Apostol Solution Ch7 Mathematical Analysis 2e

The Riemann-Stieltjes Integral

Riemann-Stieltjes integrals

7.1 Prove that



a b

d  b  a, directly from Definition 7.1.

Proof: Let f  1 on a, b, then given any partition P  a  x0, . . . , xn  b, then we

have SP, 1,  



k1 n ftkk, where tk  xk1, xk 



k1 n k  b  a. So, we know that



a b

d  b  a.

7.2

If f  R on a, b and if



_abfd  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



_abdf  fbb  faa

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



_ab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 tk inxk1, xk, we have

|SP, f,   A|  . (a) Show that if



a b

fd exists according to this definition, then it is also exists according

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 tk inxk1, xk, we have

|SP, f,   A|  . Then choosing a P withP  , then for P  P  P  . So,

we have

|SP, f,   A|  . That is,



a b

fd exists according to this definition, then it is also exists according to

(b) Let fx  x  0 for a  x  c, fx  x  1 for c  x  b,

fc  0, c  1. Show that



_abfd exists according to Definition 7.1 but does not exist

by this second definition.

Proof: Note that



a b

fd exists and equals 0 according to Definition 7.1If



_abfd exists

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 tk inxk1, xk, we have

|SP, f, |  1. We may choose a partition P  a  x0, . . . , xn  b with P   and c  xj, xj1, where j  0, . . . , n  1. Then

SP, f,   fxxj1  xj  1, where x  c, xj1

which contradicts to |SP, f, |  1.

7.4

If f  R according to Definition 7.1, prove that



_abfxdx also exists according to definition of Exercise 7.3. [Contrast with Exercise 7.3 (b).]

Hint: Let I 



_abfxdx, M  sup|fx| : x  a, b.  Given   0, choose P so that UP, f  I  /2 (notation of section 7.11). Let N be the number of subdivision points in P and let  /2MN. If P  , write

UP, f 



Mkfxk  S1  S2,

where S1 is the sum of terms arising from those subintervals of P containing no points of

P and S2 is the sum of remaining terms. Then

S1  UP, f  I  /2 and S2  NMP  NM  /2,

and hence UP, f  I  . Similarly,

LP, f  I   if P   _{for some}_.

Hence |SP, 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.

(1) Suppose that f  0 and f is continuous on a, b, and



a b

fxdx  0. Prove that

fx  0 on a, b.

Proof: Assume that there is a point c  a, b such that fc  0. Then by continuity of

f, we know that given  fc₂  0, there is a   0 such that as |x  c|  , x  a, b, we have

|fx  fc|  fc₂ which implies that

fc 2  fx if x  c  , c    a, b : I So, we have 0  fc_{2 |}I| 



Ifxdx 



a b

fxdx  0, where 0  |I|, the length of I which is absurb. Hence, we obtain that fx  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



_abgxdx  0, we always have



_abfxgxdx  0. Show that f is a constant function ona, b.

Proof: Let



a b

fxdx  I, and define gx  fx  I

ba, then we have



a b

gxdx  0, which implies that, by hypothesis,



_abfxgxdx  0 which implies that



_abfx  cgxdx  0 for any real c. So, we have



_abgx2dx  0 if letting c  _{b a}I which implies that gx  0 forall x  a, b by (1). That is, fx  I

ba ona, b.

(3) Define

hx 

0 if x  0, 1  Q

1

n if x is the rational number m/n (in lowest terms)

1 if x  0. Then h  R0, 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  R0, 1. Consider the upper sum UP, f as follows.

Given  0, there exists finitely many points x such that fx  /2. Consider a

partition P  x0  a, . . . , xn  b so that its subintervals Ij  xj1, xj for some j

containing those points and

_

|Ij|  /2. So, we have UP, f 



k1 n Mkxk 



1 



2  /2  /2  

where



₁ 



₁MjIj, 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.

UP, f,  LP, f,  ,

then we automatically have, for any finer P P,

UP, f,   LP, f,    since the refinement makes U increase and L decrease.

(4) Assume that the function fx is differentiable on a, b, but not a constant and that

fa  fb  0. Then there exists at least one point  on a, b for which |f_{| } 4

b  a2



a b

fxdx.

Proof: Consider sup_{xa,b}|f_{x| : M as follows.}

(i) If M  , then it is clear. (ii) We may assume that M  . Let x  a, ab

2 , then

fx  fx  fa  f_{yx  a  Mx  a, where y  a, x. *}

and let x  ab

2 , b, then

fx  fx  fb  f_{zx  b  Mb  x, where z  x, b. **}

So, by (*) and (**), we know that



_abfxdx 



a ab 2 fxdx 



_ab 2 b fxdx  M



a ab 2 x  adx  M



_ab 2 b b  xdx  M a  b₂ 2 which implies that

M  4

b  a2



a b

fxdx.

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  4

b  a2



a b

fxdx.

By definition of supremum, we know that there exists at least one point on a, b for

which

|f_{| } 4

b  a2



a b

fxdx.

(5) Gronwall Lemma: Let f and g be continuous non-negative function defined on a, b, and c  0. If

fx  c 



a x

gtftdt for all x  a, b, then

fx  ce



a x

gtdt

. In particular, as c  0, we have f  0 on a, b.

Fx  c 



a x gtftdt, then we have (i). Fa  c  0.

(ii). F_{x  gxfx  0  F is increasing on a, b by Mean Value Theorem}

(iii). Fx  fx on a, b  F_{x  gxFx by (ii).}

So, from (iii), we know that

Fx  Fae



a x gtdt  ce



a x gtdt by (i). For c  0, we choose cn  1/n  0, then by preceding result,

f_{x  1n e}



a x

gtdt

 0 as n  . So, we have proved all.

(6) Define

fx 



x x1

sint2_dt.

(a) Prove that |fx|  1/x if x  0.

Proof: Let x  0, then we have, by change of variable(u  t2_{), and integration by}

parts, f_{x  12}



x2 x12 sin u u du  1₂



x2 x12 _{dcos u} u  1₂ cos u_u x2 x12 



x2 x12 _{cos u} 2u3/2 du  cos_2xx2  cos x  1 2 2x  1 



x2 x12 _{cos u} 4u3/2du

which implies that,

|fx|  cos_2xx2  cos x  1 2 2x  1 



x2 x12 cos u 4u3/2 du  1_{2x  2} 1 x  1  14



x2 x12 du u3/2  1_{2x  2} 1 x  1  2x  11  12x  1/x.

Note: There is another proof by Second Mean Value Theorem to show above as

follows. Since

f_{x  12}



x2

x12 _{sin u}

u du by (a),

f_{x  12} 1_x



x2 y sin udu _{x 1}1



y x12 sin udu

 12 1x cosx2  cos y  _{x 1}1 cos y cos x  12

 12 1x  _{x 1}1 cosy  1x cosx2  _{x 1}1 cos x  12

which implies that

|f_{x|  12 }1_{x  x 1}1 |cos y| cos_xx2  cos x  1

2 x 1  12 1x  _{x 1 }1 cosx 2_ x  cos x  12 x 1  12 1x  _{x 1 }1 1x  _{x 1}1

since no x makes |cosx2_{|  cos x  1}2 _{ 1}

 1/x.

(b) Prove that 2xfx  cosx2_{  cos x  1}2 _{ rx, where |rx|  c/x and c is a}

constant.

Proof: By (a), we have

fx  cos_2xx2  cos x  1 2 2x  1 



x2 x12 cos u 4u3/2du

which implies that

2xfx  cosx2_{ } x x 1 cos x  12  x



x2 x12 _{cos u} 2u3/2du  cosx2_{  cos x  1}2 _ 1 x 1 cos x  12  x



x2 x12 cos u 2u3/2du where rx  _{x 1}1 cos x  12  x₂



x2 x12 _{cos u} u3/2 du

which implies that

|rx|  _{x 1 }1 x₂



x2 x12 _{|cos u|} u3/2 du  _{x 1 }1 x₂



x2 x12 u3/2_du  _{x 1 }1 _{x 1}1  2x .

Note: Of course, we can use the note in (a) to show it. We write it as follows. Proof: Since

f_{x  12 }1_{x } 1

x 1 cosy  1x cosx2  x 11 cos x  12 which implies that

2xfx  _{x 1 }x 1 cosy  cosx2_{ } x

x 1 cos x  12  cosx2_{  cos x  1}2 _ 1

x 1 cos x  12  x 1 x 1 cosy where

rx  _{x 1}1 cos x  12  _{x 1 }x 1 cosy which implies that

|rx|  _{x 1 }1 1 _{x 1}x  _{x 1}2

 2/x.

(c) Find the upper and lower limits of xfx, as x  .

Proof: Claim that lim supxcosx2  cos x  12  2 as follows. Taking

x  n 2 , where n  Z, then

cosx2_{  cos x  1}2 _{  cos n 8  1 . 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  Qc 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  Qc _{is dense in} R. It is equivalent toar : a  Z, where r  Qc _{is dense in0, 1 modulus 1.}

Proof: Sayar  b : a  Z, b  Z  S, and since r  Qc_{, then by Exercise 1.16,}

there are infinitely many rational numbers h/k with k  0 such that |kr  h|  1

k . Consider

x  , x   : I, where   0, and thus choosing k0 large enough so that 1/k0  .

Define L  |k0r h0|, then we have sL  I for some s  Z. So,

sL  sk0r  sh0  S. That is, we have proved that S is dense in R.

(d) Does



0  sint2_{dt converge?} Proof: Yes,



_xxsin2_tdt  12



x x sin u

u du by the process of (a)

 12 1x



_x y

sin udu 1_x



y x

sin udu by Second Mean Value Theorem  1₂ 2_{x  x}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 sin x

converges but not absolutely converges. In other words, the Dirichlet Integral converges conditionally. Proof: Consider



_xx sin x x dx  1x



_x y sin xdx 1_x



y x

sin xdx by Second Mean Value Theorem; we have



_xx sin x

x dx  2x  _x2  4x .

So, we know that Dirichlet Integral converges. Define In  ₄  2n, ₂  2n, then



₀ sin x x dx 



_I n sin x x dx 



In 2 2  4  2n dx 



n0  2 2 4   4  2n   . So, we know that Dirichlet Integral does NOT converges absolutely.

(7) Deal similarity with

fx 



x x1 sinet_dt. Show that ex_{|fx|  2} and that

ex_{fx  cose}x_{  e}1_cosex1_{  rx,}

where |rx|  min1, Cex_{, for all x and}

Proof: Since fx 



x x1 sinet_dt 



ex ex1 sin u

u du by Change of Variable (let u  et)

 cos e_ex x  cos e x1 ex1 



_ex ex1 cos u u2 du by Integration by parts *

which implies that

|fx|  cos e_ex x  cos e x1

ex1 



_ex

ex1

du

u2 since cos u is not constant 1

 1_ex  1_ex1  1_ex 1 1e

which implies that

ex_{|fx|  2.}

In addition, by (*), we have

ex_{fx  cose}x_{  e}1_cosex1_{  rx,}

rx  ex



ex

ex1

cos u

u2 du

which implies that

|rx|  1  e1 _{ 1 for all x **}

or which implies that, by Integration by parts, |rx|  ex



ex ex1 cos u u2 du  ex sin ex1 e2x1  sin e x ex  2



ex ex1 sin u u3 du  ex 1 e2x1  1_e2x  2



_ex ex1 du

u3 since sin u is not constant 1

 2ex _{for all x. ***}

By (**) and (***), we have proved that |rx|  min1, Cex_{ for all x, where C  2.}

Note: We give another proof on (7) by Second Mean Value Theorem as follows. Proof: Since fx 



ex ex1 sin u u du  1_ex



ex y sin udu 1_ex1



y ex1

sin udu by Second Mean Value Theorem

 1_ex cos ex  cos y  1_ex1 cos y  cos ex1 *

which implies that

ex_{|fx|  |cos e}x _{ cos y  e}1_{cos y  cos e}x1_|

 |cos ex _{ e}1_{cos e}x1_{  cos y1  e}1_|

 |cos ex _{ e}1_{cos e}x1_{| 1  e}1_

 1  e1_{  1  e}1_{ since no x makes |cos e}x_{|  |cos e}x1_{|  1.}

 2.

In addition, by (*), we know that

ex_{fx  cos e}x _{ e}1_{cos e}x1_{ rx}

where

rx  e1_{cos y cos y}

which implies that

|rx|  1  e1 _{ 1 for all x. **}

|rx|  ex



ex ex1 cos u u2 du  ex 1 e2x



_ex y cos udu  1 e2x1



_y ex1 cos udu  ex_{|sin y sin e}x _{ e}2_{sin e}x1 _{ sin y|}

 ex_{|sin y1  e}2_{  e}2_{sin e}x1 _{ sin e}x_|

 ex_{1  e}2_{  e}x_|e2_{sin e}x1_{| e}x_{|sin e}x_|

 ex_{1  e}2_{  e}x_{1  e}2_{ since no x makes |sin e}x1_{|  |sin e}x_{|  1}

 2ex _{for all x. ***}

So, by (**) and (***), we have proved that |rx|  min1, Cex_{, where C  2.}

(8) Suppose that f is real, continuously differentiable function ona, b,

fa  fb  0, and



_abf2_{xdx  1.} Prove that



_abxfxf_{xdx  1} 2 and that



a b f_x2_{dx }



a b x2_f2 xdx  14 . Proof: Consider



_abxfxf_{xdx }



a b xfxdfx  xf2_x| a b _



a b fxdxfx  



a b f2_{xdx }



a b xfxf_{xdx since fa  fb  0,} so we have



_abxfxf_{xdx  1} 2 . In addition, by Cauchy-Schwarz Inequality, we know that



a b f_x2_dx



a b x2_f2_{xdx }



a b xfxf_xdx 2  1₄.

Note that the equality does NOT hold since if it was, then we have f_{x  kxfx. It}

implies that

f_{x  kxfxe}kx22  0

which implies that

fekx22



 0 which implies that

fx  Cekx2

2 , a constant

C  0 since fa  0. That is, fx  0 on a, b which is absurb.

7.5 Letan be a sequence of real numbers. For x  0, define Ax 



nx an 



n1 x an,

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:



na anfn  



1 a Axf_{xdx  Aafa.} Proof: Since



₁aAxf_{xdx }



1 a

Axdfx since f has a continous derivative on 1, a  



1

a

fxdAx  Aafa  A1f1 by integration by parts

 



na anfn  Aafa by



1 a fxdAx 



n2 a anfn and A1  a1, we know that



na anfn  



1 a Axf_{xdx  Aafa.}

7.6 Use Euler’s summation formula, integration by parts in a Stieltjes integral, to derive the following identities:

(a)



_kn_1 1 ks  _ns11  s



₁ n x xs1dx if s  1. Proof:



k1 n 1 ks 



₁ n xs_{dx  1}  



1 n xdxs _{ n}s_{n  1}s_{1  1}  s



1 n _x xs1dx  n1s  1 ns1  s



₁ n _x xs1dx if s  1. (b)



_kn_1 1 k  log n 



₁ n xx x2  1. Proof:



k1 n 1 k 



1 n 1 x dx  1  



1 n xdx1_{ n}1_{n  1}1_{1  1} 



1 n x1_dx



1 n x1_dx



1 n _x x2 dx  1  log n 



1 n _{x x} x2  1.

7.7 Assume that f _{is continuous on1, 2n and use Euler’s summation formula or}

integration by parts to prove that



k1 2n 1kfk 



1 2n f_{xx  2x/2dx.} Proof:



k1 2n 1kfk  



k1 2n fk  2



k1 n f2k  



1 2n fxdx  f1  2



1 2n fxdx/2   



1 2n xdfx  2nf2n  2 



1 2n x/2dfx  f2n2n/2  f11/2 since f _{is continuous on1, 2n} 



1 2n f_{xxdx  2nf2n }



1 2n f_{xx/2dx  2nf2n} 



1 2n f_{xx  2x/2dx.}

7.8

Let1  x  x  1₂ if x  integer, and let 1  0 if x  integer. Also, let

2 



₀

x

1tdt. If f is continuous on1, n prove that Euler’s summation formula implies

that



k1 n fk 



1 n fxdx 



1 n 2xfxdx  f1  fn₂ .

Proof: Using Theorem 7.13, then we have



k1 n fk 



1 n fxdx 



1 n f_{x1xdx } f1  fn 2 



1 n fxdx 



1 n f_{xd2x } f1  fn 2 



1 n fxdx  



1 n 2xdfx  fn2n  f121  f1  fn₂ 



1 n fxdx 



1 n 2xdfx  f1  fn₂ 



1 n fxdx 



1 n 2xfxdx  f1  fn₂ since f is continuous on1, n.

7.9 Take fx  log x in Exercise 7.8 and prove that log n!  n  1₂ log n n  1 



1 n _ 2t t2 dt.

Proof: Let fx  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



1kn log k 



1 n log xdx



1 n x_{ x  12} dx_{x } log n₂ . * Since x_{ x  12}  1/2 and



a a1

x_{ x  12 dx  0 for all real a,} ** 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  1₂ log n n  C  n where C  1 



1  x_{ x  12} dx_{x ,} and n  



n  x_{ x  12} dx_{x .} So, lim_n n! en_nn1/2  eC : C1. ***

Now, using Wallis formula, we have

lim_n 2 2  4  4   2n2n

1 3  3  5  5   2n  12n  1  /2 which implies that

2n_n!4

2n!22n  11  o1  /2 which implies that, by (***),

C₁4_2n_nn1/2_en_4

C₁2 _2n2n1/2_e2n _{2n  1}1  o1  /2

which implies that

C₁2_n

22n  11  o1  /2. Let n  , we have C1  2 , and



₁



x  x  1

2dx  12 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 



px

1,

where the sum is extended over all primes p  x. The prime number theorem states that lim_xxlog x_x  1.

This is usually proved by studying a related function given by

x 



px

log p,

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 



3/2 x log tdt, x 



3/2 x 1 log tdt.

Note. The lower limit can be replaced by any number in the open interval1, 2.

Proof: Sincex 



_pxlog p, we know that by Theorem 7.9,

x 



3/2

x

log tdt,

andx 



_px1, we know that by Theorem 7.9,

x 



3/2

x ₁

log tdt. (b) If x  2, use integration by parts to show that

x  x log x 



2 x _t t dt, x  _{log x }x



2 x _t t log2tdt.

These equations can be used to prove that the prime number theorem is equivalent to the relation limx xx  1.

Proof: Use integration by parts, we know that

x 



3/2 x log tdt  



3/2 x _t t dt  log xx  log3/23/2  



3/2 x _t t dt  log xx since 3/2  0  x log x 



2 x _t t dt since



_3/2 2 _t t dt  0 by x  0 on 0, 2, and

x 



3/2 x 1 log tdt 



3/2 x _t t log2tdt  x log x  log3/23/2 



3/2 x _t t log2_tdt  x log x since3/2  0  _{log x }x



2 x _t t log2tdt since



3/2 2 _t t log2tdt  0 by x  0 on 0, 2.

7.11 If  on a, b, prove that

(a)



a b

fd 



_acfd 



_cbfd, (a  c  b)

Proof: Given  0, there is a partition P such that

UP, f,   Ia, b  . 1

Let P _{ c  P  P1 P2, where P1  a  x0, . . . , xn}

1  c and

P2  xn1  c, . . . , xn2  bthen we have

Ia, c  Ic, b  UP1, f,  UP2, f,  UP, f,  UP, f, . 2

So, by (1) and (2), we have

Ia, c  Ic, b  Ia, b *

since is arbitrary.

On the other hand, given  0, there is a partition P1 and P2such that

UP1, f,  UP2, f,  Ia, c  Ic, b  

which implies that, let P  P1 P2

UP, f,   UP1, f,  UP2, f,  Ia, c  Ic, b  . 3

Also,

Ia, b  UP, f, . 4

By (3) and (4), we have

Ia, b  Ia, c  Ic, b ** since is arbitrary.

So, by (*) and (**), we have proved it. (b)



a b

f  gd 



_abfd 



_abgd.

Proof: In any compact interval J, we have

sup

xJf  g  supxJ f supxJ g. 1

So, given  0, there is a partition Pf and Pg such that



k1 n1 Mkfk 



a b fd  /2 2 and



k1 n2 Mkgk 



a b gd  /2. 3

So, consider P  Pf  Pg, then we have, by (1),

UP, f  g,   UP, f,   UP, g,  along with

UP, f,  



k1 n1 Mkfk and UP, g,  



k1 n2 Mkgk

which implies that, by (2) and (3),

UP, f  g,  



a b fd 



a b gd  

which implies that



a b f  gd 



a b fd 



a b gd since is arbitrary. (c)



a b f  gd 



_a  b fd 



_a  b gd

Proof: Similarly by (b), so we omit the proof.

7.12

Give an example of bounded function f and an increasing function defined on

a, b such that |f|  R but for which



_abfd does not exist.

Solution: Let

fx  1 if x  0, 1  Q 1 if x  0, 1  Qc

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 g  R on a, b and define x 



_axgtdt if x  a, b. Show that:

(a) If f  on a, b, there exists a point x0 ina, b such that



_abfd  fa



_ax0

gd  fb



_x

0

b gd.

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



_abfd  fa



a x0 dx  fb



x0 b dx

which implies that, by Theorem 7.26,



_abfd  fa



a x0 gd  fb



x0 b gd.

(b) If, in addition, f is continuous ona, b, we also have



a b fxgxdx  fa



a x0 gd  fb



x0 b gd. Proof: Since



_abfd 



a b fxgxdx by Theorem 7.26, we know that, by (a),



_abfxgxdx  fa



a x0 gd  fb



x0 b gd.

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

Vx denote the total variation of  on a, x for each x in a, b, and let Va  0. Show that



_abfd 



a b

|f|dV  MVb,

where M is an upper bound for |f| ona, b. In particular, when x  x, the inequality becomes



_abfd  Mb  a.

Proof: Given  0, there is a partition P  a  x0, . . . , xn  b such that



a b fd    SP, f,  



k1 n ftkk, where tk  xk1, xk 



k1 n |ftk||xk  xk1| 



k1 n |ftk|Vxk  Vxk1  SP, |f|, V

 UP, |f|, V since V is increasing on a, b which implies that, taking infimum,



_abfd   



a b |f|dV since |f|  RV on a, b. So, we have



_abfd 



a b |f|dV * since



a b

|f|dV is clear non-negative. If M is an upper bound for |f| ona, b, then (*) implies that



_abfd 



a b

|f|dV  MVb which implies that



_abfd  Mb  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

continuous ona, b, prove that lim n



_a b fxdnx 



a b fxdx.

Proof: Use Exercise 7.14, we then have



_abfxd  nx  MVnb  0 as n  

where Vn is the total variation of  n, and M  supxa,b|fx|.

So, we have lim n



_a b fxdnx 



a b fxdx.

Remark: We do NOT need the hypothesisa  na  0 for each n  1, 2, . . . .

7.16

If f  R, f2 _{ R, g  R, and g}2 _{ R on a, b, prove that}

1 2



a b



_ab f_fx gx_{y gy} 2 dy dx 



a b f2_{xdx}



a b g2_{xdx }



a b fxgxdx 2 . When  on a, b, deduce the Cauchy-Schwarz inequality



_abfxgxdx 2 



a b f2_{xdx}



a b g2_{xdx .}

(Compare with Exercise 1.23.)

Proof: Consider 1 2



a b



_ab f_fx gx_{y gy} 2 dy dx  1₂



a b



_abfxgy  fygx2dy dx

 1₂



a b



_abf2_xg2_{y  2fxgyfygx  f}2_yg2_{xdy dx}

 1₂



a b f2_{xdx}



a b g2_{ydy} 



a b fxgxdx



a b fygydy 



a b g2_{xdx}



a b f2_{ydy} 



a b f2_{xdx}



a b g2_{ydy }



a b fxgxdx 2 , if  on a, b, then we have

0  1₂



a b



_ab f_fx gx_{y gy} 2dy dx





a b f2_{xdx}



a b g2_{ydy }



a b fxgxdx 2 which implies that



_abfxgxdx 2 



a b f2_{xdx}



a b g2_{xdx .}

Remark: (1) Here is another proof: Let A 



_abf2_{xdx, B }



a b

fxgxdx, and

C 



_abg2_{xdx. From the fact,}

0 



a b

fxz  gx2dx for any real z

 Az2 _{ 2Bz  C.} It implies that B2 _{ AC.} That is,



_abfxgxdx 2 



a b f2_{xdx}



a b g2_{xdx .}

Note: (1) The reader may recall the inner product in Linear Algebra. We often

consider Riemann Integral by defining  f, g :



a b

fxgxdx

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 :



a b

f_{xgxdx}

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  supxa,b|fx|, 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  x0  a, . . . , xn  b such that

UP, f,   LP, f,  



j1

n

Mjf  mjfj  2. 1

Write P  A  B, where A  xj : Mjf  mjf   and B  xj : Mjf  mjf  , then





B

j 



B

Mjf  mjfj  2by (1)

which implies that



B

j  . 2

For this partition P, we define the function g as follows.

gt  _xjxj_{ x} t_j1 fxj1  _xt_j x_{ x}j1_j1fxj, where xj1  t  xj.

So, it is clear that g is continuous ona, b. In every subinterval xj1, xj

|ft  gt|  _xjx_{ x}j  t_j1 ft  fxj1  _xt_j x_{ x}j1_j1ft  fxj  |ft  fxj1|  |ft  fxj|  2Mjf  mjf 3 Consider



A



xj1,xj |ft  gt|2d 



A 4Mjf  mjf2j by (3)  4



A _{Mjf  mjfj by definition of A}  4



A j by  1  4_{b  a} and



B



xj1,xj |ft  gt|2d 



B 4K2_ j  4K2_ _{by (2).} Hence,



a b |ft  gt|2d  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

p  1q  1.

Prove that the following statements. (a) If u  0 and v  0, then

uv  u_{p }p v_{q .}p Equality holds if and only if up _{ v}q_.

Proof: Let fu  up

p  v

p

q  uv be a function defined on 0, , where 1p  1q  1, p  0, q  0 and v  0, then f_{u  u}p1 _{ v. So, we know that}

f_{u  0 if u  0, v}p11 and fu  0 if u  vp11 ,

which implies that, by f vp11  0, fu  0. Hence, we know that fu  0 for all u  0.

That is, uv  up

p  v

p

q . In addition, fu  0 if and only if u  v

1

p1 if and only if up  vq.

So, Equality holds if and only if up _{ v}q_.

Note: (1) Here is another good proof by using Young’s Inequality, let fx be an strictly increasing and continuous function defined onx : x  0, with f0  0. Then we have, let a  0 and b  0,

ab 



0 a fxdx 



0 b

f1_{xdx, where f}1 _{is the inverse function of f.}

And the equality holds if and only if fa  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 fx  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



_abfp_{d  1 }



a b gq_d, then



a b fgd  1.

Proof: By Holder’s inequality, we have

fg  f_{p }p g_qq which implies that, by  on a, b, and



a b fp_{d  1 }



a b gq_d,



_abfgd 



a b _fp p d 



_a b _gq q d  1p  1q  1.

(c) If f and g are complex functions in R, where   on a, b, then



_abfgd 



a b |f|pd 1/p



a b |g|qd 1/q. *

Proof: First, we note that



_abfgd 



a b |fg|d. Also,



_ab|f|pd  Mp _



a b _|f| M p d  1

and



_ab|g|qd  Nq _



a b _|g| N q d  1. Then we have by (b),



_ab |f| M |g|N d  1

which implies that, by (*)



_abfgd  MN 



a b |f|pd 1/p



a b |g|qd 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



a b



_abfy  fxgy  gxdy dx  b  a



a b fxgxdx 



a b fxdx



a b gxdx . If  on a, b, deduce the inequality



a b fxdx



a b gxdx  b  a



a b fxgxdx

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



a

b



_abfy  fxgy  gxdy dx  1₂



a b



_abfygy  fygx  fxgy  fxgxdy dx  b  a



a b fygydy 



a b fxdx



a b gxdx which implies that, (let, f, and g  on a, b),

0  1₂



a b



_abfy  fxgy  gxdy dx and (let, and f  on a, b, g  on a, b),

0  1₂



a b



_abfy  fxgy  gxdy dx, we know that, (let, f, and g  on a, b)



_abfxdx



a b gxdx  b  a



a b fxgxdx and (let, and f  on a, b, g  on a, b)



a b fxdx



a b gxdx  b  a



a b fxgxdx.

Riemann integrals

7.18

Assume f  R on a, b. Use Exercise 7.4 to prove that the limit lim n b an



k1 n f a k b  a_n

exists and has the value



a b fxdx. Deduce that lim n



k1 n n k2 _{ n}2  4 , limn



k1 n n2 _{ k}2_1/2 _{ log 1  2 .}

Proof: Since f  R on a, b, given   0, there exists a   0 such that as P  , we have

SP, f 



a b

fxdx  . For this, we choose n large enough so that ba

n  , that is, as n  N, we have ban  .

So,

SP, f 



a b

fxdx   which implies that

b a n



k1 n f a k b  a_n 



a b fxdx  . That is, lim n b an



k1 n f a k b  a_n

exists and has the value



a b fxdx. Since



_kn_1 n k2_n2  1n



_k1 n ₁ k n 21

, we know that by above result,

lim n



k1 n n k2 _{ n}2  limn 1n



k1 n 1 k n 2 1 



0 1 _dx 1 x2  arctan 1  arctan 0  /4. Since



_kn_1n2_{ k}2_1/2 _ 1 n



_kn_1 1 1 k n 2 1/2

lim_n



k1 n n2 _{ k}2_1/2 _{ lim} n 1n



k1 n 1 1 k n 2 1/2 



0 1 dx 1  x2_1/2 



0 /4

secd, let x  tan 





0

/4

sec sec   tan _{sec  tan }d





1

1 2 _du

u , let sec   tan   u

 log 1  2 .

7.19

Define fx 



0 x et2 dt 2, gx 



0 1 ex2_t2_1 t2 _{ 1} dt.

(a) Show that g_{x  f}_{x  0 for all x and deduce that fx  gx  /4.}

Proof: Since f_{x  2}



0 x et2 dt ex2

and note that if hx, t  ex2 t21

t2_1 , we know that h is continuous on0, a  0, 1 for any

real a  0, and hx  2xex2t21is continuous on0, a  0, 1 for any real a  0, g_{x }



0 1 hxdt 



0 1 2xex2_t2_1 dt  2ex2



0 1 xext2 dt  2ex2



0 x eu2 du,

we know that g_{x  f}_{x  0 for all x. Hence, we have fx  gx  C for all x,}

constant. Since C  f0  g0 



0 1 _dt

1t2  /4, fx  gx  /4.

Remark: The reader should think it twice on how to find the auxiliary function g.

(b) Use (a) to prove that

lim_x



0

x et2

dt  1_{2 }.

Proof: Note that

hx, t  ex_t22t_{ 1}21  |ex2_t2_1 |  1 x2_t2 _{ 1} for all x  0; we know that



0 1 _ex2_t2_1 t2 _{ 1} dt  _x12



₀ 1 _dt 1 t2  0 as x  .

lim_xfx  /4 which implies that

lim_x



0

x et2

dt _{ 12 }

since limx



₀xet2dt exists by



x₀et2dt 



₀x _1tdt2  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 



k0  x2k k! for all x and



k0  x2k k! 



k0  x2k _ 1 1 x2 if |x|  1

which implies that

1 x2 _{ e}x2 if 0  x  1  1  x2_n _{ e}nx2 1 and ex2  ₁ 1  x2 if x  0  enx 2  ₁ 1  x2 n . 2

So, we have, by (1) and (2),



0 1 1  x2_n_{dx }



0 1 enx2 dx 



0  enx2 dx 



0  ₁ 1 x2 n dx. 3 Note that



₀enx2 dx  1 n



0  ex2 dx : K n . Also,



₀11  x2_n_{dx }



0 /2 sin2n1_{tdt } 2 4  6   2n  22n 1 3  5   2n  1 and



0  ₁ 1 x2 n dx 



0 /2 sin2n2_{tdt } 1 3  5   2n  3 2 4  6   2n  2 2 , so n 2 4  6   2n  22n 1 3  5   2n  1  K  n 1 3  5   2n  3 2 4  6   2n  2 2 which implies that

n 2n 1 2  4  6   2n  22n 2 1  3  5   2n  122n  1  K 2 _ n 2n 1 1  3  5   2n  3 2_{2n  1} 2  4  6   2n  22 2 2 4 By Wallis formula, we know that, by (4)

K  ₂ .



₀ex2

dx  ₂ .

Note: (Wallis formula)

lim_n 2  4  6   2n  22n2

1  3  5   2n  122n  1  2 .

Proof: As 0  x  /2, we have

sin2n1t  sin2n_{t  sin}2n1_{t, where n  N.}

So, we know that



₀/2sin2n1_{tdt }



0 /2 sin2n_{tdt }



0 /2 sin2n1_tdt

which implies that

2n2n  2   4  2 2n  12n  1   3  1   2n 12n  3   3  1 2n2n  2   4  2 2  2n  22n  4   4  22n  12n  3   3  1. So, 2n2n  2   4  2 2n  12n  3   3  1 2 1 2n 1  2  _{2n  12n  3   3  1}2n2n  2   4  2 2 1 2n. Hence, from 2n2n  2   4  2 2n  12n  3   3  1 2 1 2n  2n1 1  12n 2  0, we know that lim_n 2  4  6   2n  22n2 1  3  5   2n  122n  1  2 .

(2) Here is another exercise from Hadamard’s result. We Write it as follows. Let

f  Ck_{R with f0  0. Prove that there exists an unique function g  C}k1_{R such that}

f  xgx on R. Proof: Consider fx  fx  f0 



0 1 dfxt 



0 1 xf_xtdt  x



0 1 f_xtdt;

we know that if gx :



₀1f_{xtdt, then we have prove it.}

Note: In fact, we can do this job by rountine work. Define

gx 

fx

x if x  0

0 if x  0.

However, it is too long to write. The trouble is to make sure that g  Ck1_R.