# Uniform Limit of Analytic Functions is Analytic

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## Theorem

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

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

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

Then $f$ is analytic.

## Proof

Let $z \in U$.

By Definition of Locally Uniform Convergence, there is an $\epsilon > 0$ so that:

- $\map {B_\epsilon} z \subset U$

and $f_n$ converges uniformly on $\map {B_\epsilon} z$.

Let $\gamma$ be any simple closed curve in $\map {B_\epsilon} z$.

Then $f_n \to f$ uniformly on $\gamma$, because $\gamma \subset \map {B_\epsilon} z$.

Thus by Definite Integral of Limit of Uniformly Convergent Sequence of Integrable Functions, we have:

- $\ds \lim_{n \mathop \to \infty} \int_\gamma \map {f_n} z \rd z = \int_\gamma \map f z \rd z$

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

- $\ds \forall n \in \N: \int_\gamma \map {f_n} z \rd z = 0$

So we conclude also that:

- $\ds \int_\gamma \map f z \rd z = 0$

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

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

$\blacksquare$