Hadamard factorization theorem

In mathematics, and particularly in the field of complex analysis, the Hadamard factorization theorem asserts that every entire function with finite order can be represented as a product involving its zeroes and an exponential of a polynomial. It is named for Jacques Hadamard.

The theorem may be viewed as an extension of the fundamental theorem of algebra, which asserts that every polynomial may be factored into linear factors, one for each root. It is closely related to Weierstrass factorization theorem, which does not restrict to entire functions with finite orders.

Define the Hadamard canonical factors $${\displaystyle E_{n}(z):=(1-z)\prod _{k=1}^{n}e^{z^{k}/k}}$$Entire functions of finite order ${\displaystyle \rho }$ have Hadamard's canonical representation:¹$${\displaystyle f(z)=z^{m}e^{Q(z)}\prod _{n=1}^{\infty }E_{p}(z/a_{n})}$$where ${\displaystyle a_{k}}$ are those roots of ${\displaystyle f}$ that are not zero (${\displaystyle a_{k}\neq 0}$), ${\displaystyle m}$ is the order of the zero of ${\displaystyle f}$ at ${\displaystyle z=0}$ (the case ${\displaystyle m=0}$ being taken to mean ${\displaystyle f(0)\neq 0}$), ${\displaystyle Q}$ a polynomial (whose degree we shall call ${\displaystyle q}$), and ${\displaystyle p}$ is the smallest non-negative integer such that the series$${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{|a_{n}|^{p+1}}}}$$converges. The non-negative integer ${\displaystyle g=\max\{p,q\}}$ is called the genus of the entire function ${\displaystyle f}$. In this notation,$${\displaystyle g\leq \rho \leq g+1}$$In other words: If the order ${\displaystyle \rho }$ is not an integer, then ${\displaystyle g=[\rho ]}$ is the integer part of ${\displaystyle \rho }$. If the order is a positive integer, then there are two possibilities: ${\displaystyle g=\rho -1}$ or ${\displaystyle g=\rho }$.

Furthermore, Jensen's inequality implies that its roots are distributed sparsely, with critical exponent ${\displaystyle \alpha \leq \rho \leq g+1}$.

For example, ${\displaystyle \sin }$, ${\displaystyle \cos }$ and ${\displaystyle \exp }$ are entire functions of genus ${\displaystyle g=\rho =1}$.

Define the critical exponent of the roots of ${\displaystyle f}$ as the following:$${\displaystyle \alpha :=\limsup \limits _{r}\log _{r}N(f,r)}$$where ${\displaystyle N(f,r)}$ is the number of roots with modulus ${\displaystyle <r}$. In other words, we have an asymptotic bound on the growth behavior of the number of roots of the function:$${\displaystyle \forall \epsilon >0\quad N(f,r)\ll r^{\alpha +\epsilon },{\text{ and exists a sequence }}r_{k}{\text{ such that }}N(f,r_{k})>kr_{k}^{\alpha -\epsilon }}$$It's clear that ${\displaystyle \alpha \geq 0}$.

Theorem:² If ${\displaystyle f}$ is an entire function with infinitely many roots, then$${\displaystyle \alpha =\inf \left\{\beta :\sum _{k}|a_{k}|^{-\beta }<\infty \right\}={\frac {1}{\liminf \limits _{k}\log _{k}|a_{k}|}}}$$Note: These two equalities are purely about the limit behaviors of a real number sequence ${\displaystyle |a_{1}|\leq |a_{2}|\leq \cdots }$ that diverges to infinity. It does not involve complex analysis.

Proposition: ${\displaystyle \alpha (f)\leq \rho }$,³ by Jensen's formula.

With Hadamard factorization we can prove some special cases of Picard's little theorem.

Theorem:⁴ If ${\displaystyle f}$ is entire, nonconstant, and has finite order, then it assumes either the whole complex plane or the plane minus a single point.

Proof: If ${\displaystyle f}$ does not assume value ${\displaystyle z_{0}}$, then by Hadamard factorization, ${\displaystyle f(z)-z_{0}=e^{Q(z)}}$ for a nonconstant polynomial ${\displaystyle Q}$. By the fundamental theorem of algebra, ${\displaystyle Q}$ assumes all values, so ${\displaystyle f(z)-z_{0}}$ assumes all nonzero values.

Theorem:⁴ If ${\displaystyle f}$ is entire, nonconstant, and has finite, non-integer order ${\displaystyle \rho }$, then it assumes the whole complex plane infinitely many times.

Proof: For any ${\displaystyle w\in \mathbb {C} }$, it suffices to prove ${\displaystyle f(z)-w}$ has infinitely many roots. Expand ${\displaystyle f(z)-w}$ to its Hadamard representation ${\displaystyle f(z)-w=e^{Q(z)}z^{m}\prod _{k}E_{p}(z/a_{k})}$. If the product is finite, then ${\displaystyle \rho =g}$ is an integer.