Let $F$ be the set of functions $f$ that that are analytic in the open unit disk and satisfy $\int_{|z|=r}\,\,|f(z)||dz| \leq 1$ for all $r$, $0<r<1$. Prove that $F$ is normal.

I tried to show that $F$ is locally bounded on each disk $|z|=r$ but I couldn't, thanks for any help.

  • $\begingroup$ I gave a solution providing you know Montel's theorem. $\endgroup$ – mathcounterexamples.net Aug 6 '15 at 19:56

Let's prove that $F$ is bounded on all compact $K \subset U$ where $U$ is the open unit disk.

As $\mathbb C \setminus U$ is closed and $K$ compact, $2\delta =d(K,\mathbb C \setminus U)$ is strictly positive and $d(K,C_R)=\delta>0$ where $C_R$ is the circle centered on the origin with radius $R=1-\delta$. According to Cauchy's integral formula for $z \in K$ $$f(z)=\frac{1}{2i\pi}\int_{C_R} \frac{f(\xi)}{z-\xi} d\xi$$ Also $\vert z- \xi \vert \ge \delta$ for $\xi \in C_R$. Hence $$\vert f(z) \vert \le \frac{1}{2\pi}\int_{C_R} \frac{\vert f(\xi) \vert}{\vert z-\xi \vert} d\xi \le \frac{1}{2\pi}\int_{C_R} \frac{\vert f(\xi) \vert}{\delta} d\xi \le \frac{1}{2\pi\delta}$$

You can then get the conclusion that $F$ is normal from Montel's theorem.

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