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Let $\mathscr F$ be a family of one-to-one holomorphic functions on a simply-connected domain $D \subset \Bbb C$ such that $\mathscr F$ omits 0. Show that $\mathscr F$ is a normal family (when considered as a family of meromorphic functions).

My attempt :

Let $\mathscr F=\{f_i\}_{i \in I}$ . Then since, $\mathscr F$ be a family of holomorphic functions on $D$ , $\mathscr F$ omits $\infty \in \hat{\Bbb C}$ . Thus $\mathscr F$ omits $0,\infty \in \hat{\Bbb C}$ . So if we can find any other point in $\hat{\Bbb C}$ that $\mathscr F$ omits, we are done by Montel's theorem on Meromorphic functions .

Now using that $D$ is simply-connected domain in $\Bbb C$ and $\mathscr F$ is actually a family of holomorphic functions on $D$ omitting 0, we get for each $f_i$, a $g_i$ holomorphic on $D$ such that $f_i = {g_i}^2$ .

Now looking at a point say 1, if $\mathscr{F}$ omits 1 we are done. Otherwise, we have the possibility that some $f_i$'s assume 1 . Looking at corresponding $g_i$'s we get that $g_i =\pm 1$ , so those which take $-1$, replacing them by $-g_i$ and unaltering the others, we obtain $\mathscr G=\{g_i\}_{i\in I}$ of meromorphic functions on $D$ that omit $0,-1,\infty \in \hat{\Bbb C}$ , thus $\mathscr G$ is a normal family of meromorphic functions on $D \ldots(*)$

Now taking any sequence $\{f_n\}_{n \in \Bbb N}\subset \mathscr F$ . Look at the corresponding $\{g_n\}_{n \in \Bbb N}\subset \mathscr G$ . By $(*), \exists \{g_{n_k}\}_{k \in \Bbb N}$ normally convergent. Then does it imply that $\{g^2_{n_k}\}_{k \in \Bbb N}$ i.e. $\{f_{n_k}\}_{k \in \Bbb N}$ normally convergent (as meromorphic functions i.e. with respect to the chordal metric), which would give our desired result!

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Assume that $g_{n_k} \to g$ normally with respect to the chordal metric. In order to show that $g_{n_k}^2 \to g^2$ normally it suffices to show that every $z_0 \in D$ has a neighborhood $U \subset G$ such that $g_{n_k}^2 \to g^2$ uniformly in $U$.

If $g(z_0)$ is finite then $g$ is bounded in a neighborhood $U$ of $z_0$, and from $$ |g_{n_k}^2(z) - g^2(z)| = |g_{n_k}(z) - g(z)||g_{n_k}(z) + g(z)| $$ it follows that $g_{n_k}^2 \to g^2$ uniformly in $U$ with respect to the Euclidean metric, and therefore also with respect to the chordal metric.

If $g(z_0) = \infty$ then consider $\frac{1}{g_{n_k}}$ and $\frac 1g$ in a neighborhood of $U$.

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  • $\begingroup$ "$𝑔^2_{𝑛_𝑘}→𝑔^2$ uniformly in 𝑈 with respect to the Euclidean metric, and therefore also with respect to the chordal metric" , I don't know about this implication, can you add a bit more details regarding this. $\endgroup$
    – user422112
    Apr 15 '19 at 11:33
  • $\begingroup$ @reflexive: If $d(z, w) = \frac{|z-w|}{\sqrt{1+|z|^2}\sqrt{1+|w|^2}}$ denotes the chordal metric then $d(z, w) \le |z-w|$ for $z, w \in \Bbb C$. $\endgroup$
    – Martin R
    Apr 15 '19 at 11:35

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