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 $\rho$ have Hadamard's canonical representation:$$f(z)=z^{m}e^{Q(z)}\prod _{n=1}^{\infty }E_{p}(z/a_{n})$$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 $f(0)\neq 0$), $Q$ 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 $g=\rho$.

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

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:$$\alpha :=\limsup \limits _{r}\log _{r}N(f,r)$$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:$$\forall \epsilon >0\quad N(f,r)\ll r^{\alpha +\epsilon },{\text{ and exists a sequence }}r_{k}{\text{ such that }}N(f,r_{k})>kr_{k}^{\alpha -\epsilon }$$It's clear that $\alpha \geq 0$.

Theorem: If $f$ is an entire function with infinitely many roots, then$$\alpha =\inf \left\{\beta :\sum _{k}|a_{k}|^{-\beta }<\infty \right\}={\frac {1}{\liminf \limits _{k}\log _{k}|a_{k}|}}$$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. It does not involve complex analysis.

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

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

Define the function $g(z):=\prod _{k=1}^{\infty }E_{p}(z/a_{k})$ where $p={\text{floor}}(\alpha )$. We will study the behavior of $f(z)/g(z)$.

Bounds on the behaviour of |E_p|

In the proof, we need four bounds on $|E_{p}|$:

1. For any $r\in (0,1)$, $|1-E_{p}(z)|\leq O(|z|^{p+1})$ when $|z|\leq r$.
2. For any $r\in (0,1)$, there exists $B>0$ such that $\ln |E_{p}(z)|\geq -B|z|^{p+1}$ when $|z|\leq r$.
3. For any $r\in (0,1)$, there exists $B>0$ such that $|E_{p}(z)|\geq |1-z|e^{-B|z|^{p}}$ when $|z|\geq r$.
4. $\ln |E_{p}(z)|\leq O(|z|^{p+1})$ for all $z$, and $\ln |E_{p}(z)|\leq O(|z|^{p})$ as $|z|\to \infty$.

These are essentially proved in the similar way. As an example, we prove the fourth one.$$\ln |E_{n}(z)|=Re\left(-z^{n+1}\sum _{k=0}^{\infty }{\frac {z^{k}}{k+n+1}}\right)\leq |z|^{n+1}|g(z)|$$where $g(z):=\sum _{k=0}^{\infty }{\frac {z^{k}}{k+n+1}}$ is an entire function. Since it is entire, for any $R>0$, it is bounded in $B(0,R)$. So $\ln |E_{n}(z)|=O(|z|^{n+1})$ inside $B(0,R)$.

Outside $B(0,R)$, we have$$\ln |E_{n}(z)|=\ln |1-z|+Re(Q(z))=o(|z|)+O(|z|^{n})<O(|z|^{n+1})$$

g is well-defined

Source:

For any $R>0$, we show that the sum $\sum _{k=1}^{\infty }\ln |E_{p}(z/a_{k})|$ converges uniformly over $B(0,R)$.

Since only finitely many $|a_{k}|<KR$, we can split the sum to a finite bulk and an infinite tail:$$\sum _{k=1}^{\infty }\ln |E_{p}(z/a_{k})|=\sum _{|a_{k}|<KR}\ln |E_{p}(z/a_{k})|+\sum _{|a_{k}|\geq KR}\ln |E_{p}(z/a_{k})|$$The bulk term is a finite sum, so it converges uniformly. It remains to bound the tail term.

By bound (1) on $|E_{p}|$, $|1-E_{p}(z/a_{k})|\leq B|z/a_{k}|^{p+1}\leq B/K^{p+1}$. So if $K$ is large enough, for some $B'>0$,$$\sum _{|a_{k}|\geq KR}\ln |1-(1-E_{p}(z/a_{k}))|\leq \sum _{|a_{k}|\geq KR}B'B|z/a_{k}|^{p+1}<|z|^{p+1}B'B\sum _{k}{\frac {1}{|a_{k}|^{p+1}}}$$Since $p+1={\text{floor}}(\alpha )+1>\alpha$, the last sum is finite.

g ≤ floor(ρ)

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

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 ${\frac {1}{|a_{k}|^{\rho +\epsilon }}}$. Then since $\sum _{k}{\frac {1}{|a_{k}|^{\rho +\epsilon }}}<\infty$ 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, $deg(Q)\leq {\text{floor}}(\rho +2\epsilon )$ for any $\epsilon >0$, and so as we take $\epsilon \to 0$, we obtain $g\leq {\text{floor}}(\rho )$.

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:$$\sum _{|a_{k}|<|z_{k}|/2}\ln |E_{p}(z/a_{k})|+\sum _{|a_{k}|\geq |z_{k}|/2}\ln |E_{p}(z/a_{k})|$$The upper-bounding can be accomplished by the bounds (2), (3) on $|E_{p}|$, and the assumption that $z$ is outside every forbidden disk. Details are found in.

ρ ≤ g + 1

This is a corollary of the following:

If $f(z)=e^{Q(z)}z^{m}\prod _{k}E_{p}(z/a_{k})$ has genus $g$, then $\ln |f(z)|=O(|z|^{g+1})$.

Split the sum to three parts:$$\ln |f(z)|=Re(Q(z))+m\,Re(\ln z)+\sum _{k}\ln |E_{p}(z/a_{k})|.$$ The first two terms are $O(|z|^{p})=o(|z|^{g+1})$. The third term is bounded by bound (4) of $|E_{p}|$:$$\sum _{k}\ln |E_{p}(z/a_{k})|\leq B|z|^{q+1}\sum _{k}{\frac {1}{|a_{k}|^{q+1}}}.$$ By assumption, $p={\text{floor}}(\alpha )$, so $\sum _{k}{\frac {1}{|a_{k}|^{q+1}}}<\infty$. Hence the above sum is $O(|z|^{q+1})=O(|z|^{g+1}).$

Applications

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

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

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

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

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