As mentioned in the title, given $0 \leq r_1<r_2 \leq +\infty$, and an entire function $f : \mathbb C \to \mathbb C$, I want to prove that $$\mathfrak F= \{f(kz) \}_{k \in \mathbb C} \text{ is normal in } A=\{r_1<|z|<r_2 \} \Longleftrightarrow f \text{ is a polynomial} $$
We will start with the $\Longleftarrow$ part:
- If $f=\text{const}$, the family consists of only one element, and is obviously normal.
Suppose then that $f$ is a non-constant polynomial, and let $\{f(k_n z) \}_n \in \mathfrak F$ be any sequence.
If the sequence $\{ k_n \}$ of complex numbers isn't bounded, we may find a subsequence $\{ k_{n_l}\}$ which tends to $\infty$. Let $K \Subset A$ be any compact subset, since $f$ has a pole at $\infty$, and since $|z| \geq m_K >r_1 \geq 0$ is bounded below in $K$, the arguments $\{k_{n_l} z \}_{z \in K}$ can be made uniformly close (in $K$) to $\infty$ for large $l$, and thanks to the pole, $f(k_{n_l} z)$ tends to $\infty$ uniformly on $K$.
If the sequence $\{ k_n \}$ is bounded, we may extract a convergent subsequence $\{k_{n_l} \}$ which tends to some $k_0 \in \mathbb C$. Let $K \Subset A$ be any compact subset. All number $\{k_{n_l} z \}$ lie in the closed disk $ \overline{\Delta}(0,M)$ where $M=\max_{z \in K} |z| \cdot \sup_l |k_{n_l}|$, on which $f$ is uniformly continuous. The distances $d(k_{n_l} z,k_0 z)=|z| |k_{n_l}-k_0|$ can be made uniformly small in $K$. Hence $f(k_{n_l} z)$ converges to $f(k_0 z)$ uniformly in $K$.
In any case, we have shown that if $f$ is a polynomial, then $\mathfrak F$ is normal.
Now for the $\Longrightarrow$ part:
Suppose that $\mathfrak F$ is normal in $A$, but nonetheless $f$ isn't a polynomial. In that case $f$ has an essential singularity at $\infty$. According to Marty's theorem the expressions $$\left\{ \frac{ 2 |k| |f'(kz)|}{1+|f(kz)|^2} \right\}_{k \in \mathbb C}$$ are locally bounded in $A$. Pick some point $z_0 \in A$, and consider the compact set $K=\{z_0 \}$. Writing $k=\frac{c}{z_0}$, we see that there exists some $M$, such that $$2 \frac{|c|}{|z_0|} \frac{|f'(c)|}{1+|f(c)|^2} \leq M $$ for all $c \in \mathbb C$, or $$\frac{|f'(c)|}{1+|f(c)|^2} \leq \frac{M |z_0|}{2|c|}. $$
This means that for large $|c|$, either $|f'(c)|$ is small or $|f(c)|$ is large, and I'm pretty much stuck here... My gut tells me that I should I somehow combine the essential singularities of $f,f'$, but I couldn't do it.
I have two questions:
- Is my proof of the $\Longleftarrow$ part correct?
- Can you please help me complete the other part? (A whole new proof is also valid)
Thanks!