Weierstrass Factorization Theorem

The Weierstrass factorization theorem shows that every entire function can be expressed as an infinite product determined by its zeros, generalizing the fundamental theorem of algebra.

Statement

Theorem9.9Weierstrass factorization theorem

Let $f$ be an entire function with zeros at $a_1, a_2, \ldots$ (listed with multiplicity, $|a_1| \leq |a_2| \leq \cdots$ , none equal to zero) and a zero of order $m \geq 0$ at the origin. Then

$f(z) = z^m e^{g(z)}\prod_{n=1}^{\infty}E_{p_n}(z/a_n)$

where $g$ is entire and $p_n$ are non-negative integers chosen so the product converges. One can always take $p_n = n - 1$ .

Proof Sketch

Proof

Step 1: Convergence of the product.

The elementary factor satisfies $|1 - E_p(z)| \leq |z|^{p+1}$ for $|z| \leq 1/2$ . Therefore $\sum |z/a_n|^{p_n+1}$ converges if the $p_n$ grow fast enough relative to the $|a_n|$ . Taking $p_n = n - 1$ ensures $\sum |z/a_n|^n$ converges for each $z$ (since $|z/a_n| \to 0$ ).

Step 2: Construction.

Define $P(z) = \prod_{n=1}^\infty E_{p_n}(z/a_n)$ . This converges uniformly on compact sets (by the $M$ -test for infinite products). The product $P$ is entire with zeros precisely at $\{a_n\}$ with the correct multiplicities.

Step 3: Extracting the exponential factor.

The function $f(z)/(z^m P(z))$ is entire and nonvanishing (its zeros have been divided out). By the simply connected domain version of the logarithm, write $f(z)/(z^m P(z)) = e^{g(z)}$ for some entire $g$ . $\blacksquare$

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Canonical Products

Definition9.7Canonical product

If we take all $p_n$ equal to a fixed $p$ , the product $\prod E_p(z/a_n)$ is called a canonical product of genus $p$ . It converges if and only if $\sum |a_n|^{-(p+1)} < \infty$ . The smallest such $p$ is the genus of the canonical product.

ExampleCanonical products

$\sin(\pi z)/(\pi z)$ : $\prod_{n=1}^\infty (1 - z^2/n^2)$ is a canonical product of genus $1$ (since $\sum 1/n^2 < \infty$ but $\sum 1/n = \infty$ ).
$1/\Gamma(z)$ : $\frac{1}{\Gamma(z)} = ze^{\gamma z}\prod_{n=1}^\infty (1+z/n)e^{-z/n}$ where $\gamma$ is the Euler-Mascheroni constant. This is a canonical product of genus $1$ .
Zeros at $n^2$ : $\prod_{n=1}^\infty (1-z/n^2)$ converges with $p = 0$ since $\sum 1/n^2 < \infty$ .

Connection to Hadamard

RemarkHadamard refinement

For entire functions of finite order $\rho$ , Hadamard's theorem refines the factorization: the genus $p$ satisfies $p \leq \rho$ , and the function $g(z)$ in the exponential factor is a polynomial of degree at most $\lfloor\rho\rfloor$ . This powerful constraint connects the growth of $f$ to the distribution of its zeros.

The exponent of convergence $\lambda = \inf\{s > 0 : \sum |a_n|^{-s} < \infty\}$ satisfies $\lambda \leq \rho$ , and $p = \lfloor\lambda\rfloor$ suffices.

ExampleOrder and genus relationship

For $f(z) = \prod_{n=1}^\infty(1-z/2^n)$ : the zeros are $a_n = 2^n$ , so $\sum |a_n|^{-s} = \sum 2^{-ns}$ converges for $s > 0$ . The exponent of convergence is $\lambda = 0$ , genus $p = 0$ , and the product converges with no exponential factors needed. The order is $\rho = 0$ .