# Can all $L^2$ holomorphic functions on a domain vanish at a particular point?

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.

• There is a trivial example that has this property in one dimension, the domain $\Omega = \mathbb{C}$. – Lukas Geyer Nov 21 '14 at 0:51
• @LukasGeyer Yes, I meant examples where the space $L^2(\Omega) \cap \mathcal{O}(\Omega)$ is nontrivial. I will edit. – Steven Gubkin Nov 21 '14 at 2:03

$$\def\CC{\mathbb{C}}$$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 $$y$$.

• This is very similar to Boas's counterexample to the Lu Qi-Keng conjecture, as described by Krantz in the book "Geometric Analysis of the Bergman Kernel and Metric", p. 170. He uses the domain $|w|<(1+|z|)^{-1}$ in $\mathbb{C}^2$ and his argument that any function in the Bergman space vanishes on $w=0$ uses power series, which makes the argument a little easier, but otherwise it looks the same. – Lukas Geyer Nov 21 '14 at 17:44
• Thanks! I will have to work through all of your estimates to make sure this works, but it looks very promising. – Steven Gubkin Nov 21 '14 at 19:26

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.

For a pseudoconvex domain $\Omega\subset\mathbb{C}^n$ with imaginary components uniformly bounded (a bound $|\Im z|\leq c\log(2+|z|^2)$ actually suffices), the answer to the question in the title is no. In such a domain one can find, for given $z_0\in\mathbb{C}^n$, a plurisubharmonic function $\varphi$ such that $e^{-\varphi}$ is integrable over a neighbourhood of $z_0$, and $e^{\varphi(z)}\leq C_N(1+|z|^2)^{-N}$ for $z\in\Omega$, $N\in\mathbb{N}$. For example, take $\varphi=\log|\hat f|$ where $\hat f$ is the Fourier--Laplace transform of a test function $f\in C_c^\infty(\mathbb{R}^n)$ such that $\hat f(z_0)\neq 0$, and apply the Paley--Wiener-Schwartz Theorem. Now a theorem of Bombieri, which is Theorem 4.4.4 in Hörmander's book on complex analysis in several variables, implies the existence of $u\in A^2(\Omega)$ with $u(z_0)=1$.

I can not answer the rephrased question, but maybe my observation is of interest here.