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 \(S_{\mathbb{B}_{n}}\).