Holomorphic Functions Fixing a Point

Question

Let $U$ be a bounded connected open subset of $\mathbb{C}$ and $z_0 \in \mathbb{C}$. Show that the group of automorphisms of $U$ fixing $z_0$ is compact. My idea is to show that this group is sequentially compact, but I am unable to show it. Thanks for any help.

Answer 1

Let $G = \{ f \in \operatorname{Aut}(U) : f(z_0) = z_0\}$. Since $U$ is bounded, $G$ is a normal family, thus every sequence in $G$ has a subsequence that converges locally uniformly on $U$. It remains to be seen that the limit function again belongs to $G$.

By Cauchy's integral formula for the derivative of a holomorphic function, $D := \{ f'(z_0) : f \in G\}$ is a bounded subset of $\mathbb{C}\setminus \{0\}$. On the other hand, from $(f\circ g)'(z_0) = f'(z_0)\cdot g'(z_0)$ for $f,g \in G$ it follows that $D$ is a subgroup of $(\mathbb{C}\setminus \{0\},\,\cdot\,)$, whence $\lvert f'(z_0)\rvert = 1$ for all $f \in G$. By a theorem of Weierstraß, if a sequence $(f_n)$ of holomorphic functions on $U$ converges locally uniformly to $f$, then the sequence $(f_n')$ of derivatives converges locally uniformly to $f'$. In particular, $f_n'(z_0)$ converges to $f'(z_0)$. Thus if $f$ is the locally uniform limit of a sequence $(f_n)$ in $G$, then we have $\lvert f'(z_0)\rvert = 1$, and clearly $f(z_0) = z_0$. But this implies $f \in \operatorname{Aut}(U)$, and thus $f\in G$, as needed.

Remark: The group homomorphism $G \to S^1$ given by $f \mapsto f'(z_0)$ is injective, thus we can view $G$ as a subgroup of $S^1$, and in fact the subgroup topology induced from $S^1$ coincides with the topology of locally uniform convergence on $G$. But we don't need this fact to prove that $G$ is (sequentially) compact.