$g\circ f$ non-constant polynomials then $f$ and $g$ are polynomials?

Question

Let $f$, $g$ be non-constant entire functions. If the composition $g\circ f$ is a non-constant polynomial, can we conclude $f$ and $g$ are polynomials?

Answer 1

Assume that $f$ is entire and not a polynomial. Then $f$ has an essential singularity at $\infty$.

Since $g$ is non-constant and entire, by Picard's Little Theorem, $g(\mathbb C) = \mathbb C$ or there is $w \in \mathbb C$ such that $g(\mathbb C) = \mathbb C \setminus \{ w\}$. In either case $g(\mathbb C)$ contains a neighborhood $\mathbb C \setminus D_R$ of $\infty$ (here $D_R$ is the disc of some radius $R$). Since $f$ has an essential singularity at $\infty$, Picard's Great Theorem tells us that $f(z)$ takes every value in $\mathbb C$ infinitely often (with the possible exception of one point) on $\mathbb C \setminus D_R$. But this shows that $(f\circ g)(z)$ takes on every value in $\mathbb C$ infinitely many times (except possibly one point). Then $f\circ g$ cannot be a polynomial since a polynomial of degree $N$ can take a given value at most $N$ times. Contrapositively, if $f \circ g$ is a polynomial, then $f$ is a polynomial.

Assume that $g$ is entire and not a polynomial. Then $g$ has an essential singularity at $\infty$. For fixed $R > 0$, by Picard's Great Theorem, $g(z)$ takes every value in $\mathbb C$ infinitely often (except possibly one value) on $\mathbb C \setminus D_R$. But then in particular, there is $a \in \mathbb C$ such that $g(z) = a$ for infinitely many $z \in \mathbb C$. But then $(f \circ g)(z) = f(a)$ for infinitely many $z \in \mathbb C$, once again implying that $f\circ g$ is not a polynomial. Contrapositively, if $f\circ g$ is a polynomial, then $g$ is a polynomial.

Thus both $f$ and $g$ are polynomials.

Answer 2

Some basic properties: 1. If $h$ is entire and nonconstant, then $h(\mathbb C)$ is dense in $\mathbb C.$ (Casorati-Weierstrass and the fundamental theorem of algebra). Thus if $D$ is dense in $\mathbb C,$ then $h(D)$ is dense in $\mathbb C.$

2. If $p$ is a nonconstant polynomial, then for every $R>0,$ there exists $R'>0$ such that $p(\{|z|>R\})$ contains $\{|z|>R'\}.$ (A proof of this is sketched in the comments.)

Now in our problem $g\circ f$ is a nonconstant polynomial. Thus $(g\circ f) (\{|z|>R\})$ is not dense in $\mathbb C$ for some $R>0.$

If $f$ is not a polynomial, then $f (\{|z|>R\})$ is dense in $\mathbb C$ ((Casorati-Weierstrass). Hence $g(f (\{|z|>R\}))$ is dense in $\mathbb C$ by 1. above, conradiction. Thus $f$ is a polynomial.

By 2., this implies $f (\{|z|>R\})$ contains some $\{|z|>R'\}.$ We can assume $R'>R.$ If $g$ is not a polynomial, then it maps $\{|z|>R'\}$ to a dense subset of $\mathbb C$ (Casorati-Weierstrass). Hence $g\circ f$ maps $\{|z|>R\}$ to a dense subset of $\mathbb C.$ This is a contradiction, proving $g$ is a polynomial.