1550251515-Classical_Complex_Analysis__Gonzalez_

198 Chapter 4

Example Consider the sequence defined on C by

Clearly,

since

un(z)=(2+n~l)~, n = 0, 1,2, ...

un(z)-+ 2z, z EC, asn-+oo

\un(z)-2z\ = ~l < €

n+

provided that n > \z\/€ - 1 = N,,z. Thus we have pointwise convergence

in C. However, on a disk \z\ :::; R we have

\un(z) - 2z\ = _\_z\_ :::; ___!i_ < €

n+l n+l

whenever n > .R/€ - 1 = N,, so the convergence is uniform on \z\ :::; R,

and, more generally, on any bounded subset of C.

There is a Cauchy condition for uniform convergence, as stated in the

following theorem

Theorem 4.15 A necessary and sufficient condition for a sequence

{ un(z)} to converge uniformly on a set B is that for every € > 0 there

is an N (depending only on €) such that

\um(z) -Un(z)\ < €

for all m and n both > N and all z E B.

Proof The necessity of the condition follows at once, as in Theorem 2.25,

by a simple application of the triangle inequality.

To prove the sufficiency part, we first note that the limit function

f(z) exists, since the Cauchy condition for numerical sequences will hold

pointwise on the set B. Now, if in

\um(z) - un(z)\ < €

we keep n fixed and let m -+ oo, we get

\J(z) -un(z)\ :::; €

for n > N and all z E B. Hence un(z) ~ f(z) as n -+ oo.

Theorem 4.16 Suppose that:

1. un(z) ~ f(z) on an infinite set B.


2. zo is an accumulation point of B.