Prove that $S$ is compact. Prove that $$S:= \left \{f \in \mathcal H(\Bbb D)\ \bigg |\ f(z) = \sum\limits_{n=0}^{\infty} a_n z^n,\ \text {with}\ |a_n| \leq n,\ \text {for all}\ n \in \Bbb N \right \}$$ is compact.
How do I proceed? Any help will be highly appreciated.
Thanks in advance.
 A: Recall $\mathcal{H}(\mathbb{D})$ is a Fréchet space with the Heine-Borel property, as detailed in this neat document (which contains way more detail about this proof, along with many other subjects). Now, all that remains to show is that $S$ is closed and bounded. Closeness is immediate, but boundedness is more delicate since $\mathcal{H}(\mathbb{D})$ is not normable. We instead look at all seminorms $\nu_{K_n}$, defined by
$$\nu_{K_n}(f)=\sup_{z\in K_n}|f(z)|$$
where $K_n=\{z\in \mathbb{D}\,|\,|z|\leq 1-1/n\}$. Showing all such seminorms are bounded on $S$ will allow us to conclude the result. Now, compute
$$|f(z)|\leq\left|\sum_{n=0}^\infty a_nz^n\right|\leq\sum_{n=0}^\infty n|z|^n=\frac{|z|}{(1-|z|)^2}.$$
It follows that $\nu_{K_n}(f)\leq n(n-1)$, from which we conclude that each seminorm is indeed bounded on $S$, and hence $S$ is compact.
A: This is very likely to be a question about normal families.
Let $H=K$ be a compact subset of $\mathbb D$. Then there exist $r \in [0,1)$ such that $|z| \leq r$ for all $z \in K$. This gives $|f(z)| \leq \sum n r^{n}$ for all $z \in K$. Thus the given family is uniformly bounded on compact sets. The result now follows by Montel's Theorem.
