Bergman kernel

Let $G\subset {ℂ}^n$ be a domain (http://planetmath.org/Domain2). And let $A^2(G)$ be the Bergman space. For a fixed $z\in G$, the functional $f\mapsto f(z)$ is a boundedlinear functional. By the Riesz representation theorem (as $A^2(G)$ is a Hilbert space) there exists an element of $A^2(G)$ that represents it, andlet’s call that element $k_z\in A^2(G)$. That is we have that$f(z)=⟨f,k_z⟩$. So we can define the Bergman kernel.

Definition.

The function

$$K(z,w):=\overline{k_z(w)}$$

is called the Bergman kernel.

By definition of the inner product in $A^2(G)$ we then have thatfor $f\in A^2(G)$

$$f(z)={\int }_{G}f(w)K(z,w)𝑑V(w),$$

where $dV$ is the volume measure.

As the $A^2(G)$ space is a subspace of $L^2(G,dV)$ which is a separable Hilbert space then $A^2(G)$ also has a countable orthonormal basis, say ${\{{\phi }_j\}}_{j=1}^{\mathrm{\infty }}$.

Theorem.

We can compute the Bergman kernel as

$$K(z,w)=\sum _{j=1}^{\mathrm{\infty }}{\phi }_j(z)\overline{{\phi }_j(w)},$$

where the sum converges uniformly on compact subsets of $G\mathrm{\times }G$.

Note that integration against the Bergman kernel is just the orthogonalprojection from $L^2(G,dV)$ to $A^2(G)$. So not only is this kernel reproducing for holomorphic functions, but it will produce a holomorphic function when we just feed in any $L^2(G,dV)$ function.