Uniformly bounded sequence of holomorphic functions converges uniformly 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.
 A: 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. 
