# Zeroes of Infinite Product of Analytic Functions

## Theorem

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

Let $\sequence {f_n}$ be a sequence of analytic functions $f_n: D \to \C$.

Let $\ds \prod_{n \mathop = 1}^\infty f_n$ converge locally uniformly to $f$.

Let $z_0\in D$.

Then:

$(1): \quad$ $f$ is identically zero if and only if some $f_n$ is identically zero
$(2): \quad$ $\map {f_n} {z_0} = 0$ for finitely many $n \in \N$
$(3): \quad$ If $f$ is not identically zero, $\map {\operatorname {mult}_{z_0} } f = \ds \sum_{n \mathop = 1}^\infty \map {\operatorname {mult}_{z_0} } {f_n}$

where $\operatorname {mult}$ denotes multiplicity.

## Proof

Note that by Infinite Product of Analytic Functions is Analytic, $f$ is analytic.

Let $n_0 \in \N$ and $U\subset D$ an open neighborhood of $z_0$ such that $\ds \prod_{n \mathop = n_0}^\infty \map {f_n} z \ne 0$ for $z\in U$.

Let $f$ be identically zero on $U$.

Then $\ds \prod_{n \mathop = 1}^{n_0 - 1} \map {f_n} z$ is identically zero on $U$.

Then some $f_k$ with $k < n_0$ is identically zero on $U$.

By Uniqueness of Analytic Continuation, $f_k$ is identically zero on $D$.

Now let $f$ be not identically zero on $U$.

We have:

 $\ds \map {\operatorname {mult}_{z_0} } f$ $=$ $\ds \sum_{n \mathop = 1}^{n_0} \map {\operatorname {mult}_{z_0} } {f_n} + \map {\operatorname {mult}_{z_0} } {\prod_{n \mathop = n_0}^\infty f_n}$ Multiplicity of Product of Analytic Functions $\ds$ $=$ $\ds \sum_{n \mathop = 1}^{n_0} \map {\operatorname {mult}_{z_0} } {f_n}$ $\ds \prod_{n \mathop = n_0}^\infty \map {f_n} {z_0} \ne 0$ $\ds$ $=$ $\ds \sum_{n \mathop = 1}^\infty \map {\operatorname {mult}_{z_0} } {f_n}$ $\map {f_n} {z_0} \ne 0$ for $n \ge n_0$

$\blacksquare$