Show that $\{|f''(z_0)|: f\in \mathcal{F}\}$ is a closed set for a compact family $ \mathcal{F}$ of holomorphic functions.

Question

I have the following problem:

Let $\mathcal{F}\neq \emptyset$ be a compact family of holomorphic functions in a domain $U$, that is, every $f_n\in \mathcal{F}$ has a locally uniformly convergent subsequence whose limit also belongs to $ \mathcal{F}$.

Let $z_0\in U$. Show that there is a function $f_0\in \mathcal{F}$ such that $|f_0''(z_0)|=\sup \{|f''(z_0)|:f\in \mathcal{F}\}$.

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My attempt:

I only can prove that $\{f''(z):f\in \mathcal{F}\}$ is also compact. Since the statement is about "subsequence", I think it is far away from the desired conclusion.

Also, the limit function $f$ of $f_n''$ may not be the one which can achieve the maximum of $|f''(z_0)|$ for some $z_0\in U$.

Answer 1

Let $S:=\sup \{|f''(z_0)|:f\in \mathcal{F}\}$. Then there is a sequence $(f_n)$ in $\mathcal{F}$ such that

$(1)$ $|f_n''(z_0)| \to S$.

$(f_n)$ has a locally uniformly convergent subsequence $(f_{n_k})$ whose limit $f$ also belongs to $\mathcal{F}$ .

The Convergence Theorem of Weierstraß says:

$(f_{n_k}'')$ converges locally uniformly to $f''$. Hence

$(2)$ $|f_n''(z_0)| \to |f''(z_0)|$.

From $(1)$ and $(2)$ we get

$$S=|f''(z_0)|.$$