worst cases.)

(The reader may be tempted to go on and consider more and more compli-

cated rules along these lines. However such rules involve assumptions about

the behaviour of higher derivatives which are often unrealistic in practice.)

Exercise K.126. [8.3, T] The object of this exercise is to de¬ne the loga-

rithm and the real exponential function, so no properties of those functions

should be used. You should quote all theorems that you use, paying partic-

ular attention to those on integration.

x

We set l(x) = 1 1 dt.

t

(i) Explain why l : (0, ∞) ’ R is a well de¬ned function.

(ii) Use the change of variable formula for integrals to show that

xy

1

dt = l(y)

t

x

whenever x, y > 0. Deduce that l(xy) = l(x) + l(y).

(iii) Show that l is everywhere di¬erentiable with l (x) = 1/x.

(iv) Show that l is a strictly increasing function.

(v) Show that l(x) ’ ∞ as x ’ ∞.

(vi) Show that l : (0, ∞) ’ R is a bijective function.

(vii) In Section 5.6 we de¬ned the logarithm as the inverse of the expo-

nential function e : R ’ (0, ∞). Turn this procedure on its head by de¬ning

e as the inverse of l. Derive the main properties of e, taking particular care

to quote those theorems on inverse functions that you require. When you

have ¬nished, glance through section 5.4 to see if you have proved all the

properties of the (real) exponential given there.

(viii) By expanding (1 + t)’1 as a geometric series and integrating, obtain

a Taylor series for log(1 + x), giving the range over which your argument is

valid.

(ix) Use (viii) to ¬nd the Taylor series of log (1 + x)/(1 ’ x) , giving the

range over which your argument is valid. Show that

3 5

y’1 y’1 y’1

1 1

log y = 2 + + + ...

y+1 3 y+1 5 y+1

498 A COMPANION TO ANALYSIS

for all y > 0.

Exercise K.127. [8.3, P, S, ‘ ] Use the Taylor series for log(1 + x) and

some result on products like Exercise 5.4.4 to obtain

∞

(’1)n

2

Sn’1 xn

(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.

x

0

506 A COMPANION TO ANALYSIS

sin »t

(We adopt the convention that, if we write f (t) = , then f (0) = ».)

t

(i) Show, by using the formula for the sum of a geometric series, or

otherwise, that

n

sin((n + 1 )x)

2

1+2 cos rx = x

sin 2

r=1

for all |x| < π and deduce that

sin((n + 1 )x)

π

2

2π = dx.

x

sin 2

’π

(ii) If > 0 show that

∞

sin »x sin x

dx ’ dx,

x x

’ ’∞

as » ’ ∞.

(iii) If π ≥ > 0 show, by using the estimates from the alternating series

test, or otherwise, that

1 1

π

sin((n + 2 )x) sin((n + 2 )x)

dx ’ dx = 2π

sin x sin x

’ ’π

2 2

as n ’ ∞.

(iv) Show that

2 1

’ ’0

1

x sin 2 x

as x ’ 0.

(v) Combine the results above to show that

∞

sin x π

dx = .

x 2

0

Exercise K.144. (Big Oh and little oh.) [9.2, T] The following nota-

tions are much used in branches of mathematics like number theory, algo-

rithmics and combinatorics. Consider functions f, g : N ’ R.

We say f (n) = O(g(n)) as n ’ ∞ if there exists a constant A > 0 and

an integer n0 such that |f (n)| ¤ A|g(n)| for n ≥ n0 .

We say f (n) = „¦(g(n)) as n ’ ∞ if there exists a constant A > 0 and

an integer n0 such that |f (n)| ≥ A|g(n)| for n ≥ n0 .

507

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

We say f (n) = o(g(n)) if, given any > 0 we can ¬nd an integer n1 ( )

such that |f (n)| ¤ |g(n)| for n ≥ n1 ( ).

We say f (n) ∼ g(n) as n ’ ∞ if, f (n)/g(n) ’ 1 as n ’ ∞.

(i) If f (n) = O(g(n)) does it follow that f (n) = „¦(g(n))? Give a proof

or counterexample. If f (n) = O(g(n)) must it necessarily be false that

f (n) = „¦(g(n))? Give a proof or counterexample.

(ii) Formulate the 11 other possible questions along the lines of (i) and

resolve them. (This is not as tedious as it might seem, they all have two

sentence answers.)

(iii) In more traditional terms, what does it mean to say that f (n) = o(1)

as n ’ ∞? What does it mean to say that f (n) = O(1) as n ’ ∞?

(iv) Give an example of a pair of functions f, g : N ’ R such that

f (n), g(n) > 0 for all n and neither of the two relations f (n) = O(g(n)),

g(n) = O(f (n)) hold.

(v) Consider f, g : R ’ R. Give de¬nitions along the lines given above

for the statement ˜f (x) = O(g(x)) as x ’ ∞™ and for the statement ˜f (x) =

o(g(x)) as x ’ 0™.

(vi) Recall or do exercise K.86 and write the result in the notation of this

question.

Although the notations introduced in this question are very useful for

stating results, they are sharp edged tools which can do as much harm as

good in the hands of the inexperienced. I strongly recommend translating

any statements involving such notations back into classical form before trying

to work with them.

Exercise K.145. [9.2, P, ‘ ] Are the following statements true or false?

Give reasons.

(i) If fj , gj : N ’ R are positive functions such that f1 (n) = O(g1 (n))

and f2 (n) = O(g2 (n)), then f1 (n) + g1 (n) = O(f2 (n) + g2 (n)).

(ii) If fj , gj : N ’ R are functions such that f1 (n) = O(g1 (n)) and

f2 (n) = O(g2 (n)), then f1 (n) + g1 (n) = O(f2 (n) + g2 (n)).

(iii) If fj , gj : N ’ R are positive functions such that f1 (n) = O(g1 (n))

and f2 (n) = O(g2 (n)), then f1 (n) + g1 (n) = O(max(f2 (n), g2 (n)).

2

(iv) n! = O(2n ) as n ’ ∞.

2

(v) cos x ’ 1 ’ (sin x) = o(x4 ) as x ’ 0.

2!

(sin x)2 4

(vi) cos x ’ 1 ’ 2! ’ (sin x) = o(x6 ) as x ’ 0.

4!

Exercise K.146. [9.2, P, ‘ ] (i) If ± > ’1, show that

n

n±+1

±

r∼ .

±+1

r=1

508 A COMPANION TO ANALYSIS

(ii) State and prove a result similar to (i) for ± = ’1.

(iii) If ± < ’1, show that

∞

n±+1

±

r∼ .

(’±) ’ 1

r=n

(iv) Does there exist a modi¬cation of (iii) similar to (ii) for ± = ’1?

Give brief reasons.

Exercise K.147. [9.2, P, ‘ ] Suppose that g : R ’ R is twice di¬erentiable

and

g (x) = O(x’» ) as x ’ ∞

for some » > 0. Prove that

n+1

(g(x) ’ g(n + 1 )) dx = O(n’» )

2

n

X

as n ’ ∞. Deduce that, if » > 1 and | g(x) dx| ’ ∞ as n ’ ∞, then

1

n n

1

∼

g(r + ) g(x) dx.

2

1

r=1

Deduce the result of Exercise K.141 as a special case.

Exercise K.148. [9.2, P] The function f : [0, 1] ’ R ∪ {∞} is de¬ned as

follows. Each 0 < x < 1 is expressed as a decimal x = .a1 a2 . . . ak . . . with

aj an integer 0 ¤ aj ¤ 9. In ambiguous cases we use the non-terminating

form. We put f (x) = k if the kth digit ak is the ¬rst digit 7 to occur in the

expansion, if there is any digit 7. If there is none (and also at x = 0 and

x = 1) we put f (x) = ∞. Prove that the function fX : [0, 1] ’ R de¬ned by

fX (x) = min(X, f (x)) is integrable for all X ≥ 0 and that

1

fX (x) dx ’ 10

0

as X ’ ∞.

Does it make any di¬erence if we rede¬ne f so at those points x where

we previously had f (x) = ∞ we now have f (x) = 0?

Exercise K.149. [9.2, H] This question investigates another way of de¬n-

ing improper integrals.

509

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

(i) If f : [a, b] ’ R and R, S ≥ 0, we write

±

R if f (x) ≥ R,

fR,S (x) = f (x) if R > f (x) > ’S,

’S if ’S ≥ f (x).

If f |R,S ∈ R[a, b] for all R, S ≥ 0 and there exists an L such that, given any

> 0, there exists an R0 > 0 with

b

fR,S (x) dx ’ L <

a

b

for all R, S ≥ R0 , then we say that LV a f (x) dx exists with value L.

∞

Show that, if f |R,S ∈ R[a, b] for all R, S ≥ 0, then LV a |f (x)| dx exists

∞

if and only if LV a |f (x)| dx exists.

(ii) Suppose that f : [a, b] ’ R is continuous except at b and f (x) ’ ∞

as x ’ b. Show that the following two statements are equivalent.

(A) There exists an L such that, given any > 0, there exists an R0 > 0

with

b

fR,S (x) dx ’ L <

a

for all R, S > R0 .

(B) There exists an L such that, given any > 0, there exists a δ0 > 0

with

b’·

f (x) dx ’ L <

a

for all 0 < · < δ0 .

Show further that, if (A) and (B) hold, then L = L .

(iii) Suppose that f : [a, b] ’ R is continuous except at possibly at b.

Show that statement (A) implies statement (B) but that the converse is

false. (Think about Question K.142 and change of variable if you need a

hint.)

[Warning: If you ever need to take the contents of this question seriously

you should switch to the Lebesgue integral.]

Exercise K.150. [9.3, P] By considering both orders of integration in

X b

e’tx dt dx,

0 a

510 A COMPANION TO ANALYSIS

where b > a > 0 and X > 0, show that

X b

e’ax ’ e’bx 1 ’ e’tX

dx = dt.

x t

0 a

Show that

b

e’tX b

dt ¤ e’aX log ,

t a

a

and hence ¬nd

∞

e’ax ’ e’bx

dx.

x

0

Exercise K.151. [9.3, T] Suppose that f : R2 ’ R is continuous. By using

results on the di¬erentiation of integrals, which should be quoted exactly,

show that

t d d t

d d

f (u, v) dv du = f (u, v) du dv.

dt dt

a c c a

Deduce that

b d d b

f (u, v) dv du = f (u, v) du dv.

a c c a

In what way is this result weaker than that of Theorem 9.3.2?

Exercise K.152. [9.3, H] (i) Reread Exercise 5.3.10.

(ii) Let us de¬ne f : [0, 1] — [0, 1] ’ R by

if 2’n < x < 2’n’1 , 2’n < y < 2’n+1 and n ≥ 1, n ∈ Z,

f (x, y) = 22n

if 2’n < x < 2’n’1 , 2’n’1 < y < 2’n and n ≥ 1, n ∈ Z,

f (x, y) = ’22n+1

f (x, y) = 0 otherwise.

Sketch the sets where f (x, y) take various values. Show that the integral

1

f (x, y) dx is well de¬ned and calculate it for all values of y. Show that the

0

1

integral 0 f (x, y) dy is well de¬ned and calculate it for all values of x. Show

that the two integrals

1 1 1 1

f (x, y) dx dy and f (x, y) dy dx

0 0 0 0

are well de¬ned and calculate them. Show that the two integrals are not

equal.

511

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

(iii) Suppose that u : R ’ R is continuous function with u(t) = 0 for

1

t < 1/4 and t > 3/4, u(t) ≥ 0 for all t and 0 u(t) dt = 1. Show that the

function g : [0, 1]2 ’ R

∞

22n u(2n x ’ 1)u(2n y ’ 1) ’ 22n+1 u(2n x ’ 1)u(2n+1 y ’ 1)

g(x, y) =

n=1

1

is well de¬ned. Show that the integral 0 f (x, y) dx is well de¬ned and calcu-

1

late it for all values of y. Show that the integral 0 f (x, y) dy is well de¬ned

and calculate it for all values of x. Show that the two integrals

1 1 1 1

g(x, y) dx dy and g(x, y) dy dx

0 0 0 0

are well de¬ned but not equal. Show that g is continuous except at (0, 0),

that the function x ’ g(x, y) is an everywhere continuous function for all y

and that the function y ’ g(x, y) is an everywhere continuous function for

all x. Why does Theorem 9.3.2 fail?

(iv) The traditional counterexample is the following. De¬ne h : [0, 1] 2 ’

R by h(0, 0) = 0 and

xy(x2 ’ y 2 )

h(x, y) = 2

(x + y 2 )3 )

1

otherwise. Show that the integral 0 h(x, y) dx is well de¬ned and calculate

it for all values of y (you may ¬nd the substitution w = x2 + y 2 useful). Show

1

that the integral 0 f (x, y) dy is well de¬ned and calculate it for all values of

x. Show that the two integrals

1 1 1 1

h(x, y) dx dy and h(x, y) dy dx

0 0 0 0

are well de¬ned but not equal. Show that g is continuous except at (0, 0),

that the function x ’ g(x, y) is an everywhere continuous function for all y

and that the function y ’ g(x, y) is an everywhere continuous function for

all x.

(v) Which of examples (ii) and (iv) do you consider ˜more natural™. Which

do you feel you ˜understand better™ ? Which is ˜better™. Why? (Of course

you can refuse to answer questions like this on the grounds that they are not

mathematical but I think you will loose something.)

512 A COMPANION TO ANALYSIS

Exercise K.153. (Interchange of in¬nite integrals.) [9.3, T] (i) Sup-

pose that g : [0, ∞) ’ R. is continuous and |g(x)| ¤ A(1 + x2 )’1 for all x.

∞

Show that 0 g(x) dx exists, that

∞

g(x) dx ¤ 2A,

0

and

∞

R

g(x) dx ¤ AR’1 ,

g(x) dx ’

0 0

for R ≥ 1. [If you do not wish to use knowledge of tan’1 , simply observe

that (1 + x2 )’1 ¤ 1 for 0 ¤ x ¤ 1 and (1 + x2 )’1 ¤ x’2 for x ≥ 1.]

(ii) Suppose that f : [0, ∞)2 ’ R is continuous and |f (x, y)| ¤ Ax’2 y ’2

∞∞ ∞∞

for x, y ≥ 1. Show that 0 0 f (x, y) dx dy and 0 0 f (x, y) dy dx exist

and that

∞ ∞ ∞ ∞

f (x, y) dx dy = f (x, y) dy dx.

0 0 0 0

(iii) (Parts (iii) and (iv) run along similar lines to parts (i) and (ii) of

Exercise K.152. You do not need to have done Exercise K.152 to do this ex-

ercise but, if you have, it is worth noting the similarities and the di¬erences.)

Reread Exercise 5.3.10. Let brs be as in part (ii) of that exercise. De¬ne

F : [0, ∞)2 ’ R by

F (x, y) = brs for r ’ 1 ¤ x < r and s ’ 1 ¤ y < s.

∞ ∞ ∞ ∞

Show that F (x, y) dx dy and F (x, y) dy dx exist and that

0 0 0 0

∞ ∞ ∞ ∞

F (x, y) dx dy = F (x, y) dy dx.

0 0 0 0

(iv) By modifying the construction in part (iii), or otherwise, ¬nd a con-

tinuous function G : [0, ∞)2 ’ R such that supx2 +y2 ≥R |G(x, y)| ’ 0 as

R ’ ∞ but

∞ ∞ ∞ ∞

G(x, y) dx dy = G(x, y) dy dx.

0 0 0 0

Exercise K.154. [9.3, H] It is a weakness of the Riemann integral that

there is no Fubini theorem for general Riemann integrable functions.

Let A = [0, 1] — [0, 1] and de¬ne f : A ’ R by f (x, 0) = 1 if x is rational,

f (x, y) = 0 otherwise. By ¬nding appropriate dissections show that f is

1

Riemann integrable. Explain why F2 (y) = 0 f (t, y) dt is not de¬ned for all

y.

[See also the next exercise.]

513

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

Exercise K.155. [9.3, H] We say that a sequence e1 , e2 , . . . of points in

A is dense in A = [0, 1] — [0, 1] if, given x ∈ A, we can ¬nd a j ≥ 1 with

x ’ ej < .

(i) By considering Q — Q, or by direct construction, or otherwise, show

that we can ¬nd a sequence e1 , e2 , . . . of points which is dense in A.

(ii) By induction, or otherwise, show that we can ¬nd a sequence w1 =

(u1 , v1 ), w2 = (u2 , v2 ), . . . of points in A such that wk ’ ek < 1/k and

ui = uj , vi = vj whenever i = j. Show that the sequence w1 , w2 , . . . is dense

in A.

(iii) De¬ne f : A ’ R by f (x) = 1 if x = wj for some j, f (x) = 0

otherwise. Show that, if x is ¬xed f (x, y) = 1 for at most one value of

1

y. Conclude that F1 (x) = 0 f (x, s) ds exists and takes the value 0 for all

1 1 1 1

y. Thus 0 0 f (x, y) dy dx exists. Similarly, 0 f (x, y) dx dy exists.

0

Show, however, that f is not Riemann integrable.

Exercise K.156. [9.3, H] The object of this exercise is to construct a

bounded open set in R which does not have Riemann length. Our con-

struction will take place within the closed interval [’2, 2] and all references

to dissections and upper sums and so forth will refer to this interval.

(i) Enumerate the rationals in [0, 1] as a sequence y1 , y2 , . . . . Let Uj =

, yk + 2’k’4 ) and U = ∞ (yk ’ 2’k’4 , yk + 2’k’4 ). Explain

j ’k’4

k=1 (yk ’ 2 k=1

—

why U is open and I (IU ) ≥ 1.

(ii) We wish to show that I— (IU ) < 1, so that IU is not Riemann integrable

and U has no Riemann length. To this end, suppose, if possible, that I— (IU ) ≥

1. Explain why this means that we can ¬nd disjoint closed intervals Ir =

[ar , br ] [1 ¤ r ¤ N ] lying inside [’2, 2] such that

N N

(br ’ ar ) ≥ 15/16 yet [ar , br ] ‚ U.

r=1 r=1

(iii) Let Kj = ([’2, 2] \ Uj ) © N [ar , br ]. Explain why Kj is a closed

r=1

bounded set. By considering the total length of the intervals making up Uj ,

show that Kj = …. Use Exercise 4.3.8 to show that

N

([’2, 2] \ U ) © [ar , br ] = ….

r=1

Deduce, by reductio ad absurdum, that I— (IU ) < 1 and U has no Riemann

length.

(iv) Show that there is a closed bounded set in R which does not have

Riemann length. (This is a one line argument.)

514 A COMPANION TO ANALYSIS

(v) Show that there are bounded closed sets and bounded open sets in

R2 which do not have Riemann area.

[The ¬rst example of this type was found by Henry Smith. Smith was a

major pure mathematician at a time and place (19th century Oxford) not

particularly propitious for such a talent. He seems to have been valued more

as a good College and University man than for anything else10 .]

Exercise K.157. [9.3, H, ‘ ] (This continues with the ideas of Exercise K.156

above.)

(i) If (a, b) ⊆ [’2, 2], ¬nd a sequence of continuous functions fn : [’2, 2] ’

R such that 0 ¤ fn (x) ¤ I(a,b) (x) for all n and fn (x) ’ I(a,b) (x) as n ’ ∞

for all x ∈ [a, b].

(ii) If Uj and U are as in Exercise K.156, ¬nd a sequence of continuous

functions fn : [’2, 2] ’ R such that 0 ¤ fn (x) ¤ IUn (x) for all n and

fn (x) ’ IU (x) as n ’ ∞ for all x ∈ [a, b].

(iii) Show 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.158. [9.4, T] Let f : [a, b] ’ R be an increasing function.

(i) If tj ∈ [a, b] and t1 ¤ t2 ¤ t3 ¤ . . . , explain why f (tj ) tends to a limit.

(ii) Suppose x ∈ [a, b]. If tj , sj ∈ [a, b],

t1 ¤ t2 ¤ t3 ¤ . . . , and s1 ¤ s2 ¤ s3 ¤ . . . ,

tj ’ x, sj ’ x and tj , sj < x for all j show that limj’∞ f (tj ) = limj’∞ f (sj ).

Show that the condition tj , sj < x for all j can not be omitted.

(iii) Suppose x ∈ [a, b]. Show that, if x ∈ (a, b] and t ’ x through values

of t with a ¤ t < x, then f (t) tends to a ˜left limit™ f (x’).

(iv) State and prove the appropriate result on ˜right limits™.

(v) If we write J(x) = f (x+) ’ f (x’) for the ˜jump™ at x, show that

J(x) ≥ 0. Show that, if a ¤ x1 < x2 < · · · < xN ’1 < xN ¤ b, then

N

J(xr ) ¤ f (b) ’ f (a).

r=1

(vi) Let us write Ek = {x ∈ [a, b] : J(x) ≥ 1/k}. Use (v) to show that

Ek is ¬nite. Deduce that E = ∞ Ek is countable and conclude that an

k=1

increasing function is continuous except at a countable set of points.

10

He even gave extra teaching on Sunday afternoon, telling his students that ˜It was

lawful on the Sabbath day to pull an ass out of the ditch™.

515

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

(vi) Let us write Ek = {x ∈ [a, b] : J(x) ≥ 1/k}. Use (v) to show that

Ek is ¬nite. Deduce that E = ∞ Ek is countable and conclude that an

k=1

increasing function is continuous except at a countable set of points.

˜ ˜

(vii) If we de¬ne f (x) = f (x+) when x ∈ E, f (x) = f (x) otherwise, show

˜

that f : [a, b] ’ R is a right continuous increasing function.

(viii) If g : R ’ R is an increasing function, show that g is continuous

except at a countable set of points E, say. If we de¬ne g (x) = f (x+) when

˜

x ∈ E, g (x) = g(x) otherwise, show that g : R ’ R is a right continuous

˜ ˜

increasing function.

(ix) Not all discontinuities are as well behaved as those of increasing

functions. If g : [’1, 1] ’ R is de¬ned by g(x) = sin(1/x) for x = 0

show that there is no choice of value for g(0) which will make g left or right

continuous at 0.

Exercise K.159. [9.4, P] De¬ne H : R ’ R by H(t) = 0 if t < 0, H(t) = 1

if t ≥ 0.

(i) If f (t) = sin(1/t) for t > 0, and f (t) = 0 for t < 0, show that, whatever

value we assign to f (0), the function f is not Riemann-Stieltjes integrable

with respect to H.

(ii) If f (t) = sin(1/t) for t < 0, and f (t) = 0 for t > 0, show that f is

Riemann-Stieltjes integrable with respect to H if and only if f (0) = 0.

(iii) By re¬‚ecting on parts (i) and (ii), or otherwise, formulate and prove

a necessary and su¬cient condition for a bounded function f : R ’ R to be

Riemann-Stieltjes integrable with respect to H.

Exercise K.160. [9.4, P, ‘ ] (i) Let H be the Heaviside function discussed

in the previous exercise. Give an example of continuous function f : R ’ R

such that f (t) ≥ 0 for all t and f is not identically zero but

f (x) dH(x) = 0.

(a,b]

(ii) By re¬‚ecting on part (i) and the proof of Exercise 8.3.3, formulate a

necessary and su¬cient condition on G so that the following theorem holds.

If f : R ’ R is continuous and bounded, f (t) ≥ 0 for all t and

f (x) dG(x) = 0,

R

then f (t) = 0 for all t.

Prove that your condition is, indeed, necessary and su¬cient.

(iii) By re¬‚ecting on part (ii), formulate a necessary and su¬cient condi-

tion on G so that the following theorem holds.

516 A COMPANION TO ANALYSIS

If f : R ’ R is continuous and bounded and

f (x)g(x) dG(x) = 0,

R

whenever g : R ’ R is continuous and bounded, it follows that f (t) = 0 for

all t.

Prove that your condition is, indeed, necessary and su¬cient.

Exercise K.161. [9.4, P, ‘ ] (i) Show that, if G : R ’ R is bounded,

increasing and right continuous and f : R ’ R is bounded, increasing and

left continuous, then f is Riemann-Stieltjes integrable with respect to G.

(ii) Suppose G : R ’ R is bounded, increasing and right continuous and

f : R ’ R is bounded, increasing and right continuous. Find necessary and

su¬cient conditions for f to be Riemann-Stieltjes integrable with respect to

G and prove that your statement is correct.

Exercise K.162. (Bounded variation.) [9.4, T] We say that a function

F : [a, b] ’ R is of bounded variation if there exists a constant K such that

whenever a = x0 ¤ x1 ¤ · · · ¤ xn = b, we have

n

|f (xj ) ’ f (xj’1 )| ¤ K.

j=1

(i) Show that any bounded increasing function G : [a, b] ’ R is of

bounded variation.

(ii) Show that the sum of two functions of bounded variation is of bounded

variation. Deduce, in particular, that the di¬erence F ’ G of two increasing

functions F, G : [a, b] ’ R is of bounded variation. What modi¬cations, if

any, would you need to make to obtain a similar result for functions F, G :

R ’ R?

(iii) Show that any function of bounded variation is bounded.

(iv) Let f : [’1, 1] ’ R be de¬ned by f (x) = x sin(1/x) for x = 0,

f (0) = 0. Show that f is continuous everywhere on [’1, 1] but not of bounded

variation. [For variations on this theme consult Exercise K.165.]

Exercise K.163. [9.4, T, ‘ ] (This continues the previous question.) In this

exercise it will be useful to remember that, to prove A ¤ B, it is su¬cient

to prove that A ¤ B + for all > 0. Suppose F : [a, b] ’ R is of bounded

variation.

(i) Explain why we can de¬ne VF : [a, b] ’ R by taking VF (t) to be the

supremum of all sums

n

|f (xj ) ’ f (xj’1 )|,

j=1

517

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

where a = x0 ¤ x1 ¤ · · · ¤ xn = t and n ≥ 1.

Explain why we can de¬ne F+ : [a, b] ’ R by taking F+ (t) to be the

supremum of all sums

n

(f (yj ) ’ f (xj )),

j=0

where a = x0 ¤ y0 ¤ x1 ¤ y1 ¤ x2 ¤ y2 ¤ · · · ¤ xn ¤ yn = t and n ≥ 0.

We de¬ne F’ : [a, b] ’ R by taking F’ (t) to be the supremum of all sums

n

(f (yj’1 ) ’ f (xj )),

j=1

where a = x0 ¤ y0 ¤ x1 ¤ y1 ¤ x2 ¤ y2 ¤ · · · ¤ xn ¤ yn = t and n ≥ 1.

(ii) Show that F+ and F’ are increasing functions and

VF = F + + F ’ , F = F + ’ F ’ .

In particular, we have shown that every function of bounded variation on

[a, b] is the di¬erence of two increasing functions. We call function VF (t) the

total variation of F on [a, t].

(iii) Suppose that we have two increasing functions G+ , G’ : [a, b] ’ R

such that G+ (a) = G’ (a) = 0 and F = G+ ’ G’ . Show that G+ (t) ≥ F+ (t)

and G’ (t) ≥ F’ (t) for all t ∈ [a, b].

Show, more precisely, that G+ ’F+ and G’ ’F’ are increasing functions.

Exercise K.164. [9.4, T, ‘ ] We use the notation and hypotheses of the

previous question. We need the results of Exercise K.158 which the reader

should either do or reread before continuing.

(i) By Exercise K.158 there is a countable (possibly ¬nite or empty) set

E+ such that

lim F+ (t) ’ lim F+ (t) > 0

t’e, t>e t’e, t<e

for e ∈ E+ , and f is continuous at each x ∈ E+ . The same result holds

/

with F+ and E+ replaced by F’ and E’ . By using Exercise K.162 (iii), or

otherwise, show that E+ © E’ = ….

(ii) Use part (i) to show that F+ and F’ are right continuous if F is.

Show that F+ and F’ are continuous if F is.

(iii) Suppose that F is continuously di¬erentiable. Show that

t t

max(F (x), 0) dx, F’ (t) = ’

F+ (t) = min(F (x), 0) dx.

a a

518 A COMPANION TO ANALYSIS

[In setting out your proof, remember that a continuous function may change

sign in¬nitely often in an interval.]

Conclude that F+ and F’ are continuously di¬erentiable if F is.

Exercise K.165. [9.4, P, ‘ ] (This exercise is not required for later parts

of the sequence.) If ± and β are real let f±β : [’1, 1] ’ R be de¬ned by

f±β (x) = x± sin(xβ ) for x = 0, f±β (0) = 0.

(i) For which values of ± and β is f±β of bounded variation?

(ii) For which values of ± and β is f±β everywhere continuous?

(iii) For which values of ± and β is f±β everywhere di¬erentiable?

(iv) For which values of ± and β is f±β everywhere di¬erentiable with

continuous derivative?

Exercise K.166. [9.4, T, ‘ ] We say that a function G : R ’ R is of

bounded variation if there exists a constant K such that, whenever x0 <

x1 < · · · < xn , we have

n

|G(xj ) ’ G(xj’1 )| ¤ K.

j=1

(i) Show that a function G : R ’ R is of bounded variation if and only if

it is the di¬erence of two bounded increasing functions.

(ii) Consider a function G : R ’ R. Show that its restriction G|[a,b] is

of bounded variation for every closed interval [a, b] if and only if it is the

di¬erence of two increasing functions.

(iii) Show that if a function G : R ’ R is of bounded variation then

there exist unique bounded increasing functions G+ , G’ : R ’ R such that

G+ (t), G’ (t) ’ 0 as t ’ ’∞ and G = G+ ’ G’ with the property that, if

F+ , F’ : R ’ R are increasing functions with G = F+ ’ F’ , then F+ ’ G+

and F’ ’ G’ increasing.

(iv) Suppose that G, G+ , G’ are as in (iii). Show that G+ and G’ are

right continuous if G is. Show that G+ and G’ are continuous if G is. Show

that G+ and G’ are continuously di¬erentiable if G is.

(v) We write GV = G+ + G’ . Identify GV as the supremum of a certain

set of sums and prove that your identi¬cation is correct.

Exercise K.167. [9.4, T, ‘ ] (i) Suppose that F, G : R ’ R are bounded

increasing right continuous functions. Show that a bounded function f :

R ’ R is Riemann-Stieltjes integrable with respect to both F and G if and

only if it is Riemann-Stieltjes integrable with respect to F + G.

(ii) Let F1 , F2 , G1 , G2 : R ’ R be bounded increasing right continuous

functions with F1 ’ G1 = F2 ’ G2 . If a bounded function f : R ’ R is

519

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

Riemann-Stieltjes integrable with respect to F1 , F2 , G1 and G2 , show that

f (x) dF1 (x) ’ f (x) dF2 (x) ’

f (x) dG1 (x) = f (x) dG2 (x).

R R R R

Exercise K.168. [9.4, T, ‘ ] We use the notation and results of Exer-

cise K.166. Although we do not use the results of Exercise K.167 directly,

they show that the path chosen is a reasonable one.

If G : R ’ R is right continuous of bounded variation, then we know that

G+ and G’ are increasing bounded right continuous functions. We say that

a bounded function f : R ’ R is Riemann-Stieltjes integrable with respect

to G if it is Riemann-Stieltjes integrable with respect to both G+ and G’ .

We write

f (x) dG+ (x) ’

f (x) dG(x) = f (x) dG’ (x).

R R R

Develop the theory of this extended integral along the lines of Section 9.4,

starting at Exercise 9.4.5 (ii) and ending at Exercise 9.4.11. Note that,

though this is easy, you must be careful to make the right adjustments. Thus,

for example, the conclusion of the result corresponding to Exercise 9.4.5 (iii)

is

f (x) dG(x) ¤ K lim GV (t),

t’∞

R

and, in Exercise 9.4.11 (ii), we can choose »j positive or negative.

Exercise K.169. (The trigonometric functions via arc length.) [9.5,

T] In section 5.5 we developed the theory of the trigonometric functions via

their di¬erential equations. In that development, the trigonometric functions

come ¬rst and angle is de¬ned in terms of those functions (see Exercise 5.5.6).

Here we reverse the process.

(i) Consider the map γ : [’1, 1] ’ R2 given by

γ(y) = ((1 ’ y 2 )1/2 , y)

where we take the positive square root. Convince yourself that this represents

an arc of the unit circle. If 1 > y ≥ 0, we de¬ne the angle θ(y) subtended

at the origin by γ(0) and γ(y) to be length of the curve γ between those

two points. If 0 ≥ y > ’1, we de¬ne the angle θ(y) subtended at the origin

by γ(0) and γ(y) to be minus the length of the curve γ between those two

points.

520 A COMPANION TO ANALYSIS

Show that

y

1

θ(y) = dt

(1 ’ t2 )1/2

0

for ’1 < y < 1. Show that there is a real number ω such that θ(y) ’ ω as

y ’ 1 through values of y < 1, and θ(y) ’ ’ω as y ’ ’1 through values of

y > ’1. We de¬ne θ(1) = ω and θ(’1) = ’ω.

(ii) Show that θ is a strictly increasing continuous function on [’ω, ω].

We may thus de¬ne a function sin : [’ω, ω] ’ R by

sin t = θ ’1 (t).

Show that sin is once di¬erentiable on (’ω, ω) with sin t = (1 ’ (sin t)2 )1/2 .

Show that sin is twice di¬erentiable on (’ω, ω) with sin t = ’ sin t.

(iii) We now de¬ne sin t = sin(2ω ’ t) for t ∈ [ω, 3ω]. Show (paying

particular attention to behaviour at ω) that sin : [’ω, 3ω] ’ R is a well

de¬ned continuous function and that sin is twice di¬erentiable on (’ω, 3ω)

with sin t = ’ sin t.

(iv) Show that we can extend the de¬nition of sin to obtain a twice

di¬erentiable function sin : R ’ R with sin t = ’ sin t.

(v) Show that, if we de¬ne cos t = sin(t ’ ω), then cos2 t + sin2 t = 1 for

all t and 0 ¤ cos s ¤ 1 for all 0 ¤ s ¤ ω. Explain why this gives the ˜right

geometrical meaning™ to cos t for 0 ¤ t ¤ ω. Show also that, if we set π = 2ω,

this gives the right geometrical meaning for π in terms of the circumference

formula for the circle.

Exercise K.170. (The trigonometric functions via area.) [9.5, T, ‘ ]

We might argue that it is better to base our de¬nition of angle on area rather

than the more delicate concept of length. Consider the part of the unit circle

shown in Figure K.3 with A the point (1, 0), B the point (x, (1 ’ x2 )1/2 ) and

C the point (x, 0). Explain why the area of the area of the sector OAB is

1

x(1 ’ x2 )1/2

(1 ’ t2 )1/2 dt.

θ(x) = +

2 x

Use the de¬nition of θ just given to obtain an appropriate de¬nition of

cos t in an appropriate range [0, ω]. Show that cos has the properties we

should expect. Extend your de¬nition so that cos is de¬ned on the whole

real line and has the properties we expect.

Exercise K.171. [9.5, G, M!] (You should treat this exercise informally.)

Here are some simple examples of a phenomenon which has to be accepted

in any reasonably advanced account of length, area and volume.

521

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

Figure K.3: The area of a sector

(i) We work in R2 . Let

Dn = {(x, y) : (x ’ n)2 + y 2 ¤ (n + 1)’2 }.

Sketch E = ∞ Dn and convince yourself that the area of E ought to be

n=1

∞ ∞ ’2

n=1 area Dn = π n=1 (n + 1) , which is ¬nite, but that the length of

the boundary of E ought to be ∞ circumference Dn = 2π ∞ (n + 1)’1 ,

n=1 n=1

which is in¬nite.

(ii) Construct a similar example in R3 involving volume and area.

(iii) We work in C. Consider

∞ n

{z : |z ’ 2’n exp(2πir/n)| ¤ 2’n’3 }.

E = {0} ∪

n=1 r=1

Convince yourself that it is reasonable to say that E has ¬nite area but that

E has boundary of in¬nite length.

(iv) (Torricelli™s trumpet, often called Gabriel™s horn.) This example goes

back historically to the very beginning of the calculus. Consider the volume

of revolution E obtained by revolving the curve y = 1/x for x ≥ 1 around

the x axis. Thus

E = {(x, y, z) ∈ R3 : y 2 + z 2 ¤ x’2 , x ≥ 1}.

∞

We write f (x) = 1/x. Convince yourself that the volume of E is π 1 f (x)2 dx,

∞

which is ¬nite11 but the curved surface of E has area 2π 1 f (x)(1+f (x)2 )1/2 dx,

which is in¬nite.

11

This result is due to Torricelli. In modern terms, this was the ¬rst in¬nite integral

ever to be considered. Torricelli ¬rst evaluated the integral by ˜slicing™ and then gave a

˜proof by exhaustion™ which met Greek (and thus modern) standards of rigour The result

created a sensation because it showed that a solid could have ˜in¬nite extent™ but ¬nite

volume. Thomas Hobbes, a political philosopher who fancied himself as mathematician,

wrote of Torricelli™s result: ˜To understand this for sense, it is not required that a man

should be a geometrician or a logician but that he should be mad™[36].

522 A COMPANION TO ANALYSIS

Thus, if we have a trumpet in the form of the curved surface of E, we

would ˜clearly™ require an in¬nite amount of paint to paint its inside (since

this has in¬nite area). Equally ˜clearly™, since E has ¬nite volume, we can

¬ll the trumpet up with a ¬nite amount of paint and empty it again leaving

the inside nicely painted!

Note that by ˜winding our trumpet like a ball of knitting” we can keep it

within a bounded set (compare parts (i) and (ii)).

(v) (Thor™s drinking horn) As a variation on these themes, consider the

volumes of revolution K obtained by revolving the curve y = x’1/2 for x ≥ 1

around the x axis and K obtained by revolving the curve y = x’1/2 + x’1

for x ≥ 1 around the x axis. Thus

K = {(x, y, z) ∈ R3 : y 2 + z 2 ¤ x’1 , x ≥ 1} and

K = {(x, y, z) ∈ R3 : y 2 + z 2 ¤ (x’1/2 + x’1 )2 , x ≥ 1}.

Show that K has in¬nite volume but that K \ K has ¬nite volume. Sluse

who was the ¬rst to discover such an object wrote to Huygen™s that he could

design a ˜glass of small weight which the hardest drinker could not empty™.

Exercise K.172. [9.5, T] In Section 9.5 we discussed how to de¬ne the line

integral

f (x) ds.

γ

In this exercise we discuss the vector line integral

f (x) ds.

γ

(i) In some sense, we should have

N

f (x) ds ≈ f (xj )(xj ’ xj’1 ).

γ j=1

Discuss informally how this leads to the following variation of the de¬nition

for γ f (x) ds found on page 229.

Let x, y : [a, b] ’ R2 have continuous derivatives (with the usual conven-

tions about left and right derivatives at end points) and consider the curve

γ : [a, b] ’ R2 given by γ(t) = (x(t), y(t)). If f : R2 ’ R is continuous, then

we de¬ne

b

f (x) ds = f (x(t), y(t))γ (t) dt.

a

γ

523

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

(Here, as usual, γ (t) = (x (t), y (t)).)

(ii) Compute γ fj (x) ds for each of the six curves γ k on page 224 and

k

each of the two functions fj given by f1 (x, y) = (x, y) and f2 (x, y) = (y, x).

(iii) (What follows requires Exercise K.168.) We can extend our de¬ni-

tions to general recti¬able curves as follows. Suppose that γ : [a, b] ’ R2

is recti¬able. Show that, if we set γ(t) = (X(t), Y (t)), then the functions

X, Y : [a, b] ’ R are of bounded variation and so we may de¬ne

b b

f (x) ds = f (X(t), Y (t)) dX(t), f (X(t), Y (t)) dY (t) .

a a

γ

Show that, if γ is su¬ciently well behaved that the de¬nition given in part (i)

applies, then the de¬nitions in (i) and (iii) give the same answer.

Exercise K.173. [10.2, H, G] Consider the system described in the ¬rst

few paragraphs of Section 10.2 where n bits are transmitted and each bit has

a probability p of being transmitted wrongly independently of what happens

to any other bit.

(i) Explain why the probability that there are k errors in all is given by

nk

p (1 ’ p)n’k .

un (k) =

k

By looking at un (k + 1)/un (k), or otherwise, show that there is an integer

k0 with un (k + 1) ≥ un (k) for k ¤ k0 ’ 1 and un (k + 1) ≥ un (k) for k0 ¤ k.

Show that np ’ 1 < k0 < np + 1.

(ii) Let > 0. Use the ideas of Lemma 10.2.4 to show that

un (k)

|k’np|<n

’0

un (k)

|k’np|≥n

as n ’ ∞. Deduce that, if we write Nn for the number of errors in a message

of length n,

Pr{|Nn ’ np| ≥ n} ’ 0, †

and so, in particular,

Pr{Nn ≥ (p + )n} ’ 0,

as n ’ ∞.

(iii) If we are prepared to use a little more probability theory then, as the

reader almost certainly knows, there is a general method for obtaining the

524 A COMPANION TO ANALYSIS

result of (ii). (If the notation is unfamiliar do not proceed further.) Write

Xj = 0 if the jth bit is transmitted correctly and Xj = 1 if not. Show that

EXj = p and var Xj = p(1 ’ p). Explain why Nn = n Xj and deduce

j=1

carefully that ENn = pn and var Nn = np(1 ’ p). Now use Tchebychev™s

inequality (Lemma 9.4.14 with X = Nn ’ np) to obtain equation † in (ii).

Exercise K.174. [10.3, P, S] This exercise and the next deal with modi¬-

cations to the axioms for a metric space. Recall that they are

(A) d(x, y) = 0 if and only if x = y.

(B) d(x, y) = d(y, x) for all x, y ∈ X.

(C) d(x, y) + d(y, z) ≥ d(x, z) for all x, y, z ∈ X.

(i) Suppose that d : X 2 ’ R satis¬es axiom (A) and

(C) d(x, y) + d(z, y) ≥ d(x, z) for all x, y, z ∈ X.

Show that (X, d) is a metric space.

(ii) Say everything that there is to say about ˜anti-triangle™ spaces (X, d)

satisfying axioms (A) and (B) together with

(D) d(x, y) + d(z, y) ¤ d(x, z) for all x, y, z ∈ X.

(iii) Suppose (X, d) satis¬es axioms (A) and (B). Show that, if we set

ρ(x, y) = d(x, y) + d(y, x), then (X, ρ) is a metric space.

Exercise K.175. [10.3, T, S] This exercise is more interesting than the

previous one because it deals with a fairly frequent situation. It requires

knowledge of the notions of equivalence relation and equivalence class.

Suppose (X, d) satis¬es axioms (B) and (C) together with

(A) d(x, x) = 0 for all x ∈ X.

Show that the relation x ∼ y if d(x, y) = 0 is an equivalence relation. If

˜

we write [x] for the equivalence class of x and X = X/ ∼ for the set of

equivalence classes, show that

˜

d([x], [y]) = d(x, y)

˜˜

˜

gives a well de¬ned function on X 2 and that (X, d) is a metric space.

Exercise K.176. [10.3, T, G] This is easy but requires enough group the-

ory to understand the statement of the question.

If G is a ¬nite group with identity e generated by a subset X, de¬ne

ρ(e) = 1 and

ρ(g) = min{N : we can ¬nd x1 , x2 , . . . xN ∈ X ∪ X ’1 such that g = x1 x2 . . . xN }

whenever g = e. (Here X ’1 = {x’1 : x ∈ X}.) Show that d(g, h) = ρ(gh’1 )

de¬nes a metric on G.

525

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

Show also that d is left invariant, that is to say d(gk, hk) = d(g, h) for all

g, h, k ∈ G. Is it necessarily true that d is right invariant (i.e. d(kg, kh) =

d(g, h) for all g, h, k ∈ G)? [Hint, if required. Think about the dihedral

group.]

We say that G has diameter maxg∈G (d(e, g)) (with respect to the gener-

ating set X).

(i) Show that Sn , the permutation group on n elements is generated by

the set X1 of elements of the form (ij). Show that Sn has diameter less that

A1 n (for some constant A1 ) with respect to X1 . Interpret this result in terms

of card shu¬„ing.

(ii) Show that Sn , is generated by the set X2 of elements of the form (1j).

Show that Sn has diameter less that A2 n (for some constant A2 ) with respect

to X2 .

(iii) Show that Sn is generated by the set X3 consisting of the two elements

(12) and (123 . . . n). Show that Sn has diameter less that A3 n2 (for some

constant A3 ) with respect to X3 .

(iv) Let n ≥ 2. Consider the dihedral group Dn generated by a and b

subject only to the relations an = e, b2 = e and bab = a’1 . Show that there

exists an A4 > 0 such that Dn has diameter at least A4 n with respect to

X4 = {a, b}.

(v) How many elements does Sn have? How many does Dn have? Com-

ment very brie¬‚y indeed on the relationship between size and diameter in

parts (i) to (iv).

Exercise K.177. [10.3, P] Let (X, d) and (Y, ρ) be a metric spaces and let

X1 and X2 be subsets of X such that X1 ∪ X2 = X. Suppose f : X ’ Y is

such that f |Xj is continuous as a function from Xj to Y . Which, if any, of

the following statements are true and which may be false? (Give proofs or

counterexamples as appropriate.)

(i) The function f is automatically continuous.

(ii) If X1 and X2 are closed, then f is continuous.

(iii) If X1 and X2 are open, then f is continuous.

(iv) If X1 is closed and X2 is open, then f is continuous.

Give Rn its usual Euclidean norm. Suppose g : X ’ Rn is continuous.

Show that there is a continuous function f : X ’ Rn such that f (x) = g(x)

when g(x) < 1 and f (x) = 1 when g(x) ≥ 1.

Exercise K.178. [10.3, P] (i) Give an example of a metric space (X, d)

which contains a set A which is both open and closed but A = X and A = ….

(ii) Suppose (X, d) is a metric space, that A and B are open subsets of

X with A ∪ B = X and that we are given a function f : X ’ R. Show that,

526 A COMPANION TO ANALYSIS

if the restrictions f |A : A ’ R and f |B : B ’ R are continuous, then f is

continuous.

(iii) Consider the interval [0, 1] with the usual metric. Suppose that A

and B are open subsets of [0, 1] with A ∪ B = [0, 1] and A © B = …. De¬ne

f (x) = 1 if x ∈ A and f (x) = 0 if x ∈ B. By considering the result of (ii),

show that A = … or B = …. Deduce that the only subsets of [0, 1] which are

both open and closed are [0, 1] and ….

(iv) Suppose that A is an open subset of Rn and a ∈ A, b ∈ A. By /

considering f : [0, 1] ’ R given by f (t) = 1 if (1 ’ t)a + tb ∈ A and f (t) = 0,

otherwise, show that A is not closed. Thus the only subsets of Rn which are

both open and closed are Rn and ….

Exercise K.179. [10.3, T] Let (X, d) be a metric space. If A is a subset

of X show that the following de¬nitions are equivalent.

¯

(i) The point x ∈ A if and only if we can ¬nd xn ∈ A such that d(xn , x) ’

0 as n ’ ∞.

(ii) Let F be the collection of closed sets F such that F ⊇ A. Then

¯

A = F ∈F F .

¯ ¯

(iii) A is a closed set with A ⊇ A such that, if F is closed and F ⊇ A,

¯

then F ⊇ A.

(It is worth noting that condition (iii) has the disadvantage that it is not

¯

clear that such a set always exists.) We call A the closure of A. We also

¯

write A = Cl A.

If B is a subset of X show that the following de¬nitions are equivalent.

(i) The point x ∈ B —¦ if and only if we can ¬nd > 0 such that y ∈ B

whenever d(x, y) < .

(ii) Let U be the collection of open sets U such that U ⊆ B. Then