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 $n \ge n_0$

$\blacksquare$

Sources

• 1973: John B. Conway: Functions of One Complex Variable ... (previous) ... (next) $\text {VII}$: Compact and Convergence in the Space of Analytic Functions: $\S5$: Weierstrass Factorization Theorem: Theorem $5.9$