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I'm trying to prove the following exercise from Gamelin's Complex Analysis:

Exercise XII.$1$.$5$ (p. $319$):
Fix $M>1$, and let $\{f_n(z)\}$ be a sequence of meromorphic functions on a domain $D$. We are asked to show that $f_n(z) \rightarrow f(z)$ normally iff $f_n(z) \rightarrow f(z)$ uniformly on compact subsets of the open set $\{|f(z)|<M\}$, and $1/f_n(z) \rightarrow 1/f(z)$ uniformly on compact subsets of the open set $\{|f(z)|>1/M\}$.

Here's what I did so far (one direction):
$\Leftarrow$: It's clear that $f_n \rightarrow f$ uniformly on compact subsets of $\{|f(z)|<M\} \Leftrightarrow \frac{1}{f_n} \rightarrow \frac{1}{f}$ uniformly on compact subsets of $\{|f(z)|>1/M\}$. We show that $f_n \rightarrow f$ normally on $D$. This is where I got kind of stuck.

Any suggestions as to how I should proceed with the proof? Wouldn't those two open sets given in the exercise be compact subsets of $D$? Too bad the exercise wasn't related to spherical derivatives, because then I think all I'll have to use is Marty's Theorem in some way I guess. I'd appreciate any guidance and help on this question that anyone can provide, thanks.

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2 Answers

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One reason you might be stuck is that it is far from clear, as you say in your proof attempt, that $f_n \to f$ uniformly on compact subsets of $\{z \in D: |f(z)| < M\}|$ if and only if $\frac{1}{f_n} \to \frac{1}{f}$ uniformly on compact subsets of of $\{z \in D: |f(z)| > 1/M\}$. In fact this is not true.

As an example:

  • Take $M = 2$, and $f_n(z) = z + (z/4)^n$, and $f(z) = z$, and $D = \mathbb{C}$.
  • The set $\{z \in D: |f(z)| < M\}$ is the set $\{z \in \mathbb{C}: |z| < 2\}$, and for any $w$ in this set and $n \geq 1$ we have $$ |f_n(w) - f(w)| = |w + (w/4)^n - w| = |w|^n/4^n < 2^n/4^n = 1/2^n, $$ which (since $1/2^n \to 0$ as $n \to \infty$) shows that $f_n \to f$ uniformly on $\{z \in \mathbb{C}: |z| < 2\}$ (and in particular on all compact subsets of it).
  • The set $\{z \in D: |f(z)| > \frac{1}{M}\}$ is the set $\{z \in \mathbb{C}: |z| > 1/2\}$. The complex number $4$ is in this set, and for all $n$ we have $$ \left|\frac{1}{f_n(4)} - \frac{1}{f(4)}\right| = \left|\frac{1}{5} - \frac{1}{4}\right| = \frac{1}{20}, $$ showing that the sequence $1/f_n(4)$ does not converge to $1/f(4)$ (i.e., the sequence $\frac{1}{f_n}$ does not even converge uniformly to $\frac{1}{f}$ on the compact subset $\{4\}$ of $\{z \in D: |f(z)| > \frac{1}{M}\}$ ).

So that is one reason you might be stuck. If you thought those two conditions implied each other, then you might think that you only need to make use of one of them in your proof of the "if" direction of the given statement. But limiting yourself to only one of them, you will get stuck, because you really do need both. They don't imply each other.

Here is a sketch of the "if" direction: suppose that $f_n \to f$ uniformly on the set $\{z \in D: |f(z)| < M\}$ and that $\frac{1}{f_n} \to f$ uniformly on $\{z \in D: |f(z)| > \frac{1}{M}\}$. To prove that $f_n \to f$ normally, we need to prove that something nice happens for every compact subset of $D$. So, fix a compact subset $K$ of $D$. Complete the following outline:

  • To prove that $f_n$ converges uniformly in the spherical metric to $f$ on all of $K$, it suffices to exhibit two compact subsets $K_1$ and $K_2$ of $K$, whose union is $K$, and prove for both $i = 1, 2$ that $f_n$ converges uniformly to $f$ in the spherical metric on the set $K_i$.

  • Take $K_1 = \{z \in K: |f(z)| \leq 1\}$ and $K_2 = \{z \in K: |f(z)| \geq 1\}$.

    • Note that since $M > 1$, the set $K_1$ is a compact subset of $\{z \in D: |f(z)| < M\}$.
      • So you know by hypothesis that $f_n$ converges uniformly to $f$ on $K_1$ (this uniform convergence is in the sense of the usual metric).
      • Use this, together with the fact that the spherical metric is equivalent to the usual one on bounded sets (in particular, on $\{w \in \mathbb{C}: |w| < M\}$ ) to conclude that $f_n$ converges uniformly to $f$ on $K_1$ in the spherical metric.
    • Note that since $M > 1$, the set $K_2$ is a compact subset of $\{z \in D: |f(z)| > \frac{1}{M}\}$.
      • So you know by hypothesis that $1/f_n$ converges uniformly to $1/f$ on $K_1$ (in the usual metric).
      • Use the fact that the spherical metric $s(u,v)$ is comparable to $|1/u - 1/v|$ when $u$ and $v$ come from a set that is some positive distance from the origin, to conclude that $f_n$ converges uniformly to $f$ on $K_2$ in the spherical metric. (Here the relevant $u$'s and $v$'s are values of $f$, which by hypothesis are coming from the set $\{w \in \mathbb{C}: |w| > \frac{1}{M}\}$, which has positive distance from the origin.)

That completes the "if" direction. The "only if" is similar. If $f_n \to f$ normally, if you fix a compact subset $K$ of $\{z \in D: |f(z)| < M\}$, using the equivalence of the spherical metric to the standard one on the bounded set $\{w \in \mathbb{C}: |w| < M\}$, you can deduce that $f_n \to f$ uniformly on $K$, and if you fix a compact subset $K$ of $\{z \in D: |f(z)| > 1/M\}$, you can use the equivalence of the spherical metric to the "inverted" standard one on sets at some distance from the origin to get what you want.

[I saw Robert's hint pop up as I was writing this; it is a concise summary of the most significant idea used in the above solution. This idea is, I think, the way any book sort of "intends" you to solve the problem. But it might be difficult to implement it, if you think the two hypotheses of the "if" direction are equivalent and forget to use one of them. So I continued to write this anyway.]

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Hint: Consider any compact subset $K$ of $D$, and write $K = K_1 \cup K_2$ where $K_1 = \{z \in K: |f(z)| \le 1\}$ and $K_2 = \{z \in K: |f(z)| \ge 1\}$.

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