Let $\mathcal{F}$ be the family of all analytic mappings $f$ of $\{z: \text{Re}(z) > 0 \}$ into itself such that $f(1)=1$. Does there exist $g\in\mathcal{F}$ such that $$|g'''(4)| = \sup_{f\in\mathcal{F}}|f'''(4)|?$$
We have that $\mathcal{F}$ is normal (see here). Do we need to show that $\{f''' : f\in \mathcal{F}\}$ is normal? How would one proceed afterwards?
If suffices to know that $\mathcal{F}$ is normal. Choose a sequence $(f_n)$ in $\mathcal{F}$ with $$ \lim_{n \to \infty} |f_n(4)| = \sup_{f\in\mathcal{F}}|f'''(4)| \, . $$ $(f_n)$ has a subsequence $(f_{n_k})$ which is locally uniformly convergent, say $f_{n_k} \to g$.
Now use that locally uniformly convergences of a sequence implies locally uniformly convergences of the derivatives (a consequence of Cauchy's integral formula for the derivatives). Therefore $$ |g'''(4)| = \lim_{k \to \infty} |f_{n_k}(4)| = \sup_{f\in\mathcal{F}}|f'''(4)| \, . $$
Finally use that $g$ cannot be constant, and conclude that $g \in \mathcal{F}$.
