(log(1 + x)) = 2

n

n=2

m

where Sm = 1/n.

n=1

Exercise K.128. (Convex functions.) [8.3, T] Recall from Exercise K.39

that we call a function f : (a, b) ’ R convex if, whenever x1 , x2 ∈ (a, b) and

1 ≥ » ≥ 0 we have

»f (x1 ) + (1 ’ »)f (x2 ) ≥ f (»x1 + (1 ’ »)x2 ).

(i) Suppose that a < x1 < x2 < x3 < b. Show algebraically that

f (x2 ) ’ f (x1 ) f (x3 ) ’ f (x2 )

¤ ,

x2 ’ x 1 x3 ’ x 2

and illustrate the result graphically.

(ii) If c ∈ (a, b) show that

f (c + h) ’ f (c)

σf (c+) = inf : c<c+h<b

h

exists and

f (c + h) ’ f (c)

’ σf (c+)

h

as h ’ 0 through values h > 0. State and prove the appropriate result for

σf (c’) Show also that σf (c+) ≥ σf (c’).

(iii) Using (ii), or otherwise, show that f is continuous at c. (Thus a

convex function is automatically continuous.)

(iv) Using (ii), or otherwise, show that we can ¬nd a real B such that

f (x) ’ f (c) ≥ B(x ’ c)

for all x ∈ (a, b). (Notice that, if σf (c+) = σf (c) then f is di¬erentiable at

c, B = f (c) and the line y ’ f (c) = B(x ’ c) is the tangent. We go into this

matter further in Exercise K.129.)

499

Please send corrections however trivial to twk@dpmms.cam.ac.uk

(v) Suppose a < ± < β < b. Let g : [±, β] ’ R be a positive continuous

β β

function with ± g(t) dt = 1. Show that c = ± tg(t) dt ∈ [±, β] and prove,

using (iv), or otherwise, that

β β

f (t)g(t) dt ≥ f tg(t) dt .

± ±

(vi) Prove the result of Exercise K.39 (iii) by the method of (iv). If you

know a little probability use the method of (iv) to show that if X is a random

variable taking values in [±, β] we have

Ef (X) ≥ f (EX).

(vii) If you have done Exercise K.32 (iii) try and obtain part (iv) as a

consequence.

Exercise K.129. [8.3, T, ‘] (i) By considering the map x ’ |x|, or other-

wise show that a convex function need not be di¬erentiable everywhere.

(ii) We use the assumptions and notation of Exercise K.128 (ii). Show

that, if a < ± ¤ c1 < c2 < · · · < cN ¤ β < b, then

N

σf (cj +) ’ σf (cj ’) ¤ σf (β+) ’ σf (±’).

j=1

Deduce, by using a ˜hamburger argument™ (see page 384), that the set of

points in [±, β] where f is not di¬erentiable is countable. Deduce that f is

di¬erentiable on (a, b) except at a countable set of points.

(iii) Show that, if f, g : (a, b) ’ R are convex, then so are f + g and µf

when µ ≥ 0. If fn : (a, b) ’ R is convex for each n and f : (a, b) ’ R is such

that fn (x) ’ f (x) as n ’ ∞ for each x ∈ (a, b), show that f is convex.

(iv) Construct an f : (’1, 1) ’ R which is convex but is not di¬erentiable

at any rational point.

Exercise K.130. [8.3, H] (i) Let f : R ’ R be continuous. Let an <

c < bn and an , bn ’ c as n ’ ∞. Write In = [an , bn ], |In | = bn ’ an and

bn

f (t) dt = an f (t) dt. Show, by using the method of proof of Theorem 8.3.6

In

but not the theorem itself, that

1

f (t) dt ’ f (c).

|In | In

(ii) How might this result generalise to higher dimensions (so we consider

f : Rm ’ R) and how might we prove the generalisation? (Without more

work we cannot give a rigorous proof, but we can certainly see how the proof

ought to run.)

500 A COMPANION TO ANALYSIS

Exercise K.131. [8.3, H!] We saw in Question K.115 that there is no hope

of successful theory of integration for functions f : Q ’ Q. None the less, I

think that looking at such functions can illuminate the role played by uniform

continuity in the proof of Theorem 8.3.1.

(i) Suppose f : [0, 1] © Q ’ Q is uniformly continuous. Show that, given

any > 0, we can ¬nd a dissection D of [0, 1] © Q such that

S(f, D) ’ s(f, D) < .

In what follows we shall construct a bounded continuous function g : [0, 1] © Q ’ Q

such that

S(g, D) ’ s(g, D) ≥ 1

for every dissection D of [0, 1] © Q.

(ii) We start by working in R. We say that an interval J is ˜good™ if

J = {x ∈ [0, 1] : (x ’ q)2 ¤ 2’m }

with q rational and m a strictly positive integer. We enumerate the elements

of [0, 1] © Q as a sequence x1 , x2 , . . . of distinct elements.

Show that, setting S0 = …, we can construct inductively ¬nite collections

Sn of disjoint good sets such that

(a)n Sn = Sn’1 ∪ Kn ∪ Ln with Sn’1 , Kn and Ln disjoint.

(b)n The total lengths of the intervals in Kn ∪ Ln is less than 2’n’3 .

(c)n If J is a subinterval of [0, 1] of length at least 2’n with J © I∈Sn’1 I =

… then we can ¬nd a K ∈ Kn and an L ∈ Ln such that J ⊇ K ∪ L.

(d)n xn ∈ I∈Sn I.

(iii) Show that the total length of the intervals making up Sn is always

less than 1/4.

Suppose that

D = {a0 , a1 , . . . , ap } with 0 = a0 ¤ a1 ¤ a2 ¤ · · · ¤ ap = 1

is a dissection of [0, 1]. Show that, provided that n is su¬ciently large, the

total length of those intervals [aj’1 , aj ] with 1 ¤ j ¤ m such that we can

¬nd a K ∈ Kn and an L ∈ Ln with [aj’1 , aj ] ⊇ K ∪ L will be at least 1/2.

(iv) Show that, if q ∈ Q © [0, 1], then there is a unique n ≥ 1 such that

q∈ I.

I∈Sn \Sn’1

Explain why q lies in exactly one of K∈Kn K or L. We de¬ne g(q) = 1

L∈Ln

if q ∈ K∈Kn K and g(q) = ’1 if q ∈ L∈Ln L.

501

Please send corrections however trivial to twk@dpmms.cam.ac.uk

(v) We now work in Q. Show that g : [0, 1] © Q ’ Q is a bounded

continuous function. By using (iii), or otherwise, show that

S(g, D) ’ s(g, D) ≥ 1

for every dissection D of [0, 1] © Q.

Exercise K.132. [8.4, P] Suppose that g : R2 ’ R is a di¬erentiable

function with continuous partial derivatives. Suppose that g : R2 ’ R is a

di¬erentiable function with continuous partial derivatives. Examiners often

ask you to show that

x

G(x) = g(x, t) dt

0

is di¬erentiable and determine its derivatives.

(i) Why can you not just quote Theorem 8.4.3?

y

(ii) Set F (x, y) = 0 g(x, t) dt. Show that F has partial derivatives. Show

further that these partial derivatives are continuous.

(iii) Use the chain rule to show that G is di¬erentiable and determine its

derivatives.

(iv) Suppose that h : R ’ R is di¬erentiable. Show that

h(x)

H(x) = g(x, t) dt

0

is di¬erentiable and determine its derivatives.

Exercise K.133. [8.4, P] The following exercise is included because it uses

several of our theorems in a rather neat manner. (It is actually a conservation

of energy result.) Suppose that u : R2 ’ R has continuous partial second

derivatives, that

‚2u ‚2u ‚u ‚u

(x, t) = 2 (x, t) and that (0, t) = (1, t) = 0.

‚x2 ‚t ‚t ‚t

Show that, if

2 2

1

‚u ‚u

E(t) = (x, t) + (x, t) dx,

‚t ‚x

0

then E is a constant.

Identify explicitly the major theorems that you use. (The author required

four major theorems to do this exercise but you may have done it another

way or have a di¬erent view about what constitutes a major theorem.)

502 A COMPANION TO ANALYSIS

Exercise K.134. [8.4, T] In Example 7.1.6 we constructed an in¬nitely

di¬erentiable function E : R ’ R with E(t) = 0 for t ¤ 0 and E(t) > 0 for

t > 0. If δ > 0, sketch the function H : R ’ R given by

t

E(δ ’ x)E(x) dx.

Hδ (t) =

0

By using functions constructed along these lines, or otherwise, prove the

following mild improvement of Exercise 8.4.12.

If

1

1 ’ (f (x))4 )2 + f (x)2 dx

I(f ) =

0

(as in Exercise 8.4.12), show that there exists a sequence of in¬nitely di¬er-

entiable functions fn : [0, 1] ’ R with fn (0) = fn (1) = 0 such that

I(fn ) ’ 0.

Exercise K.135. [8.4, M] (This question should be treated as one in math-

ematical methods rather than in analysis.)

Show that the Euler-Lagrange equation for ¬nding stationary values of

an integral of type

b

g(x, y) ds

a

where the integral is an arc length integral (informally, ˜ds is the element of

arc length™) may be written

y (x)g(x, y)

g,2 (x, y) ’ g,1 (x, y)y (x) ’ = 0.

1 + y (x)2

Hence, or otherwise, show (under the assumption that the problem admits

a well behaved solution) that the curve joining 2 given points that minimises

the surface area generated by rotating the curve about the x-axis is given by

x+a

y = c cosh ,

c

where a and c are constants.

Exercise K.136. [8.4, M] (This question should be treated as one in math-

ematical methods rather than in analysis.)

503

Please send corrections however trivial to twk@dpmms.cam.ac.uk

A well behaved function y : [a, b] ’ R is to be chosen to make the integral

b

I= f (x, y, y , y ) dx

a

stationary subject to given values of y and y at a and b. Derive an analogue

of the Euler-Lagrange equation for y and solve the problem in the case where

a = 0, b = 1, y(0) = y (0) = 0, y(1) = 0, y (1) = 4 and

1

f (x, y, y , y ) = (y )2 ’ 24y.

2

Explain why your result cannot be a maximum.

Exercise K.137. [8.5, T] Use the argument of Exercise K.122 (ii) to show

that, if f : [a, b] ’ Rm is a continuous function and we write

N ’1

b’a

f (a + j(b ’ a)/N ),

SN =

N j=0

b

then SN tends to a limit as N ’ ∞. We can de¬ne this limit to be a f (t) dt.

b

(The advantage of the procedure is that we de¬ne a f (t) dt directly with-

b

out using components. It is possible to de¬ne a f (t) dt directly for all Rie-

mann integrable f : [a, b] ’ Rm by mixing the ideas of this exercise with

the ideas we used to de¬ne the one dimensional Riemann integral, but it not

clear that the work involved is worth it.)

Exercise K.138. [9.1, T] (i) De¬ne gm : [0, 1] ’ R by gm (x) = 1’m| 2 ’x| 1

1

for | 2 ’x| ¤ m’1 , gm (x) = 0 otherwise. Show that there exists an g : [0, 1] ’

R, which you should de¬ne explicitly, such that gm (x) ’ g(x) as m ’ ∞

for each x ∈ [0, 1].

(ii) If fn is as in Exercise 9.1.1, show that there exists a sequence of

continuous functions gnm : [0, 1] ’ R such that such that gnm (x) ’ fn (x) as

m ’ ∞ for each x ∈ [0, 1].

(Thus repeated taking of limits may take us out of the class of Riemann

integrable functions. In fact, the later and more di¬cult Exercise K.157

shows that it is possible to ¬nd a sequence of bounded continuous functions

hn : [0, 1] ’ R and a function h : [0, 1] ’ R which is not Riemann integrable

such that hn (x) ’ h(x) as n ’ ∞ for each x ∈ [0, 1].)

Exercise K.139. [9.2, P] Suppose that f : [0, ∞) ’ R is a non-negative,

non-increasing function. If h > 0, show that ∞ f (nh) converges if and

n=1

504 A COMPANION TO ANALYSIS

∞ ∞

only if f (t) dt converges. If f (t) dt converges, show that

0 0

∞ ∞

f (nh) ’

h f (t) dt

0

n=1

as h ’ 0 with h > 0.

Show that, given any integer N ≥ 1 and any > 0 there exists a con-

tinuous function g : [0, 1] ’ R such that 0 ¤ g(x) ¤ 1 for all x ∈ [0, 1]

and

N

1

g(nN ’1 ) = 1.

g(x) dx < , but

0 n=1

Show that there exists a continuous function G : [0, ∞) ’ R such that

G(x) ≥ 0 for all x ≥ 0 and G(x) ’ 0 as x ’ ∞ but

∞ ∞

h G(nh) G(t) dt

0

n=1

as h ’ 0 with h > 0.

Exercise K.140. [9.2, P] (i) Suppose that g : (0, ∞) ’ R is everywhere

∞

di¬erentiable with continuous derivative and 1 |g (t)| dt converges. Show

∞

that ∞ g(n) converges if and only if 1 g(t) dt converges.

n=1

(ii) Deduce Lemma 9.2.4 for the case when f is di¬erentiable.

(iii) Suppose g satis¬es the hypotheses of the ¬rst sentence of part (i). By

giving a proof or a counterexample, establish whether if the sum ∞ g(n) n=1

is convergent then it is automatically absolutely convergent.

Exercise K.141. (A weak Stirling™s formula.) [9.2, T] Show that

n+1/2 1/2

t t

log x dx ’ log n = + log 1 ’

log 1 + dt.

n n

n’1/2 0

By using the mean value theorem, or otherwise, deduce that

n+1/2

4

log x dx ’ log n ¤ .

3n2

n’1/2

(You may replace 4/(3n2 ) by An’2 with A another constant, if you wish.)

N +1/2

log x dx ’ log N ! converges to a limit. Conclude that

Deduce that 1/2

n!

e’n ’ C

(n + 1/2)(n+1/2)

as n ’ ∞, for some constant C.

505

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.142. [9.2, T] (i) Show that the change of variable theorem

∞

works for in¬nite integrals of the form 0 f (x) dx. More speci¬cally, prove

the following theorem.

Suppose that f : [0, ∞) ’ R is continuous and g : [0, ∞) ’ R is di¬er-

entiable with continuous derivative. Suppose, further, that g(t) ≥ 0 for all t,

that g(0) = 0 and g(t) ’ ∞ as t ’ ∞. Then

∞ ∞

f (s) ds = f (g(x))g (x) dx.

0 0

Explain why this result is consistent with Example 9.2.16.

(ii) Explain why the function f : [0, ∞) ’ R de¬ned by f (x) = sin x/x

for x = 0, f (0) = 1 is continuous at 0. It is traditional to write f (x) = sin x/x

and ignore the fact that, strictly speaking, sin 0/0 is meaningless. Sketch f .

nπ

sin x

If In = dx show, by using the alternating series test, that In

x

0

∞

sin x

tends to a strictly positive limit L. Deduce carefully that dx exists

x

0

with value L, say.

∞

sin tx

dx for all t ∈ R. Show using (i), or otherwise,

(iii) Let I(t) =

x

0

that I(t) = L for all t > 0, I(0) = 0, I(t) = ’L for t < 0.

(iv) Find a continuous function g : [0, π] ’ R such that g(t) ≥ 0 for all

t ∈ [0, π], g(π/2) > 0 and

g(x ’ nπ)

sin x

≥

x n

for all nπ ¤ x ¤ (n + 1)π and all integer n ≥ 1. Hence, or otherwise, show

∞

that 0 | sin x/x| dx fails to converge.

∞

sin x

(v) For which value of real ± does I(t) = dx converge? Prove

x±

0

your answer.

[There are many methods for ¬nding the constant L of part (ii). The best

known uses complex analysis. I give a rather crude but direct method in the

next exercise. Hardy found the matter su¬ciently interesting to to justify

writing two articles (pages 528“533 and 615“618 of [22]) comparing various

methods.]

Exercise K.143. [9.2, T, ‘ ] Here is a way of evaluating

∞

sin x

dx.