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 $n_{j} → ∞$.



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 $D$.

  • $\begingroup$ What's the name of that theorem? the only thing I know for normal families is it's definition!! $\endgroup$
    – Math1
    Apr 23 '20 at 23:54
  • $\begingroup$ It is called Montel's Theorem. See en.wikipedia.org/wiki/Montel%27s_theorem $\endgroup$ Apr 24 '20 at 0:01
  • $\begingroup$ @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 $|z| \leq r$. $\endgroup$ Apr 24 '20 at 0:21
  • $\begingroup$ aha...and the max goes to zero because of the uniform convergence? $\endgroup$
    – Math1
    Apr 24 '20 at 0:26
  • $\begingroup$ Yes, that is right. @Math1 $\endgroup$ Apr 24 '20 at 0:26

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