# Thread: Family of Entire Functions

1. ## Family of Entire Functions

Problem statement:

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

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 $\mathscr{F}(f)$ has a limit in the set.

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

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

Ideas:

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

Examples: The constant functions work trivially.

In the class of non-constant entire functions, polynomials fail, $e^z$ 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 $f$ must be bounded we're done. Montel's theorem gives us $\mathscr{F}(f)$ a normal family if we have $\mathscr{F}(f)$ locally uniformly bounded. This doesn't imply f bounded though.

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