# Uniform Convergence

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

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

For each fixed ${\displaystyle n}$, ${\displaystyle u_{n}(x)}$ is continuous, but the limiting function ${\displaystyle u(x)}$ 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 ${\displaystyle \Omega }$ be any set (you may think that it is a subset of ${\displaystyle \mathbb {C} }$ or ${\displaystyle \mathbb {R} }$) and let ${\displaystyle u_{n}:\Omega \to \mathbb {C} }$ be sequence of functions defined in ${\displaystyle \Omega }$.

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


Remark 2

An equivalent formulation of Uniform convergence is

${\displaystyle n>N_{0}\Longrightarrow \sup _{z\in \Omega }|u_{n}(z)-u(z)|<\varepsilon .}$

If ${\displaystyle \Omega \subset \mathbb {C} }$ and ${\displaystyle f:\Omega \to \mathbb {C} }$ is any function, we say that ${\displaystyle f}$ is continuous at ${\displaystyle z_{0}}$ in ${\displaystyle \Omega }$ if given ${\displaystyle \varepsilon >0}$, there exists ${\displaystyle \delta >0}$ such that

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

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

Theorem 1

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

Proof

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

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


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

Example 3

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

## Uniformly convergent series

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

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

Theorem 2

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

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

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

${\displaystyle \sup _{z\in \Omega }|\sum _{j=m}^{n}f_{j}(z)|<\varepsilon .}$

Finally, we state the Weierstrass ${\displaystyle M}$-test, which is a very effective way of checking uniform convergence of many of the series which arise in complex analysis.

Theorem 3

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

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

Proof

Use the Cauchy criterion. By the triangle inequality,

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