TITLE: Can all $L^2$ holomorphic functions on a domain vanish at a particular point?
QUESTION [17 upvotes]: Let $\Omega \subset \mathbb{C}^n$.  Is it possible that there is a point $p \in \Omega$ such that every $f \in  A^2(\Omega) = L^2(\Omega) \cap \mathcal{O}(\Omega)$ has a zero at $p$?  The space $A^2(\Omega)$ is called the Bergman Space of $\Omega$.
I ask since on page 56 of the second edition of Krantz's "Function Theory of Several Complex Variables", he observes that this is not possible in the bounded case:  here it is obvious that no such point can exist, since the constant functions are $L^2$ and holomorphic.  This got me thinking about the unbounded case.
Another way of phrasing this question is "Is there an unbounded domain $\Omega$ for which the Bergman kernel has a zero at some point on the diagonal?"
Such a domain would have many strange properties.

To address Lukas Geyer's comment, I would like to further add the stipulation that $A^2(\Omega)$ does not consist only of the constant function $0$.  I would like to point out that there are interesting examples of finite dimensional Bergman spaces, but the examples I have looked at do not have this property.

REPLY [6 votes]: Since it has already been answered I would like to add that the corresponding problem in dimension $1$ has negative answer (this is sometimes called Virtanen's theorem). In dimension higher that $1$ there are a lot of examples as above, one can have additional properties as for example pseudoconvexity so these domains are not ''so exotic''. On the other hand in those examples the set where all $A^2(\Omega)$ vanish is a subvariety of dimension $n-1$ in $\Omega$  I am unaware of any example in which such a set could be an isolated point. This would be really interesting to know at least from the perspective of algebraic geometry – the absence of such points corresponds to the property of base point freeness.