Infinite Product of Analytic Functions

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

Let $\left\langle{f_n}\right\rangle$ be a sequence of analytic functions $f_n: D \to \C$ that are not identically zero.

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

Then:
 * $(1): \quad f = \displaystyle \prod_{n \mathop = 1}^\infty f_n$ converges locally uniformly absolutely on $D$
 * $(2): \quad f$ is analytic
 * $(3): \quad$ For each $z \in D$, $f_n \left({z}\right) = 0$ for finitely many $n \in \N$
 * $(4): \quad$ For each $z \in D$, $\operatorname {mult}_z \left({f}\right) = \displaystyle \sum_{n \mathop = 1}^\infty \operatorname{mult}_z \left({f_n}\right)$

where $\operatorname{mult}$ denotes multiplicity.

Proof
Let $z_0 \in D$.

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

Because $\displaystyle \sum_{n \mathop = 1}^\infty \left({f_n - 1}\right)$ 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 $\displaystyle \prod_{n \mathop = n_0}^\infty f_n \left({z_0}\right) \ne 0$; in particular, $f_n \left({z_0}\right) \ne 0$ for $n \ge n_0$.

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 $D$.

The last claim follows from Zeroes of Infinite Product of Analytic Functions

Also see

 * Derivative of Infinite Product of Analytic Functions