Szegő kernel

Remarks

By the Riesz representation theorem the function $\overline{S(z, \cdot)}$ exists and is unique. So $S$ is the reproducing kernel of the Hardy space $H^2(\partial\Omega)$, in the sense of reproducing kernel Hilbert spaces. The kernel extends to $\Omega \times \Omega$, holomorphic in the first variable and antiholomorphic in the second, with $S(z, w) = \overline{S(w, z)}$.

Example: for the unit disc $\mathbb{D} \subset \C$ with arc length, $S(z, \zeta) = \dfrac{1}{2\pi(1 - z\bar\zeta)}$. Check: for $\abs{\zeta} = 1$ one has $\bar\zeta = 1/\zeta$, so $\dfrac{1}{1 - z\bar\zeta} = \dfrac{\zeta}{\zeta - z}$, and $d\sigma = \abs{d\zeta} = \dfrac{d\zeta}{i\zeta}$. The formula becomes $f(z) = \dfrac{1}{2\pi i}\displaystyle\oint_{\abs{\zeta} = 1} \dfrac{f(\zeta)}{\zeta - z}\, d\zeta$, which is Cauchy's integral formula. For the unit ball of $\C^n$ the kernel is $S(z, \zeta) = \dfrac{(n-1)!}{2\pi^n} (1 - \ip{z}{\zeta})^{-n}$, where $\ip{z}{\zeta} = \sum_j z_j \bar\zeta_j$; at $z = 0$ it is $1$ divided by the area $2\pi^n/(n-1)!$ of the sphere.

Non-example: on the unit disc, the Poisson kernel $P(z, \zeta) = \dfrac{1 - \abs{z}^2}{2\pi\abs{\zeta - z}^2}$ also satisfies $f(z) = \int_{\partial\mathbb{D}} P(z, \zeta) f(\zeta) \, d\sigma(\zeta)$ for all $f \in H^2(\partial\mathbb{D})$. It is not the Szegő kernel, because for $z \ne 0$ the function $\zeta \mapsto P(z, \zeta)$ has nonzero Fourier coefficients of negative index, so it does not lie in $H^2(\partial\mathbb{D})$.

The Bergman kernel is defined in the same way, with $L^2(\Omega)$ in place of $L^2(\partial\Omega, \sigma)$.