Composition of a holomorphic function with a normal family of holomorphic functions. Let $\Omega_1$ and $\Omega_2$ (open) $\subset \mathbb{C}$ and $\mathcal{F}$ a normal family of holomorphic functions in $\Omega_1$ such that $f(\Omega_1) \subset \Omega_2$ $\forall f \in \mathcal{F}$.
We want to prove that if $\Omega_3$ is a another open subset such that $\overline{\Omega_2} \subset \Omega_3$ and $g \in \mathcal{H}(\Omega_3)$ then $g \circ \mathcal{F}$=$\{g \circ f $ $ |$ $ f \in \mathcal{F}\}$ is a normal family.
My try:
Given a sequence of functions in $g \circ \mathcal{F}$ it'd be  
$\{$ $g \circ f_n$ $\}$. Also we know that $\mathcal{F}$ is normal so $\{$ $f_n$ $\}$ has a subsequence $\{$ $f_{n_k}$ $\}$ which is convergent to the holomorphic function $f$.
I want to show that $\{$g $\circ$ $f_{n_k}$ $\}$$\rightarrow$$g\circ f$
Given a compact subset $K \subset$ $\Omega_1$ and given $\epsilon >0$ i need to show  that $\exists$ $n_{k_0}$ st if $n_k \geq$ $n_{k_0}$ then
|$\{$g $\circ$ $f_{n_k}$ - $g \circ f$| $< \epsilon$ $\forall z \in K$
I want to use convergence of  $f_{n_k}$ and then use continuity of $g$ but i need uniform continuity of $g$ to make sure that it works for all $z \in K$
what should I do?
 A: The problem is really about controlling the images $f_n(K)$ as $n$ goes to infinity. If we can show that these sets are contained in a suitable compact subset inside $\Omega_2$, then we win by continuity of $g$.
Let's then take a sequence $g \circ f_n$ and a subsequence $f_{n_k}$ that converges uniformly to $f$ on any compact $K \subset \Omega_1$. Given such a compact set, pick an $\epsilon$ such that the $\epsilon$-neighborhood
$$
f(K)_{\epsilon} = \{ z \in \Omega_2 \mid \sup_{w \in f(K)} | z - w| \leq \epsilon \}
$$
is contained in $\Omega_3$ (which is possible because $f(K) \subset \overline{\Omega_2} \subset \Omega_3$). Note that since $f(K)$ is compact, then so is $f(K)_{\epsilon}$. By construction, since we started with a normal family, then $f_{n_k}(K) \subset f(K)_{\epsilon}$ for all $k$ big enough. But then we win, because as $g$ is continuous and $f(K)_{\epsilon}$ is compact, there exists $M > 0$ such that
$$
|g(z) - g(w)| \leq M|z-w|
$$
for all $z,w \in f(K)_{\epsilon}$. That means that
$$
|g(f_{n_k}(z)) - g(f(z))| \leq M|f_{n_k}(z)-f(z)|
$$
for all $z \in K$ for all very big $k$, and by hypothesis we can make the right-hand side as small as we want on $K$.
