A family $\mathscr{F}\subset \operatorname{Aut}(G)$ is compact Let $G\subset\mathbb{C}$ be any bounded region, and let $w\in G$. Let $\mathscr{F}$ be the set $\{f\in \operatorname{Aut}(G):f(w)=w\}$ ; where $\operatorname{Aut}(G)$ is the group of all biholomorphic maps on G.
Then we have to show that $\mathscr{F}$ is compact.
My idea :
Let $\{f_n\}_{n=1}^{\infty}$ be a sequence in $\mathscr{F}$.
Since G is bounded region, $\overline{G}$ is compact in $\mathbb{C}$, and the diameter of $G$ is finite.. Thus there exists $M>0$ such that $\lvert f(z) \rvert \leq M$ for every $z\in G$ and for every $f\in \mathscr{F}$. Hence $\mathscr{F}$ is a locally uniformly bounded family. So, by Montel's theorem, there exists a subsequence $\{f_{n_k}\}_{k=1}^{\infty}$ of $\{f_n\}_{n=1}^{\infty}$ which converges uniformly to $f$ on every compact subset of $G$. Since $\mathscr{H}(G)$, the family of all holomorphic functions on $G$, is a complete metric space, we say that $f$ is also holomorphic on $G$. It is also clear that $f(w)=w$.
Now I am stuck. How do I show that $f$ is infact a member of $\mathscr{F}$ ? If this thing can be shown then we can say, by sequential compactness, that the family $\mathscr{F}$ is indeed compact.
 A: To simplify notation, we will denote by $f_n$ the sequence of functions in $\mathcal F$ that converges to $f$ uniformly on compact subsets. By passing to a further subsequence, we can also assume that the sequence $h_n:=f_n^{-1}$ converges uniformly on compact subsets, as well, using the same Montel argument. Let us denote the limit of $h_n$ by $h$.
First we claim that $f(G) \subset G$ (this is not clear: a priori we only have $f(G) \subset \overline{G}$).
So let $p \in \Bbb C \setminus G$. Now consider the sequence of functions $g_n:G \to \Bbb C, z \mapsto f_n(z)-p$. This converges uniformly on compact subsets to $g_n:G \to \Bbb C,z \mapsto f(z)-p$. Note that for each $n$, the function $g_n$ has no zeroes on $G$ as $f_n(G) \subset G$. By Hurwitz' theorem, this implies that either $g$ has no zeroes on $G$ or $g \equiv 0$. But clearly we have $g \not\equiv 0$, because $g(w)=w-p \neq 0$. This shows that $g$ has no zeroes on $G$ which means that $f(z) \neq p$ for all $z \in G$. As $p \in \Bbb C \setminus G$ was arbitrary, this implies that $f(G) \subset G$. This argument shows that $h(G) \subset G$ as well.
Now as we have established that $f$ and $h$ are holomorphic functions $G \to G$, it makes sense to consider the compositions $f \circ h$ and $h \circ f$.
We can do this part in more generality:

Let $U \subset \Bbb C$ be a open. Let $f_n,h_n$ be sequences of functions $U \to U$ that converge uniformly on compact subsets to $f$ and $h$, respectively and assume that $f(U) \subset U$ and $h(U) \subset U$. Then the sequence $f_n \circ h_n$ converges uniformly on compact subsets to $f \circ h$

Proof Let $K$ be a compact subset of $U$. Let $z \in K$.
Then we have for every $n \in \Bbb N$:
$$|f(h(z))-f_n(h_n(z))| \leq |f(h(z)) - f_n(h(z))|+|f_n(h(z))-f_n(h_n(z))|$$
For the first term, note that $h(K)$ is a compact subset of $U$, thus $|f(h(z)) - f_n(h(z))| \to 0$ uniformly in $z \in K$.
We now claim the following: there is some compact subset $L \subset U$ such that for all $n$ sufficiently large, we have $h_n(K) \subset L$ and also $h(K) \subset L$. Because $h(K)$ is compact and $U \subset \Bbb C$ is open, there is some $\varepsilon >0$ such that $L:=\{z \in \Bbb C \mid \exists \zeta \in h(K): |z-\zeta| \leq \varepsilon\}$ is contained in $U$. $L$ is a closed and bounded subset of $\Bbb C$, so it is compact by Heine-Borel. We get that $h_n(K) \subset L$ for $n$ sufficiently large, because $h_n$ converges to $h$ uniformly on $K$.
Having found such an $L$, we can take care of the second term. We need to show that $|f_n(h(z)) - f_n(h_n(z))| \to 0$ uniformly in $z \in K$.
Using Cauchy's inequality and compactness, the set $\{|f'_n(z)|, \in \Bbb N, z \in L\}$ is bounded. Thus we can find a common Lipschitz constant $C > 0$ for $f_n$ on $L$ that does not depend on $n$. Thus we get for all $n$ sufficiently large:
$$|f_n(h(z)) - f_n(h_n(z))| \leq C |h(z)-h_n(z)|$$ which converges to $0$ uniformly on $K$ because $h_n \to h$ uniformly on $K$.
Now back to our situation: by what we have established so far, we get that $f_n \circ h_n$ converges uniformly on compact subsets to $f \circ h$. But for all $n$, we have $f_n \circ h_n=\mathrm{id}_G$. Thus $f \circ h=\mathrm{id}_G$. By the same argument, $h \circ f = \mathrm{id}_G$.
This shows that $f \in \mathrm{Aut}(G)$. Also clearly $f(w)=w$, so $f \in \mathcal F$.
Note: More generally, the group of isometries of a Riemannian manifold has compact stabilizers. For a reference, see theorem II.1.1 in Kobayashi's Transformation Groups in Differential Geometry. As we can alway put a Riemannian metric on a Riemann surface such that the group of orientation-preserving isometries is the group of holomorphic automorphisms and the group of orientation preserving isometries is closed in the group of all isometries, we get that the group of holomorphic automorphism of any Riemann surface has compact stabilizers.
