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 $n>N_{0}\Longrightarrow |u_{n}(z)-u(z)|<\varepsilon {\mbox{ for all }}z\in \Omega .$

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[edit | edit source]

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** .