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 $

This implies that $f$ 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,

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \vert f'_n(a) - f'(a) \vert\)	\(=\)	\(\displaystyle \frac 1 {2 \pi}\left\vert \int_{\partial D} \frac{f_n(z) - f(z)}{(z-a)^2}\ dz\right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \frac{r}{r^2} \sup_{z \in \partial D} \vert f_n(z) - f(z) \vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By the Estimation Lemma for complex integrals
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac 1r \sup_{z \in \partial D} \vert f_n(z) - f(z) \vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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 $S$.

$\blacksquare$