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I am trying to prove the following statement:

If $f(z)$ is entire, show that the family formed by all the functions $f(kz)$ with constant k is normal in the annulus $r_1 < |z|<r_2$ If and only if $f$ Is a polynomial.

I have been trying to write, unsuccessfully, a concise proof of this fact just using Arzela-Ascoli, Montel’s theorem, Marty’s theorem or results in this line.

Thank you for your help.

Vcwh
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    I'm unfamiliar with the notion of a normal family. What does that mean? – Arthur Apr 05 '20 at 03:33
  • There are several equivalences, but normal means that any sequence in the family has a normal convergent subsequence. Here we allow the limit to be infinity. Normal convergence means that the subsequence converges uniformly on any compact subset in the domain of definition. – Vcwh Apr 05 '20 at 09:54
  • Not exactly, that proof is intricate. I just want to find a “more concise” proof of this fact. That’s really intricate. Thank you for your comment. – Vcwh Apr 05 '20 at 15:40
  • @caffeine The proof linked is not complete as $b_m$ in the last step may not be integral so $f_{b_m}$ doesn't make sense; it is fixable and applicable here by taking integer parts and using that the image annuli grow indefinitely apart if they do not intersect; not sure what the OP means by a more concise proof – Conrad Apr 05 '20 at 18:37
  • @Conrad I do not see why they should be integral for $f_{b_m}$ to make sense: they are a countable number, and that should suffice. –  Apr 05 '20 at 19:15
  • @caffeine The hypothesis states that $f(kz)$ is normal for $k \ge1$ integer, but doesn't say anything about arbitrary $f(az)$, so apriori we do not know that $f(b_mz)$ is part of a normal family; or at least that's what i thought since otherwise the result is easy – Conrad Apr 05 '20 at 20:03
  • @Conrad the hypotesis stated is simply $k\in \mathbb{C}$. If the result is easy you could post the answer –  Apr 05 '20 at 20:12

2 Answers2

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This is not at all true. If $f(z)=z$ then the given family is not normal. Normailty demands that the functions are uniformly bounded on any compact subset of the annulus. But $\{kz: k \in \mathbb C\}$ is not bounded even on one-point sets.

Actually the given family is normal iff $f$ is a constant.

Kavi Rama Murthy
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    Sometimes uniform divergence to infinity on compact sets (of a subsequence) is also allowed in the definition of a normal family as that is uniform convergence to the constant $\infty$ in the extended plane and with that definition, the OP problem is actually correct – Conrad Apr 05 '20 at 05:34
  • This is the analytic case, as Conrad said you can consider the extended plane and therefore merophorphic functions. – Vcwh Apr 05 '20 at 09:52
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Let's prove that if $f$ is entire and there is an annulus $r < |z| <R$ s.t the family $f(az), a \in \mathbb C$ is normal (in the extended sense where we allow compact divergence to infinity too), then $f$ is a polynomial. Let $A$ be the closed annulus $r+\delta \le |z| \le R-\delta$ for small enough $2\delta <R-r$.

We can assume wlog $f$ is not constant and we will prove that $f$ takes any value finitely many times only. This clearly implies $f$ polynomial.

Pick $w$ complex and assume $f(z)=w$ has infinitely many roots $z_n$. Clearly $z_n \to \infty$ since $f$ nonconstant, so there is $a_n \to \infty$, $\frac{z_n}{a_n}=y_n \in A$. But then $f(a_nz)-w$ has a zero at $y_n \in A$, so by normality if we take a subsequence of $f(a_nz)-w \to g$ uniformly on compact subsets of the original annulus, hence in particularly on $A$, $g(y)=0$ for any $y \in A$ limit point of $y_n$ with $n$ in the respective subsequence. On the other hand if $f(a_nz)-w$ hence $f(a_nz)$ is uniformly bounded by some $M$ on $A$, it follows by maximum modulus that $f$ is bounded in the plane as $a_n \to \infty$ and that implies $f$ constant which is not possible, so any subsequence of $f(a_nz)-w$ must diverge compactly to infinity on $A$, so $g$ constructed above must be identically infinity, which contradicts the fact that $g(y)=0$. Done!

Conrad
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  • @r9m - thank you for the comment and did what you suggested - agreed with your point and while I get why people want questions to be motivated, there is lots of unnecessary repetition because of that – Conrad Apr 05 '20 at 21:36