Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Are there any generalisations of the Weierstrass Factorization Theorem, and if so where can I find information on them? I'm trying to investigate infinite products of the form

$$\prod_{k=1}^\infty f(z)^{k^a}e^{g(z)},$$

where $g\in\mathbb{Z}[z]$ and $a\in\mathbb{N}$.

share|improve this question
add comment

2 Answers

A vast generalization of Weierstraß's theorem is to Riemann surfaces.
Florack, inspired by methods due to Behnke-Stein, proved the following in 1948:

Let $X$ be a non-compact Riemann surface. Let $D$ be a closed discrete set in $X$ and to each $d\in D$ attach a complex number $a_d$.
Then there exists a holomorphic function $f\in \mathcal O(X)$ defined on all of $X$ such that $f(a_d)=c_d.$

One may think of $f$ as holomorphically interpolating some discrete data.
This result immediately implies that $X$ is a Stein manifold, a concept of fundamental importance in the theory of holomorphic manifolds: Stein manifolds are the analogues of affine varieties in algebraic geometry.

A complete proof is in Theorem 26.7 of Forster's awesome Lectures on Riemann Surfaces.
The special case where $X$ is an open subset of $\mathbb C$ is analyzed in John's answer.

share|improve this answer
add comment

The Weierstrass factorization theorem provides a way of constructing an entire function with any prescribed set of zeros, provided the set of zeros does not have a limit point in $\mathbb{C}$. I know that this generalizes to being able to construct a function holomorphic on a region $G$ with any prescribed set of zeros in $G$, provided that the set of zeros does not have a limit point in $G$.

These are theorems VII.5.14 and VII.5.15 in Conway's Functions of One Complex Variable. They lead to the (important) corollary that every meromorphic function on an open set $\Omega$ is a ratio of functions holomorphic on $\Omega$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.