# Uniform Convergence

Consider the sequence of functions $u_{n}(x)=x^{n}$ , on the unit interval $J=[0,1]$ . For each fixed $x\in J$ , the sequence $u_{n}(x)$ is convergent:

$u_{n}(x)\to u(x)=\left\{{\begin{array}{l}0{\mbox{ if }}0\leqslant x<1;\\1{\mbox{ if }}x=1.\end{array}}\right.$ as $n\to \infty$ .

For each fixed $n$ , $u_{n}(x)$ is continuous, but the limiting function $u(x)$ is not. This is an unfortunate state of affairs which is remedied by the introduction of . This is a stronger notion of convergence for a sequence of functions, sufficient to guarantee, in particular, that continuity is preserved in the passage to limits.

Let $\Omega$ be any set (you may think that it is a subset of $\mathbb {C}$ or $\mathbb {R}$ ) and let $u_{n}:\Omega \to \mathbb {C}$ be sequence of functions defined in $\Omega$ .

Definition 1
The sequence $u_{n}$ converges uniformly to the function $u:\Omega \to \mathbb {C}$ if: given $\varepsilon >0$ , there exists $N_{0}$ (depending upon $\varepsilon$ ) such that $n>N_{0}\Longrightarrow |u_{n}(z)-u(z)|<\varepsilon {\mbox{ for all }}z\in \Omega .$ Remark 2

An equivalent formulation of Uniform convergence is

$n>N_{0}\Longrightarrow \sup _{z\in \Omega }|u_{n}(z)-u(z)|<\varepsilon .$ If $\Omega \subset \mathbb {C}$ and $f:\Omega \to \mathbb {C}$ is any function, we say that $f$ is continuous at $z_{0}$ in $\Omega$ if given $\varepsilon >0$ , there exists $\delta >0$ such that

$z\in \Omega {\mbox{ and }}|z-z_{0}|<\delta \Longrightarrow |f(z)-f(z_{0})|<\varepsilon .$ Then $f$ is said to be continuous in $\Omega$ if it is continuous at every point of $\Omega$ .

Now we show that 'the uniform limit of a sequence of continuous functions is again continuous'.

Theorem 1

Let $u_{n}:\Omega \to \mathbb {C}$ be continuous for each $n$ and suppose that $u_{n}\to u$ uniformly in $\Omega$ as $n\to \infty$ . Then $u$ is continuous in $\Omega$ .

Proof

Pick $z_{0}\in \Omega$ . We need to estimate

 $|u(z)-u(z_{0})|$ for all $z\in \Omega$ close to $z_{0}$ .  For this, use


{\begin{aligned}|u(z)-u(z_{0})|&=&|u(z)-u_{n}(z)+u_{n}(z)-u_{n}(z_{0})+u_{n}(z_{0})-u(z_{0})|\\&\leqslant &|u(z)-u_{n}(z)|+|u_{n}(z)-u_{n}(z_{0})|+|u_{n}(z_{0})-u(z_{0})|.\end{aligned}} Given $\varepsilon >0$ , by uniform convergence, there exists $N_{0}$ so that
$n>N_{0}\Longrightarrow \sup _{z\in \Omega }|u_{n}(z)-u(z)|<\varepsilon /3.$ Substituting this into Uniform convergence , we obtain
$|u(z)-u(z_{0})|=\varepsilon /3+|u_{n}(z)-u_{n}(z_{0})|+\varepsilon /3.$ Now $n>N_{0}$ , and find, by continuity of $u_{n}$ at $z_{0}$ , a $\delta >0$ such that
$|u_{n}(z)-u_{n}(z_{0})|<\varepsilon /3{\mbox{ if }}z\in \Omega {\mbox{ and }}|z-z_{0}|<\delta .$ Combining Uniform convergence and Uniform convergence we see that for this same $\delta$ ,
$|u(z)-u(z_{0})|<\varepsilon {\mbox{ for all }}z\in \Omega ,|z-z_{0}|<\delta$ which proves that $u$ is continuous.

Example 3

We can see that the example we started with fails to be uniformly convergent. Indeed, with $u_{n}$ and $u$ as in Uniform convergence , if $x<1$ , we have $|u_{n}(x)-u(x)|=x^{n}$ and this tends to $1$ as $x\to 1$ . Thus $\sup _{x\in J}|u_{n}(x)-u(x)|=1$ and so $u_{n}$ does not converge uniformly to $u$ in this case.

## Uniformly convergent series

As usual, we pass from uniformly convergent sequences to uniformly convergent series via partial sums. That is, if $f_{n}:\Omega \to \mathbb {C}$ is a sequence of functions, then we say that the series $\sum _{n=0}^{\infty }f_{n}$ is uniformly convergent in $\Omega$ if the corresponding sequence of partial sums

$F_{n}=\sum _{j=0}^{n}f_{j}$ is uniformly convergent in $\Omega$ . There is a for uniform convergence of sums, with the following statement:

Theorem 2

Let $f_{n}:\Omega \to \mathbb {C}$ be a sequence of functions. The series $\sum f_{n}$ converges uniformly in $\Omega$ if and only if:

Given any $\varepsilon >0$ , there exists $N_{0}$ depending upon $\varepsilon$ , such that

for all $n>m>N_{0}$ , we have

$\sup _{z\in \Omega }|\sum _{j=m}^{n}f_{j}(z)|<\varepsilon .$ Finally, we state the Weierstrass $M$ -test, which is a very effective way of checking uniform convergence of many of the series which arise in complex analysis.

Theorem 3

Let $f_{n}:\Omega \to \mathbb {C}$ be a sequence of functions. Suppose that there exist $M_{n}$ such that

$\sup _{z\in \Omega }|f_{n}(z)|\leqslant M_{n}$ and also
$\sum _{n=1}^{\infty }M_{n}<\infty .$ Then the series $\sum f_{n}$ converges absolutely and uniformly in $\Omega$ . In particular, if each of the $f_{n}$ is continuous, then so is $\sum _{n=1}^{\infty }f_{n}$ .

Proof

Use the Cauchy criterion. By the triangle inequality,

$|\sum _{j=m}^{n}f_{j}(z)|\leqslant \sum _{j=m}^{n}|f_{j}(z)|\leqslant \sum _{j=m}^{n}M_{j}$ for every $z\in \Omega$ . The Cauchy convergence criterion for the convergent series $\sum M_{n}$ states that, given $\varepsilon >0$ , there exists $N_{0}$ such that $n>m>N_{0}$ implies $\sum _{j=m}^{n}M_{j}<\varepsilon$ . Substituting this into Uniformly convergent series we have
$\sum _{j=m}^{n}|f_{j}(z)|<\varepsilon {\mbox{ for }}n>m>N_{0}{\mbox{ and all }}z\in \Omega$ as required for the use of Theorem~ Uniformly convergent series to get absolute and uniform convergence of $\sum f_{n}$ . The continuity statement follows from Theorem Uniform convergence .