Exercise 11.2.16. Let (X, d) and (Z, ρ) be metric spaces. If (X, d) has the

Bolzano-Weierstrass property then any continuous function f : X ’ Z is

uniformly continuous.

11.3 The uniform norm

This section is devoted to one the most important norms on functions. We

shall write F to mean either R or C.

De¬nition 11.3.1. If E is a non-empty set, we write B(E) (or, more pre-

cisely, BF (E)) for the set of bounded functions f : E ’ F. The uniform

∞ on B(E) is de¬ned by f ∞ = supx∈E |f (x)|.

norm

Exercise 11.3.2. If we use the standard operations, show that B(E) is vec-

tor space over F and ∞ is a norm.

Show also that, if f, g ∈ B(E) and we write f — g(x) = f (x)g(x), then

f — g ∈ B(E) and f — g ∞ ¤ f ∞ g ∞ .

276 A COMPANION TO ANALYSIS

just de¬ned, is called the uniform norm, sup norm or ∞

The norm ∞,

norm.

Theorem 11.3.3. The uniform norm on B(E) is complete.

Proof. This follows the model set up in Section 11.1. If reader has done

Exercise 11.1.13, the argument will be completely familiar.

∞ ). For each ¬xed x ∈ E

Suppose fn is a Cauchy sequence in (B(E),

|fn (x) ’ fm (x)| ¤ fn ’ fm ∞,

so fn (x) is a Cauchy sequence in F. The general principle of convergence

tells us that fn (x) tends to a limit f (x) as n ’ ∞.

Since any Cauchy sequence is bounded we can ¬nd a K such that fn ∞ ¤

K for all n. We observe that

|f (x)| ¤ |f (x) ’ fn (x)| + |fn (x)| ¤ |f (x) ’ fn (x)| + fn ¤ |f (x) ’ fn (x)| + K ’ K

∞

as n ’ ∞. Thus |f (x)| ¤ K for all x ∈ E, and so f ∈ B(E).

Finally we need to show that f ’ fn ∞ ’ 0 as n ’ ∞. To do this we

use the usual trick of the ˜irrelevant m™, observing that

|f (x) ’ fn (x)| ¤ |f (x) ’ fm (x)| + |fm (x) ’ fn (x)| ¤ |f (x) ’ fm (x)| + fm ’ fn ∞

¤ |f (x) ’ fm (x)| + sup fp ’ fq ∞ ’ sup fp ’ fq ∞

p,q≥M p,q≥M

as m ’ ∞. Thus |f (x) ’ fn (x)| ¤ supp,q≥M fp ’ fq for all x ∈ E, and so

∞

f ’ fn ¤ sup fp ’ fq

∞ ∞

p,q≥M

for all n ≥ M . Recalling that the sequence fm is Cauchy, we see that

f ’ fn ∞ ’ 0 as n ’ ∞ and we are done.

The space B(E) is not very interesting in itself but, if E is a metric space, it

has a very interesting subspace.

De¬nition 11.3.4. If (E, d) is a non-empty metric space we write C(E) (or,

more precisely, CF (E)) for the set of bounded continuous functions f : E ’ F.

The next remark merely restates what we already know.

Lemma 11.3.5. If (E, d) is a non-empty metric space, then C(E) is a vector

subspace of B(E). Further, if f, g ∈ C(E), the pointwise product f — g ∈

C(E).

277

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

However, the next result is new and crucial.

Theorem 11.3.6. If (E, d) is a non-empty metric space, then C(E) is a

closed subset of B(E) under the uniform metric.

This has the famous ˜ /3 proof™3 .

Proof. Let fn ∈ C(E), f ∈ B(E) and fn ’ f ∞ ’ 0. We wish to show that

f is continuous on E and to do this it su¬ces to show that f is continuous

at any speci¬ed point x ∈ E.

Let > 0 and x ∈ E be given. Since fn ’ f ∞ ’ 0 as n ’ ∞, it follows,

in particular, that there exists an N with

fN ’ f < /3.

∞

Since fN is continuous at x, we can ¬nd a δ > 0 such that |fN (x) ’ fN (t)| ¤

/3 for all t ∈ E with d(x, t) < δ. It follows that

|f (x) ’ f (t)| = |(f (x) ’ fN (x)) + (fN (x) ’ fN (t)) + (fN (t) ’ f (t))|

¤ |f (x) ’ fN (x)| + |fN (x) ’ fN (t)| + |fN (t) ’ f (t))|

¤ f ’ fN ∞ + |fN (x) ’ fN (t)| + fN ’ f ∞

< /3 + /3 + /3 =

for all t ∈ E with d(x, t) < δ.

The key to the argument above is illustrated in Figure 11.1. Suppose that

f ’g ∞ < ·. Then f is trapped in a snake of radius · with g as its backbone.

In particular, if g is continuous, f cannot be ˜terribly discontinuous™.

Since C(E) is a closed subset of B(E), Theorem 11.3.3 and Lemma 11.1.4

gives us another important theorem.

Theorem 11.3.7. If (E, d) is a non-empty metric space, then the uniform

metric on C(E) is complete.

If E is a closed bounded subset of Rm with the Euclidean metric (or, more

generally, if (E, d) is a metric space with the Bolzano-Weierstrass property),

then, by Theorem 4.3.4 (or its easy generalisation to metric spaces with

the Bolzano-Weierstrass property), all continuous functions f : E ’ F are

bounded. In these circumstances, we shall write C(E) = C(E) (or, more pre-

cisely, CF (E) = CF (E)) and refer to the space C(E) of continuous functions

on E equipped with the uniform norm.

3

Or according to a rival school of thought the ˜3 proof™.

278 A COMPANION TO ANALYSIS

Figure 11.1: The uniform norm snake

Exercise 11.3.8. The question of which subsets of C(E) have the Bolzano-

Weierstrass property is quite hard and will not be tackled. To get some un-

derstanding of the problem, show by considering fn = sin nπx, or otherwise,

that {f ∈ C([0, 1]) : f ∞ ¤ 1} does not have the Bolzano-Weierstrass

property.

The reader has now met three di¬erent norms on C([a, b]) (recall Lemma 11.1.8

and Exercise 11.1.9)

= sup |f (t)|,

f ∞

t∈[a,b]

b

|f (t)| dt,

f =

1

a

1/2

b

|f (t)|2 dt

f = .

2

a

They represent di¬erent answers to the question ˜when do two continuous

functions f and g resemble each other™. If we say that f and g are close only

if f ’ g ∞ is small then, however small |f (x) ’ g(x)| is over ˜most of the

range™, if |f (x) ’ g(x)| is large anywhere, we say that f and g are far apart.

For many purposes this is too restrictive and we would like to say that f and

g are close if ˜on average™ |f (x) ’ g(x)| is small, that is to say f ’ g 1 is

279

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

small. For a communications engineer, to whom

b

2

|f (t)|2 dt

f =

2

a

is a measure of the power of a signal, the obvious measure of the closeness

of f and g is f ’ g 2 .

The problem of ¬nding an appropriate measure of dissimilarity crops up

in many di¬erent ¬elds. Here are a few examples.

(a) In weather forecasting, how do we measure how close the forecast

turns out to be to the true weather?

(b) In archaeology, ancient graves contain various ˜grave goods™. Presum-

ably graves which have many types of grave goods in common are close in

time and those with few in common are distant in time. What is the correct

measure of similarity between graves?

(c) In high de¬nition TV and elsewhere, pictures are compressed for trans-

mission and then reconstituted from the reduced information. How do we

measure how close the ¬nal picture is to the initial one?

(d) Machines ¬nd it hard to read handwriting. People ¬nd it easy. How

can we measure the di¬erence between curves so that the same words written

by di¬erent people give curves which are close but di¬erent words are far

apart?

Since there are many di¬erent problems, there will be many measures

of closeness. We should not be surprised that mathematicians use many

di¬erent metrics and norms.

We conclude the section with a generalisation of Theorem 11.3.7. The

proof provides an opportunity to review the contents of this section. In

Exercise K.224 we use the result with V = R2 .

Exercise 11.3.9. Let (E, d) be a non-empty metric space and (V, )a

complete normed space. We write CV (E) for the set of bounded continuous

functions f : E ’ V and set

f = sup f (x)

∞ V

x∈E

whenever f ∈ CV (E). Show that (CV (E), ∞) is a complete normed vector

space.

11.4 Uniform convergence

Traditionally, the material of the previous section has been presented in a

di¬erent but essentially equivalent manner.

280 A COMPANION TO ANALYSIS

De¬nition 11.4.1. (Uniform convergence.) If E is a non-empty set and

fn : E ’ F and f : E ’ F are functions, we say that fn converges uniformly

to f as n ’ ∞ if, given any > 0, we can ¬nd an n0 ( ) such that |fn (x) ’

f (x)| < for all x ∈ E and all n ≥ n0 ( ).

Remark 1: For a generalisation see Exercise K.202.

Remark 2: On page 65 we placed the de¬nition of uniform continuity in

parallel with the de¬nition of continuity. In the same way, the reader should

compare the de¬nition of uniform convergence with the notion of convergence

that we have used so far and which we shall now call pointwise convergence.

De¬nition 11.4.2. (Pointwise convergence.) If E is a non-empty set

and fn : E ’ F and f : E ’ F are functions, we say that fn converges

pointwise to f as n ’ ∞ if, given any > 0 and any x ∈ E, we can ¬nd an

n0 ( , x) such that |fn (x) ’ f (x)| < for all n ≥ n0 ( , x).

Once again, ˜uniform™ means independent of the choice of x.

Theorem 11.3.6 now takes the following form.

Theorem 11.4.3. If (E, d) is a non-empty metric space and fn : E ’ F

forms a sequence of continuous functions tending uniformly to f , then f is

continuous.

More brie¬‚y, the uniform limit of continuous functions is continuous.

Proof of Theorem 11.4.3 from Theorem 11.3.6. The problem we must face is

that the fn need not be bounded.

To get round this, choose N such that |fN (x) ’ f (x)| < 1 for all x ∈ E

and all n ≥ N . If we set gn = fn ’ fN , then

|gn (x)| ¤ |fn (x) ’ f (x)| + |fN (x) ’ f (x)| < 2,

and so gn ∈ C(E) for all n ≥ N . If we set g = f ’ fN , then g ∈ B(E), and

gn ’ g = sup |gn (x) ’ g(x)| = sup |fn (x) ’ f (x)| ’ 0.

∞

x∈E x∈E

By Theorem 11.3.6, g ∈ C(E) and so f = g + fN is continuous.

The same kind of simple argument applied to Theorem 11.3.7 gives the

so called ˜general principle of uniform convergence™.

Theorem 11.4.4. (General principle of uniform convergence.) Sup-

pose that (E, d) is a non-empty metric space and fn : E ’ F is a continu-

ous function [n ≥ 1]. The sequence fn converges uniformly to a continuous

function f if and only if, given any > 0, we can ¬nd an n0 ( ) such that

|fn (x) ’ fm (x)| < for all n, m ≥ n0 ( ) and all x ∈ E.

281

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

Figure 11.2: The witch™s hat

This theorem is also known as the GPUC by those who do not object to

theorems which sound as though they were a branch of the secret police.

Exercise 11.4.5. Prove Theorem 11.4.4 from Theorem 11.3.7.

Exercise 11.4.6. (i) Prove Theorem 11.4.3 and Theorem 11.4.4 directly.

(Naturally, the proofs will be very similar to those of Theorem 11.3.6 and

Theorem 11.3.3. If you prefer simply to look things up, proofs are given in

practically every analysis text book.)

(ii) Prove Theorem 11.3.6 from Theorem 11.4.3 and Theorem 11.3.7 from

Theorem 11.4.4.

The next two examples are very important for understanding the di¬er-

ence between pointwise and uniform convergence.

Example 11.4.7. (The witch™s hat.) De¬ne fn : [0, 2] ’ R by

fn (x) =1 ’ n|x ’ n’1 | for |x ’ n’1 | ¤ n’1 ,

fn (x) =0 otherwise.

Then the fn are continuous functions such that fn (x) ’ 0 as n ’ ∞ for

each x but fn 0 uniformly.

More brie¬‚y, pointwise convergence does not imply uniform convergence.

We sketch the witches hat in Figure 11.2.

Proof. Observe that, if x = 0, then x ≥ 2N ’1 for some positive integer N

and so, provided n ≥ N ,

fn (x) = 0 ’ 0

282 A COMPANION TO ANALYSIS

as n ’ ∞. On the other hand, if x = 0,

fn (0) = 0 ’ 0

as n ’ ∞. Thus fn (x) ’ 0 for each x ∈ [0, 2].

We observe that fn ∞ ≥ fn (n’1 ) = 1 0, so fn 0 uniformly as

n ’ 0.

Example 11.4.8. De¬ne fn : [0, 1] ’ R by fn (x) = xn . Then fn (x) ’ f (x)

as n ’ ∞ where f (x) = 0 for 0 ¤ x < 1 but f (1) = 1.

Thus the pointwise limit of continuous functions need not be continuous. We

leave the veri¬cation to the reader.

Exercise 11.4.9. Draw a diagram to illustrate Example 11.4.8 and prove

the result.

Uniform convergence is a very useful tool when dealing with integration.

Theorem 11.4.10. Let fn ∈ C([a, b]). If fn ’ f uniformly, then f ∈

C([a, b]) and

b b

fn (x) dx ’ f (x) dx.

a a

Students often miss the full force of this theorem because it is so easy to

prove. Note, however, that we need to prove that the second integral actually

exists. (You should look brie¬‚y at Exercise 9.1.1 before proceeding. If you

want an example of a sequence of continuous functions whose pointwise limit

is not Riemann integrable, consult the harder Exercise K.157.)

Proof. Since f is the uniform limit of continuous functions it is itself contin-

uous and therefore Riemann integrable. By the sup —length inequality,

b b b b

f (t) dt ’ (f (t) ’ fn (t)) dt ¤ |f (t) ’ fn (t)| dt

fn (t) dt =

a a a a

¤ (b ’ a) f ’ fn ’0

∞

as n ’ ∞.

Theorem 11.4.10 should be considered in the context of the following two

examples.

283

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

Example 11.4.11. (The tall witch™s hat.) De¬ne fn : [0, 2] ’ R by

fn (x) =n(1 ’ n|x ’ n’1 |) for |x ’ n’1 | ¤ n’1 ,

fn (x) =0 otherwise.

Then the fn are continuous functions such that fn (x) ’ 0 as n ’ ∞ but

2

fn (x) dx 0

0

as n ’ ∞.

Example 11.4.12. (Escape to in¬nity.) De¬ne fn : R ’ R by

fn (x) =n’1 (1 ’ n’1 |x|) for |x| ¤ n,

fn (x) =0 otherwise.

Then the fn are continuous functions such that fn (x) ’ 0 uniformly as

n ’ ∞ but

∞

fn (x) dx 0

’∞

as n ’ ∞.

[We gave a similar example of escape to in¬nity in Exercise 5.3.2.]

Exercise 11.4.13. (i) Draw a diagram to illustrate Example 11.4.11 and

prove the result.

(ii) Draw a diagram to illustrate Example 11.4.12 and prove the result.

One way of contrasting Theorem 11.4.10 with Example 11.4.12 is to think

of pushing a piston down a cylinder with water at the bottom. Eventually

we must stop because the water has nowhere to escape to. However, if we

place a glass sheet on top of another glass sheet, any water between them

escapes outwards.

The following example, which requires some knowledge of probability

theory, illustrates the importance of the phenomenon exhibited in Exam-

ple 11.4.12.

Exercise 11.4.14. Let X1 , X2 , . . . be independent identically distributed

random variables. Suppose further that X1 has continuous density distribu-

tion f . We know that X1 + X2 + · · · + Xn then has a continuous distribution

fn .

∞

(i) Explain why ’∞ fn (x) dx = 1.

284 A COMPANION TO ANALYSIS

(ii) In the particular case f (x) = (2π)’1/2 exp(’x2 /2) state the value of

fn (x) and show that fn (x) ’ 0 everywhere. Thus

∞ ∞

lim fn (x) dx = lim fn (x) dx.

n’∞ ’∞ n’∞

’∞

(iii) If Y is a real-valued random variable with continuous density distri-

bution gY and a > 0, show that aY has continuous density distribution gaY

given by gaY (x) = a’1 g(a’1 x). What is the density for ’aY ?

(iv) In the particular case investigated in part (ii), show that

(a) If 1/2 > ± ≥ 0, then n± fn (n± x) ’ 0 uniformly on R as n ’ ∞

and

∞ ∞

± ±

lim n± fn (n± x) dx.

lim n fn (n x) dx =

n’∞ ’∞ n’∞

’∞

(b) If ± = 1/2, then n± fn (n± x) = f (x) and

∞ ∞

± ±

lim n± fn (n± x) dx.

lim n fn (n x) dx =

n’∞ ’∞ n’∞

’∞

(c) If ± > 1/2, then n± fn (n± x) ’ 0 for each x = 0 as n ’ ∞, but

n± fn (n± 0) ’ ∞.

(v) Draw diagrams illustrating the three cases in (iv) and give a proba-

bilistic interpretation in each case.

(vi) How much further you can go, for general f , depends on your knowl-