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.




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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.



Can all $L^2$ holomorphic functions on a domain vanish at a particular point?
$\def\CC{\mathbb{C}}$Here is a less trivial example that I think works. Let $U \subset \CC^2$ be
$$\{ (x,y) : |x| \leq \min(1, 1/|y|) \}$$


There are lots of $L^2$ holomorphic functions because the change of variables $(u,v) = (x, xy)$ changes $|x|^2 dx d\bar{x} dy d \bar{y}$ to $du d\bar{u} dv d \bar{v}$, and takes $U$ to the polydisc $|u|$, $|v|<1$. So, if $f(u,v)$ is a bounded function on the polydisc, then $x f(x,xy)$ should be $L^2$ on $U$. 


I now claim that all $L^2$ functions on $U$ vanish on the line $x=0$. This requires a bunch of estimates which I didn't know, but which are probably standard for those working in the field.


Suppose that $g(z)$ is analytic on a disc $D$ around $0$ in $\CC$. Then by Cauchy-Schwarz and the triangle inequality, $\int\_D |g|^2 \geq \frac{\left| \int\_D g \right|^2}{|D|} = \frac{|D|^2 |g(0)|^2}{|D|}$. So $|D| g(0) \leq \int\_D |g|^2$. 


Let $f(x,y)$ be an $L^2$ function on $U$. Then, using Fubini, the above bound shows that 
$$\int\_{\CC} |f(0,y)|^2 \min(1,|y|^{-2})$$
is convergent, since the disc of fixed $y$ value has area $\pi y^{-2}$. Switching to polar coordinates
$$\int\_{r=0}^{\infty} r dr \min(1,\frac{1}{r^2}) \int\_{\theta=0}^{2 \pi} |f(0,r e^{i \theta})|^2 d \theta$$
converges. But, using Cauchy-Schwarz again, $$\int\_{\theta=0}^{2 \pi} |f(0,r e^{i \theta})|^2 d \theta \geq \frac{1}{2 \pi} \left| \int\_{\theta=0}^{2 \pi} f(0,r e^{i \theta}) d \theta \right|^2 = \frac{1}{2 \pi} |f(0,0)|^2.$$


So $\int\_{r=0}^{\infty} \min(r,1/r) dr |f(0,0)|$ converges, and $f(0,0) =0$. Working a bit harder, one should be able to show that $f(0,y)=0$ for any $y$.


