We denote by $D$ the open unit disk and consider the family $$\mathcal{F}=\{f: D\to D, f \text{ is analytic and } f'(0)=\tfrac{1}{2}\}.$$ Prove that there is a function $g\in \mathcal{F}$ such that $$|g(0)|=\sup\{|f(0)|: f\in\mathcal{F}\}$$
By Pick's theorem, $$|f'(0)|\leq \frac{1-|f(0)|^2}{1-0} \Leftrightarrow |f(0)|\leq \frac{\sqrt{2}}{2}.$$ So, we want to find $g\in \mathcal{F}$ such that $|g(0)|=\tfrac{\sqrt{2}}{2}$ and $g'(0)=\tfrac{1}{2}$. The function $$g(z)=\frac{z-\tfrac{\sqrt{2}}{2}}{1-\tfrac{\sqrt{2}}{2}z}$$ satisfied what we want.
Is it correct? Is there a way to solve the problem without finding $g$ explicit?
You can use a normal families argument: There exists a sequence $f_n$ in $\mathcal F$ such that
$$|f_n(0)| > (1-1/n)\sup_{f\in \mathcal F} |f(0)|.$$
Because the $f_n$ are uniformly bounded, there exists a subsequence $f_{n_k}$ converging uniformly on compact subsets of $D$ to some $g$ analytic on $D.$ Because $f_{n_k}'$ also converges uniformly on compact subsets of $D$ to $g',$ $g$ has the desired properties.