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.
On
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!
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.