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(i) Show that, if “ is a circle with centre on the real axis, there is a
T ∈ H such that T (“) is a circle with centre on the real axis passing through
the origin.
(ii) Show that if “ is a circle with centre on the real axis passing through
the origin, there is a T ∈ H such that T (“) is a line perpendicular to the real
axis.
(iii) Show that if “ is a circle with centre on the real axis or a line per-
pendicular to the real axis, there is a T ∈ H such that T (“) is the imaginary
axis.
Exercise 10.5.6 and Exercise 10.5.9 show that the following two theorems
are equivalent.
Theorem 10.5.10. The geodesic path between iy1 and iy2 [y1 , y2 real, un-
equal and strictly positive] is a straight line.
260 A COMPANION TO ANALYSIS

Theorem 10.5.11. If z1 , z2 ∈ H with z1 = z2 , the geodesic path between
them is an arc of the circle through z1 and z2 , unless z1 = z2 , in which
case, it is the straight line between the two points.
Exercise 10.5.12. Explain why Theorem 10.5.10 implies Theorem 10.5.11.
We now turn to the proof of Theorem 10.5.10.
Sketch proof of Theorem 10.5.10. Let Z : [0, 1] ’ H be a well behaved
function such that Z(0) = y1 and Z(1) = y2 with y2 > y1 . We write
Z(t) = X(t) + iY (t) with X(t) and Y (t) real and take “ to be the path
described by Z. We observe that
1 1
(Y (t)2 )1/2
1 1 2 2 1/2
(X (t) + Y (t) ) dt ≥
ds = dt
y 0 Y (t) Y (t)
“ 0
1 1
|Y (t)| Y (t)
dt = [log Y (t)]1
dt ≥
= 0
Y (t) 0 Y (t)
0
y2
1 1
= log y2 ’ log y1 = dt = ds,
t y
y1 “0

where “0 is the straight line path from y1 to y2 . We observe that the argument
above shows that
1 1
ds = ds
y y
“ “0

only if X (t)2 = 0 and Y (t) ≥ 0 for all t ∈ [0, 1] and so, by simple arguments,
X(t) = 0 and Y (t) ≥ 0 for all t ∈ [0, 1]. Thus “0 is the unique geodesic.
Remark: This is only a sketch of a proof because we have not really decided
which curves will be eligible for paths. The proof strategy of ˜project the path
onto the y-axis and compare™ ought to work for any reasonable de¬nition of
line integral and any eligible path.
Exercise 10.5.13. We work in R2 . Suppose that we want a metric de¬ned
by

d(A, B) = inf g(x, y) ds : “ joining A and B ,


which is invariant under translation and rotation. Copy the investigation
above going through the following steps.
(i) By considering points which are close and working to ¬rst order, show
that we should try g constant. Without loss of generality take g = 1.
(ii) Use the calculus of variations to suggest the form of the geodesics.
(iii) Prove that your guess is correct.
(iv) Show that our metric is the usual Euclidean metric.
261
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 10.5.14. In this section we de¬ned a metric d on H invariant
under H but did not calculate it. We could ¬nd d by using the strategy of
Exercise 10.5.9, but, in this exercise, we follow the easier path of allowing
someone else to ¬nd the answer and then verifying it.
If z, w ∈ H, we set

|z ’ w— | + |z ’ w|
ρ(z, w) = log .
|z ’ w— | ’ |z ’ w|

(i) Show that |z ’ w — | ’ |z ’ w| > 0 for all z, w ∈ H, so ρ is well de¬ned.
(ii) Show that ρ(T z, T w) = ρ(z, w) for all T ∈ H.
(iii) Show that d(iy1 , iy2 ) = ρ(iy1 , iy2 ) for all y1 , y2 > 0.
(iv) Deduce that d = ρ.
Chapter 11

Complete metric spaces

11.1 Completeness
If we examine the arguments of Section 10.3 we see that they are all mere
algebra. What must we introduce to do genuine analysis on metric spaces?
We cannot use a variant of the fundamental axiom because there is no order
on our spaces1 . Instead, we use a generalisation of the general principle of
convergence.

De¬nition 11.1.1. If (X, d) is a metric space, we say that a sequence of
points xn ∈ X is Cauchy if, given any > 0, we can ¬nd n0 ( ) such that
d(xp , xq ) < for all p, q ≥ n0 ( ).

De¬nition 11.1.2. A metric space (X, d) is complete if every Cauchy se-
quence converges.

In this context, Theorem 4.6.3 states that Rn with the Euclidean metric
is complete.
The contraction mapping theorem (Theorem 12.1.3) and its applications
will provide a striking example of the utility of this concept. However, this
section is devoted to providing background examples of spaces which are and
are not complete.
If you want to see completeness in action immediately you should do the
next example.

Exercise 11.1.3. We say that a metric space (X, d) has no isolated points
if, given y ∈ X and > 0, we can ¬nd an x ∈ X such that 0 < d(x, y) < .
1
There is an appropriate theory for objects with order (lattices) hinted at in Ap-
pendix D, but we shall not pursue this idea further.


263
264 A COMPANION TO ANALYSIS

Show by the methods of Exercise 1.6.7 that a complete non-empty metric
space with no isolated points is uncountable.
Give an example of a countable metric space. Give an example of an
uncountable metric space all of whose points are isolated.

The next lemma gives a good supply of metric spaces which are complete
and of metric spaces which are not complete.

Lemma 11.1.4. Let (X, d) be a complete metric space. If E is a subset of
X and we de¬ne dE : E 2 ’ R by dE (u, v) = d(u, v) whenever u, v ∈ E, then
(E, dE ) is complete if and only if E is closed in (X, d).

Proof. This is just a matter of de¬nition chasing.
Observe that any Cauchy sequence xn in E is a Cauchy sequence in
X and so converges to a point x in X. If E is closed, then x ∈ E and
dE (xn , x) = d(xn , x) ’ 0 as n ’ ∞. Thus (E, dE ) is complete whenever E
is closed.
Suppose now that (E, dE ) is complete. If xn ∈ E and d(xn , x) ’ 0 for
some x ∈ X, we know (by the argument of Lemma 4.6.2 if you need it)
that xn is a Cauchy sequence in X and so a Cauchy sequence in E. Since
(E, dE ) is complete, we can ¬nd a y ∈ E such that dE (xn , y) ’ 0. Now
d(xn , y) = dE (xn , y) ’ 0 so, by the uniqueness of limits, y = x and x ∈ E.
Thus E is closed in (X, d).
Thus, for example, the closed interval [a, b] is complete for the usual metric
but the open interval (a, b) is not.

Exercise 11.1.5. Let (X, d) be a metric space and E a subset of X. De¬ne
dE : E 2 ’ R as in Lemma 11.1.4.
(i) Show that, if E is not closed in (X, d), then (E, dE ) is not complete.
(ii) Give an example where E is closed in (X, d) but (E, dE ) is not com-
plete.
(iii) Give an example where (X, d) is not complete but (E, dE ) is.

The reader is warned that, at least in my opinion, it is harder than it
looks to prove that a metric space is or is not complete and it is easy to
produce plausible but unsatisfactory arguments in this context.
If we wish to show that a metric space (X, d) is incomplete, the natural
way to proceed is to ¬nd a Cauchy sequence xn and show that it does not
converge. However, we must show that xn does not converge to any point in
X and not that ˜xn does not converge to the point that it looks as though it
ought to converge to™. Among the methods available for a correct proof are
the following.
265
Please send corrections however trivial to twk@dpmms.cam.ac.uk

˜˜ ˜˜
(1) Embed (X, d) in a larger metric space (X, d) (that is, ¬nd (X, d) such
˜
˜
that X ⊇ X and d(x, y) = d(x, y) for x, y ∈ X) and show that there is an
˜
˜
x ∈ X \ X such that d(xn , x) ’ 0. (See Exercise 11.1.5.)
(2) For each ¬xed x ∈ X show that there is a δ(x) > 0 and an N (x) (both
depending on x) such that d(xn , x) > δ(x) for n > N (x).
(3) Show that the assumption d(xn , x) ’ 0 for some x ∈ X leads to a
contradiction.
Of course, no list of this sort can be exhaustive. None the less, it is good
practice to ask yourself not simply whether your proof is correct but also
what strategies it employs.
Here are a couple of examples.

Example 11.1.6. Consider s00 , the space of real sequences a = (an )∞ such
n=1
that all but ¬nitely many of the an are zero, introduced in Exercise 10.4.8.
The norm de¬ned by

|an |
a =
1
n=1

is not complete.

Proof. Set

a(n) = (1, 2’1 , 2’2 , . . . , 2’n , 0, 0, . . . ).

We observe that, if m ≥ n,
m
2’j ¤ 2’n ’ 0
d(a(n), a(m)) =
j=n+1


as n ’ ∞ and so the sequence a(n) is Cauchy.
However, if a ∈ s00 , we know that there is an N such that aj = 0 for all
j ≥ N . It follows

d(a(n), a) ≥ 2’N

whenever n ≥ N and so the sequence a(n) has no limit.

[The proof above used method 2. For an alternative proof using method 1,
see page 269. For a generalisation of the result see Exercise K.187.]
The next example needs a result which is left to the reader to prove. (You
will need Lemma 8.3.2.)
266 A COMPANION TO ANALYSIS

Exercise 11.1.7. Let b > a. Consider C([a, b]) the set of continuous func-
tions f : [a, b] ’ R. If we set
b
f ’g |f (x) ’ g(x)| dx,
=
1
a

show that is a norm.
1

Lemma 11.1.8. The normed space of Exercise 11.1.7 is not complete.

Proof. This proof uses method 3. With no real loss of generality, we take
[a, b] = [’1, 1]. Let

fn (x) = ’1 for ’1 ¤ x ¤ ’1/n,
for ’1/n ¤ x ¤ 1/n,
fn (x) = nx
for 1/n ¤ x ¤ 1.
fn (x) = 1

If m ≥ n,
1 1/n
2
fn ’ fm |fn (x) ’ fm (x)| dx ¤ ’0
= 1 dx =
1
n
’1 ’1/n

as n ’ ∞ and so the sequence fn is Cauchy.
Suppose, if possible, that there exists an f ∈ C([’1, 1]) such that f ’ fn ’0
1
as n ’ ∞. Observe that, if n ≥ N , we have
1 1 1
|f (x) ’ 1| dx = |f (x) ’ fn (x)| dx ¤ |f (x) ’ fn (x)| dx = fn ’ f ’0
1
’1
1/N 1/N

as n ’ ∞. Thus
1
|f (x) ’ 1| dx = 0
1/N


and, by Lemma 8.3.2, f (x) = 1 for x ∈ [1/N, 1]. Since N is arbitrary,
f (x) = 1 for all 0 < x ¤ 1. A similar argument shows that f (x) = ’1 for all
’1 ¤ x < 0. Thus f fails to be continuous at 0 and we have a contradiction.
By reductio ad absurdum, the Cauchy sequence fn has no limit.

Lemma 11.1.8 is important as an indication of the unsatisfactory results
of using too narrow a class of integrable functions.
The next exercise goes over very similar ground but introduces an inter-
esting set of ideas.
267
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 11.1.9. Let b > a. Consider C([a, b]) the set of continuous func-
tions f : [a, b] ’ R. If f , g ∈ C([a, b]) and we de¬ne
b
f, g = f (t)g(t) dt
a

show that (C([a, b]), , ) is an inner product space. More formally, show
that
(i) f, g ≥ 0 with equality if and only if f = 0,
(ii) f, g = g, f ,
(iii) »f, g = » f, g ,
(iv) f, g + h = f, g + f, h
for all f , g, h ∈ C([a, b]) and all » ∈ R.
Use the arguments of Lemmas 4.1.2 and 4.1.4 to show that setting
1/2
b
1/2 2
f = f, f = f (t) dt
2
a

gives a norm on C([a, b]).
Show, however, that (C([a, b]), 2) is not a complete normed space.
If we wish to show that a metric space (X, d) is complete, the natural way
to proceed is to take a Cauchy sequence xn and show that it must converge.
However, we must show that xn actually converges to a point in X and not
that ˜the sequence xn looks as though it ought to converge™. In many cases
the proof proceeds through the following steps.
(A) The sequence xn converges in some sense (but not the sense we want)
to an object x.
(B) The object x actually lies in X.
(C) The sequence xn actually converges to x in the sense we want.
Here is an example.
Example 11.1.10. The set l 1 of real sequences a with ∞ |aj | convergent
j=1
forms a vector space if we use the natural de¬nitions of addition and scalar
multiplication

(an ) + (bn ) = (an + bn ), »(an ) = (»an ).

If we set

|aj |,
a =
1
j=1

then (l1 , 1) is a complete normed space.
268 A COMPANION TO ANALYSIS

Proof. We know that the space of all sequences forms a vector space, so
we only have to show that l 1 is a subspace. Clearly 0 ∈ l 1 , and, since
N
|»aj | = |»| N |aj |, we have »a ∈ l1 whenever a ∈ l1 and » ∈ R.
j=1 j=1
1
Suppose a, b ∈ l . We have
∞ ∞
N N N
|aj + bj | ¤ |aj | + |bj | ¤ |aj | + |bj |,
j=1 j=1 j=1 j=1 j=1

so, since an increasing sequence bounded above tends to a limit, ∞ |aj +bj |
j=1
1 1
converges and a + b ∈ l . Hence l is, indeed, a subspace of the space of all
sequences. It is easy to check that 1 is a norm.
I shall label the next three paragraphs in accordance with the discussion
just before this example.
Step A Suppose a(n) is a Cauchy sequence in (l 1 , 1 ). For each ¬xed j,
|aj (n) ’ aj (m)| ¤ a(n) ’ a(m) 1 ,
so aj (n) is a Cauchy sequence in R. The general principle of convergence
tells us that aj (n) tends to a limit aj as n ’ ∞.
Step B Since any Cauchy sequence is bounded, we can ¬nd a K such that
a(n) 1 ¤ K for all n. We observe that
N N N N
|aj | ¤ |aj ’ aj (n)| + |aj (n)| ¤ |aj ’ aj (n)| + a(n) 1
j=1 j=1 j=1 j=1
N
¤ |aj ’ aj (n)| + K ’ K
j=1


as n ’ ∞. Thus N |aj | ¤ K for all N , and so |aj | converges. We
j=1 j=1
have shown that a ∈ l1 .
Step C We now observe that, if n, m ≥ M ,
N N N
|aj ’ aj (n)| ¤ |aj ’ aj (m)| + |aj (m) ’ aj (n)|
j=1 j=1 j=1
N
¤ |aj ’ aj (m)| + a(m) ’ a(n) 1
j=1
N
¤ |aj ’ aj (m)| + sup a(p) ’ a(q) 1
p,q≥M
j=1

’ sup a(p) ’ a(q) 1
p,q≥M
269
Please send corrections however trivial to twk@dpmms.cam.ac.uk

N
as m ’ ∞. Thus |aj ’ aj (n)| ¤ supp,q≥M a(p) ’ a(q) for all N , and
1
j=1
so

a ’ a(n) ¤ sup a(p) ’ a(q)
1 1
p,q≥M


for all n ≥ M . Recalling that the sequence a(m) is Cauchy, we see that
a ’ a(n) 1 ’ 0 as n ’ ∞ and we are done.

The method of proof of Step C, and, in particular, the introduction of the
˜irrelevant m™ in the ¬rst set of inequalities is very useful but requires some
thought to master.

Exercise 11.1.11. In the proof above we said ˜any Cauchy sequence is bounded™.
Give the one line proofs of the more precise statements that follow.
(i) If (X, d) is a metric space, xn is a Cauchy sequence in X and a ∈ X,
then we can ¬nd a K such that d(a, xn ) < K for all n ≥ 1.
(ii) If (V, ) is a normed vector space and xn is a Cauchy sequence in
V, then we can ¬nd a K such that xn < K for all n ≥ 1.

Alternative proof of Example 11.1.6. We wish to show that s00 , with norm

|aj |,
a =
1
j=1


is not complete. Observe that we can consider s00 as a subspace of l1 and
that the norm on l1 agrees with our norm on s00 . Now set

a(n) = (1, 2’1 , 2’2 , . . . , 2’n , 0, 0, . . . ),

and

a = (1, 2’1 , 2’2 , . . . , 2’n , 2’n’1 , 2’n’2 , . . . ).

Since a(n) ∈ s00 for all n and a(n) ’ a 1 ’ 0 as n ’ ∞, we see that s00 is
not closed in l1 and so, by Exercise 11.1.5, (s00 , 1 ) is not complete.


Here are two exercises using the method of proof of Example 11.1.10

Exercise 11.1.12. Let U be a complete vector space with norm . Show
that the set l1 (U ) of sequences

u = (u1 , u2 , u3 , . . . )
270 A COMPANION TO ANALYSIS

with ∞ uj convergent forms a vector space if we use the natural de¬ni-
j=1
tions of addition and scalar multiplication

(un ) + (vn ) = (un + vn ), »(un ) = (»un ).

Show that, if we set

u = uj ,
1
j=1


then (l1 , 1) is a complete normed space.
Exercise 11.1.13. Show that the set l ∞ of bounded real sequences forms a
vector space if we use the usual de¬nitions of addition and scalar multiplica-
tion.
Show further that, if we set

= sup |aj |,
a ∞
j≥1

then (l∞ , ∞ ) is a complete normed space.
[We shall prove a slight generalisation as theorem 11.3.3, so the reader
may wish to work through this exercise before she meets the extension.]
Exercise 11.1.14. Consider the set c0 of real sequences a = (a1 , a2 , a3 , . . . )
such that an ’ 0. Show that c0 is a subspace of l∞ and a closed subset of
(l∞ , ∞ ). Deduce that (c0 , ∞ ) is a complete normed space.

Exercise K.188 contains another example of an interesting and important
in¬nite dimensional complete normed space.
The ¬nal result of this section helps show why the operator norm is so
useful in more advanced work.
Lemma 11.1.15. Let U and V be vector spaces with norms U and V.
Suppose further that V is complete. Then the operator norm is a complete
norm on the space L(U, V ) of continuous linear maps from U to V .
Proof. Once again our argument falls into three steps.
Step A Suppose that Tn is a Cauchy sequence in (L(U, V ), ). For each
¬xed u ∈ U ,

Tn u ’ T m u = (Tn ’ Tm )u ¤ T n ’ Tm u ,
V V V

so Tn u is a Cauchy sequence in V . Since V is complete, it follows that Tn u
tends to a limit T u, say.
271
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Step B In Step A we produced a map T : U ’ V . We want to show that, in
fact, T ∈ L(U, V ). Observe ¬rst that, since Tn is linear

T (»1 u1 + »2 u2 ) ’ »1 T u1 ’ »2 T u2 V

T (»1 u1 + »2 u2 ) ’ »1 T u1 ’ »2 T u2 ’ Tn (»1 u1 + »2 u2 ) ’ »1 Tn u1 ’ »2 Tn u2
= V

= T (»1 u1 + »2 u2 ) ’ Tn (»1 u1 + »2 u2 ) ’ »1 (T u1 ’ Tn u1 ) ’ »2 (T u2 ’ Tn u2 ) V
¤ T (»1 u1 + »2 u2 ) ’ Tn (»1 u1 + »2 u2 ) V + |»1 | T u1 ’ Tn u1 V + |»2 | T u2 ’ Tn u2 V
’0

as n ’ ∞. Thus

T (»1 u1 + »2 u2 ) ’ »1 T u1 ’ »2 T u2 = 0,
V

so

T (»1 u1 + »2 u2 ) ’ »1 T u1 ’ »2 T u2 = 0,

and T is linear.
Next, observe that, since every Cauchy sequence is bounded, we can ¬nd
a K such that Tn ¤ K for all n. It follows that Tn u V ¤ K u U for each
n. Thus

¤ T u ’ Tn u ¤ T u ’ Tn u ’K u
Tu + Tn u +K u
V V V V U U

as n ’ ∞, and so T u V ¤ K u U for all u ∈ U . Thus T is continuous.
Step C Finally we need to show that T ’ Tn ’ 0 as n ’ ∞. To do this we
use the trick of ˜the irrelevant m™ introduced in the proof of Example 11.1.10.

T u ’ Tn u ¤ T u ’ Tm u Tm u ’ Tn u V
+
V V
¤ T u ’ Tm u (Tm ’ Tn )u V
+
V
¤ T u ’ Tm u T m ’ Tn u U
+
V
¤ T u ’ Tm u sup Tp ’ Tq u
+
V U
p,q≥M

’ sup Tp ’ Tq u U
p,q≥M

as m ’ ∞. Thus T u ’ Tn u ¤ supp,q≥M Tp ’ Tq for all u ∈ U ,
u
V U
and so

T ’ Tn ¤ sup Tp ’ Tq
p,q≥M

for all n ≥ M . Recalling that the sequence Tm is Cauchy, we see that
T ’ Tn ’ 0 as n ’ ∞ and we are done.
272 A COMPANION TO ANALYSIS

Remark: In this book we are mainly concerned with the case when U and V
are ¬nite dimensional. In this special case, L(U, V ) is ¬nite dimensional and,
since all norms on a ¬nite dimensional space are equivalent (Theorem 10.4.6),
the operator norm is automatically complete.

Exercise 11.1.16. In Exercise 10.4.16, U and V are not complete. Give an
example along the same lines involving complete normed spaces.


11.2 The Bolzano-Weierstrass property
In the previous section we introduced the notion of a complete metric space
as a generalisation of the general principle of convergence. The reader may
ask why we did not choose to try for some generalisation of the Bolzano-
Weierstrass theorem instead. One answer is that it is generally agreed that
the correct generalisation of the Bolzano-Weierstrass property is via the no-
tion of compactness and that compactness is best studied in the context of
topological spaces (a concept more general than metric spaces). A second an-
swer, which the reader may ¬nd more satisfactory, is given in the discussion
below which concludes in Theorem 11.2.7.
We make the following de¬nition.

De¬nition 11.2.1. A metric space (X, d) has the Bolzano-Weierstrass prop-
erty if every sequence xn ∈ X has a convergent subsequence.

Lemma 11.2.2. A metric space (X, d) with the Bolzano-Weierstrass prop-
erty is complete.

Proof. Suppose that xn is a Cauchy sequence. By de¬nition, given any > 0,
we can ¬nd n0 ( ) such that d(xp , xq ) < for all p, q ≥ n0 ( ). By the Bolzano-
Weierstrass property, we can ¬nd n(j) ’ ∞ and x ∈ X such that n(j) ’ ∞
and xn(j) ’ x as j ’ ∞.
Thus, given any > 0, we can ¬nd a J such that n(J) ≥ n0 ( /2) and
d(x, xn(J) ) < /2. Since n(J) ≥ n0 ( /2), we know that, whenever m ≥
n0 ( /2), we have d(xn(J) , xm ) < /2, and so

d(x, xm ) ¤ d(x, xn(J) ) + d(xn(J) , xm ) < /2 + /2 = .

Thus xn ’ x as n ’ ∞.

Exercise 11.2.3. We work in a metric space (X, d).
(i) Show that if xn ’ x as n ’ ∞ then the sequence xn is Cauchy. (Any
convergent sequence is Cauchy.)
273
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(ii) If xn is a Cauchy sequence and we can ¬nd n(j) ’ ∞ and x ∈ X
such that n(j) ’ ∞ and xn(j) ’ x as j ’ ∞, then xn ’ x. (Any Cauchy
sequence with a convergent subsequence is convergent.)
To show that the converse of Lemma 11.2.2 is false it su¬ces to consider
R with the usual topology. To give another counter example (in which,
additionally, the metric is bounded) we introduce a dull but useful metric
space.
Exercise 11.2.4. Let X be any set. We de¬ne d : X 2 ’ R by
d(x, y) =1 if x = y,
d(x, x) =0.
(i) Show that d is a metric.
(ii) Show that d(xn , x) ’ 0 as n ’ ∞ if and only if there exists an N
such that xn = x for all n ≥ N .
(iii) Show that (X, d) is complete.
¯
(iv) If y ∈ X, show that the closed unit ball B(y, 1) = {x : d(x, y) ¤
1} = X.
(v) If X is in¬nite, show, using (ii) or otherwise, that X does not have
the Bolzano-Weierstrass property.
(vi) Show also that every subset of X is both open and closed.
We call the metric d of the previous lemma the discrete metric.
In order to characterise metric spaces having the Bolzano-Weierstrass
property, we must introduce a further de¬nition.
De¬nition 11.2.5. We say that (X, d) is totally bounded if, given any > 0,
we can ¬nd y1 , y2 , . . . , yN ∈ X such that N B(yj , ) = X.
j=1

In other words, (X, d) is totally bounded if, given any > 0, we can ¬nd
a ¬nite set of open balls of radius covering X.
Lemma 11.2.6. If (X, d) is a metric space with the Bolzano-Weierstrass
property, then it is totally bounded.
Proof. If (X, d) is not totally bounded, then we can ¬nd an > 0 such that
no ¬nite set of open balls of radius covers X. Choose any x1 ∈ X. We
obtain x2 , x3 , . . . inductively as follows. Once x1 , x2 , . . . , xn have been ¬xed,
we observe that n B(xj , ) = X so we can choose xn+1 ∈ n B(xj , ). / j=1
j=1
Now consider the sequence xj . By construction, d(xi , xj ) ≥ for all i = j
and so, if x ∈ X, we have max(d(x, xi ), d(x, xj )) ≥ /2 for all i = j. Thus
the sequence xj has no convergent subsequence and (X, d) does not have the
Bolzano-Weierstrass property.
274 A COMPANION TO ANALYSIS

Theorem 11.2.7. A metric space (X, d) has the Bolzano-Weierstrass prop-
erty if and only if it is complete and totally bounded.
Proof. Necessity follows from Lemmas 11.2.2 and 11.2.6.
To prove su¬ciency, suppose that (X, d) is complete and totally bounded.
Let xn be a sequence in X. We wish to show that it has a convergent
subsequence.
The key observation is contained in this paragraph. Suppose that A is a
subset of X such that xn ∈ A for in¬nitely many values of n and suppose
> 0. Since X is totally bounded we can ¬nd a ¬nite set of open balls B1 ,
B2 , . . . , BM , each of radius , such that M Bm = X. It follows that
m=1
M
m=1 A © Bm = A, and, for at least one of the balls Bm , it must be true that
xn ∈ A © Bm for in¬nitely many values of n. Thus we have shown that that
there is an open ball of radius such that xn ∈ A © B for in¬nitely many
values of n.
It follows that we can construct inductively a sequence of open balls B1 ,
B2 , . . . such that Br has radius 2’r and xn ∈ r Bs for in¬nitely many
s=1
values of n [r = 1, 2, . . . ]. Pick n(1) < n(2) < n(3) < . . . such that xn(r) ∈
r ’r+1
s=1 Bs . If p, q > r, then xn(p) , xn(q) ∈ Br , and so d(xn(p) , xn(q) ) < 2 .
Thus the sequence xn(r) is Cauchy and, since X is complete, it converges.
Exercise 11.2.8. Show that the open interval (0, 1) with the usual Euclidean
metric is totally bounded but does not have the Bolzano-Weierstrass property.
Exercise 11.2.9. Use the completeness of the Euclidean norm on Rm and
Theorem 11.2.7 to show that a closed bounded subset of Rm with the usual
Euclidean norm has the Bolzano-Weierstrass property. (Thus we have an al-
ternative proof of Theorem 4.2.2, provided we do not use Bolzano-Weierstrass
to prove the completeness of Rm .)
Exercise 11.2.10. Show that a metric space (X, d) is totally bounded if and
only if every sequence in X has a Cauchy subsequence.
We shall not make a great deal of use of the concept of the Bolzano-Weierstrass
property in the remainder of this book. Thus, although the results that fol-
low are quite important, the reader should treat them merely as a revision
exercise for some of the material of Section 4.3.
Our ¬rst result is a generalisation of Theorem 4.3.1.
De¬nition 11.2.11. If (X, d) is a metric space, we say that a subset A has
the Bolzano-Weierstrass property2 if the metric subspace (A, dA ) (where dA
is the restriction of the metric d to A) has the Bolzano-Weierstrass property.
2
It is more usual to say that A is compact. However, although the statement ˜A has
the Bolzano-Weierstrass property™ turns out to be equivalent to ˜A is compact™ for metric
spaces, this is not true in more general contexts.
275
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 11.2.12. (i) Let (X, d) and (Z, ρ) be metric spaces. Show that, if
K is a subset of X with the Bolzano-Weierstrass property and f : X ’ Z is
continuous, then f (K) has the Bolzano-Weierstrass property.
(ii) Let (X, d) be a metric space with the Bolzano-Weierstrass property
and let Rp have the usual Euclidean norm. Show, by using part (i), or oth-
erwise, that, if f : X ’ Rp is a continuous function, then f (K) is closed and
bounded.

Exercise 11.2.13. State and prove the appropriate generalisation of Theo-
rem 4.3.4.

Exercise 11.2.14. (This generalises Exercise 4.3.8.) Let (X, d) be a metric
space with the Bolzano-Weierstrass property. Show that if K1 , K2 , . . . are
closed sets such that K1 ⊇ K2 ⊇ . . . , then ∞ Kj = ….
j=1

The following is a natural generalisation of De¬nition 4.5.2.

De¬nition 11.2.15. Let (X, d) and (Z, ρ) be metric spaces. We say that a
function f : X ’ Z is uniformly continuous if, given > 0, we can ¬nd a
δ( ) > 0 such that, if x, y ∈ X and d(x, y) < δ( ), we have

ρ(f (x), f (y)) < .

The next exercise generalises Theorem 4.5.5.

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-
edge of probability. If you know any of the terms Tchebychev inequality,
central limit theorem or Cauchy distribution, discuss how they apply here.
In any case, I hope I have demonstrated that when talking about things
like
X1 + X 2 + · · · + X n

we expect the interchange of limits only to work in exceptional (but therefore
profoundly interesting) cases.
Exercise 11.4.15. As I have already noted, we gave a similar example to
Example 11.4.12 in Exercise 5.3.2. We followed that by a dominated con-
vergence theorem for sums (Lemma 5.3.3). Can you formulate a similar
dominated convergence theorem for integrals? A possible version is given as
Exercise K.218.
Traditionally, Theorem 11.4.10 is always paired with the following result
which is proved by showing that it is really a disguised theorem on integra-
tion!
285
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Theorem 11.4.16. Suppose that fn : [a, b] ’ R is di¬erentiable on [a, b]
with continuous derivative fn (we take the one-sided derivative at end points
as usual). Suppose that fn (x) ’ f (x) as n ’ ∞ for each x ∈ [a, b] and
suppose that fn converges uniformly to a limit F on [a, b]. Then f is di¬er-
entiable with derivative F .

First proof. Since fn is continuous and fn ’ F uniformly, F is continuous
and Theorem 11.4.10 tells us that
t t
fn (x) dx ’ F (x) dx
c c

as n ’ ∞ for all t, c ∈ [a, b]. By the fundamental theorem of the calculus
t
(in the form of Theorem 8.3.11), we know that c fn (x) dx = fn (t) ’ fn (c),
and so
t
f (t) ’ f (c) = F (x) dx.
c

Since F is continuous, another application of the fundamental theorem of
the calculus (this time in the form of Theorem 8.3.6) tells us that f is di¬er-
entiable with

f (t) = F (t)

as required.
This proof is easy but rather roundabout. We present a second proof
which is harder but much more direct.
Second proof of Theorem 11.4.16. Our object is to show that |f (x + h) ’
f (x) ’ F (x)h| decreases faster than linearly as h ’ 0 [x, x + h ∈ [a, b]]. We
start in an obvious way by observing that

|f (x + h) ’ f (x) ’ F (x)h| ¤ |fn (x + h) ’ fn (x) ’ fn (x)h| (1)
+ | f (x + h) ’ f (x) ’ F (x)h ’ fn (x + h) ’ fn (x) ’ fn (x)h |.

The ¬rst term in the inequality can be estimated by the mean value inequality

|fn (x + h) ’ fn (x) ’ fn (x)h| ¤ |h| sup |fn (x + θh) ’ fn (x)|. (2)
0<θ<1

To estimate sup0<θ<1 |fn (x + θh) ’ fn (x)| we reverse the argument of the
˜ /3 theorem™ (Theorem 11.3.6). We know that F is continuous because it is
the uniform limit of continuous functions. Thus, given > 0 we can ¬nd a
286 A COMPANION TO ANALYSIS

δ( ) > 0 such that |F (y) ’ F (x)| < /3 whenever y ∈ [a, b] and |x ’ y| < δ( ).
Choosing an N ( ) such that fn ’ F ∞ < /3 for all n ≥ N ( ), we have
|fn (y) ’ fn (x)| ¤ |fn (y) ’ F (y)| + |F (y) ’ F (x)| + |F (x) ’ fn (x)| (3)
¤ 2 fn ’ F ∞ + |F (y) ’ F (x)| < 2 /3 + /3 =
for all y ∈ [a, b] with |x ’ y| < δ( ) and all n ≥ N ( ). Using this result, we
see that inequality (2) gives
|fn (x + h) ’ fn (x) ’ fn (x)h| ¤ |h|. (2 )
so that inequality (1) gives, in turn,
|f (x + h) ’ f (x) ’ F (x)h| (1 )
¤ | f (x + h) ’ f (x) ’ F (x)h ’ fn (x + h) ’ fn (x) ’ fn (x)h | + |h|,
for all x + h ∈ [a, b] with |h| < δ( ), and all n ≥ N ( ). Allowing n ’ ∞, we
obtain
|f (x + h) ’ f (x) ’ F (x)h| ¤ |h|, (1 )
for all x + h ∈ [a, b] with |h| < δ( ), and this is what we want.
One major advantage of the second proof is that it generalises easily to
many dimensions.
Exercise 11.4.17. Let „¦ be an open set in Rm . Suppose that fn : „¦ ’ Rp is
a di¬erentiable function on „¦ with continuous derivative Dfn . Suppose that
fn (x) ’ f (x) as n ’ ∞ for each x ∈ „¦ and suppose that there is a function
˜ : „¦ ’ L(Rm , Rp ) (that is ˜ is a function on „¦ whose values are linear
maps from Rm to Rp ) such that
sup ˜(x) ’ Dfn (x) ’ 0
x∈„¦

(that is Dfn converges uniformly to ˜ in the operator norm) as n ’ ∞.
Then f is di¬erentiable with derivative ˜.
Although Exercise 11.4.17 is not hard, it provides a useful exercise in
understanding notation. We note, for example, that Dfn is a function on
„¦ whose values are linear maps from Rm to Rp . The statement that Dfn is
continuous must therefore be interpreted as
Dfn (x + h) ’ Dfn (x) ’ 0
is the Euclidean norm on Rm and
as h 2 ’ 0 for all x ∈ „¦, where 2
is the operator norm on L(Rm , Rp ).
287
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise 11.4.18. (Easy but optional) Rewrite Exercise 11.4.17 as a theo-
rem on di¬erentiation of functions between general normed vector spaces in
accordance with De¬nition 10.4.19. Check that essentially the same proof
works in this more general case.
The reader knows how to turn results on the limits of sequences into
results on in¬nite sums and vice versa (see Exercise 4.6.7 if necessary). Ap-
plied to Theorems 11.4.10 and 11.4.16, the technique produces the following
results.
Theorem 11.4.19. (Term by term integration.) Let gj : [a, b] ’ R be
continuous. If n gj converges uniformly as n ’ ∞, then
j=1

b∞ ∞ b
gj (x) dx = gj (x) dx.
a a
j=1 j=1

Theorem 11.4.20. (Term by term di¬erentiation.) Let gj : [a, b] ’ R
be di¬erentiable with continuous derivative. If n gj (x) converges for each
j=1
x and j=1 gj converges uniformly as n ’ ∞, then ∞ gj is di¬erentiable
n
j=1
and
∞ ∞
d
gj (x) = gj (x).
dx j=1 j=1

As a typical example of the use of Theorem 11.4.16, we use it to extend
Theorem 8.4.3 on di¬erentiation under the integral to a much more useful
result.
Theorem 11.4.21. (Di¬erentiation under an in¬nite integral.) Let
(c , d ) ⊇ [c, d]. Suppose g : [0, ∞) — (c , d ) ’ R is continuous and that the
partial derivative g,2 exists and is continuous. Suppose further, that there
exists a continuous function h : [0, ∞) — (c, d) ’ R with |g,2 (x, y)| ¤ h(x)

for all (x, y) and such that 0 h(x) dx exists and is ¬nite. Then, if G(y) =

g(x, y) dx exists for all y ∈ (c, d), we have G di¬erentiable on (c, d) with
0

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

Proof. Note that H(y) = 0 g,2 (x, y) dx exists by comparison (see Lemma 9.2.4).
n
Set Gn (y) = 0 g(x, y) dx. By Theorem 8.4.3, Gn is di¬erentiable with
n
Gn (y) = g,2 (x, y) dx.
0
288 A COMPANION TO ANALYSIS

Since
∞ ∞
|Gn (y) ’ H(y)| = g,2 (x, y) dx ¤ h(x) dx ’ 0,
n n

we see that Gn converges uniformly to H on (c, d). By hypothesis, Gn (y) ’
G(y) on (c, d) so, by Theorem 11.4.16, G is di¬erentiable with derivative H
on (c, d). This is the required result.
You should be careful when using this theorem to check that the hypotheses
actually apply. Exercise K.216 illustrates what can go wrong if we do not
prevent ˜escape to in¬nity™.


11.5 Power series
The object of this section and the next is to show how the notion of uniform
convergence is used in two topics of practical importance.
We make use of a result which is too trivial to constitute a theorem but
too useful to leave unnamed.
Lemma 11.5.1. (The Weierstrass M-test.) Consider functions fn :
E ’ C. Suppose that we can ¬nd positive real numbers Mn such that

|fn (x)| ¤ Mn for all x ∈ E and all n ≥ 1. If 1 Mn converges then

1 fn converges uniformly on E.

Proof. Let > 0. By the easy part of the general principle of convergence we
can ¬nd an N0 ( ) such that m Mn ¤ for all m ≥ n ≥ N0 ( ). It follows
r=n
that
m m m
fn (x) ¤ |fn (x)| ¤ Mn ¤
r=n r=n r=n

for all m ≥ n ≥ N0 ( ) and all x ∈ E. By the general principle of uniform
convergence (Theorem 11.4.4), ∞ fn converges uniformly on E.
1

This book has been mainly about real analysis. When we talked about
functions from Rn to Rm we made a point of the fact that we cannot divide
a vector by a vector. There are, however, two exceptions to this rule. The
¬rst is that R, itself, is a vector space where division is permitted. The
second is that, if we give R2 the algebraic structure of C, we again obtain a
system in which division is permitted. This enables us to develop a theory
of di¬erentiation for functions f : C ’ C running in parallel with the theory
of one dimensional real di¬erentiation. Note that we use the usual metric
d(z1 , z2 ) = |z1 ’ z2 | throughout.
289
Please send corrections however trivial to twk@dpmms.cam.ac.uk

De¬nition 11.5.2. Let „¦ be an open set in C and let z ∈ „¦. We say that
f : „¦ ’ C is di¬erentiable at z with derivative f (z) if
f (z + h) ’ f (z)
’ f (z) ’ 0
h
as h ’ 0.
Exercise 11.5.3. Check that the de¬nition is equivalent to the statement

f (z + h) = f (z) + f (z)h + (h)|h|

for z + h ∈ „¦, where (h) ’ 0 as h ’ 0.
Exercise 11.5.4. Convince yourself that the elementary theory of complex
di¬erentiation is the same as that of real di¬erentiation. For example, you
could run through the general results leading to the following result:-
Let ar ∈ C [0 ¤ r ¤ N ], bs ∈ C [0 ¤ s ¤ M ’ 1], bM = 1 and let
P (z) = N ar z r , Q(z) = M bs z s . Then „¦ = {z ∈ C : Q(z) = 0} is
r=0 s=0
open and the quotient P/Q : C ’ C is everywhere di¬erentiable.
Exercise 11.5.4 is routine and not meant to be taken very seriously. The
next two results are important in their own right and provide an opportunity
to review the content and proof of two important earlier results.
Exercise 11.5.5. (A mean value inequality.) Suppose that „¦ is an open
set in C and that f : „¦ ’ C is di¬erentiable at all points of „¦. Suppose,
further, that the straight line segment

L = {(1 ’ t)z1 + tz2 : 0 ¤ t ¤ 1}

joining z1 and z2 lies in „¦ and that

|f (z)| ¤ K

for all z ∈ L.
Explain why we can ¬nd a real θ such that eiθ (f (z2 ) ’ f (z1 )) is real and
positive. Show that the function F : [0, 1] ’ R given by

eiθ (f ((1 ’ t)z1 + tz2 ) ’ f (z2 ))
F (t) =

is di¬erentiable on (0, 1) and ¬nd its derivative. By applying some form of
mean value theorem or mean value inequality (many versions will work) to
F , show that

|f (z2 ) ’ f (z1 )| ¤ K|z2 ’ z1 |.
290 A COMPANION TO ANALYSIS

Exercise 11.5.6. By using the ideas of the second proof of Theorem 11.4.16,
prove the following result.
Let „¦ be an open set in C. Suppose that fn : „¦ ’ C is di¬erentiable at
all points of „¦ with continuous derivative fn . Suppose that fn (z) ’ f (z) as
n ’ ∞ for each z ∈ „¦, and suppose that there is a function g : „¦ ’ C such
that

sup |g(z) ’ fn (z)| ’ 0
z∈„¦


(that is fn converges uniformly to g). Then f is di¬erentiable at all points
of „¦ with derivative g.

To use this result we need to recall the de¬nition of the radius of conver-
gence of a power series and the proof that makes the de¬nition possible. The
reader can look these things up on page 71 but it would, I think, be helpful
to redo the work as an exercise.

Exercise 11.5.7. (i) Suppose that an ∈ C. Show that if n
n=0 an z0 con-
verges for some z0 ∈ C then ∞ an z n converges for all z ∈ C with |z| < |z0 |.
n=0
(ii) Use (i) to show that either ∞ an z n converges for all z (in which
n=0
case we say the series has in¬nite radius of convergence) or there exists an
R ≥ 0 such that ∞ an z n converges for |z| < R and diverges for |z| > R
n=0
(in which case we say that the series has radius of convergence R).

We can now improve this result slightly.

Lemma 11.5.8. Suppose that an ∈ C and ∞ an z0 has radius of conver-
n
n=0
gence R. If 0 ¤ ρ < R, then ∞ an z n converges uniformly for all |z| ¤ ρ.
n=0

Proof. Choose z0 with ρ < |z0 | < R. We know that ∞ an z0 converges
n
n=0
n
and so an z0 ’ 0 as n ’ ∞. It follows that there exists an M such that
n
|an z0 | ¤ M for all n ≥ 1. Thus, if |z| ¤ ρ, we have
n
|z|n ρ
n n
|an z | = |an z0 | ¤M
|z0 |n |z0 |

an z n converges uniformly for all |z| ¤
so, by the Weierstrass M-test, n=0
ρ.

It is very important to bear in mind the following easy example.

Exercise 11.5.9. Show that ∞ z n has radius of convergence 1 but that
n=0
∞ n
n=0 z does not converge uniformly for |z| < 1.
291
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Thus a power series converges uniformly in any disc centre 0 of strictly
smaller radius than the radius of convergence but, in general, the condition
strictly smaller cannot be dropped.
The next result is also easy to prove.

Lemma 11.5.10. Suppose that an ∈ C and ∞ an z n has radius of con-
n=0
∞ n’1
vergence R. Then n=1 nan z also has radius of convergence R.

Proof. Let |w| < R. Choose w0 ∈ C with |w| < |w0 | < R and ρ ∈ R with
|w| < ρ < |w0 |. We know that ∞ an w0 converges so, arguing as before,
n
n=0
n
there exists an M such that |an w0 | ¤ M for all n ≥ 0. Thus
n’1
n’1
n nρ ρ
’1
¤ |w0 |’1 M n
n’1
|nan ρ | = |w0 | |an w0 | ’ 0.
n’1
|w0 | |w0 |

(We use the fact that, if |x| < 1, then nxn ’ 0 as n ’ ∞. There are
many proofs of this including Exercise K.9.) Thus we can ¬nd an M such
that |nan ρn’1 | ¤ M for all n ≥ 1. Our usual argument now shows that
∞ n’1
n=1 nan w converges.
We have shown that the radius of convergence of ∞ nan z n’1 is at least
n=1
R. An easier version of the same argument shows that if ∞ nan z n’1 has
n=1
radius of convergence S, then the radius of convergence of ∞ an z n is at
n=0
∞ n’1
least S. Thus the radius of convergence of n=1 nan z is exactly R.

We can now combine Exercise 11.5.6, Lemma 11.5.8 and Lemma 11.5.10
to obtain our main result on power series.

Theorem 11.5.11. Suppose that an ∈ C and ∞ an z n has radius of con-
n=0
vergence R > 0. Set „¦ = {z : |z| < R} (if R = ∞, then „¦ = C) and de¬ne
f : „¦ ’ C by

an z n .
f (z) =
n=0


Then f is everywhere di¬erentiable on „¦ and

nan z n’1 .
f (z) =
n=1


More brie¬‚y, a power series can be di¬erentiated term by term within its
circle of convergence.
292 A COMPANION TO ANALYSIS

Proof. We wish to show that, if |w| < R, then f is di¬erentiable at w with
the appropriate derivative. To this end, choose a ρ with |w| < ρ < R. By
Lemma 11.5.8, we know that

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