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 <math display="block">E_n(z) := (1-z) \prod_{k=1}^n e^{z^k/k}</math>Entire functions of finite order <math>\rho</math> have Hadamard's canonical representation:<math display="block">f(z)=z^me^{Q(z)}\prod_{n=1}^\infty E_p(z/a_n)</math>where <math>a_k</math> are those 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>Q</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>.
Furthermore, Jensen's inequality implies that its roots are distributed sparsely, with critical exponent <math>\alpha \leq \rho \leq g+1</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 \limits_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">\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) > k r_k^{\alpha - \epsilon} </math>It's clear that <math>\alpha\geq 0</math>.
Theorem: 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 \limits_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. It does not involve complex analysis.
Proposition: <math>\alpha(f) \leq \rho</math>, by Jensen's formula.
Applications
With Hadamard factorization we can prove some special cases of Picard's little theorem.
Theorem: If <math>f</math> is entire, nonconstant, and has finite order, then it assumes either the whole complex plane or the plane minus a single point.
Proof: If <math>f</math> does not assume value <math>z_0</math>, then by Hadamard factorization, <math>f(z) - z_0 = e^{Q(z)}</math> for a nonconstant polynomial <math>Q</math>. By the fundamental theorem of algebra, <math>Q</math> assumes all values, so <math>f(z) - z_0</math> assumes all nonzero values.
Theorem: