Uniform Limit of Analytic Functions is Analytic

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

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

Suppose that for each compact subset $D \subseteq S$, $\{f_n\}$ converges uniformly to some function $f:S \to \C$.

Then $f$ is analytic, and the sequence $\{f'_n\}_{n \in \N}$ converges uniformly to $f'$.

Proof
Since each $f_n$ is analytic on $S$ we have Cauchy's Integral Formula:


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

for any comapct $a \in D \subseteq S$.

Because of uniform convergence, we may pass to the limit, obtaining:


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

For such $D$ and $a$.

Thus by the Converse to Cauchy's Theorem, $f$ is analytic inside $D$.

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

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 $S$.