Infinite Product of Analytic Functions

Theorem
Let $D\subset\C$ be an open connected set.

Let $(f_n)$ be a sequence of analytic functions $f_n:D\to\C$ that are not identically zero.

Let $\displaystyle\sum_{n\mathop=1}^\infty(f_n-1)$ converge locally uniformly absolutely on $D$.

Then:
 * $f=\displaystyle\prod_{n\mathop=1}^\infty f_n$ converges locally uniformly absolutely on $D$
 * $f$ is analytic
 * For each $z\in D$, $f_n(z)=0$ for finitely many $n\in\N$
 * For each $z\in D$, $\operatorname{mult}_z(f)=\displaystyle\sum_{n\mathop=1}^\infty\operatorname{mult}_z(f_n)$

where $\operatorname{mult}$ denotes multiplicity.

Proof
Let $z_0\in D$.

Let $K\subset D$ be a compact neighborhood of $z_0$.

Because $\displaystyle\sum_{n\mathop=1}^\infty(f_n-1)$ converges locally uniformly absolutely on $D$, the series converges uniformly absolutely on $K$.

By Uniform Absolute Convergence of Infinite Product of Complex Functions we have:
 * $f=\displaystyle\prod_{n\mathop=1}^\infty f_n$ converges uniformly absolutely on $K$
 * There exists $n_0\in\N$ such that $\displaystyle\prod_{n\mathop=n_0}^\infty f_n(z_0)\neq0$; In particular, $f_n(z_0)\neq0$ for $n\geq n_0$.

By Uniform Limit of Analytic Functions is Analytic, $f$ and $\displaystyle\prod_{n\mathop=n_0}^\infty f_n$ are analytic on the interior $K^\circ$ of $K$.

Because $z_0$ was arbitrary, $f$ is analytic on $D$.

For the last claim, we have:

Also see

 * Derivative of Infinite Product of Analytic Functions