# Bergman kernel

Let $G\subset {\u2102}^{n}$ 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 bounded^{}
linear 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, and
let’s call that element ${k}_{z}\in {A}^{2}(G)$. That is we have that
$f(z)=\u27e8f,{k}_{z}\u27e9$. So we can define the Bergman kernel^{}.

###### Definition.

By definition of the inner product^{} in ${A}^{2}(G)$ we then have that
for $f\in {A}^{2}(G)$

$$f(z)={\int}_{G}f(w)K(z,w)\mathit{d}V(w),$$ |

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 orthogonal
projection 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.

## References

- 1 Steven G. Krantz. , AMS Chelsea Publishing, Providence, Rhode Island, 1992.

Title | Bergman kernel |
---|---|

Canonical name | BergmanKernel |

Date of creation | 2013-03-22 15:04:45 |

Last modified on | 2013-03-22 15:04:45 |

Owner | jirka (4157) |

Last modified by | jirka (4157) |

Numerical id | 7 |

Author | jirka (4157) |

Entry type | Definition |

Classification | msc 32A25 |

Related topic | BergmanSpace |

Related topic | BergmanMetric |