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(iii) Prove the remaining results corresponding to to those in part (iii) of
Question K.227.
Exercise K.229. [11.4, T!, ‘ ] Question K.228 does not give all the prop-
erties of x± that we are interested in. In particular, it tells us little about
the map ± ’ x± . If a > 1, let us write P (t) = Pa (t) = at .
(i) Show that P : R ’ (0, ∞) is an increasing function.
(ii) Show that P (t) ’ 1 as t ’ 0.
(iii) By using the relation P (s+t) = P (s)P (t), show that P is everywhere
continuous.
(iv) Show that, if P is di¬erentiable at 1, then P is everywhere di¬eren-
tiable.
(v) Let f (y) = n(y 1/n ’ 1) ’ (n + 1)(y 1/(n+1) ’ 1). By considering f , or
otherwise, show that f (y) ≥ 0 for all y ≥ 1. Hence, or otherwise, show that
P (1/n) ’ P (0)
tends to a limit, L say,
1/n
557
Please send corrections however trivial to twk@dpmms.cam.ac.uk

as n ’ ∞.
(vi) Show that, if 1/n ≥ x ≥ 1/(n + 1) then

P (x) ’ P (0) P (1/n) ’ P (0) n + 1
¤ — .
x 1/n n
By using this and similar estimates, or otherwise, show that
P (x) ’ P (0)
’L
x
as x ’ 0 through values of x > 0.
(vii) Show that, if x = 0

P (x) ’ P (0) P (’x) ’ P (0)
= P (x)
’x
x
and deduce that
P (x) ’ P (0)
’L
x
as x ’ 0. Conclude that P is di¬erentiable at 0 and so everywhere with
P (t) = LP (t).
(viii) If a > 0 and we set Pa (t) = at , show that Pa is di¬erentiable with
Pa (t) = La P (t) for some constant La . Show that La > 0 if a > 1. What can
you say about La for other values of a?
(ix) If a, » > 0 show that La» = »La . Deduce that there is a real number
e > 1 such that, if we write

e(t) = et ,

then e is a di¬erentiable function with e (t) = e(t).
(x) Tear o¬ the disguise of La .
We have now completed a direct attack on the problem of de¬ning x± and
obtaining its main properties. It should be clear why the indirect de¬nition
x± = exp(± log x) is preferred.
Exercise K.230. [11.5, P] Suppose ∞ an z n has radius of convergence
n=0
greater than or equal to R > 0. Write g(z) = ∞ an z n for |z| < R. Show
n=0

that, if |z0 | < R then we can ¬nd b0 , b1 , · · · ∈ C such that n
n=0 bn z
has radius of convergence greater than or equal to R ’ |z0 | and g(z) =
∞ n
n=0 bn (z ’ z0 ) for |z0 ’ z| < R ’ |z0 |. [This is quite hard work. If you
have no idea how to attack it, ¬rst work out formally what the values of the
bn must be. Now try and use Lemma 5.3.4.]
558 A COMPANION TO ANALYSIS

Show that g is complex di¬erentiable at z0 with g (z0 ) = b1 = ∞ nan z0 .n’1
n=1
Deduce that a power series is di¬erentiable and that its derivative is that ob-
tained by term by term di¬erentiation within its radius of convergence. (We
thus have an alternative proof of Theorem 11.5.11.)

Exercise K.231. [11.5, P] Here is another proof of Theorem 11.5.11. Sup-
pose ∞ an z n has radius of convergence greater than or equal to R > 0.
n=0
(i) Show that ∞ nan z n’1 and ∞ n(n ’ 1)an z n’2 also have radius
n=1 n=2
of convergence R.
(ii) Show that

n’2
n
¤ n(n ’ 1)
j j

for all n ’ 2 ≥ j ≥ 0 and deduce that

|(z + h)n ’ z n ’ nz n’1 h| ¤ n(n ’ 1)(|z| + |h|)n’2

for all z, h ∈ C.
(iii) Use (i) and (ii) to show that, if 0 < δ and |z| + |h| < R ’ δ, then
∞ ∞ ∞
n n
nan z n’1 ¤ A(δ)|h|2
an (z + h) ’ an z ’ h
n=0 n=0 n=1

where A(δ) depends only on δ (and not on h or z).
(iv) Deduce that a power series is di¬erentiable and that its derivative is
that obtained by term by term di¬erentiation within its radius of convergence.


Exercise K.232. [11.5, P] Consider the Legendre di¬erential equation

(1 ’ x2 )y ’ 2xy + l(l + 1)y = 0,

where l is a real number. Find the general power series solution. Show that,
unless l is non-negative integer, every non-trivial power series solution has
radius of convergence 1.
Show, however, that, if l = n a non-negative integer the general series
solution can be written

y(x) = APn (x) + BQn (x)

where Qn is a power series of radius of convergence 1, Pn is a polynomial of
degree n, and A and B are arbitrary constants.
559
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Verify that, when n is a non-negative integer, the function given by v(x) =
(x2 ’ 1)n satis¬es the equation

(1 ’ x2 )v (x) + 2nxv(x) = 0.

Deduce that v (n) (x) satis¬es the Legendre di¬erential equation with l = n.
[Consult Exercise K.270 if you need a hint.] Conclude that Pn is a constant
multiple of the function pn de¬ned in Exercise K.212.

Exercise K.233. [11.5, P] Obtain the general power series solution of
2
2d w dw
+ (z 2 ’ 1)w = 0.
z +z
dz 2 dz
For what values of z is your solution valid and why?
Answer the same questions for
2
2d w dw
+ (z 2 ’ 1)w = z 2 .
z +z
dz 2 dz
Exercise K.234. [11.5, P] Using Lemma 11.5.19 or otherwise, ¬nd the
power series expansion ∞ aj xj of (1 + x)1/2 for x real and |x| < 1.
j=0
Show that the complex power series ∞ aj z j has radius of convergence
j=0
1. What is the power series for the product ∞ aj z j ∞ aj z j and why?
j=0 j=0
What is its radius of convergence?
By considering
∞ ∞ ∞
aj K ’j z j
aj z j aj z j ,
j=0 j=0 j=0


or otherwise, show that, if two power series of the same radius of conver-
gence R are multiplied, the resulting power series may have any radius of
convergence with value R or greater.
By considering expressions of the form
∞ ∞ ∞
Aj z j Bj z j + Cj z j ,
j=0 j=0 j=0


or otherwise, show that, if two power series of radius of convergence R and S
are multiplied, the resulting power series may have any radius of convergence
with value min(R, S) or greater.
560 A COMPANION TO ANALYSIS

Exercise K.235. [11.5, P] By modifying the proof of Abel™s test (Lemma 5.2.4),
or otherwise, prove the following result. Let E be a set and aj : E ’ Rm a
sequence of functions. Suppose that there exists a K such that
n
aj (t) ¤ K
j=1

for all n ≥ 1 and all t ∈ E. If »j is a decreasing sequence of real positive
numbers with »j ’ 0 as j ’ ∞ then ∞ »j aj converges uniformly on E.
j=1
Deduce, in particular, that if bn ∈ R and ∞ bn converges, then ∞ bn xn
n=0 n=0
converges uniformly on [0, 1]. Explain why this implies that
∞ ∞
1
bn
n
bn x dx = .
n+1
0 n=0 n=0

Show that, provided we interpret the left hand integral as an improper
1’
integral, lim ’0, >0 0 , equation remains true under the assumption

that n=0 bn /(n + 1) converges.
Show that
111
1 ’ + ’ + · · · = log 2
234
and
111 π
1 ’ + ’ + ··· = .
357 4
Do these provide good methods for computing log 2 and π? Why?
Recall that, if a, b > 0, then aloga b = b. Show that
log2 e ’ log4 e + log8 e ’ log16 e + . . .
converges to a value to be found.
Exercise K.236. [11.5, G, S] (Only if you have done su¬cient probability,
but easy if you have.) Let X be a random variable taking positive integral
values. Show that

φX (t) = Et = X
Pr(X = n)tn
n=0

is well de¬ned for all t ∈ [’1, 1].
If EX is bounded, show that
φX (1) ’ φX (t)
’ EX
1’t
as t ’ 1 through values t < 1. Give an example of an X with EX in¬nite.
561
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.237. [11.5, T, S] Use the ideas of Exercise K.78 to prove the
following extension. Suppose that we have a sequence of functions an : „¦ ’ C
and a sequence of positive real numbers Mn such that ∞ Mj converges
j=1
absolutely and |an (z)| ¤ Mn for all z ∈ „¦ and all n. Then

N
(1 + aj (z)) ’ (1 + aj (z))
j=1 j=1


uniformly on „¦.

Exercise K.238. [11.5, P, S, ‘ ] We do not have the apparatus to exploit
Exercise K.237 fully but the following exercise is suggestive.
(i) Show that SN (z) = z N (1 ’ z 2 n’2 ) converges uniformly to S(z) =
n=1
z ∞ (1 ’ z 2 n’2 ) on any disc {z : |z| < R}. Deduce that S is a well de¬ned
n=1
continuous function on C.
(ii) By writing SN (z) = ’(N !)’2 N n=’N (n ’ z) and considering SN (z +
1)/SN (z), or otherwise, show that S(z + 1) = S(z) (that is, S is periodic
with period 1). Show also that S(z) = 0 if and only if z is an integer.
[The product z ∞ (1 ’ z 2 n’2 ) goes back to Euler who identi¬ed it with
n=1
a well known function which the reader should also be able to guess. An
advantage of in¬nite products over in¬nite sums is that we can ¬nd zeros of
the function much more easily.]

Exercise K.239. [11.5, T] (i) Suppose that an ∈ C and ∞ an z n has
n=0
radius of convergence R > 0. If aN = 0 and an = 0 for n < N , use the
inequality
q q
n N
|an z n’N |
¤ |z| |aN | ’
an z
n=0 n=N +1


(for q > n), together with standard bounds on |an z n |, to show that there
exists a δ > 0 such that
q
|aN ||z|N
n

an z
2
n=0

for all |z| < δ.
(ii) Suppose that an ∈ C and ∞ an z n has radius of convergence R > 0.
n=0
∞ n
Set f (z) = n=0 an z for |z| < R. Show, using (i), or otherwise, that, if
there exist wm ’ 0 with f (wm ) = 0 and wm = 0, then an = 0 for all n and
f = 0.
562 A COMPANION TO ANALYSIS

Exercise K.240. [11.5, P, ‘ ] This is a commentary on the previous exer-
cise. We shall consider functions from R to R.
(i) Suppose that an ∈ R and ∞ an xn has radius of convergence R > 0.
n=0
Set f (x) = ∞ an xn for |x| < R. Show that, if there exist xm ’ 0 with
n=0
f (xm ) = 0 and xm = 0, then an = 0 for all n and f = 0.
(ii) Explain why a polynomial of degree at most n with n + 1 zeros must
be zero.
(iii) Give an example of a power series of radius of convergence ∞ which
has in¬nitely many zeros but is not identically zero.
(iv) Let F be as in Example 7.1.5. Set G(x) = F (x) sin x’1 for x = 0
and G(0) = 0. Show that G is in¬nitely di¬erentiable everywhere (this will
probably require you to go through much the same argument as we used for
F ). Show that there exist xm ’ 0 with G(xm ) = 0 and xm = 0 but G is not
identically zero.
Exercise K.241. [11.5, P] This question prepares the way for Exercise K.242.
The ¬rst two parts will probably be familiar but are intended to provide a
hint for the third part.
n n n+1
(i) If n ≥ r ≥ 1, show that + = .
r’1 r r
(ii) Use induction to show that
n
n r n’r
(x + y)n = xy
r
r=0

whenever n is a positive integer and x and y are real.
(iii) Suppose that n is a positive integer and x and y are real. Verify the
next formula directly for n = 0, 1, 2, 3 and then prove it for all n.
n r’1 n’r’1 n’1
n
(x ’ j) (y ’ k) = (x + y ’ q).
r
r=0 q=0
j=0 k=0

Exercise K.242. [11.5, T, ‘ ] The following is a modi¬cation of Cauchy™s
proof of the binomial theorem. We shall work in R. We take w as a ¬xed
real number with |w| < 1.
(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
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

for all a, b ∈ A. By considering

g(x) = (1 ’ )f (x) + a0

for some a0 ∈ A and > 0 show that

inf{ f (a) ’ a : a ∈ A} = 0

and deduce that f has a ¬xed point in A.
Give an example to show that the conclusion may be false if any one of
the three conditions:- convex, closed and bounded is omitted.
The following result is useful in the theory of Markov chains (see Exer-
cise K.258). Suppose pij ≥ 0 for 1 ¤ i, j ¤ n and n pij = 1 for 1 ¤ i ¤ n.
j=1
n
Give R the norm
n
j=1 |xj |. By choosing an appropriate
1 where x 1 =
A and f show that we can ¬nd πi ≥ 0 for 1 ¤ i ¤ n with n πi = 1 such
i=1
that
n
πi pij = πj .
i=1


Exercise K.257. [12.1, P] Throughout this question (X, d) is a non-empty
metric space with the Bolzano-Weierstrass property and T : X ’ X is
a map such that d(T x, T y) ¤ d(x, y) and such that, given any x, y ∈ X
such that x = y we can ¬nd a strictly positive integer N (x, y) such that
d(T N (x,y) x, T N (x,y) y) < d(x, y).
(i) Show that the map S : X ’ R given by Sx = d(x, T x) is continuous.
Explain why this means that there is a y ∈ X such that d(y, T y) ¤ d(T x, x)
for all x ∈ X. Show that, in fact y = T y. Show that y is the unique ¬xed
point of T .
(ii) Let > 0 and write

Un = {x ∈ X : d(T n x, y) < }.
572 A COMPANION TO ANALYSIS

Show that Un ⊆ Un+1 and that Un is open for all n ≥ 1. Write U = ∞ Un
n=1
and F = X \ U . Why does F have the Bolzano-Weierstrass property? Show
that T x ∈ F whenever x ∈ F . Use part (i) to show that, if F were non-
empty, there would exist a w ∈ F with T w = w. Deduce that F is indeed
empty.
(iii) Show that, if x ∈ X, then d(T n x, y) ’ 0 as n ’ ∞.
Exercise K.258. (Finite Markov chains.) [12.1, T!, ‘ ] In a ¬nite
Markov chain with states 1, 2, . . . , m, if the system is in state i at time
n it will be in state j at time n + 1 with probability pij irrespective of any
earlier history. Thus, if it is in state i with probability qi at time n and in
state j with probability qj at time n + 1, we have
m
qj = qi pij .
i=1

The reader may ignore the previous paragraph but it explains why we are
interested in the space
m
X = {q ∈ Rm : qi = 1, qs ≥ 0 [1 ¤ s ¤ m]}
i=1

and the map T : X ’ Rm given by T q = q where qj = m qi pij for all
i=1
1 ¤ j ¤ m. Verify that if, as we shall do from now on, we take pij ≥ 0 for all
1 ¤ i, j ¤ m and m pij = 1 for all 1 ¤ i ¤ m then T q ∈ X for all q ∈ X
j=1
and so we may consider T as a map from X to X.
Finally we have to choose a metric d. In most of this book the particular
choice of metric has not been important but here it is. We take
m
d(u, v) = u ’ v |ui ’ vi |
=
1
i=1

for all u, v ∈ X.
(i) Explain why (X, d) has the Bolzano-Weierstrass property.
(ii) Show that d(T u, T v) ¤ d(u, v) for all u, v ∈ X.
(iii) Show that, if n ≥ 1,
m
(n)
n
(T q)j = qi pij
i=1

(n) (n)
for all 1 ¤ j ¤ m where pij ≥ 0 for all 1 ¤ i, j ¤ m and m pij = 1 for
j=1
all 1 ¤ i ¤ m. If you know some probability prove this both by an algebraic
and a probabilistic argument.
573
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(iv) Show that, if
(N )
pij > 0 for all 1 ¤ i, j ¤ m

for some N ≥ 1, then d(T N u, T N v) < d(u, v) for all u, v ∈ X with u = v.
(v) Conclude, using Exercise K.257, that, provided only that holds for
some N ≥ 1, there is a unique π ∈ X such that T π = π and that, if q ∈ X,
then d(T n q, π) ’ 0 as n ’ ∞. If the associated Markov chain has been
running for a long time, what can you say about the probability that it is in
state i?
(vi) Suppose that m = 2, p12 = p21 = 1 and p11 = p22 = 0. Show that
is a unique π ∈ X such that T π = π. For which q ∈ X is it true that
d(T n q, π) ’ 0 as n ’ ∞? Why does your result not contradict (v)?
(vii) Suppose that m = 2, p12 = p21 = 0 and p11 = p22 = 1. Find all the
π ∈ X such that T π = π. What can you say about the behaviour of T n q
as n ’ ∞ for each q ∈ X? Why does your result not contradict (v)?
(viii) Suppose that m = 4 and that pij = 1/2 if 1 ¤ i, j ¤ 2 or 3 ¤ i, j ¤
4 and that pij = 0 otherwise. Find all the π ∈ X such that T π = π. What
can you say about the behaviour of T n q as n ’ ∞ for each q ∈ X?
(ix) Note that, by Exercise K.256, the equation T π = π always has a
solution in X even if does not hold.
Exercise K.259. (Countable Markov chains.) [12.1, T!] In this exer-
cise and the next we consider Markov chains with an in¬nite number of states
1, 2, . . . . In algebraic terms we consider a collection of positive numbers pij
[1 ¤ i, j] such that ∞ pij = 1 for each i ≥ 1.
j=1
1
(i) Consider l the space of real sequences whose sum is absolutely con-

n=1 |xn |. Show that, if we write
vergent, with norm 1 given by x ∞ =

(T x)j = j=1 xi pij , then T is well de¬ned continuous linear map from l 1 to
l1 with operator norm T = 1. Show further that, if we write

X = {q ∈ l1 : q = 1 and qi ≥ 0 for all i ≥ 1},
1

then T q ∈ X whenever q ∈ X. (Note that you are manipulating in¬nite
sums and you must justify each step carefully. Results like Lemma 5.3.9 may
be useful.)
(ii) We shall only be interested in systems satisfying the following exten-
sion of . If k ≥ 1 there exists an N (k) such that
(N (k))
> 0 for all 1 ¤ i, j ¤ k.
pij

Show that, under this condition, given any u, v ∈ X with u = v, we can
¬nd an N ≥ 1 such that d(T N u, T N v) < d(u, v).
574 A COMPANION TO ANALYSIS

(iii) Show, under assumption , that, if the equation T π = π has a
solution π ∈ X, then it is unique. Let

Y = {h ∈ l1 : hi ≥ 0 for all i ≥ 1.}

Find all the solutions of the equation T h = h with h ∈ Y ¬rstly under the
assumption that the equation T π = π has a solution π ∈ X and secondly
under the assumption that it does not.
(iv) Show, under assumption , that, if the equation T π = π has a
solution π ∈ X, then πi > 0 for all i.
(v) Consider the following systems. In each case show we have a Markov
chain satisfying . By solving the appropriate di¬erence equations, or
otherwise, show that in case (a) the equation T π = π has a solution π ∈ X
but in cases (b) and (c) it does not.
(a) p12 = 1; pi i’1 = 1/2, pi i = pi i+1 = 1/4 for i ≥ 2; pij = 0 otherwise.
(b) p12 = 1; pi i+1 = 1/2, pi i = pi i’1 = 1/4 for i ≥ 2; pij = 0 otherwise.
(c) p12 = 1; pi i’1 = pi i = pi i+1 = 1/3 for i ≥ 2; pij = 0 otherwise.

Exercise K.260. [12.1, T!, ‘ ] This question continues the previous one.
We assume that condition holds and that the equation T π = π has a
solution π ∈ X. We will need part (ii) of Exercise K.195.
(i) Let J ≥ 1 be ¬xed. Show, using part (iv) of question K.259, that we
can ¬nd an h ∈ Y such that T h = h and hJ = 1.
(ii) We continue with the notation of (i). Explain why X does not have
the Bolzano-Weierstrass property but

XJ = {q ∈ X : hi ≥ qi for all i ≥ 1}

does. Explain why XJ is non-empty and show, carefully, that T q ∈ XJ
whenever q ∈ XJ . Deduce that, if q ∈ XJ , then T n q ’ π 1 ’ 0 as n ’ ∞.
(iii) Let e(p) ∈ l 1 be given by e(p)p = 1, e(p)i = 0 otherwise. Explain why
T n e(p)’π 1 ’ 0 as n ’ ∞. Deduce that, if q ∈ X, then T n q’π 1 ’ 0.
Interpret this result probabilistically.

Exercise K.261. [12.2, P] Let (X, d) be a complete metric space and T :
X ’ X a mapping such that T n is a contraction mapping. Show that T has
a unique ¬xed point. [Hint. Show that if w is a ¬xed point for T n then so is
T w.]
If you have done Question K.253, explain why the example asked for in
the last paragraph of that question does not contradict the result of the ¬rst
paragraph of this question.
575
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Consider the space C([a, b]) with the uniform norm. Let φ be a continuous
real-valued function on R — [a, b] which satis¬es the Lipschitz condition

|φ(x, t) ’ φ(y, t)| ¤ M for x, y ∈ R and t ∈ [a, b],

and let g ∈ C([a, b]). De¬ne T : C([a, b]) ’ C([a, b]) by
t
(T h)(t) = g(t) + φ(h(s), s) ds.
a

Show, by induction, or otherwise that
1n
(T n h)(t) ’ (T n k)(t) M (t ’ a)n h ’ k
¤
∞ ∞
n!
for all h, k ∈ [a, b].
Show that T has a ¬xed point and comment brie¬‚y on the associated
theorem on the existence of solutions of a certain di¬erential equation.

Exercise K.262. [12.2, P] Suppose that f : R ’ R is di¬erentiable with
|f (x)| < M for all x ∈ R. Show that the system of equations
n
aij f (xj ) + bi [1 ¤ i ¤ n]
xi =
j=1


with aij and bi real has a unique solution provided that
n n
1
a2 < .
ij
M2
i=1 j=1


Exercise K.263. [12.2, P, S] Consider the di¬erential equation for a func-
tion y : R ’ R

y (t) = y(t)± , y(0) = 0

where ± is real and strictly positive. For which values of ± does have only
one solution and for which values of ± does it have more than one? If has
only one solution, prove this. If has more than one solution, give at least
two solutions explicitly.

Exercise K.264. [12.2, P] Dieudonn´ is associated in most mathemati-
e
cians™ minds with the abstract approach of Bourbaki and his own text Foun-
dations of Modern Analysis [13]. He may have had this in mind when he
576 A COMPANION TO ANALYSIS

wrote the determinedly concrete In¬nitesimal Calculus [14] from which this
exercise is taken.
(i) Suppose f : R2 ’ R is a continuous function with f (t, u) < 0 when
tu > 0 and f (t, u) > 0 when tu < 0. Explain why f (0, t) = f (u, 0) = 0 for
all t and u. If y : R ’ R is a di¬erentiable function with

y (t) = f (t, y(t)) for all t ∈ R, y(0) = 0,

show, by considering the behaviour of y at a maximum and a minimum in
each interval [0, c] with c > 0, or otherwise, that y(t) = 0 for all t ≥ 0. Show
that y(t) = 0 for all t ∈ R and so has a unique solution.
(ii) Now de¬ne
±
if x ≥ t2 ,
’2t

f (t, u) = ’2x/t if t2 > x > ’t2 ,


if ’t2 > x.
2t

Verify that f is well de¬ned and continuous everywhere and satis¬es the
hypotheses of (i). Now set y0 (t) = t2 and de¬ne
t
yn+1 (t) = f (s, yn (s)) ds.
0

Show that yn (t) fails to converge as n ’ ∞ for any t = 0 although, as we
know from (i), has a unique solution.
Exercise K.265. [12.3, P] (This exercise makes repeated use of the mean
value inequality.)
(i) Consider the di¬erential equation

y (t) = |y(t)|±

with ± > 1. Suppose y is a solution with y(0) ≥ 1. Show that y(t) is strictly
increasing for t ≥ 0 and y(t) ≥ t + 1 for t ≥ 0. Let tj be de¬ned by taking
y(tj ) = 2j . Show that tj+1 ’ tj ¤ 2(1’β)j and deduce that there exists a
t— ∈ R such that tj ’ t— as j ’ ∞. Conclude that y(t) ’ ∞ as t ’ t—
through values of t < t— .
(ii) Prove the result of Example 12.3.6 by an argument along the lines of
part (i).
(iii) Show that there exists a di¬erentiable function y : [0, ∞) ’ R such
that y(0) = e and

y (t) = y(t) log y(t).
577
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.266. (Euler™s method.) [12.3, T] The next few questions
deal with Euler™s method for ¬nding approximate solutions of a di¬erential
equation. They were suggested by the typically neat treatment in [44]. You
will ¬nd the discussion more interesting if you have used Euler™s method
before or if you try one of the excellent demonstrations which now exist on
the Internet.
Suppose we wish to solve the di¬erential equation y (t) = f (t, y(t)),
y(0) = y0 numerically on a computer. (We shall look at y(t) for t ≥ 0
but the same ideas apply when t ¤ 0.) A natural approach is to seek (ap-
proximate) solutions at the points xr = rh where the strictly positive number
h is called the ˜step length™ and r is a positive integer. Explain brie¬‚y why
we expect
y (r + 1)h ≈ y(rh) + f (rh, y(rh))h.
This suggests that our approximation yr to y(rh) could be obtained by solving
the system of equations
yr+1 = yr + f (xr , yr )h. (A)
This is called Euler™s method.
Euler™s method seems reasonable and works on suitable test examples.
In this question we investigate its behaviour in general. We shall assume
stronger conditions on f than we did in Theorem 12.2.3 and elsewhere in
Sections 12.2 and 12.3. Speci¬cally, we demand that f : R2 ’ R is once
di¬erentiable and there exist constants K and L such that |f,2 (t, u)| ¤ K
and
|f,1 (t, u)| + |f,2 (t, u)f (t, u)| ¤ L
for all (t, u) ∈ R2 . Explain brie¬‚y why these conditions imply the existence
of a unique di¬erentiable y : R ’ R such that
y (t) = f (t, y(t)), y(0) = y0 . (B)
Explain why y is twice di¬erentiable and express y (t) in terms of f (t, y(t)),
f,1 (t, y(t)) and f,2 (t, y(t)). By using the 2nd mean value theorem (Exer-
cise K.49), or otherwise, show that
Lh2
|y (n + 1)h) ’ y(nh) ’ f (nh, y(nh))h| ¤ .
2
Use this result to show that
Lh2
|y (n + 1)h) ’ yn+1 | ¤ (1 + Kh)|y(nh) ’ yn | + .
2
578 A COMPANION TO ANALYSIS

Deduce that
Lh
((1 + Kh)N +1 ’ 1).
|y(N h) ’ yN | ¤
2K
[h]
In order to emphasise the dependence on h, let us write yn = yn . Show
that, if the positive number a is ¬xed and we take hN = a/N , then

LhN
[h ]
((1 + KhN )N +1 ’ 1).
|y(a) ’ yN N | ¤
2K
Observe (or recall) that that, if x ≥ 0, then
m
1 m1
x m
xr ≥ 0,
e ’ (1 + x) ≥ ’
r mr
r!
r=0

and deduce that
LhN aK
[h ]
|y(a) ’ yN N | ¤ (e ’ 1) (C)
2K
[h ]
and, in particular, yN N ’ y(a) as N ’ ∞.

Exercise K.267. [12.3, T, ‘ ] We continue with the discussion of Exer-
cise K.266. Informally, we may interpret the inequality (C) in Exercise K.266
as saying that that ˜our error estimate for Euler™s method is roughly propor-
tional to the step length™. nequality (C) is, however, only an upper estimate
for the error. Can we get a substantially better estimate? Consider the spe-
cial case f (t, u) = u, y0 = 0 and a = 1. Show that y(t) = et and yn = (1+y)n .
Show that
[h ]
y(1) ’ yN N = e ’ (1 + N ’1 )N
N
1 N1 1 hN
≥ ’ ≥ = .
r Nr
r! 2N 2
r=0

Informally, we may say ˜in this case, the error for Euler™s method decreases
no faster than the step length™.
We see that (roughly speaking) halving the step length in Euler™s method
will double the amount of calculation and only halve the error. Thus Euler™s
method is a reasonable one if we want to solve one di¬erential equation to
a reasonable degree of accuracy but unattractive if we want to solve many
di¬erential equations to a high degree of accuracy.
579
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.268. [12.3, T, ‘ ] We continue with the discussion of Exer-
cises K.266 and K.267. So far we have ignored the fact that computers only
work to a certain degree of accuracy. As a simple model to take account of
this, we replace equation (A) of Exercise K.266 by

yr+1 = yr + f (xr , yr )h +
˜ ˜ ˜ (D)
r


where the ˜error™ r satis¬es the condition | r | ¤ for some ¬xed > 0. As
before, we impose the initial condition y0 = 0.
˜
[h]
(i) As in Exercise K.266, we take yn = yn , ¬x a positive number a and
take hN = a/N . Show that

LhN
[h ]
(eaK ’ 1).
|y(a) ’ yN N | ¤ + (E)
2K hN K

What happens to the error estimate (that is the right hand side of (E)) as
N ’ ∞ (that is as the step length approaches 0)? Explain in simple terms
why we should expect this.
Lh
(ii) Which positive value of h minimises + ? (iii) Show that the
2K hK
best bound on the error that we can guarantee using (E) is proportional to
1/2
.
(iv) Just as in Exercise K.267, we ask whether we can improve substan-
tially on (iii). Once again, we consider the special case f (t, u) = u, y0 = 0
and a = 1. A little experimentation shows that we can solve (D) if we choose
r = (1 ’ rh) , giving


yr+1 = (1 + N ’1 )˜r ’ (1 ’ rN ’1 ) .
˜ y (F)

Show that, for this choice of r, we obtain

yr = (1 ’ N ’1 )r ’ r
˜

and so
hN
[h ]
y(1) ’ yN N = e ’ (1 + N ’1 )N ’ N ≥ 21/2 1/2

˜ + .
2 hN

Thus the answer to our question is no.
There exist several di¬erent ways of improving on Euler™s method but a
good understanding of Euler™s method is very helpful in understanding how
anybody came to think of these more advanced techniques.
580 A COMPANION TO ANALYSIS

Exercise K.269. [12.3, H, ‘ ] (i) Let a > 0. By considering the di¬erential
equation y (x) = g(x) and using Exercise K.266, show that there exists a
constant κa such that, if g : R ’ R is di¬erentiable, |f (t)| ¤ M for all
t ∈ [0, a], N is a strictly positive integer and N h = a, then
N ’1
a
g(t) dt ’ h g(rh) ¤ M κa h.
0 n=0

(It is easy to obtain this result directly with better values for κa . See, for
example, Exercise K.125 (v) or just think a bit.)
(ii) Suppose that a = π, N h = a with N a strictly positive integer and
G(t) = sin2 (N t). Show that
N ’1
a
G(rh) ≥ 10’2 sup |G(t)|.
G(t) dt ’ h
t∈[0,a]
0 n=0

(iii) Consider the situation described in Exercise K.266. Suppose that we
replace the condition |f,1 (t, u)|+|f,2 (t, u)f (t, u)| ¤ L by the weaker condition
|f (t, u)| ¤ L for all (t, u) ∈ R2 . If a > 0 and > 0 are ¬xed, is it possible to
¬nd an N0 (depending only on K, L, a and ) such that
[h ]
|y(a) ’ yN N | ¤
for all N ≥ N0 ?
Exercise K.270. (Leibniz rule.) [12.3, G] (i) If u, v : R ’ R are n
times di¬erentiable at x show, by induction or otherwise, that the product
uv is n times di¬erentiable at x with
n
n (n’r)
(n)
(x)v (r) (x).
(uv) (x) = u
r
r=0

(this is called the Leibniz rule.)
(ii) If y : R ’ R is de¬ned by
(1 + x2 )1/2 y(x) = log(x + (1 + x2 )1/2 ),
show that y is di¬erentiable on (0, ∞) with
(1 + x2 )y (x) + xy(x) = 1.
Show, by using induction and the Leibniz rule, that y is in¬nitely di¬eren-
tiable and ¬nd the Taylor series

y (n) (0) n
x.
n!
n=0
581
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(iii) (This part uses sledgehammers to crack a nut and may be omitted
without loss.) Find the radius of convergence of the Taylor series and, by
using results on the di¬erentiation of power series and the uniqueness of the
solution of di¬erential equations, show that, within the radius of convergence

y (n) (0) n
y(x) = x.
n!
n=0


Exercise K.271. [12.4, P, S] Let K : [0, 1]2 ’ R and g : [0, 1] ’ R be
continuous. Explain why there exists an M > 0 such that |K(s, t)| ¤ M
for all (s, t) ∈ [0, 1]2 . Suppose that |»| < M ’1 . By ¬nding an appropriate
contraction mapping T : C([0, 1]) ’ C([0, 1]) show that there exists a unique
f ∈ C([0, 1]) such that
1
f (t) = g(t) + » K(s, t)f (t) ds.
0

Exercise K.272. (The Wronskian.) [12.4, M] This should be treated
as an exercise in calculus rather than analysis. We write
« 
a11 a12 a13 a11 a12 a13
a21 a22 a23 = det a21 a22 a23 
a31 a32 a33 a31 a32 a33

(i) If u1 , u2 , . . . w3 are all di¬erentiable show that

u1 (t) u2 (t) u3 (t)
d
v1 (t) v2 (t) v3 (t)
dt
w1 (t) w2 (t) w3 (t)
u1 (t) u2 (t) u3 (t) u1 (t) u2 (t) u3 (t) u1 (t) u2 (t) u3 (t)
= v1 (t) v2 (t) v3 (t) + v1 (t) v2 (t) v3 (t) + v1 (t) v2 (t) v3 (t) .
w1 (t) w2 (t) w3 (t) w1 (t) w2 (t) w3 (t) w1 (t) w2 (t) w3 (t)

(ii) If u1 , u2 and u3 are three solutions of

y (t) + a(t)y (t) + b(t)y (t) + c(t)y(t) = 0,

we de¬ne their Wronskian W by

u1 (t) u2 (t) u3 (t)
W (t) = u1 (t) u2 (t) u3 (t) .
u1 (t) u2 (t) u3 (t)
582 A COMPANION TO ANALYSIS

Use part (i) and results about determinants to show that

W (t) = ’a(t)W (t).

(iii) Generalise parts (i) and (ii). Reread the proof of part (i) of Lemma 12.4.3
in the light of this exercise.

Exercise K.273. [12.4, T, S] The functions f, g : R ’ R are everywhere
di¬erentiable and their Wronskian f g ’ g f never vanishes. By applying
Rolle™s theorem to f /g, or otherwise, show that if f has zeros at a and b with
a < b then g must have a zero strictly between a and b. Deduce that ˜the
zeros of f and g are intertwined™.

Exercise K.274. [12.4, P] If f1 , f2 : [a, b] ’ R are once di¬erentiable we
de¬ne the Wronskian W (f1 , f2 ) by

f1 f2
W (f1 , f2 ) = det
f1 f2

Show that if f1 and f2 are linearly dependent (that is, we can ¬nd »1 , »2 ∈ R
not both zero such that »1 f1 (t) + »2 f2 (t) = 0 for all t ∈ [a, b]) then the
Wronskian W (f1 , f2 ) is identically zero. By considering functions of the type
g with g(x) = 0 for x ¤ c, g(x) = (x ’ c)2 for x ≥ c, or otherwise, show that
the converse is false.
Suppose now that f1 , f2 are twice continuously di¬erentiable. Show, by
considering the Wronskian
« 
f f 1 f2
W (f, f1 , f2 ) = det  f f1 f2 
f f1 f2

and using results on the existence and uniqueness of di¬erential equations
with given initial conditions, that the following result holds.
There exists a di¬erential equation of the form

y (x) + a1 (x)y (x) + a2 (x)y(x) = 0

whose solutions are exactly the functions of the form »1 f1 + »2 f2 with »1 and
»2 (in more sophisticated language, having f1 and f2 as a basis for the space
of solutions) if and only if W (f1 , f2 ) is non-zero everywhere on [a, b].
Generalise the results of this question to n functions f1 , f2 , . . . , fn .
583
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.275. [12.4, M] (This question should be treated as a mathe-
matical methods one.)
(i) Consider a di¬erential equation

y (x) + p(x)y (x) + q(x)y(x) = 0.

Suppose that u is a solution of . Show that if we write y(x) = u(x)v(x)
and w(x) = u (x) then takes the form of a ¬rst order di¬erential equation
x
in w. Show that y(x) = (A + B 0 w(t) dt)v(x) gives the general a solution
for .
(ii) Show that y(x) = exp x gives a solution of

y (x) ’ (2x + 1)y (x) + (x + 1)y(x) = 0

and use the method of (i) to ¬nd the general solution.
(iii) We know that y(x) = exp x gives a solution of

y (x) ’ 2y (x) + y(x).

Find the general solution by the method of (i).
(iv) Show how, if we know one solution of

y (x) + p1 (x)y (x) + p2 (x)y (x) + p3 (x)y(x) = 0

we can reduce the problem of ¬nding a general solution to that of solving a
second order linear di¬erential equation. Generalise.

Exercise K.276. (Method of variation of parameters.) [12.4, M,
‘ ] (Like Question K.275 this question should be treated as a mathematical
methods one.)
Consider a di¬erential equation

y (x) + p(x)y (x) + q(x)y(x) = 0.

whose Wronskian W (x) = y1 (x)y2 (x) ’
Let y1 and y2 be solutions of
y1 (x)y2 (x) never vanishes. Suppose that we wish to solve the di¬erential
equation

y (x) + p(x)y (x) + q(x)y(x) = f (x).

(i) In view of the success of the method of Question K.275 we might be
tempted to look for a solution of the form

y(x) = u1 (x)y1 (x) + u2 (x)y2 (x).
584 A COMPANION TO ANALYSIS

Make this substitution in .
(ii) At the end of (i) you obtained a di¬erential equation (†) for u1 and
u2 . It is a useful heuristic guide (though not always a reliable one) that two
unknowns require two equations so we add the further equation

u1 (x))y1 (x) + u2 (x)y2 (x) = 0. (††)

Use (††) to simplify (†) and then use the pair of equations to ¬nd u1 and u2 .
Now ¬nd the general solution of .
(iii) Discuss the relation of your result to the Green™s function result given
in Theorem 12.4.6.
(iv) Use the method of this question to ¬nd the general solution of
2
y (x) ’ y(x) = .
1 + ex
(v) Try and extend the ideas of this question to the solution of

y (x) + p(x)y (x) + q(x)y (x) + r(x)y = f (x).

when three solutions y1 , y2 , y3 with non-zero Wronskian (see Question K.272)
are known.
[It would be hard to think of an example more opposed to the modern view
that ˜understanding precedes doing™. But even the most convinced proponent
of modern ways must admit that it has a certain charm.]
Exercise K.277. [12.4, P] Let C(S) be the space of continuous functions
on the unit square

S = {(x, y) : 0 ¤ x ¤ 1, 0 ¤ y ¤ 1}

equipped with uniform norm K ∞ = sup(x,y)∈S |K(x, y)| and let C(I) be the
space of continuous functions on the unit interval I = [0, 1] equipped with the
uniform norm. Let L be the space of continuous linear maps L : C(I) ’ C(I)
equipped with the operator norm. If K ∈ C(S) we set
1
(TK (f ))(x) = K(x, y)f (y) dy.
0

Show that TK ∈ L.
Prove or disprove each of the following statements.
(i) If Kn ’ K in C(S), then TKn ’ TK in L.
(ii) If TKn ’ TK in L, then Kn ’ K in C(S).
(iii) The mapping K ’ TK is an injective map from C(S) to L.
(iii) The mapping K ’ TK is a surjective map from C(S) to L.
585
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.278. [12.4, T] (i) In Exercise K.132, you showed that if g :
R2 ’ R is a di¬erentiable function with continuous partial derivatives then
x
G(x) = g(x, t) dt
0

is di¬erentiable with
x
G (x) = g(x, x) + g,1 (x, t) dt
0

Review this exercise.
(ii) Suppose that a, b, f : R ’ R are continuous. Explain why there
exists a unique twice di¬erentiable y1 : R ’ R such that
y1 (t) + a(t)y1 (t) + b(t)y1 (t) = 0, y1 (0) = 0, y1 (0) = 1,
and a unique twice di¬erentiable y2 : R ’ R such that
y2 (t) + a(t)y2 (t) + b(t)y2 (t) = 0, y2 (1) = 0, y2 (1) = 1.
We make the following
key assumption: y1 (1) = 0.
˜
Show that, if we de¬ne H, H : R ’ R by
˜
H(s, t) = y1 (t)y2 (s)W (s)’1 , H(s, t) = y2 (t)y1 (s)W (s)’1 ,
˜
then H and H are twice di¬erentiable functions with continuous second par-
tial derivatives and
H,22 (s, t) + a(t)H,2 (s, t) + b(t)H(s, t) = 0, H(s, 0) = 0,
˜ ˜ ˜ ˜
H,22 (s, t) + a(t)H,2 (s, t) + b(t)H(s, t) = 0, H(s, 1) = 0.
˜ ˜
Check that H(t, t) = H(t, t), H2 (t, t) = H2 (t, t).
(iii) We de¬ne G : R2 ’ R by G(s, t) = H(s, t) for t ¤ s and G(s, t) =
˜
H(s, t) for s ¤ t. If we set
1 t 1
˜
y(t) = G(s, t)f (s) ds = H(s, t)f (s) ds + H(s, t)f (s) ds,
0 0 t

˜
show, using part (i) and the properties of H and H established in part (ii)
˜
(but not using the de¬nitions of H and H), that y is twice di¬erentiable and
satis¬es
y (t) + a(t)y + b(t)y = f (t)
together with the conditions y(0) = y(1) = 0.
586 A COMPANION TO ANALYSIS

Exercise K.279. [12.4, M] Consider the di¬erential equation

y (4) ’ k 4 y = f

for y : [0, 1] ’ R subject to boundary conditions y(0) = y (0) = 0, y(1) =
y (1) = 0 where f : [0, 1] ’ R is continuous and k is real.
By extending the discussion of the Green™s function in Section 12.4 show
that, provided that k does not take certain exceptional values to be identi¬ed,
the system has the solution
1
y(x) = G(x, t)f (t) dt
0

where
A(sinh kx ’ sin kx) + B(cosh kx ’ cos kx) if 0 ¤ x ¤ t,
G(x, t) =
C(sinh k(1 ’ x) ’ sin k(1 ’ x)) + D(cosh k(1 ’ x) ’ cos k(1 ’ x)) if t ¤ x ¤ 1,

and A, B C, D are given by
« «

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