Fall 2012 - Problem 8

Solution.

Let $A = \set{a_n}_n$ denote the distinct zeroes of $f$ and let $2m_n$ be the multiplicity of $a_n$, where $m_n \in \N$. Indeed, because $f$ is non-constant, the zeroes of $f$ must be isolated, so each compact $\cl{B\p{0, N}}$ can contain only finitely many $a_n$, or else they accumulate by subsequential compactness. Thus,

i.e., it is a countable union of finite sets, so it is at most countable.

If $\abs{A} = N$ is finite, then by the factor theorem,

where $b$ vanishes nowhere. If $A$ is infinite, then because $A$ cannot have infinitely many roots on any $\cl{B\p{0, N}}$, it follows that $\lim_{n\to\infty} \abs{a_n} = \infty$. Hence, we may apply the Weierstrass theorem, which gives existence of an entire function $p\p{z}$ such that $p$ has roots $A$, but $a_n$ has multiplicity $m_n$ instead. Moreover, $b = \frac{f}{p^2}$ vanishes nowhere, so we have the factorization

In both cases, we may write $f\p{z} = b\p{z}p^2\p{z}$, where $b$ vanishes nowhere, $p$ has the same roots as $f$ but with half the multiplicity, and both $b, p$ are entire.

Since $b$ vanishes nowhere, $b$ has a logarithm, i.e., there exists an entire function $c$ satisfying $b = e^c$. Thus, if we set

then $g$ is a composition and product of entire functions, hence entire itself. Moreover,

which completes the proof.