5.3: The Szegö Kernel

We use the same technique to create a reproducing kernel on the boundary by starting with $L^2(\partial U, d\sigma)$ instead of $L^2(U)$. We obtain a kernel where we integrate over the boundary rather than the domain itself. Let us give a quick overview, but let us not get into the details.

Let $U \subset \mathbb{C}^n$ be a bounded domain with smooth boundary. Let $C(\overline{U}) \cap \mathcal{O}(U)$ be the holomorphic functions in $U$ continuous up to the boundary. The restriction of $f \in C(\overline{U}) \cap \mathcal{O}(U)$ to $\partial U$ is a continuous function, and hence $f|_{\partial U}$ is in $L^2(\partial U,d\sigma)$, where $d\sigma$ is the surface measure on $\partial U$. Taking a closure of these restrictions in $L^2(\partial U)$ obtains the Hilbert space $H^2(\partial U)$, which is called the Hardy space. The inner product on $H^2(\partial U)$ is the $L^2(\partial U, d\sigma)$ inner product:

$$\langle f,g\rangle \overset{\text{def}}{=} \int_{\partial U} f(z) \overline{g(z)} \, d\sigma(z) .$$

Exercise $\PageIndex{1}$

Show that monomials $z^\alpha$ are a complete orthonormal system in $H^2(\partial \mathbb{B}_{n})$.

Exercise $\PageIndex{2}$

Let $U \subset \mathbb{C}^n$ be a bounded domain with smooth boundary. Prove that $H^2(\partial U)$ is infinite-dimensional.

Given an $f \in H^2(\partial U)$, write the Poisson integral

$$Pf(z) = \int_{\partial U} f(\zeta) \, P(z,\zeta) \, d \sigma(\zeta) ,$$ where $P(z,\zeta)$ is the Poisson kernel. The Poisson integral reproduces harmonic functions. As holomorphic functions are harmonic, we find that if $f \in C(\overline{U}) \cap \mathcal{O}(U)$, then $Pf = f$.

Although $f \in H^2(\partial U)$ is only defined on the boundary, through the Poisson integral, we have the values $Pf(z)$ for $z \in U$. For each $z \in U$,

$$f \mapsto Pf(z)$$ defines a continuous linear functional. Again we find a $s_z \in H^2(\partial U)$ such that $$Pf(z) = \langle f, s_z\rangle .$$ For $z \in U$ and $\zeta \in \partial U$, define $$S_U(z,\bar{\zeta}) \overset{\text{def}}{=} \overline{s_z(\zeta)} ,$$ although for a fixed $z$ this is a function only defined almost everywhere as it is an element of $L^2(\partial U,d\sigma)$. The function $S_U$ is the Szegö kernel. If $f \in H^2(\partial U)$, then $$Pf(z) = \int_{\partial U} f(\zeta) \, S_U(z,\bar{\zeta}) \, d\sigma(\zeta) .$$

As functions in $H^2(\partial U)$ extend to $\overline{U}$, then $f \in H^2(\partial U)$ may be considered a function on $\overline{U}$, where values in $U$ are given by $Pf$. Similarly, we extend $S(z,\bar{\zeta})$ to a function on $U \times \overline{U}^*$ (where the values on the boundary are defined only almost everywhere). We state without proof that if $\{ \varphi_j \}_{j\in I}$ is a complete orthonormal system for $H^2(\partial U)$, then

$$\label{eq:2} S_U(z,\bar{\zeta}) = \sum_{j \in I} \varphi_j(z)\overline{\varphi_j(\zeta)}$$ for $(z,\bar{\zeta}) \in U \times U^*$, converging uniformly on compact subsets. As before, this formula shows that $S$ is conjugate symmetric, and so it extends to $(U \times \overline{U}^*) \cup (\overline{U} \times U^*)$.

Example $\PageIndex{1}$

In Exercise 5.2.3, we computed that if $f \in C(\overline{\mathbb{D}}) \cap \mathcal{O}(\mathbb{D})$, then

$$f(z) = \frac{1}{2\pi} \int_{\partial \mathbb{D}} \frac{f(\zeta)}{1-z\bar{\zeta}} \, ds .$$ In other words, $S_{\mathbb{D}}(z,\zeta) = \frac{1}{\pi} \frac{1}{1-z\bar{\zeta}}$ .

Exercise $\PageIndex{1}$

Using the formula $\eqref{eq:2}$ compute $S_{\mathbb{B}_{n}}$.