4
$\begingroup$

Possible Duplicate:
Locally bounded Family

Let's take $F$ to be the set of all holomorphic functions on the unit disk $U$, for which each $f \in F$ has the property:

$\int\int |f(z)|^{2} dx dy \leq 1.$

where the double integral runs over $U$. Is $F$ a normal family?

$\endgroup$
5
$\begingroup$

Yes, $F$ is uniformly bounded on compact subsets of $U$, so Montel's theorem applies. Your $F$ is the closed unit ball of the Bergman space $A^2(U)$. An excellent reference on these is Duren and Schuster's Bergman spaces, which includes what you want here in the beginning of Chapter 1. (In the proof, there is a reference to page 9 of Duren's earlier book Theory of $H^p$ spaces.)

The inequality given in Theorem 1 implies that $$|f(z)|\leq \frac{1}{\sqrt{\pi}(1-|z|)}$$ for all $f\in F$ and all $z\in U$. So given a compact set $K\subset U$, $\{|f(z)|:z\in K,f\in F\}$ is bounded above by $\displaystyle{\frac{1}{\sqrt{\pi}d(K,\partial U)}}$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Yes, it is normal. To see it, write $$f(re^{i \theta}) = \sum_{n=0}^{\infty}a_nr^ne^{in\theta}$$ Calculate the integral by writing $|f|^2 = f \bar{f}$, and we obtain the inequality $$\sum_{n=0}^{\infty}\frac{|a_n|^2}{n+1} \leq \frac{1}{\pi}$$

Thus, if $f \in F$, $f(z) = \sum_{n=0}^{\infty}a_nz^n$, $|z| \leq r <1$ we have $$|f(z)| \leq \sum_{n=0}^{\infty}|a_n|r^n = \sum_{n=0}^{\infty}\frac{|a_n|}{\sqrt{n+1}} \sqrt{n+1} r^n \leq (1/\pi)^{1/2} (\sum_{n=0}^{\infty} (n+1)r^{2n})^{1/2}$$

and so $F$ is uniformly bounded on compact subsets, thus normal by Montel's theorem.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.