# Uniform boundedness imply uniform convergence?

If ${f_n}$ is a uniformly bounded sequence of holomorphic functions in $\Omega$ such that ${f_n(z)}$ converges for every $z \in \Omega$, is the convergence is uniform on every compact subset of $\Omega$?

I'm trying to prove using Dominated Convergence Theorem with Cauchy's Integral Formula. For any compact $K \subset \Omega$, let $\gamma = \partial K$. Then for any $z \in K$, by the CIF, $$|f_n(z) - f_m(z)| \leq \frac{1}{2\pi i}\int_{\gamma}\left|\frac{f_n(w)-f_m(w)}{w-z}\right|\:dz.$$

Now since we are talking in compact sets, $|f_n(z)| \leq B \in L^1$, so by applying Dominated Convergence Theorem, and by taking the limit as $m \to \infty$, $$|f_n(z)-f(z)| \leq \frac{1}{2\pi i}\int_{\gamma}\left|\frac{f_n(w)-f(w)}{w-z}\right|\:dz.$$

But by the pointwise convergence, I am trying to claim that RHS can be made sufficiently small for large enough $n$. Am I on the right track?

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Can you use something like Ascoli's theorem? I am wondering if you can prove the equicontinuity of $\{f_n\}$ by Cauchy's formula. – Siminore Sep 14 '12 at 11:39
I did run across Arzela-Ascoli's Theorem, but we haven't technically proven it yet, so I can't use it. But I assume I can mimic the proof of the A-A Theorem. – aviness Sep 14 '12 at 11:44

## 1 Answer

Here is a powerful convergence theorem.

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Wow that is an intense convergence theorem! – aviness Sep 14 '12 at 11:46
I still believe you can solve your problem more easily. But powerful theorems are often useful. – Siminore Sep 14 '12 at 11:48
You were right on the money with Ascoli's theorem. Montel's theorem (stated in a somewhat cryptic fashion on Wikipedia) can be proved using Arzelà-Ascoli. This is done e.g. in Ahlfors. – t.b. Sep 14 '12 at 11:53