An unbounded domain with trivial Bergman space

The conclusion $A^2({\mathbb{C}}^m)=\{0\}$ and $K_{{\mathbb{C}}^m}\equiv 0$ is already proved in The Bergman space $A^2(\mathrm{\Omega })$ and the Bergman kernel, so this example uses that earlier result without repeating its large-polydisc argument.

Let $a=(z,w)\in \mathbb{D}\times \mathbb{C}$ and set $\rho =(1-\mathrm{\mid }z\mathrm{\mid })\mathrm{/}2>0$. For every $R>0$, the closed polydisc ${\bar{\mathrm{\Delta }}}_{(\rho ,R)}(a)$ lies in $\mathbb{D}\times \mathbb{C}$, since $\mathrm{\mid }z\mathrm{\mid }+\rho =(1+\mathrm{\mid }z\mathrm{\mid })\mathrm{/}2<1$ and the second coordinate is unrestricted. These polydiscs show that the product set is open. Let $f$ be the unique holomorphic representative of any class in $A^2(\mathbb{D}\times \mathbb{C})$, as in [F6]. The mean bound [F3] and integral monotonicity [F4] give $\mathrm{\mid }f(a){\mathrm{\mid }}^2\le ({\pi }^2{\rho }^2{R}^2{)}^{-1}{\int }_{{\mathrm{\Delta }}_{(\rho ,R)}(a)}\mathrm{\mid }f{\mathrm{\mid }}^{2}d{\lambda }_4\le \mathrm{\parallel }[f]{\mathrm{\parallel }}_{L^2(\mathbb{D}\times \mathbb{C})}^2\mathrm{/}({\pi }^2{\rho }^2{R}^2)$. Letting $R\to \mathrm{\infty }$ gives $f(a)=0$. Since $a$ was arbitrary, every holomorphic representative is identically zero, hence $A^2(\mathbb{D}\times \mathbb{C})=\{0\}$.

By [F7], each point evaluation on the zero space $A^2(\mathbb{D}\times \mathbb{C})$ has the unique representer $k_w=0$, so $K_{\mathbb{D}\times \mathbb{C}}(z,w)=0$ for all $z,w$. Its diagonal is zero, making the logarithm in the kernel-derived Bergman metric undefined.

By [F5], the constant function $1$ is in $A^2(\mathbb{D})$. Its evaluation at $0$ is $1$, so its Riesz representer $k_0$ is nonzero by [F7]. Reproduction with $f=k_0$ gives $K_{\mathbb{D}}(0,0)=k_0(0)=⟨k_0,k_0⟩=\mathrm{\parallel }{k}_0{\mathrm{\parallel }}^2>0$, proving the claimed contrast with the two unbounded examples.