TITLE: Show that F is a normal family
QUESTION [2 upvotes]: G is a domain in $C$, where $C$ is the complex plane and $G \neq C$ and $M>0$ we have that:
$F=\{f\in H(G): \int_{G}|f(z)|^2 dm_2(z)\leq M\}$
How do I show that F is a normal family? - Some of the other answers are missing some calculations and so I don't quite follow it.

REPLY [7 votes]: Fix $f \in F$, $z_0 \in G$. We can find $R > 0$ so that $\overline{D}(z_0, 2R) \subset G$. Let $z \in D(z_0, R)$, $0 \le r \le R$. By the mean value property:
$$
f(z) = \frac{1}{2\pi} \int_0^{2\pi} f(z + re^{i\theta}) \,d\theta
$$
Multiply both sides by $r$ and integrate with respect to $r$ from $0$ to $R$ to get:
$$
f(z) = \frac{1}{\pi R^2} \int_0^R \int_0^{2\pi} f(z + re^{i\theta})r \,d\theta dr
$$
By Cauchy-Schwarz:
\begin{align}
\left|f(z)\right| &\le \frac{1}{\pi R^2} \int_0^R \int_0^{2\pi} \left|f(z + re^{i\theta})r\right| \,d\theta dr \\
&\le \frac{1}{\pi R^2} \left\{\int_0^R \int_0^{2\pi} \left|f(z + re^{i\theta})\right|^2 r \,d\theta dr\right\}^{1/2} \left\{\int_0^R \int_0^{2\pi} r \,d\theta dr\right\}^{1/2}
\end{align}
The last expression is bounded by a bound that is independent of $f$. It only depends on $M$ and $R$. Hence it is valid for all $z \in D(z_0, R)$. It follows that all $f \in F$ are uniformly bounded in a neighborhood of $z_0$ for any $z_0 \in G$. Let $K$ be a compact subset of $G$. Cover $K$ with a finite set of such neighborhoods. It follows that $F$ is uniformly bounded on each compact subset of $G$. Thus $F$ is normal.
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