Montel's Theorem

Theorem
Let $U \subseteq \C$ be an open subset of the complex numbers.

Let $\map \HH U$ be the space of holomorphic mappings on $U$.

Then a family of mappings $\FF \subseteq \map \HH U$ is normal $\FF$ is locally bounded.

Normal implies locally bounded
By the Arzelà-Ascoli Theorem, every normal family is locally bounded.

Locally bounded implies normal
By the Arzelà-Ascoli Theorem, any locally bounded and locally uniformly equicontinuous family is normal.

Hence it suffices to show that $\FF$ is locally uniformly equicontinuous.

As a Family of Lipschitz Continuous Functions with same Lipschitz Constant is Uniformly Equicontinuous, it suffices to show that every point $z_0 \in U$ has a neighbourhood $N$ such that all functions from $\FF$ have the same Lipschitz constant on $N$.

To this end, take any $z_0 \in U$.

By the assumption that $\FF$ is locally bounded, we may choose $R > 0$ such that for all $z \in B := \map B {x; R}$ (the disk of radius $R$ around $x$), we have $\size {\map f z} < M$.

Now take any $r \in \tuple {0, R}$ and let $B' = \map B {x; r}$.

We then have for any $z, z' \in B'$ and $w \in B$:


 * $\size {\paren {w - z} \paren {w - z'} } \ge \paren {R - r}^2$

This allows us to derive:

Therefore, $M \paren {R - r}^{-2}$ is a Lipschitz constant for all functions in $\FF$ on $B'$.