Problem statement:

Let $\displaystyle f$ be an entire function and set $\displaystyle \mathscr{F}(f) = \{f(2^n z) : n \in \{0, 1, 2, \dots \} \}$. Determine all entire functions $\displaystyle f$ such that $\displaystyle \mathscr{F}(f)$ is a compact normal family on $\displaystyle \mathbb{C}$.

Defs:I'm including these as I'm clear on the definition of normal and compact, but unsure of the definition for "compact normal". Please comment if you think I have a definition in error.

Compact: I'm assuming the problem's author implies that we have sequential compactness, i.e., every sequence from $\displaystyle \mathscr{F}(f)$ has a limit in the set.

Normal: Every sequence from $\displaystyle \mathscr{F}(f)$ has a subsequence that converges uniformly on any compact subset of $\displaystyle \mathbb{C}$.

So compact normal means that any sequence from $\displaystyle \mathscr{F}(f)$ has a subsequence that converges uniformly on any compact subset of $\displaystyle \mathbb{C}$ to a function in $\displaystyle \mathscr{F}(f)$.

Ideas:

Any function $\displaystyle f_n \in \mathscr{F}(f)$ is entire since it is the composition of two entire functions, $\displaystyle f$ and $\displaystyle 2^n z$.

Examples: The constant functions work trivially.

In the class of non-constant entire functions, polynomials fail, $\displaystyle e^z$ fails.

I don't see how to make the general argument. Liouville's theorem identifies that any non-constant functions are unbounded. So if we can prove $\displaystyle f$ must be bounded we're done. Montel's theorem gives us $\displaystyle \mathscr{F}(f)$ a normal family if we have $\displaystyle \mathscr{F}(f)$ locally uniformly bounded. This doesn't imply f bounded though.

I'm sure that I'm overlooking something ... thank you for your help.