# Existence of a subsequence and analytic function [closed]

Let $$\{f_{n}\} _n{_\in \mathbb{_N}}$$ be a sequence of functions which are analytic on the open unit disc $$D$$ and such that $$|f_{n}(z)| ≤ 1$$ for all $$n$$ and all $$z ∈ D$$. Prove that there is a subsequence$$\{f_{n}{_j}\}$$ and an analytic function $$f$$ on $$D$$ satisfying the following property:

For every $$r$$, $$0 < r < 1, max_{|z|≤r} |f(z) −f_{n}{_j}(z)| → 0$$ as $$n_{j} → ∞$$

Show by example that it is false in general that $$sup_{z∈D} |f(z) − f_{n}{_j}(z| → 0$$ as $$n_{j} → ∞$$.

Thanks!!

The first part comes from a basic theorem on normal families which can be found in any book on Complex Analysis. For the second part take $$f_n(z)=z^{n}$$. Use the fact that $$(1-\frac 1 {n_k})^{n_k} \to \frac 1 e$$ to see that the convergence is not uniform on $$D$$.
• @Math1 Boundedness implies that $(f_n)$ is a normal family. Hence a subsequence converging uniformly on compact sets exists. Since $\{x: |z| \leq r\}$ is a compact set for each $r <1$ the convergence is uniform for $|z| \leq r$. Apr 24 '20 at 0:21