A compact subset of Holomorphic function space with uniform convergence topology Let $F=\left\{f:f\left(z\right)=\sum^{\infty}_{n=0}a_{n}z^{n},\left|a_{n}\right|\leq n,n=0,1,...\right\}$
(a).Prove that every $f\in F$ defines a holomorphic function on $D$.
(b).Prove that $F$ is a compact subset of the set of all holomorphic functions on $D$ in the topology of uniform convergence on compact subsets of $D$.
For the question one ,we can use Cauchy-Hadamard formula to get that this power series converges on the unit disk.
As for the second question,I feel helpless and don't know which theorems in complex analysis I can resort to.
 A: I assume $D$ means the unit disk. For the first part, you are right. You can also use the root test.
For the second part, it suffices to show that every sequence $(f_n)$ of functions in $F$ converges along a subsequence (uniformly on compact subsets) to a function $f\in F$. In turn, it suffices to show that the $f_n$ are uniformly bounded on all compact subsets of $D$, and then apply Montel's theorem.
So, indeed, suppose that $K \subset D$ is compact. Then we have that $K \subset \{z:\; |z|<R\}$ for some $R<1$. For $z \in K$ and $n \in \Bbb N$, we have that $|f_n(z)| \leq \sum n R^n = \frac{R}{(1-R)^2}< \infty$. This shows that the $f_n$ are locally uniformly bounded, which proves convergence along a subsequence $f_{n_k}$ to some analytic function $f$.
We still need to show that this limit function $f$ is actually in $F$. We will prove something stronger, namely that $F$ is a closed subset of the space of holomorphic functions. In other words, if $g_n \to g$ uniformly on compact sets and if $g_n \in F$, then $g \in F$. Suppose $g_n(z) = \sum_k a^{(n)}_k z^k$ with $|a^{(n)}_k| \leq k$, and suppose that $g(z) = \sum_k a_k z^k$. Then I claim that for each fixed $k \geq 0$ we have that $a^{(n)}_k \to a_k$ as $n \to \infty$, which would imply that $|a_k| \leq k$ as desired. Indeed, this is an easy consequence of Cauchy's integral formula: $$a^{(n)}_k = \frac{1}{2 \pi i} \int_{\gamma} \frac{g_n(z)}{z^{k+1}} dz \;\stackrel{n \to \infty}{\longrightarrow} \;\frac{1}{2 \pi i}\int_{\gamma} \frac{g(z)}{z^{k+1}} dz = a_k$$ where $\gamma$ is any smooth curve winding once around $0$ in $D$, and we can pass the limit under the integral by uniform convergence on $\gamma$ of the $g_n$ to $g$.
A: Let $\{f_n\}\subset F$. Then $f_n=\sum_{k=0}^\infty a_{n,k}z^k$, with $\lvert a_{n,k}\rvert\le k$. 
The sequence $\{a_{n,1}\}_{n\in\mathbb N}$ is bounded, and hence it possesses a convergent subsequence $a_{j_n^1,1}\to b_1$, and clearly $\lvert b_1\rvert\le 1$.
The sequence $\{a_{j_n^1,2}\}_{n\in\mathbb N}$ is bounded, and hence it possesses a convergent subsequence $a_{j_n^2,2}\to b_2$, and clearly $\lvert b_2\rvert\le 2$.
Recursively, we obtain a convergent subsequence $a_{j_n^k,k}\to b_k$, of the subsequence $a_{j_n^{k-1},k}$ and $\lvert b_k\rvert\le k$.
Observe now that $a_{j_n^n,k}\to b_k$, for all $k\in\mathbb N$, and let $g(z)=\sum_{n=0}^\infty b_nz^n$. Clearly $g\in F$.
Claim. $\{\,f_{j_n^n}\}$ is uniformly convergent to $g$ on any compact $K\subset\mathbb D$.
Proof. If $K$ compact and $K\subset\mathbb D$, then $K\subset \{z\in\mathbb D: \lvert z\rvert\le R\}$, for some $R<1$. Then let $\varepsilon>0$ and $N\in\mathbb N$, such that 
$$
\sum_{k=N+1}^\infty k R^k= \frac{(N+1)R^{N}}{1-R}+\frac{R^{N+1}}{(1-R)^2}<\frac{\varepsilon}{4}.
$$ 
Also, let $n_0\in\mathbb N$, so that
$$
\lvert a_{j_n^n,k}-b_k\rvert<\frac{\varepsilon}{2(N+1)},\quad \text{for}\,\,\,k=0,\ldots,N\,\,\,\text{and}\,\,\,n\ge n_0.
$$
Then for $n\ge n_0$, we have
$$
\max_{z\in K}\lvert\, f_{j_n^n}(z)-g_n(z)\rvert\le\sum_{k=0}^\infty\lvert a_{j^n_n,k}-b_k\rvert R^k \le \\ \sum_{k=0}^N\lvert a_{j^n_n,k}-b_k\rvert R^k
+\sum_{k=N+1}^\infty\lvert a_{j^n_n,k}-b_k\rvert R^k
\frac{\varepsilon}{2}+\sum_{k=N+1}^\infty2k R^k <\varepsilon.
$$
