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.

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    $\begingroup$ What other answers? $\endgroup$ – Jonas Meyer May 18 '13 at 13:26
  • $\begingroup$ This looks like a duplicate of this question $\endgroup$ – robjohn May 18 '13 at 14:09

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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  • $\begingroup$ I think what you mean here is that: let $z_0 \in G$, and $R > 0$ such that $\bar{D}(z_0,2R) \subset G$. Now, for every $z\in D(z_0,R)$, we have $\bar{D}(z,R)\subset G$, hence as you show, $|f(z)|$ is bounded by a bound that is independent of $f$ and $z$. Then conclude as yours. Or I misunderstand something? $\endgroup$ – Du Phan Dec 5 '13 at 16:29
  • $\begingroup$ @AymanHourieh: If the bound depends on $z$, how can it happen that each $f$ is bounded in a neighborhood of $z$? I think that uniformly bounded on compact subsets means that the bound not depends on specific $z$, nor $f$; instead, just depends on which compact subset one works with. $\endgroup$ – Du Phan Dec 5 '13 at 18:53
  • $\begingroup$ @AymanHourieh: I agree that we just need the family is uniformly bounded on compact subsets. But as you did, pick $z$, $R$, then claimed that $|f(z)|$ is bounded; how about another $w$ in $D(z,R)$? For example, the domain is the whole plane. One has to prove that $f$ is bounded in a neighborhood of $0$ by a bound not depend on $f$, but here one just has $|f(0)|$ is bounded by a bound not depend on $f$. $\endgroup$ – Du Phan Dec 6 '13 at 1:10
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    $\begingroup$ @DuPhan I see what you mean now. I should have read my answer more carefully (it's bean a while since I posted it!). Yes, that was implicit in the answer. I've edited it to add what said in your first comment. Thanks! $\endgroup$ – Ayman Hourieh Dec 6 '13 at 9:36

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