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.

Formal statement

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

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

Critical exponent

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

Theorem (): If $f$ is an entire function with infinitely many roots, then$$\alpha =\inf \left\{\beta :\sum _{k}|a_{k}|^{-\beta }<\infty \right\}={\frac {1}{\liminf _{k}\log _{k}|a_{k}|}}$$Note: These two equalities are purely about the limit behaviors of a real number sequence $|a_{1}|\leq |a_{2}|\leq \cdots$ that diverges to infinity.

( Theorem 12.3.4): $\alpha (f)\leq \rho$. This is proved by Jensen's formula.

Proof

Since $f(z)/z^{m}$ is also an entire function with the same order $\rho$ and genus, we can WLOG assume $m=1$.

If $f$ has only finitely many roots, then $f(z)=e^{P(z)}\prod _{k=1}^{n}E_{0}(z/a_{k})$ with the function $e^{P(z)}$ of order $\rho$. Thus by an application of the Borel–Carathéodory theorem, $P$ is a polynomial of degree $deg(P)\leq floor(\rho )$, and so we have $\rho =deg(P)=g$.

Otherwise, $f$ has infinitely many roots. This is the tricky part and requires splitting into two cases. First show that $g\leq floor(\rho )$, then show that $\rho \leq g+1$.

Proving $g\leq floor(\rho )$

As usual in analysis, we fix some small $\epsilon >0$.

Define the function $g(z):=\prod _{k=1}^{\infty }E_{p}(z/a_{k})$ where $p=floor(\rho )$, then the goal is to show that $f(z)/g(z)$ is of order $\leq \rho +\epsilon$. This does not exactly work, however, due to bad behavior of $f(z)/g(z)$ near $a_{k}$. Consequently, we need to pepper the complex plane with "forbidden disks", one around each $a_{k}$, each with radius ${\frac {1}{|a_{k}|^{\alpha +\epsilon }}}$. Then since $\sum _{k}{\frac {1}{|a_{k}|^{\alpha +\epsilon }}}<\infty$ by the previous result on $\alpha$, we can pick an increasing sequence of radii $R_{1}<R_{2}<\cdots$ that diverge to infinity, such that each circle $\partial B(0,R_{n})$ avoids all these forbidden disks.

Thus, if we can prove a bound of form $\ln \left|{\frac {f(z)}{g(z)}}\right|=O(|z|^{\rho +\epsilon })=o(|z|^{\rho +2\epsilon })$ for all large $z$ that avoids these forbidden disks, then by the same application of Borel–Carathéodory theorem, $deg(P)\leq floor(\rho +2\epsilon )$ for any $\epsilon >0$, and so as we take $\epsilon \to 0$, we obtain $g\leq floor(\rho )$.

Since $\ln \left|f(z)\right|=o(|z|^{\rho +\epsilon })$ by the definition of $\rho$, it remains to show that $-\ln \left|g(z)\right|=O(|z|^{\rho +\epsilon })$, that is, there exists some constant $B>0$ such that $$\sum _{k=1}^{\infty }\ln |E_{p}(z/a_{k})|\geq -B|z|^{\rho +\epsilon }$$for all large $z$ that avoids these forbidden disks.

As usual in analysis, this infinite sum can be split into two parts: a finite bulk and an infinite tail term, each of which is to be separately handled. There are finitely many $a_{k}$ with modulus $<|z_{k}|/2$ and infinitely many $a_{k}$ with modulus $\geq |z_{k}|/2$. So we have to bound:$$\sum _{|a_{k}|<|z_{k}|/2}\ln |E_{p}(z/a_{k})|+\sum _{|a_{k}|\geq |z_{k}|/2}\ln |E_{p}(z/a_{k})|$$This bound can be accomplished by the three bounds on $|E_{p}|$:

• There exists $B>0$ such that $\ln |E_{p}(z)|\geq -B|z|^{p+1}$ when $|z|\leq 1/2$.
• There exists $B>0$ such that $|E_{p}(z)|\geq |1-z|e^{-B|z|^{p}}$ when $|z|\geq 1/2$.