Showing a certain operator on a set of holomorphic functions attains its supremum 
Let $G \subseteq \mathbb{C}$ be open and connected, and fix $a_1, a_2 \in G$. Let $\mathcal{F} := \{f: G \rightarrow \mathbb{C}: f$ is holomorphic and $|f(z)| \leq 1 \}$.
Define $\Phi: \mathcal{F} \rightarrow \mathbb{C}$ by $$\Phi (f) := |f(a_1)| + |f'(a_2)|.$$  Show that $\Phi$ has a maximum, i.e., that there is $f_0 \in \mathcal{F}$ such that $\Phi(f_0) = \sup \{\Phi(f): f\in \mathcal{F}\}$.

It should be possible to write the proof with just complex analysis tools and the basic functional analysis needed to prove Arzelà-Ascoli.  Let $\rho$ denote the distance for $C(G,\mathbb{C})$, as defined in pg. 142-148 of Functions of One Complex Variable I, 2nd edition, by Conway.  We can endow $\mathcal{F}$ with $\rho$.
One idea is to show $\Phi$ is continuous and $\mathcal{F}$ is compact.  If these two things are true, the conclusion would follow immediately.
Another idea is to construct $f_0$ somehow.  Since $|f(a_1)| \leq 1$ $\forall f \in \mathcal{F}$, we could focus on making $|f_0'(a_2)|$ maximal.
Also, it'd be helpful to understand what the operator $\Phi$ is actually measuring.
 A: 
One idea is to show $\Phi$ is continuous and $\mathcal{F}$ is compact. If these two things are true, the conclusion would follow immediately.

These things are true, if we take the standard topology - the topology of locally uniform convergence - on $\mathscr{O}(G)$.
With the standard topology, the subspace topology induced by $C(G)$, the space $\mathscr{O}(G)$ of holomorphic functions on the domain $G$ is a Fréchet-Montel space. That is, a Fréchet space in which every closed and bounded subset is compact. The latter is basically the content of Montel's little theorem, a locally bounded family of holomorphic functions is normal. The set
$$\mathcal{F} = \left\{ f \in \mathscr{O}(G) : \bigl(\forall z\in G\bigr)\bigl( \lvert f(z)\rvert \leqslant 1\bigr)\right\}$$
is clearly bounded, and since the inequality is non-strict, also closed. Hence is is compact.
The map $f \mapsto \lvert f(a_1)\rvert$ is even continuous on $C(G)$, hence also on the subspace $\mathscr{O}(G)$.
The map $f \mapsto \lvert f'(a_2)\rvert$ is continuous on $\mathscr{O}(G)$ due to the integral formula for the derivatives, which gives a bound for the values of $f'$ on any compact subset $K\subset G$ in terms of a bound for the values of $f$ on any compact neighbourhood $K' \subset G$ of $K$.
So $\Phi$ is the sum of two continuous functions, hence continuous.
However, I expect that the exercise was intended to be solved with less functional analytic arguments.
By the classical Montel theorem, the family $\mathcal{F}$ is normal, since it is (uniformly, even) bounded.
Then consider a sequence $(f_n)$ of functions in $\mathcal{F}$ with
$$\lim_{n\to \infty} \Phi(f_n) = \sup \{ \Phi(f) : f \in \mathcal{F}\}.$$
By the normality of $\mathcal{F}$, we can assume that $f_n$ converges locally uniformly on $G$ to a function $f$. By a theorem of Weierstraß, the locally uniform limit of holomorphic functions is holomorphic, and clearly $\lvert f(z)\rvert \leqslant 1$ for all $z\in G$, so $f\in \mathcal{F}$.
By another theorem of Weierstraß, if a sequence of holomorphic functions $(f_n)$ converges locally uniformly, then the sequence $(f_n')$ of derivatives also converges locally uniformly, and of course to the derivative of $\lim f_n$.
Thus we have $f(a_1) = \lim\limits_{n\to\infty} f_n(a_1)$ and $f'(a_2) = \lim\limits_{n\to\infty} f_n'(a_2)$, and therefore
$$\Phi(f) = \lim_{n\to\infty} \Phi(f_n) = \sup \{\Phi(g) : g \in \mathcal{F}\},$$
i.e. the supremum is attained.
