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at the simple continued fraction of Jd has period length two if and only if d : a2 + b where a and b are integers irrationals, then ) ca1-a2)' : d'r - d2 (c''c.z)' : ot't'or2 . 1 1 . Which of the following quadratic irrationals have purely periodic continued fractions l + . 6 a ) 2 + ,/-B b) c ) 4 + ' , m c) d) e) (tt - ,/-toltg e + ,f?l)/z (tz + -'.ft-g)l: t 12. Suppose that a : G+JF)/c, where 4,b, and c are integers, b ) 0, and b is noi u perfecl square. Show that is a reduced quatratic irrational if and only if 390 Decimal Fractions and Continued Fractions 13. Show that if quadratic irrational. ir-u reduced quadratic jrrational, then _ l/a, is also a reduced 1 14' Let k be a positive integer. Show that there are infinitely mgy positive integers D, such that the simple continued fraction expansion of ,/6 h., , period of Show that if p : (tar + l)2 * 2a1,-1 * r, where / is a nonnegative integer, then rD has a period of length k + l.) ( , 15' Let k be a continued fraction of JOp has a period of length 6ft. lgsitiu: iF:r. Let Dk - (3k+t)2 + 3 Show that the simple 10.4 Computer Projects Write computer programs to do the following: 1' Find the quadratic irrational that is the value of a periodic simple continued fraction. 2' Find the periodic simple continued fraction expansion of a quadratic irrational. 11 some Nonlinear Diophantine Equations 11.1 Pythagorean TriPles The Pythagorean theorem tells us that the sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypothenrur.. Conversely, any triangle for which the sum of the squares of the lengths of the two shortest sides equals the square of the third side is a right triangle. Consequently, to find all right triangles with integral side lengths, we need to find all triples of positive integ ers x ,y ,z satisfying the diophantine equation ( Triples of Pythagorean triPles. positive integers satisfying this equation are called Example. The triples 3,4,5; 6,8,10; and 5,12,,13 are Pythagorean triples b ecause 32 + 42 : 5'.62 + 82: 102, and 52 + 122 : 132. Unlike most nonlinear diophantine equations, it is possible to explicitly Before developing the result describe all the integral solutions of (ll.l). describing all Pythagorean triples, we need a definition. Definition. A Pythagorean triple x,!,2 is called primitive if (x,y,z) : l. Example. The Pythagorean triptes 3,4,5 and 5,I2,I3 are primitive' whereas 3 9 1 392 Som e N onli near D iophant ine E quati ons the Pythagorean triple 6,g,10 is not. Let x,!,2 be a pythagorean triple with (x,y,z) : d . Then, there are l . integers xr, t,zr with x : dxi,y : dyt ,, J i r, Furthermore, because " i-r' r,,r1, 21): " "A we have so that ? . Hence, xt,!t,21 is a primitive pythagorean triple, and the original triple x,!,2 is simply an integral multiple of this primitive pytgagorean triple. Also, note that any integral multiple of a primitive (or for that matter any) Pythagorean triple is again a pythagorean triple. If x1 ])t,zt is a primitive Pythagorean triple, then we have and hence , so that dx 1,dy 1,dz 1 is a Pythagorean triple. Consequently, all Pythagorean triples can be found by forming integral multiples of primitive Pythagorean triples. To find all primitive pythago*un triples, we need some lemmata. The first lemma tells us that any two integers of a primitive Pythagorean triple are relatively prime primitive Pythagorean triple, then Proof. suppose x ,! ,z is a primitive pythagorean triple and (x ,y) > l. Then, there is a primep such th a,t p^l (xy), sothat p I x and p I y. S inc that p I z (using problem 32 of Section 3.2). This is a contradiction since (x ,y ,z) : In a similar manner we can easilv show that (x ,z) : (y ,z. Therefore , (x g) : l. D l. 1 1 .1 Pythagorean TriPles 393 Next, we establish a lemma about the parity of the integers of a primitive Pythagorean triPle. Lemma 11.2. If x,y,z is a primitive Pythagorean triple, then x is even and y is odd or x is odd and Y is even' Proof. Let x ,!,z be a Primitive that (x ,y\ : 1, so that x and both be odd. If x and Y were we would have Pythagorean triple. By Lemma 1 l '1, we know y cannot both be even. Also x and y cannot both odd, then (from problem 2 of Section 2'1) so that ) x - = vz = I (mod 4), (mod 4). This is impossible (again from problem and y is odd, or vice versa. E 2 of Section 2.1). Therefore, x is even The final lemma that we need is a consequence of the fundamental theorem of arithmetic. It tells us that two relatively prime integers that multiply together to give a square must both be squares' Lemma 11.3. If r,s, and t are positive integers such that (r,s) : : t2, then there are integers z and n such that r : m2 and s : n2. ; I and Proof. ,upptr. l b e and If r :1 or s : that r ) l, then the lemma is obviously true, so we may I and s ) 1. Let the prime-power factorizations of r,,s, and , : :,i\ p:,it p:" t : ql' ql' quo'. Since (r,s ) : distinct. Since rs : t2, we have l, the primes occurring in the factorizations of r and s are pi'pi' pi"pi,+ipi,n pl,' : q?"q'ru' qiur' From the fundamental theorem of arithmetic, the prime-powers occurring on 394 Som e No nli near D iophant ine E quati ons the two sides of the above equation are the same. Hence, each pi must be equal to Qi for some j with matching exponents, so that a; : 2bi. consequently, every exponent a; is even, and therefore ai/2 is an integer. we see that r - m2 and , : 12, where m and n arethe integers and a./2 a-/ : pi,r('pi,C' a / 2 P r " ! We can now prove the desired result that describes all primitive Pythagorean triples. Theorem ll.l. The positive integers x,l,z form a primitive pythagorean triple, with y even, if and only if there are relatively prime positive integers 172 and n, |/t ) n, with m odd and n even or m even and, n odd, such that x : m 2 - n 2 'r7-'#ir' Prot{. Let x ,y ,z be a primitive Pythagorean triple. Lemma I 1.2 tells us that x is odd and y is even, or vice versa. Since we have assumed that y is even, x and z are both odd. Hence, z*x and z-x are both even, so that there are p ositive integers r ands with r : (z+i/2and s : (z -il /2 . lz+x] f ,-"1 Ir)' lr): I , .l t ' J:" . Using Lemma I 1.3, we see that there are integers la and n such that r : m2 and, s : n2. Writing x,y,a ndz in te rmsof m and n w e hav PYthagorean TriPle s 395 x : m2-n2', y :2 mn, and z : *'+r', and w e know that (x ,y,z ) : we see also that (m ,n) : 1, since any common divisor of m and n must also l ' Oi "iO" We also note that rn and n cannot both be odd, for if they were' then x y ' and z would all be even, contradicting the condition (x,y ,z) : l ' Since I and m and n cannot both be odd, we see m is even and n is odd, (m,n) : or vice versa. This shows that every primitive Pythagorean triple has the appropriate form. To see that everY triPle , where m and n m * n (mod 2), forms a primitive Pythagorean triple, first note that are positive integers, m ) n, (m,n) : 1, and m a-2m2n2+n4 ) * 4m2n2 : ^ m2+n2)2 : 2 2 . To see that these values of x,y, and z are mutually relatively .prime, assume ! . ) note that p * 2, since x is odd (because x: m2-n2 where mz and n2 have o fp o rit" parity). Also, note th at because p I,x and p l t, p I G+ i :2m2 1. (* ,i) triple. This concludes the proof Therefore, (r,y,z) : l, and xoy,z i s a pri mi tiv e P ythagor ean The following example illustrates the use of Theorem I I .l to produce Pythagorean triPles. Hence, Theorem 1 I .1 tells us that is a primitive Pythagorean triple. 396 Som e No nli near D iophant ine E quati ons We list the primitive pythagorean rn :< 6 in Table I l.l. triples generated using Theorem I l.l with 24 20 40 r 2 60 5 l 3 l 7 2 5 29 4 l 3 7 6 t Table 11.1. Some Primitive pythagorean Triples. I l.l Problems l . F i n d a l l il primitive Pythagorean triples x,l,z with z < 40) Pythagorean triples x,!,2 with z < 40. Show that if x,!,2 divisible by 3. is a primitive pythagorean triple, then either x or y is Show that if x ,!,z is a Pythagorean triple, then divisible by 5. exactly one of x,y , and, z is Show that if x,l,z divisible by 4. is a Pythagorean triple, then at least one of x,y, and z is Show that every positive integer greater than Pythagorean triple. three is part of at least one recursivelv bv 11.2 Fermat's Last Theorem 397 ' Show that xnln,zn is a Pythagorean triple the Pythagorean triples given in problem 6. Find all solutions in positive integers of the diophantine equation x2 I 2y2 : t2' t2g. Find all solutions in positive integers of the diophantine equation x2 * 3y2: 10. Find all solutions in positive integers of the diophantine equation . 1 3 . 1 4 . 1 5 . Find all Pythagorean triples containing the integer 12. Find formulae for the integers of all Pythagorean triples x,l,z with z - y*l Find formulae for the integers of all Pythagorean triples x,l,z with z - y * 2' Show that the number of Pythagorean triples x,-y,z (with x2 + y2 : z2) with a fixed integer x is (rk2)-l)/2if Find all solutions in positive integers of the diophantine equation *' * py' : 22 is odd, and (r!2l4-1)/2 if x is even. 11.1 Computer Projects Write programs to do the following: l. Find all Pythagorean triples xJ,z with xy,and z less than a given bound. 2. Find all Pythagorean triples containing a given integer' ll.2 Fermat's Last Theorem In the previous section, we showed that the diophantine equation x2 + y2 : z2 has infinitely many solutions in nonzero integers x, !, z . What happens when we replace the exponent two in this equation with an integer grrut.. than two? Next to the discussion of the equation xz + y2 : z2 in his copy of the works of Diophantus, Fermat wrote in the margin: "However, it is impossible to write a cube as the sum of two cubes, a fourth power as the sum of two fourth powers and in general any power the sum of two similar powers. For this I have discovered a truly wonderful proof, but the margin is too small to contain it." 398 Some Nonl inear D iophant ine E quati ons Since Fermat made |
this statement many people have searched for a proof of this assertion without success. Even trrouitr no ,or...t proof has yet been discovered, the foilowing conjecture is knowi as Fermat,s rast theorem. Fermat's Last Theorem. The diophantine equation x ' + l n : z n has no solutions in nonzero integers x, r, z when n is an integer with n D 3. Currently' we know that Fermat's last theorem is true for all positive integers n with 3 ( n <125000. In th is section , we w il l show that the spec ial cas e of Fermat's last theorem with n: 4 is true. That is, we will ,ho* that the diophantine equation x a + ! 4 : 2 4 has no solutions in nonzero integers x, !, z. Note that if we could also show that the diophantine equations x P + Y P : 7 P has no solutions in nonzero integers x,!,2 whenever p is an odd prime, then we would know that Fermat's last theorem is true (see probl em 2 at the end of this section). The proof we will give of the special case of n - 4 uses the method of infnite descent devised by Fermat. This method is an offshoot of the well-ordering property, and shows that a diophantine equation has no solutions by showing that for every solution there is a "smaller', solution. contradicting the well-ordering property. Using the method of infinite descent we will show that the diophantine equation xa + !4 : 22. has no solutions in nonzero integers x, !, and z. This is stronger than showing that Fermat's last theorem is true for n: 4, because : Theorem 11.2. The diophantine equation **',ro,r: t' has no solutions in nonzer" ,",.*1, Proof. Assume that the above equation has a solution in nonzero integers x,l,z. Since we may replace any number of the variables with their negatives 1 1.2 Fermat's Last Theorem 399 without changing the validity of the equation' we positive integers' We may also suppose that (x,y) : 1' To see x : dx 1 and y = dY,, with (xvYt) : 1' where x1 since xa + Y4 : '2 ' vte have may assume that x,Y,z are this, let (x,Y) : d. Then and y 1 itro Positive integers' so that ( ' Hence do | ,', and, by problem 32 of Section 2'2' we know Therefore , z : d'r r, where z 1is a positive integer' Thus ? , so that x f + y l : t ? . Thi s gives a solution of xa + ya2 in positive in tegers x : xt '! : l r'z : zr xo, lo, and zsare positive integers with (xe,-/o) : 1 ' We will show that there i s a nother solution in positi ve integers x : xr,! : l t, z: zt w ith (x r'y l) : 1' su ch that 21 1 zs , is a Pythagorean triple. Furthermore, we have so that x&, y&, ,o l-fi, r&> - i, ro. if p is a prime such that p I x3 and p I y&' then p I xs contradicting the fact that (xq,lrq): l. Hence, *3,yE, zs is a ;;';'l'ro, prim-itive iythagorean triple, and by Theorem- 11.1, we know that there afe positive integers z and n with (z ,n), m # rl (mod 2) ' and x & : m 2 - n 2 !& : Zmn z o : m 2 + n 2 , where we have interchanged x62 and yfr, if necessary' to make yfr the even integer of this Pair. 400 Some Nonl inear D iophant ine E quati ons From the equation for xfr, we see that x & + n 2 : m 2 . Since (m,n) : l, it foilows that x,s,n,m is a primitive pythagorean tripre. Again using Theorem I I .1, we see that there are fositive integers r and s with ( . Since m is odd and (m,n) : b ecause y&: ? positive integer, so that l, we know th at (m, 2d : l . W e note that (2d m, Lemma ll.3 te lls us that ther e are posi ti ve i ntegers z1 . si nce (r,s): y1 such that r : xl and s : y? . Note th at si nce (r, s) : that (xl,-yr) : I , Lemma 11.3 te lls us th at ther e are pos iti ve i ntegers x1 erd l , i t easi ry fol ow s l. Hence. x { + y f : - 2z l where x t,! t,z 1 ?re positive integers with r r,y1) : l. Moreover, we have . To complete the proof, assume that xa * y4 : z2 has at least one integral solution' By the well-ordering property, we know that among the solutions in positive integers, there is a solution with the smallest value is of the variable z However, we have shown that from this solution we can find another solution with a smaller value of the variable z, leading to a contradiction. This completes the proof by the method of infinite descent. n Readers interested in the history of Fermat's last theorem and how investigations relating to this conjecture led to the genesis of the theory of algebraic numbers are encouraged to consult the books of Edwards Il4l and Ribenboim Irt]. A great deal of research relating to Fermat's last theorem is underway. Recently, the German mathematician Faltings established a result that shows that for a fixed positive integer n, n > 3, the diophantine equation xn + yn : z' has at most a finite number of solutions where x g, and, z are integers and (x,-y) : l. 1 1.3 Pell's Equation 401 ll.2 Problems l. show that if x,! ,z is a Pythagorean triple and n is an integer n ) 2' then .. Show that Fermat's last theorem is a consequence of Theorem I l '2' and the assertion that xP * yp : zP has no solutions in nonzero integers when p is an odd prime. 3. Using Fermat's little theorem, show that if p is prime and a) b) if xp-l * yn-t : zP-r, then p | *yt if xP + lP : zP, then p | (x+Y-z). . 4. Show that the diophantine equation xo-yo: integers using the method of infinite descent' z2 has no solutions in nonzero never a Perfect square. with integer sides is 6. Show that the diophantine equation xa + 4ya - z2 has no solutions in nonzero integers. i. Show that the diophantine equation x' - 8y4 : z2 has no solutions in nonzero 8 . 9 . 1 0 . integers. Show that the diophantine equation xa + 3ya : z4 has infinitely many solutions' Show that in a Pythagorean triple there is at most one perfect square' Show that the diophantine equation xz + y2: z3 has infinitely many integer integers that solutions by showing x : 3k2-1, | - k(k2-3), z : k2 * I form a solution. for each positive integer k the tt.2 Computer Proiects l . Write a computer program to search for solutions of diophantine equations such . 11.3 Pell's Equation In this section, we study diophantine equations of the form ( , where d and n are fixed integers. When d <0 and n (0, there are no solutions of (11.2). When d < 0 and n ) 0, there can be at most a finite 402 Some Nonlinear Diophantine Equations number of solutions, since the equation x2 - dyr: n implies that l"l < fi il* Also, note that when d is a perfect ,quur., say d : D2, lrl < JM Hence, any solution of Qt.D, when d is a perfect square, corresponds to a simultaneous solution of the equations ::'d=; , where a and b are integers such that n : ab. In this case, there are only a finite number of solutions, since there is at most one solution in integers of these two equations for each factorization n : ab For the rest of this section, we are interested in the diophantine equation x2 - dy':n, where d and n are integers and d is a positive integer which is not a perfect square. As the following theorem shows, the simpL continued fraction of -,/v is very useful for the study of this equation. Theorem 11.3. Let d and n be integers such that d > 0, d is not a perfect square, and lrl < r/7. .lf x2 - dyI: n, then xfy is a convergent of the simple continued fraction of ^/7. Proof. First consider the case where n ) A. Since x2 _ dyr: n,wesee that ( t r . : ) G +y./7) G -y,/V) : n From (tt.:), we see that x - y.,/7 ) 0, so that x > yrT. consequently, * _ , / 7 v > 0 , and since 0 1 n < ,8, we see that ta G -,/7v) Y W v x2-dY2 y G + y,/7) : 1 1 .3 Pell's Equation 403 - f r\ t \ q I 1 : l YQYJA) fi Zy'rld ) rr2 L! 1 Since 0 < convergent When n Theorem 10.18 tells us that x ly must be a .,17 < +, -r 2 v ' slmple contlnueo x _ v 1 fraction of JL of the ( 0. we divide both sides of x2 - dy' : n by -d, to obtain v2 - ,fr*': -3 is a By a similar argument to that given when n ) 0 o we see that y /x convergent of the simple continuid fraction expansion of ll.r/7' Therefore' must be a from problem 7 of Slction 10'3, we know tB converyent of the simple continued fraction of './d : *l!,:1l,j.,/x) l/(l/{cl ) ' u we have shown that solutions of the diophantine equation x2 - dy': n, are gifn by the convergents of the simple continued The next theorem will help us use these ^1"1 . *h;; fraction expansion of convefgents to find solutions of this diophantine equation' .n, fi. Theorem 11.4. Let d be a positive integer that is not ^ perfect square' [47.1, P*+r --!*Q! and dk : (io + ',/hlQr, il; t* L :0 ,1,2,... where ao: Jd ' Further more' Iet (;"- pt*'JlQ*, denote tie kth convergent of the simple continued fraction expansion of - 'o'' oo: O;';-r: ;J; . Before we prove Theorem 1 1.4, we prove a useful lemma numbers and d is a positive integer that is not a perfect square. Then proof. Since r * s,/7 : t * u,/7, *"see that if s # u then ,/7 - r-t u - s 444 Som e Nonti near D iophant ine E quati ons By Theorem 10.1, (r-t)/(u-s) irrational. Hence, s : u, and consequently r : t. A is rational, and by Theorem r0.2 Jv i , We can now prove Theorem I 1.4. Proof. Since ^E : o,0: Ias;ar, e2,...,ek, otk+tL, Theorem 10.9 tells us that tj - v s ott+tp* I p*_t , ' Since dk+t : (pt *, + ,/7)/er+r J V : Therefore, we see that we have + ,8)p* * e*+pr,_t (P**t (P**, + ,/V)qr * et +rQ*_t dqt t (Pt+flt, I Qt +rQtr-r)fi : (pr,+tpr, * e*+rpt,-r) + p*fi. From Lemma 11.4, we that dqr, Pt+fl* f Qt+rQn-t: pk When we mul tiply t t. by qt and the second by pt, subtract the first simplify, we obtain find : P*+tPt, * Q*+et -r and first of these two equations from the second, and then , where we have used rheorem 10.10 to complete the proof. tr The special case of the diophantine equation x2 _ dy, : , with n : I is called Pell's equation. we will use Theorems ll.3 and rr.4 to find all solutions of Pell's equation and the related equation x2 - dy, : -t. Theorem 1l'5' Let d be a positive integer that is not a perfect square. Let px/qt denote the kth convergent of the simple continued fraction of .8, k : 1,2,3,"' and let n be the period length of this continued fraction. Then, y.!"n ,r, the diophantine equation equation x2 - dy' : - l has no solutions. when |
n is odd, the positive ' : even, the positive solutions of * : Pyoof. Theorem 1r.3 tells us that if xo,ro is a positive solution of simple continued fraction of ,/7 . On the other hand, from Theorem I 1.4 we know that t l , 1 1 .3 Pell's Equation 405 . Because the period cf the continued expansion oL"/j Qjn : Qo:I for 7 : 1,2,3,"', ('int" J'l : "tf ' Hence' is n, we know that pk-, - d q?^-t : (- l)i'Qni : (- I )/n ' This equation , . . . shows that when n is even Pin-t, Qin-t is a solution of and Pz(j -Dr-r,Qz(i-Dn -, is a sol ution of x2 - dy' : for -l To show that the diophantine equations have no solutions other than those already implies that n lk and that Q1 # -l for 7 : We first note that if Qt*t: l, th en found, we will show that Qpal,k+l: P1ra1 * 'ftr' the continued fraction expansiOn of a1a1 is purely Since ok+l : periodic. Hence, Theoiem !0.20 tells us that -1 1 a*+r: Pk+r - ''17 < O' Thi s implies that Pk+t:lr/71, so th at dk : c"o , a nd nl k' la1ra,.a1r1z,...l -'Sin"" ct; has a purely periodic simple continued fraction -G. dj : -pi expansion, we know that - and . From the first of these inequalities, we see that Pi > -r/7 second, we see that Pi < -l -fi. contradictory, we see that Qt # -1- and, from the Since these two inequalities for p1 are Since we have found all solutions of x2-dy2: x and y arc positive integers, we have completed the proof. n I and x2-dy2: -1, where We illustrate the use of Theorem 11.5 with the following examples' Example. Since the simple continued fraction of .,8 is tl;f ,f 'f ,f ,el the 406 Some No nli near D iophant ine E quati ons positive solutions of th e diop hantine equa tion x2 i : l'2'3"" *T]: p1_o1/e.roi-r continued fraction expansion of pe:649, I are pni _t,et. 'j _t, is the (roi-l)th ctnvergent or ,r," simple least po-ritiu" sorution is {e : 180. The positi ve soluti ons of the di ophanti ne equati on x2-13y2 : -I are Prci-o,Qtoi-oi : 1,2,3,. ..; the l east pos iti ve sol uti on l 3yr: .,m. The Example. Since the continued fraction of -,.fr is t3;Wl, the positive solutions of x2 - t4 y2 _: I are pai-1,e4j-r, j : r,. 2,3,.. . w here p+ i -tbqi- r i s the 7th convergent of the simple continued fraction expansion of Vl4. The l east positive sohltion is pt: 15, Qt: 4. The di ophanti ne equati on xz - l4y2 : -1 has no rotuiionr, since the period length of the simple continued fraction expansion af ,/la is even. We conclude this section with the following theorem that shows how to find all the positive solutions of pell's equation x2-- dyt : I from the least positive solution, without finding subsequent convergents of the continued fraction expansion of ,/7. Theorem 11.6. L9t xg1 be the least positive solution of the diophantine equation x2 - dyL : l, where d is a positive integer that is not a perfect square. Then all positive solutions xk,lk are given by Note that xp and y1, are determined by the use of Lemma Proof. We need to show that x1r,y1, is a solution for k : every solution is of this form. and that To show that x1,/r -.!! a solution, tst note that by taking conjugates, it lr,,/T)k, because from Lemma 10.4, the follows that x1, - ytrfi: conjugate of a power is the power of the conjugate. Now, note that (x r- xt - dyt : (xp + yr,fi)G,, - yr,fix? - ayilo : 1 . Hence xk,lt is a solution for fr : To show that every positive solution is equal to is a positive solution integer ft, assume that X,y k : 1,2,3,.... Then th ere is an integer r such t hat x*,lt< for some positive different from x*,lk for 1 1 .3 Pell's Equation 407 (xl + yJ7)" < x + Y./7 ( (x t * v]/a)n*t' When we multiply this inequality by (x t * y rfi)-"' we obtain ' since x? - dy? :1 implies that x t - !t,[i : (x1 * yt,[d)-t. Now let and note that : (*? - + yf/7)'8 - dy?)'8' - dYz) - Y,l7)Gt - - t . We see that s,/ is a solution of x2 - dy': i .; we see that 0 < (s + tJa)-r < 1. Hence l, and furthermore, we know that ,fr'.'"*;;';r",lV.--Mor.oner, since we know that s + t-,/7 > 1, 1 - r : +t(s t r,/7> +(s - r.'.ff)l > o /- and , : 1[(s + t-./7) - (s - t',17)] > o. 2Jd Th is means that s,/ is a positi ve solution, so th at s 2 x1, and t' 2 y1, by the choice of x1,y1 as the smallest-positive solution' But this contradicts the Therefore X,I' must be xpy1, for some inequality s * f ../7 < xr * ytfi. choice of /c. tr To illustrate the use of Theorem I1.6, we have the following example' Example. From a previous example we know that the least positive solution of - - positive solutions are given by xt, yp where * * yr,./n : (649 + tgo\[Lte . For instance, we have 408 Some Nonlinear Diophantine Equations x z * y 2,8 : 842361 + 233640.,/ : 233640 is . the least positive solution of ll.3 Problems l ' Find all the solutions of each of the foilowing diophantine equations ' Find all the solutions of each of the following diophantine equations ) 4xz - 9/2 : loo. 3' For which of the x2 - 3ly' : n havea solution following values of n does the diophantine equation a ) l b) c ) 2 -1 d) d 4 f) -3 -s? 4. Find the least positive solution of the diophantine equations a) b ) x2 - 29y2 : -. Find the three smallest positive . For each of the equation x2 - drz solutions of the diophantine equation following values : -l has solutions of d determine whether the diophantine ) 4r h) s0. 7. The least positive solution of lt- the diophantine equation xz - 6lyz : 1 is 2261i398A. Find the least positive solution other than xt:17663190493 Pell's Equation 409 8. S!g* that if pr/qt is a converggnt of the simple continued fraction expansion of Jd then lp? - dq?l < | + zJd. 9. Show that if d is a positive integer divisible by a prime of the form 4ft * 3, then the diophantine equation x2 - dy': -l has no solutions. Let d and n be positive integers. il Show that if r,s is a solution of the diophantine equation x2 - dyz : the diophantine equation x2 - dy' : , X,Y Xr + dYs, Xs t Yr is alsoa solution of x2 - dy': is a solution of r. I and then b) Show that the diophantine equation x2 - dyz: n either has no solutions, or infinitelv many solutions. I l. Find those right triangles having legs with lengths that are consecutive integers. (Hint: use Theorem 11.1 to write the lengths of the legs as x -.r2 - 12 and y :2st, where s and t are positive integers such that (s,t) : / and 12. Show that each of the following diophantine equations has no solutions . 11.3 Computer Projects Write programs to do the following: 1. Find those integers n with x2 - dyz: rz has no solutions. lrl < Ji such that the diophantine equation 2. Find the least positive solutions of the diophantine equations x2 - dy': and 3. Find the solutions of Pell's equation from the least positive solution (see Theorem I 1 . 6 ) . Appendix 412 Tabfe 1. Factor Table. Appendix The least prime fac1o1,of .::h.odd positive integer less than 10000 and not divisible by five is given in the table. ThJinitial digits of tile integei are listed to the side and the last digit is at the top of the column. primes are indicated with a dash.. 20 2 l 22 23 24 25 26 27 28 29 30 3 l 32 3 3 34 3 5 4A 4 l 42 43 44 45 46 47 48 49 50 5 1 52 53 54 55 56 57 58 59 60 6 l 62 63 64 65 66 67 68 69 70 7 l 72 73 74 75 80 8 l 82 83 84 85 86 87 88 89 90 9 r 92 93 94 95 96 97 98 99 100 l 0 l t02 103 rc4 105 106 107 r08 109 l l 0 l l l lt2 l l 3 rt4 - t20 t2l r22 123 t24 125 r26 t27 128 t29 1 3 0 1 3 1 r32 r 3 3 134 1 3 5 136 r37 r 3 8 1 3 9 140 t4l r42 r43 144 145 t46 147 148 r49 150 r 5 l rs2 1 5 3 r54 t 5 5 Appendix 4 1 3 Table 1. (Continued). 160 r6l t62 r63 t64 r65 r66 r67 r6 8 r69 170 17l 172 173 174 175 176 177 1 7 8 179 1 80 l 8 l r82 r83 1 84 1 8 5 1 86 t87 1 8 8 1 8 9 r90 l 9 l 76 77 78 79 2m 201 202 203 204 205 206 207 208 209 210 2tl 2t2 2r3 2t4 2t5 2r6 217 2t8 2r9 220 22r 222 223 224 225 226 227 228 229 230 231 29 t3 rt7 1 1 8 l l 9 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 ? 5 S 256 257 2s8 259 260 26r 262 263 264 265 266 267 268 269 270 27r t57 1 5 8 1 5 9 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 29s 296 297 298 299 300 301 302 303 304 305 306 307 308 309 - 414 192 r93 r94 195 r96 r97 198 r99 320 321 322 323 324 325 326 327 328 329 330 3 3 1 332 333 334 335 336 337 3 3 8 339 340 341 342 343 344 345 346 347 232 233 234 235 236 237 238 239 :oo 3 6 1 362 363 364 365 366 367 368 369 370 37r 372 373 374 375 376 377 378 379 380 3 8 1 382 383 384 385 386 387 Table 1. (Continued). Appendix - 272 273 274 275 276 277 278 279 400 40r 402 403 404 405 406 40'7 408 409 4r0 4tr 412 413 414 4t5 416 4t7 4 1 8 419 420 421 422 423 424 42s 426 427 37 47 53 59 23 19 43 - 3t2 3 r 3 3r4 3 1 5 316 317 3 1 8 3le440 44r 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 4s9 460 461 462 463 464 465 466 467 Appendix Table 1. (Continued). 4 1 5 348 349 3 50 3 5 1 352 3s3 354 3 55 3 56 357 3 5 8 3 59 480 4 8 1 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 s00 5 01 502 503 tn- 3 47 29 37 13 - 388 389 390 39r 392 393 394 39s 396 397 398 399 5zo 521 522 523 524 525 526 527 s28 s29 530 5 3 1 532 5 3 3 534 5 3 5 s36 537 5 3 8 s39 540 54r 542 543 - Tt1 qt* - 428 429 430 431 432 433 434 435 436 437 438 439 560 5 6 1 562 563 564 565 566 s67 568 s69 570 571 572 573 5',74 ) t ) 576 5',77 578 579 580 5 8 1 582 583 :-Tl r 7t - 468 469 470 471 4',72 473 474 475 476 477 478 479_ 600 601 602 603 604 605 606 607 608 609 6 1 0 6 1 1 612 6 1 3 614 6 1 5 6r6 617 6 1 8 6 1 9 620 621 622 623 7 1' 323 Table 1. (Continued). Appendix 504 505 506 507 508 509 5 1 0 5 l l 512 5 1 3 514 5 1 5 516 517 5 1 8 5 1 9 640 64r 642 643 644 645 646 647 648 649 650 651 652 6s3 654 655 656 657 658 659 - 544 545 546 547 548 549 550 551 552 553 s54 555 556 s57 s58 559 680 6 8 1 682 683 684 685 686 687 688 689 690 69r 692 693 694 695 696 697 698 699 - 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 720 721 722 723 724 725 726 727 728 129 730 7 3 r 732 733 734 735 736 737 738 739 il 7 1929 624 625 626 627 628 629 630 6 3 1 632 633 634 635 636 637 638 639 760 761 762 763 764 765 766 76 |
7 768 769 770 771 772 773 774 775 776 777 7 7 8 779 Ap pendix Table 1. (Continued)' 660 661 662 663 664 665 666 667 668 669 670 671 672 673 674 675 676 677 678 679 800 8 01 802 803 804 805 806 807 808 809 8 1 0 8 1 1 812 8 1 3 8 1 4 8 1 5 700 701 '702 703 704 705 706 707 708 709 7r0 7 t l 7 t 2 713 7 t 4 7 1 5 7 t 6 717 7 1 8 7t9 840 841 842 843 844 845 846 847 848 849 850 8 5 1 852 8s3 854 855 740 741 742 743 744 745 746 747 748 749 750 7 5 1 752 753 7 5 4 7 5 5 756 757 7 5 8 759 880 8 8 1 882 883 884 885 886 887 888 889 890 891 892 893 894 895 780 7 8 1 782 783 784 785 786 787 788 789 790 791 792 793 794 795 796 797 798 799 920 92r 922 923 924 925 926 927 928 929 930 9 3 1 932 933 934 935 417 418 Table 1. (Continued). 820 821 822 823 824 825 826 827 828 829 830 8 3 1 832 833 834 835 836 837 838 839 960 961 962 963 964 965 966 967 968 969 856 857 858 859 860 86r 862 863 864 865 866 867 868 869 870 871 872 873 874 875 876 877 878 879 970 971 972 973 974 975 976 977 978 979 896 897 898 899 900 901 902 903 904 905 906 907 908 909 910 9tl 912 913 9r4 9 1 5 916 917 9 r 8 919 980 9 8 1 982 983 984 985 986 987 988 989 7l 29 936 937 938 939 940 94r 942 943 944 945 946 947 948 949 950 951 952 953 954 955 956 957 958 959 990 991 992 993 994 99s 996 997 998 999 Appendix Reprinted with permission from u. Dudley, Elementary Number Theory, Second Edition, copyrighto 1969 and l97g by w. H. Freeman and company. All rights reserved. Appendix 4 1 9 Table 2. Values of Some Arithmetic Functions' I 2 3 4 5 6 ''l l 8 l 9 2A 2 l 22 2 3 24 25 26 2'I 28 29 30 3 l 3 2 3 3 34 40 4 l 42 4 3 44 45 46 4"1 48 49 20 t 2 l 8 t 2 28 I 30 l 6 20 l 6 24 t 2 36 l 8 24 l 6 40 t 2 4 2 20 24 22 46 l 6 42 28 t 4 24 24 3 l l 8 3 9 20 42 3 2 36 24 60 3 l 42 40 48 54 48 9 l 3 8 60 56 90 42 96 44 84 7 8 7 2 48 124 5 7 420 Table 2. (Continued). A ppendix 50 5 l 5 2 5 3 54 5 5 56 5 7 58 5 9 60 6 r 62 6 3 64 65 66 67 68 69 7A 7 l 72 7 3 74 7 5 7 6 1 1 78 79 80 8 l 8 2 8 3 84 8 5 86 8 7 8 8 89 90 9 l 92 9 3 94 9 5 96 9',| 9 8 99 1 0 0 20 3 2 24 52 l 8 40 24 3 6 28 58 l 6 60 30 3 6 3 2 48 20 66 3 2 44 24 70 24 72 3 6 40 3 6 60 24 78 3 2 54 40 8 2 24 64 42 56 40 8 8 24 7 2 44 60 46 72 3 2 96 42 60 40 93 7 2 9 8 5 4 120 72 1 2 0 80 90 60 1 6 8 62 96 1 0 4 127 84 144 68 r26 96 t44 7 2 r 9 5 74 n 4 t24 140 96 1 6 8 80 1 8 6 t2r r26 84 224 1 0 8 t32 120 1 8 0 90 234 n 2 r 6 8 128 t44 t20 252 9 8 t 7 l r 5 6 217 Appendix 421 Table 3. Primitive Roots Modulo Primes The least primitive root r modulo p for each prim e p, p < 1000 is given in the table 29 3 1 3',1 4 l 4 3 47 89 97 l 0 l 1 0 3 1 07 1 09 1 1 3 127 1 3 1 r37 1 3 9 t49 l 5 l 1 5 7 1 6 3 r67 r73 179 r97 199 2tl 223 227 229 233 239 241 251 257 263 269 271 277 28r 283 293 307 3 1 1 3 1 3 317 3 3 1 33',1 347 349 3 s 3 359 367 3 7 3 379 3 8 3 389 397 401 409 4 1 9 421 43r 433 439 443 449 457 46r 463 467 479 487 49r 499 s03 s09 521 523 5 4 1 547 5 ) / 5 6 3 569 57r 577 5 8 7 5 9 3 599 601 607 6 1 3 617 6r9 63r 641 643 647 6 5 3 659 6 01 6 73 677 6 8 3 691 701 709 7 1 9 727 733 739 743 7 5 r 7 5 1 76r 769 773 787 797 809 8 1 1 8 2 r 823 827 829 863 877 8 8 1 8 8 3 887 907 9 l l 919 929 937 94r 947 9 5 3 967 97r 977 983 9 9 1 997 Appendix Table 4. Indices Numbers 422 29 3 r 3 i 4 l 43 47 5 3 59 6 l 67 7 l 73 79 83 89 97 I r( I to l t 2 l 8 22 28 30 36 40 42 46 52 58 60 66 70 72 78 82 88 96 'ilil,Y,l 'i l;i l^ilrrl trlfr|JIl,lil'i p l 9 23 29 3 l 37 4 l 43 47 53 59 6 l 67 7 l 73 79 83 89 97 Numbers t 7 1 8 l 1 9 20 2 l 22 23 24 25 26 27 28 29 30 3 l 32 33 l 0 7 2 l 7 7 3 3 38 t 6 l 0 40 47 64 49 2 l 2 l 56 6 89 I Indices rr I oa 26 t 7 3 l 29 t 5 25 7 26 l 6 60 37 63 72 25 t 2 24 5 24 8 25 34 3 7 3 7 49 8 24 1 7 40 1 7 70 29 t 4 69 l 3 t 7 29 22 t 4 36 6 3 l t 0 55 62 27 39 54 80 82 5 20 27 l 5 36 t 6 5 3 9 l 5 57 28 l 5 46 26 60 57 77 8 l 3 29 l 3 40 28 20 53 9 42 44 30 l 3 75 49 76 l 9 5 t 2 l 7 l 7 29 25 46 4 l 20 45 67 38 78 39 59 l 6 l 0 l 0 4 8 2 42 t 2 44 30 56 2 46 54 52 2 l 5 t 4 23 l l 39 l 3 57 29 5 5 60 l 5 67 l 8 87 9 9 28 34 3 3 3 49 59 47 l l l l 56 3 8 3 l 46 20 l 8 3 l 27 23 t 7 2 l 32 57 6 l 69 t 4 85 60 f 5 5 5 30 40 20 5 80 74 5 l 0 9 44 I Reprinted with permission from J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill Book Company. Copyright O 1939. Appendix 423 Table 4. (Continued). p 3'l 4 l 43 47 s 3 59 6 l 67 7 l 7 8 79 83 89 97 p 5 3 5 9 6 l 67 7 l '73 7 9 83 89 97 p 67 7 l 78 79 83 89 97 p I r 9 23 34 l l 4 l 48 65 5 5 29 25 5 7 22 27 l 9 2 l l 8 3 3 9 24 l l 3 8 29 34 37 3 5 63 32 t 8 2 l 4 30 36 44 l 4 l 4 64 28 l 0 64 34 t 6 3 2 42 30 5 5 39 22 2A 64 t 9 20 27 l l 22 70 3 6 48 'l 46 5 8 65 65 3 5 67 24 95 20 22 9 50 9 2s l 8 46 25 74 30 30 Numbers Numbers dices I n I 1 3 | ')) | 33 I 43 1 e l 48 I43 1 8 l 27 1 r i I 6 r l 43 1 7 r I 76 1 26 1 28 1 5 8 1 4 l 29 48 34 27 l 0 l 3 64 7 72 45 23 40 1 6 5 8 29 2 l 54 30 6 l 73 l 5 44 23 20 50 9 3 l 59 23 54 84 2 l 54 l 0 43 50 3 8 l 7 76 65 l 4 23 3 6 3 8 46 2 66 28 l 6 74 62 50 5 l 52 53 54 ) ) 56 ) t 5 8 5 9 60 6 l 62 63 64 65 43 r 3 45 3 l 62 l 0 50 5 5 68 36 27 3 2 5 3 5 t 5 27 22 46 7 63 76 47 42 2 l 5 l 3 42 79 5 5 93 22 3 3 5 7 23 5 3 '7'7 59 78 l 0 3 5 t 9 52 l 4 26 5 3 l 9 \) 3 l 37 8 5 9 56 5 2 5 l 66 8 7 2 l 52 26 t 9 5 7 65 l l 4 l 3 7 30 3 2 49 42 68 3 3 37 3 6 5 5 29 36 45 4 43 t 5 1 3 7 5 47 3 l 36 J 5 3 l 34 43 67 66 67 68 69 't0 'tl 7 2 7 3 74 7 5 Numbers 30 56 66 23 7 l l 9 l 5 43 't6 Inc Lices 48 45 5 5 70 83 75 69 5 8 45 66 69 64 6 36 48 24 6 8 t 2 34 67 60 1 8 22 5 26 I I 7 8 79 80 8 l 3 3 63 69 7 3 t 5 1 3 94 47 50 48 45 56 5 7 6 l J I 29 5 8 3 8 6 l 4 l 52 2'7 50 5 8 5 l 3 5 42 4 l 36 79 66 44 5 l 3 3 62 l t 3 6 t 4 65 50 50 Inc lices 39 68 40 J J 3 l 46 4 l 42 4 88 44 69 20 28 23 z l 27 29 4'l 44 5 3 72 40 49 67 5 3 43 32 77 2 l Numbers 82 83 84 8 5 86 87 88 89 90 9 1 92 93 94 95 96 8 3 89 97 4 l 3 7 23 6 l t 7 26 '73 76 90 45 3 8 60 83 44 92 lces I n d 20 1 22 82 48 Appendix Table 4. (Continued). Indices llNumbers t0 | z0 | 40 | 2t | +z ! zs I so I +r I zt I + 15 | t I t+ | za I se I zq I ss I t: I ee | + Indices . 2l 3 l 3'' 4 1 43 4'l 5 3 59 6 l 67 7 l 73 79 83 89 97 p t 7 l 8 t 9 20 2 l 22 I 23 24 25 26 27 28 29 30 3r 32 33 I 1 9 23 29 3 l 3 7 4 1 43 47 5 3 59 6 l 67 7 l 7 3 79 83 89 9? l 0 l 5 2 l 22 l 8 26 26 3 8 J 3 3 44 20 62 20 48 l 5 6 83 I 6 l 3 4 36 3 3 35 2 6 27 40 8 27 65 30 t 8 2t t 2 23 5 3 3 40 l 4 3 24 28 47 26 37 l 8 32 3 7 73 93 l 4 t 7 l 5 29 3 5 42 l 5 48 56 3 3 s2 46 t 7 t 7 74 4 l 77 I I ) t 4 2 1 5 40 28 43 5 3 5 3 7 38 t 2 5 l 65 34 94 26 l 2 3 5 34 l 9 l 0 t 2 t 4 54 t 3 56 62 3 7 60 54 38 l 0 I I ) 30 34 46 3 3 47 l 0 7 53 60 74 47 I 3 82 20 2 l 0 t 6 t 6 42 l 3 3 5 20 l 4 t 6 8 64 l l 39 22 l l 6 20 t 4 ) 22 26 l l 40 28 4 l 40 34 22 28 t 3 I Numbers 36 57 3 8 l 1 5 l 3 0 l 60 4 s l 2 3 l 4 6 l 25 2 t l 5 1 3 5 1 32 s r l : e l : + l 24 6 e l 4 e l 6 8 l 46 s | 'o I zo I 40 t + 1 + + l + t l 40 z + l t s l u l 79 22 l 8 3 2 l 5 1 6 52 22 l 9 56 3 54 23 44 84 65 22 59 50 l l 47 59 80 3 l 1 3 7 I 3 42 t7 4 3 5 t 4 t'l 3 9 35 3 r 43 5 3 66 42 57 7 l t 2 78 Ap pe ndix Table 4. (Continued). 425 p 34 3 5 1 3 6 37 3 8 39 40 4 l 42 4t 4 4 i | 4 5 i . 4 64',1 48 49 Indices 17 4 l 43 47 5 3 5 9 6 l 6',1 7 l 7 3 79 83 89 97 28 20 3 l 34 9 27 45 65 l 0 3 5 l 3 59 36 2 1 9 3 8 t 29 l 8 54 29 63 70 29 39 3 5 l 9 l 0 I 23 z l 4 3 6 49 5 8 59 & 72 3 8 70 5 7 50 1 5 20 20 l 9 39 5 5 5 l 22 68 3 5 57 82 56 8 1 7 6 3 8 l 9 49 3 5 t 2 48 26 3 l 68 86 I 8 30 23 3 8 37 3 l 3 2 l 78 62 26 42 I 24 9 46 l 7 l 3 6 20 32 76 4 l 78 l 6 I Numbers rl 52 2 5 l 5 0 l 3 e t' 67 7 1 1 6 3 1 2 3 42 ? 62 e l 4 5 l 3 1 5 8 29 45 39 34 26 t 2 69 l 4 70 82 56 80 I 37 25 9 52 24 57 70 52 8 l 79 1 2 M 50 l 8 43 48 44 5 8 77 79 5 9 60 lndices 20 3 t 1 l 0 45 34 67 'll 1 3 19 8 3 89 97 p I 60 49 1 0 6 l 5 5 70 67 7 l 7 3 19 83 89 97 p 82 I 2r 95 83 89 97 40 3 l 4 47 48 61 50 69 72 72 27 6 28 27 5 2 43 7 l 5 5 3 8 69 I t 2 5 6 54 9 69 5 5 27 25 54 24 5 l 4 l 63 5 3 I 54 7 5 76 48 4 l t 5 l 5 46 z l 25 47 89 37 2 l 30 34 l l 63 50 < ) 5 7 l 5 42 60 25 )) 3 1 l 7 6',7 9 l 30 23 5 3 3 3 56 t 4 34 23 67 I 46 39 1 8 6 l 42 68 69 44 Indices I mbers 68 3 3 28 46 8 3 68 1 7 36 3 7 45 36 l 6 6 l 34 39 3 9 56 72 48 47 5 3 29 26 I 22 30 65 62 23 87 65 26 30 39 76 59 29 57 l l 7 8 50 4 6 l 66 3 3 7 3 6 l 20 I 3 8 20 63 5 3 44 60 43 l 5 1 5 I 22 3 45 75 66 6 46 84 40 l 2 49 32 4 l 24 5 8 63 44 48 85 24 5 3 l 3 7',| 23 Indices rl 2 r I{umbt It' l s 2 2 42 94 95 96 88 52 N 66 lrs umbr l r s Table 5. Simple Continued Fractions for Square Roots of positive lntegers d J7 d J7 | , I r,,ll o I f3:LAGt n I r+;1i;l)"rl I I 20 I Ia:2,81 i I zt I t4;iJJJJst ,, I rs:s, ror o I Is:z,rol 42 ) [6:2,t21 | . 60 6 l 6 2 6 3 6 5 66 6 1 6 8 6 9 70 7 l 7 2 7 3 11 I / ) | 76 1 77 1 7 8 I 7 e l 80 I 8 2 I 8 3 I 34 1 t 5 | - - l 16 ' tt,T,zl,tqi I t't;zr+s:+. ro ts;zT;, rJ,lot - l8;22,-l ,1 .t ,z,z,t ol [ ta;ffi. tq:1.1..-r'r.sJJJmr [ [\ry,zJal [ rg;mr [ lg;iJTl 4 2 6 Answers to Selected Problems Section l.l 1 . a ) 20 b) s5 c) :as d) 2046 . 2. 6, 24, 120,720, 5040 , 4032 0, 36288 0, 36288 00 Section 1. Section 1. ! Section 1.t . (8705736) r6 8 . (l I C) rc, (28 95) ro 428 A nsw ers to S elec ted probl ems ) 3 gross, I I dozen, and 6 eggs Section 1.5 a) prime b) prime c) prime d) composite e) prime f) composite 16. a) 1,3,7,9,13,15,21,2 5,3 1 ,33 ,37,43,49, 51, 63,67,69, 73, 75,7g,g7.93.99 Section 2.. I if a is odd and b is even or vice versa, 2 otherwise 15. 66,70,105; 66,70,165; or 42,70,165 Section 2, Section 2.. 249,331 il 24 b) 210 c) r+o d) I l2l I e) soo+o il 3426s7 308, 490 . 25. 29. 30 Answers to Selected Problems 429 Section 2. Section 2.) no sorution il ,x'ZI cb1 y =-zi^\13:::il;4,-"44r. 39 French francs, I I Swiss francs 3. 17 apples, 23 oranges 8-'lPt 0f = ( ) no solution nickels, dimes, quarters first-class, l9 second-class) no solution l-n l : 1 0 . I l . t2. 1 3 . Section 3.) b c) o d) 12 d + f .+ 12. a) 4 o'clock b) 6 o'clock c) 4 o'clock 430 Section 3.2 A nsw ers to S elec ted probl ems ( 7,0), (7,2).(7,4),(l.0 (mod g) no soluti |
on Section 3.) b) no soluti on c) (x ,y) = (0,2) , (1,3) , (2,4 ), (:, 0) or (2,5), (3,2), (4,6), (5,3) , (6,0) (mod 7) b) no sol uti on . a) x = 28 (mod 30) b) no solution ) x : 44 (mod g40) e) no solution Section 3. rl ls rl fr rJ ls l ) l5 ) An swers to Selected Problems 4 3 1 Section 4.. a) by 3, not by 9 b) by 3, and 9 c) by 3' and 9 d) not bv 3 4. a) no b) Yes c) no d) no 5. a) those with their number of digits divisible by 3, and by 9 b) those with an even number of digits c) those with their numbcr of digits divisible by 6 (same ior 7 and for 13) d. a) incorrect b) incorrect c) passes casting out nines check d) no' for example part (c) is incorrect, but passes check Section 4.2 2. a) Friday e) Saturday i) Monday d) Thursday c) Monday b) Friday f) Saturday g) Tuesday h) Thursday k) Friday j) Sunday l) Wednesday Section 4. ') 6 1 I b\,c J . 4 5 3 ,4 t b)' () ') t l' - - l ' ) l - l -- ----- 1 3 ,| b)'c --vc ) 3 , Section 4. 432 A nsw ers to S elec ted probl ems Section 5. ) Section 5. Section 5. ) : Section 6.1 il mod 4) factor; n has at reast two odd prime factors; or n has a prime factor . Section 6.. perfect squares 4' those positive integers that have only even powers of odd primes in their prime- i l 2 s 6 power factorization ' a) primes b) squares of primes c) products to two distinct primes or cubes of primes Section 6. An swers to Selected Problems 433 ), c) Prime Section 7. * bc 'r d (mod 26 Section 7. . RL OQ NZ OF XM CQ KE QI VD AZ ] 124 25 ) . digraphic Hill cipher with enciphering matrix Itj 163 ol ol 0 l r l I' l r l Section 7. Section 7. . EAT CHO COLA TE CAKE 434 A nsw ers to S elec ted probl ems 5' a) 0371 0354 0858 0858 0087 1359 0354 0000 0087 1543 I 7g7 053s b) 001 g 0977 0274 0872 082r 0073 084s 07400000 0008 0r48 0803 04r5 ffi8 #l 3l1i'u* 0000 0734 0152 0647 0972 6' d 0042 0056 0481 0481 0763 0000 0051 0000 0294 0262 0995 0495 05:|' ag72 . 0872 I 152 1537 0169 00 00 0972 l5l5 0937 1297 1208 2273 l5l5 00 00 Section 7.5 l. a) yes b) no c) yes d) no . 6242382306332274 g. (44,37,7 4,7 2,50,24 Section 7, Section 8.. il 23.89 18. d 2209 Section 8.. il q b) the modulus is not prime 6 . 1 i l. b Section 8. ) : Section 8. . . , Section 8.. 120, 240 e) no soluti on I 6 8. 252. 504 e) t go f) 38808 0 g ) 8o+ o h) I 254 I l 328000 Section 8. : Section 8.. a) use sPread s : 3 b) Section 9.1 e) 30 i) 20 436 Section 9.2 Answers to Setected problems Section 9. Section l0.l 6. a) .lb) .ar6 c) .92nr6 d) .5 e) xOq i. a) (:s)g b) (.2)s c) (.r+o:), ai'f.'i6 ) 10. b :2s'3s'5"7"', where s1,.92,s3, and sa are nonnegative 12. a) l,o b) 2,0 c) 1,4 d) 2,1 e) l, 365 f) 2.4 343 9 0 70 20 'i' 6'T';'t't't';,r,7,T';, ;';,;';,;,; f) .000999 e) (.05 2)6 f) (.02 721350564)R integers, not a1 zero Section 10./2 d s/3 h) 8/ Section 10. 99532 , , + Section 10.4 l. ") IU,t,t,+1 b) t3;:,61 c) ta;l":,r.sl a) to;FrZt . a) (z: +.,/Til/rc b) (-l +,/+sl/z c) (s + .,Fazlto 4. b) [ lo;20] , 117:frl, I4t:il) Answers to Selected Problems 437 5. c) [q;j,J8], tt o:z2o|lte;Tt4I?q,2,+t1 6. d to:ffil, 17 :7,t41 l6;l,t 5,1,321 Section I l. ; 21,28,35, 35,12,37 ; I 5,36,39; 24,32,40 8 . x : ,: ' - (2^2-nz),! - - : r,, : +Q.m2+n2) where m and n a(e positive integers, *>it,li, and n is even 9. , - I l-{^z-3n2),y : mn,, - * ) Section 11.3 , , r | ) ? \ f,(^2+3n2) r - - ^ ^ where m and n are positive integers 52Q, y : 273 ; x : 4620799, y : 829920; x : 427035667968011. x : 6239'765965'120528801, ! : 19892016576262330040 Bibliography BOOKS Number Theory l ' w. W. Adams and L. J. Goldstein, Intoduction Prentice-Hall, Englewood Cliffs, New Jersey, 1g76. to Number Theory, 2. G. E. Andrews, Number Theory, w. B. Saunders, philadelphia, lg7l. 3. T. A. Apostol, Introduction Verlag, New York, 1976. to Analytic Number Theory, Springer- 4. R' G. Archibald, An Introduction to the Theory of Numbers, Merrill, Columbus, Ohio, 1970. 5. I. A. Barnett, Elements of Number Theory, prindle, weber, and Schmidt, Boston, 1969. 6. A. H. Beiler, Recreations in the Theory of Numbers, 2nd ed., Dover, New York, 1966. 7. E. D. Bolker, Elementary Number Theory, Benjamin, New york, 1970. 8. Z. I. Borevich and I. R. Shafarevich, Number Theory, Academic press, New York, 1966. 9. D. M. Burton, Elementary Number Theory, Allyn and Bacon, Boston, t 9 7 6 . 10. R. D. Carmichael, The Theory of Numbers and Diophantine Analysis, Dover, New York, 1959 (reprint of th e ori ginal 1914 and l 9l5 edi tio ns). I l. H. Davenport, The Higher Arithmetic, 5th ed., Cambridge University Press, Cambridge, 1982. 12. L. E, Dickson, History of the Theory of Numbers, three volumes, chelsea, New York, 1952 (reprint of the l9l9 original). 13. L. E. Dickson, Introduction to the Theory of Numbers, Dover, New York 1957 (reprint of the original 1929 edition). 4 3 8 Bi bl iograPhY 439 1 4 . H. M. Edwa rds, Ferm at's Last Theorem, Spri nger-v erlag , N ew Y ork' 1911 . 1 5 . 1 6 . t'7 . 1 8 . 1 9 . 20. Grossw ald,, Topics from the Theory of Numbers, 2nd ed., Birkhausero . K. Guy, l.)nsolved Problems in l,{umber Theory, springer-verlag. H. Hardy and E. M. Wright, An In tro duct ion to the Theory of 1,,{umbers, 5th ed., Oxford University Press, Oxford, 1919' L. Hua, Introduction to Number Theory, Springer-verlag, New York l 9 8 2 . K. Ireland and M. L Rosen, A Classical Introduction to Modern IYumber Theory, Springer-Verl ag, New Yor k, 1982' E. Landau, Ele mentary Num ber Theory, Chel sea, N ew Y ork, 1958' W. J. LeVeque, Fundamentals of Number Theory, Addispn-Wesley, Reading, Massachusetts, 1977 . w. J. LeVeque, Reviews in Number TheOry, six volumes, American Mathemat ical Socie ty, Wash ington, D.C., 1974' C. T. Long, Elementary Introduction to Number Theory, 2nd ed., Heath, Lexington, Massachusetts , 1972. 25. G. B. Matthews, Theory of Numbers, Chelsea, New York (no date)' 26. I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers, 4th ed., Wiley, New York, 1980 . 2l. O. Ore, An Invitation to Number Theory, Random House, New York' t 9 6 7 . 28. O. Ore, Number Theory and its History, McGraw-Hill, New York, I 9 4 8 . 29. A. J. Pettofrezzo and D. R. Byrkit, Elements of Number Theory, Prentice-Hall, Englewood Cliffs, New Jersey, 1970' 30. H. Rademacher, Lectures on Elementary [t{umber Theory, Blaisdell, New York 1964, reprint Krieger, 1977 . 31. P. Ribenboim, 1-J Lectures on Fermat's Last Theorem, Springer-Verlag, New York, 1919 . 40. 440 B ibl iogr aphy 32. J. Roberts, Elementary Number Theory, MIT press, cambridge, Massachusetts, 1977. 33. D. shanks, solved and unsolved problems in Number Theory, 2nd ed., Chelsea, New york. 197g. J. E. Shockley, Introduction to Number Theory, Holt, Rinehart, and Winston, 1967. w. Sierpifski, Elementary Theory of Numbers, polski Akademic Nauk, Warsaw, 1964. w. Sierpifiski, A selection of problems in the Theory of Numbers, Pergammon Press, New york, 1964. w. Sierpirlski, 250 problems in Elementory Number Theory, polish Scientific Publishers, Warsaw, 1g70. H. M. Stark, An Introduction to Number Theory, Markham, chicago, 1970; reprint MIT press, cambridge, Massachuseits, r9ig. B. M. Stewart, The Theory of Numbers, 2nd, ed., Macmiilan, New York, 1964. J. v. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, New York. lg3g. 4l' C' Vanden Eyden, Number Theory, International Textbook, Scranton, Pennsylvania, 1970. 42. I. M. vinogradov. Elements of Number Theory, Dover, New york, t954. Number Theory with Computer Science 4 3 . 44. 4 5 . A. M. Kirch, Elementary Number Theory: A computer Approach, Intext, New York, 1974. D. G. Malm, A computer Laboratory Manual for Number Theory, COMPress, Wentworth, New Hampshire, 1979. D. D. spencer, computers in Number Theory, computer science press, Rockville, Maryland, 1982. Bi bli ograPhY CryptographY 441 46. B. Boswo rth, codes, ciphers, and computers, Hayden, Rochelle Park, New JerseY, 1982. 47. D. E. R. Denning, Cryptography and Data Security, Addison.Wesley, Reading, Massachusetts, 1982' 48. w. F. Friedman, Elements of Cryptanalysis, Aegean Park Press, Laguna Hills, California, 1978 computer Engineering, Univ. calif. Santa Barbara, 1982. 50. D. Kahn, The Codebreakers, the Story of Secret Writing' Macmillan . New York' 1967. A. G. Konheim, Cryptography: A Primer, Wiley' New York' 1981' S. Kullback, s/atis tical Methods in cryptanalysis, Aegean Park Press, Laguna Hills, California, 1976. C. H. Meyer and S. M. Matyas' Cryptography: A New Dimension Computer Data Security, Wiley, New York, 1982' A. sinkov, Elementary cryptanalysis, Mathematical Association America, Washington, D.C., 1966' of t n Computer Science 55. K. Hwan g, Computer Arithmetic: Principles, Architecture and Design' WileY, New York, 1979. 56. D. E. Knuth, Art volume Algorithms Massachusetts, l98l . 'of computer Programming: €d., semi-Numertcal Addison wesley, Reading 2, 2nd 57. D. E. Knuth, Art of computer Programming: sorting and searching, volume 3, Addison-wesley, Reading, Massachusetts, 1973. 58. L. Kronsj o, Algorithms: Their complexity and Efficiency, wiley, New York, 1979. 59. N. S. Szab5 and R. J. Tanaka, Residue Arithmetic and its Applications to Computer Technology, McGraw-Hill' 1967' 442 General Bibliography 60. 6 1 . 6 2 . H. Anton, Elementary Linear Algebra, 3rd ed., Wiley, New York, 1981. E. Landau, Foundations of Analysfs, 2nd ed., Chelsea, New York, 1960. W. Rudin, Principles of Mathematical Analysis, 2nd ed., McGraw-Hill, New York 1964. ARTICLES Numben Theory 63. Ll M. Adleman, C. Pomerancq and R. S. Rumely, "On distinguishing prlime numbers from composite numbers," Annals of Mathematics . 64. J. Ewing, t 286243-l is prime," The Mathematical Intelligencer, Volume . 65. J.lE. Freund, "Round Robin Mathematicso" American Mathematical tullonthly, Volume 63 (1 956), ll2-114. 66. R. K. Guy, "How the Ftfth Manitoba Coderence on Numerical Mathematics, Utilitas, Winnepeg, Manitoba, 197 5, 49-89. factor a number" Proceedings of to I ot. l 6 8 . 69. 7 0 . 7 1 . A. K. Head , "Mul tiplication mod ulo n, " B IT, V olume 20 (tgS O), 115I 1 6 . P |
. Hagis, Jr., "Sketch of a proof that an odd perfect number relatively factors," Mathematics of least eleven prime prime Computations, Volume 46 0983), 399-404. to 3 has at J. C. Lagarias and A. M. Odlyzko, "New algorithms for computing n(ff)," Bell Laboratories Technical Memorandum TM-82-1 I 218-57. H. P. Lawther, Jr., "An application of number theory to the splicing of telephone cables," American Mathematical Monthly,Yolume 42 (tggS), 8 l - 9 1 . H.1 W. Lenstra, Jr., "Primality testing," Studieweek Getaltheorie en Co[nputers, 1-5 September 1980, Stichting Mathematisch Centrum, Arfrsterdam. Holland. Bi bl iograPhY 443 72. G. L. Miller, "Riemann's hypothesis and tests for primality "' Proceedings of thq seventh Annual Ac:M symposium on the Theory of computing, 234-239. C. pomerance, "Recent developments in primality In tell igencer, volume 3 (lg g l ), 97-105 . urrir*"rical i ' \lq. C. pomerance, "The search for primes," Scientific American' Volume testing"' The 1,7315. ,,probabilistic algorithms for lesting primality," Journal of M. o. Rabin, Number Theory, Vol ume 12 0980)' 128-138' ./6. R. Rumely, ,,Recent advances in primality testing," Notices of the American Mathematical Sociely, Volume 30 (1983), 4,75-47,7, 77. D. Slowinski, "searching for the 2'7th Mersenne prime"' Journal of 7 8 . 7 9 . Recreational Mathematics, Vol ume I I (19 18/9), 258-261 ' R. Solovay and V. Strassen' "A fast Monte Carlo SIAM Journal for Computing, Volume 6 09ll)' . C. Williams, "The influence of computers in number theory," Computers and Mathematics ' test for PrimalitY," 84-85 and erratum, the develoPment of with APPlications, g0. H. c. williams, "Primality testing on a computer", Ars combinatorica ' CryptograPhY 81. L. M. Adleman, "A subexponential algorithm for the discrete logarithm problem with applications to cryptogiaphy," Proceedings of the 2ath Annual Sy*:,porium on the Fonia'tioit of Computer Science' 1979' 55' 60. g2. M. Blum, "coin-flipping by telephone - a protocol for solving impossible problems," IEEE Proceedings' Spring Compcon" 133-137' 83. w. Diffie and M. Hellman, "New directions in cryptography"' IEEE Transactions on Idormation Theory, Volume 22 (l976),644-655' g4. D. R. Floyd, "Annotated bibliographical in conventional and public key cryptograpnr,. Cryptologia, Volume 7 (1983) ' 12'24' 444 ibl iogr aphy J. Gordon, "Use of intractable problems Privacy, Volume 2 (19g0), l7g-fg4. rn cryptography," Information M. E. Hellman, "The mathematics Scientffic American, Volume 241 (1979) L. S. Hill, "Concerning certain linear cryptography," American Mathematical l 3 5 - 1 5 4 . of public-key cryptography," t 4 6 - t 5 7 . transformation apparatus of Monthl y, V olume 3g (1931) . 8 8 . 8 9 . 90. Lempel, "cryptology in transition," computing surveys, volume ll Q979), 285-303. in Idormatioi R. J. Lipton, "How to cheat at mental poker,,, and ,,An improved power encryption method," unpublished reports, Department of computer Science, University of California, Berklir'y, 1979. R. c. Merkle and M. E. Hellman, "Hiding information and signatures in trapdoor knapsacks," IEEE Transactiins Theory, Volume 24 (19 79), 525-530. s. Pohlig and M. Hellman, "An improved argorithm for computing logarithms over GF(p) and its .ryptog.upt i. significance,,' IEEE Transactions on Information Theory, volume 24 (rgj"$, roC_iio. M. o. Rabin,. "Digitalized signatures and public-key functions as intractable as factorization," MIT Laboratory for computer science Technical Report LCS/TR-212, cambridge, Massachusetts, rg7g. R. L. Rivest, A. Shamir, and L. M. Adleman, "A method for obtaining digital signatures a1d public-key cryptosystems," communications of the ACM, Volume 2t (1979), tZO-126. A. shamir, uA polynomial time algorithm for breaking the basic Merkle-Hellman cryptosystem," proceedings of the 2ird Annual symposium of the Foundations of computeiscie,nce, r45-r52. A. Shamir, "How to share a secret," communications of the ACM, Volume 22 0979 ), 612-6t3. A. Shamir, R. L. Rivest, and L. M. Adleman, "Mentar poker,,, The Mathematical Gardner, ed. D. A. Klarner, wadsworth International, Belmont, California, 198 l, 37-43. List of SYmbols Summation, 5 Factorial, 8 Product, 9 Binomial coeficient, l0 7 2 5 3 (of n integers), Divides, 19 Does not divide, 19 Greatest integer, 20 Base b exPansion, 27 Computer word size, 33 Big-O notation, 38 Number of Primes, 47 Greatest common divisor, Greatest common divisor Fibonacci number, 60 Least common multiPle, Minimum, 72 Maximum, 72 Exactly divide, 76 Least common multiple (of n integers), Fermat number, 81 Congruent, 9l Not congruent, 91 Inverse, 104 Congruent (matrices), I l9 Inverse (of matrix), l2l Identity lnatrix, l2l Adjoint, 122 Hashing function, 141 Euler's phi-function, l6l 5 5 7 7 t 2 n t II l*) ov) ,r(.x min(xy) ma x(x,y) p ' l l n ta 1,a2, .. ., anl F(mod nr adj Ca* ) ord.a ind,a I ( n ) X6(n ) |t I s-l l p ) r ) lLl Fn Ck : Pr/qr [ ;a r,...,o * - ,,ffifr|' Q , List of Symbols Summation over divisors, 170 Dirichlet product, 172 Miibius function. 173 Sum of divisors function, I74 Number of divisors function , 17 s Mersenne number. l g2 Enciphering transformation, ZI2 Deciphering transformation, 212 Order of a modulo m. Z3Z Index of a to the base r, 252 Minimal universal exponent, 269 Maximal +l - exponent, 2g0 Legendre symbol, 289 Jacobi symbol, 314 Base D expansion, 341 Periodic base 6 expansion, 343 Farey series of order n, 349 Finite simple continued fraction, 351 Convergent of a continued fraction, 354 Infinite simple continued fraction, 362 Periodic continued fraction, 3i4 Conjugate, 377 lndex Absolute least residues, 93 Abundant integer, 185 Additive function, 17 4 Affine transformation, l9l division, 19 Euclidean, 58 for addition, 33 for division, 3'7,41 for matrix multiPlication, 43 for modular exPonentiation, 97 for modular multiPlication, 100 for multiplication, 35,39 for subtraction, 34 least-remainder, 67 Amicable pair, 185 Approximation, best rational, 37 | by rationals, 369 Arithmetic function, 166,418 Arithmetic, fundamental theorem of, 2,69 Arithmetic progression, primes in Balanced ternary exPansion, 30 Base, 27 Base D expansion, 27,341 Best rational aPProximation, 371 Big-O notation, 38,39 Binary notation, 27 Binomial coeffficient, l0 Binomial theorem, 12 Biorhythms, I l4 Bit operation, 38 Bits, 27 Block cipher, 198 Borrow, 35 Caesar. Julius, 189 Caesar cipher, 189 Calendar, 134 International Fixed, 138 Cantor expansion, 30 Card shuffiing, 152 Carmichael number, 155'272 Carry, 34 Casting out nines, 134 Character ciPher, 189 Chinese, ancient, 2,107, Chinese remainder theorem, 107 Cicada, periodic, 5' block, 198 Caesar, 189 character, 189 digraphic, 198 exponentiation, 205 H i l l , 1 9 8 iterated knapsack, 224 knapsack, 221 monographic, 189 polygraphic, 198 product, 19'l public-key, 2,212 Rabin, 215 R S A , 2 1 2 substitution, 189 transposition, 204 Vigndre, 197 Ciphertext, 188 Clustering, 142 Coconut problem, 101 Coefficients, binomial, 10 Coin flipping, 298 Collatz conjecture, 24 Collision. 142 Common key, 208 Common ratio, 5 Complete system of residues, 93 Completely additive function, 174 448 Index Completely multiplicative function, 166 Composite, 1,45 Computational complexity, 3g of addition, 39 of Euclidean algorithm, 62 of matrix multiplication, 43 of multiplication, 39 of subtraction, 39 Computer arithmetic, 33,109 Computer files, 141,227 Computer word size, 33,109 Congruence, 2,gl linear, 102 of matrices, I l9 Congruence class. 92 Conjecture, Ccllatz, 24 Goldbach, 50 Conjugate, 377 Continued fraction, 350 infinite, 362 periodic, 374 425 purely periodic, 3g3 Convergent, 354 Coversion of bases, Zg Covering set of congruences, I l5 Cryptography, 188 Cryptology, 188 Cubic residue, 262 Database, 227 Day of the week, 134 Decimal notation, 27 Deciphering, 186 Deciphering key, 213 Decryption, 188 Deficient integer, 185 Descent, proof by, 398 Diabolic matrix, 127 Digraphic cipher, 198 Diophantine equations, 86,391 linear, 86 Diophantus, 86 Dirichlet, G. Lejeune, 74 Dirichlet product, 172 Dirichlet's theorem on primes in arithmetic progression, 74 Divisibilitytests, lZ9 Division algorithm, l9 Divisor, l8 Double hashing, 143 Draim factorization, g4 Duodecimal notation, 44 Electronic poker, 209,304 Enciphering, 188 Encryption, 188 Equation, diophantine, 86 Pell's, 404 Eratosthenes, I Eratosthenes, sieve of, 2,46 Euclid, I Euclidean algorithm, 5g Euler. L.. I Eul er phi -funct ion, l 6l,l 67 Euler pseudoprime, 325 Euler's criterion. 290 Euler's factorization method, g5 Euler's theorem, 161 Exactly divide. i6 Expansion, base b, 27 Cantor, 30 continued fraction, 350 periodic base b, 343 periodic continued function, 374 terminating, 341 t l-exponent, 280 Exponentiation cipher, 205 Factor, l8 Factor table, 4ll Factorial function, 8 Factorization, 69,79 Draim, 84 Euler, 85 l n d e x Fermat. 80 prime, 68 prime-power, 69 speed of, 80,215 Faltings, G., 400 Farey series, 349 Fermat, P. de, 1,397 Fermat factorization, 80 Fermat prime, 8l Fermat quotient, 152 Fermat's last theorem, 398 Fermat's little theorem, 148 Fibonacci, 60 Fibonacci numbers, 60 generalized, 68 Fibonacci pseudo-random number generator, 219 Frequencies, of letters, 193 of digraphs, 202 of polygraphs, 203 Function. additive, 17 4 arithmetic, 166 completely additive, l7 4 completely multiPlicative, 166 Euler phi, 161 factorial, 8 greatest integer, 20 Liouville's , 17 4 Mobius, l'73 multiplicative, 166 number of divisors. 175 sum of divisors. 174 Fundamental Theorem of Arithmetic, 69 Game of Euclid, 67 Gauss, C. G., 2,47 Gauss' generalization of Wilson's theorem, 152 Gauss'lemma, 293 Generalized Riemann hypothesis, 158 Generalized Fibonacci numbers, 68 Geometric progression, 5 Goldbach, C., 50 Goldbach's conjecture, 50 Greatest common divisor, 53 Greatest integer function, 20 Greeks, ancient, 2 Hadamard, J., 48 Hanoi, tower of, l' double, 143 quadratic, 304 Hashing function, 141 Hexadecimal notation, 27 Hilbert prime, 76 Hill ciphe |
r, 198 l2l Identity matrix modulo z, Inclusion-exclusion, principle of, 17,51 Incongruent, 9l Index of an integer, 252,421 Index of summation, 5 Index system, 262 Induction, mathematical, 4 Infinite simple continued fraction, Infinitude of primes, 45,82 Integer, 362 deficient, 185 palindromic, 133 powerful, 16 t73 Inverse of an arithmetic function, Inverse modulo lrr, 104 Inverse of a matrix modulo nr, l2I Involutory matrix, 126,244 Irrational number, 336,36'l Jacobi symbol, 314 Kaprekar constant, 3l K e y , l 4 l common, 208 deciphering, 213 enciphering, 212 mastero 228 public, 212 shared, 208 450 I n d e x Knapsack cipher, 221 Knapsack problem, 219 k-perfect number, 186 Kronecker symbol, 324 k th power residue, 256 Lagrange, J., 147 Lagrange interpolation, 242 Lagrange's theorem (on continued functions), 378 Lagrange's theorem (on polynomial congruences) , 219 Lam6, G., 62 Lam6's theorem, 62 Law of quadratic reciprocity, 297,314 Least common multiple, 72 Least nonnegative residue, 93 Least-remainder algorithm, 67 Legendre symbol, 289 Lemma, Gauss'. 293 Linear combination, 54 greatest common divisor as a, 54,63 Linear congruence, 102 Linear congruential method, 275 Liouville's function, 114 Logarithms modulo p, 207 Lowest terms, 336 Lucas-Lehmer test, 183 Lucky numbers, 52 Magic square, 127 Master key, 228 Mathematical induction. 4 Matrix, involutory, 126 Matrix multiplication, 43 Maximal t1-exponent, 280 Mayans, 1,25 Mersenne, M., 182 Mersenne number. 182 Mersenne prime, 182 Method of infinite descent, 398 Middle-square method, 275 Miller's test, 156 Minimal universal exponent, 269 Mobius function, 173 Mobius inversion formula, 173 Modular exponentiation, 97 algorithm for, 97 Monographic cipher, 189 Monkeys, l0l Multiple precision, 33 Multiplication, 35,39 matrix, 43 Multiplicative function, 166 Multiplicative knapsack problem, 226 Mutually relatively prime, 56 N i m . 3 l Notation, big-O, 38 binary, 27 decimal, 27 duodecimal, 44 hexadecimal, 27 octal, 27 product Number, Carmichael, 155,2'12 Fermat, 8l Fibonacci, 60 generalized Fibonacci, 68 irrational. 336 k-perfect, 186 lucky, 52 Mersenne, 182 perfect, 180 rational, 336 superperfect, 186 Number of divisors function. 17 5 Octal notation, 27 Operation, bit, 38 Order of an integer, 232 Pairwise relatively prime, 56 Palindromic integer, 133 Partial remainder, 37 Partial quotient, 351 Pascal's triangle, 12 Pell's equation, 404 Pepin's test, 3l I Perfect number, 180 Period, l n d e x 4 5 1 of a base b exPansion, 343 of a continued fraction, 37 4 Periodic base b exPansion, 343 Periodic cicada, 5'l Periodic continued fraction, 374 Poker. 209,304 Polygraphic ciPher, 198 Powerful integer, 76 Prepperiod, 343 Primality test, 153,263 probabilistic, 158,334 Hilbert, 76 in arithmetic Progressions, 74 infinitude of, 45 Mersenne, 182 Wilson, 152 Prime number theorem, 47 Prime-power factorization, 69 Primitive root, 234,243 42O Primitive Pythagorean triPle, 391 Principle of inclusion-exclusion, l7 Principle of mathematical induction, second, 8 Probabilistic primality test, 158'334 Probing sequence, 143 Problem, knapsack, 219 multiplicative knaPsack, 226 Product, Dirichlet, 172 Product ciPher, 192 Property, reflexive, 92 symmetric, 92 transitive, 92 well-ordering, 4 Pseudoconvergent, 37 4 Pseudoprime, 2,153 Euler, 325 strong, 157 Pseudo-random numbers, 27 5 Pseudo-random number generator' Fibonacci, 279 linear congruential, 27 5 middle'square, 27 5 pure multiPlicative, 277 Purely periodic continued fraction' 383 Pythagoras, 1 Pythagorean triPle, 391 Pythagorean theorem, 391 Quadratic hashing, 304 Quadratic irrational, 37 5 Quadratic nonresidue, 288 Quadratic reciProcitY law, 297,304 Quadratic residue, 288 Quotient, l9 Fermat, l52 Rabbits, 68 Rabin's probabilistic PrimalitY Rational number, 336 Read subkeY, 227 Recursive definition, 8 Reduced residue system, 162 Reduced quadratic irrational, 384 Reflexive proPertY, 92 Regular polygon, constructabilitY, 83 Relatively prime, 53 mutually, 56 pairwise, 56 Residue, cubic, 262 k th power , 256 least nonnegative, 93 quadratic, 288 Residues, absolute least, 93 complete sYstem of, 93 reduced, 162 Root of a polynomial modulo rn, 238 Round-robin tournament, 139 RSA cipher system, 212,274 Second princiPle of 4s2 l n d e x Fermat's last, 398 Fermat's little. 148 Lagrange's (on continued fractions), 378 Lagrange's (on polynomial congruences), 239 Lam6's, 62 Wilson's, 147 Threshold scheme, 228,243 Tower of Hanoi. 17 Transitive property, Transposition cipher, Triangle, 204 92 Pascal's, l2 Pythogrean, 391 Twin primes, 50 Universal exponent, 269 Vall6e-Poussin, C. de la, 48 Vignrire ciphers, 197 Weights, problem of, 30 Well-ordering property, 4 Wilson, J., 147 Wilson prime, 152 Wilson's theorem, 147 Gauss' generalization of, 152 Write subkey, 22'l mathematical induction. 8 Seed, 276 Shadows, 228 Shift transformation. l9l Shifting, 35 Sieve of Eratosthenes, 2,46 Signature, 216 Signed message, 216,218 Solovay-Strassen probabilistic primality test, 334 Splicing of telephone cables, 284 Spread of a splicing scheme, 284 Square-free integer, 7 5 Strong pseudoprime, 157 Subkey, read, 227 write, 227 Substitution cipher, 189 Succinct certificate of primality, 266 Sum of divisors function, 174 Summation notation, 5 Super-increasingsequence, 22O Superperfect number, 186 Symbol, Jacobi. 314 Kronecker, 324 Legendre, 289 Symmetric property, 92 System of residues, complete, 93 reduced, 162 System of congruences, 107,1 l6 Telephone cables, 284 Terminating expansion, 341 Test, divisibility, 129 Lucas-Lehmer, 183 Miller's, 156 Pepin's, 3l I primality, 153,263 probalistic primality, 158,334 Theorem, binomial, 12 Chinese remainder. 107 Dirichlet's, 74 Eulerns, l6l |
ons, the likelihood ratio test is still a good idea. In fact, you will show in the example sheets that for a discrete distribution, as long as a likelihood ratio test of exactly size α exists, the same result holds. Example. Suppose X1, · · · , Xn are iid N (µ, σ2 0 is known. We want to find the best size α test of H0 : µ = µ0 against H1 : µ = µ1, where µ0 and µ1 0), where σ2 21 2 Hypothesis testing IB Statistics are known fixed values with µ1 > µ0. Then Λx(H0; H1) = (2πσ2 0)−n/2 exp (2πσ2 0)−n/2 exp µ1 − µ0 σ2 0 n¯x + = exp − 1 2σ2 0 − 1 2σ2 0 n(µ2 (xi − µ1)2 (xi − µ0)2 0 − µ2 1) 2σ2 0 . This is an increasing function of ¯x, so for any k, Λx > k ⇔ ¯x > c for some c. Hence we reject H0 if ¯x > c, where c is chosen such that P( ¯X > c | H0) = α. Under H0, ¯X ∼ N (µ0, σ2 0/n), so Z = Since ¯x > c ⇔ z > c′ for some c′, the size α test rejects H0 if √ n( ¯X − µ0)/σ0 ∼ N (0, 1). √ z = n(¯x − µ0) σ0 > zα. For example, suppose µ0 = 5, µ1 = 6, σ0 = 1, α = 0.05, n = 4 and x = (5.1, 5.5, 4.9, 5.3). So ¯x = 5.2. From tables, z0.05 = 1.645. We have z = 0.4 and this is less than 1.645. So x is not in the rejection region. We do not reject H0 at the 5% level and say that the data are consistent with H0. Note that this does not mean that we accept H0. While we don’t have sufficient reason to believe it is false, we also don’t have sufficient reason to believe it is true. This is called a z-test. In this example, LR tests reject H0 if z > k for some constant k. The size of such a test is α = P(Z > k | H0) = 1 − Φ(k), and is decreasing as k increasing. Our observed value z will be in the rejected region iff z > k ⇔ α > p∗ = P(Z > z | H0). Definition (p-value). The quantity p∗ is called the p-value of our observed data x. For the example above, z = 0.4 and so p∗ = 1 − Φ(0.4) = 0.3446. In general, the p-value is sometimes called the “observed significance level” of x. This is the probability under H0 of seeing data that is “more extreme” than our observed data x. Extreme observations are viewed as providing evidence against H0. 2.2 Composite hypotheses For composite hypotheses like H : θ ≥ 0, the error probabilities do not have a single value. We define Definition (Power function). The power function is W (θ) = P(X ∈ C | θ) = P(reject H0 | θ). We want W (θ) to be small on H0 and large on H1. Definition (Size). The size of the test is α = sup θ∈Θ0 W (θ). 22 2 Hypothesis testing IB Statistics This is the worst possible size we can get. For θ ∈ Θ1, 1 − W (θ) = P(Type II error | θ). Sometimes the Neyman-Pearson theory can be extended to one-sided alter- natives. For example, in the previous example, we have shown that the most powerful size α test of H0 : µ = µ0 versus H1 : µ = µ1 (where µ1 > µ0) is given by √ C = x : n(¯x − µ0) σ0 > zα . The critical region depends on µ0, n, σ0, α, and the fact that µ1 > µ0. It does not depend on the particular value of µ1. This test is then uniformly the most powerful size α for testing H0 : µ = µ0 against H1 : µ > µ0. Definition (Uniformly most powerful test). A test specified by a critical region C is uniformly most powerful (UMP) size α test for test H0 : θ ∈ Θ0 against H1 : θ ∈ Θ1 if (i) supθ∈Θ0 W (θ) = α. (ii) For any other test C ∗ with size ≤ α and with power function W ∗, we have W (θ) ≥ W ∗(θ) for all θ ∈ Θ1. Note that these may not exist. However, the likelihood ratio test often works. Example. Suppose X1, · · · , Xn are iid N (µ, σ2 wish to test H0 : µ ≤ µ0 against H1 : µ > µ0. 0) where σ0 is known, and we First consider testing H ′ 0 : µ = µ0 against H ′ Neyman-Pearson test of size α of H ′ 0 against H ′ √ n(¯x − µ0) σ0 C = x : > zα . 1 : µ = µ1, where µ1 > µ0. The 1 has We show that C is in fact UMP for the composite hypotheses H0 against H1. For µ ∈ R, the power function is W (µ) = Pµ(reject H0) √ √ = Pµ = Pµ n( ¯X − µ0) σ0 n( ¯X − µ) σ0 √ = 1 − Φ zα + > zα > zα + n(µ0 − µ) σ0 n(µ0 − µ) σ0 √ To show this is UMP, we know that W (µ0) = α (by plugging in). W (µ) is an increasing function of µ. So W (µ) = α. sup µ≤µ0 So the first condition is satisfied. 0 vs H ′ For the second condition, observe that for any µ > µ0, the Neyman-Pearson 1 has critical region C. Let C ∗ and W ∗ belong to any size α test of H ′ other test of H0 vs H1 of size ≤ α. Then C ∗ can be regarded as a test of H ′ 0 vs H ′ 1 of size ≤ α, and the Neyman-Pearson lemma says that W ∗(µ1) ≤ W (µ1). This holds for all µ1 > µ0. So the condition is satisfied and it is UMP. 23 2 Hypothesis testing IB Statistics We now consider likelihood ratio tests for more general situations. Definition (Likelihood of a composite hypothesis). The likelihood of a composite hypothesis H : θ ∈ Θ given data x to be Lx(H) = sup θ∈Θ f (x | θ). So far we have considered disjoint hypotheses Θ0, Θ1, but we are not interested in any specific alternative. So it is easier to take Θ1 = Θ rather than Θ \ Θ0. Then Λx(H0; H1) = Lx(H1) Lx(H0) = supθ∈Θ1 f (x | θ) supθ∈Θ0 f (x | θ) ≥ 1, with large values of Λ indicating departure from H0. Example. Suppose that X1, · · · , Xn are iid N (µ, σ2 0 known, and we wish to test H0 : µ = µ0 against H1 : µ ̸= µ0 (for given constant µ0). Here Θ0 = {µ0} and Θ = R. 0), with σ2 For the numerator, we have supΘ f (x | µ) = f (x | ˆµ), where ˆµ is the mle. We know that ˆµ = ¯x. Hence Λx(H0; H1) = (2πσ2 (2πσ2 0)−n/2 exp 0)−n/2 exp − 1 2σ2 0 − 1 2σ2 0 (xi − ¯x)2 . (xi − µ0)2 Then H0 is rejected if Λx is large. To make our lives easier, we can use the logarithm instead: 2 log Λ(H0; H1) = 1 σ2 0 (xi − µ0)2 − (xi − ¯x)2 = n σ2 0 (¯x − µ0)2. So we can reject H0 if we have √ n(¯x − µ0) σ0 > c for some c. √ We know that under H0, Z = generalised likelihood test rejects H0 if n( ¯X − µ0) σ0 ∼ N (0, 1). So the size α √ n(¯x − µ0) σ0 > zα/2. Alternatively, since n( ¯X − µ0)2 σ2 0 ∼ χ2 1, we reject H0 if n(¯x − µ0)2 σ2 0 > χ2 1(α), (check that z2 α/2 = χ2 1(α)). Note that this is a two-tailed test — i.e. we reject H0 both for high and low values of ¯x. 24 2 Hypothesis testing IB Statistics The next theorem allows us to use likelihood ratio tests even when we cannot find the exact relevant null distribution. First consider the “size” or “dimension” of our hypotheses: suppose that H0 imposes p independent restrictions on Θ. So for example, if Θ = {θ : θ = (θ1, · · · , θk)}, and we have – H0 : θi1 = a1, θi2 = a2, · · · , θip = ap; or – H0 : Aθ = b (with A p × k, b p × 1 given); or – H0 : θi = fi(φ), i = 1, · · · , k for some φ = (φ1, · · · , φk−p). We say Θ has k free parameters and Θ0 has k − p free parameters. We write |Θ0| = k − p and |Θ| = k. Theorem (Generalized likelihood ratio theorem). Suppose Θ0 ⊆ Θ1 and |Θ1| − |Θ0| = p. Let X = (X1, · · · , Xn) with all Xi iid. If H0 is true, then as n → ∞, 2 log ΛX(H0; H1) ∼ χ2 p. If H0 is not true, then 2 log Λ tends to be larger. We reject H0 if 2 log Λ > c, where c = χ2 p(α) for a test of approximately size α. We will not prove this result here. In our example above, |Θ1| − |Θ0| = 1, 1 exactly for all n in that and in this case, we saw that under H0, 2 log Λ ∼ χ2 particular case, rather than just approximately. 2.3 Tests of goodness-of-fit and independence 2.3.1 Goodness-of-fit of a fully-specified null distribution So far, we have considered relatively simple cases where we are attempting to figure out, say, the mean. However, in reality, more complicated scenarios arise. For example, we might want to know if a dice is fair, i.e. if the probability of getting each number is exactly 1 6 . Our null hypothesis would be that p1 = p2 = · · · = p6 = 1 6 , while the alternative hypothesis allows any possible values of pi. In general, suppose the observation space X is partitioned into k sets, and let pi be the probability that an observation is in set i for i = 1, · · · , k. We want to test “H0 : the pi’s arise from a fully specified model” against “H1 : the pi’s are unrestricted (apart from the obvious pi ≥ 0, pi = 1)”. Example. The following table lists the birth months of admissions to Oxford and Cambridge in 2012. Sep Oct Nov Dec 457 470 515 470 Jan Feb Mar Apr May 437 466 473 381 457 Jun 396 Jul Aug 394 384 Is this compatible with a uniform distribution over the year? Out of n independent observations, let Ni be the number of observations in ith set. So (N1, · · · , Nk) ∼ multinomial(k; p1, · · · , pk). For a generalized likelihood ratio test of H0, we need to find the maximised likelihood under H0 and H1. 25 2 Hypothesis testing IB Statistics Under H1, like(p1, · · · , pk) ∝ pn1 1 · · · pnk k . So the log likelihood is l = constant + ni log pi. We want to maximise this subject to pi = 1. Using the Lagrange multiplier, we will find that the mle is ˆpi = ni/n. Also |Θ1| = k − 1 (not k, since they must sum up to 1). Under H0, the values of pi are specified completely, say pi = ˜pi. So |Θ0| = 0. Using our formula for ˆpi, we find that 2 log Λ = 2 log ˆpn1 1 · · · ˆpnk 1 · · · ˜pnk ˜pn1 k k = 2 ni log ni n˜pi (1) Here |Θ1|−|Θ0| = k −1. So we reject H0 if 2 log Λ > χ2 size α test. k−1(α) for an approximate Under H0 (no effect of month of birth), ˜pi is the proportion of births in month i in 1993/1994 in the whole population — this is not simply proportional to the number of days in each month (or even worse, 1 12 ), as there is for example an excess of September births (the “Christmas effect”). Then 2 log Λ = 2 ni log ni n˜pi = 44.86. P(χ2 11 > 44.86) = 3 × 10−9, which is our p-value. Since this is certainly less than 0.001, we can reject H0 at the 0.1% level, or can say the result is “significant at the 0.1% level”. The traditional levels for comparison are α = 0.05, 0.01, 0.001, roughly corresponding to “evidence”, “strong evidence” and “very strong evidence”. A similar common situation has H0 : pi = pi(θ) for some parameter θ and H1 as before. Now |Θ0| is the number of independent parameters to be estimated under H0. Under H0, we find mle ˆθ by maximizing ni log pi(θ), and then 2 log Λ = 2 log nk ˆp1 n1 · · · ˆpk p1(ˆθ)n1 · · · pk(ˆθ)nk = 2 ni log . ni npi(ˆθ) (2) The degrees of freedom are k − 1 |
− |Θ0|. 2.3.2 Pearson’s chi-squared test Notice that the two log likelihoods are of the same form. In general, let oi = ni (observed number) and let ei = n ˜pi or npi(ˆθ) (expected number). Let δi = oi − ei. Then 2 log Λ = 2 oi log oi ei = 2 (ei + δi) log = 2 (ei + δi) δi ei = 2 δi + δ2 i ei − − δ2 i 2ei 26 + O(δ3 i ) 1 + δi ei δ2 i 2e2 i + O(δ3 i ) 2 Hypothesis testing IB Statistics We know that δi = 0 since ei = oi. So ≈ = δ2 i ei (oi − ei)2 ei . This is known as the Pearson’s chi-squared test. Example. Mendel crossed 556 smooth yellow male peas with wrinkled green peas. From the progeny, let (i) N1 be the number of smooth yellow peas, (ii) N2 be the number of smooth green peas, (iii) N3 be the number of wrinkled yellow peas, (iv) N4 be the number of wrinkled green peas. We wish to test the goodness of fit of the model 16 , 1 Suppose we observe (n1, n2, n3, n4) = (315, 108, 102, 31). H0 : (p1, p2, p3, p4) = 9 16 , 3 16 , 3 16 . We find (e1, e2, e3, e4) = (312.75, 104.25, 104.25, 34.75). The actual 2 log Λ = 0.618 and the approximation we had is (oi−ei)2 = 0.604. ei Here |Θ0| = 0 and |Θ1| = 4 − 1 = 3. So we refer to test statistics χ2 Since χ2 3(0.05) = 7.815, we see that neither value is significant at 5%. So there is no evidence against Mendel’s theory. In fact, the p-value is approximately P(χ2 3 > 0.6) ≈ 0.90. This is a really good fit, so good that people suspect the numbers were not genuine. 3(α). Example. In a genetics problem, each individual has one of the three possible genotypes, with probabilities p1, p2, p3. Suppose we wish to test H0 : pi = pi(θ), where p1(θ) = θ2, p2 = 2θ(1 − θ), p3(θ) = (1 − θ)2. for some θ ∈ (0, 1). We observe Ni = ni. Under H0, the mle ˆθ is found by maximising ni log pi(θ) = 2n1 log θ + n2 log(2θ(1 − θ)) + 2n3 log(1 − θ). We find that ˆθ = 2n1+n2 2n . Also, |Θ0| = 1 and |Θ1| = 2. After conducting an experiment, we can substitute pi(ˆθ) into (2), or find the corresponding Pearson’s chi-squared statistic, and refer to χ2 1. 2.3.3 Testing independence in contingency tables Definition (Contingency table). A contingency table is a table in which observations or individuals are classified according to one or more criteria. Example. 500 people with recent car changes were asked about their previous and new cars. The results are as follows: 27 2 Hypothesis testing IB Statistics New car Large Medium Small Large Medium Small 56 50 18 52 83 51 42 67 81 This is a two-way contingency table: Each person is classified according to the previous car size and new car size. Consider a two-way contingency table with r rows and c columns. For i = 1, · · · , r And j = 1, · · · , c, let pij be the probability that an individual selected from the population under consideration is classified in row i and column j. (i.e. in the (i, j) cell of the table). Let pi+ = P(in row i) and p+j = P(in column j). Then we must have p++ = i j pij = 1. Suppose a random sample of n individuals is taken, and let nij be the number of these classified in the (i, j) cell of the table. j nij and n+j = i nij. So n++ = n. Let ni+ = We have (N11, · · · , N1c, N21, · · · , Nrc) ∼ multinomial(rc; p11, · · · , p1c, p21, · · · , prc). We may be interested in testing the null hypothesis that the two classifications are independent. So we test – H0: pij = pi+p+j for all i, j, i.e. independence of columns and rows. – H1: pij are unrestricted. Of course we have the usual restrictions like p++ = 1, pij ≥ 0. Under H1, the mles are ˆpij = nij n . n and ˆp+j = n+j Under H0, the mles are ˆpi+ = ni+ n . Write oij = nij and eij = nˆpi+ ˆp+j = ni+n+j/n. Then 2 log Λ = 2 r c i=1 j=1 oij log oij eij ≈ r c i=1 j=1 (oij − eij)2 eij . using the same approximating steps for Pearson’s Chi-squared test. We have |Θ1| = rc − 1, because under H1 the pij’s sum to one. Also, i pi+ = 1 and |Θ0| = (r − 1) + (c − 1) because p1+, · · · , pr+ must satisfy p+1, · · · , p+c must satisfy j p+j = 1. So |Θ1| − |Θ0| = rc − 1 − (r − 1) − (c − 1) = (r − 1)(c − 1). Example. In our previous example, we wish to test H0: the new and previous car sizes are independent. The actual data is: New car Large Medium Small Total Large Medium Small 56 50 18 52 83 51 42 67 81 Total 124 186 190 150 120 150 500 28 2 Hypothesis testing IB Statistics while the expected values given by H0 is New car Large Medium Small Total Large 37.2 Medium 49.6 37.2 Small Total 124 55.8 74.4 55.8 186 57.0 76.0 57.0 190 150 120 150 500 Note the margins are the same. It is quite clear that they do not match well, but we can find the p value to be sure. = 36.20, and the degrees of freedom is (3 − 1)(3 − 1) = 4. (oij − eij)2 eij 4(0.05) = 9.488 and χ2 From the tables, χ2 So our observed value of 36.20 is significant at the 1% level, i.e. there is strong evidence against H0. So we conclude that the new and present car sizes are not independent. 4(0.01) = 13.28. 2.4 Tests of homogeneity, and connections to confidence intervals 2.4.1 Tests of homogeneity Example. 150 patients were randomly allocated to three groups of 50 patients each. Two groups were given a new drug at different dosage levels, and the third group received a placebo. The responses were as shown in the table below. Improved No difference Worse Total Placebo Half dose Full dose Total 18 20 25 63 17 10 13 40 15 20 12 47 50 50 50 150 Here the row totals are fixed in advance, in contrast to our last section, where the row totals are random variables. For the above, we may be interested in testing H0 : the probability of “improved” is the same for each of the three treatment groups, and so are the probabilities of “no difference” and “worse”, i.e. H0 says that we have homogeneity down the rows. In general, we have independent observations from r multinomial distributions, each of which has c categories, i.e. we observe an r × c table (nij), for i = 1, · · · , r and j = 1, · · · , c, where (Ni1, · · · , Nic) ∼ multinomial(ni+, pi1, · · · , pic) independently for each i = 1, · · · , r. We want to test H0 : p1j = p2j = · · · = prj = pj, 29 2 Hypothesis testing IB Statistics for j = 1, · · · , c, and Using H1, for any matrix of probabilities (pij), H1 : pij are unrestricted. and like((pij)) = r i=1 ni+! ni1! · · · nic! i1 · · · pnic pni1 ic , log like = constant + r c i=1 j=1 nij log pij. Using Lagrangian methods, we find that ˆpij = nij ni+ . Under H0, log like = constant + By Lagrangian methods, we have ˆpj = n+j n++ Hence n+j log pj. c j=1 . 2 log Λ = r c i=1 j=1 nij log ˆpij ˆpj = 2 r c i=1 j=1 nij log nij ni+n+j/n++ , which is the same as what we had last time, when the row totals are unrestricted! We have |Θ1| = r(c − 1) and |Θ0| = c − 1. So the degrees of freedom is r(c − 1) − (c − 1) = (r − 1)(c − 1), and under H0, 2 log Λ is approximately χ2 (r−1)(c−1). Again, it is exactly the same as what we had last time! We reject H0 if 2 log Λ > χ2 If we let oij = nij, eij = ni+n+j n++ (r−1)(c−1)(α) for an approximate size α test. , and δij = oij − eij, using the same approxi- mating steps as for Pearson’s chi-squared, we obtain 2 log Λ ≈ (oij − eij)2 eij . Example. Continuing our previous example, our data is Improved No difference Worse Total Placebo Half dose Full dose Total 18 20 25 6 3 17 10 13 40 15 20 12 47 50 50 50 150 The expected under H0 is Improved No difference Worse Total Placebo Half dose Full dose Total 21 21 21 63 15.7 15.7 15.7 47 50 50 50 150 13.3 13.3 13.3 40 30 2 Hypothesis testing IB Statistics We find 2 log Λ = 5.129, and we refer this to χ2 as the mean of χ2 chance. 4. Clearly this is not significant, 4 is 4, and is something we would expect to happen solely by We can calculate the p-value: from tables, χ2 4(0.05) = 9.488, so our observed value is not significant at 5%, and the data are consistent with H0. We conclude that there is no evidence for a difference between the drug at the given doses and the placebo. For interest, (oij − eij)2 eij = 5.173, giving the same conclusion. 2.4.2 Confidence intervals and hypothesis tests Confidence intervals or sets can be obtained by inverting hypothesis tests, and vice versa Definition (Acceptance region). The acceptance region A of a test is the complement of the critical region C. Note that when we say “acceptance”, we really mean “non-rejection”! The name is purely for historical reasons. Theorem (Duality of hypothesis tests and confidence intervals). Suppose X1, · · · , Xn have joint pdf fX(x | θ) for θ ∈ Θ. (i) Suppose that for every θ0 ∈ Θ there is a size α test of H0 : θ = θ0. Denote the acceptance region by A(θ0). Then the set I(X) = {θ : X ∈ A(θ)} is a 100(1 − α)% confidence set for θ. (ii) Suppose I(X) is a 100(1 − α)% confidence set for θ. Then A(θ0) = {X : θ0 ∈ I(X)} is an acceptance region for a size α test of H0 : θ = θ0. Intuitively, this says that “confidence intervals” and “hypothesis acceptance/rejection” are the same thing. After gathering some data X, we can produce a, say, 95% confidence interval (a, b). Then if we want to test the hypothesis H0 : θ = θ0, we simply have to check whether θ0 ∈ (a, b). On the other hand, if we have a test for H0 : θ = θ0, then the confidence interval is all θ0 in which we would accept H0 : θ = θ0. Proof. First note that θ0 ∈ I(X) iff X ∈ A(θ0). For (i), since the test is size α, we have P(accept H0 | H0 is true) = P(X ∈ A(θ0) | θ = θ0) = 1 − α. And so P(θ0 ∈ I(X) | θ = θ0) = P(X ∈ A(θ0) | θ = θ0) = 1 − α. For (ii), since I(X) is a 100(1 − α)% confidence set, we have P (θ0 ∈ I(X) | θ = θ0) = 1 − α. So P(X ∈ A(θ0) | θ = θ0) = P(θ ∈ I(X) | θ = θ0) = 1 − α. 31 2 Hypothesis testing IB Statistics Example. Suppose X1, · · · , Xn are iid N (µ, 1) random variables and we want a 95% confidence set for µ. One way is to use the theorem and find the confidence set that belongs to the hypothesis test that we found in the previous example. We find a test of size 0.05 n(¯x − µ0)| > 1.96 of H0 : µ = µ0 against H1 : µ ̸= µ0 that rejects H0 when | (where 1.96 is the upper 2.5% point of N (0, 1)). √ Then I(X) = {µ : X ∈ A(µ)} = {µ : | n( ¯X − µ)| < 1.96 |
}. So a 95% √ √ √ confidence set for µ is ( ¯X − 1.96/ n, ¯X + 1.96/ n). 2.5 Multivariate normal theory 2.5.1 Multivariate normal distribution So far, we have only worked with scalar random variables or a vector of iid random variables. In general, we can have a random (column) vector X = (X1, · · · , Xn)T , where the Xi are correlated. The mean of this vector is given by µ = E[X] = (E(X1), · · · , E(Xn))T = (µ1, · · · , µn)T . Instead of just the variance, we have the covariance matrix cov(X) = E[(X − µ)(X − µ)T ] = (cov(Xi, Xj))ij, provided they exist, of course. We can multiply the vector X by an m × n matrix A. Then we have E[AX] = Aµ, and The last one comes from cov(AX) = A cov(X)AT . (∗) cov(AX) = E[(AX − E[AX])(AX − E[AX])T ] = E[A(X − EX)(X − EX)T AT ] = AE[(X − EX)(X − EX)T ]AT . If we have two random vectors V, W, we can define the covariance cov(V, W) to be a matrix with (i, j)th element cov(Vi, Wj). Then cov(AX, BX) = A cov(X)BT . An important distribution is a multivariate normal distribution. Definition (Multivariate normal distribution). X has a multivariate normal distribution if, for every t ∈ Rn, the random variable tT X (i.e. t · X) has a normal distribution. If E[X] = µ and cov(X) = Σ, we write X ∼ Nn(µ, Σ). Note that Σ is symmetric and is positive semi-definite because by (∗), tT Σt = var(tT X) ≥ 0. So what is the pdf of a multivariate normal? And what is the moment generating function? Recall that a (univariate) normal X ∼ N (µ, σ2) has density fX (x; µ, σ2) = √ 1 2πσ exp − 1 2 (x − µ)2 σ2 , 32 2 Hypothesis testing IB Statistics with moment generating function MX (s) = E[esX ] = exp µs + . σ2s2 1 2 Hence for any t, the moment generating function of tT X is given by MtT X(s) = E[estT X] = exp tT µs + tT Σts2 . 1 2 Hence X has mgf MX(t) = E[etT X] = MtT X(1) = exp tT µ + tT Σt . 1 2 (†) Proposition. (i) If X ∼ Nn(µ, Σ), and A is an m × n matrix, then AX ∼ Nm(Aµ, AΣAT ). (ii) If X ∼ Nn(0, σ2I), then |X|2 σ2 = Instead of writing |X|2/σ2 ∼ χ2 X 2 i XT X σ2 ∼ χ2 σ2 = n. n, we often just say |X|2 ∼ σ2χ2 n. Proof. (i) See example sheet 3. (ii) Immediate from definition of χ2 n. Proposition. Let X ∼ Nn(µ, Σ). We split X up into two parts: X = where Xi is a ni × 1 column vector and n1 + n2 = n. X1 X2 , Similarly write µ1 µ2 where Σij is an ni × nj matrix. µ = , Σ = Σ11 Σ12 Σ21 Σ22 , Then (i) Xi ∼ Nni (µi, Σii) (ii) X1 and X2 are independent iff Σ12 = 0. Proof. (i) See example sheet 3. (ii) Note that by symmetry of Σ, Σ12 = 0 if and only if Σ21 = 0. From (†), MX(t) = exp(tT µ+ 1 2 tT Σt) for each t ∈ Rn. We write t = . t1 t2 Then the mgf is equal to MX(t) = exp 2 µ2 + tT 1 µ1 + tT tT 1 2 i µi + 1 From (i), we know that MXi(ti) = exp(tT MX1 (t1)MX2 (t2) for all t if and only if Σ12 = 0. tT 2 Σ22t2 + 1 Σ11t1 + 1 2 tT 1 Σ12t2 + tT 2 Σ21t1 . 1 2 2 tT i Σiiti). So MX(t) = 33 2 Hypothesis testing IB Statistics Proposition. When Σ is a positive definite, then X has pdf fX(x; µ, Σ) = 1 |Σ|2 1 √ 2π n exp − 1 2 (x − µ)T Σ−1(x − µ) . Note that Σ is always positive semi-definite. The conditions just forbid the case |Σ| = 0, since this would lead to dividing by zero. 2.5.2 Normal random samples We wish to use our knowledge about multivariate normals to study univariate normal data. In particular, we want to prove the following: Theorem (Joint distribution of ¯X and SXX ). Suppose X1, · · · , Xn are iid Xi, and SXX = (Xi − ¯X)2. Then N (µ, σ2) and ¯X = 1 n (i) ¯X ∼ N (µ, σ2/n) (ii) SXX /σ2 ∼ χ2 (iii) ¯X and SXX are independent. n−1. Proof. We can write the joint density as X ∼ Nn(µ, σ2I), where µ = (µ, µ, · · · , µ). n (the other Let A be an n × n orthogonal matrix with the first row all 1/ √ rows are not important). One possible such matrix is A = 2×1 3×2 1√ n 1√ 1√ ... 1√ n(n−1) 2×1 3×2 1√ n −1√ 1√ ... 1√ n(n−1) 1√ n 0 −2√ ... 1√ 3×2 n(n−1) 1√ n 0 0 ... 1√ n(n−1√ n 0 0 ... −(n−1) √ n(n−1) Now define Y = AX. Then Y ∼ Nn(Aµ, Aσ2IAT ) = Nn(Aµ, σ2I). We have √ Aµ = ( nµ, 0, · · · , 0)T . √ So Y1 ∼ N ( independent, since the covariance matrix is every non-diagonal term 0. nµ, σ2) and Yi ∼ N (0, σ2) for i = 2, · · · , n. Also, Y1, · · · , Yn are But from the definition of A, we have Y1 = 1 √ n n i=1 √ n ¯X. Xi = 34 2 Hypothesis testing IB Statistics √ So n ¯X ∼ N ( √ nµ, σ2), or ¯X ∼ N (µ, σ2/n). Also = YT Y − Y 2 1 = XT AT AX − Y 2 1 = XT X − n ¯X 2 n X 2 i − n ¯X 2 i=1 n (Xi − ¯X)2 = = i=1 = SXX . So SXX = ∼ σ2χ2 n−1. Finally, since Y1 and Y2, · · · , Yn are independent, so are ¯X and SXX . 2.6 Student’s t-distribution Definition (t-distribution). Suppose that Z and Y are independent, Z ∼ N (0, 1) and Y ∼ χ2 k. Then T = Z Y /k is said to have a t-distribution on k degrees of freedom, and we write T ∼ tk. The density of tk turns out to be fT (t) = Γ((k + 1)/2) Γ(k/2) 1 √ πk 1 + t2 k −(k+1)/2 . This density is symmetric, bell-shaped, and has a maximum at t = 0, which is rather like the standard normal density. However, it can be shown that P(T > t) > P(Z > t), i.e. the T distribution has a “fatter” tail. Also, as k → ∞, tk approaches a normal distribution. Proposition. If k > 1, then Ek(T ) = 0. If k > 2, then vark(T ) = k If k = 2, then vark(T ) = ∞. In all other cases, the values are undefined. In particular, the k = 1 case has k−2 . undefined mean and variance. This is known as the Cauchy distribution. Notation. We write tk(α) be the upper 100α% point of the tk distribution, so that P(T > tk(α)) = α. Why would we define such a weird distribution? The typical application is to study random samples with unknown mean and unknown variance. √ Let X1, · · · , Xn be iid N (µ, σ2). Then ¯X ∼ N (µ, σ2/n). So Z = n( ¯X−µ) σ ∼ N (0, 1). Also, SXX /σ2 ∼ χ2 n−1 and is independent of ¯X, and hence Z. So √ n( ¯X − µ)/σ SXX /((n − 1)σ2) ∼ tn−1, 35 2 Hypothesis testing IB Statistics or √ n( ¯X − µ) SXX /(n − 1) ∼ tn−1. We write ˜σ2 = SXX confidence interval for µ is found from n−1 (note that this is the unbiased estimator). Then a 100(1−α)% 1 − α = P −tn−1 α 2 ≤ √ n( ¯X − µ) ˜σ ≤ tn−1 . α 2 This has endpoints ¯X ± ˜σ √ n tn−1 . α 2 36 3 Linear models IB Statistics 3 Linear models 3.1 Linear models Linear models can be used to explain or model the relationship between a response (or dependent) variable, and one or more explanatory variables (or covariates or predictors). As the name suggests, we assume the relationship is linear. Example. How do motor insurance claim rates (response) depend on the age and sex of the driver, and where they live (explanatory variables)? It is important to note that (unless otherwise specified), we do not assume normality in our calculations here. Suppose we have p covariates xj, and we have n observations Yi. We assume n > p, or else we can pick the parameters to fix our data exactly. Then each observation can be written as Yi = β1xi1 + · · · + βpxip + εi. ((*)) for i = 1, · · · , n. Here – β1, · · · , βp are unknown, fixed parameters we wish to work out (with n > p) – xi1, · · · , xip are the values of the p covariates for the ith response (which are all known). – ε1, · · · , εn are independent (or possibly just uncorrelated) random variables with mean 0 and variance σ2. We think of the βjxij terms to be the causal effects of xij and εi to be a random fluctuation (error term). Then we clearly have – E(Yi) = β1xi1 + · · · βpxip. – var(Yi) = var(εi) = σ2. – Y1, · · · , Yn are independent. Note that (∗) is linear in the parameters β1, · · · , βp. Obviously the real world can be much more complicated. But this is much easier to work with. Example. For each of 24 males, the maximum volume of oxygen uptake in the blood and the time taken to run 2 miles (in minutes) were measured. We want to know how the time taken depends on oxygen uptake. We might get the results Oxygen Time Oxygen Time Oxygen Time 42.3 918 36.2 1045 53.3 743 53.1 805 49.7 810 47.2 803 42.1 892 41.5 927 56.9 683 50.1 962 46.2 813 47.8 844 42.5 968 48.2 858 48.7 755 42.5 907 43.2 860 53.7 700 47.8 770 51.8 760 60.6 748 49.9 743 53.3 747 56.7 775 37 3 Linear models IB Statistics For each individual i, we let Yi be the time to run 2 miles, and xi be the maximum volume of oxygen uptake, i = 1, · · · , 24. We might want to fit a straight line to it. So a possible model is Yi = a + bxi + εi, where εi are independent random variables with variance σ2, and a and b are constants. The subscripts in the equation makes it tempting to write them as matrices: Y = Y1 ... Yn , X = x11 ... xn1 x1p · · · ... . . . · · · xnp = ε1 ... εn β1 ... βp Then the equation becomes We also have – E(ε) = 0. – cov(Y) = σ2I. Y = Xβ + ε. (2) We assume throughout that X has full rank p, i.e. the columns are independent, and that the error variance is the same for each observation. We say this is the homoscedastic case, as opposed to heteroscedastic. Example. Continuing our example, we have, in matrix form Y = Then Y1 ... Y24 , X = x1 1 ... ... 1 x24 , = Xβ + ε. ε1 ... ε24. Definition (Least squares estimator). In a linear model Y = Xβ + ε, the least squares estimator ˆβ of β minimizes S(β) = ∥Y − Xβ∥2 = (Y − Xβ)T (Y − Xβ) = n i=1 (Yi − xijβj)2 with implicit summation over j. If we plot the points on a graph, then the least square estimators minimizes the (square of the) vertical distance between the points and the line. To minimize it, we want for all k. So ∂S ∂βk β= ˆβ = 0 −2xik(Yi − xij ˆβj) = 0 38 3 Linear models IB Statistics for each k (with implicit summation over i and j), i.e. xikxij ˆβj = xikYi for all k. Putting this back in matrix form, we have Proposition. The least squares estimator satisfies X T X ˆβ = X T Y. (3) We could also have derived this by completing the square of (Y − Xβ)T (Y − Xβ), but that would be more complicated. In order to find ˆβ, our life would be much easier if X T X has an inverse. Fortunately, it always does. We assumed that X is of full rank p. Then tX T Xt = (Xt)T (Xt) = ∥Xt∥2 > 0 for t ̸= 0 in Rp (the last inequality is since if there were a t such that |
∥Xt∥ = 0, then we would have produced a linear combination of the columns of X that gives 0). So X T X is positive definite, and hence has an inverse. So ˆβ = (X T X)−1X T Y, (4) which is linear in Y. We have E(ˆβ) = (X T X −1)X T E[Y] = (X T X)−1X T Xβ = β. So ˆβ is an unbiased estimator for β. Also cov(ˆβ) = (X T X)−1X T cov(Y)X(X T X)−1 = σ2(X T X)−1, (5) since cov Y = σ2I. 3.2 Simple linear regression What we did above was so complicated. If we have a simple linear regression model then we can reparameterise it to Yi = a + bxi + εi. Yi = a′ + b(xi − ¯x) + εi, where ¯x = xi/n and a′ = a + b¯x. Since (xi − ¯x) = 0, this leads to simplified calculations. (6) In matrix form, 1 ... 1 Since (xi − ¯x) = 0, in X T X, the off-diagonals are all 0, and we have (x1 − ¯x) ... (x24 − ¯x Sxx , 39 3 Linear models IB Statistics where Sxx = (xi − ¯x)2. Hence (X T X)−1 = 1 n 0 0 1 Sxx So ˆβ = (X T X −1)X T Y = , ¯Y SxY Sxx where SxY = Yi(xi − ¯x). Hence the estimated intercept is ˆa′ = ¯y, and the estimated gradient is ˆb = = = Sxy Sxx = r × i yi(xi − ¯x) i(xi − ¯x)2 i(yi − ¯y)(xi − ¯x) i(xi − ¯x)2 Syy Sxx . i(yi − ¯y)2 × Syy Sxx (∗) We have (∗) since ¯y(xi − ¯x) = 0, so we can add it to the numerator. Then the other square root things are just multiplying and dividing by the same things. So the gradient is the Pearson product-moment correlation coefficient r times the ratio of the empirical standard deviations of the y’s and x’s (note that the gradient is the same whether the x’s are standardised to have mean 0 or not). Hence we get cov(ˆβ) = (X T X)−1σ2, and so from our expression of (X T X)−1, var(ˆa′) = var( ¯Y ) = σ2 n , var(ˆb) = σ2 Sxx . Note that these estimators are uncorrelated. Note also that these are obtained without any explicit distributional assump- tions. Example. Continuing our previous oxygen/time example, we have ¯y = 826.5, Sxx = 783.5 = 28.02, Sxy = −10077, Syy = 4442, r = −0.81, ˆb = −12.9. Theorem (Gauss Markov theorem). In a full rank linear model, let ˆβ be the least squares estimator of β and let β∗ be any other unbiased estimator for β which is linear in the Yi’s. Then var(tT ˆβ) ≤ var(tT β∗). for all t ∈ Rp. We say that ˆβ is the best linear unbiased estimator of β (BLUE). Proof. Since β∗ is linear in the Yi’s, β∗ = AY for some p × n matrix A. Since β∗ is an unbiased estimator, we must have E[β∗] = β. However, since β∗ = AY, E[β∗] = AE[Y] = AXβ. So we must have β = AXβ. Since this holds for any β, we must have AX = Ip. Now cov(β∗) = E[(β∗ − β)(β∗ − β)T ] = E[(AY − β)(AY − β)T ] = E[(AXβ + Aε − β)(AXβ + Aε − β)T ] 40 3 Linear models IB Statistics Since AXβ = β, this is equal to = E[Aε(Aε)T ] = A(σ2I)AT = σ2AAT . Now let β∗ − ˆβ = (A − (X T X)−1X T )Y = BY, for some B. Then BX = AX − (X T X)−1X T X = Ip − Ip = 0. By definition, we have AY = BY + (X T X)−1X T Y, and this is true for all Y. So A = B + (X T X)−1X T . Hence cov(β∗) = σ2AAT = σ2(B + (X T X)−1X T )(B + (X T X)−1X T )T = σ2(BBT + (X T X)−1) = σ2BBT + cov(ˆβ). Note that in the second line, the cross-terms disappear since BX = 0. So for any t ∈ Rp, we have var(tT β∗) = tT cov(β∗)t = tT cov(ˆβ)t + tT BBT tσ2 = var(tT ˆβ) + σ2∥BT t∥2 ≥ var(tT ˆβ). Taking t = (0, · · · , 1, 0, · · · , 0)T with a 1 in the ith position, we have var( ˆβi) ≤ var(β∗ i ). Definition (Fitted values and residuals). ˆY = X ˆβ is the vector of fitted values. These are what our model says Y should be. R = Y − ˆY is the vector of residuals. These are the deviations of our model from reality. The residual sum of squares is RSS = ∥R∥2 = RT R = (Y − X ˆβ)T (Y − X ˆβ). We can give these a geometric interpretation. Note that X T R = X T (Y − ˆY) = X T Y − X T X ˆβ = 0 by our formula for β. So R is orthogonal to the column space of X. Write ˆY = X ˆβ = X(X T X)−1X T Y = P Y, where P = X(X T X)−1X T . Then P represents an orthogonal projection of Rn onto the space spanned by the columns of X, i.e. it projects the actual data Y to the fitted values ˆY. Then R is the part of Y orthogonal to the column space of X. The projection matrix P is idempotent and symmetric, i.e. P 2 = P and P T = P . 41 3 Linear models IB Statistics 3.3 Linear models with normal assumptions So far, we have not assumed anything about our variables. In particular, we have not assumed that they are normal. So what further information can we obtain by assuming normality? Example. Suppose we want to measure the resistivity of silicon wafers. We have five instruments, and five wafers were measured by each instrument (so we have 25 wafers in total). We assume that the silicon wafers are all the same, and want to see whether the instruments consistent with each other, i.e. The results are as follows 130.5 130.4 113.0 128.0 121.2 112.4 138.2 120.5 117.5 110.5 Wafer 3 118.9 116.7 128.9 114.9 118.5 4 5 125.7 132.6 103.4 114.9 100.5 134.0 104.2 118.1 98.8 120.9 Let Yi,j be the resistivity of the jth wafer measured by instrument i, where i, j = 1, · · · , 5. A possible model is Yi,j = µi + εi,j, where εij are independent random variables such that E[εij] = 0 and var(εij) = σ2, and the µi’s are unknown constants. This can be written in matrix form, with Y = Then We have Y1,1 ... Y1,5 Y2,1 ... Y2,5 ... Y5,1 ... Y5, ... ... 1 0 0 1 ... ... 0 1 ... ... 0 0 ... ... ... 0 0 ... 0 ... 1 ... 1 , β = µ1 µ2 µ3 µ4 µ5 , ε = ε1,1 ... ε1,5 ε2,1 ... ε2,5 ... ε5,1 ... ε5,5 Y = Xβ + ε. X T X = 5 0 ... 0 0 5 ... ... 5 42 3 Linear models IB Statistics Hence So we have (X T X)− ... 0 0 1 5 ... ... 1 5 ˆµ = (X T X)−1X Y1 ... ¯Y5 The residual sum of squares is RSS = 5 5 i=1 j=1 (Yi,j − ˆµi)2 = 5 5 (Yi,j − ¯Yi)2 = 2170. i=1 j=1 This has n − p = 25 − 5 = 20 degrees of freedom. We will later see that ¯σ = RSS/(n − p) = 10.4. Note that we still haven’t used normality! We now make a normal assumption: Y = Xβ + ε, ε ∼ Nn(0, σ2I), rank(X) = p < n. This is a special case of the linear model we just had, so all previous results hold. Since Y = Nn(Xβ, σ2I), the log-likelihood is l(β, σ2) = − n 2 log 2π − n 2 log σ2 − 1 2σ2 S(β), where S(β) = (Y − Xβ)T (Y − Xβ). If we want to maximize l with respect to β, we have to maximize the only term containing β, i.e. S(β). So Proposition. Under normal assumptions the maximum likelihood estimator for a linear model is ˆβ = (X T X)−1X T Y, which is the same as the least squares estimator. This isn’t coincidence! Historically, when Gauss devised the normal distribution, he designed it so that the least squares estimator is the same as the maximum likelihood estimator. To obtain the MLE for σ2, we require i.e. ∂l ∂σ2 ˆβ,ˆσ2 = 0, − n 2ˆσ2 + S(ˆβ) 2ˆσ4 = 0 43 3 Linear models IB Statistics So ˆσ2 = 1 n S(ˆβ) = 1 n (Y − X ˆβ)T (Y − X ˆβ) = 1 n RSS. Our ultimate goal now is to show that ˆβ and ˆσ2 are independent. Then we can apply our other standard results such as the t-distribution. First recall that the matrix P = X(X T X)−1X T that projects Y to ˆY is idempotent and symmetric. We will prove the following properties of it: Lemma. (i) If Z ∼ Nn(0, σ2I) and A is n × n, symmetric, idempotent with rank r, then ZT AZ ∼ σ2χ2 r. (ii) For a symmetric idempotent matrix A, rank(A) = tr(A). Proof. (i) Since A is idempotent, A2 = A by definition. So eigenvalues of A are either 0 or 1 (since λx = Ax = A2x = λ2x). Since A is also symmetric, it is diagonalizable. So there exists an orthogonal Q such that Λ = QT AQ = diag(λ1, · · · , λn) = diag(1, · · · , 1, 0, · · · , 0) with r copies of 1 and n − r copies of 0. Let W = QT Z. So Z = QW. Then W ∼ Nn(0, σ2I), since cov(W) = QT σ2IQ = σ2I. Then ZT AZ = WT QT AQW = WT ΛW = r i=1 w2 i ∼ χ2 r. (ii) rank(A) = rank(Λ) = tr(Λ) = tr(QT AQ) = tr(AQT Q) = tr A Theorem. For the normal linear model Y ∼ Nn(Xβ, σ2I), (i) ˆβ ∼ Np(β, σ2(X T X)−1) (ii) RSS ∼ σ2χ2 n−p, and so ˆσ2 ∼ σ2 n χ2 n−p. (iii) ˆβ and ˆσ2 are independent. The proof is not particularly elegant — it is just a whole lot of linear algebra! Proof. 44 3 Linear models IB Statistics – We have ˆβ = (X T X)−1X T Y. Call this CY for later use. Then ˆβ has a normal distribution with mean (X T X)−1X T (Xβ) = β and covariance (X T X)−1X T (σ2I)[(X T X)−1X T ]T = σ2(X T X)−1. So ˆβ ∼ Np(β, σ2(X T X)−1). – Our previous lemma says that ZT AZ ∼ σ2χ2 r. So we pick our Z and A so that ZT AZ = RSS, and r, the degrees of freedom of A, is n − p. Let Z = Y − Xβ and A = (In − P ), where P = X(X T X)−1X T . We first check that the conditions of the lemma hold: Since Y ∼ Nn(Xβ, σ2I), Z = Y − Xβ ∼ Nn(0, σ2I). Since P is idempotent, In − P also is (check!). We also have rank(In − P ) = tr(In − P ) = n − p. Therefore the conditions of the lemma hold. To get the final useful result, we want to show that the RSS is indeed ZT AZ. We simplify the expressions of RSS and ZT AZ and show that they are equal: ZT AZ = (Y − Xβ)T (In − P )(Y − Xβ) = YT (In − P )Y. Noting the fact that (In − P )X = 0. Writing R = Y − ˆY = (In − P )Y, we have RSS = RT R = YT (In − P )Y, using the symmetry and idempotence of In − P . Hence RSS = ZT AZ ∼ σ2χ2 n−p. Then ˆσ2 = RSS n ∼ σ2 n χ2 n−p. – Let V = ˆβ R = DY, where D = C In − P is a (p + n) × n matrix. Since Y is multivariate, V is multivariate with cov(V ) = Dσ2IDT = σ2 = σ2 = σ2 CC T (In − P )C T CC T (In − P )C T 0 In − P CC T 0 C(In − P )T (In − P )(In − P )T C(In − P ) (In − P ) 45 3 Linear models IB Statistics Using C(In − P ) = 0 (since (X T X)−1X T (In − P ) = 0 since (In − P )X = 0 — check!). Hence ˆβ and R are independent since the off-diagonal covariant terms are 0. So ˆβ and RSS = RT R are independent. So ˆβ and ˆσ2 are independent. From (ii), E(RSS) = σ2(n − p). So ˜σ2 = RSS n−p is an unbiased estimator of σ2. ˜σ is often known as the residual standard error on n − p degrees of freedom. 3.4 The F distribution Definition (F distribution). Suppose U and V are independent with U ∼ χ |
2 m and V ∼ χ2 V /n is said to have an F -distribution on m and n degrees of freedom. We write X ∼ Fm,n n. The X = U/m Since U and V have mean m and n respectively, U/m and V /n are approxi- mately 1. So F is often approximately 1. It should be very clear from definition that Proposition. If X ∼ Fm,n, then 1/X ∼ Fn,m. We write Fm,n(α) be the upper 100α% point for the Fm,n-distribution so that if X ∼ Fm,n, then P(X > Fm,n(α)) = α. Suppose that we have the upper 5% point for all Fn,m. Using these information, it is easy to find the lower 5% point for Fm,n since we know that P(Fm,n < 1/x) = P(Fn,m > x), which is where the above proposition comes useful. Note that it is immediate from definitions of tn and F1,n that if Y ∼ tn, then Y 2 ∼ F1,n, i.e. it is a ratio of independent χ2 1 and χ2 n variables. Inference for β 3.5 We know that ˆβ ∼ Np(β, σ2(X T X)−1). So ˆβj ∼ N (βj, σ2(X T X)−1 jj ). The standard error of ˆβj is defined to be SE( ˆβj) = ˜σ2(X T X)−1 jj , where ˜σ2 = RSS/(n − p). Unlike the actual variance σ2(X T X)−1 error is calculable from our data. jj , the standard Then ˆβj − βj SE( ˆβj) = ˆβj − βj ˜σ2(X T X)−1 jj = ( ˆβj − βj)/ σ2(X T X)−1 jj RSS/((n − p)σ2) By writing it in this somewhat weird form, we now recognize both the numerator and denominator. The numerator is a standard normal N (0, 1), and the 46 3 Linear models IB Statistics denominator is an independent But a standard normal divided by χ2 is, by definition, the t distribution. So n−p/(n − p), as we have previously shown. χ2 ˆβj − βj SE( ˆβj) ∼ tn−p. So a 100(1 − α)% confidence interval for βj has end points ˆβj ± SE( ˆβj)tn−p( α 2 ). In particular, if we want to test H0 : βj = 0, we use the fact that under H0, ˆβj SE( ˆβj ) ∼ tn−p. 3.6 Simple linear regression We can apply our results to the case of simple linear regression. We have Yi = a′ + b(xi − ¯x) + εi, where ¯x = xi/n and εi are iid N (0, σ2) for i = 1, · · · , n. Then we have ˆa′ = ¯Y ∼ N a′, σ2 n ˆb = SxY Sxx ∼ N b, σ2 Sxx ˆYi = ˆa′ + ˆb(xi − ¯x) RSS = i (Yi − ˆYi)2 ∼ σ2χ2 n−2, and (ˆa′, ˆb) and ˆσ2 = RSS/n are independent, as we have previously shown. Note that ˆσ2 is obtained by dividing RSS by n, and is the maximum likelihood estimator. On the other hand, ˜σ is obtained by dividing RSS by n − p, and is an unbiased estimator. Example. Using the oxygen/time example, we have seen that ˜σ2 = RSS n − p = 67968 24 − 2 = 3089 = 55.62. So the standard error of ˆβ is SE(ˆb) = ˜σ2(X T X)−1 22 = 3089 Sxx = 55.6 28.0 = 1.99. So a 95% interval for b has end points ˆb ± SE(ˆb) × tn−p(0.025) = 12.9 ± 1.99 ∗ t22(0.025) = (−17.0, −8.8), using the fact that t22(0.025) = 2.07. Note that this interval does not contain 0. So if we want to carry out a size 0.05 test of H0 : b = 0 (they are uncorrelated) vs H1 : b ̸= 0 (they are correlated), the test statistic would be ˆb 1.99 = −6.48. Then we reject H0 because this is less than −t22(0.025) = −2.07. = −12.9 SE(ˆb) 47 3 Linear models IB Statistics 3.7 Expected response at x∗ After performing the linear regression, we can now make predictions from it. Suppose that x∗ is a new vector of values for the explanatory variables. The expected response at x∗ is E[Y | x∗] = x∗T β. We estimate this by x∗T ˆβ. Then we have x∗T (ˆβ − β) ∼ N (0, x∗T cov(ˆβ)x∗) = N (0, σ2x∗T (X T X)−1x∗). Let τ 2 = x∗T (X T X)−1x∗. Then x∗T (ˆβ − β) ˜στ ∼ tn−p. Then a confidence interval for the expected response x∗T β has end points x∗T ˆβ ± ˜στ tn−p . α 2 Example. Previous example continued: Suppose we wish to estimate the time to run 2 miles for a man with an oxygen take-up measurement of 50. Here x∗T = (1, 50 − ¯x), where ¯x = 48.6. The estimated expected response at x∗T is x∗T ˆβ = ˆa′ + (50 − 48.5) × ˆb = 826.5 − 1.4 × 12.9 = 808.5, which is obtained by plugging x∗T into our fitted line. We find τ 2 = x∗T (X T X)−1x∗ = 1.42 783.5 So a 95% confidence interval for E[Y | x∗ = 50 − ¯x] is x∗2 Sxx 1 24 1 n + + = x∗T ˆβ ± ˜στ tn−p α 2 = 808.5 ± 55.6 × 0.21 × 2.07 = (783.6, 832.2). = 0.044 = 0.212. Note that this is the confidence interval for the predicted expected value, NOT the confidence interval for the actual obtained value. The predicted response at x∗ is Y ∗ = x∗β + ε∗, where ε∗ ∼ N (0, σ2), and Y ∗ is independent of Y1, · · · , Yn. Here we have more uncertainties in our prediction: β and ε∗. A 100(1 − α)% prediction interval for Y ∗ is an interval I(Y) such that P(Y ∗ ∈ I(Y)) = 1 − α, where the probability is over the joint distribution of Y ∗, Y1, · · · , Yn. So I is a random function of the past data Y that outputs an interval. First of all, as above, the predicted expected response is ˆY ∗ = x∗T β. This is an unbiased estimator since ˆY ∗ − Y ∗ = x∗T (ˆβ − β) − ε∗, and hence E[ ˆY ∗ − Y ∗] = x∗T (β − β) = 0, To find the variance, we use that fact that x∗T (ˆβ − β) and ε∗ are independent, and the variance of the sum of independent variables is the sum of the variances. So var( ˆY ∗ − Y ∗) = var(x∗T (ˆβ)) + var(ε∗) = σ2x∗T (X T X)−1x∗ + σ2. = σ2(τ 2 + 1). 48 3 Linear models IB Statistics We can see this as the uncertainty in the regression line σ2τ 2, plus the wobble about the regression line σ2. So We therefore find that ˆY ∗ − Y ∗ ∼ N (0, σ2(τ 2 + 1)). ˆY ∗ − Y ∗ √ τ 2 + 1 ˜σ ∼ tn−p. So the interval with endpoints x∗T ˆβ ± ˜σ τ 2 + 1tn−p α 2 is a 95% prediction interval for Y ∗. We don’t call this a confidence interval — confidence intervals are about finding parameters of the distribution, while the prediction interval is about our predictions. Example. A 95% prediction interval for Y ∗ at x∗T = (1, (50 − ¯x)) is x∗T ± ˜σ τ 2 + 1tn−p α 2 = 808.5 ± 55.6 × 1.02 × 2.07 = (691.1, 925.8). Note that this is much wider than our our expected response! This is since there are three sources of uncertainty: we don’t know what σ is, what ˆb is, and the random ε fluctuation! Example. Wafer example continued: Suppose we wish to estimate the expected resistivity of a new wafer in the first instrument. Here x∗T = (1, 0, · · · , 0) (recall that x is an indicator vector to indicate which instrument is used). The estimated response at x∗T is We find x∗T ˆµ = ˆµ1 = ¯y1 = 124.3 τ 2 = x∗T (X T X)−1x∗ = 1 5 . So a 95% confidence interval for E[Y ∗ 1 ] is x∗T ˆµ ± ˜στ tn−p α 2 = 124.3 ± 10.4 √ 5 × 2.09 = (114.6, 134.0). Note that we are using an estimate of σ obtained from all five instruments. If we had only used the data from the first instrument, σ would be estimated as 5 j=1 y1,j − ¯y1 5 − 1 ˜σ1 = = 8.74. The observed 95% confidence interval for µ1 would have been ¯y1 ± ˜σ1√ 5 t4 α 2 = 124.3 ± 3.91 × 2.78 = (113.5, 135.1), which is slightly wider. Usually it is much wider, but in this special case, we only get little difference since the data from the first instrument is relatively tighter than the others. A 95% prediction interval for Y ∗ 1 at x∗T = (1, 0, · · · , 0) is x∗T ˆµ ± ˜σ τ 2 + 1tn−p α 2 = 124.3 ± 10.42 × 1.1 × 2.07 = (100.5, 148.1). 49 3 Linear models IB Statistics 3.8 Hypothesis testing 3.8.1 Hypothesis testing In hypothesis testing, we want to know whether certain variables influence the result. If, say, the variable x1 does not influence Y , then we must have β1 = 0. So the goal is to test the hypothesis H0 : β1 = 0 versus H1 : β1 ̸= 0. We will tackle a more general case, where β can be split into two vectors β0 and β1, and we test if β1 is zero. We start with an obscure lemma, which might seem pointless at first, but will prove itself useful very soon. Lemma. Suppose Z ∼ Nn(0, σ2In), and A1 and A2 are symmetric, idempotent n × n matrices with A1A2 = 0 (i.e. they are orthogonal). Then ZT A1Z and ZT A2Z are independent. This is geometrically intuitive, because A1 and A2 being orthogonal means they are concerned about different parts of the vector Z. Proof. Let Xi = AiZ, i = 1, 2 and Then W = W1 W2 = A1 A2 Z. W ∼ N2n 0 0 , σ2 A1 0 0 A2 since the off diagonal matrices are σ2AT 1 A2 = A1A2 = 0. So W1 and W2 are independent, which implies WT 1 W1 = ZT AT 1 A1Z = ZT A1A1Z = ZT A1Z and are independent WT 2 W2 = ZT AT 2 A2Z = ZT A2A2Z = ZT A2Z Now we go to hypothesis testing in general linear models: Suppose X n×p = X0 n×p0 X1 n×(p−p0) and B = p0. , where rank(X) = p, rank(X0) = β0 β1 We want to test H0 : β1 = 0 against H1 : β1 ̸= 0. Under H0, X1β1 vanishes and Under H0, the mle of β0 and σ2 are Y = X0β0 + ε. ˆˆβ0 = (X T RSS0 ˆˆσ2 = n 0 X0)−1X T 1 n = 0 Y (Y − X0 ˆˆβ0)T (Y − X0 ˆˆβ0) and we have previously shown these are independent. 50 3 Linear models IB Statistics Note that our poor estimators wear two hats instead of one. We adopt the convention that the estimators of the null hypothesis have two hats, while those of the alternative hypothesis have one. So the fitted values under H0 are ˆˆY = X0(X T 0 X0)−1X T 0 Y = P0Y, where P0 = X0(X T 0 X0)−1X T 0 . The generalized likelihood ratio test of H0 against H1 is ΛY(H0, H1) = = = exp − 1 exp − 1 2ˆσ2 (Y − X ˆβ)T (Y − X ˆβ) ˆˆβ0) ˆˆβ0)T (Y − X 2ˆˆσ2 (Y − X 1√ 1√ 2π ˆσ2 2π ˆˆσ2 n/2 ˆˆσ2 ˆσ2 RSS0 RSS n/2 = 1 + RSS0 − RSS RSS n/2 . We reject H0 when 2 log Λ is large, equivalently when RSS0−RSS is large. RSS Using the results in Lecture 8, under H0, we have 2 log Λ = n log 1 + RSS0 − RSS RSS , which is approximately a χ2 random variable. This is a good approximation. But we can get an exact null distribution, and p1−p0 get an exact test. We have previously shown that RSS = YT (In − P )Y, and so RSS0 − RSS = YT (In − P0)Y − YT (In − P )Y = YT (P − P0)Y. Now both In − P and P − P0 are symmetric and idempotent, and therefore rank(In − P ) = n − p and rank(P − P0) = tr(P − P0) = tr(P ) − tr(P0) = rank(P ) − rank(P0) = p − p0. Also, (In − P )(P − P0) = (In − P )P − (In − P )P0 = (P − P 2) − (P0 − P P0) = 0. (we have P 2 = P by idempotence, and P P0 = P0 since after projecting with P0, we are already in the space of P , and applying P has no effect) Finally, YT (In − P )Y = (Y − X0β0)T (In − P )(Y − X0β0) YT (P − P0)Y = (Y − X0β0)T (P − P0)(Y − X0β0) 51 3 Linear models IB Statistics since (In − P )X0 = (P − P0)X0 = 0. If we let Z = Y − X0β0, |
A1 = In − P , A2 = P − P0, and apply our previous lemma, and the fact that ZT AiZ ∼ σ2χ2 r, then RSS = YT (In − P )Y ∼ χ2 RSS0 − RSS = YT (P − P0)Y ∼ χ2 n−p p−p0 and these random variables are independent. So under H0, F = YT (P − P0)Y/(p − p0) YT (In − P )Y/(n − p) = (RSS0 − RSS)/(p − p0) RSS/(n − p) ∼ Fp−p0,n−p, Hence we reject H0 if F > Fp−p0,n−p(α). RSS0 − RSS is the reduction in the sum of squares due to fitting β1 in addition to β0. Source of var. d.f. sum of squares mean squares F statistic Fitted model Residual Total p − p0 n − p n − p0 RSS0 − RSS RSS RSS0 RSS0−RSS p−p0 RSS n−p (RSS0−RSS)/(p−p0) RSS/(n−p) The ratio RSS0−RSS by β1, and denoted R2. RSS0 is sometimes known as the proportion of variance explained 3.8.2 Simple linear regression We assume that where ¯x = xi/n and εi are N (0, σ2). Yi = a′ + b(xi − ¯x) + εi, Suppose we want to test the hypothesis H0 : b = 0, i.e. no linear relationship. We have previously seen how to construct a confidence interval, and so we could simply see if it included 0. Alternatively, under H0, the model is Yi ∼ N (a′, σ2), and so ˆa′ = ¯Y , and the fitted values are ˆYi = ¯Y . The observed RSS0 is therefore RSS0 = (yi − ¯y)2 = Syy. i The fitted sum of squares is therefore RSS0 − RSS = i (yi − ¯y)2 − (yi − ¯y − ˆb(xi − ¯x))2 = ˆb2 (xi − ¯x)2 = ˆb2Sxx. Source of var. d.f. Fitted model Residual Total 1 n − 2 n − 1 sum of squares RSS0 − RSS = ˆb2SXX RSS = i(yi − ˆy)2 RSS0 = i(yi − ¯y)2 mean squares F statistic ˆb2Sxx ˜σ2. F = ˆb2Sxx ˜σ2 52 3 Linear models IB Statistics Note that the proposition of variance explained is ˆb2Sxx/Syy = where r is the Pearson’s product-moment correlation coefficient S2 xy SxxSyy = r2, r = Sxy SxxSyy . ∼ tn−2, where SE(ˆb) = ˜σ/ √ Sxx. We have previously seen that under H0, So we let ˆb SE(ˆb) √ ˆb = ˆb SE(ˆb) is precisely the same as checking whether Sxx ˜σ . t = α 2 Checking whether |t| > tn−2 t2 = F > F1,n−2(α), since a F1,n−2 variable is t2 n−2. Hence the same conclusion is reached, regardless of whether we use the t-distribution or the F statistic derived form an analysis of variance table. 3.8.3 One way analysis of variance with equal numbers in each group Recall that in our wafer example, we made measurements in groups, and want to know if there is a difference between groups. In general, suppose J measurements are taken in each of I groups, and that Yij = µi + εij, where εij are independent N (0, σ2) random variables, and the µi are unknown constants. Fitting this model gives RSS = I J i=1 j=1 (Yij − ˆµi)2 = I J (Yij − ¯Yi.)2 i=1 j=1 on n − I degrees of freedom. Suppose we want to test the hypothesis H0 : µi = µ, i.e. no difference between groups. are ˆYij = ¯Y . Under H0, the model is Yij ∼ N (µ, σ2), and so ˆµ = ¯Y , and the fitted values The observed RSS0 is therefore RSS0 = i,j (yij − ¯y..)2. The fitted sum of squares is therefore RSS0 − RSS = i j (yij − ¯y..)2 − (yij − ¯yi.)2 = J (¯yi. − ¯y..)2. i Source of var. d.f. Fitted model Residual Total sum of squares J i(¯yi − ¯y..)2 j(yij − ¯yi.)2 j(yij − ¯y..) mean squares J i (¯yi.−¯y..)2 I−1 ˜σ2 F statistic J i (¯yi.−¯y..)2 (I−1)˜σ2 53 |
s+1 1/2 N, N t = M, M t s 1/2 , s (Ni2−n − N(i−1)2−n )2 1/2 (Cauchy–Schwarz) where all equalities are u.c.p. Claim. For all 0 ≤ s < t, we have t s |dM, N u| ≤ M, M t s N, N t s. Indeed, for any subdivision s = t0 < t1 < · · · tn = t, we have n i=1 |M, N ti ti−1| ≤ n M, M ti ti−1 N, N ti ti−1 i=1 n i=1 ≤ M, M ti ti−1 1/2 n N, N ti ti−1 1/2 . i=1 (Cauchy–Schwarz) Taking the supremum over all subdivisions, the claim follows. Claim. For all bounded Borel sets B ⊆ [0, ∞), we have B |dM, N u| ≤ B 23 dM u B dN u. 2 Semi-martingales III Stochastic Calculus and Applications We already know this is true if B is an interval. If B is a finite union of integrals, then we apply Cauchy–Schwarz. By a monotone class argument, we can extend to all Borel sets. Claim. The theorem holds for H = k =1 h1B, K = n =1 k1B for B ⊆ [0, ∞) bounded Borel sets with disjoint support. We have |HsKs| |dM, N s| ≤ ≤ ≤ n =1 n |hk| |hk| B |d ¯M , N s| 1/2 1/2 dN s dM s =1 n =1 h2 B B dM s 1/2 n =1 B k2 B 1/2 dN s To finish the proof, approximate general H and K by step functions and take the limit. 2.6 Semi-martingale Definition (Semi-martingale). A (continuous) adapted process X is a (continuous) semi-martingale if X = X0 + M + A, where X0 ∈ F0, M is a continuous local martingale with M0 = 0, and A is a continuous finite variation process with A0 = 0. This decomposition is unique up to indistinguishables. Definition (Quadratic variation). Let X = X0 + M + A and X = X be (continuous) semi-martingales. Set , X = M, M . This definition makes sense, because continuous finite variation processes do not have quadratic variation. Exercise. We have X, Y (n) t = 2nt i=1 (Xi2−n − X(i−1)2−n )(Yi2−n − Y(i−1)2−n ) → X, Y u.c.p. 24 3 The stochastic integral III Stochastic Calculus and Applications 3 The stochastic integral 3.1 Simple processes We now have all the background required to define the stochastic integral, and we can start constructing it. As in the case of the Lebesgue integral, we first define it for simple processes, and then extend to general processes by taking a limit. Recall that we have Definition (Simple process). The space of simple processes E consists of functions H : Ω × [0, ∞) → R that can be written as Ht(ω) = n i=1 Hi−1(ω)1(ti−1,ti](t) for some 0 ≤ t0 ≤ t1 ≤ · · · ≤ tn and bounded random variables Hi ∈ Fti . Definition (H · M ). For M ∈ M2 and H ∈ E, we set t 0 H dM = (H · M )t = n i=1 Hi−1(Mti∧t − Mti−1∧t). If M were of finite variation, then this is the same as what we have previously seen. Recall that for the Lebesgue integral, extending this definition to general functions required results like monotone convergence. Here we need some similar In fact, we get results that put bounds on how large the integral can be. something better than a bound. Proposition. If M ∈ M2 c and H ∈ E, then H · M ∈ M2 ∞ c and H · M 2 M2 = E H 2 s dM s . 0 (∗) Proof. We first show that H · M ∈ M2 for c. By linearity, we only have to check it X i We have to check that E(X i is when t > ti−1. t = Hi−1(Mti∧t − Mti−1∧t) t | Fs) = 0 for all t > s, and the only non-trivial case E(X i t | Fs) = Hi−1E(Mti∧t − Mti−1∧t | Fs) = 0. We also check that X iM2 ≤ 2H∞M M2. So it is bounded. So H · M ∈ M2 c. To prove (∗), we note that the X i are orthogonal and that X it = H 2 i−1(M ti∧t − M ti−1∧t). So we have H ·M, H ·M = X i, X i = H 2 i−1(M ti∧t −M ti−1∧t) = t 0 H 2 s dM s. 25 3 The stochastic integral III Stochastic Calculus and Applications In particular, H · M 2 M2 = EH · M ∞ = E ∞ 0 H 2 s dM s . Proposition. Let M ∈ M2 c and H ∈ E. Then H · M, N = H · M, N for all N ∈ M2. In other words, the stochastic integral commutes with the bracket. Proof. Write H · M = X i = Hi−1(Mti∧t − Mti−1∧t) as before. Then X i, N t = Hi−1Mti∧t − Mti−1∧t, N = Hi−1(M, N ti∧t − M, N ti−1∧t). 3.2 Itˆo isometry We now try to extend the above definition to something more general than simple processes. Definition (L2(M )). Let M ∈ M2 lence classes of) previsible H : Ω × [0, ∞) → R such that c. Define L2(M ) to be the space of (equiva- HL2(M ) = HM = E ∞ 0 H 2 s dM s 1/2 < ∞. For H, K ∈ L2(M ), we set (H, K)L2(M ) = E ∞ 0 HsKs dM s . In fact, L2(M ) is equal to L2(Ω × [0, ∞), P, dP dM ), where P is the previsible σ-algebra, and in particular is a Hilbert space. c. Then E is dense in L2(M ). Proposition. Let M ∈ M2 Proof. Since L2(M ) is a Hilbert space, it suffices to show that if (K, H) = 0 for all H ∈ E, then K = 0. So assume that (K, H) = 0 for all H ∈ E and t Xt = Ks dM s, 0 Then X is a well-defined finite variation process, and Xt ∈ L1 for all t. It suffices to show that Xt = 0 for all t, and we shall show that Xt is a continuous martingale. Let 0 ≤ s < t and F ∈ Fs bounded. We let H = F 1(s,t] ∈ E. By assumption, we know 0 = (K, H) = E t F s Ku dM u = E(F (Xt − XS)). Since this holds for all Fs measurable F , we have shown that So X is a (continuous) martingale, and we are done. E(Xt | Fs) = Xs. 26 3 The stochastic integral III Stochastic Calculus and Applications Theorem. Let M ∈ M2 c. Then (i) The map H ∈ E → H ·M ∈ M2 c extends uniquely to an isometry L2(M ) → M2 c, called the Itˆo isometry. (ii) For H ∈ L2(M ), H · M is the unique martingale in M2 c such that H · M, N = H · M, N for all N ∈ M2 (as above) and the RHS is the finite variation integral. c, where the integral on the LHS is the stochastic integral (iii) If T is a stopping time, then (1[0,T ]H) · M = (H · M )T = H · M T . Definition (Stochastic integral). H · M is the stochastic integral of H with respect to M and we also write (H · M )t = t 0 Hs dMs. It is important that the integral of martingale is still a martingale. After proving Itˆo’s formula, we will use this fact to show that a lot of things are in fact martingales in a rather systematic manner. For example, it will be rather effortless to show that B2 t − t is a martingale when Bt is a standard Brownian motion. Proof. (i) We have already shown that this map is an isometry when restricted to E. So extend by completeness of M2 c and denseness of E. (ii) Again the equation to show is known for simple H, and we want to show it is preserved under taking limits. Suppose H n → H in L2(M ) with H n ∈ L2(M ). Then H n · M → H · M in M2 c. We want to show that H · M, N ∞ = lim n→∞ H · M, N = lim n→∞ H n · M, N ∞ in L1. H n · M, N in L1. for all N ∈ M2 c. To show the first holds, we use the Kunita–Watanabe inequality to get E|H · M − H n · M, N ∞| ≤ E (H · M − H n · M ∞)1/2 (EN ∞)1/2 , and the first factor is H · M − H n · M M2 → 0, while the second is finite since N ∈ M2 c. The second follows from E |((H − H n) · M, N )∞| ≤ H − H nL2(M )N M2 → 0. So we know that H · M, N ∞ = (H · M, N )∞. We can then replace N by the stopped process N t to get H · M, N t = (H · M, N )t. To see uniqueness, suppose X ∈ M2 c is another such martingale. Then we have X − H · M, N = 0 for all N . Take N = X − H · M , and then we are done. 27 3 The stochastic integral III Stochastic Calculus and Applications (iii) For N ∈ M2 c, we have (H · M )T , N t = H · M, N t∧T = H · M, N t∧T = (H1[0,T ] · M, N )t for every N . So we have shown that (H · M )T = (1[0,T ]H · M ) by (ii). To prove the second equality, we have , N t∧T = ((H1[0,T ] · M, N )t. Note that (ii) can be written as (−) 0 Hs dMs, N = t t 0 Hs dM, N s. Corollary. H · M, K · N = H · (K · M, N ) = (HK) · M, N . In other words, (−) 0 Hs dMs, (−) 0 Ks dNs = t t 0 HsKs dM, N s. Corollary. Since H · M and (H · M )(K · N ) − H · M, K · N are martingales starting at 0, we have t E 0 H dMs = 0 t E 0 Hs dMs t Ks dNs = 0 t 0 HsKs dM, N s. Corollary. Let H ∈ L2(M ), then HK ∈ L2(M ) iff K ∈ L2(H · M ), in which case (KH) · M = K · (H · M ). Proof. We have ∞ E 0 K 2 s H 2 s dMs ∞ = E 0 K 2 s H · M s , so KL2(H·M ) = HKL2(M ). For N ∈ M2 c, we have (KH) · M, N t = (KH · M, N )t = (K · (H · M, N ))t = (K · H · M, N )t. 3.3 Extension to local martingales We have now defined the stochastic integral for continuous martingales. We next go through some formalities to extend this to local martingales, and ultimately to semi-martingales. We are not doing this just for fun. Rather, when we later prove results like Itˆo’s formula, even when we put in continuous (local) martingales, we usually end up with some semi-martingales. So it is useful to be able to deal with semi-martingales in general. 28 3 The stochastic integral III Stochastic Calculus and Applications Definition (L2 bc(M )). Let L2 bc(M ) be the space of previsible H such that t H 2 s dM s < ∞ a.s. for all finite t > 0. 0 Theorem. Let M be a continuous local martingale. (i) For every H ∈ L2 bc(M ), there is a unique continuous local martingale H · M with (H · M )0 = 0 and H · M, N = H · M, N for all N, M . (ii) If T is a stopping time, then (1[0,T ]H) · M = (H · M )T = H · M T . (iii) If H ∈ L2 and then loc(M ), K is previsible, then K ∈ L2 loc(H · M ) iff HK ∈ L2 loc(M ), K · (H · M ) = (KH) · M. (iv) Finally, if M ∈ M2 the previous one. c and H ∈ L2(M ), then the definition is the same as Proof. Assume M0 = 0, and that t H = 0 when this fails). Set t 0 H 2 Sn = inf t ≥ 0 : 0 (1 + H 2 s ) dM s ≥ n . s dM s < ∞ for all ω ∈ Ω (by setting These Sn are stopping times that tend to infinity. Then M Sn , M Sn t = M, M t∧Sn ≤ n. So M Sn ∈ M2 c. Also, ∞ 0 Hs dM Sn s = Sn 0 H 2 s dM s ≤ n. So H ∈ L2(M Sn ), and we have already defined what H · M Sn is. Now notice that H · M Sn = (H · M Sm )Sn for m ≥ n. So it makes sense to define H · M = lim n→∞ This is the unique process such that (H · M )Sn = H · M Sn . We see that H · M is a continuous adapted local martingale with reducing sequence Sn. H · M Sn . Claim. H · M, N = H · M, N . Sn ∧ S Indeed, assume that N0 = 0. Set S c. Then n. Observe that N S H · M, N Tn = H · M Sn , N S n ∈ M2 n = inf{t ≥ 0 : |Nt| ≥ n}. Set Tn = n = H · M Sn , N S n = H · M, N Tn. Taking the limit n → ∞ gives the desired result. The proofs of the other claims are the same as before, since they only use the characterizing property H · M, N = H · M, N . 29 3 The stochastic integral III Stochastic Calculus and Applications 3.4 |
Extension to semi-martingales Definition (Locally boounded previsible process). A previsible process H is locally bounded if for all t ≥ 0, we have |Hs| < ∞ a.s. sup s≤t Fact. (i) Any adapted continuous process is locally bounded. (ii) If H is locally bounded and A is a finite variation process, then for all t ≥ 0, we have t 0 |Hs| |dAs| < ∞ a.s. Now if X = X0 + M + A is a semi-martingale, where X0 ∈ F0, M is a continuous local martingale and A is a finite variation process, we want to define Hs dXs. We already know what it means to define integration with respect to dMs and dAs, using the Itˆo integral and the finite variation integral respectively, and X0 doesn’t change, so we can ignore it. Definition (Stochastic integral). Let X = X0 + M + A be a continuous semimartingale, and H a locally bounded previsible process. Then the stochastic integral H · X is the continuous semi-martingale defined by and we write Proposition, (H · X)t = T 0 Hs dXs. (i) (H, X) → H · X is bilinear. (ii) H · (K · X) = (HK) · X if H and K are locally bounded. (iii) (H · X)T = H1[0,T ] · X = H · X T for every stopping time T . (iv) If X is a continuous local martingale (resp. a finite variation process), then so is H · X. (v) If H = n then i=1 Hi−11(ti−1,ti] and Hi−1 ∈ Fti−1 (not necessarily bounded), (H · X)t = n i=1 Hi−1(Xti∧t − Xti−1∧t). Proof. (i) to (iv) follow from analogous properties for H · M and H · A. The last part is also true by definition if the Hi are uniformly bounded. If Hi is not bounded, then the finite variation part is still fine, since for each fixed ω ∈ Ω, Hi(ω) is a fixed number. For the martingale part, set Tn = inf{t ≥ 0 : |Ht| ≥ n}. 30 3 The stochastic integral III Stochastic Calculus and Applications Then Tn are stopping times, Tn → ∞, and H1[0,Tn] ∈ E. Thus (H · M )t∧Tn = n i=1 Hi−1T[0,Tn](Xti∧t − Xti−1∧t). Then take the limit n → ∞. Before we get to Itˆo’s formula, we need a few more useful properties: Proposition (Stochastic dominated convergence theorem). Let X be a continuous semi-martingale. Let H, Hs be previsible and locally bounded, and let K be previsible and non-negative. Let t > 0. Suppose (i) H n (ii) |H n s → Hs as n → ∞ for all s ∈ [0, t]. s | ≤ Ks for all s ∈ [0, t] and n ∈ N. (iii) t s dM s < ∞ and t 0 K 2 okay if K is locally bounded). 0 Ks |dAs| < ∞ (note that both conditions are Then t 0 H n s dXs → t 0 Hs dXs in probability. Proof. For the finite variation part, the convergence follows from the usual dominated convergence theorem. For the martingale part, we set Tm = inf t ≥ 0 : K 2 s dM s ≥ m . t 0 So we have E Tm∧t H n 0 s dMs − Tn∧t Hs dMs 0 2 ≤ E Tn∧t 0 (H n s − Hs)2 dM s → 0. using the usual dominated convergence theorem, since Tn∧t K 2 s dM s ≤ m. 0 Since Tn ∧ t = t eventually as n → ∞ almost surely, hence in probability, we are done. Proposition. Let X be a continuous semi-martingale, and let H be an adapted bounded left-continuous process. Then for every subdivision 0 < t(m) 1 < i − t(m) · · · < t(m) nm of [0, t] with maxi |t(m) i−1| → 0, then 0 < t(m) t 0 Hs dXs = lim m→∞ nm i=1 Ht(m) i−1 (Xt(m) i − Xt(m) i−1 ) in probability. Proof. We have already proved this for the Lebesgue–Stieltjes integral, and all we used was dominated convergence. So the same proof works using stochastic dominated convergence theorem. 31 3 The stochastic integral III Stochastic Calculus and Applications 3.5 Itˆo formula We now prove the equivalent of the integration by parts and the chain rule, i.e. Itˆo’s formula. Compared to the world of usual integrals, the difference is that the quadratic variation, i.e. “second order terms” will crop up quite a lot, since they are no longer negligible. Theorem (Integration by parts). Let X, Y be a continuous semi-martingale. Then almost surely, XtYt − X0Y0 = t 0 Xs dYs + t 0 Ys dXs + X, Y t The last term is called the Itˆo correction. Note that if X, Y are martingales, then the first two terms on the right are martingales, but the last is not. So we are forced to think about semi-martingales. Observe that in the case of finite variation integrals, we don’t have the correction. Proof. We have XtYt − XsYs = Xs(Yt − Ys) + (Xt − Xs)Ys + (Xt − Xs)(Yt − Ys). When doing usual calculus, we can drop the last term, because it is second order. However, the quadratic variation of martingales is in general non-zero, and so we must keep track of this. We have Xk2−n Yk2−n − X0Y0 = k (Xi2−nYi2−n − X(i−1)2−n Y(i−1)2−n) i=1 n i=1 = X(i−1)2−n (Yi2−n − Y(i−1)2−n ) + Y(i−1)2−n (Xi2−n − X(i−1)2−n ) + (Xi2−n − X(i−1)2−n )(Yi2−n − Y(i−1)2−n ) Taking the limit n → ∞ with k2−n fixed, we see that the formula holds for t a dyadic rational. Then by continuity, it holds for all t. The really useful formula is the following: Theorem (Itˆo’s formula). Let X 1, . . . , X p be continuous semi-martingales, and let f : Rp → R be C 2. Then, writing X = (X 1, . . . , X p), we have, almost surely, f (Xt) = f (X0) + p t i=1 0 ∂f ∂xi (Xs) dX i s + 1 2 p t i,j=1 0 ∂2f ∂xi∂xj (Xs) dX i, X js. In particular, f (X) is a semi-martingale. The proof is long but not hard. We first do it for polynomials by explicit computation, and then use Weierstrass approximation to extend it to more general functions. 32 3 The stochastic integral III Stochastic Calculus and Applications Proof. Claim. Itˆo’s formula holds when f is a polynomial. It clearly does when f is a constant! We then proceed by induction. Suppose Itˆo’s formula holds for some f . Then we apply integration by parts to g(x) = xkf (x). where xk denotes the kth component of x. Then we have g(Xt) = g(X0) + t 0 X k s df (Xs) + t 0 f (Xs) dX k s + X k, f (X)t We now apply Itˆo’s formula for f to write t 0 X k s df (Xs) = p t i=1 0 X k s ∂f ∂xi (Xs) dX i s + 1 2 p t i,j=1 0 X k s ∂2f ∂xi∂xj (Xs) dX i, X js. We also have So we have X k, f (X)t = p t i=1 0 ∂f ∂xi (Xs) dX k, X is. g(Xt) = g(X0) + p t i=1 0 ∂g ∂xi (Xs) dX i s + 1 2 p t i,j=1 0 ∂2g ∂xi∂xj (Xs) dX i, X js. So by induction, Itˆo’s formula holds for all polynomials. Claim. Itˆo’s formula holds for all f ∈ C 2 if |Xt(ω)| ≤ n and t all (t, ω). 0 |dAs| ≤ n for By the Weierstrass approximation theorem, there are polynomials pk such that sup |x|≤k |f (x) − pk(x)| + max i ∂f ∂xi − ∂p ∂xi + max i,j ∂2f ∂xi∂xj − ∂pk ∂xi∂xj ≤ 1 k . By taking limits, in probability, we have f (Xt) − f (X0) = lim k→∞ (pk(Xt) − pk(X0)) t 0 ∂f ∂xi (Xs) dX i s = lim k→∞ ∂pk ∂xi (Xs) dX i s by stochastic dominated convergence theorem, and by the regular dominated convergence, we have t 0 ∂f ∂xi∂xj dX i, X js = lim k→∞ t 0 ∂2pk ∂xi∂xj dX i, X j. 33 3 The stochastic integral III Stochastic Calculus and Applications Claim. Itˆo’s formula holds for all X. Let Tn = inf t ≥ 0 : |Xt| ≥ n or |dAs| ≥ n t 0 Then by the previous claim, we have f (X Tn t ) = f (X0) + + 1 2 = f (X0) + + 1 2 0 t∧Tn 0 t∧Tn i,j p i=1 i,j 0 p t 0 t i=1 s ) d(Xi)Tn s ∂f ∂xi (X Tn ∂2f ∂xi∂xj (X Tn s ) d(Xi)Tn , (X j)Tns ∂f ∂xi (Xs) d(Xi)s ∂2f ∂xi∂xj (Xs) d(Xi), (X j)s. Then take Tn → ∞. Example. Let B be a standard Brownian motion, B0 = 0 and f (x) = x2. Then In other words, B2 t = 2 t 0 BS dBs + t. B2 t − t = 2 t 0 Bs dBs. In particular, this is a continuous local martingale. Example. Let B = (B1, . . . , Bd) be a d-dimensional Brownian motion. Then we apply Itˆo’s formula to the semi-martingale X = (t, B1, . . . , Bd). Then we find that f (t, Bt) − f (0, B0) − t 0 ∂ ∂s + 1 2 ∆ f (s, Bs) ds = d t i=1 0 ∂ ∂xi f (s, Bs) dBi s is a continuous local martingale. There are some syntactic tricks that make stochastic integrals easier to manipulate, namely by working in differential form. We can state Itˆo’s formula in differential form df (Xt) = p i=1 ∂f ∂xi dX i + 1 2 p i,j=1 ∂2f ∂xi∂xj dX i, X j, which we can think of as the chain rule. For example, in the case case of Brownian motion, we have 1 2 f (Bt) dt. df (Bt) = f (Bt) dBt + 34 3 The stochastic integral III Stochastic Calculus and Applications Formally, one expands f using that that “(dt)2 = 0” but “(dB)2 = dt”. The following formal rules hold: t Zt − Z0 = Hs dXs ⇐⇒ dZt = Ht dXt Zt = X, Y t = 0 t 0 dX, Y t ⇐⇒ dZt = dXt dYt. Then we have rules such as Ht(Kt dXt) = (HtKt) dXt Ht(dXt dYt) = (Ht dXt) dYt d(XtYt) = Xt dYt + Yt dXt + dXt dYt df (Xt) = f (Xt) dXt + 1 2 f (Xt) dXt dXt. 3.6 The L´evy characterization A more major application of the stochastic integral is the following convenient characterization of Brownian motion: Theorem (L´evy’s characterization of Brownian motion). Let (X 1, . . . , X d) be continuous local martingales. Suppose that X0 = 0 and that X i, X jt = δijt for all i, j = 1, . . . , d and t ≥ 0. Then (X 1, . . . , X d) is a standard d-dimensional Brownian motion. This might seem like a rather artificial condition, but it turns out to be quite useful in practice (though less so in this course). The point is that we know that , and in particular if we are integrating things with respect to Brownian motions of some sort, we know Btt = t, and so we are left with some explicit, familiar integral to do. Proof. Let 0 ≤ s < t. It suffices to check that Xt − Xs is independent of Fs and Xt − Xs ∼ N (0, (t − s)I). Claim. E(eiθ·(Xt−Xs) | Fs) = e− 1 2 |θ|2(t−s) for all θ ∈ Rd and s < t. This is sufficient, since the right-hand side is independent of Fs, hence so is the left-hand side, and the Fourier transform characterizes the distribution. To check this, for θ ∈ Rd, we define Yt = θ · Xt = d i=1 θiX i t . Then Y is a continuous local martingale, and we have Y t = Y, Y t = d i,j=1 θjθk X j, X kt = |θ|2t. by assumption. Let Zt = eiYt+ 1 2 Y t = eiθ·Xt+ 1 2 |θ|2t. 35 3 The stochastic integral III Stochastic Calculus and Applications By Itˆo’s formula, with X = iY + 1 2 Y t and f (x) = ex, we get dZt = Zt idYt − 1 2 dY t + 1 2 dY t = iZt dYt. So this implies Z is a continuous local martingale. Moreover, since Z is bounded on bounded intervals of t, we know Z is in fact a martingale, and Z0 = 1. Then by definition of a martingale, we have E(Zt | Fs) = Zs, And unwrapping the definition of Zt shows that the result follows. In general, th |
e quadratic variation of a process doesn’t have to be linear in t. It turns out if the quadratic variation increases to infinity, then the martingale is still a Brownian motion up to reparametrization. Theorem (Dubins–Schwarz). Let M be a continuous local martingale with M0 = 0 and M ∞ = ∞. Let Ts = inf{t ≥ 0 : M t > s}, the right-continuous inverse of M t. Let Bs = MTs and Gs = FTs . Then Ts is a (Ft) stopping time, M Ts = s for all s ≥ 0, B is a (Gs)-Brownian motion, and Mt = BM t. Proof. Since M is continuous and adapted, and M ∞ = ∞, we know Ts is a stopping time and Ts < ∞ for all s ≥ 0. Claim. (Gs) is a filtration obeying the usual conditions, and G∞ = F∞ Indeed, if A ∈ Gs and s < t, then A ∩ {Tt ≤ u} = A ∩ {Ts ≤ u} ∩ {Tt ≤ u} ∈ Fu, using that A ∩ {Ts ≤ u} ∈ Fu since A ∈ Gs. Then right-continuity follows from that of (Ft) and the right-continuity of s → Ts. Claim. B is adapted to (Gs). In general, if X is c´adl´ag and T is a stopping time, then XT 1{T <∞} ∈ FT . Apply this is with X = M , T = Ts and FT = Gs. Thus Bs ∈ Gs. Claim. B is continuous. Here this is actually something to verify, because s → Ts is only right continuous, not necessarily continuous. Thus, we only know Bs is right continuous, and we have to check it is left continuous. Now B is left-continuous at s iff Bs = Bs− , iff MTs = MTs− . Now we have Ts− = inf{t ≥ 0 : M t ≥ s}. If Ts = Ts−, then there is nothing to show. Thus, we may assume Ts > Ts−. Then we have M Ts = M Ts− . Since M t is increasing, it means M Ts is constant in [Ts−, Ts]. We will later prove that 36 3 The stochastic integral III Stochastic Calculus and Applications Lemma. M is constant on [a, b] iff M being constant on [a, b]. So we know that if Ts > Ts−, then MTs = MTs−. So B is left continuous. We then have to show that B is a martingale. Claim. (M 2 − M )Ts is a uniformly integrable martingale. To see this, observe that M Ts ∞ = M Ts = s, and so M Ts is bounded. So (M 2 − M )Ts is a uniformly integrable martingale. We now apply the optional stopping theorem, which tells us E(Bs | Gr) = E(M Ts ∞ | Gs) = MTt = Bt. So Bt is a martingale. Moreover, E(B2 s − s | Gr) = E((M 2 − M )Ts | FTr ) = M 2 Tr − M Tr = B2 r − r. So B2 t − t is a martingale, so by the characterizing property of the quadratic variation, Bt = t. So by L´evy’s criterion, this is a Brownian motion in one dimension. The theorem is only true for martingales in one dimension. In two dimensions, this need not be true, because the time change needed for the horizontal and vertical may not agree. However, in the example sheet, we see that the holomorphic image of a Brownian motion is still a Brownian motion up to a time change. Lemma. M is constant on [a, b] iff M being constant on [a, b]. Proof. It is clear that if M is constant, then so is M . To prove the converse, by continuity, it suffices to prove that for any fixed a < b, {Mt = Ma for all t ∈ [a, b]} ⊇ {M b = M a} almost surely. We set Nt = Mt − Mt ∧ a. Then N t = M t − M t∧a. Define Tε = inf{t ≥ 0 : N t ≥ ε}. Then since N 2 − N is a local martingale, we know that E(N 2 t∧Tε ) = E(N t∧Tε) ≤ ε. Now observe that on the event {M b = M a}, we have N b = 0. So for t ∈ [a, b], we have E(1{M b=M a}N 2 t ) = E(1{M b=M a N 2 t∧Tε ) = E(N t∧Tε) = 0. 3.7 Girsanov’s theorem Girsanov’s theorem tells us what happens to our (semi)-martingales when we change the measure of our space. We first look at a simple example when we perform a shift. 37 3 The stochastic integral III Stochastic Calculus and Applications Example. Let X ∼ N (0, C) be an n-dimensional centered Gaussian with positive definite covariance C = (Cij)n i,j=1. Put M = C −1. Then for any function f , we have Ef (X) = det M 2π 1/2 Rn f (x)e− 1 2 (x,M x) dx. Now fix an a ∈ Rn. The distribution of X + a then satisfies Ef (X + a) = det M 2π 1/2 Rn f (x)e− 1 2 (x−a,M (x−a)) dx = E[Zf (X)], where Z = Z(x) = e− 1 Thus, if P denotes the distribution of X, then the measure Q with 2 (a,M a)+(x,M a). dQ dP = Z is that of N (a, C) vector. Example. We can extend the above example to Brownian motion. Let B be a Brownian motion with B0 = 0, and h : [0, ∞) → R a deterministic function. We then want to understand the distribution of Bt + h. Fix a finite sequence of times 0 = t0 < t1 < · · · < tn. Then we know that i=1 is a centered Gaussian random variable. Thus, if f (B) = f (Bt1 , . . . , Btn ) (Bti)n is a function, then E(f (B)) = c · Rn − 1 2 n i=1 f (x)e (xi−xi−1)2 ti−ti−1 dx1 · · · dxn. Thus, after a shift, we get E(f (B + h)) = E(Zf (B)), Z = exp − 1 2 n i=1 (hti − hti−1)2 ti − ti−1 + n i=1 (hti − hti−1)(Bti − Bti−1) ti − ti−1 . In general, we are interested in what happens when we change the measure by an exponential: Definition (Stochastic exponential). Let M be a continuous local martingale. Then the stochastic exponential (or Dol´eans–Dade exponential ) of M is E(M )t = eMt− 1 2 M t The point of introducing that quadratic variation term is Proposition. Let M be a continuous local martingale with M0 = 0. Then E(M ) = Z satisfies i.e. dZt = Zt dM, Zt = 1 + t Zs dMs. 0 In particular, E(M ) is a continuous local martingale. Moreover, if M is uniformly bounded, then E(M ) is a uniformly integrable martingale. 38 3 The stochastic integral III Stochastic Calculus and Applications There is a more general condition for the final property, namely Novikov’s condition, but we will not go into that. Proof. By Itˆo’s formula with X = M − 1 dZt = Ztd Mt − 1 2 2 M , we have 1 2 dM t + ZtdM t = Zt dMt. Since M is a continuous local martingale, so is Zs dMs. So Z is a continuous local martingale. Now suppose M ∞ ≤ b < ∞. Then P Mt ≥ a sup t≥0 = P sup t≥0 Mt ≥ a, M ∞ ≤ b ≤ e−a2/2b, where the final equality is an exercise on the third example sheet, which is true for general continuous local martingales. So we get E exp Mt sup t ∞ 0 ∞ = = P(exp(sup Mt) ≥ λ) dλ P(sup Mt ≥ log λ) dλ 0 ≤ 1 + ∞ 1 e−(log λ)2/2b dλ < ∞. Since M ≥ 0, we know that E(M )t ≤ exp (sup Mt) , sup t≥0 So E(M ) is a uniformly integrable martingale. Theorem (Girsanov’s theorem). Let M be a continuous local martingale with M0 = 0. Suppose that E(M ) is a uniformly integrable martingale. Define a new probability measure dQ dP = E(M )∞ Let X be a continuous local martingale with respect to P. Then X − X, M is a continuous local martingale with respect to Q. Proof. Define the stopping time Tn = inf{t ≥ 0 : |Xt − X, M t| ≥ n}, and P(Tn → ∞) = 1 by continuity. Since Q is absolutely continuous with respect to P, we know that Q(Tn → ∞) = 1. Thus it suffices to show that X Tn − X Tn , M is a continuous martingale for any n. Let Y = X Tn − X Tn , M , Z = E(M ). Claim. ZY is a continuous local martingale with respect to P. 39 3 The stochastic integral III Stochastic Calculus and Applications We use the product rule to compute d(ZY ) = Yt dZt + Zt dYt + dY, Zt = Y Zt dMt + Zt(dX Tn − dX Tn, M t) + Zt dM, X Tn = Y Zt dMt + Zt dX Tn So we see that ZY is a stochastic integral with respect to a continuous local martingale. Thus ZY is a continuous local martingale. Claim. ZY is uniformly integrable. Since Z is a uniformly integrable martingale, {ZT : T is a stopping time} is uniformly integrable. Since Y is bounded, {ZT YT : T is a stopping time} is also uniformly integrable. So ZY is a true martingale (with respect to P). Claim. Y is a martingale with respect to Q. We have EQ(Yt − Ys | Fs) = EP(Z∞Yt − Z∞Ys | Fs) = EP(ZtYt − ZsYs | Fs) = 0. Note that the quadratic variation does not change since X − X, M = Xt = lim n→∞ 2nt i=1 (Xi2−n − X(i−1)2−n )2 a.s. along a subsequence. 40 4 Stochastic differential equations III Stochastic Calculus and Applications 4 Stochastic differential equations 4.1 Existence and uniqueness of solutions After all this work, we can return to the problem we described in the introduction. We wanted to make sense of equations of the form ˙x(t) = F (x(t)) + η(t), where η(t) is Gaussian white noise. We can now interpret this equation as saying dXt = F (Xt) dt + dBt, or equivalently, in integral form, Xt − X0 = T 0 F (Xs) ds + Bt. In general, we can make the following definition: Definition (Stochastic differential equation). Let d, m ∈ N, b : R+ × Rd → Rd, σ : R+ × Rd → Rd×m be locally bounded (and measurable). A solution to the stochastic differential equation E(σ, b) given by dXt = b(t, Xt) dt + σ(t, Xt) dBt consists of (i) a filtered probability space (Ω, F, (Ft), P) obeying the usual conditions; (ii) an m-dimensional Brownian motion B with B0 = 0; and (iii) an (Ft)-adapted continuous process X with values in Rd such that Xt = X0 + t 0 σ(s, Xs) dBs + t 0 b(s, Xs) ds. If X0 = x ∈ Rd, then we say X is a (weak) solution to Ex(σ, b). It is a strong solution if it is adapted with respect to the canonical filtration of B. Our goal is to prove existence and uniqueness of solutions to a general class of SDEs. We already know what it means for solutions to be unique, and in general there can be multiple notions of uniqueness: Definition (Uniqueness of solutions). For the stochastic differential equation E(σ, b), we say there is – uniqueness in law if for every x ∈ Rd, all solutions to Ex(σ, b) have the same distribution. – pathwise uniqueness if when (Ω, F, (Ft), P) and B are fixed, any two solutions X, X with X0 = X 0 are indistinguishable. These two notions are not equivalent, as the following example shows: 41 4 Stochastic differential equations III Stochastic Calculus and Applications Example (Tanaka). Consider the stochastic differential equation where dXt = sgn(Xt) dBt, X0 = x, sgn(x) = +1 x > 0 −1 x ≤ 0 . This has a weak solution which is unique in law, but pathwise uniqueness fails. To see the existence of solutions, let X be a one-dimensional Brownian motion with X0 = x, and set Bt = t 0 sgn(Xs) dXs, which is well-defined because sgn(Xs) is previsible and left-continuous. Then we have x + t 0 sgn(Xs) dBs = x + t 0 sgn(Xs)2 dXs = x + Xt − X0 = Xt. So it remains to show that B is a Brownian motion. We already know that B is a continuous local martingale, so by L´evy’s characterization, it suffices t |
o show its quadratic variation is t. We simply compute B, Bt = t 0 dXs, Xs = t. So there is weak existence. The same argument shows that any solution is a Brownian motion, so we have uniqueness in law. Finally, observe that if x = 0 and X is a solution, then −X is also a solution with the same Brownian motion. Indeed, −Xt = t 0 sgn(Xs) dBs = t 0 sgn(−Xs) dBs + 2 t 0 1Xs=0 dBs, where the second term vanishes, since it is a continuous local martingale with quadratic variation t 0 1Xs=0 ds = 0. So pathwise uniqueness does not hold. In the other direction, however, it turns out pathwise uniqueness implies uniqueness in law. Theorem (Yamada–Watanabe). Assume weak existence and pathwise uniqueness holds. Then (i) Uniqueness in law holds. (ii) For every (Ω, F, (Ft), P) and B and any x ∈ Rd, there is a unique strong solution to Ex(a, b). We will not prove this, since we will not actually need it. The key, important theorem we are now heading for is the existence and uniqueness of solutions to SDEs, assuming reasonable conditions. As in the case of ODEs, we need the following Lipschitz conditions: 42 4 Stochastic differential equations III Stochastic Calculus and Applications Definition (Lipschitz coefficients). The coefficients b : R+ × Rd → Rd, σ : R+ × Rd → Rd×m are Lipschitz in x if there exists a constant K > 0 such that for all t ≥ 0 and x, y ∈ Rd, we have |b(t, x) − b(t, y)| ≤ K|x − y| |σ(t, x) − σ(t, y)| ≤ |x − y| Theorem. Assume b, σ are Lipschitz in x. Then there is pathwise uniqueness for the E(σ, b) and for every (Ω, F, (Ft), P) satisfying the usual conditions and every (Ft)-Brownian motion B, for every x ∈ Rd, there exists a unique strong solution to Ex(σ, b). Proof. To simplify notation, we assume m = d = 1. We first prove pathwise uniqueness. Suppose X, X are two solutions with t)2] = 0. We will actually put some 0. We will show that E[(Xt − X X0 = X bounds to control our variables. Define the stopping time S = inf{t ≥ 0 : |Xt| ≥ n or |X t| ≥ n}. By continuity, S → ∞ as n → ∞. We also fix a deterministic time T > 0. Then whenever t ∈ [0, T ], we can bound, using the identity (a + b)2 ≤ 2a2 + 2b2, E((Xt∧S − X t∧S)2) ≤ 2E t∧S 0 (σ(s, Xs) − σ(s, X s)) dBs 2 + 2E t∧S 0 (b(s, Xs) − b(s, X s)) ds 2 . We can apply the Lipschitz bound to the second term immediately, while we can simplify the first term using the (corollary of the) Itˆo isometry t∧S E 0 (σ(s, Xs) − σ(s, X s)) dBs 2 = E t∧S 0 (σ(s, Xs) − σ(s, X s))2 ds . So using the Lipschitz bound, we have E((Xt∧S − X t∧S)2) ≤ 2K 2(1 + T )E t∧S 0 |Xs − X s|2 ds ≤ 2K 2(1 + T ) t 0 E(|Xs∧S − X s∧S|2) ds. We now use Gr¨onwall’s lemma: Lemma. Let h(t) be a function such that for some constant c. Then h(t) ≤ c t 0 h(s) ds h(t) ≤ h(0)ect. 43 4 Stochastic differential equations III Stochastic Calculus and Applications Applying this to t∧S)2), we deduce that h(t) ≤ h(0)ect = 0. So we know that h(t) = E((Xt∧S − X E(|Xt∧S − X t∧S|2) = 0 for every t ∈ [0, T ]. Taking n → ∞ and T → ∞ gives pathwise uniqueness. We next prove existence of solutions. We fix (Ω, F, (Ft)t) and B, and define F (X)t = X0 + t 0 σ(s, Xs) dBs + t 0 b(s, Xs) ds. Then X is a solution to Ex(a, b) iff F (X) = X and X0 = x. To find a fixed point, we use Picard iteration. We fix T > 0, and define the T -norm of a continuous adapted process X as XT = E sup t≤T 1/2 . |Xt|2 In particular, if X is a martingale, then this is the same as the norm on the space of L2-bounded martingales by Doob’s inequality. Then B = {X : Ω × [0, T ] → R : XT < ∞} is a Banach space. Claim. F (0)T < ∞, and F (X) − F (Y )2 T ≤ (2T + 8)K 2 T 0 X − Y 2 t dt. We first see how this claim implies the theorem. First observe that the claim implies F indeed maps B into itself. We can then define a sequence of processes X i t by X 0 t = x, X i+1 = F (X i). Then we have X i+1 − X i2 T ≤ CT T 0 So we find that X i − X i−12 t dt ≤ · · · ≤ X 1 − X 02 T CT i i! . ∞ i=1 X i − X i−12 T < ∞ for all T . So X i converges to X almost surely and uniformly on [0, T ], and F (X) = X. We then take T → ∞ and we are done. To prove the claim, we write F (0)T ≤ |X0| + t 0 b(s, 0) ds + 44 t 0 σ(s, 0) dBs T . 4 Stochastic differential equations III Stochastic Calculus and Applications The first two terms are constant, and we can bound the last by Doob’s inequality and the Itˆo isometry: t 0 σ(s, 0) dBs T ≤ 2E T 0 σ(s, 0) dBs 2 = 2 T 0 σ(s, 0)2 ds. To prove the second part, we use F (X) − F (Y )2 ≤ 2E sup t≤T t 0 b(s, X − s) − b(s, Ys) ds 2 + 2E sup t≤T t 0 (σ(s, Xs) − σ(s, Ys)) dBs 2 . We can bound the first term with Cauchy–Schwartz by T E T 0 |b(s, Xs) − b(s, Ys)| dt, and the second term with Doob’s inequality by T E 0 |σ(s, Xs) − σ(s, Ys)|2 ds ≤ 4K 2 T 0 X − Y 2 t dt. 4.2 Examples of stochastic differential equations Example (The Ornstein–Uhlenbeck process). Let λ > 0. Then the Ornstein– Uhlenbeck process is the solution to dXt = −λXt dt + dBt. The solution exists by the previous theorem, but we can also explicitly find one. By Itˆo’s formula applied to eλtXt, we get So we find that d(eλtXt) = eλtdXt + λeλtXt dt = dBt. Xt = e−λtX0 + t 0 e−λ(t−s) dBs. Observe that the integrand is deterministic. So we can in fact interpret this as an Wiener integral. Fact. If X0 = x ∈ R is fixed, then (Xt) is a Gaussian process, i.e. (Xti)n i=1 is jointly Gaussian for all t1 < · · · < tn. Any Gaussian process is determined by the mean and covariance, and in this case, we have EXt = e−λtx, cov(Xt, Xs) = e−λ|t−s| − e−λ|t+s| 1 2λ Proof. We only have to compute the covariance. By the Itˆo isometry, we have t s E((Xt − EXt)(Xs − EXs)) = E e−λ(t−u) dBu e−λ(s−u) dBu 0 0 = e−λ(t+s) t∧s 0 eλu du. 45 4 Stochastic differential equations III Stochastic Calculus and Applications In particular, Xt ∼ N e−λtx, 1 − e−2λt 2λ → N 0, . 1 2λ Fact. If X0 ∼ N (0, 1 covariance, i.e. the covariance depends only on time differences: 2λ ), then (Xt) is a centered Gaussian process with stationary 1 2λ The difference is that in the deterministic case, the EXt cancels the first cov(Xt, Xs) = e−λ|t−s|. e−λtX0 term, while in the non-deterministic case, it doesn’t. This is a very nice example where we can explicitly understand the long-time behaviour of the SDE. In general, this is non-trivial. Dyson Brownian motion Let HN be an inner product space of real symmetric N × N matrices with inner product N Tr(HK) for H, K ∈ HN . Let H 1, . . . , H dim(HN ) be an orthonormal basis for HN . Definition (Gaussian orthogonal ensemble). The Gaussian Orthogonal Ensemble GOEN is the standard Gaussian measure on HN , i.e. H ∼ GOEN if H = dim Hn r=1 H iX i where each X i are iid standard normals. We now replace each X i by a Ornstein–Uhlenbeck process with λ = 1 2 . Then GOEN is invariant under the process. Theorem. The eigenvalues λ1(t) ≤ · · · ≤ λN (t) satisfies dλi t = − λi 2 + 1 N j=i 1 λi − λj dt + 2 N β dBi. Here β = 1, but if we replace symmetric matrices by Hermitian ones, we get β = 2; if we replace symmetric matrices by symplectic ones, we get β = 4. This follows from Itˆo’s formula and formulas for derivatives of eigenvalues. Example (Geometric Brownian motion). Fix σ > 0 and t ∈ R. Then geometric Brownian motion is given by dXt = σXt dBt + rXt dt. We apply Itˆo’s formula to log Xt to find that Xt = X0 exp σBt + r − t . σ2 2 46 4 Stochastic differential equations III Stochastic Calculus and Applications Example (Bessel process). Let B = (B1, . . . , Bd) be a d-dimensional Brownian motion. Then satisfies the stochastic differential equation Xt = |Bt| dXt = d − 1 2Xt dt + dBt if t < inf{t ≥ 0 : Xt = 0}. 4.3 Representations of solutions to PDEs Recall that in Advanced Probability, we learnt that we can represent the solution to Laplace’s equation via Brownian motion, namely if D is a suitably nice domain and g : ∂D → R is a function, then the solution to the Laplace’s equation on D with boundary conditions g is given by u(x) = Ex[g(BT )], where T is the first hitting time of the boundary ∂D. A similar statement we can make is that if we want to solve the heat equation ∂u ∂t = ∇2u with initial conditions u(x, 0) = u0(x), then we can write the solution as u(x, t) = Ex[u0( √ 2Bt)] This is just a fancy way to say that the Green’s function for the heat equation is a Gaussian, but is a good way to think about it nevertheless. In general, we would like to associate PDEs to certain stochastic processes. Recall that a stochastic PDE is generally of the form dXt = b(Xt) dt + σ(Xt) dBt for some b : Rd → R and σ : Rd → Rd×m which are measurable and locally bounded. Here we assume these functions do not have time dependence. We can then associate to this a differential operator L defined by L = 1 2 i,j aij∂i∂j + i bi∂i. where a = σσT . Example. If b = 0 and σ = √ 2I, then L = ∆ is the standard Laplacian. The basic computation is the following result, which is a standard application of the Itˆo formula: Proposition. Let x ∈ Rd, and X a solution to Ex(σ, b). Then for every f : R+ × Rd → R that is C 1 in R+ and C 2 in Rd, the process M f t = f (t, Xt) − f (0, X0) − t 0 ∂ ∂s + L f (s, Xs) ds is a continuous local martingale. 47 4 Stochastic differential equations III Stochastic Calculus and Applications We first apply this to the Dirichlet–Poisson problem, which is essentially to solve −Lu = f . To be precise, let U ⊆ Rd be non-empty, bounded and open; f ∈ Cb(U ) and g ∈ Cb(∂U ). We then want to find a u ∈ C 2( ¯U ) = C 2(U ) ∩ C( ¯U ) such that −Lu(x) = f (x) for x ∈ U u(x) = g(x) for x ∈ ∂U. If f = 0, this is called the Dirichlet problem; if g = 0, this is called the Poisson problem. We will have to impose the following technical condition on a: Definition (Uniformly elliptic). We say a : ¯U → Rd×d is uniformly elliptic if there is a constant c > 0 such that for all ξ ∈ Rd and x ∈ ¯U , we have ξT a(x)ξ ≥ c|ξ|2. If a is symmetric (which it is in our case), this is the same as asking for the smallest eigenvalue of a to be bounded away from 0. It would be very nice if we can write down a solution to the Dirichlet–Poisson problem using a solut |
ion to Ex(σ, b), and then simply check that it works. We can indeed do that, but it takes a bit more time than we have. Instead, we shall prove a slightly weaker result that if we happen to have a solution, it must be given by our formula involving the SDE. So we first note the following theorem without proof: Theorem. Assume U has a smooth boundary (or satisfies the exterior cone condition), a, b are H¨older continuous and a is uniformly elliptic. Then for every H¨older continuous f : ¯U → R and any continuous g : ∂U → R, the Dirichlet–Poisson process has a solution. The main theorem is the following: Theorem. Let σ and b be bounded measurable and σσT uniformly elliptic, U ⊆ Rd as above. Let u be a solution to the Dirichlet–Poisson problem and X a solution to Ex(σ, b) for some x ∈ Rd. Define the stopping time Then ETU < ∞ and TU = inf{t ≥ 0 : Xt ∈ U }. u(x) = Ex g(XTU ) + f (Xs) ds . TU 0 In particular, the solution to the PDE is unique. Proof. Our previous proposition applies to functions defined on all of Rn, while u is just defined on U . So we set Un = x ∈ U : dist(x, ∂U ) > 1 n , Tn = inf{t ≥ 0 : Xt ∈ Un}, and pick un ∈ C 2 let b (Rd) such that u|Un = un|Un . Recalling our previous notation, M n t = (M un )Tn t = un(Xt∧Tn ) − un(X0) − 48 t∧Tn 0 Lun(Xs) ds. 4 Stochastic differential equations III Stochastic Calculus and Applications Then this is a continuous local martingale that is bounded by the proposition, and is bounded, hence a true martingale. Thus for x ∈ U and n large enough, the martingale property implies u(x) = un(x) = E u(Xt∧Tn) − Lu(Xs) ds t∧Tn 0 = E u(Xt∧Tn ) + f (Xs) ds . t∧Tn 0 We would be done if we can take n → ∞. To do so, we first show that E[TU ] < ∞. Note that this does not depend on f and g. So we can take f = 1 and g = 0, and let v be a solution. Then we have E(t ∧ Tn) = E − t∧Tn 0 Lv(Xs) ds = v(x) − E(v(Xt∧Tn)). Since v is bounded, by dominated/monotone convergence, we can take the limit to get E(TU ) < ∞. Thus, we know that t ∧ Tn → TU as t → ∞ and n → ∞. Since TU E 0 |f (Xs)| ds ≤ f ∞E[TU ] < ∞, the dominated convergence theorem tells us t∧Tn E f (Xs) ds → E TU f (Xs) ds . 0 0 Since u is continuous on ¯U , we also have E(u(Xt∧Tn )) → E(u(Tu)) = E(g(Tu)). We can use SDE’s to solve the Cauchy problem for parabolic equations as b (Rd), we well, just like the heat equation. The problem is as follows: for f ∈ C 2 want to find u : R+ × Rd → R that is C 1 in R+ and C 2 in Rd such that = Lu ∂u ∂t u(0, · ) = f on R+ × Rd on Rd Again we will need the following theorem: Theorem. For every f ∈ C 2 b (Rd), there exists a solution to the Cauchy problem. Theorem. Let u be a solution to the Cauchy problem. Let X be a solution to Ex(σ, b) for x ∈ Rd and 0 ≤ s ≤ t. Then In particular, Ex(f (Xt) | Fs) = u(t − s, Xs). u(t, x) = Ex(f (Xt)). 49 4 Stochastic differential equations III Stochastic Calculus and Applications In particular, this implies Xt is a continuous Markov process. Proof. The martingale has ∂ g(s, x) = u(t − s, x). Then ∂t + L, but the heat equation has ∂ ∂t − L. So we set ∂ ∂s + L g(s, x) = − ∂ ∂t u(t − s, x) + Lu(t − s, x) = 0. So g(s, Xs) − g(0, X0) is a martingale (boundedness is an exercise), and hence u(t − s, Xs) = g(s, Xs) = E(g(t, Xt) | Fs) = E(u(0, Xt) | Fs) = E(f (Xt) | Xs). There is a generalization to the Feynman–Kac formula. Theorem (Feynman–Kac formula). Let f ∈ C 2 suppose that u : R+ × Rd → R satisfies b (Rd) and V ∈ Cb(Rd) and = Lu + V u ∂u ∂t u(0, · ) = f on R+ × Rd on Rd, where V u = V (x)u(x) is given by multiplication. Then for all t > 0 and x ∈ Rd, and X a solution to Ex(σ, b). Then u(t, x) = Ex f (Xt) exp t 0 V (Xs) ds . If L is the Laplacian, then this is Schr¨odinger equation, which is why Feynman was thinking about this. 50 Index Index (H · A)t, 9 H · X, 30 L2(M ), 26 L2 bc(M ), 29 X T , 11 E, 10 M2, 16 Bessel process, 47 bounded variation, 7 bracket, 22 BV, 7 c`adl`ag, 7 c`adl`ag adapted process, 9 Cauchy problem, 49 covariation, 22 Dirichlet problem, 48 Dirichlet–Poisson problem, 48 Dol´eans–Dade exponential, 38 Doob’s inequality, 16 Dubins–Schwarz theorem, 36 Dyson Brownian motion, 46 Feynman–Kac formula, 50 finite variation, 8 Finite variation process, 9 Gaussian orthogonal ensemble, 46 Gaussian space, 4 Gaussian white noise, 4 Geometric Brownian motion, 46 Girsanov’s theorem, 39 Hahn decomposition, 6 integration by parts, 32 Itˆo correction, 32 Itˆo isometry, 27 Itˆo’s formula, 32 Kunita–Watanabe, 23 III Stochastic Calculus and Applications Lebesgue–Stieltjes integral, 8 Lipschitz coefficients, 43 local martingale, 12 locally bounded previsible process, 30 Ornstein–Uhlenbeck process, 45 pathwise uniqueness, 41 Poisson problem, 48 previsible σ-algebra, 10 previsible process, 10 quadratic variation, 18 semi-martingale, 24 reduces, 12 semi-martingale, 24 signed measure, 6 simple process, 10, 25 stochastic differential equation, 41 stochastic dominated convergence theorem, 31 stochastic exponential, 38 stochastic integral, 27, 30 stopped process, 11 strong solution, 41 Tanaka equation, 42 total variation, 6, 7 total variation process, 9 u.c.p., 17 uniformly elliptic, 48 uniformly on compact sets in probability, 17 uniqueness in law, 41 usual conditions, 11 Vitali theorem, 13 weak solution, 41 L´evy’s characterization of Brownian motion, 35 Yamada–Watanabe, 42 51 |
dzk = i 2 j,k hjk dzk ∧ ¯zj. So we have hkj = hjk. The non-degeneracy condition ω∧n = 0 is equivalent to det hjk = 0, since ωn = n i 2 n! det(hjk) dz1 ∧ d¯z1 ∧ · · · ∧ dzn ∧ d¯zn. So hjk is a non-singular Hermitian matrix. Finally, we take into account the compatibility condition ω(v, Jv) > 0. If we write then we have So we have v = aj j ∂ ∂zj + bj ∂ ∂ ¯zj , Jv = i j aj ∂ ∂zj − bj . ∂ ∂ ¯zj ω(v, Jv) = i 2 hjk(−iajbk − iajbk) = hjkajbk > 0. So the conclusion is that hjk is positive definite. Thus, the conclusion is Theorem. A K¨ahler form ω on a complex manifold M is a ∂- and ¯∂-closed form of type (1, 1) which on a local chart is given by ω = i 2 j,k hjk dzj ∧ d¯zk where at each point, the matrix (hjk) is Hermitian and positive definite. Often, we start with a complex manifold, and want to show that it has a K¨ahler form. How can we do so? First observe that we have the following proposition: 22 2 Complex structures III Symplectic Geometry Proposition. Let (M, ω) be a complex K¨ahler manifold. If X ⊆ M is a complex submanifold, then (X, i∗ω) is K¨ahler, and this is called a K¨ahler submanifold . In particular, if we can construct K¨ahler forms on Cn and CPn, then we have K¨ahler forms for a lot of our favorite complex manifolds, and in particular complex projective varieties. But we still have to construct some K¨ahler forms to begin with. To do so, we use so-called strictly plurisubharmonic functions. Definition (Strictly plurisubharmonic (spsh)). A function ρ ∈ C∞(M, R) is strictly plurisubharmonic (spsh) if locally, is positive definite. ∂2ρ ∂zj ∂ ¯zk Proposition. Let M be a complex manifold, and ρ ∈ C∞(M ; R) strictly plurisubharmonic. Then ω = ∂ ¯∂ρ i 2 is a K¨ahler form. We call ρ the K¨ahler potential for ω. Proof. ω is indeed a 2-form of type (1, 1). Since ∂2 = ¯∂2 = 0 and ∂ ¯∂ = − ¯∂∂, we know ∂ω = ¯∂ω = 0. We also have ω = i 2 j,k ∂2ρ ∂zj∂ ¯zk dzj ∧ d¯zk, and the matrix is Hermitian positive definite by assumption. Example. If M = Cn ∼= R2n, we take Then we have ρ(z) = |z|2 = zk ¯zk. hjk = ∂2ρ ∂zj∂ ¯zk = δjk, so this is strictly plurisubharmonic. Then i 2 i 2 ω = = = dxk ∧ dyk, dzj ∧ d¯zk d(xj + iyj) ∧ d(xk − iyk) which is the standard symplectic form. So (Cn, ω) is K¨ahler and ρ = |z|2 is a (global) K¨ahler potential for ω0. There is a local converse to this result. Proposition. Let M be a complex manifold, ω a closed real-valued (1, 1)-form and p ∈ M , then there exists a neighbourhood U of p in M and a ρ ∈ C∞(U, R) such that ω = i∂ ¯∂ρ on U. 23 2 Complex structures III Symplectic Geometry Proof. This uses the holomorphic version of the Poincar´e lemma. When ρ is K¨ahler, such a ρ is called a local K¨ahler potential for ω. Note that it is not possible to have a global K¨ahler potential on a closed K¨ahler manifold, because if ω = i 2 ∂ ¯∂ρ, then ω = d ¯∂ρ i 2 is exact, and we know symplectic forms cannot be exact. Example. Let M = Cn and ρ(z) = log(|z|2 + 1). It is an exercise to check that ρ is strictly plurisubharmonic. Then ωF S = i 2 ∂ ¯∂(log(|z2| + 1)) is another K¨ahler form on Cn, called the Fubini–Study form. The reason this is interesting is that it allows us to put a K¨ahler structure on CPn. Example. Let M = CPn. Using homogeneous coordinates, this is covered by the open sets Uj = {[z0, . . . , zn] ∈ CPn | zj = 0}. with the chart given by ϕj : Uj → Cn z0 zj [z0, . . . , zn] → , . . . , zj−1 zj , zj+1 zj , . . . , . zn zj One can check that ϕ∗ form on CPn, making it a K¨ahler manifold. j ωF S = ϕ∗ kωF S. Thus, these glue to give the Fubini–Study 2.4 Hodge theory So what we have got so far is that if we have a complex manifold, then we can decompose Ωk(M ; C) = Ωp,q, and using ¯∂ : Ωp,q → Ωp,q+1, we defined the Dolbeault cohomology groups p+q=k H p,q Dolb(M ) = ker ¯∂ im ¯∂ . It would be nice if we also had a decomposition H k dR ∼= H p,q Dolb(M ). p+q=k This is not always true. However, it is true for compact K¨ahler manifolds: 24 2 Complex structures III Symplectic Geometry Theorem (Hodge decomposition theorem). Let (M, ω) be a compact K¨haler manifold. Then H k dR ∼= H p,q Dolb(M ). p+q=k To prove the theorem, we will first need a “real” analogue of the theorem. This is an analytic theorem that lets us find canonical representatives for each cohomology class. We can develop the same theory for Dolbeault cohomology, and see that the canonical representatives for Dolbeault cohomology are the same as those for de Rham cohomology. We will not prove the analytic theorems, but just say how these things piece together. Real Hodge theory Let V be a real oriented vector space m with inner product G. Then this induces an inner product on each Λk = k(V ), denoted · , · , defined by v1 ∧ · · · ∧ vk, w1 ∧ · · · ∧ wk = det(G(vi, wj))i,j Let e1, . . . , en be an oriented orthonormal basis for V . Then {ej1 ∧ · · · ∧ ejk : 1 ≤ j1 < · · · < jk ≤ m} is an orthonormal basis for Λk. Definition (Hodge star). The Hodge ∗-operator ∗ : Λk → Λm−k is defined by the relation α ∧ ∗β = α, β e1 ∧ · · · ∧ em. It is easy to see that the Hodge star is in fact an isomorphism. It is also not hard to verify the following properties: Proposition. – ∗(e1 ∧ · · · ∧ ek) = ek+1 ∧ · · · ∧ em – ∗(ek+1 ∧ · · · ∧ em) = (−1)k(m−k)e1 ∧ · · · ∧ ek. – ∗∗ = α = (−1)k(m−k)α for α ∈ Λk. In general, let (M, g) be a compact real oriented Riemannian manifold of dimension m. Then g gives an isomorphism T M ∼= T ∗M , and induces an inner product on each T ∗ p M , which we will still denote gp. This induces an inner product on each kT ∗ p M , which we will denote · , · again. The Riemannian metric and orientation gives us a volume form Vol ∈ Ωm(M ), defined locally by Volp(e1 ∧ · · · ∧ em), where e1, . . . , em is an oriented basis of T ∗ on Ωk(M ), p M . This induces an L2-inner product α, βL2 = α, β Vol. Now apply Hodge ∗-operator to each (V, G) = (T ∗ get p M, gp) and p ∈ M . We then M 25 2 Complex structures III Symplectic Geometry Definition (Hodge star operator). The Hodge ∗-operator on forms ∗ : Ωk(M ) → Ωm−k(M ) is defined by the equation α ∧ (∗β) = α, β Vol. We again have some immediate properties. Proposition. (i) ∗ ∗ α = (−1)k(m−k)α for α ∈ Ωk(M ). (ii) ∗1 = Vol We now introduce the codifferential operator Definition (Codifferential operator). We define the codifferential operator δ : Ωk → Ωk−1 to be the L2-formal adjoint of d. In other words, we require dα, βL2 = α, δβL2 for all α ∈ Ωk and β ∈ Ωk+1. We immediately see that Proposition. δ2 = 0. Using the Hodge star, there is a very explicit formula for the codifferential (which in particular shows that it exists). Proposition. Proof. δ = (−1)m(k+1)+1∗ d ∗ : Ωk → Ωk−1. dα, βL2 = = M M dα ∧ ∗β d(α ∧ ∗β) − (−1)k M α ∧ d(∗β) = (−1)k+1 = (−1)k+1 M M α ∧ d(∗β) (Stokes’) (−1)(m−k)kα ∧ ∗ ∗ d(∗β) = (−1)k+1+(m−k)k M α, ∗ d ∗β. We can now define the Laplace–Beltrami operator Definition (Laplace–Beltrami operator). We define the Laplacian, or the Laplace–Beltrami operator to be ∆ = dδ + δd : Ωk → Ωk. Example. If M = Rm with the Euclidean inner product, and f ∈ Ω0(M ) = C∞(M ), then ∆f = − ∂2f ∂x2 i . n i=1 26 2 Complex structures III Symplectic Geometry It is an exercise to prove the following Proposition. (i) ∆∗ = ∗∆ : Ωk → Ωm−k (ii) ∆ = (d + δ)2 (iii) ∆α, βL2 = α, ∆βL2 . (iv) ∆α = 0 iff dα = δα = 0. In particular, (iv) follows from the fact that ∆α, α = dα, dα + δα, δα = dα2 L2 + δα2 L2. Similar to IA Vector Calculus, we can define Definition (Harmonic form). A form α is harmonic if ∆α = 0. We write Hk = {α ∈ Ωk(m) | ∆α = 0} for the space of harmonic forms. Observe there is a natural map Hk → H k dR(M ), sending α to [α]. The main result is that Theorem (Hodge decomposition theorem). Let (M, g) be a compact oriented Riemannian manifold. Then every cohomology class in H k dR(M ) has a unique harmonic representation, i.e. the natural map Hk → H k dR(M ) is an isomorphism. We will not prove this. Complex Hodge theory We now see what we can do for complex K¨ahler manifolds. First check that Proposition. Let M be a complex manifold, dimC M = n and (M, ω) K¨ahler. Then (i) ∗ : Ωp,q → Ωn−p,n−q. (ii) ∆ : Ωp,q → Ωp,q. define the L2-adjoints ¯∂∗ = ± ∗ ¯∂∗ and ∂∗ = − − ∗∂∗ with the appropriate signs as before, and then We can then define d = ∂ + ¯∂, δ = ∂∗ + ¯∂∗. ∆∂ = ∂∂∗ + ∂∗∂ : Ωp,q → Ωp,q ∆ ¯∂ = ¯∂ ¯∂∗ + ¯∂∗ ¯∂ : Ωp,q → Ωp,q. Proposition. If our manifold is K¨ahler, then ∆ = 2∆∂ = 2∆ ¯∂. 27 2 Complex structures III Symplectic Geometry So if we have a harmonic form, then it is in fact ∂ and ¯∂-harmonic, and in particular it is ∂ and ¯∂-closed. This give us a Hodge decomposition Hk C = Hp,q, p+q=k where Hp,q = {α ∈ Ωp,q(M ) : ∆α = 0}. Theorem (Hodge decomposition theorem). Let (M, ω) be a compact K¨ahler manifold. The natural map Hp,q → H p,q Dolb is an isomorphism. Hence H k dR(M ; C) ∼= Hk C = Hp,q ∼= H p,q Dolb(M ). p+q=k p+q=k What are some topological consequences of this? Recall the Betti numbers are defined by bk = dimR H k dR(M ) = dimC H k dR(M ; C). We can further define the Hodge numbers hp,q = dimC H p,q Dolb(M ). Then the Hodge theorem says bk = hp,q. p+q=k Moreover, since H p,q Dolb(M ) = H q,p Dolb(M ). So we have Hodge symmetry, hp,q = hq,p. Moreover, the ∗ operator induces an isomorphism So we have H p,q Dolb ∼= H n−p,n−q Dolb . hp,q = hn−p,n−q. There is called central symmetry, or Serre duality. Thus, we know that Corollary. Odd Betti numbers are even. Proof. b2k+1 = hp,q = 2 p+q=2k+1 hp,2k+1−p . k p=0 Corollary. h1,0 = h0,1 = 1 2 b1 is a topological invariant. We have also previously seen that for a general compact K¨ahler manifold, we have Proposition. Even Betti numbers are positive. 28 2 Complex structures III Symplectic Geometry Recall that we proved this by arguing that [ωk] = 0 ∈ H 2k(M ). In fact, ωk ∈ H k,k Dolb(M ), and so Proposition. hk,k = 0. We usually organize these hp,q in the form of a Hodge diamond , e.g. h2,0 h2,1 h1,0 h2,2 h1,1 h0,0 h1,2 h0,1 h0,2 29 3 Hamiltonian vector fields III Symplectic Geometry 3 Hamiltonian vector fields Symplectic geometry was first studied by physicists, who mod |
eled their systems by a symplectic manifold. The Hamiltonian function H ∈ C∞(M ), which returns the energy of a particular configuration, generates a vector field on M which is the equation of motion of the system. In this chapter, we will discuss how this process works, and further study what happens when we have families of such Hamiltonian functions, giving rise to Lie group actions. 3.1 Hamiltonian vector fields Definition (Hamiltonian vector field). Let (M, ω) be a symplectic manifold. If H ∈ C∞(M ), then since ˜ω : T M → T ∗M is an isomorphism, there is a unique vector field XH on M such that ιXH ω = dH. We call XH the Hamiltonian vector field with Hamiltonian function H. Suppose XH is complete (e.g. when M is compact). This means we can integrate XH , i.e. solve ∂ρt ∂t (p) = XH (ρt(p)), ρ0(p) = p. These flow have some nice properties. Proposition. If XH is a Hamiltonian vector field with flow ρt, then ρ∗ In other words, each ρt is a symplectomorphism. Proof. It suffices to show that ∂ t ω = 0. We have ∂t ρ∗ t ω = ω. d dt (ρ∗ t ω) = ρ∗ t (LXH ω) = ρ∗ t (dιXH ω + ιXH dω) = ρ∗ t (ddH) = 0. Thus, every function H gives us a one-parameter subgroup of symplectomor- phisms. Proposition. ρt preserves H, i.e. ρ∗ t H = H. Proof. d dt t H = ρ∗ ρ∗ t (LXH H) = ρ∗ t (ιXH dH) = ρ∗ t (ιXH ιXH ω) = 0. So the flow lines of our vector field are contained in level sets of H. Example. Take (S2, ω = dθ ∧ dh). Take to be the height function. Then XH solves H(h, θ) = h So ιXH (dθ ∧ dh) = dh. XH = ∂ ∂θ , ρt(h, θ) = (h, θ + t). As expected, the flow preserves height and the area form. 30 3 Hamiltonian vector fields III Symplectic Geometry We have seen that Hamiltonian vector fields are symplectic: Definition (Symplectic vector field). A vector field X on (M, ω) is a symplectic vector field if LX ω = 0. Observe that LX ω = ιX dω + dιX ω = dιX ω. So X is symplectic iff ιX ω is closed, and is Hamiltonian if it is exact. Thus, locally, every symplectic vector field is Hamiltonian and globally, the obstruction lies in H 1 dR(M ). Example. Take (T 2, ω = dθ1 ∧ dθ2). Then Xi = ∂ ∂θi Hamiltonian, since ιXiω = dθ2−i is closed but not exact. are symplectic but not Proposition. Let X, Y be symplectic vector fields on (M, ω). Then [X, Y ] is Hamiltonian. Recall that if X, Y are vector fields on M and f ∈ C∞(M ), then their Lie bracket is given by [X, Y ]f = (XY − Y X)f. This makes χ(M ), the space of vector fields on M , a Lie algebra. In order to prove the proposition, we need the following identity: Exercise. ι[X,Y ]α = [LX , ιY ]α = [ιX , LY ]α. Proof of proposition. We need to check that ι[X,Y ]ω is exact. By the exercise, this is ι[X,Y ]ω = LX ιY ω − ιY LX ω = d(ιX ιY ω) + ιX dιY ω + ιY dιX ω − ιY ιY dω. Since X, Y are symplectic, we know dιY ω = dιX ω = 0, and the last term always vanishes. So this is exact, and ω(Y, X) is a Hamiltonian function for [X, Y ]. Definition (Poisson bracket). Let f, g ∈ C∞(M ). We then define the Poisson bracket {f, g} by This is defined so that {f, g} = ω(Xf , Xg). X{f,g} = −[Xf , Xg]. 31 3 Hamiltonian vector fields III Symplectic Geometry Exercise. The Poisson bracket satisfies the Jacobi identity, and also the Leibniz rule {f, gh} = g{f, h} + {f, g}h. Thus, if (M, ω) is symplectic, then (C∞(M ), { · , · }) is a Poisson algebra. This means it is a commutative, associative algebra with a Lie bracket that satisfies the Leibniz rule. Further, the map C∞(M ) → χ(M ) sending H → XH is a Lie algebra (anti-)homomorphism. Proposition. {f, g} = 0 iff f is constant along integral curves of Xg. Proof. LXg f = ιXg df = ιXg ιXf ω = ω(Xf , Xg) = {f, g} = 0. Example. If M = R2n and ω = ω0 = dxj ∧ dyj, and f ∈ C∞(R2n), then Xf = i ∂f ∂yi ∂ ∂xi − ∂f ∂xi ∂ ∂yi . If ρ0(p) = p, then ρt(p) = (x(t), y(t)) is an integral curve for Xf iff dxi dt = ∂f ∂yi , ∂yi ∂t = − ∂f ∂xi . In classical mechanics, these are known as Hamilton equations. 3.2 Integrable systems In classical mechanics, we usually have a fixed H, corresponding to the energy. Definition (Hamiltonian system). A Hamiltonian system is a triple (M, ω, H) where (M, ω) is a symplectic manifold and H ∈ C∞(M ), called the Hamiltonian function. Definition (Integral of motion). A integral of motion/first integral /constant of motion/conserved quantity of a Hamiltonian system is a function f ∈ C∞(M ) such that {f, H} = 0. For example, H is an integral of motion. Are there others? Of course, we can write down 2H, H 2, H 12, eH , etc., but these are more-or-less the same as H. Definition (Independent integrals of motion). We say f1, . . . , fn ∈ C∞(M ) are independent if (df1)p, . . . , (dfn)p are linearly independent at all points on some dense subset of M . Definition (Commuting integrals of motion). We say f1, . . . , fn ∈ C∞ commute if {fi, fj} = 0. If we have n independent integrals of motion, then we clearly have dim M ≥ n. In fact, the commuting condition implies: Exercise. Let f1, . . . , fn be independent commuting functions on (M, ω). Then dim M ≥ 2n. 32 3 Hamiltonian vector fields III Symplectic Geometry The idea is that the (dfi)p are not only independent, but span an isotropic subspace of T M . If we have the maximum possible number of independent commuting first integrals, then we say we are integrable. Definition (Completely integrable system). A Hamiltonian system (M, ω, H) of dimension dim M = 2n is (completely) integrable if it has n independent commuting integrals of motion f1 = H, f2, . . . , fn. Example. If dim M = 2, then we only need one integral of motion, which we can take to be H. Then (M, ω, H) is integrable as long as the set of non-critical points of H is dense. Example. The physics of a simple pendulum of length 1 and mass 1 can be modeled by the symplectic manifold M = T ∗S1, where the S1 refers to the angular coordinate θ of the pendulum, and the cotangent vector is the momentum. The Hamiltonian function is H = K + V = kinetic energy + potential energy = 1 2 ξ2 + (1 − cos ω). We can check that the critical points of H are (θ, ξ) = (0, 0) and (π, 0). So (M, ω, H) is integrable. Example. If dim M = 4, then (M, ω, H) is integrable as long as there exists an integral motion independent of H. For example, on a spherical pendulum, we have M = T ∗S2, and H is the total energy. Then the angular momentum is an integral of motion. What can we do with a completely integrable system? Suppose (M, ω, H) is completely integrable system with dim M = 2n and f1 = H, f2, . . . , fn are commuting. Let c be a regular value of f = (f1, . . . , fn). Then f −1(c) is an n-dimensional submanifold of M . If p ∈ f −1(c), then Since we know Tp(f −1(c)) = ker(df )p. dfp = (df1)p ... (dfn)p = ιXf1 ... ιXfn ω , ω Tp(f −1(c)) = ker(dfp) = span{(Xf1)p, . . . (Xfn )p}, Moreover, since ω(Xfi, Xfj ) = {fi, fj} = 0, we know that Tp(f −1(c)) is an isotropic subspace of (TpM, ωp). If Xf1, . . . , Xfn are complete, then following their flows, we obtain global coordinates of (the connected components of) f −1(c), where q ∈ f −1(c) has coordinates (ϕ1, . . . , ϕm) (angle coordinates) if q is achieved from the base point p by following the flow of Xfi for ϕi seconds for each i. The fact that the fi are Poisson commuting implies the vector fields commute, and so the order does not matter, and this gives a genuine coordinate system. By the independence of Xfi, the connected components look like Rn−k × T k, where T k = (S1)k is the k torus. In the extreme case k = n, we simply get a torus, which is a compact connected component. 33 3 Hamiltonian vector fields III Symplectic Geometry Definition (Liouville torus). A Liouville torus is a compact connected component of f −1(c). It would be nice if the (ϕi) are part of a Darboux chart of M , and this is true. Theorem (Arnold–Liouville thoerem). Let (M, ω, H) be an integrable system with dim M = 2n and f1 = H, f2, . . . , fn integrals of motion, and c ∈ R a regular value of f = (f1, . . . , fn). (i) If the flows of Xfi are complete, then the connected components of f −1({c}) are homogeneous spaces for Rn and admit affine coordinates ϕ1, . . . , ϕn (angle coordinates), in which the flows of Xfi are linear. (ii) There exists coordinates ψ1, . . . , ψn (action coordinates) such that the ψi’s are integrals of motion and ϕ1, . . . , ϕn, ψ1, . . . , ψn form a Darboux chart. 3.3 Classical mechanics As mentioned at the beginning, symplectic geometry was first studied by physicists. In this section, we give a brief overview of how symplectic geometry arises in physics. Historically, there have been three “breakthroughs” in classical mechanics: (i) In ∼ 1687, Newton wrote down his three laws of physics, giving rise to Newtonian mechanics. (ii) In ∼ 1788, this was reformulated into the Lagrangian formalism. (iii) In ∼ 1833, these were further developed into the Hamiltonian formalism. Newtonian mechanics In Newtonian mechanics, we consider a particle of mass m moving in R3 under the potential V (x). Newton’s second law then says the trajectory of the particle obeys m Hamiltonian mechanics. d2x dt2 = −∇V (x). To do Hamiltonian mechanics, a key concept to introduce is the momentum: Definition (Momentum). The momentum of a particle is y = m dx dt . We also need the energy function Definition (Energy). The energy of a particle is |y|2 + V (x). H(x, y) = 1 2m 34 3 Hamiltonian vector fields III Symplectic Geometry We call R3 the configuration space and T ∗R3 the phase space, parametrized by (x, y). This has a canonical symplectic form ω = dxi ∧ dyi. Newton’s second law can be written as dyi dt = − ∂V ∂xi . Combining with the definition of y, we find that (x, y) evolves under. dxi dt dyi dt = ∂H ∂yi = − ∂H ∂xi So physical motion is given by Hamiltonian flow under H. H is called the Hamiltonian of the system. Lagrangian mechanics Lagrangian mechanics is less relevant to our symplectic picture, but is nice to know about nevertheless. This is formulated via a variational principle. In general, consider a system with N particles of masses m1, . . . , mN moving in R3 under a potential V ∈ C∞(R3N ). The Ham |
iltonian function can be defined exactly as before: H(x, y) = k 1 2mk |yk|2 + V (x), where x(t) = (x1, . . . , xn) and each xi is a 3-vector; and similarly for y with dxt yk = mk dt . Then in Hamiltonian mechanics, we say (x, y) evolves under Hamiltonian flow. Now fix a, b ∈ R and p, q ∈ R3N . Write P for the space of all piecewise differentiable paths γ = (γ1, . . . , γn) : [a, b] → R3N . Definition (Action). The action of a path γ ∈ P is Aγ = b a k mk 2 dγk dt (t) 2 − V (γ(t)) dt. The integrand is known as the Lagrangian function. We will see that γ(t) = x(t) is (locally) a stationary point of Aγ iff mk d2xt dt = − ∂V ∂xk , i.e. if and only if Newton’s second law is satisfied. The Lagrangian formulation works more generally if our particles are constrained to live in some submanifold X ⊆ R3n. For example, if we have a pendulum, then the particle is constrained to live in S1 (or S2). Then we set P to be the maps γ : [a, b] → X that go from p to q. The Lagrangian formulation is then exactly the same, except the minimization problem is now performed within this P. 35 3 Hamiltonian vector fields III Symplectic Geometry More generally, suppose we have an n-dimensional manifolds X, and F : T X → R is a Lagrangian function. Given a curve γ : [a, b] → X, there is a lift ˜γ : [a, b] → T X t → γ(t), (t) . dγ dt The action is then Aγ = b a (˜γ∗F )(t) = b a F γ(t), (t) dt. dγ dt To find the critical points of the action, we use calculus of variations. Pick a chart (x1, . . . , xn) for X and naturally extend to a chart (x1, . . . , xn, v1, . . . , vn) for T X. Consider a small perturbation of our path γε(t) = (γ1(t) + εc1(t), . . . , γn(t) + εcn(t)) for some functions c1, . . . , cn ∈ C∞([a, b]) such that ci(a) = ci(b) = 0. We think of this as an infinitesimal variation of γ. We then find that dAγε dε ε=0 = 0 ⇔ ∂F ∂xk = d dt ∂F ∂vF for k = 1, . . . , n. These are the Euler–Lagrange equations. Example. In X = R3N and F (x1, . . . , xn, v1, . . . , vn) = mk 2 k |vk|2 − V (x1, . . . , xn). Then the Euler–Lagrange equations are − ∂V ∂xki = mk d2xki dt2 . Example. On a Riemannian manifold, if we set F : T X → R be (x, v) → |v|2, then we obtain the Christoffel equations for a geodesic. In general, there need not be solutions to the Euler–Lagrange equation. However, if we satisfy the Legendre condition det ∂2F ∂vi∂vj = 0, then the Euler–Lagrange equations become second order ODEs, and there is a unique solution given γ(0) and ˙γ(0). If furthermore this is positive definite, then the solution is actually a locally minimum. 3.4 Hamiltonian actions In the remainder of the course, we are largely interested in how Lie groups can act on symplectic manifolds via Hamiltonian vector fields. These are known as Hamiltonian actions. We first begin with the notion of symplectic actions. Let (M, ω) be a symplectic manifold, and G a Lie group. 36 3 Hamiltonian vector fields III Symplectic Geometry Definition (Symplectic action). A symplectic action is a smooth group action ψ : G → Diff(M ) such that each ψg is a symplectomorphism. In other words, it is a map G → Symp(M, ω). Example. Let G = R. Then a map ψ : G → Diff(M ) is a one-parameter group of transformations {ψt : t ∈ R}. Given such a group action, we obtain a complete vector field Xp = dψt dt (p) t=0 . Conversely, given a complete vector field X, we can define ψt = exp tX, and this gives a group action by R. Under this correspondence, symplectic actions correspond to complete sym- plectic vector fields. Example. If G = S1, then a symplectic action of S1 is a symplectic action of R which is periodic. In the case where G is R or S1, it is easy to define what it means for an action to be Hamiltonian: Definition (Hamiltonian action). An action of R or S1 is Hamiltonian if the corresponding vector field is Hamiltonian. Example. Take (S2, ω = dθ ∧ dh). Then we have a rotation action ψt(θ, h) = (θ + t, h) generated by the vector field ∂ Hamiltonian S1 action. ∂θ . Since ι ∂ ∂θ ω = dh is exact, this is in fact a Example. Take (T 2, dθ1 ∧ dθ2). Consider the action ψt(θ1, θ2) = (θ1 + t, θ2). This is generated by the vector field ∂ ∂θ1 not exact. So this is a symplectic action that is not Hamiltonian. . But ι ∂ ∂θ1 ω = dθ2, which is closed but How should we define Hamiltonian group actions for groups that are not R or S1? The simplest possible next case is the torus G = T n = S1 × · · · × S1. If we have a map ψ; T n → Symp(M, ω), then for this to be Hamiltonian, it should definitely be the case that the restriction to each S1 is Hamiltonian in the previous sense. Moreover, for these to be compatible, we would expect each Hamiltonian function to be preserved by the other factors as well. For the general case, we need to talk about the Lie group of G. Let G be a Lie group. For each g ∈ GG, there is a left multiplication map Lg : G → G a → ga. 37 3 Hamiltonian vector fields III Symplectic Geometry Definition (Left-invariant vector field). A left-invariant vector field on a Lie group G is a vector field X such that (Lg)∗X = X for all g ∈ G. We write g for the space of all left-invariant vector fields on G, which comes with the Lie bracket on vector fields. This is called the Lie algebra of G. If X is left-invariant, then knowing Xe tells us what X is everywhere, and specifying Xe produces a left-invariant vector field. Thus, we have an isomorphism g ∼= TeG. The Lie algebra admits a natural action of G, called the adjoint action. To construct this, note that G acts on itself by conjugation, ϕg(a) = gag−1. This fixes the identity, and taking the derivative gives Adg : g → g, or equivalently, Ad is a map Ad : G → GL(g). The dual g∗ admits the dual action of G, called the coadjoint action. Explicitly, this is given by Ad∗ g(ξ), x = ξ, Adg x. An important case is when G is abelian, i.e. a product of S1’s and R’s, in which case the conjugation action is trivial, hence the (co)adjoint action is trivial. Returning to group actions, the correspondence between complete vector fields and R/S1 actions can be described as follows: Given a smooth action ψ : G → Diff(M ) and a point p ∈ M , there is a map Differentiating this at e gives G → M g → ψg(p). g ∼= TeG → TpM X → X # p . We call X # the vector field on M generated by X ∈ g. In the case where G = S1 or R, we have g ∼= R, and the complete vector field corresponding to the action is the image of 1 under this map. We are now ready to define Definition (Hamiltonian action). We say ψ : G → Symp(M, ω) is a Hamiltonian action if there exists a map µ : M → g∗ such that (i) For all X ∈ g, X # is the Hamiltonian vector field generated by µX , where µX : M → R is given by µX (p) = µ(p), X. (ii) µ is G-equivariant, where G acts on g∗ by the coadjoint action. In other words, µ ◦ ψg = Ad∗ g ◦µ for all g ∈ G. 38 3 Hamiltonian vector fields III Symplectic Geometry µ is then called a moment map for the action ψ. In the case where G is abelian, condition (ii) just says µ is G-invariant. Example. Let M = Cn, and We let acting by ω = 1 2 j dzj ∧ d¯zj = i rj drj ∧ dθj. T n = {(t1, . . . , tn) ∈ Cn : |tk| = 1 for all k}, ψ(t1,...,tn)(z1, . . . , zn) = (tk1 1 z1, . . . , tkn n zn) where k1, . . . , kn ∈ Z. We claim this action has moment map µ : Cn → (tn)∗ ∼= Rn 1 2 It is clear that this is invariant, and if X = (a1, . . . , an) ∈ tn ∈ Rn, then (k1|z1|2, . . . , kn|zn|2). (x1, . . . , zn) → − X # = k1a1 ∂ ∂θ1 + · · · + knan ∂ ∂θn . Then we have dµX = d − 1 2 kjajr2 j = − kjajrj drj = ιX# ω. Example. A nice example of a non-abelian Hamiltonian action is coadjoint orbits. Let G be a Lie group, and g the Lie algebra. If X ∈ g, then we get a vector field gX # generated by X via the adjoint action, and also a vector field gX on g∗ generated by the co-adjoint action. If ξ ∈ g∗, then we can define the coadjoint orbit through ξ Oξ = {Ad∗ g(ξ) : g ∈ G}. What is interesting about this is that this coadjoint orbit is actually a symplectic manifold, with symplectic form given by ωξ(, [X, Y ]. Then the coadjoint action of G on Oξ has moment map Oξ → g∗ given by the inclusion. 3.5 Symplectic reduction Given a Lie group action of G on M , it is natural to ask what the “quotient” of M looks like. What we will study is not quite the quotient, but a symplectic reduction, which is a natural, well-behaved subspace of the quotient that is in particular symplectic. We first introduce some words. Let ψ : G → Diff(M ) be a smooth action. 39 3 Hamiltonian vector fields III Symplectic Geometry Definition (Orbit). If p ∈ M , the orbit of p under G is Op = {ψg(p) : g ∈ G}. Definition (Stabilizer). The stabilizer or isotropy group of p ∈ M is the closed subgroup Gp = {g ∈ G : ψg(p) = p}. We write gp for the Lie algebra of Gp. Definition (Transitive action). We say G acts transitively if M is one orbit. Definition (Free action). We say G acts freely if Gp = {e} for all p. Definition (Locally free action). We say G acts locally freely if gp = {0}, i.e. Gp is discrete. Definition (Orbit space). The orbit space is M/G, and we write π : M → M/G for the orbit projection. We equip M/G with the quotient topology. The main theorem is the following: Theorem (Marsden–Weinstein, Meyer). Let G be a compact Lie group, and (M, ω) a symplectic manifold with a Hamiltonian G-action with moment map µ : M → g∗. Write i : µ−1(0) → M for the inclusion. Suppose G acts freely on µ−1(0). Then (i) Mred = µ−1(0)/G is a manifold; (ii) π : µ−1(0) → Mred is a principal G-bundle; and (iii) There exists a symplectic form ωred on Mred such that i∗ω = π∗ωred. Definition (Symplectic quotient). We call Mred the symplectic quotient/reduced space/symplectic reduction of (M, ω) with respect to the given G-action and moment map. What happens if we do reduction at other levels? In other words, what can we say about µ−1(ξ)/G for other ξ ∈ g∗? If we want to make sense of this, we need ξ to be preserved under the coadjoint action of G. This is automatically satisfied if G is abelian, and in this case, we simply have µ−1(ξ) = ϕ−1(0), where ϕ = µ − ξ is another moment map. So this is not more general. If ξ is not |
preserved by G, then we can instead consider µ−1(ξ)/Gξ, or equivalently take µ−1(Oξ)/G. We check that µ−1(ξ)/Gξ ∼= µ−1(Oξ)/G ∼= ϕ−1(0)/G, where ϕ : M × Oξ → g∗ (ρ, η) → µ(p) − η is a moment map for the product action of G on (M × Oξ, ω × ωξ). So in fact there is no loss in generality for considering just µ−1(0). 40 3 Hamiltonian vector fields III Symplectic Geometry Proof. We first show that µ−1(0) is a manifold. This follows from the following claim: Claim. G acts locally freely at p iff p is a regular point of µ. We compute the dimension of im dµp using the rank-nullity theorem. We know dµpv = 0 iff dµp(v), X = 0 for all X ∈ g. We can compute dµp(v), X = (dµX )p(v) = (ιX# p ω)(v) = ωp(X # p , v). Moreover, the span of the X # p is exactly TpOp. So ker dµp = (TpOp)ω. Thus, dim(im dµp) = dim Op = dim G − dim Gp. In particular, dµp is surjective iff Gp = 0. Then (i) and (ii) follow from the following theorem: Theorem. Let G be a compact Lie group and Z a manifold, and G acts freely on Z. Then Z/G is a manifold and Z → Z/G is a principal G-bundle. Note that if G does not act freely on µ−1(0), then by Sard’s theorem, generically, ξ is a regular value of µ, and so µ−1(ξ) is a manifold, and G acts locally freely on µ−1(ξ). If µ−1(ξ) is preserved by G, then µ−1(ξ)/G is a symplectic orbifold. It now remains to construct the symplectic structure. Observe that if p ∈ µ−1(0), then TpOp ⊆ Tpµ−1(0) = ker dµp = (TpOp)ω. So TpOp is an isotropic subspace of (TpM, ω). We then observe the following straightforward linear algebraic result: Lemma. Let (V, Ω) be a symplectic vector space and I an isotropic subspace. Then Ω induces a canonical symplectic structure Ωred on I Ω/I, given by Ωred([u], [v]) = Ω(u, v). Applying this, we get a canonical symplectic structure on (TpOp)ω TpOp = Tpµ−1(0) TpOp = T[p]Mred. This defines ωred on Mred, which is well-defined because ω is G-invariant, and is smooth by local triviality and canonicity. It remains to show that dωred = 0. By construction, i∗ω = π∗ωred. So π∗(dωred) = dπ∗ωred = di∗ω = i∗dω = 0 Since π∗ is injective, we are done. Example. Take (M, ω) = Cn, ω0 = i 2 dzk ∧ d¯zk = dxk ∧ dyk = rk drk ∧ dθk . 41 3 Hamiltonian vector fields III Symplectic Geometry We let G = S1 act by multiplication eit · (z1, . . . , zn) = (eitz1, . . . , eitzn). This action is Hamiltonian with moment map µ : Cn → R z → − |z|2 2 + 1 2 . The + 1 is a moment map, we compute 2 is useful to make the inverse image of 0 non-degenerate. To check this 1 2 d = d − |z|2 r2 k 1 2 = − rk drk. On the other hand, if then we have So we have X = a ∈ g ∼= R, X # = a ∂ ∂θ1 + · · · + . ∂ ∂θn ιX# ω = −a rk drk = dµX . It is also clear that µ is S1-invariant. We then have µ−1(0) = {z ∈ Cn : |z|2 = 1} = S2n−1. Then we have Mred = µ−1(0)/S1 = CPn−1. One can check that this is in fact the Fubini–Study form. So (CPn−1, ωF S) is the symplectic quotient of (Cn, ω0) with respect to the diagonal S1 action and the moment map µ. Example. Fix k, ∈ Z relatively prime. Then S1 acts on C2 by eit(z1, z2) = (eiktz1, eitz2). This action is Hamiltonian with moment map µ : C2 → R (z1, z2) → − 1 2 (k|z1|2 + |z2|2). There is no level set of µ where the action is free, since – (z, 0) has stabilizer Z/kZ – (0, z) has stabilizer Z/Z – (z1, z2) has trivial stabilizer if z1, z2 = 0. 42 3 Hamiltonian vector fields III Symplectic Geometry On the other hand, the action is still locally free on C2 \ {(0, 0)} since the stabilizers are discrete. The reduced spaces µ−1(ξ)/S1 for ξ = 0 are orbifolds, called weighted or twisted projective spaces. The final example is an infinite dimensional one by Atiyah and Bott. We will not be able to prove the result in full detail, or any detail at all, but we will build up to the statement. The summary of the result is that performing symplectic reduction on the space of all connections gives the moduli space of flat connections. Let G → P π→ B be a principal G-bundle, and ψ : G → Diff(P ) the associated free action. Let dψ : g → χ(P ) X → X # be the associated infinitesimal action. Let X1, . . . , Xk be a basis of the Lie algebra g. Then since ψ is a free action, X # k are all linearly independent at each # Define the vertical tangent space Vp = span{(X # 1 )p, . . . , (X # k )p} ⊆ TpP = ker(dπp) We can then put these together to get V ⊆ T P , the vertical tangent bundle. Definition (Ehresmann Connection). An (Ehresmann) connection on P is a choice of subbundle H ⊆ T P such that (i) P = V ⊕ H (ii) H is G-invariant. Such an H is called a horizontal bundle. There is another way of describing a connection. A connection form on P is a g-valued 1-form A ∈ Ω1(P ) ⊗ g such that (i) ιX# A = X for all X ∈ g (ii) A is G-invariant for the action g · (Ai ⊗ Xi) = (ψg−1)∗Ai ⊗ Adg Xi. Lemma. Giving an Ehresmann connection is the same as giving a connection 1-form. Proof. Given an Ehresmann connection H, we define Ap(v) = X, where v = X # p + hp ∈ V ⊕ H. Conversely, given an A, we define Hp = ker Ap = {v ∈ TpP : ivAp = 0}. 43 3 Hamiltonian vector fields III Symplectic Geometry We next want to define the notion of curvature. We will be interested in flat connections, i.e. connections with zero curvature. To understand curvature, if we have a connection T P = V ⊕ H, then we get further decompositions T ∗P = V ∗ ⊕ H ∗, Λ2(T ∗P ) = (2V ∗) ⊕ (V ∗ ∧ H ∗) ⊕ (2H ∗). So we end up having Ω1(P ) = Ω1 Ω2(P ) = Ω2 vert(P ) ⊕ Ω1 hor(P ) vert ⊕ Ωmixed ⊕ Ω2 hor If A = k i=1 Ai ⊗ Xi ∈ Ω1 ⊗ g, then dA ∈ Ω2 ⊗ g. We can then decompose this as dA = (dA)vert + (dA)mix + (dA)hor. The first part is uninteresting, in the sense that it is always given by (dA)vert(X, Y ) = [X, Y ], the second part always vanishes, and the last is the curvature form curv A ∈ Ω2 hor ⊗ g. Definition (Flat connection). A connection A is flat if curv A = 0. We write A for the space of all connections on P . Observe that if A1, A0 ∈ A, then A1 − A0 is G-invariant and ιX# (A1 − A0) = X − X = 0 for all X ∈ g. So A1 − A0 ∈ (Ω1 hor ⊗ g)G. So A = A0 + (Ω1 hor ⊗ g)G, and TA0 A = (Ω1 hor ⊗ g)G. Suppose B is a compact Riemann surface, and G a compact or semisimple Lie group. Then there exists an Ad-invariant inner product on g. We can then define ω : (Ω1 hor ⊗ g)G × (Ω1 hor ⊗ g)G → R, sending → biXi aiXi, ai ∧ bj(Xi, Xj). B i,j This is easily seen to be bilinear, anti-symmetric and non-degenerate. It is also closed, if suitably interpreted, since it is effectively constant across the affine space A. Thus, A is an “infinite-dimensional symplectic manifold”. 44 3 Hamiltonian vector fields III Symplectic Geometry To perform symplectic reduction, let G be the gauge group, i.e. the group of G-equivariant diffeomorphisms f : P → P covering the identity. G acts on A by V ⊕ H → V ⊕ Hf , where Hf is the image of H by df . Atiyah and Bott proved that the action of G on (A, ω) is Hamiltonian with moment map µ : A → Lie(G)∗ = (Ω2 hor ⊗ g)G A → curv A Performing symplectic reduction, we get M = µ−1(0)/G, the moduli space of flat connections, which has a symplectic structure. It turns out this is in fact a finite-dimensional symplectic orbifold. 3.6 The convexity theorem We focus on the case G = T n. It turns out the image of the moment map µ has a very rigid structure. Theorem (Convexity theorem (Atiyah, Guillemin–Sternberg)). Let (M, ω) be a compact connected symplectic manifold, and µ : M → Rn a moment map for a Hamiltonian torus action. Then (i) The levels µ−1(c) are connected for all c (ii) The image µ(M ) is convex. (iii) The image µ(M ) is in fact the convex hull of µ(M G). We call µ(M ) the moment polytope. Here we identify G ∼= T n Rn/Zn, which gives us an identification g ∼= Rn and g∗ ∼= (Rn)∗ ∼= Rn. Example. Consider (M = CPn, ωF S). T n acts by letting t = (t1, . . . , tn) ∈ T n ∼= U(1)n send ψt([z0 : · · · zn]) = [z0 : t1z1 : · · · : tnzn]. This has moment map µ([z0 : · · · zn]) = − 1 2 (|z1|2, . . . , |zn|2) |z0|2 + · · · + |zn|2 . The image of this map is µ(M ) = x ∈ Rn : xk ≤ 0, x1 + · · · + xn ≥ − . 1 2 For example, in the CP2 case, the moment image lives in R2, and is just 45 3 Hamiltonian vector fields III Symplectic Geometry − 1 2 − 1 2 The three vertices are µ([0 : 0 : 1]), µ([0 : 1 : 0]) and µ([1 : 0 : 0]). We now want to prove the convexity theorem. We first look at (iii). While it seems like a very strong claim, it actually follows formally from (ii). Lemma. (ii) implies (iii). Proof. Suppose the fixed point set of the action has k connected components Z = Z1 ∪ · · · ∪ Zk. Then µ is constant on each Zj, since X #|Zj = 0 for all X. Let µ(Zj) = ηj ∈ Rn. By (ii), we know the convex hull of {η1, . . . ηk} is contained in µ(M ). To see that µ(M ) is exactly the convex hull, observe that if X ∈ Rn has rationally independent components, so that X topologically generates T , then p is fixed by T iff X # p = 0. Thus, µX attains its maximum on one of the Zj. p = 0, iff dµX Now if ξ is not in the convex hull of {ηj}, then we can pick an X ∈ Rn with rationally independent components such that ξ, X > ηj, X for all j, since the space of such X is open and non-empty. Then ξ, X > sup p∈ Zj µ(p), X = sup p∈M µ(p), X. So ξ ∈ µ(M ). With a bit more work, (i) also implies (ii). Lemma. (i) implies (ii). Proof. The case n = 1 is immediate, since µ(M ) is compact and connected, hence a closed interval. In general, to show that µ(M ) is convex, we want to show that the intersection of µ(M ) with any line is connected. In other words, if π : Rn+1 → Rn is any projection and ν = π ◦ µ, then π−1(c) ∩ µ(M ) = µ(ν−1(c)) is connected. This would follow if we knew ν−1(c) were connected, which would follow from (i) if ν were a moment map of an T n action. Unfortunately, most of the time, it is just the moment map of an Rn action. For it to come from a T n action, we need π to be represented by an integer matrix. Then T = {πT t : t ∈ T n = Rn/Zn} ⊆ T n+1 is a subtorus, and one readily checks that ν is the moment map for the T action. 46 3 Hamiltonian vector fields III Symplectic Geometry Now for any p0, p1 ∈ M , we can find p 1 arbitrarily close to p0, p1 and |
a line of the form π−1(c) with π integral. Then the line between p 0 and p 1 is contained in µ(M ) by the above argument, and we are done since µ(M ) is compact, hence closed. 0, p It thus remains to prove (i), where we have to put in some genuine work. This requires Morse–Bott theory. Let M be a manifold, dim M = m, and f : M → R a smooth map. Let Crit(f ) = {p ∈ M : dfp = 0} be the set of critical points. For p ∈ Crit(f ) and (U, x1, . . . , xm) a coordinate chart around p, we have a Hessian matrix Hpf = ∂2f ∂xi∂xj Definition (Morse(-Bott) function). f is a Morse function if at each p ∈ Crit(f ), Hpf is non-degenerate. f is a Morse–Bott function if the connected components of Crit(f ) are submanifolds and for all p ∈ Crit(p), Tp Crit(f ) = ker(Hpf ). If f is Morse, then the critical points are isolated. If f is Morse–Bott, then the Hessian is non-degenerate in the normal bundle to Crit(f ). If M is compact, then there is a finite number of connected components of Crit(f ). So we have Crit(f ) = Z1 ∪ · · · ∪ Zk, and the Zi are called the critical submanifolds. For p ∈ Zi, the Hessian Hpf is a quadratic form and we can write TpM = E− p ⊕ TpZi ⊕ E+ p , where E± that dim E± index of Zi to be dim E− p are the positive and negative eigenspaces of Hpf respectively. Note p are locally constant, hence constant along Zi. So we can define the p , and the coindex to be dim E+ p . We can then define a vector bundle E− → Zi, called the negative normal bundle. Morse theory tells us how the topology of M − c = {p ∈ M : f (p) ≤ c} changes with c ∈ R. Theorem (Morse theory). (i) If f −1([c1, c2]) does not contain any critical point. Then f −1(c1) ∼= f −1(c2) and Mc1 ∼= Mc2 (where ∼= means diffeomorphic). (ii) If f −1([c1, c2]) contains one critical manifold Z, then M − c2 M − c1 ∪ D(E−), is, up to homotopy equivalence, where D(E−) is the disk bundle of E−. In particular, if Z is an isolated point, M − c2 p -cell to M − obtained by adding a dim E− . c1 The key lemma in this proof is the following result: 47 3 Hamiltonian vector fields III Symplectic Geometry Lemma. Let M be a compact connected manifold, and f : M → R a Morse–Bott function with no critical submanifold of index or coindex 1. Then (i) f has a unique local maximum and local minimum (ii) All level sets of f −1(c) are connected. Proof sketch. There is always a global minimum since f is compact. If there is another local minimum at c, then the disk bundle is trivial, and so in M − c+ε M − c−ε ∪ D(E−) for ε small enough, the union is a disjoint union. So Mc+ε has two components. Different connected components can only merge by crossing a level of index 1, so this cannot happen. To handle the maxima, consider −f . More generally, the same argument shows that a change in connectedness must happen by passing through a index or coindex 1 critical submanifold. To apply this to prove the convexity theorem, we will show that for any X, µX is a Morse–Bott function where all the critical submanifolds are symplectic. In particular, they are of even index and coindex. Lemma. For any X ∈ Rn, µX is a Morse–Bott function where all critical submanifolds are symplectic. Proof sketch. Note that p is a fixed point iff X # point. So the critical points are exactly the fixed points of X. p = 0 iff dµX p = 0 iff p is a critical (TpM, ωp) models (M, ω) in a neighbourhood of p by Darboux theorem. Near a fixed point T n, an equivariant version of the Darboux theorem tells us there is a coordinate chart U where (M, ω, µ) looks like ω|U = dxi ∧ dyi 1 2 i µ|U = µ(p) − (x2 i + y2 i )αi, where αi ∈ Z are weights. Then the critical submanifolds of µ are given by {xi = yi = 0 : αi = 0}, which is locally a symplectic manifold and has even index and coindex. Finally, we can prove the theorem. Lemma. (i) holds. Proof. The n = 1 case follows from the previous lemmas. We then induct on n. Suppose the theorem holds for n, and let µ = (µ1, . . . , µn) : M → Rn+1 be a moment map for a Hamiltonian T n+1-action. We want to show that for all c = (c1, . . . , cn) ∈ Rn+1, the set µ−1(c) = µ−1 1 (c1) ∩ · · · ∩ µ−1 n+1(cn+1) is connected. 48 3 Hamiltonian vector fields III Symplectic Geometry The idea is to set 1 (c1) ∩ · · · µ−1 and then show that µn+1|N : N → R is a Morse–Bott function with no critical submanifolds of index or coindex 1. N = µ−1 1 (cn), We may assume that dµ1, . . . , dµn are linearly independent, or equivalently, dµX = 0 for all X ∈ Rn. Otherwise, this reduces to the case of an n-torus. To make sense of N , we must pick c to be a regular value. Density arguments imply that C = X=0 Crit(µX ) = Crit µX . X∈Zn+1\{0} Since Crit µX is a union of codimension ≥ 2 submanifolds, its complement is dense. Hence by the Baire category theorem, C has dense complement. Then a continuity argument shows that we only have to consider the case when c is a regular value of µ, hence N is a genuine submanifold of codimension n. By the induction hypothesis, N is connected. We now show that µn+1|N : N → R is Morse–Bott with no critical submanifolds of index or coindex 1. Let x be a critical point. Then the theory of Lagrange multipliers tells us there are some λi ∈ R such that dµn+1 + n n=1 λidµi = 0 x Thus, µ is critical in M for the function µY = µn+1 + n i=1 λiµi, where Y = (λ1, . . . , λn, 1) ∈ Rn+1. So by the claim, µY is Morse–Bott with only even indices and coindices. Let W be a critical submanifold of µY containing x. Claim. W intersects N transversely at x. If this were true, then µY |N has W ∩ N as a non-degenerate critical submanifold of even index and coindex, since the coindex doesn’t change and W is even-dimensional. Moreover, when restricted to N , λiµi is a constant. So µn+1|N satisfies the same properties. To prove the claim, note that TxN = ker dµ1|x ∩ · · · ∩ ker dµn|x. With a moments thought, we see that it suffices to show that dµ1, . . . , dµn remain linearly independent when restricted to TxW . Now observe that the Hamiltonian vector fields X # n |x are independent since dµ1|x, . . . dµn|x are, and they live in TxW since their flows preserve W . 1 |x, . . . , X # Since W is symplectic (by the claim), for all k = (k1, . . . kn), there exists v ∈ TxW such that In other words, ω kiX # i |x, v = 0. kidµi (v) = 0. 49 3 Hamiltonian vector fields III Symplectic Geometry It is natural to seek a non-abelian generalization of this, and it indeed exists. Let (M, ω) be a symplectic manifold, and G a compact Lie group with a Hamiltonian action on M with moment map µ : M → g∗. From Lie group theory, there is a maximal torus T ⊆ G with Lie algebra t, and the Weyl group W = N (T )/T is finite (where N (T ) is the normalizer of T ). Then under the coadjoint action, we have g∗/G ∼= t∗/W, Pick a Weyl chamber t∗ µ induces a moment map µ+ : M → t∗ says + of t∗, i.e. a fundamental domain of t∗ under W . Then +, and the non-abelian convexity theorem Theorem (Kirwan, 1984). µ+(M ) ⊆ t∗ + is a convex polytope. We shall end with an application of the convexity theorem to linear algebra. λ the set of all 2 × 2 Hermitian λ be? Let λ = (λ1, λ2) ∈ R2 and λ1 ≥ λ2, and H2 matrices with eigenvalues (λ1, λ2). What can the diagonal entries of A ∈ H2 We can definitely solve this by brute force, since any entry in H looks like a z b ¯z A = where a, b ∈ R and z ∈ C. We know tr a = a + b = λ1 + λ2 det a = ab − |z|2 = λ1λ2. The first implies b = λ1 + λ2 − a, and all the second condition gives is that ab > λ1λ2. b a This completely determines the geometry of H2 λ, since for each allowed value of a, there is a unique value of b, which in turn determines the norm of z. Topologically, this is a sphere, since there is a S1’s worth of choices of z except at the two end points ab = λ1λ2. What has this got to do with Hamiltonian actions? Observe that U(2) acts transitively on H2 λ by conjugation, and T 2 = eiθ1 0 0 eiθ2 ⊆ U(2). This contains a copy of S1 given by eiθ1 0 S1 = 0 e−iθ1 ⊆ T 2. 50 3 Hamiltonian vector fields III Symplectic Geometry We can check that eiθ 0 0 e−iθ a z b ¯z eiθ 0 0 e−iθ −1 = a eiθz eiθz b Thus, if ϕ : H2 λ → R2 is the map that selects the diagonal elements, then ϕ−1(a, b) = a |z|e−iθ |z|eiθ b is one S1-orbit. This is reminiscent of the S1 action of S2 quotienting out to a line segment. We can think more generally about Hn λ, the n × n Hermitian matrices with eigenvalues λ1 ≥ · · · ≥ λn, and ask what the diagonal elements can be. Take ϕ : Hn λ → Rn be the map that selects the diagonal entries. Then the image lies on the plane tr A = λi. This certainly contain the n! points whose coordinates are all possible permutation of the λi, given by the diagonal matrices. Theorem (Schur–Horn theorem). ϕ(Hn from the diagonal matrices. λ) is the convex hull of the n! points To view this from a symplectic perspective, let M = Hn λ, and U(n) acts transitively by conjugation. For A ⊆ M , let GA be the stabilizer of A. Then Thus, Hn λ = U(n)/GA. TAHn λ ∼= iHn gA where Hn is the Hermitian matrices. The point of this is to define a symplectic form. We define ωA : iHn × iHn → R (X, Y ) → i tr([X, Y ]A) = i tr(X(Y A − AY )) So ker ωA = {Y : [A, Y ] = 0} = gA. So ωA induces a non-degenerate form on TAHn form on Hn λ. λ. In fact, this gives a symplectic Let T n ⊆ U(n) be the maximal torus consisting of diagonal entries. We can show that the T n action is Hamiltonian with moment map ϕ. Since T n fixes exactly the diagonal matrices, we are done. 3.7 Toric manifolds What the convexity theorem tells us is that if we have a manifold M with a torus action, then the image of the moment map is a convex polytope. How much information is retained by a polytope? Of course, if we take a torus that acts trivially on M , then no information is retained. 51 3 Hamiltonian vector fields III Symplectic Geometry Definition (Effective action). An action G on M is effective (or faithful ) if every non-identity g ∈ G moves at least one point of M . But we can still take the trivial torus T 0 that acts trivially, and it will still be effective. Of course, no information is retained in the pol |
ytope as well. Thus, we want to have as large of a torus action as we can. The following proposition puts a bound on “how much” torus action we can have: Proposition. Let (M, ω) be a compact, connected symplectic manifold with moment map µ : M → Rn for a Hamiltonian T n action. If the T n action is effective, then (i) There are at least n + 1 fixed points. (ii) dim M ≥ 2n. We first state without proof a result that is just about smooth actions. Fact. An effective action of T n has orbits of dimension n. This doesn’t mean all orbits are of dimension n. It just means some orbit has dimension n. Proof. (i) If µ = (µ1, . . . , µn) : M → Rn and p is a point in an n-dimensional orbit, then {(dµi)p} are linearly independent. So µ(p) is an interior point (if p is not in the interior, then there exists a direction X pointing out of µ(M ). So (dµX )p = 0, and dµX gives a non-trivial linear combination of the dµi’s that vanishes). So if there is an interior point, we know µ(M ) is a non-degenerate polytope in Rn. This mean it has at least n + 1 vertices. So there are at least n + 1 fixed points. (ii) Let O be an orbit of p in M . Then µ is constant on O by invariance of µ. So TpO ⊆ ker(dµp) = (TpO)ω. So all orbits of a Hamiltonian torus action are isotropic submanifolds. So dim O ≤ 1 2 dim M . So we are done. In the “optimal” case, we have dim M = 2n. Definition ((Symplectic) toric manifold). A (symplectic) toric manifold is a compact connected symplectic manifold (M 2n, ω) equipped with an effective T n action of an n-torus together with a choice of corresponding moment map µ. Example. Take (CPn, ωF S), where the moment map is given by µ([z0 : z1 : · · · : zn]) = − 1 2 (|z1|2, . . . , |zn|2) |z0|2 + |z1|2 + · · · + |zn|2 . Then this is a symplectic toric manifold. 52 3 Hamiltonian vector fields III Symplectic Geometry Note that if (M, ω, T µ) is a toric manifold and µ = (µ0, . . . , µn) : M → Rn, then µ1, . . . , µn are commuting integrals of motion {µi, µj} = ω(. So we get an integrable system. The punch line of the section is that there is a correspondence between toric manifolds and polytopes of a suitable kind. First, we need a suitable notion of equivalence of toric manifolds. Definition (Equivalent toric manifolds). Fix a torus T = R2n/(2πZ)n, and fix an identification t∗ ∼= t ∼= Rn. Given two toric manifolds (Mi, ωi, T, µi) for i = 1, 2, We say they are (i) equivalent if there exists a symplectomorphism ϕ : M1 → M2 such that ϕ(x · p) = x · ϕ(p) and µ2 ◦ ϕ = µ1. (ii) weakly equivalent if there exists an automorphism λ : T → T and ϕ : M1 → M2 symplectomorphism such that ϕ(x, p) = λ(x) · ϕ(p). We also need a notion of equivalence of polytopes. Recall that Aut(T ) = GL(n, Z), and we can define Definition. AGL(n, Z) = {x → Bx + c : B ∈ GL(n, Z), c ∈ Rn}. Finally, not all polytopes can arise from the image of a moment map. It is not hard to see that the following are some necessary properties: Definition (Delzant polytope). A Delzant polytope in Rn is a compact convex polytope satisfying (i) Simplicity: There exists exactly n edges out meeting at each vertex. (ii) Rationality: The edges meeting at each vertex P are of the form P + tui for t ≥ 0 and ui ∈ Zn. (iii) Smoothness: For each vertex, the corresponding ui’s can be chosen to be a Z-basis of Zn. Observe that all polytopes arising as µ(M ) satisfy these properties. We can equivalently define rationality and smoothness as being the exact same conditions on the outward-pointing normals to the facets (co-dimension 1 faces) meeting at P . Example. In R, there is any Delzant polytope is a line segment. This corresponds to the toric manifold S2 = CP1 as before, and the length of the polytope corresponds to the volume of CP1 under ω. Example. In R2, this is Delzant polytope: 53 3 Hamiltonian vector fields III Symplectic Geometry On the other hand, this doesn’t work: since on the bottom right vertex, we have det To fix this, we can do −1 −1 2 0 = −2 = ±1. Of course, we can also do boring things like rectangles. There is in fact a classification of all Delzant polytopes in R2, but we shall not discuss this. Example. The rectangular pyramid in R3 is not Delzant because it is not simple. The tetrahedron is. Theorem (Delzant). There are correspondences symplectic toric manifolds up to equivalence symplectic toric manifolds up to weak equivalence ←→ ←→ Delzant polytopes Delzant polytopes modulo AGL(n, Z) Proof sketch. Given a Delzant polytope ∆ in (Rn)∗ with d facets, we want to construct (M∆, ω∆, T∆, µ∆) with µ∆(M∆) = ∆. The idea is to perform the construction for the “universal” Delzant polytope with d facets, and then obtain the desired M∆ as a symplectic reduction of this universal example. As usual, the universal example will be “too big” to be a genuine symplectic toric manifold. Instead, it will be non-compact. If ∆ has d facets with primitive outward-point normal vectors v1, . . . , vd (i.e. they cannot be written as a Z-multiple of some other Z-vector), then we can write ∆ as ∆ = {x ∈ (Rn)∗ : x, vi ≤ λi for i = 1, . . . , d} for some λi. 54 3 Hamiltonian vector fields III Symplectic Geometry There is a natural (surjective) map π : Rd → Rn that sends the basis vector ed of Rd to vd. If λ = (λ1, . . . , λd), and we have a pullback diagram ∆ Rd λ (Rn)∗ π∗ (Rd)∗ where Rd λ = {X ∈ (Rd)∗ : X, ei ≤ λi for all i}. In more down-to-earth language, this says π∗(x) ∈ Rd λ ⇐⇒ x ∈ ∆, which is evident from definition. Now there is a universal “toric manifold” with µ(M ) = Rd λ, namely (Cd, ω0) with the diagonal action (t1, . . . , td) · (z1, . . . , zd) = (eit1z1, . . . , eitnzn), using the moment map φ(z1, . . . , zd) = − 1 2 (|z1|2, . . . , |zd|2) + (λ1, . . . , λd). We now want to pull this back along π∗. To this extent, note that π sends Zd to Zn, hence induces a map T d → T n with kernel N . If n is the Lie algebra of N , then we have a short exact sequence 0 −→ (Rn)∗ π∗ −→ (Rd)∗ i∗ −→ n∗ −→ 0. Since im π∗ = ker i∗, the pullback of Cd along π∗ is exactly Z = (i∗ ◦ φ)−1(0). It is easy to see that this is compact. Observe that i∗ ◦ φ is exactly the the moment map of the induced action by N . So Z/N is the symplectic reduction of Cd by N , and in particular has a natural symplectic structure. It is natural to consider Z/N instead of Z itself, since Z carries a T d action, but we only want to be left with a T n action. Thus, after quotienting out by N , the T d action becomes a T d/N ∼= T n action, with moment map given by the unique factoring of through (Rn)∗. The image is exactly ∆. Z → Cd → (Rd)∗ 55 4 Symplectic embeddings III Symplectic Geometry 4 Symplectic embeddings We end with a tiny chapter on symplectic embeddings, as promised in the course description. Definition (Symplectic embedding). A symplectic embedding is an embedding s ϕ : M1 → M2 such that ϕ∗ω2 = ω1. The notation we use is (M1, ω1) → (M2, ω2). A natural question to ask is, if we have two symplectic manifolds, is there a symplectic embedding between them? For concreteness, take (Cn, ω0) ∼= (R2n, ω0), and consider the subsets B2n(r) and Z 2n(R) = B2(R)×R2n−2 (where the product is one of symplectic manifolds). If r ≤ R, then there is a natural inclusion of B2n(r) into Z 2n(R)? If we only ask for volume-preserving embeddings, then we can always embed B2n(r) into Z 2n(R), since Z 2n(R) has infinite volume. It turns out, if we require the embedding to be symplectic, we have Theorem (Non-squeezing theorem, Gromov, 1985). There is an embedding B2(n) → Z 2n(R) iff r < R. When studying symplectic embeddings, it is natural to consider the following: Definition (Symplectic capacity). A symplectic capacity is a function c from the set of 2n-dimensional manifolds to [0, ∞] such that (i) Monotonicity: if (M1, ω1) → (M2, ω2), then c(M1, ω1) ≤ c(M2, ω2). (ii) Conformality: c(M, λω) = λc(M, ω). (iii) Non-triviality: c(B2n(1), ω0) > 0 and c(Z 2n(1), ω0) < ∞. If we only have (i) and (ii), this is called a generalized capacity. Note that the volume is a generalized capacity, but not a symplectic capacity. Proposition. The existence of a symplectic capacity is equivalent to Gromov’s non-squeezing theorem. Proof. The ⇒ direction is clear by monotonicity and conformality. Conversely, if we know Gromov’s non-squeezing theorem, we can define the Gromov width WG(M, ω) = sup{πr2 | (B2n(r), ω0) → (M, ω)}. This clearly satisfies (i) and (ii), and (iii) follows from Gromov non-squeezing. Note that Darboux’s theorem says there is always an embedding of B2n(r) into any symplectic manifold as long as r is small enough. 56 Index Index ∗, 25, 26 J-anti-holomorphic, 20 J-anti-holomorphic tangent vectors, 18 J-holomorphic, 20 J-holomorphic tangent vectors, 18 Lg, 37 T 0,1, 19 T 1,0, 19 T0,1, 19 T1,0, 18, 19 Λp,q, 19 Ωk(M, C), 19 Ωp,q, 19 δ, 26 J (V, Ω), 17, 18 Hk, 27 Hp,q, 28 Oξ, 39 ˜ω, 9 n-periodic point, 11 action, 35, 36 action coordinates, 34 adjoint action, 38 almost complex structure, 17 compatible, 17 integrable, 17 angle coordinates, 33 anti-holomorphic tangent vector, 18 Arnold–Liouville theorem, 34 Betti numbers, 28 calculus of variations, 36 canonical almost-complex structure, 17 canonical symplectic form, 7 central symmetry, 28 coadjoint action, 38 coadjoint orbit, 39 codifferential operator, 26 coindex, 47 commuting integrals of motion, 32 compatible almost complex structure, 17 compatible complex structure, 16 57 III Symplectic Geometry completely integrable system, 33 complex anti-linear cotangent vectors, 19 complex linear cotangent vectors, 19 complex structure, 16 configuration space, 6, 35 connection, 43 connection form, 43 conormal bundle, 8 conormal space, 8 conserved quantity, 32 constant of motion, 32 Convexity theorem, 45 cotangent bundle, 6 critical submanifolds, 47 curvature form, 44 Darboux theorem, 6 deformation equivalent, 5 Delzant polytope, 53 Dolbeault cohomology groups, 21 effective action, 52 Ehresmann connection, 43 energy, 34 Euler–Lagrange equations, 36 exactly homotopic, 15 first integral, 32 flat connection, 44 forms of type (p, q), 19 free |
action, 40 Fubini–Study form, 24 gauge group, 45 generalized capacity, 56 generating function, 8, 10 Hamilton equations, 32 Hamiltonian, 35 Hamiltonian action, 37, 38 Hamiltonian function, 30, 32 Hamiltonian system, 32 Hamiltonian vector field, 30 harmonic form, 27 Hessian matrix, 47 Index III Symplectic Geometry Hodge decomposition theorem, 25, 27, 28 Hodge diamond, 29 Hodge numbers, 28 Hodge star, 26 hodge star, 25 Hodge symmetry, 28 holomorphic tangent vector, 18 horizontal bundle, 43 independent integrals of motion, 32 index, 47 integrable almost complex structures, 17 integral of motion, 32 isotopic, 5 isotropic subspace, 4 isotropy group, 40 K¨ahler form, 21 K¨ahler manifold, 21 K¨ahler potential, 23 K¨ahler submanifold, 23 Lagrangian function, 35, 36 Lagrangian neighbourhood theorem, 13 Lagrangian submanifold, 8 Lagrangian subspace, 4 Laplace–Beltrami operator, 26 Laplacian, 26 left-invariant vector field, 38 Legendre condition, 36 Lie algebra, 38 Lie bracket, 31 Liouville torus, 34 local K¨ahler potential, 24 locally free action, 40 moment map, 39 moment polytope, 45 momentum, 34 Morse function, 47 Morse theory, 47 Morse–Bott function, 47 Moser’s equation, 6 Moser’s trick, 5 relative, 6 negative normal bundle, 47 Newlander–Nirenberg, 20 58 Newton’s second law, 34 Nijenhuis torsion, 21 non-degenerate bilinear map, 3 non-degenerate function, 15 non-squeezing theorem, 56 orbit, 40 orbit space, 40 periodic point, 11 phase space, 6, 35 Poincar´e’s last geometric theorem, 13 Poisson algebra, 32 Poisson bracket, 31 polar decomposition, 16, 17 product form, 9 rank, 3 reduced space, 40 relative Moser’s trick, 6 Schur–Horn theorem, 51 Serre duality, 28 spsh, 23 stabilizer, 40 strictly plurisubharmonic, 23 strongly isotopic, 5 symplectic action, 37 symplectic basis, 4 symplectic capacity, 56 symplectic embedding, 56 symplectic form, 4 symplectic manifold, 4 symplectic quotient, 40 symplectic reduction, 40 symplectic subspace, 4 symplectic toric manifold, 52 symplectic vector field, 31 symplectic vector space, 3 symplectomorphic, 5 symplectomorphism, 4 symplectomorphism generated by f , 10 tautological 1-form, 7 toric manifold, 52 transitive action, 40 twisted product form, 9 twisted projective spaces, 43 Index III Symplectic Geometry vertical tangent bundle, 43 vertical tangent space, 43 Weyl chamber, 50 Weyl group, 50 weighted projective space, 43 Whitney extension theorem, 14 59 |
art of our usual picture of the reals is the sense that some numbers are “bigger” than others or more to the “right” than others. We express this by using inequalities x < y or x y. The order structure is closely related to the field structure. For example, when we use inequalities in elementary courses we frequently use the fact that if x < y and 0 < z, then xz < yz (i.e., that inequalities can be multiplied through by positive numbers). ≤ This structure, too, can be axiomatized and reduced to a small set of rules. Once again, these same rules can be found in other applications of mathematics. When these rules are added to the field axioms the result is called an ordered field. The real number system is an ordered field, satisfying the four additional axioms. Here a < b is now a statement that is either true or false. (Before a + b and a b were not statements, but elements of R.) · O1 For any a, b ∈ O2 For any a, b, c O3 For any a, b O4 For any a, b ∈ ∈ Exercises R exactly one of the statements a = b, a < b or b < a is true. R if a < b is true and b < c is true, then a < c is true. ∈ R if a < b is true, then a + c < b + c is also true for any c R if a < b is true, then a c < b · · c is also true for any c ∈ R. ∈ R for which c > 0. 1.4.1 Using just the axioms, prove that ad + bc < ac + bd if a < b and c < d. 1.4.2 Show for every n IN that n2 n. ≥ 1.4.3 Using just the axioms, prove the arithmetic-geometric mean inequality: ∈ ≤ R with a > 0 and b > 0. (Assume, for the moment, the existence of square roots.) √ab a + b 2 for any a, b See Note 3 ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 1.5. Bounds 1.5 Bounds 11 Let E be some set of real numbers. There may or may not be a number M that is bigger than every number in the set E. If there is, we say that M is an upper bound for the set. If there is no upper bound, then the set is said to be unbounded above or to have no upper bound. This is a simple enough idea, but it is critical to an understanding of the real numbers and so we shall look more closely at it and give some precise definitions. Definition 1.1: (Upper Bounds) Let E be a set of real numbers. A number M is said to be an upper bound for E if x M for all x E. ≤ ≤ ∈ ∈ Definition 1.2: (Lower Bounds) Let E be a set of real numbers. A number m is said to be a lower bound for E if m x for all x E. It is often important to note whether a set has bounds or not. A set that has an upper bound and a lower bound is called bounded. A set can have many upper bounds. Indeed every number is an upper bound for the empty set . A set may have no upper bounds. We can use the phrase “E is unbounded above” if there are no upper bounds. For some sets the most natural upper bound (from among the infinitely many to choose) is just the largest member of the set. This is called the maximum. Similarly, the most natural lower bound for some sets is the smallest member of the set, the minimum. ∅ Definition 1.3: (Maximum) Let E be a set of real numbers. If there is a number M that belongs to E and is larger than every other member of E, then M is called the maximum of the set E and we write M = max E. Definition 1.4: (Minimum) Let E be a set of real numbers. If there is a number m that belongs to E and is smaller than every other member of E, then m is called the minimum of the set E and we write m = min E. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 12 Properties of the Real Numbers Chapter 1 Example 1.5: The interval x : 0 has a maximum and a minimum. The maximum is 1 and 1 is also an upper bound for the set. (If a set has a maximum, then that number must certainly be an upper bound for the set.) Any number larger than 1 ◭ is also an upper bound. The number 0 is the minimum and also a lower bound. [0, 1] = ≤ ≤ x { } 1 Example 1.6: The interval (0, 1) = x : 0 < x < 1 { } has no maximum and no minimum. At first glance some novices insist that the maximum should be 1 and the minimum 0 as before. But look at the definition. The maximum must be both an upper bound and also a member of the set. Here 1 and 0 are upper and lower bounds, respectively, but do not belong to the ◭ set. Example 1.7: The set IN of natural numbers has a minimum but no maximum and no upper bounds at ◭ all. We would say that it is bounded below but not bounded above. 1.6 Sups and Infs Let us return to the subject of maxima and minima again. If E has a maximum, say M , then that maximum could be described by the statement M is the least of all the upper bounds of E, that is to say, M is the minimum of all the upper bounds. The most frequent language used here is “M is the least upper bound.” It is possible for a set to have no maximum and yet be bounded above. In any example that comes to mind you will see that the set appears to have a least upper bound. Example 1.8: The open interval (0, 1) has no maximum, but many upper bounds. Certainly 2 is an upper bound and so is 1. The least of all the upper bounds is the number 1. Note that 1 cannot be described as ◭ a maximum because it fails to be in the set. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 1.6. Sups and Infs 13 Definition 1.9: (Least Upper Bound/Supremum) Let E be a set of real numbers that is bounded above and nonempty. If M is the least of all the upper bounds, then M is said to be the least upper bound of E or the supremum of E and we write M = sup E. Definition 1.10: (Greatest Lower Bound/Infimum) Let E be a set of real numbers that is bounded below and nonempty. If m is the greatest of all the lower bounds of E, then m is said to be the greatest lower bound of E or the infimum of E and we write M = inf E. To complete the definition of inf E and sup E it is most convenient to be able write this expression even for E = ∅ or for unbounded sets. Thus we write 1. inf = and sup = . ∅ ∅ 2. If E is unbounded above, then sup E = −∞ ∞ . ∞ 3. If E is unbounded below, then inf E = . −∞ The Axiom of Completeness Any example of a nonempty set that you are able to visualize that has an upper bound will also have a least upper bound. Pages of examples might convince you that all nonempty sets bounded above must have a least upper bound. Indeed your intuition will forbid you to accept the idea that this could not always be the case. To prove such an assertion is not possible using only the axioms for an ordered field. Thus we shall assume one further axiom, known as the axiom of completeness. Completeness Axiom A nonempty set of real numbers that is bounded above has a least upper bound (i.e., if E is nonempty and bounded above, then sup E exists and is a real number). This now is the totality of all the axioms we need to assume. We have assumed that R is a field with two operations of addition and multiplication, that R is an ordered field with an inequality relation “<”, and finally that R is a complete ordered field. This is enough to characterize the real numbers and the phrase “complete ordered field” refers to the system of real numbers and to no other system. (We shall not prove this statement; see Exercise 1.11.3 for a discussion.) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 14 Exercises Properties of the Real Numbers Chapter 1 1.6.1 Show that a set of real numbers E is bounded if and only if there is a positive number r so that all x E. ∈ 1.6.2 Find sup E and inf E and (where possible) max E and min E for the following examples of sets: < r for x | | (a) E = IN (b) E = Z (c) E = Q (d) E = R (e) E = (f) E = (g) E = (h) E = (i) E = 3, 2, 5, 7 } {− x : x2 < 2 x : x2 x − 1/n : n n√n : n { { { ∈ ∈ { } − IN } IN } 1 < 0 } 1.6.3 Under what conditions does sup E = max E? 1.6.4 Show for every nonempty, finite set E that sup E = max E. See Note 4 1.6.5 For every x R define ∈ { called the greatest integer function. Show that this is well defined and sketch the graph of the function. ≤ ∈ } [x] = max n Z : n x 1.6.6 Let A be a set of real numbers and let B = between min A and max B. x : x A } ∈ {− . Find a relation between max A and min B and 1.6.7 Let A be a set of real numbers and let B = between inf A and sup B. x : x A } ∈ {− . Find a relation between sup A and inf B and 1.6.8 Let A be a set of real numbers and let B = and sup B. x + r : x { A } ∈ for some number r. Find a relation between sup A ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 1.6. Sups and Infs 15 1.6.9 Let A be a set of real numbers and let B = xr : x sup A and sup B. (What happens if r is negative?) { A } ∈ for some positive number r. Find a relation between 1.6.10 Let A and B be sets of real numbers such that A B. Find a relation among inf A, inf B, sup A, and sup B. ⊂ 1.6.11 Let A and B be sets of real numbers and write C = A 1.6.12 Let A and B be sets of real numbers and write C = A B. Find a relation among sup A, sup B, and sup C. B. Find a relation among sup A, sup B, and sup C. ∪ ∩ 1.6.13 Let A and B be sets of real numbers and write Find a relation among sup A, sup B, and sup C, y B . } ∈ ∈ 1.6.14 Let A and B be sets of real numbers and write { Find a relation among inf A, inf B, and inf C. C = x + y : x A, y B . } ∈ ∈ 1.6.15 Let A be a set of real numbers and write A2 = the infs and sups of the two sets? x2 : x { A } ∈ . Are there any relations you can find between 1.6.16 Let E be a set of real numbers. Show that x is not an upper bound of E if and only if there exists a number E such that e > x. e ∈ 1.6.17 Let A be a set of real numbers. Show that a real number x is the supremum of A if and only if a a ∈ A and for every positive number ε there is an element a′ A such that x − ∈ ε < a′. x for all ≤ 1.6.18 Formulate a condition analogous to the preceding exercise for an infimum. 1.6.19 Using the completeness axiom, show that every nonempty set E of real numbers that is bounded below has a greatest low |
er bound (i.e., inf E exists and is a real number). 1.6.20 A function is said to be bounded if its range is a bounded set. Give examples of functions f : R R that are bounded and examples of such functions that are unbounded. Give an example of one that has the property that → is finite but max { f (x) : x R } ∈ sup { does not exist. f (x) : x R } ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 16 Properties of the Real Numbers Chapter 1 1.6.21 The rational numbers Q satisfy the axioms for an ordered field. Show that the completeness axiom would not be satisfied. That is show that this statement is false: Every nonempty set E of rational numbers that is bounded above has a least upper bound (i.e., sup E exists and is a rational number). 1.6.22 Let F be the set of all numbers of the form x + √2y, where x and y are rational numbers. Show that F has all the properties of an ordered field but does not have the completeness property. 1.6.23 Let A and B be nonempty sets of real numbers and let δ(A, B) is often called the “distance” between the sets A and B. δ(A, B) = inf a : a A, b {| − b | ∈ B . } ∈ (a) Let A = IN and B = R (b) If A and B are finite sets, what does δ(A, B) represent? IN. Compute δ(A, B) \ (c) Let B = [0, 1]. What does the statement δ( { (d) Let B = (0, 1). What does the statement δ( { } x } x , B) = 0 mean for the point x? , B) = 0 mean for the point x? 1.7 The Archimedean Property There is an important relationship holding between the set of natural numbers IN and the larger set of real numbers R. Because we have a well-formed mental image of what the set of reals “looks like,” this property is entirely intuitive and natural. It hardly seems that it would require a proof. It says that the set of natural numbers IN has no upper bound (i.e., that there is no real number x so that n x for all n = 1, 2, 3, . . . ). ≤ At first sight this seems to be a purely algebraic and order property of the reals. In fact it cannot be proved without invoking the completeness property of Section 1.6. The property is named after the famous Greek mathematician known as Archimedes of Syracuse (287 B.C.–212 B.C.).1 1 Archimedes seems to be the archetypical absent-minded mathematician. The historian Plutarch tells of his death at the hand of an invading army: “As fate would have it, Archimedes was intent on working out some problem by a diagram, and ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 1.7. The Archimedean Property 17 Theorem 1.11 (Archimedean Property of R) The set of natural numbers IN has no upper bound. Proof. The proof is obtained by contradiction. If the set IN does have an upper bound, then it must have a least upper bound. Let x = sup IN, supposing that such does exist as a finite real number. Then n x for all natural numbers n but n 1 cannot be true for all natural numbers n. Choose some natural 1. Then m + 1 is also an natural number and m + 1 > x. But that cannot be so number m with m > x since we defined x as the supremum. From this contradiction the theorem follows. ≤ − − ≤ x The archimedean theorem has some consequences that have a great impact on how we must think of the real numbers. 1. No matter how large a real number x is given, there is always a natural number n larger. 2. Given any positive number y, no matter how large, and any positive number x, no matter how small, one can add x to itself sufficiently many times so that the result exceeds y (i.e., nx > y for some n IN). ∈ 3. Given any positive number x, no matter how small, one can always find a fraction 1/n with n a natural number that is smaller (i.e., so that 1/n < x). Each of these is a consequence of the archimedean theorem, and the archimedean theorem in turn can be derived from any one of these. Exercises 1.7.1 Using the archimedean theorem, prove each of the three statements that follow the proof of the archimedean theorem. having fixed both his mind and eyes upon the subject of his speculation, he did not notice the entry of the Romans nor that the city was taken. In this transport of study a soldier unexpectedly came up to him and commanded that he accompany him. When he declined to do this before he had finished his problem, the enraged soldier drew his sword and ran him through.” For this biographical detail and many others on all the mathematicians in this book consult http://www-history.mcs.st-and.ac.uk/history. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 18 Properties of the Real Numbers Chapter 1 1.7.2 Suppose that it is true that for each x > 0 there is an n using this assumption. IN so that 1/n < x. Prove the archimedean theorem ∈ 1.7.3 Without using the archimedean theorem, show that for each x > 0 there is an n See Note 5 1.7.4 Let x be any real number. Show that there is a m Show that m is unique. m ≤ Z so that ∈ x < m + 1. IN so that 1/n < x. ∈ 1.7.5 The mathematician Leibniz based his calculus on the assumption that there were “infinitesimals,” positive real numbers that are extremely small—smaller than all positive rational numbers certainly. Some calculus students also believe, apparently, in the existence of such numbers since they can imagine a number that is “just next to zero.” Is there a positive real number smaller than all positive rational numbers? 1.7.6 The archimedean property asserts that if x > 0, then there is a natural number N so that 1/N < x. The proof requires the completeness axiom. Give a proof that does not use the completeness axiom that works for x rational. Find a proof that is valid for x = √y, where y is rational. 1.7.7 In Section 1.2 we made much of the fact that there is a number whose square is 2 and so √2 does exist as a real number. Show that exists as a real number and that α2 = 2. See Note 6 α = sup { x ∈ R : x2 < 2 } 1.8 Inductive Property of IN Since the natural numbers are included in the set of real numbers there are further important properties of IN that can be deduced from the axioms. The most important of these is the principle of induction. This is the basis for the technique of proof known as induction, which is often used in this text. For an elementary account and some practice, see Section A.8 in the appendix. We first prove a statement that is equivalent. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 1.8. Inductive Property of IN 19 Theorem 1.12 (Well-Ordering Property) Every nonempty subset of IN has a smallest element. Proof. Let S If α ⊂ ∈ IN and S = . Then α = inf S must exist and be a real number since S is bounded below. S, then we are done since we have found a minimal element. ∅ Suppose not. Then, while α is the greatest lower bound of S, α is not a minimum. There must be an element of S that is smaller than α + 1 since α is the greatest lower bound of S. That element cannot be α since we have assumed that α S. Thus we have found x S with 6∈ ∈ α < x < α + 1. Now x is not a lower bound of S, since it is greater than the greatest lower bound of S, so there must be yet another element y of S such that α < y < x < α + 1. But now we have reached an impossibility, for x and y are in S and both natural numbers, but 0 < x which cannot happen. From this contradiction the proof now follows. y < 1, − Now we can state and prove the principle of induction. IN so that 1 ∈ ⊂ S and, for every natural number n, if Theorem 1.13 (Principle of Induction) Let S n S then so also is n + 1. Then S = IN. S. We claim that E = ∈ Proof. Let E = IN Suppose not (i.e., suppose E because 1 in S. By hypothesis it follows that α = (α we have obtained a contradiction, proving our theorem. S by hypothesis. Thus α = − ∈ ∅ \ ∅ and then it follows that S = IN proving the theorem. ). By Theorem 1.12 there is a first element α of E. Can α = 1? No, 1 is also a natural number and, since it cannot be in E it must be 1) + 1 must be in S. But it is in E. This is impossible and so − Exercises 1.8.1 Show that any bounded, nonempty set of natural numbers has a maximal element. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 20 Properties of the Real Numbers Chapter 1 1.8.2 Show that any bounded, nonempty subset of Z has a maximum and a minimum. 1.8.3 For further exercises on proving statements using induction as a method, see Section A.8. 1.9 The Rational Numbers Are Dense There is an important relationship holding between the set of rational numbers Q and the larger set of real numbers R. The rational numbers are dense. They make an appearance in every interval; there are no gaps, no intervals that miss having rational numbers. For practical purposes this has great consequences. We need never actually compute with arbitrary real numbers, since close by are rational numbers that can be used. Thus, while π is irrational, in routine computations with a practical view any nearby fraction might do. At various times people have used 3, 22/7, and 3.14159, for example. For theoretical reasons this fact is of great importance too. It allows many arguments to replace a consideration of the set of real numbers with the smaller set of rationals. Since every real is as close as we please to a rational and since the rationals can be carefully described and easily worked with, many simplifications are allowed. Definition 1.14: (Dense Sets) A set E of real numbers is said to be dense (or dense in R) if every interval (a, b) contains a point of E. Theorem 1.15: The set Q of rational numbers is dense. Proof. Let x < y and consider the interval (x, y). We must find a rational number inside this interval. By the archimedean theorem, Theorem 1.11, there is a natural number This means that ny > nx + 1. n > 1 − y . x ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 21 Z ∈ Section 1.9. The Rational Numbers Are Dense Let m be chosen as the integer |
just less than nx + 1; more precisely (using Exercise 1.7.4), find m so that m Now some arithmetic on these inequalities shows that nx + 1 < m + 1. ≤ and then m 1 − ≤ nx < ny thus exhibiting a rational number m/n in the interval (x, y). Exercises 1.9.1 Show that the definition of “dense” could be given as A set E of real numbers is said to be dense if every interval (a, b) contains infinitely many points of E. 1.9.2 Find a rational number between √10 and π. 1.9.3 If a set E is dense, what can you conclude about a set A ⊃ 1.9.4 If a set E is dense, what can you conclude about the set R 1.9.5 If two sets E1 and E2 are dense, what can you conclude about the set E1 ∩ 1.9.6 Show that the dyadic rationals (i.e., rational numbers of the form m/2n for m E? E? \ E2? Z, n ∈ ∈ IN) are dense. 1.9.7 Are the numbers of the form IN dense? What is the length of the largest interval that contains no such number? m/2100 ± for m ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 22 Properties of the Real Numbers Chapter 1 1.9.8 Show that the numbers of the form for m, n See Note 7 ∈ IN are dense. 1.10 The Metric Structure of R m√2/n ± In addition to the algebraic and order structure of the real numbers, we need to make measurements. We need to describe distances between points. These are the metric properties of the reals, to borrow a term from the Greek for measure (metron). As usual, the distance between a point x and another point y is either x y or y x depending on which π. To is positive. Thus the distance between 3 and describe this in general requires the absolute value function which simply makes a choice between positive and negative. 4 is 7. The distance between π and √10 is √10 − − − − Definition 1.16: (Absolute Value) For any real number x write and = x if if x < 0 . (Beginners tend to think of the absolute value function as “stripping off the negative sign,” but the example shows that this is a limited viewpoint.) π | − √10 | = √10 π − Properties of the Absolute Value Since the absolute value is defined directly in terms of inequalities (i.e., 0 or x < 0), there are a number of properties that can be proved directly from properties of the choice x ≥ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 1.10. The Metric Structure of R 23 inequalities. These properties are used routinely and the student will need to have a complete mastery of them. Theorem 1.17: The absolute value function has the following properties: 1. For any x ∈ 2. For any x, y 3. For any x, y 4. For any x, y R, ∈ ∈ ∈ −| R, R, R, x x = | ≤ xy | | . | | and . | − Distances on the Real Line Using the absolute value function we can define the distance function or metric. Definition 1.18: (Distance) The distance between two real numbers x and y is d(x, y) = x | − y . | We hardly ever use the notation d(x, y) in elementary analysis, preferring to write even while we are thinking of this as the distance between the two points. Thus if a sequence of points x1, x2, x3, . . . is growing ever closer to a point c, we should perhaps describe d(xn, c) as getting smaller and smaller, thus emphasizing that the distances are shrinking; more often we would simply write and expect you to interpret this as a distance. xn − − x y c | | | | Properties of the Distance Function The main properties of the distance function are just interpretations of the absolute value function. Expressed in the language of a distance function, they are geometrically very intuitive: 1. d(x, y) 0 ≥ (all distances are positive or zero). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 24 Properties of the Real Numbers Chapter 1 2. d(x, y) = 0 if and only if x = y (different points are at positive distance apart). 3. d(x, y) = d(y, x) (distance is symmetric, that is the distance from x to y is the same as from y to x)). 4. d(x, y) d(x, z) + d(z, y) ≤ (the triangle inequality, that is it is no longer to go directly from x to y than to go from x to z and then to y). In Chapter 13 we will study general structures called metric spaces, where exactly such a notion of distance satisfying these four properties is used. For now we prefer to rewrite these properties in the language of the absolute value, where they lose some of their intuitive appeal. But it is in this form that we are likely to use them. = 0 if and only if a = 0. 1. 2. 3. 4. a | | ≥ 0the triangle inequality). Exercises 1.10.1 Show that 1.10.2 Show that max { x, y } 1.10.3 Show that the inequalities = max x | are equivalent. y /2 + (x + y)/2. What expression would give min { | x, y ? } < ε and − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 1.11. Challenging Problems for Chapter 1 25 1.10.4 Show that if α < x < β and α < y < β, then about the interval (α, β). x | − y | < β − α and interpret this geometrically as a statement 1.10.5 Show that y | − | is also called the triangle inequality. || ≤ | − x x || y | assuming the triangle inequality (i.e., that | ). This inequality | 1.10.6 Under what conditions is it true that 1.10.7 Under what conditions is it true that .10.8 Show that x1 + x2 + | for any numbers x1, x2, . . . , xn. + xn| ≤ | x1| + + x2| | · · · + xn.10.9 Let E be a set of real numbers and let A = sups of the two sets? x {| | : x E } ∈ . What relations can you find between the infs and 1.10.10 Find the inf and sup of the set x : { 2x + π | | < √2 . } 1.11 Challenging Problems for Chapter 1 1.11.1 The complex numbers C are defined as equal to the set of all ordered pairs of real numbers subject to these operations: and (a1, b1) + (a2, b2) = (a1 + a2, b1 + b2) (a1, b1) (a2, b2) = (a1a2 − · b1b2, a1b2 + a2b1). (a) Show that C is a field. (b) What are the additive and multiplicative identity elements? (c) What are the additive and multiplicative inverses of an element (a, b)? (d) Solve (a, b)2 = (1, 0) in C. (e) We identify R with a subset of C by identifying the elements x how this can be interpreted as saying that “R is a subfield of C.” ∈ R with the element (x, 0) in C. Explain ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 26 Properties of the Real Numbers Chapter 1 (f) Show that there is an element i z = x + iy for x, y R. ∈ C with i2 = ∈ 1 so that every element z − C can be written as ∈ (g) Explain why the equation x2 + x + 1 = 0 has no solution in R but two solutions in C. 1.11.2 Can an order be defined on the field C of Exercise 1.11.1 in such a way so to make it an ordered field? 1.11.3 The statement that every complete ordered field “is” the real number system means the following. Suppose that F is a nonempty set with operations of addition “+” and multiplication “ ” and an order relation “<” · that satisfies all the axioms of an ordered field and also the axiom of completeness. Then there is a one-to-one onto function f : R F that has the following properties: → (a) f (x + y) = f (x) + f (y) for all x, y (b) f (x (c) f (x) < f (y) if and only if x < y for x, y f (y) for all x, y y) = f (x) ∈ R. R. ∈ · · R. ∈ Thus, in a certain sense, F and R are essentially the same object. Attempt a proof of this statement. [Note R refers to the addition in the reals whereas f (x) + f (y) refers to the addition in the set that x + y for x, y F .] ∈ 1.11.4 We have assumed in the text that the set IN is obviously contained in R. After all, 1 is a real number (it’s in the axioms), 2 is just 1 + 1 and so real, 3 is 2 + 1 etc. In that way we have been able to prove the material of Section 1.8. But there is a logical flaw here. We would need induction really to define IN in this way (and not just say “etc.”). Here is a set of exercises that would remedy that for students with some background in set manipulations. S. Show that R is inductive. (a) Define a set S (b) Show that there is a smallest inductive set by showing that the intersection of the family of all inductive R to be inductive if 1 S implies that x + 1 S and if x ⊂ ∈ ∈ ∈ sets is itself inductive. (c) Define IN to be that smallest inductive set. (d) Prove Theorem 1.13 now. (That is, show that any set S with the property stated there is inductive and conclude that S = IN.) (e) Prove Theorem 1.12 now. (That is, with this definition of IN prove the well-ordering property.) 1.11.5 Use this definition of “dense in a set” to answer the following questions: ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 27 A set E of real numbers is said to be dense in a set A if every interval (a, b) that contains a point of A also contains a point of E. (a) Show that dense in the set of all reals is the same as dense. (b) Give an example of a set E dense in IN but with E IN = (c) Show that the irrationals are dense in the rationals. (A real number is irrational if it is not rational, ∩ ∅ . that is if it belongs to R but not to Q.) (d) Show that the rationals are dense in the irrationals. (e) What property does a set E have that is equivalent to the assertion that R E is dense in E? \ 1.11.6 Let G be a subgroup of the real numbers under addition (i.e., if x and y are in G, then x + y x − ∈ G). Show that either G is a dense subset of R or else there is a real number α so that nα : n = 0, 3, . . . G = 1, 2, . { ± ± ± } G and ∈ See Note 8 Notes 1Exercise 1.3.3. Let F be the set of all numbers of the form x + y√2 where x, y Q. Again to be sure that nine properties of a field hold it is enough to check, here, that a + b and a · ∈ b are in F if both a and b are. 2Exercise 1.3.5. As a first step define what x2 and 2x really mean. In fact, define 2. (It would be defined as 2 = 1 + 1 since 1 and addition are defined in the field axioms.) Then multiply (x + 1) (x + 1) using only the rules given here. Since your proof uses only the field axioms, it must be valid in any situation in which these axioms are true, not just for R. · 3Exercise 1.4.3. Su |
ppose a > 0 and b > 0 and a = b. Establish that √a = √b. Establish that (√a − √b)2 > 0. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 28 NOTES Carry on. What have you proved? Now what if a = b? 4Exercise 1.6.4. You can use induction on the size of E, that is, prove for every natural number n that if E has n elements, then 5Exercise 1.7.3. Suppose not, then the set sup E = max E. 1/n : n = 1, 2, 3, . . . { } has a positive lower bound, etc. You will have to use the existence of a greatest lower bound. 6Exercise 1.7.7. Not that easy to show. Rule out the possibilities α2 < 2 and α2 > 2 using the archimedean property to assist. 7Exercise 1.9.8. To find a number in (x, y), find a rational in (x//√2, y//√2). Conclude from this that the set of all (irrational) numbers of the form m√2/n is dense. ± 8Exercise 1.11.6. 0 } dense. Case 2: If α > 0 show that If G = { , then take α = 0. If not, let α = inf G G = { nα : n = 0, 1, 2, ± ± ± 3, . . . ). Case 1: If α = 0 show that G is (0, ∞ ∩ . } For case 1 consider an interval (r, s) with r < s. We wish to find a member of G in that interval. To keep the argument simple just consider, for the moment, the situation in which 0 < r < s. Choose g G with 0 < g < s r. The set ∈ − is nonempty (why?) and so there is a minimal element m in M (why?). Now check that (m the interval (r, s). − 1)g is in G and inside M = n { ∈ IN : ng s } ≥ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Chapter 2 SEQUENCES 2.1 Introduction Let us start our discussion with a method for solving equations that originated with Newton in 1669. To solve an equation f (x) = 0 the method proposes the introduction of a new function F (x) = x f (x) f ′(x) . − We begin with a guess at a solution of f (x) = 0, say x1 and compute x2 = F (x1) in the hopes that x2 is closer to a solution than x1 was. The process is repeated so that x3 = F (x2), x4 = F (x3), x5 = F (x4), . . . and so on until the desired accuracy is reached. Processes of this type have been known for at least 3500 years although not in such a modern notation. We illustrate by finding an approximate value for √2 this way. We solve the equation f (x) = x2 by computing the function 2 = 0 − f (x) f ′(x) and using it to improve our guess. A first (very crude) guess of x1 = 1 will produce the following list of values for our subsequent steps in the procedure. We have retained 60 digits in the decimal expansions to F (x) = x − 2x = x − − 2 x2 29 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 30 show how this is working: Sequences Chapter 2 x1 = 1.00000000000000000000000000000000000000000000000000000000000 x2 = 1.50000000000000000000000000000000000000000000000000000000000 x3 = 1.41666666666666666666666666666666666666666666666666666666667 x4 = 1.41421568627450980392156862745098039215686274509803921568628 x5 = 1.41421356237468991062629557889013491011655962211574404458490 x6 = 1.41421356237309504880168962350253024361498192577619742849829 x7 = 1.41421356237309504880168872420969807856967187537723400156101. To compare, here is the value of the true solution √2, computed in a different fashion to the same number of digits: √2 = 1.41421356237309504880168872420969807856967187537694807317668. Note that after only four steps the procedure gives a value differing from the true value only in the sixth decimal place, and all subsequent values remain this close. A convenient way of expressing this is to write that xn − | By the seventh step, things are going even better and we can claim that xn − 47 for all n < 10− < 10− √2 4. 7. ≥ ≥ | | | 5 for all n √2 It is inconceivable that anyone would require any further accuracy for any practical considerations. The 47, which is a tiny number. Even so, as mathematicians we can error after the sixth step cannot exceed 10− ask what may seem an entirely impractical sort of question. Can this accuracy of approximation continue forever? Is it possible that, if we wait long enough, we can find an approximation to √2 with any degree of accuracy? ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.2. Sequences 31 Expressed more formally, if we are given a positive number ε (we call it epsilon to suggest that it measures an error) no matter how small, can we find a stage in this procedure so that the value computed and all subsequent values are closer to √2 than ε? In symbols, is there an integer n0 (which will depend on just how small ε is) that is large enough so that If this is true then this sequence has a remarkable property. It is not merely in its first few terms a convenient way of computing √2 to some accuracy; the sequence truly represents the number √2 itself, and it cannot represent any other number. We shall say that the sequence converges to √2 and write xn − | √2 | < ε for all n n0? ≥ xn = √2. lim n →∞ This is the beginning of the theory of convergence that is central to analysis. If mathematicians had never considered the ultimate behavior of such sequences and had contented themselves with using only the first few terms for practical computations, there would have been no subject known as analysis. These ideas lead, as you might imagine, to an ideal world of infinite precision, where sequences are not merely useful gadgets for getting good computations but are precise tools in discussing real numbers. From the theory of sequences and their convergence properties has developed a vast world of beautiful and useful mathematics. For the student approaching this material for the first time this is a critical test. All of analysis, both pure and applied, rests on an understanding of limits. What you learn in this chapter will offer a foundation for all the rest that you will have to learn later. 2.2 Sequences A sequence (of real numbers, of sets, of functions, of anything) is simply a list. There is a first element in the list, a second element, a third element, and so on continuing in an order forever. In mathematics a finite list is not called a sequence; a sequence must continue without interruption. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 32 Sequences Chapter 2 For a more formal definition notice that the natural numbers are playing a key role here. Every item in the sequence (the list) can be labeled by its position; label the first item with a “1,” the second with a “2,” and so on. Seen this way a sequence is merely then a function mapping the natural numbers IN into some set. We state this as a definition. Since this chapter is exclusively about sequences of real numbers, the definition considers just this situation. Definition 2.1: By a sequence of real numbers we mean a function f : IN R. → Thus the sequence is the function. Even so, we usually return to the list idea and write out the sequence f as f (1), f (2), f (3), . . . , f (n), . . . with the ellipsis (i.e., the three dots) indicating that the list is to continue in this fashion. The function values f (1), f (2), f (3), . . . are called the terms of the sequence. When it is not confusing we will refer to such a sequence using the expression (with the understanding that the index n ranges over all of the natural numbers). f (n) } { If we need to return to the formality of functions we do, but try to keep the intuitive notion of a sequence as an unending list in mind. While computer scientists much prefer the function notation, mathematicians have become more accustomed to a subscript notation and would rather have the terms of the preceding sequence rendered as In this chapter we study sequences of real numbers. Later on we will encounter the same word applied to other lists of objects (e.g., sequences of intervals, sequences of sets, sequences of functions. In all cases the word sequence simply indicates a list of objects). f1, f2, f3, . . . , fn, . . . or . fn} { ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.2. Sequences 33 x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 Figure 2.1. An arithmetic progression. 2.2.1 Sequence Examples In order to specify some sequence we need to communicate what every term in the sequence is. For example, the sequence of even integers 2, 4, 6, 8, 10, . . . could be communicated in precisely that way: “Consider the sequence of even integers.” Perhaps more direct would be to give a formula for all of the terms in the sequence: “Consider the sequence whose nth term is xn = 2n.” Or we could note that the sequence starts with 2 and then all the rest of the terms are obtained by adding 2 to the previous term: “Consider the sequence whose first term is 2 and whose nth term is 2 added to the (n 1)st term,” that is, − xn = 2 + xn 1. − Often an explicit formula is best. Frequently though, a formula relating the nth term to some preceding term is preferable. Such formulas are called recursion formulas and would usually be more efficient if a computer is used to generate the terms. Arithmetic Progressions The simplest types of sequences are those in which each term is obtained from the preceding by adding a fixed amount. These are called arithmetic progressions. The sequence is the most general arithmetic progression. The number d is called the common difference. c, c + d, c + 2d, c + 3d, c + 4d, . . . , c + (n 1)d, . . . − Every arithmetic progression could be given by a formula xn = c + (n 1)d − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 34 or a recursion formula x1 = c xn = xn Sequences Chapter 2 1 + d. Note that the explicit formula is of the form xn = f (n), where f is a linear function, f (x) = dx + b for some b. Figure 2.1 shows the points of an arithmetic progression plotted on the line. If, instead, you plot the points (n, xn) you will find that they all lie on a straight line with slope d. − Geometric Progressions. A var |
iant on the arithmetic progression is obtained by replacing the addition of a fixed amount by the multiplication by a fixed amount. These sequences are called geometric progressions. The sequence c, cr, cr2, cr3, cr4, . . . , crn 1, . . . − is the most general geometric progression. The number r is called the common ratio. Every geometric progression could be given by a formula or a recursion formula xn = crn − 1 x1 = c xn = rxn 1. − Note that the explicit formula is of the form xn = f (n), where f is an exponential function f (x) = brx for some b. Figure 2.2 shows the points of a geometric progression plotted on the line. Alternatively, plot the points (n, xn) and you will find that they all lie on the graph of an exponential function. If c > 0 and the common ratio r is larger than 1, the terms increase in size, becoming extremely large. If 0 < r < 1, the terms decrease in size, getting smaller and smaller. (See Figure 2.2.) Iteration The examples of an arithmetic progression and a geometric progression are special cases of a process called iteration. So too is the sequence generated by Newton’s method in the introduction to this chapter. Let f be some function. Start the sequence x1, x2, x3, . . . by assigning some value in the domain of f , say x1 = c. All subsequent values are now obtained by feeding these values through the function repeatedly: c, f (c), f (f (c)), f (f (f (c))), f (f (f (f (c)))), . . . . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.2. Sequences 35 x8 x7 x6 x5 x4 x3 x2 x1 Figure 2.2. A geometric progression. As long as all these values remain in the domain of the function f , the process can continue indefinitely and defines a sequence. If f is a function of the form f (x) = x + b, then the result is an arithmetic progression. If f is a function of the form f (x) = ax, then the result is a geometric progression. A recursion formula best expresses this process and would offer the best way of writing a computer program to compute the sequence: Sequence of Partial Sums. If a sequence x1 = c xn = f (xn 1). − x1, x2, x3, x4, . . . is given, we can construct a new sequence by adding the terms of the old one: s1 = x1 s2 = x1 + x2 s3 = x1 + x2 + x3 s4 = x1 + x2 + x3 + x4 and continuing in this way. The process can also be described by a recursion formula: The new sequence is called the sequence of partial sums of the old sequence sequences in considerable depth in the next chapter. xn} { . We shall study such s1 = x1 , sn = sn 1 + xn. − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 36 Sequences Chapter 2 For a particular example we could use xn = 1/n and the sequence of partial sums could be written as Is there a more attractive and simpler formula for sn? The answer is no. sn = 1 + 1/2 + 1/3 + + 1/n. · · · Example 2.2: The examples, given so far, are of a general nature and describe many sequences that we will encounter in analysis. But a sequence is just a list of numbers and need not be defined in any manner quite so systematic. For example, consider the sequence defined by an = 1 if n is divisible by three, an = n 2n if n is two more than a multiple of three. The first if n is one more than a multiple of three, and an = few terms are evidently − What would be the next three terms? Exercises 4, 1, 4, 1, − − 32, 1, . . . . ◭ 2.2.1 Let a sequence be defined by the phrase “consider the sequence of prime numbers 2, 3, 5, 7, 11, 13 . . . ”. Are you sure that this defines a sequence? 2.2.2 On IQ tests one frequently encounters statements such as “what is the next term in the sequence 3, 1, 4, 1, 5, . . . ?”. In terms of our definition of a sequence is this correct usage? (By the way, what do you suppose the next term in the sequence might be?) See Note 9 2.2.3 Give two different formulas (for two different sequences) that generate a sequence whose first four terms are 2, 4, 6, 8. See Note 10 2.2.4 Give a formula that generates a sequence whose first five terms are 2, 4, 6, 8, π. 2.2.5 The examples listed here are the first few terms of a sequence that is either an arithmetic progression or a geometric progression. What is the next term in the sequence? Give a general formula for the sequence. (a) 7, 4, 1, . . . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 37 Section 2.3. Countable Sets (b) .1, .01, .001, . . . (c) 2, √2, 1, . . . 2.2.6 Consider the sequence defined recursively by x1 = √2 , xn = √2 + xn−1. Find an explicit formula for the nth term. 2.2.7 Consider the sequence defined recursively by x1 = √2 , xn = √2xn−1. Find an explicit formula for the nth term. 2.2.8 Consider the sequence defined recursively by x1 = √2 , xn = 2 + xn−1. Show, by induction, that xn < 2 for all n. p 2.2.9 Consider the sequence defined recursively by Show, by induction, that xn < xn+1 for all n. p 2.2.10 The sequence defined recursively by x1 = √2 , xn = 2 + xn−1. is called the Fibonacci sequence. It is possible to find an explicit formula for this sequence. Give it a try. See Note 11 f1 = 1 , f2 = 1 , fn+2 = fn + fn+1 2.3 Countable Sets " Enrichment section. May be omitted. A sequence of real numbers, formally, is a function whose domain is the set IN of natural numbers and whose range is a subset of the reals R. What sets might be the range of some sequence? To put it another ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 38 Sequences Chapter 2 way, what sets can have their elements arranged into an unending list? Are there sets that cannot be arranged into a list? The arrangement of a collection of objects into a list is sometimes called an enumeration. Thus another way of phrasing this question is to ask what sets of real numbers can be enumerated? The set of natural numbers is already arranged into a list in its natural order. The set of integers (including 0 and the negative integers) is not usually presented in the form of a list but can easily be so presented, as the following scheme suggests: 0, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, . . . . − Example 2.3: The rational numbers can also be listed but this is quite remarkable, for at first sight no reasonable way of ordering them into a sequence seems likely to be possible. The usual order of the rationals in the reals is of little help. − − − − − − | | To find such a scheme define the “rank” of a rational number m/n in its lowest terms (with n m 1) to + n. Now begin making a finite list of all the rational numbers at each rank; list these from smallest be to largest. For example, at rank 1 we would have only the rational number 0/1. At rank 2 we would have only the rational numbers 1/2, 1/2, 2/1. Carry on in this fashion through all the ranks. Now construct the final list by concatenating these shorter lists in order of the ranks: 1/1, 1/1. At rank 3 we would have only the rational numbers 2/1, ≥ − − − The range of this sequence is the set of all rational numbers. 0/1, 1/1, 1/1, 2/1, 1/2, 1/2, 2/1, . . . . − − − ◭ Your first impression might be that few sets would be able to be the range of a sequence. But having seen in Example 2.3 that even the set of rational numbers Q that is seemingly so large can be listed, it might then appear that all sets can be so listed. After all, can you conceive of a set that is “larger” than the rationals in some way that would stop it being listed? The remarkable fact that there are sets that cannot be arranged to form the elements of some sequence was proved by Georg Cantor (1845–1918). This proof is essentially his original proof. (Note that this requires some familiarity with infinite decimal expansions; the exercises review what is needed.) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.3. Countable Sets 39 Theorem 2.4 (Cantor) No interval (a, b) of real numbers can be the range of some sequence. It is enough to prove this for the interval (0, 1) since there is nothing special about it (see Proof. Exercise 2.3.1). The proof is a proof by contradiction. We suppose that the theorem is false and that there sn} is a sequence obtain a contradiction by showing that this cannot be so. We shall use the sequence c in the interval (0, 1) so that sn 6 decimal fraction. If we write this sequence out in decimal notation it might look like Each of the points s1, s2, s3 . . . in our sequence is a number between 0 and 1 and so can be written as a so that every number in the interval (0, 1) appears at least once in the sequence. We to find a number = c for all n. sn} { { s1 = 0.x11x12x13x14x15x16 . . . s2 = 0.x21x22x23x24x25x26 . . . s3 = 0.x31x32x33x34x35x36 . . . etc. Now it is easy to find a number that is not in the list. Construct c = 0.c1c2c3c4c5c6 . . . by choosing ci to be either 5 or 6 whichever is different from xii. This number cannot be equal to any of the listed numbers s1, s2, s3 . . . since c and si differ in the ith position of their decimal expansions. This gives us our contradiction and so proves the theorem. Definition 2.5: (Countable) A nonempty set S of real numbers is said to be countable if there is a sequence of real numbers whose range is the set S. In the language of this definition then we can see that (1) any finite set is countable, (2) the natural numbers and the integers are countable, (3) the rational numbers are countable, and (4) no interval of real numbers is countable. By convention we also say that the empty set is countable. ∅ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 40 Exercises Sequences Chapter 2 2.3.1 Show that, once it is known that the interval (0, 1) cannot be expressed as the range of some sequence, it follows that any interval (a, b), [a, b), (a, b], or [a, b] has the same property. See Note 12 2.3.2 Some novices, on reading the proof of Cantor’s theorem, say “Why can’t you just put the number c that you found at t |
he front of the list.” What is your rejoinder? 2.3.3 A set (any set of objects) is said to be countable if it is either finite or there is an enumeration (list) of the set. Show that the following properties hold for arbitrary countable sets: (a) All subsets of countable sets are countable. (b) Any union of a pair of countable sets is countable. (c) All finite sets are countable. 2.3.4 Show that the following property holds for countable sets: If S1, S2, S3, . . . is a sequence of countable sets of real numbers, then the set S formed by taking all elements that belong to at least one of the sets Si is also a countable set. See Note 13 2.3.5 Show that if a nonempty set is contained in the range of some sequence of real numbers, then there is a sequence whose range is precisely that set. 2.3.6 In Cantor’s proof presented in this section we took for granted material about infinite decimal expansions. This is entirely justified by the theory of sequences studied later. Explain what it is that we need to prove about infinite decimal expansions to be sure that this proof is valid. See Note 14 2.3.7 Define a relation on the family of subsets of R as follows. Say that A if there is a function B, where A and B are subsets of R, ∼ that is one-to-one and onto. (If A this is an equivalence relation, that is, show that ∼ B we would say that A and B are “cardinally equivalent.”) Show that f : A B → ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.4. Convergence 41 (a) A ∼ (b) If A (c) If A A for any set A. A. B then B C then A B and B ∼ ∼ ∼ ∼ C. ∼ 2.3.8 Let A and B be finite sets. Under what conditions are these sets cardinally equivalent (in the language of Exercise 2.3.7)? 2.3.9 Show that an infinite set of real numbers that is countable is cardinally equivalent (in the language of Exercise 2.3.7) to the set IN. Give an example of an infinite set that is not cardinally equivalent to IN. 2.3.10 We define a real number to be algebraic if it is a solution of some polynomial equation anxn + an−1xn−1 + where all the coefficients are integers. Thus √2 is algebraic because it is a solution of x2 number π is not algebraic because no such polynomial equation can ever be found (although this is hard to prove). Show that the set of algebraic numbers is countable. A real number that is not algebraic is said to be transcendental. For example, it is known that e and π are transcendental. What can you say about the existence of other transcendental numbers? See Note 15 + a1x + a0 = 0, 2 = 0. The · · · − 2.4 Convergence The sequence 1 4 is getting closer and closer to the number 0. We say that this sequence converges to 0 or that the limit of the sequence is the number 0. How should this idea be properly defined, , , , , The study of convergent sequences was undertaken and developed in the eighteenth century without any precise definition. The closest one might find to a definition in the early literature would have been something like A sequence L. sn} { converges to a number L if the terms of the sequence get closer and closer to ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 42 Sequences Chapter 2 Apart from being too vague to be used as anything but a rough guide for the intuition, this is misleading in other respects. What about the sequence .1, .01, .02, .001, .002, .0001, .0002, .00001, .00002, . . .? Surely this should converge to 0 but the terms do not get steadily “closer and closer” but back off a bit at each second step. Also, the sequence .1, .11, .111, .1111, .11111, .111111, . . . is getting “closer and closer” to .2, but we would not say the sequence converges to .2. A smaller number (1/9, which it is also getting closer and closer to) is the correct limit. We want not merely “closer and closer” but somehow a notion of “arbitrarily close.” The definition that captured the idea in the best way was given by Augustin Cauchy in the 1820s. He found a formulation that expressed the idea of “arbitrarily close” using inequalities. In this way the notion of limit is defined by a straightforward mathematical statement about inequalities. Definition 2.6: (Limit of a Sequence) Let converges to a number L and write be a sequence of real numbers. We say that sn} { or provided that for every number ε > 0 there is an integer N so that sn → L as n → ∞ whenever n N . ≥ sn − | L | < ε A sequence that converges is said to be convergent. A sequence that fails to converge is said to diverge. We are equally interested in both convergent and divergent sequences. sn} { lim n →∞ sn = L ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.4. Convergence 43 Note. In the definition the N depends on ε. If ε is particularly small, then N might have to be chosen large. In fact, then N is really a function of ε. Sometimes it is best to emphasize this and write N (ε) rather than N . Note, too, that if an N is found, then any larger N would also be able to be used. Thus the definition requires us to find some N but not necessarily the smallest N that would work. While the definition does not say this, the real force of the definition is that the N can be determined no matter how small a number ε is chosen. If ε is given as rather large there may be no trouble finding the N value. If you find an N that works for ε = .1 that same N would work for all larger values of ε. Example 2.7: Let us use the definition to prove that n2 2n2 + 1 1 2 lim n →∞ It is by no means clear from the definition how to obtain that the limit is the number L = 1 definition is not intended as a method of finding limits. It assigns a precise meaning to the statement about the limit but offers no way of computing that limit. Fortunately most of us remember some calculus devices that can be used to first obtain the limit before attempting a proof of its validity. 2 . Indeed the = . n2 2n2 + 1 lim n →∞ = lim n →∞ 1 2 + 1/n2 = 1 (2 + 1/n2) 1 = 2 + limn (1/n2) limn →∞ 1 2 . = →∞ Indeed this would be a proof that the limit is 1/2 provided that we could prove the validity of each of these steps. Later on we will prove this and so can avoid the ε, N arguments that we now use. Let any positive ε be given. We need to find a number N [or N (ε) if you prefer] so that every term in the sequence on and after the N th term is closer to 1/2 than ε, that is, so that n2 2n2 + 1 − 1 2 < ε ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 44 Sequences Chapter 2 for n = N , n = N + 1, n = N + 2, . . . . It is easiest to work backward and discover just how large n should be for this. A little work shows that this will happen if or 1 2(2n2 + 1) < ε 4n2 + 2 > 1 ε . The smallest n for which this statement is true could be our N . Thus we could use any integer N with N 2 > 1 4 1 ε − 2 . There is no obligation to find the smallest N that works and so, perhaps, the most convenient one here might be a bit larger, say take any integer N larger than 1 2√ε N > . The real lesson of the example, perhaps, is that we wish never to have to use the definition to check any limit computation. The definition offers a rigorous way to develop a theory of limits but an impractical method of computation of limits and a clumsy method of verification. Only rarely do we have to do a computation of this sort to verify a limit. Uniqueness of Sequence Limits Let us take the first step in developing a theory of limits. This is to ensure that our definition has defined limit unambiguously. Is it possible that the definition allows for a sequence L1 for a to converge to two different limits? If we have established that sn → different number L1? L is it possible that sn → ◭ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.4. Convergence 45 Theorem 2.8 (Uniqueness of Limits) Suppose that are both true. Then L1 = L2. sn = L1 and lim n →∞ sn = L2 lim n →∞ Proof. Let ε be any positive number. Then, by definition, we must be able to find a number N1 so that L1| sn − < ε | whenever n whenever n ≥ ≥ are true. N1. We must also be able to find a number N2 so that sn − N2. Take m to be the maximum of N1 and N2. Then both assertions L2| < ε | sm − | L1| < ε and sm − | L2| < ε This allows us to conclude that so that L1 − | L2| ≤ | L1 − sm| + sm − | L2| < 2ε L2| But ε can be any positive number whatsoever. This could only be true if L1 = L2, which is what we wished to show. L1 − | < 2ε. Exercises 2.4.1 Give a precise ε, N argument to prove that limn→∞ 1 n = 0. 2.4.2 Give a precise ε, N argument to prove the existence of lim n→∞ 2n + 3 3n + 4 . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 46 Sequences Chapter 2 2.4.3 Show that a sequence 2.4.4 Show that a sequence { sn} sn} { converges to a limit L if and only if the sequence converges to a limit L if and only if the sequence 2.4.5 Show that Definition 2.6 is equivalent to the following slight modification: converges to zero. { sn − L } converges to sn} {− L. − We write limn→∞ sn = L provided that for every positive integer m there is a real number N so that < 1/m whenever n N . L | 2.4.6 Compute the limit sn − | ≥ and verify it by the definition. See Note 16 2.4.7 Compute the limit See Note 17 lim n→∞ 1 + 2 + 3 + n2 · · · + n lim n→∞ 12 + 22 + 32 + n3 + n2 . · · · 2.4.8 Suppose that is a convergent sequence. Prove that limn→∞ 2sn exists. { 2.4.9 Prove that limn→∞ n does not exist. sn} 2.4.10 Prove that limn→∞( − 2.4.11 The sequence sn = ( 1)n does not exist. integer N so that an integer N so that < ε whenever n 0 ≥ < ε whenever n − sn − 1 | | sn − | | N ? ≥ 1)n does not converge. For what values of ε > 0 is it nonetheless true that there is an N ? For what values of ε > 0 is it nonetheless true that there is 2.4.12 Let be a sequence that assumes only integer values. Under what conditions can such a sequence 2.4.13 Let be a sequence and obtain a new s |
equence (sometimes called the “tail” of the sequence) by writing where M is some integer (perhaps large). Show that tn = sM +n for n = 1, 2, 3, . . . sn} { converges if and only if tn} { converges. sn} { converge? sn} { ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.5. Divergence 47 2.4.14 Show that the statement “ { inequality sn} converges to L” is false if and only if there is a positive number c so that the holds for infinitely many values of n. sn − | L | > c 2.4.15 If 2.4.16 If { sn} sn} { √L. is a sequence of positive numbers converging to 0, show that √sn} { also converges to zero. is a sequence of positive numbers converging to a positive number L, show that √sn} { converges to 2.5 Divergence A sequence that fails to converge is said to diverge. Some sequences diverge in a particularly interesting way, and it is worthwhile to have a language for this. The sequence sn = n2 diverges because the terms get larger and larger. We are tempted to write This conflicts with our definition of limit and so needs its own definition. We do not say that this sequence .” “converges to ” but rather that it “diverges to ∞ ∞ n2 or → ∞ lim n →∞ n2 = . ∞ Definition 2.9: (Divergence to to and write ∞ or ) Let sn} { ∞ be a sequence of real numbers. We say that sn} { diverges sn = lim n →∞ ∞ → ∞ provided that for every number M there is an integer N so that sn → ∞ as n whenever n N . ≥ sn ≥ M ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 48 Sequences Chapter 2 Note. The definition does not announce this, but the force of the definition is that the choice of N is possible no matter how large M is chosen. There may be no difficulty in finding an N if the M given is not big. Example 2.10: Let us prove that n2 + 1 n + 1 → ∞ using the definition. If M is any positive number we need to find some point in the sequence after which all terms exceed M . Thus we need to consider the inequality After some arithmetic we see that this is equivalent to n2 + 1 n + 1 ≥ M. Since . n n + 1 < 1 we see that, as long as n that M + 1 this will be true. Thus take any integer N M + 1 and it will be true ≥ ≥ n2 + 1 n + 1 ≥ N . (Any larger value of N would work too.) M ◭ for all n ≥ Exercises 2.5.1 Formulate the definition of a sequence diverging to 2.5.2 Show, using the definition, that limn→∞ n2 = ∞ n3+1 n2+1 = 2.5.3 Show, using the definition, that limn→∞ . ∞ . −∞ . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.6. Boundedness Properties of Limits 49 2.5.4 Prove that if sn → ∞ 2.5.5 Prove that if sn → ∞ 2.5.6 Prove that if xn → ∞ See Note 18 . sn → −∞ then − then (sn)2 → ∞ then the sequence sn = xn also. xn+1 is convergent. Is the converse true? 2.5.7 Suppose that a sequence the converse true? sn} { of positive numbers satisfies limn→∞ sn = 0. Show that limn→∞ 1/sn = . Is ∞ 2.5.8 Suppose that a sequence Show that sn → ∞ 2.5.9 The sequence sn = ( . integer N so that sn > M whenever n 1)n does not diverge to N ? − ≥ sn} { of positive numbers satisfies the condition sn+1 > αsn for all n where α > 1. . For what values of M is it nonetheless true that there is an ∞ 2.5.10 Show that the sequence np + α1np−1 + α2np−2 + + αp · · · diverges to ∞ , where here p is a positive integer and α1, α2, . . . , αp are real numbers (positive or negative). 2.6 Boundedness Properties of Limits A sequence is said to be bounded if its range is a bounded set. Thus a sequence a number M so that every term in the sequence satisfies sn} { is bounded if there is sn| ≤ | M. For such a sequence, every term belongs to the interval [ M, M ]. − It is fairly evident that a sequence that is not bounded could not converge. This is important enough to state and prove as a theorem. Theorem 2.11: Every convergent sequence is bounded. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 50 Proof. Suppose that sn → L. Then for every number ε > 0 there is an integer N so that Sequences Chapter 2 N . In particular we could take just one value of ε, say ε = 1, and find a number N so that whenever n whenever n ≥ ≥ N . From this we see that sn − | L | < 1 sn − | L < ε | N . This number for all n L | | no indication of the values for | ≥ Thus if we write = L + L sn − | sn − sn| | + 1 would be an upper bound for all the numbers s1 sN + < L − | | | , | . | s2| M = max 1| s2| | , s1| {| , . . . , sN L , 1| − | | | + 1 } we must have for every value of n. This is an upper bound, proving the theorem. sn| ≤ | M except that we have sn| | As a consequence of this theorem we can conclude that an unbounded sequence must diverge. Thus, even though it is a rather crude test, we can prove the divergence of a sequence if we are able somehow to show that it is unbounded. The next example illustrates this technique. Example 2.12: We shall show that the sequence 1 2 1 3 diverges. The easiest proof of this is to show that it is unbounded and hence, by Theorem 2.11, could not converge. sn = 1 + 1 n 1 4 · · · + + + + We watch only at the steps 1, 2, 4, 8, . . . and make a rough lower estimate of s1, s2, s4, s8, . . . in order to show that there can be no bound on the sequence. After a bit of arithmetic we see that s1 = 1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.6. Boundedness Properties of Limits s2 = 1 + 1 2 s4 = s8 = + and, in general, that s2n 1 + n/2 ≥ for all n = 0, 1, 2, . . . . Thus the sequence is not bounded and so must diverge. 51 ◭ Example 2.13: As a variant of the sequence of the preceding example consider the sequence 1 2p + 1 3p + 1 4p + tn = 1 + 1 np + · · · where p is any positive real number. The case p = 1 we have just found diverges. For p < 1 the sequence is larger than it is for p = 1 and so the case is even stronger for divergence. For p > 1 the sequence is smaller and we cannot see immediately whether it is bounded or unbounded; in fact, with some effort we can show that such a sequence is bounded. What can we conclude? Nothing yet. An ◭ unbounded sequence diverges. A bounded sequence may converge or diverge. Exercises 2.6.1 Which statements are true? (a) If (b) If (c) If { sn} sn} { sn} { is unbounded then it is true that either limn→∞ sn = is unbounded then limn→∞ | and ∞ are both bounded then so is sn| = . . sn + tn} { tn} { or else limn→∞ sn = . −∞ ∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Sequences Chapter 2 . 52 (d) If (e) If (f) If (g) If (h) If 2.6.2 If sn} { are both unbounded then so is { { and and and tn} { tn} tn} { sn} { sn} sn} { sn} { sn} { is bounded prove that is unbounded then is bounded then so is are both bounded then so is { are both unbounded then so is 1/sn} . is bounded. is convergent. { 1/sn} { sn/n } { sn + tn} . sntn} { sntn} { . 2.6.3 State the converse of Theorem 2.11. Is it true? 2.6.4 State the contrapositive of Theorem 2.11. Is it true? 2.6.5 Suppose that is a sequence of positive numbers converging to a positive limit. Show that there is a sn} { positive number c so that sn > c for all n. See Note 19 2.6.6 As a computer experiment compute the values of the sequence 1 3 for large values of n. Is there any indication in the numbers that you see that this sequence fails to converge or must be unbounded? sn = .7 Algebra of Limits Sequences can be combined by the usual arithmetic operations (addition, subtraction, multiplication, and division). Indeed most sequences we are likely to encounter can be seen to be composed of simpler sequences combined together in this way. In Example 2.7 we suggested that the computations n2 2n2 + 1 lim n →∞ = lim n →∞ 1 2 + 1/n2 = 1 (2 + 1/n2) limn →∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.7. Algebra of Limits 53 = 1 2 + limn 1/n2 = 1 2 →∞ could be justified. Note how this sequence has been obtained from simpler ones by ordinary processes of arithmetic. To justify such a method we need to investigate how the limit operation is influenced by algebraic operations. Suppose that Then we would expect and sn → S and tn → T. Csn → sn + tn → tn → sn − sntn → CS S + T T S − ST Each of these statements must be justified, however, solely on the basis of the definition of convergence, not on intuitive feelings that this should be the case. Thus we need to develop what could be called the “algebra of limits.” sn/tn → S/T. Theorem 2.14 (Multiples of Limits) Suppose that Then { is a convergent sequence and C a real number. Proof. Let S = limn what positive number ε is given, we can find an integer N so that, for all n →∞ →∞ sn. In order to prove that limn Csn = CS we need to prove that, no matter N , ≥ Csn = C lim n →∞ sn} lim n →∞ sn . Csn − | CS | < ε. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 54 Note that | by properties of absolute values. This gives us our clue. | | Csn − CS = C Sequences Chapter 2 sn − | | S | Suppose first that C = 0 and let ε > 0. Choose N so that if n ≥ N . Then if n ≥ N we must have Csn − | CS This is precisely the statement that sn − | S | < ε/ C | | = C sn − | | | S | | < C | | (ε/ C | | ) = ε. and the theorem is proved in the case C = 0. The case C = 0 is obvious. (Now we should probably delete our first paragraph since it does not contribute to the proof; it only serves to motivate us in finding the correct proof.) Csn = CS lim n →∞ Theorem 2.15 (Sums/Differences of Limits) Suppose that the sequences gent. Then sn} { and tn} { are conver- and lim n →∞ (sn + tn) = lim →∞ n sn + lim n →∞ tn lim n →∞ (sn − tn) = lim →∞ n sn − lim n →∞ tn. Proof. Let S = limn sn and T = limn →∞ tn. In order to prove that we need to prove that no matter what positive number ε is given we can find an integer N so that →∞ lim n →∞ (sn + tn) = S + T (sn + tn) | − (S + T ) < ε | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElement |
ary Real Analysis, 2nd Edition (2008)6 6 55 Section 2.7. Algebra of Limits if n ≥ N . Note that (sn + tn) (S + T ) | − sn − S + | tn − | T | | ≤ | by the triangle inequality. Thus we can make this expression smaller than ε by making each of the two expressions on the right smaller than ε/2. This provides the method. Suppose that ε > 0. Choose N1 so that N1 and also choose N2 so that sn − | S | < ε/2 N2. Then if n is greater than both N1 and N2 both of these inequalities will be true. Set tn − | | T < ε/2 if n if n ≥ ≥ and note that if n N we must have ≥ N = max N1, N2} { − This is precisely the statement that | (sn + tn) (S + T ) sn − | ≤ | S | + tn − | T | < ε/2 + ε/2 = ε. and the first statement of the theorem is proved. The second statement is similar and is left as an exercise. (Once again, for a more formal presentation, we would delete the first paragraph.) (sn + tn) = S + T lim n →∞ Theorem 2.16 (Products of Limits) Suppose that (sntn) = lim n →∞ sn} sn and tn} tn { lim n →∞ { lim n →∞ are convergent sequences. Then . Proof. Let S = limn prove that no matter what positive number ε is given we can find an integer N so that, for all n sntn − tn. In order to prove that limn sn and T = limn < ε. ST →∞ →∞ →∞ | | (sntn) = ST we need to N , ≥ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 56 Sequences Chapter 2 It takes some experimentation with different ways of writing this to find the most useful version. Here is an inequality that offers the best approach: sntn − | ST = sn(tn − | | T ) + snT ST | − T + tn − sn| | | ≤ | , T is constant, and T tn − | sn − | | T | S | | . cannot be too big. To control the we need to recall that convergent sequences are bounded (Theorem 2.11) and get a bound from We can control the size of size of there. With these preliminaries explained the rest of the proof should seem less mysterious. (Now this paragraph can be deleted for a more formal presentation.) sn − sn| sn| and S | | | | Suppose that ε > 0. Since converges it is bounded and hence, by Theorem 2.11, there is a positive number M so that sn| ≤ | M for all n. Choose N1 so that sn} { (1) if n ≥ N1. [We did not use ε/(2T ) since there is a possibility that T = 0.] Also, choose N2 so that if n ≥ N2. Set N = max { sn − | 2M | | T < tn − and note that if n N1, N2} ≥ tn − sn| | sntn − | T + M ST | ≤ | ε 2M | | ≤ T N we must have sn − | | < ε | | This is precisely the statement that and the theorem is proved. sntn = ST lim n →∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.7. Algebra of Limits 57 Theorem 2.17 (Quotients of Limits) Suppose that further that tn 6 = 0 for all n and that the limit sn} { and tn} { are convergent sequences. Suppose = 0. Then lim n →∞ tn 6 lim n →∞ sn tn = limn limn →∞ →∞ sn tn . Proof. Rather than prove the theorem at once as it stands let us prove just a special case of the theorem, namely that →∞ tn. We need to show that no matter what positive number ε is given we can find an lim n →∞ 1 tn = 1 limn . tn Let T = limn →∞ integer N so that 1 tn − 1 T < ε if n ≥ N . To work with this inequality requires us to consider tn − T = | T tn| | | 1 tn − 1 T | | tn| It is only the in the denominator that offers any trouble since if it is too small we cannot control the size of the fraction. This explains the first step in the proof that we now give, which otherwise might have seemed strange. | . Suppose that ε > 0. Choose N1 so that if n ≥ N1 and also choose N2 so that tn − | T | < /2 T | | tn − | T | < ε T | | 2/2 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 58 if n ≥ and so N2. From the first inequality we see that T | T tn| ≤ | tn| − < /2 T | | | − | Sequences Chapter 2 if n ≥ N1. Set N = max N1, N2} { and note that if n | /2 T tn| ≥ | | N we must have ≥ tn − T = | T tn| | | = ε. 1 T | | 1 tn − ε < 2/2 T | | 2/2 T | | (1/tn) = 1/T . This is precisely the statement that limn →∞ We now complete the proof of the theorem by applying the product theorem along with what we have just proved to obtain as required. Exercises lim n →∞ sn tn = sn lim n →∞ lim n →∞ 1 tn = limn limn →∞ →∞ sn tn 2.7.1 By imitating the proof given for the first part of Theorem 2.15 show that tn. lim n→∞ (sn − tn) = lim n→∞ sn − lim n→∞ 2.7.2 Show that limn→∞ (sn)2 = (limn→∞ sn)2 using the theorem on products and also directly from the definition of limit. 2.7.3 Explain which theorems are needed to justify the computation of the limit that introduced this section. lim n→∞ n2 2n2 + 1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.7. Algebra of Limits 59 2.7.4 Prove Theorem 2.16 but verifying and using the inequality T ) sntn − | | in place of the inequality (1). Which proof do you prefer? S)(tn − (sn − | ≤ | ST + S(tn − | T ) | + T (sn − | S) | 2.7.5 Which statements are true? (a) If (b) If (c) If (d) If (e) If (f) If (g) If { sn} sn} { sn} { sn} sn} { sn} (sn)2 { { { and and and and . { { tn} tn} { sn + tn} { sntn} { are both divergent then so is are both divergent then so is sn + tn} . sntn} { are both convergent then so is { tn} are both convergent then so is 1/sn} (sn)2 . } sn} { is convergent so too is { { { . . . tn} . is convergent so too is is convergent so too is } 2.7.6 Note that there are extra hypotheses in the quotient theorem (Theorem 2.17) that were not in the product theorem (Theorem 2.16). Explain why both of these hypotheses are needed. 2.7.7 A careless student gives the following as a proof of Theorem 2.16. Find the flaw: “Suppose that ε > 0. Choose N1 so that N1 and also choose N2 so that if n if n ≥ ≥ N2. If n N = max { ≥ sn − | S | < ε + 1 2 T | | tn − | T | < 2 ε sn| | + 1 ST | ≤ | ε sn| | + 1 2 tn − sn sn − | | S | < ε. 2 T | | + 1 then N1, N2} sntn − | sn| ≤ | Well, that works!” ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 60 Sequences Chapter 2 2.7.8 Why are Theorems 2.15 and 2.16 no help in dealing with the limits and What else can you do? lim n→∞ √n + 1 √n − √n lim n→∞ √n + 1 − √n ? 2.7.9 In calculus courses one learns that a function f : R → so that < ε for all Use this to prove that limn→∞(sn)2 = (limn→∞ sn)2. f (x) | f (y. Show that if f is continuous at y and sn → R is continuous at y if for every ε > 0 there is a δ > 0 f (y). y then f (sn) → 2.8 Order Properties of Limits In the preceding section we discussed the algebraic structure of limits. It is a natural mathematical question to ask how the algebraic operations are preserved under limits. As it happens, these natural mathematical questions usually are important in applications. We have seen that the algebraic properties of limits can be used to great advantage in computations of limits. There is another aspect of structure of the real number system that plays an equally important role as the algebraic structure and that is the order structure. Does the limit operation preserve that order structure the same way that it preserves the algebraic structure? For example, if for all n, can we conclude that sn ≤ tn lim n →∞ In this section we solve this problem and several others related to the order structure. These results, lim n →∞ sn ≤ tn? too, will prove to be most useful in handling limits. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.8. Order Properties of Limits 61 Theorem 2.18: Suppose that sn} { and tn} { for all n. Then are convergent sequences and that sn ≤ tn lim n →∞ sn ≤ lim n →∞ tn. Proof. Let S = limn sn and T = limn →∞ →∞ tn and suppose that ε > 0. Choose N1 so that sn − < ε/2 S | | N1 and also choose N2 so that if n if n ≥ ≥ N2. Set N = max 0 This shows that N1, N2} { tn − ≤ sn = T tn − | and note that if n S + (tn − − ≥ T ) + (S T < ε/2 | N we must have sn) < T − − S + ε/2 + ε/2. This statement is true for any positive number ε. It would be false if T positive or zero (i.e., T S as required). S is negative and hence T − S is − ≥ ε < T S. − − Note. There is a trap here that many students have fallen into. Since the condition sn ≤ tn implies would it not follow “similarly” that the condition sn < tn implies lim n →∞ sn ≤ lim n →∞ tn Be careful with this. It is false. See Exercise 2.8.1. lim n →∞ sn < lim →∞ n tn? ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 62 Sequences Chapter 2 Corollary 2.19: Suppose that sn} { for all n. Then is a convergent sequence and that sn ≤ ≤ α β α lim n →∞ sn ≤ ≤ β. Proof. Consider that the assumption here can be read as αn ≤ sequences. Now apply the theorem. sn ≤ βn where αn} { and βn} { are constant Note. Again, don’t forget the trap. The condition α < sn < β for all n implies that It would not imply that α lim n →∞ sn ≤ ≤ β. α < lim n →∞ sn < β. The Squeeze Theorem The next theorem is another useful variant on these themes. Here an unknown sequence is sandwiched between two convergent sequences, allowing us to conclude that that sequence converges. This theorem is often taught as “the squeeze theorem,” which seems a convenient label. Theorem 2.20 (Squeeze Theorem) Suppose that and tn} { are convergent sequences, that sn} { sn = lim n →∞ tn lim n →∞ and that for all n. Then xn} { is also convergent and sn ≤ xn ≤ tn lim n →∞ xn = lim →∞ n sn = lim →∞ n tn. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.8. Order Properties of Limits Proof. Let L be the limit of the two sequences. Choose N1 so that if n if n ≥ ≥ N1 and also choose N2 so that N2. Set N = max N1, N2} { . Note that sn − sn − | L | < ε tn − | L | < ε L xn − L L tn − ≤ ≤ for all n and so if n ≥ N . From this we see that or, to put it in a more familiar form, proving the statement of the theorem. ε < sn − − L ≤ xn − L ≤ tn − L < ε ε < xn − − L < ε xn − | L | < ε 63 Example 2.21: Let θ be some real number and consider the computation of While |
this might seem hopeless at first sight since the values of sin nθ are quite unpredictable, we recall that none of these values lies outside the interval [ 1, 1]. Hence sin nθ n . lim n →∞ 1 n ≤ − − sin nθ 1 n . n ≤ The two outer sequences converge to the same value 0 and so the inside sequence (the “squeezed” one) must converge to 0 as well. ◭ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 64 Sequences Chapter 2 Absolute Values A further theorem on the theme of order structure is often needed. The absolute value, we recall, is defined directly in terms of the order structure. Is absolute value preserved by the limit operation? Theorem 2.22 (Limits of Absolute Values) Suppose that sequence is also a convergent sequence and sn|} {| sn} { is a convergent sequence. Then the Proof. Let S = limn →∞ = lim →∞ | n sn| sn and suppose that ε > 0. Choose N so that sn − lim n →∞ S < ε sn | | . if n ≥ N . Observe that, because of the triangle inequality, this means that for all n ≥ N . By definition as required. sn| − | || S || ≤ | sn − S | < ε lim →∞ | n sn| = S | | Maxima and Minima Since maxima and minima can be expressed in terms of absolute values, there is a corollary that is sometimes useful. Corollary 2.23 (Max/Min of Limits) Suppose that sequences sn} { and are convergent sequences. Then the are also convergent and and tn} { sn, tn}} max sn, tn}} { { and min { { lim n →∞ max sn, tn} { = max lim n →∞ sn, lim n →∞ { tn} lim n →∞ min sn, tn} { = min lim n →∞ sn, lim n →∞ { . tn} ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.8. Order Properties of Limits 65 Proof. The first of these follows from the identity max sn, tn} { = sn + tn 2 + | sn − 2 tn| and the theorem on limits of sums and the theorem on limits of absolute values. In the same way the second assertion follows from min sn, tn} { = sn + tn 2 sn − 2 | − tn| . Exercises 2.8.1 Show that the condition sn < tn does not imply that lim n→∞ sn < lim n→∞ tn. (If the proof of Theorem 2.18 were modified in an attempt to prove this false statement, where would the modifications fail?) See Note 20 2.8.2 If is a sequence all of whose values lie inside an interval [a, b] prove that sn} { sn/n is convergent. { } 2.8.3 A careless student gives the following as a proof of the squeeze theorem. Find the flaw: “If limn→∞ sn = limn→∞ tn = L, then take limits in the inequality L. This can only be true if limn→∞ xn = L.” sn ≤ xn ≤ tn . What can you conclude? to get L 2.8.4 Suppose that sn ≤ 2.8.5 Suppose that limn→∞ limn→∞ xn ≤ ≤ 2.8.6 Suppose that sn} { sn tn for all n and that sn → ∞ n > 0 Show that sn → ∞ tn} { and . and that sn → ∞ . What can you conclude? lim n→∞ sn tn = α are sequences of positive numbers, that ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 66 Sequences Chapter 2 2.8.7 Suppose that sn} { and tn} { are sequences of positive numbers, that lim n→∞ sn tn = ∞ 2.8.8 Suppose that and that tn → ∞ sn} { . What can you conclude? and tn} { are sequences of positive numbers, that lim n→∞ sn tn = ∞ 2.8.9 Let and that sn} { is bounded. What can you conclude? sn} { be a sequence of positive numbers. Show that the condition lim n→∞ sn+1 sn < 1 implies that sn → See Note 21 0. 2.8.10 Let sn} { be a sequence of positive numbers. Show that the condition lim n→∞ sn+1 sn > 1 implies that sn → ∞ See Note 22 . 2.9 Monotone Convergence Criterion In many applications of sequence theory we find that the sequences that arise are going in one direction: The terms steadily get larger or steadily get smaller. The analysis of such sequences is much easier than for general sequences. Definition 2.24: (Increasing) We say that a sequence is increasing if sn} { s1 < s2 < s3 < · · · < sn < sn+1 < . . . . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.9. Monotone Convergence Criterion 67 Definition 2.25: (Decreasing) We say that a sequence s1 > s2 > s3 > · · · is decreasing if sn} > sn > sn+1 > . . . . { Often we encounter sequences that “increase” except perhaps occasionally successive values are equal rather than strictly larger. The following language is usually1 used in this case. Definition 2.26: (Nondecreasing) We say that a sequence sn ≤ Definition 2.27: (Nonincreasing) We say that a sequence sn ≥ s3 ≥ · · · ≥ s3 ≤ · · · ≤ sn} { sn+1 ≤ sn} { sn+1 ≥ s2 ≤ s2 ≥ s1 ≤ s1 ≥ is nondecreasing if . . . . is nonincreasing if . . . . Thus every increasing sequence is also nondecreasing but not conversely. A sequence that has any one of these four properties (increasing, decreasing, nondecreasing, or nonincreasing) is said to be monotonic. Monotonic sequences are often easier to deal with than sequences that can go both up and down. The convergence issue for a monotonic sequence is particularly straightforward. We can imagine that an increasing sequence could increase up to some limit, or we could imagine that it could increase indefinitely and diverge to + . It is impossible to imagine a third possibility. We express this as a theorem that will become our primary theoretical tool in investigating convergence of sequences. ∞ 1 In some texts you will find that a nondecreasing sequence is said to be increasing and an increasing sequence is said to be strictly increasing. The way in which we intend these terms should be clear and intuitive. If your monthly salary occasionally rises but sometimes stays the same you would not likely say that it is increasing. You might, however, say “at least it never decreases” (i.e., it is nondecreasing). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) is convergent if and only if Theorem 2.28 (Monotone Convergence Theorem) Suppose that sn} sn} 1. If is bounded. More specifically, is nondecreasing then either is bounded and converges to sup { { sn} { Sequences Chapter 2 sn} { is a monotonic sequence. Then sn} { or else sn} { is unbounded 2. If is nonincreasing then either sn} { is bounded and converges to inf sn} { or else sn} { is unbounded 68 { sn} and sn → ∞ sn} { . and sn → −∞ . If the sequence is unbounded then it diverges. This is true for any sequence, not merely monotonic Proof. sequences. sup us prove the first of these cases. Thus the proof is complete if we can show that for any bounded monotonic sequence sn} Let be assumed to be nondecreasing and bounded, and let in case the sequence is nondecreasing, or it is inf the limit is in case the sequence is nonincreasing. Let sn} sn} { { { sn} { sn} L for all n and if β < L there must be some term sm say, with sm > β. Let ε > 0. We know L = sup { . Then sn ≤ that there is an m so that for all n sm > L m. But we already know that every term sn ≤ sn ≥ ≥ ε − L. Putting these together we have that or for all n m. By definition then sn → ≥ L ε < sn ≤ − L < L + ε sn − | L as required. L | < ε How would we normally apply this theorem? Suppose a sequence increasing (or maybe just nondecreasing). Then to establish that { { sn} were given that we recognize as sn} converges we need only show that ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.9. Monotone Convergence Criterion 69 the sequence is bounded above, that is, we need to find just one number M with for all n. Any crude upper estimate would verify convergence. sn ≤ M Example 2.29: Let us show that the sequence sn = 1/√n converges. This sequence is evidently decreasing. Can we find a lower bound? Yes, all of the terms are positive so that 0 is a lower bound. Consequently, the sequence must converge. If we wish to show that 1 √n lim n →∞ = 0 we need to do more. But to conclude convergence we needed only to make a crude estimate on how low ◭ the terms might go. Example 2.30: Let us examine the sequence 1 3 This sequence is evidently increasing. Can we find an upper bound? If we can then the series does converge. If we cannot then the series diverges. We have already (earlier) checked this sequence. It is unbounded and ◭ so limn sn = 1 + sn = 1 n 1 4 1 2 · · · + + + + . . →∞ ∞ Example 2.31: Let us examine the sequence √2, 2 + √2 , s2 + 2 + r q 2 + √2, . . . . Handling such a sequence directly by the limit definition seems quite impossible. This sequence can be defined recursively by x1 = √2 xn = 2 + xn 1. − p ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 70 Sequences Chapter 2 The computation of a few terms suggests that the sequence is increasing and so should be accessible by the methods of this section. We prove this by induction. That x1 < x2 is just an easy computation (do it). Let us suppose that 1 < xn for some n and show that it must follow that xn < xn+1. But − xn where the middle step is the induction hypothesis (i.e., that xn sequence is increasing. p − 1 < xn). It follows by induction that the Now we show inductively that the sequence is bounded above. Any crude upper bound will suffice. It is xn = 2 + xn − 1 < √2 + xn = xn+1 clear that x1 < 10. If xn 1 < 10 then − xn = 2 + xn − 1 < √2 + 10 < 10 and so it follows, again by induction, that all terms of the sequence are smaller than 10. We conclude from the monotone convergence theorem that this sequence is convergent. p But to what? (Certainly it does not converges to 10 since that estimate was extremely crude.) That is not so easy to sort out, it seems. But perhaps it is, since we know that the sequence converges to something, say L. In the equation (xn)2 = 2 + xn 1, − obtained by squaring the recursion formula given to us, we can take limits as n does xn L and (xn)2 L2. Hence → ∞ . Since xn → L so too 1 → − → L2 = 2 + L. The only possibilities for L in this quadratic equation are L = and we know that it is either 1 or 2. We can clearly rule out 2. were positive. Hence xn → Exercises − 1 and L = 2. We know the limit L exists 1 as all of the numbers in our sequence ◭ − − 2.9 |
.1 Define a sequence sn} { recursively by setting s1 = α and sn = (sn−1)2 + β 2sn−1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.9. Monotone Convergence Criterion 71 where α, β > 0. (a) Show that for n = 1, 2, 3, . . . √β)2 (sn − 2sn = sn+1 − β. (b) Show that sn > √β for all n = 2, 3, 4, . . . unless α = √β. What happens if α = √β? (c) Show that s2 > s3 > s4 > . . . sn > . . . except in the case α = √β. (d) Does this sequence converge? To what? (e) What is the relation of this sequence to the one introduced in Section 2.1 as Newton’s method? 2.9.2 Define a sequence tn} { recursively by setting t1 = 1 and tn = tn−1 + 1. p Does this sequence converge? To what? 2.9.3 Consider the sequence s1 = 1 and sn = 2 s2 n−1 Our conclusion is that limn→∞ sn = 3√2. Do you have any criticisms of this argument? . We argue that if sn → L then L = 2 L2 and so L3 = 2 or L = 3√2. 2.9.4 Does the sequence converge? 2.9.5 Does the sequence converge2n 1) − (2n2n2n − · 1) 1 · n2 2.9.6 Several nineteenth-century mathematicians used, without proof, a principle in their proofs that has come to be known as the nested interval property: ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 72 Sequences Chapter 2 Given a sequence of closed intervals [a1, b1] [a2, b2] [a3, b3] . . . ⊃ ⊃ ⊃ arranged so that each interval is a subinterval of the one preceding it and so that the lengths of the intervals shrink to zero, then there is exactly one point that belongs to every interval of the sequence. Prove this statement. Would it be true for a descending sequence of open intervals (a1, b1) (a3, b3) (a2, b2) . . .? ⊃ ⊃ ⊃ 2.10 Examples of Limits The theory of sequence limits has now been developed far enough that we may investigate some interesting limits. Each of the limits in this section has some cultural interest. Most students would be expected to know and recognize these limits as they arise quite routinely. For us they are also an opportunity to show off our methods. Mostly we need to establish inequalities and use some of our theory. We do not need to use an ε, N argument since we now have more subtle and powerful tools at hand. Example 2.32: (Geometric Progressions) Let r be a real number. What is the limiting behavior of the sequence forming a geometric progression? If r > 1 then it is not hard to show that 1, r, r2, r3, r4, . . . , rn, . . . If r 1 the sequence certainly diverges. If r = 1 this is just a constant sequence. ≤ − The interesting case is rn . → ∞ To prove this we shall use an easy inequality. Let x > 0 and n an integer. Then, using the binomial theorem (or induction if you prefer), we can show that lim n →∞ rn = 0 1 < r < 1. if − (1 + x)n > nx. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.10. Examples of Limits Case (i): Let 0 < r < 1. Then (where x = 1/r 1 > 0) and so − r = 1 1 + x 0 < rn = 1 (1 + x)n < 1 nx → 0 73 as n . By the squeeze theorem we see that rn 0 as required. → ∞ Case (ii): If 1 < r < 0 then r = − − → t for 0 < t < 1. Thus tn. tn rn By case (i) we know that tn → ≤ 0. By the squeeze theorem we see that rn ≤ − 0 again as required. ◭ → Example 2.33: (Roots) An interesting and often useful limit is To show this we once again derive an inequality from the binomial theorem. If n n√n = 1. lim n →∞ 2 and x > 0 then ≥ For n ≥ 2 write (where xn = n√n 1 > 0) and so − (1 + x)n > n(n 1)x2/2. − n√n = 1 + xn n = (1 + xn)n > n(n 1)x2 n/2 − or as n n < n . By the squeeze theorem we see that xn → 0 < x2 As a special case of this example note that → ∞ 2 0 1 → − 0 and it follows that n√n 1 as required. → n√C 1 → ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 74 as n → ∞ for any positive constant C. This is true because if C > 1 then 1 < n√C < n√n Sequences Chapter 2 for large enough n. By the squeeze theorem this shows that n√C n√C = 1 1/C → 1 n by the first case since 1/C > 1. p 1. If, however, 0 < C < 1 then → Example 2.34: (Sums of Geometric Progressions) For all values of x in the interval ( lim n →∞ 1 + x + x2 + x3 + + xn = · · · 1 1 − . x While at first a surprising result, this is quite evident once we check the identity · · · which just requires a straightforward multiplication. Thus − (1 x) 1 + x + x2 + x3 + + xn = 1 xn+1, − = lim n →∞ where we have used the result we proved previously, namely that lim n →∞ 1 + x + x2 + x3 + + xn · · · 1 xn+ One special case of this is useful to remember. Set x = 1/2. Then xn+1 → 0 if x < 1. | | 1 + lim n →∞ 1 2 + 1 22 + 1 23 + + · · · 1 2n = 2. ◭ ◭ 1, 1) the limit − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.10. Examples of Limits 75 Example 2.35: (Decimal Expansions) What meaning is assigned to the infinite decimal expansion x = 0.d1d2d3d4 . . . dn . . . where the choices of integers 0 di ≤ converted into fractions and so the infinite process can be avoided. But if the pattern does not repeat, a different interpretation must be made. 9 can be made in any way? Repeating decimals can always be ≤ The most obvious interpretation of this number x is to declare that it is the limit of the sequence But how do we know that the limit exists? Our theory provides an immediate answer. Since this sequence is nondecreasing and every term is smaller than 1, by the monotone convergence theorem the sequence converges. This is true no matter what the choices of the decimal digits are. ◭ 0.d1d2d3d4 . . . dn. lim n →∞ Example 2.36: (Expansion of ex) Let x > 0 and consider the two closely related sequences sn = 1 + x + x2 2! + x3 3! + + · · · xn n! and tn = 1 + n . x n The relation between the two sequences becomes more apparent once the binomial theorem is used to expand the latter. In more advanced mathematics it is shown that both sequences converge to ex. Let us be content to prove that The sequence is clearly increasing since each new term is the preceding term with a positive number added to it. To show convergence then we need only show that the sequence is bounded. This takes some arithmetic, but not too much. sn} { lim n →∞ sn = lim →∞ n tn. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 76 Sequences Chapter 2 Choose an integer N larger than 2x. Note then that that and that Thus xN +1 (N + 1)! < xN +2 (N + 2)! < 1 2 1 4 xN N ! xN N ! xN +3 (N + 3)! < 1 8 xN N ! . 1 + x + sn ≤ x2 2! + + xN − 1 (N 1)! + xN N ! xN 1 − − + (N 1)! − · · · x2 2 xN N ! 1 + x + + · · · ≤ Here we have used the limit for the sum of a geometric progression from Example 2.34 to make an upper estimate on how large this sum can get. Note that the N is fixed and so the number on the right-hand side of this inequality is just a number, and it is larger than every number in the sequence It follows now from the monotone convergence theorem that the binomial theorem to obtain sn} converges. To handle { { . , first apply tn} { sn} 2/n) tn = 1 + x + 1 − 1/n 2! (1 1/n)(1 x2 + − − 3! is increasing and that it is smaller than the convergent sequence · · · ≤ x3 + sn. converges. Moreover, . It follows, sn} { From this we see that tn} again from the monotone convergence theorem, that { tn ≤ { lim n →∞ tn} lim n →∞ sn. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.10. Examples of Limits 77 If we can obtain the opposite inequality we will have proved our assertion. Let m be a fixed number and let n > m. Then, from the preceding expansion, we note that tn > 1 + x + (/n 2! 1/n)(1 x2 + (1 − 2/n) + − · · · m! 1/n)(1 3! + (1 2/n) x3 − [m − − 1]/n) xm. We can hold m fixed and allow n → ∞ in this inequality and obtain that lim n →∞ tn ≥ sm lim n →∞ tn ≥ lim n →∞ sn ◭ for each m. From this it now follows that and we have completed our task. Exercises 2.10.1 Since we know that this suggests the formula Do you have any criticisms? See Note 23 1 + x + x2 + x3 + + xn · · · → + 16 + = 1 · · · 1 − = 1. − 2 2.10.2 Let α and β be positive numbers. Discuss the convergence behavior of the sequence αβn βαn . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 78 2.10.3 Define Show that 2 < e < 3. 2.10.4 Show that 2.10.5 Check the simple identity and use it to show that 2.11 Subsequences The sequence 1, Sequences Chapter 2 e = lim n→∞ 1 + 1 n n . lim n→∞ 1 + n 1 2n = √e lim n→∞ 1 + n 2 n = e2. appears to contain within itself the two sequences 1, 2, 2, 3, − − 3, 4, − − 4, 5, − 5, . . . and 1, 2, 3, 4, 5, . . . 1, 2, 3, 4, 5, . . . . − − − − − In order to have a language to express this we introduce the term subsequence. We would say that the latter two sequences are subsequences of the first sequence. Often a sequence is best studied by looking at some of its subsequences. But what is a proper definition of this term? We need a formal mathematical way of expressing the vague idea that a subsequence is obtained by crossing out some of the terms of the original sequence. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.11. Subsequences 79 Definition 2.37: (Subsequences) Let s1, s2, s3, s4, . . . be any sequence. Then by a subsequence of this sequence we mean any sequence where sn1, sn2, sn3, sn4, . . . n1 < n2 < n3 < . . . is an increasing sequence of natural numbers. Example 2.38: We can consider to be a subsequence of sequence 1, 2, 3, 4, 5, . . . − because it contains just the first, third, fifth, etc. terms of the original sequence. Here n1 = 1, n2 = 3, n3 = 5, 1, 2, 2, 3, 3, 4, 4, 5, 5, . . . In many applications of sequences it is the subsequences that need to be studied. For example, what can we say about the existence of monotonic subsequences, or bounded subsequences, or divergent subsequences, or convergent subsequences? The answers to these questions have important uses. Existence of Monotonic Subsequenc |
es Our first question is easy to answer for any specific sequence, but harder to settle in general. Given a sequence can we always select a subsequence that is monotonic, either monotonic nondecreasing or monotonic nonincreasing? Theorem 2.39: Every sequence contains a monotonic subsequence. Proof. We construct first a nonincreasing subsequence if possible. We call the mth element xm of the xn sequence a turn-back point if all later elements are less than or equal to it, in symbols if xm ≥ xn} { ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 80 Sequences Chapter 2 for all n > m. If there is an infinite subsequence of turn-back points xm1, xm2, xm3, xm4, . . . then we have found our nonincreasing subsequence since xm1 ≥ xm3 ≥ This would not be possible if there are only finitely many turn-back points. Let us suppose that xM is the last turn-back point so that any element xn for n > M is not a turn-back point. Since it is not there must be an element further on in the sequence greater than it, in symbols xm > xn for some m > n. Thus we can choose xm1 > xM +1 with m1 > M + 1, then xm2 > xm1 with m2 > m1, and then xm3 > xm2 with m3 > m2, and so on to obtain an increasing subsequence xm4 ≥ xm2 ≥ . . . . xM +1 < xm1 < xm2 < xm3 < xm4 < . . . as required. Existence of Convergent Subsequences Having answered this question about the existence of monotonic subsequences, we can also now answer the question about the existence of convergent subsequences. This might, at first sight, seem just a curiosity, but it will give us later one of our most important tools in analysis. The theorem is traditionally attributed to two major nineteenth-century mathematicians, Karl Theodor Wilhelm Weierstrass (1815-1897) and Bernhard Bolzano (1781–1848). These two mathematicians, the first German and the second Czech, rank with Cauchy among the founders of our subject. Theorem 2.40 (Bolzano-Weierstrass) Every bounded sequence contains a convergent subsequence. Proof. By Theorem 2.39 every sequence contains a monotonic subsequence. Here that subsequence would be both monotonic and bounded, and hence convergent. Other (less important) questions of this type appear in the exercises. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.11. Subsequences Exercises 81 2.11.1 Show that, according to our definition, every sequence is a subsequence of itself. How would the definition have to be reworded to avoid this if, for some reason, this possibility were to have been avoided? xn} . { snk + tmk } is itself a subsequence of tn} xn} is a subsequence of { 2.11.2 Show that every subsequence of a subsequence of a sequence is a subsequence of then is it true that 2.11.3 If tmk } sn} and is a { { { { { snk } subsequence of snk } { sn + tn} ? { 2.11.4 If sn} 2.11.5 Describe all sequences that have only finitely many different subsequences. is a subsequence of a subsequence of is { { } } { ? (snk )2 (sn)2 2.11.6 Establish which of the following statements are true. (a) A sequence is convergent if and only if all of its subsequences are convergent. (b) A sequence is bounded if and only if all of its subsequences are bounded. (c) A sequence is monotonic if and only if all of its subsequences are monotonic. (d) A sequence is divergent if and only if all of its subsequences are divergent. 2.11.7 Establish which of the following statements are true for an arbitrary sequence (a) If all monotone subsequences of a sequence (b) If all monotone subsequences of a sequence sn} sn} { sn} (c) If all convergent subsequences of a sequence { sn} (d) If all convergent subsequences of a sequence { { converges to 0. . sn} is bounded. are convergent, then are convergent, then is convergent. converge to 0, then converges to 0. { { sn} sn} { sn} sn} { { converge to 0 and is bounded, then sn} { 2.11.8 Where possible find subsequences that are monotonic and subsequences that are convergent for the following sequences (a) (b) (c) 1)nn ( { − sin (nπ/8) } } n sin (nπ/8) { { } ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 82 (d) (e) (f) n+1 n sin (nπ/8) 1)n 1 + ( { rn} { − consists of all rational numbers in the interval (0, 1) arranged in some order. } Sequences Chapter 2 2.11.9 Describe all subsequences of the sequence Describe all convergent subsequences. Describe all monotonic subsequences. 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, . . . . 2.11.10 If snk } { is a subsequence of sn} { show that nk ≥ k for all k = 1, 2, 3, . . . . 2.11.11 Give an example of a sequence that contains subsequences converging to every natural number (and no other numbers). 2.11.12 Give an example of a sequence that contains subsequences converging to every number in [0, 1] (and no other numbers). 2.11.13 Show that there cannot exist a sequence that contains subsequences converging to every number in (0, 1) and no other numbers. See Note 24 2.11.14 Show that if 2.11.15 If a sequence { sn} xn} { has no convergent subsequences, then sn| → ∞ | as n . → ∞ has the property that lim n→∞ converges to L. x2n = lim n→∞ x2n+1 = L show that the sequence xn} { 2.11.16 If a sequence has the property that xn} { lim n→∞ . ∞ 2.11.17 Let α and β be positive real numbers and define a sequence by setting s1 = α, s2 = β and sn+2 = 1 show that the sequence x2n = lim n→∞ diverges to x2n+1 = xn} ∞ { 2 (sn+sn+1) are monotonic and convergent. Does for all n = 1, 2, 3, . . . . Show that the subsequences the sequence converge? To what? sn} { s2n} { and s2n−1} { ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 2.11.20 Suppose a sequence xn} { L. Show that xn} { 2.11.21 Let Section 2.11. Subsequences 83 2.11.18 Without appealing to any of the theory of this section prove that every unbounded sequence has a strictly monotonic subsequence (i.e., either increasing or decreasing). 2.11.19 Show that if a sequence xn} precisely the same behavior. { converges to a finite limit or diverges to then every subsequence has ±∞ has the property that every subsequence has a further subsequence convergent to xn} { converges to L. be a bounded sequence and let x = sup { Prove that there is a subsequence convergent to x. xn : n IN } ∈ . Suppose that, moreover, xn < x for all n. 2.11.22 Let xn} { be a bounded sequence, let y = inf } Suppose that, moreover, y < xn < x for all n. Prove that there is a pair of convergent subsequences and so that ∈ ∈ { { } xnk } { xn : n IN and x = sup xn : n IN . xmk } { lim k→∞ | xnk − xmk | = x y. − 2.11.23 Does every divergent sequence contain a divergent monotonic subsequence? 2.11.24 Does every divergent sequence contain a divergent bounded subsequence? 2.11.25 Construct a proof of the Bolzano-Weierstrass theorem for bounded sequences using the nested interval property and not appealing to the existence of monotonic subsequences. 2.11.26 Construct a direct proof of the assertion that every convergent sequence has a convergent, monotonic subsequence (i.e., without appealing to Theorem 2.39). 2.11.27 Let be a bounded sequence that we do not know converges. Suppose that it has the property that every one of its convergent subsequences converges to the same number L. What can you conclude? be a bounded sequence that diverges. Show that there is a pair of convergent subsequences xnk } { xmk | be a sequence. A number z with the property that for all ε > 0 there are infinitely many terms ε, z + ε) is said to be a cluster point of the sequence. Show that z is a xnk − lim k→∞ | > 0. of the sequence in the interval (z cluster point of a sequence if and only if there is a subsequence − xnk } { converging to z. xn} { xn} { xmk } { xn} { 2.11.28 Let and 2.11.29 Let so that ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 84 Sequences Chapter 2 2.12 Cauchy Convergence Criterion What property of a sequence characterizes convergence? As a “characterization” we would like some necessary and sufficient condition for a sequence to converge. We could simply write the definition and consider that that is a characterization. Thus the following technical statement would, indeed, be a sn} characterization of the convergence of a sequence L so that is convergent if and only if N with the property that A sequence ε > 0 { . sn} { whenever n N . ≥ | ∃ sn − < ε L | ∀ ∃ In mathematics when we ask for a characterization of a property we can expect to find many answers, some more useful than others. The limitation of this particular characterization is that it requires us to find the number L which is the limit of the sequence in advance. Compare this with a characterization of convergence of a monotonic sequence . sn} { A monotonic sequence sn} { is convergent if and only if it is bounded. This is a wonderful and most useful characterization. But it applies only to monotonic sequences. A correct and useful characterization, applicable to all sequences, was found by Cauchy. This is the content of the next theorem. Note that it has the advantage that it describes a convergent sequence with no reference whatsoever to the actual value of the limit. Loosely it asserts that a sequence converges if and only if the terms of the sequence are eventually arbitrarily close together. Theorem 2.41 (Cauchy Criterion) A sequence exists an integer N with the property that whenever n N and m N . ≥ ≥ sn − | sn} { sm| < ε is convergent if and only if for each ε > 0 there ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.12. Cauchy Convergence Criterion 85 Proof. This property of the theorem is so important that it deserves some terminology. A sequence is said to be a Cauchy sequence if it satisfies this property. Thus the theorem states that a sequence is convergent if and only if it is a Cauchy sequence. The terminology is most significant in more advanced situations where being a Cauchy sequence i |
s not necessarily equivalent with being convergent. Our proof is a bit lengthy and will require an application of the Bolzano-Weierstrass theorem. The proof in one direction, however, is easy. Suppose that is convergent to a number L. Let ε > 0. Then there must be an integer N so that ε 2 N . Thus if both m and n are larger than N , sk − | < L | whenever k ≥ sn} { ε 2 ε 2 { L L + sn} sn − which shows that sm| ≤ | sn − | is a Cauchy sequence. Now let us prove the opposite (and more difficult) direction. For the first step we show that every Cauchy sequence is bounded. Since the proof of this can be obtained by copying and modifying the proof of Theorem 2.11, we have left this as an exercise. (It is not really interesting that Cauchy sequences are bounded since after the proof is completed we know that all Cauchy sequences are convergent and so must, indeed, be bounded.) sm| = ε − + < | | For the second step we apply the Bolzano-Weierstrass theorem to the bounded sequence to obtain a convergent subsequence . snk } { we can show that sn → The final step is a feature of Cauchy sequences. Once we know that snk → L also. Let ε > 0 and choose N so that L and that { is Cauchy, sn} { sn} for all m, n ≥ N . Choose K so that sn − | sm| < ε/2 snk − | L | < ε/2 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 86 Sequences Chapter 2 for all k k K. Suppose that n K. For this value sm = snk ≥ ≥ N . Set m equal to any value of nk that is larger than N and so that ≥ By definition, sn} { sn − converges to L and so the proof is complete. snk − | sn − | snk | | ≤ | + L L | < ε/2 + ε/2 = ε. Example 2.42: The Cauchy criterion is most useful in theoretical developments rather than applied to concrete examples. Even so, occasionally it is the fastest route to a proof of convergence. For example, consider the sequence xn} { defined by setting x1 = 1, x2 = 2 and then, recursively, 1 + xn 2 xn = xn − − 2 . Each term after the second is the average of the preceding two terms. The distance between x1 and x2 is 1, that between x2 and x3 is 1/2, between x3 and x4 is 1/4, and so on. We see then that after the N stage all the distances are smaller than 2− N +1, that is, that for all n N and m N xm| ≤ This is exactly the Cauchy criterion and so this sequence converges. Note that the Cauchy criterion offers no information on what the sequence is converging to. You must come up with another method to find out. xn − 1 . − | ◭ ≥ ≥ 1 2N Exercises 2.12.1 Show directly that the sequence sn = 1/n is a Cauchy sequence. 2.12.2 Show directly that any multiple of a Cauchy sequence is again a Cauchy sequence. 2.12.3 Show directly that the sum of two Cauchy sequences is again a Cauchy sequence. 2.12.4 Show directly that any Cauchy sequence is bounded. 2.12.5 The following criterion is weaker than the Cauchy criterion. Show that it is not equivalent: ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.13. Upper and Lower Limits 87 For all ε > 0 there exists an integer N with the property that whenever n N . ≥ See Note 25 sn+1 − | sn| < ε 2.12.6 A careless student believes that the following statement is the Cauchy criterion. For all ε > 0 and all positive integers p there exists an integer N with the property that sn+p − sn| < ε | whenever n N . ≥ 2.12.7 Show directly that if Is this statement weaker, stronger, or equivalent to the Cauchy criterion? sn|} {| is a Cauchy sequence then so too is sn} converges whenever converges. { sn} { . From this conclude that sn|} {| 2.12.8 Show that every subsequence of a Cauchy sequence is Cauchy. (Do not use the fact that every Cauchy sequence is convergent.) 2.12.9 Show that every bounded monotonic sequence is Cauchy. (Do not use the monotone convergence theorem.) 2.12.10 Show that the sequence in Example 2.42 converges to 5/3. See Note 26 2.13 Upper and Lower Limits " Advanced section. May be omitted. If limn xn = L then, according to our definition, numbers α and β on either side of L, that is, α < L < β, have the property that →∞ for all sufficiently large n. In many applications only half of this information is used. α < xn and xn < β ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 88 Sequences Chapter 2 Example 2.43: Here is an example showing how half a limit is as good as a whole limit. Let sequence of positive numbers with the property that xn} { be a n√xn = L < 1. lim n →∞ 0. To see this pick numbers α and β so that Then we can prove that xn → There must be an integer N so that α < L < β < 1. α < n√xn < β < 1 for all n ≥ N . Forget half of this and focus on Then we have for all n N and it is clear now why xn → ≥ 0. n√xn < β < 1. xn < βn ◭ This example suggests that the definition of limit might be weakened to handle situations where less is needed. This way we have a tool to discuss the limiting behavior of sequences that may not necessarily converge. Even if the sequence does converge this often offers a tool that can be used without first finding a proof of convergence. We break the definition of sequence limit into two half-limits as follows. Definition 2.44: (Lim Sup) A limit superior of a sequence , denoted as xn} { is defined to be the infimum of all numbers β with the following property: There is an integer N so that xn < β for all n N . ≥ lim sup xn, n →∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.13. Upper and Lower Limits 89 Definition 2.45: (Lim Inf ) A limit inferior of a sequence , denoted as xn} { lim inf n →∞ xn, is defined to be the supremum of all numbers α with the following property: There is an integer N so that α < xn for all n N . ≥ Note. In interpreting this definition note that, by our usual rules on infs and sups, the values and are allowed. If there are no numbers β with the property of the definition, then the sequence is simply −∞ ∞ unbounded above. The infimum of the empty set is taken as lim supn xn = →∞ ∞ ⇔ the sequence { and so has no upper bound. ∞ xn} On the other hand, if every number β has the property of the definition this means exactly that our sequence must be diverging to . The infimum of the set of all real numbers is taken as and so −∞ lim supn xn = →∞ −∞ ⇔ the sequence xn} → −∞ . { −∞ The same holds in the other direction. A sequence that is unbounded below can be described by saying lim inf n xn = →∞ −∞ . A sequence that diverges to can be described by saying lim inf n ∞ xn = . ∞ →∞ We refer to these concepts as “upper limits” and “lower limits” or “extreme limits.” They extend our theory describing the limiting behavior of sequences to allow precise descriptions of divergent sequences. Obviously, we should establish very quickly that the upper limit is indeed greater than or equal to the lower limit since our language suggests this. Theorem 2.46: Let xn} { be a sequence of real numbers. Then lim inf n →∞ xn ≤ lim sup xn. n →∞ Proof. ∞ number β larger than lim supn If lim supn xn = →∞ or if lim inf n −∞ xn and any number α smaller than lim inf n we have nothing to prove. If not then take any xn. By definition then xn = →∞ →∞ →∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 90 Sequences Chapter 2 there is an integer N so that xn < β for all n be true that α < β. But β is any number larger than lim supn xn. lim sup ≥ α →∞ N and an integer M so that α < xn for all n xn. Hence M . It must ≥ Similarly, α is any number smaller than lim inf n as required. How shall we use the limit superior of a sequence →∞ ≤ →∞ n →∞ xn. Hence lim inf n →∞ xn ≤ n lim sup xn ? If xn} { xn = L lim sup n →∞ then every number β > L has the property that xn < β for all n large enough. This is because L is the b infimum of such numbers β. On the other hand, any number b < L cannot have this property so xn ≥ for infinitely many indices n. Thus numbers slightly larger than L must be upper bounds for the sequence eventually. Numbers slightly less than L are not upper bounds eventually. To express this a little more precisely, the number L is the limit superior of a sequence exactly when the following holds: xn} { For every ε > 0 there is an integer N so that xn < L + ε for all n infinitely many n N . ≥ N and xn > L ε for − ≥ The next theorem gives another characterization which is sometimes easier to apply. This version also better explains why we describe this notion as a “lim sup” and “lim inf.” Theorem 2.47: Let xn} { and be a sequence of real numbers. Then lim sup n →∞ xn = lim →∞ n sup { xn, xn+1, xn+2, xn+3, . . . } lim inf n →∞ xn = lim →∞ n inf { xn, xn+1, xn+2, xn+3, . . . . } ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.13. Upper and Lower Limits 91 Proof. Let us prove just the statement for lim sups as the lim inf statement can be proved similarly. Write Then xn ≤ But yn} { yn = sup { yn for all n and so, using the inequality promised in Exercise 2.13.5, xn, xn+1, xn+2, xn+3, . . . } . n is a nonincreasing sequence and so →∞ lim sup xn ≤ lim sup yn. n →∞ lim sup n →∞ yn = lim →∞ n yn. Let us now show the reverse inequality. If lim supn xn = ∞ →∞ then the sequence is unbounded above. lim sup n →∞ xn ≤ lim n →∞ sup { xn, xn+1, xn+2, xn+3, . . . . } sup { xn, xn+1, xn+2, xn+3, . . . = } ∞ lim sup n →∞ xn = lim →∞ n sup { xn, xn+1, xn+2, xn+3, . . . } From this it follows that Thus for all n and so, in this case, must certainly be true. If lim sup xn < n ∞ →∞ xn. By definition then there is an integer N so that xn < β then take any number β larger than lim supn for all n N . It follows that →∞ ≥ But β is any number larger than lim supn xn. Hence lim n →∞ sup { xn, xn+1, xn+2, xn+3, . . . β. } ≤ →∞ lim n →∞ sup { xn, xn+1, xn+2, xn+3, . . . lim sup xn. } ≤ n →∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 92 Sequences Chapter 2 We have proved both inequalities, the |
equality follows, and the theorem is proved. The connection between limits and extreme limits is close. If a limit exists then the upper and lower limits must be the same. Theorem 2.48: Let lim supn xn} { xn = lim inf n be a sequence of real numbers. Then xn and these are finite. In this case →∞ →∞ xn} { is convergent if and only if lim sup n →∞ xn = lim inf n →∞ xn = lim →∞ n xn. Proof. Let ε > 0. If lim supn If it is also true that lim inf n Putting these together we have →∞ →∞ xn = L then there is an integer N1 so that xn < L + ε for all n ε for all n xn = L then there is an integer N2 so that xn > L − for all By definition then limn Conversely, if limn →∞ →∞ L − ε < xn < L + ε n N = max { N1, N2} . ≥ xn = L. xn = L then for some N , N . Thus L − ε < xn < L + ε for all n ≥ xn ≤ Since ε is an arbitrary positive number we must have lim inf n →∞ ≤ − L ε lim sup n →∞ xn ≤ L + ε. as required. L = lim inf n →∞ xn = lim sup n →∞ xn N1. N2. ≥ ≥ In the exercises you will be asked to compute several lim sups and lim infs. This is just for familiarity with the concepts. Computations are not so important. What is important is the use of these ideas ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.13. Upper and Lower Limits 93 in theoretical developments. More critical is how these limit operations relate to arithmetic or order properties. The limit of a sum is the sum of the two limits. Is this true for lim sups and lim infs? (See Exercise 2.13.9.) Do not skip these exercises. Exercises 2.13.1 Complete Example 2.43 by showing that if is a sequence of positive numbers with the property that lim supn→∞ n√xn < 1 then xn → xn} 0. Show that if { lim inf n→∞ . What can you conclude if lim supn→∞ n√xn > 1 n√xn > 1? What can you conclude if lim inf n→∞ n√xn < then xn → ∞ 1? 2.13.2 Compute lim sups and lim infs for the following sequences (a) (b) (c) (d) (e) (f) } 1)nn ( { − sin (nπ/8) } { n sin (nπ/8) { [(n + 1) sin (nπ/8)]/n { 1 + ( { rn} { 1)n } } − consists of all rational numbers in the interval (0, 1) arranged in some order. } 2.13.3 Give examples of sequences of rational numbers an} { with (a) upper limit √2 and lower limit √2, and lower limit √2, − (b) upper limit + ∞ (c) upper limit π and lower limit e. 2.13.4 Show that lim supn→∞( 2.13.5 If two sequences − and an} { xn) = bn} { lim sup n→∞ (lim inf n→∞ xn). − satisfy the inequality an ≤ lim sup n→∞ an ≤ bn and bn for all sufficiently large n, show that bn. lim inf n→∞ an ≤ lim inf n→∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 94 Sequences Chapter 2 2.13.6 Show that limn→∞ xn = ∞ if and only if lim sup n→∞ xn = lim inf n→∞ xn = . ∞ 2.13.7 Show that if lim supn→∞ an = L for a finite real number L and ε > 0, then an > L + ε for only finitely 2.13.8 Show that for any monotonic sequence − many n and an > L ε for infinitely many n. xn} { lim sup n→∞ (including the possibility of infinite limits). an} { (an + bn) and { lim sup n→∞ lim sup n→∞ Give an example to show that the equality need not occur. bn} ≤ 2.13.9 Show that for any bounded sequences xn = lim inf n→∞ xn = lim n→∞ xn an + lim sup n→∞ bn. 2.13.10 What is the correct version for the lim inf of Exercise 2.13.9? 2.13.11 Show that for any bounded sequences { lim sup n→∞ an} and (anbn) bn} (lim sup n→∞ of positive numbers bn). an)(lim sup n→∞ { ≤ Give an example to show that the equality need not occur. 2.13.12 Correct the careless student proof in Exercise 2.8.3 for the squeeze theorem by replacing lim with limsup and liminf in the argument. 2.13.13 What relation, if any, can you state for the lim sups and lim infs of a sequence an} { and one of its has no convergent subsequences, what can you state about the lim sups and lim infs of subsequences 2.13.14 If a sequence the sequence? ? { ank } an} { 2.13.15 Let S denote the set of all real numbers t with the property that some subsequence of a given sequence converges to t. What is the relation between the set S and the lim sups and lim infs of the sequence See Note 27 { an} { ? an} ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.14. Challenging Problems for Chapter 2 2.13.16 Prove the following assertion about the upper and lower limits for any sequence numbers: an+1 n√an ≤ an ≤ Give an example to show that each of these inequalities may be strict. lim sup n→∞ n√an ≤ lim inf n→∞ lim inf n→∞ lim sup n→∞ an+1 an . 95 an} { of positive real 2.13.17 For any sequence an} { write sn = (a1 + a2 + lim inf n→∞ an ≤ · · · lim inf n→∞ + an)/n. Show that lim sup sn ≤ n→∞ sn ≤ lim sup n→∞ an. Give an example to show that each of these inequalities may be strict. 2.14 Challenging Problems for Chapter 2 2.14.1 Let α and β be positive numbers. Show that lim n→∞ n αn + βn = max α, β { } . 2.14.2 For any convergent sequence an} { p write sn = (a1 + a2 + lim n→∞ could converge even if · · · an = lim n→∞ + an)/n, the sequence of averages. Show that sn. an} diverges. { Give an example to show that sn} { 2.14.3 Let a1 = 1 and define a sequence recursively by Show that limn→∞ an n = 1/2. an+1 = √a1 + a2 + + an. · · · 2.14.4 Let x1 = θ and define a sequence recursively by For what values of θ is it true that xn → 2.14.5 Let be a sequence of numbers in the interval (0, 1) with the property that an} { for all n = 2, 3, 4, . . . . Show that this sequence is convergent. xn 1 + xn/2 . xn+1 = 0? an < an−1 + an+1 2 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 96 Sequences Chapter 2 2.14.6 For any convergent sequence an} { write sn = n (a1a2 . . . an), the sequence of geometric averages. Show that limn→∞ an = limn→∞ sn. Give an example to show that sn} could converge even if diverges. an} p { { 2.14.7 If what can you conclude about the sequence 2.14.8 A function f is defined by lim n→∞ α sn − sn + α = 0 ? sn} { f (x) = lim n→∞ n x2 1 − 1 + x2 at every value x for which this limit exists. What is the domain of the function? 2.14.9 A function f is defined by at every value x for which this limit exists. What is the domain of the function? f (x) = lim n→∞ 1 xn + x−n 2.14.10 Suppose that f : R R is a positive function with a derivative f ′ that is everywhere continuous and negative. Apply Newton’s method to obtain a sequence → x1 = θ , xn+1 = xn − f (xn) f ′(xn) . for any starting value θ. [This problem assumes some calculus background.] 2.14.11 Let f (x) = x3 3x + 3. Apply Newton’s method to obtain a sequence Show that xn → ∞ − x1 = θ , xn+1 = xn − f (xn) f ′(xn) . Show that for any positive integer p there is a starting value θ such that the sequence period p. See Note 28 xn} { is periodic with ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 2.14. Challenging Problems for Chapter 2 97 2.14.12 Determine all subsequential limit points of the sequence xn = cos n. See Note 29 2.14.13 A sequence sn} { for all n = 2, 3, 4, . . . . is said to be contractive if there is a positive number 0 < r < 1 so that sn| ≤ sn+1 − | sn−1| sn − | r (a) Show that the sequence defined by s1 = 1 and sn = (4 + sn−1)−1 for n = 2, 3, . . . is contractive. (b) Show that every contractive sequence is Cauchy. (c) Show that a sequence can satisfy the condition sn+1 − | sn−1| sn − | sn| < for all n = 2, 3, 4, . . . and not be contractive, nor even convergent. (d) Is every convergent sequence contractive? See Note 30 2.14.14 The sequence defined recursively by is called the Fibonacci sequence. Let f1 = 1, f2 = 1 fn+2 = fn + fn+1 rn = fn+1/fn be the sequence of ratios of successive terms of the Fibonacci sequence. (a) Show that r1 < r3 < r5 < r2n−1 → (b) Show that r2n − (c) Deduce that the sequence · · · 0. rn} to the roots of the equation x2 { < r6 < r4 < r2. converges. Can you find a way to determine that limit? (This is related x − − 1 = 0.) 2.14.15 A sequence of real numbers xn} { Show that limn→∞ xn = 1. See Note 31 has the property that (2 xn)xn+1 = 1. − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 98 2.14.16 Let an} { be an arbitrary sequence of positive real numbers. Show that lim sup n→∞ a1 + an+1 an n e. ≥ See Note 32 2.14.17 Suppose that the sequence whose nth term is is convergent. Show that See Note 33 sn} { is also convergent. sn + 2sn+1 2.14.18 Show that the sequence √7, converges and find its limit. See Note 34 √77, s7 q − r 7 + 7 q − √7, . . . 2.14.19 Let a1 and a2 be positive numbers and suppose that the sequence an+2 = √an + √an+1. an} { is defined recursively by Show that this sequence converges and find its limit. See Note 35 Notes 9Exercise 2.2.2. For the next term in the sequence some people might expect a 1. Most mathematicians would expect a 9. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 10Exercise 2.2.3. Here is a formula that generates the first five terms of the sequence 0, 0, 0, 0, c, . . . . 11Exercise 2.2.10. The formula is f (n) = c(n − 1)(n − 2)(n 4! − 3)(n 4) . − fn = 1 √5 ( n 1 + √5 2 ! − 1 √5 − 2 ! n . ) It can be verified by induction. 12Exercise 2.3.1. Find a function f : (a, b) is a sequence that is claimed to have all of (a, b) as its range. → (0, 1) one-to-one onto and consider the sequence f (sn), where 99 sn} { 13Exercise 2.3.4. We can consider that the elements of each of the sets Si can be listed, say, and so on. Now try to think of a way of listing all of these items, that is, making one big list that contains them all. S1 = S2 = x11, x12, x13, . . . { x21, x22, x23, . . . { } } 14Exercise 2.3.6. We need (i) every number has a decimal expansion; (ii) the decimal expansion is unique except in the case of expansions that terminate in a string of zeros or nines [e.g., 1/2 = 0.5000000 = .49999999 . . . ], thus if a and b are numbers such that in the nth decimal place one has a 5 (or a 6) and the other does not then either = b, or perhap |
s one ends in a string of zeros and the other in a string of nines; and (iii) every string of 5’s and 6’s a defines a real number with that decimal expansion. · · · 15Exercise 2.3.10. Try to find a way of ranking the algebraic numbers in the same way that the rational numbers were ranked. 16Exercise 2.4.6. You will need the identity (n + 1)/2. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 100 17Exercise 2.4.7. You will need to find an identity for the sum of the squares similar to the identity 1+2+3+ n(n + 1)/2. NOTES +n = · · · 18Exercise 2.5.6. To establish a correct converse, reword: If all xn > 0 and xn this is true. The converse of the statement in the exercise is false (e.g., xn = 1/n). xn+1 → 1, then xn → ∞ . Prove that 19Exercise 2.6.5. Use the same method as used in the proof of Theorem 2.11. 20Exercise 2.8.1. Give a counterexample. Perhaps find two sequences so that sn < 0 < tn for all n and yet limn→∞ sn = limn→∞ tn = 0. 21Exercise 2.8.9. Take any number r strictly between 1 and that limit. Show that for some N , sn+1 < rsn if N . Deduce that n ≥ and Carry on. sN +2 < r2sN sN +3 < r3sN . 22Exercise 2.8.10. Take any number r strictly between 1 and that limit. Show that for some N , sn+1 > rsn if N . Deduce that n ≥ and Carry on. sN +2 > r2sN sN +3 > r3sN . 23Exercise 2.10.1. In terms of our theory of convergence this statement has no meaning since (as you should show) the sequence diverges. Even so, many great mathematicians, including Euler, would have accepted and used this formula. The fact that it is useful suggests that there are ways of interpreting such statements other than as convergence assertions. 24Exercise 2.11.13. If a sequence contains subsequences converging to every number in (0, 1) show that it also contains a subsequence converging to 0. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 101 25Exercise 2.12.5. Consider the sequence 26Exercise 2.12.10. Compare to sn = 1 + 1/2 + 1/3 + + 1/n − − . . . which is the sum of a geometric progression. 27Exercise 2.13.15. Consider separately the cases where the sequence is bounded or not. 28Exercise 2.14.11. A sequence xn} of p will work. (Note that if { is periodic with period p if xn+p = xn for all values of n and no smaller value xn} { is periodic with period p, then xn = xn+p = xn+2p = xn+3p = . . . .) 29Exercise 2.14.12. Clearly, no number larger than 1 or less than interval [ Now consider the set of numbers − 1, 1] is the set of all such limit points. If x [ − ∈ 1 could be such a limit. Show that in fact the 1, 1] there must be a number y so that cos y = x (why?). − Using Exercise 1.11.6 or otherwise, show that this is dense. Hence there are pairs of integers n, m so that G = n + 2mπ : n, m { Z . } ∈ From this deduce that and so x | − cos n | < ε. 30Exercise 2.14.13. For (a) show that y | − n + 2mπ < ε. | cos y | cos(n + 2mπ) | − < ε sn+1 − | sn| ≤ 1 17 | sn − sn−1| for all n = 2, 3, 4, . . . . For (b) you will need to use the fact that the sum of geometric progressions is bounded, in fact that 1 + r + r2 + + rn < (1 r)−1 · · · − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 102 if 0 < r < 1. Express for m > n, sm − | sn+2 − | and then use the contractive hypothesis. Note that + sn| sn+1 − sn| ≤ | sm − + + | · · · sn+1| NOTES sm−1| For (d) you might have to wait for the study of series in order to find an appropriate example of a convergent sequence that is not contractive. s4 − | s3| ≤ r s3 − | s2| ≤ r2 s2 − | . s1| 31Exercise 2.14.15. This is from the 1947 Putnam Mathematical Competition. 32Exercise 2.14.16. This is from the 1949 Putnam Mathematical Competition. 33Exercise 2.14.17. This is from the 1950 Putnam Mathematical Competition. 34Exercise 2.14.18. This is from the 1953 Putnam Mathematical Competition. 35Exercise 2.14.19. Problem posed by A. Emerson in the Amer. Math. Monthly, 85 (1978), p. 496. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Chapter 3 INFINITE SUMS " This chapter on infinite sums and series may be skipped over in designing a course or covered later as the need arises. The basic material in Sections 3.4, 3.5, and parts of 3.6 will be needed, but not before the study of series of functions in Chapter 9. All of the enrichment or advanced sections may be omitted and are not needed in the sequel. 3.1 Introduction The use of infinite sums goes back in time much further, apparently, than the study of sequences. The sum 16 + 1 32 + 1 64 + · · · = 2 has been long known. It is quite easy to convince oneself that this must be valid by arithmetic or geometric “reasoning.” After all, just start adding and keeping track of the sum as you progress: 1 15 16 , . . . . Figure 3.1 makes this seem transparent. But there is a serious problem of meaning here. A finite sum is well defined, an infinite sum is not. Neither humans nor computers can add an infinite column of numbers. 103 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 104 Infinite Sums Chapter 3 Figure 3.1. 1 + 1/2 + 1/4 + 1/8 + 1/16 + = 2. · · · The meaning that is commonly assigned to the preceding sum appears in the following computations + · · · = lim n →∞ 2n = lim n 2 →∞ 1 2n+1 1 − = 2. This reduces the computation of an infinite sum to that of a finite sum followed by a limit operation. Indeed this is exactly what we were doing when we computed 1, 1 1 a compelling reason for thinking of the sum as 2. 16 , . . . and felt that this was 8 , 1 15 2 , 1 3 4 , 1 7 In terms of the development of the theory of this textbook this seems entirely natural and hardly surprising. We have mastered sequences in Chapter 2 and now pass to infinite sums in Chapter 3 using the methods of sequences. Historically this was not the case. Infinite summations appear to have been studied and used long before any development of sequences and sequence limits. Indeed, even to form the notion of an infinite sum as previously, it would seem that we should already have some concept of sequences, but this is not the way things developed. It was only by the time of Cauchy that the modern theory of infinite summation was developed using sequence limits as a basis for the theory. We can transfer a great deal of our expertise in sequential limits ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.2. Finite Sums 105 to the problem of infinite sums. Even so, the study in this chapter has its own character and charm. In many ways infinite sums are much more interesting and important to analysis than sequences. 3.2 Finite Sums We should begin our discussion of infinite sums with finite sums. There is not much to say about finite sums. Any finite collection of real numbers may be summed in any order and any grouping. That is not to say that we shall not encounter practical problems in this. For example, what is the sum of the first 10100 prime numbers? No computer or human could find this within the time remaining in this universe. But there is no mathematical problem in saying that it is defined; it is a sum of a finite number of real numbers. There are a number of notations and a number of skills that we shall need to develop in order to succeed at the study of infinite sums that is to come. The notation of such summations may be novel. How best to write out a symbol indicating that some set of numbers a1, a2, a3, . . . , an} { has been summed? Certainly a1 + a2 + a3 + + an · · · is too cumbersome a way of writing such sums. The following have proved to communicate much better: where I is the set 1, 2, 3, . . . , n } { or ai I Xi ∈ ai or n ai. n i X1 ≤ ≤ Here the Greek letter Σ, corresponding to an uppercase “S,” is used to indicate a sum. Xi=1 It is to Leonhard Euler (1707–1783) that we owe this sigma notation for sums (first used by him in 1 are also his. These alone indicate the level 1755). The notations f (x) for functions, e and π, i for √ − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 106 Infinite Sums Chapter 3 of influence he has left. In his lifetime he wrote 886 papers and books and is considered the most prolific writer of mathematics that has lived. The usual rules of elementary arithmetic apply to finite sums. The commutative, associative, and distributive rules assume a different look when written in Euler’s notation: ai + bi = (ai + bi) , I Xi ∈ I Xi ∈ I Xi ∈ cai = c ai, I Xi ∈ I Xi ∈ and ai I Xi ∈ ! × J Xj ∈ = bj I Xi ∈ J Xj ∈ = aibj aibj . ! J Xj ∈ I Xi ∈ Each of these can be checked mainly by determining the meanings and seeing that the notation produces the correct result. Occasionally in applications of these ideas one would like a simplified expression for a summation. The best known example is perhaps (n + 1) 2 , Xk=1 which is easily proved. When a sum of n terms for a general n has a simpler expression such as this it is usual to say that it has been expressed in closed form. Novices, seeing this, usually assume that any summation with some degree of regularity should allow a closed form expression and that it is always important to get a closed form expression. If not, what can you do with a sum that cannot be simplified? One of the simplest of sums · · · does not allow any convenient formula, expressing the sum as some simple function of n. This is typical=1 1 k ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.2. Finite Sums 107 It is only the rarest of summations that will allow simple formulas. Our work is mostly in estimating such expressions; we hardly ever succeed in computing them exactly. Even so, there are a few special cases that should be remembered and which make our task in some cases much easier. If a sum can be rewritten in the special form below, a simple computation (canceling Telescoping |
Sums. s1, s2, etc.) gives the following closed form: s1) + (s3 − s0) + (s2 − s2) + (s4 − It is convenient to call such a sum “telescoping” as an indication of the method that can be used to compute it. 1) = sn − + (sn − (s1 − s3) + s0. · · · sn − Example 3.1: For a specific example of a sum that can be handled by considering it as telescoping, consider the sum n Xk=1 1 k(k + 1n 1 1) − . n · A closed form is available since, using partial fractions, each term can be expressed as Thus n 1 k(k + 1(k + 1) = n Xk= . − Xk=1 The exercises contain a number of other examples of this type. ◭ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 108 Infinite Sums Chapter 3 Geometric Progressions. constant factor times the previous term), then a closed form for any such sum is available: If the terms of a sum are in a geometric progression (i.e., if each term is some 1 + r + r2 + · · · + rn − 1 + rn = 1 − 1 rn+1 r − . (1) = 1; if r = 1 the sum is easily seen to be just n + 1. The formula in (1) can be proved This assumes that r by converting to a telescoping sum. Consider instead (1 1 + rn) = (1 r)(1 + r + r2 + + rn (1 − − r) times the preceding sum: r2) + r) + (r + (rn rn+1). − · · · Now add this up as a telescoping sum to obtain the formula stated in (1). − − − · · · Any geometric progression assumes the form A + Ar + Ar2 + · · · + Arn = A(1 + r + r2 + + rn) · · · and formula (1) (which should be memorized) is then applied. Summation By Parts. Sums are frequently given in a form such as n akbk for sequences ak} { and bk} { . If a formula happens to be available for Xk=1 sn = a1 + a2 + + an, · · · then there is a frequently useful way of rewriting this sum (using s0 = 0 for convenience): n n akbk = (sk − 1)bk sk − = s1(b1 − Xk=1 b2) + s2(b2 − Xk=1 b3) + + sn 1(bn − 1 − − bn) + snbn. · · · ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 3.2. Finite Sums 109 Usually some extra knowledge about the sequences can then be used to advantage. The { computation is trivial (it is all contained in the preceding equation which is easily checked). Sometimes this summation formula is referred to as Abel’s transformation after the Norwegian mathematician Niels Henrik Abel (1802–1829), who was one of the founders of the rigorous theory of infinite sums. It is the analog for finite sums of the integration by parts formula of calculus. bk} { sk} and Abel’s most important contributions are to analysis but he is forever immortalized in group theory (to which he made a small contribution) by the fact that commutative groups are called “Abelian.” Exercises 3.2.1 Prove the formula k = n(n + 1) 2 . n Xk=1 3.2.2 Give a formal definition of number of elements of I. See Note 36 P i∈I ai for any finite set I and any function a : I R that uses induction on the → Your definition should be able to handle the case I = . ∅ 3.2.3 Check the validity of the formulas given in this section for manipulating finite sums. Are there any other formulas you can propose and verify? 3.2.4 Is the formula valid? See Note 37 ai = ai + ai Xi∈I∪J Xi∈I Xi∈J 3.2.5 Let I = (i, j) : 1 { ≤ i ≤ m, 1 j n } ≤ ≤ . Show that m n aij = aij. X(i,j)∈I i=1 X j=1 X ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 110 Infinite Sums Chapter 3 3.2.6 Give a formula for the sum of n terms of an arithmetic progression. (An arithmetic progression is a list of numbers, each of which is obtained by adding a fixed constant to the previous one in the list.) For the purposes of infinite sums (our concern in this chapter) such a formula will be of little use. Explain why. 3.2.7 Obtain formulas (or find a source for such formulas) for the sums n of the pth powers of the natural numbers where p = 1, 2, 3, 4, . . . . Again, for the purposes of infinite sums such formulas will be of little use. kp = 1p + 2p + 3p + + np · · · Xk=1 3.2.8 Explain the (vague) connection between integration by parts and summation by parts. See Note 38 3.2.9 Obtain a formula for 3.2.10 Obtain a formula for P 3.2.11 Obtain the formula n k=1( − 1)k. 2 + 2√2 + 4 + 4√2 + 8 + 8√2 + + 2m. · · · sin θ + sin 2θ + sin 3θ + sin 4θ + − 2 sin θ/2 How should the formula be interpreted if the denominator of the fraction is zero? See Note 39 + sin nθ = · · · cos θ/2 cos(2n + 1)θ/2 . 3.2.12 Obtain the formula 3.2.13 If cos θ + cos 3θ + cos 5θ + cos 7θ + + cos(2n 1)θ = − · · · sin 2nθ 2 sin θ . sn = )n+1 1 n show that 1/2 sn ≤ ≤ 1 for all n. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.2. Finite Sums 3.2.14 If sn = 111 show that s2n ≥ 1 + n/2 for all n. 3.2.15 Obtain a closed form for 3.2.16 Obtain a closed form for n Xk=1 n 3.2.17 Let ak} { and bk} { be sequences with bk} { for all k. Show that for all n. 1 k(k + 2)(k + 4) . αr + β k(k + 1)(k + 2) Xk=1 decreasing and a1 + a2 + | · · · + ak| ≤ . K Kb1 ≤ akbk n Xk=1 3.2.18 If r is the interest rate (e.g., r = .06) over a period of years, then P (1 + r)−1 + P (1 + r)−2 + + P (1 + r)−n · · · is the present value of an annuity of P dollars paid every year, starting next year and for n years. Give a shorter formula for this. (A perpetuity has nearly the same formula but the payments continue forever. See Exercise 3.4.12.) 3.2.19 Define a finite product (product of a finite set of real numbers) by writing What elementary properties can you determine for products? Yk=1 n ak = a1a2a3 . . . an. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 112 Infinite Sums Chapter 3 3.2.20 Find a closed form expression for k3 1 − k3 + 1 . n Yk=1 3.3 Infinite Unordered sums " Advanced section. May be omitted. We now pass to the study of infinite sums. We wish to interpret ai I Xi ∈ for an index set I that is infinite. The study of finite sums involves no analysis, no limits, no ε’s, in short none of the processes that are special to analysis. To define and study infinite sums requires many of our skills in analysis. To begin our study imagine that we are given a collection of numbers ai indexed over an infinite set I R) and we wish the sum of the totality of these numbers. If the set I has (i.e., there is a function a : I some structure, then we can use that structure to decide how to start adding the numbers. For example, if a is a sequence so that I = IN, then we should likely start adding at the beginning of the sequence: → a1, a1 + a2, a1 + a2 + a3, a1 + a2 + a3 + a4, . . . and so defining the sum as the limit of this sequence of partial sums. Another set I would suggest a different order. For example, if I = Z (the set of all integers), then a popular method of adding these up would be to start off: a0, a 1 + a0 + a1, − 1 + a0 + a1 + a2 + a0 + a1 + a2 + a3, . . . a − 3 + a − − once again defining the sum as the limit of this sequence. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.3. Infinite Unordered sums 113 It seems that the method of summation and hence defining the meaning of the expression ai I Xi ∈ for infinite sets I must depend on the nature of the set I and hence on the particular problems of the subject one is studying. This is true to some extent. But it does not stop us from inventing a method that will apply to all infinite sets I. We must make a definition that takes account of no extra structure or ordering for the set I and just treats it as a set. This is called the unordered sum and the notation I ai is always meant to indicate that an unordered sum is being considered. The key is just how to pass from finite sums to infinite sums. Both of the previous examples used the idea of taking some finite sums (in a systematic way) and then passing to a limit. P i ∈ Definition 3.2: Let I be an infinite set and a a function a : I ai = c R. Then we write → and say that the sum converges if for every ε > 0 there is a finite set I0 ⊂ I0 ⊂ I, ⊂ J I Xi ∈ I so that, for every finite set J, J Xi ∈ A sum that does not converge is said to diverge. < ε. c ai − Note that we never form a sum of infinitely many terms. The definition always computes finite sums. Example 3.3: Let us show, directly from the definition, that i 2−| | = 3. Z Xi ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 114 If we first sum by rearranging the terms into the sum 2−| i | i N X− ≤ ≤ N we can see why the sum is likely to be 3. Let ε > 0 and choose N so that 2− for a finite geometric progression we have 1 + 2(2− 1 + 2− 2 + + 2− N ) · · · Infinite Sums Chapter 3 N < ε/4. From the formula i N X− ≤ ≤ Z with K finite and N 2−| i | − 3 = 2 (2− 1 + 2− | > N for all k | ∈ k | Also, if K ⊂ K, then k 2−| | < 2(2− N ) < ε/2 2 + · · · + 2− N ) 1 | − = 2(2− N ) < ε/2. again from the formula for a finite geometric progression. Let K Xk ∈ If I0 ⊂ J ⊂ Z with J finite then I0 = Z : i { ∈ N − i N . } ≤ ≤ as required. 2−| i | 3 − = J Xi ∈ 2−| i | 3 − i N X− ≤ ≤ N + i 2−| | < ε I0 J Xi \ ∈ ◭ 3.3.1 Cauchy Criterion " Advanced section. May be omitted. In most theories of convergence one asks for a necessary and sufficient condition for convergence. We saw in studying sequences that the Cauchy criterion provided such a condition for the convergence of a ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.3. Infinite Unordered sums 115 sequence. There is usually in any theory of this kind a type of Cauchy criterion. Here is the Cauchy criterion for sums. Theorem 3.4: A necessary and sufficient condition that the sum there is a finite set I0 so that i ∈ P I ai converges is that for every ε > 0 for every finite set J ⊂ I that contains no elements of I0 (i.e., for all finite sets J ai < ε J Xi ∈ I \ ⊂ I0). Proof. As usual in Cauchy criterion proofs, one direction is easy to prove. Suppose that converges. Then for every ε > 0 there is a finite set I0 so that I ai = C i ∈ P C < ε/2 ai − I. Let J be a finite subset of I |
K Xi ∈ I0 and consider taking a sum over K = I0 ∪ \ J. for every finite set I0 ⊂ Then K ⊂ and By subtracting these two inequalities and remembering that C ai − I0 Xi ∪ ∈ J C ai − I0 Xi ∈ < ε/2 < ε/2. ai + ai = ai I0 Xi ∪ ∈ J J Xi ∈ I0 Xi ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 116 Infinite Sums Chapter 3 (since I0 and J are disjoint) we obtain This is exactly the Cauchy criterion. Conversely, suppose that the sum does satisfy the Cauchy criterion. Then, applying that criterion to J Xi ∈ ai < ε. ε = 1, 1/2, 1/3,. . . we can choose a sequence of finite sets In} { so that for every finite set J I \ ⊂ ai < 1/n J Xi ∈ In. We can arrange our choices to make I1 ⊂ I2 ⊂ I3 ⊂ . . . so that the sequence of sets is increasing. Let Then for any m > n, cn = ai In Xi ∈ It follows from this that number c. Let ε > 0 and choose an integer N larger than 2/ε and so that n > N and any finite set J with IN ⊂ cn} I, ⊂ J { is a Cauchy sequence of real numbers and hence converges to some real cN − | c | < ε/2. Then, for any cn − | cm| = < 1/n. Im Xi \ ∈ In ai c ai − IN Xi ∈ ≤ ai − cN J Xi ∈ + cN − | c | + IN J Xi \ ∈ ai < 0 + ε/2 + 1/N < ε. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.3. Infinite Unordered sums By definition, then, and the theorem is proved. ai = c I Xi ∈ 117 All But Countably Many Terms in a Convergent Sum Are Zero. Our next theorem shows that having “too many” numbers to add up causes problems. If the set I is not countable then most of the ai that we are to add up should be zero if the sum is to exist. This shows too that the theory of sums is in an essential way limited to taking sums over countable sets. It is notationally possible to have a sum f (x) [0,1] Xx ∈ but that sum cannot be defined unless f (x) is mostly zero with only countably many exceptions. Theorem 3.5: Suppose that I. i ∈ P I ai converges. Then ai = 0 for all i ∈ I except for a countable subset of Proof. We shall use Exercise 3.3.2, where it is proved that for any convergent sum there is a positive integer M so that all the sums ai I. Let m be an integer. We ask how many elements ai are there such that ai > 1/m? for any finite set I0 ⊂ It is easy to see that there are at most M m of them since if there were any more our sum would exceed M . Similarly, there are at most M m terms such that that is not − zero can be given a “rank” m depending on whether ai > 1/m. Thus each element of ai : i I0 Xi ∈ M ≤ ∈ } { I 1/m < ai ≤ 1/(m − 1) or 1/m < ai ≤ − 1/(m 1). − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 118 Infinite Sums Chapter 3 As there are only finitely many elements at each rank, this gives us a method for listing all of the nonzero elements in and so this set is countable. I ai : i { ∈ } The elementary properties of unordered sums are developed in the exercises. These sums play a small role in analysis, a much smaller role than the ordered sums we shall consider in the next sections. The methods of proof, however, are well worth studying since they are used in some form or other in many parts of analysis. These exercises offer an interesting setting in which to test your skills in analysis, skills that will play a role in all of your subsequent study. Exercises 3.3.1 Show that if i∈I ai converges, then the sum is unique. See Note 40 P 3.3.2 Show that if i∈I ai converges, then there is a positive number M so that all the sums P for any finite set I0 ⊂ See Note 41 I. M ≤ ai Xi∈I0 3.3.3 Suppose that all the terms in the sum all the sums P i∈I ai are nonnegative and that there is a positive number M so that for any finite set I0 ⊂ See Note 42 I. Show that P ai ≤ M Xi∈I0 i∈I ai must converge. 3.3.4 Show that if i∈I ai converges so too does i∈J ai for every subset J I. ⊂ P P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.3. Infinite Unordered sums 119 3.3.5 Show that if i∈I ai converges and each ai ≥ 0, then P ai = sup ai : J I, J finite . ⊂ ) ( Xi∈J 3.3.6 Each of the rules for manipulation of the finite sums of Section 3.2 can be considered for infinite unordered sums. Formulate the correct statement and prove what you think to be the analog of these statements that we know hold for finite sums: Xi∈I ai + bi = (ai + bi) Xi∈I Xi∈I Xi∈I cai = c ai Xi∈I bj = Xi∈I aibj = aibj. Xi∈J Xi∈I Xj∈J Xj∈J Xi∈I ai × Xi∈I ai + ai = ai + ai 3.3.7 Prove that under appropriate convergence assumptions. Xi∈I∪J Xi∈I∩J Xi∈I Xi∈J 3.3.8 Let σ : I → J one-to-one and onto. Establish that under appropriate convergence assumptions. 3.3.9 Find the sum See Note 43 aj = aσ(i) Xj∈J Xi∈I 1 2i . Xi∈IN ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 120 3.3.10 Show that diverges. Are there any infinite subsets J ⊂ 1 i Xi∈IN IN such that 1 i Xi∈J Infinite Sums Chapter 3 i∈I [ai]− converge and that [ai]− P converges? 3.3.11 Show that i∈I ai converges if and only if both P and ai = Xi∈I i∈I [ai]+ and [ai]+ − Xi∈I P Xi∈I See Note 44 3.3.12 Compute = ai| | Xi∈I [ai]+ + [ai]−. Xi∈I Xi∈I 2−i−j. What kind of ordered sum would seem natural here (in the way that ordered sums over IN and Z were considered in this section)? X(i,j)∈IN×IN See Note 45 3.4 Ordered Sums: Series For the vast majority of applications, one wishes to sum not an arbitrary collection of numbers but most commonly some sequence of numbers: a1 + a2 + a3 + . . . . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.4. Ordered Sums: Series 121 The set IN of natural numbers has an order structure, and it is not in our best interests to ignore that order since that is the order in which the sequence is presented to us. The most compelling way to add up a sequence of numbers is to begin accumulating: a1, a1 + a2, a1 + a2 + a3, a1 + a2 + a3 + a4, . . . and to define the sum as the limit of this sequence. This is what we shall do. If you studied Section 3.3 on unordered summation you should also compare this “ordered” method with the unordered method. The ordered sum of a sequence is called a series and the notation is used exclusively for this notion. ∞ ak Xk=1 Definition 3.6: Let ak} { be a sequence of real numbers. Then we write and say that the series converges if the sequence ak = c ∞ Xk=1 n sn = ak Xk=1 (called the sequence of partial sums of the series) converges to c. If the series does not converge it is said to be divergent. This definition reduces the study of series to the study of sequences. We already have a highly developed theory of convergent sequences in Chapter 2 that we can apply to develop a theory of series. Thus we can rapidly produce a fairly deep theory of series from what we already know. As the theory develops, ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 122 Infinite Sums Chapter 3 however, we shall see that it begins to take a character of its own and stops looking like a mere application of sequence ideas. 3.4.1 Properties The following short harvest of theorems we obtain directly from our sequence theory. The convergence or divergence of a series ∞k=1 ak depends on the convergence or divergence of the sequence of partial sums P n sn = ak Xk=1 and the value of the series is the limit of the sequence. To prove each of the theorems we now list requires only to find the correct theorem on sequences from Chapter 2. This is left as Exercise 3.4.2. Theorem 3.7: If a series ∞k=1 ak converges, then the sum is unique. P Theorem 3.8: If both series ∞k=1 ak and ∞k=1 bk converge, then so too does the series and Theorem 3.9: If the series c and P P ∞ (ak + bk) Xk=1 ∞ Xk=1 (ak + bk) = ∞ ak + ∞ bk. Xk=1 Xk=1 ∞k=1 ak converges, then so too does the series ∞k=1 cak for any real number P ∞ Xk=1 cak = c ∞ ak. Xk=1 P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.4. Ordered Sums: Series 123 Theorem 3.10: If both series ∞k=1 ak and ∞k=1 bk converge and ak ≤ bk for each k, then P P ∞ Xk=1 ak ≤ ∞ bk. Xk=1 Theorem 3.11: Let M ≥ 1 be any integer. Then the series converges if and only if the series ak = a1 + a2 + a3 + a4 + . . . ∞ Xk=1 converges. aM +k = aM +1 + aM +2 + aM +3 + aM +4 + . . . ∞ Xk=1 ∞p ai a “tail” for the series Note. If we call ∞1 ai, then we can say that this last theorem asserts that it is the behavior of the tail that determines the convergence or divergence of the series. Thus in questions of convergence we can easily ignore the first part of the series—however many terms we like. Naturally, the actual sum of the series will depend on having all the terms. P P 3.4.2 Special Series Telescoping Series Any series for which we can find a closed form for the partial sums we should probably be able to handle by sequence methods. Telescoping series are the easiest to deal with. If the sequence of partial sums of a series can be computed in some closed form , then the series sn} { can be rewritten in the telescoping form (s1) + (s2 − s1) + (s3 − and the series studied by means of the sequence s2) + (s4 − sn} { . s3) + + (sn − sn − · · · 1) + . . . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 124 Infinite Sums Chapter 3 Example 3.12: Consider the series ∞ 1 k(k + 1 Xk=1 with an easily computable sequence of partial sums. Xk=1 = lim n 1 →∞ − 1 n + 1 = 1 ◭ Do not be too encouraged by the apparent ease of the method illustrated by the example. In practice we can hardly ever do anything but make a crude estimate on the size of the partial sums. An exact expression, as we have here, would be rarely available. Even so, it is entertaining and instructive to handle a number of series by such a method (as we do in the exercises). Geometric Series Geometric series form another convenient class of series that we can handle simply by sequence methods. From the elementary formula we |
see immediately that the study of such a series reduces to the computation of the limit 1 + r + r2 + · · · + rn − 1 + rn = 1 − 1 rn+1 r − (r = 1) 1 1 < r < 1 (which is usually expressed as − 1 rn+1 r − 1 = lim n →∞ 1 r | − r | < 1) and invalid for all other values of r. ∞ rk − 1 = 1 + r + r2 + 1 = · · · 1 r Xk=1 − 1 the series diverges. It is well worthwhile to memorize this fact and (2) which is valid for Thus, for r − < 1 the series | | and is convergent and for formula (2) for the sum of the series. | ≥ r | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 3.4. Ordered Sums: Series 125 Harmonic Series As a first taste of an elementary looking series that presents a new challenge to our methods, consider the series which is called the harmonic series. Let us show that this series diverges + . . . , ∞ Xk=1 →∞ This series has no closed form for the sequence of partial sums and so there seems no hope of merely computing limn sn to determine the convergence or divergence of the harmonic series. But we can make estimates on the size of sn even if we cannot compute it directly. The sequence of partial sums increases at each step, and if we watch only at the steps 1, 2, 4, 8, . . . and make a rough lower estimate of s1, s2, s4, s8, and so 1 + n/2 for all n (see Exercise 3.2.14). From this we see that limn . . . we see that s2n the series diverges. sn} sn = →∞ ∞ ≥ { Alternating Harmonic Series A variant on the harmonic series presents immediately a new challenge. Consider the series which is called the alternating harmonic series. ∞ 1) − + . . . , Xk=1 The reason why this presents a different challenge is that the sequence of partial sums is no longer increasing. Thus estimates as to how big that sequence get may be of no help. We can see that the sequence is bounded, but that does not imply convergence for a non monotonic sequence. Once again, we have no closed form for the partial sums so that a routine computation of a sequence limit is not available. By computing the partial sums s2, s4, s6, . . . we see that the subsequence 1} computing the partial sums s1, s3, s5, . . . we see that the subsequence observations show us that s2n { − is increasing. By s2n} { is decreasing. A few more 1/2 = s2 < s4 < s6 < < s5 < s3 < s1 = 1. · · · (3) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 126 Infinite Sums Chapter 3 Our theory of sequences now allows us to assert that both limits exist. Finally, since lim n →∞ s2n and lim →∞ n s2n 1 − s2n − s2n − 1 = − 1 2n → 0 we can conclude that limn 2 and 1 because of the inequalities (3) but exactly what it is would take much further analysis.] Thus we have proved that the alternating harmonic series converges (which is in contrast to the divergence of the harmonic series). sn exists. [It is somewhere between 1 →∞ p-Harmonic Series The series ∞ 1 kp = 1 + 1 2p + 1 3p + . . . Xk=1 ∞ is called the p-harmonic series. The methods we have used in the study of for any parameter 0 < p < the harmonic series can be easily adapted to handle this series. As a first observation note that if 0 < p < 1, then 1 kp > Thus the p-harmonic series for 0 < p < 1 is larger than the harmonic series itself. Since the latter series has 1. unbounded partial sums it is easy to argue that our series does too and, hence, diverges for all 0 < p What about p > 1? Now the terms are smaller than the harmonic series, small enough it turns out that the series converges. To show this we can group the terms in the same manner as before for the harmonic series and obtain 1 k ≤ . 1 2p + 1 3p 1 + + 1 7p 1 4p + 2 2p + 1 5p + 4 4p + 1 6p + 8 8p + + · · · ≤ 1 1 8p + 1 21 − · · · p − 1 + ≤ + 1 15p + . . . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.4. Ordered Sums: Series 127 since we recognize the latter series as a convergent geometric series with ratio 21 an upper bound for the partial sums of the series − p. In this way we obtain for all p > 1. Since the partial sums are increasing and bounded above, the series must converge. 1 kp ∞ Xk=1 Size of the Terms have ultimately small terms. If Certainly for the geometric series that idea precisely described the situation: It should seem apparent from the examples we have seen that a convergent series must ∞k=1 ak converges, then it seems that ak must tend to 0 as k gets large. P ∞ 1 rk − Xk=1 < 1, which is exactly when the terms tend to zero and diverges when converges if exactly when the terms do not tend to zero. r | | r | | ≥ 1, which is A reasonable conjecture might be that this is always the situation: A series only if ak → get small; they simply don’t get small fast enough. Thus the correct observation is simple and limited. ∞k=1 ak converges if and . But we have already seen the harmonic series diverges even though its terms do 0 as k → ∞ P If ∞k=1 ak converges, then ak → 0 as k . → ∞ P To check this is easy. If sn} { an = lim →∞ is the sequence of partial sums of a convergent series sn − lim n →∞ 1) = lim n →∞ The converse, as we just noted, is false. To obtain convergence of a series it is not enough to know that the terms tend to zero. We shall see, though, that many of the tests that follow discuss the rate at which the terms tend to zero. lim n →∞ (sn − C = 0. 1 = C P sn sn − − − n ∞k=1 ak = C, then ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 128 Exercises 3.4.1 Let sn} { the series be any sequence of real numbers. Show that this sequence converges to a number S if and only if Infinite Sums Chapter 3 s1 + ∞ Xk=2 (sk − sk−1) converges and has sum S. 3.4.2 State which theorems from Chapter 2 would be used to prove Theorems 3.7–3.11. 3.4.3 If ∞ k=1(ak + bk) converges, what can you say about the series 3.4.4 If P P ∞ ∞ ak and bk? Xk=1 ∞ k=1(ak + bk) diverges, what can you say about the series Xk=1 ∞ ∞ ak and bk? 3.4.5 If the series 3.4.6 If the series Xk=1 ∞ k=1(a2k + a2k−1) converges, what can you say about the series ∞ k=1 ak converges, what can you say about the series Xk=1 P ∞ k=1 ak? P P ∞ (a2k + a2k−1)? 3.4.7 If both series ∞ k=1 ak and Xk=1 ∞ k=1 bk converge, what can you say about the series 3.4.8 How should we interpret P P ∞ ∞ ∞ ak+1, ak+6 and ak−4? ∞ k=1 akbk? P Xk=−5 3.4.9 If sn is a strictly increasing sequence of positive numbers, show that it is the sequence of partial sums of some Xk=5 Xk=0 series with positive terms. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.4. Ordered Sums: Series 129 Figure 3.2. What is the area of the black region? 3.4.10 If is a subsequence of , is there anything you can say about the relation between the convergence ank } { behavior of the series an} { ∞ k=1 ak and its “subseries” ∞ k=1 ank ? See Note 46 P P 3.4.11 Express the infinite repeating decimal .123451234512345123451234512345 . . . as the sum of a convergent geometric series and compute its sum (as a rational number) in this way. 3.4.12 Using your result from Exercise 3.2.18, obtain a formula for a perpetuity of P dollars a year paid every year, starting next year and for every after. You most likely used a geometric series; can you find an argument that avoids this? 3.4.13 Suppose that a bird flying 100 miles per hour (mph) travels back and forth between a train and the railway station, where the train and the bird start off together 1 mile away and the train is approaching the station at a fixed rate of 60 mph. How far has the bird traveled when the train arrives? You most likely did not use a geometric series; can you find an argument that does? 3.4.14 What proportion of the area of the square in Figure 3.2 is black? ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Infinite Sums Chapter 3 130 3.4.15 Does the series converge or diverge? See Note 47 3.4.16 Show that for all r > 1. See Note 48 3.4.17 Obtain a formula for the sum 3.4.18 Obtain a formula for the sum 3.4.19 Obtain a formula for the sum ∞ log Xk= r2 + 1 + 4 r4 + 1 + 8 r8 + 2 + 1 + 1 √2 + 1 2 + 1 2√2 + . . . . 1 k(k + 2)(k + 4) . αr + β k(k + 1)(k + 2) . ∞ Xk=1 ∞ Xk=1 3.4.20 Find all values of x for which the the following series converges and determine the sum1 + x)2 + x (1 + x)3 + x (1 + x)4 + . . . . 3.4.21 Determine whether the series ∞ converges or diverges where a and b are positive real numbers. 1 a + kb Xk=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.4. Ordered Sums: Series 131 3.4.22 We have proved that the harmonic series diverges. A computer experiment seems to show otherwise. Let sn be the sequence of partial sums and, using a computer and the recursion formula compute s1, s2, s3, . . . and stop when it appears that the sequence is no longer changing. This does happen! Explain why this is not a contradiction. sn+1 = sn + 1 n + 1 , 3.4.23 Let M be any integer. In Theorem 3.11 we saw that the series ∞ k=1 aM +k converges. What is the exact relation between the sums of the two series? ∞ k=1 ak converges if and only if the series P 3.4.24 Write up a formal proof that the p-harmonic series P ∞ converges for p > 1 using the method sketched in the text. See Note 49 1 kp Xk=1 3.4.25 With a short argument using what you know about the harmonic series, show that the p-harmonic series for 0 < p ≤ 1 is divergent. 3.4.26 Obtain the divergence of the improper calculus integral by comparing with the harmonic series. See Note 50 ∞ | sin x x | dx 0 Z 3.4.27 We have seen that the condition an → series k=1 ak. Is the condition nan → are going to zero faster than 1/k. ∞ P 0 is a necessary, but not sufficient, condition for convergence of the 0 either necessary or sufficient for the convergence? This says terms 3.4.28 Let p be an integer greater or equal to 2 and let x be a real number in the interval [0, 1). Construct a sequence of integers kn} { as follows |
: Divide the interval [0, 1) into p intervals of equal length [0, 1/p), [1/p, 2/p), . . . , [(p 1)/p, 1) − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 132 Infinite Sums Chapter 3 1. Then k1 is chosen so that x belongs to the k1th interval. and label them from left to right as 0, 1, . . . , p 1)/p, k1/p) in which x lies, dividing it into p intervals Repeat the process applying it now to the interval [(k1 − of equal length and choose k2 so that x belongs to the k2th interval of the new subintervals. Continue this . Show that process inductively to define the sequence − kn} { ∞ x = ki pi . i=1 X [This is called the p-adic representation of the number x.] See Note 51 3.5 Criteria for Convergence How do we determine the convergence or divergence of a series? The meaning of convergence or divergence is directly given in terms of the sequence of partial sums. But usually it is very difficult to say much about that sequence. Certainly we hardly ever get a closed form for the partial sums. For a successful theory of series we need some criteria that will enable us to assert the convergence or divergence of a series without much bothering with an intimate acquaintance with the sequence of partial sums. The following material begins the development of these criteria. 3.5.1 Boundedness Criterion If a series forms a monotonic sequence. It is strictly increasing if all terms are positive. ∞k=1 ak consists entirely of nonnegative terms, then it is clear that the sequence of partial sums We have a well-established fundamental principle for the investigation of all monotonic sequences: P A monotonic sequence is convergent if and only if it is bounded. Applied to the study of series, this principle says that a series ∞k=1 ak consisting entirely of nonnegative terms will converge if the sequence of partial sums is bounded and will diverge if the sequence of partial sums is unbounded. P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.5. Criteria for Convergence 133 This reduces the study of the convergence/divergence behavior of such series to inequality problems: Is there or is there not a number M so that for all integers n? sn = n Xk=1 ak ≤ M This is both good news and bad. Theoretically it means that convergence problems for this special class of series reduce to another problem: one of boundedness. That is good news, reducing an apparently difficult problem to one we already understand. The bad news is that inequality problems may still be difficult. Note. A word of warning. The boundedness of the partial sums of a series is not of as great an interest for series where the terms can be both positive and negative. For such series the boundedness of the partial sums does not guarantee convergence. 3.5.2 Cauchy Criterion One of our main theoretical tools in the study of convergent sequences is the Cauchy criterion describing (albeit somewhat technically) a necessary and sufficient condition for a sequence to be convergent. If we translate that criterion to the language of series we shall then have a necessary and sufficient condition for a series to be convergent. Again it is rather technical and mostly useful in developing a theory rather than in testing specific series. The translation is nearly immediate. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 134 Definition 3.13: The series Infinite Sums Chapter 3 Xk=1 is said to satisfy the Cauchy criterion for convergence provided that for every ε > 0 there is an integer N so that all of the finite sums m ∞ ak for any ak Xk=n Now we have a principle that can be applied in many theoretical situations: A series ∞k=1 ak converges if and only if it satisfies the Cauchy criterion for convergence. Note. It may be useful to think of this conceptually. The criterion asserts that convergence is equivalent to the fact that blocks of terms P M Xk=N added up and taken from far on in the series must be small. Loosely we might describe this by saying that a convergent series has a “small tail.” Note too that if the series converges, then this criterion implies that for every ε > 0 there is an integer ak N so that for every n N . ≥ < ε ∞ ak Xk=n ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.5. Criteria for Convergence 3.5.3 Absolute Convergence 135 If a series consists of nonnegative terms only, then we can obtain convergence or divergence by estimating the size of the partial sums. If the partial sums remain bounded, then the series converges; if not, the series diverges. No such conclusion can be made for a series ∞k=1 ak of positive and negative numbers. Boundedness of the partial sums does not allow us to conclude anything about convergence or divergence since the sequence of partial sums would not be monotonic. What we can do is ask whether there is any relation between the two series P ∞ ak and ∞ ak| | Xk=1 Xk=1 where the latter series has had the negative signs stripped from it. We shall see that convergence of the series of absolute values ensures convergence of the original series. Divergence of the series of absolute values gives, however, no information. This gives us a useful test that will prove the convergence of a series ∞k=1 ak by investigating instead the related series ∞k=1 | ak| without the negative signs. P Theorem 3.14: If the series P ∞k=1 | ak| converges, then so too does the series ∞k=1 ak. Proof. The proof takes two applications of the Cauchy criterion. If ε > 0 there is an integer N so that all of the finite sums P for any N n < m < ≤ ∞ . But then m Xk=n < ε ak| | m m < ε. ak| | ≤ Xk=n ak Xk=n P ak| converges, then for every ∞k=1 | P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 136 Infinite Sums Chapter 3 It follows, by the Cauchy criterion applied to the series ∞k=1 ak, that this series is convergent. Note. Note that there is no claim in the statement of this theorem that the two series have the same sum, just that the convergence of one implies the convergence of the other. P For theoretical reasons it is important to know when the series of absolute values converges. Such series are “more” than convergent. They are convergent in a way that allows more manipulations than would otherwise be available. They can be thought of as more robust; a series that converges, but whose absolute series does not converge is in some ways fragile. This leads to the following definitions. P ∞k=1 | ak| Definition 3.15: A series verges. ∞k=1 ak is said to be absolutely convergent if the related series P ∞k=1 | ak| con- P Definition 3.16: A series but the series ∞k=1 | ak| diverges. P ∞k=1 ak is said to be nonabsolutely convergent if the series ∞k=1 ak converges P Note that every absolutely convergent series is also convergent. We think of it as “more than convergent.” P Fortunately, the terminology preserves the meaning even though the “absolutely” refers to the absolute value, not to any other implied meaning. This play on words would not be available in all languages. Example 3.17: Using this terminology, applied to series we have already studied, we can now assert the following: Any geometric series 1 + r + r2 + r3 + . . . is absolutely convergent if r 1. | | ≥ < 1 and divergent if r | | and The alternating harmonic series + . . . is nonabsolutely convergent. ◭ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.5. Criteria for Convergence 137 Exercises 3.5.1 Suppose that converse hold? the converse hold? 3.5.3 Suppose that both series P P ∞ k=1 ak is a convergent series of positive terms. Show that ∞ k=1 a2 k is convergent. Does the 3.5.2 Suppose that ∞ k=1 ak is a convergent series of positive terms. Show that P ∞ k=1 √akak+1 is convergent. Does P ∞ ∞ ak and bk Xk=1 are absolutely convergent. Show that then so too is the series Xk=1 ∞ k=1 akbk. Does the converse hold? 3.5.4 Suppose that both series ∞ ak and ∞ P bk are nonabsolutely convergent. Show that it does not follow that the series Xk=1 Xk=1 ∞ k=1 akbk is convergent. 3.5.5 Alter the harmonic series ∞ k=1 1/k by deleting all terms in which the denominator contains a specified digit P (say 3). Show that the new series converges. See Note 52 P 3.5.6 Show that the geometric series criterion. P ∞ n=1 rn is convergent for r | | < 1 by using directly the Cauchy convergence 3.5.7 Show that the harmonic series is divergent by using directly the Cauchy convergence criterion. 3.5.8 Obtain a proof that every series using the Cauchy criterion. See Note 53 P ∞ k=1 ak for which ∞ k=1 | ak| P converges must itself be convergent without 3.5.9 Show that a series ∞ k=1 ak is absolutely convergent if and only if two at least of the series P ∞ ∞ ∞ ak , [ak]+ and [ak]− Xk=1 Xk=1 Xk=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 138 Infinite Sums Chapter 3 converge. (If two converge, then all three converge.) 3.5.10 The sum rule for convergent series ∞ ∞ ∞ (ak + bk) = ak + bk can be expressed by saying that if any two of these series converges so too does the third. What kind of statements can you make for absolute convergence and for nonabsolute convergence? Xk=1 Xk=1 Xk=1 3.5.11 Show that a series xn} 3.5.12 A sequence { ∞ k=1 ak is absolutely convergent if and only if every subseries ∞ k=1 ank converges. P of real numbers is said to be of bounded variation if the series ∞ P converges. xk − | xk−1| Xk=2 (a) Show that every sequence of bounded variation is convergent. (b) Show that not every convergent sequence is of bounded variation. (c) Show that all monotonic convergent sequences are of bounded variation. (d) Show that any linear combination of two sequences of bounded variation is of bounded variation. (e) Is the product of of two sequences of bounded variation also of bounded variation? 3.5.13 Establish |
the Cauchy-Schwarz inequality: For any finite sequences the inequality must hold. a1, a2, . . . , an} { and b1, b2, . . . , bn} { n (ak)2 ≤ Xk=1 1 2 n ! (bk)2 Xk=1 1 2 ! n Xk=1 akbk ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 139 3.5.14 Using the Cauchy-Schwarz inequality (Exercise 3.5.13), show that if ∞ n=1 an converges, then the series an} { is a sequence of nonnegative numbers for which P also converges for any p > 1 convergence? √an np ∞ n=0 X 2 . Without the Cauchy-Schwarz inequality what is the best you can prove for 3.5.15 Suppose that ∞ n=1 a2 n converges. Show that P lim sup n→∞ a1 + √2a2 + √3a3 + √4a4 + n + √nan · · · < . ∞ See Note 54 3.5.16 Let x1, x2, x3 be a sequence of positive numbers and write x1 + x2 + x3 + and If sn → S and tn → See Note 55 T , show that ST tn = 1. ≥ sn = + xn · · · + · · · + 1 xn . 1 x1 + 1 x2 n + 1 x3 n 3.6 Tests for Convergence In many investigations and applications of series it is important to recognize that a given series converges, converges absolutely, or diverges. Frequently the sum of the series is not of much interest, just the convergence behavior. Over the years a battery of tests have been developed to make this task easier. There are only a few basic principles that we can use to check convergence or divergence and we have already discussed these in Section 3.5. One of the most basic is that a series of nonnegative terms is ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 140 Infinite Sums Chapter 3 convergent if and only if the sequence of partial sums is bounded. Most of the tests in the sequel are just clever ways of checking that the partial sums are bounded without having to do the computations involved in finding that upper bound. 3.6.1 Trivial Test The first test is just an observation that we have already made about series: If a series then ak → over a series such as ∞k=1 ak converges, 0. We turn this into a divergence test. For example, some novices will worry for a long time P 1 k√k ∞ Xk=1 applying a battery of tests to it to determine convergence. The simplest way to see that this series diverges is to note that the terms tend to 1 as k . Perhaps this is the first thing that should be considered for any series. If the terms do not get small there is no point puzzling whether the series converges. It does not. → ∞ 3.18 (Trivial Test) If the terms of the series ∞k=1 ak do not converge to 0, then the series diverges. Proof. We have already proved this, but let us prove it now as a special case of the Cauchy criterion. For all ε > 0 there is an N so that P n for all n N and so, by definition, ak → ≥ 0. 3.6.2 Direct Comparison Tests = an| | < ε ak Xk=n A series ∞k=1 ak with all terms nonnegative can be handled by estimating the size of the partial sums. Rather than making a direct estimate it is sometimes easier to find a bigger series that converges. This larger series provides an upper bound for our series without the need to compute one ourselves. P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 141 Note. Make sure to apply these tests only for series with nonnegative terms since, for arbitrary series, this information is useless. 3.19 (Direct Comparison Test I) Suppose that the terms of the series are each smaller than the corresponding terms of the series Xk=1 ∞ ak that is, that 0 for all k. If the larger series converges, then so does the smaller series. ∞ bk, Xk=1 ak ≤ bk ≤ Proof. If 0 ak ≤ ≤ bk for all k, then n n Thus the number B = It follows that ∞k=1 ak must converge. P ak ≤ bk ≤ ∞ bk. Xk=1 ∞k=1 bk is an upper bound for the sequence of partial sums of the series Xk=1 Xk=1 ∞k=1 ak. P P Note. In applying this and subsequent tests that demand that all terms of a series satisfy some requirement, we should remember that convergence and divergence of a series ∞k=1 ak depends only on the behavior of ak for large values of k. Thus this test (and many others) could be reformulated so as to apply only for k greater than some integer N . P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 142 Infinite Sums Chapter 3 3.20 (Direct Comparison Test II) Suppose that the terms of the series are each larger than the corresponding terms of the series Xk=1 ∞ ak that is, that 0 for all k. If the smaller series diverges, then so does the larger series. ∞ ck, Xk=1 ck ≤ ≤ ak Proof. This follows from Test 3.19 since if the larger series did not diverge, then it must converge and so too must the smaller series. Here are two examples illustrating how these tests may be used. Example 3.21: Consider the series ∞ k + 5 k3 + k2 + k + 1 . Xk=1 While the partial sums might seem hard to estimate at first, a fast glance suggests that the terms (crudely) are similar to 1/k2 for large values of k and we know that the series ∞k=1 1/k2 converges. Note that k + 5 k3 + k2 + k + 1 = 1 + 5/k P k2(1 + 1/k + 1/k2 + 1/k3) ≤ C k2 for some choice of C (e.g., C = 6 will work). We now claim that our given series converges by a direct comparison with the convergent series ∞k=1 C/k2. (This is a p-harmonic series with p = 2.) ◭ P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 143 Example 3.22: Consider the series ∞ k + 5 k2 + k + 1 . Xk=1 r Again, a fast glance suggests that the terms (crudely) are similar to 1/√k for large values of k and we know that the series ∞k=1 1/√k diverges. Note that P k + 5 k2 + k + 1 = 1 + 5/k k(1 + 1/k + 1/k2) ≥ C k for some choice of C (e.g., C = 1 comparison with the divergent series 4 will work). We now claim that our given series diverges by a direct ∞k=1 √C/√k. (This is a p-harmonic series with p = 1/2.) ◭ The examples show both advantages and disadvantages to the method. We must invent the series that is to be compared and we must do some amount of inequality work to show that comparison. The next tests replace the inequality work with a limit operation, which is occasionally easier to perform. P 3.6.3 Limit Comparison Tests ∞k=1 ak with all terms nonnegative can be handled by comparing with a larger We have seen that a series convergent series or a smaller divergent series. Rather than check all the terms of the two series being compared, it is convenient sometimes to have this checked automatically by the computation of a limit. In this section, since the tests involve a fraction, we must be sure not only that all terms are nonnegative, but also that we have not divided by zero. P 3.23 (Limit Comparison Test I) Let each ak ≥ be compared to the terms of the series ak bk P lim k →∞ ∞k=1 bk by computing < ∞ 0 and bk > 0. If the terms of the series and if the latter series converges, then so does the former series. ∞k=1 ak can P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 144 Infinite Sums Chapter 3 Proof. The proof is easy. If the stated limit exists and is finite then there are numbers M and N so that for all k test, we find that the series N . This shows that ak ≤ ∞k=N ak converges by comparison with the convergent series N . Consequently, applying the direct comparison ∞k=N M bk. ≥ ≥ ak bk M bk for all k < M P 3.24 (Limit Comparison Test II) Let each ak > 0 and ck > 0. If the terms of the series be compared to the terms of the series ∞k=1 ck by computing P and if the latter series diverges, then so does the original series. P ak ck lim k →∞ > 0 Proof. Since the limit exists and is not zero there are numbers ε > 0 and N so that ak ck > ε ∞k=1 ak can P We repeat our two examples, Example 3.21 and 3.22, where we previously used the direct comparison ∞k=N ak diverges by comparison with the divergent for all k ≥ N . This shows that, for all k N , ≥ ak ≥ Consequently, by the direct comparison test the series series ∞k=N εck. εck. P P test to check for convergence. Example 3.25: We look again at the series k + 5 k3 + k2 + k + 1 , ∞ Xk=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 145 comparing it, as before, to the convergent series ∞k=1 1/k2. This now requires computing the limit k2(k + 5) P k3 + k2 + k + 1 , lim k →∞ which elementary calculus arguments show is 1. Since it is not infinite, the original series can now be claimed to converge by a limit comparison. ◭ Example 3.26: Again, consider the series ∞ k + 5 k2 + k + 1 Xk=1 r by comparing with the divergent series ∞k=1 1/√k. We are required to compute the limit which elementary calculus arguments show is 1. Since it is not zero, the original series can now be claimed ◭ to diverge by a limit comparison. P √k lim k →∞ r k + 5 k2 + k + 1 , 3.6.4 Ratio Comparison Test Again we wish to compare two series directly comparing the size of the terms we compare the ratios of the terms. The inspiration for this test P ∞k=1 bk is a geometric rests on attempts to compare directly a series with a convergent geometric series. If series with common ratio r, then evidently ∞k=1 bk composed of positive terms. Rather than ∞k=1 ak and P bk+1 bk This suggests that perhaps a comparison of ratios of successive terms would indicate how fast a series might be converging. = r. P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 146 Infinite Sums Chapter 3 3.27 (Ratio Comparison Test) If the ratios satisfy bk+1 bk for all k (or just for all k sufficiently large) and the series the series ak+1 ak ≤ ∞k=1 ak is also convergent. P ∞k=1 bk, with the larger ratio is convergent, then Proof. As usual, we assume all terms are positive in both series. If the ratios satisfy P for k > N , then they also satisfy ak+1 ak ≤ bk+1 bk ak+1 bk+1 ≤ ak bk , which means that the sequence above, say by C, and so ak/bk} { is decreasing for k > N . In particula |
r, that sequence is bounded Thus an application of the direct comparison test shows that the series ak ≤ Cbk. 3.6.5 d’Alembert’s Ratio Test ∞k=1 ak converges. P ∞k=1 rk for The ratio comparison test requires selecting a series for comparison. Often a geometric series some 0 < r < 1 may be used. How do we compute a number r that will work? We would wish to use bk = rk with a choice of r so that P bk+1 bk One useful and easy way to find whether there will be such an r is to compute the limit of the ratios. ak+1 ak ≤ rk+1 rk = r. = ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 147 3.28 (Ratio Test) If terms of the series ∞k=1 ak are all positive and the ratios satisfy then the series ∞k=1 ak is convergent. Proof. The proof is easy. If P then there is a number β < 1 so that P lim k →∞ ak+1 ak < 1 ak+1 ak lim k →∞ < 1, ak+1 ak < β for all sufficiently large k. Thus the series convergent geometric series ∞k=1 βk. P ∞k=1 ak converges by the ratio comparison test applied to the Note. The ratio test can also be pushed to give a divergence answer: If P ak+1 ak lim k →∞ > 1 (4) then the series answer as useful as the convergence test. From (4) it follows that there must be an N and β so that ∞k=1 ak is divergent. But it is best to downplay this test or you might think it gives an P for all k ≥ N . Then and ak+1 ak > β > 1 aN +1 > βaN , aN +2 > βaN +1 > β2aN , aN +3 > βaN +2 > β3aN . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 148 Infinite Sums Chapter 3 We see that the terms ak of the series are growing large at a geometric rate. Not only is the series diverging, but it is diverging in a dramatic way. We can summarize how this test is best applied. If terms of the series ∞k=1 ak are all positive, compute ak+1 ak lim k →∞ = L. P 1. If L < 1, then the series 2. If L > 1, then the series 3. If L = 1, then the series ∞k=1 ak is convergent. ∞k=1 ak is divergent; moreover, the terms ak → ∞ ∞k=1 ak may diverge or converge, the test being inconclusive. . P P Example 3.29: The series P ∞ (k!)2 (2k)! Xk=0 is particularly suited for an application of the ratio test since the ratio is easily computed and a limit taken: If we write ak = (k!)2/(2k)!, then ak+1 ak (2k)! (k!)2 = Consequently, this is a convergent series. More than that, it is converging faster than any geometric series ((k + 1)!)2 (2k + 2)! (2k + 2)(2k + 1) → (k + 1)2 1 4 = . Xk=0 for any positive ε. (To make this expression “converging faster” more precise, see Exercise 3.12.5.) ∞ k + ε 1 4 ◭ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 3.6.6 Cauchy’s Root Test 149 P There is yet another way to achieve a comparison with a convergent geometric series. We suspect that a ∞k=1 rk but do not know how to compute the series value of r that might work. The limiting values of the ratios P ∞k=1 ak can be compared to some geometric series ak+1 ak provide one way of determining what r might work but often are difficult to compute. Instead we recognize that a comparison of the form would mean that Crk ak ≤ k√ak ≤ For large k the term k√C is close to 1, and this motivates our next test, usually attributed to Cauchy. k√Cr. 3.30 (Root Test) If terms of the series ∞k=1 ak are all nonnegative and if the roots satisfy then that series converges. Proof. This is almost trivial. If for all k N , then ≥ P lim k →∞ k√ak < 1, (ak)1/k < β < 1 ak < βk and so ∞k=1 ak converges by direct comparison with the convergent geometric series ∞k=1 βk. Again we can summarize how this test is best applied. The conclusions are nearly identical with those P P for the ratio test. Compute (ak)1/k = L. lim k →∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 150 Infinite Sums Chapter 3 1. If L < 1, then the series 2. If L > 1, then the series 3. If L = 1, then the series ∞k=1 ak is convergent. ∞k=1 ak is divergent; moreover, the terms ak → ∞ ∞k=1 ak may diverge or converge, the test being inconclusive. . P P Example 3.31: In Example 3.29 we found the series P to be handled easily by the ratio test. It would be extremely unpleasant to attempt a direct computation using the root test. On the other hand, the series (k!)2 (2k)! ∞ Xk=0 for x > 0 can be handled by either of these tests. You should try the ratio test while we try the root test: Xk=0 ∞ kxk = x + 2x2 + 3x3 + 4x4 + . . . and so convergence can be claimed for all 0 < x < 1 and divergence for all x > 1. The case x = 1 is inconclusive for the root test, but the trivial test shows instantly that the series diverges for x = 1. ◭ lim k →∞ 1/k kxk = lim k →∞ k√kx = x 3.6.7 Cauchy’s Condensation Test " Enrichment section. May be omitted. Occasionally a method that is used to study a specific series can be generalized into a useful test. Recall that in studying the sequence of partial sums of the harmonic series it was convenient to watch only at the steps 1, 2, 4, 8, . . . and make a rough lower estimate. The reason this worked was simply that the terms in the harmonic series decrease and so estimates of s1, s2, s4, s8, . . . were easy to obtain using just that fact. This turns quickly into a general test. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 151 3.32 (Cauchy’s Condensation Test) If the terms of a series monotonically to zero, then that series converges if and only if the related series ∞k=1 ak are nonnegative and decrease P converges. 2ja2j ∞ Xj=1 Proof. series. Computing first the sum of 2p+1 Since all terms are nonnegative, we need only compare the size of the partial sums of the two 1 terms of the original series, we have − a1 + (a2 + a3) + · · · + (a2p + a2p+1 + + 2pa2p. a1 + 2a2 + · · · ≤ · · · + a2p+1 1) − And, with the inequality sign in the opposite direction, we compute the sum of 2p terms of the original series to obtain a1 + a2 + (a3 + a4) + + (a2p−1+1 + a2p−1+2 + + a2pa1 + 2a2 + + 2pa2p) . · · · If either series has a bounded sequence of partial sums so too then does the other series. Thus both converge or else both diverge. Example 3.33: Let us use this test to study the p-harmonic series: for p > 0. The terms decrease to zero and so the convergence of this series is equivalent to the convergence of the series 1 kp ∞ Xk=1 p ∞ 2j Xj=1 1 2j ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 152 Infinite Sums Chapter 3 and this series is a geometric series ∞ p 21 − j . This converges precisely when 21 1 or p exactly the convergence behavior of the p-harmonic series for all values of p. (For p just by the trivial test.) p < 1 or p > 1 and diverges when 21 ≥ − − p 1. Thus we know ≤ 0 we have divergence ◭ ≤ Xj=1 It is worth deriving a simple test from the Cauchy condensation test as a corollary. This is an improvement on the trivial test. The trivial test requires that limk ak = 0 for a convergent series ∞k=1 ak. This next test, which is due to Abel, shows that slightly more can be said if the terms form a must go to zero faster than 1/k . monotonic sequence. The sequence P ak} { →∞ } ∞k=1 ak decrease monotonically, then { Corollary 3.34: If the terms of a convergent series Proof. By the Cauchy condensation test we know that kak = 0. P lim k →∞ 2ja2j = 0. lim j →∞ kak ≤ 0 as required. 2 2ja2j , If 2j k 2j+1, then ak ≤ ≤ ≤ a2j and so which is small for large j. Thus kak → 3.6.8 Integral Test " Enrichment section. May be omitted. To determine the convergence of a series ∞k=1 ak of nonnegative terms it is often necessary to make some kind of estimate on the size of the sequence of partial sums. Most of our tests have done this P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 153 automatically, saving us the labor of computing such estimates. Sometimes those estimates can be obtained by calculus methods. The integral test allows us to estimate the partial sums instead between series and infinite integrals, which is of much importance in analysis. n 1 f (x) dx in certain circumstances. This is more than a convenience; it also shows a close relation R n k=1 f (k) by computing P 3.35 (Integral Test) Let f be a nonnegative decreasing function on [1, X 1 f (x) dx can be computed for all X > 1. If R X ) such that the integral ∞ exists, then the series P ∞k=1 f (k) converges. If then the series ∞k=1 f (k) diverges. lim X 1 →∞ Z f (x) dx < ∞ X lim X →∞ Z 1 f (x) dx = , ∞ Proof. Since the function f is decreasing we must have P Applying these inequalities for k = 2, 3, 4, . . . , n we obtain k+1 f (x) dx f (kx) dx. k Z n+1 n f (x) dx ≤ 1 Z Xk=1 f (k) ≤ n f (1) + f (x) dx. 1 Z (5) The series converges if and only if the partial sums are bounded. But we see from the inequalities (5) that if the limit of the integral is finite, then these partial sums are bounded. If the limit of the integral is infinite, then these partial sums are unbounded. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 154 Infinite Sums Chapter 3 Note. The convergence of the integral yields the convergence of the series. There is no claim that the sum of the series however, see Exercise 3.6.21. ∞k=1 f (k) and the value of the infinite integral ∞1 f (x) dx are the same. In this regard, R P Example 3.36: According to this test the harmonic series ∞k=1 1 k can be studied by computing lim X 1 →∞ Z For the same reasons the p-harmonic series X dx x = lim X →∞ P log X = . ∞ 1 kp ∞ Xk=1 1 for p > 1 can be studied by computing X dx xp = lim X p lim X →∞ − In both cases we obtain the same conclusion as before. The harmonic series diverges and, for p > 1, the ◭ p-harmonic series converges. − →∞ 3.6.9 Kummer’s Tests " Advanced section. May be omitted. The ratio test requires merely taking the limit of |
the ratios ak+1 ak but often fails. We know that if this tends to 1, then the ratio test says nothing about the convergence or divergence of the series ∞k=1 ak. Kummer’s tests provide a collection of ratio tests that can be designed by taking different choices of . The choices Dk = 1, Dk = k and Dk = k ln k are used in the following tests. Ernst Eduard sequence Dk} { P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 155 Kummer (1810–1893) is probably most famous for his contributions to the study of Fermat’s last theorem; his tests arose in his study of hypergeometric series. 3.37 (Kummer’s Tests) The series any sequence of positive numbers and compute ∞k=1 ak can be tested by the following criteria. Let Dk} { denote P ak ak+1 − ∞k=1 ak converges. On the other hand, if L = lim inf k →∞ Dk Dk+1 . If L > 0 the series P for all sufficiently large k and if the series Dk ak ak+1 − Dk+1 0 ≤ diverges, then the series ∞k=1 ak diverges. 1 Dk ∞ Xk=1 Proof. there must exist an integer N so that for all k P If L > 0, then we can choose a positive number α < L. By the definition of a liminf this means N , ≥ ak ak+1 − Dk+1 . Rewriting this, we find that α < Dk We can write this inequality for k = N, N + 1, N + 2, . . . , N + p to obtain αak+1 < Dkak − Dk+1ak+1. αaN +1 < DN aN − αaN +2 < DN +1aN +1 − DN +1aN +1 DN +2aN +2 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 156 Infinite Sums Chapter 3 and so on. Adding these (note the telescoping sums), we find that α(aN +1 + aN +2 + + aN +p+1) · · · < DN +1aN +1 − (The final inequality just uses the fact that all the terms here are positive.) From this inequality we can determine that the partial sums of the series DN +p+1aN +p+1 < DN +1aN +1. usual criterion, this proves that this series converges. ∞k=1 ak are bounded. By our The second part of the theorem requires us to establish divergence. Suppose now that P Dk ak ak+1 − Dk+1 ≤ 0 for all k ≥ N . Then Thus the sequence Dkak} { Dkak ≤ is increasing after k = N . In particular, Dk+1ak+1. N and so Dkak ≥ C for some C and all k ≥ C Dk It follows by a direct comparison with the divergent series ak ≥ . C/Dk that our series also diverges. Note. In practice, for the divergence part of the test, it may be easier to compute P If L < 0, then we would know that L = lim sup k →∞ Dk ak ak+1 − Dk+1 . Dk ak ak+1 − Dk+1 0 ≤ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 157 for all sufficiently large k and so, if the series ∞k=1 1 Dk diverges, then the series ∞k=1 ak diverges. Example 3.38: What is Kummer’s test if the sequence used is the simplest possible Dk = 1 for all k? In this case it is simply the ratio test. For example, suppose that P P Then, replacing Dk = 1, we have ak+1 ak lim k →∞ = r. ak ak+1 − lim k →∞ Dk Dk+1 1 1 < 0 we have divergence while if 1/r = lim k →∞ ak ak+1 − Thus, by Kummer’s test, if 1/r are just the cases r > 1 and r < 1 of the ratio test. − = 1 r − 1. 1 > 0 we have convergence. These ◭ − 3.6.10 Raabe’s Ratio Test A simple variant on the ratio test is known as Raabe’s test. Suppose that " Adv. so that the ratio test is inconclusive. Then instead compute ak ak+1 lim k →∞ = 1 k lim k →∞ ak ak+1 − 1 . The series ∞k=1 ak converges or diverges depending on whether this limit is greater than or less than 1. P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 158 Infinite Sums Chapter 3 3.39 (Raabe’s Test) The series ∞k=1 ak can be tested by the following criterion. Compute Then P L = lim k →∞ k ak ak+1 − 1 . 1. If L > 1, the series ∞k=1 ak converges. 2. If L < 1, the series P ∞k=1 ak diverges. 3. If L = 1, the test is inconclusive. P Proof. This is Kummer’s test using the sequence Dk = k. Example 3.40: Consider the series ∞ kk ekk! . Xk=0 An attempt to apply the ratio test to this series will fail since the ratio will tend to 1, the inconclusive case. But if instead we consider the limit as called for in Raabe’s test, we can use calculus methods (L’Hˆopital’s rule) to obtain a limit of 1 2 . Consequently, this series diverges. ◭ k lim k →∞ kk ekk! ek+1(k + 1)! (k + 1)k+1 − 1 3.6.11 Gauss’s Ratio Test " Advanced section. May be omitted. Raabe’s test can be replaced by a closely related test due to Gauss. We might have discovered while ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence using Raabe’s test that 159 = L. k lim k →∞ ak ak+1 − 1 This suggests that in any actual computation we will have discovered, perhaps by division, that ak ak+1 = 1 + L k + terms involving 1 k2 etc. The case L > 1 corresponds to convergence and the case L < 1 to divergence, both by Raabe’s test. What if L = 1, which is considered inconclusive in Raabe’s test? Gauss’s test offers a different way to look at Raabe’s test and also has an added advantage that it handles this case that was left as inconclusive in Raabe’s test. 3.41 (Gauss’s Test) The series ∞k=1 ak can be tested by the following criterion. Suppose that where φ(k) (k = 1, 2, 3, . . . ) forms a bounded sequence. Then P ak ak+1 = 1 + L k + φ(k) k2 1. If L > 1 the series ∞k=1 ak converges. 2. If L ≤ 1 the series P ∞k=1 ak diverges. Proof. As we noted, for L > 1 and L < 1 this is precisely Raabe’s test. Only the case L = 1 is new! Let us assume that P ak ak+1 = 1 + + 1 k xk k2 where xk} { is a bounded sequence. To prove this case (that the series diverges) we shall use Kummer’s test with the sequence Dk = k log k. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 160 We consider the expression which now assumes the form Dk ak ak+1 − Dk+1 , k log k ak ak+1 − (k + 1) log(k + 1) Infinite Sums Chapter 3 We need to compute the limit of this expression as k should try) to see that the limit is 1. For this use the facts that → ∞ . It takes only a few manipulations (which you = k log k 1 + (k + 1) log(k + 1). 1 k + xk k2 − − (log k)/k 0 → (k + 1) log(1 + 1/k) 1 → and as k . → ∞ We are now in a position to claim, by Kummer’s test, that our series part of the test requires us to check that the series ∞ ∞k=1 ak diverges. To apply this P 1 k log k Xk=2 diverges. Several tests would work for this. Perhaps Cauchy’s condensation test is the easiest to apply, although the integral test can be used too [see Exercise 3.6.2(c)]. Note. In Gauss’s test you may be puzzling over how to obtain the expression ak ak+1 = 1 + L k + φ(k) k2 . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 161 In practice often the fraction ak/ak+1 is a ratio of polynomials and so usual algebraic procedures will supply this. In theory, though, there is no problem. For any L we could simply write Thus the real trick is whether it can be done in such a way that the φ(k) do not grow too large. φ(k) = k2 ak ak+1 − 1 + L k . Also, in some computations you might prefer to leave the ratio as ak+1/ak the way it was for the ratio test. In that case Gauss’s test would assume the form ak+1 ak = 1 L k + φ(k) k2 . − (Note the minus sign.) The conclusions are exactly the same. Example 3.42: The series 1 + mx + x2+ − 1) m(m 2! 1) . . . (m k! − m(m − 1)(m 3! 2) − x3 + m(m k + 1) − xk + . . . is called the binomial series. When m is a positive integer all terms for k > m are zero and the series reduces to the binomial formula for (1 + x)m. Here now m is any real number and the hope remains that the formula might still be valid, but using a series rather than a finite sum. This series plays an important role in many applications. Let us check for absolute convergence at x = 1. We can assume that m since that case is trivial. = 0 If we call the absolute value of the k + 1–st term ak so then a simple calculation shows that for large values of k ak+1 = m(m − 1) . . . (m k! − k + 1) ak+1 ak = 1 m + 1 k . − , ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 162 Infinite Sums Chapter 3 Here we are using the version ak+1/ak rather than the reciprocal; see the preceding note. There are no higher-order terms to worry about in Gauss’s test here and so the series ak converges if m + 1 > 1 and diverges if m + 1 < 1. Thus the binomial series converges absolutely for x = 1 if m > 0. For m = 0 the series certainly converges since all terms except for the first one are identically zero. For m < 0 we know so far only that it does not converge absolutely. A closer analysis, for those who might care to try, will show that the series is nonabsolutely convergent for 1 < m < 0 and divergent for m P 1. ◭ − ≤ − 3.6.12 Alternating Series Test We pass now to a number of tests that are needed for studying series of terms that may change signs. The ∞i=1 ai, where the ai are both negative and positive, is to apply one simplest first step in studying a series from our battery of tests to the series . If any test shows that this converges, then we know that ∞i=1 | our original series converges absolutely. This is even better than knowing it converges. ai| But what shall we do if the series is not absolutely convergent or if such attempts fail? One method P P applies to special series of positive and negative terms. Recall how we handled the series ∞ ( 1) − + . . . Xk=1 (called the alternating harmonic series). We considered separately the partial sums s2, s4, s6, . . . and s1, s3, s5, . . . . The special pattern of + and signs alternating one after the other allowed us to see that each subsequence and that applies to a wide class of series, similar to the alternating harmonic series. was monotonic. All the features of this argument can be put into a test s2n} 1} − s2n − { { ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Conv |
ergence 163 3.43 (Alternating Series Test) The series ∞ ( 1)k − 1ak, − Xk=1 whose terms alternate in sign, converges if the sequence decreases monotonically to zero. Moreover, the value of the sum of such a series lies between the values of the partial sums at any two consecutive stages. ak} { The proof is just exactly the same as for the alternating harmonic series. Since the ak are Proof. nonnegative and decrease, we compute that a2 = s2 ≤ a1 − s3 ≤ These subsequences then form bounded monotonic sequences and so s6 ≤ · · · ≤ s5 ≤ s4 ≤ s1 = a1. exist. Finally, since lim n →∞ s2n and lim →∞ n s2n 1 − sn = L exists. From the proof it is clear that the value L lies in each of the we can conclude that limn intervals [s2, s1], [s2, s3], [s4, s3], [s4, s5], . . . and so, as stated, the sum of the series lies between the values of the partial sums at any two consecutive stages. →∞ s2n − s2n − 1 = a2n → − 0 3.6.13 Dirichlet’s Test " Advanced section. May be omitted. Our next test derives from the summation by parts formula n Xk=1 akbk = s1(b1 − b2) + s2(b2 − b3) + + sn 1(bn − 1 − − · · · bn) + snbn ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 164 Infinite Sums Chapter 3 that we discussed in Section 3.2. We can see that if there is some special information available about the n sequences k=1 akbk can be proved. The test gives here, then the convergence of the series one possibility for this. The next section gives a different variant. sn} bn} and { { The test is named after Lejeune Dirichlet1 (1805–1859) who is most famous for his work on Fourier P series, in which this test plays an important role. bn} 3.44 (Dirichlet Test) If { are bounded, then the series ∞k=1 akbk converges. is a sequence decreasing to zero and the partial sums of the series ∞k=1 ak P n k=1 ak. By our assumptions on the series P ∞k=1 ak there is a positive number M so Proof. Write sn = sn| ≤ that | P M for all n. Let ε > 0 and choose N so large that bn < ε/(2M ) if n The summation by parts formula shows that for akbk = anbn + an+1bn+1 + | + ambm| · · · = Xk=n 1bn + sn(bn − sn sn(bn − 1bn| bn+1) | M (bn + [bn − Notice that we have needed to use the fact that |− sn ≤ |− + ≤ − − bn+1) + + sm · · · + + | · · · | bm] + bm) − sm 1(bm − 1(bm 1 − 1 − − − 2M bn < ε. ≤ bm) + smbm| smbm| + bm) | | bk 1 − − bk ≥ 0 for each k. This is precisely the Cauchy criterion for the series convergence. ∞k=1 akbk and so we have proved P 1 One of his contemporaries described him thus: “He is a rather tall, lanky-looking man, with moustache and beard about to turn grey with a somewhat harsh voice and rather deaf. He was unwashed, with his cup of coffee and cigar. One of his failings is forgetting time, he pulls his watch out, finds it past three, and runs out without even finishing the sentence.” (From http://www-history.mcs.st-and.ac.uk/history. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence Example 3.45: The series + . . . converges by the alternating series test. What other pattern of + and have convergence? Let ak = ± 1. If the partial sums n 165 signs could we insert and still − remain bounded, then, by Dirichlet’s test, the series n Xk=1 must converge. Thus, for example, the pattern Xk=1 ak ak k − − − would produce a convergent series (that is not alternating). − − − + + + + + + + − + + − − . . . ◭ 3.6.14 Abel’s Test " Advanced section. May be omitted. The next test is another variant on the same theme as the Dirichlet test. There the series was proved to be convergent by assuming a fairly weak fact for the series sums) and a strong fact for the second. ∞k=1 akbk ∞k=1 ak (i.e., bounded partial (i.e., monotone convergence to 0). Here we strengthen the first and weaken bk} P P { 3.46 (Abel Test) If the series bn} { ∞k=1 akbk converges. is a convergent monotone sequence and the series ∞k=1 ak is convergent, then P Proof. P Suppose first that bk is decreasing to a limit B. Then bk − B decreases to zero. We can apply ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 166 Infinite Sums Chapter 3 Dirichlet’s test to the series ∞ ak(bk − B) Xk=1 to obtain convergence, since if ∞k=1 ak is convergent, then it has a bounded sequence of partial sums. But this allows us to express our series as the sum of two convergent series: P ∞ akbk = ∞ Xk=1 Xk=1 ak(bk − B) + B ∞ ak. Xk=1 If the sequence bk is instead increasing to some limit then we can apply the first case proved to the an} be a sequence of positive numbers. If limn→∞ n2an = 0, what (if anything) can be said about the ∞ ∞ n=1 an. (If we drop n=1 an. If limn→∞ nan = 0, what (if anything) can be said about the series { series the assumption about the sequence being positive does anything change?) P 3.6.2 Which of these series converge? P an} { series ∞k=1 ak( bk). − − P Exercises 3.6.1 Let (a) (b) (c) (d) ∞ n=1 X ∞ n=1 X ∞ n=2 X ∞ n=1 X n(n + 1) (n + 2)2 3n(n + 1)(n + 2) n3√n 1 ns log n a1/n 1 − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 167 1 n(log n)t 1 ns(log n)t ∞ n=2 X ∞ n=2 X ∞ (e) (f) (g) n2 1 n 1 − n=1 X 3.6.3 For what values of x do the following series converge? (a) (b) P xn ∞ n=2 log n ∞ n=2(log n)xn ∞ n=1 e−nx P (c) (d) 1 + 2x + 32x2 P 2! + 43x3 3! + . . . . See Note 56 3.6.4 Let ak be a sequence of positive numbers and suppose that lim k→∞ kak = L. What can you say about the convergence of the series ∞ k=1 ak if L = 0? What can you say if L > 0? 3.6.5 " Let ak} { (a) lim sup k→∞ √kak > 0 be a sequence of nonnegative numbers. Consider the following conditions: P (b) lim sup k→∞ √kak < ∞ (c) lim inf k→∞ √kak > 0 (d) lim inf k→∞ √kak < ∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 168 Which condition(s) imply convergence or divergence of the series are inconclusive as to convergence or divergence? Supply examples. P See Note 57 Infinite Sums Chapter 3 ∞ k=1 ak? Supply proofs. Which conditions 3.6.6 Suppose that ∞ n=1 √an also be convergent? 3.6.7 Give examples of series both convergent and divergent that illustrate that the ratio test is inconclusive when ∞ n=1 an is a convergent series of positive terms. Must the series P P the limit of the ratios L is equal to 1. 3.6.8 Give examples of series both convergent and divergent that illustrate that the root test is inconclusive when the limit of the roots L is equal to 1. 3.6.9 " Apply both the root test and the ratio test to the series where α, β are positive real numbers. α + αβ + α2β + α2β2 + α3β2 + α3β3 + . . . 3.6.10 " Show that the limit comparison test applied to series with positive terms can be replaced by the following version. If and if ∞ k=1 bk converges, then so does P lim sup k→∞ ∞ k=1 ak. If P lim inf k→∞ ak bk ak ck < ∞ > 0 and if ∞ k=1 ck diverges, then so does ∞ k=1 ak. 3.6.11 " Show that the ratio test can be replaced by the following version. Compute P P lim inf k→∞ ak+1 ak = L and lim sup k→∞ ak+1 ak = M. (a) If M < 1, then the series (b) If L > 1, then the series (c) If L 1 ≤ ≤ M , then the series P ∞ k=1 ak is convergent. ∞ k=1 ak is divergent; moreover, the terms ak → ∞ P . ∞ k=1 ak may diverge or converge, the test being inconclusive. P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.6. Tests for Convergence 169 3.6.12 " Show that the root test can be replaced by the following version. Compute lim sup k→∞ k√ak = L. (a) If L < 1, then the series (b) If L > 1, then the series (c) If L = 1, then the series ∞ k=1 ak is convergent. ∞ k=1 ak is divergent; moreover, some subsequence of the terms akj → ∞ ∞ k=1 ak may diverge or converge, the test being inconclusive. . P P 3.6.13 " Show that for any sequence of positive numbers P lim inf k→∞ ak+1 ak ≤ lim inf k→∞ lim sup k→∞ k√ak ≤ lim sup k→∞ ak+1 ak . ak} { k√ak ≤ What can you conclude about the relative effectiveness of the root and ratio tests? 3.6.14 " Give examples of series for which one would clearly prefer to apply the root (ratio) test in preference to the ratio (root) test. How would you answer someone who claims that “Exercise 3.6.13 shows clearly that the ratio test is inferior and should be abandoned in favor of the root test?” 3.6.15 " Let an} { be a sequence of positive numbers and write Show that if lim inf Ln > 1, then diverges. P Ln = log 1 an log n . an converges. Show that if Ln ≤ 1 for all sufficiently large n, then an P 3.6.16 Apply the test in Exercise 3.6.15 to obtain convergence or divergence of the following series (x is positive): (a) (b) (c) ∞ n=2 xlog n ∞ n=2 xlog log n ∞ n=2(log n)− log n P P 3.6.17 Prove the alternating series test directly from the Cauchy criterion. P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 170 Infinite Sums Chapter 3 3.6.18 Determine for what values of p the series 1 4p + . . . is absolutely convergent and for what values it is nonabsolutely convergent. ( − Xk=1 1)k−1 1 kp = 1 1 3p − 1 2p + − ∞ 3.6.19 How many terms of the series ∞ ( − 1)k−1 k2 Xk=1 must be taken to obtain a value differing from the sum of the series by less than 10−10? 3.6.20 If the sequence xn} { converges. is monotonically decreasing to zero then prove that the series x1 − 1 2 (x1 + x2) + 1 3 (x1 + x2 + x3) 1 4 − (x1 + x2 + x3 + x4) + . . . 3.6.21 This exercise attempts to squeeze a little more information out of the integral test. In the notation of that test consider the sequence n n+1 Show that the sequence is increasing and that 0 ∞ k=1 f (k) and en} ∞ 1 f (x) dx? { en ≤ ≤ f (1). What is the exact relation between en = f (k) Xk=1 f (x) dx − 1 Z 3.6.22 Show that P R for some number γ, .5 < γ < 1. See Note 58 3.6.23 Show that n+1 n lim n→∞ Xk= dx ! lim n→∞ 2n Xk=n+1 1 k = log 2. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Secti |
on 3.6. Tests for Convergence 171 3.6.24 Let F be a positive function on [1, ) with a positive, decreasing and continuous derivative F ′. ∞ ∞ k=1 F ′(k) converges if and only if (a) Show that P converges. F ′(k) F (k) ∞ Xk=1 (b) Suppose that ∞ k=1 F ′(k) diverges. Show that P converges if and only if p > 1. See Note 59 F ′(k) [F (k)]p ∞ Xk=1 3.6.25 This collection of exercises develops some convergence properties of power series; that is, series of the form A full treatment of power series appears in Chapter 10. Xk=0 ∞ akxk = a0 + a1x + a2x2 + a3x3 + . . . . (a) Show that if a power series converges absolutely for some value x = x0 then the series converges absolutely for all x | . x0| | ≤ | (b) Show that if a power series converges for some value x = x0 then the series converges absolutely for all < x | | (c) Let . x0| | ∞ R = sup t : ( aktk converges . ) Xk=0 ∞ k=0 akxk must converge absolutely for all Show that the power series > R. [The number R is called the radius of convergence of the series. The explanation for the word x | “radius” (which conjures up images of circles) is that for complex series the set of convergence is a disk.] < R and diverge for all x | P | | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 172 Infinite Sums Chapter 3 (d) Give examples of power series with radius of convergence 0, , 1, 2, and √2. ∞ (e) Explain how the radius of convergence of a power series may be computed with the help of the ratio test. (f) Explain how the radius of convergence of a power series may be computed with the help of the root test. (g) " " Establish the formula for the radius of convergence of the power series R = 1 lim supk→∞ k ak| | ∞ k=0 akxk. p (h) Give examples of power series ∞ k=0 akxk with radius of convergence R so that the series converges P absolutely at both endpoints of the interval [ at the right-hand endpoint but diverges at the left-hand endpoint of [ are there? R, R]. Give another example so that the series converges R, R]. What other possibilities P − − 3.6.26 The series 1 + mx + m(m − 1)(m 3! 2) − x3 + m(m k + 1) − xk + . . . x2+ − 1) m(m 2! 1) . . . (m k! − is called the binomial series. Here m is any real number. (See Example 3.42.) (a) Show that if m is a positive integer then this is precisely the expansion of (1 + x)m by the binomial theorem. (b) Show that this series converges absolutely for any m and for all (c) Obtain convergence for x = 1 if m > 1. − 1 if m > 0. (d) Obtain convergence for x = − 3.7 Rearrangements " Enrichment section. May be omitted. < 1. x | | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.7. Rearrangements 173 Any finite sum may be rearranged and summed in any order. This is because addition is commutative. We might expect the same to occur for series. We add up a series ∞k=1 ak by starting at the first term and adding in the order presented to us. If the terms are rearranged into a different order do we get the same result? P Example 3.47: The most famous example of a series that cannot be freely rearranged without changing the sum is the alternating harmonic series. We know that the series 1 1 2 is convergent (actually nonabsolutely convergent) with a sum somewhere between 1/2 and 1. rearrange this so that every positive term is followed by two negative terms, thus, 1 4 1 2 − we shall arrive at a different sum. Grouping these and adding, we obtain 1 10 − 12 + . . . + . . . 10 − 1 12 + . . . whose sum is half the original series. Rearranging the series has changed the sum! If we ◭ For the theory of unordered sums there is no such problem. If an unordered sum a number c, then so too does any rearrangement. Exercise 3.3.8 shows that if σ : I onto, then aj = aσ(i). I Xi ∈ I Xi ∈ J aj converges to j I is one-to-one and ∈ → P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 174 Infinite Sums Chapter 3 We had hoped for the same situation for series. If σ : IN IN is one-to-one and onto, then → ∞ ak = ∞ aσ(k) may or may not hold. We call Xk=1 Xk=1 ∞k=1 aσ(k) a rearrangement of the series ∞k=1 ak. We propose now to characterize those series that allow unlimited rearrangements, and those that are more fragile (as is the alternating harmonic series) and cannot permit rearrangement. P P 3.7.1 Unconditional Convergence A series is said to be unconditionally convergent if all rearrangements of that series converge and have the same sum. Those series that do not allow this but do converge are called conditionally convergent. Here the “conditional” means that the series converges in the arrangement given, but may diverge in another arrangement or may converge to a different sum in another arrangement. We shall see that conditionally convergent series are extremely fragile; there are rearrangements that exhibit any behavior desired. There are rearrangements that diverge and there are rearrangements that converge to any desired number. Our first theorem asserts that any absolutely convergent series may be freely rearranged. All absolutely convergent series are unconditionally convergent. In fact, the two terms are equivalent although we must wait until the next section to prove that. unconditionally convergent absolutely convergent ⇔ Theorem 3.48 (Dirichlet) Every absolutely convergent series is unconditionally convergent. Proof. convergence and absolute convergence mean the same thing. Let us prove this first for series ∞k=1 ak whose terms are all nonnegative. For such series P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.7. Rearrangements 175 Let ∞k=1 aσ(k) be any rearrangement. Then for any M P by choosing an N large enough so that M N ∞ ak aσ(k) ≤ ak ≤ Xk=1 Xk=1 1, 2, 3, . . . , N { } σ(1), σ(2), σ(3), . . . , σ(M ) Xk=1 { . } includes all the integers By the bounded partial sums criterion this shows that ∞k=1 ak. But this same argument would show that ∞k=1 aσ(k) and consequently all rearrangements converge to the same sum. We now allow the series P ∞k=1 ak to have positive and negative values. Write ∞k=1 aσ(k) is convergent and to a sum smaller than ∞k=1 ak is convergent and to a sum smaller than P P P P ∞ ak = ∞ [ak]+ Xk=1 (cf. Exercise 3.5.8) where we are using the notation Xk=1 [ak]− − ∞ Xk=1 [X]+ = max X, 0 { } and [X]− = max X, 0 {− } and remembering that X = [X]+ [X]− and − X | | = [X]+ + [X]−. Any rearrangement of the series on the left-hand side of this identity just results in a rearrangement in the two series of nonnegative terms on the right. We have just seen that this does nothing to alter the convergence or the sum. Consequently, any rearrangement of our series will have the same sum as required to prove the assertion of the theorem. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 176 Infinite Sums Chapter 3 3.7.2 Conditional Convergence A convergent series is said to be conditionally convergent if it is not unconditionally convergent. Thus such a series converges in the arrangement given, but either there is some rearrangement that diverges or else there is some rearrangement that has a different sum. In fact, both situations always occur. We have already seen (Example 3.47) how the alternating harmonic series can be rearranged to have a different sum. We shall show that any nonabsolutely convergent series has this property. Our previous rearrangement took advantage of the special nature of the series; here our proof must be completely general and so the method is different. The following theorem completes Theorem 3.48 and provides the connections: and conditionally convergent nonabsolutely convergent ⇔ unconditionally convergent absolutely convergent ⇔ Note. You may wonder why we have needed this extra terminology if these concepts are identical. One reason is to emphasize that this is part of the theory. Conditional convergence and nonabsolutely convergence may be equivalent, but they have different underlying meanings. Also, this terminology is used for series of other objects than real numbers and for series of this more general type the terms are not equivalent. Theorem 3.49 (Riemann) Every nonabsolutely convergent series is conditionally convergent. In fact, every nonabsolutely convergent series has a divergent rearrangement and can also be rearranged to sum to any preassigned value. Proof. enough if we observe that both series Let P ∞k=1 ak be an arbitrary nonabsolutely convergent series. To prove the first statement it is ∞ [ak]+ and ∞ [ak]− Xk=1 Xk=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.7. Rearrangements 177 must diverge in order for since the series is assumed to be convergent. ∞k=1 ak to be nonabsolutely convergent. We need to observe as well that ak → (skipping any zero or negative Write p1, p2, p3, for the sequence of positive numbers in the sequence P 0 ones) and write q1, q2, q3, . . . for the sequence of terms that we have skipped. We construct a new series ak} { · · · where we have chosen 0 = n0 < n1 < n2 < n3 < . . . so that p1 + p2 + + pn1 + q1 + pn1+1 + pn1+2 + + pn2 + q2 + pn2+1 + . . . · · · pnk+1 + pnk+2 + + pnk+1 > 2k · · · for each k = 0, 1, 2, . . . . Since the terms of our original series and so is a rearrangement. Since the terms qk → with the goal of producing ever larger partial sums for the new series and so, evidently, this new series diverges to ∞k=1 pk diverges, this is possible. The new series so constructed contains all 0, they will not interfere P . The second requirement of the theorem is to produce a convergent rearrangement, convergent to a given number α. We proceed in much the same way but with rather more caution. We leave this to the exercises. ∞ 3.7.3 Comparison of " Advanced section. May be omitted. ∞i=1 ai and IN ai i ∈ The unordered sum of a sequence of real numbers, written as, P P has an apparent c |
onnection with the ordered sum ai, IN Xi ∈ ∞ ai. Xi=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 178 Infinite Sums Chapter 3 We should expect the two to be the same when both converge, but is it possible that one converges and not the other? The answer is that the convergence of IN ai is equivalent to the absolute convergence of ∞i=1 ai. i ∈ Theorem 3.50: A necessary and sufficient condition for absolutely convergent and in this case P ∞ ai = P IN ai to converge is that the series ∞i=1 ai is P i ∈ P ai. IN Xi ∈ Proof. We shall use a device we have seen before a few times: For any real number X write Xi=1 [X]+ = max X, 0 { } and [X]− = max X, 0 {− } and note that X = [X]+ [X]− and − X | | = [X]+ + [X]−. The absolute convergence of the series and the convergence of the sum in the statement in the theorem now reduce to considering the equality of the right-hand sides of and ai = [ai]+ IN Xi ∈ IN Xi ∈ [ai]− − IN Xi ∈ This reduces our problem to considering just nonnegative series (sums). ∞ ai = ∞ [ai]+ ∞ [ai]−. Xi=1 Xi=1 − Xi=1 Thus we may assume that each ai ≥ 0. For any finite set I IN it is clear that ⊂ ai ≤ ∞ ai. Xi=1 I Xi ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.7. Rearrangements It follows that if ∞i=1 ai converges, then (by Exercise 3.3.3) so too does P Similarly, if N is finite, ai ≤ ∞ ai. Xi=1 IN Xi ∈ N IN ai and i ∈ P It follows that if i ∈ P IN ai converges, then, by the boundedness criterion, so too does ∞i=1 ai and P Together these two assertions and the equations (6) and (7) prove the theorem for the case of nonnegative series (sums). ai ≤ ai. IN Xi ∈ ai ≤ ai. IN Xi ∈ Xi=1 ∞ Xi=1 179 (6) (7) Exercises 3.7.1 Let Show that − + . . . . 3s .7.2 For what values of x does the following series converge and what is the sum? 1 + x2 + x + x4 + x6 + x3 + x8 + x10 + x5 + . . . 3.7.3 For what series is the computation ∞ ∞ ∞ ak = a2k + a2k−1 Xk=1 Xk=1 Xk=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 180 Infinite Sums Chapter 3 valid? Is this a rearrangement? 3.7.4 For what series is the computation valid? Is this a rearrangement? 3.7.5 For what series is the computation ∞ valid? Is this a rearrangement? Xk=1 ∞ ∞ ak = (a2k + a2k−1) Xk=1 Xk=1 ak = a2 + a1 + a4 + a3 + a6 + a5 + . . . 3.7.6 Give an example of an absolutely convergent series for which is it much easier to compute the sum by rearrangement than otherwise. 3.7.7 For what values of α and β does the series converge.7.8 Let a series be altered by the insertion of zero terms in a completely arbitrary manner. Does this alter the convergence of the series? 3.7.9 Suppose that a convergent series contains only finitely many negative terms. Can it be safely rearranged? 3.7.10 Suppose that a nonabsolutely convergent series has been rearranged and that this rearrangement converges. Does this rearranged series converge absolutely or nonabsolutely? 3.7.11 Is there a divergent series that can be rearranged so as to converge? Can every divergent series be rearranged , can it be rearranged to diverge to ∞ k=1 ak diverges, but does not diverge to so as to converge? If or ∞ −∞ ? ∞ P 3.7.12 How many rearrangements of a nonabsolutely convergent series are there that do not alter the sum? ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.8. Products of Series 3.7.13 Complete the proof of Theorem 3.49 by showing that for any nonabsolutely convergent series series and any α there is a rearrangement of the series so that 181 ∞ k=1 ak P See Note 60 3.7.14 Improve Theorem 3.49 by showing that for any nonabsolutely convergent series series aσ(k) = α. ∞ Xk=1 there is a rearrangement of the series so that n −∞ ≤ α β ≤ ≤ ∞ α = lim inf n→∞ Xk=1 aσ(k) ≤ lim sup n→∞ aσ(k) = β. n Xk=1 ∞ k=1 ak and any P 3.8 Products of Series " Enrichment section. May be omitted. The rule for the sum of two convergent series2 in Theorem 3.8 ∞ (ak + bk) = ∞ ak + ∞ bk is entirely elementary to prove and comes directly from the rule for limits of sums of sequences. If An and Bn represent the sum of n + 1 terms of the two series, then Xk=0 Xk=0 Xk=0 lim n →∞ ∞ Xk=0 (ak + bk) = lim →∞ n (An + Bn) = lim →∞ n An + lim →∞ n Bn 2 In the formula for a product of series in this section we prefer to label the series starting with 0. This does not change the series in any way. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 182 Infinite Sums Chapter 3 ∞ = ak + ∞ bk. At first glance we might expect to have a similar rule for products of series, since (An × An × Bn n Bn) = lim →∞ lim n →∞ lim n →∞ Xk=0 Xk=0 Xk=0 But what is AnBn? If we write out this product we obtain Xk=0 ∞ = ak × ∞ bk. AnBn = (a0 + a1 + a2 + + an) (b0 + b1 + b2 + n · · · n · · · + bn) = aibj. From this all we can show is the curious observation that Xi=0 Xj=1 lim n →∞ n n Xi=0 Xj=1 aibj = ∞ Xk=0 ak × ∞ bk. Xk=0 What we would rather see here is a result similar to the rule for sums: Can this result be interpreted as “series + series = series.” “series × series = series?” We need a systematic way of adding up the terms aibj in the double sum so as to form a series. The terms are displayed in a rectangular array in Figure 3.3. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.8. Products of Series 183 × b0 b1 b2 b3 b4 b5 . . . a0 a0b0 a0b1 a0b2 a0b3 a0b4 a0b5 . . . a1 a1b0 a1b1 a1b2 a1b3 a1b4 a1b5 . . . a2 a2b0 a2b1 a2b2 a2b3 a2b4 a2b5 . . . a3 a3b0 a3b1 a3b2 a3b3 a3b4 a3b5 . . . a4 a4b0 a4b1 a4b2 a4b3 a4b4 a4b5 . . . a5 a5b0 a5b1 a5b2 a5b3 a5b4 a5b5 . . . . . . . . . . . . . . . . . . . . . . . . Figure 3.3. The product of the two series ∞ 0 ak and ∞ 0 bk. P P If we replace the series here by a power series, this systematic way will become much clearer. How should we add up · · · + anxn b0 + b1x + b2x2 + a0 + a1x + a2x2 + + bnxn (which with x = 1 is the same question we just asked)? The now obvious answer is a0b0 + (a0b1 + a1b0)x + (a0b2 + a1b1 + a2b0)x2 +(a0b3 + a1b2 + a2b1 + a3b0)x3 + . . . . Notice that this method of grouping the terms corresponds to summing along diagonals of the rectangle in Figure 3.3. · · · This is the source of the following definition. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Infinite Sums Chapter 3 184 Definition 3.51: The series is called the formal product of the two series provided that ∞ ck Xk=0 ak and ∞ Xk=0 ∞ bk Xk=0 ck = k Xi=0 aibk i. − Our main goal now is to determine if this “formal” product is in any way a genuine product; that is, if ∞ ck = ∞ ∞ bk. ak × Xk=0 Xk=0 Xk=0 The reason we expect this might be the case is that the series expansion of ∞k=0 ck contains all the terms in the (a0 + a1 + a2 + a3 + . . . ) (b0 + b1 + b2 + b3 + . . . ) . P A good reason for caution, however, is that the series arrangement and we know that series can be sensitive to rearrangement. ∞k=0 ck contains these terms only in a particular P 3.8.1 Products of Absolutely Convergent Series It is a general rule in the study of series that absolutely convergent series permit the best theorems. We can rearrange such series freely as we have seen already in Section 3.7.1. Now we show that we can form products of such series. We shall have to be much more cautious about forming products of nonabsolutely convergent series. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.8. Products of Series 185 Theorem 3.52 (Cauchy) Suppose that ∞k=0 ck is the formal product of two absolutely convergent series Then ∞k=0 ck converges absolutely too and Xk=0 P ∞ ak and ∞ bk. Xk=0 P Proof. We write By definition and so ck = ∞ Xk=0 ∞ Xk=0 ak × ∞ bk. Xk=0 A = ∞ Xk=0 ak, A′ = ∞ Xk=0 , An = ak| | n ak, Xk=0 n B = ∞ Xk=0 bk, B′ = ∞ Xk=0 bk| | , and Bn = bk. Xk=0 ck = k Xi=0 aibk i − N N k N N ck| ≤ | Xk=0 i| ≤ Since the latter two series converge, this provides an upper bound A′B′ for the sequence of partial sums and hence the series ck| P aibj taken over all i N k=1 | Let us recall that the formal product of the two series is just a particular rearrangement of the terms 0. Consider any arrangement of these terms. This must form an absolutely ∞k=0 ck converges absolutely. Xi=0 Xi=0 Xi=0 Xk=0 0, j P − | | | ai|! bi|! ≤ A′B′. ai| · | bk ≥ ≥ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 186 Infinite Sums Chapter 3 convergent series by the same argument as before since A′B′ will be an upper bound for the partial sums of the absolute values . Thus all rearrangements will converge to the same value by Theorem 3.48. aibj| We can rearrange the terms aibj taken over all i | 0 in the following convenient way “by squares.” Arrange always so that the first (m + 1)2 (m = 0, 1, 2, . . . ) terms add up to AmBm. For example, one such arrangement starts off 0, j ≥ ≥ a0b0 + a1b0 + a0b1 + a1b1 + a2b0 + a2b1 + a0b2 + a1b2 + a2b2 + . . . . (A picture helps considerably to see the pattern needed.) We know this arrangement converges and we know it must converge to AmBm = AB. lim m →∞ In particular, the series required. P ∞k=0 ck which is just another arrangement, converges to the same number AB as It is possible to improve this theorem to allow one (but not both) of the series to converge nonabsolutely. The conclusion is that the product then converges (perhaps nonabsolutely), but different methods of proof will be needed. As usual, nonabsolutely convergent series are much more fragile, and the free and easy moving about of the terms in this proof is not allowed. 3.8.2 Products of Nonabsolutely Convergent Series Let us give a famous example, due to Cauchy, of a pair of convergent series whose product diverges. We know that the alternating series is convergent, but not absolutely convergent since the related absolute series is a p-ha |
rmonic series with p = 1 2 . ∞ ( 1)k − 1 √k + 1 Xk=0 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.8. Products of Series Let ∞ ck be the formal product of this series with itself. By definition the term ck is given by Xk=0 1)k ( − " 1 (k + 1) 1 · + 1 · 2 + (k) 1 (k 3 · 1) − + + · · · 1 (k + 1) . 1 # · There are k + 1 terms in the sum for ck and each term is larger than 1/(k + 1) so we see that Since the terms of the product series ∞k=0 ck do not tend to zero, this is a divergent series. p p p p 187 ck| ≥ | 1. This example supplies our observation: The formal product of two nonabsolutely convergent series need not converge. In particular, there may be no convergent series to represent the product P ∞ Xk=0 ak × ∞ bk Xk=0 for a pair of nonabsolutely convergent series. For absolutely convergent series the product always converges. We should not be too surprised at this result. The theory begins to paint the following picture: Absolutely convergent series can be freely manipulated in most ways and nonabsolutely convergent series can hardly be manipulated in general in any serious manner. Interestingly, the following theorem can be proved that shows that even though, in general, the product might diverge, in cases where it does converge it converges to the “correct” value. Theorem 3.53 (Abel) Suppose that ∞k=0 ck is the formal product of two nonabsolutely convergent series ∞k=0 bk and suppose that this product ∞k=0 ck is known to converge. Then ∞k=0 ak and P P P ck = ∞ Xk=0 ∞ P ak × Xk=0 ∞ bk. Xk=0 Proof. The proof requires more technical apparatus and will not be given until Section 3.9.2. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 188 Exercises 3.8.1 Form the product of the series ∞ k=0 akxk with the geometric series Infinite Sums Chapter 3 and obtain the formula P ∞ 1 Xk=0 For what values of x would this be valid? − 1 x = 1 + x + x2 + x3 + . . . 1 − 1 x ∞ akxk = (a0 + a1 + a2 + + ak)xk. · · · 3.8.2 Show that for appropriate values of x. 3.8.3 Using the fact that show that where σk = 1 + 1/2 + 1/3 + Xk=0 (1 − ∞ x)2 = (k + 1)xk Xk=0 1)k ( − k + 1 ∞ Xk=0 = log 2, ∞ Xk=0 + 1/(k + 1). 1)kσk ( − k + 2 = (log 2)2 2 3.8.4 Verify that ex+y = exey by proving that · · · ∞ Xk=0 ∞ ∞ (x + y)k k! = xk k! yk k! . Xk=0 Xk=0 3.8.5 For what values of p and q are you able to establish the convergence of the product of the two series 1)k ( − (k + 1)p 1)k ( (k + 1)q ? − and ∞ ∞ Xk=0 Xk=0 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.9. Summability Methods 3.9 Summability Methods " Advanced section. May be omitted. 189 A first course in series methods often gives the impression of being obsessed with the issue of convergence or divergence of a series. The huge battery of tests in Section 3.6 devoted to determining the behavior of series might lead one to this conclusion. Accordingly, you may have decided that convergent series are useful and proper tools of analysis while divergent series are useless and without merit. In fact divergent series are, in many instances, as important or more important than convergent ones. Many eighteenth century mathematicians achieved spectacular results with divergent series but without a proper understanding of what they were doing. The initial reaction of our founders of nineteenth-century analysis (Cauchy, Abel, and others) was that valid arguments could be based only on convergent series. Divergent series should be shunned. They were appalled at reasoning such as the following: The series s = 1 1 + 1 − − 1 + . . . can be summed by noting that s = 1 and so 2s = 1 or s = 1 the series is clearly divergent. 2 . But the sum 1 ( proves to be a useful value for the “sum” of this series even though There are many useful ways of doing rigorous work with divergent series. One way, which we now study, is the development of summability methods. Suppose that a series ∞k=0 ak diverges and yet we wish to assign a “sum” to it by some method. Our standard method thus far is to take the limit of the sequence of partial sums. We write P n sn = ak sn. If the series diverges, this means precisely that this sequence does and the sum of the series is limn not have a limit. How can we use that sequence or that series nonetheless to assign a different meaning to the sum? →∞ Xk=0 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 190 Infinite Sums Chapter 3 3.9.1 Ces`aro’s Method " Advanced section. May be omitted. An infinite series ∞k=0 ak has a sum S if the sequence of partial sums P n sn = ak Xk=0 converges to S. If the sequence of partial sums diverges, then we must assign a sum by a different method. We will still say that the series diverges but, nonetheless, we will be able to find a number that can be considered the sum. We can replace limn →∞ sn, which perhaps does not exist, by + sn s0 + s1 + s2 + n + 1 if this exists and use this value for the sum of the series. This is an entirely natural method since it merely takes averages and settles for computing a kind of “average” limit where an actual limit might fail to exist. lim n →∞ = C · · · For a series ∞k=0 ak often we can use this method to obtain a sum even when the series diverges. Definition 3.54: If P sn} { then the new sequence is the sequence of partial sums of the series s0 + s1 + s2 + n + 1 + sn = C · · · lim n →∞ ∞k=0 ak and P σn = s0 + s1 + s2 + n + 1 · · · + sn is called the sequence of averages or Ces`aro means and we write ak = C [Ces`aro]. ∞ Xk=0 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.9. Summability Methods 191 Thus the symbol [Ces`aro] indicates that the value is obtained by this method rather than by the usual method of summation (taking limits of partial sums). The method is named after Ernesto Ces`aro (1859–1906). Our first concern in studying a summability method is to determine whether it assigns the “correct” value to a series that already converges. Does Xk=0 Any method of summing a series is said to be regular or a regular summability method if this is the case. Xk=0 ∞ ak = A ∞ ak = A [Ces`aro]? ⇒ Theorem 3.55: Suppose that a series also true. ∞k=0 ak converges to a value A. Then ∞k=0 ak = A [Ces`aro] is P Proof. This is an immediate consequence of Exercise 2.13.17. For any sequence P sn} { write In that exercise we showed that σn = s1 + s2 + n + sn . · · · n If you skipped that exercise, here is how to prove it. Let →∞ lim inf n →∞ sn ≤ lim inf n →∞ σn ≤ lim sup σn ≤ lim sup sn. n →∞ β > lim sup n sn. →∞ (If there is no such β, then lim supn for some N . Thus for all n ≥ N . Fix N , allow n σn ≤ → ∞ sn = and there is nothing to prove.) Then sn < β for all n N ≥ ∞ →∞ 1 n , and take limit superiors of each side to obtain N + 1)β n (s1 + s2 + + sN 1) + · · · (n − − lim sup n →∞ σn ≤ β. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 192 It follows that σn ≤ The other inequality is similar. In particular, if limn equal, proving the theorem. lim sup →∞ n →∞ lim sup sn. n →∞ sn exists so too does limn Infinite Sums Chapter 3 →∞ σn and they are Example 3.56: As an example let us sum the series The partial sums form the sequence 1, 0, 1, 0, . . . , which evidently diverges. Indeed the series diverges merely by the trivial test: The terms do not tend to zero. Can we sum this series by the Ces`aro summability method? The averages of the sequence of partial sums is clearly tending to 1 2 . Thus we can write )k = − 1 2 ∞ ( Xk=0 [Ces`aro] ◭ even though the series is divergent. 3.9.2 Abel’s Method " Advanced section. May be omitted. We require in this section that you recall some calculus limits. We shall need to compute a limit F (x) lim 1 x − → for a function F defined on (0, 1) where the expression x indicates a left-hand limit. In Chapter 5 we present a full account of such limits; here we need remember only what this means and how it is computed. ∞k=0 ak diverges and yet we wish to assign a “sum” to it by some other method. Suppose that a series → − 1 If the terms of the series do not get too large, then the series P F (x) = ∞ akxk Xk=0 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.9. Summability Methods 193 will converge (by the ratio test) for all 0 x < 1. The value we wish for the sum of the series would appear to be F (1), but for a divergent series inserting the value 1 for x gives us nothing we can use. Instead we compute ≤ lim 1 x − → F (x) = lim 1 − → x and use this value for the sum of the series. Definition 3.57: We write akxk = A ∞ Xk=0 if akxk = A [Abel] ∞ Xk=0 akxk = A. lim 1 x − → ∞ Xk=0 Here the symbol [Abel] indicates that the value is obtained by this method rather than by the usual method of summation (taking limits of partial sums). As before, our first concern in studying a summability method is to determine whether it assigns the “correct” value to a series that already converges. Does We are asking, in more correct language, whether Abel’s method of summability of series is regular. ∞ ak = A ∞ ak = A [Abel]? Xk=0 ⇒ Xk=0 Theorem 3.58 (Abel) Suppose that a series ∞k=0 ak converges to a value A. Then ∞ akxk = A. P lim 1 x − → Xk=0 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 194 Infinite Sums Chapter 3 Proof. Our first step is to note that the convergence of the series In particular, the terms are bounded and so the root test will prove that the series absolutely for all < 1 at least. Thus we can define ∞k=0 ak requires that the terms ak → 0. ∞k=0 akxk converges P x | | P x < 1. F (x) = ∞ akxk Xk=0 for 0 ≤ Let us form the product of the series for F (x) with the geometric series (cf. Exercise 3.8.1). Since both series are absolutely convergent for any 0 1 − 1 = 1 + x + x2 + x3 + . . . x x < 1, we obtain ≤ Writing and |
using the fact that we obtain F (x) x 1 − ∞ = (a0 + a1 + a2 + Xk=0 + ak)xk. · · · sk = (a0 + a1 + a2 + + ak) · · · sk → A = ∞ ak, Xk=0 F (x) = (1 x) − Let ε > 0 and choose N so large that ∞ Xk=0 skxk = A (1 − − x) A)xk. (sk − ∞ Xk=0 sk − | A | < ε/2 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 195 Section 3.9. Summability Methods for k > N . Then the inequality N holds for all 0 ≤ x < 1. The sum here is just a finite sum, and taking limits in finite sums is routine: F (x) | A − | ≤ (1 − x) sk − | A xk + ε/2 | Xk=0 N Xk=0 (1 lim 1 x − → − x) A)xk = 0. (sk − Thus for x < 1 but sufficiently close to 1 we can make this smaller than ε/2 and conclude that We have proved that and the theorem is proved. Example 3.59: Let us sum the series F (x) | A | − < ε. F (x) = A lim 1 x − → by Abel’s method. We form 1) + . . . ∞ ( Xk=0 obtaining the formula by recognizing this as a geometric series. Since F (x) = ∞ ( 1)kxk = − 1 1 + x Xk=0 F (x) = lim 1 x − → 1 2 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 196 we have proved that ∞ Infinite Sums Chapter 3 1)k = ( − 1 2 [Abel]. Recall that we have already obtained in Example 3.56 that Xk=0 Xk=0 ∞ 1)k = ( − 1 2 [Ces`aro] so these two different methods have assigned the same sum to this divergent series. You might wish to explore whether the same thing will happen with all series. ◭ As an interesting application we are now in a position to prove Theorem 3.53 on the product of series. Theorem 3.60 (Abel) Suppose that ∞k=0 ck is the formal product of two convergent series ∞k=0 bk and suppose that P P ∞k=0 ck is known to converge. Then P ∞k=0 ak and P Xk=0 Proof. The proof just follows on taking limits as x Xk=0 ∞ ck = ∞ ak × ∞ bk. Xk=0 in the expression 1 → − akxk ∞ ckxk = ∞ Xk=0 Xk=0 ∞ bkxk. × Xk=0 Abel’s theorem, Theorem 3.58, allows us to do this. How do we know, however, that this identity is true < 1 and, by Theorem 3.52, for all 0 absolutely convergent series can be multiplied in this way. x < 1? All three of these series are absolutely convergent for x | ≤ | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.10. More on Infinite Sums 197 Exercises 3.9.1 Is the series Ces`aro summable? 3.9.2 Is the series Ces`aro summable? 3.9.3 Is the series Abel summable.9.4 Show that a divergent series of positive numbers cannot be Ces`aro summable or Abel summable. 3.9.5 Find a proof from an appropriate source that demonstrates the exact relation between Ces`aro summability and Abel summability. 3.9.6 In an appropriate source find out what is meant by a Tauberian theorem and present one such theorem appropriate to our studies in this section. See Note 61 3.10 More on Infinite Sums " Advanced section. May be omitted. How should we form the sum of a double sequence numbers? In many applications of analysis such sums are needed. A variety of methods come to mind: ajk} { where both j and k can range over all natural 1. We might simply form the unordered sum ajk. IN X(j,k) ∈ × IN ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 198 Infinite Sums Chapter 3 2. We could construct “partial sums” in some systematic method and take limits just as we do for ordinary series: These are called square sums and are quite popular. If you sketch a picture of the set of points lim N →∞ N N ajk. Xj=1 Xk=1 ≤ in the plane the square will be plainly visible. { (j, k) : 1 j ≤ N, 1 k N } ≤ ≤ 3. We could construct partial sums using rectangular sums: Xj=1 Here the limit is a double limit, requiring both M and N to get large. If you sketch a picture of the set of points Xk=1 →∞ M N ajk. lim M,N in the plane you will see the rectangle. (j, k) : 1 { j ≤ ≤ M, 1 k N } ≤ ≤ 4. We could construct partial sums using circular sums: Once again, a sketch would show the circles. 5. We could “iterate” the sums, by summing first over j and then over k: lim R →∞ Xj2+k2 ≤ R2 ajk. ∞ ∞ ajk Xj=1 Xk=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.10. More on Infinite Sums or, in the reverse order, 199 ∞ ∞ ajk. Xj=1 Our experience in the study of ordinary series suggests that all these methods should produce the same sum if the numbers summed are all nonnegative, but that subtle differences are likely to emerge if we are required to add numbers both positive and negative. Xk=1 In the exercises there are a number of problems that can be pursued to give a flavor for this kind of theory. At this stage in your studies it is important to grasp the fact that such questions arise. Later, when you have found a need to use these kinds of sums, you can develop the needed theory. The tools for developing that theory are just those that we have studied so far in this chapter. Exercises 3.10.1 Decide on a meaning for the notion of a double series ∞ ajk and prove that if all the numbers ajk are nonnegative then this converges if and only if Xj,k=1 converges and that the values assigned to (8) and (9) are the same. X(j,k)∈IN×IN ajk 3.10.2 Decide on a meaning for the notion of an absolutely convergent double series and prove that such a series is absolutely convergent if and only if Xj,k=1 ∞ ajk ajk X(j,k)∈IN×IN (8) (9) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 200 converges. Infinite Sums Chapter 3 3.10.3 Show that the methods given in the text for forming a sum of a double sequence the numbers are nonnegative. 3.10.4 Show that the methods given in the text for forming a sum of a double sequence general. ajk} { ajk} { are equivalent if all are not equivalent in 3.10.5 What can you assert about the convergence or divergence of the double series 3.10.6 What is the sum of the double series 3.11 Infinite Products " Enrichment section. May be omitted. ∞ Xj,k=1 ∞ Xj,k=0 1 j k4 ? xjyk j! k! ? In this chapter we studied, quite extensively, infinite sums. There is a similar theory for infinite products, a theory that has much in common with the theory of infinite sums. In this section we shall briefly give an account of this theory, partly to give a contrast and partly to introduce this important topic. Similar to the notion of an infinite sum is the notion of an infinite product ∞ n=1 X ∞ n=1 Y an = a1 + a2 + a3 + a4 + . . . pn = p1 × p2 × p3 × p4 × . . . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.11. Infinite Products 201 with a nearly identical definition. Corresponding to the concept of “partial sums” for the former will be the notion of “partial products” for the latter. The main application of infinite series is that of series representations of functions. The main application of infinite products is exactly the same. Thus, for example, in more advanced material we will find a representation of the sin function as an infinite series sin x = x 1 3! − x3 + 1 5! x5 − 1 7! x7 + . . . and also as an infinite product sin x = x2 π2 1 − x2 4π2 1 − x2 9π2 1 − x2 16π2 1 − . . . . The most obvious starting point for our theory would be to define an infinite product as the limit of the sequence of partial products in exactly the same way that an infinite sum is defined as the limit of the sequence of partial sums. But products behave differently from sums in one important regard: The number zero plays a peculiar role. This is why the definition we now give is slightly different than a first guess might suggest. Our goal is to define an infinite product in such a way that a product can be zero only if one of the factors is zero (just like the situation for finite products). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 202 Infinite Sums Chapter 3 Definition 3.61: Let bk} { be a sequence of real numbers. We say that the infinite product converges if there is an integer N so that all bk 6 ∞ bk Yk=1 = 0 for k > N and if lim M →∞ M bk Yk=N +1 exists and is not zero. For the value of the infinite product we take ∞ Yk=1 bk = b1 × b2 × . . . bN × lim M →∞ M bk. Yk=N +1 This definition guarantees us that a product of factors can be zero if and only if one of the factors is zero. This is the case for finite products, and we are reluctant to lose this. Theorem 3.62: A convergent product if and only if one of the factors is zero. bk = 0 ∞ Yk=1 Proof. This is built into the definition and is one of its features. We expect the theory of infinite products to evolve much like the theory of infinite series. We recall that 0. Naturally, the product analog requires the terms to tend n k=1 ak could converge only if ak → a series to 1. P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.11. Infinite Products Theorem 3.63: A product ∞ bk 203 that converges necessarily has bk → Proof. This again is a feature of the definition, which would not be possible if we had not handled the zeros in this way. Choose N so that none of the factors bk is zero for k > N . Then 1 as k → ∞ . Yk=1 as required. bn = lim n n k=N +1 bk 1 n k=N +1 bk − →∞ Q = 1 Q As a result of this theorem it is conventional to write all infinite products in the special form ∞ (1 + ak) Yk=1 in a convergent product. Also, our assumption about the 1 only for finitely many values of k. The expressions (1 + ak) are called the “factors” → ∞ and remember that the terms ak → zeros allows for ak = of the product and the ak themselves are called the “terms.” A close linkage with series arises because the two objects 0 as k − ∞ ak and ∞ (1 + ak), the series and the product, have much the same kind of behavior. Xk=1 Yk=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 204 Theorem 3.64: A product Infinite Sums Chapter 3 ∞ (1 + ak) where all the terms ak are positive is convergent if and only if the series Yk=1 ∞k=1 ak converges. Proof. Here we use our usual criterion that has |
served us through most of this chapter: A sequence that is monotonic is convergent if and only if it is bounded. P Note that a1 + a2 + a3 + + an ≤ · · · (1 + a1)(1 + a2)(1 + a3) (1 + an) × · · · × so that the convergence of the product gives an upper bound for the partial sums of the series. It follows that if the product converges so must the series. In the other direction we have (1 + a1)(1 + a2)(1 + a3) × · · · × (1 + an) ≤ ea1+a2+a3+ ··· +an and so the convergence of the series gives an upper bound for the partial products of the infinite product. It follows that if the series converges, so must the product. Exercises 3.11.1 Give an example of a sequence of positive numbers exists, but so that the infinite product lim n→∞ nonetheless diverges. so that bk} { b1b2b3 . . . bn bk ∞ n=1 Y ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.11. Infinite Products 3.11.2 Compute 205 ∞ Yk=1 1 k2 1 − . 3.11.3 In Theorem 3.64 we gave no relation between the value of the product ∞ (1 + ak) and the value of the series ∞ k=1 ak where all the terms ak are positive. What is the best you can state? Yk=1 3.11.4 For what values of p does the product P ∞ 1 + n=1 Y 1 kp (1 + x4) (1 + x8) (1 + x16) . . . × × × × converge? 3.11.5 Show that converges to 1/(1 ∞ (1 + x2k ) = (1 + x2) Yk=1 x2) for all − − 1 < x < 1 and diverges otherwise. 3.11.6 Find a Cauchy criterion for the convergence of infinite products. 3.11.7 A product Yk=1 is said to converge absolutely if the related product ∞ ∞ (1 + ak) converges. (1 + Yk=1 ) ak| | (a) Show that an absolutely convergent product is convergent. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 206 Infinite Sums Chapter 3 (b) Show that an infinite product ∞ (1 + ak) converges absolutely if and only if the series of its terms Yk=1 ∞ k=1 ak converges absolutely. (c) For what values of x does the product converge absolutely? (d) For what values of x does the product converge absolutely? (e) For what values of x does the product converge absolutely? (f) Show that converges but not absolutely. ∞ Yk=1 1 + x k P ∞ Yk=1 1 + x k2 ∞ 1 + xk Yk=1 ∞ Yk=1 1 + 1)k ( − k 3.11.8 Develop a theory that allows for the order of the factors in a product to be rearranged. 3.12 Challenging Problems for Chapter 3 3.12.1 If an is a sequence of positive numbers such that the following three series? ∞ n=1 an diverges what (if anything) can you say about P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.12. Challenging Problems for Chapter 3 207 (a) (b) (c) ∞ n=1 ∞ n=1 ∞ n=1 P P an 1+an an 1+nan an 1+n2an 3.12.2 Prove the following variant on the Dirichlet Test 3.44: If P (cf. Exercise 3.5.12) that converges to zero and the partial sums of the series the series ∞ k=1 akbk converges. 3.12.3 Prove this variant on the Cauchy condensation test: If the terms of a series P decrease monotonically to zero, then that series converges if and only if the series bn} { is a sequence of bounded variation ∞ k=1 ak are bounded, then P ∞ k=1 ak are nonnegative and P converges. (2j + 1)aj2 ∞ j=1 X 3.12.4 Prove this more general version of the Cauchy condensation test: If the terms of a series nonnegative and decrease monotonically to zero, then that series converges if and only if the related series ∞ k=1 ak are P converges. Here m1 < m2 < m3 < m4 < . . . is assumed to be an increasing sequence of integers and mj ≤ C (mj − mj+1 − mj−1) for some positive constant and all j. 3.12.5 For any two series of positive terms write ∞ ∞ ∞ j=1 X (mj+1 − mj)amj ak bk Xk=1 Xk=1 0 as k if ak/bk → (a) If both series converge, explain why this might be interpreted by saying that → ∞ . faster than P ∞ k=1 bk. ∞ k=1 ak is converging P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 208 (b) If both series diverge, explain why this might be interpreted by saying that slowly than ∞ k=1 bk. Infinite Sums Chapter 3 ∞ k=1 ak is diverging more P (c) For convergent series is there any connection between P and (d) For what values of p, q is (e) For what values of r, s is (f) Arrange the divergent series ∞ ∞ ak Xk=1 ∞ ak ≤ 1 kp rk Xk=1 ∞ Xk=1 ∞ Xk=1 bk Xk=1 ∞ bk? Xk=1 ∞ Xk=1 ∞ Xk=1 1 kq ? sk? ∞ into the correct order. Xk=2 1 k , ∞ Xk=2 1 k log k , ∞ Xk=2 1 k log(log k) , ∞ Xk=2 1 k log(log(log k)) . . . (g) Arrange the convergent series ∞ Xk=2 Xk=2 into the correct order. Here p > 1. 1 kp , ∞ ∞ ∞ 1 k(log k)p , 1 k log k(log(log k))p , Xk=2 Xk=2 1 k log k(log(log k))(log(log(log k)))p . . . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 3.12. Challenging Problems for Chapter 3 209 (h) Suppose that P ∞ k=1 bk is a divergent series of positive numbers. Show that there is a series ∞ ∞ (i) Suppose that that also diverges (but more slowly). ∞ k=1 ak is a convergent series of positive numbers. Show that there is a series ∞ ∞ P ak bk Xk=1 Xk=1 ak bk Xk=1 Xk=1 that also converges (but more slowly). (j) How would you answer this question? Is there a “mother” of all divergent series diverging so slowly that all other divergent series can be proved to be divergent by a comparison test with that series? See Note 62 3.12.6 This collection of exercises develops some convergence properties of trigonometric series; that is, series of the form ∞ a0/2 + (ak cos kx + bk sin kx) . (10) (b) For what values of x does (a) For what values of x does ∞ k=1 ∞ P k=1 ∞ k=1 ( P series (10) for all values of x. (c) Show that the condition | P See Note 63 Xk=1 converge? converge? bk| | ) < ∞ sin kx k2 sin kx k + ak| ensures the absolute convergence of the trigonometric 3.12.7 Let ak} { be a decreasing sequence of positive real numbers with limit 0 such that bk = ak − 2ak+1 + ak+2 ≥ 0. Prove that ∞ k=1 kbk = a1. See Note 64 P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 210 3.12.8 Let ak} { be a monotonic sequence of real numbers such that ∞ k=1 ak converges. Show that Infinite Sums Chapter 3 converges. See Note 65 ∞ Xk=1 P ak+1) k(ak − 3.12.9 Show that every positive rational number can be obtained as the sum of a finite number of distinct terms of the harmonic series See Note 66 .12.10 Let x1 ≥ terms of the series. Show that P is an interval if and only if ∞ k=1 xk be a convergent series of positive numbers that is monotonically nonincreasing; that is, x2 ≥ P . . . . Let P denote the set of all real numbers that are sums of finitely or infinitely many x3 ≥ for every integer n. See Note 67 xn ≤ ∞ xk Xk=n+1 3.12.11 Let p1, p2, p3, be a sequence of distinct points that is dense in the interval (0, 1). The points p1, p2, p3, . . . , pn−1 decompose the interval [0, 1] into n closed subintervals. The point pn is an interior point of one of those intervals and decomposes that interval into two closed subintervals. Let an and bn be the lengths of those two intervals. Prove that ∞ See Note 68 akbk(ak + bk) = 3. Xk=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 3.12.12 Let an} { be a sequence of positive number such that the series ∞ k=1 ak converges. Show that 211 (ak)n/(n+1) P ∞ Xk=1 also converges. See Note 69 converges. See Note 71 an} { sequence { See Note 72 3.12.13 Let ak} { be a sequence of positive numbers and suppose that a2k + a2k+1 ak ≤ ∞ k=1 ak diverges. for all k = 1, 2, 3, 4, . . . . Show that See Note 70 P 3.12.14 If ak} { is a sequence of positive numbers for which ∞ k=1 ak diverges, determine all values of p for which ak (a1 + a2 + + ak)p P · · · ∞ Xk=1 3.12.15 Let be a sequence of real numbers converging to zero. Show that there must exist a monotonic ∞ k=1 akbk is absolutely convergent. bn} ∞ k=1 bk diverges and the series such that the series P P Notes 36Exercise 3.2.2. Define it for I with n + 1 elements and show well defined. P i∈I ai for I with zero or one elements. Suppose it is defined for I with n elements. Define ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 212 NOTES 37Exercise 3.2.4. The answer is yes if I and J are disjoint. Otherwise the correct formula would be ai + ai = ai + ai. Xi∈I∪J Xi∈I∩J Xi∈I Xi∈J 38Exercise 3.2.8. Try to interpret the “difference” ∆sk = sk+1 − 39Exercise 3.2.11. Use a telescoping sum method. Even if you cannot remember your trigonometric identities you can work backward to see which one is needed. Check the formula for values of θ with sin θ/2 = 0 and see that it can be interpreted by taking limits. sk = ak+1 as the analog of a derivative. 40Exercise 3.3.1. This is similar to the statement that convergent sequences have unique limits. Try to imitate that proof. 41Exercise 3.3.2. This is similar to the statement that convergent sequences are bounded. Try to imitate that proof. 42Exercise 3.3.3. This is similar to the statement that monotone, bounded sequences are convergent. Try to imitate that proof. 43Exercise 3.3.9. Compare with the sum given in the introduction to this chapter. 44Exercise 3.3.11. Here we are using, as elsewhere and and note that [X]+ = max X, 0 { } [X]− = max {− X, 0 } X = [X]+ [X]− and − X | | = [X]+ + [X]−. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 213 45Exercise 3.3.12. Note that the index set is Thus we can study unordered sums of double sequences { in the form I = IN IN. × aij} aij. X(i,j)∈IN×IN 46Exercise 3.4.10. Handle the case where each ak ≥ 47Exercise 3.4.15. Using properties of the log function, you can view this series as a telescoping one. 0 separately from the general case. 48Exercise 3.4.16. Consider that 1 − r 1 r + 1 = r2 1 − 2 − . 1 49Exercise 3.4.24. Establish the inequalities 1 kp ≤ ∞ Xk=1 2k−1 (2k−1)p 2n−1 Xk=1 ∞ = (21−p)j = j=0 X 2p−1 2p−1 − . 1 Conclude that the partial sums of the p-harmonic series for p > 1 are increasing and bounded. Explain now why the seri |
es must converge. 50Exercise 3.4.26. As a first step show that 2kπ+3π/4 2kπ+π/4 Z | sin x x | dx ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 214 NOTES (Remember that in calculus an integral 51Exercise 3.4.28. Establish that ∞ 0 R 1 √2 ≥ 2kπ+3π/4 2kπ+π/4 Z 1 x dx. is interpreted as limX→∞ X 0 .) R n ki pi 52Exercise 3.5.5. Add up the terms containing p digits in the denominator. Note that our deletions leave only 1 pn . i=1 X ≤ − x 9p−1 of them. The total sum is bounded by 8 × 8(1/1 + 9/10 + 92/100 + . . . ) = 80. 53Exercise 3.5.8. Instead consider the series where and and note that ∞ ∞ [ak]+ and [ak]− Xk=1 Xk=1 [X]+ = max X, 0 } { [X]− = max {− X, 0 } X = [X]+ [X]− and − X | | = [X]+ + [X]−. 54Exercise 3.5.15. Use the Cauchy-Schwarz inequality. 55Exercise 3.5.16. Use the Cauchy-Schwarz inequality. 56Exercise 3.6.3. The answer for (d) is x < 1/e. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 215 57Exercise 3.6.5. Only one condition is sufficient to supply divergence. Give a proof for that one and counterexamples for the three others. Here is an idea that may help: Let ak = 0 for all values of k except if k = 2m for some m in which case ak = 1/√k. Note that lim supk→∞ √kak = 1 in this case and that ∞ k=1 ak will converge. 58Exercise 3.6.22. The exact value of γ, called Euler’s constant, is not needed in the problem; it is approximately P .5772156. 59Exercise 3.6.24. The integral test should occur to you while thinking of this problem. Start by checking that converges if and only if exists. Find similar statements for the other series. F ′(k) ∞ Xk=1 lim X→∞ F (X) 60Exercise 3.7.13. Imitate the proof of the first part of Theorem 3.49 but arrange for the partial sums to go larger than α before inserting a term qk. You must take the first opportunity to insert qk when this occurs. 61Exercise 3.9.6. The name “Tauberian theorem” was coined by Hardy and Littlewood after a result of Alfred Tauber (1866–1942?). The date of his death is unknown; all that is certain is that he was sent by the Nazis to Theresienstadt concentration camp on June 28, 1942. 62Exercise 3.12.5. For (h) consider the series the series given. ∞ k=1(sk+1 − P sk)/sk+1 where sk is the sequence of partial sums of 63Exercise 3.12.6. For (b) use Abel’s method and the computation in Exercise 3.2.11. Further treatment of some aspects of trigonometric series may be found in Section 10.8. 64Exercise 3.12.7. This is from the 1948 Putnam Mathematical Competition. 65Exercise 3.12.8. This is from the 1952 Putnam Mathematical Competition. 66Exercise 3.12.9. This is from the 1954 Putnam Mathematical Competition. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 216 NOTES 67Exercise 3.12.10. This is from the 1955 Putnam Mathematical Competition. 68Exercise 3.12.11. This is from the 1964 Putnam Mathematical Competition. 69Exercise 3.12.12. This is from the 1988 Putnam Mathematical Competition. 70Exercise 3.12.13. This is from the 1994 Putnam Mathematical Competition. 71Exercise 3.12.14. Problem posed by A. Torchinsky in Amer. Math. Monthly, 82 (1975), p. 936. 72Exercise 3.12.15. Problem posed by Jan Mycielski in Amer. Math. Monthly, 83 (1976), p. 284. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Chapter 4 SETS OF REAL NUMBERS 4.1 Introduction Modern set theory and the world it has opened to mathematics has its origins in a problem in analysis. A young Georg Cantor in 1870 began to attack a problem given to him by his senior colleague Edward Heine, who worked at the same university. (We shall see Heine playing a key role in some ideas of this chapter too.) The problem was to determine if the equation ∞ 1 2 a0 + (ak cos kx + bk sin kx) = 0 (1) Xk=1 must imply that all the coefficients of the series, the are zero. Cantor solved this using { the methods of his time. It was a good achievement, but not the one that was to make him famous. What he did next was to ask, as any good mathematician would, whether his result could be generalized. Suppose that the series (1) converges to zero for all x except possibly for those in a given set E. If this set E is very small, then perhaps, the coefficients of the series should also have to be all zero. and the ak} bk} { The nature of these exceptional sets (nowadays called sets of uniqueness) required a language and techniques that were entirely new. Previously a number of authors had needed a language to describe sets 217 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 218 Sets of Real Numbers Chapter 4 that arose in various problems. What was used at the time was limited, and few interesting examples of sets were available. Cantor went beyond these, introducing a new collection of ideas that are now indispensable to analysis. We shall encounter in this chapter many of the notions that arose then: accumulation points, derived sets, countable sets, dense sets, nowhere dense sets. Incidentally, Cantor never did finish his problem of describing the sets of uniqueness, as the development of the new set theory was more important and consumed his energies. In fact, the problem remains unsolved, although much interesting information about the nature of sets of uniqueness has been discovered. The theory of sets that Cantor initiated has proved to be fundamental to all of mathematics. Very quickly the most talented analysts of that time began applying his ideas to the theory of functions, and by now this material is essential to an understanding of the subject. This chapter contains the most basic material. In Chapter 6 we will need some further concepts. 4.2 Points In our studies of analysis we shall often need to have a language that describes sets of points and the points that belong to them. That language did not develop until late in the nineteenth century, which is why the early mathematicians had difficulty understanding some problems. For example, consider the set of solutions to an equation f (x) = 0 where f is some well-behaved function. In the simplest cases (e.g., if f is a polynomial function) the solution set could be empty or a finite number of points. There is no difficulty there. But in more general settings the solution set could be very complicated indeed. It may have points that are “isolated,” points appearing in clusters, or it may contain intervals or merely fragments of intervals. You can see that we even lack the words to describe the possibilities. The ideas in this section are all very geometric. Try to draw mental images that depict all of these ideas to get a feel for the definitions. The definitions themselves should be remembered but may prove hard to remember without some associated picture. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.2. Points 219 a x c − x x+c b Figure 4.1. Every point in (a, b) is an interior point. The simplest types of sets are intervals. We call a closed interval, and [a, b] = x : a { x b } ≤ ≤ an open interval. The other sets that we often consider are the sets IN of natural numbers, Q of rational numbers, and R of all real numbers. Use these in your pictures, as well as sets obtained by combining them in many ways. (a, b.2.1 Interior Points Every point inside an open interval I = (a, b) has the feature that there is a smaller open interval centered at that point that is also inside I. Thus if x (a, b) then for any positive number c that is small enough ∈ − (x c, x + c) (a, b). ⊂ Indeed the arithmetic to show this is easy (and a picture makes it transparent). Let c be any positive number that is smaller than the shortest distance from x to either a or b. Then (x Figure 4.1.) c, x + c) (a, b). (See − ⊂ Note. Often we use the following suggestive language. An open interval that contains a point x is said to be a neighborhood of x. Thus each point in (a, b) possesses a neighborhood, indeed many neighborhoods, that lie entirely inside the set I. On occasion the point x itself is excluded from the neighborhood: We say ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 220 Sets of Real Numbers Chapter 4 that an interval (c, d) is a neighborhood of x if x belongs to the interval and we say that the set (c, d) is a deleted neighborhood. This is just the interval with the point x removed. x } \ { We can distinguish between points that are merely in a set and points that are more deeply inside the set. The word chosen to convey this image of “inside” is interior. Definition 4.1: (Interior Point) Let E be a set of real numbers. Any point x that belongs to E is said to be an interior point of E provided that some interval (x − c, x + c) E. ⊂ Thus an interior point of the set E is not merely in the set E; it is, so to speak, deep inside the set, at a positive distance at least c away from every point that does not belong to E. Example 4.2: The following examples are immediate if a picture is sketched. 1. Every point x of an open interval (a, b) is an interior point. 2. Every point x of a closed interval [a, b], except the two endpoints a and b, is an interior point. 3. The set of natural numbers IN has no interior points whatsoever. 4. Every point of R is an interior point. 5. No point of the set of rational numbers Q is an interior point. [This is because any interval (x c, x+c) must contain both rational numbers and irrational numbers and, hence, can never be a subset of Q.] − In each case, we should try to find the interval (x such interval. c, x + c) inside the set or explain why there can be no ◭ − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.2. Points 4.2.2 Isolated Points 221 Most sets that we consider will have infinitely many points. Certainly any interval (a, b) or [a, b] has infinitely many points. The |
set IN of natural numbers also has infinitely many points, but as we look closely at any one of these points we see that each point is all alone, at a certain distance away from every other point in the set. We call these points isolated points of the set. Definition 4.3: (Isolated Point) Let E be a set of real numbers. Any point x that belongs to E is said to be an isolated point of E provided that for some interval (x c, x + c) (x − c, x + c) E = { ∩ − x . } Thus an isolated point of the set E is in the set E but has no close neighbors who are also in E. It is at some positive distance at least c away from every other point that belongs to E. Example 4.4: As before, the examples are immediate if a picture is sketched. 1. No point x of an open interval (a, b) is an isolated point. 2. No point x of a closed interval [a, b] is an isolated point. 3. Every point belonging to the set of natural numbers IN is an isolated point. 4. No point of R is isolated. 5. No point of Q is isolated. In each case, we should try to find the interval (x that there is none. c, x + c) that meets the set at no other point or show ◭ − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 222 Sets of Real Numbers Chapter 4 4.2.3 Points of Accumulation Most sets that we consider will have infinitely many points. While the isolated points are of interest on occasion, more than likely we would be interested in points that are not isolated. These points have the property that every containing interval contains many points of the set. Indeed we are interested in any point x with the property that the intervals (x c, x + c) meet the set E at infinitely many points. This could happen even if x itself does not belong to E. We call these points accumulation points of the set. An accumulation point need not itself belong to the set. − Definition 4.5: (Accumulation Point) Let E be a set of real numbers. Any point x (not necessarily in E) is said to be an accumulation point of E provided that for every c > 0 the intersection contains infinitely many points. (x − c, x + c) E ∩ Thus an accumulation point of E is a point that may or may not itself belong to E and that has many close neighbors who are in E. Note. The definition requires that for all c > 0 the intersection contains infinitely many points of E. In checking for an accumulation point it may be preferable merely to check that there is at least one point in this intersection (other than possibly x itself). If there is always at least one point, then there must in fact be infinitely many (Exercise 4.2.18). (x c, x + c) E − ∩ Example 4.6: Yet again, the examples are immediate if a picture is sketched. 1. Every point of an open interval (a, b) is a an accumulation point of (a, b). Moreover, the two endpoints a and b are also accumulation points of (a, b) [although they do not belong themselves to (a, b))]. 2. Every point of a closed interval [a, b] is an accumulation point of (a, b). No point outside can be. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.2. Points 3. No point at all is an accumulation point of the set of natural numbers IN. 4. Every point of R is an accumulation point. 5. Every point on the real line, both rational and irrational, is an accumulation point of the set Q. 223 ◭ 4.2.4 Boundary Points The intervals (a, b) and [a, b] have what appears to be an “edge”. The points a and b mark the boundaries between the inside of the set (i.e., the interior points) and the “outside” of the set. This inside/outside language with an idea of a boundary between them is most useful but needs a precise definition. Definition 4.7: (Boundary Point) Let E be a set of real numbers. Any point x (not necessarily in E) c, x + c) contains at least one point of is said to be a boundary point of E provided that every interval (x E and also at least one point that does not belong to E. − This definition is easy to apply to the intervals (a, b) and [a, b] but harder to imagine for general sets. For these intervals the only points that are immediately seen to satisfy the definition are the two endpoints that we would have naturally said to be at the boundary. Example 4.8: The examples are not all transparent but require careful thinking about the definition. 1. The two endpoints a and b are the only boundary points of an open interval (a, b). 2. The two endpoints a and b are the only boundary points of a closed interval [a, b]. 3. Every point in the set IN of natural numbers is a boundary point. 4. No point at all is boundary point of the set R. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 224 Sets of Real Numbers Chapter 4 5. Every point on the real line, both rational and irrational, is a boundary point of the set Q. (Think for a while about this one!) ◭ Exercises 4.2.1 Determine the set of interior points, accumulation points, isolated points, and boundary points for each of the following sets: } } 1, 1/2, 1/3, 1/4, 1/5, . . . (a) { 1, 1/2, 1/3, 1/4, 1/5, . . . 0 (b) } ∪ { { (2, 3) (1, 2) (c) (0, 1) ∪ ∪ (1/4, 1/2) (d) (1/2, 1) ∪ x : (e) π x { | − | x : x2 < 2 (f) { } (g) R IN Q (h) R ∪ ∪ < 1 } (3, 4) (1/8, 1/4) \ \ ∪ · · · ∪ (n, n + 1) (1/16, 1/8.2.2 Give an example of each of the following or explain why you think such a set could not exist. (a) A nonempty set with no accumulation points and no isolated points (b) A nonempty set with no interior points and no isolated points (c) A nonempty set with no boundary points and no isolated points 4.2.3 Show that every interior point of a set must also be an accumulation point of that set, but not conversely. 4.2.4 Show that no interior point of a set can be a boundary point, that it is possible for an accumulation point to be a boundary point, and that every isolated point must be a boundary point. 4.2.5 Let E be a nonempty set of real numbers that is bounded above but has no maximum. Let x = sup E. Show that x is a point of accumulation of E. Is it possible for x to also be an interior point of E? Is x a boundary point of E? ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.2. Points 225 4.2.6 State and solve the version of Exercise 4.2.5 that would use the infimum in place of the supremum. 4.2.7 Let A be a set and B = R 4.2.8 Let A be a set and B = R A. Show that every boundary point of A is also a boundary point of B. \ A. Show that every boundary point of A is a point of accumulation of A or else a point of accumulation of B, perhaps both. \ 4.2.9 Must every boundary point of a set be also an accumulation point of that set? 4.2.10 Show that every accumulation point of a set that does not itself belong to the set must be a boundary point of that set. 4.2.11 Show that a point x is not an interior point of a set E if and only if there is a sequence of points xn} { converging to x and no point xn ∈ E. 4.2.12 Let A be a set and B = R 4.2.13 Let A be a set and B = R A. Show that every interior point of A is not an accumulation point of B. A. Show that every accumulation point of A is not an interior point of of B. \ \ 4.2.14 Give an example of a set that has the set IN as its set of accumulation points. 4.2.15 Show that there is no set which has the interval (0, 1) as its set of accumulation points. 4.2.16 Show that there is no set which has the set Q as its set of accumulation points. 4.2.17 Give an example of a set that has the set as its set of accumulation points. E = 0 { 1, 1/2, 1/3, 1/4, 1/5, . . . } ∪ { } 4.2.18 Show that a point x is an accumulation point of a set E if and only if for every ε > 0 there are at least two points belonging to the set E (x ∩ − ε, x + ε). 4.2.19 Suppose that the set xn} { is a convergent sequence converging to a number L and that xn 6 = L for all n. Show that has exactly one point of accumulation, namely L. Of what importance was the assumption that xn 6 all n for this exercise? = L for x : x = xn for some n { } ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) xn} { xn} { xn} { xn} { 226 4.2.20 Let E be a set and a sequence of distinct elements of E. Suppose that limn→∞ xn = x. Show that x is Sets of Real Numbers Chapter 4 a point of accumulation of E. 4.2.21 Let E be a set and a sequence of points, not necessarily elements of E. Suppose that limn→∞ xn = x and that x is an interior point of E. Show that there is an integer N so that xn ∈ E for all n N . ≥ 4.2.22 Let E be a set and a sequence of elements of E. Suppose that lim n→∞ xn = x and that x is an isolated point of E. Show that there is an integer N so that xn = x for all n N . ≥ 4.2.23 Let E be a set and a sequence of distinct points, not necessarily elements of E. Suppose that limn→∞ xn = x and that x2n ∈ E and x2n+1 6∈ E for all n. Show that x is a boundary point of E. 4.2.24 If E is a set of real numbers, then E′, called the derived set of E, denotes the set of all points of accumulation of E. Give an example of each of the following or explain why you think such a set could not exist. (a) A nonempty set E such that E′ = E (b) A nonempty set E such that E′ = ∅ (c) A nonempty set E such that E′ = ∅ (d) A nonempty set E such that E′, E′′ (e) A nonempty set E such that E′, E′′, E′′′, . . . are all different (f) A nonempty set E such that (E but E′′ = = but E′′′ = = (E E′)′ E′) ∅ ∅ ∅ ∪ ∪ 4.2.25 Show that there is no set with uncountably many isolated points. See Note 73 4.3 Sets We now begin a classification of sets of real numbers. Almost all of the concepts of analysis (limits, derivatives, integrals, etc.) can be better understood if a classification scheme for sets is in place. By far the most important notions are those of closed sets and open sets. This is the basis for much advanced mathematics and leads to the subject known as topology, which is fundamental to an understanding of ClassicalRealAnalysis.comThomson*Bruckner*BrucknerE |
lementary Real Analysis, 2nd Edition (2008)6 6 6 Section 4.3. Sets 227 many areas of mathematics. On the real line we can master open and closed sets and describe precisely what they are. 4.3.1 Closed Sets In many parts of mathematics the word “closed” is used to indicate that some operation stays within a system. For example, the set of natural numbers IN is closed under addition and multiplication (any sum or product of two of them is yet another) but not closed under subtraction or division (2 and 3 are natural numbers, but 2 3 and 3/2 are not). This same word was employed originally to indicate sets of real numbers that are “closed” under the operation of taking points of accumulation. If all points of accumulation turn out to be in the set, then the set is said to be closed. This terminology has survived and become, perhaps, the best known usage of the word “closed.” − Definition 4.9: (Closed) Let E be a set of real numbers. The set E is said to be closed provided that every accumulation point of E belongs to the set E. Thus a set E is not closed if there is some accumulation point of E that does not belong to E. In particular, a set with no accumulation points would have to be closed since there is no point that needs to be checked. Example 4.10: The examples are immediate since we have previously described all of the accumulation points of these sets. 1. The empty set ∅ is closed since it contains all of its accumulation points (there are none). 2. The open interval (a, b) not closed because the two endpoints a and b are accumulation points of (a, b) and yet they do not belong to the set. 3. The closed interval [a, b] is closed since only points that are already in the set are accumulation points. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 228 Sets of Real Numbers Chapter 4 4. The set of natural numbers IN is closed because it has no points of accumulation. 5. The real line R is closed since it contains all of its accumulation points, namely every point. 6. The set of rational numbers Q is not closed. Every point on the real line, both rational and irrational, is an accumulation point of Q, but the set fails to contain any irrationals. ◭ If a set is not closed it is because it neglects to contain points that “should” be The Closure of a Set there since they are accumulation points but not in the set. On occasions it is best to throw them in and consider a larger set composed of the original set together with the offending accumulation points that may not have belonged originally to the set. Definition 4.11: (Closure) Let E be any set of real numbers and let E′ denote the set of all accumulation points of E. Then the set is called the closure of the set E. E = E E′ ∪ For example, (a, b) = [a, b], [a, b] = [a, b], IN = IN, and Q = R. Each of these is an easy observation since we know what the points of accumulation of these sets are. 4.3.2 Open Sets Originally, the word “open” was used to indicate a set that was not closed. In time it was realized that this is a waste of terminology, since the class of “not closed sets” is not of much general interest. Instead the word is now used to indicate a contrasting idea, an idea that is not quite an opposite—just at a different extreme. This may be a bit unfortunate since now a set that is not open need not be closed. Indeed some sets can be both open and closed, and some sets can be both not open and not closed. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.3. Sets 229 Definition 4.12: (Open) Let E be a set of real numbers. Then E is said to be open if every point of E is also an interior point of E. Thus every point of E is not merely a point in the set E; it is, so to speak, deep inside the set. For each point x0 of E there is some positive number δ and all points outside E are at least a distance δ away from x0. Note that this means that an open set cannot contain any of its boundary points. Example 4.13: These examples are immediate since we have seen them before in the context of interior points in Section 4.2.1. 1. The empty set is open since it contains no points that are not interior points of the set. (This is the first example of a set that is both open and closed.) ∅ 2. The open interval (a, b) is open since every point x of an open interval (a, b) is an interior point. 3. The closed interval [a, b] is not open since there are points in the set (namely the two endpoints a and b) that are in the set and yet are not interior points. 4. The set of natural numbers IN has no interior points and so this set is not open; all of its points fail to be interior points. 5. Every point of R is an interior point and so R is open. (Remember, R is also closed so it is both open and closed. Note that R and ∅ are the only examples of sets that are both open and closed.) 6. No point of the set of rational numbers Q is an interior point and so Q definitely fails to be open. ◭ If a set is not open it is because it contains points that “shouldn’t” be there since The Interior of a Set they are not interior. On occasions it is best to throw them away and consider a smaller set composed entirely of the interior points. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 230 Sets of Real Numbers Chapter 4 Definition 4.14: (Interior) Let E be any set of real numbers. Then the set int(E) denotes the set of all interior points of E and is called the interior of the set E. For example, int((a, b)) = (a, b), int([a, b]) = (a, b), int(IN) = observation since we know what the interior points of these sets are. ∅ , and int(Q) = . Each of these is an easy ∅ Component Intervals of Open Sets Think of the most general open set G that you can. A first feeble suggestion might be any open interval G = (a, b). We can do a little better. How about the union of two of these If these are disjoint, then we would tend to think of G as having two “components.” It is easy to see that every point is an interior point. We need not stop at two component intervals; any number would work: G = (a, b) (c, d)? ∪ G = (a1, b1) (a2, b2) (a3, b3) (an, bn). ∪ The argument is the same and elementary. If x is a point in this set, then x is an interior point. Indeed we can form the union of a sequence of such open intervals and it is clear that we shall obtain an open set. For a specific example consider ∪ · · · ∪ ∪ ( , 3) (1/2, 1) (1/8, 1/4) (1/32, 1/16) (1/128, 1/64) . . . . ∪ − −∞ ∪ At this point our imagination stalls and it is hard to come up with any more examples that are not obtained by stringing together open intervals in exactly this way. This suggests that, perhaps, all open sets have this structure. They are either open intervals or else a union of a sequence of open intervals. This theorem characterizes all open sets of real numbers and reveals their exact structure. ∪ ∪ ∪ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.3. Sets 231 Theorem 4.15: Let G be a nonempty open set of real numbers. Then there is a unique sequence (finite or infinite) of disjoint, open intervals called the component intervals of G such that (a1, b1), (a2, b2), (a3, b3), . . . , (an, bn), . . . G = (a1, b1) (a2, b2) (a3, b3) ∪ · · · ∪ (an, bn) . . . . ∪ ∪ ∪ G. We know that there must be some interval (a, b) containing the point x Proof. Take any point x and contained in the set G. This is because G is open and so every point in G is an interior point. We need to take the largest such interval. The easiest way to describe this is to write ∈ and Note that α < x < β. Then α = inf { t : (t, x) G } ⊂ β = sup { t : (x, t) G . } ⊂ Ix = (α, β) is called the component of G containing the point x. (It is possible here for α = or β = .) ∞ −∞ One feature of components that we require is this: If x and y belong to the same component, then Ix = Iy If x and y do not belong to the same component, then Ix and Iy have no points in common. This is easily checked (Exercise 4.3.21). There remains the task of listing the components as the theorem requires. If the collection is finite, then this presents no difficulties. If it is infinite we need a clever strategy. Ix : x { G } ∈ Let r1, r2, r3, . . . be a listing of all the rational numbers contained in the set G. We construct our list of components of G by writing for the first step (a1, b1) = Ir1. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 232 Sets of Real Numbers Chapter 4 The second component must be disjoint from this first component. We cannot simply choose Ir2 since if r2 belongs to (a1, b1), then in fact (a1, b1) = Ir1 = Ir2. Instead we travel along the sequence r1, r2, r3, . . . until we reach the first one, say rm2, that does not already belong to the interval (a1, b1). This then serves to define our next interval: (a2, b2) = Irm2 If there is no such point, then the process stops. This process is continued inductively resulting in a sequence of open intervals: . (a1, b1), (a2, b2), (a3, b3), . . . , (an, bn), . . . , which may be infinite or finite. At the kth stage a point rmk is selected so that rmk does not belong to any component thus far selected. If this cannot be done, then the process stops and produces only a finite list of components. The proof is completed by checking that (i) every point of G is in one of these intervals, (ii) every point in one of these intervals belongs to G, and (iii) the intervals in the sequence must be disjoint. For (i) note that if x first number rk in the list that belongs to this component. But then x this interval Irk at some stage. Thus x does belong to one of these intervals. ∈ ∈ G, then there must be rational numbers in the component Ix. Indeed there is a Irk and so we must have chosen For (ii) note that if x is in G, then Ix ⊂ For (iii) consider some pair of intervals in the sequence we have constructed. The late |
r one chosen was required to have a point rmk that did not belong to any of the preceding choices. But that means then that the new component chosen is disjoint from all the previous ones. G. Thus every point in one of the intervals belongs to G. This completes the checking of the details and so the proof is done. Exercises 4.3.1 Is it true that a set, all of whose points are isolated, must be closed? See Note 74 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.3. Sets 233 4.3.2 If a set has no isolated points must it be closed? Must it be open? 4.3.3 A careless student, when asked, incorrectly remembers that a set is closed “if all its points are points of accumulation.” Must such a set be closed? 4.3.4 A careless student, when asked, incorrectly remembers that a set is open “if it contains all of its interior points.” Is there an example of a set that fails to have this property? Is there an example of a nonopen set that has this property? 4.3.5 Determine which of the following sets are open, which are closed, and which are neither open nor closed. (a) ( , 0) (0, ) −∞ 1, 1/2, 1/3, 1/4, 1/5, . . . ∞ ∪ 1, 1/2, 1/3, 1/4, 1/5, . . . } (b) { (c) 0 } ∪ { { (d) (0, 1) ∪ (e) (1/2, 1 : x2 < 2 IN Q \ (f) { (g) { (h) R (i) R } \ (1, 2) (2, 3) (3, 4) (n, n + 1) (1/8, 1/4) (1/16, 1/81/4, 1/2) ∪ ∪ < 1 } 4.3.6 Show that the closure operation has the following properties: E2. E2. E2, then E1 ⊂ E2 = E1 ∪ E1 ∩ E2 ⊂ (a) If E1 ⊂ (b) E1 ∪ (c) E1 ∩ (d) Give an example of two sets E1 and E2 such that E2 6 E1 ∩ E2. = E1 ∩ E2. (e) E = E. 4.3.7 Show that the interior operation has the following properties: ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 234 Sets of Real Numbers Chapter 4 (a) If E1 ⊂ (b) int(E1 ∩ (c) int(E1 ∪ (d) Give an example of two sets E1 and E2 such that E2, then int(E1) E2) = int(E1) int(E1) E2) ⊂ int(E2). int(E2). int(E2). ⊃ ∪ ∩ (e) int(int(E)) = int(E). int(E1 ∪ E2) = int(E1) int(E2). ∪ 4.3.8 Show that if the set E′ of points of accumulation of E is empty, then the set E must be closed. 4.3.9 Show that the set E′ of points of accumulation of any set E must be closed. 4.3.10 Show that the set int(E) of interior points of any set E must be open. 4.3.11 Show that a set E is closed if and only if E = E. 4.3.12 Show that a set E is open if and only if int(E) = E. 4.3.13 If A is open and B is closed, what can you say about the sets A \ 4.3.14 If A and B are both open or both closed, what can you say about the sets A B and B \ A? B and B A? \ \ 4.3.15 If E is a nonempty bounded, closed set, show that max E and min E both exist. If E is a bounded, open set, show that neither max { E } nor min E { } exist (although sup { E } { } { } and inf do). E { } 4.3.16 Show that if a set of real numbers E has at least one point of accumulation, then for every ε > 0 there exist points x, y ∈ E so that 0 < x | y | − < ε. 4.3.17 Construct an example of a set of real numbers E that has no points of accumulation and yet has the property that for every ε > 0 there exist points x, y x | be a sequence of real numbers. Let E denote the set of all numbers z that have the property that E so that 0 < < ε. − ∈ y | 4.3.18 Let xn} { there exists a subsequence convergent to z. Show that E is closed. xnk } { 4.3.19 Determine the components of the open set R IN. \ 4.3.20 Let F = set , 1/2, 1/3, 1/4, 1/5, . . . } . Show that F is closed and determine the components of the open ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 4.3. Sets 235 4.3.21 In the proof of Theorem 4.15 show that if x and y belong to the same component, then Ix = Iy, while if x and y do not belong to the same component, then Ix and Iy have no points in common. 4.3.22 In the proof of Theorem 4.15, after obtaining the collection of components , there remained the task of listing them. In classroom discussions the following suggestions were made as to how the components might be listed: Ix : x G ∈ { } (a) List the components from largest to smallest. (b) List the components from smallest to largest. (c) List the components from left to right. (d) List the components from right to left. For each of these give an example of an open set with infinitely many components for which this strategy would work and also an example where it would fail. 4.3.23 In searching for interesting examples of open sets, you may have run out of ideas. Here is an example of a construction due to Cantor that has become the source for many important examples in analysis. We describe the component intervals of an open set G inside the interval (0, 1). At each “stage” n we shall describe 2n−1 components. At the first stage, stage 1, take (1/3, 2/3) and at stage 2 take (1/9, 2/9) and (7/9, 8/9) and so on so that at each stage we take all the middle third intervals of the intervals remaining inside (0, 1). The set G is the open subset of (0, 1) having these intervals as components. (a) Describe exactly the collection of intervals forming the components of G. (b) What are the endpoints of the components. How do they relate to ternary expansions of numbers in [0, 1]? (c) What is the sum of the lengths of all components? (d) Sketch a picture of the set G by illustrating the components at the first three stages. (e) Show that if x, y G, x < y, but x and y are not in the same component, then there are infinitely many components of G in the interval (x, y). ∈ See Note 75 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 236 Sets of Real Numbers Chapter 4 4.4 Elementary Topology The study of open and closed sets in any space is called topology. Our goal now is to find relations between these ideas and examine the properties of these sets. Much of this is a useful introduction to topology in any space; some is very specific to the real line, where the topological ideas are easier to sort out. The first theorem establishes the connection between the open sets and the closed sets. They are not quite opposites. They are better described as “complementary.” Theorem 4.16 (Open vs. Closed) Let A be a set of real numbers and B = R A is open if and only if B is closed. \ A its complement. Then If A is open and B fails to be closed then there is a point z that is a point of accumulation of B Proof. and yet is not in B. Thus z must be in A. But if z is a point in an open set it must be an interior point. Hence there is an interval (z δ, z + δ) contained entirely in A; such an interval contains no points of B. Hence z cannot be a point of accumulation of B. This is a contradiction and so we have proved that B must be closed if A is open. − Conversely, if B is closed and A fails to be open, then there is a point z A that is not an interior point of A. Hence every interval (z δ, z + δ) must contain points outside of A, namely points in B. By definition this means that z is a point of accumulation of B. But B is closed and so z, which is a point in A, should really belong to B. This is a contradiction and so we have proved that A must be open if B is closed. − ∈ Theorem 4.17 (Properties of Open Sets) Open sets of real numbers have the following properties: 1. The sets ∅ and R are open. 2. Any intersection of a finite number of open sets is open. 3. Any union of an arbitrary collection of open sets is open. 4. The complement of an open set is closed. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.4. Elementary Topology 237 Proof. The first assertion is immediate and the last we have already proved. The third is easy. Thus it is enough for us to prove the second assertion. Let us suppose that E1 and E2 are open. To show that E1 ∩ is in both of the sets E1 and E2 and both are open there are intervals E2 is also open we need to show that every point is an interior point. Let z E2. Then, since z E1 ∩ ∈ and Let δ = min { δ1, δ2} . We must then have (z − δ1, z + δ1) E1 ⊂ (z − δ2, z + δ2) E2. ⊂ − which shows that z is an interior point of E1 ∩ (z E1 ∩ δ, z + δ) E2. Since z is any point, this proves that E1 ∩ E2, ⊂ Having proved the theorem for two open sets, it now follows for three open sets since E2 ∩ E3 = (E1 ∩ E1 ∩ E2) E3. ∩ That any intersection of an arbitrary finite number of open sets is open now follows by induction. E2 is open. Theorem 4.18 (Properties of Closed Sets) Closed sets of real numbers have the following properties: 1. The sets ∅ and R are closed. 2. Any union of a finite number of closed sets is closed. 3. Any intersection of an arbitrary collection of closed sets is closed. 4. The complement of a closed set is open. Proof. second one. Let us suppose that E1 and E2 are closed. To show that E1 ∪ show that every accumulation point belongs to that set. Let z be an accumulation point of E1 ∪ Except for the second assertion these are easy or have already been proved. Let us prove the E2 is also closed we need to E2 that ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 238 Sets of Real Numbers Chapter 4 does not belong to the set. Since z is in neither of the closed sets E1 and E2, this point z cannot be an accumulation point of either. Thus some interval (z − Consequently, that interval contains no points of E1 ∪ contradicting our assumption. Since z is any accumulation point, this proves that E1 ∪ δ, z + δ) contains no points of either E1 or E2. E2 and z is not an accumulation point after all, E2 is closed. Having proved the theorem for two closed sets, it now follows for three closed sets since E2 ∪ E3 = (E1 ∪ E1 ∪ E2) E3. ∪ That any union of an arbitrary finite number of closed sets is closed now follows by induction. Exercises 4.4.1 Explain why it is that the sets and R are open and also closed. ∅ 4.4.2 Show that a union of an arbitrary collection of open sets is open. 4.4.3 Show that an intersection of an arbitrary collection of |
closed sets is closed. 4.4.4 Give an example of a sequence of open sets G1, G2, G3, . . . whose intersection is neither open nor closed. Why does this not contradict Theorem 4.17? 4.4.5 Give an example of a sequence of closed sets F1, F2, F3, . . . whose union is neither open nor closed. Why does this not contradict Theorem 4.18? 4.4.6 Show that the set E can be described as the smallest closed set that contains every point of E. See Note 76 4.4.7 Show that the set int(E) can be described as the largest open set that is contained inside E. See Note 77 4.4.8 A function f : R so that is open. Let E be an arbitrary closed set. Is it possible to construct a function f : R points at which f is not bounded is precisely the set E? R is said to be bounded at a point x0 provided that there are positive numbers ε and M ε, x0 + ε). Show that the set of points at which a function is bounded R so that the set of < M for all x f (x) | | (x0 − → → ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.5. Compactness Arguments 239 4.4.9 This exercise continues Exercise 4.3.23. Define the Cantor ternary set K to be the complement of the open set G of Exercise 4.3.23 in the interval [0, 1]. (a) If all the open intervals up to the nth stage in the construction of G are removed from the interval [0, 1], there remains a closed set Kn that is the union of a finite number of closed intervals. How many intervals? (b) What is the sum of the lengths of these closed intervals that make up Kn? (c) Show that K = (d) Sketch a picture of the set K by illustrating the sets K1, K2, and K3. T (e) Show that if x, y K, x < y, then there is an open subinterval I ∞ n=1 Kn. (f) Give an example of a number z (0, 1) that is not an endpoint of a component of G. ∈ K ∈ ∩ (x, y) containing no points of K. ⊂ 4.4.10 Express the closed interval [0, 1] as an intersection of a sequence of open sets. Can it also be expressed as a union of a sequence of open sets? 4.4.11 Express the open interval (0, 1) as a union of a sequence of closed sets. Can it also be expressed as an intersection of a sequence of closed sets? 4.5 Compactness Arguments " Parts of this section could be cut in a short course. For a minimal approach to compactness arguments, you may wish to skip over all but the Bolzano-Weierstrass property. For all purposes of elementary real analysis this is sufficient. Proofs in the sequel that require a compactness argument will be supplied with one that uses the Bolzano-Weierstrass property and, perhaps, another that can be omitted. In analysis we frequently encounter the problem of arguing from a set of “local” assumptions to a “global” conclusion. Let us focus on just one problem of this type and see the kind of arguments that can be used. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 240 Sets of Real Numbers Chapter 4 Local Boundedness of a Function Suppose that a function f is locally bounded at each point of a set E. By this we mean that for every point x δ, x + δ) and f is bounded on the points in E that belong to that interval. Can we conclude that f is bounded on the whole of the set E? E there is an interval (x − ∈ Thus we have been given a local condition at each point x in the set E. There must be numbers δx and Mx so that The global condition we want, if possible, is to have some single number M that works for all t is, E; that ∈ f (t) | | ≤ Mx for all t E in the interval (x δx, x + δx). ∈ − Two examples show that this depends on the nature of the set E. f (t) | | ≤ M for all t E. ∈ Example 4.19: The function f (x) = 1/x is locally bounded at each point x in the set (0, 1) but is not bounded on the set (0, 1). It is clear that f cannot be bounded on (0, 1) since the statement 1 t ≤ M for all t (0, 1) ∈ cannot be true for any M . But this function is locally bounded at each point x here. Let x δx = x/2 and Mx = 2/x. Then (0, 1). Take ∈ f (t) = = Mx 1 t ≤ 2 x if x/2 = x δx < t < x + δx. − What is wrong here? What is there about this set E = (0, 1) that does not allow the conclusion? The point 0 is a point of accumulation of (0, 1) that does not belong to (0, 1), and so there is no assumption that f is bounded at that point. We avoid this difficulty if we assume that E is closed. ◭ Example 4.20: The function f (x) = x is locally bounded at each point x in the set [0, bounded on the set [0, ). It is clear that f cannot be bounded on [0, ) since the statement ∞ ) but is not ∞ f (t) = t ≤ M for all t [0, ) ∞ ∈ ∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.5. Compactness Arguments 241 cannot be true for any M . But this function is locally bounded at each point x here. Let x δx = 1 and Mx = x + 1. Then ∈ [0, ∞ ). Take f (t) = t ≤ x + 1 = Mx if x 1 < t < x + 1. − What is wrong here? What is there about this set E = [0, ) that does not allow the conclusion? This set is closed and so contains all of its accumulation points so that the difficulty we saw in the preceding example does not arise. The difficulty is that the set is too big, allowing larger and larger bounds as we move to the right. We could avoid this difficulty if we assume that E is bounded. ◭ ∞ Indeed, as we shall see, we have reached the correct hypotheses now for solving our problem. The version of the theorem we were searching for is this: Theorem Suppose that a function f is locally bounded at each point of a closed and bounded set E. Then f is bounded on the whole of the set E. Arguments that exploit the special features of closed and bounded sets of real numbers are called compactness arguments. Most often they are used to prove that some local property has global implications, which is precisely the nature of our boundedness theorem. We now solve our problem using various different compactness arguments. Each of these arguments will become a formidable tool in proving theorems in analysis. Many situations will arise in which some local property must be proved to hold globally, and compactness will play a huge role in these. 4.5.1 Bolzano-Weierstrass Property A closed and bounded set has a special feature that can be used to design compactness arguments. This property is essentially a repeat of a property about convergent subsequences that we saw in Section 2.11. Theorem 4.21 (Bolzano-Weierstrass Property) A set of real numbers E is closed and bounded if and only if every sequence of points chosen from the set has a subsequence that converges to a point that belongs to E. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 242 Sets of Real Numbers Chapter 4 Suppose that E is both closed and bounded and let xn} be a sequence of points chosen from E. Proof. { must be bounded too. We apply the Bolzano-Weierstrass theorem Since E is bounded this sequence that converges. If xnk → for sequences (Theorem 2.40) to obtain a subsequence z then since all the points of the subsequence belong to E either the sequence is constant after some term or else z is a point E. This proves the theorem in one direction. of accumulation of E. In either case we see that z xnk } xn} { { In the opposite direction we suppose that a set E, which we do not know in advance to be either closed or bounded, has the Bolzano-Weierstrass property. Then E cannot be unbounded. For example, if E is unbounded then there is a sequence of points xn} sequence converges, contradicting the assumption. of E with xn → ∞ and no subsequence of that −∞ or { Also, E must be closed. If not, there is a point of accumulation z that is not in E. This means that there is a sequence of points to z and, since z xn} { E, we again have a contradiction. in E converging to z. But any subsequence of xn} { would also converge ∈ This theorem can also be interpreted as a statement about accumulation points. 6∈ Corollary 4.22: A set of real numbers E is closed and bounded if and only if every infinite subset of E has a point of accumulation that belongs to E. Let us use the Bolzano-Weierstrass property to prove our theorem about local boundedness. Theorem Suppose that a function f is locally bounded at each point of a closed and bounded set E. Then f is bounded on the whole of the set E. Proof. (Bolzano-Weierstrass compactness argument) To use this argument we will need to construct a sequence of points in E that we can use. Our proof is a proof by contradiction. If f is not bounded on E chosen from E so that there must be a sequence of points xn} { f (xn) > n | for all n. If such a sequence could not be chosen, then at some stage, N say, there are no more points with f (xN ) > N and N is an upper bound. | | | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.5. Compactness Arguments 243 By compactness (i.e., by Theorem 4.21) there is a convergent subsequence E. By the local boundedness assumption there is an open interval (z xnk } converging to a point δ, z + δ) and a number Mz so { − z ∈ that | ≤ whenever t is in E and inside that interval. But for all sufficiently large values of k, the point xnk must belong to the interval (z δ, z + δ). The two statements | f (t) Mz − f (xnk ) > nk and f (xnk ) Mz | | ≤ | | cannot both be true for all large k and so we have reached a contradiction, proving the theorem. 4.5.2 Cantor’s Intersection Property " Enrichment section. May be omitted. A famous compactness argument, one that is used often in analysis, involves the intersection of a descending sequence of sets; that is, a sequence with E2 ⊃ E1 ⊃ What conditions on the sequence will imply that E3 ⊃ E4 ⊃ . . . . Example 4.23: An example shows that some conditions are needed. Suppose that for each n En = (0, 1/n). Then ∈ IN we let ∞ n=1 \ = En 6 ? ∅ E3 ⊃ is a descending sequence of sets with empty intersection. The same is true of the sequence En} are bounded (but not closed) while the sets in ◭ ). Observe that the set |
s in the sequence are closed (but not bounded). E1 ⊃ E2 ⊃ . . . , { En} so { Fn = [n, the sequence ∞ Fn} { ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 244 Sets of Real Numbers Chapter 4 In a paper in 1879 Cantor described the following theorem and the role it plays in analysis. He pointed out that variants on this idea had been already used throughout most of that century, notably by Lagrange, Legendre, Dirichlet, Cauchy, Bolzano, and Weierstrass. be a sequence of nonempty closed and bounded subsets of real numbers such that E2 ⊃ En} { . . . . Let E = E3 ⊃ For each i ∞n=1 En. Then E is not empty. Theorem 4.24: Let E1 ⊃ Proof. Ei. The sequence bounded set E1. Therefore, because of Theorem 4.21, denote that limit. Fix an integer m. Because the sets are descending, xik ∈ k ∈ IN. But Em is closed, from which it follows that z IN choose xi ∈ T ∈ ∈ xi} { xi} { Em. This is true for all m is bounded since every point lies inside the . Let z has a convergent subsequence xik } { Em for all sufficiently large E. IN, so z ∈ ∈ Corollary 4.25 (Cantor Intersection Theorem) Suppose that subsets of real numbers such that En} { is a sequence of nonempty closed If then the intersection consists of a single point. E1 ⊃ E2 ⊃ E3 ⊃ . . . . diameter En → 0, E = ∞ En n=1 \ Proof. Here the diameter of a nonempty, closed bounded set E would just be max E min E, which exists and is finite for such a set (see Exercise 4.3.15). Since we are assuming that the diameters shrink to zero it follows that, at least for all sufficiently large n, En must be bounded. − That E and y x ∈ = R, y follows from Theorem 4.24. It remains to show that E contains only one point. Let x ∅ = x. Since diameter En → IN such that diameter Ei < ∈ . Since − ∈ x y | | 0, there exists i x E and E = as required. E Ei, y cannot be in Ei. Thus y / ∈ ∈ { } ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 Section 4.5. Compactness Arguments 245 Now we prove our theorem about local boundedness by using the Cantor intersection property to frame an argument. Theorem Suppose that a function f is locally bounded at each point of a closed and bounded set E. Then f is bounded on the whole of the set E. (Cantor intersection compactness argument). To use this argument we will need to construct Proof. a sequence of closed and bounded sets shrinking to a point. Our proof is again a proof by contradiction. Suppose that f is not bounded on E. Since E is bounded we may assume that E is contained in some interval [a, b]. Divide that interval in a)/2. At least one of these intervals contains half, forming two subintervals of the same length, namely (b points of E and f is unbounded on that interval. Call it [a1, b1]. − Now do the same to the interval [a1, b1]. Divide that interval in half, forming two subintervals of the same length, namely (b a)/4. At least one of these intervals contains points of E and f is unbounded on that interval. Call it [a2, b2]. Continue this process inductively, producing a descending sequence of a)/2n, contains points of E, and f is intervals [an, bn] unbounded on E so that the nth interval [an, bn] has length (b [an, bn]. − − { By the Cantor intersection property there is a single point z E contained in all of these intervals. But c, z + c) so that f is bounded on the points of by our local boundedness assumption there is an interval (z E in that interval. For any large enough value of n, though, the interval [an, bn] would be contained inside c, z + c). This would be impossible and so we have reached a contradiction, proving the the interval (z theorem. − − ∈ } ∩ 4.5.3 Cousin’s Property " Enrichment section. May be omitted. Another compactness argument dates back to Pierre Cousin in the last years of the nineteenth century. This exploits the order of the real line and considers how small intervals may be pieced together to give ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 246 Sets of Real Numbers Chapter 4 larger intervals. The larger interval [a, b] is subdivided a = x0 < x1 < < xn = b · · · and then expressed as a finite union of nonoverlapping subintervals said to form a partition: [i=1 This again provides us with a compactness argument since it allows a way to argue from the local to the global. n [a, b] = [xi 1, xi]. − Lemma 4.26 (Cousin) Let x length smaller than δ. Then there exists a partition [a, b] there exists δ = δ(x) > 0 such that ∈ C C be a collection of closed subintervals of [a, b] with the property that for each [a, b] that contain x and have contains all intervals [c, d] ⊂ of [a, b] such that [xi − 1, xi] ∈ C for i = 1, . . . , n. a = x0 < x1 < < xn = b · · · This lemma makes precise the statement that if a collection of closed intervals contains all “sufficiently small” ones for [a, b], then it contains a partition of [a, b]. We shall frequently see the usefulness of such a partition. This is the most elementary of a collection of tools called covering theorems. Roughly, a cover of a set is a family of intervals covering the set in the sense that each point in the set is contained in one or more of the intervals. We formalize the assumption in Cousin’s lemma in this language: Definition 4.27: (Cousin Cover) A collection lemma is called a Cousin cover of [a, b]. C of closed intervals satisfying the hypothesis of Cousin’s Proof. (Proof of Cousin’s lemma) Let us, in order to obtain a contradiction, suppose that not contain a partition of the interval [a, b]. Let c be the midpoint of that interval and consider the two does C ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.5. Compactness Arguments 247 subintervals [a, c] and [c, b]. If partitions together we can obtain a partition of [a, b], which we have supposed is impossible. contains a partition of both intervals [a, c] and [c, b], then by putting those C Let I1 = [a, b] and let I2 be either [a, c] or [c, b] chosen so that we can continue in this fashion, obtaining a shrinking sequence of intervals I1 ⊃ length of In is (b contains no partition of In. a)/2n 1 and − C contains no partition of I2. Inductively . . . so that the I3 ⊃ I2 ⊃ By the Cantor intersection theorem (Theorem 4.25) there is a single point z in all of these intervals. For sufficiently large n, the interval In contains z and has length smaller than δ(z). Thus, by definition, is In} In ∈ C itself a partition. But this contradicts the way in which the sequence was chosen and this contradiction completes our proof. does indeed contain a partition of that interval In since the single interval . In particular, C { − C Now we reprove our theorem about local boundedness by using Cousin’s property to frame an argument. Theorem Suppose that a function f is locally bounded at each point of a closed and bounded set E. Then f is bounded on the whole of the set E. Proof. [a, b]. Let us say that an interval [c, d] (Cousin compactness argument) The set E is bounded and so is contained in some interval [a, b] is “black” if the following statement is true: ⊂ There is a number M (which may depend on [c, d]) so that the interval [c, d]. | f (t) | ≤ M for all t ∈ E that are in The collection of all black intervals is a Cousin cover of [a, b]. This is because of the local boundedness assumption on f . Consequently, by Cousin’s lemma, there is a partition of the interval [a, b] consisting of black intervals. The function f is bounded in E on each of these finitely many black intervals and so, since there are only finitely many of them, f must be bounded on E in [a, b]. But [a, b] includes all of E and so the proof is complete. 4.5.4 Heine-Borel Property " Advanced section. May be omitted. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 248 Sets of Real Numbers Chapter 4 Another famous compactness property involves covers too, as in the Cousin lemma, but this time covers consisting of open intervals. This theorem has wide applications, including again extensions of local properties to global ones. You may find this compactness argument more difficult to work with than the others. On the real line all of the arguments here are equivalent and, in most cases, any one will do the job. Why not use the simpler ones then? The answer is that in more general spaces than the real line these other versions may be more useful. Time spent learning them now will pay off in later courses. The property we investigate is named after two mathematicians, ´Emile Borel (1871–1956) and Heinrich Eduard Heine (1821–1881), whose names have become closely attached to these ideas. We begin with some definitions. Definition 4.28: (Open Cover) Let A there exists at least one interval U ⊂ such that x R and let be a family of open intervals. If for every x is called an open cover of A. U , then U A ∈ ∈ U ∈ U Definition 4.29: (Heine-Borel Property) A set A open cover of A can be reduced to a finite subcover. That is, if finite subset of such that ⊂ , U U1, U2, . . . , Un} { U R is said to have the Heine-Borel property if every is an open cover of A, then there exists a A U1 ∪ U2 ∪ · · · ∪ ⊂ Un. Example 4.30: Any finite set has the Heine-Borel property. Just take one interval from the cover for each ◭ element in the finite set. Example 4.31: The set IN does not have the Heine-Borel property. Take, for example, the collection of open intervals (0, n) : n = 1, 2, 3, . . . While this forms an open cover of IN, no finite subcollection could also be an open cover. Example 4.32: The set A = the collection of open intervals { 1/n : n IN } ∈ does not have the Heine-Borel property. Take, for example, (1/n, 2) : n = 1, 2, 3, . . . While this forms an open cover of A, no finite subcollection could also be an open cover. ◭ ◭ { { . } . } ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.5. Compactness Arguments 249 O |
bserve in these examples that IN is closed (but not bounded) while A is bounded (but not closed). We shall prove, in Theorem 4.33, that a set A has the Heine-Borel property if and only if that set is both closed and bounded. Theorem 4.33 (Heine-Borel) A set A and bounded. ⊂ R has the Heine-Borel property if and only if A is both closed Proof. Suppose A otherwise there is nothing to prove. Let [a, b] be the smallest closed interval containing A; that is, is an open cover for A. We may assume A R is both closed and bounded, and ⊂ U = , ∅ a = inf x : x A and b = sup x : x { ∈ } { A . } ∈ ∈ ∈ For each x A. We shall apply Cousin’s lemma to the interval [a, b], so we need to first A and b Observe that a define an appropriate Cousin cover of [a, b]. A, since ∈ is an open cover of A, there exists an open interval Ux ∈ U t, x + t) U Since Ux is open, there exists δ(x) > 0 for which (x points in A. Now consider points in V = [a, b] and all t of Ux. (0, δ(x)). This defines δ(x) for A. We must define δ(x) for points of V . Since A is closed V for of [a, b] as follows: An interval [c, d] is a member V . C Observe that an interval of type (i) can contain points of V , but an interval of type (ii) cannot contain A, V is open (why?); thus for each x (0, δ(x)). We can therefore obtain a Cousin cover { ∈ if there exists x V there exists δ(x) > 0 such that (x [a, b] such that either (i) x C A and x Ux or (ii) x Ux for all t such that x V and x t, x + t) [c, d] [c, d] } ⊂ a, b ⊂ − ⊂ ⊂ − ⊂ ∈ ∈ ∈ ∈ ∈ ∈ ∈ ∈ \ points of A. Figure 4.2 illustrates examples of both types of intervals. In that figure [c, d] interval of type (i) in It is clear that C forms a Cousin cover of [a, b]. From Cousin’s lemma we infer the existence of a V is an interval of type (ii) in Ux is an ⊂ ; [c′, d′] ⊂ C . < xn = b with [xi 1, xi] is partition a = x0 < x1 < 1, xi] . We either contained in V (in which case it is disjoint from A) or is contained in some member Ui ∈ U now “throw away” from the partition those intervals that contain only points of V , and the union of the remaining closed intervals covers all of A. Each interval of this finite collection is contained in some open interval U from the cover for i = 1, . . . , n. Each of the intervals [xi . More precisely, let ∈ and [xi − 1, xi] . Ui} ⊂ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 250 Then so Sets of Real Numbers Chapter 4 A V ] ( ( [ c x Ux ) ] d - [ ] c′ d′ Figure 4.2. The two types of intervals in the proof of Theorem 4.33. A ⊂ [i S ∈ [xi − 1, xi] ⊂ Ui, S [i ∈ Ui : i { S } ∈ is the required subcover of A. To prove the converse, we must show that if A is not bounded or if A is not closed, then there exists an open cover of A with no finite subcover. Suppose first that A is not bounded. Consider the family of open intervals is an open cover of A. (Indeed it is an open cover of all of R.) But it is also clear that contains Clearly no finite subcover of A since a finite subcover will cover only a bounded set and we have assumed that A is unbounded. U U = ( n, n) : n U { − IN . } ∈ Now suppose A is not closed. Then there is a point of accumulation z of A that does not belong to A. Consider the family of open intervals = U , z − −∞ 1 n : n ∈ IN ∪ z + 1 n , : n ∈ IN . ∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.5. Compactness Arguments 251 U is an open cover of A. (Indeed it is an open cover of all of R Clearly U contains no finite subcover of A since a finite subcover contains no points of the interval (z c, z + c) for some small positive c and yet, since z is an accumulation point of A, this interval must contain infinitely many points of A. .) But it is also clear that \ { − } z Once again, we return to our sample theorem, which shows how a local property can be used to prove a global condition, this time using a Heine-Borel compactness argument. Theorem Suppose that a function f is locally bounded at each point of a closed and bounded set E. Then f is bounded on the whole of the set E. (Heine-Borel compactness argument). As f is locally bounded at each point of E, for every < Mx for E there exists an open interval Ux containing x and a positive number Mx such that f (t) | | Ux ∩ E. Let ∈ Proof. x ∈ all t Then U such that Let is an open cover of E. By the Heine-Borel theorem there exists = Ux : x U { E . } ∈ Ux1, Ux2, . . . , Uxn} { E Ux1 ∪ Ux2 ∪ · · · ∪ ⊂ Uxn. Let x ∈ E. Then there exists i, 1 i ≤ ≤ we conclude that f is bounded on E. | M = max n, for which x { . Mx1, Mx2, . . . , Mxn} Uxi. Since ∈ M Mxi ≤ | ≤ f (x) Our ability to reduce to a finite subcover in the proof of this theorem was crucial. You may wish to use the function f (x) = 1/x on (0, 1] to appreciate this statement. U ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 252 4.5.5 Compact Sets Sets of Real Numbers Chapter 4 We have seen now a wide range of techniques called compactness arguments that can be applied to a set that is closed and bounded. We now introduce the modern terminology for such sets. Definition 4.34: A set of real numbers E is said to be compact if it has any of the following equivalent properties: 1. E is closed and bounded. 2. E has the Bolzano-Weierstrass property. 3. E has the Heine-Borel property. In spaces more general than the real line there may be analogues of the notions of closed, bounded, convergent sequences, and open covers. Thus there can also be analogues of closed and bounded sets, the Bolzano-Weierstrass property, and the Heine-Borel property. In these more general spaces the three properties are not always equivalent and it is the Heine-Borel property that is normally chosen as the definition of compact sets there. Even so, a thorough understanding of compactness arguments on the real line is an excellent introduction to these advanced and important ideas in other settings. If we return to our sample theorem we see that now, perhaps, it should best be described in the language of compact sets: Theorem Suppose that E is compact. Then every function f : E on E is bounded on the whole of the set E. Conversely, if every function f : E → locally bounded on E is bounded on the whole of the set E, then E must be compact. → R that is locally bounded R that is In real analysis there are many theorems of this type. The concept of compact set captures exactly when many local conditions can have global implications. Exercises 4.5.1 Give an example of a function f : R R that is not locally bounded at any point. → ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.5. Compactness Arguments See Note 78 253 ) does not have the Bolzano-Weierstrass property. 4.5.2 Show directly that the interval [0, ∞ 4.5.3 " Show directly that the interval [0, 4.5.4 " Show directly that the set [0, 1] 4.5.5 Develop the properties of compact sets: (i) is the union of a pair of compact sets compact? (ii) is the union of a finite sequence of compact sets compact? (iii) is the union of a sequence of compact sets compact? Do the same for intersections. See Note 79 ∞ Q does not have the Heine-Borel property. ) does not have the Heine-Borel property. ∩ 4.5.6 Show directly that the union of two sets with the Bolzano-Weierstrass property must have the Bolzano- Weierstrass property. 4.5.7 " Show directly that the union of two sets with the Heine-Borel property must have the Heine-Borel property. 4.5.8 We defined an open cover of a set E to consist of open intervals covering E. Let us change that definition to allow an open cover to consist of any family of open sets covering E. What changes are needed in the proof of Theorem 4.33 so that it remains valid in this greater generality? See Note 80 R is said to be locally increasing at a point x0 if there is a δ > 0 so that 4.5.9 A function f : R → whenever f (x) < f (x0) < f (y) δ < x < x0 < y < x0 + δ. x0 − Show that a function that is locally increasing at every point in R must be increasing; that is, that f (x) < f (y) for all x < y. See Note 81 4.5.10 Let f : E → R have this property: For every e E there is an ε > 0 so that ∈ f (x) > ε if x E (e ε, e + ε). ∈ ∩ − Show that if the set E is compact then there is some positive number c so that f (e) > c ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 254 Sets of Real Numbers Chapter 4 for all e ∈ E. Show that if E is not closed or is not bounded, then this conclusion may not be valid. 4.5.11 Prove the following variant of Lemma 4.26: C be a collection of closed subintervals of [a, b] with the property that for each x contains all intervals [c, d] ⊂ has the property that if [α, β] and [β, γ] both belong to [a, b] there [a, b] that contain x and have length then so Let exists δ = δ(x) > 0 such that smaller than δ. Suppose that too does [α, γ]. Then [a, b] belongs to C C ∈ C . C 4.5.12 Use the version of Cousin’s lemma given in Exercise 4.5.11 to give a simpler proof of the sample theorem on local boundedness. 4.5.13 " Give an example of an open covering of the set Q of rational numbers that does not reduce to a finite subcover. 4.5.14 Suppose that E is closed and K is compact. Show that E “closed + bounded” definition and also using the Bolzano-Weierstrass property). ∩ K is compact. Do this in two ways (using the 4.5.15 Prove that every function f : E R that is locally bounded on E is bounded on the whole of the set E only if the set E is compact, by supplying the following two constructions: → (a) Show that if the set E is not bounded, then there is an unbounded function f : E locally bounded on E. R so that f is → (b) Show that if the set E is not closed, then there is an unbounded function f : E bounded on E. R so that f is locally → 4.5.16 " Suppose that E is closed and K is compact. Show that E K is compact using the Heine-Borel property. ∩ 4.5.17 Suppose that E is compact. Is the set of boundary points |
of E also compact? 4.5.18 " Prove Lindel¨off’s covering theorem: be a collection of open intervals such that every point of a set E belongs to at least one of the Let intervals. Then there is a sequence of intervals I1, I2, I3, . . . chosen from C that also covers E. C See Note 82 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.6. Countable Sets 255 4.5.19 Describe briefly the distinction between the covering theorem of Lindel¨off (expressed in Exercise 4.5.18) and that of Heine-Borel. See Note 83 4.5.20 " We have seen that the following four conditions on a set A R are equivalent: ⊂ (a) A is closed and bounded. (b) Every infinite subset of A has a limit point in A. (c) Every sequence of points from A has a subsequence converging to a point in A. (d) Every open cover of A has a finite subcover. Prove directly that (b) (c), (b) ⇒ (d) and (c) ⇒ (d). ⇒ See Note 84 4.5.21 Let f be a function that is locally bounded on a compact interval [a, b]. Let is bounded on [a, x] . } (a) Show that S = . ∅ (b) Show that if z = sup S, then a < z b. ≤ (c) Show that z S. ∈ (d) Show that z = b by showing that z < b is impossible. Using these steps, construct a proof of the sample theorem on local boundedness. 4.6 Countable Sets As part of our discussion of properties of sets in this chapter let us review a special property of sets that relates, not to their topological properties, but to their size. We can divide sets into finite sets and infinite sets. How do we divide infinite sets into “large” and “larger” infinite sets? ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 256 Sets of Real Numbers Chapter 4 We did this in our discussion of sequences in Section 2.3. (If you skipped over that section now is a good time to go back.) If an infinite set E has the property that the elements of E can be written as a list (i.e., as a sequence) e1, e2, e3, . . . , en, . . . , then that set is said to be countable. Note that this property has nothing particularly to do with the other properties of sets encountered in this chapter. It is yet another and different way of classifying sets. The following properties review our understanding of countable sets. Remember that the empty set, any finite set, and any infinite set that can be listed are all said to be countable. An infinite set that cannot be listed is said to be uncountable. Theorem 4.35: Countable sets have the following properties: 1. Any subset of a countable set is countable. 2. Any union of a sequence of countable sets is countable. 3. No interval is countable. Exercises 4.6.1 Give examples of closed sets that are countable and closed sets that are uncountable. 4.6.2 Is there a nonempty open set that is countable? 4.6.3 If a set is countable, what can you say about its complement? 4.6.4 Is the intersection of two uncountable sets uncountable? 4.6.5 Show that the Cantor set of Exercise 4.3.23 is infinite and uncountable. See Note 85 4.6.6 Give (if possible) an example of a set with (a) Countably many points of accumulation ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.7. Challenging Problems for Chapter 4 257 (b) Uncountably many points of accumulation (c) Countably many boundary points (d) Uncountably many boundary points (e) Countably many interior points (f) Uncountably many interior points 4.6.7 A set is said to be co-countable if it has a countable complement. Show that the intersection of finitely many co-countable sets is itself co-countable. 4.6.8 Let E be a set and f : R R be an increasing function [i.e., if x < y, then f (x) < f (y)]. Show that E is countable if and only if the image set f (E) is countable. (What property other than “increasing” would work here?) → 4.6.9 Show that every uncountable set of real numbers has a point of accumulation. See Note 86 4.6.10 Let F be a family of (nondegenerate) intervals; that is, each member of is an interval (open, closed or neither) but is not a single point. Suppose that any two intervals I and J in the family have no point in common. Show that the family See Note 87 F can be arranged in a sequence I1, I2, . . . . F 4.7 Challenging Problems for Chapter 4 4.7.1 Cantor, in 1885, defined a set E to be dense-in-itself if E illustrative examples. E′. Develop some facts about such sets. Include ⊂ 4.7.2 One of Cantor’s early results in set theory is that for every closed set E there is a set S with E = S′. Attempt a proof. 4.7.3 Can the closed interval [0, 1] be expressed as the union of a sequence of disjoint closed subintervals each of length smaller than 1? ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 258 Sets of Real Numbers Chapter 4 4.7.4 In many applications of open sets and closed sets we wish to work just inside some other set A. It is convenient to have a language for this. A set E R that is open. A set E set G is closed. Answer the following questions. ⊂ ⊂ ⊂ A is said to be closed relative to A if E = A F for some set F A is said to be open relative to A if E = A G for some R that ∩ ⊂ ∩ (a) Let A = [0, 1] describe, if possible, sets that are open relative to A but not open as subsets of R. (b) Let A = [0, 1] describe, if possible, sets that are closed relative to A but not closed as subsets of R. (c) Let A = (0, 1) describe, if possible, sets that are open relative to A but not open as subsets of R. (d) Let A = (0, 1) describe, if possible, sets that are closed relative to A but not closed as subsets of R. 4.7.5 Let A = Q. Give examples of sets that are neither open nor closed but are both relative to Q. 4.7.6 Show that all the subsets of IN are both open and closed relative to IN. 4.7.7 Introduce for any set E ⊂ R the notation ∂E = x : x is a boundary point of E . } (a) Show for any set E that ∂E = E E). (b) Show that for any set E the set ∂E is closed. (c) For what sets E is it true that ∂E = (d) Show that ∂E E for any closed set E. ⊂ { (R ∩ \ ? ∅ (e) If E is closed, show that ∂E = E if and only if E has no interior points. (f) If E is open, show that ∂E can contain no interval. 4.7.8 Let E be a nonempty set of real numbers and define the function (a) Show that f (x) = 0 for all x (b) Show that f (x) = 0 if and only if x E. (c) Show for any nonempty closed set E that ∈ ∈ f (x) = inf x {| . x { ∈ R : f (x) > 0 = (R E). \ } ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 4.7. Challenging Problems for Chapter 4 259 4.7.9 Let f : R → R have this property: For every x0 ∈ − < δ. Show that for all x, y f (x) | R there is a δ > 0 so that f (x0) | R, x x | = y, x0| < − whenever 0 < x | x0| − 4.7.10 Let f : E → R have this property: For every e ∈ f (y) | f (x there is an ε > 0 so that ∈ f (x) > ε if x E (e ε, e + ε). ∈ ∩ − Show that if the set E is compact, then there is some positive number c so that for all e ∈ E. Show that if E is not closed or is not bounded, then this conclusion may not be valid. 4.7.11 (Separation of Compact Sets) Let A and B be nonempty sets of real numbers and let f (e) > c δ(A, B) is often called the “distance” between the sets A and B. δ(A, B) = inf a {| − b | : a A, b ∈ B . } ∈ (a) Prove δ(A, B) = 0 if A (b) Give an example of two closed, disjoint sets in R for which δ(A, B) = 0. (c) Prove that if A is compact, B is closed, and A , then δ(A, B) > 0. B = = B ∩ ∅ . ∩ ∅ See Note 88 4.7.12 Show that every closed set can be expressed as the intersection of a sequence of open sets. 4.7.13 Show that every open set can be expressed as the union of a sequence of closed sets. 4.7.14 A collection of sets Sα : α nonempty intersection. { A } ∈ is said to have the finite intersection property if every finite subfamily has a (a) Show that if Sα : α { A } ∈ is a family of compact sets that has the finite intersection property, then = Sα 6 . ∅ \α∈A ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 260 NOTES (b) Give an example of a collection of closed sets yet Sα : α { A } ∈ Sα = . ∅ \α∈A that has the finite intersection property and 4.7.15 A set S R is said to be disconnected if there exist two disjoint open sets U and V each containing a point of S so that S ⊂ U ⊂ ∪ V . A set that is not disconnected is said to be connected. (a) Give an example of a disconnected set. (b) Show that every compact interval [a, b] is connected. (c) Show that R is connected. (d) Show that every nonempty connected set is an interval. 4.7.16 Show that the only subsets of R that are both open and closed are and R. ∅ 4.7.17 Given any uncountable set of real numbers E show that it is possible to extract a sequence ∞ k=1 ak/k diverges. terms of E so that the series ak} { of distinct P Notes 73Exercise 4.2.25. Let containing x and containing no other point of the set. Pick the least integer n so that qn ∈ with the isolated points in a set. be an enumeration of the rationals. If x is isolated, then there is an open interval Ix Ix. This associates integers qn} { 74Exercise 4.3.1. Consider the set 1/n : n { IN . } ∈ 75Exercise 4.3.23. The ternary expansion of a number x [0, 1] is given as ∈ x = 0.a1a2a3a4 · · · = ai/3i ∞ i=1 X ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 261 0, 1, 2 . (Thus this is merely the “base 3” version of a decimal expansion.) Observe that 1/3 and 2/3 where the ai ∈ { } can be expressed as 0.0222222 . . . and 0.200000 . . . in ternary. Observe that each number in the interval (1/3, 2/3), that is the first stage component of G, must be written as 0.1a2a3a4 . . . in ternary. How might this lead to a description of the points in G? 76Exercise 4.4.6. Consider the intersection of the family of all closed sets that contain the set E. 77Exercise 4.4.7. Consider the union of the family of all open sets that are contained in the set E. 78Exercise 4.5.1. Try this one: Define f (x) = 0 for x |
irrational and f (x) = q if x = p/q where p/q is a rational with p, q integers and with no common factors. 79Exercise 4.5.5. Take compact to mean closed and bounded. Show that a finite union or arbitrary intersection of compact sets is again compact. Check that an arbitrary union of compact sets need not be compact. Show that any closed subset of a compact set is compact. Show that any finite set is compact. 80Exercise 4.5.8. For a course in functions of one variable open covers can consist of intervals. In more general settings there may be nothing that corresponds to an “interval;” thus the more general covering by open sets is needed. Your task is just to look through the proof and spot where an “open interval” needs to be changed to an “open set.” 81Exercise 4.5.9. Cousin’s lemma offers the easiest proof, although any other compactness argument would work. Take the family of all intervals [c, d] for which f (c) < f (d) and check that the hypotheses of that lemma hold on any interval [x, y]. 82Exercise 4.5.18. Let { rational endpoints. For each x choice k(x). Thus C = ∈ Vα : α A E there is a Vα and a k so that the interval Nk satisfies x be the open cover. Let N1, N2, . . . be a listing of all open intervals with Vα. Call this ∈ } Nk ⊂ ∈ = { N Nk(x) : x E } ∈ is a countable open cover of E (but not the countable open cover that we want). But corresponding to each member of that is a member of forms a cover of E. that contains it. Using that correspondence we construct the countable subcollection of N C C 83Exercise 4.5.19. Lindel¨off’s theorem asserts that an open cover of any set of reals can be reduced to a countable ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 262 NOTES subcover. The Heine-Borel theorem asserts that an open cover of any compact set of reals can be reduced to a finite subcover. ⇒ 84Exercise 4.5.20. For (b) (d) and for (c) (d). Suppose that there is an open cover of A but no finite subcover. ⇒ Step 1: You may assume that the open cover is just a sequence of open sets. (This is because of Exercise 4.5.18.) Step 2: You may assume that the open cover is an increasing sequence of open sets G1 ⊂ . . . (just take G3 ⊂ A but not in the union of the first terms in the sequence you were given). Step 3: Now choose points xi to be in Gi ∩ A that is an accumulation point of the any previous Gj for j < i. Step 4: Now apply (b) [or (c)] to get a point z points xi. This would have to be a point in some set GN (since these cover A) but for n > N none of the points xn can belong to GN . G2 ⊂ ∈ 85Exercise 4.6.5. This result may seem surprising at first since the Cantor set, at first sight, seems to contain only the endpoints of the open intervals that are removed at each stage, and that set of endpoints would be countable. (That view is mistaken; there are many more points.) Show that a point x in [0, 1] belongs to the Cantor set if and only if it can be written as a ternary expansion x = 0.c1c2c3 . . . (base 3) in such a way that only 0’s and 2’s occur. This is now a simple characterization of the Cantor set (in terms of string of 0’s and 2’s) and you should be able to come up with some argument as to why it is now uncountable. 86Exercise 4.6.9. You will need the Bolzano-Weierstrass theorem (Theorem 4.21). But this uncountable set E might be unbounded. How could we prove that an uncountable set would have to contain an infinite bounded subset? Consider ∞ E = n=1 [ E [ − ∩ n, n]. 87Exercise 4.6.10. Select a rational number from each member of the family and use that to place them in an order. 88Exercise 4.7.11. For part (b) look ahead to part (c): Any such example must have A and B unbounded. For < 1/n. As A is compact part (c) assume δ(A, B) = 0. Then there must be points xn ∈ there is a convergent subsequence xnk converging to a point z in A. What is happening to ynk ? (Be sure to use here the fact that B is closed.) A and yn ∈ xn − | B with yn| ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Chapter 5 CONTINUOUS FUNCTIONS 5.1 Introduction to Limits The definition of the limit of a function lim x0 x → is given in calculus courses, but in many classes it is not explored to any great depth. Computation of limits is interesting and offers its challenges, but for a course in real analysis we must master the definition itself and derive its consequences. f (x) Our viewpoint is larger than that in most calculus treatments. There it is common to insist, in order δ, x0 + δ) that for a limit to be defined, that the function f must be defined at least in some interval (x0 − contains the point x0 (with the possible exception of x0 itself). Here we must allow a function f that is defined only on some set E and study limits for points x0 that are not too remote from E. We do not insist that x0 be in the domain of f but we do require that it be “close.” This requirement is expressed using our language from Chapter 4. We must have x0 a point of accumulation of E. Except for this detail about the domain of the function the definition we use is the usual ε-δ definition from calculus. Readers familiar with the sequence limit definitions of Chapter 2 will have no trouble handling this definition. It is nearly the same in general form as the ε-N definition for sequences, and many of the proofs use similar ideas. 263 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 264 Continuous Functions Chapter 5 5.1.1 Limits (ε-δ Definition) The definition of a sequence limit, limn if n is sufficiently large. The definition of a function limit →∞ sn, made precise the statement that sn is arbitrarily close to L f (x) lim x0 x → is intended, in much the same way, to make precise the statement that f (x) is arbitrarily close to L if x is sufficiently close to x0. One feature of the definition must be to exclude the value at the point x0 from consideration; it should be irrelevant to the value of the limit. It is possible (likely even) that f (x0) = L, but whether this is true or false should not be any influence on the existence of the limit. Thus the definition assumes the following form. The requirement that x0 be a point of accumulation of E may seem strange at first sight, but we will see that it is needed in order for the definition to have some meaning. Without it any number would be the limit and the theory of limits would be useless. Definition 5.1: (Limit) Let f : E accumulation of E. Then we write → R be a function with domain E and suppose that x0 is a point of if for every ε > 0 there is a δ > 0 so that f (x) = L lim x0 x → whenever x is a point of E differing from x0 and satisfying f (x) | L < ε − | x | − x0| < δ. Note. The condition on x can be written as or as or, yet again, as 0 < x | − x0| < δ x (x0 − ∈ δ, x0 + δ) , x = x0 x0 − δ < x < x0 + δ , x = x0. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 Section 5.1. Introduction to Limits 265 Figure 5.1. Graphical interpretation of the ε-δ limit definition. The exclusion of x = x0 should be seen as an advantage here. An inequality is required to be true for all x satisfying some condition, and we are allowed not to have to check x = x0. It may happen to be true that < ε when x = x0 but it is irrelevant to the definition. For example, you will recall that the limit f (x) | used to define a derivative − L | f ′(x0) = lim x0 → x f (x) x − − f (x0) x0 must require that the value for x = x0 be excluded; the expression is not defined when x = x0. See Figure 5.1 for a graphical interpretation of the definition. In the picture a particular value of ε is illustrated and for that value the figure shows a choice of δ that works. Every smaller value of δ would have worked, too. The definition requires doing this, however, for every positive ε, and the figure cannot convey that. We now present some examples illustrating how to prove the existence of a limit directly from the definition. These are to be considered as exercises in understanding the definition. We would rarely use the definition to compute a limit, and we hope seldom to use the definition to verify one; we will use the definition to develop a theory that will verify limits for us. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 266 Continuous Functions Chapter 5 Example 5.2: Any function f (x) = ax + b will have the easily predicted limit lim x0 x → f (x) = lim x0 → x (ax + b) = ax0 + b. If you sketch a picture similar to that of Figure 5.1 you see easily that the choice of δ is monitored by the slope of the line y = ax + b. The steeper the slope, the smaller the δ has to be taken in comparison with ε. Let us do this for the linear function f (x) = 10x 11. We expect that (10x lim 5 x → − − 11) = 10(5) 11 = 39. − Let us prove this. We need a condition ensuring that the expression is smaller than ε. Some arithmetic converts this to (10x | 11) 39 | − − Now it is clear that, if we insist that (10x 11) | − x | − 39 = 10x 50 = 10 − 5 | | | < ε/10, we will have − | x . 5 | − | | | (10x | 11) 39 < ε. − − | That completes the proof. Better, though, would be to write it in a more straightforward manner that obscures how we did it but gets to the point of the proof more simply: Let ε > 0. Let δ = ε/10. Then for all x with 5 < δ we have x | = − 10 | x (10x | 11) 39 − − 11) = 39 as required. | | | | < 10δ = ε. 5 | − By definition, limx 5(10x → − An alert reader of our short proof will know that the choice of δ as ε/10 took some time to compute and is ◭ not just an inspired second sentence of the proof. Example 5.3: Let us use the definition to verify the existence of x2. lim x0 x → ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.1. Introduction to Limits 267 Again the definition gives no hints as how to compute the limit; it can be used only to verify the correctness 3 x2 = 9. We need a condition en |
suring that of a limit statement. To keep it simple let us show that limx the expression → is smaller than ε. Some arithmetic converts this to = x2 9 If we insist that x2 /M, where M is bigger than any value of how big might x + 3 values of x), then this is not too big. For example, if x stays inside (2, 4), then enough computations to allow us to write up a proof. < ε exactly as we need. But just be? If we remember that we are interested only in values of x close to 3 (not huge < 7. These are , then we will have | | | | x2 Let ε > 0. Let δ = ε/7 or δ = 1, whichever is smaller (i.e., δ = min it follows that { ε/7, 1 } ). Then if x | − 3 | < δ and hence that By definition, limx → x2 x − = 9 3 | | 3 x2 = 9 as required(ε/7) = ε. The finished proof is shorter and lacks all the motivating steps that we just went through. ◭ In spite of these examples and the necessity in elementary courses such as this to work through similar examples, the main goal of our definition is to build up a theory of limits that can then be used to justify other methods of computation and lead to new discoveries. On occasions we must, however, return to the definition to handle an unusual case. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Continuous Functions Chapter 5 268 Exercises 5.1.1 Prove the existence of the limit limx→x0 (4 See Note 89 12x). − 5.1.2 Prove the validity of the limit limx→x0 (ax + b) = ax0 + b. See Note 90 5.1.3 Prove the existence of the limit limx→−4 x2. See Note 91 5.1.4 Prove the validity of the limit limx→x0 x2 = x2 0. See Note 92 5.1.5 Suppose in the definition of the limit that the phrase “x0 be a point of accumulation of the domain of f ” is deleted. Show that then the limit statement limx→−2 √x = L would be true for every number L. 5.1.6 Recall that in the definition of limx→x0 f (x) there is a requirement that x0 be a point of accumulation of the domain of f . Which values of x0 would be excluded from consideration in the limit 5.1.7 Which values of x0 would be excluded from consideration in the limit p lim x→x0 x2 2? − lim x→x0 arcsin x + 2 | ? | 5.1.8 Prove the validity of the limit limx→x0 √x = √x0. See Note 93 5.1.9 Prove that the limit limx→0 1 x fails to exist. 5.1.10 Prove that the limit limx→0 sin(1/x) fails to exist. 5.1.11 Using the definition, show that if limx→x0 f (x) = L, then lim x→x0 | f (x) | = . L | | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.1. Introduction to Limits 269 5.1.12 Suppose that x0 is a point of accumulation of both A and B and that f : A → that f and g must agree in the sense that f (x) = g(x) if x is in both A and B. R and g : B → R. We insist (a) What conditions on A and B ensure that if the limit limx→x0 f (x) exists so too must the limit limx→x0 g(x)? (b) What conditions on A and B ensure that if lim x→x0 f (x) and lim x→x0 g(x) both exist then they must be equal. See Note 94 5.1.2 Limits (Sequential Definition) The theory of function limits can be reduced to the theory of sequence limits. This is a popular device in mathematics. Some new theory turns out to be contained in an old theory. This allows easy proofs of results since the old theory has all the tools needed for constructing proofs in the new subject. If our goal were merely to prove all the properties of limits, this would allow us to skip over ε-δ proofs. But since we are trying in this elementary course to learn many methods of analysis, we shall not escape from learning to use ε-δ arguments. Even so, this is an interesting tool for us to use. We can call upon our sequence experience to discover new facts about function limits. Definition 5.4: (Limit) Let f : E accumulation of E. Then we write → R be a function with domain E and suppose that x0 is a point of if for every sequence en} { f (x) = L lim x0 x → of points of E with en 6 lim n →∞ = x0 and en → f (en) = L. x0 as n , → ∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 270 Continuous Functions Chapter 5 Note. If x0 is not a point of accumulation of E, then there would be no sequence with en 6 definition. = x0 for all n and en → x0 as n → ∞ en} { . Thus once again this is an essential ingredient of our limit of points of E Before we can use this definition we need to establish that it is equivalent to the ε-δ definition. We prove that now. Proof. (Definitions 5.1 and 5.4 are equivalent) Suppose first that f (x) = L lim x0 x → according to Definition 5.1 and that to x0. Let ε > 0. There must be a positive number δ so that , en 6 en} { = x0, is a sequence of points in the domain of f converging f (x) | L < ε − | f (en) | L < ε − | x if 0 < n < δ. But en → N . Putting these together, we find that x0 and en 6 x0| − | = x0 so there is number N such that 0 < en − | x0| < δ for all ≥ if n N . This proves that f (en) ≥ Conversely, suppose that L is not the limit of f (x) as x converges to L. This verifies that Definition 5.1 implies Definition 5.4. x0 according to Definition 5.1. We must find a sequence of points in the domain of f and converging to x0 such that f (en) does not converge to L. Because L is not the limit, there must be some ε0 > 0 so that for any δ > 0 there will be points x in the domain of f with 0 < < δ and yet the inequality en} x → } { { x0| | − − fails. Applying this to δ = 1, 1/2, 1/3, 1/4 . . . we obtain a sequence of points xn with xn in the domain of f and | | f (x) L < ε0 and yet 0 < xn − | x0| < 1/n f (xn) | L − | ≥ ε0. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.1. Introduction to Limits 271 This is precisely the sequence we wanted since Definition 5.4 implies Definition 5.1. { f (xn) cannot converge to L. Thus we have shown that } Since the two definitions are equivalent, we can use either a sequential argument or an ε-δ argument in our discussions of limits. Example 5.5: Suppose we wish to prove that limx x0 f (x) = L implies that → f (x) = √L. lim x0 x → p We could convert this into an ε-δ statement, which will involve us in some unpleasant inequality work. Or we can see that, alternatively, we need to prove that if we know f (xn) conclude f (xn) (Exercise 2.4.16). L, then we can √L. But we did study just such problems in our investigation of sequence limits → → ◭ p Exercises 5.1.13 Prove the existence of the limit limx→x0(4 5.1.14 Prove the validity of the limit 12x) by converting to a statement about sequences. − lim x→x0 by converting to a statement about sequences. 5.1.15 Prove the validity of the limit limx→x0 x2 = x2 x 5.1.16 Show that limx→0 | See Note 95 0 by converting to a statement about sequences. /x does not exist by using the sequential definition of limit. | (ax + b) = ax0 + b 5.1.17 Prove that the limit limx→0 1 x fails to exist by converting to a statement about sequences. 5.1.18 Prove that the limit limx→0 sin(1/x) fails to exist by converting to a statement about sequences. 5.1.19 Let x0 be an accumulation point of the domain E of a function f . Prove that the limit limx→x0 f (x) fails to exist if and only if there is a sequence of distinct points divergent. en} { of E converging to x0 but with f (en) } { ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 272 Continuous Functions Chapter 5 5.1.20 Let f be the characteristic function of the rational numbers; that is, f is defined for all real numbers by setting f (x) = 1 if x is a rational number and f (x) = 0 if x is not a rational number. Determine where, if possible, the limit limx→x0 f (x) exists. 5.1.21 Using the sequential definition, show that if limx→x0 f (x) = L, then 5.1.22 Find hypotheses under which you can prove that if limx→x0 f (x) = L, then lim x→x0 | f (x) | = . L | | f (x) = √L. lim x→x0 p See Note 96 5.1.3 Limits (Mapping Definition) " Enrichment section. May be omitted. The essential idea behind a limit is that values of x close to x0 get mapped by f into values close to L. We have been able to express this idea by using inequalities that express this closeness: δ-close for the x values and ε-close for the f (x) values. This is essentially a mapping property that can be expressed by arbitrary open sets. f (x) = L lim x0 x → The following definition is equivalent to both Definitions 5.1 and 5.4. Definition 5.6: (Limit) Let f : E accumulation of E. Then we write → R be a function with domain E and suppose that x0 is a point of if for every open set V containing the point L there is an open set U containing the point x0 and every point x = x0 of U that is in the domain of f is mapped into a point in V ; that is, f (x) = L lim x0 x → f : E U x0} → \ { ∩ V. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 5.1. Introduction to Limits 273 Once again, we must show that this definition is equivalent to the ε-δ definition. We prove that now. Proof. (Definitions 5.1 and 5.6 are equivalent) Suppose first that f (x) = L lim x0 x → according to Definition 5.1. Let V be an open set containing the point L. Then, since L is an interior point of V there is a positive number ε with Choose δ > 0 so that (L − ε, L + ε) V. ⊂ L | < δ whenever x is a point in E (the domain of f ). Let U be the open set (x0 − δ, x0 + δ). = x0 of U that is in the domain of f is mapped into f (x) | if 0 < Then the inequality we have shows that every point x a point in V . This is precisely Definition 5.6. x0| < ε − − x | Conversely, suppose that limx x0 f (x) = L according to Definition 5.6. Let ε > 0. Choose ε, L + ε). By our definition there must be an open set U containing the point x0 and every point = x0 of U that is in the domain of f is mapped into a point in V . Since x0 is an interior point of U there − V = (L x must be a positive number δ so that → This mapping property implies that (x0 − δ, x0 + δ) U. ⊂ < δ. This is exactly our ε-δ definition of Definition 5.1. f (x) | L < ε − | if 0 < x x0| | − Since |
all three of our definitions are equivalent we can use either a sequential argument, a mapping argument, or an ε-δ argument in our discussions of limits. Exercises x 5.1.23 Show that limx→0 | 5.1.24 Prove directly that the sequential definition of limit is equivalent to the mapping definition. /x does not exist using the mapping definition of limit. | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 274 5.1.4 One-Sided Limits Continuous Functions Chapter 5 It is possible for a function to fail to have a limit at a point and yet appear to have limits on one side. If we ignore what is happening on the right for a function, perhaps it will have a “left-hand limit.” This is easy to achieve. Let f be defined everywhere near a point x0 and define a new function g(x) = f (x) for all x < x0. This new function g is defined on a set to the left of x0 and knows nothing of the values of f on the right. Thus the limit can be thought of as a left-hand limit for f . It would be written as g(x) lim x0 x → where the “x0− notation ” is the indication that a left-hand limit is used, not an ordinary limit. Similarly, the f (x) lim x0 → x − lim x → denotes a right-hand limit with the “x0+” indicating the limit on the positive or right side of x0. f (x) x0+ Since these one-sided limits are really just ordinary limits for a different function, they must satisfy all the theory of ordinary limits with no further fuss. We can use them quite freely without worrying that they need a different definition or a different theory. Even so, it is convenient to translate our usual definitions into one-sided limits just to have an expression for them. We give the right-hand version. You can supply a left-hand version. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.1. Introduction to Limits 275 Definition 5.7: (Right-Hand Limit) Let f : E is a point of accumulation of E ). Then we write → (x0, ∩ ∞ R be a function with domain E and suppose that x0 f (x) = L lim → x0+ x f (x) | L | − < ε if for every ε > 0 there is a δ > 0 so that whenever x0 < x < x0 + δ and x E. ∈ An equivalent sequential version can be established. Definition 5.8: (Right-Hand Limit) Let f : E is a point of accumulation of E ). Then we write → (x0, ∩ ∞ f (x) = L lim → x0+ x R be a function with domain E and suppose that x0 if for every decreasing sequence en} { of points of E with en > x0 and en → f (en) = L. lim n →∞ x0 as n , → ∞ Exercises 5.1.25 Show directly that Definitions 5.7 and 5.8 are equivalent. 5.1.26 Under appropriate additional assumptions about the domain of the function f show that limx→x0 f (x) = L if and only if both are valid. 5.1.27 If the two limits lim x→x0+ f (x) = L and lim x→x0− f (x) = L lim x→x0+ f (x) = L1 and lim x→x0− f (x) = L2 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 276 Continuous Functions Chapter 5 L2 exist and are different, then the function is said to have a jump discontinuity at that point. The value L1 − is called the magnitude of the jump. Give an example of a function with a jump of magnitude 3 at the value 3. x0 = 2. Give an example with a jump of magnitude − 5.1.28 Compute the one-sided limits of the function at any point x0. See Note 97 f (x) = x x | | 5.1.29 Compute, if possible, the one-sided limits of the function f (x) = e1/x at 0. See Note 98 5.1.30 According to our definitions, is there any distinction between the assertions lim x→0 √x = 0 and lim x→0+ √x = 0? What is the meaning of limx→0− √x = 0? See Note 99 5.1.5 Infinite Limits We can easily check that the limits 1 x fail to exist. A glance at the graph of the function f (x) = 1/x suggests that we should write instead and lim 0 x − lim 0+ x → 1 x → 1 x lim 0+ x → = ∞ and lim 0 x − → 1 x = −∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.1. Introduction to Limits 277 as a way of conveying more information about what is happening rather than saying merely that the limits do not exist. In this we are following our custom in the study of divergent sequences. Some sequences merely diverge, some diverge to function limit definition, we should arrive at the following definition. or to −∞ ∞ . If we look back at the definition for sequences and compare it with our R be a function with domain E and suppose that x0 is a Definition 5.9: (Infinite Limit) Let f : E point of accumulation of E (x0, ). Then we write → ∩ ∞ lim → if for every M > 0 there is a δ > 0 so that f (x) x x0+ f (x) = ∞ M whenever ≥ x0 < x < x0 + δ and x E. ∈ Similarly, we can define x x0+ f (x) = lim → if for every m < 0 there is a δ > 0 so that f (x) limits on the left are similarly defined and denoted limx and limx two-sided limits are defined in the same manner, but with a two-sided condition. Note. Just as for sequences, we do not say that the limit of a function exists unless that limit is finite. Thus, for example, we would say that the limit limx . are used to describe certain situations, but they are not and A limit is a real number. The symbols interpreted as numbers themselves. 0+ 1/x does not exist, and that in fact limx m whenever x0 < x < x0 + δ and x E. The infinite . Also, ∈ f (x) = 0+ 1/x = → −∞ f (x) = −∞ −∞ ∞ ∞ ∞ ≤ x0 x0 → → → − − Exercises 5.1.31 Give an equivalent formulation for infinite limits using a sequential version. 5.1.32 Formulate a definition for the statement that limx→x0− f (x) = lim x→x0− f (x) = ∞ . Using your definition, show that ∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Continuous Functions Chapter 5 278 if and only if lim x→(−x0)+ f ( − x) = . ∞ 5.1.33 Where does the function f (x) = have infinite limits? Give proofs using the definition. 5.1.34 Formulate a definition for the statements 1 √x2 1 − lim x→∞ f (x) = L and lim x→−∞ f (x) = L. See Note 100 5.1.35 Formulate a definition for the statements 5.1.36 Let f : (0, ) ∞ → R. Show that lim x→∞ f (x) = and lim x→−∞ f (x) = . ∞ ∞ lim x→∞ f (x) = L if and only if lim x→0+ f (1/x) = L. 5.1.37 What are the limits limx→∞ xp for various real numbers p? 5.1.38 Show that one of the limits limx→0+ f (x) and limx→0− f (x) of the function at 0 is infinite and one is finite. What can you say about the limits lim x→∞ f (x) and lim x→−∞ f (x)? f (x) = e1/x See Note 101 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.2. Properties of Limits 5.2 Properties of Limits 279 The computation of limits in calculus courses depended on a theory of limits. For most simple computations it was enough to know how to handle functions that were put together by adding, subtracting, multiplying, or dividing other functions. Later, more subtle problems required advanced techniques (e.g., L’Hˆopital’s rule). Here we develop the rudiments of a theory of function limits. We start with the uniqueness property, the boundedness property and continue to the algebraic properties. In this we are following much the same path we did when we began our study of sequential limits. Indeed the definitions of sequential limits and function limits are so similar that the theories are necessarily themselves quite similar. 5.2.1 Uniqueness of Limits When we write the statement we wish to be assured that it is not also true for some other numbers different from L. f (x) = L lim x0 x → Theorem 5.10 (Uniqueness of Limits) Suppose that Then the number L is unique: No other number has this same property. f (x) = L. lim x0 x → Proof. We suppose that and f (x) = L lim x0 x → f (x) = L1 lim x0 x → ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 280 Continuous Functions Chapter 5 are both true. To prove the theorem we must show that L = L1. If we convert this to a statement about = x0 and all points in the domain of f must have x0 with xn 6 sequences this asserts that any sequence xn → L f (xn) → and also must have For these limits to exist the point x0 must be a point of accumulation for the domain of f and so there exists at least one such sequence. But we have already established for sequence limits that this is impossible (Theorem 2.8) unless L = L1. f (xn) L1. → Exercises 5.2.1 Give an ε-δ proof of Theorem 5.10. See Note 102 5.2.2 Explain why the proof fails if the part of the limit definition that asserts x0 is to be a point of accumulation of the domain of f were omitted. 5.2.2 Boundedness of Limits We recall that convergent sequences are bounded. There is a similar statement for functions. If a function limit exists the function cannot be too large; the statement must be made precise, however, since it is really only valid close to the point where the limit is taken. For example, you will recall from our discussion of local boundedness in Section 4.5 that the function f (x) = 1/x is unbounded and yet locally bounded at each point other than at 0. In the same way we will see that the existence of the limit for every value of x0 6 = 0 also requires that local boundedness property. 1 x = 1 x0 lim x0 x → ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.2. Properties of Limits 281 Theorem 5.11 (Boundedness of Limits) Suppose that the limit exists. Then there is an interval (x0 − for every value of x in that interval that is in the domain of f . f (x) = L lim x0 x → c, x0 + c) and a number M such that f (x) | | ≤ M Proof. There is a δ > 0 so that whenever x is a point in the domain of f differing from x0 and satisfying domain of f , then this means that f (x) L < 1 | − | x | − x0| < δ. If x0 is not in the f (x) | | = | f (x) − L + L f (x, x0 + δ) that are in the domain of f . This would complete the proof since we can take for all x in (x0 − L M = | + 1. | If x0 is in the domain of f , then take instead Then x0) . | | for all x in (x0 − | ≤ δ, x0 + δ) that are in the domain of f . | f (x) M A similar stat |
ement can be made about boundedness away from zero. This shows that if a function has a nonzero limit, then close by to the point the function stays away from zero. The proof uses similar ideas and is left for the exercises. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 282 Continuous Functions Chapter 5 Theorem 5.12 (Boundedness Away from Zero) If the limit f (x) lim x0 x → exists and is not zero, then there is an interval (x0 − | ≥ f (x) | m > 0 for every value of x = x0 in that interval and that belongs to the domain of f . c, x0 + c) and a positive number m such that Exercises 5.2.3 Prove Theorem 5.11 using the sequential definition of limit instead. See Note 103 5.2.4 Use Theorem 5.11 to show that limx→0 1 x cannot exist. 5.2.5 Prove Theorem 5.12 using an ε-δ argument. 5.2.6 Prove Theorem 5.12 using a sequential argument. 5.2.7 Prove Theorem 5.12 by deriving it from Theorem 5.11 and the fact (proved later) that if then 5.2.3 Algebra of Limits lim x→x0 f (x) = L = 0 lim x→x0 1 f (x) = 1 L . Functions can be combined by the usual arithmetic operations (addition, subtraction, multiplication and division). Indeed most functions we are likely to have encountered in a calculus course can be seen to be composed of simpler functions combined together in this way. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 Section 5.2. Properties of Limits 283 Example 5.13: The computations 2x3 + 4 3x2 + 1 lim 3 x → 2(limx 3(limx = = limx limx 3 x3) + 4 3 x2) + 1 → → = 3(2x3 + 4) 3(3x2 + 1) 33 + 4 32 + 1 2 3 → → × × should return fond memories of calculus homework assignments. But how are these computations properly ◭ justified? Because of our experience with sequence limits, we can anticipate that there should be an “algebra of function limits” just as there was an algebra of sequence limits. The proofs can be obtained either by imitating the proofs we constructed earlier for sequences or by using the fact that function limits can be reduced to sequential limits. There is an extra caution here. An example illustrates. Example 5.14: We know that limx 0 √ → 0 √x = 0. Does it follow that There is only one point in the domain of the function and so no limit statement is possible. f (x) = √x + √ x − ◭ The extra hypothesis throughout the following theorems appears in order to avoid examples like this. We must assume that the domain of f , call it dom(f ), and the domain of g, call it dom(g), must have enough points in common to define the limit at the point x0 being considered. In most simple applications the domains of the functions do not cause any troubles. For proofs we have a number of strategies available. We can reduce these limit theorems to statements about sequences and then appeal to the theory of sequential limits that we developed in Chapter 2. x = 0 and limx √x + √ x → = 0? − lim 0 x → − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 284 Continuous Functions Chapter 5 Alternatively, we can construct ε-δ proofs by modeling them after the similar statements that we proved for sequences. We do not need any really new ideas. The proofs have, accordingly, been left to the exercises. Theorem 5.15 (Multiples of Limits) Suppose that the limit exists and that C is a real number. Then f (x) lim x0 x → Theorem 5.16 (Sums and Differences) Suppose that the limits Cf (x) = C lim x0 x → lim x0 x → f (x) . exist and that x0 is a point of accumulation of dom(f ) and lim x0 x → f (x) and lim x0 x → g(x) lim x0 x → dom(g). Then ∩ (f (x) + g(x)) = lim x0 → x f (x) + lim x0 → x g(x) (f (x) lim x0 x → g(x)) = lim x0 → x − f (x) lim x0 x → − g(x). Theorem 5.17 (Products of Limits) Suppose that the limits exist and that x0 is a point of accumulation of dom(f ) ∩ f (x) and lim x0 x → g(x) lim x0 x → dom(g). Then f (x)g(x) = lim x0 x → lim x0 x → f (x) lim x0 x → g(x) . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.2. Properties of Limits 285 Theorem 5.18 (Quotients of Limits) Suppose that the limits lim x0 x → exist and that the latter is not zero and that x0 is a point of accumulation of dom(f ) f (x) and lim x0 x → g(x) f (x) g(x) = lim x0 x → limx limx → → x0 f (x) x0 g(x) . dom(g). Then ∩ Exercises 5.2.8 Let f and g be functions with domains dom(f ) and dom(g). What are the domains of the functions listed below obtained by combining these functions algebraically or by a composition? − (a) f + g (b) f g (c) f g (d) f /g (e) f g ◦ (f) √f + g (g) √f g 5.2.9 What exactly is the trouble that arises in the theorems of this section that required us to assume “that x0 is a point of accumulation of dom(f ) See Note 104 ∩ dom(g)?” 5.2.10 Is it true that if both limx→x0 f (x) and limx→x0 g(x) fail to exist, then lim x→x0 (f (x) + g(x)) must also fail to exist? 5.2.11 In the statement of Theorem 5.18 don’t we also have to assume that g(x) is never zero? See Note 105 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 286 Continuous Functions Chapter 5 5.2.12 A careless student gives the following as a proof of Theorem 5.17. Find the flaw: “Suppose that ε > 0. Choose δ1 so that f (x if 0 < x | x0| − < δ1 and also choose δ2 so that if 0 < x | x0| − < δ2 Define δ = min { f (x)g(x) | Well, that shows f (x)g(x) f (x) | ≤ | LM if f (x) → g(x) M < | − . If 0 < δ1, δ2} LM − | x | f (x) − | | ε 2 f (x) | | x0| g(x) M − | ≤ | ε f (x .” → L and g(x) + 1 < δ, then we have f (x. 5.2.13 Prove Theorem 5.15 by using an ε-δ proof and by using the sequential definition of limit. 5.2.14 Prove Theorem 5.16 by using an ε-δ proof and by using the sequential definition of limit. 5.2.15 Prove Theorem 5.18 by using the sequential definition of limit. 5.2.16 Prove Theorem 5.17 by correcting the flawed ε-δ proof in Exercise 5.2.12 and by using the sequential definition of limit. Which method is easier? . 5.2.4 Order Properties Just as we saw that sequence limits preserve both the algebraic structure and the order structure, so we will find that function limits have the same properties. We have just completed the algebraic properties. We turn now to the order properties. If f (x) ≤ g(x) for all x, then we expect to conclude that f (x) lim x0 x → lim x0 x → ≤ g(x). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.2. Properties of Limits 287 We now prove this and several other properties that relate directly to the order structure of the real numbers. Theorem 5.19: Suppose that the limits f (x) and lim x0 x → exist and that x0 is a point of accumulation of dom(f ) g(x) lim x0 x → dom(g). If for all x dom(f ) ∩ ∈ dom(g), then Proof. Let us give an indirect proof. Let ∩ g(x) f (x) ≤ f (x) lim x0 x → lim x0 x → ≤ g(x). and suppose, contrary to the theorem, that L > M . Choose ε so small that M + ε < L ε; that is, choose − L = lim x0 x → f (x) and M = lim x0 → x g(x) By the definition of limits there are numbers δ1 and δ2 so that ε < (L M )/2. − if x = x0 is within δ1 of x0 and in the domain of f and f (x) > L ε − g(x) < M + ε = x0 is within δ2 of x0 and is in the domain of g. But the conditions in the theorem assure us that if x there must be at least one point, x = z say, that satisfies both conditions. That would mean g(zz). − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 288 Continuous Functions Chapter 5 This is impossible as it contradicts the fact that all the values of f (x) are less than the values g(x). This contradiction completes the proof. Note. There is a trap here that we encountered in our discussions of sequence limits. We remember that the condition sn < tn does not imply that sn < lim →∞ In the same way the condition f (x) < g(x) does not imply lim n →∞ n tn. Be careful with this, too. Corollary 5.20: Suppose that the limit lim x0 x → f (x) < lim x0 → x g(x). exists and that α f (x) ≤ ≤ β for all x in the domain of f . Then f (x) lim x0 x → α lim x0 x → ≤ f (x) β. ≤ Note. Again, don’t forget the trap. The condition α < f (x) < β for all x implies at best that It would not imply that α lim x0 x → ≤ f (x) β. ≤ α < lim x0 x → f (x) < β. The next theorem is another useful variant on these themes. Here an unknown function is sandwiched between two functions whose limit behavior is known, allowing us to conclude that a limit exists. This theorem is often taught as “the squeeze theorem” just as the version for sequences in Theorem 2.20 was labeled. Here we need the functions to have the same domain. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.2. Properties of Limits 289 Theorem 5.21 (Squeeze Theorem) Suppose that f , g, h : E of the common domain E. Suppose that the limits → R and that x0 is a point of accumulation exist and that f (x) = L and lim x0 x → g(x) = L lim x0 x → ≤ E except perhaps at x = x0. Then limx f (x) h(x) g(x) ≤ x0 h(x) = L. → for all x ∈ Proof. The easiest proof is to use a sequential argument. This is left as Exercise 5.2.19. Example 5.22: Let us prove that the limit x sin(1/x) = 0 lim 0 x → is valid. Certainly the expression sin(1/x) seems troublesome at first. But we notice that the inequalities are valid for all x (except x = 0 where the function is undefined). Since x −| | ≤ x sin(1/x) x | ≤ | Theorem 5.21 supplies our result. lim x → 0 | x | = lim x → 0 −| x | = 0 ◭ A final theorem on the theme of order structure is often needed. The absolute value, we recall, is defined directly in terms of the order structure. Is absolute value preserved by the limit operation? As the proof does not require any new ideas, it is left as Exercise 5.2.21. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 290 Continuous Functions Chapter 5 Theorem 5.23 (Limits of Absolute Values) Suppose that the limit exists. Then f (x) = L l |
im x0 x → lim x → x0 | f (x) = L . | | | Since maxima and minima can be expressed in terms of absolute values, there is a corollary that is sometimes useful. Corollary 5.24 (Max/Min of Limits) Suppose that the limits exist and that x0 is a point of accumulation of dom(f ) and lim x0 x → max { f (x) = L and lim x0 x → g(x) = M lim x0 x → dom(g). Then = max L, M { } } ∩ f (x), g(x) lim x0 x → min { f (x), g(x) = min L, M . } { } Proof. The first of the these follows from the identity max { f (x), g(x) f (x) + g(x) 2 = } f (x) + | g(x) | − 2 and the theorem on limits of sums and the theorem on limits of absolute values. In the same way the second assertion follows from min { f (x), g(x) f (x) + g(x) 2 | − = } f (x) − 2 g(x) | . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 291 Section 5.2. Properties of Limits Exercises 5.2.17 Show that the condition f (x) < g(x) does not imply that lim x→x0 f (x) < lim x→x0 g(x). 5.2.18 Give a sequential type proof for Theorem 5.19. 5.2.19 Give a sequential type proof for Theorem 5.21. 5.2.20 Give an ε-δ proof of Theorem 5.23. 5.2.21 Give a proof of Theorem 5.23 by converting it to a statement about sequences. 5.2.22 Extend Corollary 5.24 to the case of more than two functions; that is, determine lim x→x0 max { f1(x), f2(x), . . . , fn(x) } . 5.2.5 Composition of Functions You will have observed a pattern that is attractive in the study of limits. These examples suggest the pattern: 2 , f (x) [f (x)]2 = lim x0 x → f (x) = lim x0 x → lim x0 x → q p ef (x) = elimx→x0 f (x). lim x0 x → lim x0 x → f (x), The first is easy to prove since [f (x)]2 = f (x)f (x) and we can use the product rule. The square root example is harder but could be proved using an ε-δ argument and requires only the assumption that x0 f (x) < 0. limx x0 f (x) is positive. It could be false if limx x0 f (x) = 0 and definitely is false if limx The third will require some familiarity with the exponential function and is harder still, though always → → → true. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 292 Continuous Functions Chapter 5 The general pattern is the following. Some function F is composed with f , and the limit computation we wish to use is Can this be justified? More correctly, what are the conditions under which it can be justified? F (f (x)) = F lim x0 x → lim x0 x → f (x) . Let us analyze this using a sequence argument since that often simplifies function limits. We suppose xn → x0. We have then our supposition that f (xn) L. Can we conclude → F (f (xn)) F (L)? → This is exactly what we are doing when we try to use lim x0 x → The property of the function F that we desire is simple: F (f (x)) = F lim x0 x → f (x) . Think of zn = f (xn); then zn → z0 then F (zn) F (z0). If zn → → L and the required property is This is the same as requiring that F (zn) F (L) whenever zn → L. → Thus we have proved the following theorem, which completely answers our question about justifying the preceding operations. F (z) = F (L). lim L z → ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.2. Properties of Limits 293 Theorem 5.25: Let F be a function defined in a neighborhood of the point L and such that If then F (z) = F (L). lim L z → f (x) = L lim x0 x → F (f (x)) = F lim x0 x → lim x0 x → f (x) = F (L). The condition on the function F that is called continuity at the point L and is the subject of Section 5.4. F (z) = F (L) lim L z → Exercises 5.2.23 Show that lim x→x0 f (x) = lim x→x0 f (x) could be false if limx→x0 f (x) = 0 and definitely is false if limx→x0 f (x) < 0. p q 5.2.24 Give a formal proof of Theorem 5.25 using the sequential method sketched in the text. 5.2.25 Give a formal proof of Theorem 5.25 using an ε-δ method. 5.2.26 Give a formal proof of Theorem 5.25 using the mapping idea. 5.2.27 Give an example of a limit for which even though both of the limits in the statement do exist. lim x→x0 F (f (x)) = F lim x→x0 f (x) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 294 5.2.28 Show that lim x→x0 | f (x) | = under some appropriate assumption by applying Theorem 5.25. See Note 106 lim x→x0 f (x) Continuous Functions Chapter 5 5.2.29 Show that lim x→x0 under some appropriate assumption by applying Theorem 5.25. See Note 107 = f (x) | | p lim x→x0 f (x) s 5.2.30 Obtain Corollary 5.24 as an application of Theorem 5.25. 5.2.6 Examples There are a number of well-known examples of limits that every student should know. Partly this is because there will be an expectation in later courses that these should have been seen. But, more important, an abundance of examples is needed to gain some insight into when limits exist and when they do not and how they behave. For any function f defined near a point x0 there are several possibilities we should look for. 1. Does the limit limx x0 f (x) exist? → 2. If the limit does exist, is the limit the most likely value, namely (Such functions are said to be continuous at the point x0.) f (x) = f (x0)? lim x0 x → ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.2. Properties of Limits 295 3. If the limit fails to exist, then could it be that the one-sided limits do exist but happen to be unequal; that is, x f (x) lim → x0+ = lim x0 x → − f (x)? (Such a function is said to have a jump discontinuity at the point x0.) The case that is most familiar, namely where f (x) = f (x0), lim x0 x → is described by the language of continuity. Our study of continuity comes in the next section. But let us be aware now of when a function has this property. Polynomials All polynomial functions have entirely predictable limits. If + anxn, p(x) = a0 + a1x + a2x2 + · · · then p(x) = p(x0) lim x0 x → x0 a0 = a0 and the fact that limx at every value. (In the language we shall use, these functions are continuous.) To prove this we can use the x0 x = x0. These are trivial to prove. Then the polynomial fact that limx is built up from this by additions and multiplications. The theorems of Section 5.2.3 can be used to x0(x)(x2) = x3 complete the verification [e.g., limx the product rule applied again]. 0 by the product rule, limx x0 x3 = limx x0 x2 = x2 0 by → → → → → Rational Functions A rational function is a function of the form R(x) = p(x) q(x) where p and q are polynomials (i.e., a ratio of polynomials and hence the name). Since we can take limits R(x) = lim x0 x → limx limx → → x0 p(x) x0 q(x) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 296 Continuous Functions Chapter 5 freely, excepting only the case where the denominator is zero, we have found that R(x) = R(x0) lim x0 x → except at those points where q(x0) = 0. At those points, it is possible that the limit exists. Note, however, that it cannot equal R(x0) since R(x0) is not defined. It is also possible that the right-hand and left-hand limits are infinite. There are some examples in the exercises to illustrate these possibilities. Exponential Functions The exponential function ex can be proved to have the limiting value that we would expect, namely To prove this depends on how we have defined the exponential function in the first place. There are many ways in which we can develop such a theory. It is usual to wait for more theoretical apparatus and then define the exponential function in an appropriate way that allows that to be exploited. Recall that we mentioned in Example 2.36 that ex = ex0. lim x0 x → x3 3! Sums like this are called power series. As part of the theory of power series we will discover precisely when they are continuous. Then it is possible to define the exponential function as a power series and claim continuity immediately. ex = 1 + x + xn n! + . . . . x2 2! · · · + + + Most of the elementary functions of the calculus (trigonometric functions, inverse trigonometric functions, etc.) can be handled in this way. We do not pause here to worry about limits of such functions. Characteristic Function of the Rationals The characteristic function of a set E of real numbers is the function that assigns value 1 at points in E and value 0 at points outside E. Some authors call it an indicator function since it does, indeed, indicate when points are or are not in the set. For an interesting example of a function that would have been considered bizarre in the early days of calculus, consider the characteristic function of the rationals: χQ (x) = 1 if x Q ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.2. Properties of Limits and It is an easy exercise to check that χQ (x) = 0 if x Q. 6∈ lim → x0+ x χQ (x) and lim x0 → x − χQ (x) both fail to exist. Dirichlet Function The Dirichlet function is defined on [0, 1] by 297 f (x) = 0, 1/q, if x is irrational or x = 0 if x = p/q, p, q IN, p/q in lowest terms. ∈ To examine the limiting behavior of this function, we need to observe that while there are many points where this function is positive (all rationals) there are not many points where it assumes a value greater than some positive number ε. Indeed if we count them we will see that for any positive integer q the set of points contains at most q(q are only finitely many such points. − 1)/2 points. The exact number is not important; all we need to observe is that there Sq = x { ∈ [0, 1] : f (x) 1/q ≥ } Thus let ε > 0 and choose any integer q large enough so that 1/q < ε. For any point x0 we can choose δ, x0) and (x0, x0 + δ) contain no points of the finite set Sq. δ, x0) or (x0, x0 + δ) satisfies δ > 0 in such a way that both intervals (x0 − That must mean that every point x in (x0 − 0 f (x) < 1/q < ε. ≤ Thus it follows that at every point x0. In particular, the equation f (x) = 0 lim x0 x → f (x) = f (x0) lim x0 x → ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Ed |
ition (2008) 298 Continuous Functions Chapter 5 will hold at every irrational point x0 but must fail at every rational point. In the language of continuity we have proved that this function is continuous at every irrational point but discontinuous at every rational point. A curious function: It appears to be continuous at nearly every point and discontinuous at nearly every point. Nineteenth-century mathematicians were quite intrigued by such functions and called them pointwise discontinuous, a term that seems not to have survived. (We shall return to this example occasionally. For example, Exercise 7.5.4 asks for an account of the local extrema of this function.) Nondecreasing Functions with Jumps The simplest example of a function with a discontinuity is perhaps This function fails to have a limit at x = 0 since limx H(x) = 0. In the language → introduced earlier in this section we would say that H has a jump (or a jump discontinuity) at the point 0. c) has a jump at x = c. Moreover, if The discontinuity can be placed at any point. The function H(x 0+ H(x) = 1 and limx ≥ → − 0 < ck is a finite sequence of distinct points, then the function − c1 < c2 < c3 < · · · H(x) = 0 if x < 0 0 1 if x k F (x) = H(x ci) − Xi=1 is a nondecreasing function with jumps at each of the points c1, c2, c3, . . . , ck. At every other point x0 it is the case that limx x0 F (x) = F (x0). An interesting question now occurs. We have succeeded in constructing a function that is nondecreasing → and has jumps at a prescribed finite set of points. Can we construct such a function if we wish to have jumps at a given infinite set of points? This is a question to which we will return. Step Functions A function f is a step function if it assumes finitely many values, say b1, b2, . . . , bN and for each 1 N the set i ≤ ≤ f − 1(bi) = x : f (x) = bi} , { ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.2. Properties of Limits 299 b5 b4 b3 b2 b1 Figure 5.2. Graph of a step function. which represents the set of points at which f assumes the value bi, is a finite union of intervals and singleton point sets. Another way to think about this is that a function of the form is a step function if all the Ai are intervals or singleton sets. (See Figure 5.2 for an illustration.) Xi=1 Step functions play an important role in integration theory. They offer a crude way of approximating M f (x) = aiχAi (x) functions. The function H(x) = 0 if x < 0 0 1 if x that we have just seen is a simple step function since it assumes just two values, 0 and 1, where 0 is assumed on the interval ( , 0) and 1 is assumed on [0, ≥ ). ∞ The discontinuities of a step function are easy to detect. −∞ Distance of a Closed Set to a Point Let C be a closed set and define a function by writing d(x, C) = inf x y | − {| : y C . } ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 300 Continuous Functions Chapter 5 This function gives a meaning to the distance between a set C and a point x. If x0 ∈ and if x0 6∈ This function is continuous at every point; that is, this function has the property that C, then d(x0, C) > 0. C, then d(x0, C) = 0, We can interpret (1) geometrically: If two points x1 and x2 are close together, then they are at roughly the same distance from the closed set C. d(x, C) = d(x0, C). lim x0 x → (1) The Characteristic Function of the Cantor Set Let K be the Cantor set and let χK be its characteristic function; that is, let χK = 1 if x K and χK (x) = 0 otherwise. This function has the property that ∈ if x0 is not in the Cantor set and the limit exists at no point in the Cantor set. For an easy proof of this you will have to review the properties of the Cantor set and its complement in Exercises 4.3.23 and 4.4.9. lim x0 x → χK (x) = 0 Exercises 5.2.31 Give a proof that includes all necessary details that the limit for all polynomials p. 5.2.32 Suppose that you know that Prove that limx→x0 ex = ex0 for all x0. See Note 108 lim x→x0 p(x) = p(x0) ex = e2. lim x→2 5.2.33 Suppose that you know that lim x→0 Prove that limx→x0 sin x = sin x0 for all x0. cos x = 1 and lim x→0 sin x = 0. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.2. Properties of Limits See Note 109 301 5.2.34 In the text we constructed a nondecreasing function with jumps at each of the points c1, c2, c3, . . . , ck and continuous everywhere else. Construct an increasing function with this property. See Note 110 5.2.35 Let f : [a, b] → R be a step function. Show that there is a partition so that f is constant on each interval (xi−1, xi), i = 1, 2, . . . , n. a = x0 < x1 < x2 < < xn−1 < xn = b · · · 5.2.36 Suppose that M f (x) = aiχAi where the Ai are intervals. Show that f is a step function; that is, that f assumes finitely many values, and for each b in the range of f the set f −1(b) is a finite union of intervals or singleton sets. Where are the discontinuities of such a function? See Note 111 i=1 X 5.2.37 Show that the characteristic function of the rationals can also be defined by the formula 5.2.38 Show that χQ (x) = lim m→∞ lim n→∞ | n. cos(m!πx) | lim x→x0+ χQ (x) and lim x→x0− χQ (x) both fail to exist, where χQ is the characteristic function of the rationals. What would be the answer to the corresponding question for the characteristic function of the irrationals? 5.2.39 Describe the graph of the function χQ . What kind of a sketch would convey this set? 5.2.40 Give an example of a set E such that the characteristic function χE of E has limits at every point. Can you describe the most general set E with this property? 5.2.41 Give an example of a set E such that the characteristic function χE of E has one-sided limits at every point. Can you describe the most general set E with this property? ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 302 5.2.42 Show that Continuous Functions Chapter 5 at every point x0 where d(x, C) is the distance from x to the closed set C as defined in this section. lim x→x0 d(x, C) = d(x0, C) 5.2.43 Sketch the graph of the function d(x, C) for several closed sets C (e.g., , IN, [0, 1], 0 { } 0 { 1, 1/2, 1/3, 1/4, . . . }∪{ , } 5.2.44 Sketch the graph of the characteristic function χK of the Cantor set (Exercises 4.3.23 and 4.4.9) and show and [0, 1] [2, 3]). ∪ that lim x→x0 at all points x not in the Cantor set and that this limit fails to exist at all points in Cantor set. See Note 112 χK (x) = 0 5.3 Limits Superior and Inferior If limits fail to exist we need not abandon all hope of discussing the limiting behavior. We saw this situation in our study of sequence limits in Section 2.13. Even if sn fails to exist, it is possible that the two extreme limits diverges so that limn sn} →∞ { " Adv. provide some meaningful information. These two concepts always exist (possibly as situation occurs for functions. The theory is nearly identical in many respects. lim inf n →∞ sn and lim sup →∞ n sn or ∞ −∞ ). A similar Definition 5.26: (Lim Sup and Lim Inf ) Let f : E that x0 is a point of accumulation of E. Then we write → R be a function with domain E and suppose lim sup x0 x → f (x) = inf δ>0 sup { f (x) : x (x0 − ∈ δ, x0 + δ) E, x = x0} ∩ and lim inf x0 x → f (x) = sup δ>0 inf { f (x) : x (x0 − ∈ δ, x0 + δ) E, x = x0} ∩ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 Section 5.3. Limits Superior and Inferior 303 As this section is for more advanced readers we have left the development of this concept to the exercises. Exercises 5.3.1 Show from the definition that 5.3.2 Compute each of the following. (a) lim supx→0 sin x−1 (b) lim supx→0 x sin x−1 (c) lim supx→0 x−1 sin x−1 lim sup x→x0 f (x) lim inf x→x0 ≥ f (x). 5.3.3 Formulate an equivalent definition for lim supx→0 f (x) expressed in terms of sequential limits; that is, in x0. Show that your definition is equivalent to that in the text. terms of limits of f (xn) for xn → 5.3.4 Give an example of a function f so that lim inf x→0 f (x) = 0 and lim sup f (x) = 1. x→0 5.3.5 What changes, if any, are there if the definition of lim sup had been written as lim sup x→x0 f (x) = inf δ>0 sup { f (x) : x (x0 − ∈ δ, x0 + δ) E ? } ∩ See Note 113 5.3.6 Formulate a definition for the one-sided concepts 5.3.7 Give an example of a function f with the properties lim sup x→x0+ f (x) and lim sup x→x0− f (x). lim inf x→0+ f (x) = 0, lim sup x→0+ f (x) = , ∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 304 and Continuous Functions Chapter 5 lim inf x→0− f (x) = , −∞ lim sup x→0− f (x) = 1. 5.3.8 Show that limx→0 f (x) exists if and only if all four of and are equal and finite. 5.3.9 Show that limx→0 f (x) = if and only if ∞ and lim inf x→0+ f (x), lim sup x→0+ f (x), lim inf x→0− f (x), lim sup x→0− f (x) lim inf x→0+ f (x) = lim sup x→0+ f (x) = lim inf x→0− f (x) = lim sup x→0− f (x.3.10 For the function χQ , the characteristic function of the rationals, determine the values of each of the limits lim inf x→x0+ χQ (x), lim sup x→x0+ χQ (x), lim inf x→x0− χQ (x), and lim sup x→x0− χQ (x) at any point x0. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.4. Continuity 305 5.3.11 Give an example of a function f such that x0 : lim sup x→x0− f (x) > lim sup x→x0+ f (x) } { is infinite. 5.4 Continuity The earliest use of the term “continuity” is somewhat clouded by misconceptions of the nature of a function. If a function was given by a single formula then it was considered in the eighteenth century to be “continuous.” If, however, the function had a “break” in the formula—defined differently in one interval than in another—it was considered as “discontinuous.” As the subject developed these notions continued to obscure the really important ideas. Augustin Cauchy (1789–1857) was the first to give the modern definition a |
nd to focus attention on the concept that has now assumed such an important role in analysis. 5.4.1 How to Define Continuity " Enrichment section. May be omitted. Before we proceed to the present day definition, let us consider another notion. Even as late as the middle of the nineteenth century, some mathematicians believed this notion should form the basis for a definition of continuity. This concept is suggested by the phrase “the graph has no jumps.” While some instructors of calculus courses might use such phrases to convey a sense of continuity to students, the phrase is not a precise one, nor does it fully convey all we wish a continuous function to be. This notion is related to continuity, however, and has some importance in its own right. We’ll begin with a brief discussion of it. Here is one attempt at making our phrase precise. (See Figure 5.3.) Definition 5.27: (Intermediate Value Property) Let f be defined on an interval I. Suppose that for = f (b), and for each d between f (a) and f (b), there exists c between a and b for each a, b which f (c) = d. We then say that f has the intermediate value property (IVP) on I. I with f (a) ∈ Functions with this property are called Darboux functions after Jean Gaston Darboux (1842–1917), who ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 306 Continuous Functions Chapter 5 f(a) d f(b) Figure 5.3. At some point between a and b the function assumes any given value d between f (a) and f (b). a c b showed in 1875 that for every differentiable function F on an interval I, the derivative F ′ has the IVP on I. He is also particularly famous for his 1875 account of the Riemann integral using upper and lower sums; often reference is made to the “Darboux integral,” meaning this version of the classical Riemann integral. Example 5.28: Let F (x) = 1 sin x− 0 if x = 0 if x = 0. The graph of F is shown in Figure 5.4. You may wish to verify that F has the IVP. In particular, F assumes every value in the interval [ 1, 1] infinitely often in every neighborhood of x = 0. ◭ − We haven’t yet made precise the phrase “the graph has no jumps,” but the IVP seems to convey that idea well enough. Since this property is so easy to describe and appears to have content that is easy to visualize, why not take it as the definition of continuity? Before attempting to answer that question, let us offer a competing phrase to capture the idea of continuity: “If x is near x0, then f (x) is near f (x0).” As stated, this phrase is not precise, but we can make it precise using the limit concept. This phrase could be interpreted really as asserting that f (x0) = lim x0 → x f (x). (2) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 5.4. Continuity 307 Figure 5.4. Graph of the function F (x) = sin x −1 on [ π/8, π/8]. − → According to this criterion our function F of Example 5.28 would not be continuous at x0 = 0, because F (0) = 0, but limx 0 F (x) does not exist. We shall see presently that the definition based on limits allows the development of a useful theory. We’ll see that the class of continuous functions [as defined using equation (2)] is closed under addition and multiplication, and that such functions have many other desirable properties. For example, the class is closed under certain kinds of limits of sequences, and every continuous function on [a, b] is integrable. On the other hand (as is shown in the exercises), none of the analogous statements is valid for the class of functions defined by IVP. Thus a theory of continuity based on the limit concept allows a rich structure and enjoys wide applicability, whereas one based on the IVP is rather limited. In addition, the fundamental notion of limit extends to much more general settings than R. In contrast, extensions of IVP, while possible, are peripheral to mathematical analysis. Exercises 5.4.1 Refer to Example 5.28. Let G(x) = F (x) − 1 if x = 0 if x = 0. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 308 Continuous Functions Chapter 5 Show that G has the IVP, yet F + G does not. Thus the class of functions with IVP is not closed under addition. 5.4.2 Give an example to show that the class of functions with IVP is not closed under multiplication. 5.4.3 Let H(x) = x−1 sin x−1 0 if x = 0 if x = 0. Show that H has IVP on [0, 1] but is not bounded there. (This shows that the IVP does not imply boundedness; we shall see that, in contrast, any continuous function on [0, 1] would have to be bounded.) 5.4.4 Give an example of a function f with IVP on [0, 1] that is bounded but achieves no absolute maximum on [0,1]. 5.4.5 Let K be the Cantor set of Exercises 4.3.23 and 4.4.9 and let complementary to K in (0,1). (ak, bk) } { be the sequence of intervals (a) Write down a set of equations defining a function f that vanishes at every two-sided point of accumulation of K, is continuous on each interval [ak, bk], and for which (See Figure 5.5 for an illustration of one possible choice.) lim x→ak+ f (x) = − 1 and lim x→bk− f (x) = 1. (b) Verify that f has the intermediate value property. (c) Verify that f is not continuous in the sense that f (x0) = limx→x0 f (x) fails at certain points. (Which points?) 5.4.6 " We construct a function with IVP whose graph may be more difficult to visualize. Let I0 = (0, 1). Each IN and x = .a1a2 . . . in I0, let I0 has a unique decimal expansion not ending in a string of 9’s. For each n x ∈ fn(x) = a1(x) + a2(x) + n + an(x) . · · · ∈ Thus fn(x) represents the average of the first n digits of x. For each x (a) Show that f : I0 → (b) Describe how to construct x I0 such that f (x) = π. [0, 10]. ∈ I0, let f (x) = lim supn fn(x). ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 5.4. Continuity 309 1 1 Figure 5.5. Function defined on the complement of the Cantor set, as described in Exercise 5.4.5. The first three stages are shown. (c) Describe how to construct x (.01, .02) such that f (x) = π. ∈ (d) Show that for each interval (a, b) I0 and each d Thus, f assumes every value in [0,10] in every interval in I0. In particular, f has IVP. [0, 10] there exists c ⊂ ∈ ∈ (a, b) such that f (c) = d. (e) Let A = { (f) Show that x : f (x) = x } need not have IVP. − . Let g(x) = 0 if x A, g(x) = f (x) for x / ∈ ∈ A. Show that g(x) has IVP. g(x) + x does not have IVP. Thus the sum of a function with IVP with the identity function 5.4.2 Continuity at a Point Let us look at Cauchy’s concept of continuous function. We begin by defining continuity at a point, more specifically continuity at an interior point of the domain of a function f . This way we are assured that if we are interested in what is happening at the point x0 then f is defined in a neighborhood of x0; that c, x0 + c) for a positive number c. This simplifies some of the is, that f is defined in some interval (x0 − computations. Definition 5.29: (Continuous) Let f be defined in a neighborhood of x0. The function f is continuous at x0 provided limx x0 f (x) = f (x0). → ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 310 Continuous Functions Chapter 5 ∈ ∈ U , then f (x) This means that for each neighborhood V of f (x0) there is a neighborhood U of x0 such that f (U ) ⊂ V . We can, of course, state the definition in terms of δ’s and ε’s: f is that is, if x continuous at x0 if for each ε > 0 there exists δ > 0 such that In the exercises we ask you to verify that the three formulations, involving the language of limits, of neighborhoods, and of δ’s and ε’s, are equivalent. We believe that this is an important exercise for readers who do not yet feel comfortable with the limit concept. Feeling comfortable with the various forms that continuity takes is essential to feeling comfortable with many of the arguments that appear in the sections and chapters that follow. < ε whenever f (x0) x0| f (x) < δ. V : − − x | | | Observe that a function f can fail to be continuous at x0 in three ways: 1. f is not defined at x0. 2. limx → x0 f (x) fails to exist. 3. f is defined at x0 and limx x0 f (x) exists, but → f (x0) = lim x0 x → f (x). We leave it to you to provide simple examples of each of these possibilities. Example 5.30: Let f : (0, continuous at x0. ) ∞ R be defined by f (x) = 1/x. We show that if x0 ∈ → (0, ∞ ), then f is Let’s first try the “neighborhood” definition of continuity. Let V be a neighborhood of f (x0), say V = (A, B). Thus A < f (x0) < B. We must find a neighborhood U = (a, b) of x0 such that f (U ) V . A picture suggests what to do: Let a = 1/B, b = 1/A. (See Figure 5.6.) But we must be a bit careful here. Nothing in our neighborhood definition of continuity allows us to assume A > 0, so b might not be defined (if A = 0), or might not be in the domain of f (if A < 0). This presents, however, only a minor nuisance. Thus we assume A > 0 in our proof. ⊂ So, let us assume A > 0, a = 1/B, b = 1/A. Then, since A < f (x0) < B, we have 1 A > x0 = 1 f (x0) = a, 1 B b = > ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 5.4. Continuity 311 1/x V B f(x0) A a x0 U b Figure 5.6. Graphical interpretation of the neighborhood definition of continuity for the function f (x) = x a = 1/B and b = 1/A. −1. Note that so x0 ∈ (a, b) = U. Furthermore, if c ∈ U , then a < c < b and B > 1/c = f (c) > A, so f (c) ∈ V . This shows that f (U ) V as was required. ⊂ ◭ (0, ), then f is continuous at x0. Example 5.31: Let us take the same example f : (0, that if x0 ∈ We see now how a proof based on the δ-ε definition might look. As with our first proof, we shall provide many details of the proof. After you feel conversant with limits and continuity, you may wish to streamline the proofs somewhat by leaving out details that “any reader finds obvious.” x ) and let ε > 0. We wish to find δ > 0 such that if R, defined by f (x) = 1/x. We sh |
ow again < δ and x > 0, then ∞ ∞ → ) Let x0 ∈ 1/x0| − 1/x | (0, < ε. Rewriting this last inequality as ∞ | − x0| x0| suggests we try δ = εxx0. But δ should depend only on ε and x0, not on x. There is no δ > 0 for which the inequality < δ implies the inequality < εxx0 − x x | (3) x0| | − x | − x0| < εxx0 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 312 for all x ). We can remove this problem by first requiring x to stay away from 0. Continuous Functions Chapter 5 (0, ∈ ∞ For example, we first require that Then x | − x0| < 1 2 x0. 1 2 1 2 x0 < x and εx2 0 < εxx0. (4) (5) (6) The inequalities (3), (4), and (6) suggest taking δ = min For this δ , we compute easily that if x 1 2 x2 0ε 1 2 x2 0 the last inequality being obtained by using (6) on the numerator ◭ = | < | < δ, then x0| − xx0| x | − | 1 x − x0| 1 x0 1 2 x0, 1 2 x2 0ε . = ε, x x0| − and (5) on the denominator xx0| | . Exercises 5.4.7 Prove that the function f (x) = x2 is continuous at every point of R using the δ-ε form of continuity, 5.4.8 Prove that the function f (x) = x | | is continuous at every point of R using the δ-ε form of continuity, 5.4.9 Show that the three formulations of continuity appearing at the beginning of this section are equivalent. 5.4.10 In the δ-ε verification of continuity of the function 1/x we obtained a δ that did the job. We made no claim that this δ is the largest possible δ we could have chosen. Show that for ε = 1 and x0 ∈ works must satisfy δ < x2 0/(1 + x0). (0, 1) any δ that ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.4. Continuity 313 5.4.11 (Sequence Definition of Continuity) Prove that f is continuous at x0 if and only if limn→∞ f (xn) = f (x0) x0. How would you expect this characterization of continuity at x0 to be modified for every sequence xn} → if x0 is not an interior point of its domain? { 5.4.12 Give three examples of a function f that fails to be continuous at a point x0. The first should be discontinuous merely because f is not defined at x0. The second should be discontinuous because limx→x0 f (x) fails to exist. The third should have neither of these defects but should nonetheless be discontinuous. 5.4.13 A function f is said to be symmetrically continuous at a point x if − Show that if f is continuous at a point, then it must be symmetrically continuous there and that the converse does not hold. − lim h→0+ [f (x + h) f (x h)] = 0. 5.4.3 Continuity at an Arbitrary Point To this point we have discussed continuity of a function at an interior point of its domain. How should we modify our notions if x0 is not an interior point? Continuity at Endpoints For example, if f : [a, b] Since the function is defined only on the interval [a, b] and we have defined continuity in terms of limits, it seems that we should require, as before, that for any interior point x0 ∈ R, how can we define continuity of f at a or at b? (a, b) → while at the endpoints continuity would be defined by a one-sided limit, f (x) = f (x0) lim x0 x → f (x) = f (a) and lim a+ x → f (x) = f (b). lim b x − → We can also reformulate our definition in a way that recognizes that f is defined only on [a, b]. In our V . neighborhood definition we interpret U as a relative neighborhood of x0: We require that f (U Here by a relative neighborhood we mean that part of an ordinary neighborhood that is inside the set where f is defined. [a, b]) ⊂ ∩ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 314 Continuous Functions Chapter 5 Similarly, for the δ-ε definition, our requirement becomes that f (x) | − f (x0) | < ε whenever defined. x | − x0| < δ and x ∈ [a, b]. Again we are merely restricting our attention to the set where f is Continuity on an Arbitrary Set These reformulations would work for any set A, not just an interval. Thus we assume that so that f is a function with domain A and x0 is an arbitrary point of A, which need not be an interior point nor even a point of accumulation (it might be isolated in A). f : A R → There are four versions of the definition. As before, you should check to see that they are indeed equivalent. Some extra care is needed because x0 could be any point of A and may be isolated in A. Definition 5.32: (ε-δ Version) Let f be defined on a set A and let x0 be any point of A. The function f is continuous at x0 provided for every ε > 0 there is a δ > 0 so that for every x A for which x | − x0| < δ. ∈ f (x) | − f (x0) | < ε Definition 5.33: (Limit Version) Let f be defined on a set A and let x0 be any point of A. The function f is continuous at x0 provided either that x0 is isolated in A or else that x0 is a point of accumulation of that set and f (x) = f (x0). lim x0 x → Definition 5.34: (Neighborhood Version) Let f be defined on a set A and let x0 be any point of A. The function f is continuous at x0 provided that for every open set V containing f (x0) there is an open set U containing x0 so that f (U A) V . ∩ ⊂ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.4. Continuity 315 In other words, the neighborhood version asserts that there is a set U A open relative to A that f A is relatively open relative to A if B is the intersection of some ∩ maps into V . We recall that a set B open set (here U ) with A. ⊂ Definition 5.35: (Sequential Version) Let f be defined on a set A and let x0 be any point of A. The function f is continuous at x0 provided that for every sequence of points belonging to A and converging to x0, it follows that f (xn) f (x0). xn} { → Exercises 5.4.14 Prove the equivalence of the four definitions for the continuity of a function defined on an arbitrary set A. 5.4.15 Let f : IN R by writing f (n) = 1/n2. Is f continuous at any point in its domain? → See Note 114 5.4.16 Using each of the four versions of continuity, show that any function is automatically continuous at any point of its domain that is isolated. 5.4.17 Let f be defined on the set containing the points 0, 1, 1/2, 1/4, 1/8, . . . , 1/2n only. What values can you assign at these points that will make this function continuous everywhere where it is defined? See Note 115 5.4.18 Let f be defined on the set containing the points 0, 1, 1/2, 1/4, 1/8, . . . , can you assign at these points that will make this function continuous everywhere where it is defined? ± ± ± ± ± 1/2n, . . . only. What values 5.4.19 If f is continuous at a point x0 then is it necessarily true that lim x→x0 f (x) = f (x0)? At what points in the domain of f can you say this? See Note 116 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 316 Continuous Functions Chapter 5 R is said to be Lipschitz if there is a positive number M so that y 5.4.20 A function f : [a, b] for all x, y [a, b]. Show that a Lipschitz function must be continuous. Is the converse true? [Rudolf Otto Sigismund Lipschitz (1832–1903) is probably best remembered for this condition, now forever attached to his name, which he used in formulating an existence theorem for differential equations of the form y′ = f (x, y).] See Note 117 f (x) | f (y.4.4 Continuity on a Set " Enrichment section. May be omitted. Continuity is defined at points. A function such as f (x) = x2 could be said to be continuous at every 0 for every real number. In many cases the function real number x0, meaning only that limx considered is continuous at every point in its domain. We say simply that f is continuous. But we must remember this is an assertion about every single point where f is defined. x0 x2 = x2 → Definition 5.36: Let f : A point of A. → R. Then f is continuous (or continuous on A) if f is continuous at each If we wish to prove directly from this definition that f is continuous, we must show that f is continuous A. It is sometimes easier to use the global characterization of continuity that follows. at every x0 ∈ Theorem 5.37: Let f : A V f − 1(V ) = A : f (x) x → } R. Then f is continuous if and only if for every open set V is open (relative to A). R, the set ⊂ ∈ ∈ { Suppose first that f is continuous. Let V be open, let x0 ∈ Proof. (α, β) V and so that x0 ∈ that α < f (x) < β for all x ∈ exists δ > 0 such that if 1(V ) and choose α < β so that 1((α, β)). Then α < f (x0) < β. We will find a neighborhood U of x0 such α). Since f is continuous at x0, there A. Let ε = min(β f (x0), f (x0) f − U f − − ⊂ ∩ then A x ∈ ∩ − (x0 − δ, x0 + δ), f (x) | − f (x0) | < ε. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.4. Continuity Thus and so f (x) < β. Similarly, f (x) − f (x0) < β f (x0), − 317 − and so f (x) > α. Thus the relative neighborhood U = (x0 − hence also a subset of f − in f − 1(V ) is open relative to A. 1(V ). That is, f − 1((α, β)) and A is a subset of f − ∩ 1(V ) has a relative neighborhood 1(V ). We have shown that each member of f − δ, x0 + δ) − f (x) f (x0) > α f (x0), To prove the converse, suppose f satisfies the condition that for each open interval (α, β) with α < β, A. We must show that f is continuous at x0. Let ε > 0, 1((α, β)) is open relative to A. Let x0 ∈ ε. Our hypothesis implies that f − 1((α, β)) is open relative to A. Thus the set f − β = f (x0) + ε, α = f (x0) − f − 1((α, β)) = (ai, bi) A, ∩ the union being a finite or countable union of pairwise disjoint open intervals. One of these intervals, say (aj, bj), contains x0. Let [ For x | − x0| < δ and x ∈ A we find Because β = f (x0) + ε and α = f (x0) − This shows that f is continuous at x0. δ = min(x0 − aj, bj − x0). α < f (x) < β. ε we must have f (x) | − f (x0) | < ε. We spelled out the details of the proof of Theorem 5.37. This may have caused it to appear rather lengthy. But the proof is nothing more than writing down in a rigorous way what some intuitive pictures indicate. You might find that the neighborhood notion of continuity is a more natural one to use for proving the theorem. W |
e leave this as Exercise 5.4.23. As a corollary let us point out that we can replace open sets by open intervals; thus to check continuity of a function f it is enough to show that f − 1((α, β)) is open for every interval (α, β). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 318 Continuous Functions Chapter 5 Corollary 5.38: Let f : A is open (relative to A). → R. Then f is continuous if and only if for every interval (α, β), f − 1((α, β)) 1((α, β)) Proof. We verify that the conditions (i) f − is relatively open for all α < β are equivalent. But this is immediate. If (i) is satisfied, then (ii) is also, since the requirement (ii) is just a special case of (i). On the other hand, if (ii) is satisfied and 1(V ) is relatively open for all open V R and (ii) f − ⊂ then V = (αi, βi), [ f − 1(V ) = f − 1((αi, βi)). Each of the sets f − of open sets. 1((αi, βi)) is open by hypothesis, so f − [ 1(V ) is also open because it is a union of a family Example 5.39: Let f (x) = 1/x (x > 0). We find f − 1((α, β)) = 1 β , 1 α . Since (1/β, 1/α) is open it would follow that f is continuous on (0, ). ∞ Exercises 5.4.21 Prove that the function f (x) = x2 is continuous on R by using Theorem 5.37. 5.4.22 Prove that the function f (x) = x | | is continuous on R by using Theorem 5.37. 5.4.23 Prove Theorem 5.37 using the neighborhood definition of continuity. 5.4.24 Let f be continuous in a neighborhood U of the point x0. If f (x) < β for all x β. Show by example that we cannot conclude f (x0) < β. f (x0) ≤ ◭ U x0} \ { ∈ , prove that ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.4. Continuity 319 5.4.25 Let f, g be defined on R. Suppose f (0) = 0 and f is continuous at x = 0. Suppose g is bounded in some neighborhood of zero. Prove that f g is continuous at x = 0. Apply this to the function f (x) = x sin(1/x) (f (0) = 0) at x = 0. 5.4.26 Let x0 ∈ R. Following are four δ-ε conditions on a function f : R imply continuity of f at x0? Which, if any, are implied by continuity at x0? → R. Which, if any, of these conditions (a) For every ε > 0 there exists δ > 0 such that if (b) For every ε > 0 there exists δ > 0 such that if (c) For every ε > 0 there exists δ > 0 such that if (d) For every ε > 0 there exists δ > 0 such that if < δ, then f (x0) | < ε, then f (x0) | R. Following are four δ-ε conditions on a function f : R imply continuity of f at x0? Which, if any, are implied by continuity at x0? f (x) | < δ, then f (x) | < ε, then x − | f (x) | x | − f (x) | x0| − x0| − → f (x0) | − x0| x | − f (x0) − | x0| x | − < ε. < ε. < δ. < δ. 5.4.27 Let x0 ∈ R. Which, if any, of these conditions (a) There exists ε > 0 such that for each δ > 0, if (b) There exists ε > 0 such that for each δ > 0, if (c) There exists ε > 0 such that for each δ > 0, if (d) There exists ε > 0 such that for each δ > 0, if x | − f (x) | x | − f (x) | x0| − x0| − < δ, then f (x0) | < ε, then f (x0) | f (x) | < δ, then f (x) | < ε, then f (x0) − | x0| x − | f (x0) − | x0| x | − < ε. < ε. < δ. < δ. 5.4.28 For each of the eight conditions of Exercises 5.4.26 and 5.4.27, describe in words which functions satisfy the condition. (Some of these conditions characterize familiar classes of functions, including the empty class.) ⊂ ◦ 5.4.29 Let A then g R, f : A f is continuous at x0. Apply this to prove that if f is continuous at x0, then R. Prove that if f is continuous at x0 ∈ A and g is continuous at f (x0), is continuous at x0. R, g : f (A) → → 5.4.30 Using the notions of unilateral or one-sided limits, define left continuity of a function f at a point x0. Do the same for right continuity. If f is defined in a neighborhood of x0, prove that f is continuous at x0 if and only if f is both left continuous and right continuous at x0. 5.4.31 Let f : R R. Prove that f is continuous if and only if for every closed set K → in R. State carefully and prove the analogous result if f : A of R. → R, the set f −1(K) is closed R, where A is an arbitrary nonempty subset ⊂ 5.4.32 Suppose f has the IVP on (a, b) and is discontinuous at x0 ∈ x : f (x) = y is infinite. { } (a, b). Prove that there exists y R such that ∈ f | | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 320 Continuous Functions Chapter 5 5.5 Properties of Continuous Functions We now present some of the most basic of the properties of continuous functions. The first theorem is an algebraic one; it asserts that the family of continuous functions defined on a set has many of the properties of an algebra: elements may be added, subtracted, multiplied, and (under some conditions) divided. Theorem 5.40: Let f, g : A f + g and f g are continuous at x0. Furthermore, if g(x0) R. Suppose f and g are continuous at x0 ∈ = 0, then f /g is continuous at x0. R and let c → ∈ A. Then cf , Proof. The results follow immediately from the limit definition of continuity and the usual algebraic properties of limits. Corollary 5.41: Every polynomial is continuous on R. Proof. The functions f (x) = 1 and g(x) = x are continuous on R. The corollary follows from Theorem 5.40. Corollary 5.42: Every rational function is continuous at each point in its domain (i.e., at each x which the denominator does not vanish). ∈ R at One of our most important properties allows us to compose two continuous functions. Be careful, though, with the conditions on the domains as they cannot be overlooked. Theorem 5.43: Let f : A point x0 ∈ → → A and that g is continuous at the point y0 = f (x0) B. Then the composition function R, g : B R and suppose that f (A) B. Suppose that f is continuous at a ⊂ ∈ f : A g ◦ R → is continuous at x0. Proof. This follows from Theorem 5.25. A global version follows as a corollary. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 5.6. Uniform Continuity 321 Corollary 5.44: Let f : A continuous on B, then the composition function R, g : B → → R and suppose that f (A) B. If f is continuous on A and g is ⊂ is continuous on A. Exercises f : A g ◦ R → 5.5.1 If f and g are functions such that f + g is continuous, does it follow that at least one of f or g must be continuous? 5.5.2 If is continuous, does it follow that f is continuous? f | | 5.5.3 If ef (x) is continuous, does it follow that f is continuous? 5.5.4 If f (f (x)) is continuous, does it follow that f is continuous? 5.6 Uniform Continuity Let us take a closer look at the meaning of continuity of a function f on an interval I. The definition I and for every ε > 0, there exists δ > 0 such that if x asserts that for each x0 ∈ I and x0| < δ, then − ∈ x | − Now carefully consider the following statement: | f (x) f (x0) | < ε. For every ε > 0, there exists δ > 0 such that if x, x0 ∈ I and x | − x0| < δ, then f (x) | − f (x0) | < ε. This may appear at first sight to be just a restatement of the meaning of continuity expressed in the first paragraph. If you cannot detect the difference, then you are in good company: Cauchy did not see any difference and used the property just quoted incorrectly to prove that a continuous function on an interval [a, b] must be integrable. We need to focus on the fact that the number δ depends not only on f and on ε, but also on x0; that is, δ = δ(f, ε, x0). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 322 Continuous Functions Chapter 5 Example 5.45: Consider the function f (x) = 1/x on the interval I = (0, 1). We found in Exercise 5.4.10 that if we take ε = 1, we can choose but we cannot choose a larger value. Thus if x0 → I. to “work” for all x0 ∈ δ(f, 1, x0) = x2 0 1 + x0 0, then δ(f, 1, x0) , 0. No number δ is sufficiently small ◭ → It is often important to be able to select δ independently of x0. When this is possible, we say that f is uniformly continuous on I. Definition 5.46: (Uniformly Continuous) Let f be defined on a set A ⊂ continuous (on A) if for every ε > 0 there exists δ > 0 such that if x, y f (x) f (y) < ε. | − | R. We say that f is uniformly < δ, then A and x y ∈ | − | As an illustration of the usefulness of uniform continuity, we note that if f is uniformly continuous on a bounded interval I, then f is bounded on I. Theorem 5.47: If a function f is uniformly continuous on a bounded interval I, then f is bounded on I. f (x) Proof. Here we suppose that I is one of (a, b), [a, b], [a, b), or (a, b]. To check that f is bounded, choose δ so that < xn = b y such that | intervals [xi < 1 whenever x, y < δ for i = 1, . . . , n. Our definition of δ implies that f is bounded on each of the < δ. There is a finite set a = x0 < x1 < − xi − I and · · · − ∈ x | | | f (y) | xi 1| − I. Let 1, xi] ∩ − mi = inf { Mi = sup { m = min { M = max f (x) : xi 1 ≤ − f (x) : xi 1 ≤ − m1, . . . , mn} M1, . . . , Mn} { . x x ≤ ≤ xi , x xi , x ∈ I , } I } ∈ Then, for every x I, m ∈ ≤ f (x) ≤ M , so f is bounded on I. , ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.6. Uniform Continuity 323 Observe that if we tried to present a similar argument for the function f (x) = 1/x on the interval I = (0, 1), the continuity of f would allow us to conclude that every x bounded, but we would be unable to obtain a finite number of such intervals that cover I. ∈ I is in an interval on which f is In our illustration that uniform continuity on I implies boundedness, we did not specify whether I contained one or more of its endpoints. Our next objective is to show that when I = [a, b] is a closed interval, then every function f that is continuous on I is uniformly continuous on I. (Note also the more general version given in Exercise 5.6.14.) This result will be of importance in many places. In particular, the important result we will later prove, that a continuous function f on [a, b] is integrable, depends on the uniform continuity of f . Cauchy certainly recognized this fact but failed to distinguish |
between continuity and uniform continuity. Theorem 5.48: Let f be continuous on [a, b]. Then f is uniformly continuous. Proof. Our proof invokes a compactness argument. We recall from our investigations of compactness in Section 4.5 that there are several equivalent formulations possible. We shall use the Bolzano-Weierstrass property. Exercise 5.6.13 you are asked to prove it using the Heine-Borel property.) (Exercise 5.6.2 asks for another proof of this same theorem using Cousin’s lemma. In We use an indirect proof. If f is not uniformly continuous, then there are sequences xn} { and yn} { so that xn − yn → 0 but f (xn) f (yn) > c − | | for some positive c. (The verification of this step is left as Exercise 5.6.12.) Now apply the Bolzano-Weierstrass property to obtain a convergent subsequence yn → ynk → both converge to the same limit z, which f (z) and f (ynk ) xnk } the limit of this new sequence { the corresponding subsequence ynk } { must be a point in the interval [a, b]. By the continuity of f , f (xnk ) f (xn) . Observe that xnk − yn} of the sequence > c for all n, this means from our study of sequence limits that . Write z as and xnk } f (z). Since 0 since xn − xnk } { 0. Thus f (ynz) | − f (z) | ≥ c > 0 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 324 Continuous Functions Chapter 5 and this is impossible. This contradiction proves the theorem. Boundedness of Continuous Functions As an application of Theorem 5.48 we can now prove that any continuous function on a closed bounded interval [a, b] is bounded. Indeed such a function must be uniformly continuous there, and we have already seen in Theorem 5.47 that a uniformly continuous function on a bounded interval is bounded. Thus we have the following useful theorem. Theorem 5.49: Let f be continuous on [a, b]. Then f is bounded. Exercises 5.6.1 Adjust the proof of Theorem 5.48 to show that if f is continuous on a compact set K, then f is uniformly continuous on K. See Note 118 5.6.2 Give another proof of Theorem 5.48 but this time using Cousin’s lemma. See Note 119 5.6.3 Because of Theorem 5.47 any function that is continuous on (0, 1) but unbounded cannot be uniformly continuous there. Give an example of a continuous function on (0, 1) that is bounded, but not uniformly continuous. 5.6.4 Let x1, x2, . . . , xn be real numbers, each in the domain of some function f . Show that f is uniformly 5.6.5 Let X = continuous on the set X = { x1, x2, . . . , xn, . . . uniformly continuous on X? See Note 120 { . x1, x2, . . . , xn} . What property must X have so that every function continuous on X is } 5.6.6 Suppose f is uniformly continuous on each of the sets X1, X2, . . . , Xn and let X = n i=1 Xi. Show that f need not be continuous on X. Show that, even if f is continuous on X, f need not be uniformly continuous on X. S ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.6. Uniform Continuity 325 5.6.7 Suppose f is uniformly continuous on each of the compact sets X1, X2, . . . , Xn. n i=1 Xi. Show that this need not be the case if the sets Prove that f is uniformly continuous on the set X = Xk are not closed and need not be the case if the sets Xk are not bounded. See Note 121 S 5.6.8 Let f be a uniformly continuous function on a set E. Show that if is a Cauchy sequence in E then is a Cauchy sequence in f (E). Show that this need not be true if f is continuous but not uniformly xn} { f (xn) { } continuous. 5.6.9 A function f : E for all x, y See Note 122 ∈ R is said to be Lipschitz if there is a positive number M so that M E. Show that such a function must be uniformly continuous on E. Is the converse true? f (x) | f (y.6.10 Explain how Exercise 5.6.4 can be deduced from Exercise 5.6.6 or from Exercise 5.6.7. See Note 123 5.6.11 Give an example of a function f that is continuous on R and a sequence of compact intervals X1, X2, . . . , Xn, ∞ i=1 Xi. . . . on each of which f is uniformly continuous, but for which f is not uniformly continuous on X = See Note 124 5.6.12 Show that if f is not uniformly continuous on an interval [a, b] then there are sequences 0 but > c for some positive c. from that interval so that xn − yn → f (xn) | − f (yn) | See Note 125 S yn} { chosen xn} { and 5.6.13 " Prove Theorem 5.48 using the Heine-Borel property. See Note 126 5.6.14 Prove the following more general and complete version of Theorem 5.48. Suppose that f : E E. Conversely, if every continuous function f : E compact. → R is continuous. If E is compact, then f must be uniformly continuous on R is uniformly continuous, then E must be → ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 326 Continuous Functions Chapter 5 5.6.15 Prove Theorem 5.49 without using the fact that such a function is uniformly continuous. Use Cousin’s lemma. See Note 127 5.6.16 Prove Theorem 5.49 without using the fact that such a function is uniformly continuous. Use the Bolzano-Weierstrass property. See Note 128 5.6.17 " Prove Theorem 5.49 without using the fact that such a function is uniformly continuous. Use the Heine-Borel property. See Note 129 5.7 Extremal Properties A familiar kind of problem that we study in elementary calculus involves locating extrema of continuous functions defined on an interval [a, b]. The technique entails checking values of the function at points where its derivative is zero, at the endpoints of the interval, and at any points of nondifferentiability. For such a process to work, we must be sure the function has a maximum (or minimum) on the interval. We verify this now. Theorem 5.50: Let f be continuous on [a, b]. Then f possesses both an absolute maximum and an absolute minimum. x . By Theorem 5.48, f is uniformly continuous on [a, b]. Thus, by f (x) : a . If there exists x0 such that f (x0) = M , then f achieves a maximum value M . Proof. Let M = sup Theorem 5.49, M < [a, b]. We show this is impossible. Suppose, then, that f (x) < M for all x [a, b], f (x) f (x)). For each x [a, b]. From the definition of M we see that Let g(x) = 1/(M = M ; as a consequence, g is continuous and g(x) > 0 for all ∈ inf M { − f (x) : x [a, b] } ∈ = 0, ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 5.7. Extremal Properties so sup 327 = . ∞ 1 f (x) M − : x ∈ [a, b] This means that g is not bounded on [a, b]. This is impossible because, as we saw in Section 5.6, a continuous function defined on a closed interval must be bounded. A similar proof would show that f has an absolute minimum on A. Example 5.51: Does this theorem extend to more general situations? If we replace the interval [a, b] by some other set does the conclusion remain true? The example 1 x shows that the closed interval cannot be replaced by an open one. On the other hand, the example f (x) = (0, 1) for x ∈ shows that the bounded closed interval [a, b] cannot be replaced by an unbounded closed one. f (x) = x for x [0, ) ∞ ∈ ◭ From this example the suggestion that we need a closed and bounded set (i.e., a compact set) seems to offer itself. Indeed that is the correct generalization of Theorem 5.50. Theorem 5.52: Let f be continuous on a closed and bounded set A. Then f possesses an absolute maximum and an absolute minimum on A. Exercises 5.7.1 Give an example of an everywhere discontinuous function that possesses a unique point at which there is an absolute maximum and a unique point at which there is an absolute minimum. 5.7.2 Show that a continuous function maps compact sets to compact sets. See Note 130 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 328 Continuous Functions Chapter 5 5.7.3 Prove Theorem 5.50 using a Bolzano-Weierstrass argument. See Note 131 5.7.4 Give an example of a function defined only on the rationals and continuous at each point in its domain and yet does not have an absolute maximum. 5.7.5 Let f : R → R be a continuous function with the property that lim x→∞ f (x) = lim x→−∞ f (x) = 0. Show that f has either an absolute maximum or an absolute minimum but not necessarily both. See Note 132 5.7.6 Let f : R for all x ∈ R be a continuous function that is periodic in the sense that for some number p, f (x + p) = f (x) → R. Show that f has an absolute maximum and an absolute minimum. 5.8 Darboux Property We have already observed that the IVP (Darboux property) is not the same as continuity. It is true, however, that if f is continuous on [a, b], then f has the Darboux property. We state Theorem 5.53 in a form that suggests use of Cousin’s lemma. (Readers that prefer to use the Bolzano-Weierstrass theorem should see the hint for Exercise 5.8.3.) Expressed this way the theorem asserts that if the graph has no point on some horizontal line y = c, then the graph must be entirely above or below that line. Another way to say this (see Exercise 5.8.8) is that the function must assume every value between any two of its values. Theorem 5.53: Let f be continuous on [a, b] and let c f (x) > c for all x [a, b] or f (x) < c for all x [a, b]. ∈ ∈ R. If for every x ∈ ∈ [a, b], f (x) = c, then either Again, as in the proof of Theorem 5.48, we will use a compactness argument. We shall use Proof. Cousin’s lemma (Lemma 4.26). In the exercises you are asked to prove this same theorem using the Bolzano-Weierstrass property and the Heine-Borel property. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 5.8. Darboux Property 329 J or f (x) > c for all ∈ denote the collection of closed intervals J such that f (x) < c for all x forms a Cousin cover of [a, b]. ∈ [a, b]. Thus, if f (x) < c, then f (t) < c for all t − | | c = ε > 0, so there exists δ > 0 such that x C Let J. We verify that ∈ If x [a, b], then C f (x) and t ∈ for all t such that for i = 1, . . . , n, [xi [x − ∈ [x ∈ − f (t) f (x) < δ δ/2, x + δ/2], while if f (x) > c, |
then f (t) > c < xn = b < ε whenever − − /2, x + δ/2]. By Cousin’s lemma there exists a partition of [a, b], a = x0 < x1 < Suppose now that f (a) < c. The argument is similar if f (a) > c. Since [a, x1] = [x0, x1] for all x this way, we see that f (x) < c for all x [x0, x1]. Analogously, since [x1, x2] [a, b]. ∈ , and f (x1) < c, f (x) < c for x ∈ ∈ C , f (x) < c ∈ C [x1, x2]. Proceeding in 1, xi] − . ∈ C ∈ You may wish to look at Exercise 5.8.8 for other wordings of this theorem that suggest IVP as “connectedness.” Exercises 5.8.1 Show that a nondecreasing function with the Darboux property must be continuous. 5.8.2 Show that a continuous function maps compact intervals to compact intervals. See Note 133 5.8.3 Prove Theorem 5.53 using the Bolzano-Weierstrass property of sequences rather than Cousin’s lemma. See Note 134 5.8.4 " Prove Theorem 5.53 using the Heine-Borel property. See Note 135 5.8.5 Prove Theorem 5.53 using the following “last point” argument: suppose that f (a) < c < f (b) and let z be the last point in [a, b] where f (z) ≤ Show that f (z) = c. See Note 136 c, that is, let z = sup x { ∈ [a, b] : f (x) c . } ≤ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 330 5.8.6 A function f : [a, b] function f mapping [a, b] into itself has at least one fixed point. → ∈ [a, b] is said to have a fixed point c Continuous Functions Chapter 5 [a, b] if f (c) = c. Show that every continuous See Note 137 5.8.7 Let f : [a, b] → Show that if the sequence zn} 5.8.8 Show that Theorem 5.53 can be reworded in the following ways: { [a, b] be continuous. Define a sequence recursively by z1 = x1, zn = f (zn−1) where x1 ∈ is convergent, then it must converge to a fixed point of f . [a, b]. (a) Let f be defined and continuous on an interval I, let a, b I with f (a) = f (b). Let d lie between f (a) and f (b). Then there exists c between a and b such that f (c) = d. ∈ (b) A continuous function defined on an interval I maps subintervals of I onto either single points or else subintervals of R. [Singleton points are often considered to be (degenerate) intervals.] See Note 138 5.8.9 A continuous function maps compact intervals to compact intervals. Is it true that continuous functions map closed sets to closed sets? Is it true that continuous functions map open sets to open sets? 5.8.10 State forms of Theorem 5.53 and its rewordings in Exercise 5.8.8 for continuous functions defined on intervals that need not be closed and/or bounded. 5.9 Points of Discontinuity In our discussion of continuous functions we have mentioned discontinuities only as a contrast to the notion of continuity. In many applications of mathematics the functions that arise will have discontinuities and it is well to study such functions. We first ask for a language of discontinuity points. Then we investigate an important class of functions, the monotonic functions, and determine just how badly discontinuous they could be. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 5.9. Points of Discontinuity 5.9.1 Types of Discontinuity 331 Let x0 be a point of the domain of some function f . If x0 is a point of discontinuity, then this means that either the limit limx x0 f (x) fails to exist or else that limit does exist but → f (x0) = lim x0 x → f (x). Note that when we discuss discontinuity points we are discussing only points at which the function is defined. (Some calculus texts might call x0 a point of discontinuity even if f (x0) fails to be defined. This is not our usage here.) Note, too, that a discontinuity point cannot occur at an isolated point of the domain of the function. Removable Discontinuities We can separate these cases into situations of increasing severity. The weakest possibility is that limx discontinuity of f . The word “removable” suggests that were we merely to assign a new value to f (x0) we would no longer have a discontinuity. x0 f (x) does indeed exist but fails to equal f (x0). We call this a removable → Jump Discontinuities A little more serious case of discontinuity occurs if the limit does not exist, but it fails to exist only because f (x) lim x0 x → f (x) and lim → x0+ x f (x), lim x0 → x − the two one-sided limits, exist but disagree. In that case, no matter what value f (x0) assumes, this is a point of discontinuity. We call this a jump discontinuity of f . The difference between the two limits is a measure of the “size” of the discontinuity and is called the jump. f (x) lim → x0+ x − x lim x0 → − f (x) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 332 Continuous Functions Chapter 5 Essential Discontinuities Finally, the most intractable kind of discontinuity would be the situation in which x0 f (x) does not exist, and at least one of the two right-hand and left-hand limits (perhaps both) limx → f (x) and lim → x0+ x f (x) lim x0 → x − also does not exist. Again, no matter what value f (x0) assumes, this is a point of discontinuity. We call this an essential discontinuity of f . Example 5.54: Let f (x) = 0 for all x = 0 and let f (0) = 2. It is clear that 0 is a removable discontinuity of f . Perhaps this example seems entirely artificial. A more natural example would be the function given by the following formula: f (x) = x + 1 x2 1 − (x = ± 1), f (1) = c1, f ( 1) = c2. − This function is clearly continuous at every point other than x = at try to determine which one is removable and which one is essential. 1 but may have two discontinuities, one 1 and one at 1. One of these is not, however, a serious discontinuity since it is removable. You should ◭ ± − Example 5.55: Let f (x) be defined as the linear function x + 1 for x < 0 and a different linear function 2x 0. Then there is a discontinuity at 0 since 1 for x − ≥ lim 0+ x → f (x) = lim 0+ x → (2x 1) = 1 − − lim 0 x − → f (x) = lim 0 − → x (x + 1) = 1. 2. A picture would show exactly what this jump represents. ◭ but In this case the size of the jump is − Exercises 5.9.1 Show that a function that has the Darboux property cannot have either removable or jump discontinuities. 5.9.2 What kind of discontinuities does the Dirichlet function (see Section 5.2.6) have? ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 Section 5.9. Points of Discontinuity 333 5.9.3 What kind of discontinuities does the characteristic function of the Cantor set (see Section 5.2.6) have? 5.9.4 Let the function f : R R have just one point of discontinuity and assume only rational values. What kind of discontinuity point must that be? → 5.9.5 Classify the discontinuities of the rational function 5.9.6 Give an example of a function continuous at 0 but with an essential discontinuity at each other point. f (x) = x + 1 x2 1 − (x = ± 1), f (1) = c1, f ( − 1) = c2. 5.9.7 Give an example of a function f with a jump discontinuity and yet (f )2 is continuous everywhere. 5.9.8 Give an example of a function f with an essential discontinuity everywhere and yet (f )2 is continuous everywhere. 5.9.9 Define a function F by the formula xn 1 + xn . What is the domain of this function? Classify all discontinuities. F (x) = lim n→∞ 5.9.2 Monotonic Functions In general, there is not too much to say about the continuity of an arbitrary function. It is possible for a function to be discontinuous everywhere. But if the function is monotonic this is not possible. We start with some definitions, needed here and again later in many places. Definition 5.56: (Nondecreasing) Let f be real valued on an interval I. If f (x1) and x2 are points in I with x1 < x2, we say f is nondecreasing on I. ≤ f (x2) whenever x1 Definition 5.57: (Increasing) Let f be real valued on an interval I. If the strict inequality f (x1) < f (x2) holds whenever x1 and x2 are points in I with x1 < x2, we say f is increasing on I. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 334 Continuous Functions Chapter 5 In the opposite direction we define nonincreasing and decreasing.1 Definition 5.58: (Nonincreasing) Let f be real valued on an interval I. If f (x1) and x2 are points in I with x1 < x2, we say f is nonincreasing on I. ≥ f (x2) whenever x1 Definition 5.59: (Decreasing) Let f be real valued on an interval I. If the strict inequality f (x1) > f (x2) holds whenever x1 and x2 are points in I with x1 < x2, we say f is decreasing on I. A function that is either nonincreasing or nondecreasing is said to be monotonic. Sometimes, to emphasize that there is a strict inequality, we say that a function that is increasing or decreasing is strictly monotonic. The class of monotonic functions has a particularly interesting structure as regards continuity. Such functions can never have essential discontinuities. This is because if f is monotonic nondecreasing or monotonic nonincreasing, then at any point both one-sided limits limx x0+ f (x) and limx f (x) exist. x0 → → − → x0 f (x) and limx x0+ f (x) both exist. Theorem 5.60: Let f be monotonic on an interval I. If x0 is interior to I, then the one-sided limits limx − → Suppose f is nondecreasing on I; the proof for the case that f is nonincreasing will then follow Proof. f is nondecreasing. To prove Theorem 5.60 let x0 be interior to I and let by noting that in this case is xk} { a nondecreasing sequence of numbers bounded from above by f (x0). Thus by the monotone convergence principle be an increasing sequence of points in I such that limk xk = x0. Then the sequence approaches a limit L. f (xk) →∞ − } { f (xk) { } For xk < x < x0, Let ε > 0. Since f (xk) L, there exists N ∈ → f (xk) f (x) L. ≤ IN such that ≤ 1Some authors prefer the terms “increasing” and “strictly increasing” for what we would call nondecreasing and increasing. See the comments on page 435. L − f (xk) < ε ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.9. Points of Discon |
tinuity whenever k ≥ N . For all x satisfying xN ≤ L − It follows that x ≤ f (x) x0 we thus have L − ≤ f (xN ) < ε. lim x0 − → so f has a left-sided limit at x0. A similar argument shows that f also has a right-sided limit at x0. f (x) = L, x 335 Monotonic Functions Have Jump Discontinuities Recall that a function f is said to have a jump at x0 if f has limits from the left and from the right at x0, but these limits are different. Thus, if f is monotonic nondecreasing, say, then clearly x Thus the only possibility of a discontinuity at the point x0 is if the jump x0+ − x ≤ f (x) f (x0) ≤ lim → f (x). lim x0 → J(x0) = lim x0+ x → f (x) − x lim x0 → − f (x) is positive. Thus monotonic functions do not have removable discontinuities nor do they have essential discontinuities. They have only jump discontinuities. Monotonic Functions Have Countably Many Discontinuities We can go further than this. We can ask about the set of points at which there can be a discontinuity point. We ask how large this set can be. The answer is “not very.” Theorem 5.61: Let f be monotonic on an interval [a, b]. Then the set of points of discontinuity of f in that interval is countable. In particular, f must be continuous at the points of a set dense in [a, b]. Proof. We consider again the case that f is nondecreasing since the case that f is nonincreasing follows f . If f is nondecreasing and discontinuous at a point x0 in (a, b), then the by considering the function open interval − I(x0) = lim x0 → x − f (x), lim → x0+ x f (x) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 336 Continuous Functions Chapter 5 either contains no points in the range of f or contains only the single point f (x0) in the range. (To check this statement, see Exercise 5.9.12.) Thus, each point of discontinuity x0 of f in I corresponds to an interval I(x0). For two different points of discontinuity x1 and x2, the intervals I(x1) and I(x2) are disjoint (because f is nondecreasing). But any collection of disjoint intervals in R can be arranged into a sequence (Exercise 4.6.10) and so there can be only countably many points of discontinuity of f . It is easy to construct monotonic functions with infinitely many points of discontinuity. For example, if f (x) = n on [n, n + 1), then f has jumps at all the integers. It is natural to ask which countable sets can be the set of discontinuities for some monotonic f . For example, does there exist an increasing function that is discontinuous at every rational number in R? (Exercise 5.9.14 provides an answer.) Example 5.62: Our theorem shows that a monotonic function has a countable set of points at most where it can be discontinuous. It is easy to find examples of monotonic functions with a prescribed set of discontinuities if the set given to us is finite. Could any countable set be given and we then find a monotonic function that has exactly that set as its points of discontinuity? The answer, remarkably, is yes. Let C be a countable subset of (a, b). List the elements as c1, c2, c3, . . . . Define the function for a x ≤ ≤ b as f (x) = 1 2n . cn<x X This function is hard to visualize since it depends on the order of the terms. Clearly, f (a) = 0 and f (b) = 1. The other values are much less clear. But we can see that there is a jump of magnitude 1/2 at the point c1, a jump of magnitude 1/4 at the point c2, a jump of magnitude 1/8 at the point c3, and so on. The function is strictly increasing on any subinterval in which C is dense and would be constant in any interval ◭ that contains no points of C. It can be shown that the only discontinuities occur at the points of C. Exercises 5.9.10 Construct a function with a jump discontinuity of magnitude everywhere else. 5 at the point x = 1 and continuous − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.9. Points of Discontinuity 5.9.11 Find a monotonic function on [0, 1] with discontinuities at 1/3, 2/3, and 3/4 only. 5.9.12 Suppose f is increasing on an interval I. Let x0 be an interior point of I. Prove that limx→x0− f (x) f (x0) ≤ limx→x0+ f (x). 5.9.13 Verify the claims made in Example 5.62 about the function f there. See Note 139 337 ≤ 5.9.14 Using Example 5.62, show that there is a (strictly) increasing function on [0, 1] that is discontinuous at each rational number in (0, 1) and continuous at each irrational number. 5.9.15 Show that there is no monotonic function on [0, 1] that is discontinuous precisely at each irrational number in (0, 1). See Note 140 5.9.16 Show that if f : [a, b] continuous and increasing on the interval on which it is defined. See Note 141 → R is continuous and increasing, then the inverse function f −1 exists and is also 5.9.17 Let f be a continuous function on an open interval (a, b). Suppose that f has no local maximum or local minimum at any point. Show that f must be monotonic. 5.9.18 Suppose that f : R R and that f (x) + αx is monotonic for every α R. Show that f (x) = ax + b for some ∈ → be a sequence of monotonic functions defined on the interval [0, 1]. Suppose that f (x) = lim n→∞ fn(x) exists for each 0 problem by “continuous,” the exercise would be invalid: show this, too.) 1. Show that f is monotonic. (If the word “monotonic” is replaced throughout this ≤ ≤ x 5.9.20 Can the range of an increasing function on the interval [0, 1] consist only of rational numbers? Can it consist only of irrational numbers? a, b. 5.9.19 Let fn} { ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 338 Continuous Functions Chapter 5 5.9.3 How Many Points of Discontinuity? " Advanced section. May be omitted. We have already answered the question as to how many points of discontinuity a monotonic function may have. The set of such points must be countable. We know too that all of these are jump discontinuities; a monotonic function has no removable discontinuities and no essential discontinuities. What is the situation for an arbitrary function? There are three questions. How many removable discontinuities are possible? How many jump discontinuities are possible? How many essential discontinuities are possible? Example 5.63: One example that we have seen before shows that there can be a great many essential discontinuities. Let f be the characteristic function of the rational numbers; that is, f (x) is 1 if x is a rational number and is 0 if x is irrational. Clearly, and lim sup x0 x → f (x) = 1 f (x) = 0 lim inf x0 x → at every point x0. In particular, the limit does not exist anywhere and so every point is an essential discontinuity. ◭ Surprisingly, though, this is not the case for the removable discontinuities or the jump discontinuities. No function can have an uncountable number of such discontinuities. Theorem 5.64: Let f be a real function defined on an interval [a, b]. The sets of points in [a, b] at which f has a removable discontinuity and at which f has a jump discontinuity are both countable. Proof. Let J be the set of points at which there is a jump discontinuity. Every point of J is in one of the two sets: J+ = x { ∈ (a, b) : lim x+ → y f (y) > lim x − → y f (y) } ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.9. Points of Discontinuity 339 or − We shall show that J+ is countable. J = x { ∈ (a, b) : lim x+ → y f (y) < lim x − → y f (y) . } If x ∈ J+, then y lim x+ → f (y) > lim x − → y f (y) and so there is for any such x at least one rational number r so that Moreover, there then must exist some integer m (depending on x and r) so that lim x+ → y f (y) > r > lim x − → y f (y). 1//m. f (z) > r > f (y) whenever x − Let Jrn, where r is a rational and n a positive integer, denote the set of all points x with the property that f (y) < r < f (z) whenever x 1//n. − We claim that this set is countable. If not, then it must have a point of accumulation and, in particular, there would have to be at least three points a < b < c, with c a < 1/n, all belonging to Jrn. But by the way that Jrn was defined this means, since a and c Jrn, that f (b) < r and r < f (b) are both true. Since this is impossible, all points in Jrn are isolated and hence Jrn is countable. The union − ∈ n=1 [ is a countable union of countable sets and is thus also countable. But this set contains every point of J+ and so that set is also countable. Similarly, it is true that J is countable and hence the set of points with jump discontinuities is countable. Q [r ∈ − ∞ Jrn ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 340 Continuous Functions Chapter 5 That the set of points at which the function has a removable discontinuity is also countable is left as an exercise. The ideas of the proof here can be used to prove it in a similar fashion. Notice especially this technique of inserting a rational number between two unequal numbers. Incidentally, this theorem throws a new light on the theorem about the discontinuity points of In that proof we used the properties of monotonic functions to show that the monotonic functions. collection of discontinuity points was countable. But we know easily that the only such points are the jump discontinuities and any function, monotonic or not, has only countably many of these points by our theorem here. Thus we have another way of looking at Theorem 5.61. Exercises 5.9.21 Give an example of a function with a dense set of removable discontinuities. 5.9.22 Give an example of a function with a dense set of jump discontinuities. 5.9.23 Prove the remaining statement of Theorem 5.64 that is not proved in the text. 5.10 Challenging Problems for Chapter 5 5.10.1 Suppose that f is a function defined on the real line with the property that f (x + y) = f (x) + f (y) for all x, y. Suppose that f is continuous at 0. Show that f must be continuous everywhere. See Note 142 5.10.2 Suppose that f is a function defined on the real li |
ne with the property that f (x + y) = f (x) + f (y) for all x, y. Suppose that f is continuous at 0. Show that f (x) = Cx for all x and some number C. See Note 143 5.10.3 Suppose that f is a function defined on the real line with the property that f (x + y) = f (x)f (y) for all x, y. Suppose that f is continuous at 0. Show that f must be continuous everywhere. See Note 144 5.10.4 Generalize Theorem 5.61 to prove that if a function f (not necessarily monotonic) has left-sided limits and right-sided limits at every point of an open interval I, then f must be continuous except on a countable set. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 5.10. Challenging Problems for Chapter 5 341 5.10.5 Determine necessary and sufficient conditions on a pair of sets A and B so that they will have the property that there exists a continuous function f : R x B. See Note 145 ∈ → R such that f (x) = 0 for all x A and f (x) = 1 for all ∈ 5.10.6 Let f : [1, ∞ ) be continuous, positive and increasing with f (x) → ∞ as x → ∞ . Show that is convergent if and only if the series Xk=1 converges (where f −1 denotes the inverse function). 1 f (k) ∞ Xk=1 ∞ f −1(k) k2 5.10.7 (Extensions of continuous functions) If f : A R, g : B R, A B, and f (x) = g(x) for all x then the function g is said to be an extension of the function f . Prove each of the following: → → ⊂ A, ∈ (a) A function that is continuous on a closed set A can be extended to a function that is continuous on R. (b) A function that is uniformly continuous on a set A can be extended to a function that is uniformly continuous on A. (c) A function that is uniformly continuous on an arbitrary nonempty subset of R can be extended to a function that is uniformly continuous on all of R. (d) Give an example of a function f that is continuous on (0,1) but that cannot be extended to a function continuous on [0,1]. 5.10.8 " For an arbitrary function f : R is countable. → R show that x0 : lim sup x→x0− { f (x) > lim sup x→x0+ f (x) } 5.10.9 " Give an example of a function f : R → f (x0) > lim sup x→x0 R such that there are infinitely many points x0 at which either f (x). f (x) or f (x0) < lim inf x→x0 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 342 NOTES 5.10.10 " For an arbitrary function f : R → is countable. R show that the set of points x0 at which f (x0) does not lie between lim inf x→x0 f (x) f (x) and lim sup x→x0 5.10.11 " Let y be a real number or of numbers in E and converging to a point c with xn 6 y then y is called a cluster value of f at = c and with f (xn) c. Show that every cluster value at c lies between lim inf x→c f (x) and lim supx→c f (x). Show that both lim inf x→c f (x) and lim supx→c f (x) are themselves cluster values of f at c. R be a function. If there is a sequence and let f : E xn} ±∞ → → { 5.10.12 Is there a continuous function f : R equation f (x) = y? 5.10.13 Is there a continuous function f : R R such that for every real y there are precisely two solutions to the R such that for every real y there are precisely three solutions to the → → R, then the set x : f is right continuous at x but not left continuous at x } { equation f (x) = y? 5.10.14 Prove that if f : R → is countable. See Note 146 Notes 89Exercise 5.1.1. Model your answer after Example 5.2. 90Exercise 5.1.2. Consider the cases a = 0 and a = 0 separately. If it is easier for you, break into the three cases a > 0, a < 0, and a = 0. 91Exercise 5.1.3. Model your answer after Example 5.3. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 NOTES 343 92Exercise 5.1.4. Consider the cases x0 = 0 and x0 6 device of restricting x to be close to x0 by assuming that x | 93Exercise 5.1.8. Don’t forget to exclude x0 < 0 from your answer since it is not a point of accumulation of the < 1 at least. x0| − = 0 separately. Use the factoring trick in Example 5.3 and the domain of this function. Consider the cases x0 = 0 and x0 > 0 separately. 94Exercise 5.1.12. If B Can you find other conditions? If x0 is a point of accumulation of A deduced, assuming that both exist. ∩ ⊂ A, then the existence of limx→x0 g(x) can be deduced from the existence of limx→x0 f (x). B, then the equality of the two limits can be 95Exercise 5.1.16. Either find a single sequence with xn 6 = 0 so that the limit xn → 0 lim n→∞ | xn| /xn does not exist or else find two such sequences with different limits. 96Exercise 5.1.22. You could assume (i) that L > 0 or (ii) that f (x) statement about sequences. 97Exercise 5.1.28. At x0 6 98Exercise 5.1.29. On one side the limit is zero and on the other the limit fails to exist. (Look ahead to Exercise 5.1.38, which means that the limit does not exist.) You may use the elementary = 0 the two one-sided limits are equal. What are they? At x0 = 0 they differ. where you are asked to show that the limit is inequality ∞ 0 for all x in its domain. Then convert to a ≥ 0 < z < ez (which is valid for all z > 0) in your argument. Consider the sequences 1/n 0 and 1/n − → 0. → 99Exercise 5.1.30. Check the definition: There would be no distinction. The limit lim x→0− √x, however, would be meaningless since 0 is not a point of accumulation of the domain of the square root function on the left. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 344 NOTES 100Exercise 5.1.34. Use the definitions in this section as a model. You will need a replacement for the “x0 is a point of accumulation” of the domain condition. If you cannot think of anything better, then simply use the assumption that f is defined in some interval (a, ). ∞ 101Exercise 5.1.38. On one side at 0 the limit is zero and on the other the limit is . See Exercise 5.1.29. ∞ 1/n, x0 + 1/n) there would be a point xn for 102Exercise 5.2.1. Model your proof after Theorem 2.8 for sequences. 103Exercise 5.2.3. If the theorem were false, then in every interval (x0 − which > n. f (xn) | | 104Exercise 5.2.9. If x0 is not a point of accumulation of then the statement dom(f ) ∩ dom(g), lim x→x0 f (x) + g(x) = L does not have any meaning even though the two statements about lim x→x0 f (x) and lim x→x0 g(x) may have. 105Exercise 5.2.11. What exactly is the domain of the function f (x)/g(x)? Show that x0 would be a point of accumulation of that domain provided that g(x) C as x = 0. x0 and C → → 106Exercise 5.2.28. It is enough to assume that limx→x0 f (x) exists and to apply Theorem 5.25 with F (x) = sure to explain why this function F has the properties expressed in that theorem. x | . Be | 107Exercise 5.2.29. It is enough to assume that limx→x0 f (x) exists and is positive and then apply Theorem 5.25 with F (x) = √x. Alternatively, assume that f (x) 0 for all x in a neighborhood of x0. Again be sure to explain why this function F has the properties expressed in that theorem. ≥ 108Exercise 5.2.32. Use the property of exponentials that ea+b = eaeb and the product rule for limits. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 NOTES 345 109Exercise 5.2.33. Use a trigonometric identity for sin(x x0 + x0) and the sum and products rule for limits. − 110Exercise 5.2.34. Take the function H(x) of the text and consider instead H(x) + x. 111Exercise 5.2.36. This would be trivial if the sets Ai were disjoint. So it is the case where these are not disjoint that you need to address. If x0 is not in the Cantor set K, then it is in some open interval complementary to that set. x0 and 112Exercise 5.2.44. Use that to prove the existence of the limit. If x0 is in the Cantor set, then there must be sequences xn → yn → 113Exercise 5.3.5. Consider separately the cases x0 ∈ would the lim sup be larger according to this revised definition? K. Use that to prove the nonexistence of the limit. E. Under what circumstances in the latter case x0 with xn ∈ K and yn 6∈ E and x0 6∈ 114Exercise 5.4.15. One of the definitions treats isolated points in a special way. Note that each point in the domain of f is isolated. 115Exercise 5.4.17. You must arrange for f (0) to be the limit of the sequence of values f (2−n). No other condition is necessary. 116Exercise 5.4.19. At an isolated point x0 of the domain the limit limx→x0 f (x) has no meaning. But if x0 is not an isolated point in the domain of f it must be a point of accumulation and then limx→x0 f (x) is defined and it must be equal to f (x0). 117Exercise 5.4.20. For the converse consider the function f (x) = √x on [0, 1]. 118Exercise 5.6.1. Let a = inf K and b = sup K and apply Cousin’s lemma to the interval [a, b] by taking the same collection nearly, namely consist of all closed subintervals [t, s] such that C f (t′) | − f (s′) | < ε/2 for all t′, s′ K ∈ ∩ [t, s]. You will have to find a different choice of δ to make your argument work. 119Exercise 5.6.2. As usual in applications of Cousin’s lemma, we should define first our collection of closed consist subintervals so as to have a desired property that can be extended to the whole interval [a, b]. Let ε > 0. Let C ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 346 NOTES of all closed subintervals [t, s] such that [t, s]. We check that satisfies the hypotheses of Lemma 4.26. C [a, b] there exists δ(x) > 0 such that if f (t′) | − f (s′) | < ε/2 for all t′, s′ ∈ For each x ∈ then t ∈ [a, b] (x ∩ − δ(x), x + δ(x)), f (t) | f (x) | − < ε/4. [a, b] (x ∩ − δ(x), x + δ(x)), f (s′) | ≤ | < f (t′) ε 4 + − ε 4 + f (x) | − f (s′) | f (x) | ε . 2 = [a, b] (x ∩ − δ(x), x + δ(x)) It follows that if t′ and s′ are in the set then f (t′) | − Consequently, every interval [t, s] inside belongs to . C Thus Lemma 4.26 may be applied and there exists a partition such that if, for some i = 1, . . . , n, then Let a = x0 < x1 < < xn = b · · · xi−1 ≤ x, y ≤ xi, f (x) | f (y) | − < ε/2. δ = min i=1,...,n | xi − . xi−1| ClassicalRealAnalysis.comThomson*Bruckner*B |
rucknerElementary Real Analysis, 2nd Edition (2008) NOTES 347 If x < y and x | y | − < δ, then either there exists i for which in which case or there exists i such that in which case xi−1 ≤ x, y ≤ xi, f (x) | − f (y) < ε/2, xi−1 ≤ x ≤ xi ≤ y ≤ xi+1, f (y) | + − f (xi) | f (x) | ≤ | f (x) | − f (y) ε 2 < − + f (xi) | ε = ε. 2 Since this argument applies to any positive ε, we have proved that f is uniformly continuous on [a, b]. 120Exercise 5.6.5. If X is compact (closed and bounded) then this property should hold. If the set X has a point of accumulation that does not belong to X, then it is possible to give an example of a continuous function defined on X that is not uniformly continuous on X. Finally consider the situation in which the set is closed but not bounded: are there points xi and xj arbitrarily close together? 121Exercise 5.6.7. You need consider only two compact sets X1, X2. Since they are compact, there is a positive distance between them that you can use to help define your δ. For not closed consider X1 = (0, 1) and X2 = (1, 2) and define f appropriately. For not bounded use and and define f appropriately. X1 = { 1, 2, 3, . . . } X2 = { 1, 2 + 1/2, 3 + 1/3, 4 + 1/4, . . . 122Exercise 5.6.9. For the converse consider the function f (x) = √x on [0, 1]. By Theorem 5.48 we know that this function is uniformly continuous on [0, 1]. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 348 NOTES 123Exercise 5.6.10. Show that any function defined on a set X containing just one element is uniformly continuous. Then consider the sequence Xi = , i = 1, 2, . . . , n. xi} { 124Exercise 5.6.11. For the sequence of intervals you might choose [1, 2], [2, 3], [3, 4], . . . . (Why would you not be able to choose [1/2, 1], [1/4, 1/2], [1/8, 1/4], . . . ?) 125Exercise 5.6.12. This can be obtained merely by negating the formal statement that f is uniformly continuous on [a, b]. 126Exercise 5.6.13. Using the local continuity property, claim that there are open intervals Ix containing any point x so that for any y subcover. ∈ Ix. Now apply the Heine-Borel property to this open cover. Obtain uniform continuity from the finite f (y) | f (x) | − < ε 127Exercise 5.6.15. Let C lemma to find a partition of [a, b] using intervals in be the collection of all closed intervals I . C [a, b] so that f is bounded on I. Use Cousin’s ⊂ 128Exercise 5.6.16. Use an indirect proof. Show that if f is not bounded then there is a sequence [a, b] so that xn} { of points in f (xn) | | > n for all n. Now apply the Bolzano-Weierstrass property to obtain subsequences and get a contradiction. 129Exercise 5.6.17. Using the local continuity property, claim that there are open intervals Ix containing any point x so that for any y subcover. ∈ Ix. Now apply the Heine-Borel property to this open cover. Obtain boundedness of f from the finite f (y) | f (x) | − < 1 130Exercise 5.7.2. That is, prove that the image set f (K) = f (x) : x { ∈ K } ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 349 is compact if K is compact and f is a continuous function defined at every point of K. Give a direct proof that uses the fact that a set is compact if and only if every sequence in the set has a subsequence convergent to a point in the set. Start with a sequence of points in K with f (xn) = yn etc. in f (K), explain why there must be a sequence xn} yn} { { 131Exercise 5.7.3. Let Explain why you can choose a sequence of points x b . ≤ from [a, b] so that ≤ } f (x) : a M = sup { xn} { f (xn) > M 1/n. − Now apply the Bolzano-Weierstrass theorem and use the continuity of f . 132Exercise 5.7.5. x > N and x < N . − If f (x0) = c > 0, then there is an interval [ N, N ] so that x0 ∈ [ − N, N ] and f (x) | | < c/2 for all − 133Exercise 5.8.2. That is, prove that the image set f ([c, d]) is a compact interval for any interval [c, d] if f is a continuous function defined at every point of [c, d]. Apply Theorem 5.52 and Theorem 5.53. from [a, b] so that f (xn) > c, f (yn) < c and 134Exercise 5.8.3. Suppose that the theorem is false and explain, then, why there should exist sequences yn| xn − yn} | { 135Exercise 5.8.4. Suppose that the theorem is false and explain, then, why there should exist at each point x an open interval Ix centered at x so that either f (t) > c for all t [a, b] or else f (t) < c for all t < 1/n. { and xn} [a, b] ∈ [a, b]. Ix ∩ ∈ Ix ∩ ∈ 136Exercise 5.8.5. You may take c = 0. Show that if f (z) > 0, then there is an interval [z δ, z] on which f is positive. Show that if f (z) < 0, then there is an interval [z, z + δ] on which f is negative. Explain why each of these two cases is impossible. − 137Exercise 5.8.6. The function must be onto. Hence there is a point x1 with f (x1) = a and a point x2 with f (x2) = b. Now convince yourself that there is a point on the graph of the function that is also on the line y = x. 138Exercise 5.8.8. Condition (a) is the intermediate value property (IVP) according to Definition 5.27, while (b) can be interpreted as saying that connectedness is preserved by continuous functions. This latter interpretation requires a careful definition of connectedness in R. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 350 NOTES 139Exercise 5.9.13. You wish to show that (i) f is discontinuous at every point in C, indeed has a jump discontinuity at each such point; (ii) f is continuous at every point not in C; (iii) f is nondecreasing; (iv) f is increasing on any interval in which C is dense; and (v) f is constant on any interval containing no point of C. The most direct and easiest proof that f is continuous at every point not in C would be to use “uniform conver- gence” but that is in a later chapter. Here you will have to use an ε-δ argument. 140Exercise 5.9.15. How large can the set of discontinuity points be? 141Exercise 5.9.16. The function f −1 is defined on the interval J = [f (a), f (b)]. Explain first why it exists (not all functions must have an inverse). Prove that it is increasing. Prove that it is continuous (using the fact that it is increasing). 142Exercise 5.10.1. The equation f (x + y) = f (x) + f (y) is called a functional equation. You are told about this function only that it satisfies such a relationship and has a nice property at one point. Now you must show that this implies more. Show first that f (0) = 0 and that f (x y) = f (x) f (y). − − 143Exercise 5.10.2. This continues Exercise 5.10.1. Show first that f (r) = rf (1) for all r = m/n rational. Then make use of the continuity of f that you had already established in the other exercise. 144Exercise 5.10.3. Show that either f is always zero or else f (0) = 1. Establish 145Exercise 5.10.5. Consider the intersection f (x − y) = f (x)/f (y). B. A ∩ 146Exercise 5.10.14. You will need to use the fact that x : lim sup − x→x 0 f (x) > lim sup x→x+ 0 { f (x) } is countable. See Exercise 5.10.8. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Chapter 6 MORE ON CONTINUOUS FUNCTIONS AND SETS " This chapter can be considered enrichment material containing also several more advanced topics and may be skipped in its entirety. You can proceed directly to the study of derivatives and integrals with no loss in the continuity of the material. 6.1 Introduction In this chapter we go much more deeply into the analysis of continuous functions. For this we need some new set theoretic ideas and methods. 6.2 Dense Sets [This section reviews material from Section 1.9.] Consider the set Q of rational numbers and let (a, b) be an open interval in R. How do we show that there is a member of Q in the interval (a, b); that is, that (a, b) Q = ? ∅ ∩ 351 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 352 More on Continuous Functions and Sets Chapter 6 Suppose first that 0 < a. Since b is a positive integer q such that − a > 0, the archimedean property (Theorem 1.11) implies that there Thus q(b − a) > 1. qb > 1 + qa. The archimedean property also implies that the set of integers m { IN : m > qa } ∈ is nonempty. Thus, according to the well-ordering principle, there is a smallest integer p in this set and for this p, it is true that p qa < p. It follows that 1 − ≤ which implies a < p r = p/q in the interval (a, b). q < b. We have shown that, under the assumption a > 0, there exists a rational number qa < p ≤ 1 + qa < qb, r = 0. If a < b < 0, then 0 < r − − The same is true under the assumption a < 0. To see this observe first that if a < 0 < b, we can take a, so the argument of the previous paragraph shows that there exists b < Q such that The preceding discussion proves that every open interval contains a rational number. We often express a. In this case a < b < r < r < b. − − − ∈ this fact by saying that the set of rational numbers is a dense set. Definition 6.1: A set of real numbers A is said to be dense (in R) if for each open interval (a, b) the set A (a, b) is nonempty. ∩ It is important to have a more general concept, that of a set A being dense in a set B. Definition 6.2: Let A and B be subsets of R. If every open interval that intersects B also intersects A, we say that A is dense in B. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.2. Dense Sets 353 Thus Definition 6.1 states the special case of Definition 6.2 that occurs when B = R. We should note B in their version of Definition 6.2. We find it more convenient not to that some authors require that A impose this restriction. Thus, for example, in our language Q is dense in R ⊂ Q. B (Exercise 6.2.1). \ It is easy to verify that A is dense in B if and only if A ⊃ Exercises 6.2.1 Verify that A is dense in B if and only if A B. ⊃ 6.2.2 Prove that every set A is dense in its closure A. 6.2.3 Prove that if A is dense in B and C B, then A is dense in C. ⊂ 6.2.4 Prove that if A B a |
nd A is dense in B, then A = B. Is the statement correct without the assumption that ⊂ B? A ⊂ 6.2.5 Is R Q dense in Q? \ 6.2.6 The following are several pairs (A, B) of sets. In each case determine whether A is dense in B. (a) A = IN, B = IN (b) A = IN, B = Z (c) A = INd) A = 2n , m Z, n ∈ ∈ IN , B = Q 6.2.7 Let A and B be subsets of R. Prove that A is dense in B if and only if for every b an} { of points from A such that limn→∞ an = b. 6.2.8 Let B be the set of all irrational numbers. Prove that the set B there exists a sequence ∈ is a countable subset of B that is dense in B. A = q + √2 : q { Q } ∈ 6.2.9 Let f : R true that → R be a strictly increasing continuous function. Does f map dense sets to dense sets; that is, is it f (E) = f (x) : x { E } ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 354 More on Continuous Functions and Sets Chapter 6 is dense if E is dense? See Note 147 6.2.10 Prove that every set B ⊂ 6.3 Nowhere Dense Sets R contains a countable set A that is dense in B. We might view a set A that is dense in R as being somehow large: Inside every interval, no matter how small, we find points of A. There is an opposite extreme to this situation: A set is said to be nowhere dense, and hence is in some sense small, if it is not dense in any interval at all. The precise definition of this important concept of smallness follows. Definition 6.3: The set A an open subinterval J such that A ⊂ J = . ∅ ∩ R is said to be nowhere dense in R provided every open interval I contains We can state this another way: A is nowhere dense provided A contains no open intervals. (See Exercise 6.3.4.) Example 6.4: It is easy to construct examples of nowhere dense sets. 1. Any finite set 2. IN 1/n : n 3. { IN } ∈ Each of these sets is nowhere dense, as you can verify. ◭ Each of the sets in Example 6.4 is countable and hence also small in the sense of cardinality. It is hard to imagine an uncountable set that is nowhere dense but, as we shall see in Section 6.5, such sets do exist. We establish a simple result showing that any finite union of nowhere dense sets is again nowhere dense. It is not true that a countable union of nowhere dense sets is again nowhere dense. Indeed countable unions of nowhere dense sets will be important in our subsequent study. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.3. Nowhere Dense Sets 355 Theorem 6.5: Let A1, A2, . . . , An be nowhere dense in R. Then A1 ∪ · · · ∪ Proof. i = 1, 2, . . . , n. Let I be any open interval in R. We seek an open interval J Since A1 is nowhere dense, there exists an open interval I1 ⊂ nowhere dense in R, so there exists an open interval I2 ⊂ we obtain open intervals ⊂ I such that I1 ∩ I2 = ∅ I1 such that A2 ∩ I such that J Ai = for ∅ ∩ A1 = . Now A2 is also . Proceeding in this way ∅ An is also nowhere dense in R. such that for i = 1, ..., n, Ai ∩ for i = 1, . . . , n. Thus Ii = ∅ I1 ⊃ I2 ⊃ I3 ⊃ · · · ⊃ In . It follows from the fact that In ⊂ Ii for i = 1, . . . , n that Ai ∩ In = ∅ n [i=1 Ai ! ∩ In = n [i=1 (Ai ∩ In) = n [i=1 = , ∅ ∅ as was to be proved. Exercises 6.3.1 Give an example of a sequence of nowhere dense sets whose union is not nowhere dense. See Note 148 6.3.2 Which of the following statements are true? (a) Every subset of a nowhere dense set is nowhere dense. (b) If A is nowhere dense, then so too is c) If A is nowhere dense, then so too is cA = (d) If A is nowhere dense, then so too is A′, the set of derived points of A. (e) A nowhere dense set can have no interior points. A ∈ { } { ct : t A for every number c. } ∈ for every positive number c. (f) A set that has no interior points must be nowhere dense. (g) Every point in a nowhere dense set must be isolated. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 356 More on Continuous Functions and Sets Chapter 6 (h) If every point in a set is isolated, then that set must be nowhere dense. See Note 149 6.3.3 If A is nowhere dense, what can you say about R A? If A is dense, what can you say about R A? \ \ 6.3.4 Prove that a set A is empty. ⊂ R is nowhere dense if and only if A contains no intervals; equivalently, the interior of A 6.3.5 What should the statement “A is nowhere dense in the interval I” mean? Give an example of a set that is nowhere dense in [0, 1] but is not nowhere dense in R. 6.3.6 Let A and B be subsets of R. What should the statement “A is nowhere dense in the B” mean? Is IN nowhere dense in [0, 10]? Is IN nowhere dense in Z? Is nowhere dense in IN? 4 { } 6.3.7 Prove that the complement of a dense open subset of R is nowhere dense in R. 6.3.8 Let f : R R be a strictly increasing continuous function. Show that f maps nowhere dense sets to nowhere dense sets; that is, → is nowhere dense if E is nowhere dense. f (E) = f (x) : x { E } ∈ 6.4 The Baire Category Theorem " Advanced section. May be omitted. In this section we shall establish the Baire category theorem, which gives a sense in which nowhere dense sets can be viewed as “small:” A union of a sequence of nowhere dense sets cannot fill up an interval. If we interpret Cantor’s theorem (Theorem 2.4) as asserting that a union of a sequence of finite sets cannot fill up an interval, then we see the Baire category theorem as a far-reaching generalization. We motivate this important theorem by way of a game idea that is due to Stefan Banach (1892–1945) and Stanislaw Mazur (1905–1981). Although the origins of the theorem are due to Ren´e Baire, after whom the theorem is named, the game approach helps us see why the Baire category theorem might be true. This Banach-Mazur game is just one of many mathematical games that are used throughout mathematics to develop interesting concepts. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.4. The Baire Category Theorem 6.4.1 A Two-Player Game " Advanced section. May be omitted. 357 A. Player I1. The players \ We introduce the Baire category theorem via a game between two players (A) and (B). Player (A) is given a subset A of R, and player (B) is given the complementary set B = R (A) first selects a closed interval I1 ⊂ alternate moves, a move consisting of selecting a closed interval inside the previously chosen interval. R; then player (B) chooses a closed interval I2 ⊂ The play of the game thus determines a descending sequence of closed intervals where player (A) chooses those with odd index and player (B) those with even index. If I1 ⊃ I2 ⊃ I3 ⊃ · · · ⊃ In ⊃ . . . then player (A) wins; otherwise player (B) wins. A ∩ ∞ n=1 \ = In 6 , ∅ The goal of player (A) is evidently to make sure that the intersection contains a point of A; the goal of player (B) is to ensure that the intersection is empty or contains only points of B. We expect that player (A) should win if his set A is large while player (B) should win if his set is large. It is not, however, immediately clear what “large” might mean for this game. Example 6.6: If the set A given to player (A) contains an open interval J, then (A) should choose any J. No matter how the game continues, player (A) wins. Another way to say this: If the set interval I1 ⊂ ◭ given to player (B) is not dense, he loses. Example 6.7: For a more interesting example, let player (A) be dealt the “large” set of all irrational numbers, so that player (B) is dealt the rationals. (Both players have been dealt dense sets now.) Let A consist of the irrational numbers. Player (A) can win by following the strategy we now describe. Let q1, q2, q3, . . . be a listing of all of the rational numbers; that is, Q = { q1, q2, q3, . . . . } ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 358 More on Continuous Functions and Sets Chapter 6 Player (A) chooses the first interval I1 as any closed interval such that q1 / ∈ I1. Inductively, suppose I1, I2, . . . , I2n have been chosen according to the rules of the game so that it is now time for player (A) to choose I2n+1. The set is finite, so there exists a closed interval I2n+1 ⊂ q1, q2, . . . , qn} I2n such that { is empty. Player (A) chooses such an interval. I2n+1 ∩ { q1, q2, . . . , qn} Since for each n IN, qn / ∈ ∈ sequence of closed intervals, I2n+1, the set = ∞n=1 In 6 ∅ . Thus A T ∩ ∞n=1 In 6 ∅ ∞n=1 In contains no rational numbers, but, as a descending ◭ , and (A) wins. = T In these two examples, using informal language, we can say that player (A) has a strategy to win: No T matter how player (B) proceeds, player (A) can “answer” each move and win the game. In both examples player (A) had a clear advantage: The set A was larger than the set B. But in what sense is it larger? It is not the fact that A is uncountable while B is countable that matters here. It is something else: The fact that given an interval I2n, player (A) can choose I2n+1 inside I2n in such a way that I2n+1 misses the set . Let us try to see in the second example a general strategy that should work for player (A) in some cases. . Suppose instead that B is the union of a sequence of IN, there The set B was the union of the singleton sets qn} { “small” sets Qn. Then the same “strategy” will prevail if given any interval J and given any n exists an interval I J such that ∈ q1, q2, . . . , qn} { I (Q1 ∪ Q2 ∪ · · · ∪ ∩ . Qn) = ∞n=1 Qn. Thus, if B = ∅ ∞n=1 In will be nonempty, and will miss the set The set ∞n=1 Qn, player (A) has a winning strategy. It is in this sense that the set B is “small.” The set A is “large” because the set B is “small”. If we look carefully at the requirement on the sets Qk, we see it is just that each of these sets is nowhere dense in R. S T S Thus the key to player (A) winning rests on the concept of a nowhere dense set. But note that it rests on the set B being the union of a sequence of nowhere dense sets. ⊂ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Editi |
on (2008) Section 6.4. The Baire Category Theorem 359 6.4.2 The Baire Category Theorem " Advanced section. May be omitted. We can formulate our result from our discussion of the game in several ways: 1. R cannot be expressed as a countable union of nowhere dense sets. 2. The complement of a countable union of nowhere dense sets is dense. The second of these provides a sense in which countable unions of nowhere dense sets are “small:” No matter which countable collection of nowhere dense sets we choose, their union leaves a dense set uncovered. To formulate the Baire category theorem we need some definitions. This is the original language of Baire and it has survived; he simply places sets in two types or categories. Into the first category he places the sets that are to be considered small and into the second category he puts the remaining (not small) sets. Definition 6.8: Let A be a set of real numbers. 1. A is said to be of the first category if it can be expressed as a countable union of nowhere dense sets. 2. A is said to be of the second category if it is not of the first category. 3. A is said to be residual in R if the complement R A is of the first category. \ The following properties of first category sets and their complements, the residual sets, are easily proved and left as exercises. Lemma 6.9: A union of any sequence of first category sets is again a first category set. Lemma 6.10: An intersection of any sequence of residual sets is again a residual set. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 360 More on Continuous Functions and Sets Chapter 6 Theorem 6.11 (Baire Category Theorem) Every residual subset of R is dense in R. Proof. A = X ∩ The discussion in Section 6.4.1 constitutes a proof. Suppose that player (A) is dealt a set [a, b] where X is residual. Then there is a sequence of nowhere dense sets so that Qn} { Then player (A) wins by choosing any interval I1 ⊂ strategy of Section 6.4.1. In particular, X must contain a point of the interval [a, b], and hence a point of any interval. n=1 [ [a, b] that avoids Q1 and continues following the X = R \ ∞ Qn. Theorem 6.11 provides a sense of largeness of sets that is not shared by dense sets in general. The intersection of two dense sets might be empty, but the intersection of two, or even countably many, residual sets must still be dense. Exercises 6.4.1 Show that the union of any sequence of first category sets is again a first category set. See Note 150 6.4.2 Show that the intersection of any sequence of residual sets is again a residual set. See Note 151 6.4.3 Rewrite the proof of Theorem 6.11 without using the games language. See Note 152 6.4.4 Give an example of two dense sets whose intersection is not dense. Does this contradict Theorem 6.11? See Note 153 6.4.5 Suppose that (c′, d′) See Note 154 ⊂ S (c, d) so that An0 is dense in (c′, d′). ∞ n=1 An contains some interval (c, d). Show that there is a set, say An0 , and a subinterval ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.4. The Baire Category Theorem 6.4.3 Uniform Boundedness " Advanced section. May be omitted. 361 There are many applications of the Baire category Theorem in analysis. For now, we present just one of functions is bounded. This means that for each ∈ F } . We can describe this situation Mx for all f is pointwise bounded. Does this imply that the collection is uniformly bounded; that is, application, dealing with the concept of uniform boundedness. Suppose we have a collection defined on R with the property that for each x R there exists a number Mx ≥ x 0 such that by saying that that there is a single number M so that and every x M for all f R, f (x) {| | ≤ f (x) f (x) ∈ F ∈ | R, fn(x) = 0 for all other values x. Let Example 6.12: Let q1, q2, q3, . . . be an enumeration of Q. For each n fn(qk) = k if n for all f k for all f ∈ F R bounded. The bounds can be taken to be 0 if x \ But since Q is dense in R, none of the functions fn is bounded on any interval. (Verify this.) Thus a collection of functions may be pointwise bounded but not uniformly bounded on any interval. ∈ Q, f (x) = 0 R . Then if x IN \ ∈ } R, the set is : f f (x) | Q) and we can take Mqk = k. ∈ { . Thus, for each x Q (Mx = 0 if x IN we define a function fn by , and if x = qk, fn : n ∈ F} f (x) ∈ F | ≤ ∈ R {| = ≤ F ∈ ∈ \ | ◭ The functions fn in Example 6.12 are everywhere discontinuous. Our next theorem shows that if we had taken a collection be bounded on closed intervals (as Theorem 5.49 guarantees), but there would be an interval I on which the entire collection is I. uniformly bounded ; that is, there exists a constant M such that of continuous functions, then not only would each f M for all f and each x f (x) ∈ F F Theorem 6.13: Let constant Mx > 0 such that M > 0 such that f (x) F | | ≤ f (x) Mx for each f | ≤ | M for each f and x ∈ F ∈ ∈ F ∈ F be a collection of continuous functions on R such that for each x | ≤ | ∈ R there exists a . Then there exists an open interval I and a constant I. ∈ Proof. For each n hypothesis, each f IN, let An = n for all f is continuous and so it is easy to check that each of the sets . By hypothesis, R = ∈ F} f (xn=1 An. Also, by S x : { | f (x) | ≤ n } ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 362 More on Continuous Functions and Sets Chapter 6 must be closed (e.g., Exercise 5.4.31). Thus An = x : { | f (x) | ≤ n } \f ∈F is an intersection of closed sets and is therefore itself closed. This expresses the real line R as a union of the sequence of closed sets . It now follows from the Baire category theorem that at least one of the sets, say An0, must be dense in some open interval I. Since An0 is closed and dense in the interval I, An0 must contain I. This means that f (x) and all x n0 for each f I. An} { | | ≤ ∈ F ∈ Exercises 6.4.6 Let fn} { be a sequence of continuous functions on an interval [a, b] such that lim n→∞ fn(x) = f (x) exists at every point x bounded on some subinterval of [a, b]. ∈ [a, b]. Show that f need not be continuous nor even bounded, but that f must be 6.4.7 Let fn} { be a sequence of continuous functions on [0, 1] and suppose that lim n→∞ fn(x) = 0 x 1. Show that there must be an interval [c, d] for all 0 fn(x) | See Note 155 | ≤ ≤ ≤ 1 for all x [c, d]. ∈ [0, 1] so that, for all sufficiently large n, ⊂ 6.4.8 Give an example of a sequence of functions on [0, 1] with the property that lim n→∞ fn(x) = 0 for all 0 ≤ fn(x) > 1. x ≤ 1 and yet for every interval [c, d] [0, 1] and every N there is some x [c, d] and n > N with ∈ ⊂ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.5. Cantor Sets 6.5 Cantor Sets " Advanced section. May be omitted. 363 We say that a set is perfect if it is a nonempty closed set with no isolated points. The only examples that might come to mind are sets that are finite unions of intervals. It might be difficult to imagine a perfect subset of R that is also nowhere dense. In this section we obtain such a set, the very important classical Cantor set. We also discuss some of its variants. Such sets have historical significance and are of importance in a number of areas of mathematical analysis. 6.5.1 Construction of the Cantor Ternary Set " Advanced section. May be omitted. We begin with the closed interval [0, 1]. From this interval we shall remove a dense open set G. The remaining set K = [0, 1] G will then be closed and nowhere dense in [0,1]. We construct G in such a way that K has no isolated points and is nonempty. Thus K will be a nonempty, nowhere dense perfect subset of [0,1]. \ It is easiest to understand the set G if we construct it in stages. Let G1 = G1. is what remains when the middle third of the interval [0,1] is removed. This is , and let K1 = [0, 1] \ 3 1 3 , 2 We repeat this construction on each of the two component intervals of K1. Let G2 = ∪ 0, 1 3 Thus K1 = the first stage of our construction. (G1 ∪ let K2 = [0, 1] G2). Thus and ∪ K2 = 0 . ∪ ∪ ∪ This completes the second stage. We continue inductively, obtaining two sequences of sets, For each n IN ∈ 1. Gn is a union of 2n − 1 pairwise disjoint open intervals. 2. Kn is a union of 2n pairwise disjoint closed intervals. Kn} { and Gn} { with the following properties: ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 364 More on Continuous Functions and Sets Chapter 6 K1 K2 K3 Figure 6.1. The third stage in the construction of the Cantor ternary set. 3. Kn = [0, 1] (G1 ∪ G2 ∪ · · · ∪ \ Gn). 4. Each component of Gn+1 is the “middle third” of some component of Kn. 5. The length of each component of Kn is 1/3n. Figure 6.1 shows K1, K2, and K3. Now let and let G = ∞ Gn n=1 [ Then G is open and the set K (our Cantor set) is closed. K = [0, 1] G = \ Kn. ∞ n=1 \ To see that K is nowhere dense, it is enough, since K is closed, to show that K contains no open intervals (Exercise 6.3.4). Let J be an open interval in [0, 1] and let λ be its length. Choose n IN such that 1/3n < λ. By property 5, each component of Kn has length 1/3n < λ, and by property 2 the components ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.5. Cantor Sets of Kn are pairwise disjoint. Thus Kn cannot contain J, so neither can K = the closed set K contains no intervals and is therefore nowhere dense. 365 ∞1 Kn. We have shown that T x1 − It remains to show that K has no isolated points. Let x0 ∈ K. We show that x0 is a limit point of K. To do this we show that for every ε > 0 there exists x1 ∈ < ε. Choose n such x0| that 1/3n < ε. There is a component L of Kn that contains x0. This component is a closed interval of length 1/3n < ε. The set Kn+1 ∩ L has two components L0 and L1, each of which contains points of K. < ε. The point x0 is in one of the components, say L0. Let x1 be any point of K This verifies |
that x0 is a limit point of K. Thus K has no isolated points. K such that 0 < L1. Then 0 < x0 − x1| ∩ | | The set K is called the Cantor set. Because of its construction, it is often called the Cantor middle third set. In a moment we shall present a purely arithmetic description of the Cantor set that suggests another common name for K, the “Cantor ternary set”. But first, we mention a few properties of K and of its complement G that may help you visualize these sets. First note that G is an open dense set in [0, 1]. Write G = ∞k=1(ak, bk). (The component intervals (ak, bk) of G can be called the intervals complementary to K in (0, 1). Each is a middle third of a component interval of some Kn.) Observe that no two of these component intervals can have a common endpoint. If, for example, bm = an, then this point would be an isolated point of K, and K has no isolated points. S Next observe that for each k IN, the points ak and bk are points of K. But there are other points of K as well. In fact, we shall see presently that K is uncountable. These other points are all limit points of the endpoints of the complementary intervals. The set of endpoints is countable, but the closure of this set is uncountable as we shall see. Thus, in the sense of cardinality, “most” points of the Cantor set are not endpoints of intervals complementary to K. ∈ Each component interval of the set Gn has length 1/3n; thus the sum of the lengths of these component intervals is 1 − 2n 3n = 1 2 2 3 n . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 366 More on Continuous Functions and Sets Chapter 6 It follows that the lengths of all component intervals of G forms a geometric series with sum 1 2 2 3 n = 1. ∞ n=1 X (This also gives us a clue as to why K cannot contain an interval: After removing from the unit interval a sequence of pairwise disjoint intervals with length-sum one, no room exists for any intervals in the set K that remains.) Exercises 6.5.1 Let E be the set of endpoints of intervals complementary to the Cantor set K. Prove that E = K. 6.5.2 Let G be a dense open subset of R and let be its set of component intervals. Prove that H = R is perfect if and only if no two of these intervals have common endpoints. (ak, bk) } { G \ 6.5.3 Let K be the Cantor set and let (ak, bk) } IN, let ck = (ak + bk)/2 (the midpoint of the interval (ak, bk)) and let N = be the sequence of intervals complementary to K in [0, 1]. For each . Prove each of the ck : k IN { { ∈ } k following: ∈ = cj, there exists k (a) Every point of N is isolated. (b) If ci 6 (c) Show that there is an order-preserving mapping φ : Q N ]. This may seem surprising since Q “neighbor” in N ). ∈ φ(x) < φ(y) ∈ isolated points. IN such that ck is between ci and cj (i.e., no point in N has an immediate (0, 1) (0, 1), then (0, 1) has no isolated points while N has only N [i.e., if .5.4 It is common now to say that a set E of real numbers is a Cantor set if it is nonempty, bounded, perfect, and nowhere dense. Show that the union of a finite number of Cantor sets is also a Cantor set. 6.5.5 Show that every Cantor set is uncountable. 6.5.6 " Let A and B be subsets of R. A function h that maps A onto B, is one-to-one, and with both h and h−1 continuous is called a homeomorphism between A and B. The sets A and B are said to be homeomorphic. Prove that a set C is a Cantor set if and only if it is homeomorphic to the Cantor ternary set K. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.5. Cantor Sets 367 6.5.2 An Arithmetic Construction of K " Enrichment section. May be omitted. We turn now to a purely arithmetical construction for the Cantor set. You will need some familiarity with ternary (base 3) arithmetic here. Each x ∈ [0, 1] can be expressed in base 3 as x = .a1a2a3 . . . , where ai = 0, 1 or 2, i = 1, 2, 3, . . . . Certain points have two representations, one ending with a string of zeros, the other in a string of twos. For example, .1000 = .0222 . . . both represent the number 1/3 (base ten). Now, if x G1 must have ‘1’ in the first position of its ternary ∈ expansion. Similarly, if (1/3, 2/3), a1 = 1, thus each x · · · ∈ 1 9 it must have a 1 in the second position of its ternary expansion (i.e., a2 = 1). In general, each point in Gn must have an = 1. It follows that every point of G = ∞1 Gn must have a 1 someplace in its ternary expansion. G2 = 7 9 8 9 2 9 ∪ ∈ x , , , Now endpoints of intervals complementary to K have two representations, one of which involves no 1’s. The remaining points of K never fall in the middle third of a component of one of the sets Kn, and so have ternary expansions of the form S We can therefore describe K arithmetically as the set x = .a1a2 . . . ai = 0 or 2. x = .a1a2a3 . . . (base three) : ai = 0 or 2 for each i { . IN } ∈ As an immediate result, we see that K is uncountable. In fact, K can be put into 1-1 correspondence with [0,1]: For each x = .a1a2a3 . . . (base 3), ai = 0, 2, ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 368 More on Continuous Functions and Sets Chapter 6 in the set K, let there correspond the number y = .b1b2b3 . . . (base 2), bi = ai/2. This provides a 1-1 correspondence between K (minus endpoints of complementary intervals) and [0, 1] (minus the countable set of numbers with two base 2 representations). By allowing these two countable sets to correspond to each other, we obtain a 1-1 correspondence between K and [0, 1]. Note. We end this section by mentioning that variations in the constructions of K can lead to interesting situations. For example, by changing the construction slightly, we can remove intervals in such a way that with G′ = ∞ [k=1 (a′k, b′k) ∞ (b′k − (instead of 1), while still keeping K′ = [0, 1] G′ nowhere dense and perfect. The resulting set K′ created problems for late nineteenth-century mathematicians trying to develop a theory of measure. The “measure” of G′ should be 1/2; the “measure” of [0,1] should be 1. Intuition requires that the measure of the nowhere dense set K′ should be 1 2 . How can this be when K′ is so “small?” a′k) = 1/2 Xk=1 \ 2 = 1 1 − Exercises 6.5.7 Find a specific irrational number in the Cantor ternary set. See Note 156 6.5.8 Show that the Cantor ternary set can be defined as K = x ( ∈ [0, 1] : x = ∞ n=1 X in 3n for in = 0 or 2 ) . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.5. Cantor Sets 6.5.9 Let ∞ D = x ( ∈ [0, 1] : x = n=1 X 369 jn 3n for jn = 0 or 1 ) . Show that D + D = K + K = [0, 2]. { x + y : x, y D } ∈ = [0, 1]. From this deduce, for the Cantor ternary set K, that 6.5.10 A careless student makes the following argument. Explain the error. “If G = (a, b), then G = [a, b]. Similarly, if G = It follows that an open set G and its closure G differ by at most a countable set.” ∞ i=1(ai, bi) is an open set, then G = S ∞ i=1[ai, bi]. S See Note 157 6.5.3 The Cantor Function " Advanced section. May be omitted. The Cantor set allows the construction of a rather bizarre function that is continuous and nondecreasing on the interval [0, 1]. It has the property that it is constant on every interval complementary to the Cantor set and yet manages to increase from f (0) = 0 to f (1) = 1 by doing all of its increasing on the Cantor set itself. It has sometimes been called “the devil’s staircase.” Define the function f in the following way. On (1/3, 2/3), let f = 1/2; on (1/9, 2/9), let f = 1/4; on 1 open intervals appearing at the nth stage, define (7/9, 8/9), let f = 3/4. Proceed inductively. On the 2n f to satisfy the following conditions: − 1. f is constant on each of these intervals. 2. f takes the values on these intervals. 1 2n , 3 2n , . . . , 1 2n − 2n 3. If x and y are members of different nth-stage intervals with x < y, then f (x) < f (y). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 370 More on Continuous Functions and Sets Chapter 6 This description defines f on G = [0, 1] \ K. Extend f to all of [0, 1] by defining f (0) = 0 and, for 0 < x 1, ≤ In order to check that this defines the function that we want, we need to check each of the following. f (x) = sup f (t) : t { G, t < x } . ∈ 1. f (G) is dense in [0, 1]. 2. f is nondecreasing on [0, 1]. 3. f is continuous on [0, 1]. 4. f (K) = [0, 1]. These have been left as exercises. Figure 6.2 illustrates the construction. The function f is called the Cantor function. Observe that f “does all its rising” on the set K. The Cantor function allows a negative answer to many questions that might be asked about functions and derivatives and, hence, has become a popular counterexample. For example, let us follow this kind of reasoning. If f is a continuous function on [0, 1] and f ′(x) = 0 for every x (0, 1) then f is constant. (This is proved in most calculus courses by using the mean value theorem.) Now suppose that we know less, that f ′(x) = 0 for every x (0, 1) excepting a “small” set E of points at which we know nothing. If E is finite it is still easy to show that f must be constant. If E is countable it is possible, but a bit more difficult, to show that it is still true that f must be constant. The question then arises, just how small a set E can appear here; that is, what would we have to know about a set E so that we could say f ′(x) = 0 for every x (0, 1) ∈ The Cantor function is an example of a function constant on every interval complementary to the Cantor set K (and so with a zero derivative at those points) and yet is not constant. The Cantor set, since it is nowhere dense, might be viewed as extremely small, but even so it is not insignificant for this problem. E implies that f is constant? ∈ ∈ \ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.5. Cantor Sets 371 Figure 6.2. The third stage in the construction of the Cantor |
function. Exercises 6.5.11 In the construction of the Cantor function complete the verification of details. (a) Show that f (G) is dense in [0, 1]. (b) Show that f is nondecreasing on [0, 1]. (c) Infer from (a) and (b) that f is continuous on [0, 1]. (d) Show that f (K) = [0, 1] and thus (again) conclude that K is uncountable. 6.5.12 Find a calculus textbook proof for the statement that a continuous function f on an interval [a, b] that has a ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 372 More on Continuous Functions and Sets Chapter 6 zero derivative on (a, b) must be constant. Improve the proof to allow a finite set of points on which f is not known to have a zero derivative. 6.6 Borel Sets " Advanced section. May be omitted. In our study of continuous functions we have seen that the classes of open sets and closed sets play a significant role. But the class of sets that are of importance in analysis goes beyond merely the open and closed sets. E. Borel (1871–1956) recognized that for many operations of analysis we need to form countable intersections and countable unions of classes of sets. The collection of Borel sets was introduced exactly to allow these operations. We recall that a countable union of closed sets may not be closed (or open) and that a countable intersection of open sets, also, may not be open (or closed). In this section we introduce two additional types of sets of importance in analysis, sets of type sets of type they are precisely the right classes of sets to solve some fundamental questions about real functions. Gδ and F σ. These classes form just the beginning of the large class of Borel sets. We shall find that 6.6.1 Sets of Type Gδ " Advanced section. May be omitted. Recall that the union of a collection of open sets is open (regardless of how many sets are in the collection), but the intersection of a collection of open sets need not be open if the collection has infinitely many sets. For example, Similarly, if q1, q2, q3, . . . is an enumeration of Q, then ∞ n=1 \ 1 n , 1 n = 0 . } { − the set of irrational numbers. The set 0 { } ∞ (R qk} \ { ) = R Q, \ \k=1 is closed (not open), and R Q is neither open nor closed. The \ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.6. Borel Sets 373 Q is a countable intersection of open sets. Such sets are of sufficient importance to give them a set R \ name. Definition 6.14: A subset H of R is said to be of type intersection of open sets, that is, if there exist open sets G1, G2, G3, . . . such that H = Gδ set) if it can be expressed as a countable Gδ (or a ∞k=1 Gk. Example 6.15: A closed interval [a, b] or a half-open interval (a, b] is of type T Gδ since and ∞ [a, b] = n=a, b] = a, b + n=1 \ 1 n . ◭ Theorem 6.16: Every open set and every closed set in R is of type Gδ. Proof. Let G be an open set in R. It is clear that G is of type as a countable union of closed sets. Express G in the form Gδ. We also show that G can be expressed ∞ G = (ak, bk) [k=1 where the intervals (ak, bk) are pairwise disjoint. Now for each k ∈ such that the sequence dkj } j decreases to ak, the sequence IN. Thus ckj } { { IN there exist sequences dkj } increases to bk and ckj < dkj for each ckj } and { { ∈ (ak, bk) = [ckj , dkj ]. ∞ [j=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 374 More on Continuous Functions and Sets Chapter 6 We have expressed each component interval of G as a countable union of closed sets. It follows that ∞ ∞ G = [ckj , dkj ] = ∞ [ckj , dkj ] [k=1 is also a countable union of closed sets. Now take complements. This shows that R G can be expressed as a countable intersection of open sets (by using the de Morgan laws). Since every closed set F can be written [j,k=1 [j=1 \ for some open set G, we have shown that any closed set is of type F = R G \ Gδ. We observed in Section 6.4 that a dense set can be small in the sense of category. For example, Q is a Gδ must be large in the sense of category. first category set. Our next result shows that a dense set of type Theorem 6.17: Let H be of type Gδ and be dense in R. Then H is residual. Proof. Write H = ∞ Gk with each of the sets Gk open. Since H is dense by hypothesis and H open sets Gk is also dense. Thus R \ result now follows from Lemma 6.10. Gk is nowhere dense for every k ∈ \k=1 Gk for each k IN, each of the ⊂ ∈ IN, and so each Gk is residual. The Exercises 6.6.1 Which of the following sets are of type Gδ? (a) IN 1 n (b) : n IN ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.6. Borel Sets 375 (c) The set (d) A finite union of intervals (that need not be open or closed) of midpoints of intervals complementary to the Cantor set Cn : n IN ∈ { } 6.6.2 Prove Theorem 6.17 for the interval [a, b] in place of R. 6.6.3 Prove that a set E of type Gδ in R is either residual or else there is an interval containing no points of E. 6.6.2 Sets of Type Fσ " Advanced section. May be omitted. Just as the countable intersections of open sets form a larger class of sets, the countable unions of closed sets form a larger class of sets. The complements of open sets are closed. By dealing with complements of notion of a set of type F σ. Gδ sets, so also the Gδ sets we arrive at the dual Definition 6.18: A subset E of R is said to be of type countable union of closed sets; that is, if there exist closed sets F1, F2, F3, . . . such that E = F σ set) if it can be expressed as a F σ (or an ∞k=1 Fk. Using the de Morgan laws, we verify easily that the complement of a F σ and vice versa (Exercise 6.6.4). This is closely related to the fact that a set is open if and only if its complement is closed. Gδ set is an S Example 6.19: The set of rational numbers, Q is a set of type as F σ. This is clear since it can be expressed ∞ Q = rn} { n=1 [ is any enumeration of the rationals. The singleton sets where rn} rn} { Q is not of type (because a countable set is first category). In particular, Q is not of type type Gδ also. It follows from Theorem 6.17 that a dense set of type are clearly closed. But note that Gδ must be uncountable Q is not of \ ◭ Gδ (and therefore R { F σ). Theorem 6.20: A set is of type Gδ if and only if its complement is of type F σ. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 376 More on Continuous Functions and Sets Chapter 6 Example 6.21: A half-open interval (a, b] is both of type ∞ (a, b] = a, b + n=1 \ = 1 n n=1 [ Gδ and of type : . ◭ Note. The only subsets of R that are both open and closed are the empty set and R itself. There are, however, many sets that are of type Gδ and also of type F σ. See Exercise 6.6.1. We can now enlarge on Theorem 6.16. There we showed that all open sets and all closed sets are in the class Gδ. We now show they are also in the class F σ. Theorem 6.22: Every open set and every closed set in R is both of type F σ and Gδ. In the proof of Theorem 6.16 we showed explicitly how to express any open set as an Proof. F σ. Thus Gδ (the latter being trivial). The part pertaining to closed sets open sets are of type F σ as well as of type now follows by considering complements and using the de Morgan laws. The complement of a closed set is F σ set is a open and therefore the complement of an Gδ set. Exercises 6.6.4 Verify that a subset A of R is an F σ ( 6.6.5 Which of the following sets are of type Gδ) if and only if R F σ? A is a Gδ ( F σ). \ (a) IN 1 n (b) : n (c) The set IN Cn : n ∈ { ∈ } (d) A finite union of intervals (that need not be open or closed) IN of midpoints of intervals complementary to the Cantor set ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.7. Oscillation and Continuity 377 6.6.6 " Prove that a set of type 6.6.7 " Let fn} Show that { F σ in R is either first category or contains an open interval. be a sequence of real functions defined on R and suppose that fn(x) f (x) at every point x. → x : f (x) > α = } { x : fn(x) { α + 1/m . } ≥ ∞ ∞ ∞ m=1 [ r=1 [ n=r \ If each function fn is continuous, what can you assert about the set See Note 158 6.7 Oscillation and Continuity " Advanced section. May be omitted. x : f (x) > α ? } { the set of discontinuity points of a function. To discuss this set we shall need the notions of and we need to introduce a new tool, the oscillation of a function. In this section we return to a problem that we began investigating in Section 5.9 about the nature of Gδ sets We begin with an example of a function f that is discontinuous at every rational number and continuous F σ and at every irrational number. Example 6.23: Let q1, q2, q3, . . . be an enumeration of Q. Define a function f by f (x) = 1 k , 0, if x = qk R if x ∈ \ Q. \ Q is dense in R, f can be continuous at a point x only if f (x) = 0; that is, only if x Since R f is discontinuous at every x let ε > 0. Choose k there exists δ > 0 such that x at x0. Since x0 was an arbitrary irrational point, we see that f is continuous at every irrational. R Q, let x0 ∈ IN such that 1/k < ε. Since the set q1, q2, . . . , qk is a finite set not containing x0, < δ, then either x0| 1 k < ε. This verifies the continuity of f ◭ | Q or x = qj for some j > k. In either case Q. To check that f is continuous at each point of R δ for each i = 1, . . . , k. Thus if x Q. Thus Q and ∈ qi − x0| ≥ R and f (x0) f (x \ \ \ | | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 378 More on Continuous Functions and Sets Chapter 6 Our example shows that it is possible for a function to be continuous at every irrational number and discontinuous at every rational number. Is it possible for the opposite to occur? Does there exist a function f continuous on Q and discontinuous on R continuity of some function f defined on an interval. Q? More generally, what sets can be the set |
of points of \ We answer this question in this section. The principal tool is that of oscillation of a function at a point. 6.7.1 Oscillation of a Function " Advanced section. May be omitted. In order to describe a point of discontinuity we need a way of measuring that discontinuity. For monotonic functions the jump was used previously for such a measure. For general, nonmonotonic, functions a different tool is used. Definition 6.24: Let f be defined on a nondegenerate interval I. We define the oscillation of f on I as the quantity ωf (I) = sup I | x,y ∈ f (x) f (y) . | − Let’s see how oscillation relates to continuity. Suppose f is defined in a neighborhood of x0, and f is continuous at x0. Then To see this, let ε > 0. Since f is continuous at x0, there exists δ0 > 0 such that inf δ>0 ωf ((x0 − δ, x0 + δ)) = 0. if x | − x0| < δ0. If then f (x) | − f (x0) | < ε/2 x0 − δ0 < x1 ≤ x2 < x0 + δ0, f (x1) | − f (x2) f (x1) f (x0) + | | − f (x0) − f (x2. (1) (2) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.7. Oscillation and Continuity Since (2) is valid for all x1, x2 ∈ (x0 − f (x1) δ0, x0 + δ0), we have : x0 − f (x2) δ0 < x1 ≤ | − sup {| x2 < x0 + δ0} ≤ ε. 379 (3) But (3) implies that if 0 < δ < δ0, then Since ε was arbitrary, the result follows. The converse is also valid. Suppose (1) holds. Let ε > 0. Choose δ > 0 such that ωf ([x0 − δ, x0 + δ]) ε. ≤ ωf (x0 − δ, x0 + δ) < ε. Then sup so f (x0) < ε whenever f (x) | We summarize the preceding as a theorem. − − x | | {| f (x) x0| : x f (x0) (x0 − − < δ. This implies continuity of f at x0. δ, x0 + δ) < ε, ∈ } | Theorem 6.25: Let f be defined on an interval I and let x0 ∈ I. Then f is continuous at x0 if and only if inf δ>0 ωf ((x0 − δ, x0 + δ)) = 0. The quantity in the statement of the theorem is sufficiently important to have a name. Definition 6.26: Let f be defined in a neighborhood of x0. The quantity ωf (x0) = inf δ>0 ωf ((x0 − δ, x0 + δ)) is called the oscillation of f at x0. Theorem 6.25 thus states that a function f is continuous at a point x0 if and only if ωf (x0) = 0. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 380 More on Continuous Functions and Sets Chapter 6 Returning to the function that introduced this section, we see that ωf (x) = 1/k, 0, if x = qk R if x ∈ \ Q. Let’s now see how the concept of oscillation relates to the set of points of continuity of a function. Theorem 6.27: Let f be defined on a closed interval I (which may be all of R). Let γ > 0. Then the set is open and the set is closed. x : ωf (x) < γ { } x : ωf (x) { γ } ≥ A. We wish to find a neighborhood U of x0 such that U A; ⊂ (α, γ). From Definition 6.26 we infer the existence of a number δ > 0 Proof. Let A = { that is, such that ωf (x) < γ for all x Let ωf (x0) = α < γ and let β x : ωf (x) < γ and let x0 ∈ U . ∈ } ∈ such that for u, v (x0 − ∈ δ, x0 + δ). Let f (u) | − f (v) | ≤ β U = (x0 − δ, x0 + δ) U . Since U is open, there exists δ1 < δ such that and let x ∈ Then (x − δ1, x + δ1) U. ⊂ ωf (x) sup sup ≤ ≤ {| {| f (t) f (u) f (s) | f (v) | − − : t, s ∈ : u, v (x − U δ1, x + δ1) β < γ, } ∈ } ≤ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.7. Oscillation and Continuity so x ∈ A. This proves A is open. It follows then that the complement of A in I, the set must be closed. x : ωf (x) { , γ } ≥ 381 We use the oscillation in the next subsection to answer a question about the nature of the set of points of continuity of a function. Exercises 6.7.1 Suppose that f is bounded on an interval I. Prove that ωf (I) = sup x∈I f (x) inf x∈I − f (x). 6.7.2 A careless student believes that the oscillation can be written as Show that this is not true, even for bounded functions. ωf (x0) = lim sup x→x0 f (x) lim inf x→x0 − f (x). 6.7.3 Prove that See Note 159 ωf (x0) = lim δ→0+ ωf ((x0 − δ, x0 + δ)). 6.7.4 Calculate ωf (0) for each of the following functions. (a) f (x) = (b) f (x) = (c) f (x) = x, 4, 0, 1, n, 0, if x = 0 if x = 0 Q Q if x ∈ if x / ∈ if x = 1 n otherwise ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 382 More on Continuous Functions and Sets Chapter 6 (d) f (x) = (e) f (x) = (f) f (x) = sin 1 x , 0, sin 1 x , 7, x sin 1 1 x , 0, if x = 0 if x = 0 = 0 if x if x = 0 if x = 0 if x = 0 6.7.5 In the proof of Theorem 6.27 we let ωf (x0) = α < γ and let β the proof have worked if we had used β = γ? ∈ (α, γ). Why was the β introduced? Would 6.7.2 The Set of Continuity Points " Advanced section. May be omitted. Given an arbitrary function, how can we describe the nature of the set of points where f is continuous? Can it be any set? Given a set E, how can we know whether there is a function that is continuous at every point of E and discontinuous at every point not in E? We saw in Example 6.23 that a function exists whose set of continuity points is exactly the irrationals. Can a function exist whose set of continuity points is exactly the rationals? By characterizing the set of such points we can answer this and other questions about the structure of functions. We now prove the main result of this section using primarily the notion of oscillation introduced in Section 6.7.1. Theorem 6.28: Let f be defined on a closed interval I (which may be all of R). Then the set Cf of points of continuity of f is of type F σ. Conversely, if H is a set of type Gδ, and the set Df of points of discontinuity of f is of type Gδ, then there exists a function f defined on R such that Cf = H. Proof. To prove the first part, let f : I is of type Gδ. For each k ∈ IN, let R. We show that the set → Cf = { x : ωf (x) = 0 } Bk = x : ωf (x) 1 k . ≥ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 6 Section 6.7. Oscillation and Continuity By Theorem 6.27, each of the sets Bk is closed. Thus the set B = ∞ Bk [k=1 F σ. By Theorem 6.25, Df = B. Therefore, Cf = I is of type the set Cf is a Gδ. To prove the converse, let H be any subset of R of type \ B. Since the complement of an F σ is a Gδ. Then H can be expressed in the form 383 Gδ, H = ∞ Gk \k=1 with each of the sets Gk being open. We may assume without loss of generality that G1 = R and that Gi ⊃ Let IN. (Verify this.) be sequences of positive numbers, each converging to zero, with and Gi+1 for each i ∈ βk} αk} { { for all k ∈ IN. Define a function f : R R by → αk > βk > αk+1, f (x) = 0 αk βk if x if x if x H (Gk \ (Gk \ ∈ ∈ ∈ Gk+1) Gk+1) Q (R ∩ ∩ Q). \ We show that f is continuous at each point of H and discontinuous at each point of R Let x0 ∈ H and let ε > 0. Choose n such that αn < ε. Since H. \ we see that x0 ∈ Gn. The set Gn is open, so there exists δ > 0 such that (x0 − δ, x0 + δ) ⊂ Gn. From the x0 ∈ H = Gk, ∞ \k=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 384 More on Continuous Functions and Sets Chapter 6 definition of f on Gn, we see that for all x (x0 − ∈ δ, x0 + δ). Thus − < δ, so f is continuous at x0. | f (x) if | x0| x − Now let x0 ∈ f (x) 0 ≤ ≤ αn < ε f (x0) f (xx) | < ε R IN such that x0 belongs to the set Gk \ or f (x0) = βk. Let us suppose that f (x0) = αk. If x0 is an interior point of Gk \ point of H. Then there exists k ∈ \ Gk+1. Thus f (x0) = αk Gk+1, then x0 is a limit so f is discontinuous at x0. x : x { (Gk \ ∈ Gk+1) (R Q) } \ = x : f (x) = βk} { , ∩ The argument is similar if x0 is a boundary point of Gk \ close to x0 there are points of the set Gk+1. Again, assume f (x0) = αk. Arbitrarily At these points, f takes on values in the set R (Gk \ \ Gk+1). =k [i The only limit point of this set is zero and so S is closed. In particular, αk is not a limit point of this set and does not belong to the set. Let ε be half the distance from the point αk to the closed set S; that is, let =k [j S = 0 { } ∪ αi ∪ βj. ε = 1 2 d(αk, S). Arbitrarily close to x0 there are points x such that f (x) S. For such a point, f (x) | f (x0) = | | − αk| − > ε, ∈ f (x) so f is discontinuous at x0. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 Section 6.8. Sets of Measure Zero 385 Observe that Theorem 6.28 answers a question we asked earlier: Is there a function f continuous on Q and discontinuous at every point of R \ Q? The answer is negative, since Q is not of type Gδ. Exercises 6.7.6 In the second part of the proof of Theorem 6.28 we provided a construction for a function f with Cf = H, Gδ. Exhibit explicitly sets Gk that will give rise to a function f such that Q. Can you do this in such a way that the resulting function is the one we obtained at the beginning where H is an arbitrary set of type Cf = R \ of this section? 6.7.7 In the proof of Theorem 6.28 we took ε = 1 2 d(αk, S). Show that this number equals 1 2 min i6=k { min αi − {| , αk| βi − | βk|}} . 6.8 Sets of Measure Zero " Advanced section. May be omitted. In analysis there are a number of ways in which a set might be considered as “small.” For example, the Cantor set is not small in the sense of counting: It is uncountable. It is small in another different sense: It is nowhere dense, that is there is no interval at all in which it is dense. Now we turn to another way in which the Cantor set can be considered small: It has “zero length.” Example 6.29: Suppose we wish to measure the “length” of the Cantor set. Since the Cantor set is rather bizarre, we might look instead at the sequence of intervals that have been removed. There is no difficulty a. What is the total length of the in assigning a meaning of length to an interval; the length of (a, b) is b intervals removed in the construction of the Cantor set? From the interval [0, 1] we remove first a middle third interval of length 1/3, then two middle third intervals of length 1/9, and so on so that at the nth n. The sum of the lengths of all intervals so removed is stage we remove 2n 1 intervals each of length 3 |
− − − 1/3 + 2(1/9) + 4(1/27) + = · · · 1/3 (1 + 2/3 + (2/3)2 + (2/3)3 + . . . ) = 1. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 386 More on Continuous Functions and Sets Chapter 6 From the interval [0, 1] we appear to have removed all of the length. What is left over, the Cantor set, must have length zero. This method of computing lengths has some merit but it is not the one we wish to adopt here. Another approach to “measuring” the length of the Cantor set is to consider the length that remains at each stage. At the first stage the Cantor set is contained inside the union ∪ which has length 2(1/3). At the next stage it is contained inside a union of four intervals, with total length 4(1/9). Similarly, at the nth stage the Cantor set is contained inside the union of 2n intervals each of length 3− Thus, as before, it seems we should assign zero length to the Cantor set. n. The sum of the lengths of all these intervals is (2/3)n, and this tends to zero as n gets large. ◭ [0, 1/3] [2/3, 1], We convert the second method of the example into a definition of what it means for a set to be of measure zero. “Measure” is the technical term used to describe the “length” of sets that need not be intervals. In the example we used closed intervals while in our definition we have employed open intervals. There is no difference (see Exercise 6.8.13). In the example we covered the Cantor set with a finite sequence of intervals while in our definition we have employed an infinite sequence. For the Cantor set there is no difference but for other sets (sets that are not bounded or are not closed) there is a difference. Definition 6.30: Let E be a set of real numbers. Then E is said to have measure zero if for every ε > 0 there is a finite or infinite sequence of open intervals covering the set E so that (a1, b1), (a2, b2), (a3, b3), (a4, b4), . . . Note. In the definition of measure zero sets is there a change if we insist on an infinite sequence of (bk − ak) ≤ ε. ∞ Xk=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.8. Sets of Measure Zero 387 intervals, disallowing finite sequences? Suppose that the sequence (a1, b1), (a2, b2), (a3, b3), (a4, b4), . . . (aN , bN ) of open intervals covers the set E so that N Then to satisfy the definition we could add in some further intervals that do not amount in length to more than ε/2. For example, take (bk − ak) < ε/2. Xk=1 (aN +p, bN +p) = (0, ε/2p+1) for p = 1, 2, 3, . . . . Then ∞ (bk − ak) = (bk − ∞ ak) + ε/2p+1 < ε. Xk=1 p=1 X N Xk=1 Thus the definition would not be changed if we had required infinite coverings. Here are some examples of sets of measure zero. Example 6.31: Every finite set has measure zero. The empty set is easily handled. If and ε > 0, then the sequence of intervals E = x1, x2, . . . , xN } { ε 2N covers the set E and the sum of all the lengths is ε. xi − ε 2N , xi + i = 1, 2, 3, . . . , N ◭ Example 6.32: Every infinite, countable set has measure zero. If E = { x1, x2, . . . } ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 388 More on Continuous Functions and Sets Chapter 6 and ε > 0, then the sequence of intervals covers the set E and, since the sum of all the lengths is ε. xi − ε 2i+1 , xi + ε 2i+1 i = 1, 2, 3, . . . ε 2k+1 ∞ 2 Xk=1 = ε2− k = ε, Xk=1 ◭ Example 6.33: The Cantor set has measure zero. Let ε > 0. Choose n so that (2/3)n < ε. Then the nth stage intervals in the construction of the Cantor set give us 2n closed intervals each of length (1/3)n. This covers the Cantor set with 2n closed intervals of total length (2/3)n, which is less than ε. If the closed intervals trouble you (the definition requires open intervals), see Exercise 6.8.13 or argue as follows. Since (2/3)n < ε there is a positive number δ so that (2/3)n + δ < ε. Enlarge each of the closed intervals to form a slightly larger open interval, but change the length of each only enough so that the sum of the lengths of all the 2n closed intervals does not increase by more than δ. The resulting collection of open intervals also covers the Cantor set, and the sum of the length of these ◭ intervals is less than ε. One of the most fundamental of the properties of sets having measure zero is how sequences of such sets combine. We recall that the union of any sequence of countable sets is also countable. We now prove that the union of any sequence of measure zero sets is also a measure zero set. Theorem 6.34: Let E1, E2, E3, . . . be a sequence of sets of measure zero. Then the set E formed by taking the union of all the sets in the sequence is also of measure zero. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.8. Sets of Measure Zero 389 Proof. Let ε > 0. We shall construct a cover of E consisting of a sequence of open intervals of total length less than ε. Since E1 has measure zero, there is a sequence of open intervals (a11, b11), (a12, b12), (a13, b13), (a14, b14), . . . covering the set E1 and so that the sum of the lengths of these intervals is smaller than ε/2. Since E2 has measure zero, there is a sequence of open intervals (a21, b21), (a22, b22), (a23, b23), (a24, b24), . . . covering the set E2 and so that the sum of the lengths of these intervals is smaller than ε/4. In general, for each k = 1, 2, 3, . . . there is a sequence of open intervals (ak1, bk1), (ak2, bk2), (ak3, bk3), (ak4, bk4), . . . covering the set Ek and so that the sum of the lengths of these intervals is smaller than ε/2k. The totality of all these intervals can be arranged into a single sequence of open intervals that covers every point in the . The sum of the lengths of all the intervals in the large sequence is smaller than union of the sequence Ek} { ε/2 + ε/4 + ε/8 + = ε. · · · It follows that E has measure zero. Let us return to the situation for the Cantor set once again. For each ε > 0 we were able to choose a finite cover of open intervals with total length less than ε. This is not the case for all sets of measure zero. For example, the set of all rational numbers on the real line is countable and hence also of measure zero. Any finite collection of intervals must fail to cover that set, in fact cannot come close to covering all rational numbers. For what sets is it possible to select finite coverings of small length? The answer is that this is possible for compact sets of measure zero. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 390 More on Continuous Functions and Sets Chapter 6 Theorem 6.35: Let E be a compact set of measure zero. Then for every ε > 0 there is a finite collection of open intervals (a1, b1), (a2, b2), (a3, b3), (a4, b4), . . . , (aN , bN ) that covers the set E and so that N Xk=1 Proof. Since E has measure zero, it is certainly possible to select a sequence of open intervals (bk − ak) < ε. that covers the set E and so that (a1, b1), (a2, b2), (a3, b3), (a4, b4), . . . (bk − But how can we reduce this collection to a finite one that also covers the set E? If you studied the Heine-Borel theorem (Theorem 4.33), then you know how. ak) < ε. Xk=1 ∞ We shall present here a proof that uses the Bolzano-Weierstrass theorem instead. We claim that we can find an integer N so that all points of E are in one of the intervals (a1, b1), (a2, b2), (a3, b3), (a4, b4), . . . , (aN , bN ). This will prove the theorem. We prove this by contradiction. If this is not so, then for each integer k = 1, 2, 3, . . . we must be able to find a point xk ∈ E but xk is not in any of the intervals (a1, b1), (a2, b2), (a3, b3), (a4, b4), . . . (ak, bk). is bounded because E is bounded. By the Bolzano-Weierstrass theorem the sequence The sequence { . Let z be the limit of the convergent subsequence. Since E is closed z has a convergent subsequence is in E. The original sequence of intervals covers all of E and so there must be an interval (aM , bM ) that xnj } xk} { ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 6.8. Sets of Measure Zero 391 contains z. For large values of j the points xnj also belong to (aM , bM ). But this is impossible since xnj cannot belong to the interval (aM , bM ) for nj ≥ M . Since this is a contradiction, the proof is done. Exercises 6.8.1 Show that every subset of a set of measure zero also has measure zero. 6.8.2 If E has measure zero, show that the translated set also has measure zero.8.3 If E has measure zero, show that the expanded set also has measure zero for any c > 0. cE = cx : x { E } ∈ 6.8.4 If E has measure zero, show that the reflected set also has measure zero. E = − {− x : x E } ∈ 6.8.5 Without referring to the proof of Theorem 6.34, show that the union of any two sets of measure zero also has measure zero. 6.8.6 If E1 ⊂ 6.8.7 Show that any interval (a, b) or [a, b] is not of measure zero. E2 and E1 has measure zero but E2 has not, what can you say about the set E2 \ E1? 6.8.8 Give an example of a set that is not of measure zero and does not contain any interval [a, b]. 6.8.9 A careless student claims that if a set E has measure zero, then it is obviously true that the closure E must also have measure zero. Is this correct? 6.8.10 If a set E has measure zero what can you say about interior points of that set? 6.8.11 Suppose that a set E has the property that E E also have measure zero? ∩ [a, b] has measure zero for every compact interval [a, b]. Must ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 392 More on Continuous Functions and Sets Chapter 6 6.8.12 Show that the set of real numbers in the interval [0, 1] that do not have a 7 in their infinite decimal expansion is of measure zero. 6.8.13 In Definition 6.30 show that closed intervals may be used without changing the definition. 6.8.14 Describe completely the class of sets E with the following property: Fo |
r every ε > 0 there is a finite collection of open intervals (a1, b1), (a2, b2), (a3, b3), (a4, b4), . . . , (aN , bN ) that covers the set E and so that (These sets are said to have zero content.) N (bk − Xk=1 ak) < ε. 6.8.15 Show that a set E has measure zero if and only if there is a sequence of intervals (a1, b1), (a2, b2), (a3, b3), (a4, b4), . . . so that every point in E belongs to infinitely many of the intervals and ∞ k=1(bk − ak) converges. 6.8.16 By altering the construction of the Cantor set, construct a nowhere dense closed subset of [0, 1] so that the P sum of the lengths of the intervals removed is not equal to 1. Will this set have measure zero? 6.9 Challenging Problems for Chapter 6 6.9.1 Show that a function is discontinuous except at the points of a first category set if and only if it is continuous at a dense set of points. 6.9.2 Let f : R R be a continuous function. Assume that for every positive number ε the sequence → converges to zero as n . Prove that → ∞ lim x→∞ f (x) = 0. f (nε) } { See Note 160 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 393 6.9.3 Let fn be a sequence of continuous functions defined on an interval [a, b] such that limn→∞ fn(x) = 0 for each [a, b]. Show that for any ε > 0 there is an interval [c, d] [a, b] and an integer N so that x ∈ ⊂ < ε fn(x) | | for every n ≥ N and every x ∈ [c, d]. Show that this need not be true for [c, d] = [a, b]. 6.9.4 Let fn be a sequence of continuous functions defined on an interval [a, b] such that limn→∞ fn(x) = x ∈ [a, b]. Show that for any M > 0 there is an interval [c, d] fn(x) > M ⊂ [a, b] and an integer N so that for each ∞ for every n ≥ N and every x ∈ [c, d]. Show that this need not be true for [c, d] = [a, b]. Notes 147Exercise 6.2.9. To make this true, assume that f is onto or else show that if E is dense then f (E) is dense in the set (interval) f (R). 148Exercise 6.3.1. dense, but If q1, q2, q3, . . . is an enumeration of the rationals, then each of the sets , i qi} { ∈ IN, is nowhere n { i=1 [ = Q qi} is not nowhere dense. (Indeed it is dense.) 149Exercise 6.3.2. All of (a)–(e) and (h) are true. Find counterexamples for (f) and (g). The proofs that the others are true follow routinely from the definition. 150Exercise 6.4.1. Suppose that ∞ An = Ank [k=1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 394 NOTES with each of the sets Ank nowhere dense. Then expresses that union as a first category set. ∞ ∞ ∞ Ank = Ank n=1 [ [k=1 [n,k=1 151Exercise 6.4.2. Let a first category set An. For each n write Bn} { be a sequence of residual subsets of R. Thus each of the sets Bn is the complement of with each of the sets Ank nowhere dense. Then ∞ An = Ank [k=1 ∞ \ [k=1 Ank. Bn = R Now use De Morgan’s laws. 152Exercise 6.4.3. Suppose that X is residual, that is, X = R ∞ \ n=1 [ Qn where each Qn is nowhere dense. Show that for any interval [a, b] there is a point in X appropriate descending sequence of closed subintervals of [a, b]. ∩ [a, b] by constructing an 153Exercise 6.4.4. Make sure your sets are dense but not both residual (e.g., Q and R Q). \ 154Exercise 6.4.5. This follows, with the correct interpretation, directly from the Baire category theorem. 155Exercise 6.4.7. Consider the sequence AN = x { ∈ [0, 1] : fn(x) | | ≤ 1, all n N . } ≥ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES Check that AN = [0, 1]. ∞ [N =1 395 156Exercise 6.5.7. is uncountable and the rationals are countable. Your job is to find just one. It is clear that there must be many irrational numbers in the Cantor ternary set, since that set 157Exercise 6.5.10. Consider G = (0, 1) C where C is the Cantor ternary set. \ 158Exercise 6.6.7. Often to prove a set identity such as this the best way is to start with a point x that belongs to the set on the right and then show that point must be in the set on the left. After that is successful start with a point x that belongs to the set on the left. For example, if f (x) > α, then f (x) ≥ α + 1/m for some integer m. But fn(x) and so there must be an integer R so that fn(x) > α + 1/m for all n f (x) → R, etc. ≥ This exercise shows how unions and intersections of sequences of open and closed sets might arise in analysis. Note that the sets would be closed if the functions fn are continuous. Thus it would follow that the set x : fn(x) α + 1/m ≥ } { } F σ. This says something interesting about a function f that is the limit of a sequence of continuous { x : f (x) > α must be of type . fn} functions { 159Exercise 6.7.3. You need to recall Theorem 5.60, which asserts that monotone functions have left- and right-hand limits. 160Exercise 6.9.2. This is from the 1964 Putnam Mathematical Competition. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Chapter 7 DIFFERENTIATION 7.1 Introduction Calculus courses succeed in conveying an idea of what a derivative is, and the students develop many technical skills in computations of derivatives or applications of them. We shall return to the subject of derivatives but with a different objective. Now we wish to see a little deeper and to understand the basis on which that theory develops. Much of this chapter will appear to be a review of the subject of derivatives with more attention paid to the details now and less to the applications. Some of the more advanced material will be, however, completely new. We start at the beginning, at the rudiments of the theory of derivatives. 7.2 The Derivative Let f be a function defined on an interval I and let x0 and x be points of I. Consider the difference quotient determined by the points x0 and x: representing the average rate of change of f on the interval with endpoints at x and x0. f (x) x − − f (x0) x0 , (1) 396 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.2. The Derivative 397 f f(x) f(x0) x0 x Figure 7.1. The chord determined by (x, f (x)) and (x0, f (x0)). In Figure 7.1 this difference quotient represents the slope of the chord (or secant line) determined by the points (x, f (x)) and (x0, f (x0)). This same picture allows a physical interpretation. If f (x) represents the f (x0) distance a point moving on a straight line has moved from some fixed point in time x, then f (x) represents the (net) distance it has moved in the time interval [x0, x], and the difference quotient (1) represents the average velocity in that time interval. − Suppose now that we fix x0, and allow x to approach x0. We learn in elementary calculus that if exists, then the limit represents the slope of the tangent line to the graph of the function f at the point (x0, f (x0)). In the setting of motion, the limit represents instantaneous velocity at time x0. f (x) x lim x0 x → − − f (x0) x0 The derivative owes its origins to these two interpretations in geometry and in the physics of motion, but now completely transcends them; the derivative finds applications in nearly every part of mathematics and the sciences. We shall study the structure of derivatives, but with less concern for computations and applications than we would have seen in our calculus courses. Now we wish to understand the notion and see why it has the properties used in the many computations and applications of the calculus. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 398 Differentiation Chapter 7 7.2.1 Definition of the Derivative We begin with a familiar definition. Definition 7.1: Let f be defined on an interval I and let x0 ∈ f ′(x0), is defined as f (x0) x0 f (x) x x f ′(x0) = lim x0 → provided either that this limit exists or is infinite. If f ′(x0) is finite we say that f is differentiable at x0. If f is differentiable at every point of a set E I, we say that f is differentiable on E. When E is all of I, we ⊂ simply say that f is a differentiable function. − − (2) , I. The derivative of f at x0, denoted by Note. We have allowed infinite derivatives and they do play a role in many studies, but differentiable always refers to a finite derivative. Normally the phrase “a derivative exists” also means that that derivative is finite. Example 7.2: Let f (x) = x2 on R and let x0 ∈ x2 = x f (x) x f (x0) x0 − − R. If x R, x = x0, then ∈ (x = x2 0 x0 − − − x0)(x + x0) x0) (x − Since x = x0, the last expression equals x + x0, so establishing the formula, f ′(x0) = 2x0 for the function f (x) = x2. f (x) x lim x0 x → − − f (x0) x0 = lim x0 x → (x + x0) = 2x0, . ◭ Let us take a moment to clarify the definition when the interval I contains one or both of its endpoints. Suppose I = [a, b]. For x0 = a (or x0 = b), the limit in (2) is just a one-sided, or unilateral, limit. The function f is defined only on [a, b] so we cannot consider points outside of that interval. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 Section 7.2. The Derivative 399 This brings us to another point. It can happen that a function that is not differentiable at a point x0 x0 does satisfy the requirement of (2) from one side of x0. This means that the limit in (2) exists as x from that side. We present a formal definition. → Definition 7.3: Let f be defined on an interval I and let x0 ∈ denoted by f ′+(x0) is the limit I. The right-hand derivative of f at x0, x → provided that one-sided limit exists or is infinite. Similarly, the left-hand derivative of f at x0, f ′ − the limit x0+ (x0), is f ′+(x0) = lim f (x) x f (x0) x0 , f ′ − (x0) = lim x0 x → − f (x) x f (x0) x0 . − − − − Observe that, if x0 is an interior point of I, then f ′(x0) exists if and only if f ′+(x0) = f ′ − Exercise 7.2.8) (x0). (See Example 7.4: Let f (x) = quotient (1) becomes x | | on R. Let us consider the differentiability of f at x0 = 0. The difference Thus while f (x) x − − f (0) 0 = | x x | = 1, 1, − if x > 0 if x < 0. f ′+(0) = lim x00) = lim x0 |
The function has different right-hand and left-hand derivatives at x0 = 0 so is not differentiable at x0 = 0. ◭ f ′ − = − 1. → − x | x | x ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 400 Differentiation Chapter 7 Figure 7.2. A function trapped between x2 and x2. − Example 7.5: (A “trapping principle”) Let f be any function defined in a neighborhood I of zero. Suppose f satisfies the inequality I. Thus, the graph of f is “trapped” between the parabolas y = x2 and y = for all x f (0) = 0. The difference quotient computed for x0 = 0 becomes − ∈ x2 f (x) x2. In particular, | ≤ | from which we calculate so Thus f (x) x − − f (0) 0 = f (x) x , f (x) x f (x) x lim 0 x → x2 x ≤ = x | | ≤ lim 0 | x → = 0. x | f (x) x lim 0 x → = 0. As a result, f ′(0) = 0. Figure 7.2 illustrates the principle. ◭ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.2. The Derivative 401 Higher-Order Derivatives When a function f is differentiable on I, it is possible that its derivative f ′ is also differentiable. When this is the case, the function f ′′ = (f ′)′ is called the second derivative of the function f . Inductively, we can define derivatives of all orders: f (n+1) = (f (n))′ (provided f (n) is differentiable). When n is small, it is customary to use the convenient notation f ′′ for f (2), f ′′′ for f (3) etc. Notation It is useful to have other notations for the derivative of a function f . Common notations are dx and dy df dx (when the function is expressed in the form y = f (x)). Another notation that is useful is Df . These alternate notations along with slight variations are useful for various calculations. You are no doubt familiar with such uses—the convenience of writing dy dx = dy du du dx when using the chain rule, or viewing D as an operator in solving linear differential equations. Notation can be important at times. Consider, for example, how difficult it would be to perform a simple arithmetic calculation such as the multiplication (104)(90) using Roman numerals (CIV)(XC)! Exercises 7.2.1 You might be familiar with a slightly different formulation of the definition of derivative. If x0 is interior to I, then for h sufficiently small, the point x0 + h is also in I. Show that expression (2) then reduces to Repeat Examples 7.2 and 7.4 using this formulation of the derivative. See Note 161 f ′(x0) = lim h→0 f (x0 + h) h − f (x0) . 7.2.2 Let c R. Calculate the derivatives of the functions g(x) = c and k(x) = x directly from the definition of derivative. ∈ 7.2.3 Check the differentiability of each of the functions below at x0 = 0. (a) f (x) = x x | | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 402 Differentiation Chapter 7 (b) f (x) = x sin x−1 (f (0) = 0) (c) f (x) = x2 sin x−1 (f (0) = 0) x2, 0, if x rational if x irrational (d) f (x) = 7.2.4 Let f (x) = x2, ax, if x 0 ≥ if x < 0 (a) For which values of a is f differentiable at x = 0? (b) For which values of a is f continuous at x = 0? (c) When f is differentiable at x = 0, does f ′′(0) exist? 7.2.5 For what positive values of p is the function f (x) = x | 7.2.6 A function f has a symmetric derivative at a point if p differentiable at 0? | f ′ s(x) = lim h→0 f (x + h) f (x h) − − 2h exists. Show that f ′ is not differentiable at x. See Note 162 s(x) = f ′(x) at any point at which the latter exists but that f ′ s(x) may exist even when f 7.2.7 Find all points where f (x) = √1 See Note 163 − cos x is not differentiable and at those points find the one-sided derivatives. 7.2.8 Prove that if x0 is an interior point of an interval I, then f ′(x0) exists or is infinite if and only if +(x0) = f ′ f ′ 7.2.9 Let a function f : R −(x0). and elsewhere f (x) = 0. Find a condition on that sequence so that f ′(0) exists. → R be defined by setting f (1/n) = cn for n = 1, 2, 3, . . . where cn} { is a given sequence 7.2.10 Let a function f : R R be defined by setting f (1/n2) = cn for n = 1, 2, 3, . . . where sequence and elsewhere f (x) = 0. Find a condition on that sequence so that f ′(0) exists. → cn} { is a given ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.2. The Derivative 403 7.2.11 Give an example of a function with an infinite derivative at some point. Give an example of a function f with f ′ at some point x0. 7.2.12 If f ′(x0) > 0 for some point x0 in the interior of the domain of f show that there is a δ > 0 so that −(x0) = +(x0) = −∞ ∞ and f ′ f (x) < f (x0) < f (y) whenever x0 − See Note 164 δ < x < x0 < y < x0 + δ. Does this assert that f is increasing in the interval (x0 − δ, x0 + δ)? 7.2.13 Let f be increasing and differentiable on an interval. Does this imply that f ′(x) this imply that f ′(x) > 0 on that interval? See Note 165 0 on that interval? Does ≥ 7.2.14 Suppose that two functions f and g have the following properties at a point x0: f (x0) = g(x0) and ≤ g(x) for all x in an open interval containing the point x0. If both f ′(x0) and g′(x0) exist show that f (x) they must be equal. How does this compare to the trapping principle used in Example 7.5, where it seems much more is assumed about the function f . See Note 166 7.2.15 Suppose that f is a function defined on the real line with the property that f (x + y) = f (x)f (y) for all x, y. Suppose that f is differentiable at 0 and that f ′(0) = 1. Show that f must be differentiable everywhere and that f ′(x) = f (x). See Note 167 7.2.2 Differentiability and Continuity A continuous function need not be differentiable (Example 7.4) but the converse is true. Every differentiable function is continuous. Theorem 7.6: Let f be defined in a neighborhood I of x0. If f is differentiable at x0, then f is continuous at x0. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 404 Differentiation Chapter 7 Proof. It suffices to show that lim x0 → x (f (x) − f (x0)) = 0. For x = x0, f (x) − f (x0) = f (x) x − − f (x0) x0 (x − x0). Now and lim x0 x → (x − x0) = 0. We then obtain lim x0 x → f (x) x − − f (x0) x0 = f ′(x0) (f (x) lim x0 x → − f (x0)) = (f ′(x0))(0) = 0 by the product rule for limits. We can use this theorem in two ways. If we know that a function has a discontinuity at a point, then we know immediately that there is no derivative there. On the other hand, if we have been able to determine by some means that a function is differentiable at a point then we know automatically that the function must also be continuous at that point. Exercises 7.2.16 Construct a function on the interval [0, 1] that is continuous and is not differentiable at each point of some infinite set. See Note 168 7.2.17 Suppose that a function has both a right-hand and a left-hand derivative at a point. What, if anything, can you conclude about the continuity of that function at that point? 7.2.18 Suppose that a function has an infinite derivative at a point. What, if anything, can you conclude about the continuity of that function at that point? See Note 169 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 7.2. The Derivative 405 7.2.19 Show that if a function f has a symmetric derivative f ′ continuous at x0 in the sense that limh→0[f (x0 + h) See Note 170 s(x0) (see Exercise 7.2.6), then f must be symmetrically h)] = 0. Must f in fact be continuous? f (x0 − − 7.2.20 If f ′(x0) = 7.2.21 Find an example of an everywhere differentiable function f so that f ′ is not everywhere continuous. , does it follow that f must be continuous at x0 on one side at least? ∞ 7.2.22 Show that a function f that satisfies an inequality of the form | ≤ for some constant M and all x, y must be everywhere continuous but need not be everywhere differentiable. − − | f (x) | f (y) M y x | p 7.2.23 The Dirichlet function (see Section 5.2.6) is discontinuous at each rational number. By Theorem 7.6 it follows that this function has no derivative at any rational number. Does it have a derivative at any irrational number? 7.2.3 The Derivative as a Magnification " Enrichment section. May be omitted. We offer now one more interpretation of the derivative, this time as a magnification factor. In elementary calculus one often makes use of the geometric content of the graph of a function f . In particular, we can view the derivative in terms of slopes of tangent lines to the graph. But the graph of f is a subset of two-dimensional space, while the range of f is a subset of one-dimensional space and, as such, has some additional geometric content. Suppose f is differentiable on an interval I, and let J be a closed sub-interval of I. The range of f on J will also be a closed interval, because f is differentiable and hence continuous on J, and continuous functions map closed intervals onto closed intervals (Exercise 5.8.2). If we compare the length of the of the interval f (J) the expression interval J to the length f (J) J | | | | represents the amount that the interval J has been expanded (or contracted) under the mapping f . | f (J) J | | | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 406 Differentiation Chapter 7 For example, if f (x) = x2 and J = [2, 3], then | f (J) J | | | [4, 9] | [2, 3. Thus the interval [2, 3] has been expanded by f to an interval of 5 times its size. If we look only at small intervals then the derivative offers a clue to the size of the magnification factor. If J is a sufficiently small interval having x0 as an endpoint, then the ratio | , the approximation becoming “exact in the limit.” Thus f ′(x0) | factor” of small intervals containing the point x0. In our illustration with the function f (x) = x2, the magnification factor at x0 = 2 is f ′(2) = 4. Small intervals about x0 are magnified by a factor of about 4. At the other endpoint x0 = 3, small intervals about x0 are magnified by a factor of about 6. f ′(x0) | | f (J) is approximately can be viewed as a “ma |
gnification / | J | | | In Exercise 7.2.26 we ask you to prove a precise statement covering the preceding discussion. Exercises 7.2.24 What is the ratio f (J) | | J | | for the function f (x) = x2 if J = [2, 2.001], J = [2, 2.0001], J = [2, 2.00001]? 7.2.25 In this section we have interpreted f ′(x0) as a magnification factor. If f ′(x0) = 0, does this mean that small intervals containing the point x0 are magnified by a factor of 0 when mapped by f ? 7.2.26 Let f be differentiable on an interval I and let x0 be an interior point of I. Make precise the following statement and prove it: lim J→x0 f (J) | | J | | = f ′(x0) . | | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.3. Computations of Derivatives 407 7.3 Computations of Derivatives Example 7.2 provides a calculation of the derivative of the function f (x) = x2. The calculation involved direct evaluation of the limit of an appropriate difference quotient. For the function f (x) = x2, this evaluation was straightforward. But limits of difference quotients can be quite complicated. You are familiar with certain rules that are useful in calculating derivatives of functions that are “built up” from functions whose derivatives are known. In this section we review some of the calculus rules that are commonly used to compute derivatives. We need first to prove the algebraic rules: The sum rule, the product rule, and the quotient rule. Then we turn to the chain rule. Finally, we look at the power rule. Our viewpoint here is not to practice the computation of derivatives but to build up the theory of derivatives, making sure to see how it depends on work on limits that we proved earlier on. The various rules we shall obtain in this section should be viewed as aids for computations of derivatives. An understanding of these rules is, of course, necessary for various calculations. But they in no way can substitute for an understanding of the derivative. And they might not be useful in calculating certain derivatives. (For example, derivatives of the functions of Exercise 7.2.3 cannot be calculated at x0 = 0 by using these rules.) Nonetheless, it is true that one often has a function that can be expressed in terms of several functions via the operations we considered in this section, functions whose derivatives we know. In those cases, the techniques of this section might be useful. 7.3.1 Algebraic Rules Functions can be combined algebraically by multiplying by constants, by addition and subtraction, by multiplication, and by division. To each of these there is a calculus rule for computing the derivative. We recall that the limit of a sum (a difference, a product, a quotient) is the sum (difference, product, quotient) of the limits. Perhaps we might have thought the same kind of rule would apply to derivatives. The derivative of the sum is indeed the sum of the derivatives, but the derivative of the product is not ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 408 Differentiation Chapter 7 the product of the derivatives. Nor do quotients work in such a simple way. The reasons for the form of the various rules can be found by writing out the definition of the derivative and following through on the computations. Theorem 7.7: Let f and g be defined on an interval I and let x0 ∈ then f + g and f g are differentiable at x0. If g(x0) following formulas are valid: I. If f and g are differentiable at x0 = 0, then f /g is differentiable at x0. Furthermore, the (i) (cf )′(x0) = cf ′(x0) for any real number c. (ii) (f + g)′(x0) = f ′(x0) + g′(x0). (iii) (f g)′(x0) = f (x0)g′(x0) + g(x0)f ′(x0). (iv) f g ′ (x0) = g(x0)f ′(x0) f (x0)g′(x0) − (g(x0))2 (if g(x0) = 0). Proof. Parts (i) and (ii) follow easily from the definition of the derivative and appropriate limit theorems. To verify part (iii), let h = f g. Then for each x h(x) − h(x0) = f (x)[g(x) I we have g(x0)] + g(x0)[f (x) f (x0)] − ∈ − h(x) x − − h(x0) x0 = f (x) g(x) x g(x0) x0 + g(x0) f (x) x f (x0) x0 . (3) − − − − x0, f (x) f (x0) since f being differentiable is also continuous. By the definition of the derivative we also know that → → so As x and g(x) x f (x) x − − − − g(x0) x0 → f (x0) x0 → g′(x0) f ′(x0) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 Section 7.3. Computations of Derivatives 409 as x → x0. We now see from equation (3) that verifying part (iii). lim x0 x → h(x) x − − h(x0) x0 = f (x0)g′(x0) + g(x0)f ′(x0), Finally, to establish part (iv) of the theorem, let h = f /g. Straightforward algebraic manipulations show that h(x) x h(x0) x0 = − − f (x0) x0 1 g(x)g(x0) g(x0) f (x) x − − − x0. Since f and g are continuous at x0, f (x) f (x) g(x) x f (x0) and g(x) g(x0) x0 → − − Now let x the theorem follows from equation (4), the definition of derivative, and basic limit theorems. g(x0). Thus part (iv) of → → . (4) Example 7.8: To calculate the derivative of h(x) = (x3 + 1)2 we have several ways to proceed. 1. Apply the definition of derivative. You may wish to set up the difference quotient and see that a calculation of its limit is a formidable task. 2. Write h(x) = x6 + 2x3 + 1 and apply the formula d dx xn = nxn 1 (Exercise 7.3.5) and the rule for − sums. Thus we get 3. Use the product rule to obtain h′(x) = 6x5 + 6x2. Then, again, use the formula d dx xn = nxn − h′(x) = (x3 + 1) (x3 + 1) + (x3 + 1) d dx 1 and the rule for sums to continue: (x3 + 1). d dx h′(x) = (x3 + 1)3x2 + (x3 + 1)3x2 = 6x5 + 6x2. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 410 Differentiation Chapter 7 ◭ Exercises 7.3.1 Give the details needed in the proof of Theorem 7.7 for the sum rule for derivatives; that is, (f + g)′(x0) = f ′(x0) + g′(x0). 7.3.2 The table shown in Figure fig–table2 gives the values of two functions f and g at certain points. Calculate (f + g)′(1), (f g)′(1) and (f /g)′(1). What can you assert about (f /g)′(3)? Is there enough information to calculate f ′′(3)? x 1 2 3 4 5 f (x) 3 4 6 -1 2 ′ f (x) 3 4 1 0 5 g(x) 2 4 1 1 3 ′ g (x) 2 0 0 1 3 Figure 7.3. Values of f and g at several points. 7.3.3 Obtain the rule d dx 1 f (x) = f ′(x) f (x)2 − from Theorem 7.7 and also directly from the definition of the derivative. 7.3.4 Obtain the rule for d dx (f (x))2 = 2f (x)f ′(x) from Theorem 7.7 and also directly from the definition of the derivative. 7.3.5 Obtain the formula for n = 1, 2, 3, . . . by induction. d dx xn = nxn−1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.3. Computations of Derivatives 411 See Note 171 7.3.6 State and prove a theorem that gives a formula for f ′(x0) when f = f1 + f2 + and each of the functions f1, . . . , fn is differentiable at x0. 7.3.7 State and prove a theorem that gives a formula for f ′(x0) when · · · + fn and each of the functions f1, . . . , fn is differentiable at x0. f = f1f2 . . . fn 7.3.8 Show that (f g)′′(x0) = f ′′(x0)g(x0) + 2f ′(x0)g′(x0) + f (x0)g′′(x0) under appropriate hypotheses. 7.3.9 Extend Exercise 7.3.8 by obtaining a similar formula for (f g)′′′(x0). 7.3.10 Obtain a formula for (f g)(n)(x0) valid for n = 1, 2, 3, . . . . See Note 172 7.3.2 The Chain Rule There is another, nonalgebraic, interpretation of Example 7.8 that you may recall from calculus courses. Example 7.9: We can view the function h(x) = (x3 + 1)2 as a composition of the function f (x) = x3 + 1 and g(u) = u2. Thus You are familiar with the chain rule that is useful in calculating derivatives of composite functions. In this case the calculation would lead to h(x) = g f (x). ◦ h′(x) = g′(f (x))f ′(x) = g′(x3 + 1)3x2 = 2(x3 + 1)3x2 = 6x5 + 6x2. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 412 Differentiation Chapter 7 In elementary calculus you might have preferred to obtain dy dx = dy du du dx = 2(x3 + 1)(3x2) = 6x5 + 6x2 by making the substitution u = x3 + 1, y = u2. The chain rule is the familiar calculus formula d dx g(f (x)) = g′(f (x))f ′(x) ◭ for the differentiation of the composition of two functions g students often memorize this in the form dy dx by using the new variables y = g(u) and u = f (x). = dy du du dx f under appropriate assumptions. Calculus ◦ Let us first try to see why the chain rule should work. Then we’ll provide a precise statement and proof of the chain rule. Perhaps the easiest way to “see” the chain rule is by interpreting the derivative as a magnification factor. Let f be defined in a neighborhood of x0 and let g be defined in a neighborhood of f (x0). If f is differentiable at x0, then f maps each small interval J containing x0 onto an interval f (J) containing f (x0) with interval f (J) containing f (x0) onto an interval g(f (J)) with f maps J onto the interval h(J) = g(f (J)) and Thus h = g . If, also, g is differentiable at f (x0), then g will map a small | . f (J) g′(f (x0)) | approximately approximately g(f (J)) | f ′(x0) f (J(J) J | | | = | g(f (J)) | f (J) | f (J) J | | | | | and this is approximately equal to In short, the magnification factors f ′(x0) | | g′(f (x0)) f ′(x0) | and || g′(f (x0)) | | . | multiply to give the magnification factor h′(x0) . | | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.3. Computations of Derivatives 413 h(J) * g f (J) f J 1 2 3 4 5 6 Figure 7.4. f maps J to f (J) and g maps that to h(J). Here h = g f , x0 = 1, and J = [.9, 1.1]. ◦ Example 7.10: Let us relate this discussion to our example h(x) = (x3 + 1)2. Here f (x) = x3 + 1, g(x) = x2. At x0 = 1 we obtain f (x0) = 2, f ′(x0) = 3, g(f (x0)) = 4 and g′(f (x0)) = 4. The function f maps small intervals about x0 = 1 onto ones about three times as long, and in turn, the function g maps those intervals onto ones about four times as long, so the total magnification factor for the function ◭ h = g f is about 12 at x0 = 1 (Fig. 7.4). ◦ Proof of the Chain Rule disc |
ussion we could begin by writing If we wished to formulate a proof of the chain rule based on the preceding g(f (x)) x − − g(f (x0)) x0 = g(f (x)) f (x) − − g(f (x0)) f (x0) f (x) x − − f (x0) x0 (5) which compares to our formula | | = | h(J) J | | g(f (J)) | f (J) | f (J) J | . | | | | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 414 Differentiation Chapter 7 If we let x → x0 in (5), we would expect to get the desired result f )′(x0) = g′(f (x0))f ′(x0). (g ◦ And this argument would be valid if f were, for example, increasing. But in order for equation (5) to be = f (x0). When computing the limit of a difference quotient, we can valid, we must have x assume x = f (x0). Yet the chain rule applies nonetheless. = x0, but we can’t assume, without additional hypotheses, that if x = x0 then f (x) = x0 and f (x) The proof is clearer if we separate these two cases. In the simpler case the function does not repeat the value f (x0) in some neighborhood of x0. In the harder case the function repeats the value f (x0) in every neighborhood of x0. Exercise 7.3.11 shows that in that case we must have f ′(x0) = 0 and so the chain rule reduces to showing that the composite function g f also has a zero derivative. ◦ Theorem 7.11 (Chain Rule) Let f be defined on a neighborhood U of x0 and let g be defined on a neighborhood V of f (x0) for which ∈ Suppose f is differentiable at x0 and g is differentiable at f (x0). Then the composite function h = g differentiable at x0 and ⊂ f (x0) f (U ) V. f is ◦ h′(x0) = (g ◦ f )′(x0) = g′(f (x0))f ′(x0). Proof. Consider any sequence of distinct points xn different from x0 and converging to x0. If we can show that the sequence converges to g′(f (x0))f ′(x0) for every such sequence then we have obtained our required formula. Sn = g(f (xn)) g(f (x0)) x0 − xn − Note that if f (xn) = f (x0), then we can write yn = f (xn), y0 = f (x0) and display Sn as Sn = g(yn) − yn − g(y0) y0 f (xn) − xn − f (x0) x0 . (6) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 6 6 6 6 Section 7.3. Computations of Derivatives 415 Seen in this form it becomes obvious that Sn → g′(y0)f ′(x0) = g′(f (x0))f ′(x0) except for the problem that we cannot (as we remarked before beginning our proof) assume that in all cases f (xn) = f (x0). Thus we consider two cases. In the first case we assume that for any sequence of distinct points xn converging to x0 there cannot be infinitely many terms with f (xn) = f (x0). In that case the chain rule formula is evidently valid. In the second case we assume that there does exist a sequence of distinct points xn converging to x0 with f (xn) = f (x0) for infinitely many terms. In that case (Exercise 7.3.11) we must have f ′(x0) = 0 and so, to establish the chain rule, we need to prove that h′(x0) = 0. But in this case for any sequence xn converging to x0 either Sn = 0 [when f (xn) = f (x0)] or else Sn can be written in the form of equation (6) [when f (xn) 0 and the proof is complete. = f (x0)]. It is then clear that Sn → Exercises 7.3.11 Show that if for each neighborhood U of x0 there exists x f ′(x0) does not exist or else f ′(x0) = 0. See Note 173 U , x = x0 for which f (x) = f (x0), then either ∈ 7.3.12 Give an explicit example of functions f and g such that the “proof” of the chain rule based on equation (5) fails. See Note 174 7.3.13 The heuristic discussion preceding Theorem 7.11 dealt with of f ′(x0) and g′(f (x0)) affect the discussion. In particular, how can we modify the discussion to get the correct sign for h′(x0)? , not with h′(x0). Explain how the signs h′(x0) | | 7.3.14 Most calculus texts use a proof of Theorem 7.11 based on the following ideas. Define a function G in the neighborhood V of f (x0) by G(v) = g(f (x0))]/[v [g(v) − g′(f (x0)), f (x0)], − if v = f (x0) if v = f (x0). (7) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 6 6 416 Differentiation Chapter 7 (a) Show that G is continuous at f (x0). (b) Show that G(v)(v − (c) Prove that limx→x0 f (x0)) = g(v) h(x)−h(x0) x−x0 − = g′(f (x0))f ′(x0). g(f (x0)) for every v V , regardless of whether or not f (x0) = v. ∈ 7.3.15 State and prove a theorem that gives a formula for f ′(x0) when f2 ◦ fn−1 ◦ · · · ◦ f = fn ◦ (Be sure to state all the hypotheses that you need.) f1. g)′(1) 7.3.16 The table in Figure 7.3.16 gives the values of two functions f and g at certain points. Calculate (f ◦ d (f 2)(1) dx f )′(1). Is there enough information to calculate (f f )′(1)? f )′(3)? How about g)′(3) and/or (g and (f and (x) 3 4 6 -1 2 ′ f (x) 3 4 1 0 5 g(x) 2 4 1 1 3 ′ g (x) 2 0 0 1 3 Figure 7.5. Values of f and g at several points. 7.3.3 Inverse Functions Suppose that a function f : I J has an inverse. This simply means that there is a function g (called the inverse of f ) that reverses the mapping: If f (a) = b then g(b) = a. We can assume that I and J are intervals. Thus f maps the interval I onto the interval J and the inverse function g then maps J back to I. Not all functions have an inverse, but we are supposing that this one does. → Suppose too that f is differentiable at a point x0 ∈ I. Then we would expect from geometric considerations that that the inverse function g should be differentiable at the image point z0 = f (x0) ∈ J. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.3. Computations of Derivatives 417 This is entirely elementary. The connection between a function f and its inverse g is given by or f (g(x)) = x for all x g(f (x)) = x for all x Using the chain rule on the second of these immediately gives J ∈ I. ∈ and hence we have the connection g′(f (x))f ′(x) = 1 g′(f (x)) = 1 f ′(x) , which a geometrical argument could also have found. Example 7.12: Suppose that the exponential function ex has been developed and that we have proved that it is differentiable for all values of x and we have the usual formula d dx ex = ex. Then, provided we can be sure there is an inverse, a formula for the derivative of that inverse can be found. Let L(x) be the inverse function of f (x) = ex. Then, since we know that f ′(x) = f (x) or, replacing f (x) by another letter, say z, we have L′(f (x)) = 1 f ′(x) = 1 f (x) L′(z) = 1 z . This must be valid for every value z in the domain of L, that is, for every value in the range of f . You should recognize the derivative of the function ln z here. Even so, we would still need to justify the existence ◭ of the inverse function before we could properly claim to have proved this formula. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 418 Differentiation Chapter 7 We would like a better way to handle inverse functions than presented here. Our observations here allow us to compute the derivative of an inverse but do not assure us that an inverse will exist. For a theorem that allows us merely to look at the derivative and determine that an inverse exists and has a derivative, see Theorem 7.32. Exercises 7.3.17 Find a formula for the derivative of the function sin−1 x assuming that the usual formula for has been found. See Note 175 d dx sin x = cos x 7.3.18 Find a formula for the derivative of the function tan−1 x assuming that the usual formula for d dx tan x = sec2 x has been found. 7.3.19 Give a geometric interpretation of the relationship between the slope of the tangent at a point (x0, y0) on the graph of y = f (x) and the slope of the tangent at the point (y0, x0) on the graph of y = g(x) where g is the inverse of f . See Note 176 7.3.20 What facts about the function f (x) = ex would need to be established in order to claim that there is indeed an inverse function? What is the domain and range of that inverse function? 7.3.4 The Power Rule The power rule is the formula d dx which is the basis for many calculus problems. We have already shown (in Exercise 7.3.5) that xp = pxp − 1 d dx xn = nxn − 1 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.3. Computations of Derivatives 419 for n = 1, 2, 3, . . . and for every value of x. This is easy enough to extend to negative integers. Just interpret for n = 1, 2, 3, . . . and for every value of x = 0, d dx x− n = d dx 1 xn and, using the quotient rule, we find that again the power rule formula is valid for p = any value of x other than 0. 1, − 2, − − 3, . . . and The formula also works for p = 0 since we interpret x0 as the constant 1 (although for x = 0 we prefer not to make any claims). Is the formula indeed valid for every value of p, not just for integer values? Example 7.13: We can verify the power rule formula for p = 1/2; that is, we prove that d dx √x = d dx x1/2 = 1 2 x1/2 1 = − 1 2√x . First we must insist that x > 0 otherwise √x and the fraction in our formula would not be defined. Now interpret √x as the inverse of the square function f (x) = x2. Specifically let f (x) = x2 for x > 0 and g(x) = √x for x > 0 and note that f (g(x)) = g(f (x)) = x. Thus and so, since f ′(x) = 2x and f ′(g(x))g′(x) = 1 we obtain 2√xg′(x) = 1 and finally that d dx f (g(x)) = d dx x = 1 as required if the power rule formula is valid. Is the power rule g′(x) = 1 2√x d dx xp = pxp − 1 ◭ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 420 Differentiation Chapter 7 valid for all rational values of p? We can handle the case p = m/n for integer m and n by essentially the same methods. We state this as a theorem whose proof is left as an exercise. For irrational p there is also a discussion in the exercises. Theorem 7.14: Let f (x) = x m n for x > 0 and integers m, n. Then f ′(x) = m n − 1. x m n Example 7.15: Every polynomial is differentiable on R and its derivative can be calculated via term by term differentiation; that is, d dx (a0 + a1x + a2x2 + · · · + anxn) = a1 + 2a2x + + nanxn 1. − · · · This follows from the power rule formul |
a and the rule for sums. Note that the derivative of a polynomial ◭ is again a polynomial. Example 7.16: A rational function is a function R(x) that can be expressed as the quotient of two polynomials, R(x) = p(x) q(x) . This would be defined at every point at which the denominator q(x) is not equal to zero. Every rational function is differentiable except at those points at which the denominator vanishes. This follows from the previous example, which showed how to differentiate a polynomial, and from the quotient rule. Thus d dx p(x) q(x) = p′(x)q(x) p(x)q′(x) − q2(x) . Notice that the derivative is another rational function with the same domain since both numerator and denominator are again polynomials. ◭ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.4. Continuity of the Derivative? 421 Exercises 7.3.21 Prove Theorem 7.14. See Note 177 7.3.22 Show that the power formula is available for all values of p once the formula d dx ex = ex is known. See Note 178 7.3.23 Let p(x) = a0 + a1x + a2x2 + Compute the sequence of values p(0), p′(0), p′′(0), p′′′(0), . . . . See Note 179 + anxn. · · · 7.3.24 Determine the coefficients of the polynomial by using the formulas that you obtained in Exercise 7.3.23. See Note 180 p(x) = (1 + x)n = a0 + a1x + a2x2 + + anxn · · · 7.4 Continuity of the Derivative? We have already observed (Theorem 7.6) that if a function f is differentiable on an interval I, then f is also continuous on I. This statement should not be confused with the (incorrect) statement that the derivative, f ′, is continuous. Example 7.17: Consider the function f defined on R by 1, f (x) = x2 sin x− 0, if x = 0 if x = 0. sin x− Since that f ′(0) = 0. For x | ≤ | 1 for all x = 0, x2 for all x = 0, we can calculate, as in elementary calculus, that f (x) | ≤ ∈ | 1 R. We can now conclude (e.g., from Example 7.5) f ′(x) = − cos x− 1 + 2x sin x− 1. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 6 422 Differentiation Chapter 7 This function f ′ is continuous at every point x0 6 need only consider an appropriate sequence xn → sequence = 0. At x0 = 0 it is discontinuous. To see this we 0 and see what happens to f ′(xn). For example, try the xn = 1 πn . Since cos 1 xn = cos (πn) and these numbers are alternately +1 and discontinuous at 0. − 1 it is clear that f ′(xn) cannot converge. Consequently, f ′ is ◭ Observe that the function f provides an example of a function that is differentiable on all of R, yet f ′ is discontinuous at a point. It is possible to modify this function to obtain a differentiable function g whose derivative g′ is discontinuous at infinitely many points, and even at all the points of the Cantor set (see Exercise 7.4.2). You might wonder, then, if anything positive could be said about the properties of a derivative f ′. It is possible for the derivative of a differentiable function to be discontinuous on a dense set1: An example is given later in Section 9.7. We will also show, in Section 7.9, that the function f ′, while perhaps discontinuous, nonetheless shares one significant property of continuous functions: It has the intermediate value property (Darboux property). Exercises 7.4.1 Give a simple example of a function f differentiable in a deleted neighborhood of x0 such that limx→x0 f ′(x) does not exist. 7.4.2 " Let P be a Cantor subset of [0, 1] (i.e., P is a nonempty, nowhere dense perfect subset of [0, 1]) and let be the sequence of intervals complementary to P in (0, 1). (See Section 6.5.1.) (an, bn) } { 1 It is not possible for a derivative to be discontinuous at every point. See Corollary 9.40. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.5. Local Extrema 423 (a) On each interval [an, bn] construct a differentiable function such that n)+(an) = (f ′ n)−(bn) = 0, fn(an) = fn(bn) = (f ′ lim sup x→a+ n f ′(x) = lim sup f ′ n(x) = 1, x→b f ′(x) = lim inf − n x→b − n f ′ n(x) = 1, − lim inf x→a+ n f ′ n(x) | | is bounded by 1 in each interval [an, bn]. and fn(x) | − | ≤ (b) Let g be defined on [0, 1] by − (x an)2(x bn)2 and g(x) = Sketch a picture of the graph of g. fn(x), 0, if x if x ∈ ∈ (an, bn), n = 1, 2, . . . P . (c) Prove that g is differentiable on [0, 1]. (d) Prove that g′(x) = 0 for each x (e) Prove that g′ is discontinuous at every point of P . P . ∈ 7.5 Local Extrema We have seen in Section 5.7 that a continuous function defined on a closed interval [a, b] achieves an absolute maximum value and an absolute minimum value on the interval. These are called extreme values or extrema. There must be points where the maximum and minimum are attained, but how do we go about finding such points? One way is to find all points that may not be themselves extrema, but are local extreme points. A function defined on an interval I is said to have a local maximum at x0 in the interior f (x0) for all x in the smaller interval. A of I, if there exists δ > 0 such that [x0 − local minimum is similarly defined. I and f (x) δ, x0 + δ] ≤ ⊂ A familiar process studied in elementary calculus is sometimes useful for locating these extrema when the function is differentiable on (a, b): We look for critical points (i.e., points where the derivative is zero). We begin with the theorem that forms the basis for this process. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 424 Differentiation Chapter 7 Theorem 7.18: Let f be defined on an interval I. If f has a local extremum at a point x0 in the interior of I and f is differentiable at x0, then f ′(x0) = 0. Proof. similar. Then there exists δ > 0 such that Suppose f has a local maximum at x0 in the interior of I, the proof for a local minimum being for all x [x0 − ∈ δ, x0 + δ]. Thus and If f ′(x0) exists, then [x0 − δ, x0 + δ] ⊂ I and f (x) f (x0) ≤ f (x) x f (x) x − − − − f (x0) x0 ≤ f (x0) x0 ≥ 0 for x ∈ (x0, x0 + δ) 0 for x (x0 − ∈ δ, x0). f ′(x0) = lim x0+ x → f (x) x − − f (x0) x0 = lim x0 x → − f (x) x − − (8) (9) (10) f (x0) x0 . By (8), the first of these limits is at most zero; by (9), the second is at least zero. By (10), these limits are equal and are therefore equal to zero. It follows from Theorem 7.18 that a function f that is continuous on [a, b] must achieve its maximum at one (or more) of these types of points: 1. Points x0 ∈ 2. Points x0 ∈ 3. The points a or b (a, b) at which f ′(x0) = 0 (a, b) at which f is not differentiable We leave it to you to provide simple examples of each of these possibilities. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.5. Local Extrema 425 The usual process for locating extrema in elementary calculus thus involves locating points at which f has a zero derivative and comparing the values of f at those points and the points of nondifferentiability (if any) and at the endpoints a and b. In the setting of elementary calculus the situation is usually relatively simple: The function is differentiable, the set on which f ′(x) = 0 is finite (or contains an interval), and the equation f ′(x) = 0 is easily solved. Much more complicated situations can occur, of course. The following exercises provide some examples and theorems that indicate just how complicated the set of extrema can be. Exercises 7.5.1 Give an example of a differentiable function on R for which f ′(0) = 0 but 0 is not a local maximum or minimum of f . 7.5.2 Let f (x) = x4(2 + sin x−1), 0, if x = 0 if x = 0. (a) Prove that f is differentiable on R. (b) Prove that f has an absolute minimum at x = 0. (c) Prove that f ′ takes on both positive and negative values in every neighborhood of 0. 7.5.3 " Let K be the Cantor set and let be the sequence of intervals complementary to K in [0, 1]. For each k, let ck = (ak + bk)/2. Define f on [0, 1] to be zero on K, 1/k at ck, linear and continuous on each of the intervals. (See Figure 7.6.) (ak, bk) } { (a) Write equations that represent f on the intervals [ak, ck] and [ck, bk]. (b) Show that f is continuous on [0, 1]. (c) Verify that f has minimum zero, achieved at each x (d) Verify that f has a local maximum at each of the points ck. (e) Modify f to a differentiable function with the same set of extrema. K. ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 426 Differentiation Chapter 7 1 Figure 7.6. Part of the graph of the function in Exercise 7.5.3. 1 7.5.4 Find all local extrema of the Dirichlet function (see Section 5.2.6) defined on [0, 1] by 7.5.5 Show that the functions in Exercises 7.5.3 and 7.5.4 have infinitely many maxima, all of them strict. Show f (x) = 0, 1/q, if x is irrational or x = 0 if x = p/q, p, q IN, p/q in lowest terms. ∈ that the sets of points at which these functions have a strict maximum is countable. 7.5.6 Prove that if f : R See Note 181 → R, then { x : f achieves a strict maximum at x is countable. } 7.5.7 Let f : R at x. → R have the following property: For each x ∈ R, f achieves a local maximum (not necessarily strict) (a) Give an example of such an f whose range is infinite. (b) Prove that for every such f , the range is countable. See Note 182 7.5.8 There are continuous functions f : R → means that there is no interval on which the function is increasing, decreasing, or constant. For such functions, the set of maxima as well as the set of minima is dense in R. Construction of such functions is given later in Section 13.14.2. Show that such a function f maps its set of extrema onto a dense subset of the range of f . R, even differentiable functions, that are nowhere monotonic. This ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.6. Mean Value Theorem 7.6 Mean Value Theorem 427 There is a close connection between the values of a function and the values of its derivative. In one direction this is trivial since the derivative is defined in terms of the values |
of the function. The other direction is more subtle. How does information about the derivative provide us with information about the function? One of the keys to providing that information is the mean value theorem. Suppose f is continuous on an interval [a, b] and is differentiable on (a, b). Consider a point x in (a, b). For y ∈ (a, b), y = x, the difference quotient f (y) y − − f (x) x represents the slope of the chord determined by the points (x, f (x)) and (y, f (y)). This slope may or may not be a good approximation to f ′(x). If y is sufficiently near x, the approximation will be good; otherwise it may not be. The mean value theorem asserts that somewhere in the interval determined by x and y there will be a point at which the derivative is exactly the slope of the given chord. It is the existence of such a point that provides a connection between the values of the function [in this case the value (f (y) and y). x)] and the value of the derivative (in this case the value at some point between x f (x))/(y − − 7.6.1 Rolle’s Theorem We begin with a preliminary theorem that provides a special case of the mean value theorem. This derives its name from Michel Rolle (1652–1719) who has little claim to fame other than this. Indeed Rolle’s name was only attached to this theorem because he had published it in a book in 1691; the method itself he did not discover. Perhaps his greatest real contribution is the invention of the notation n√x for the nth root of x. Theorem 7.19 (Rolle’s Theorem) Let f be continuous on [a, b] and differentiable on (a, b). If f (a) = f (b) then there exists c (a, b) such that f ′(c) = 0. ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 428 Differentiation Chapter 7 f a c1 c2 b Figure 7.7. Rolle’s theorem [note that f (a) = f (b)]. Proof. If f is constant on [a, b], then f ′(x) = 0 for all x (a, b), so c can be taken to be any point of (a, b). Suppose then that f is not constant. Because f is continuous on the compact interval [a, b], f achieves a maximum value M and a minimum value m on [a, b] (Theorem 5.50). Because f is not constant, one of (a, b) such that f (c) = M . the values M or m is different from f (a) and f (b), say M > f (a). Choose c Since M > f (a) = f (b), c (a, b). By Theorem 7.18, f ′(c) = 0. = a and c = b, so c ∈ ∈ Observe that Rolle’s theorem asserts that under our hypotheses, there is a point at which the tangent to the graph of the function is horizontal, and therefore has the same slope as the chord determined by the points (a, f (a)) and (b, f (b)). (See Figure 7.7.) There may, of course, be many such points; Rolle’s theorem just guarantees the existence of at least one such point. Observe also that we did not require that f be differentiable at the endpoints a and b. The 1, f (0) = 0, on the interval [0, 1/π]. This function theorem applies to such functions as f (x) = x sin x− is not differentiable at zero, but it does have an infinite number of points between 0 and 1/π where the derivative is zero. ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 Section 7.6. Mean Value Theorem 429 Exercises 7.6.1 Apply Rolle’s theorem to the function f (x) = √1 the endpoints of the interval. x2 on [ − − 7.6.2 Use Rolle’s theorem to explain why the cubic equation 1, 1]. Observe that f fails to be differentiable at x3 + αx2 + β = 0 cannot have more than one solution whenever α > 0. 7.6.3 If the nth-degree equation p(x) = a0 + a1x + a2x2 + + anxn = 0 · · · has n distinct real roots, then how many distinct real roots does the (n See Note 183 1)st degree equation p′(x) = 0 have? − 7.6.4 Suppose that f ′(x) > c > 0 for all x 7.6.5 Suppose that f : R be some point ξ so that f ′′(ξ) = 0. See Note 184 → [0, ). Show that limx→∞ f (x) = ∈ ∞ . ∞ R and both f ′ and f ′′ exist everywhere. Show that if f has three zeros, then there must 7.6.6 Let f be continuous on an interval [a, b] and differentiable on (a, b) with a derivative that never is zero. Show that f maps [a, b] one-to-one onto some other interval. See Note 185 7.6.7 Let f be continuous on an interval [a, b] and twice differentiable on (a, b) with a second derivative that never is zero. Show that f maps [a, b] two-one onto some other interval; that is, there are at most two points in [a, b] mapping into any one value in the range of f . See Note 186 7.6.2 Mean Value Theorem If we drop the requirement in Rolle’s theorem that f (a) = f (b), we now obtain the result that there is a c (a, b) such that ∈ f ′(c) = f (b) b − − f (a) a . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 430 Differentiation Chapter 7 a c b Figure 7.8. Mean value theorem [f ′ (c) is slope of the chord]. Geometrically, this states that there exists a point c the function at (c, f (c)) is parallel to the chord determined by the points (a, f (a)) and (b, f (b)). (See Figure 7.8.) (a, b) for which the tangent to the graph of ∈ This is the mean value theorem, also known as the law of the mean or the first mean value theorem (because there are other mean value theorems). Theorem 7.20 (Mean Value Theorem) Suppose that f is a continuous function on the closed interval [a, b] and differentiable on (a, b). Then there exists c (a, b) such that f ′(c) = ∈ f (b) b f (a) a . − − Proof. We prove this theorem by subtracting from f a function whose graph is the straight line determined by the chord in question and then applying Rolle’s theorem. Let L(x) = f (a) + f (b) b − − f (a) a (x − a). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.6. Mean Value Theorem We see that L(a) = f (a) and L(b) = f (b). Now let g(x) = f (x) L(x). − 431 (11) Then g is continuous on [a, b], differentiable on (a, b), and satisfies the condition g(a) = g(b) = 0. By Rolle’s theorem, there exists c f ′(c) = L′(c). But ∈ (a, b) such that g′(c) = 0. Differentiating (11), we see that so as was to be proved. L′(c) = f ′(c) = f (b) b f (b) b − − − − f (a) a , f (a) a , Rolle’s theorem and the mean value theorem were easy to prove. The proofs relied on the geometric content of the theorems. We suggest that you take the time to understand the geometric interpretation of these theorems. Exercises 7.6.8 A function f is said to satisfy a Lipschitz condition on an interval [a, b] if for all x, y in the interval. Show that if f is assumed to be continuous on [a, b] and differentiable on (a, b) then this condition is equivalent to the derivative f ′ being bounded on (a, b). See Note 187 f (x) | − f (y.6.9 Suppose f satisfies the hypotheses of the mean value theorem on [a, b]. Let S be the set of all slopes of chords determined by pairs of points on the graph of f and let (a) Prove that S D. ⊂ f ′(x) : x D = { (a, b) } . ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 432 Differentiation Chapter 7 (b) Give an example to show that D can contain numbers not in S. See Note 188 7.6.10 Interpreting the slope of a chord as an average rate of change and the derivative as an instantaneous rate of change, what does the mean value theorem say? If a car travels 100 miles in 2 hours, and the position s(t) of the car at time t satisfies the hypotheses of the mean value theorem, can we be sure that there is at least one instant at which the velocity is 50 mph? 7.6.11 Give an example to show that the conclusion of the mean value theorem can fail if we drop the requirement that f be differentiable at every point in (a, b). Give an example to show that the conclusion can fail if we drop the requirement of continuity at the endpoints of the interval. 7.6.12 Suppose that f is differentiable on [0, Determine See Note 189 ∞ ) and that lim x→∞ f ′(x) = C. lim x→∞ [f (x + a) − f (x)]. 7.6.13 Suppose that f is continuous on [a, b] and differentiable on (a, b). If what can you conclude about the right-hand derivative of f at a? See Note 190 lim x→a+ f ′(x) = C 7.6.14 Suppose that f is continuous and that lim x→x0 f ′(x) exists. What can you conclude about the differentiability of f ? What can you conclude about the continuity of f ′? See Note 191 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.6. Mean Value Theorem 433 7.6.15 Let f : [0, ) ∞ → R so that f ′ is decreasing and positive. Show that the series is convergent if and only if f is bounded. See Note 192 f ′(i) ∞ i=1 X 7.6.16 Prove a second-order version of the mean value theorem. Let f be continuous on [a, b] and twice differentiable on (a, b). Then there exists c that (a, b) such ∈ f (b) = f (a) + (b a)f ′(a) + (b − − a)2 f ′′(c) 2! . See Note 193 7.6.17 Determine all functions f : R → for every x = y. R that have the property that x + y 2 f (x) x f ′ = f (y) y − − 7.6.18 A function is said to be smooth at a point x if f (x + h) + f (x h2 Show that a smooth function need not be continuous. Show that if f ′′ is continuous at x, then f is smooth at x. See Note 194 lim h→0 2f (x) = 0. h) − − 7.6.3 Cauchy’s Mean Value Theorem " Enrichment section. May be omitted. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 434 Differentiation Chapter 7 We can generalize the mean value theorem to curves given parametrically. Suppose f and g are continuous on [a, b] and differentiable on (a, b). Consider the curve given parametrically by As t varies over the interval [a, b], the point (x, y) traces out a curve C joining the points (g(a), f (a)) and (g(b), f (b)). If g(a) = g(b), the slope of the chord determined by these points is x = g(t) , y = f (t) [a, b]). (t ∈ − − Cauchy’s form of the mean value theorem asserts that there is a point (x, y) on C at which the tangent is parallel to the chord in question. We state and prove this theorem. f (b) g(b) f (a) g(a) . Theorem 7.21 (Cauchy Mean Value Theorem) Let f and g be continuous on [a, b] and differentiable on (a, b). Then ther |
e exists c (a, b) such that ∈ [f (b) − f (a)]g′(c) = [g(b) g(a)]f ′(c). − (12) Proof. Let φ(x) = [f (b) Then φ is continuous on [a, b] and differentiable on (a, b). Furthermore, f (a)]g(x) [g(b) g(a)]f (x). − − − By Rolle’s theorem, there exists c ∈ (a, b) for which φ′(c) = 0. It is clear that this point c satisfies (12). φ(a) = f (b)g(a) − f (a)g(b) = φ(b). Exercises 7.6.19 Use Cauchy’s mean value theorem to prove any simple version of L’Hˆopital’s rule that you can remember from calculus. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 7.7. Monotonicity 435 7.6.20 Show that the conclusion of Cauchy’s mean value can be put into determinant form as 7.6.21 Formulate and prove a generalized version of Cauchy’s mean value whose conclusion is the existence of a f (a) f (b) f ′(c) g(a) g(b) g′(c) = 0. 1 1 0 f (a) f (b) f ′(c) h(a) h(b) g(a) g(b) g′(c) h′(c) = 0. point c such that See Note 195 7.7 Monotonicity 0 on an interval I, then f is nondecreasing on I. We use this In elementary calculus one learns that if f ′ ≥ and related results for a variety of purposes: sketching graphs of functions, locating extrema, etc. In this section we take a closer look at what’s involved. We recall some definitions. Definition 7.22: Let f be real valued on an interval I. 1. If f (x1) ≤ f (x2) whenever x1 and x2 are points in I with x1 < x2, we say f is nondecreasing on I. 2. If the strict inequality f (x1) < f (x2) holds, we say f is increasing. A similar definition was given for nonincreasing and decreasing functions. Note. Some authors prefer the terms “increasing” and “strictly increasing” for what we would call nondecreasing and increasing. This has the unfortunate result that constant functions are then considered to be both increasing and decreasing. According to our definition we must say that they are both nondecreasing and nonincreasing, which sounds more plausible—if something stays constant it is neither going up nor going down). The disadvantage of our usage is the discomfort you may at first feel in using ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 436 Differentiation Chapter 7 the terms (which disappears with practice). It is always safe to say “strictly increasing” for increasing even though it is redundant according to the definition. By a monotonic function we mean a function that is increasing, decreasing, nondecreasing, or nonincreasing. The theorems involving monotonicity of functions that one encounters in elementary calculus usually are stated for differentiable functions. But a monotonic function need not be differentiable, or even continuous. Example 7.23: For example, if x, x + 1, then f is increasing on R, but is not continuous at x = 0. (For more on discontinuities of monotonic functions, see Section 5.9.2.) for x < 0 0, for x f (x) = ≥ ◭ Let us now address the role of the derivative in the study of monotonicity. We prove a familiar theorem that is the basis for many calculus applications. Note that the proof is an easy consequence of the mean value theorem. Theorem 7.24: Let f be differentiable on an interval I. 0 for all x I, then f is nondecreasing on I. (i) If f ′(x) ≥ (ii) If f ′(x) > 0 for all x ∈ I, then f is increasing on I. ∈ (iii) If f ′(x) 0 for all x ≤ (iv) If f ′(x) < 0 for all x I, then f is nonincreasing on I. I, then f is decreasing on I. ∈ ∈ I, then f is constant on I. (v) If f ′(x) = 0 for all x Proof. To prove (i), let x1, x2 ∈ such that ∈ I with x1 < x2. By the mean value theorem (7.20) there exists c (x1, x2) ∈ f (x2) f (x1) = f ′(c)(x2 − − x1). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.7. Monotonicity 437 If f ′(c) 0, then f (x2) 0 for all x Parts (ii), (iii) and (iv) have similar arguments, and (v) follows immediately from parts (i) and (iii). I, f is nondecreasing on I. f (x1). Thus, if f ′(x) ≥ ≥ ≥ ∈ Exercises 7.7.1 Establish the inequality ex See Note 196 1 1−x for all x < 1. ≤ 7.7.2 Suppose that f and g are differentiable functions such that f ′ = g and g′ = number C with the property that f . Show that there exists a − [f (x)]2 + [g(x)]2 = C for all x. 7.7.3 Suppose f is continuous on (a, c) and a < b < c. Suppose also that f is differentiable on (a, b) and on (b, c). Prove that if f ′ < 0 on (a, b) and f ′ > 0 on (b, c), then f has a minimum at b. See Note 197 7.7.4 The hypotheses of Theorem 7.24 require that f be differentiable on all of the interval I. You might think that a positive derivative at a single point also implies that the function is increasing, at least in a neighborhood of that point. This is not true. Consider the function f (x) = x/2 + x2 sin x−1, 0, if x = 0 if x = 0. (a) Show that the function g(x) = x2 sin x−1 (g(0) = 0) is everywhere differentiable and that g′(0) = 0. (b) Show that g′ is discontinuous at x = 0 and that g′ takes on values close to (c) Show that f ′ takes on both positive and negative values in every neighborhood of zero. (d) Show that f ′(0) = 1 2 > 0 but that f is not increasing in any neighborhood of zero. (e) Prove that if a function F is differentiable on a neighborhood of x0 with F ′(x0) > 0 and F ′ is continuous 1 arbitrarily near 0. ± at x0, then F is increasing on some neighborhood of x0. (f) Why does the example f (x) given here not contradict part (e)? ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 438 Differentiation Chapter 7 7.7.5 Let f be differentiable on [0, on [0, ∞ ). Show that ∞ ) and suppose that f (0) = 0 and that the derivative f ′ is an increasing function f (x) x < f (y) y for all 0 < x < y. See Note 198 7.7.6 Suppose that f , g : R → R and both have continuous derivatives and the determinant is never zero. Show that between any two zeros of f there must be a zero of g. See Note 199 φ(x) = f (x) f ′(x) g(x) g′(x) 7.8 Dini Derivates " Advanced section. May be omitted. We observed in Example 7.4 that the function f (x) = x = 0 but does have the one-sided derivatives f ′+(0) = 1 and f ′ − continuous functions that don’t have even one-sided derivatives at a point. (0) = − x | | does not have a derivative at the point 1. It is not difficult to construct Example 7.25: Consider the function f (x) = (See Figure 7.25). Since cos x− 1 | | ≤ 1 for all x cos x− 1 , | x | 0, if x = 0 if x = 0. = 0, so f is continuous at x = 0. It is clear that f is continuous at all other points in R, so f is a continuous function. f (x) = 0 = f (0) lim 0 x → ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 Section 7.8. Dini Derivates 439 Figure 7.9. Graph of f (x) = x cos x −1 | . | The oscillatory behavior of f is such that the sets x : cos x− 1 = 1 and x : cos x− 1 = 0 both have zero as a two-sided limit point. Thus each of the sets x : f (x) = { x : f (x) = 0 } x | |} and { has zero as two-sided limit point. Inspection of the difference quotient reveals that lim sup 0+ x → f (x) x f (0) 0 − − = 1, while lim inf 0+ x → f (x) x f (0) 0 − − = 0, so f ′+ does not exist at x = 0. Similarly, f ′ − have a derivative, or a one-sided derivative, don’t exist at x = 0. (0) does not exist. The limits that are required to exist for f to ◭ Example 7.26: A function defined on an interval I may fail to have a derivative, even a one-sided derivative, at every point. Let g(x) = 0, 1, if x is rational, if x is irrational. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 440 Differentiation Chapter 7 Since g is everywhere discontinuous on both sides, g has no derivative and no one-sided derivative at any ◭ point. There are, also, continuous functions that fail to have a one-sided derivative, finite or infinite, at even a single point. Such functions are difficult to construct, the first construction having been given by Besicovitch in 1925. Now the derivative, when it exists, plays an important role in analysis, and it is useful to have a substitute when it doesn’t exist. Many good substitutes have been developed for certain situations. Perhaps the simplest such substitutes are the Dini derivates. These exist at every point for every function defined on an open interval. They are named after the Italian mathematician Ulisse Dini (1845–1918). Definition 7.27: Let f be real valued in a neighborhood of x0. We define the four Dini derivates of f at x0 by 1. [Upper right Dini derivate] 2. [Lower right Dini derivate] 3. [Upper left Dini derivate] 4. [Lower left Dini derivate] D+f (x0) = lim sup x0+ x → D+f (x0) = lim inf x0+ x → D−f (x0) = lim sup − x0 → x f (x) x − − f (x0) x0 f (x) x − − f (x0) x0 f (x) x − − f (x0) x0 D − f (x0) = lim inf − x0 → x f (x) x − − f (x0) x0 . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.8. Dini Derivates Example 7.28: For the function f (x) = cos x− 1 x | | , f (0) = 0, we calculate that D+f (0) = 1, D+f (0) = 0, D−f (0) = 0, D f (0) = − 1. − Elsewhere f ′(x) exists and all four Dini derivatives have that value. Example 7.29: The function has at every rational x g(x) = 0, 1, if x is rational, if x is irrational. For x irrational there are similar values for the Dini derivates (see Exercise 7.8.1a). D+g(x) = 0, D+g(x) = , D−g(x) = , D − ∞ g(x) = 0. −∞ 441 ◭ ◭ It is easy to check that a function f has a derivative at a point x0 if and only if all four Dini derivates are equal at that point, and a one-sided derivative at x0 if the two Dini derivates from that side are equal (see Exercise 7.8.2). We end this section with an illustration of the way in which knowledge about a Dini derivate can substitute for that of the ordinary derivative. We prove a theorem about monotonicity. You are familiar with the fact that if f is differentiable on an interval [a, b] and f ′(x) > 0 for all x [a, b], then f is an increasing function on [a, b]. (We provided a formal proof in Section 7.7.) ∈ Here is a generalization |
of that theorem. Theorem 7.30: Let f be continuous on [a, b]. If D+f (x) > 0 at each point x on [a, b]. ∈ [a, b), then f is increasing Proof. Let us first show that f is nondecreasing on [a, b]. We prove this by contradiction. If f fails to be nondecreasing on [a, b], there exist points c and d such that a b and f (c) > f (d). Let y be any point in the interval (f (d), f (c)). c < d ≤ ≤ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 442 Differentiation Chapter 7 Since f is continuous on [a, b], it possesses the intermediate value property. Thus from Theorem 5.53 [or (c, d) more precisely from the version of that theorem given as Exercise 5.8.8(a)] there exists a point t such that f (t) = y. Thus the set ∈ is nonempty. Let x : f (x) = y) [c, d] } ∩ { x0 = sup x : c { x ≤ ≤ d and f (x) = y . } Now, f (d) < y and f is continuous, from which it follows that x0 < d. Thus f (x) < y for x Furthermore, the set is closed (because f is continuous), so f (x0) = y. 0.This contradicts our hypothesis that D+f (x) > 0 for all x x : f (x) = y But this implies that D+f (x0) ∈ { This contradiction completes the proof that f is nondecreasing. } ≤ (x0, d]. [a, b). ∈ Now we wish to show that it is in fact increasing. If not, then there must be some subinterval in which the function is constant. But at every point interior to that interval we would have f ′(x) = 0 and so it would be impossible for D+f (x) > 0 at such points. Exercises 7.8.1 Calculate the four Dini derivates for each of the following functions at the given point. (a) for x = π. g(x) = 1, 0, if x is rational if x is irrational (b) h(x) = x sin x−1 (h(0) = 0) at x = 0 (c) f (x) = x sin x−1 (f (0) = 5) at x = 0 (d) at x = 0 and at x = 1 u(x) = x2, 0, if x is rational if x is irrational ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.8. Dini Derivates 443 7.8.2 Prove that f has a derivative at x0 if and only if D+f (x0) = D+f (x0) = D−f (x0) = D−f (x0). In that case, f ′(x0) is the common value of the Dini derivates at x0. (We assume that f is defined in a neighborhood of x0.) 7.8.3 (Derived Numbers) The Dini derivates are sometimes called “extreme unilateral derived numbers.” Let with limk→∞ xk = x0 ]. Then λ is a derived number for f at x0 if there exists a sequence xk} { λ , [ −∞ such that ∈ ∞ λ = lim k→∞ f (xk) − xk − f (x0) x0 . (a) For the function f (x) = 1, 1] is a derived number for f at x = 0. Show that the two extreme derived numbers from the right are 0 and 1, and the two from the left are , f (0) = 0, show that every number in the interval [ − x cos x−1 1 and 0. (b) Show that a function has a derivative at a point if and only if all derived numbers at that point coincide. (c) Let f : R R. Prove that if f is continuous on R, then the set of derived numbers of f at x0 consists of either one or two closed intervals (that might be degenerate or unbounded). Give examples to illustrate the various possibilities. R and let x0 ∈ → − 7.8.4 Let f, g : R R. → (a) Prove that D+(f + g)(x) (b) Give an example to illustrate that the inequality in (a) can be strict. (c) State and prove the analogue of part (a) for the lower right derivate D+f . D+f (x) + D+g(x). ≤ 7.8.5 Generalize Theorem 7.18 to the following: If f achieves a local maximum at x0, then D+f (x0) 0 and D−f (x0) 0. ≥ ≤ Illustrate the result with a function that is not differentiable at x0. 7.8.6 Prove a variant of Theorem 7.30 that assumes that, for all x in [a, b) except for x in some countable set, the Dini derivate D+f (x) > 0. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 444 Differentiation Chapter 7 7.8.7 Prove a variant of Theorem 7.30: If f is continuous and D+f (x) on [a, b]. See Note 200 0 for all x ∈ ≥ [a, b), then f is nondecreasing 7.8.8 Prove yet another (more subtle) variant of Theorem 7.30: If f is continuous and D+f (x) > 0 for all x except for x in some countable set, then f is increasing on [a, b]. [a, b) ∈ 7.8.9 Prove that no continuous function can have D+f (x) = f : R → See Note 201 R such that D+f (x) = for all x R. ∈ ∞ for all x ∈ ∞ R. Give an example of a function 7.8.10 Show that the set x : D+f (x) < D−f (x) cannot be uncountable. Give an example of a function f such that D+f < D−f on an infinite set. See Note 202 7.9 The Darboux Property of the Derivative Suppose f is differentiable on an interval [a, b]. We argued in the proof of Rolle’s theorem (7.19) that if (a, b) at which f achieves an extremum. At this point c we have f (a) = f (b), then there exists a point c f ′(c) = 0. ∈ A different hypothesis can lead to the same conclusion. Suppose f is differentiable on [a, b] and f ′(a) < 0 < f ′(b) (or f ′(b) < 0 < f ′(a)). Once again, the extreme value f achieves must occur at a point c in the interior of [a, b], (why?), and at this point we must have f ′(c) = 0. This observation is a special case of the following theorem first proved by Darboux in 1875. Theorem 7.31: Let f be differentiable on an interval I. Suppose a, b be any number between f ′(a) and f ′(b). Then there exists c ∈ (a, b) such that f ′(c) = γ. I, a < b, and f ′(a) = f ′(b). Let γ ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 7.9. The Darboux Property of the Derivative 445 Proof. Let g(x) = f (x) The discussion preceding the statement of the theorem shows that there exists c For this c we have γx. If f ′(a) < γ < f ′(b), then g′(a) = f ′(a) − − ∈ γ < 0 and g′(b) = f ′(b) γ > 0. (a, b) such that g′(c) = 0. − f ′(c) = g′(c) + γ = γ, completing the proof for the case f ′(a) < f ′(b). The proof when f ′(a) > f ′(b) is similar. You might have noted that Theorem 7.31 is exactly the statement that the derivative of a differentiable function has the Darboux property (i.e., the intermediate value property) that we established for continuous functions in Section 5.8. The derivative f ′ of a differentiable function f need not be continuous, of course. The result does imply, however, that f ′ cannot have jump discontinuities and cannot have removable discontinuities. Both the mean value theorem and Theorem 7.31 give information about the range of the derivative f ′ of a differentiable function f . The mean value theorem implies that the range of f ′ includes all slopes of chords determined by the graph of f on the interval of definition of f . Theorem 7.31 tells us that this range is actually an interval. This interval may be unbounded and, if bounded, may or may not contain its endpoints. (See Exercise 7.9.1.) Derivative of an Inverse Function Theorem 7.31 allows us to establish a familiar theorem about differentiating inverse functions. Theorem 7.32: Suppose f is differentiable on an interval I and for each x I, f ′(x) = 0. Then ∈ (i) f is one-to-one on I, (ii) f − 1 is differentiable on J = f (I), (iii) (f − 1)′(f (x)) = 1 f ′(x) for all x I. ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 446 Proof. By Theorem 7.31 either f ′(x) > 0 for all x increasing or decreasing on I, and is thus one-to-one, establishing (i). I or f ′(x) < 0 for all x ∈ Differentiation Chapter 7 I. In either case, f is either ∈ To verify (ii) and (iii), observe first that f − (see Exercise 5.9.16). Let y0 ∈ value 1/(f ′(x0)). For x J and let x0 = f − I, write y = f (x), so x = f − 1(y). Consider the difference quotient ∈ 1 is continuous, since f is continuous and strictly monotonic 1)′(y0) exists and has 1(y0). We wish to show that (f − − − 1 is continuous. Thus x0, because the function f − − − = f − 1(y) y 1(y0) f − y0 x f (x) x0 f (x0) . lim y0 y → f − 1(y) y f − y0 − − 1(y0) = lim x0 x → 1 f (x0) x0 − − f (x) x = 1 f ′(x0) . As y y0, x → → Exercises 7.9.1 Let f be differentiable on [a, b] and let (f ′) can be R (f ′) denote the range of f ′ on [a, b]. Give examples to illustrate that R (a) a closed interval (b) an open interval (c) a half-open interval (d) an unbounded interval See Note 203 7.9.2 Give an example of a differentiable function f such that f ′(x0) = lim x→x0 f ′(x). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 7.9. The Darboux Property of the Derivative 447 Show that if f is defined and continuous in a neighborhood of x0 and if the limit lim x→x0 f ′(x) exists and is finite, then f is differentiable at x0 and f ′ is continuous at x0. 7.9.3 Most classes of functions we have encountered are closed under the operations of addition and multiplication (e.g., polynomials, continuous functions, differentiable functions). The class of derivatives is closed under addition, but behaves badly with respect to multiplication. Consider, for example, the pair of functions F and G defined on R by F (x) = x2 sin G(x) = x2 cos 1 x3 , (F (0) = 0), and 1 x3 , (G(0) = 0). Verify each of the following statements: (a) F and G are differentiable on R. (b) The functions F G′ and GF ′ are bounded functions. (c) F (x)G′(x) F ′(x)G(x) = − 3, 0, if x = 0 if x = 0. (d) At least one of the functions F G′ or GF ′ must fail to be a derivative. Thus, even the product of a differentiable function F with a derivative G′ need not be a derivative. See Note 204 7.9.4 Show, in contrast to Exercise 7.9.3, that if a function f has a continuous derivative on R and g is differentiable, then f g′ is a derivative. See Note 205 7.9.5 Let f be a differentiable function on an interval [a, b]. Show that f ′ is continuous if and only if the set Eα = x : f ′(x) = α { } is closed for each real number α. See Note 206 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 448 Differentiation Chapter 7 f Figure 7.10. Concave up/down/up. R be a continuous function that is differentiable on (0, 1) and with f (0) = 0 and f (1) = 1. 7.9.6 Let f : [0, 1] → Show there must exist distinct numbers ξ1 and ξ2 in that interval such that 7.9.7 Prove or disprove that if f : R → R |
is differentiable and monotonic, then f ′ must be continuous on R. f ′(ξ1)f ′(ξ2) = 1. 7.10 Convexity In elementary calculus one studies functions that are concave-up or concave-down on an interval. A knowledge of the intervals on which a function is concave-up or concave-down is useful for such purposes as sketching the graph of the equation y = f (x) and studying extrema of the function (Fig. 7.10). In the setting of elementary calculus the functions usually have second derivatives on the intervals 0 on I, and involved. In that setting we define a function as being concave-up on an interval I if f ′′ ≥ 0 on I. Definitions involving the first derivative, but not the second, can also be concave-down if f ′′ ≤ given: f is concave-up on I if f ′ is increasing on I, concave-down if f ′ is decreasing on I. Equivalently, f is concave-up if the graph of f lies “above” (more precisely “not below”) each of its tangent lines, concave-down if the graph lies below (not above) each of its tangent lines. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.10. Convexity 449 The geometric properties we wish to capture when we say a function is concave-up or concave-down do not depend on differentiability properties. The condition is that the graph should lie below (or above) all its chords. The following definitions make this concept precise. We shall follow the common practice of using the terms “convex” and “concave” in place of the terms “concave-up” and “concave-down.” Definition 7.33: Let f be defined on an interval I. If for all x1, x2 ∈ f (αx1 + (1 αf (x1) + (1 α)x2) − ≤ α)f (x2) − I and α ∈ [0, 1] the inequality (13) is satisfied, we say that f is convex on I. If the reverse inequality in (13) applies, we say that f is concave on I. If the inequalities are strict for all α (0, 1) we say f is strictly convex or strictly concave on I. ∈ For example, the function f (x) = x | that the graph of f has no line segments in it. Note that the function f (x) = x = 0. is convex, but not strictly convex on R. Strict convexity implies is not differentiable at x | | | The geometric condition defining convexity does imply a great deal of regularity of a function. Our first objective is to address this issue. We begin with some simple geometric considerations. Suppose f is convex on an open interval I. Let x1 and x2 be points in I with x1 < x2. The chord determined by the points (x1, f (x1)) and (x2, f (x2)) defines a linear function M on [x1, x2]: If x = αx1 + (1 α)x2, then − M (x) = αf (x1) + (1 α)f (x2). − The definition of “convex” states that for all x ∈ [x1, x2] and that Now let z ∈ (x1, x2) Then f (x) ≤ M (x) M (x1) = f (x1) and M (x2) = f (x2). f (z) z − − f (x1) x1 M (z) z ≤ M (x1) x1 = − − M (x2) − x2 − M (z) z ≤ f (x2) − x2 − f (z) z (14) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 450 Differentiation Chapter 7 f M M(z) f(z) x1 z x2 Figure 7.11. Comparison of the three slopes in the inequalities (14). (Fig. 7.11). Thus, the chord determined by f and the points x1 and x2 has a slope between the slopes of the chord determined by x1 and z and the chord determined by z and x2. The inequalities (14) have a number of useful consequences: 1. For fixed x I, ∈ (f (x + h) f (x))/h − is a nondecreasing function of h on some interval (0, δ). Thus f (x + h) h − f (x) = inf h>0 f (x + h) h − f (x) lim 0+ h → exists or possibly is lower bound, since −∞ . That it is in fact finite can be shown by using (14) again to get a finite (f (x′) f (x))/(x′ x) (f (x + h) f (x))/h − − ≤ − I with x′ < x. Thus f has a right-hand derivative f ′+(x) at x. Similarly, f has a finite for any x′ ∈ left-hand derivative at x. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.10. Convexity 451 2. If x, y ∈ I and x < y, then From observation 1 we infer that f ′+(x) ≤ f ′+(y). (f (x + h) f (x))/h (f (y + h) f (y))/h − ≤ − whenever h > 0 and x + h, y + h are in I. Thus f ′+ is a nondecreasing function. Similarly, f ′ − nondecreasing function. is a 3. It is also clear from (14) that for all x I. ∈ (x) f ′ − ≤ f ′+(x) 4. f is continuous on I. To see this, observe that since both one-sided derivatives exist at every point the function must be continuous on both sides, hence continuous. We summarize the preceding discussion as a theorem. Theorem 7.34: Let f be convex on an open interval I. Then (i) f has finite left and right derivatives at each point of I. Each of these one-sided derivatives is a nondecreasing function of x on I, and (x) f ′ − ≤ f ′+(x) for all x I. ∈ (15) (ii) f is continuous on I. Note. If f is convex on a closed interval [a, b], some of the results do not apply at the endpoints a and b. (See Exercise 7.10.8.) Note, too, that the corresponding results are valid for concave functions on I, the one-sided derivatives now being nonincreasing functions of x and the inequality in (15) being reversed. We can now obtain the characterizations of convex functions familiar from elementary calculus. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 452 Differentiation Chapter 7 Corollary 7.35: Let f be defined on an open interval I. (i) If f is differentiable on I, then f is convex on I if and only if f ′ is nondecreasing on I. (ii) If f is twice differentiable on I, then f is convex on I if and only if f ′′ ≥ 0 on I. We leave the verification of Corollary 7.35 as Exercise 7.10.9. Exercises 7.10.1 Show that a function f is convex on an interval I if and only if the determinant f (x1) f (x2) f (x3) is nonnegative for any choices of x1 < x2 < x3 in the interval I. 1 x1 1 x2 1 x3 7.10.2 If f and g are convex on an interval I, show that any linear combination αf + βg is also convex provided α and β are nonnegative. 7.10.3 If f and g are convex functions, can you conclude that the composition g See Note 207 f is also convex? ◦ 7.10.4 Let f be convex on an open interval (a, b). Show that then there are only two possibilities. Either (i) f is nonincreasing or nondecreasing on the entire interval (a, b) or else (ii) there is a number c so that f is nonincreasing on (a, c] and nondecreasing on [c, b). 7.10.5 Suppose f is convex on an open interval I. Prove that f is differentiable except on a countable set. See Note 208 7.10.6 Suppose f is convex on an open interval I. Prove that if f is differentiable on I, then f ′ is continuous on I. 7.10.7 Let f be convex on an open interval that contains the closed interval [a, b]. Let M = max +(a), f ′ f ′ −(b) . } { ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.10. Convexity 453 Show that for all x, y ∈ [a, b]. f (x) | − f (y.10.8 Theorem 7.34 pertains to functions that are convex on an open interval. Discuss the extent to which the results of the theorem hold when f is convex on a closed interval [a, b]. In particular, determine whether continuity of f at the endpoints of the interval follows from the definition. Must f ′ +(a) and f ′ −(b) be finite? 7.10.9 Prove Corollary 7.35. 7.10.10 Let f be convex on an open interval (a, b). Must f be bounded above? Must f be bounded below? See Note 209 7.10.11 Let f be convex on an open interval (a, b). Show that f does not have a strict maximum value. 7.10.12 Let f be defined and continuous on an open interval (a, b). Show that f is convex there if and only if there do not exist real numbers α and β such that the function f (x) + αx + β has a strict maximum value in (a, b). 7.10.13 " Let A = a1, a2, a3, . . . { } be any countable set of real numbers. Let 1 X Prove that f is convex on R, differentiable on the set R See Note 210 \ f (x) = ∞ x | ak| − 10k . A, and nondifferentiable on the set A. 7.10.14 " (Inflection Points) In elementary calculus one studies inflection points. The definitions one finds try to capture the idea that at such a point the sense of concavity changes from strict “up to down” or vice versa. Here are three common definitions that apply to differentiable functions. In each case f is defined on an open interval (a, b) containing the point x0. The point x0 is an inflection point for f if there exists an open interval I (a, b) such that on I ⊂ (Definition A) f ′ increases on one side of x0 and decreases on the other side. (Definition B) f ′ attains a strict maximum or minimum at x0. (Definition C) The tangent line to the graph of f at (x0, f (x0)) lies below the graph of f on one side of x0 and above on the other side. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 454 Differentiation Chapter 7 (a) Prove that if f satisfies Definition A at x0, then it satisfies Definition B at x0. (b) Prove that if f satisfies Definition B at x0, then it satisfies Definition C at x0. (c) Give an example of a function satisfying Definition B at x0, but not satisfying Definition A. (d) Give an example of an infinitely differentiable function satisfying Definition C at x0, but not satisfying Definition B. (e) Which of the three definitions states that the sense of concavity of f is “up” on one side of x0 and “down” on the other? See Note 211 7.10.15 (Jensen’s Inequality) Let f be a convex function on an interval I, let x1, x2, . . . , xn be points of I and let α1, α2, . . . , αn be positive numbers satisfying Show that See Note 212 αk = 1. n Xk=1 n n f αkxk Xk=1 ! ≤ Xk=1 αkf (xk) . 7.10.16 Show that the inequality is strict in Jensen’s inequality (Exercise 7.10.15) except in the case that f is linear on some interval that contains the points x1, x2, . . . , xn. 7.11 L’Hˆopital’s Rule " Enrichment section. May be omitted. Suppose that f and g are defined in a deleted neighborhood of x0 and that f (x) = A and lim x0 x → g(x) = B. lim x0 x → ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.11. L’Hˆopital’s Rule 455 According to our usual theory of limits, we then have f (x) g(x) = lim x0 |
x → limx limx → → x0 f (x) x0 g(x) = A B , unless B = 0. But what happens if B = 0, which is often the case? A number of possibilities exist: If B = 0 and = 0, then the limit does not exist. The most interesting case remains: If both A and B are zero, then A the limiting behavior depends on the rates at which f (x) and g(x) approach zero. Example 7.36: Consider 6x 3x Look at this simple example geometrically. For x straight line y = 6x approaches zero at twice the rate that the line y = 3x does. = lim x 0 → lim 0 x → = 2. 6 3 = 0, the height 6x is twice that of the height 3x. The ◭ Example 7.37: Now consider the slightly more complicated limit f (x) g(x) lim 0 x → = lim 0 x → 6x + x2 3x + 5x3 . If we divide the numerator and denominator by x = 0, we see that the limit is the same as lim 0 x → This last limit can be calculated by our usual elementary methods as equaling 6/3 = 2. Here, for x = 0 near zero, the height f (x) = 6x + x2 is approximately 6x, while the height of g(x) = 3x + 5x3 is approximately 3x, that is, the desired ratio is approximately 2. Again, the numerator approaches zero at about twice the rate that the denominator does. 6 + x 3 + 5x2 . We can be more precise by calculating these rates exactly. Let f (x) = 6x + x2 and g(x) = 3x + 5x3. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 6 6 6 456 Differentiation Chapter 7 3.0 1.5 f g 6x x3 1.0 Figure 7.12. Comparison of the rates in Example 7.37. Then f ′(x) = 6 + 2x, f ′(0) = 6 g′(x) = 3 + 5x2, g′(0) = 3. This makes precise our statement that the numerator approaches zero twice as fast as the denominator does. (See Figure 7.12 where there is an illustration showing the graphs of the functions f and g compared ◭ to the lines y = 6x and y = 3x.) Let us try to generalize from these two examples. Suppose f and g are differentiable in a neighborhood of x = a and that f (a) = g(a) = 0. Consider the following calculations and what conditions on f and g are required to make them valid. f (x) g(x) = f (x) g(x) f (a) g(a) = − − − − f (x) x g(x) x f (a) a g(a) a x a → −→ f ′(a) g′(a) = lim a x → f ′(x) g′(x) − − If these calculations are valid, they show that under these assumptions (f (a) = g(a) = 0 and both f ′(a) . (16) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.11. L’Hˆopital’s Rule 457 and g′(a) exist) we should be able to claim that f (x) g(x) lim a x → = lim a x → f ′(x) g′(x) . You should check the various conditions that must be met to justify the calculations: g(x) cannot equal zero at any point of the neighborhood in question (other than a); nor can g(x) = g(a), (for x = a); f (a) and g(a) must equal zero (for the first equality), and f ′/g′ must be continuous at x = a (for the last equality). The calculations (16) provide a simple proof of a rudimentary form for a method of computing limits known as L’Hˆopital’s rule. We say “rudimentary” because some of the conditions we assumed are not needed for the conclusion 7.11.1 L’Hˆopital’s Rule: 0 " Enrichment section. May be omitted. 0 Form lim a x → f (x) g(x) = lim a x → f ′(x) g′(x) . Our first theorem provides a version of the rule identical with our introductory remarks but under weaker assumptions. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 458 Differentiation Chapter 7 Theorem 7.38 (L’Hˆopital’s Rule: 0 deleted neighborhood N of x = a. Suppose 0 Form) Suppose that the functions f and g are differentiable in a (i) limx → (ii) limx a f (x) = 0, a g(x) = 0, → (iii) For every x ∈ f ′(x) g′(x) exists. (iv) limx a → N , g′(x) = 0, and Then lim a → x f (x) g(x) = lim a x → f ′(x) g′(x) . Proof. Our hypotheses do not require f and g to be defined at x = a. But we can in any case define (or redefine) f and g at x = a by f (a) = g(a) = 0. Because of assumptions (i) and (ii), this results in continuous functions defined on the full neighborhood N of the point x = a. We can now apply Cauchy’s form of the mean value theorem (7.21). a } ∪ { Suppose x ∈ N and a < x. By Theorem 7.21 there exists c = cx in (a, x) such that [f (x) − f (a)]g′(cx) = [g(x) g(a)]f ′(cx). − Since f (a) = g(a) = 0, (17) becomes f (x)g′(cx) = g(x)f ′(cx). Equation (18) is valid for x > a in N . We would like to express (18) in the form f (x) g(x) = f ′(cx) g′(cx) . (17) (18) (19) To justify (19) we show that g(x) is never zero in N . (That g′(cx) is never zero in N is N , x > a, we have g(x) = 0, then by Rolle’s theorem there would x : x > a } ∩ { our hypothesis (iii).) If for some x ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 7.11. L’Hˆopital’s Rule 459 exist a point 19) holds. (a, x) such that g′(t) = 0, contradicting hypothesis (ii). Thus equation (19) is valid for all ∈ (x, a) such that . A similar argument shows that if x } N , x < a, then there exists cx ∈ ∈ Now as x → a, cx also approaches a, since cx is between a and x. Thus f (x) g(x) lim a x → = lim a x → f ′(cx) g′(cx) = lim a x → f ′(x) g′(x) , since the last limit exists by hypothesis (iv). Note. Observe that we did not require f to be defined at x = a, nor did we require that f ′/g′ be continuous at x = a. It is also important to observe that L’Hˆopital’s rule does not imply that, under hypotheses (i), (ii), and (iii) of Theorem 7.38, if limx a f ′(x)/g′(x) must also exist. Exercise 7.11.5 provides an example to illustrate this. a f (x)/g(x) exists, then limx → → Example 7.39: Let us use L’Hˆopital’s rule to evaluate Let f (x) = ln(1 + x), g(x) = x. Then ln(1 + x)/x. lim 0 x → Thus lim 0 x → f (x) = lim 0 → x g(x) = 0, f ′(x) = 1 1 + x , and g′(x) = 1. ln(1 + x) x lim 0 x → = lim 0 x → 1 1 + x = 1. ◭ We refer to this theorem as the “ 0 0 form” for obvious reasons. There is also a version of the form ∞ ∞ (see Theorem 7.42). In addition, other modifications are possible. The point a can be replaced with a = or a = , (Theorem 7.41), and the results are valid for one-sided limits. (Our proof of Theorem 7.38 ∞ −∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 460 Differentiation Chapter 7 actually established that fact since we considered the case x > a and x < a separately.) Various other “indeterminate forms,” ones for which the limit depends on the rates at which component parts approach their separate limits, can be manipulated to make use of L’Hˆopital’s rule possible. Here is an example in which the forms “1∞” and “1−∞” come into play. Observe that the function whose limit we wish to calculate is of the form f (x)g(x) where f (x) and g(x) as x a . → −∞ → − 1 as x → → a but g(x) → ∞ as x a+ → Example 7.40: Evaluate limx 0(1 + x)2/x, write limx → → 0(1 + x)2/x. This expression is of the form 1∞ (when x > 0). To calculate y = (1 + x)2/x, z = ln y = 2 x ln(1 + x). Now the numerator and denominator of the function z satisfy the hypotheses of L’Hˆopital’s rule. Thus Since limx → 0 z = 2, limx → z = lim 0 x → 2 ln(1 + x) x 2 1 + x = 2. = lim 0 x → lim 0 x → 0 y = e2. 7.11.2 L’Hˆopital’s Rule as x " Enrichment section. May be omitted. → ∞ We proved Theorem 7.38 under the assumption that a . In this case we are, of course, dealing with one-sided limits. As before, the relation R, but the theorem is valid when a = ∈ a = + ∞ ◭ or −∞ f ′(x) g′(x) lim x →∞ = L implies something about relative rates of growth of the functions f (x) and g(x) as x base a proof of the versions of L’Hˆopital’s rule that have a = transformation. −∞ (or ∞ ) on Theorem 7.38 by a simple . We can → ∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.11. L’Hˆopital’s Rule 461 Theorem 7.41: Let f, g be differentiable on some interval ( , b). Suppose −∞ (i) limx →−∞ f (x) = 0, (ii) limx →−∞ g(x) = 0, (iii) For every x ( , b), g′(x) = 0, and −∞ ∈ f ′(x) g′(x) exists. (iv) limx →−∞ Then A similar result holds when we replace x lim →−∞ by f (x) g(x) = lim x →−∞ f ′(x) g′(x) . in the hypotheses. ∞ −∞ Proof. Let x = 1/t. Then, as t 0+, x → − F (t) = f → −∞ 1 t − and vice-versa. Define functions F and G by and G(t) = g 1 t . − Both functions F and G are defined on some interval (0, δ). We verify easily that lim 0+ → t F (t) = lim 0+ t → G(t) = 0 and that Using Theorem 7.38, we infer F ′(t) G′(t) lim 0+ → t = lim x →−∞ f ′(x) g′(x) . F ′(t) G′(t) lim t 0+ → = lim 0+ t → F (t) G(t) = lim 0( − − = lim x →−∞ f (x) g(x) . (20) (21) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 462 The result follows from (20) and (21) Differentiation Chapter 7 " Enrich. 7.11.3 L’Hˆopital’s Rule: ∞ ∞ Form When f (x) takes the form given in Theorem 7.42. Note, however, that we don’t require f (x) or even that f (x) approaches any limit. a we obtain the indeterminate form ∞ ∞ and g(x) → ∞ → ∞ as x → . L’Hˆopital’s theorem then in our hypotheses, → ∞ Theorem 7.42: Let f and g be differentiable on a deleted neighborhood N of x = a. Suppose that (i) limx a g(x) = → (ii) For every x ∈ . ∞ N g′(x) = 0. a f ′(x)/g′(x) exists. (iii) limx → Then The analogous statements are valid if a = f (x) = lim g(x) a x → or if limx → . f ′(x) g′(x) a g(x) = lim a x → ±∞ . −∞ Proof. We prove the main part of Theorem 7.42 under the assumption that f ′(x)/g′(x) lim a x → is finite. The case that the limit is infinite as well as variants are left as Exercises 7.11.6 and 7.11.7. It suffices to consider the case of right-hand limits, the proof for left-hand limits being similar. Let We will show that if p < L < q, then there exists δ > 0 such that L = lim a+ x → f ′(x)/g′(x). p < f (x)/g(x) < q ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 7.11. L’Hˆopital’s Rule 463 for x ∈ (a, a + δ). Since p and q are arbitrary (subject to the restriction p < L < q), we can then conclude f (x)/g(x) = L lim a+ x → as required. Choose r (L, q). By (iii) and the definition of L |
there exists δ1 such that f ′(x)/g′(x) < r whenever (a, a + δ1). If a < x < y < a + δ1, then we infer from Theorem 7.21, Cauchy’s form of the mean value ∈ x theorem, and our assumption (ii) that there exists c ∈ (x, y) such that ∈ f (x) g(x) − − f (y) g(y) = f ′(c) g′(c) < r. (22) Fix y in (22). Since limx a+ g(x) = , there exists δ2 > 0 such that a + δ2 < y and such that → g(x) > g(y) and g(x) > 0 if a < x < a + δ2. We then have ∞ (g(x) − g(y))/g(x) > 0 for x ∈ (a, a + δ2), so we can multiply both sides of the inequality (22) by (g(x) g(y))/g(x), obtaining − f (x) g(x) < r r g(y) g(x) − + f (y) g(x) for x ∈ (a, a + δ2). (23) Now let x a+. Then g(x) a+ by assumption (i). Since r, g(y), and f (y) are constants, the second and third terms on the right side of (23) approach zero. It now follows from the inequality r < q that there exists δ3 ∈ (0, δ2) such that → ∞ as x → → f (x) g(x) < q whenever a < x < a + δ3. (24) In a similar fashion we find a δ4 > 0 such that f (x) g(x) > p whenever a < x < a + δ4. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 464 Differentiation Chapter 7 If we let δ = min(δ3, δ4), we have shown that p < f (x) g(x) < q whenever x (a, a + δ). ∈ Since p and q were arbitrary numbers satisfying p < L < q, our conclusion f (x) g(x) lim a+ x → = L = lim a+ x → f ′(x) g′(x) follows. Exercises 7.11.1 Consider the function f (x) = (3x 2x)/x defined everywhere except at x = 0. − (a) What value should be assigned to f (0) in order that f be everywhere continuous? (b) Does f ′(0) exist if this value is assigned to f (0)? (c) Would it be correct to calculate f ′(0) by computing instead f ′(x) by the usual rules of the calculus and finding limx→0 f ′(x). See Note 213 7.11.2 Suppose that f and g are defined in a deleted neighborhood of x0 and that lim x→x0 f (x) = A = 0 and lim x→x0 g(x) = 0. Show that See Note 214 lim x→x0 f (x) g(x) = . ∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 7.11. L’Hˆopital’s Rule 465 7.11.3 Discuss the limiting behavior as x (a) (c) 1 x 1 sin x → 0 for each of the following functions. (b) (d) 1 x2 1 x sin x−1 7.11.4 Evaluate each of the following limits. (a) lim x→0 (b) lim t→0 (c) lim u→1 ex cos x − x t sin t t3 − u5 + 5u 2u5 + 8u 6 10 − − 7.11.5 Let f (x) = x2 sin x−1, g(x) = x. Show that but that does not exist. lim x→0 f (x) g(x) = 0 f ′(x) g′(x) lim x→0 7.11.6 The proof we provided for Theorem 7.42 required that limx→a f ′(x)/g′(x) be finite. Prove that the result holds if this limit is infinite. 7.11.7 Prove the part of Theorem 7.42 dealing with a = or limx→a g(x) = . −∞ ±∞ 7.11.8 Evaluate the following limits. (a) lim x→∞ (b) lim x→∞ x3 ex ln x x ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 466 Differentiation Chapter 7 (c) (d) lim x→0+ lim x→0+ x ln x xx 7.11.9 This exercise gives information about the relative rates of increase of certain types of functions. Prove that for each positive number p, lim x→∞ ln x xp = lim x→∞ xp ex = 0. 7.11.10 Give an example of functions f and g defined on R such that lim x→∞ g(x) = , lim sup x→∞ ∞ f (x) = , lim inf x→∞ ∞ f (x) = −∞ and Theorem 7.42 applies. See Note 215 7.12 Taylor Polynomials " Enrichment section. May be omitted. Suppose f is continuous on an open interval I and c f closely when x is sufficiently close to the point c, but may or may not provide a good approximation elsewhere. If f is differentiable on I, then we see from the mean value theorem (Theorem 7.20) that for each x = c) there exists z between x and c such that I (x I. The constant function g(x) = f (c) approximates ∈ ∈ f (x) = f (c) + f ′(z)(x c). − f (c) provides the size of the error obtained in approximating The expression R0(x) = f ′(z)(x the function f by a constant function P0(x) = f (c). We can think of this as approximation by a zero-degree polynomial. c) = f (x) − − We do not expect a constant function to be a good approximation to a given continuous function in general. But our acquaintance with Taylor series (as presented in elementary calculus courses) suggests that if a function is sufficiently differentiable, it can be approximated well by polynomials of sufficiently high degree. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 Section 7.12. Taylor Polynomials 467 Suppose we wish to approximate f by a polynomial Pn of degree n. In order for the polynomial Pn to have a chance to approximate f well in a neighborhood of a point c, we should require Pn(c) = f (c), P ′n(c) = f ′(c), . . . , P (n) n (c) = f (n)(c). In that case we at least guarantee that Pn “starts out” with the correct value, the correct rate of change, etc. to give it a chance to approximate f well in some neighborhood I of c. The test however is this. Write Is it true that the “error” or “remainder” Rn(x) is small when x f (x) = Pn(x) + Rn(x). I? ∈ In order to answer this sort of question, it would be useful to have workable forms for this error term Rn(x). We present two forms for the remainder. The first is due to Joseph-Louis Lagrange (1736–1813), who obtained Theorem 7.43 in 1797. He used integration methods to prove the theorem. We provide a popular and more modern proof based on the mean value theorem. Theorem 7.43 (Lagrange) Let f possess at least n + 1 derivatives on an open interval I and let c Let I. ∈ Pn(x) = f (c) + f ′(c)(x c) + − f ′′(c) 2! (x − c)2 + + · · · f (n)(c) n! c)n (x − and let Rn(x) = f (x) − Pn(x). Then for each x Rn(x) = I there exists z between x and c (z = c if x = c) such that ∈ f (n+1)(z) (n + 1)! c)n+1. (x − Proof. Fix x ∈ I. Then there is a number M (depending on x, of course) such that We wish to show that M = (f (n+1)(z))/(n + 1)! for some z between x and c. f (x) = Pn(x) + M (x c)n+1. − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 468 Differentiation Chapter 7 Consider the function g defined on I by g(t) = f (t) − = Rn(t) Pn(t) M (t − c)n+1 M (t − c)n+1. − − Now Pn is a polynomial of degree at most n, so P (n+1) n g(n+1)(t) = f (n+1)(t) − (t) = 0 for all t ∈ (n + 1)!M for all t Also, since f (k)(c) = P (k) n (c) for k = 1, 2, . . . , n, we readily see that g(k)(c) = 0 for k = 0, 1, 2, . . . , n. I. Thus I. ∈ (25) (26) Suppose now that x > c, the case x < c having a similar proof, and the case x = c being obvious. We have chosen M in such a way that g(x) = 0 and, by (26), we see that g(c) = 0. Thus g satisfies the (c, x) such that hypotheses of Rolle’s theorem on the interval [c, x]. Therefore there exists a point z1 ∈ g′(z1) = 0. Now apply Rolle’s theorem to g′ on the interval [c, z1], obtaining a point z2 ∈ Continuing in this way we use (26) and Rolle’s theorem repeatedly to obtain a point zn ∈ (c, z1) such that g′′(z2) = 0. 1) such that g(n)(zn) = 0. Finally, we apply Rolle’s theorem to the function g(n) on the interval [c, zn]. We obtain a point z (c, zn) such that g(n+1)(z) = 0. From (25) we deduce (c, zn − ∈ f (n+1)(z) = (n + 1)!M, completing the proof. Note. The function Pn is called the nth Taylor polynomial for f . You will recognize Pn as the nth partial sum of the Taylor series studied in elementary calculus. (See also Chapter 10.) The function Rn is called the remainder or error function between f and Pn. If Pn is to be a good approximation to f , then Rn must be small in absolute value. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.12. Taylor Polynomials 469 Observe that Pn(c) = f (c) and that n (c) = f (k)(c) for k = 0, 1, 2, . . . , n. P (k) Observe also that the mean value theorem is the special case of Theorem 7.43 obtained by taking n = 0: on the interval [c, x] there is a point z with f (x) − f (c) = f ′(z)(x c). − Lagrange’s result expresses the error term Rn in a particular way. It provides a sense of the error in approximating f by Pn. Note that we do not get an exact statement of the error term since it is given in terms of the value f (n+1)(z) at some point z. But if we know a little bit about the function f (n+1) on the interval in question, we might be able to say that this error is not very large. Example 7.44: Suppose we wish to approximate the function f (x) = sin x on the interval [ Taylor polynomial of degree 3, with c = 0. Here − a, a] by a f ′(x) = cos x , f ′′(x) = sin x , f ′′′(x) = − − cos x and f (4)(x) = sin x. Thus P3(x) = cos(0)x R3(x) = sin z 4! x2 sin(0) 2! x4 for some z in [ − − a, a]. − cos(0) 3! x3 = x x3 3! − and The exact error depends on which z makes this all true. But since that | sin z | ≤ 1 for all z, we get immediately so P3 approximates f to within a4/24 on the interval [ be sufficient for the purposes at hand. For large a, a higher-degree polynomial can produce the desired a, a]. For a small, the approximation should − R3(x) | | ≤ a4/4! = a4/24, ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 470 accuracy, since Rn(x) | | ≤ xn+1 | | (n + 1)! . Differentiation Chapter 7 ◭ Various other forms for the error term Rn are useful. The integral form is one of them. We state this form without proof. We assume that you are familiar with the integral as studied in calculus courses. Theorem 7.45 (Integral Form of Remainder) Suppose that the function f possesses at least n + 1 derivatives on an open interval I and that f (n+1) is Riemann integrable on every closed interval contained in I. Let c I. Then ∈ Rn(x) = x 1 n! c Z f (n+1)(t)(x − t)n dt for all x I. ∈ We shall see this form of the remainder again in Chapter 10 when we study Taylor series. Exercises 7.12.1 Exhibit the Taylor polynomial about x = 0 of degree n for the function f (x) = ex. Find n so that Rn(x) | | ≤ .0001 for all x [0, 2]. ∈ 7.12.2 Show that if f is a polynomial of degree n, then it is its own Taylor polynomial of degree n with c = 0. 7.12.3 Calculate the Taylor polynomial of degree 5 with c = 1 for th |
e functions f (x) = x5 and g(x) = ln x. 7.12.4 Let f (x) = 1 1, and n = 2. Show that x+2 , c = − where, for some z between x and 1 x + 2 1, − = 1 − (x + 1) + (x + 1)2 + R3 R3 = (x + 1)3 (2 + z)4 . − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.13. Challenging Problems for Chapter 7 471 7.12.5 Let f (x) = ln(1 + x), c = 0, and (x > 1). Show that − f (x) = x 1 2 x2 + 1 3 x3 + + ( − · · · 1)n−1 xn n + Rn − where Rn = 1)n ( − n + 1 n+1 x 1 + z for some z between 0 and x. Estimate Rn on the interval [0, 1/10]. 7.12.6 Just because a function possesses derivatives of all orders on an interval I does not guarantee that some Taylor polynomial approximates f in a neighborhood of some point of I. Let f (x) = e− 1 x2 , 0, if x = 0 if x = 0. (a) Show that f has derivatives of all orders and that f (k)(0) = 0 for each k = 0, 1, 2, . . . . (b) Write down the polynomial Pn with c = 0. (c) Write down Lagrange’s form for the remainder of order n. Observe its magnitude and take the time to understand why Pn is not a good approximation for f on any interval I, no matter how large n is. 7.13 Challenging Problems for Chapter 7 7.13.1 (Straddled derivatives) Let f : R → R. Prove that f is differentiable at x0 if and only if R and let x0 ∈ lim u→x0−, v→x0+ f (v) v f (u) u − − 7.13.2 (Unstraddled Derivatives) Let f : R exists (finite), and, in this case, f ′(x0) equals this limit. R and let x0 ∈ lim u→x0, v→x0, u6=v → f (v) v f (u) u − − exists. R. We say f is strongly differentiable at x0 if (a) Show that a differentiable function need not be strongly differentiable everywhere. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 472 Differentiation Chapter 7 (b) Show that a strongly differentiable function must be differentiable. (c) If f is strongly differentiable at a point x0 and differentiable in a neighborhood of x0, show that f ′ must be continuous there. 7.13.3 Let p be a polynomial of the nth degree that is everywhere nonnegative. Show that for all x. See Note 216 p(x) + p′(x) + p′′(x) + + p(n)(x) 0 ≥ · · · 7.13.4 Suppose that f is continuous on [0, 1], differentiable on (0, 1), and f (0) = 0 and f (1) = 1. For every integer n show that there must exist n distinct points ξ1, ξ2, . . . , ξn in that interval so that Xk=1 7.13.5 Show that there exists precisely one real number α with the property that for every function f differentiable on [0, 1] and satisfying f (0) = 0 and f (1) = 1 there exists a number ξ in (0, 1) (which depends, in general, on f ) so that n 1 f ′(ξk) = n. f ′(ξ) = αξ. 7.13.6 Let f be a continuous function. Show that the set of points where f is differentiable but not strongly differentiable (as defined in Exercise 7.13.2) is of the first category. 7.13.7 Let f be a continuous function on an open interval I. Show that f is convex on I if and only if See Note 217 7.13.8 (Wronskians) The Wronskian of two differentiable functions f and g is the determinant x + y 2 f ≤ f (x) + f (y) 2 . Prove that if W (f, g) does not vanish on an interval I and f (x1) = f (x2) = 0 for points x1 < x2 in I, then (x1, x2) such that g(x3) = 0. [The functions f (x) = sin x, g(x) = cos x furnish an example.] there exists x3 ∈ W (f, g) = f (x) f ′(x) g(x) g′(x) . ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 7.13. Challenging Problems for Chapter 7 473 See Note 218 7.13.9 " Let f be a continuous function on an open interval I. Show that f is convex if and only if lim sup h→0 f (x + h) + f (x h2 − h) − 2f (x) 0 ≥ for every x See Note 219 ∈ I. 7.13.10 " Let f be continuous on an interval (a, b). (a) Prove that the four Dini derivates of f and the difference quotient f (y)−f (x) y−x same bounds. (x = y ∈ (a, b)) have the (b) Prove that if one of the Dini derivates is continuous at a point x0, then f is differentiable at x0. (c) Show by example that the statements in the first two parts can fail for discontinuous functions. 7.13.11 " (Denjoy-Young-Saks Theorem) The theorem with this name is a far-reaching theorem relating the four Dini derivates D+f , D+f , D−f and D−f . It was proved independently by an English mathematician, Grace Chisolm Young (1868–1944), and a French mathematician, Arnaud Denjoy (1884–1974), for continuous functions in 1916 and 1915 respectively. Young then extended the result to a larger class of functions called measurable functions. Finally, the Polish mathematician Stanislaw Saks (1897–1942) proved the theorem for all real-valued functions in 1924. Here is their theorem. Theorem (Denjoy-Young-Saks) Let f be an arbitrary finite function defined on [a, b]. Then except for a set of measure zero every point x (1) A1 on which f has a finite derivative. (2) A2 on which D+f = D−f (finite), D−f = (3) A3 on which D−f = D+f (finite), D+f = (4) A4 on which D−f = D+f = [a, b] is in one of four sets: and D−f = D+f = . −∞ . −∞ ∞ ∞ ∈ and D+f = and D−f = . −∞ ∞ (a) Sketch a picture illustrating points in the sets A2, A3 and A4. To which set does x = 0 belong when sin x−1, f (0) = 0? f (x) = x | p | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008)6 474 Differentiation Chapter 7 (b) Use the Denjoy-Young-Saks theorem to prove that an increasing function f has a finite derivative except on a set of measure zero. (c) Use the Denjoy-Young-Saks theorem to show that if all derived numbers of f are finite except on a set of measure zero, then f is differentiable except on a set of measure zero. (d) Use the Denjoy-Young-Saks theorem to show that, for every finite function f , the set has measure zero. x : f ′(x) = { ∞} 7.13.12 Let f be a continuous function on an interval [a, b] with a second derivative at all points in (a, b). Let a < x < b. Show that there exists a point ξ f (x)−f (a) ∈ (a, b) so that f (b)−f (a) b−a x−a − − x b = 1 2 f ′′(ξ). See Note 220 7.13.13 Let f : R R be a differentiable function with f (0) = 0 and suppose that → Show that f is identically zero. See Note 221 f ′(x) | f (x) | for all x R. ∈ | ≤ | 7.13.14 Let f : R → R have a third derivative that exists at all points. Suppose that exists and that Show that See Note 222 lim x→∞ f (x) lim x→∞ f ′′′(x) = 0. lim x→∞ f ′(x) = lim x→∞ f ′′(x) = 0. 7.13.15 Let f be defined on an interval I of length at least 2 and suppose that f ′′ exists there. If f ′′(x) | See Note 223 | ≤ 1 for all x I show that ∈ f ′(x) | | ≤ 2 on the interval. f (x) | | ≤ 1 and ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 7.13.16 Let f : R → R be infinitely differentiable and suppose that n2 n2 + 1 1 n = f for all n = 1, 2, 3, . . . . Determine the values of 475 f ′(0), f ′′(0), f ′′′(0), f (4)(0), . . . . R have a third derivative that exists at all points. Show that there must exist at least one point f (ξ)f ′(ξ)f ′′(ξ)f ′′′(ξ) 0. ≥ See Note 224 7.13.17 Let f : R → ξ for which See Note 225 Notes 161Exercise 7.2.1. Write x = x0 + h. 162Exercise 7.2.6. Write f (x + h) f (x − − f (x)] + [f (x) = [f (x + h) − h) − f (x h)]. − 163Exercise 7.2.7. Use 1 cos x = 2 sin2 x/2. − When you take the square root be sure to use the absolute value. 164Exercise 7.2.12. Just use the definition of the derivative. Give a counterexample with f (0) = 0 and f ′(0) > 0 but so that f is not increasing in any interval containing 0. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 476 NOTES 165Exercise 7.2.13. Even for polynomials, p(x) increasing does not imply that p′(x) > 0 for all x. For example, take p(x) = x3. That has only one point where the derivative is not positive. Can you do any better? 166Exercise 7.2.14. Actually the assumptions are different. Here we assume f ′(x0) does exist, whereas in the trapping principle we had to assume more inequalities to deduce that it exists. 167Exercise 7.2.15. Review Exercise 5.10.3 first. 168Exercise 7.2.16. Advanced (very advanced) methods would allow you to find a function continuous on [0, 1] that is differentiable at no point of that interval. For the purpose of this exercise just try to find one that is not differentiable at 1/2, 1/3, 1/4, . . . . (Novices constructing examples often feel they need to give a simple formula for functions. Here, for example, you can define the function on [1/2, 1], then on [1/4, 1/2], then on [1/8, 1/4], and so on . . . and then finally at 0.) 169Exercise 7.2.18. Find two examples of functions, one continuous and one discontinuous at 0, with an infinite derivative there. 170Exercise 7.2.19. Imitate the proof of Theorem 7.6. Find a counterexample to the question. 171Exercise 7.3.5. Use Theorem 7.7 (the product rule) and for the induction step consider d dx xn = d dx [x][xn−1]. 172Exercise 7.3.10. This formula is known as Leibniz’s rule (which should indicate its age since Leibniz, one of the founders of the calculus, was born in 1646). It extends both Exercises 7.3.8 and 7.3.9. The formula is (f g)(n)(x0) Xk=0 173Exercise 7.3.11. Consider a sequence xn → x0 with xn 6 = x0 and f (xn) = f (x0). n = n! k!(n k)! − f (k)(x0)g(n−k)(x0). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 174Exercise 7.3.12. Let f (x) = x2 sin x−1 477 (f (0) = 0) and take x0 = 0. Utilize the fact that 0 is a limit point of the set x : f (x) = 0 . } { 175Exercise 7.3.17. 1/ cos x. This needs some work. Use If I(x) is the inverse function then I(sin x) = x. The chain rule gives derivative as I ′(sin x) = and obtain cos x = 1 p I ′(sin x) = sin2 x − 1 sin2 x . 1 − Now replace the sin x by some other variable. Caution: While doing this exercise make sure that you know how the p arcsin function sin−1 x is actually defined. It is not the inverse of the function sin x since that function has no inverse. 176Exercise 7.3.19. Draw a good picture. The graph of y = g(x) is the reflection in the line y = x of the graph of y = f (x). What is the slope of the reflected tange |
nt line? 177Exercise 7.3.21. Use the idea in the example. If f (x) = x1/m, then [f (x)]m = x and use the chain rule. If then and use the chain rule. 178Exercise 7.3.22. Once you know that you can determine that F (x) = xn/m, [F (x)]m = xn d dx ex = ex d dx ln x = 1/x using inverse functions. Then consider xp = ep(ln x). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 478 NOTES 179Exercise 7.3.23. The formula you should obtain is ak = p(k)(0) k! for k = 0, 1, 2, . . . . 180Exercise 7.3.24. need to compute p(0), p′(0), p′′(0), p′′′(0), . . . to do this. If you succeed, then you have proved the binomial theorem using derivatives. Of course, you 181Exercise 7.5.6. Define sets An consisting of all x for which f (t) < f (x) for all 0 < ∞ n=1 An is the set in question. x | t | − < 1 n and observe that 182Exercise 7.5.7. Modify the hint in Exercise 7.5.6. S 183Exercise 7.6.3. Use Rolle’s theorem to show that if x1 and x2 are distinct solutions of p(x) = 0, then between them is a solution of p′(x) = 0. 184Exercise 7.6.5. Use Rolle’s theorem twice. See Exercise 7.6.7 for another variant on the same theme. 185Exercise 7.6.6. Since f is continuous we already know (look it up) that f maps [a, b] to some closed bounded interval [c, d]. Use Rolle’s theorem to show that there cannot be two values in [a, b] mapping to the same point. 186Exercise 7.6.7. cf. Exercise 7.6.5. 187Exercise 7.6.8. First show directly from the definition that the Lipshitz condition will imply a bounded derivative. Then use the mean value theorem to get the converse, that is, apply the mean value theorem to f on the interval [x, y] for any a x < y b. ≤ ≤ 188Exercise 7.6.9. Note that an increasing function f would allow only positive numbers in S. 189Exercise 7.6.12. Apply the mean value theorem to f on the interval [x, x + a] to obtain a point ξ in [x, x + a] with f (x + a) − f (x) = af ′(ξ). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 479 190Exercise 7.6.13. Use the mean value theorem to compute lim x→a+ f (x) x f (a) a . − − 191Exercise 7.6.14. This is just a variant on Exercise 7.6.13. Show that under these assumptions f ′ is continuous at x0. 192Exercise 7.6.15. Use the mean value theorem to relate to ∞ i=1 X (f (i + 1) f (i)) − ∞ f ′(i). Note that f is increasing and treat the former series as a telescoping series. i=1 X 193Exercise 7.6.16. The proof of the mean value theorem was obtained by applying Rolle’s theorem to the function For this mean value theorem apply Rolle’s theorem twice to a function of the form g(x) = f (x) f (a) − f (b) b − f (a) a (x − a). − − h(x) = f (x) f (a) − − f ′(a)(x a) − − α(x − a)2 for an appropriate number α. 194Exercise 7.6.18. Write f (x + h) + f (x h) − − 2f (x) = [f (x + h) − f (x)] + [f (x h) − − f (x)] ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 480 NOTES and apply the mean value theorem to each term. 195Exercise 7.6.21. Let φ(x) be Interpret as a monotonicity statement about the function f (a) f (b) f (x) g(a) h(a) g(b) h(b) g(x) h(x) and imitate the proof of Theorem 7.21. 196Exercise 7.7.1. f (x) = (1 x)ex. − 197Exercise 7.7.3. We do not assume differentiability at b. For example, this would apply to the function f (x) = with b = 0. x | , | 198Exercise 7.7.5. Interpret this as a monotonicity property for the function F (x) = f (x)/x. We need to show that F ′ is positive. Show that this is true if f ′(x) > f (x)/x for all x. But how can we show this? Apply the mean value theorem to f on the interval [0, x] (and don’t forget to use the hypothesis that f ′ is an increasing function). 199Exercise 7.7.6. that f (x)/g(x) is monotone (increasing or decreasing) on [a, b]. If not, there is an interval [a, b] with f (a) = f (b) = 0 and neither f nor g vanish on (a, b). Show 200Exercise 7.8.7. Let ε > 0 and consider f (x) + εx. 201Exercise 7.8.9. Figure out a way to express R as a countable union of disjoint dense sets An and then let f (x) = n F is not an increasing for all x function, and apply Theorem 7.30. An. For an example subtract an appropriate linear function F from f such that f − ∈ 202Exercise 7.8.10. then the function defined by the series In connection with this exercise we should make this remark. If A = ak} { is any countable set, has D+f (x) < D−f (x) for all x ∈ A. This can be verified using the results in Chapter 9 on uniform convergence. −| x − 2k ak| ∞ Xk=0 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 481 203Exercise 7.9.1. For the third part use the function F (x) = x2 sin x−1, F (0) = 0 to show that there exists a x3 on an appropriate interval. differentiable function f such that f ′(x) = cos x−1, f (0) = 0. Consider g(x) = f (x) − 204Exercise 7.9.3. If either F G′ or GF ′ were a derivative, so would the other be since (F G)′ = F G′ + GF ′. In that case F G′ GF ′ is also a derivative. But now show that this is impossible [because of (c)]. − 205Exercise 7.9.4. Use f g′ = (f g)′ f ′g. You need to know the fundamental theorem of calculus to continue. − 206Exercise 7.9.5. If f ′ is continuous, then it is easy to check that Eα is closed. In the opposite direction suppose that every Eα is closed and f ′ is not continuous. Then show that there must be a number β and a sequence of points β and f ′(z) < β. Apply the Darboux property of the derivative to xn} { show that this cannot happen if Eβ is closed. Deduce that f ′ is continuous. converging to a point z and yet f ′(xn) ≥ 207Exercise 7.10.3. you should be able to prove that g this might not be true. ◦ If f is convex on an interval I and g is convex and also nondecreasing on the interval f (I), then f is also convex. Show also that if the monotonicity assumption on g is dropped 208Exercise 7.10.5. Show that at every point of continuity of f ′ nuities does the (nondecreasing) function f ′ + have? + the function is differentiable. How many disconti- 209Exercise 7.10.10. Give an example of a convex function on the interval (0, 1) that is not bounded above; that answers the first question. For the second question use Exercise 7.10.4 to show that f must be bounded below. 210Exercise 7.10.13. The methods of Chapter 9 would help here. There we learn in general how to check for the differentiability of functions defined by series. For now just use the definitions and compute carefully. 211Exercise 7.10.14. For (d) let (sin 1/x)2, e−1/x2 (sin 1/x)2, e−1/x2 0, − for x > 0 for x = 0, for x < 0 . f (x) = ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 482 NOTES The three definitions in the exercise are not equivalent even for infinitely differentiable functions. They are, however, equivalent for analytic functions; that is, functions represented by power series (a topic we cover in Chapter 10). Since the scope of elementary calculus is more or less limited to functions that are analytic on the intervals on which the functions are concave up or down, we might argue that on that level, the definition to take is the one that is simplest to develop. We should mention, however, that there are differentiable functions that are not concave-up or concave-down on any interval! 212Exercise 7.10.15. Order the terms so that And write x1 ≤ x2 ≤ · · · ≤ xn. n p = αkxk. Choose a number M between f ′ −(p) and f ′ Xk=1 +(p). Check that Check that x1 ≤ p ≤ xn. f (xk) M (xk − ≥ p) + f (p) for k = 1, 2, . . . , n. Now use these inequalities to obtain Jensen’s inequality. 213Exercise 7.11.1. Use L’Hˆopital’s rule to find that f (0) should be ln(3/2). Use the definition of the derivative and L’Hˆopital’s rule twice to compute f ′(0) = [(ln 3)2 (ln 2)2]/2. − Exercise 7.6.13 shows that the technique in (c) part does in fact compute the derivative provided only that you can show that this limit exists. 214Exercise 7.11.2. Treat the cases A > 0 and A < 0 separately. 215Exercise 7.11.10. We must have lim x→∞ f ′(x) = 0 in this case. (Why?) ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) NOTES 483 216Exercise 7.13.3. Consider the function and note, in particular, the relation between H, H ′ and p. H(x) = p(x) + p′(x) + p′′(x) + + p(n)(x) · · · 217Exercise 7.13.7. Such functions are called midpoint convex. By the definition of convexity we need to show that if x1, x2 ∈ [0, 1], then the inequality I and α ∈ f (αx1 + (1 α)x2) − ≤ αf (x1) + (1 α)f (x2) − is satisfied. Use the midpoint convexity condition to show that this is true whenever α is a fraction of the form p/2q [0, 1]. Without continuity this argument fails for integers p and q. Now use continuity to show that it holds for all α and, indeed, there exist discontinuous midpoint convex functions that fail to be convex. [For an extensive account of what is known about such conditions, see B. S. Thomson, Symmetric Properties of Real Functions, Marcel Dekker, (New York, 1994).] ∈ 218Exercise 7.13.8. If g does not vanish on (x1, x2), then Rolle’s theorem applied to the quotient f /g provides a contradiction. Incidentally, Josef de Wronski (1778–1853), whose name was attached firmly to this concept in 1882 in a multivolume History of Determinants, was a rather curious figure whom you are unlikely to encounter in any other context. One biographer writes about him: For many years Wronski’s work was dismissed as rubbish. However, a closer examination of the work in more recent times shows that, although some is wrong and he has an incredibly high opinion of himself and his ideas, there is also some mathematical insights of great depth and brilliance hidden within the papers. 219Exercise 7.13.9. Consider the function H(x) = f (x) + cx2 + ax + b for c > 0 and various choices of lines y = ax + b and make use of Exercise 7.10.14. 220Exercise 7.13.12. This is from the 1939 Putnam Mathematical Competition. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Rea |
l Analysis, 2nd Edition (2008) 484 NOTES 221Exercise 7.13.13. This is from the 1946 Putnam Mathematical Competition. 222Exercise 7.13.14. This is from the 1958 Putnam Mathematical Competition. 223Exercise 7.13.15. This is from the 1962 Putnam Mathematical Competition. 224Exercise 7.13.16. This is from the 1992 Putnam Mathematical Competition. 225Exercise 7.13.17. This is from the 1998 Putnam Mathematical Competition. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Chapter 8 THE INTEGRAL " For a short course the integral as conceived by Cauchy can be introduced and the material on Riemann’s integral omitted or abridged. The study of the Riemann integral introduces new techniques and ideas that may not be needed for some courses. 8.1 Introduction Calculus students learn two processes, both of which are described as “integration.” The following two examples should be familiar: and x3 dx = x4/4 + C Z 2 x3 dx = 24/4 14/4 = 16/4 − 1/4 = 15/4. − 1 Z The first is called an indefinite integral or antiderivative and the second a definite integral. The use of nearly identical notation, terminology, and methods of computation does a lot to confuse the underlying meanings. Many calculus students would be hard pressed to make a distinction. 485 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 486 The Integral Chapter 8 Indeed even for many eighteenth-century mathematicians these two different procedures were not much distinguished. It was a great discovery that the computation of an area could be achieved by finding an antiderivative. It is attributed to Newton, but vague ideas along this line can be found in the thinking of earlier authors. For these mathematicians a definite integral was defined directly in terms of the antiderivative. This is most unfortunate for the development of a rigorous theory, as recognized by Cauchy. He saw clearly that it was vital that the meaning of the definite integral be separated from the indefinite integral and given a precise definition independent of it. For this he turned to the geometry of the Greeks, who had long ago described a method for computing areas of regions enclosed by curves. This method, the so-called method of exhaustion, involves computing the areas of simpler figures (squares, triangles, rectangles) that approximate the area of the region. We return to the example 2 x3 dx interpreted as an area. The region is that bounded on the left and right by the lines x = 1 and x = 2, below by the line y = 0, and above by the curve y = x3. (See Figure 8.1.) 1 Z Using the method of exhaustion, we may place this figure inside a collection of rectangles by dividing the interval [1, 2] into n equal sized subintervals each of length 1/n. This means selecting the points and constructing rectangles with the height of the rectangles determined by right hand endpoints. The total area of these rectangles exceeds the true area and is precisely 1, 1 + 1/n, 1 + 2/n, . . . , 1 + (n 1)/n − The method of exhaustion requires a lower estimate as well and the true area of the region must be greater Xk=1 n (1 + k/n)3(1/n). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.1. Introduction 487 8 1 x3 1 2 Figure 8.1. Region bounded by x = 1, x = 2, y = x3, and y = 0. than n (See Figure 8.2 for an illustration with n = 4.) (1 + (k − 1)/n)3(1/n). Xk=1 The method of exhaustion requires us to show that as n increases both approximations, the upper one and the lower one, approach the same number. Cauchy saw that, because of the continuity of the function f (x) = x3, these limits would be the same. More than that, any choice of points ξk from the interval [1 + (k 1)/n, 1 + (k)/n] would have the property that − would exist. lim n →∞ n Xk=1 3(1/n) ξk This procedure, borrowed heavily from the Greeks, will work for any continuous function and thus it ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 488 The Integral Chapter 8 8 1 1 2 Figure 8.2. Method of exhaustion (n = 4). offered to Cauchy a way to define the integral for any function f , continuous on an interval [a, b], without any reference whatsoever to notions of derivatives or antiderivatives. The key ingredients here are first dividing the interval [a, b] by a finite sequence of points b a f (x) dx Z thus forming a collection of nonoverlapping subintervals called a partition of [a, b] a = x0 < x1 < x2 < x3 < < xn − · · · 1 < xn = b, (it is not important that they have equal size, just that they get small). Then we form the sums n [x0, x1], [x1, x2], . . . , [xn 1, xn], − f (ξk)(xk − with respect to this partition. The only constraint on the choice of the points ξk is that each is taken from Xk=1 (1) xk 1) − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.2. Cauchy’s First Method 489 the appropriate interval [xk It is an unfortunate trick of fate that the sums (1) that originated with Cauchy are called Riemann sums because of Riemann’s later (much later) use of them in defining his integral. 1, xk] of the partition; these are often called the associated points. − In this chapter we start with Cauchy’s methods of integration and proceed to Riemann. The important thing for you to keep track of is how this theory develops in a manner that assigns meaning to the integral of various classes of functions in a way distinct from how we would compute an integral in a calculus course. 2 1 x3 dx = 15/4 in the familiar way, rather than as a limit of Riemann It is much easier to compute that sums; but the meaning of this statement is properly given in this more difficult way. R 8.2 Cauchy’s First Method Cauchy’s first goal in defining an integral was to give meaning to the integral for continuous functions. The integral is defined as the limit of Riemann sums. Before such a definition is valid we must show that the limit exists. Thus the first step is the proof of the following theorem. Theorem 8.1 (Cauchy) Let f be a continuous function on an interval [a, b]. Then there is a number I, called the definite integral of f on [a, b], such that for each ε > 0 there is a δ > 0 so that n whenever [x0, x1], [x1, x2], . . . , [xn than δ and each ξk is a point in the interval [xk − 1, xk]. − 1, xn] is a partition of the interval [a, b] into subintervals of length less f (ξk)(xk − xk 1) − − I < ε Xk=1 Once the theorem is proved, then we can safely define the definite integral of a continuous function as that number I guaranteed by the theorem. Loosely speaking, we say that the integral is defined as a limit of Riemann sums (1). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 490 The Integral Chapter 8 Definition 8.2: Let f be a continuous function on an interval [a, b]. Then we define to be that number I whose existence is proved in Theorem 8.1. b a f (x) dx Z Now we must prove Theorem 8.1. Proof. For any particular partition (let us call it π) of the interval [a, b] write [x0, x1], [x1, x2], . . . , [xn 1, xn] − and n Xk=1 n M (π) = m(π) = max { f (x) : x [xk − 1, xk] (xk − } xk 1) − ∈ min { f (x) : x [xk − 1, xk] (xk − } xk − 1). ∈ Xk=1 Here M (π) and m(π) depend on the partition π. These are called the upper sums and lower sums for the partition. Note that any Riemann sum over this partition must lie somewhere between the lower sum and the upper sum. Since f is continuous on [a, b] it is uniformly continuous there (Theorem 5.48). Thus for every ε > 0 there is a δ > 0 that depends on ε so that | x < δ. Since we shall need to find a different δ for many choices of ε, let us write it as δ(ε). Thus if the partition we are using has the property that every interval is shorter than δ(ε), we must − − − y | | a b if | f (x) f (y) < ε have max { f (x) : x [xk − 1, xk] ∈ } − min { f (x) : x [xk − 1, xk] } ∈ < b ε − a ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.2. Cauchy’s First Method 491 It follows that for such partitions 0 Select a sequence of partitions M (π) ≤ , each one containing all the points of the previous partition, and πn} such that every interval in the nth partition πn is shorter than δ(1/n). If M (πn) and m(πn) denote the corresponding sums for the nth partition of our sequence of partitions, then m(π) < ε. − { One more technical point needs to be raised. As we add points to a partition the upper sums cannot m(πn+1). (The details M (πn+1) while m(πn) increase nor can the lower sums decrease. Thus M (πn) just require some inequality work and are left as Exercise 8.2.17.) ≥ ≤ 0 ≤ M (πn) m(πn) < 1/n. − If m(πn) = M (πn) at some stage in the process then we are done and can choose this number as I. Otherwise the intervals [m(πn), M (πn)] form a descending sequence with lengths shrinking to zero. By Cantor’s intersection property (see Section 4.5.2) there is a number I so that m(πn) has the property of the theorem. I and M (πn) . We shall show that I I as n → ∞ → → Now let ε > 0 and choose any partition π with the property that every interval is shorter than δ(ε/2). By what we have seen, the interval [m(π), M (π)] has length smaller than ε/2. Any Riemann sum over the partition π must evidently belong to the interval Let N > 2/ε. Suppose for a moment that the intervals [m(π), M (π)]. [m(π), M (π)] and [m(πN ), M (πN )] intersect at some point. In that case the Riemann sum over the partition π and the value I, which is inside the interval [m(πN ), M (πN )], must be closer together than ε/2 + 1/N , which is smaller than ε. As this is precisely what we want to prove, we are done. It remains to check that the two intervals [m(π), M (π)] and [m(πN ), M (πN )] ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 492 The Integral Chapter 8 intersect at some point. To find a point common to these two intervals combine the |
two partitions π and πN to form a partition containing all points in either partition. The Riemann sum over such a partition belongs to the interval [m(π), M (π)] and also to the interval [m(πN ), M (πN )]. This completes the proof. (That I is unique is left as Exercise 8.2.2.) A special case of this definition and this theorem allows us to compute an integral as a limit of a sequence. In practice this is seldom the best way to compute it, but it is interesting and useful in some parts of the theory. Corollary 8.3: Let f be a continuous function on an interval [a, b] Then b a Z f (x) dx = lim →∞ + Xk=0 k n (b − a) . 8.2.1 Scope of Cauchy’s First Method " Enrichment section. May be omitted. It is natural to ask whether this method of Cauchy for describing the integral of a continuous function would apply to a larger class of functions. But Cauchy did not ask this question. His goal was to assign a meaning for continuous functions, a class of functions that was large enough for most applications. The only limitation he might have seen was that this method would fail for functions having infinite singularities (i.e., discontinuity points where the function is unbounded). Thus he was led to the method we discuss in Section 8.4 as Cauchy’s second method. Cauchy and other mathematicians of his time were sufficiently confused as to the meaning of the word “function” that they might never have asked such a question. But we can. And many years later Riemann did too, as we shall see in Section 8.6. In the exercises you are asked to prove the following: The first method of Cauchy will fail if applied to an unbounded function f on an interval [a, b]. The first method of Cauchy succeeds if applied to any function f that is bounded on an interval [a, b] and has only finitely many discontinuities there. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.2. Cauchy’s First Method 493 The first statement shows that the method used here to define an integral is severely limited. It can never be used for unbounded functions. Since we have restricted it here to continuous functions that is no problem; any function continuous on an interval [a, b] is bounded there. The second statement shows that the method is not, however, limited only to continuous functions even though that was Cauchy’s intention. Later we will use the method to define Riemann’s integral which applies to a large class of (bounded) functions that are permitted to have many, even infinitely many, points of discontinuity. Exercises 8.2.1 To complete the computations in the introduction to this chapter, show that n (1 + (k)/n)3(1/n) = 15/4. lim n→∞ Xk=1 This computation alone should be enough to convince you that the definition is intended theoretically and hardly ever used to compute integrals. See Note 226 8.2.2 Show that the number I in the statement of Theorem 8.1 is unique; that is, that there cannot be two numbers that would be assigned to the symbol b a f (x) dx. 8.2.3 If f is constant and f (x) = α for all x in [a, b] show that R 8.2.4 If f is continuous and f (x) ≥ b f (x) dx = α(b a Z 0 for all x in [a, b] show that a). − b a Z f (x) dx 0. ≥ 8.2.5 If f is continuous and m f (x) ≤ ≤ M for all x in [a, b] show that b m(b a) − ≤ a Z f (x) dx M (b a). − ≤ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 494 8.2.6 Calculate length. 1 b 0 xp dx (for whatever values of p you can manage) by partitioning [0, 1] into subintervals of equal R 8.2.7 Calculate a xp dx (for whatever values of p you can manage) by partitioning [a, b] into subintervals [a, aq], [aq, aq2], . . . , [aqn−1, b] where aqn = b. (Note that the subintervals are not of equal length, but that the lengths form a geometric progression.) R The Integral Chapter 8 8.2.8 Use the method of the preceding exercise to show that and check it by the usual calculus method. 2 dx x2 = 1 2 1 Z 8.2.9 Compute the Riemann sums for the integral b a x−2 dx (a > 0) taken over a partition of the interval [a, b] and with associated points ξi = √xixi−1. What can you conclude from this? See Note 227 [x0, x1], [x1, x2], . . . , [xn−1, xn] R 8.2.10 Compute the Riemann sums for the integral b a x−1/2 dx (a > 0) taken over a partition R of the interval [a, b] and with associated points [x0, x1], [x1, x2], . . . , [xn−1, xn] What can you conclude from this? √xi + √xi−1 2 2 . ξi = 8.2.11 Show that 8.2.12 Express lim n→∞ n 1 (n + 1)2 + 1 (n + 2)2 + 1 (n + 3)2 + + · · · 1 (2n)2 = 1 2 . as a definite integral where f is continuous on [0, 1]. lim n→∞ 1 n n f Xk=1 k n ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.2. Cauchy’s First Method 8.2.13 Calculate lim n→∞ e1/n + e2/n + · · · + e(n−1)/n + en/n n 495 by expressing this limit as a definite integral of some continuous function and then using calculus methods. 8.2.14 Prove that the conclusion of Theorem 8.1 is false if f is discontinuous at any point in the interval and is not bounded. 8.2.15 Prove that the conclusion of Theorem 8.1 is true if f is continuous at all but a finite number of points in the interval [a, b] and is bounded. 8.2.16 Prove that the conclusion of Theorem 8.1 is true for the function f defined on the interval [0, 1] as follows: f (0) = 0 and f (x) = 2−n for each How many points of discontinuity does f have in the interval [0, 1]? What is the value of the number I in this case? 2−n−1 < x 2−n (n = 0, 1, 2, 3, . . . ). ≤ 8.2.17 For a bounded function f and any partition π [x0, x1], [x1, x2], . . . , [xn−1, xn] of the interval [a, b] write and n Xk=1 n M (f, π) = m(f, π) = f (x) : x sup { [xk−1, xk] } (xk − ∈ xk−1) f (x) : x inf { [xk−1, xk] } (xk − ∈ xk−1) Xk=1 These are called the upper sums and lower sums for the partition for the function f and were used in the proof of Theorem 8.1. (a) Show that if π2 contains all of the points of the partition π1, then m(f, π1) m(f, π2) ≤ ≤ M (f, π2) ≤ M (f, π1). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 496 The Integral Chapter 8 (b) Show that if π1 and π2 are arbitrary partitions and f is any bounded function, then (c) Show that if π is any arbitrary partition and f is any bounded function on [a, b] then where C = sup f and c = inf f . c(b a) − ≤ m(f, π) ≤ M (f, π) C(b a) − ≤ m(f, π1) ≤ M (f, π2). (d) Show that with any choice of associated points the Riemann sum over a partition π is in the interval [m(f, π), M (f, π)]. (e) Show that, if f is continuous, every value in the interval between m(f, π) and M (f, π) is equal to some particular Riemann sum over the partition π with an appropriate choice of associated points ξk. (f) Show that if f is not continuous the preceding assertion may be false. 8.3 Properties of the Integral The integral has thus far been defined just for continuous functions. We ask what properties it must have. Later we shall have to extend the scope of the integral to much broader classes of functions. It will be important to us then that the collection of elementary properties here will still be valid. These properties exhibit the structure of the integral. They are the most vital tools to use in handling integrals both for theoretical and practical matters. Since we are restricted to continuous functions in this section, the proofs are simple. As we enlarge the scope of the integral the proofs may become more difficult, and subtle differences in assertions may arise. Note. All functions f and g appearing in the statements are assumed to be continuous on the intervals [a, b], [b, c] , [a, c] in the statements. Thus the integrals all have meaning. This means we do not have to prove that any of these integrals exist: They do. It is the stated identity that needs to be proved in each case. To prove the identity, we consider a sequence of partitions πn chosen so that the points in the partition are closer together than 1/n. Let us use the notation S(πn, f ) to denote a Riemann sum taken over this partition for the function f with associated points chosen (say) at the left-hand endpoint of the ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.3. Properties of the Integral corresponding intervals. Then We shall use this idea in the proofs. lim n →∞ b S(πn, f ) = f (x) dx. a Z 497 8.4 (Additive Property) Let f be continuous on [a, c] and suppose that a < b < c. Then b a Z c c f (x) dx + f (x) dx = f (x) dx. b Z a Z Proof. For our sequence of partitions we choose πn to be a partition of [a, c] chosen so that the points in the partition are closer together than 1/n and so that the point b is one of the points. Each partition πn splits into two parts; π′n and π′′n where the former is a partition of [a, b] and where the latter is a partition of [b, c]. Note that S(πn, f ) = S(π′n, f ) + S(π′′n, f ) by elementary arithmetic. If we let n statement we wish to prove. → ∞ in this identity we obtain immediately the identity in the 8.5 (Linear Property) Let f and g be continuous on [a, b]. Then, for all α, β b a Z [αf (x) + βg(x)] dx = α f (x) dx + β g(x) dx. b b a Z a Z R, ∈ Proof. Again consider a sequence of partitions of [a, b], πn chosen so that the points in the partition are closer together than 1/n. If S(πn, f ) denotes a Riemann sum taken over this partition for the function f , then b lim n →∞ S(πn, f ) = f (x) dx. a Z ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 498 The Integral Chapter 8 In the same way for g we would have But it is easy to check that lim n →∞ b S(πn, g) = g(x) dx. a Z S(πn, αf + βg) = αS(πn, f ) + βS(πn, g) and taking n to prove that → ∞ in this identity gives exactly the statement in the property. Note that we do not have This follows from Theorem 8.1 because the integrand is continuous. S(πn, αf + βg) → b a [αf (x) + βg(x)] dx. Z 8.6 (Monotone Property) Let f and g be continuous on [a, b]. Then, i |
f f (x) b a Z f (x) dx ≤ a Z b g(x) dx. g(x) for all a x ≤ ≤ b, ≤ Proof. Consider a sequence of partitions πn chosen so that the points in the partition are closer together than 1/n. If S(πn, f ) denotes a Riemann sum taken over this partition for the function f , then In the same way for g we would have S(πn, f ) = lim n →∞ lim n →∞ g(x) for all x we must have S(πn, g) = But since f (xx) dx. g(x) dx. S(πn, f ) ≤ S(πn, g). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.3. Properties of the Integral Taking limits as n → ∞ in this inequality yields the property. 8.7 (Absolute Property) Let f be continuous on [a, b]. Then or, equivalently, b − a | Z f (x) dx | ≤ b a Z b f (x) dx f (x) dx | ≤ a | Z Proof. This follows immediately from the monotone property because b a Z f (x) dx b ≤ a | Z f (x) dx. | f (x) −| f (x) f (x) . | ≤ | | ≤ 499 Fundamental Theorem of Calculus The next two properties are known together as the fundamental theorem of calculus. They establish the close relationship between differentiation and integration and offer, to the calculus student, a useful method for the computation of integrals. This method reduces the computational problem of integration (i.e., computing a limit of Riemann sums) to the problem of finding an antiderivative. 8.8: (Differentiation of the Indefinite Integral) Let f be continuous on [a, b]. Then the function has a derivative on [a, b] and F ′(x) = f (x) at each point. a Z x F (x) = f (t) dt Proof. Let h > 0 and x [a, b). We compute ∈ F (x + h) F (x) − − hf (x) = x+h x Z (f (t) − f (x)) dt ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 500 The Integral Chapter 8 provided only that x + h b. Thus, using Exercise 8.3.1, we have ≤ F (x + h) F (x) − − hf (x) | ≤ h max f (t) f (x) : t | ∈ − {| [x, x + h] } F (x + h) h F (x) − − max f (t) f (x) : t [x, x + h] } . ∈ | − {| and hence that | As f is continuous at x ≤ f (x) f (x) as h 0+ and this inequality shows that the right-hand derivative of F at x A similar argument would show that the left-hand derivative of F at x → max {| f (t) − : t | ∈ [x, x + h] 0 } → proves the property. [a, b) is exactly f (x). ∈ (a, b] is exactly f (x). This ∈ 8.9: (Integral of a Derivative) If the function F has a continuous derivative on [a, b], then b a Z F ′(x) dx = F (b) F (a). − Proof. Given any ε > 0 there is a δ > 0 so that any Riemann sum for the continuous function F ′ over a partition of [a, b] into intervals of length less than δ is within ε of b a F ′(x) dx. If is such a partition then observe that, if we choose the associated points ξk ∈ theorem in such a way that [xk − 1, xk] by the mean value [x0, x1], [x1, x2], . . . , [xn 1, xn] R − then we will have F (xk) F (xk 1) = F ′(ξk)(xk − − xk 1) − − F (b) − F (a) = n Xk=1 F (xk) F (xk 1) = − − n Xk=1 F ′(ξk)(xk − xk − 1). ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.3. Properties of the Integral 501 Since the right side of the identity is within ε of this is true for any ε > 0 and hence it follows that these must be equal; that is, that b b a F ′(x) dx so too must be the value F (b) R − F (a). But F ′(x) dx = F (b) F (a). − a Z Exercises 8.3.1 If f is continuous on an interval [a, b] and show that M = max f (x) | : x ∈ [a, b] |} {| 8.3.2 (Mean Value Theorem for Integrals) If f is continuous show that there is a point ξ in (a, b) so that M (b a). − ≤ b a Z f (x) dx b f (x) dx = f (ξ)(b a). − a Z 8.3.3 If f is continuous and m f (x) ≤ ≤ M for all x in [a, b] show that b b for any continuous, nonnegative function g. a Z m g(x) dx f (x)g(x) dx M ≤ ≤ a Z b a Z g(x) dx 8.3.4 If f is continuous and nonnegative on an interval [a, b] and show that f is identically equal to zero there. b a Z f (x) dx = 0 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 502 The Integral Chapter 8 8.3.5 (Second Mean Value Theorem for Integrals) If f and g are continuous on an interval [a, b] and g is nonnegative, show that there is a number ξ (a, b) such that ∈ b b a Z f (x)g(x) dx = f (ξ) g(x) dx. a Z 8.3.6 If f is continuous on an interval [a, b] and for every continuous function g on [a, b] show that f is identically equal to zero there. a Z b f (x)g(x) dx = 0 8.3.7 (Integration by parts) Suppose that f , g, f ′ and g′ are continuous on [a, b]. Establish the integration by parts formula f (x)g′(x) dx = [f (b)g(b) b a Z f (a)g(a)] − b − a Z f ′(x)g(x) dx. 8.3.8 (Integration by substitution) State conditions on f and g so that the integration by substitution formula is valid. b a Z f (g(x))g′(x) dx = f (s) ds g(b) g(a) Z 8.3.9 State conditions on f , g and h so that the integration by substitution formula is valid. b a Z f (g(h(x)))g′(h(x))h′(x) dx = f (s) ds g(h(b)) g(h(a)) Z 8.3.10 If f and g are continuous on an interval [a, b] show that b Z a f (x)g(x) dx 2 ! ≤ Z a b [f (x)]2 dx b [g(x)]2 dx . ! ! Z a See Note 228 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.4. Cauchy’s Second Method 8.4 Cauchy’s Second Method 503 Defining an integral only for continuous functions, as we did in the preceding section, is far too limiting. Even in the early nineteenth century the need for considering more general functions was apparent. For Cauchy this meant handling functions that have discontinuities. But Cauchy would not have felt any need to handle badly discontinuous functions, indeed he may not even have considered such objects as functions. In our terminology we could say that Cauchy was interested in extending his integral from continuous functions to functions possessing isolated discontinuities (i.e., the set of discontinuity points contains only isolated points). We have already noted in Section 8.2.1 that bounded functions with finitely many discontinuities present no difficulties. Cauchy’s first method can be applied to them. It is the case of unbounded functions that offers real resistance. What should we mean by the integral While the integrand has only one discontinuity (at x = 0) the function is unbounded and Cauchy’s first method cannot be applied. If the integral did make sense, then we would expect that the function 1 dx √x ? 0 Z would be defined and continuous everywhere on the interval [0, 1] and the value F (0) would equal our integral. But here F (x) is not defined at x = 0 although it is defined for all x in (0, 1] since the integrand is continuous on any interval [x, 1] for x > 0. If we compute it we see that F (δ) = 1 δ dx √x Z While we cannot take F (0) itself (it is not defined), we can take the limit, 1 dx √x = 2 − 2√δ. F (δ) = δ Z lim 0+ → δ F (δ) = lim 0+ δ → 1 dx √x δ Z = 2, ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 504 The Integral Chapter 8 8 4 2 1 ab c 1 Figure 8.3. Computation of 1 0 x −1/2 dx = 2. R as a perfectly reasonable value for the integral. Indeed if we consider this as a problem in determining the area of the unbounded region in Figure 8.3 we can see graphically why the answer should be 2. Note that in the figure 0 < a < b < c < 1 and these numbers have the values a = 1/64, b = 1/16, and c = 1/4. The integrals have values c 1 b 1/2 dx = 1, 1/2 dx = 1/2, and 1/2 dx = 1/4 c Z and so we would expect x− x− b Z as indeed this method does give. x− 1/2 dx = 1 + 1/2 + 1/4 + 1 0 Z x− a Z = 2 · · · This is precisely Cauchy’s second method. If you understand this example, you understand the method. Any general write-up of the method might obscure this simple idea. We need some language, however. The procedure of taking a limit to obtain the final value of the integral may or may not work. If the limit does exist, we say that the integral converges, or is a convergent integral, and we say that the function f is integrable by Cauchy’s second method or simply integrable if the context is clear. Otherwise the integral is said to be divergent. We give a formal definition valid just in the case that the function has one point of unboundedness, and ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.4. Cauchy’s Second Method 505 that point occurs at the left-hand endpoint of the interval. For more than one point, or for a point not at an endpoint, the definition is best generalized by splitting the integral into separate integrals, each of which can be handled one at a time in this fashion. (See Exercise 8.4.3.) Definition 8.10: Let f be a continuous function on an interval (a, b] that is unbounded in every interval (a, a + δ). Then we define to be b a Z f (x) dx b f (x) dx lim 0+ → δ a+δ Z if this limit exists, and in this case the integral is said to be convergent. If both integrals converge the integral is said to be absolutely convergent. a Z b f (x) dx and b a | Z f (x) dx | The role of the extra condition of absolute convergence is much like its role in the study of infinite series. You will recall that absolutely convergent series are more “robust” in the sense that they can be rather freely manipulated, unlike the nonabsolutely convergent series, which are rather fragile. The same is true here of absolutely convergent integrals. Note that the integral is both convergent and absolutely convergent merely because the integrand is nonnegative. 0 Z 1 x− 1/2 dx Exercises 8.4.1 Formulate a definition of the integral right-hand endpoint. Supply an example. b a f (x) dx for a function continuous on [a, b) and unbounded at the R ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 506 The Integral Chapter 8 8.4.2 Formulate a definition of the integral in every interval containing c. Supply an example. R 8.4.3 How would an integral of the form b a f (x) dx for a function continuous on [a, c) and on (c, b] and unbounded be interpreted, where f is continuous? 3 0 Z x(x | p f (x) 1)(x 2)(x 3) | − − − dx 8.4.4 Let f and g |
be continuous on (a, b] and such that f (x) | absolutely convergent, show that so also is the integral g(x) | ≤ | | b a f (x) dx. for all a < x b. If the integral b a g(x) dx is ≤ R 8.4.5 For what continuous functions f must the integral 1 R f (x) converge? See Note 229 −1 Z √1 x2 − dx 8.4.6 Let f be a bounded function, continuous on (a, b] and that is discontinuous at the endpoint a. Show that if the second method of Cauchy is applied to f then the result is the same as applying the first method to the entire interval [a, b] [regardless of the value assigned to f (a)]. 8.4.7 Suppose that f is continuous on [ − 1, 1] except for an isolated discontinuity at x = 0. If the limit −δ 1 lim δ→0+ Z −1 f (x) dx + f (x) dx δ Z ! exists does it follow that f is integrable on [ − 1, 1]? 8.4.8 As a project determine which of the properties of the integral in Section 8.3 (which apply only to continuous functions on an interval [a, b]) can be extended to functions that are integrable by Cauchy’s second method on [a, b]. Give proofs. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.5. Cauchy’s Second Method (Continued) 507 8.5 Cauchy’s Second Method (Continued) The same idea that Cauchy used to assign meaning to the integral of unbounded functions he also used to handle functions on unbounded intervals. How should we interpret the integral We might try first to form a partition of the unbounded interval [1, Riemann sums. A much simpler idea is to adapt Cauchy’s second method to this in the obvious way. ) and seek some kind of limit of ∞ ∞ dx x2 ? 1 Z dx x2 = lim In Figure 8.4 we show graphically how to compute the area that is represented by dx x2 = lim 1 →∞ 1 →∞ Z 1 X = 1. − ∞ Z X X 1 X and so we would expect x− 2 dx = 1/2, 2 1 Z x− 2 dx = 1/4, 4 2 Z x− 2 dx = 1/8 8 4 Z 2 dx. Note that ∞1 x− R as indeed this method does give. (See Figure 8.4.) ∞ x− 2 dx = 1/2 + 1/4 + 1/8 + = 1 · · · 1 Z This is precisely Cauchy’s second method applied to unbounded intervals. Again, if you understand this example, you understand the method. We give a formal definition valid just for an infinite interval of the form [a, similar. The case ( of which can be handled in this fashion. (See Exercise 8.5.2.) −∞ , + ∞ ∞ ) is best split up into the sum of two integrals, from ( ). The case ( , a] and [a, , b] is ), each −∞ ∞ −∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 508 The Integral Chapter 8 1 1 2 4 8 b Figure 8.4. Computation of ∞ 1 x −2 dx = 1. R Definition 8.11: Let f be a continuous function on an interval [a, ). Then we define ∞ to be ∞ f (x) dx a Z X f (x) dx lim X a →∞ Z if this limit exists, and in this case the integral is said to be convergent. If both integrals converge the integral is said to be absolutely convergent. f (x) dx and ∞ a Z ∞ f (x) dx | | a Z Again, the role of the extra condition of absolute convergence is much like its role in the study of infinite 2 dx is convergent and also absolutely convergent merely because the series. Note that the integral integrand is nonnegative. ∞1 x− R Exercises 8.5.1 Formulate a definition of the integral convergent and divergent integrals of this type. R b −∞ f (x) dx for a function continuous on ( , b]. Supply examples of −∞ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.5. Cauchy’s Second Method (Continued) 8.5.2 Formulate a definition of the integral of convergent and divergent integrals of this type. See Note 230 R 8.5.3 For what values of p is the integral 8.5.4 Show that ∞ 1 x−p dx convergent? R ∞ ∞ −∞ f (x) dx for a function continuous on ( 509 ). Supply examples ∞ , −∞ xne−x dx = n!. 0 Z ) such that limx→∞ f (x) = α. Show that if the integral ∞ 1 f (x) dx ∞ 1 f (x) dx converges. Can you conclude that R R ). Show that the integral ∞ 1 f (x) dx converges if and R ) so that the integral ∞ 1 f (x) dx converges but the series R ) so that the integral ∞ 1 f (x) dx diverges but the series 8.5.5 Let f be a continuous function on [1, converges, then α must be 0. ∞ 8.5.6 Let f be a continuous function on [1, limx→∞ f (x) = 0? ) such that the integral ∞ 8.5.7 Let f be a continuous, decreasing function on [1, ∞ n=1 f (n) converges. only if the series ∞ 8.5.8 Give an example of a function f continuous on [1, P ∞ n=1 f (n) diverges. ∞ 8.5.9 Give an example of a function f continuous on [1, P ∞ n=1 f (n) converges. 8.5.10 Show that P is convergent but not absolutely convergent. See Note 231 0 Z ∞ ∞ R sin x x dx 8.5.11 (Cauchy Criterion for Convergence) Let f : [a, ) R be a continuous function. Show that the → ∞ a f (x) dx converges if and only if for every ε > 0 there is a number M so that, for all M < c < d, ∞ integral R d f (x) dx < ε. c Z ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 510 The Integral Chapter 8 8.5.12 (Cauchy Criterion for Absolute Convergence) Let f : [a, R be a continuous function. Show that ∞ the integral a f (x) dx converges absolutely if and only if for every ε > 0 there is a number M so that, for all M < c < d, R → ∞ ) d f (x) | | dx < ε. c Z 8.5.13 As a project determine which of the properties of the integral in Section 8.3 (which apply only to continuous functions on a finite interval) can be extended to integrals on an infinite interval [a, ]. Give proofs. ∞ 8.6 The Riemann Integral Thus far in our discussion of the integral we have defined the meaning of the symbol first for all continuous functions, by Cauchy’s first method, and then for functions that may have a finite number of discontinuities at which the function is unbounded, by Cauchy’s second method. b a f (x) dx Z Let us return to Cauchy’s first method. We ask just how far this method can be applied. It can be applied to all continuous functions; that was the content of Theorem 8.1. It can be applied to all bounded functions with finitely many discontinuities (Exercise 8.2.15). It can be applied to some bounded functions with infinitely many discontinuities (Exercise 8.2.16). Rather than search for broader classes of functions to which this method applies, we adopt the viewpoint that was taken by Riemann. We simply define the class of all functions to which Cauchy’s first method can be applied and then seek to characterize that class. This represents a much more modern point of view than Cauchy would have taken with his much more limited idea of what a function is. Note that we need only turn Theorem 8.1 into a definition. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.6. The Riemann Integral 511 Definition 8.12: Let f be a function on an interval [a, b]. Suppose that there is a number I such that for all ε > 0 there is a δ > 0 so that n whenever [x0, x1], [x1, x2], . . . , [xn than δ and each ξk is a point in the interval [xk and we write − − 1, xn], is a partition of the interval [a, b] into subintervals of length less 1, xk]. Then f is said to be Riemann integrable on [a, b] f (ξk)(xk − xk 1) − − I < ε Xk=1 b f (x) dx for that number I. We can call the set of points a Z the partition of the interval [a, b] or, equivalently, if it is more convenient, the set of intervals π = { x0, x1, x2, . . . , xn 1, xn} − [x0, x1], [x1, x2], . . . , [xn 1, xn] − can be called the partition. The points ξk that are chosen from each interval [xk associated points of the partition. Notice that in the definition the associated points can be freely chosen inside the intervals of the partition. 1, xk] are called the − Loosely a function f is Riemann integrable if the limit of the Riemann sums for f exists over that interval. The program now is to determine what classes of functions are Riemann integrable and to obtain characterizations of Riemann integrability. We shall investigate this in the remainder of this section. We need also to find out whether the properties of the integral that hold for continuous functions now continue to hold for all Riemann integrable functions. We shall consider that in the next section. Two observations are immediate from our earlier work: All continuous functions are Riemann integrable. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 512 The Integral Chapter 8 All Riemann integrable functions are bounded. In light of this last statement we see that the Riemann integral is somewhat limited in that it will not do anything to handle unbounded functions. For that we must still return to Cauchy’s second method. But, as we shall see, the Riemann integral will handle many bounded functions that are badly discontinuous (but not too badly). As research progressed in the nineteenth century the Riemann integral became the standard tool for discussing integrals of bounded functions. For unbounded functions Cauchy’s second method continued to be employed, although other methods emerged. By the early twentieth century the Riemann integral was abandoned by all serious analysts in favor of Lebesgue’s integral. The Riemann integral survives in texts such as this mainly because of the technical difficulties of Lebesgue’s better, but more difficult, methods. 8.6.1 Some Examples All Riemann integrable functions are bounded. All continuous functions are Riemann integrable. In order to obtain some insight into the question as to what functions are Riemann integrable we present some examples, first of a bounded function that is not integrable and then of a badly discontinuous function that is integrable. Example 8.13: Here is an example of a function that is bounded but “too discontinuous” to be Riemann integrable. On the interval [0, 1] let f be the function equal to 1 for x rational and to 0 for x irrational. Let be any partition. If we choose associated points ξk ∈ sum n [xk − n 1, xk] so that ξk is rational, then the Riemann [x0, x1], [x1, x2], . . . , [xn 1, xn] − f (ξk)(xk − xk − 1) = (xk − xk − 1) = 1 Xk=1 Xk=1 (2) Classical |
RealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 513 (3) ◭ Section 8.6. The Riemann Integral while if we choose associated points ηk ∈ [xk n − 1, xk] so that ηk is irrational Because of (2) and (3), the integral f (ηk)(xk − xk − 1) = 0. Xk=1 1 0 f (x) dx cannot exist. R Example 8.14: Recall the Dirichlet function (Section 5.2.6) which provides an example of a function that is discontinuous at every rational number and continuous at every irrational. We show that this function is Riemann integrable. On the interval [0, 1] let f (0) = 0, at rational numbers f (p/q) = 1/q for 0 < p/q 1 (assuming that p/q has been expressed in its lowest terms) and to f (x) = 0 for x irrational. ≤ Let ε > 0. Let q0 be any positive integer larger than 2/ε. We count the number of points x in [0, 1] at which f (x) > 1/q0. There are finitely many of these, say M of them. Choose δ1 sufficiently small so that any two of these points are further apart than 2δ1. Choose δ < δ1 so that (for reasons that become clear only after all our computations are done) M δ < ε/2. This will allow us to use the inequality M δ + 1/q0 < ε. (4) Let [x0, x1], [x1, x2], . . . , [xn 1, xn] − [xk be any partition chosen so that each of the intervals is shorter than δ. For any choice of associated points ξk ∈ (i) f (ξk) = 0 if ξk = 0 or if ξk is irrational, or 1, xk] we note that either − (ii) f (ξk) > 1/q0 if ξk is one of the M points counted previously, or (iii) 0 < f (ξk) ≤ 1/q0 if ξk is any other rational point. We can estimate the Riemann sum n Xk=1 f (ξk)(xk − xk 1) − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 514 The Integral Chapter 8 by considering separately these three cases. Case (i) evidently contributes nothing to this sum. Case (ii) can contribute at most M δ to this sum since each interval in the partition can contain at most one of the points of type (ii) and there are only M such points. Finally, case (iii) can contribute in total no more than 1/q0. Thus, using the inequality (4), we have n 0 ≤ f (ξk)(xk − xk 1) − ≤ M δ + 1/q0 < ε. Xk=1 This proves that the integral dense set of discontinuities) it is startling that it is nonetheless integrable. 1 0 f (x) dx = 0. Considering just how discontinuous this function is (it has a ◭ R 8.6.2 Riemann’s Criteria " Advanced section. May be omitted. What bounded functions then are Riemann integrable? The answer is that such functions must be “mostly” continuous. The example of the very discontinuous function in Example 8.13 suggests this. On the other hand, Example 8.14 shows that the discontinuities of a Riemann integrable function might even be dense. Riemann first analyzed this by using the oscillation of the function f on an interval. We recall (Definition 6.24) that this is defined as ωf ([c, d]) = sup [c,d] x ∈ f (x) − x inf [c,d] ∈ f (x). This measures how much the function f changes in the interval [c, d]. For a continuous function this is just the difference between the maximum and minimum values of f on [c, d] and will be small if the interval [c, d] is small enough. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.6. The Riemann Integral 515 Theorem 8.15 (Riemann) A function f defined on an interval [a, b] is Riemann integrable if and only if for every ε > 0 there is a δ > 0 so that n whenever [x0, x1], [x1, x2], . . . , [xn than δ. − ωf ([xk 1, xk])(xk − − xk 1) < ε − Xk=1 1, xn], is a partition of the interval [a, b] into subintervals of length less Proof. If f is Riemann integrable on [a, b] with integral I, then for any ε > 0 there must be a δ > 0 so that any two Riemann sums taken over a partition with intervals smaller than δ are both within ε/4 of I. In particular, we have n n whenever [x0, x1], [x1, x2], . . . , [xn than δ. Here ξk and ηk are any choices from [xk − 1, xk]. We rewrite this as − 1, xn], is a partition of the interval [a, b] into subintervals of length less f (ξk)(xk − xk 1) − − f (ηk)(xk − xk − < ε/2 Xk=1 1) Now notice that [f (ξk) f (ηk)](xk − − xk − ε/2 < ε. ≤ (5) 1) sup [xk−1,xk] (f (ξ) − f (η)) = ωf ([xk 1, xk]). − Thus we see that the criterion follows immediately on taking sups over these choices of ξk and ηk in the inequality (5). The other direction of the theorem can be interpreted as a “Cauchy criterion” and proved in a manner similar to all our other Cauchy criteria so far in the text (indeed similar to the proof of Theorem 8.1). We omit the details. Theorem 8.15 offers an interesting necessary and sufficient condition for integrability. It is awkward to Xk=1 n Xk=1 η,ξ ∈ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 516 The Integral Chapter 8 use the sufficiency criterion here since it demands that we check that all small partitions have a certain property. The following variant is a little easier to apply since we need find only one partition for each positive ε. Theorem 8.16: A function f on an interval [a, b] is Riemann integrable if and only if for every ε > 0 there is at least one partition [x0, x1], [x1, x2], . . . , [xn 1, xn], of the interval [a, b] so that − n Xk=1 ωf ([xk 1, xk])(xk − − xk − 1) < ε. Proof. By Theorem 8.15 we see that if f is Riemann integrable there would have to exist such a partition. In the opposite direction we must show that the condition here implies integrability. Certainly this condition implies that f is bounded (or else this sum would be infinite) and so we may assume that f (x) | [c, d]: M for all x. This gives us a useful, if crude, estimate on the size of the oscillation on any interval | ≤ ωf ([c, d]) 2M. ≤ Let ε > 0. We shall find a number δ so that the criterion of Theorem 8.15 is satisfied. Let [x0, x1], 1, xn] be the partition whose existence is given. We use that to find our δ. Choose δ [x1, x2], . . . , [xn sufficiently small so that − Now let 2M nδ < ε. [y0, y1], [y1, y2], . . . , [ym 1, ym], − be any partition of the interval [a, b] into subintervals of length less than δ. These intervals are of two types: Type (i) are those that are contained entirely inside intervals of our original partition, and type (ii) are those that include as interior points one of the points xk for k = 1, 2, . . . , n 1. In any case there are 1 of these intervals and each is of length less than δ. Thus, using just a crude estimate on each of only n − − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.6. The Riemann Integral these terms, the intervals of type (ii) contribute to the sum m no more than (2M )nδ. The sum taken over all the type (i) intervals must be smaller than ωf ([yk 1, yk])(yk − − yk − 1) Xk=1 Thus the total sum m m Xk=1 ωf ([xk 1, xk])(xk − − xk − 1) < ε. ωf ([yk 1, yk])(yk − − yk − 1) < 2M nδ + ε < 2ε. Xk=1 It follows by the criterion in Theorem 8.15 that f is Riemann integrable as required. 517 8.6.3 Lebesgue’s Criterion " Advanced section. May be omitted. Theorem 8.16 is beautiful and seemingly characterizes the class of Riemann integrable functions in a meaningful way. But at the time of Riemann there was only an imperfect understanding of sets of real numbers and so it did not occur to Riemann that the property of Riemann integrability for a bounded function f depended exclusively on the nature of the set of points of discontinuity of f . Indeed the condition n ωf ([xk 1, xk])(xk − − xk − 1) < ε Xk=1 on the oscillation of the function suggests that something more subtle than just this is happening. In 1901 Henri Lebesgue completed this theorem by using the notion of a set of measure zero. Recall (from Section 6.8) that a set E of real numbers is of measure zero if for every ε > 0 there is a sequence of intervals ci) < ε. The exact characterization of (ci, di) { Riemann integrable functions is precisely this: They are bounded (as we already well know) and they are covering all points of E and with total length ∞i=1(di − } P ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 518 The Integral Chapter 8 continuous at all points except perhaps at the points of a set of measure zero. (In modern language they are said to be continuous almost everywhere or continuous a.e.) Theorem 8.17 (Riemann-Lebesgue) A function f on an interval [a, b] is Riemann integrable if and only if f is bounded and the set of points in [a, b] at which f is not continuous is a set of measure zero. Proof. The necessity is not difficult to prove but is the least important part for us. The sufficiency is more important and harder to prove. Throughout the proof we require a familiarity with the notion of the oscillation ωf (x) of a function f at a point x as discussed in Section 6.7. Recall that this value is positive if and only if f is discontinuous at x. Let us suppose that f is Riemann integrable. Certainly f is bounded. Fix e > 0 and consider the set N (e) of points x such that the oscillation of f at x is greater than e; that is, so that Any interval (c, d) that contains a point x ∈ Let ε > 0 and use Theorem 8.15 to find intervals ωf (x) > e. N (e) will certainly have ωf ([c, d]) e. ≥ forming a partition of the interval [a, b] and such that n [x0, x1], [x1, x2], . . . , [xn 1, xn], − ωf ([xk 1, xk](xk − − xk − 1) < εe/2. Xk=1 Select from this collection just those intervals that contain a point from N (e) in their interior. The total length of these intervals cannot exceed (eε)/(2e) since for each such interval [xk ωf ([xk 1, xk] we must have 1, xk]) e. − Thus we have succeeded in covering the set N (e) by a sequence of open intervals (xk 1, xk) of total − ≥ length less than ε/2, except for an oversight. One or more of the points xi} { might be in the set N (e), and − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.6. The Riemann Integral 519 we have neglected to cover it. Since there are only finitely many such points, we ca |
n add a few sufficiently short intervals to our collection. Thus we have proved that for any ε > 0 the set N (e) can be covered by a collection of open intervals of total length less than ε. It follows that N (e) has measure zero. But the set of points of discontinuity of f is the union of the sets N (1), N (1/2), N (1/4), N (1/8), . . . . Since each of these is measure zero, it follows from Theorem 6.34 that the set of points of discontinuity of f has measure zero too as required. This proves the theorem in one direction. In the other suppose that f is bounded, say that M for all x and that the set E of points in [a, b] at which f is not continuous is a set of measure zero. Let ε > 0. By Theorem 8.16 we need to find at least one partition f (x) | ≤ | of the interval [a, b] so that n [x0, x1], [x1, x2], . . . , [xn 1, xn] − Let E1 denote the set of points x in [a, b] at which the oscillation is greater than or equal to ε/(2(b ωf ([xk 1, xk])(xk − − xk − 1) < ε. Xk=1 that is, for which ωf (x) ε/(2(b a)). − ≥ a)); − This set is closed (see Theorem 6.27) and, being a subset of E, it must have measure zero. Now closed sets of measure zero can be covered by a finite number of small open intervals of total length smaller than ε/(4M + 1). (See Theorem 6.35.) We can assume that these open intervals do not have endpoints in common. Note that, at points in the intervals that remain, the oscillation of f is smaller than ε/(2(b these intervals may be subdivided into smaller intervals on which the oscillation is at least that small (Exercise 8.6.6). a)). Consequently, − Thus we may construct a partition [x0, x1], [x1, x2], . . . , [xn 1, xn], − ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 520 The Integral Chapter 8 of the interval [a, b] consisting of two kinds of closed intervals: (i) the first kind cover all the points of E1 and have total length smaller than ε/(4M + 1) and (ii) the remaining kind contain no points of E1 and the oscillation of f on each of these intervals is smaller than ε/(2(b a)); that is, The sum ωf ([xk − 1, xk]) < ε/(2(b n − a)). − splits into two sums depending on the intervals of type (i) or type (ii). The former sum contributes no more than ωf ([xk 1, xk])(xk − − xk 1) − Xk=1 while the latter sum contributes no more than (2M ) × ε/(4M + 1) < ε/2 Altogether, then, and the proof is complete. ε/(2(b a)) (b × − − a) = ε/2. n Xk=1 ωf ([xk 1, xk])(xk − − xk − 1) < ε 8.6.4 What Functions Are Riemann Integrable? " Advanced section. May be omitted. Theorem 8.17 exactly characterizes those functions that are Riemann integrable as the class of bounded functions that do not have too many points of discontinuity. We should recognize immediately that certain types of functions that we are used to working with are also integrable. We express these as corollaries to our theorem. (Recall that step functions were defined in Section 5.2.6.) Corollary 8.18: Every step function on an interval [a, b] is Riemann integrable there. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.6. The Riemann Integral 521 Proof. A step function is bounded and has only finitely many discontinuities. Thus the set of discontinuities has measure zero. Consequently, the corollary follows from Theorem 8.17. Corollary 8.19: Every bounded function with only countably many points of discontinuity in an interval [a, b] is Riemann integrable there. Proof. The corollary follows directly from Theorem 8.17 since countable sets have measure zero. Corollary 8.20: Every function monotonic on an interval [a, b] is Riemann integrable there. Proof. A monotonic function is bounded and has only countably many discontinuities. Consequently, this corollary follows from the preceding corollary. Corollary 8.21: If a function f is Riemann integrable on an interval [a, b] then the function Riemann integrable on that interval. f | | is also Proof. The corollary follows directly from Theorem 8.17 since if f is Riemann integrable on [a, b] it must has be bounded and continuous at every point except a set of measure zero. Exercise 8.6.7 shows that precisely the same properties. f | | Exercises 8.6.1 Show directly from Theorem 8.16 that the characteristic function of the rationals is not Riemann integrable on any interval. See Note 232 8.6.2 Show that the product of two Riemann integrable functions is itself Riemann integrable. 8.6.3 If f is Riemann integrable on an interval and f is never zero, does it follow that 1/f is Riemann integrable there? What extra hypothesis could we invoke to make this so? See Note 233 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 522 The Integral Chapter 8 8.6.4 If f is Riemann integrable on an interval [a, b] show that for every ε > 0 there are a pair of step functions so that See Note 234 L(x) f (x) ≤ ≤ U (x) b a Z (U (x) − L(x)) dx < ε. 8.6.5 Let f be a function on an interval [a, b] with the property that for every ε > 0 there are a pair of step functions L(x) f (x) ≤ ≤ U (x) so that Show that f is Riemann integrable. b a Z (U (x) − L(x)) dx < ε. 8.6.6 Suppose that the oscillation ωf (x) of a function f is smaller than η at each point x of an interval [c, d]. Show that there must be a partition [x0, x1], [x1, x2], . . . , [xn−1, xn], of [c, d] so that the oscillation on each member of the partition. See Note 235 ωf ([xk−1, xk]) < η 8.6.7 Show that the set of points at which a function F is discontinuous includes all points at which discontinuous but not conversely. Deduce Corollary 8.21 as a result of this observation from Theorem 8.17. is F | | 8.6.8 Deduce Corollary 8.18 directly from Theorem 8.15 rather than from Theorem 8.17. 8.6.9 Deduce Corollary 8.19 directly from Theorem 8.15. 8.6.10 Deduce Corollary 8.20 directly from Theorem 8.15. 8.6.11 Show that the converse of Corollary 8.21 does not hold. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.7. Properties of the Riemann Integral 523 8.7 Properties of the Riemann Integral " The proofs in this section make use of the Lebesgue criterion for integrability. You may skip the proofs and just see how the properties are essentially unchanged from Section 8.3 for Cauchy’s original integral. The Riemann integral is an extension of Cauchy’s first integral from continuous functions to a larger class of bounded functions—those that are bounded and continuous except at the points of a small set (a set of measure zero). We have enlarged the class of functions to which the notion of an integral may be applied. Have we lost any of our crucial properties of Section 8.3? These properties express how we expect integration to behave; it would be distressing to lose any of them. In some cases they remain completely unchanged. In some cases they need to be modified slightly. But our goal was never simply to integrate as many functions as possible; it is to preserve the theory of the integral and to apply that theory sufficiently broadly to handle all necessary applications. If we lose our basic properties we have lost too much. Fortunately the Riemann integral keeps all of the basic properties of the integral of continuous functions. The few differences should be carefully noted. Note especially how some of the properties must be rephrased. 8.22 (Additive Property) If f is Riemann integrable on both intervals [a, b] and [b, c] then it is Riemann integrable on [a, c] and b a Z c c f (x) dx + f (x) dx = f (x) dx. b Z a Z Proof. The proof of the identity need not change from the way we handled it for continuous functions (check this). It is the first assertion in the statement that must be verified. We prove that f is Riemann integrable on [a, c]. By Theorem 8.17 if f is Riemann integrable on both of these intervals it is bounded on both and the set of points of discontinuity in each interval has measure zero. It follows that f is bounded on [a, c]. Also, its set of points of discontinuity in [a, c] is the union of the set of points of discontinuity in [a, b] and [b, c] ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 524 The Integral Chapter 8 together with (possibly) the point b itself. Thus the set of points of discontinuity in [a, c] is also of measure zero. Consequently, by Theorem 8.17, f is Riemann integrable. 8.23 (Linear Property) If f and g are both Riemann integrable on [a, b], then so too is any linear combination αf + βg and b a Z [αf (x) + βg(x)] dx = α f (x) dx + β g(x) dx. b b a Z a Z Again the proof of the identity does not change from the way we handled it for continuous Proof. functions (check this). It is the first assertion in the statement that needs to be verified. We must prove that αf + βg is Riemann integrable on [a, b]. The points of discontinuity of the function function αf + βg are either points of discontinuity of f or else they are points of discontinuity of g. If both functions f and g are Riemann integrable, then they are both bounded and continuous except at the points of a set of measure zero. It follows that αf + βg is bounded and continuous except at the points of a set of measure zero. Hence, by Theorem 8.17, αf + βg is Riemann integrable. 8.24 (Monotone Property) If f and g are both Riemann integrable on [a, b], then, if f (x) all a b, x g(x) for ≤ ≤ ≤ b a Z f (x) dx ≤ a Z b g(x) dx. Proof. The proof for continuous functions works equally well here. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.7. Properties of the Riemann Integral 525 8.25 (Absolute Property) If f is Riemann integrable on [a, b], then so too is or, equivalently, b − a | Z f (x) dx | ≤ b a Z b f (x) dx f (x) dx | ≤ a | Z and f | | Proof. The proof for continuous functions works equally well here because we have already shown, in Corollary 8.21 that if f is Riemann integrable on [a, b], then so too is f |
. | | b a Z f (x) dx b ≤ a | Z f (x) dx. | Fundamental Theorem of Calculus The next two properties, 8.26 and 8.27, are important. They show how the processes of integration and differentiation are inverses of each other. Together they are known as the fundamental theorem of calculus for the Riemann integral. You should note, however, a weakness in this theory. If we compute F ′ we cannot immediately conclude from 8.27 that F (a). We need first to check that F ′ is Riemann integrable. This may not always be easy. Worse yet, it may be false, even for bounded derivatives (see the discussion in Section 9.7). It was this failure of the Riemann integral to integrate all derivatives that Lebesgue claimed was his motivation to look for a more general theory of integration. b a F ′(x) dx = F (b) R − 8.26: (Differentiation of the Indefinite Integral) If f is Riemann integrable on [a, b] then the function x F (x) = f (t) dt is continuous on [a, b] and F ′(x) = f (x) at each point x at which the function f is continuous. a Z Proof. Once again, the proof for continuous functions works equally well here. Note, however, that we are no longer trying to prove that F ′(x) = f (x) at every point x, only at those points x where f is continuous. It is left to you to check the proof and verify that it works here, unchanged. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 526 The Integral Chapter 8 8.27 (Integral of a Derivative) Suppose that the function F is differentiable on [a, b]. Provided it is also true that F ′ is Riemann integrable on [a, b], then b a Z F ′(x) dx = F (b) F (a). − Proof. Again the proof for continuous functions works here. Exercises 8.7.1 Give a set of conditions under which the integration by substitution formula holds. See Note 236 b a Z f (φ(t))φ′(t) dt = f (x) dx φ(b) φ(a) Z 8.7.2 Give a set of conditions under which the integration by parts formula holds. b a Z f (t)g′(t) dt = f (b)g(b) f (a)(g(a) − b − a Z f ′(t)g(t) dt 8.7.3 Suppose that f is Riemann integrable on [a, b] and define the function x F (x) = f (t) dt. a Z (a) Show that F satisfies a Lipschitz condition on [a, b]; that is, that there exists M > 0 such that for every x, y ∈ [a, b], (b) If x is a point at which f is not continuous is it still possible that F ′(x) = f (x)? (c) Is it possible that F ′(x) exists but is not equal to f (x)? F (y) | − F (x) M y | x . | − | ≤ ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.7. Properties of the Riemann Integral 527 (d) Is it possible that F ′(x) fails to exist? 8.7.4 The function x F (x) = sin(1/t) dt 0 Z has a derivative at every point where the integrand is continuous. Does it also have a derivative at x = 0? 8.7.5 Improve Property 8.27 by assuming that F is continuous on [a, b] and allowing that F ′ exists at all points of [a, b] with finitely many exceptions. 8.7.6 " Do much better than the preceding exercise and improve Property 8.27 by assuming that F is continuous on [a, b] and allowing that F ′ exists at all points of [a, b] with countably many exceptions. 8.7.7 If f and g are Riemann integrable on an interval [a, b] show that b Z a f (x)g(x) dx 2 ! ≤ Z a b [f (x)]2 dx b [g(x)]2 dx . ! ! Z a This extends the Cauchy-Schwarz inequality of Exercise 8.3.10. 8.7.8 Show that the integration by parts formula of Exercise 8.3.7 extends to the case where f and g are continuous and f ′ and g′ are Riemann integrable. 8.7.9 " (More on the Fundamental Theorem of Calculus) Let f be bounded on [a, b] and continuous a.e. on [a, b]. Suppose that F is defined on [a, b] and that F ′(x) = f (x) for all x in [a, b] except at the points of some set of measure zero. (a) Is it necessarily true that F (x) − x F (a) = f (t) dt for every x [a, b]? ∈ a Z (b) Same question as in (a) but assume also that F is continuous. (c) Same question, but this time assume that F is a Lipschitz function. You may assume the nonelementary fact that a Lipschitz function H with H ′ = 0 a.e. must be constant. (d) Give an example of a Lipschitz function F such that F is differentiable, F ′ is bounded, but F ′ is not integrable. See Note 237 ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 528 The Integral Chapter 8 8.8 The Improper Riemann Integral " Enrichment section. May be omitted. The Riemann integral applies only to bounded functions. What should we mean by the integral 1 0 Z Since the integrand is unbounded on [0, 1], it is not Riemann integrable even though the integrand is continuous at all but one point. There is not much else for us to do but to back track by several decades and return to Cauchy’s second method; namely, we compute 1 F (δ) = lim 0+ What we should probably do now is to create a new hybrid integral by combining the Riemann integral lim 0+ → = 2. δ Z → δ δ dx √x with Cauchy’s second method. This is often called the improper Riemann integral. As before we give a definition that considers only one point of unboundedness (at the left endpoint of the interval) with the understanding that the ideas can be applied to any finite number of such points. Definition 8.28: Let f be a function on an interval (a, b] that is Riemann integrable on [a + δ, b] and that is unbounded in the interval (a, a + δ) for every 0 < δ < b a. Then we define dx √x ? to be b a Z − f (x) dx b f (x) dx lim 0+ → δ a+δ Z if this limit exists, and in this case the integral is said to be convergent. If both integrals converge the integral is said to be absolutely convergent. a Z b f (x) dx and b a | Z f (x) dx | ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.8. The Improper Riemann Integral 529 In the same way we also extend the Riemann integral from bounded intervals to unbounded ones. How should we interpret the integral This cannot exist as a Riemann integral since the definition is clearly restricted to finite intervals and would not allow any easy interpretation for infinite intervals. As before we use Cauchy’s second method to obtain ∞ dx x2 ? 1 Z dx x2 = lim We give a formal definition valid just for an infinite interval of the form [a, dx x2 = lim 1 →∞ 1 →∞ Z 1 X = 1. − ∞ Z X X 1 X ∞ ) is best split up into the sum of two integrals, from ( similar. The case ( −∞ of which can be handled in this fashion. , + ∞ ). The case ( , a] and [a, , b] is ), each −∞ ∞ −∞ Definition 8.29: Let f be a function on an interval [a, . Then we define [a, b] for a < b < ) that is Riemann integrable on every interval ∞ ∞ to be ∞ f (x) dx a Z X f (x) dx lim X a →∞ Z if this limit exists, and in this case the integral is said to be convergent. If both integrals converge the integral is said to be absolutely convergent. f (x) dx and ∞ a Z ∞ f (x) dx | | a Z Both of these definitions extend the Riemann integral to a more general concept. Note that in any applications using an improper Riemann integral of either type, we are obliged to announce whether the integral is convergent or divergent, and frequently whether it is absolutely or nonabsolutely convergent. ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) 530 The Integral Chapter 8 It might seem that this theory would be important to master and represents the final word on the subject of integration. By the end of the nineteenth century it had become increasingly clear that this theory of the Riemann integral itself was completely inadequate to handle the bounded functions that were arising in many applications. The extra step here, using Cauchy’s second method, designed to handle unbounded functions, also proved far too restrictive. The modern theory of integration was developed in the first decades of the twentieth century. The methods are different and even the language needs many changes. Thus, the material in these last few sections has largely a historical interest. Some mathematicians claim it has only that, others that learning this material is a good preparation for learning the more advanced material. Exercises 8.8.1 For what values of p, q are the integrals ordinary Riemann integrals, convergent improper Riemann integrals, or divergent improper Riemann integrals? 1 sin x xp dx and 1 (sin x)q x dx 0 Z 0 Z 8.9 More on the Fundamental Theorem of Calculus " Enrichment section. May be omitted. The Riemann integral does not integrate all bounded derivatives and so the fundamental theorem of calculus for this integral assumes the awkward form provided F is differentiable on [a, b] and the derivative F ′ is Riemann integrable there. b a F ′(x) dx = F (b) F (a) − Z The emphasized phrase is unfortunate. It means we have a limited theory and it also means that, in practice, we must always check to be sure that a derivative F ′ is integrable before proceeding to integrate ClassicalRealAnalysis.comThomson*Bruckner*BrucknerElementary Real Analysis, 2nd Edition (2008) Section 8.9. More on the Fundamental Theorem of Calculus 531 it. In Section 9.7 we shall show how to construct a function F that is everywhere differentiable on an interval and whose derivative F ′ is bounded but not itself Riemann integrable on any subinterval. Let us take another look at the integrability of derivatives to see if we can discover what goes wrong. We take a completely naive approach and start with the definition of the derivative itself. If F ′ = f everywhere, then, at each point ξ and for every ε > 0, there is a δ > 0 so that F (x′′) F (x′) f (ξ)(x′′ | − − x′) < ε(x′′ x′) − | (6) − x′ < δ. A careless student might argue that one can recover F (b) x′′ and 0 < x′′ − ≤ ξ for x′ ≤ follows. Let F (a) as a limit of Riemann sums for f as − be a partition of [a, b], and let ξi ∈ F (a) = F (b) [xi − n − where R = Xi=1 n Xi=1 a = x0 < x1 < x2 < < xn = b · · · 1, xi]. Then (F (xi − 1) − F (xi)) = n Xi=1 f (ξi)(xi − xi − 1) + R (F (xi) F (xi − 1) f (ξi)(xi − − xi − 1)) . − Thus F (b) that, if the partition is finer than the number δ |
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