This seems to be the correct place for complex analysis. Here's the problem I'm working on:

Let $\displaystyle \Delta (a,r)$ be the disk $\displaystyle \{z \in \mathbb{C}:|z-a|<r \}$, and let $\displaystyle \Delta$ be the unit disk $\displaystyle \Delta (0,1)$. Let $\displaystyle f$ be an analytic function on $\displaystyle \Delta$.

(a) Let $\displaystyle a \in \Delta$. Show that for any $\displaystyle r$, $\displaystyle 0<r<1-|a|$, we have:

$\displaystyle f(a)=\frac{1}{\pi{r}^2}\int_{\Delta (a,r)}f(z)\ dxdy$

(b) Show that if $\displaystyle M=\int_{\Delta}|f(z)|\ dxdy$, then:

$\displaystyle |f(a)| \le \frac{M}{\pi(1-|a|)^2}$, $\displaystyle a \in \Delta$

(c) Let $\displaystyle \mathcal{F}$ be a family of analytic functions on $\displaystyle \Delta$ for which there exists $\displaystyle M>0$ such that:

$\displaystyle \int_{\Delta}|f(z)|\ dxdy \le M\text{ for all }f \in \mathcal{F}$.

Show that $\displaystyle \mathcal{F}$ is a normal family.

For (c), a normal family is one in which every sequence in $\displaystyle \mathcal{F}$ has a subsequence that converges uniformly on compact subsets.

Unless I'm missing something, (b) follows from (a) by just taking the absolute value:

$\displaystyle |f(a)| \le \frac{1}{\pi{r}^2}\int_{\Delta (a,r)}|f(z)|\ dxdy$

$\displaystyle \le \frac{1}{\pi{r}^2}\int_{\Delta}|f(z)|\ dxdy$

$\displaystyle = \frac{1}{\pi{r}^2}M$

and since this must be true for all $\displaystyle r<1-|a|$, taking the limit as $\displaystyle r$ goes to $\displaystyle 1-|a|$ gives

$\displaystyle |f(a)| \le \frac{M}{\pi{(1-|a|)}^2}$

I'm still lost on (a) and (c), though.

Thanks,

Hollywood