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== Hadamard factorization theorem ==
{{Main page|Hadamard factorization theorem}}
Define the Hadamard canonical factors <math display="block">E_n(z) := (1-z) \prod_{k=1}^n e^{z^k/k}</math>Entire functions of finite [[Entire function|order]] have [[Jacques Hadamard|Hadamard]]'s canonical representation<ref name="conway" />:<math display="block">f(z)=z^me^{P(z)}\prod_{n=1}^\infty E_p(z/a_k)</math>where <math>a_k</math> are those [[Zero of a function|roots]] of <math>f</math> that are not zero (<math>a_k \neq 0</math>), <math>m</math> is the order of the zero of <math>f</math> at <math>z = 0</math> (the case <math>m = 0</math> being taken to mean <math>f(0) \neq 0</math>), <math>P</math> a polynomial (whose degree we shall call <math>q</math>), and <math>p</math> is the smallest non-negative integer such that the series<math display="block">\sum_{n=1}^\infty\frac{1}{|a_n|^{p+1}}</math>converges. The non-negative integer <math>g=\max\{p,q\}</math> is called the genus of the entire function <math>f</math>. In this notation, <math display="block">g \leq \rho \leq g + 1</math>
In other words: If the order <math>\rho</math> is not an integer, then <math>g = [ \rho ]</math> is the integer part of <math>\rho</math>. If the order is a positive integer, then there are two possibilities: <math>g = \rho-1</math> or <math>g = \rho </math>.
For example, <math>\sin</math>, <math>\cos</math> and <math>\exp</math> are entire functions of genus <math>g = \rho = 1</math>.
=== Critical exponent ===
Define the '''critical exponent''' of the roots of <math>f</math> as the following:<math display="block">\alpha := \limsup_r \log_r N(f, r)</math>where <math>N(f, r)</math> is the number of roots with modulus <math>< r</math>. In other words, we have an asymptotic bound on the growth behavior of the number of roots of the function:<math display="block">r^{\alpha - \epsilon} \ll N(f, r) \ll r^{\alpha + \epsilon}\quad \forall \epsilon > 0</math>It's clear that <math>\alpha\geq 0</math>.
Theorem (<ref name=":0">{{Cite web |last=Dupuy |first=Taylor |title=Hadamard’s Theorem and Entire Functions of Finite Order — For Math 331 |url=https://www.uvm.edu/~tdupuy/complexspring2017/hadamard.pdf}}</ref>): If <math>f</math> is an entire function with infinitely many roots, then<math display="block">\alpha = \inf\left\{\beta : \sum_k |a_k|^{-\beta} < \infty\right\} = \frac{1}{\liminf_k \log_k |a_k|}</math>Note: These two equalities are purely about the limit behaviors of a real number sequence <math>|a_1| \leq |a_2| \leq \cdots</math> that diverges to infinity.
(<ref>{{Cite web |last=Kupers |first=Alexander |date=April 30, 2020 |title=Lectures on complex analysis |url=https://www.utsc.utoronto.ca/people/kupers/wp-content/uploads/sites/50/2020/12/complex-analysis-2019.pdf |website=Lecture notes for Math 113.}}</ref> Theorem 12.3.4): <math>\alpha(f) \leq \rho</math>. This is proved by [[Jensen's formula]].
=== Proof sketch ===
Since <math>f(z) / z^m</math> is also an entire function with the same order <math>\rho</math> and genus, we can WLOG assume <math>m = 1</math>.
If <math>f</math> has only finitely many roots, then <math>f(z) = e^{P(z)}\prod_{k=1}^n E_0(z/a_k)</math> with the function <math>e^{P(z)}</math> of order <math>\rho</math>. Thus by [[Borel–Carathéodory theorem|an application of the Borel–Carathéodory theorem]], <math>P</math> is a polynomial of degree <math>deg(P) \leq floor(\rho)</math>, and so we have <math>\rho = deg(P) = g</math>.
Otherwise, <math>f</math> has infinitely many roots. This is the tricky part and requires splitting into two cases. First show that <math>g \leq floor(\rho)</math>, then show that <math>\rho \leq g+1</math>.
==== Proving <math>g \leq floor(\rho)</math><ref name=":0" /> ====
As usual in analysis, we fix some ''small'' <math>\epsilon > 0</math>.
Define the function <math>g(z) := \prod_{k=1}^\infty E_p(z/a_k)</math> where <math>p = floor(\rho)</math>, then the goal is to show that <math>f(z)/g(z)</math> is of order <math>\leq \rho+\epsilon</math>. This does not exactly work, however, due to bad behavior of <math>f(z)/g(z)</math> near <math>a_k</math>. Consequently, we need to pepper the complex plane with "forbidden disks", one around each <math>a_k</math>, each with radius <math>\frac{1}{|a_k|^{\alpha + \epsilon}}</math>. Then since <math>\sum_k \frac{1}{|a_k|^{\alpha + \epsilon}} < \infty </math> by the previous result on <math>\alpha</math>, we can pick an increasing sequence of radii <math>R_1 < R_2 < \cdots</math> that diverge to infinity, such that each circle <math>\partial B(0, R_n)</math> avoids all these forbidden disks.
Thus, if we can prove a bound of form <math>\ln \left| \frac{f(z)}{g(z)}\right| = O(|z|^{\rho + \epsilon}) = o(|z|^{\rho + 2\epsilon})</math> for all ''large'' <math>z</math> {{r|g=nb|name=|r=That is, we fix some yet-to-be-determined constant <math>R</math>, and use "<math>z</math> is large" to mean <math>|z| > R</math>.}} that avoids these forbidden disks, then by the same application of Borel–Carathéodory theorem, <math>deg(P) \leq floor(\rho + 2\epsilon)</math> for any <math>\epsilon > 0</math>, and so as we take <math>\epsilon \to 0</math>, we obtain <math>g \leq floor(\rho)</math>.
Since <math>\ln \left| f(z)\right| = o(|z|^{\rho + \epsilon})</math> by the definition of <math>\rho</math>, it remains to show that <math>-\ln \left| g(z)\right| = O(|z|^{\rho + \epsilon})</math>, that is, there exists some constant <math>B > 0</math> such that <math display="block">\sum_{k=1}^\infty \ln |E_p(z/a_k)| \geq -B |z|^{\rho + \epsilon}</math>for all large <math>z</math> 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 <math>a_k</math> with modulus <math>< |z_k|/2</math> and infinitely many <math>a_k</math> with modulus <math>\geq |z_k|/2</math>. So we have to bound:<math display="block">\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)| </math>This bound can be accomplished by the three bounds on <math>|E_p|</math>:
* There exists <math>B>0</math> such that <math>\ln |E_p(z)| \geq -B |z|^{p+1}</math> when <math>|z| \leq 1/2</math>.
* There exists <math>B>0</math> such that <math>|E_p(z)| \geq |1-z| e^{-B |z|^{p}}</math> when <math>|z| \geq 1/2</math>.
==== Proving <math>\rho \leq g + 1</math><ref>{{Cite web |last=Garrett |first=Paul |date=September 15, 2019 |title=Weierstrass and Hadamard products |url=https://www-users.cse.umn.edu/~garrett/m/mfms/notes_2019-20/05c_Hadamard_products.pdf}}</ref> ====
==See also==
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