MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 1 down vote accepted

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)

share|cite|improve this answer

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 $X$.

share|cite|improve this answer
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

Your Answer


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.