Uniform Limit of Analytic Functions is Analytic

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

Let $\{f_n\}_{n \in \N}$ be a sequence of analytic functions $U \to \C$ converging pointwise to $f \colon U \to \C$.

Suppose that for each compact subset $K \subseteq U$, $\{f_n\}$ converges uniformly on $K$.

Then $f$ is analytic, and the sequence $\{f'_n\}_{n \in \N}$ converges uniformly to $f'$ on all compact subsets of $U$.

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 $\varepsilon > 0$ so that $B_\varepsilon \left( z \right) \subset U$ and $f_n$ converges uniformly on $B_\varepsilon \left( z \right)$.

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

Since $f_n \to f$ uniformly on $\gamma$ (because $\gamma \subset B_\varepsilon \left( z \right)$), 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
 * $\displaystyle \int_\gamma f_n \left( z \right) \mathrm d z = 0$

for all $n$, and 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 $B_\varepsilon \left( z \right)$.

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

For the statement for the derivative, let $D$ be a disk of radius $r$ about $a$, contained in $U$.

We have Cauchy's Integral Formula for Derivatives


 * $\displaystyle f'_n(a) = \frac 1 {2 \pi i} \int_{\partial D} \frac{f_n(z)}{(z-a)^2} \ dz $

and


 * $\displaystyle f'(a) = \frac 1 {2 \pi i} \int_{\partial D} \frac{f(z)}{(z-a)^2} \ dz $

Therefore,

Now the $f_n$ tend uniformly to $f$, and we can bound $r$ away from zero.

It follows that $f_n' \to f'$ uniformly in each compact disk contained in $U$.