Suppose that $f : \mathbb{C} \to \mathbb{C}$ is entire. Let $\Omega = \{ z : \frac{1}{2} < |z| < 2\}$. Furthermore, suppose that $ \mathcal{F} = \{f(kz) : k \in \mathbb{C}\}$ is a normal family of analytic functions defined on $\Omega$; the family $\mathcal{F}$ is normal on $\Omega$ iff each sequence from $\mathcal{F}$ has a subsequence converging uniformly on compact subsets of $\Omega$. Prove that $f$ is a polynomial.

I'm studying for my qualifying exam in complex analysis, and this question appears on an old exam. Usually, when I want to show that some entire $f$ is a polynomial, I just try to show that there exists some $N$ so that $f^{(k)}(0) = 0$, all $k \ge N$, but I'm having trouble getting that strategy to work here. I've been looking at the sequence $f(\frac{z}{n})$ trying to make something work, but have not had any success.

I would prefer to receive a hint or two, and then I'll try to post my own solution. Thanks in advance!

  • $\begingroup$ Do you know whether or not the subsequence is required to converge uniformly to a finite function on compact sets, or if it is allowed to tend uniformly to $\infty$? $\endgroup$ – bryanj Jun 18 '13 at 2:42
  • $\begingroup$ The idea to look at $f(z/n)$ is good. The annuli $\{0.6n \le |z|\le 1.9n\}$ cover almost all of the plane, outside of some disk. So, if $f(z/n)$ converge uniformly on $\{0.6\le |z|\le 1.9\}$ (either to a finite function or to $\infty$), you will get a strong statement concerning the behavior of $f(z)$ as $|z|\to\infty$. $\endgroup$ – ˈjuː.zɚ79365 Jun 18 '13 at 13:28
  • $\begingroup$ @bryanj thanks for your comment, yes it's okay for the functions to tend uniformly to $\infty$. $\endgroup$ – JZS Jun 18 '13 at 16:33
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    $\begingroup$ I'm learning it, like you, so take this with a grain of salt. Suppose the limit (of the subsequence of $f(nz)$) function $g$ is bounded. Then we can find (using the subsequence) a collection of annuli of large radii with $|f| < M$ for all these annuli. Use the Cauchy Integral Formula to show (similar to the proof of Liouville's Theorem) that $f$ is constant (and so a polynomial). Im still working on the case when $g = \infty$. $\endgroup$ – bryanj Jun 19 '13 at 17:01
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    $\begingroup$ jtms88: You are right that there's an issue, but @bryanj found a way to solve it. Here is a condensed version: If the limit of subsequence is finite, the maximum principle fills the gaps between annuli. And if there is no subsequence with finite limit, then the entire sequence goes to $\infty$, and you have $|f(z)|\to \infty$ as $|z|\to\infty$. $\endgroup$ – ˈjuː.zɚ79365 Jun 19 '13 at 23:29

I'd like to attempt an answer. I will have no problem with down-votes or with deleting it, since I am not 100% confident in it.

Let $A$ be the annulus $\frac{2}{3} \le |z| \le \frac{3}{2}$. For a sequence $a_n \in \mathbb{C}$, define $f_{a_n}(z) = f(a_n z)$ and $B_{a_n} = \{a_n z : z \in A \}$.

$B_{a_n}$ is an annulus with inner radius $|a_n|\frac{2}{3}$ and outer radius $|a_n|\frac{3}{2}$.

Let $a_n = n$, n positive integer. Let $f_{n_k}$ be the subsequence of $f_n$ guaranteed by the statement. Then either $f_{n_k} \to \infty$ uniformly on $A$ or $f_{n_k} \to g$ uniformly on $A$ for an analytic function $g$.

In the second case, $g$ is analytic so $g(A)$ is bounded. Using uniform convergence, you can see that There is an $M$ such that $f_{n_k}(A) < M$ for all $k$ sufficiently large. This means $f(z) < M$ for all $z \in B_{n_k}$, with $k$ sufficiently large. Mimicking the proof of Liouville's theorem that uses the Cauchy Integral Formula, this shows that $f(z)$ is constant.

In the first case, for any $M$ you can achieve $|f_{n_k}(z)| > M$ for $k$ sufficiently large. If in addition, for $k$ sufficiently large, $B_{n_k} \cap B_{n_{k+1}}$ is non-emtpy, then this shows $f(z) \to \infty$ as $z \to \infty$, and therefore $f$ must be a polynomial.

If on the other hand there are infinitely many $k$ with $B_{n_k} \cap B_{n_{k+1}}$ empty, then we need to look at what happens in these "gaps". Note that the outer radius of $B_{n_k}$ is in this case strictly less than the inner radius of $B_{n_{k+1}}$. This "gap" is an open annulus, but $f$ is analytic on the closure. Suppose such a gap does not contain a zero of $f$. Then the Minimum Modulus Theorem says that the modulus of $f$ in the gap is at least as great as the minimum modulus on the boundary, and so in the gap $f|(z)| > M$.

Therefore if only finitely many gaps contain zeros, we again see that $f(z) \to \infty$ as $z \to \infty$.

Finally, suppose there are infinitely $n_k$ with $B_{n_k} \cap B_{n_{k+1}}$ empty, and infinitely many of these "gaps" contain a zero of $f$. For the $m$-th gap pick a zero $b_m$, with $|b_m| < |b_{m+1}|$ and $|b_m| \to \infty$. These $b_m$ define a new sequence $f_{b_m}$ of functions on $A$, a subsequence of which converges uniformly. But in this case it cannot converge uniformly to $\infty$, because each function in the sequence has a zero at $z = 1$. Then as in the proof of the second case, $f$ must be constant.

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    $\begingroup$ One thing that worries me about this is that it seems like the conclusion of the problem would remain true even if the original domain $\Omega$ were not required to be an annulus, but instead was only required to be an open set containing some circle $|z| = r$. $\endgroup$ – bryanj Jun 19 '13 at 21:24
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    $\begingroup$ Sure, it would. If an open set contains some circle $|z|=r$, then it contains an annulus of the form $r_1<|z|<r_2$, and we are in the setting of the original problem, only with different numbers in place of $1/2$ and $2$. $\endgroup$ – ˈjuː.zɚ79365 Jun 19 '13 at 23:33
  • $\begingroup$ @bryanj - nice job, clear all the way through and it does the job I think. Thanks! $\endgroup$ – JZS Jun 20 '13 at 13:08
  • $\begingroup$ I'm sorry but I have several questions, and it wouod be great if you could answer them: (1) why is g(A) bounded in the first case? (2) why does $f(z) \to \infty$ as $z \to \infty$ imply that $f$ is a polynomial? $\endgroup$ – Cute Brownie Dec 7 '18 at 9:32

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