I am studying for a qualifying exam and I have been stuck on the following problem from a previous exam:
Given a bounded open set $U$ with $w_0\in U$, let $\mathcal{F}$ be a family of one-to-one holomorphic functions $f:D(0,1) \to U$ satisfying $f(0)=w_0$. Prove that there is a function $f\in \mathcal{F}$ that maximizes the area of $f(D(0,1/2))$. The question also asks to prove $\mathcal{F}$ is a normal family, which is clear because $U$ is bounded.
My thoughts so far are to use Schwarz' lemma and find a maximum value of $\{|f(z)| :f\in \mathcal{F} \text{ and } |z|=1/2 \}$, since the maximum modulus principle gives that the maximum value of each function occurs on the boundary. However, I am not seeing how to apply this, so any help would be appreciated.
Note that the area of the image $f(D(0,1/2))$ for $f\in \mathcal F$ is $$\mathcal D(f)=\int_{D(0,1/2)} |f'(z)|^2\, dA.$$
Because $U$ is bounded, this value is bounded above and therefore has a supremum, $M$. We need to show some function actually attains this supremum. Let $\{g_i\}$ be a sequence of functions such that the areas of their images tend to $M$. Since $\mathcal F$ is normal, there is a subsequence of $\{g_i\}$ converging uniformly on compact subsets to a holomorphic function $g$. Without loss of generality, we may suppose the subsequence is $\{g_i\}$. Then, using the uniform convergence to switch limits and integration (and I leave it to you to justify this), $$\mathcal D(g) = \mathcal D (\lim g_i) = \lim \mathcal D(g_i) =M,$$ so $g$ attains the supremum.