Apostol Solution Ch7 Mathematical Analysis 2e
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The Riemann-Stieltjes Integral Riemann-Stieltjes integrals b
7.1 Prove that d b a, directly from Definition 7.1. a
Proof: Let f 1 on a, b, then given any partition P a x 0 , . . . , x n b, then we have n
SP, 1,
ft k k , where t k x k1 , x k k1 n
k k1
b a. b
So, we know that d b a. a
7.2 If f R on a, b and if ab fd 0 for every f which is monotonic on a, b,
prove that must be constant on a, b.
Proof: Use integration by parts, and thus we have b
a df fbb faa Given any point c a, b, we may choose a monotonic function f defined as follows. f
0 if x c 1 if x c.
So, we have b
a df c b. So, we know that is constant on a, b.
7.3 The following definition of a Riemann-Stieltjes integral is often used in the
literature: We say that f is integrable with respect to if there exists a real number A having the property that for every 0, there exists a 0 such that for every partition P of a, b with norm P and for every choice of t k in x k1 , x k , we have |SP, f, A| . b (a) Show that if fd exists according to this definition, then it is also exists according a to Definition 7.1 and the two integrals are equal. Proof: Since refinement will decrease the norm, we know that if there exists a real number A having the property that for every 0, there exists a 0 such that for every partition P of a, b with norm P and for every choice of t k in x k1 , x k , we have |SP, f, A| . Then choosing a P with P , then for P P P . So, we have |SP, f, A| . b
That is, fd exists according to this definition, then it is also exists according to a Definition 7.1 and the two integrals are equal.
(b) Let fx x 0 for a x c, fx x 1 for c x b, b fc 0, c 1. Show that fd exists according to Definition 7.1 but does not exist a by this second definition. b
b
Proof: Note that fd exists and equals 0 according to Definition 7.1If fd exists a a according to this definition, then given 1, there exists a 0 such that for every partition P of a, b with norm P and for every choice of t k in x k1 , x k , we have |SP, f, | 1. We may choose a partition P a x 0 , . . . , x n b with P and c x j , x j1 , where j 0, . . . , n 1. Then SP, f, fxx j1 x j 1, where x c, x j1 which contradicts to |SP, f, | 1.
7.4 If f R according to Definition 7.1, prove that ab fxdx also exists according to
definition of Exercise 7.3. [Contrast with Exercise 7.3 (b).] b Hint: Let I fxdx, M sup|fx| : x a, b. Given 0, choose P so that a UP , f I /2 (notation of section 7.11). Let N be the number of subdivision points in P and let /2MN. If P , write UP, f
M k fx k S 1 S 2 ,
where S 1 is the sum of terms arising from those subintervals of P containing no points of P and S 2 is the sum of remaining terms. Then S 1 UP , f I /2 and S 2 NMP NM /2, and hence UP , f I . Similarly, LP, f I if P for some . Hence |SP, f I| if P min, . Proof: The hint has proved it. Remark: There are some exercises related with Riemann integrals, we write thme as references. b (1) Suppose that f 0 and f is continuous on a, b, and fxdx 0. Prove that a fx 0 on a, b. Proof: Assume that there is a point c a, b such that fc 0. Then by continuity of fc f, we know that given 2 0, there is a 0 such that as |x c| , x a, b, we have fc |fx fc| 2 which implies that fc fx if x c , c a, b : I 2 So, we have b fc 0 |I| fxdx fxdx 0, where 0 |I|, the length of I 2 I a which is absurb. Hence, we obtain that fx 0 on a, b. (2) Let f be a continuous function defined on a, b. Suppose that for every continuous function g defined on a, b which satisfies that
b
a gxdx 0, we always have b
a fxgxdx 0. Show that f is a constant function on a, b. b
Proof: Let fxdx I, and define gx fx a
I ba
, then we have
b
a gxdx 0, which implies that, by hypothesis, b
a fxgxdx 0 which implies that b
a fx cgxdx 0 for any real c. So, we have b
a gx 2 dx 0 if letting c
I ba which implies that gx 0 forall x a, b by (1). That is, fx
I ba
on a, b.
(3) Define 0 if x 0, 1 Q hx
1 n
if x is the rational number m/n (in lowest terms) 1 if x 0.
Then h R0, 1. Proof: Note that we have shown that h is continuous only at irrational numbers on 0, 1 Q. We use it to show that h is Riemann integrable, i.e., h R0, 1. Consider the upper sum UP, f as follows. Given 0, there exists finitely many points x such that fx /2. Consider a partition P x 0 a, . . . , x n b so that its subintervals I j x j1 , x j for some j containing those points and |I j | /2. So, we have n
UP, f
M k x k k1
1
2
/2 /2 where 1 1 M j I j , and 2 , is the sum of others. So, we have shown that f satisfies the Riemann condition with respect to x x. Note: (1) The reader can show this by Theorem 7.48 (Lebesgue’s Criterion for Riemann Integrability). Also, compare Exercise 7.32 and Exercise 4.16 with this. (2) In Theorem 7.19, if we can make sure that there is a partition P such that
UP , f, LP , f, , then we automatically have, for any finer P P , UP, f, LP, f, since the refinement makes U increase and L decrease. (4) Assume that the function fx is differentiable on a, b, but not a constant and that fa fb 0. Then there exists at least one point on a, b for which 4 b a 2
|f |
b
a fxdx.
Proof: Consider sup xa,b |f x| : M as follows. (i) If M , then it is clear. (ii) We may assume that M . Let x a, ab , then 2 fx fx fa f yx a Mx a, where y a, x. , b, then and let x ab 2
*
fx fx fb f zx b Mb x, where z x, b. So, by (*) and (**), we know that
**
ab 2
b
b
a fxdx a
fxdx ab fxdx 2
M
ab 2
a
b
x adx M ab b xdx 2
M ab 2
2
which implies that b
4 fxdx. b a 2 a Note that by (*) and (**), the equality does NOT hold since if it was, then we had f x M on a, b which implies that f is a constant function. So, we have M
b
4 fxdx. b a 2 a By definition of supremum, we know that there exists at least one point on a, b for which M
|f |
4 b a 2
b
a fxdx.
(5) Gronwall Lemma: Let f and g be continuous non-negative function defined on a, b, and c 0. If x
fx c gtftdt for all x a, b, a
then x
a gtdt
fx ce In particular, as c 0, we have f 0 on a, b. Proof: Let c 0 and define
.
x
Fx c gtftdt, a
then we have (i). Fa c 0. (ii). F x gxfx 0 F is increasing on a, b by Mean Value Theorem (iii). Fx fx on a, b F x gxFx by (ii). So, from (iii), we know that x
x
a gtdt
a gtdt
Fx Fae ce by (i). For c 0, we choose c n 1/n 0, then by preceding result, x
gtdt fx 1 0 as n . ne a So, we have proved all. (6) Define fx
x1
x
sint 2 dt.
(a) Prove that |fx| 1/x if x 0. Proof: Let x 0, then we have, by change of variable(u t 2 ), and integration by parts, fx 1 2
x1 2
x
1 2 1 2
2
sin u du u
x1 2
x
2
dcos u u
cos u u
x1 2 x2
x1 2 x2
cos x 1 2 cosx 2 2x 2x 1
cos u du 2u 3/2
x1 2 x2
cos u du 4u 3/2
which implies that, |fx|
cos x 1 2 cosx 2 2x 2x 1
x1 2
x
2
x1 2
cos u du 4u 3/2
du 1 1 1 2x 4 x2 2x 1 u 3/2 1 1 1 1 2x 2x 2x 1 2x 1 1/x. Note: There is another proof by Second Mean Value Theorem to show above as follows. Since fx 1 2
x1 2
x
2
sin u du by (a), u
we know that, by Second Mean Value Theorem,
fx 1 2 1 2 1 2 which implies that
1 x
y
x2 sin udu x 1 1
x1 2
y
sin udu
1 cosx 2 cos y 1 cos y cos x 1 2 x x1 1x 1 cosy 1x cosx 2 1 cos x 1 2 x1 x1 cos x 1 2 cosx 2 1 1 x |cos y| x x1 x1
|fx| 1 2
cos x 1 2 cosx 2 x x1
1 1 x x1
1 2
1 1 1 1x 1 x 2 x1 x1 2 since no x makes |cosx | cos x 1 2
1
1/x. (b) Prove that 2xfx cosx 2 cos x 1 2 rx, where |rx| c/x and c is a constant. Proof: By (a), we have x1 2 cos x 1 2 cosx 2 cos u du fx 2 2x 2x 1 4u 3/2 x which implies that 2xfx
cosx 2
cosx 2
x1 2
x cos x 1 2 x x1 x2 cos x 1
2
cos u du 2u 3/2 x1 2
1 cos x 1 2 x x1 x2
cos u du 2u 3/2
where rx 1 cos x 1 2 x 2 x1
x1 2
x
2
cos u du u 3/2
which implies that |rx| 1 x 2 x1
x1 2
x
2
x1 2
1 x 2 x2 x1 1 1 x1 x1 2 x.
|cos u| du u 3/2 u 3/2 du
Note: Of course, we can use the note in (a) to show it. We write it as follows. Proof: Since fx 1 2 which implies that
1x
1 x1
cosy 1x cosx 2 1 cos x 1 2 x1
x 1 cosy cosx 2 x cos x 1 2 x1 x1 2 x 1 cosy 1 2 cosx cos x 1 cos x 1 2 x1 x1
2xfx
where 1 cos x 1 2 x1
rx
x 1 cosy x1
which implies that 1 1 x x1 x1 2 x1 2/x.
|rx|
(c) Find the upper and lower limits of xfx, as x . Proof: Claim that lim sup x cosx 2 cos x 1 2 2 as follows. Taking x n 2 , where n Z, then cosx 2 cos x 1 2 cos n 8 1 . If we can show that n 8 is dense in 0, 2 modulus 2. It is equivalent to show that n 2 is dense in 0, 1 modulus 1. So, by lemma ar : a Z, where r Q c is dense in 0, 1 modulus 1, we have proved the claim. In other words, we have proved the claim. Note: We use the lemma as follows. ar b : a Z, b Z, where r Q c is dense in R. It is equivalent to ar : a Z, where r Q c is dense in 0, 1 modulus 1. Proof: Say ar b : a Z, b Z S, and since r Q c , then by Exercise 1.16, there are infinitely many rational numbers h/k with k 0 such that |kr h| 1k . Consider x , x : I, where 0, and thus choosing k 0 large enough so that 1/k 0 . Define L |k 0 r h 0 |, then we have sL I for some s Z. So, sL sk 0 r sh 0 S. That is, we have proved that S is dense in R.
(d) Does sint 2 dt converge? 0
Proof: Yes, x
x sin 2 tdt
x
1 2
x
1 2
x sin udu x1 2 2 x x 1 x
sin u du by the process of (a) u y
x
y sin udu
by Second Mean Value Theorem
1 2 2x which implies that the integral exists.
Note: (i) We can show it without Second Mean Value Theorem by the method of (a). However Second Mean Value Theorem is more powerful for this exercise. (ii) Here is the famous Integral named Dirichlet Integral used widely in the STUDY of Fourier Series. We write it as follows. Show that the Dirichlet Integral 0 sinx x dx
1
converges but not absolutely converges. In other words, the Dirichlet Integral converges conditionally. Proof: Consider x
x
sin x dx 1 x x
y
x sin xdx x1
x
y sin xdx by Second Mean Value Theorem;
we have x
x
sin x dx 2 2 4 . x x x x So, we know that Dirichlet Integral converges. Define I n 4 2n, 2 2n, then 0 sinx x dx In sinx x dx
I
2 2 4
n
2n 2 2
dx
4
n0
4 . 2n
So, we know that Dirichlet Integral does NOT converges absolutely. (7) Deal similarity with fx
x1
x
sine t dt.
Show that e x |fx| 2 and that e x fx cose x e 1 cose x1 rx, where |rx| min1, Ce x , for all x and Proof: Since x1
fx
x
e
e x1 x
sine t dt sin u du by Change of Variable (let u e t ) u e x1
ex cos e x1 cos e x e x1 e x which implies that
cos u du by Integration by parts u2 e x1
e x cos e x1 du since cos u is not constant 1 |fx| cos x1 ex x u2 e e 1 1 1 1 e1x x1 e ex e which implies that e x |fx| 2. In addition, by (*), we have e x fx cose x e 1 cose x1 rx, where
*
rx
e x
e x1
e
x
cos u du u2
which implies that |rx| 1 e 1 1 for all x or which implies that, by Integration by parts, |rx| e x
ex
ex
e x1
e
x
cos u du u2 e x1
sin e x1 sin e x 2 ex e 2x1 ex 1
**
e 2x1
e x1
12x 2 e ex
du u3
sin u du u3 since sin u is not constant 1
2e x for all x. By (**) and (***), we have proved that |rx| min1, Ce x for all x, where C 2.
***
Note: We give another proof on (7) by Second Mean Value Theorem as follows. Proof: Since fx
e x1
e
x
sin u du u y
e x1
1 sin udu by Second Mean Value Theorem e1x sin udu x1 e ex y 1 cos y cos e x1 e1x cos e x cos y x1 e which implies that e x |fx| |cos e x cos y e 1 cos y cos e x1 | |cos e x e 1 cos e x1 cos y1 e 1 | |cos e x e 1 cos e x1 | 1 e 1 1 e 1 1 e 1 since no x makes |cos e x | |cos e x1 | 1. 2. In addition, by (*), we know that e x fx cos e x e 1 cos e x1 rx where rx e 1 cos y cos y which implies that |rx| 1 e 1 1 for all x. In addition, from the proof of the process in (7), we know that
*
**
|rx|
ex
e x1
e
x
1 e 2x
ex
cos u du u2 y
e
e x1
y
1
cos udu
cos udu e 2x1 e x |sin y sin e x e 2 sin e x1 sin y| e x |sin y1 e 2 e 2 sin e x1 sin e x | e x 1 e 2 e x |e 2 sin e x1 | e x |sin e x | x
e x 1 e 2 e x 1 e 2 since no x makes |sin e x1 | |sin e x | 1 2e x for all x. So, by (**) and (***), we have proved that |rx| min1, Ce x , where C 2. (8) Suppose that f is real, continuously differentiable function on a, b, fa fb 0, and b
a f 2 xdx 1. Prove that b
a xfxf xdx
1 2
and that b
b
a f x 2 dx a x 2 f 2 xdx
1. 4
Proof: Consider b
b
a xfxf xdx a xfxdfx b
xf 2 x| ba fxdxfx a
b
b
a
a
f 2 xdx xfxf xdx since fa fb 0, so we have b
a xfxf xdx
1 . 2 In addition, by Cauchy-Schwarz Inequality, we know that b
a
b
b
f x 2 dx x 2 f 2 xdx
a
2
1. 4 a Note that the equality does NOT hold since if it was, then we have f x kxfx. It implies that f x kxfxe
xfxf xdx
kx 2 2
0
which implies that fe
kx 2 2
0
which implies that fx Ce which implies that
kx 2 2
, a constant
***
C 0 since fa 0. That is, fx 0 on a, b which is absurb. 7.5 Let a n be a sequence of real numbers. For x 0, define x
Ax
an an, nx
n1
where x is the largest integer in x and empty sums are interpreted as zero. Let f have a continuous derivative in the interval 1 x a. Use Stieltjes integrals to derive the following formula: a
a n fn 1 Axf xdx Aafa. na
Proof: Since a
a
1 Axf xdx 1 Axdfx since f has a continous derivative on 1, a a
fxdAx Aafa A1f1 by integration by parts 1
a n fn Aafa by na
a
a
1 fxdAx a n fn and A1 a 1 , n2
we know that a
a n fn 1 Axf xdx Aafa. na
7.6 Use Euler’s summation formula, integration by parts in a Stieltjes integral, to derive the following identities: n n 1 (a) k1 k1s n s1 s xx s1 dx if s 1. 1
Proof: n
k1s k1
n
1 x s dx 1 n
xdx s n s n 1 s 1 1 1 n
x dx n 1s s1 1 x n x 1s1 s s1 dx if s 1. n 1 x
s
n
(b) k1 Proof:
1 k
log n
n xx x2
1
1.
n
1k
k1
n
1
1 dx 1 x n
xdx 1 n 1 n 1 1 1 1 1
n
n
n
1 x 1 dx 1 x 1 dx 1
log n
x dx 1 x2
x x 1. x2
n 1
7.7 Assume that f is continuous on 1, 2n and use Euler’s summation formula or integration by parts to prove that 2n
2n
1 k fk 1
f xx 2x/2dx.
k1
Proof: 2n
2n
n
1 fk fk 2 f2k k
k1
k1 2n
1
k1
2n
1
fxdx f1 2
fxdx/2
2n
2n
1
1
xdfx 2nf2n 2 x/2dfx f2n2n/2 f11/2 since
2n
1
f
is continuous on 1, 2n 2n
f xxdx 2nf2n f xx/2dx 2nf2n 1
2n
1
f xx 2x/2dx.
7.8x Let 1 x x
2 1 tdt. If 0 that
f
1 2
if x integer, and let 1 0 if x integer. Also, let is continuous on 1, n prove that Euler’s summation formula implies
n
n
n
fk 1 fxdx 1 2 xf xdx k1
f1 fn . 2
Proof: Using Theorem 7.13, then we have n
f1 fn 2
n
n
n
n
n
2 xdf x f n 2 n f 1 2 1
fk 1 fxdx 1 f x 1 xdx k1
1 fxdx 1 f xd 2 x
1 fxdx
n
1
f1 fn 2 f1 fn 2
f1 fn 2 n n f1 fn fxdx 2 xf xdx since f is continuous on 1, n. 2 1 1
n
n
1 fxdx 1 2 xdf x
7.9 Take fx log x in Exercise 7.8 and prove that n t log n! n 1 log n n 1 2 2 dt. 2 t 1 Proof: Let fx log x, then by Exercise 7.8, it is clear. So, we omit the proof. Remark: By Euler’s summation formula, we can show that n n log k 1 log xdx 1 x x 12 dxx log2 n .
*
1kn
Since x x 1 2
1/2
and a1
a
x x 1 dx 0 for all real a, 2 we thus have the convergence of the improper integral 1 x x 12 dx by Second Mean Value Theorem. So, by (*), we have log n! n 1 log n n C n 2 where , C 1 x x 1 dx 2 x 1
**
and n
n
x x 1 2
dx . x
So, n! e C : C 1 . e n n n1/2 Now, using Wallis formula, we have 2 2 4 4 2n2n /2 lim n 1 3 3 5 5 2n 12n 1 which implies that 2 n n! 4 1 o1 /2 2n! 2 2n 1 which implies that, by (***), 4 C 41 2 n n n1/2 e n 1 o1 /2 C 21 2n 2n1/2 e 2n 2n 1 lim n
which implies that C 21 n 1 o1 /2. 22n 1 Let n , we have C 1
2 , and x x 1
1 2
dx
1 2
log 2 1.
Note: In (***), the formula is called Stirling formula. The reader should be noted that Wallis formula is equivalent to Stirling formula.
***
7.10 If x 1, let x denote the number of primes p x, that is, x 1, px
where the sum is extended over all primes p x. The prime number theorem states that log x x x 1. lim x This is usually proved by studying a related function given by x
log p, px
where again the sum is extended over all primes p x. Both function and are step functions with jumps at the primes. This exercise shows how the Riemann-Stieltjes integral can be used to relate these two functions. (a) If x 2, prove that x and x can be expressed as the following Riemann-Stieltjes integrals: x x 1 dt. x log tdt, x log t 3/2 3/2 Note. The lower limit can be replaced by any number in the open interval 1, 2. Proof: Since x px log p, we know that by Theorem 7.9, x
x
3/2 log tdt,
and x px 1, we know that by Theorem 7.9, x
x
3/2
1 dt. log t
(b) If x 2, use integration by parts to show that x x log x
x 2
t t dt,
x t x dt. 2 log x 2 t log t These equations can be used to prove that the prime number theorem is equivalent to x the relation lim x x 1.
x
Proof: Use integration by parts, we know that x x t x log tdt t dt log xx log3/23/2 3/2 3/2 x t t dt log xx since 3/2 0 3/2 x log x and
x 2
t t dt since
2
3/2
t t dt 0 by x 0 on 0, 2,
x
x
3/2
x
3/2
1 dt log t
t x 3/2 dt 2 log x log3/2 t log t
t x dt since 3/2 0 2 log x t log t x t 2 t x dt since dt 0 by x 0 on 0, 2. 2 2 log x 2 t log t 3/2 t log t
x
3/2
7.11 If on a, b, prove that b c b (a) fd fd fd, (a c b) a
a
c
Proof: Given 0, there is a partition P such that UP, f, Ia, b . Let P c P P 1 P 2 , where P 1 a x 0 , . . . , x n 1 c and P 2 x n 1 c, . . . , x n 2 bthen we have Ia, c Ic, b UP 1 , f, UP 2 , f, UP , f, UP, f, . So, by (1) and (2), we have Ia, c Ic, b Ia, b since is arbitrary. On the other hand, given 0, there is a partition P 1 and P 2 such that UP 1 , f, UP 2 , f, Ia, c Ic, b which implies that, let P P 1 P 2 UP, f, UP 1 , f, UP 2 , f, Ia, c Ic, b . Also, Ia, b UP, f, . By (3) and (4), we have Ia, b Ia, c Ic, b since is arbitrary. So, by (*) and (**), we have proved it. b
b
1
2 *
3 4 **
b
(b) f gd fd gd. a a a Proof: In any compact interval J, we have supf g sup f sup g. xJ
xJ
xJ
1
So, given 0, there is a partition P f and P g such that n1
b
M k f k a fd /2
2
k1
and n2
b
M k g k a gd /2. k1
So, consider P P f P g , then we have, by (1), UP, f g, UP, f, UP, g, along with
3
n1
UP, f,
M k f k and UP, g, k1
n2
M k g k k1
which implies that, by (2) and (3), b
b
a fd a gd
UP, f g, which implies that b
a
f gd
b
a
b
fd gd a
since is arbitrary. b
b
b
(c) f gd fd gd a a a
Proof: Similarly by (b), so we omit the proof.
7.12 Give an example of bounded function f and an increasing function defined on b
a, b such that |f| R but for which a fd does not exist. Solution: Let 1 if x 0, 1 Q
fx
1 if x 0, 1 Q c
and x x on 0, 1. Then it is clear that f R on a, b and |f| R on a, b. 7.13 Let be a continuous function of bounded variation on a, b. Assume that x g R on a, b and define x gtdt if x a, b. Show that: a (a) If f on a, b, there exists a point x 0 in a, b such that b b x fd fa 0 gd fb gd. a
a
x0
Proof: Since is a continuous function of bounded variation on a, b, and g R on a, b, we know that x is a continuous function of bounded variation on a, b, by Theorem 7.32. Hence, by Second Mean-Value Theorem for Riemann-Stieltjes integrals, we know that b
x0
a fd fa a
b
dx fb dx x0
which implies that, by Theorem 7.26, b
x0
a fd fa a
b
gd fb gd. x0
(b) If, in addition, f is continuous on a, b, we also have b
x0
a fxgxdx fa a
b
gd fb gd. x0
Proof: Since b
b
a fd a fxgxdx by Theorem 7.26, we know that, by (a),
b
x0
a fxgxdx fa a
b
gd fb gd. x0
Remark: We do NOT need the hypothesis that f is continuous on a, b.
7.14 Assume that f R on a, b, where is of bounded variation on a, b. Let
Vx denote the total variation of on a, x for each x in a, b, and let Va 0. Show that b
a fd
b
a |f|dV MVb,
where M is an upper bound for |f| on a, b. In particular, when x x, the inequality becomes b
a fd
Mb a.
Proof: Given 0, there is a partition P a x 0 , . . . , x n b such that b
a fd SP, f, n
ft k k , where t k x k1 , x k k1 n
|ft k ||x k x k1 | k1 n
|ft k |Vx k Vx k1 k1
SP, |f|, V UP, |f|, V since V is increasing on a, b which implies that, taking infimum, b
b
a fd a |f|dV since |f| RV on a, b. So, we have b
a fd
b
a |f|dV
b
since |f|dV is clear non-negative. If M is an upper bound for |f| on a, b, then (*) implies a that b
a fd
b
a |f|dV MVb
which implies that b
a fd
Mb a
if x x.
7.15 Let n be a sequence of functions of bounded variation on a, b. Suppose
there exists a function defined on a, b such that the total variation of n on a, b tends to 0 as n . Assume also that a n a 0 for each n 1, 2, . . . . If f is
*
continuous on a, b, prove that lim n
b
b
a fxd n x a fxdx.
Proof: Use Exercise 7.14, we then have b
a fxd n x
MV n b 0 as n
where V n is the total variation of n , and M sup xa,b |fx|. So, we have lim n
b
b
a fxd n x a fxdx.
Remark: We do NOT need the hypothesis a n a 0 for each n 1, 2, . . . . 7.16 If f R, f 2 R, g R, and g 2 R on a, b, prove that 1 2
b
a a b
a
fx gx fy gy
b
b
a
f 2 xdx
2
dy dx
g 2 xdx
b
a fxgxdx
2
.
When on a, b, deduce the Cauchy-Schwarz inequality b
a
2
fxgxdx
b
b
a f 2 xdx a g 2 xdx
.
(Compare with Exercise 1.23.) Proof: Consider 1 2
b
b
a a b
b
b
b
fx gx fy gy
2
dy dx
1 2
a a fxgy fygx 2 dy
1 2
a a f 2 xg 2 y 2fxgyfygx f 2 yg 2 xdy
1 2
a f 2 xdx a g 2 ydy
b
dx
b
b
b
fxgxdx fygydy a b
b
a
a
a
g 2 xdx f 2 ydy
b
b
b
a f 2 xdx a g 2 ydy a fxgxdx
if on a, b, then we have
2
,
dx
0 1 2
b
a
b
2
fx gx fy gy
b
a a
f 2 xdx
b
a
dy dx
g 2 ydy
b
a fxgxdx
2
which implies that 2
b
a fxgxdx
b
b
a f 2 xdx a g 2 xdx
b
b
a
a
.
Remark: (1) Here is another proof: Let A f 2 xdx, B fxgxdx, and b
C g 2 xdx. From the fact, a
0
b
a fxz gx 2 dx for any real z
Az 2 2Bz C. It implies that B 2 AC. That is, b
a fxgxdx
2
b
b
a f 2 xdx a g 2 xdx
.
Note: (1) The reader may recall the inner product in Linear Algebra. We often consider Riemann Integral by defining f, g :
b
a fxgxdx
where f and g are real continuous functions defined on a, b. This definition is a real case. For complex case, we need to preserve its positive definite. So, we define f, g :
b
a fxg xdx
where f and g are complex continuous functions defined on a, b, and g means its conjugate. In addition, in this sense, we have the triangular inequality: f g f h f h, where f
f, f .
(2) Suppose that f R on a, b where on a, b and given 0, then there
exists a continuous function g on a, b such that f g .
Proof: Let K sup xa,b |fx|, and given 0, we want to show that f g . Since f R on a, b where on a, b, given 1 0, there is a partition P x 0 a, . . . , x n b such that n
UP, f, LP, f,
M j f m j f j 2 . j1
Write P A B, where A x j : M j f m j f and B x j : M j f m j f , then
1
j B
M j f m j f j 2 by (1) B
which implies that
j .
2
B
For this partition P, we define the function g as follows. tx x t gt x j j x j1 fx j1 x j xj1 fx j , where x j1 t x j . j1 So, it is clear that g is continuous on a, b. In every subinterval x j1 , x j t x j1 xj t |ft gt| x j x j1 ft fx j1 x j x j1 ft fx j |ft fx j1 | |ft fx j | 2M j f m j f Consider
x A
j1 ,x j
2 |ft gt| d
4M j f m j f 2 j by (3) A
4 M j f m j f j by definition of A A
4 j by 1
A 4 b
a
and
x B
j1 ,x j
2 |ft gt| d
4K 2 j B
4K 2 by (2). Hence, b
a |ft gt| 2 d 4 b a 4K 2 2 if we choose is small enough so that 4 b a 4K 2 2 . That is, we have proved that f g . P.S.: The exercise tells us a Riemann-Stieltjes integrable function can be approximated (approached) by continuous functions. (3)There is another important result called Holder’s inequality. It is useful in Analysis and more general than Cauchy-Schwarz inequality. In fact, it is the case p q 2 in Holder’s inequality. We consider the following results. Let p and q be positive real numbers such that 1 1 1. q p Prove that the following statements. (a) If u 0 and v 0, then
3
p p uv up vq .
Equality holds if and only if u p v q . p
p
Proof: Let fu up vq uv be a function defined on 0, , where p 0, q 0 and v 0, then f u u p1 v. So, we know that f u 0 if u 1
1 q
1,
1
v p1 ,
and f u 0 if u
0, fu 0. Hence, we know that fu 0 for all u 0.
which implies that, by f v p1 up
1
0, v p1
1 p
1
vp
That is, uv p q . In addition, fu 0 if and only if u v p1 if and only if u p v q . So, Equality holds if and only if u p v q . Note: (1) Here is another good proof by using Young’s Inequality, let fx be an strictly increasing and continuous function defined on x : x 0, with f0 0. Then we have, let a 0 and b 0, ab
a
b
0 fxdx 0 f 1 xdx, where f 1 is the inverse function of f.
And the equality holds if and only if fa b. Proof: The proof is easy by drawing the function f on x y plane. So, we omit it. So, by Young’s Inequality, let fx x , where 0, we have the Holder’s inequality. (2) The reader should be noted that there are many proofs of (a), for example, using the concept of convex function, or using A. P. G. P. along with continuity. (b) If f, g R on a, b where on a, b, f, g 0 on a, b, and b
b
a f p d 1 a g q d, then b
a fgd 1. Proof: By Holder’s inequality, we have gq fp fg p q b
b
a
a
which implies that, by on a, b, and f p d 1 g q d, b
b
b fp gq 1 1 d p q d p q 1. a
a fgd a
(c) If f and g are complex functions in R, where on a, b, then b
a
b
a
fgd
1/p
|f| p d
b
a
|g| q d
1/q
Proof: First, we note that b
a fgd
b
a |fg|d.
Also, b
b
a |f| p d M p a
|f| M
p
d 1
.
*
and b
b
a |g| q d N q a
|g| N
q
d 1.
Then we have by (b), b
a
|f| |g| d 1 M N
which implies that, by (*) b
a fgd
MN
1/p
b
a |f| p d
b
a |g| q d
1/q
.
(d) Show that Holder’s inequality is also true for the ”improper” integrals. Proof: It is clear by (c), so we omit the proof. 7.17 Assume that f R, g R, and f g R on a, b. Show that 1 2
b
b
a a fy fxgy gxdy b
b a fxgxdx a
dx
b
b
a fxdx a gxdx
.
If on a, b, deduce the inequality b
b
a fxdx a gxdx
b
b a fxgxdx a
when both f and g are increasing (or both are decreasing) on a, b. Show that the reverse inequality holds if f increases and g decreases on a, b. Proof: Since 1 2 1 2
b
b
b
b
a a fy fxgy gxdy
dx
a a fygy fygx fxgy fxgxdy b
b a fygydy a
b
dx
b
a fxdx a gxdx
which implies that, (let , f, and g on a, b), b
b
b
b
0 1 fy fxgy gxdy dx 2 a a and (let , and f on a, b, g on a, b), 0 1 fy fxgy gxdy dx, 2 a a we know that, (let , f, and g on a, b) b
b
a fxdx a gxdx
b
b a fxgxdx a
and (let , and f on a, b, g on a, b) b
b
a fxdx a gxdx Riemann integrals
b
b a fxgxdx. a
7.18 Assume f R on a, b. Use Exercise 7.4 to prove that the limit ba lim n n
n
f
a akb n
k1
b
exists and has the value fxdx. Deduce that n
lim n
k1
a
n , lim 4 n k2 n2
n
n 2 k 2 1/2 log
1 2 .
k1
Proof: Since f R on a, b, given 0, there exists a 0 such that as P , we have b
SP, f fxdx . a
For this , we choose n large enough so that So,
ba n
, that is, as n N, we have
b
SP, f fxdx a
which implies that ba n
n
f
a akb n
k1
b
fxdx . a
That is, ba lim n n
n
f
a akb n
k1
b
exists and has the value fxdx. a n n n 1 Since k1 k 2 n 2 n k1 n
lim n
k1
1 k n
2
1
, we know that by above result,
n 1 lim n n k2 n2
n
1 n
n k1
1
0
k1
1 2 1 k n
dx 1 x2 arctan 1 arctan 0 /4. 1 , we know that by above result, 2 1/2
Since k1 n 2 k 2 1/2
1
n
k n
ba n
.
n
lim n
n
1 n 2 k 2 1/2 lim n n
k1
1
0
0
0
1
1 1 nk
k1
2
1/2
dx 1/2 1 x 2
/4
/4
sec d, let x tan sec sec tan d sec tan
1 2
du , let sec tan u u
log 1 2 .
7.19 Define x
2
0 e t dt
fx
2
1
0
, gx
e x 2 t 2 1 dt. t2 1
(a) Show that g x f x 0 for all x and deduce that fx gx /4. Proof: Since f x 2
x
0 e t dt 2
e x
2
x 2 t 2 1
and note that if hx, t e t 2 1 , we know that h is continuous on 0, a 0, 1 for any real a 0, and h x 2xe x 2 t 2 1 is continuous on 0, a 0, 1 for any real a 0, g x
1
0 h x dt 1
0 2xe x t 1 dt 2
2e x
2
2
1
0 xe xt dt 2
x
2e x 2 e u 2 du, 0
0 for all x. Hence, we have fx gx C for all x, we know that 1 dt constant. Since C f0 g0 1t 2 /4, fx gx /4. g x
f x
0
Remark: The reader should think it twice on how to find the auxiliary function g. (b) Use (a) to prove that lim x
x
0 e t dt 2
1 . 2
Proof: Note that 2
2
x t 1 hx, t e 2 t 1
|e x 2 t 2 1 |
1 for all x 0; x 2 t 2 1
we know that 1
0 So, by (a), we get
e x 2 t 2 1 dt t2 1
1 x2
1
0
dt 1 t2
0 as x .
lim fx /4 x which implies that x
0 e t dt
lim x x
x
0
0
2
2 2 since lim x e t dt exists by e t dt
x
dt 0 1t 2
1 2 arctan x /2 as x .
Remark: (1) There are many methods to show this. But here is an elementary proof with help of Taylor series and Wallis formula. We prove it as follows. In addition, the reader will learn some beautiful and useful methods in the future. For example, use the application of Gamma function, and so on. Proof: Note that two inequalities,
1
x2
e x2
xk!
2k
for all x
k0
and
k0
x 2k k!
x 2k
1 if |x| 1 1 x2
k0
which implies that 1 x 2 e x 2 if 0 x 1 1 x 2 n e nx 2
1
and 1 if x 0 e nx 2 1 x2
2 e x
1 1 x2
n
.
2
So, we have, by (1) and (2), 1
1
0 1 x 2 n dx 0 e nx dx 0 e nx dx 0 2
2
1 1 x2
n
dx.
3
Note that
0 e nx dx 2
1 n
0 e x dx : 2
K . n
Also, /2
1
0 1 x 2 n dx 0
sin 2n1 tdt
2 4 6 2n 22n 1 3 5 2n 1
and
0
1 1 x2
n
dx
/2
0
sin 2n2 tdt
1 3 5 2n 3 , 2 4 6 2n 2 2
so 2 4 6 2n 22n 1 3 5 2n 3 K n 1 3 5 2n 1 2 4 6 2n 2 2 which implies that 2 4 6 2n 22n 2 1 3 5 2n 3 2 2n 1 n n 2 K 2n 1 1 3 5 2n 1 2 2n 1 2n 1 2 4 6 2n 2 2 By Wallis formula, we know that, by (4) . K 2 That is, we have proved that Euler-Possion Integral n
2
2
4
. 2
0 e x dx 2
Note: (Wallis formula) lim n
2 4 6 2n 22n 2 . 2 2 1 3 5 2n 1 2n 1
Proof: As 0 x /2, we have sin 2n1 t sin 2n t sin 2n1 t, where n N. So, we know that /2
0
sin 2n1 tdt
/2
0
/2
0
sin 2n tdt
sin 2n1 tdt
which implies that 2n 12n 3 3 1 2n 22n 4 4 2 2n2n 2 4 2 . 2 2n2n 2 4 2 2n 12n 3 3 1 2n 12n 1 3 1 So, 2n2n 2 4 2 2n 12n 3 3 1 Hence, from
2
2n2n 2 4 2 2n 12n 3 3 1
1 2n 1 2
2n2n 2 4 2 2n 12n 3 3 1
2
1 1 2n 2n 1
2
1 . 2n
1 0, 2n 2
we know that lim n
2 4 6 2n 22n 2 . 2 2 1 3 5 2n 1 2n 1
(2) Here is another exercise from Hadamard’s result. We Write it as follows. Let f C k R with f0 0. Prove that there exists an unique function g C k1 R such that f xgx on R. Proof: Consider fx fx f0 1
0 dfxt
0 xf xtdt
1
1
x f xtdt; 0
we know that if gx :
1 0
f xtdt,
then we have prove it.
Note: In fact, we can do this job by rountine work. Define gx
fx x
if x 0
0 if x 0.
However, it is too long to write. The trouble is to make sure that g C k1 R. x
7.20 Assume g R on a, b and define fx gtdt if t a, b. Prove that the a
x
integral |gt|dt gives the total variation of f on a, x. a x
Proof: Since |gt|dt exists, given 0, there exists a partition a P 1 x 0 a, . . . , x n x such that LP, |g|
x
a |gt|dt .
1
So, for this P 1 , we have n
|fx k fx k1 | k1
n
xk
x k1
n
gtdt
k1
|c k x k x k1 | by Mean Value Theorem
2
k1
where inf xx k1 ,x k |gx| c k sup xx k1 ,x k |gx|. Hence, we know that, by (1) and (2), n
x
|fx k fx k1 | a |gt|dt k1
which implies that x
a |gt|dt
V f a, b
since is arbitrary. x Conversely, since |gt|dt exists, given 0, there exists a partition P 2 such that a
UP 2 , |g|
x
a |gt|dt /2.
3
Also, for the same , there exists a partition P 3 t 0 a, . . . , t m x such that m
V f a, b /2
|ft k ft k1 |. k1
Let P P 2 P 3 s 0 a, . . . , s p x, then by (3) and (4), we have UP, |g|
x
a |gt|dt /2
and p
V f a, b /2
|fs k fs k1 | k1 p
sk
s k1 p
gtdt
k1
|c k x k x k1 | k1
UP, |g| which imply that V f a, b
x
a |gt|dt
since is arbitrary. Therefore, from above discussion, we have proved that V f a, b
x
a |gt|dt.
7.21 If f f 1 , . . . , f n be a vector-valued function with a continuous derivative f on
4
a, b. Prove that the curve described by f has length f a, b
b
a f tdt.
Proof: Since f f 1 , . . . , f n is continuous on a, b, we know that 2
n f j t j1
1/2
f t is uniformly continuous on a, b. So, given 0, there exists a 1 0 such that as |x y| 1 , where x, y a, b, we have . 1 |f x f y| 3b a Since f t R on a, b, for the same , there exists 2 0 such that as P 1 2 , where P 1 x 0 a, . . . , x n bwe have b
SP 1 , f f tdt /3, where SP 1 , f a
n
f t j x j
2
j1
and f a, b exists by Theorem 6.17, for the same , there exists a partition P 2 s 0 a, . . . , s m b such that m
f a, b /3
fs k fs k1 k1
m
n
k1
j1
1/2
f j s k f j s k1
2
.
3
Let min 1 , 2 and P P 2 so that P , where P y 0 a, . . . , y q bthen by (1)-(3), we have (i) As |x y| , where x, y a, b, we have . 4 |f x f y| 3b a (ii) As P , we have SP, f
b a
q
f tdt
/3, where
SP, f
f tj y j
5
j1
(iii) As P , we have f a, b /3
m
n
k1
j1
q
n
k1
j1
q
n
k1 q
j1
1/2
f j s k f j s k1
2
1/2
f j y k f j y k1 2 f j z k 2
y j , by Mean Value Theorem
f z k y j k1
f a, b By (ii) and (iii), we have
6
q
gz k y j SP, g
k1
q
q
k1 q
j1
gz k y j gtj y j |gz k gtj |y j k1 q
3b a y j k1
/3.
7
Hence, (5)-(7) implies that b
a f tdt f a, b
.
Since is arbitrary, we have proved that f a, b
b
a f tdt.
7.22 If f n1 is continuous on a, x, define x I n x 1 x t n f n1 tdt. n! a (a) Show that I k1 x I k x
f k ax a k , k 1, 2, . . . , n. k!
Proof: Since, for k 1, 2, . . . , n, x I k x 1 x t k f k1 tdt k! a x 1 x t k df k t k! a x x 1 x t k f k t a k x t k1 f k tdt k! a k x k f ax a 1 x t k1 f k tdt k! k 1! a
f k ax a k I k1 x, k!
we know that I k1 x I k x
f k ax a k , for k 1, 2, . . . , n. k!
(b) Use (a) to express the remainder in Taylor’s formula (Theorem 5.19) as an integral. Proof: Since fx fa I 0 x, we know that
fx fa I 0 x n
fa I k1 x I k x I n x k1
n
k0
f k ax a k 1 n! k!
So, by Taylor’s formula, we know that f n1 cx a n1 R n x 1 n! n 1! where R n x is the remainder term.
x
a x t n f n1 tdt by (a).
x
a x t n f n1 tdt, for some c a, x.
Remark: 1. The reader should be noted that with help of Mean Value Theorem, we have x n1 cx a n1 1 x t n f n1 tdt f . n! a n 1!
*
2. Use Integration by parts repeatedly; we can show (*). Of course, there is other proofs such as Mathematical Induction. Proof: Since
uv n1 dt uv n u v n1 u v n2 . . . 1 n u n v 1 n1 u n1 vdt, letting vt x t n and ut ft, then n
fx
f k0
k a
k!
x a k 1 n!
x
a x t n f n1 tdt.
Note: The reader should give it a try to show it. Since it is not hard, we omit the detail. 3. The remainder term as an integral is useful; the reader should see the textbook in Ch9, pp242-244. 4. There is a good exercise related with an application of Taylor’s Remainder. We write it as a reference. Let u t ftut 0, where ft is continuous and non-negative on 0, c If u is defined and not a zero function on 0, c and b
a b ta tft b a for all a, b 0, c, where a b.
*
Then u at most has one zero on 0, c. Proof: First, we note that u has at most finitely many zeros in the interval 0, c by uniqueness theorem on O.D.E. So, let ua ub 0, where a, b 0, c with a b, and no point y a, b such that uy 0. Consider a, b and by Taylor’s Theorem with Remainder Term as an integral, we have x
ux ua u ax a x tu tdt x
a
u ax a x tu tdt a x
u ax a x tutftdt. a
Note that ux is positive on a, b ( Or, ux is negative on a, b ) So, we have
**
|ux| |u a|x a. By (**), b
0 ub u ab a b tutftdt a
which implies that u ab a
b
a b tutft
which implies that by (***), and note that u a 0, ba
b
a b tt aftdt
which contradicts to (*). So, u at most has one zero on 0, c. Note: (i) In particular, let ft e t , we have (*) holds. Proof: Since b
a b tt ae t dt e a 2 b a e b 2 b a by integration by parts twice, we have, (let b a x), e a 2 b a e b 2 b a b a e a 2 x e xa 2 x x xe a 1 e ax 2e x x 2 0 since a b and e x 1 x. (ii) In the proof of exercise, we use the uniqueness theorem: If px and qx are continuous on 0, a, then y pxy qxy 0, where y0 y 0 , and y 0 y 0 has one and only one solution. In particular, if y0 y 0 0, then y 0 on 0, a is the only solution. We do NOT give a proof; the reader can see the book, Theory of Ordinary Differential Equation by Ince, section 3.32, or Theory of Ordinary Differential Equation by Coddington and Levison, Chapter 6. However, we need use the uniqueness theorem to show that u (in the exercise) has at most finitely many zeros in 0, c. Proof: Let S x : ux 0, x 0, c. If #S , then by Bolzano-Weierstrass Theorem, S has an accumulation point p in 0, c. Then up 0 by continuity of u. In addition, let r n p, and ur n 0, then ux up ur n up u p lim lim 0. xp rn p xp n (Note that if p is the endpoint of 0, c, we may consider x p or x p ). So, by uniqueness theorem, we then have u 0 on 0, c which contradicts to the hypothesis, u is not a zero function on 0, c. So, #S . 7.23 Let f be continuous on 0, a. If x 0, a, define f 0 x fx and let x f n1 x 1 x t n ftdt, n 0, 1, 2, . . . n! 0 (a) Show that the nth derivative of f n exists and equals f. Proof: Consider, by Chain Rule,
***
f n
1 n 1!
x
0 x t n1 ftdt f n1 for all n N,
we have f n n f. That is, nth derivative of f n exists and equals f. Remark: (1) There is another proof by Mathematical Induction and Integration by parts. It is not hard; we omit the proof. (2) The reader should note that the exercise tells us that given any continuous function f on a, b, there exists a function g n on a, b such that g n n f, where n N. In fact, the function x g n 1 x t n ftdt, n 0, 1, 2, . . . n! a (3) The reader should compare the exercise with 7.22. At the same time, look at two integrands in both exercises. (b) Prove the following theorem of M. Fekete: The number of changes in sign of f in 0, a is not less than the number of changes in sign in the ordered set of numbers fa, f 1 a, . . . , f n a. Hint: Use mathematical induction. Proof: Let Tf denote the number of changes in sign of f on 0, a and S n f the number of changes in sign in the ordered set of numbers fa, f 1 a, . . . , f n a. We prove Tf S n f for each n by Mathematical Induction as follows. Note that S n f n. As n 1, if S 1 f 0, then there is nothing to prove it. If S 1 f 1, it means that faf 1 a 0. Without loss of generality, we may assume that fa 0, so f 1 a 0 which implies that a 0 f 1 a 1 ftdt 0! 0 which implies that there exists a point y 0, a such that fy 0. Hence, Tf S 1 f holds for any continuous functions defined on 0, a. Assume that n k holds for any continuous functions defined on 0, a, As n k 1, . we consider the ordered set of numbers fa, f 1 a, . . . , f k a, f k1 a. Note that f n1 a f 1 n a for all n N, so by induction hypothesis, Tf 1 S k f 1 Suppose S k f 1 p, and f 1 0 0, then f 1 f at least has p zeros by Rolle’s Theorem. Hence, * Tf Tf 1 S k f 1 p We consider two cases as follows. (i) faf 1 a 0 :With help of (*), Tf S k f 1 S k1 f.
(ii) faf 1 a 0 :Claim that Tf S k f 1 p as follows. Suppose NOT, it means that Tf Tf 1 p by (*). Say fa 1 fa 2 . . . fa p 0, where 0 a 1 a 2 . . . a p 1. and f 1 b 1 f 1 b 2 . . . f 1 b p 0, where 0 b 1 b 2 . . . b p 1. By faf 1 a 0, we know that fxf 1 x 0 where x 0, c, c mina 1 , b 1 which is impossible since fxf 1 x fxf 1 x f 1 0 by f 1 0 0 fxf 1 y, where y 0, x 0, c fxfy 0 since fx and fy both positive or negative. So, we obtain that Tf S k f 1 p. That is, Tf S k f 1 1 S k1 f. From above results, we have proved it by Mathmatical Induction. (c) Use (b) to prove the following theorem of Feje’r: The number of changes in sign of f in 0, a is not less than the number of changes in sign in the ordered set a
a
a
0
0
0
f0, ftdt, tftdt, . . . , t n ftdt. Proof: Let gx fa x, then, define g 0 x gx, and for n 0, 1, 2, . . . , a g n1 a 1 a t n gtdt n! 0 a 1 u n fudu by change of variable (u a t). n! 0 So, by (b), the number of changes in sign of g in 0, a is not less than the number of changes in sign in the ordered set ga, g 1 a, . . . , g n1 a. That is, the number of changes in sign of g in 0, a is not less than the number of changes in sign in the ordered set a
a
a
0
0
0
f0, ftdt, tftdt, . . . , t n ftdt. Note that the number of changes in sign of g in 0, a equals the number of changes in sign of f in 0, a, so we have proved the Feje’r Theorem.
7.24 Let f be a positive continuous function in a, b. Let M denote the maximum
value of f on a, b. Show that
lim n
b
a
fx n dx
1/n
M.
Proof: Since f is a positive continuous function in a, b, there exists a point c a, b such that fc M sup xa,b fx 0. Then given M 0, there is a 0 such that as x Bc, a, b : I, we have 0 M fx M . Hence, we have
1/n
I f n xdx
|I| 1/n M
1/n
b
a fx n dx
b a 1/n M
which implies that M lim inf n 1/n
b
So, lim n inf fx n dx a
b
lim n sup fx n dx
1/n
a
b
1/n
a fx dx n
M.
M since is arbitrary. Similarly, we can show that b
M. So, we have proved that lim n fx n dx
1/n
a
M.
Remark: There is good exercise; we write it as a reference. Let fx and gx are continuous and non-negative function defined on a, b. Then b
a fx gxdx
lim n
n
1/n
max fx. xa,b
x
Since the proof is similar, we omit it. (The reader may let x gtdt). a
7.25 A function f of two real variables is defined for each point x, y in the unit square 0 x 1, 0 y 1 as follows: 1 if x is rational,
fx, y 1
1
0
0_
2y if x is irrational.
(a) Compute fx, ydx and fx, ydx in terms of y. Proof: Consider two cases for upper and lower Riemann-Stieltjes integrals as follows. (i) As y 0, 1/2 : Given any partition P x 0 0, . . . , x n 1, we have sup fx, y 1, and inf fx, y 2y. xx j1 ,x j
1
1
0
0_
xx j1 ,x j
Hence, fx, ydx 1, and fx, ydx 2y. (ii) As y 1/2, 1 : Given any partition P x 0 0, . . . , x n 1, we have sup fx, y 2y, and inf fx, y 1. xx j1 ,x j
1
xx j1 ,x j
1
Hence, fx, ydx 2y, and fx, ydx 1. 0
0_
1
t
(b) Show that fx, ydy exists for each fixed x and compute fx, ydy in terms of x 0 0 and t for 0 x 1, 0 t 1. Proof: If x Q 0, 1, then fx, y 1. And if x Q c 0, 1, then fx, y 2y. So, for each fixed x, we have 1
1
0 fx, ydy 0 1dy 1 if x Q 0, 1 and 1
1
0 fx, ydy 0 2ydy 1 if x Q c 0, 1. In addition, t
t
0 fx, ydy 0 1dy t if x Q 0, 1
and t
t
0 fx, ydy 0 2ydy t 2 if x Q c 0, 1. 1
1
0
0
(c) Let Fx fx, ydy. Show that Fxdx exists and find its value. Proof: By (b), we have Fx 1 on 0, 1. 1
So, Fxdx exists and 0
1
0 Fxdx 1. 7.26 Let f be defined on 0, 1 as follows: f0 0; if 2 n1 x 2 n , then
fx 2 n for n 0, 1, 2, . . .
1
(a) Give two reasons why fxdx exists. 0
Proof: (i) fx is monotonic decreasing on 0, 1. (ii) x : f is discontinuous at x measure zero.
has
Remark: We compute the value of the integral as follows. Solution: Consider the interval I n 2 n , 1 where n N, then we have f R on I n for each n, and n
1
2
n
fxdx
2 k1
2 k1 n
k
2 k1
k1 n
fxdx 2 k1
2
dx
k
2 k1 2 k k1
2
1
So, th integral fxdx 0
1 4
1 14 1 14
2 1 3 2 3
1 4
n
n
2 as n . 3
.
Note: In the remark, we use the following fact. If f R on a, b, then b
b
a fxdx lim n a
fxdx n
where a n is a sequence with a n a, and a n a for all n. Proof: Since a n a, given 0, there is a positive integer N such that as n N, we have |a n a| /M, where M sup |fx| xa,b
So,
b
b
a fxdx a
fxdx n
an
a
fxdx
M|a n a| . b
b
That is, fxdx lim n fxdx. a
an
x
(b) Let Fx ftdt. Show that for 0 x 1 we have 0 Fx xAx 1 Ax 2 , 3 where Ax 2 log x/ log 2 and where y is the greatest integer in y. 1
1
Proof: First, we note that Fx ftdt ftdt 0 x consider the value of the integral
2 3
1
ftdt. So, it suffices to x
1
x ftdt. Given any x 0, 1, then there exists a positive N such that 2 N1 x 2 N . So, 1
1
x ftdt 2
N
ftdt
2 1 3
1 4
2 1 3 3
1 4
2 N x N
N
ftdt 2 N 2 N x by Remark in (a)
1 2
N
x.
So, N x 1 1 3 4 2 N x 1 2 2N 3 xAx 1 Ax 2 3 where Ax 2 N . Note that 2 N1 x 2 N , we have N log 1/2 x , where y is the Gauss symbol.
Fx
1 2
N
Hence, Ax 2 log x/ log 2 . Remark: (1) The reader should give it a try to show it directly by considering 0, x, where 0 x 1. (2) Here is a good exercise. We write it as a reference. Suppose that f is defined on 0, 1 by the following fx
1 2n
if x
j 2n
where j is an odd integer and 0 j 2 n , n 1, 2, . . . , 0 otherwise.
Show that f R on 0, 1 and has the value of the integral 0. Proof: In order to show this, we consider the Riemann’s condition with respect to x x as follows. Given a partition n j P x 0 0 20n , x 1 21n , x 2 22n , . . . , x j 2 n , . . . , x 2 n 22 n 1 , then the upper sum
2n
M k x k
UP, f
k1
1n 2
2n
Mk k1
1n 2n n 1 2 2 So, f satisfies the Riemann’s condition on 0, 1.
0 as n .
Note: (1) The reader should give it a try to show that the set of discontinuities of f has measure zero. Thus by Theorem 7.48 (Lebesgue’s Criterion for Riemann Integral), we know that f R on 0, 1. In addition, by the fact, the lower Riemann integral equals the Riemann integral, we know that its integral is zero. (2) For the existence of Riemann integral, we summarize to be the theorem: Let f be a bounded function on a, b. Then the following statements are equivalent: (i) f R on a, b. (ii) f satisfies Riemann’s condition on a, b. b b (iii) fxdx fxdx a a (iv) the set of discontinuities of f on I has measure zero. P.S.: The reader should see the textbook, pp 391; we have the general discussion.
7.27 Assume f has a derivative which is monotonic decreasing and satisfies f x m 0 for all x in a, b. Prove that b
a cos fxdx
2. m
Hint: Multiply and divide the integrand by f x and use Theorem 7.37(ii). Proof: Since f x m 0, and b
b
a cos fxdx a
1 f
is monotonic increasing on a, b, we consider
cos fx f xdx f x 1
f b
b
c cos fxf xdx, by Theorem 7.37(ii) fb
1 cos udu, by Change of Variable f b fc sin fb sin fc f b
which implies that b
a cos fxdx
2. m
7.28 Given a decreasing seq uence of real numbers Gn such that Gn 0 as
n . Define a function f on 0, 1 in terms of Gn as follows: f0 1; if x is irrational, then fx 0; if x is rational m/n (in lowest terms), then fm/n Gn. Compute the oscillation f x at each x in 0, 1 and show that f R on 0, 1.
Proof: Let x 0 Q c 0, 1. Since lim n Gn 0, given 0, there exists a positive integer K such that as n K, we have |Gn| . So, there exists a finite number
of positive integers n such that Gn . Denote S x : |fx| , then #S . Choose a 0 such that x 0 , x 0 0, 1 does NOT contain all points of S. Note that fx 0 0. Hence, we know that f is continuous at x 0 . That is, f x 0 for all x Q c 0, 1. Let x 0 0, then it is clear that f 0 1 f0 0. So, f is not continuous at 0. Q 0, 1. Let x 0 Q 0, 1, say x 0 M (in lowest terms). Since Gn is monotonic N decreasing, there exists a finite number of positive integers n such that Gn GN. Denote T x : |fx| GN, then #T . Choose a 0 such that x 0 , x 0 0, 1 does NOT contain all points of T. Let h 0, , then supfx fy : x, y x 0 h, x 0 h 0, 1 fx 0 GN. So, f x 0 GN. That is, f x fx for all x Q 0, 1. Remark: (1) If we have proved f is continuous on Q c 0, 1, then f is automatically Riemann integrable on 0, 1 since D Q 0, 1 Q, the set of discontinuities of f has measure zero. (2) Here is a good exercise. We write it as a reference. Given a function f defined on a, b, then the set of continuities of f on a, b is G set. Proof: Let C denote the set of continuities of f on a, b, then C x : fx 0 and x : fx
k1 x : f x 1/k 1/k is open. We know that C is a G set.
Note: (i) We call S a G set if S n1 O n , where O n is open for each n. (ii) Given y x a, b : f x 1/k : I, then f y 1/k. Hence, there exists a d 0, such that f By, d 1/k, where By, d a, b For z By, d, consider a smaller so that Bz, By, d. Hence, f Bz, 1/k which implies that f z 1/k. Hence, By, d I. That is, y is an interior point of I. That is, I is open since every point of I is interior.
7.29 Let f be defined as in Exercise 7.28 with Gn 1/n. Let gx 1 if 0 x 1, g0 0. Show that the composite function h defined by hx gfx is not Riemann-integrable on 0, 1, although both f R and g R on 0, 1. Proof: By Exercise 7.28, we know that hx
0 if x Q c 0, 1 1 if x Q 0, 1
which is discontinuous everywhere on 0, 1. Hence, the function h (Dirichlet Function) is not Riemann-integrable on 0, 1. 7.30 Use Lebesgue’s theorem to prove Theorem 7.49. (a) If f is of bounded variation on a, b, then f R on a, b.
Proof: Since f is of bounded variation on a, b, by Theorem 6.13 (Jordan Theorem), f f 1 f 2 , where f 1 and f 2 are increasing on a, b. Let D i denote the set of discontinuities of f i on a, b, i 1, 2. Hence, D, the set of discontinuities of f on a, b is D D1 D2. Since |D 1 | |D 2 | 0, we know that |D| 0. In addition, f is of bounded on a, b since f is bounded variation on a, b. So, by Theorem 7.48, f R on a, b. (b) If f R on a, b, then f R on c, d for every subinterval c, d a, b, |f| R and f 2 R on a, b. Also, f g R on a, b whenever g R on a, b. Proof: (i) Let D a,b and D c,d denote the set of discontinuities of f on a, b and c, d, respectively. Then D c,d D a,b . Since f R on a, b, and use Theorem 7.48, |D a,b | 0 which implies that |D c,d | 0. In addition, since f is bounded on a, b, f is automatically is bounded on c, d for every compact subinterval c, d. So, by Theorem 7.48, f R on c, d. (ii) Let D f and D |f| denote the set of discontinuities of f and |f| on a, b, respectively, then D |f| D f . Since f R on a, b, and use Theorem 7.48, |D f | 0 which implies that |D |f| | 0. In addition, since f is bounded on a, b, it is clear that |f| is bounded on a, b. So, by Theorem 7.48, |f| R on a, b. (iii) Let D f and D f 2 denote the set of discontinuities of f and f 2 on a, b, respectively, then D f2 D f . Since f R on a, b, and use Theorem 7.48, |D f | 0 which implies that D f 2 0. In addition, since f is bounded on a, b, it is clear that f 2 is bounded on a, b. So, by Theorem 7.48, f 2 R on a, b. (iv) Let D f and D g denote the set of discontinuities of f and g on a, b, respectively. Let D fg denote the set of discontinuities of fg on a, b, then D fg D f D g Since f, g R on a, b, and use Theorem 7.48, |D f | |D g | 0 which implies that |D fg | 0. In addition, since f and g are bounded on a, b, it is clear that fg is bounded on a, b. So, by Theorem 7.48, fg R on a, b. (c) If f R and g R on a, b, then f/g R on a, b whenever g is bounded away from 0. Proof: Let D f and D g denote the set of discontinuities of f and g on a, b, respectively. Since g is bounded away from 0, we know that f/g is well-defined and f/g is also bounded on a, b. Consider D f/g , the set of discontinuities of f/g on a, b is D f/g D f D g . Since f R and g R on a, b, by Theorem 7.48, |D f | |D g | 0 which implies that |D f/g | 0. Since f/g is bounded on a, b with |D f/g | 0, f/g R on a, b by Theorem 7.48. Remark: The condition that the function g is bounded away from 0 CANNOT omit. For example, say gx x on 0, 1 and g0 1. Then it is clear that g R on 0, 1, but
1/g R on 0, 1. In addition, the reader should note that when we ask whether a function is Riemann-integrable or not, we always assume that f is BOUNDED on a COMPACT INTERVAL a, b. (d) If f and g are bounded functions having the same discontinuities on a, b, then f R on a, b if, and only if, g R on a, b. Proof: () Suppose that f R on a, b, then D f , the set of discontinuities of f on a, b has measure zero by Theorem 7.48. From hypothesis, D f D g , the set of discontinuities of g on a, b, we know that g R on a, b by Theorem 7.48. () If we change the roles of f and g, we have proved it.
(e) Let g R on a, b and assume that m gx M for all x a, b. If f is
continuous on m, M, the composite function h defined by hx fgx is Riemann-integrable on a, b.
Proof: Note that h is bounded on a, b. Let D h and D g denote the set of dsicontinuities of h and g on a, b, respectively. Then Dh Dg. Since g R on a, b, then |D g | 0 by Theorem 7.48. Hence, |D h | 0 which implies that h R on a, b by Theorem 7.48. Remark: (1) There has a more general theorem related with Riemann-Stieltjes Integral. We write it as a reference. (Theorem)Suppose g R on a, b, m gx M for all x a, b. If f is continuous on m, M, the composite function h defined by hx fgx R on a, b. Proof: It suffices to consider the case that is increasing on a, b. If a b, there is nothing to prove it. So, we assume that a b. In addition, let K sup xa,b |hx|. We claim that h satisfies Riemann condition with respect to on a, b. That is, given 0, we want to find a partition P such that n
UP, h, LP, h,
M k h m k h k . k1
Since f is uniformly continuous on m, M, for this 0, there is a 2K1 0 such that as |x y| where x, y m, M, we have . 1 |fx fy| 2b a Since g R on a, b, for this 0, there exists a partition P such that n
UP, g, LP, g,
M k g m k g k 2 . k1
Let P A B, where A x j : M k g m k g and B x j : M k g m k g , then k B
M k h m k h k 2 by (2) B
which implies that
k . B
2
So, we have UA, h, LA, h,
M k h m k h k A
/2 by (1) and UB, h, LB, h,
M k h m k h k B
2K k B
2K /2. It implies that UP, h, LP, h, UA, h, LA, h, UB, h, LB, h, . That is, we have proved that h satisfies the Riemann condition with respect to on a, b. So, h R on a, b. Note: We mention that if we change the roles of f and g, then the conclusion does NOT hold. Since the counterexample is constructed by some conclusions that we will learn in Real Analysis, we do NOT give it a proof. Let C be the standard Cantor set in 0, 1 and C the Cantor set with positive measure in 0, 1. Use similar method on defining Cantor Lebesgue Function, then there is a continuous function f : 0, 1 0, 1 such that fC C . And Choose g X C on 0, 1. Then h g f X C which is NOT Riemann integrable on 0, 1. (2) The reader should note the followings. Since these proofs use the exercise 7.30 and Theorem 7.49, we omit it. (i) If f R on a, b, then |f| R on a, b, and f r R on a, b, where r 0, . (ii) If |f| R on a, b, it does NOT implies f R on a, b. And if f 2 R on a, b, it does NOT implies f R on a, b. For example, f
1 if x Q a, b 1 Q c a, b
(iii) If f 3 R on a, b, then f R on a, b. 7.31 Use Lebesgue’s theorem to prove that if f R and g R on a, b and if fx m 0 for all x in a, b, then the function h defined by hx fx gx is Riemann-integrable on a, b. Proof: Consider hx exph log f, then by Theorem 7.49, f R log f R h log f R exph log f h R.
7.32 Let I 0, 1 and let A 1 I 13 , 23 be the subset of I obtained by removing
those points which lie in the open middle third of I; that is, A 1 0, 13 23 , 1. Let A 2 be the subset of A 1 obtained by removing he open middle third of 0, 13 and of 23 , 1. Continue this process and define A 3 , A 4 , . . . . The set C n1 A n is called the Cantor set. Prove that (a) C is compact set having measure zero. Proof: Write C n1 A n . Note that every A n is closed, so C is closed. Since A 1 is closed and bounded, A 1 is compact and C A 1 , we know that C is compact by Theorem 3.39. n In addition, it is clear that |A n | 23 for each n. Hence, |C| lim n |A n | 0, which implies that C has a measure zero.
(b) x C if, and only if, x n1 a n 3 n , where each a n is either 0 or 2. Proof: ()Let x C n1 A n , then x A n for all n. Consider the followings. (i) Since x A 1 0, 13 23 , 1, then it implies that a 1 0 or 2. (ii) Since x A 2 0, 19 29 , 39 69 , 79 89 , 99 , then it implies that a 2 0 or 2. Inductively, we have a n 0 or 2. So, x n1 a n 3 n , where each a n is either 0 or 2. () If x n1 a n 3 n , where each a n is either 0 or 2, then it is clear that x A n for each n. Hence, x C. (c) C is uncountable. Proof: Suppose that C is countable, write C x 1 , x 2 , . . . . We consider unique ternary expansion: if x n1 a n 3 n , then x : a 1 , . . . , a n , . . . . From this definition, by (b), we have x k x k1 , x k2 , . . . x kk, . . . where each component is 0 or 2. Choose y y 1 , y 2 , . . . where yj
2 if x jj 0 0 if x jj 2.
By (b), y C. It implies that y x k for some k which contradicts to the choice of y. Hence, C is uncountable. Remark: (1) In fact, C C means that C is a perfect set. Hence, C is uncountable. The reader can see the book, Principles of Mathematical Analysis by Walter Rudin, pp 41-42.
(2) Let C x : x n1 a n 3 n , where each a n is either 0 or 2 . Define a new function : C 0, 1 by
x
a2n n/2 , n1
then it is clear that is 1-1 and onto. So, C is equivalent to 0, 1. That is, C is uncountable. (d) Let fx 1 if x C, fx 0 if x C. Prove that f R on 0, 1. Proof: In order to show that f R on 0, 1, it suffices to show that, by Theorem 7.48, f is continuous on 0, 1 C since it implies that D C, where D is the set of discontinuities of f on 0, 1. Let x 0 0, 1 C, and note that C C , so there is a 0 such that
x 0 , x 0 C , where x 0 , x 0 0, 1. Then given 0, there is a 0 such that as x x 0 , x 0 , we have |fx fx 0 | 0 . Remark: (1) C C :Given x C n1 A n , and note that every endpoints of A n belong to C. So, x is an accumulation point of the set y : y is the endpoints of A n . So, C C . In addition, C C since C is closed. Hence, C C . (2) In fact, we have f is continuous on 0, 1 C and f is not continuous on C. Proof: In (d), we have proved that f is continuous on 0, 1 C, so it remains to show that f is not continuous on C. Let x 0 C, if f is continuous at x 0 , then given 1/2, there is a 0 such that as x x 0 , x 0 0, 1, we have |fx fx 0 | 1/2 which is absurb since we can choose y x 0 , x 0 0, 1 and y C by the fact C does NOT contain an open interval since C has measure zero. So, we have proved that f is not continuous on C. Note: In a metric space M, a set S M is called nonwhere dense if intclS . Hence, we know that C is a nonwhere dense set. Supplement on Cantor set. From the exercise 7.32, we have learned what the Cantor set is. We write some conclusions as a reference. (1) The Cantor set C is compact and perfect. (2) The Cantor set C is uncountable. In fact, #C #R. (3) The Cantor set C has measure zero. (4) The Cantor set C is nonwhere dense. (5) Every point x in C can be expressed as x n1 a n 3 n , where each a n is either 0 or 2. (6) X C : 0, 1 0, 1 the characteristic function of C on 0, 1 is Riemann integrable. The reader should be noted that Cantor set C in the exercise is 1 dimensional case. We can use the same method to construct a n dimensional Cantor set in the set x 1 , . . . , x n : 0 x j 1, j 1, 2, . . n. In addition, there are many researches on Cantor set. For example, we will learn so called Space-Filling Curve on the textbook, Ch9, pp 224-225. In addition, there is an important function called Cantor-Lebesgue Function related with Cantor set. The reader can see the book, Measure and Integral (An Introduction to Real Analysis) written by Richard L. Wheeden and Antoni Zygmund, pp 35.
7.33 The exercise outlines a proof (due to Ivan Niven) that 2 is irrational. Let n
fx x n 1 x /n!. Prove that:
(a) 0 fx 1/n! if 0 x 1. Proof: It is clear. (b) Each kth derivative f k 0 and f k 1 is an integer. Proof: By Leibnitz Rule,
f k x
1 n!
k
kj n n j 1x nj
1 kj n n k j 11 x nkj
j0
which implies that 0 if k n f k 0
.
1 if k n kn 1 kn n 2n k 1 if k n
So, f k 0 Z for each k N. Similarly, f k 1 Z for each k Z. Now assume that 2 a/b, where a and b are positive integers, and let n
Fx
bn
1 k f 2k x 2n2k . k0
Prove that: (c) F0 and F1 are integers. Proof: By (b), it is clear. (d) 2 a n fx sin x
d dx
F x sin x Fx cos x
Proof: Note that n
F x
2 Fx
bn
n
1 k f 2k2 x 2n2k
2bn
k0
1 k f 2k x 2n2k k0
n1
b n 1 k f 2k2 x 2n2k b n 1 n f 2n2 x k0
n
2 b n 1 k f 2k x 2n2k 2 b n fx 2n k1
n1
b n 1 k f 2k2 x 2n2k 1 k1 f 2k2 x 2n2k
k0 b n 1 n f 2n2 x
2 b n fx 2n
2 a n fx since f is a polynomial of degree 2n. So, d F x sin x Fx cos x dx sin xF x 2 Fx 2 a n fx sin x. 1
(e) F1 F0 a n fx sin xdx. 0
Proof: By (d), we have 1
2 a n fx sin xdx F x sin x Fx cos x| 10 0
F 1 sin F1 cos F 0 sin 0 F0 cos 0 F1 F0 which implies that
1
F1 F0 a n fx sin xdx. 0
(f) Use (a) in (e) to deduce that 0 F1 F0 1 if n is sufficiently large. This contradicts (c) and show that 2 (and hence ) is irrational. Proof: By (a), and sin x 0, 1 on 0, , we have 1 1 n n 0 a n fx sin xdx a sin xdx 2a 0 as n . n! 0 n! 0 So, as n is sufficiently large, we have, by (d), 0 F1 F0 1 which contradicts (c). So, we have proved that 2 (and hence ) is irrational. Remark: The reader should know that is a transcendental number. (Also, so is e). It is well-known that a transcendental number must be an irrational number. In 1900, David Hilbert asked 23 problems, the 7th problem is that, if 0, 1 is an algebraic number and is an algebraic number but not rational, then is it true that is a transcendental number. The problem is completely solved by Israil Moiseevic Gelfand in 1934. There are many open problem now on algebraic and transcendental numbers. For example, It is an open problem: Is the Euler Constant lim 1 1 . . . 1 n log n 2 a transcendental number. 7.34 Given a real-valued function , continuous on the interval a, b and having a finite bounded derivative on a, b. Let f be defined and bounded on a, b and assume that both integrals b
b
a fxdx and a fx xdx exists. Prove that these integrals are equal. (It is not assumed that is continuous.) Proof: Since both integrals exist, given 0, there exists a partition P x 0 a, . . . , x n b such that b
SP, f, fxdx /2 a
where n
SP, f,
ft j j for t j x j1 , x j j1 n
ft j s j x j by Mean Value Theorem, where s j x j1 , x j
*
j1
and b
SP, f fx xdx /2 a
where n
SP, f
ft j t j x j for t j x j1 , x j j1
So, let t j s j , then we have
**
SP, f, SP, f . Hence, b
b
a fxdx a fx xdx b
b
a
a
SP, f, fxdx SP, f fx xdx . So, we have proved that both integrals are equal.
7.35 Prove the following theorem, which implies that a function with a positive
integral must itself be positive on some interval. Assume that f R on a, b and that b 0 fx M on a, b, where M 0. Let I fxdx, let h 12 I/M b a, and a assume that I 0. Then the set T x : fx h contains a finite number of intervals, the sum of whose lengths is at least h. n Hint. Let P be a partition of a, b such that every Riemann sum SP, f k1 ft k x k satisfies SP, f I/2. Split SP, f into two parts, SP, f kA kB , where A k : x k1 , x k T, and B k : k A. If k A, use the inequality ft k M; if k B, choose t k so that ft k h. Deduce that kA x k h. Proof: It is clear by Hint, so we omit the proof. Remark: There is another proof about that a function with a positive integral must itself be positive on some interval. Proof: Suppose NOT, it means that in every subinterval, there is a point p such that fp 0. So, n
LP, f
m j x j 0 since m j 0 j1
for any partition P. Then it implies that sup LP, f P
b
a fxdx 0
which contradicts to a function with a positive integral. Hence, we have proved that a function with a positive integral must itself be positive on some interval. Supplement on integration of vector-valued functions. (Definition) Given f 1 , . . . , f n real valued functions defined on a, b, and let f f 1 , . . . , f n : a, b R n . If on a, b. We say that f R on a, b means that f j R on a, b for j 1, 2, . . . , n. If this is the case, we define b
b
b
a fd a f 1 d, . . . , a f n d
.
From the definition, the reader should find that the definition is NOT stranger for us. When we talk f f 1 , . . . , f n R on a, b, it suffices to consider each f j R on a, b for j 1, 2, . . . , n. For example, if f R on a, b where on a, b, then f R on a, b. Proof: Since f R on a, b, we know that f j R on a, b for j 1, 2, . . . , n. Hence,
n
f 2j R on a, b k1
which implies that, by Remark (1) in Exercise 7.30, 1/2
n
f
f 2j
R on a, b
R on a, b.
k1
Remark: In the case above, we have b
a fd
b
a fd.
Proof: Consider b
n
b
b
b
a f 1 d, . . . , a f n d, a f 1 d, . . . , a f n d
y 2
b
b
a f j d a f j d
,
j1 b
which implies that, (let y j f j d, y y 1 , . . . , y n ), a
n
2
y
b
y j a f j d j1 n
b
a f j y j d j1
n
b
a fjyj
d
j1
b
a fyd b
y fd a
which implies that y
b
a fd.
Note: The equality holds if, and only if, ft kty.
Existence theorems for integral and differential equations The following exercises illustrate how the fixed-point theorem for contractions. (Theorem 4.48) is used to prove existence theorems for solutions of certain integral and differential equations. We denote by Ca, b the metric space of all real continuous functions on a, b with the metric df, g f g max |fx gx|, xa,b
and recall the Ca, b is a complete metric space. 7.36 Given a function g in Ca, b, and a function K is continuous on the rectangle Q a, b a, b, consider the function T defined on Ca, b by the equation b
Tx gx Kx, ttdt, a
where is a given constant. (a) Prove that T maps Ca, b into itself. Proof: Since K is continuous on the rectangle Q a, b a, b, and x Ca, b, we know that b
a Kx, ttdt Ca, b. Hence, we prove that Tx Ca, b. That is, T maps Ca, b into itself. (b) If |Kx, y| M on Q, where M 0, and if || M 1 b a 1 , prove that T is a contraction of Ca, b and hence has a fixed point which is a solution of the integral b equation x gx Kx, ttdt. a
Proof: Consider T 1 x T 2 x
b
Kx, t 1 t 2tdt a b
|| |Kx, t 1 t 2t|dt a
b
||M | 1 t 2t|dt a
Since || (*), we know that
M 1 b
||Mb a 1 t 2t. a , then there exists c such that || c M 1 b a 1 . Hence, by 1
T 1 x T 2 x 1 t 2t where 0 cMb a : 1. So, T is a contraction of Ca, b and hence has a fixed b point which is a solution of the integral equation x gx Kx, ttdt. a
7.37 Assume f is continuous on a rectangle Q a h, a h b k, b k, where h 0, k 0. (a) Let be a function, continuous on a h, a h, such that x, x Q for all x in a h, a h. If 0 c h, prove that satisfies the differential equation y fx, y on a c, a c and the initial condition a b if, and only if, satisfies the integral equation x
x b ft, tdt on a c, a c. a
Proof: ()Since x a c, a c
t
ft, t on a c, a c and a b, we have, x
x a tdt a x
a ft, tdt on a c, a c. a
()Assume x
x b ft, tdt on a c, a c, a
then x ft, x on a c, a c. (b) Assume that |fx, y| M on Q, where M 0, and let c minh, k/M. Let S
*
denote the metric subspace of Ca c, a c consisting of all such that |x b| Mc on a c, a c. Prove that S is closed subspace of Ca c, a c and hence that S is itself a complete metric space. Proof: Since Ca c, a c is complete, if we can show that S is closed, then S is complete. Hence, it remains to show that S is closed. Given f S , then there exists a sequence of functions f n such that f n f under the sup norm . . So, given 0, there exists a positive integer N such that as n N, we have max |f n x fx| . xac,ac
Consider |fx b| |fx f N x| |f N x b| fx f N x f N x b Mc which implies that |fx b| Mc for all x since is arbitrary. So, f S. It means that S is closed. (c) Prove that the function T defined on S by the equation x
Tx b ft, tdt a
maps S into itself. Proof: Since x
|Tx b|
a ft, tdt
a |ft, t|dt
x
x aM Mc we know that Tx S. That is, T maps S into itself. (d) Now assume that f satisfies a Lipschitz condition of the form |fx, y fx, z| A|y z| for every pair of points x, y and x, z in Q, where A 0. Prove that T is a contraction of S if h 1/A. Deduce that for h 1/A the differential equation y fx, y has exactly one solution y x on a c, a c such that a b. Proof: Note that h 1/A, there exists such that h 1/A. Since T 1 x T 2 x
x
a |ft, 1 t ft, 2 t|dt x
A | 1 t 2 t|dt by |fx, y fx, z| A|y z| a
Ax a 1 t 2 t Ah 1 t 2 t 1 t 2 t where 0 A : 1. Hence, T is a contraction of S. It implies that there exists one and
only one S such that
x
x b ft, tdt a
which implies that x fx, x. That is, the differential equation y fx, y has exactly one solution y x on a c, a c such that a b.
Supplement on Riemann Integrals 1. The reader should be noted that the metric space Ra, b, d is NOT complete, where df, g
b
a |fx gx|dx.
We do NOT give it a proof. The reader can see the book, Measure and Integral (An Introduction to Real Analysis) by Richard L. Wheeden and Antoni Zygmund, Ch5. 2. The reader may recall the Mean Value Theorem: Let f be a continuous function on a, b. Then b
a fxdx fx 0 b a where x 0 a, b. In fact, the point x 0 can be chosen to be interior of a, b. That is, x 0 a, b. Proof: Let M sup xa,b fx, and m inf xa,b fx. If M m, then it is clear. So, we may assume that M m as follows. Suppose NOT, it means that x 0 a or b. Note that, fx 1 m fx 0 : r M fx 2 by continuity of f on a, b. Then we claim that fx 0 m or M. If NOT, i.e., fx 1 r fx 2 it means that there exists a point p x 1 , x 2 such that fp r by Intermediate Value Theorem. It contradicts to p a or b. So, we have proved the claim. If fa m, then b
b
a fxdx mb a 0 a fx mdx which implies that, by fx m 0 on a, b, fx m forall x a, b. So, it is impopssible. Similarly for other cases. Remark: (1) The reader can give it a try to consider the Riemann-Stieltjes Integral as follows. Let be a continuous and increasing function on a, b. If f is continuous on a, b, then b
a fxdx fcb a where c a, b. Note: We do NOT omit the continuity of on a, b since
0 if x 0
fx x on 0, 1; x
1 if x 0, 1
.
(2) The reader can see the textbook, exercise 14.13 pp 404. Exercise: Show that 2
/2
0
dx 1
1 2
2
sin x
. 2
Proof: It is clear by the choice of x 0 0, /2. 3. Application on Integration by parts for Riemann-integrable function. It is well-known that
fxdx xfx xdfx. If fx has the inverse function gy x, then (*) implies that
fxdx xfx gydy. For example,
arcsin xdx x arcsin x sin ydy. 4. Here is an observation on Series, Differentiation and Integration. We write it as a table to make the reader think it twice. Series Differentiation Integration
: Summation by parts Cesaro Sum ? : fg f g fg Mean Value Theorem Chain Rule : Integration by parts Mean Value Theorem Change of Variable .
*
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