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? 
 A: 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
$$
K\subset D(0,r)=\{z\in\mathbb{D}:|z|<r\}.
$$ 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.
