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Consider an open connected set $\Omega\subset \mathbb{C}$, and $f_n\subset H(\Omega)$. Suppose $f(z)=\lim_{n\to\infty}f_n(z)$ exists and $|f_n(z)|\leq M$ for all $z\in \Omega$. Show that $$\lim_{n\to\infty}\sup_{z\in K}|f_n(z)-f(z)|=0$$ for any compact $K\subset \Omega$.

Note: If correct, I am sure this is a well known result; but I don't see why it's correct, and I haven't found a reference, either.

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    $\begingroup$ The result is a consequence of Montel's theorem. $\endgroup$ – saz Sep 10 '13 at 12:38
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    $\begingroup$ Montel's Theorem only gives normality (which shows that a subsequence of $\{f_n\}$ converges uniformly), but we need to prove something stronger - that the original sequence converges uniformly. $\endgroup$ – pre-kidney Sep 10 '13 at 15:20
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    $\begingroup$ Since pointwise convergence is assumed, any subsequence of $f_n$ converges to $f$, hence $f_n \to f$ uniformly. $\endgroup$ – saz Sep 10 '13 at 15:42
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Since the sequence of maps $f_n$ is uniformly bounded, the Cauchy integral formula will give you a uniform bound on derivatives $|f_n'(z)|\le C_K$ on every compact $K\subset \Omega$. Hence, the family of restricted maps $f_n|K$ is equicontinuous and it follows from the Arzela-Ascoli theorem that for every compact our sequence of functions contains a uniformly convergent subsequence. Since the original sequence converges point-wise, every convergent subsequence has the same limit. Now, it is a general fact of pointset topology that if $x_n$ is a sequence in compact topological space and every convergent subsequence in $x_n$ has the same limit, then the sequence $x_n$ also converges. We apply this fact to the sequence of restrictions of our functions to a given compact subset $K$ in $\Omega$, where $X$ is the space of $L$-Lipschitz functions on $K$ with topology of uniform convergence. It now follows that our sequence of functions converges uniformly on each compact subset $K$ implying the claim.

The same argument works with Montel's theorem, see the comments.

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  • $\begingroup$ Arzela-Ascoli only yields a subsequence of the $\{f_n\}$ with this property (and proves nothing about the original sequence $\{f_n\}$) so this approach doesn't seem to yield much. $\endgroup$ – pre-kidney Sep 10 '13 at 15:08
  • $\begingroup$ Since the original sequence converges pointwise, all subsequential limits will be the same. Thus, AA theorem implies the result. $\endgroup$ – Moishe Kohan Sep 10 '13 at 15:28
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    $\begingroup$ Maybe I'm being dense, but I'm not following your explanation. Can you formulate this implication in more precise mathematical language (or provide a rigorous proof of what you are describing?) Also, it would be nice if this was added to your original answer (because it is a necessary step in the proof.) $\endgroup$ – pre-kidney Sep 10 '13 at 15:59
  • $\begingroup$ Ok, I will give more details. $\endgroup$ – Moishe Kohan Sep 10 '13 at 17:55
  • $\begingroup$ Thank you for your explanation! $\endgroup$ – pre-kidney Sep 13 '13 at 6:15

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