# Weierstrass factorization with $L^2$ estimates?

Let $\Omega$ be a bounded domain in $\mathbb{C}$. Let $X$ be a discrete set of points whose boundary is in the boundary of $\Omega$. Can I find an $L^2$ holomorphic function which vanishes on $X$? Can I solve the problem in weighted $L^2$ spaces?

If there are counterexamples, are precise conditions on the set $X$ known to ensure the existence of an $L^2$ solution?

I have been learning about Hormander's approach to the $\bar{\partial}$-problem, and this seems like a natural question to ask from that perspective, but I have not been able to find any work done on this.

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## 2 Answers

The right place to start is the seminal work of Seip, Kristian Seip, Beurling type density theorems in the unit disk, Invent. math. 1993, Vol 113, 1, pp 21-39 (look at the last sections). You will see that much is known in the case of the disc, but I fairly doubt that a complete characterization is known for arbitrary domains (especially in the weighted case).

Also a good reference would be Ohsawa, T. On the extension of L2 holomorphic functions. V. Effects of generalization. Nagoya Math. J. 161 (2001), 1–21. (again look just at the last chapters)

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You have to specify what you mean by $L^2$. Is this $L^2$ with respect to Lebesgue measure (area) in $D$? Whatever you mean by $L^2$, the answer is "no". The reason is Jensen's formula. It says that a function which has too many zeros must grow fast.

If you want to solve it in weighted $L^2$ space, then your weight must be related to the growth rate of the set $X$. If instead you want to fix the weight in $L^2$, the conditions on $X$ will come from the Jensen formula. If you are interested in $L^2$ without weight, look in the books about Bergman space. There you can find the exact conditions on $X$.

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Ah, I had not thought to use Jensen's formula. I will see if I can cook up a concrete counterexample. I was thinking of L^2 wrt lebesgue measure. I will wait a day or so to accept your answer. Thanks. – Steve Apr 17 '13 at 22:41