I have the following question and I have no idea how I can solve it. Let $F$ be family of functions $f$ in $A(D)$ where $D=\{x\in\mathbb{C}, \vert x\vert<1\}$ is a unit disc and $A(D)$ denotes the set of analytic functions on $D$.

If: $$\begin{cases} f(0) &=2i\\ \vert f(z)\vert&>1 \end{cases}$$ for all $z\in D$, then prove that $F$ is a normal family in $A(D)$.

I tried to use montel's but this family of functions is not locally bounded and the same issue with Ascoli-Arzela theorem.

Could you please give me a hint. Thanks.

  • $\begingroup$ Are $A(D)=\text{adh}(D)$ and $D=\{x\in\mathbb{C}, \vert x\vert < 1\}$? $\endgroup$ – MoebiusCorzer Dec 8 '15 at 16:11
  • $\begingroup$ Sorry for the typo. By A(D) which is the set of all analytic functions. $\endgroup$ – user160492 Dec 8 '15 at 16:32
  • $\begingroup$ I've corrected my wrong editing :) As an information, the set of analytic functions on some given set $D$ is often denoted $C^{\omega}(D)$, if I'm not mistaken (but it is maybe only in the real case) $\endgroup$ – MoebiusCorzer Dec 8 '15 at 16:33

We define $$ G:= \bigg\{ \frac{1}{f}\ \ \bigg|\ \ f \in F \bigg\} $$

Easy to see $G \subset A(D)$ is a normal family (since uniformly bounded).

Let $f_n$ be a sequence in $F$.

Define $g_n := \dfrac{1}{f_n}$ for all $n \in \mathbb N$.

Now, since $G$ is normal, there is a subsequence of $g_n$ (let us say $g_{n_k}$) and a function $g$ such that $g_{n_k} \to g$ uniformly on $D$.

Check that $f_{n_k} := \frac{1}{g_{n_k}}$ converges uniformly to $f := \frac{1}{g}$ on $D$. (use the fact that $g_n$ are bounded)

  • $\begingroup$ Is there any use of the condition $f(0)=2i$? $\endgroup$ – skylark Jun 13 at 15:04

Hint: Consider the family of all reciprocals of functions in $\mathcal{F}$.

Or, use the stronger theorem also called Montel's theorem which states that any family of analytic functions on a region omitting two complex values is normal.


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