# Prove that there exists $h\in V$, such that $|f'(\frac{1}{2})|\leq|h'(\frac{1}{2})|$ for all $f\in V$.

Let $$V=\{f\in\mathcal{O}(\mathbb{D}):f(z)=\sum_{n=0}^{\infty}a_{n}z^{n}\text{ with }|a_{n}|\leq n^{2}\text{ for all }n\}$$. Prove that there exists $$h\in V$$, such that $$|f'(\frac{1}{2})|\leq|h'(\frac{1}{2})|$$ for all $$f\in V$$.

I want to use Montel's Theorem:

Let $$\mathcal{F}$$ be a family of holomorphic functions on Ω. If $$\mathcal{F}$$ is uniformly bounded on every compact subset of Ω, then $$\mathcal{F}$$ is equicontinuous on every compact subset of Ω, and hence $$\mathcal{F}$$ is a normal family.

My initial thought is to first prove that the set $$V$$ is uniformly bounded on every compact subset, but I'm not quite sure how to show that. Also, How do I use the Montel's theorem to prove above? I guess my question is how is showing the existence of $$h$$ relate to $$V$$ is a normal family?

In fact, the problem does not require Montel's theorem necessarily: simply let $$h(z) = \sum_{n=0}^\infty n^2 z^n,$$ and observe that $$|f'(\frac{1}{2})|\leq |h'(\frac{1}{2})|$$ for all $$f\in V$$.
However, I'll present how Montel's theorem implies that such supremum is attained in $$V$$ as a maximum. For any compact set $$K\subset\mathbb{D}$$, observe that there exists $$r\in (0,1)$$ such that $$K\subset D(0,r)=\{z\in\mathbb{D}:|z| Therefore, this leads to a simple estimate that $$|f(z)|=|\sum_{j=0}^\infty a_j z^j|\leq \sum_{j=0}^\infty j^2 r^j <\infty,\quad\forall z\in K.$$ This shows that $$V$$ is a normal family. Now, let $$f_n\in V$$ be a sequence such that $$|f'_n(\frac{1}{2})| \to \sup_{f\in V}|f'(\frac{1}{2})|.$$ Since $$V$$ is normal, by passing to a subsequence, we may assume that $$f_n \to h$$ locally uniformly on $$\mathbb{D}$$. By Cauchy's integral formula, this of course implies that each $$k$$-th derivative $$f_n^{(k)}$$ converges locally uniformly to $$h^{(k)}$$. Since $$V$$ is closed, we have $$h\in V$$ and that $$\sup_{f\in V}|f'(\frac{1}{2})| = \lim_{n\to\infty}\sup_{f\in V}|f'(\frac{1}{2})| = |h'(\frac{1}{2})|,$$as desired.
• The precise definition is given in en.wikipedia.org/wiki/Normal_family. If $V$ is a normal family, then It happens that every sequence in $V$ has a locally uniformly convergent subsequence. Montel's theorem provides a sufficient condition for this. So $\sup|f'(c)|$ is actually attained in $V$, if $V$ is closed, by a limit $h\in V$ of approaching sequence $f_n\in V$ such that $|f'_n(c)| \to \sup|f'(c)|$. – Song Dec 11 '18 at 20:25