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(i) Use the ratio test to show that ∞ n! n’1 (x ’ j)w n converges. By
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)
569
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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

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