In an exercise I received there is the following question:
Let $R>0$.
Let $\mathcal{F}$ be the family of functions $f \colon \{z:|z|<R\} \to \mathbb{C}$ that
- are analytic
- avoid the values $0,1$
- all satisfy $f(0) = a_0$ for some constant $a_0 \in \mathbb{C}$
For all $0\leq r < R$ define $$Q(r) := \sup_{f \in \mathcal{F}}\ [\max_{|z|\leq r}|f(z)|]$$ I have to prove that $Q$ is continuous, unbounded and strictly increasing, and that $Q$ is in fact a maximum, in addition to being a supremum.
The strictly increasing part I managed to prove, using the fact that if $f$ is not constant, $f$'s maxima over growing sub-disks of $\{z:|z|<R\}$ must strictly increase, otherwise it is a contradiction to the maximum principle.
In addition, Schottky's theorem guarantees that $Q(r)$ is finite for all $0\leq r < R$.
But how to address the other things to prove? I tried the following direction:
By Montel's theorem, $\mathcal{F}$ is a normal family. Let $0\leq r_0 < R$. By the definition of supremum, there exists a sequence $\{f_n\} \in \mathcal{F}$ such that $\max_{|z|\leq r} |f_n(z)| \to Q(r)$. By normality, there is a convergent subsequence $\{f_{n_k}\}$. On the compact subset $|z| \leq r_0$, this sequence converges uniformly to some analytic function $g$, and the uniform converges yields $\max_{|z|\leq r} |g(z)| = Q(r)$. But I do not know whether $g \in \mathcal{F}$ (if it does, this solves the maximum part).
Please help.