# $\{g_{n_k}\}_{k \in \Bbb N}$ normally convergent $\implies \{g^2_{n_k}\}_{k \in \Bbb N}$ normally convergent (as meromorphic functions)?

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!

• @MartinR injectivity of $\mathscr F$
– user422112
Apr 15 '19 at 11:14
• this has been asked before math.stackexchange.com/questions/3146837/… Apr 15 '19 at 12:38

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 $$U$$.
• "$𝑔^2_{𝑛_𝑘}→𝑔^2$ 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.
• @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 $z, w \in \Bbb C$. Apr 15 '19 at 11:35