Let $\mathcal F$ be the family of analytic functions on the unit disk $\,\mathbb D=\{z: \lvert z\rvert<1 \},$ such that $f[\mathbb D] \subset \mathbb C\!\smallsetminus\!(-\infty, 0]$. Show that $\mathcal F$ is a normal family.

(Normal Family= relatively compact= every sequence contains a subsequence which converges uniformly on every compact subset of $\mathbb D$)

EDIT. I see that this is immediate from Montel's theorem, but can it be proved somewhat easily without that?


Hint. Note that $\mathbb C\setminus (-\infty,0]$ is a simply connected domain, and hence there exists a conformal map $$\varphi : \mathbb C\setminus (-\infty,0]\to\mathbb D.$$ Next observe that the family $$ \mathcal G=\{\varphi\circ f : f\in\mathcal F\} $$ is bounded and hence a normal family. Next, if $\{\varphi\circ f_n\}$ converges on compact subsets of $\mathbb D$, then so does the sequence $$ f_n=\varphi^{-1}\circ(\varphi\circ f_n), \quad n\in\mathbb N. $$

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  • $\begingroup$ wonderful. thanks. $\endgroup$ – Ashley May 30 '15 at 18:34

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