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I want to prove that if $ f_n $ are analytic functions on a domain $ U \subset \mathbb{C} $ and $ f_n $ tends locally uniformly to $ f $ on $U$, then $f$ is analytic on $U$.

My thoughts:

I'd like to show that $ f $ is necessarily continuous on $ U $, and that for any simple closed curve $ \gamma $, $ \int_\gamma f(z) dz = 0 $. Then by Morera's theorem, $f$ is analytic.

Now $f_n $ tends locally uniformly to $ f$ on $U$ means that for any $ x \in U $, $ \exists r > 0 $ such that $ f_n $ tends uniformly on $ B(x;r) $ to $f$. I know that continuity is preserved under uniform convergence, and so for any $ x \in U $, $ f$ is continuous on some ball around $x$, and so is continuous at $x$. Thus $ f $ is continuous on $ U $.

Let $ \gamma $ be a simple, closed curve in $ U $. Then for any $ x $ on $ \gamma $, $\exists $ $ r > 0 $ such that $ f_n $ converges uniformly to $f$ on $ B(x;r_x) $. Let $ N_x $ be the associated 'magic uniform convergence number' (for lack of a better term) for the sequence on this ball. Now, the collection of these balls for all $ x $ on $\gamma$ forms a cover of $\gamma$, and by compactness there is a finite subcover, say $ \{ B(x_i;r_i), 1 \leq i \leq n \} $. Let $ N = \max_i N_{x_i} $. Then $ N $ gives us uniform convergence of the sequence $ f_n $ to $ f$ on $\gamma$.

Thus we can say $ \int_\gamma f(z) dz = \lim_{n\to \infty} \int_\gamma f_n(z) dz = 0 $, and so we're done.

Is this proof valid?


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yes, it's perfectly valid – user8268 May 18 '11 at 14:11
great, thanks :) – user938272 May 18 '11 at 14:14

As user8268 said (a while ago), the reasonning is valid.

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I see this is an old question and that it already has been answered. Let me just point out that you can restrict the argument involving Morera's Theorem to the same disk you used for continuity. In this disk the integral around any closed loop will be zero and therefore $f$ will be analytic throughout the disk. Since this applies for any $x \in U$ $f$ is analytic on $U$.

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