Hadamard factorization theorem

Remarks

The proof first bounds the number $n(r)$ of zeros in the disc $\abs{z} \le r$ by Jensen's formula: $n(r) = O(r^{\rho + \varepsilon})$ for every $\varepsilon > 0$. This gives the convergence of $\sum_n \abs{a_n}^{-(p+1)}$, so the Weierstrass product with the fixed index $p$ converges. The quotient of $f$ by $z^m$ times the product is $e^{g}$ with $g$ entire. Lower bounds for the product away from its zeros show that $g$ is a polynomial of degree at most $\rho$. For example, $\sin(\pi z)$ has order $1$, so $p = 1$ and $\sin(\pi z) = \pi z \prod_{n \ne 0} (1 - z/n) e^{z/n}$. The finite order hypothesis is necessary: $f(z) = e^{e^z}$ has no zeros and infinite order, and $f = e^{g}$ with $g(z) = e^z$, which is not a polynomial.

Going deeper

For a student

A useful consequence: an entire function of finite order that is not an integer has infinitely many zeros. Indeed, with finitely many zeros the factorization gives a polynomial times $e^{g}$ with $g$ a polynomial, whose order is an integer. Another consequence is that an entire function of finite order without zeros is $e^{g}$ with $g$ a polynomial. The theorem is a basic tool for the Riemann zeta function, through the product formula of the function $\xi$, and for the Laguerre–Pólya class.