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B —¦ = U ∈U U .
State a condition (iii) along the lines of (iii) and show that it is equivalent
to (i) and (ii) .
We call B —¦ the interior of B. We also write B —¦ = Int B.
Show that Cl A = X \ Int(X \ A) and Int A = X \ Cl(X \ A). Show that
A = Cl A if and only if A is closed and A = Int A if and only if A is open.
Suppose we work in R with the usual distance. Find the interior and
closure of the following sets Q, Z, {1/n : n ∈ Z, n ≥ 1}, [a, b], (a, b) and
[a, b) where b > a.
Let (X, d) and (Y, ρ) be metric spaces. Show that f : X ’ Y is continu-
¯
ous if and only if f (A) ⊆ f (A) for all subsets A of X.
(An unimportant warning.) Find a metric space (X, d) and an x ∈ X
such that
Cl{y : d(x, y) < 1} = {y : d(x, y) ¤ 1} and Int{y : d(x, y) ¤ 1} = {y : d(x, y) < 1}.
527
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.180. [10.3, P, ‘ ] We use the notation of Exercise K.179. Show
that, if U is open,

Cl(Int(Cl U )) = Cl U.

Show that, starting from a given A, it is not possible to produce more
than 7 distinct sets by repeated application of the operators Cl and Int.
Give an example of a set A in R with the usual metric which gives rise
to 7 distinct sets in this way.

Exercise K.181. [10.4, T] The results of this question are simple but give
some insight into the nature of the norm.
(i) Let be a norm on a real or complex vector space. Show that the
closed unit ball

¯
“ = B(0, 1) = {x ∈ V : x ¤ 1}

has the following properties.
(A) “ is convex, that is to say, if x, y ∈ “ and 1 ≥ » ≥ 0, then »x + (1 ’
»)y ∈ “.
(B) “ is centrally symmetric, that is to say, if x ∈ “, then ’x ∈ “.
(C) “ is absorbing, that is to say, if x ∈ V , then we can ¬nd a strictly
positive real number » with »x ∈ “.
(ii) Show, conversely, that if “ is a closed, convex, centrally symmetric,
absorbing set then “ is the closed unit ball of a norm.
(iii) If A and B are norms on V and K > 0 state and prove a
necessary and su¬cient condition in terms of the closed unit balls of the two
norms for the inequality

¤K x
x A B


to hold for all x ∈ V .

Exercise K.182. [10.4, P] Consider the real vector space l ∞ of real se-
quences a = (a1 , a2 , . . . ) with norm a = supn≥1 |an |. Show that the set

E = {a : there exists an N with an = 0 for all n ≥ N }

is a subspace of l∞ but not a closed subset.
Show that any ¬nite-dimensional subspace of any normed vector space is
closed.
528 A COMPANION TO ANALYSIS

) be a real normed space and let R
Exercise K.183. [10.4, T] Let (E,
have its usual norm. Let T : E ’ R be a linear map.
(i) Show that, if T is not the zero map, the null space

N = T ’1 (0) = {e ∈ E : T e = 0}

is a subspace of E such that, if y ∈ E then every x ∈ E can be written in
/
exactly one way as x = »y + e with » ∈ R and e ∈ N . Comment very brie¬‚y
on what happens when T is the zero map.
(ii) Suppose that T is not the zero map and N is as in (i). Show that N
is closed if and only if there exists a δ > 0 such that »y + e > |»|δ for all
» ∈ R and e ∈ N .
(iii) Show that T is continuous if and only if its null space T ’1 (0) is closed.
(iv) Find T ’1 (0) and check directly that the conclusions of (iii) are satis-
¬ed if we take E = s00 and take T as in part (ii) of Exercise 10.4.14. (There
are ¬ve norms to check.)
Exercise K.184. [10.4, H] This question and the one that follows are in-
cluded more for the bene¬t of the author than the reader. I was worried
whether the proof of Theorem 10.4.6 actually required the use of results
from analysis. In this question I outline a simple example supplied by Imre
Leader which should convince all but the most stubborn that the proof should
indeed use analysis.
Consider the map N : Q2 ’ R given by N (x, y) = |x + y21/2 |. Show that
(1) N (x) ≥ 0 for all x ∈ Q2 .
(2) If N (x) = 0, then x = 0.
(3) If » ∈ Q and x ∈ Q2 , then N (»x) = |»|N (x).
(4) (The triangle inequality) If x, y ∈ Q2 , then N (x + y) ¤ N (x) + N (y).
Let (x, y) = max(|x|, |y|). Show that given any δ > 0 we can ¬nd an
x ∈ Q2 such that

N (x) < δ x .

Explain why, if N : Q2 ’ R is any function satisfying conditions (1)
to (4), there exists a K > 0 such that

N (x) < K x

for all x ∈ Q2 .
Exercise K.185. [10.4, H!] In this question we give another example of
Imre Leader which proves conclusively that Theorem 10.4.6 is indeed a result
of analysis. It is quite complicated and I include it to provide pleasure for
529
Please send corrections however trivial to twk@dpmms.cam.ac.uk

the expert12 rather than enlightenment for the beginner. Our object is to
prove the following result.
There is a map N : Q3 ’ Q such that
(1) N (x) ≥ 0 for all x ∈ Q3 .
(2) If N (x) = 0, then x = 0.
(3) If » ∈ Q and x ∈ Q3 , then N (»x) = |»|N (x).
(4) (The triangle inequality) If x, y ∈ Q3 , then N (x + y) ¤ N (x) + N (y).
(5) If we write x ∞ = max1¤j¤3 |xj |, then, given any δ > 0, we can ¬nd
a x ∈ Q3 such that

N (x) < δ x ∞.

(i) Why does this result show that Theorem 10.4.6 is a result of analysis?
(ii) (The proof begins here. It will be helpful to recall the ideas and
notation of Question K.181.) We work in R2 with the usual Euclidean norm
. Suppose u1 , u2 , . . . are elements of R2 none of which are scalar multiples
of any other and such that 1/2 ¤ uj ¤ 2 for all j ≥ 1. Show, by using
induction, or otherwise, that we can ¬nd an an ∈ Q such that, writing
vn = an un , we have
(a) 1/2 < vn < 2, and
(b) vn ∈ “n’1
/
¯
where “0 is the closed disc B(0, 1/2) of radius 1/2 and centre 0 and, if n ≥ 1

“n = {»x + µvn : x ∈ “n’1 , |»| + |µ| = 1, », µ ∈ R}.

(For those who know the jargon, “n is the convex hull of “n’1 ∪ {vn , ’vn }.)
(iii) Continuing with the ideas of (ii), write “ for the closure of ∞ “n .
n=0
Show that “ is a closed centrally symmetric convex set with
¯ ¯
B(0, 1/2) ⊆ “ ⊆ B(0, 2).

Show further that, if µ ∈ R, then µvn ∈ “ if and only if |µ| ¤ 1. Use the
— on R such
2
results of Question K.181 to deduce the existence of a norm
that un — ∈ Q for all n and

2x ≥ x ≥ x /2


for all x ∈ R2 .
(iv) Let

E = {(a + b21/2 , c + b31/2 ) : a, b, c ∈ Q}.
Who can start by considering the question of whether we can replace Q3 by Q2 in the
12

next paragraph.
530 A COMPANION TO ANALYSIS

By considering a suitable sequence un ∈ E show that there exists a norm
— on R such that e — ∈ Q for all e ∈ E and
2


2x ≥ x ≥ x — /2



for all x ∈ R2 .
(v) De¬ne N : Q3 ’ Q by N (x, y, z) = (x + y21/2 , z + y31/2 ) — . Show
that N has properties (1) to (5) as required.

Exercise K.186. [11.1, P] Suppose that (X, d) is a complete metric space.
We say that a subset F of X has bounded diameter if there exists a K such
that d(x, y) < K for all x, y ∈ F and we then say that F has diameter

sup{d(x, y) : x, y ∈ F }.

Suppose that we have sequence of closed subsets Fj of bounded diameter
with F1 ⊇ F2 ⊇ . . . . Show that, if the diameter of Fn tends to 0, then

j=1 Fj contains exactly one point.
Show that ∞ Fj may be empty if
j=1
(a) (X, d) is not complete, or
(b) the Fj do not have bounded diameter, or
(c) the Fj have bounded diameters but the diameters do not tend to 0.

Exercise K.187. [11.1, P] Consider the space s00 introduced in Exer-
cise 10.4.8. Recall that s00 the space of real sequences a = (an )∞ such
n=1
that all but ¬nitely many of the an are zero. Our object in this question is
to show that no norm on s00 can be complete. To this end, let us suppose
that is a norm on s00 .
(i) Write En = {a ∈ s00 : aj = 0 for all j ≥ n}. Show that En is closed.
(ii) If b ∈ En+1 \ En , show that there exists a δ > 0 such that b ’ a > δ
for all a ∈ En .
(iii) If h ∈ En+1 \ En and a ∈ En , show that a + »h ∈ En+1 \ En for all
» = 0.
(iv) Show that, given x(n) ∈ En and δn > 0, we can ¬nd x(n + 1) ∈ En+1
such that x(n+1)’x(n) < δn /4, and δn+1 < δn /4 such that x(n+1)’a >
δn+1 for all a ∈ En .
(v) Starting with x(1) = 0 and δ1 = 1, construct a sequence obeying the
conclusions of part (iv) for all n ≥ 2. Show that the sequence x(n) is Cauchy.
(vi) Continuing with the notation of (v), show that, if the sequence x(n)
converges to some y ∈ s00 then y ’ x(n + 1) < δn+1 /2, for each n ≥ 1.
Conclude that y ∈ En for any n and deduce the required result by reductio
/
ad absurdum.
531
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.188. (The space l 2 .) [11.1, T] (i) Let aj , bj ∈ R. Use the
triangle inequality for the Euclidean norm on Rm and careful handling of
limits to show that, if ∞ a2 and ∞ b2 converge, so does ∞ (aj + bj )2 .
j=1 j j=1 j j=1
Show further that, in this case,
1/2 1/2 1/2
∞ ∞ ∞
(aj + bj )2 a2 b2
¤ + .
j j
j=1 j=1 j=1


(ii) Show that the set l 2 of real sequences a with 2
j=1 aj convergent
forms a vector space if we use the natural de¬nitions of addition and scalar
multiplication
(an ) + (bn ) = (an + bn ), »(an ) = (»an ).
Show that, if we set

a2 ,
a =
2 j
j=1

then (l2 , 2 ) is a complete normed space.
(iii) The particular space (l 2 , 2 ) has a further remarkable property to
which we devote the next paragraph.
Show using the Cauchy“Schwarz inequality for Rm , or otherwise, that if
a, b ∈ l2 , then ∞ |aj bj | converges. We may thus de¬ne
j=1


a·b= aj b j .
j=1

Show that this inner product satis¬es all the conclusions of Lemma 4.1.1.
Exercise K.189. (H¨lder™s inequality.) [11.1, T] Throughout this ques-
o
tion p and q are real numbers with p > 1 and p’1 + q ’1 = 1.
(i) Show that q > 1.
(ii) Use the convexity of ’ log (see Question K.39) to show that, if x and
y are strictly positive real numbers, then
xp y q
xy ¤ +.
p q
Observe that this equality remains true if we merely assume that x, y ≥ 0.
(iii) Suppose that f, g : [a, b] ’ R are continuous. Show that
b b b
1 1
p
|g(t)|q dt.
|f (t)g(t)| dt ¤ |f (t)| dt +
p q
a a a
532 A COMPANION TO ANALYSIS

Deduce that, if F, G : [a, b] ’ R are continuous and
b b
p
|G(t)|q dt = 1,
|F (t)| dt =
a a

then
b
|F (t)G(t)| dt ¤ 1.
a

(iv) By considering κF and µG with F and G as in (iii), or otherwise,
show that, if f, g : [a, b] ’ R are continuous, then
1/p 1/q
b b b
|f (t)|p dt |g(t)|q dt
|f (t)g(t)| dt ¤ .
a a a

This inequality is known as H¨lder™s inequality for integrals.
o
(v) Show that the Cauchy-Schwarz inequality for integrals is a special
case of H¨lder™s inequality.
o

Exercise K.190. [11.1, T, ‘ ] (i) Suppose that f : [a, b] ’ R is a continuous
function such that
1/q
b b
|g(t)|q dt
|f (t)g(t)| dt ¤ A
a a

for some constant A and all continuous functions g : [a, b] ’ R. By taking
g(t) = f (t)± for a suitable ±, or otherwise, show that
1/p
b
|f (t)|p dt ¤ A.
a

This result is known as the ˜reverse H¨lder inequality™.
o
(ii) By ¬rst applying H¨lder™s inequality to the right hand side of the
o
inequality
b b b
|(f (t) + h(t))g(t)| dt ¤ |f (t)g(t)| dt + |h(t)g(t)| dt
a a a

and then using the reverse H¨lder inequality, show that
o
1/p 1/p 1/p
b b b
p p p
|f (t) + h(t)| dt ¤ |f (t)| dt |h(t)| dt
+
a a a
533
Please send corrections however trivial to twk@dpmms.cam.ac.uk

for all continuous functions f, h : [a, b] ’ R. (This inequality is due to
Minkowski.)
(iii) Show that, if we set
1/p
b
|f (t)|p dt
f = ,
p
a

then (C([a, b]), p ) is a normed space. Show, however, that (C([a, b]), p)
is not complete.
(iv) Rewrite H¨lder™s inequality and the reverse H¨lder inequality using
o o
the notation introduced in (iii).
Exercise K.191. The l p spaces.) [11.1, T, ‘ ] Throughout this question
p and q are real numbers with p > 1 and p’1 + q ’1 = 1.
(i) By imitating the methods of Exercise K.189, or otherwise, show that,
if aj , bj ∈ R, then
1/p 1/q
n n n
|aj |p |bj |q
|aj ||bj | ¤ .
j=1 j=1 j=1
∞ ∞
|aj |p and |bj |q converge, then so does
Deduce carefully that, if j=1 j=1

j=1 |aj ||bj | and
1/p 1/q
∞ ∞ ∞
|aj |p |bj |q
|aj ||bj | ¤ .
j=1 j=1 j=1

This inequality is known as H¨lder™s inequality for sums.
o
(ii) By imitating the methods of Exercise K.189, or otherwise, show that
the collection lp of real sequences a = (a1 , a2 , . . . ) forms a normed vector
space under the appropriate algebraic operations (to be speci¬ed) and norm
1/p

|aj |p
a = .
p
j=1

(iii) By using the ideas of Example 11.1.10, or otherwise, show that
p
(l , p ) is complete.
(iv) Use the parallelogram law (see Exercise K.297 (i)) to show that the
p
l norm is not derived from an inner product if p = 2. Do the same for
(C([a, b]), p ).

(v) We de¬ned (l1 , 1 ) in Exercise 11.1.10 and (l , ∞ ) in Exer-
cise 11.1.13. Investigate H¨lder™s inequality and the reverse H¨lder inequality
o o
for p = 1, q = ∞ and for p = ∞, q = 0. Carry out a similar investigation for
(C([a, b]), 1 ) (see Lemma 11.1.8) and (C([a, b]), ∞ ).
534 A COMPANION TO ANALYSIS

Exercise K.192. (The Hausdor¬ metric.) [11.1, T] Let (X, d) be a
metric space and E a non-empty subset of X. For each x ∈ X, de¬ne

d(x, E) = inf{d(x, y) : y ∈ E}.

Show that the map f : X ’ R given by f (x) = d(x, E) is continuous.
Now consider the set E of non-empty closed bounded sets in Rn with the
usual Euclidean metric d. If K, L ∈ E de¬ne

ρ (K, L) = sup{d(k, L) : k ∈ K}.

and

ρ(K, L) = ρ (K, L) + ρ (L, K).

Prove that, if K, L, M ∈ E, k ∈ K and l ∈ L then

d(k, M ) ¤ d(k, l) + ρ (L, M )

and deduce, or prove otherwise, that

ρ (K, M ) ¤ ρ (K, L) + ρ (L, M ).

Hence show that (E, ρ) is a metric space.
Suppose now that (X, d) is R with the usual metric. Suppose K is closed
and bounded in Rn (again with the usual Euclidean metric) and fn : K ’ X
is a sequence of continuous functions converging uniformly to some function
f . Show that ρ(fn (K), f (K)) ’ 0 as n ’ ∞. How far can you generalise
this result?
Exercise K.193. [11.1, T, ‘ ] The metric ρ de¬ned in the previous question
(Exercise K.192) is called the Hausdor¬ metric. It is clearly a good way of
comparing the ˜similarity™ of two sets. Its utility is greatly increased by the
observation that (E, ρ) is complete. The object of this question is to prove
this result.
(i) If Kn ∈ E and K1 ⊇ K2 ⊇ K3 ⊇ . . . explain why K = ∞ Kj ∈ E j=1
and show that ρ(Kn , K) ’ 0 as n ’ ∞.
(ii) Suppose Ln ∈ E and ρ(Ln , Lm ) ¤ 4’n for all 1 ¤ n ¤ m. If we set
¯
Kj = Lj + B(0, 2’j ) = {l + x : l ∈ Lj , x ¤ 2’j },

show that Kn ∈ E and K1 ⊇ K2 ⊇ K3 ⊇ . . . . Show also that, if we write
K = ∞ Kj we have ρ(Ln , K) ’ 0 as n ’ ∞.
j=1
(iii) Deduce that (E, ρ) is complete.
535
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.194. [11.2, T] Let (V, ) be a normed vector space and E a
closed subspace (that is a vector subspace which is also a closed subset). If
E = V there must exist a z ∈ V \ E. By considering

inf{ z ’ y : y ∈ E},

> 0 we can ¬nd an x ∈ V such that
or otherwise, show that given any
x = 1 but

x ’ e > 1 ’ for all e ∈ E.

Show that if V is in¬nite dimensional we can ¬nd a sequence xj ∈ V such
that

xj = 1 for all j, but xi ’ xj > 1/2 for all i = j.
¯
Deduce that if (V, ) is a normed space then the closed unit ball B =
{x : x ¤ 1} has the Bolzano-Weierstrass property if and only if V is ¬nite
dimensional.
Exercise K.195. [11.2, P] (i) Consider l ∞ the space of bounded real se-
∞ given by x ∞ = supn≥1 |xn |. Let κn be a positive
quences with norm
real number [n ≥ 1]. Show that the set

E = {x ∈ l∞ : |xn | ¤ κn for all n ≥ 1}

has the Bolzano-Weierstrass property if and only if κn ’ 0 as n ’ ∞.
(ii) Consider l1 the space of real sequences whose sum is absolutely con-

n=1 |xn |. Let κn be a positive
vergent, with norm 1 given by x ∞ =
real number [n ≥ 1]. Show that the set

E = {x ∈ l∞ : 0 ¤ xn ¤ κn for all n ≥ 1}

has the Bolzano-Weierstrass property if and only if κn converges.
n=1

Exercise K.196. [11.2, T] This exercise takes up ideas discussed in Exer-
cises K.29 to K.36 but the reader only needs to have looked at Exercise K.29.
We work in a metric space (X, d). Consider the following two possible
properties of (X, d).
(A) If a collection K of closed sets has the ¬nite intersection property,
then K∈K K = ….
(B) If a collection U of open sets is such that U ∈U U = X, then we can
¬nd an n ≥ 1 and U1 , U2 , . . . , Un ∈ U such that n Un = X.
j=1
(i) Show that (X, d) has property (A) if and only if it has property (B).
536 A COMPANION TO ANALYSIS

(ii) Suppose that (X, d) has the Bolzano-Weierstrass property and that
U is a collection of open sets such that U ∈U U = X. Show that there exists
a δ > 0 such that, given any x ∈ X, we can ¬nd a U ∈ U with B(x, δ) ⊆ U .
(iii) Use (ii) and Lemma 11.2.6 to show that, if (X, d) has the Bolzano-
Weierstrass property, then it has property (B).

Exercise K.197. [11.2, T, ‘ ] This exercise continues Exercise K.196.
(i) Suppose that xn ∈ X for all n but that the sequence xn has no
convergent subsequence. Show that given any x ∈ X we can ¬nd a δx > 0
and an Nx ≥ 1 such that xn ∈ B(x, δx ) for all n ≥ Nx .
/
(ii) By considering sets of the form Ux = B(x, δx ), or otherwise, show
that the space X described in (i) can not have property (B).
(iii) Deduce that a metric space (X, d) has property (B) if and only if it
has the Bolzano-Weierstrass property.
(iv) Use (iii) to prove parts (ii) and (iv) of Exercise K.36.
If (X, d) has property (B) we say that it is compact. We have shown that
compactness is identical with the Bolzano-Weierstrass property for metric
spaces. However, the de¬nition of compactness makes no explicit use of the
notions of ˜distance™ (that is, metric) or ˜convergence™ and can be generalised
to situations where these concepts cease to be useful.

Exercise K.198. [11.3, P] If f : [0, 1] ’ R is continuous we write
1
= sup |f (t)|, and f |f (t)| dt.
f =
∞ 1
t∈[0,1] 0


Consider the space C 1 ([0, 1]) of continuously di¬erentiable functions. We set

f = f ∞+ f
A 1
f =f∞
B
f = f ∞+ f ∞
C
= |f (0)| + f
f D 1


Which of these formulae de¬ne norms? Of those that are norms, which are
complete? Consider those which are norms together with the norms ∞
and 1 . Which are Lipschitz equivalent and which not? Prove all your
answers.

Exercise K.199. [11.3, T] Weierstrass™s example (see page 199) becomes
very slightly less shocking if we realise that our discussion uses an inappropri-
ate notion of when one function is close to another. In the next two exercises
we discuss a more appropriate notion. It will become clear why much of
537
Please send corrections however trivial to twk@dpmms.cam.ac.uk

the early work in what we now call the theory of metric spaces was done by
mathematicians interested in the calculus of variations.
(i) Find a sequence of functions fn : [’1, 1] ’ R which have continuous
derivative (with the usual convention about derivative at the end points)
such that fn ’ |x| uniformly on [’1, 1] as n ’ ∞.
(ii) If b > a let us write C 1 ([a, b]) for the space of functions on [a, b]
which have continuous derivative. Show that C 1 ([a, b]) is not closed in
1
(C([a, b]), ∞ ) and deduce that (C ([a, b]), ∞ ) is not complete.
(iii) If f ∈ C 1 ([a, b]) let us write

f =f +f ∞.
— ∞


Show that (C 1 ([a, b]), —) is a complete normed space.
(iv) Consider

A = {f ∈ C 1 ([0, 1]) : f (0) = f (1) = 0}.

Show that A is a closed subset of (C 1 ([a, b]), and that A is a vector subspace
of C 1 ([a, b]). Conclude that (A, — ) is a complete normed space. Show also
that (A, ∞ ) is not complete.

Exercise K.200. [11.3, T, ‘ ] We continue with the discussion and notation
of Exercise K.199. The example we gave on page 199 involved studying the
function I : A ’ R given by
1
(1 ’ (f (x))4 )2 + f (x)2 dx.
I(f ) =
0


(i) Show that (if we give R the usual metric) the function I : A ’ R is
not continuous if we give A the norm ∞ but is continuous if we give A
the norm —.
(ii) Do (or recall) Exercise 8.3.4 which shows that

inf{I(f ) : f ∈ A} = 0

but that I(f ) > 0 for all f ∈ A. This is the main point of the Weierstrass
counterexample and is una¬ected by our discussion.
(iii) Write f0 = 0. By using Exercise 8.4.12, or otherwise, show that we
can ¬nd fn ∈ A such that fn ’ f0 ∞ ’ 0 and Ifn ’ 0.
(iv) Show that if fn ∈ A and Ifn ’ 0 then the sequence fn does not
converge in (A, — ).
538 A COMPANION TO ANALYSIS

Exercise K.201. [11.3, T, ‘ ] Exercise K.200 shows that the question ˜Is
f0 = 0 a local minimum for I?™ should be asked using the norm — rather
than the norm ∞ but leaves the question unanswered.
(i) By considering functions of the form f (x) = sin 2πx or otherwise
show that given any δ > 0 we can ¬nd u, v ∈ A with u — , v — < δ and
Iu > 0 > Iv. Conclude that f0 = 0 is neither a local maximum nor a local
minimum of J when we use the norm — . (Thus the Euler-Lagrange search
in Exercise 8.4.11 does not look at su¬ciently many ways of approaching f0 .)
(ii) Find an in¬nitely di¬erentiable function P such that P (0) = 0, P (t) >
0 for t = 0, P (1) = 0 and P (1) > 0 (so P has a strict local minimum at 1).
[The simplest such function that I can think of is a polynomial.]
(iii) De¬ne J : A ’ R by
1
P (1 ’ f (x)2 ) + f (x)2 dx.
J(f ) =
0

Show that

inf{J(f ) : f ∈ A} = 0

but that J(f ) > 0 for all f ∈ A. Show that J has a strict local minimum
at f0 = 0 when we use the norm — (that is to say, there exists a δ > 0
such that J(f ) > J(f0 ) whenever f ’ f0 — < δ). Show that we can ¬nd a
sequence fn ∈ A such that fn ’ 0 uniformly and J(fn ) ’ 0.
(iv) Find an in¬nitely di¬erentiable function G : R2 ’ R such that,
de¬ning K : A ’ R by
1
K(f ) = G(f (x), f (x)) dx,
0

we have

sup{K(f ) : f ∈ A} = 1, inf{K(f ) : f ∈ A} = 0

but 1 > K(f ) > 0 for all f ∈ A.
Exercise K.202. [11.4, P] (i) If E is a non-empty set and (Y, ρ) a complete
metric space, we write B(E) for the set of bounded functions f : E ’ Y .
The uniform distance d∞ on B(E) is de¬ned by d∞ (f, g) = supx∈E ρ(f, g).
Show that (B(E), d∞ ) is a complete metric space.
(ii) Generalise and prove Theorem 11.3.6 and Theorem 11.3.7.
(iii) Let E be a non-empty set and (Y, ρ) a complete metric space. If fn :
E ’ Y and f : E ’ Y are functions we say that fn converges uniformly to f
539
Please send corrections however trivial to twk@dpmms.cam.ac.uk

as n ’ ∞ if, given any > 0, we can ¬nd an n0 ( ) such that ρ(fn (x), f (x)) <
for all x ∈ E and all n ≥ n0 ( ).
Generalise and prove Theorem 11.4.3 and Theorem 11.4.4.
(iv) Suppose that we now drop the condition that (Y, ρ) is complete.
Which of the results above continue to hold and which fail? [Hint. Take E
to consist of one point.]
Exercise K.203. [11.4, P] (i) Show that the integral

e’ux xγ’1 dx
φγ (u) =
0

converges for all u > 0 and all γ > 0. By di¬erentiating under the integral
sign, integrating by parts and solving the resulting di¬erential equation show
that
φγ (u) = Aγ (0)uγ
where Aγ is a constant depending on γ. You should justify each step of your
argument.
(ii) Obtain the result of (i) in the case when γ is an integer by integrating
by parts.
(iii) Show that the integral

2 /2
e’ux e’x
ψ(u) = dx
0

converges for all real u. By di¬erentiating under the integral sign, integrating
by parts and solving the resulting di¬erential equation show that
2 /2
ψ(u) = Ae’u
for some constant A.
(iv) Obtain the result of (iii) by a change of variable. [However, the
method of (iii) carries over to the case when u is complex and the method
of (iv) does not.]
Exercise K.204. [11.4, P] A function f : R ’ R is in¬nitely di¬erentiable
everywhere and there exists a function g : R ’ R such that f (n) (x) ’ g(x)
uniformly on each bounded interval as n ’ ∞. By considering the equation
x
f (n+1) (t) dt = f (n) (t) ’ f (n) (0),
a

or otherwise, ¬nd a di¬erential equation for g and show that g(x) = Cex for
some C. Must it be true that f (x) = Cex ? Give reasons.
540 A COMPANION TO ANALYSIS

Exercise K.205. [11.4, P] Consider a sequence of functions gn : [a, b] ’ R
and a continuous function g : [a, b] ’ R. Suppose that gn (x) ’ g(x) as
n ’ ∞ for each x ∈ [a, b].
Show that gn ’ g uniformly on [a, b] if and only if, given any sequence
xn ∈ [a, b] with xn ’ x, we have gn (xn ) ’ g(x) as n ’ ∞.
If we replace [a, b] by (a, b) does the ˜if™ part remain true? Does the
˜only if™ part remain true? Would your answers be di¬erent if we insisted, in
addition, that the gn were continuous? Give proofs and counterexamples as
appropriate.

Exercise K.206. [11.4, P] (i) A continuous function f : [0, 1] ’ R has
f (1) = 1 and satis¬es

0 ¤ f (x) < 1 for all 0 ¤ x < 1

Show that
1’δ
(f (x))n dx ’ 0
n
0

as n ’ ∞. Deduce that, if the left derivative f (1) exists and is non-zero,
then
1
1
(f (x))n dx ’
n
f (1)
0

as n ’ ∞.
(ii) A di¬erentiable function g : [0, 1] ’ R satis¬es

0 ¤ g(x) ¤ 1 for all 0 ¤ x ¤ 1.
1
(g(x))n dx remains bounded as n ’ ∞, then g(x) < 1 for
Show that, if n 0
all 0 < x < 1.

Exercise K.207. (Weierstrass approximation theorem.) [11.4, T] (i)
If 1 > · > 0, de¬ne Sn : [’1, 1] ’ R by Sn (t) = (1 + · ’ t2 )n . By considering
the behaviour of
’1
1
Tn (t) = Sn (x) dx Sn (t),
’1

or otherwise, show that, given any > 0, we can ¬nd a sequence of polyno-
mials Un such that
(a) Un (t) ≥ 0 for all t ∈ [’1, 1] and all n ≥ 1,
541
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(b) Un (t) ’ 0 uniformly on [’1, ’ ] ∪ [ , 1] as n ’ ∞,
1
Un (t) dt = 1 for all n ≥ 1.
(c)
’1
t
(ii) Sketch the functions given by Vn (t) = Un (s) ds and Wn (t) =
’1
t
Vn (s) ds.
’1
(iii) Show that if g : [’1, 1] ’ R is given by g(t) = 0 for t ∈ [’1, 0] and
g(t) = t for t ∈ [0, 1], then there exists a sequence of real polynomials Pn
with Pn (t) ’ g(t) uniformly on [’1, 1].
(iv) By considering functions of the form cPn (at + b), show that, if ± ∈
[’1, 1] and if g± : [’1, 1] ’ R is given by g± (t) = 0 for t ∈ [’1, ±] and
f (t) = t ’ ± for t ∈ [±, 1], then there exists a sequence of real polynomials
Qn with Qn (t) ’ g± (t) uniformly on [’1, 1].
(v) If h : [’1, 1] ’ R is piecewise linear, show that there exists a sequence
of real polynomials Rn with Rn (t) ’ h(t) uniformly on [’1, 1].
(vi) If F : [’1, 1] ’ R is continuous, show that there exists a sequence of
real polynomials Sn with Sn (t) ’ F (t) uniformly on [’1, 1].
(vi) If f : [a, b] ’ R is continuous show that there exists a sequence of
real polynomials Pn with Pn (t) ’ f (t) uniformly on [a, b].
(vii) If f : [a, b] ’ C is continuous, show that there exists a sequence of
polynomials Pn with Pn (t) ’ f (t) uniformly on [a, b].
Exercise K.208. [11.4, P, S, ‘ ] (This is easy if you see how to apply
the result of Exercise K.207 appropriately.) Show that, given a continuous
function f : [0, 1] ’ R, we can ¬nd a sequence of polynomials Pn such that
Pn (x2 ) ’ f (x) uniformly on [0, 1] as n ’ ∞.
Show, however that there exists a continuous function g : [’1, 1] ’ R
such that we cannot ¬nd a sequence of polynomials Qn such that Qn (x2 ) ’
g(x) uniformly on [’1, 1] as n ’ ∞.
Exercise K.209. [11.4, P, T] This proof of Weierstrass™s approximation
theorem is due to Bernstein and requires some elementary probability theory.
The random variable Xn (t) is the total number of successes in n indepen-
dent trials in each of which the probability of success is t. Show, by using
Tchebychev™s inequality, or otherwise, that
Xn (t) 1
’t ≥δ ¤
Pr .
nδ 2
n

Suppose that f : [0, 1] ’ R is continuous. Let pn (t) = Ef (Xn (t)/n) the
expectation of f (Xn (t)/n). Show that pn ’ f uniformly. (You will need
542 A COMPANION TO ANALYSIS

to use the fact that a continuous function on a closed interval is uniformly
continuous and bounded. You should indicate explicitly where you use these
two facts.)
Show also that pn is a polynomial of degree at most n (pn is called a
Bernstein polynomial). Deduce the result of Exercise K.207.
If you know about convex functions (for example, if you have done Exer-
cise K.39) show that, if f is convex, then pn (x) ≥ f (x) for all x ∈ [0, 1].
Exercise K.210. [11.4, P] (i) (Dini™s theorem) Let I be a bounded interval
on the real line and let fn : I ’ R [n ≥ 1] and f : I ’ R be functions such
that fn (x) ’ f (x) as n ’ ∞ for each x ∈ I. Show that if all four of the
following conditions hold
(a) fn is continuous on I for each n,
(b) f is continuous on I,
(c) for each x, the sequence fn (x) is decreasing,
(d) I is closed,
then fn ’ f uniformly on I.
(ii) Exhibit counterexamples to show that the result is false if any one of
the four conditions is omitted.
(iii) Let
1
p0 (t) = 0 and pn (t) = pn’1 (t) + (t2 ’ pn’1 (t)).
2
Show, by induction, or otherwise, that 0 ¤ pn’1 (t) ¤ pn (t) ¤ |t| for all
t ∈ [’1, 1]. Use part (i) to show that pn (t) ’ |t| uniformly on [’1, 1].
(iv) Sketch the function h : R ’ R given by h(x) = |x| + |x ’ a| with
a > 0.
(v) Use the result of (iii) to give an alternative proof of Exercise K.207 (iii).
(vi) Generalise the result of (i) as far as you can. (For example, you could
take I to be a subset of Rn or a metric space.)
Exercise K.211. (Orthogonal polynomials.) [11.4, T] Let a < b and
let w : [a, b] ’ R be a strictly positive continuous function.
(i) Show that
b
f, g = f (t)g(t)w(t) dt
a

de¬nes an inner product on the space C([a, b]) of continuous functions from
[a, b] to R. We write f = f, f 1/2 as usual.
(ii) Prove there is a sequence s0 , s1 , s2 , . . . of polynomials with real
coe¬cients such that
543
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(a) sn has exact degree n,
(b) each sn has leading coe¬cient 1, and
(c) sn , sm = 0 for n = m.
(iii) Show that, if P is a polynomial of degree at most n then there exists
one and only one solution to the equation
n
P= aj s j
j=0


with aj ∈ R. Deduce, in particular, that sn , Q = 0 whenever Q is a
polynomial of degree at most n ’ 1.
(iv) Use the fact that the polynomials are uniformly dense in C([a, b])
(Weierstrass™s theorem, Exercise K.207) to show that, if f ∈ C([a, b]),
n
: aj ∈ R for 0 ¤ j ¤ n
aj s j ’ f ’0
inf
j=0

as n ’ ∞.
n
(v) Show that,if f ∈ C([a, b]), then aj sj ’ f has a unique mini-
j=0
mum, attained when

f, sj
ˆ
aj = f (j) = .
sj , s j

Show that
n
ˆ
f (j)sj ’ f ’ 0
j=0

as n ’ ∞.
(vi) By considering the expression
b
sn (t)q(t)w(t) dt,
a

where

q(t) = (t ’ t1 )(t ’ t2 ) . . . (t ’ tk )

and t1 , t2 , . . . , tk are the points of (a, b) at which sn changes sign, prove that
sn must have exactly n distinct zeros, all of which lie in the open interval
(a, b).
544 A COMPANION TO ANALYSIS

Exercise K.212. (Legendre polynomials.) [11.4, T, ‘ ] Let p0 (x) = 1
and
dn
pn (x) = n (x2 ’ 1)n
dx
for n ≥ 1. By integrating by parts, or otherwise, show that
1
pn (x)pm (x) dx = δnm γn
’1


where γn is to be found explicitly. (Recall that δnm = 0 if n = m, δnn = 1.)
Let [a, b] = [’1, 1], w(x) = 1. Show that there exists a cn (to be found
explicitly) such that, if we write sn = cn pn , then the sn obey the conditions
of Exercise K.211 (ii). We call the pn Legendre polynomials.
Show that
1 1
2n+1 (n!)2
k n
t Pn (t) dt = 0, for 0 ¤ k ¤ n ’ 1 and t Pn (t) dt = .
(2n + 1)!
’1 ’1


Exercise K.213. (Gaussian quadrature.) [11.4, T, ‘ ] Suppose x1 , x2 , . . . , xn
are distinct points in [’1, 1]. Show that, if we write

x ’ xi
ej (x) = ,
xj ’ x i
i=j


then any polynomial P of degree at most n ’ 1 satis¬es
n
P (t) = P (xj )ej (t)
j=1


for all t ∈ [’1, 1]. Deduce that we can ¬nd real numbers a1 , a2 , . . . , an such
that
n
1
P (t) dt = aj P (xj )
’1 j=1


for all polynomials of degree n ’ 1 or less.
Gauss realised that this idea could be made much more e¬ective if we
take the x1 , x2 , . . . , xn to be the zeros of the Legendre polynomials (see
Exercise K.212 and part (iv) of Exercise K.211) and this we shall do from
now on. Show that if P is a polynomial of degree 2n ’ 1 or less we can write
545
Please send corrections however trivial to twk@dpmms.cam.ac.uk

P = pn Q + R where Q and R are polynomials of degree n ’ 1. By using this
result and equation show that
n
1
P (t) dt = aj P (xj )
’1 j=1

for all polynomials of degree 2n ’ 1 or less. (For another example of a choice
involving zeros of an orthogonal polynomial, see Exercise K.48.)
By considering polynomials of the form i=j (x ’ xi )2 show that aj > 0
for each j. By considering the constant polynomial 1 show that n aj = 2.
j=1
Let us write
n 1
Gn f = aj f (xj ), and If = f (t) dt
’1
j=1

whenever f ∈ C([’1, 1]). Show that, if f, g ∈ C([’1, 1]), then

|Gn (f ) ’ Gn (g)| ¤ 2 f ’ g and |I(f ) ’ I(g)| ¤ 2 f ’ g
∞, ∞.

Use the fact that the polynomials are uniformly dense in C([’1, 1]) (Weier-
strass™s theorem, Exercise K.207) to show that
1
Gn (f ) ’ f (t) dt
’1

as n ’ ∞ whenever f ∈ C([’1, 1]).
Exercise K.214. [11.4, P] Consider the orthogonal polynomials sn of Ex-
ercise K.211. By imitating the arguments of that exercise, show that, if » is
real, then sn ’ »sn’1 has at least n ’ 1 distinct real roots and has no multiple
roots. Deduce that sn ’ »sn’1 has n distinct real roots.
(i) If P and Q are real polynomials with a common real root at x show
that there exists a real » such that P ’ »Q has a multiple root at x. Deduce
that sn and sn’1 have no common root.
(ii) If P and Q are real polynomials such that P has real roots at x and
y with x < y and Q has no real root in [x, y] show that there exists a real
» such that P ’ »Q has a multiple root at some point w ∈ (x, y). Deduce
that, between any two roots of sn , there is a root of sn’1 .
Exercise K.215. [11.4, T] (i) De¬ne fn : [’1, 1] ’ R by fn (x) = (x2 +
n’2 )1/2 . Show that fn converges pointwise to a limit F , to be found, and fn
converges uniformly to a limit f , to be found. Show that f is not everywhere
di¬erentiable.
546 A COMPANION TO ANALYSIS

(ii) By considering the integral of appropriate witch™s hats, or otherwise,
¬nd di¬erentiable functions fn : [’1, 1] ’ R such that fn (x) = 0 for x ¤ 0
and fn (x) = 1 for x ≥ 1/n. Show that fn (x) ’ 0 for each x and there is an
f such that fn (x) ’ f (x) for each x as n ’ ∞ but that f is not continuous.
(iii) Why do these results not contradict Theorem 11.4.16?

Exercise K.216. [11.4, H] Suppose that u, v : R ’ R are in¬nitely di¬er-

entiable, that v(0) = 0, that u(t) ≥ 0 for all t ∈ R and that 0 u(x) dx = 1.
In this question we shall be interested in the function g : R2 ’ R de¬ned by

g(x, y) = yv(y)u(xy)

(i) Just to get an idea of what is going on, sketch g in the particular case
when v(y) = y and u(x) = 0 for x < 1 and x > 2.
(ii) Show that g has partial derivatives of all orders (thus g is in¬nitely
di¬erentiable). Show also that

g(x, y) dx = v(y)
0

for all y.

(iii) Show that if G(y) = g(x, y) dx then G (0) = v (0), but
0


g,2 (x, 0) dx = 0.
0

We can certainly choose v with v (0) = 0 (for example v(y) = y). Why does
this not contradict Theorem 11.4.21?

Exercise K.217. [11.4, H] In this exercise we modify the ideas we used in
Exercise K.216 to obtain an example relevant to Theorem 8.4.3.
Suppose that u, v : R ’ R have continuous derivatives, that v(t)t1/2 ’ 0
as t ’ 0, that u(t) ≥ 0 for all t ∈ R, that u(t) = 0 if t ¤ 1 or t ≥ 2 and

that 0 u(x) dx = 1. In this question we shall be interested in the function
g : R2 ’ R de¬ned by

g(x, y) = |y|’1/2 v(y)u(x|y|’1/2 ) for y = 0, g(x, 0) = 0.

(i) Just to get an idea of what is going on, sketch g in the particular case
when v(y) = y
(ii) Show that g is everywhere continuous.
(iii) Show that g has partial derivatives g,1 and g,2 everywhere and that
these derivatives are continuous except, possibly, at (0, 0).
547
Please send corrections however trivial to twk@dpmms.cam.ac.uk

1
(iv) Show that G(y) = g(x, y) dx then G (0) = v (0), but
0


g,2 (x, 0) dx = 0.
0

We can certainly choose v with v (0) = 0 (for example v(y) = y). Why does
this not contradict Theorem 8.4.3?

Exercise K.218. (A dominated convergence theorem for integrals.)
[11.4, P] Suppose that g : [0, ∞) ’ R is a continuous function such that

g(t) dt converges. If fn : [0, ∞) ’ R is a continuous function with
0
|fn (t)| ¤ g(t) for all t ∈ [0, ∞) and fn ’ f uniformly on [0, ∞) as n ’ ∞,
∞ ∞
show that all the integrals 0 fn (t) dt [n ≥ 1] and 0 f (t) dt converge and
∞ ∞
fn (t) dt ’ f (t) dt.
0 0

[Hints. If you cannot do this at once, you can adopt one or more of the
following strategies. (1) Consider the special case h(t) = (1 + t2 )’1 . (2) Ask
how the hypotheses prevent us using Example 11.4.12 as a counterexample.
(3) Reread the proof of Lemma 5.3.3 where a similar result is obtained.]

Exercise K.219. [11.4, P] (i) Suppose that d1 , d2 , . . . are metrics on a
space X. Show that

2’j dj (x, y)
d(x, y) =
1 + dj (x, y)
j=1

is a metric on X. Show that if each dj is complete, then so is d.
(ii) Consider C ∞ , the space of in¬nitely di¬erentiable functions on [0, 1].
Show that

2’j supt∈[0,1] |f (j) (t) ’ g (j) (t)|
d(f, g) =
1 + supt∈[0,1] |f (j) (t) ’ g (j) (t)|
j=0


is a complete metric on C ∞ . (Be careful, supt∈[0,1] |f (j) (t) ’ g (j) (t)| is not a
metric on C ∞ if j = 0.)
(j)
Show that d(fn , f ) ’ 0 as n ’ ∞ if and only if fn (x) ’ f (x) uniformly
on [0, 1] for each j.
Let gn (x) = n’n sin(n + 1)2 πx. Show that d(gn , 0) ’ 0 but
(j)
sup sup |gn (x)| = ∞
j≥0 x∈[0,1]
548 A COMPANION TO ANALYSIS

(j)
(more formally, the set {|gn (x)| : x ∈ [0, 1], j ≥ 0} is unbounded) for all n.
(iii) Consider the space C(D) of continuous (but not necessarily bounded)
functions f : D ’ C, where D is the open unit disc given by

D = {z ∈ C : |z| < 1}.

Find a complete metric d such that d(fn , f ) ’ 0 as n ’ ∞ if and only if

|f (z) ’ g(z)| ’ 0
sup
|z|¤1’n’1


for all n ≥ 1. Prove that you have, indeed, found such a metric.

Exercise K.220. (Pointwise convergence cannot be given by a met-
ric.) [11.4, H] The object of this exercise is to show that the notion of a
metric is not su¬cient to cover all kinds of convergence. More speci¬cally,
we shall show that there does not exist a metric d on the space C([0, 1])
of continuous functions such that d(fn , f ) ’ 0 as n ’ ∞ if and only if
fn (x) ’ f (x) for all x ∈ [0, 1]. The proof is quite complicated and simpler
proofs can be obtained once we have studied topology, so the reader may
prefer to omit it.
The proof is by contradiction, so we assume that d is a metric such that
d(fn , f ) ’ 0 as n ’ ∞ if and only if fn (x) ’ f (x) for all x ∈ [0, 1].
(i) Let gn : [0, 1] ’ R be a ˜witch™s hat™ given by

gn (x) =1 ’ n(x ’ n’1 ’ 2’1 ) for |x ’ n’1 ’ 2’1 | ¤ n’1 ,
gn (x) =0 otherwise.

Show that d(gn , 0) ’ 0 as n ’ ∞.
(ii) By using the ideas of (i), or otherwise, show that, given any closed
interval I ⊆ [0, 1] and any > 0, we can ¬nd a g ∈ C([0, 1]) and a closed
interval J ⊆ I such that d(g, 0) ¤ but g(x) ≥ 1/2 on I.
(iii) Show that we can ¬nd a sequence of closed intervals I1 ⊇ I2 ⊇ I3 ⊇
. . . and continuous functions fn such that d(fn , 0) ¤ 2’n but fn (x) ≥ 1/2 on
In .
(iv) Use Exercise 4.3.8 to derive a contradiction.

Exercise K.221. [11.4, T, S] (This sketches some background for the next
exercise.) Let V be a vector space. Suppose that is a norm on V and d a
metric on V such that d(xn , x) ’ 0 as n ’ ∞ if and only if xn ’ x ’ 0.
Show that
(a) d(a + xn , a + x) ’ 0 if and only if d(xn , x) ’ 0.
(b) d(»xn , 0) ’ 0 whenever d(xn , 0) ’ 0.
549
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(c) d(»n x, 0) ’ 0 whenever »n ’ 0.
These remarks make it easy to ¬nd metrics which do not have the con-
vergence properties of norms. As an example, check which of properties (a),
(b) and (c) hold for the discrete metric and deduce that the discrete metric
does not have the convergence properties of a norm. Do the same for the
British Railway non-stop metric of Exercise 10.3.7.
Show however, that the metric of part (ii) of Exercise K.219 has all the
properties (a), (b) and (c). In the next exercise we show that, even so, it
does not have the convergence properties of a norm.
Exercise K.222. (An interesting metric not derivable from a norm.)
[11.4, H, ‘ ] In this question we shall show that there does not exist a norm

D on the space C ([0, 1]) of continuous functions such that fn D ’ 0 as
(j)
n ’ ∞ if and only if fn (x) ’ 0 uniformly on [0, 1] for each j. This shows
that the metric d de¬ned in part (ii) of Exercise K.219 cannot be replaced
by a norm.
The proof proceeds by contradiction. We assume that D has the prop-
erties given in the previous paragraph.
(i) By reductio ad absurdum, or otherwise, show that, for each integer j ≥
0, there exists an j > 0 such that f D ≥ j whenever supx∈[0,1] |f (j) (x)| ≥ 1.
(ii) Deduce that g D ≥ j supx∈[0,1] |g (j) (x)| for all g ∈ C ∞ ([0, 1]) and all
j ≥ 0.
(iii) Show that, by choosing δk and Nk appropriately, and setting fk (x) =
δk sin πNk x, or otherwise, that we can ¬nd fk ∈ C ∞ ([0, 1]) such that
(j)
sup |fk (x)| ¤ 2’k when 0 ¤ j ¤ k ’ 1,
x∈[0,1]
(k)
sup |fk (x)| ≥ 2j j .
j
x∈[0,1]

(j)
(iv) Show that fn D ’ ∞ as n ’ ∞, but fn (x) ’ 0 uniformly on
[0, 1] for each j. This contradiction gives the desired conclusion.
Exercise K.223. (A continuous nowhere di¬erentiable function.) [11.4,
T](i) Suppose that f : [0, 1] ’ R is di¬erentiable with continuous derivative.
Explain why we can ¬nd a K, depending on f , such that
|f (x) ’ f (y)| ¤ K|x ’ y|
for all x, y ∈ [0, 1].
(ii) Suppose f : [0, 1] ’ R is di¬erentiable with continuous derivative.
By considering
g(x) = f (x) + sin N x
550 A COMPANION TO ANALYSIS

with N very large, show that, given any > 0 and any · > 0, we can ¬nd a
g : [0, 1] ’ R with the following properties.
(a) g is di¬erentiable with continuous derivative.
(b) g ’ f ∞ ¤ .
(c) Given any x ∈ [0, 1], we can ¬nd a y ∈ [0, 1] such that |x ’ y| < ·
but |g(x) ’ g(y)| > /2.
(iii) Set g0 = 0. Show that we can ¬nd a sequence of functions gn :
[0, 1] ’ R such that, whenever n ≥ 1,
(a)n gn is di¬erentiable with continuous derivative.
(b)n gn ’ gn’1 ∞ ¤ 2’3n .
(c)n Given any x ∈ [0, 1], we can ¬nd a y ∈ [0, 1] such that |x’y| < 2’4n
but |gn (x) ’ gn (y)| > 2’3n’1 .
(iv) Show that gn converges uniformly to a continuous function g. Show
further that

2’3j = 2’3n /7.
g ’ gn ¤

j=n+1


Use this fact, together with (c)n , to show that, given any x ∈ [0, 1], we can
¬nd a y ∈ [0, 1] such that |x ’ y| < 2’4n but

|g(x) ’ g(y)| > 2’3n’1 ’ 2’3n+1 /7 = 3 — 2’3n’1 /7.

Deduce that
3 — 2n’1
g(x) ’ g(y)

sup
x’y 7
y∈[0,1], y=x


for all n ≥ 1 and all x. Conclude that g is nowhere di¬erentiable.
[At the beginning of the 20th century continuous, nowhere di¬erentiable func-
tions were considered to be playthings of pure mathematicians of uncertain
taste. Classes of continuous nowhere di¬erentiable functions now form the
main subject of study in ¬nancial mathematics and parts of engineering.]

Exercise K.224. (A continuous space ¬lling curve.) [11.4, T] This
exercise uses Exercise 11.3.9. We work in R2 and R with the usual Euclidean
norms.
(i) Let (x0 , y0 ), (x1 , y1 ) ∈ [0, 1]2 . Show, by specifying f piece by piece,
or otherwise, that we can ¬nd a continuous function f : [0, 1] ’ [0, 1]2 such
that f (0) = (x0 , y0 ), f (1) = (x1 , y1 ) and (rN ’1 , sN ’1 ) ∈ f ([0, 1]) for each
0 ¤ r, s ¤ N . (In other words the curve described by f starts at (x0 , y0 ),
ends at (x1 , y1 ) and passes through all points of the form (rN ’1 , sN ’1 ).)
551
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(ii) Suppose that h : [0, 1] ’ [0, 1]2 is continuous and 0 = t0 < t1 < t2 <
· · · < tM = 1. Show that given any > 0 we can ¬nd an · > 0 such that

t0 + · < t 1 ’ · < t 1 + · < t 2 ’ · < t 2 + · < t 3 ’ · < t 3 + · < · · · < t M ’ ·

and h(t) ’ h(s) < whenever t, s ∈ [0, 1] and |t ’ s| ¤ 2·.
(iii) Let g0 : [0, 1] ’ [0, 1]2 be a continuous function such that g0 (0) =
(0, 0), g0 (1/3) = (0, 1), g0 (2/3) = (1, 0) and g0 (1) = (1, 1) and let E0 =
{0, 1/3, 2/3, 1}. Show that we can ¬nd a sequence of functions gn : [0, 1] ’
[0, 1]2 and ¬nite sets En ‚ [0, 1] such that, if n ≥ 1,
(a)n gn ’ gn’1 ∞ ¤ 2’n+4 .
(b)n If a ∈ En then gn (a) = (r2’n , s2’n ) for some integers r and s with
0 ¤ r, s ¤ 2n .
(c)n If r and s are integers with 0 ¤ r, s ¤ 2n then there exists an
a ∈ En with gn (a) = (r2’n , s2’n ).
(d)n If r and s are integers with 0 ¤ r, s ¤ 2n’1 and a ∈ En’1 satis¬es
gn’1 (a) = (r2’n+1 , s2’n+1 ) then, if u and v are integers with 0 ¤ u, v ¤ 2n
and

|u2’n ’ r2n’1 |, |v2’n ’ s2n’1 | ¤ 2’n ,

we can ¬nd b ∈ En such that |a ’ b| ¤ 2’n and gn (b) = (r2’n , s2’n ).
[Conditions (a)n and (d)n are the important ones. The other two are included
to help keep track of what is going on. Note that gn will not be injective for
n ≥ 1 and that En will contain several points x with gn (x) = (r2’n , s2’n ).]
(iv) Show that gn converges uniformly to a continuous function g :
[0, 1] ’ [0, 1]2 .
(v) Using the fact that every x = [0, 1] can be written x = ∞ κj 2’j
j=1
with κj ∈ {0, 1}, or otherwise, use condition (d)n to show that if (x, y) ∈ [0, 1]
we can ¬nd tn ∈ En [n ≥ 0] such that |tn ’ tn’1 | ¤ 2’n for n ≥ 1 and
gn (tn ) ’ (x, y) ’ 0 as n ’ 0. Explain why we can ¬nd a t ∈ [0, 1] with
tn ’ t as n ’ ∞ and use the inequality

g(t) ’ (x, y) ¤ g(t) ’ g(tn ) + g(tn ) ’ gn (tn ) + gn (tn ) ’ (x, y)
¤ g(t) ’ g(tn ) + g ’ gn ∞ + gn (tn ) ’ (x, y)

to show that g(t) = (x, y). Conclude that g : [0, 1] ’ [0, 1]2 is a continuous
surjective map.
[I cannot claim any practical use for space ¬lling curves, but the popularity
of fractals means that mathematicians do come across continuous maps h :
[0, 1] ’ [0, 1]2 where the image h([0, 1]) is ˜unexpectedly large™.]
552 A COMPANION TO ANALYSIS

Exercise K.225. (The devil™s staircase.) [11.4, T] This result is less
spectacular than the previous two and probably harder to explain. If you do
not understand my presentation, the example can be found in most books
on measure theory. (The set E de¬ned in (iii) is called the Cantor set or the
˜Cantor middle third set™.)
(i) We work on [0, 1]. Write E(0) = (3’1 , 2.3’1 ) for the ˜middle third™
open interval of [0, 1]. We have [0, 1] \ E(0) = [0, 3’1 ] ∪ [2.3’1 , 1] and we
write E0 = (3’2 , 2.3’2 ) for the middle third open interval of [0, 3’1 ] and
E1 = (2.3’1 + 3’2 , 2.3’1 + 2.3’2 ) for the middle third interval of [2.3’1 , 1].
More generally, we set
n n
2 (j)3’j + 3’n’1 , 2 (j)3’j + 2.3’n’1 ),
E (1), (2),..., (n) = (
j=1 j=1

where (j) = 0 or (j) = 1. Sketch

E(n) = E (1), (2),..., (n)
(j)∈{0,1}

for n = 1, 2, 3 and verify that E(n) is the union of 2n intervals each of
length 3’n’1 and such that the intervals are the middle third open intervals
of the closed intervals which make up [0, 1] \ n’1 Ek . (You should worry
k=0
more about understanding the pattern than about rigour.)
(ii) We take fn : [0, 1] ’ R to be the simplest piecewise linear function
which satis¬es fn (0) = 0, fn (1) = 1 and
fn (x) = 2’1 for x ∈ E(0),
m
fn (x) = 2’(m+1) + 2’j (j)
for x ∈ E (1), (2),..., (m) and 1 ¤ m ¤ n.
j=1

I sketch f2 in Figure K.4. Sketch f3 for yourself. Explain why fn is an
increasing function. (iii) Show that fn ’ fm ∞ ¤ 2’n , and deduce that fn
converges uniformly to a continuous function f . Show that f is increasing
and that f (0) = 0, f (1) = 1 and
f (x) = 2’1 for x ∈ E(0),
m
f (x) = 2’(m+1) + 2’j (j)
x ∈ E (1), (2),..., (m) and all m ≥ 1.
j=1

Explain why f is di¬erentiable on E(m) with f (x) = 0 for each x ∈ E(m).
Conclude that f is di¬erentiable on E = ∞ E(m) with f (x) = 0 for each
m=0
x ∈ E.
553
Please send corrections however trivial to twk@dpmms.cam.ac.uk




Figure K.4: The second approximation to the devil™s staircase

(iv) Show that the total length of the intervals making up n E(m)
m=0
is 1 ’ (2/3) . If IE : [0, 1] ’ R is given by IE (x) = 1 if x ∈ E and
n+1

IE (x) = 0 otherwise, show, by ¬nding appropriate dissections, that IE is
Riemann integrable and
1
IE (x) dx = 1.
0

Informally, we have shown that ˜f has zero derivative on practically the whole
of [0, 1] but still manages to increase from 0 to 1™ !
(v) This part takes us back to Section 9.4, and, in particular, to the
promise made on page 223 to exhibit a real-valued random variable which
is not a simple mix of discrete and continuous. Our argument is informal
(partly because we have not de¬ned what we mean by a ˜mix of discrete and
continuous™). De¬ne F : R ’ R by F (x) = 0 if x < 0, F (x) = f (x) if
0 ¤ x ¤ 1 and F (x) = 1 otherwise. Explain why F is continuous. Let X be
the real-valued random variable with Pr{X ¤ x} = F (x).
Suppose, if possible, that X is a ˜mix of discrete and continuous™. Since
F is continuous, X must, in fact, be a continuous random variable and so
x
F (x) = Pr{X ¤ x} = g(t) dt
’∞

for some well behaved g. If g is Riemann integrable, then it is bounded on
554 A COMPANION TO ANALYSIS

[0, 1] with |g(x)| ¤ K for all x ∈ [0, 1]. Deduce that, on this assumption,

|F (x) ’ F (y)| ¤ K(y ’ x)

for 0 ¤ x ¤ y ¤ 1. By showing that this last inequality is false for some x
and y, obtain a contradiction.
Thus X must be a novel type of random variable.
[Graphs very similar to that of F turn up in the theory of di¬erential equa-
tions, in particular for processes intended to mimic transition to turbulence
in ¬‚uid ¬‚ows.]
Exercise K.226. [11.4, H] (i) By considering fn (x) = An sin Bn (x + Cn )
where An , Bn and Cn are to be stated explicitly, show that we can ¬nd an
in¬nitely di¬erentiable function fn : R ’ R such that (writing g (0) (x) =
g(x))

|fn (x)| ¤ 2’n’1 for all n > r ≥ 0 and all x ∈ R
(r)


yet
n’1
(n) 2 (n)
≥ (n!) + 1 + |fr (0)|.
fn (0)
r=0

Explain why fn tends uniformly to an in¬nitely di¬erentiable function
f : R ’ R with f (n) (0) ≥ (n!)2 . Show that the Taylor series

f (n) (0) n
x
n!
n=0

diverges for all x = 0.
(ii) The result of (i) can be extended as follows. Show that we can ¬nd a
sequence y1 , y2 . . . of rational numbers such that each rational number occurs
in¬nitely often in the sequence. (In other words, given a rational number q
and an integer N , we can ¬nd an n ≥ N with yn = q.)
By considering fn (x) = An sin Bn (x + Cn ) where An , Bn and Cn are to
be stated explicitly, show that we can ¬nd an in¬nitely di¬erentiable function
fn : R ’ R such that

|fn (x)| ¤ 2’n’1 for all n > r ≥ 0 and all x ∈ R
(r)


yet
n’1
(n) 2 (n)
≥ (n!) + 1 + |fr (yn )|.
fn (yn )
r=0
555
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Deduce the existence of an in¬nitely di¬erentiable function f : R ’ R
whose Taylor series

f (n) (q)
(q + h)n
n!
n=0

diverges for all h = 0 and all rational numbers q.
(iii) The result of (ii) is even stronger than at ¬rst appears. Use the ideas
of Exercise K.230 to show that, if a function F : R ’ R has derivatives of
all orders at x and

F (n) (x) n
F (x + h) = h
n!
n=0

for all |h| < R, then F is in¬nitely di¬erentiable at t for all |t ’ x| < R and

F (n) (t) n
F (t + k) = k
n!
n=0

whenever |k| < R ’ |t ’ x|.
Conclude that the function f obtained in (ii) can not have a valid Taylor
expansion about any point.
Exercise K.227. [11.4, T!] The object of this question and the next two is
to de¬ne x± and to obtain its properties directly rather than by the indirect
de¬nition used in Section 5.7. In this ¬rst question we deal with the case ±
rational. We assume the properties of xn when n is an integer.
(i) Carefully quoting the results that you use, show that, if n is a strictly
positive integer, then, given any x > 0, there exists a unique r1/n (x) > 0 with
r1/n (x)n = x. Show further (again carefully quoting the results that you use)
that the function r1/n : (0, ∞) ’ (0, ∞) is di¬erentiable with

r1/n (x)
r1/n (x) = .
nx
(ii) Show that if m, m , n and n are strictly positive integers with m/n =
m /n then

(r1/n (x))m = (r1/n (x))m

for all x > 0. Explain why this means that we can de¬ne rm/n (x) = r1/n (x)m
for all m, n positive integers.
(iii) If ±, β ∈ Q and ±, β > 0 prove the following results.
556 A COMPANION TO ANALYSIS

(a) r±+β (x) = r± (x)rβ (x) for all x > 0.
(b) r±β (x) = r± (rβ (x)) for all x > 0.
(c) r± (xy) = r± (x)r± (y) for all x, y > 0.
(d) r± (1) = 1.
(e) The function r± : (0, ∞) ’ (0, ∞) is di¬erentiable with
±r± (x)
r± (x) = .
x
Show further that, if n is a strictly positive integer,
(f) rn (x) = xn for all x > 0.
(iv) Show carefully how to de¬ne r± for all ± ∈ Q and obtain results
corresponding to those in part (iii).
Exercise K.228. [11.4, T!, ‘ ] We continue the arguments of Question K.227.
(i) Suppose that x > 0, ± ∈ R and ±n ∈ Q with ±n ’ ± as n ’ ∞.
Show that the sequence r±n (x) is Cauchy and so tends to a limit y, say.
Suppose that βn ∈ Q with βn ’ ± as n ’ ∞. Show that rβn (x) ’ y as
n ’ ∞. We can thus write r± (x) = y.
Show also that r± (x) > 0.
(ii) Suppose that ± ∈ R, ± = 0 and ±n ∈ Q with ±n ’ ± as n ’ ∞.
Show that, if 0 < a < b, we have r±n ’ r± uniformly on [a, b]. Deduce,
quoting any results that you use, that the function r± : (0, ∞) ’ (0, ∞) is
di¬erentiable with
±r± (x)
r± (x) = .
x

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