Proof of Hadamard's Factorization Theorem

Hadamard's factorization theorem refines the Weierstrass factorization for entire functions of finite order, constraining the exponential factor to be a polynomial.

Statement

Theorem9.11Hadamard factorization theorem (restated)

Let $f$ be entire of finite order $\rho$ with zeros $a_1, a_2, \ldots$ (none at the origin) and a zero of order $m$ at the origin. Let $p = \lfloor\rho\rfloor$ . Then

$f(z) = z^m e^{Q(z)}\prod_{n=1}^{\infty}E_p(z/a_n)$

where $Q(z)$ is a polynomial of degree at most $p$ , and the product converges absolutely.

Key Ingredients

Theorem9.12Jensen's formula

If $f$ is holomorphic on $|z| \leq R$ , $f(0) \neq 0$ , and $a_1, \ldots, a_k$ are the zeros of $f$ in $|z| < R$ (counted with multiplicity), then

$\log|f(0)| = -\sum_{j=1}^k \log\frac{R}{|a_j|} + \frac{1}{2\pi}\int_0^{2\pi}\log|f(Re^{i\theta})|\,d\theta.$

RemarkJensen's formula and zero counting

Jensen's formula relates the growth of $f$ (measured by $\log|f|$ on circles) to the number and distribution of zeros. Specifically, the counting function $n(r) = \#\{|a_k| \leq r\}$ satisfies

$N(r) = \int_0^r \frac{n(t)}{t}\,dt \leq \log M(r) + O(1)$

where $M(r) = \max_{|z|=r}|f(z)|$ . For $f$ of order $\rho$ : $N(r) = O(r^{\rho+\varepsilon})$ , which implies $\sum |a_n|^{-(\rho+\varepsilon)} < \infty$ for every $\varepsilon > 0$ .

Proof Outline

Proof

Step 1: The canonical product converges.

Since $\sum |a_n|^{-(\rho+\varepsilon)} < \infty$ for $\varepsilon > 0$ and $p = \lfloor\rho\rfloor \geq \rho - 1$ , we have $p + 1 > \rho$ , so $\sum |a_n|^{-(p+1)} < \infty$ . This ensures the canonical product $P(z) = \prod E_p(z/a_n)$ converges absolutely and uniformly on compact sets.

Step 2: $P(z)$ has the same zeros as $f$ .

By construction, $P$ is entire with zeros exactly at $\{a_n\}$ (with correct multiplicities). Therefore $h(z) = f(z)/(z^m P(z))$ is entire and nonvanishing, so $h(z) = e^{g(z)}$ for some entire function $g$ .

Step 3: Growth estimate on $g$ .

The key step is showing $g$ is a polynomial of degree $\leq p$ . The growth of $P(z)$ can be estimated: $\log|P(z)| = O(r^{\rho+\varepsilon})$ using properties of the elementary factors. Therefore

$\text{Re}(g(z)) = \log|h(z)| = \log|f(z)| - m\log|z| - \log|P(z)| = O(r^{\rho+\varepsilon}).$

Step 4: A real part bound implies polynomial.

If $g(z) = \sum c_n z^n$ is entire with $\text{Re}(g(z)) \leq C r^{\rho+\varepsilon}$ , then by Cauchy's estimates and the Borel-Caratheodory theorem (which bounds $|g|$ in terms of $\text{Re}(g)$ ):

$|c_n| R^n \leq 2\max_{|z|=R}\text{Re}(g(z)) + 2|g(0)| \leq C'R^{\rho+\varepsilon}$

so $|c_n| \leq C'R^{\rho+\varepsilon-n}$ . For $n > \rho$ , letting $R \to \infty$ gives $c_n = 0$ . Therefore $g = Q$ is a polynomial of degree at most $\lfloor\rho\rfloor = p$ . $\blacksquare$

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Applications

ExampleEntire functions of order 1

Every entire function of order $1$ with zeros $\{a_n\}$ satisfying $\sum 1/|a_n|^{1+\varepsilon} < \infty$ has the form

$f(z) = z^m e^{\alpha z + \beta}\prod_n\left(1 - \frac{z}{a_n}\right)e^{z/a_n}.$

This applies to $\sin(\pi z)$ , $1/\Gamma(z)$ , and the Riemann xi function $\xi(s)$ .

RemarkSignificance

Hadamard's theorem bridges analysis and algebra: it shows entire functions of finite order are essentially determined by their zeros plus a polynomial, much like polynomials themselves. This has profound applications in number theory (the prime number theorem uses the Hadamard factorization of the Riemann zeta function).