1

n=0 j=0

thinking about your proof and the Weierstrass M-test, improve this result to

show that ∞ n! n’1 (x ’ j)w n converges uniformly for |x| ¤ M whenever

1

n=0 j=0

M is ¬xed. We write

∞ n’1

1

(x ’ j)w n .

f (x) =

n!

n=0 j=0

563

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

Explain why f is continuous.

(ii) Use Exercise 5.4.4 (¬rst proved by Cauchy) and Exercise K.241 to

show that

f (x)f (y) = f (x + y)

for all x and y.

(iii) Show that f (0) = 1 and deduce that f (x) = 0 for all x. Deduce,

quoting any theorem you need, that f (x) > 0 for all x.

(iv) Explain why we can de¬ne g : R ’ R by g(x) = log f (x) and why g

is continuous. Show that

g(x) + g(y) = g(x + y)

for all x and y. Use Exercise K.90 (i) (¬rst proved by Cauchy) to show that

g(x) = ax and so f (x) = bx for some real positive number b.

(v) Find b by considering f (1). Deduce that

∞ n’1

1

x

(x ’ j)w n .

(1 + w) =

n!

n=0 j=0

[One problem faced by anyone teaching elementary analysis in France is that

every theorem seems to be called Cauchy™s theorem. A glance at the exercise

above shows why13 .]

Exercise K.243. [11.5, T] (Part (iii) of this question makes repeated use

of Exercise K.76.) Exercise 11.5.24 raises the question of when we can tell

whether a di¬erential equation of the form u (t) = f (t, u(t)) has a power

series solution. The following theorem of Cauchy (see the concluding remark

of the previous question) gives a rather natural condition.

Suppose that cn,m ∈ R [n, m ≥ 0] and that there exists a ρ > 0 and an

K > 0 such that

|cn,m | ¤ Kρn+m for all n, m ≥ 0.

(Exercise K.74 (c) shows why we use this condition.) Then we can ¬nd a

δ > 0 with ρ > δ and an ∈ R such that ∞ an tn converges to u(t), say, for

n=0

13

Moreover, Cauchy™s contributions to the rigorising of analysis form only a fragment

of his contribution to pure and applied mathematics. The following quotation from a

materials engineer can be echoed in many ¬elds. ˜[Cauchy™s paper of 1822] was perhaps

the most important event in the history of elasticity since Hooke. After this, that science

showed promise of becoming a practical tool for engineers rather than a happy hunting-

ground for a few somewhat eccentric philosophers.™ ([18], Chapter 3)

564 A COMPANION TO ANALYSIS

all |t| < δ and such that, writing

∞ ∞

cn,m xn y m

f (x, y) =

n=0 m=0

for |x|, |y| < ρ’1 we have u(0) = 0, |u(t)| < ρ and

u (t) = f (t, u(t))

for all |t| ¤ δ.

(i) The object of this question is to prove Cauchy™s theorem but in the

¬rst two parts of the question we are merely interested in seeing what is

going on so we work in a non-rigorous manner. Assuming that a power

series solution ∞ an tn with the required properties actually exists show

n=0

by formal manipulation that a0 = 0 and kak should be the coe¬cient of xk’1

in

m

∞ ∞ ∞

cn,m xn ar x r .

n=0 m=0 r=0

Explain why that coe¬cient is a ¬nite sum only depending on a0 , a1 , . . . ,

ak’1 . Show that, if we de¬ne A0 = 0, and de¬ne A1 , A2 , . . . , formally by

means of the equation

m

∞ ∞ ∞ ∞

d

An tn Kρn+m tn Ar tr

= ,

dt n=0 n=0 m=0 r=0

then |ak | ¤ Ak .

(ii) We continue with the ideas of (i). Show that can be rewritten (at

least, formally) as

dw K

=

(1 ’ ρt)(1 ’ ρw)

dt

where w(t) = ∞ An tn . Solve equation for w(0) = 0.

n=0

(iii) We now reverse the non-rigorous argument but this time proceed

rigorously. Show that equation has a solution w(t) with w(0) = 0 and

the property that we can ¬nd Aj and · > 0 such that ∞ An tn has radius

n=0

∞ n

of convergence at least · and w(t) = n=0 An t for all |t| < ·. Show that

for |t| < ·. Deduce that, if ak is given by equating

the An satisfy equation

coe¬cients in the equation

m

∞ ∞ ∞ ∞

an x n = cn,m xn ar x r ,

n=0 n=0 m=0 r=0

565

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

then |ak | ¤ Ak and ∞ an tn has radius of convergence at least ·. Show

n=0

that if we set u(t) = ∞ an tn for |t| < ·, then u(0) = 0 and there exists

n=0

a δ with 0 < δ ¤ · such that |u(t)| < ρ. Show that δ and u satisfy the

conclusions of Cauchy™s theorem.

Exercise K.244. (Convolution.) [11.6, T] Let us write CP (R) for the

set of continuous function f : R ’ R which are periodic with period 2π.

(i) Show that, if f, g ∈ CP (R), then the equation

π

1

f — g(t) = f (t ’ s)g(s) ds

2π ’π

gives a well de¬ned, continuous, 2π periodic function f — g : R ’ R. (Thus

f — g ∈ CP (R).)

(ii) Show that, if f, g ∈ CP (R), then

2π+a

1

f — g(t) = f (t ’ s)g(s) ds

2π a

for all t ∈ R and a ∈ R.

(iii) Show that if f, g, h ∈ CP (R) and » ∈ R then

(»f ) — g = »(f — g), f — (g + h) = f — g + f — h,

f — g = g — f and f — (g — h) = (f — g) — h.

(iv) Suppose that f, g ∈ CP (R) and f is di¬erentiable with f ∈ CP (R).

Show that f — g is di¬erentiable and

(f — g) = f — g.

(v) Suppose that f, un ∈ CP (R), that un (t) ≥ 0 for all t ∈ R, that

un (t) = 0 for π/n ¤ |t| ¤ π and

π

1

un (t) dt = 1

2π ’π

show that un — f ’ f uniformly as n ’ ∞.

(vii) Suppose that f ∈ CP (R) and > 0. Show, by using parts (iv) and (v)

that there exists a twice di¬erentiable function g with g, g , g ∈ CP (R) such

that f ’ g ∞ ¤ .

(viii) Let us write CP (C) for the set of continuous function f : R ’ C

which are periodic with period 2π. Show. in as much detail as you consider

desirable, that parts (i) to (vi) hold with CP (R) replaced by CP (C)

566 A COMPANION TO ANALYSIS

Exercise K.245. [11.6, T, ‘ ] We continue with the notation of the previous

question.

(i) Suppose that that f, g ∈ CP (C). Show that

ˆg

f — g(n) = f (n)ˆ(n)

for all integers n. (We saying that ˜taking Fourier coe¬cients converts con-

volution to multiplication™.)

(ii) Show that if f ∈ CP (C), then

ˆ

|f (n)| ¤ f ∞.

(iii) Show that if f, g ∈ CP (C), then

ˆ

|f (n)| ¤ |ˆ(n)| + f ’ g

g ∞.

(iv) Use part (iii) of this exercise, part (vii) of Exercise K.244 and Exer-

ˆ

cise 11.6.10 to show that, if f ∈ CP (C), then f (n) ’ 0 as |n| ’ ∞. (This is

a version of the important Riemann-Lebesgue lemma.)

(v) Suppose, if possible, that there exists an e ∈ CP (C) such that e—f = f

for all f ∈ CP (C). Find e and use (iv) to show that no such e can exist.

ˆ

(Informally, convolution on CP (C) has no unit.)

Exercise K.246. [11.6, T, ‘ ] This question uses the version of the Riemann-

Lebesgue lemma proved in Exercise K.245 (iv). If f ∈ CP (R) we write

n

ˆ

Sn (f, t) = f (n) exp(int).

j=’n

(i) Suppose that f1 , g1 ∈ CP (R) and f1 (t) = g1 (t) sin t for all t. Show

ˆ

that f1 (j) = g1 (j + 1) ’ g1 (j ’ 1) and deduce that Sn (f1 , 0) ’ 0 as n ’ ∞.

ˆ ˆ

(ii) Suppose that f2 ∈ CP (R), f2 (nπ) = 0 and f2 is di¬erentiable at nπ for

all integer n. Show that there exists a g2 ∈ CP (R) such that f2 (t) = g2 (t) sin t

for all t and deduce that Sn (f2 , 0) ’ 0 as n ’ ∞.

(iii) Suppose that f3 ∈ CP (R), f3 (0) = 0 and f3 is di¬erentiable at 0.

ˆ

Write f4 (t) = f3 (t/2). Compute f4 (j) in terms of the Fourier coe¬cients of

f3 . Show that Sn (f4 , 0) ’ 0 and deduce that Sn (f3 , 0) ’ 0 as n ’ ∞.

(iv) Suppose that f ∈ CP (R), and f is di¬erentiable at some point x.

Show that Sn (f, x) ’ f (x) as n ’ ∞.

Exercise K.247. [12.1, H] The result of this question is a special case of

part of Lemma 13.1.4. However, precisely because it is a special case, some

readers may ¬nd it a useful introduction to the ideas of Section 13.1.

567

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

(i) Let δ > 0 and let f : C ’ C be an everywhere di¬erentiable function

such that |f (z) ’ 1| ¤ 1/2 for all |z| ¤ δ. Set X = {z ∈ C : |z| ¤ δ}. If

|w| ¤ δ/2 and we de¬ne T z = z ’ f (z) + w for z ∈ X show, by using the

mean value inequality (Exercise 11.5.5) that T z ∈ X whenever z ∈ X. Thus

T is a function T : X ’ X.

(ii) We continue with the hypotheses and notation of part (i). Show that

T is a contraction mapping and deduce that the equation f (z) = w has

exactly one solution with |z| ¤ δ.

(iii) Let F : C ’ C be an everywhere di¬erentiable function with con-

tinuous derivative. Suppose further that F (0) = 0 and F (0) = 1. Show

that there exists a δ > 0 such that, if |w| ¤ δ/2, the equation F (z) = w has

exactly one solution with |z| ¤ δ.

(iv) Let g : C ’ C be an everywhere di¬erentiable function with contin-

uous derivative. Show that, if g (0) = 0, there exists ·1 . ·2 > 0 such that, if

|ω ’ g(0)| ¤ ·1 , the equation g(z) = w has exactly one solution with |z| ¤ ·2 .

(v) If we omit the condition g (0) = 0 in (iv) does it remain true that

there exists an · > 0 such that, if |ω ’ g(0)| ¤ ·, the equation g(z) = w has

a solution? Give a proof or counterexample.

(vi) If we omit the condition g (0) = 0 in (iv) but add the condition

g (0) = 0 does it remain true that there exist ·1 . ·2 > 0 such that, if |ω ’

g(0)| ¤ ·1 , the equation g(z) = w has at most one solution with |z| ¤ ·2 ?

Give a proof or counterexample.

Exercise K.248. [12.1, P, ‘] We work in C. Consider the following state-

ment.

There exists a δ > 0 such that, if |w| ¤ δ, the equation fj (z) = w has a

solution.

Give a proof or counterexample in each of the following cases.

(i) f1 (z) = z — .

(ii) f2 (z) = z + z — .

(iii) f3 (z) = z + |z|.

(iv) f4 (z) = z + |z|2 .

Is our statement true for all F in the following cases? Give a proof or

counterexample.

(v) f5 (z) = F (z) + |z| where F : C ’ C is an everywhere di¬erentiable

function with continuous derivative and F (0) = 0 and F (0) = 1.

(vi) f6 (z) = F (z) + |z|2 where F : C ’ C is an everywhere di¬erentiable

function with continuous derivative and F (0) = 0 and F (0) = 1.

Exercise K.249. (Newton-Raphson.) [12.1, T] (i) Suppose that f :

568 A COMPANION TO ANALYSIS

R ’ R is a twice di¬erentiable function such that

f (x)f (x)

¤»

f (x)2

for all x and some |»| < 1 Show that the mapping

f (x)

Tx = x ’

f (x)

is a contraction mapping and deduce that f has a unique root y.

(ii) Suppose that F : R ’ R is a twice di¬erentiable function such that

F (x)F (x)

¤»

F (x)2

for all |x| ¤ a and some |»| < 1 and that F (0) = 0. Consider the mapping

F (x)

Tx = x ’ .

F (x)

Show that T n x ’ 0.

Suppose that

sup|t|¤a |F (t)| sup|t|¤a |F (t)|

= M.

inf |t|¤a |F (t)|2

By using the mean value theorem twice, show that, if |x| ¤ a, then

|T x| ¤ M x2 .

(iii) If you know what the Newton-Raphson method is, comment on the

relevance of the results of (i) and (ii) to that method. Comment in particular

on the speed of convergence.

Exercise K.250. [12.1, G, P, S] (This is a short question and involves no

analysis.) Suppose f, g : X ’ X are such that f g = gf . Show that if f

has a unique ¬xed point then g has a ¬xed point. Can g have more than one

¬xed point? (Give a proof or counterexample.)

Show that, if we merely know that f has ¬xed points, it does not follow

that g has any.

Exercise K.251. [12.1, P] If (X, d) is a complete metric space and T :

X ’ X is a surjective map such that

d(T x, T y) ≥ Kd(x, y)

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

for all x, y ∈ X and some K > 1, show that T has a unique ¬xed point.

By considering the map T : R ’ R de¬ned by T (x) = 1 + 4n + 2x

for 0 ¤ x < 1 and n an integer, or otherwise, show that the condition T

surjective cannot be dropped.

Exercise K.252. [12.1, P] We work in Rm with the usual distance. Let E

be a closed non-empty subset of Rm and let T be a map T : E ’ E.

(i) Suppose T (a) ’ T (b) < a ’ b for all a, b ∈ E with a = b. We

saw in Example 12.1.4 that T need not have a ¬xed point. Show that, if T

has a ¬xed point, it is unique.

(ii) Suppose T (a) ’ T (b) > a ’ b for all a, b ∈ E with a = b. Show

that T need not have a ¬xed point but, if T has a ¬xed point, it is unique.

(iii) Suppose T (a) ’ T (b) = a ’ b for all a, b ∈ E. Show that T

need not have a ¬xed point and that, if T has a ¬xed point, it need not be

unique.

(iv) Suppose now that E is non-empty, closed and bounded and

T (a) ’ T (b) < a ’ b

for all a, b ∈ E with a = b. By considering inf x∈E x ’ T (x) , or otherwise,

show that T has a ¬xed point.

(v) Suppose that E = Rm and

T (a) ’ T (b) < a ’ b

for all a, b ∈ Rm with a = b. Show that, if there exists a point c ∈ Rm and

a K > 0 such that T n c < k for all positive integer n, then T has a ¬xed

point.

Exercise K.253. [12.1, G, P, S] (This is easy if you see what is going on.)

Let (X, d) be a metric space. Show that the set of isometries f : X ’ X

forms a group G(X) under composition.

By choosing X to be an appropriate closed bounded subset of R2 with

the usual metric, or otherwise, prove the following statements about G(X)

when X has the Bolzano-Weierstrass property.

(i) G(X) need not be Abelian.

(ii) There exists an X such that G(X) has only one element.

(iii) There exists an X such that G(X) has in¬nitely many elements.

If n ≥ 2 give an example of an (X, d) with the Bolzano-Weierstrass prop-

erty and a T : X ’ X such that T m has no ¬xed points for 1 ¤ m ¤ n ’ 1

but T n does.

570 A COMPANION TO ANALYSIS

Exercise K.254. [12.1, P] (i) Let (X, d) be a metric space with the Bolzano-

Weierstrass property. Suppose f : X ’ X is expansive in the sense that

d(f (x), f (y)) ≥ d(x, y)

for all x, y ∈ X. (Note that we do not assume that f is continuous.) If

a, b ∈ X we write a0 = a and an+1 = f (an ) and de¬ne a sequence bn

similarly. Show that we can ¬nd a sequence of integers n(1) < n(2) < . . .

such that d(an(k) , a0 ) ’ 0 and d(bn(k) , b0 ) ’ 0 as n ’ ∞. Deduce that

d(a, b) = d(f (a), f (b)) for all a, b ∈ X (that is, f is an isometry).

Find a complete metric space (Y, ρ) and maps g, h : Y ’ Y such that

ρ(g(x), g(y)) ≥ 2ρ(x, y), ρ(h(x), h(y)) ≥ 2ρ(x, y)

and g is continuous but h is not.

(ii) Let (X, d) be a metric space with the Bolzano-Weierstrass property.

Suppose f : X ’ X is an isometry in the sense that

d(f (x), f (y)) = d(x, y)

for all x, y ∈ X. If a ∈ X we write a0 = a and an+1 = f (an ). Show (as

in (i)) that we can ¬nd a sequence of integers n(1) < n(2) < . . . such that

d(an(k) , a0 ) ’ 0 as n ’ ∞ and deduce that f is bijective.

Find a complete metric space (Y, ρ) and a map g : Y ’ Y such that

ρ(g(x), g(y)) = ρ(x, y)

for all x, y ∈ Y , but g is not surjective.

Exercise K.255. [12.1, P] Let (X, d) be a metric space with the Bolzano-

Weierstrass property and (Y, ρ) a metric space. Suppose f : X ’ Y and

g : Y ’ X are such that ρ(f (x), f (x )) ≥ d(x, x ) for all x, x ∈ X and

d(g(y), g(y )) ≥ ρ(y, y ) for all y, y ∈ Y . By considering g —¦ f and us-

ing Exercise K.254 show that ρ(f (x), f (x )) = d(x, x ) for all x, x ∈ X,

d(g(y), g(y )) = ρ(y, y ) for all y, y ∈ Y and f and g are bijective. Is it

necessarily true that f and g are inverse functions?

Show that the result of the ¬rst paragraph may fail if we simply have

f : X ’ f (X) bijective with f and f |’1 continuous and g : Y ’ g(Y )

f (X)

’1

bijective with g and g|g(Y ) continuous.

Show that the result may fail, even if (X, d) and (Y, ρ) are complete, if

(X, d) does not have the Bolzano-Weierstrass property.

571

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

Exercise K.256. [12.1, P] Let (V, ) be a complete normed space. We

say that a subset A ⊆ V is convex if whenever a, b ∈ A and 0 ¤ » ¤ 1 we

have

»a + (1 ’ »)b ∈ A.

Suppose that A is closed bounded convex subset of V and f : A ’ A satis¬es

f (a) ’ f (b) ¤ a ’ b