# Uniform Limit of Analytic Functions is Analytic

## Theorem

Let $U$ be an open subset of $\C$.

Let $\left\{ {f_n}\right\}_{n \mathop \in \N}$ be a sequence of analytic functions $f_n : U \to \C$.

Let $\left\{ {f_n}\right\}$ converge locally uniformly to $f$ on $U$.

Then $f$ is analytic.

## Proof

By Equivalence of Local Uniform Convergence and Compact Convergence, $f_n$ converges to $f$ locally uniformly on $U$.

Then for any $z \in U$, there is an $\epsilon > 0$ so that:

$B_\epsilon \left({z}\right) \subset U$

and $f_n$ converges uniformly on $B_\epsilon \left({z}\right)$.

Let $\gamma$ be any simple closed curve in $B_\epsilon \left({z}\right)$.

Since $f_n \to f$ uniformly on $\gamma$ (because $\gamma \subset B_\epsilon \left({z}\right)$), we have:

$\displaystyle \lim_{n \to \infty} \int_\gamma f_n \left({z}\right) \, \mathrm d z = \int_\gamma f \left({z}\right) \, \mathrm d z$

Since each $f_n$ is analytic, we have that:

$\displaystyle \forall n \in \N: \int_\gamma f_n \left({z}\right) \, \mathrm d z = 0$

So we conclude also that

$\displaystyle \int_\gamma f \left({z}\right) \, \mathrm d z = 0$

Since $\gamma$ was arbitrary, we have by Morera's Theorem that $f$ is analytic in $B_\epsilon \left({z}\right)$.

Since $z$ was arbitrary, $f$ is analytic on all of $U$.

$\blacksquare$