Uniform Limit of Analytic Functions is Analytic
 | This article needs to be linked to other articles. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
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 $U$.
$\blacksquare$