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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
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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

N
an z n ’ f (z)
n=0


uniformly for |z| ¤ ρ. By Lemma 11.5.8, we know that there exists a function
g : „¦ ’ C such that N nan z n’1 ’ g(z) as N ’ ∞ for all z ∈ „¦. Using
n=1
Lemma 11.5.8 again, we have

N
nan z n’1 ’ g(z)
n=1


uniformly for |z| ¤ ρ. Since

N N
d
an z n nan z n’1 ,
=
dz n=0 n=1


Exercise 11.5.6 now tells us that f is di¬erentiable in {z : |z| < ρ} with

nan z n’1 .
f (z) =
n=1

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