5.2: The Bergman Kernel
- Page ID
- 74242
Let \(U \subset \mathbb{C}^n\) be a domain. Define Bergman space of \(U\): \[A^2(U) \overset{\text{def}}{=} \mathcal{O}(U) \cap L^2(U) .\] That is, \(A^2(U)\) denotes the space of holomorphic functions \(f \in \mathcal{O}(U)\) such that \[||f||_{A^2(U)}^2 \overset{\text{def}}{=} ||f||_{L^2(U)}^2 = \int_U |f(z)|^2 dV < \infty .\] \(A^2(U)\) is an inner product space with the \(L^2(U)\) inner product \[\langle f, g\rangle \overset{\text{def}}{=} \int_U f(z) \overline{g(z)} \, dV .\] We will prove that \(A^2(U)\) is complete, in other words, it is a Hilbert space. We first prove that the \(A^2(U)\) norm bounds the uniform norm on compact sets.
Let \(U \subset \mathbb{C}^n\) be a domain and \(K \subset \subset U\) compact. Then there exists a constant \(C_K\), such that \[||f||_K = \sup_{z \in K} |f(z)| \leq C_K ||f||_{A^2(U)} \qquad \text{ for all } f\in A^2(U) .\] Consequently, \(A^2(U)\) is complete.
- Proof
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As \(K\) is compact there exists an \(r > 0\) such that \(\overline{\Delta_r(z)} \subset U\) for all \(z \in K\). Take any \(z \in K\), and apply Exercise 1.2.8 and Cauchy–Schwarz: \[\begin{align}\begin{aligned} |f(z)| &= \left| \frac{1}{V\bigl(\Delta_r(z)\bigr)} \int_{\Delta_r(z)} f(\xi) \, dV(\xi)\right| \\ & \leq \frac{1}{\pi^n r^{2n}} \sqrt{\int_{\Delta_r(z)} 1^2 \, dV(\xi)} \sqrt{\int_{\Delta_r(z)} |f(\xi)|^2 \, dV(\xi)} \\ & = \frac{1}{\pi^{n/2} r^n} ||f||_{A^2(\Delta_r(z))} \leq \frac{1}{\pi^{n/2} r^n} ||f||_{A^2(U)} . \end{aligned}\end{align}\] Taking supremum over \(z \in K\) proves the estimate. Therefore, if \(\{ f_j \}\) is a sequence of functions in \(A^2(U)\) converging in \(L^2(U)\) to some \(f \in L^2(U)\), then it converges uniformly on compact sets, and so \(f \in \mathcal{O}(U)\). Consequently, \(A^2(U)\) is a closed subspace of \(L^2(U)\), and hence complete.
For a bounded domain, \(A^2(U)\) is always infinite-dimensional, see exercise below. There exist unbounded domains for which either \(A^2(U)\) is trivial (just the zero function) or even finite-dimensional. When \(n=1\), \(A^2(U)\) is either trivial, or infinite-dimensional.
Show that if a domain \(U \subset \mathbb{C}^n\) is bounded, then \(A^2(U)\) is infinite-dimensional.
- Show that \(A^2(\mathbb{C}^n)\) is trivial (it is just the zero function).
- Show that \(A^2(\mathbb{D} \times \mathbb{C})\) is trivial.
- Find an example of an unbounded domain \(U\) for which \(A^2(U)\) is infinite-dimensional. Hint: Think in one dimension for simplicity.
Show that \(A^2(\mathbb{D})\) can be identified with \(A^2(\mathbb{D} \setminus \{ 0 \})\), that is, every function in the latter can be extended to a function in the former.
Let \(U \subset \mathbb{C}^n\) be a domain, \(f \in \mathcal{O}(U)\), and \(X = f^{-1}(0)\). Show that every function in \(A^2(U \setminus X)\) is a restriction of a function in \(A^2(U)\), that is, \(A^2(U) \cong A^2(U \setminus X)\).
The lemma says that point evaluation is a bounded linear functional. That is, fix \(z \in U\) and take \(K= \{ z \}\), then the linear operator \[f \mapsto f(z)\] is a bounded linear functional. By the Riesz–Fisher theorem, there exists a \(k_z \in A^2(U)\), such that \[f(z) = \langle f, k_z\rangle .\] Define the Bergman kernel for \(U\) as \[K_U(z,\bar{\zeta}) \overset{\text{def}}{=} \overline{k_z(\zeta)} .\] The function \(K_U\) is defined as \((z,\bar{\zeta})\) vary over \(U \times U^*\), where we write \[U^* = \{ \zeta \in \mathbb{C}^n : \bar{\zeta} \in U \}.\]
Then for all \(f \in A^2(U)\) we have \[\label{eq:1} f(z) = \int_U f(\zeta) K_U(z,\bar{\zeta}) \, dV(\zeta) .\] This last equation is sometimes called the reproducing property of the kernel.
Note that the Bergman kernel depends on \(U\), which is why we write it as \(K_U(z,\bar{\zeta})\).
The Bergman kernel \(K_U(z,\bar{\zeta})\) is holomorphic in \(z\), antiholomorphic in \(\zeta\), and \[\overline{K_U(z,\bar{\zeta})} = K_U(\zeta,\bar{z}) .\]
- Proof
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As each \(k_z\) is in \(A^2(U)\), it is holomorphic in \(\zeta\). Hence, \(K_U\) is antiholomorphic in \(\zeta\). If we prove \(\overline{K_U(z,\bar{\zeta})} = K_U(\zeta,\bar{z})\), then we prove \(K_U\) is holomorphic in \(z\).
As \(\overline{K_U(z,\bar{\zeta})} = k_z(\zeta)\) is in \(A^2(U)\), then \[\begin{align}\begin{aligned} \overline{K_U(z,\bar{\zeta})} & = \int_{U} \overline{K_U(z,\bar{w})} K_U(\zeta,\bar{w}) dV(w) \\ & = \overline{ \left( \int_{U} \overline{K_U(\zeta,\bar{w})} K_U(z,\bar{w}) dV(w) \right) } = \overline{ \overline{ K_U(\zeta,\bar{z}) }} = K_U(\zeta,\bar{z}) . \end{aligned}\end{align}\]
Therefore, thinking of \(\bar{\zeta}\) as the variable, \(K_U\) is a holomorphic function of \(2n\) variables.
Let us compute the Bergman kernel (and the Szegö kernel of the next section while we’re at it) explicitly for the unit disc \(\mathbb{D} \subset \mathbb{C}\). Let \(f \in \mathcal{O}(\mathbb{D}) \cap C(\overline{D})\), that is, \(f\) is holomorphic in \(\mathbb{D}\) and continuous up to the boundary. Let \(z \in \mathbb{D}\). Then \[f(z) = \frac{1}{2\pi i} \int_{\partial \mathbb{D}} \frac{f(\zeta)}{\zeta-z} \, d\zeta .\] On the unit circle \(\zeta \bar{\zeta} = 1\). Let \(ds\) be the arc-length measure on the circle, parametrized as \(\zeta = e^{is}\). Then \(d\zeta = i e^{is} ds\), and \[f(z) = \frac{1}{2\pi i} \int_{\partial \mathbb{D}} \frac{f(\zeta)}{\zeta-z} \, d\zeta = \frac{1}{2\pi i} \int_{\partial \mathbb{D}} \frac{f(\zeta)}{1-z\bar{\zeta}} \bar{\zeta} \, d\zeta = \frac{1}{2\pi} \int_{\partial \mathbb{D}} \frac{f(\zeta)}{1-z\bar{\zeta}} \, ds .\] The integral is now a regular line integral of a function whose singularity, which used to be inside the unit disc, disappeared (we “reflected it” to the outside). The kernel \(\frac{1}{2\pi} \frac{1}{1-z\bar{\zeta}}\) is called the Szegö kernel, which we will briefly mention next. We apply Stokes to the second integral above: \[\begin{align}\begin{aligned} \frac{1}{2\pi i} \int_{\partial \mathbb{D}} \frac{f(\zeta)}{1-z\bar{\zeta}} \bar{\zeta} \, d\zeta &= \frac{1}{2\pi i} \int_{\mathbb{D}} f(\zeta) \frac{\partial}{\partial \bar{\zeta}} \left[ \frac{\bar{\zeta}}{1-z\bar{\zeta}} \right] \, d\bar{\zeta} \wedge d\zeta \\ &= \frac{1}{\pi} \int_{\mathbb{D}} \frac{f(\zeta)}
Callstack:
at (Bookshelves/Analysis/Tasty_Bits_of_Several_Complex_Variables_(Lebl)/05:_Integral_Kernels/5.02:_The_Bergman_Kernel), /content/body/div[6]/div/p/span[12], line 1, column 5
In an exercise you found that \(A^2(\mathbb{C}^n) = \{ 0 \}\). Therefore, \(K_{\mathbb{C}^n}(z,\bar{\zeta}) \equiv 0\).
The Bergman kernel for a more general domain is diffcult (usually impossible) to compute explicitly. We do have the following formula however.
Suppose \(U \subset \mathbb{C}^n\) is a domain, and \(\{ \varphi_j (z) \}_{j\in I}\) is a complete orthonormal system for \(A^2(U)\). Then \[K_U(z,\bar{\zeta}) = \sum_{j \in I} \varphi_j(z) \overline{\varphi_j(\zeta)} ,\] with uniform convergence on compact subsets of \(U \times U^*\).
- Proof
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For any fixed \(\zeta \in U\), the function \(z \mapsto K_U(z,\bar{\zeta})\) is in \(A^2(U)\) and so expand this function in terms of the basis and use the reproducing property of \(K_U\) \[K_U(z,\bar{\zeta}) = \sum_{j \in I} \left(\int_U K_U(w,\bar{\zeta}) \overline{\varphi_j(w)} \, dV(w) \right) \varphi_j(z) = \sum_{j \in I} \overline{\varphi_j(\zeta)} \varphi_j(z) .\] The convergence is in \(L^2\) as a function of \(z\), for a fixed \(\zeta\). Let \(K \subset \subset U\) be a compact set. Via Lemma \(\PageIndex{1}\), \(L^2\) convergence in \(A^2(U)\) is uniform convergence on compact sets. Therefore, for a fixed \(\zeta\) the convergence is uniform in \(z \in K\). In particular, we get pointwise convergence. So, \[\sum_{j \in I} \left| \varphi_j(z) \right| ^2 = \sum_{j \in I} \varphi_j(z) \overline{\varphi_j(z)} = K_U(z,\bar{z}) \leq C_K < \infty ,\] where \(C_K\) is the supremum of \(K_U(z,\bar{\zeta})\) on \(K \times K^*\). Hence for \((z,\bar{\zeta}) \in K \times K^*\), \[\sum_{j \in I} \left| \varphi_j(z) \overline{\varphi_j(\zeta)} \right| \leq \sqrt{ \sum_{j \in I} \left| \varphi_j(z) \right| ^2 } \sqrt{ \sum_{j \in I} \left| \varphi_j(\zeta) \right| ^2 } \leq C_K < \infty .\] And so the convergence is uniform on \(K \times K^*\).
- Show that if \(U \subset \mathbb{C}^n\) is bounded, then \(K_U(z,\bar{z}) > 0\) for all \(z \in U\).
- Why can this fail if \(U\) is unbounded? Find a (trivial) counterexample.
Show that given a domain \(U \subset \mathbb{C}^n\), the Bergman kernel is the unique function \(K_U(z,\bar{\zeta})\) such that
- for a fixed \(\zeta\), \(K_U(z,\bar{\zeta})\) is in \(A^2(U)\),
- \(\overline{K_U(z,\bar{\zeta})} = K_U(\zeta,\bar{z})\),
- the reproducing property \(\eqref{eq:1}\) holds.
Let \(U \subset \mathbb{C}^n\) be either the unit ball or the unit polydisc. Show that \(A^2(U) \cap C(\overline{U})\) is dense in \(A^2(U)\). In particular, this exercise says we only need to check the reproducing property on functions continuous up to the boundary to show we have the Bergman kernel.
Let \(U, V \subset \mathbb{C}^n\) be two domains and \(f \colon U \to V\) a biholomorphism. Prove \[K_U(z,\bar{\zeta}) = \det D f(z) \, \overline{\det D f (\zeta)} \, K_V\bigl(f(z),\overline{f(\zeta)}\bigr) .\]
Show that the Bergman kernel for the polydisc is \[K_{\mathbb{D}^n}(z,\bar{\zeta}) = \frac{1}{\pi^n} \prod_{j=1}^n \frac{1}{{(1-z_j\bar{\zeta}_j)}^2}.\]
Show that for some constants \(c_\alpha\), the set of all monomials \(\frac{z^\alpha}{c_\alpha}\) gives a complete orthonormal system of \(A^2(\mathbb{B}_{n})\). Hint: To show orthonormality compute the integral using polar coordinates in each variable separately, that is, let \(z_j = r_j e^{i\theta_j}\) where \(\theta \in [0,2\pi]^n\) and \(\sum_j r_j^2 < 1\). Then show completeness by showing that if \(f \in A^2(\mathbb{B}_{n})\) is orthogonal to all \(z^\alpha\), then \(f = 0\). Finding \(c_\alpha = \sqrt{\frac{\pi^n \alpha!}{(n+|\alpha|)!}}\) requires the classical \(\beta\) function of special function theory.
Using the previous exercise, show that the Bergman kernel for the unit ball is
\[K_{\mathbb{B}_{n}} (z,\bar{\zeta})=\frac{n!}{\pi ^{n}}\frac{1}{(1-\langle z,\zeta \rangle )^{n+1}},\] where \(\langle z, \zeta\rangle \) is the standard inner product on \(\mathbb{C}^n\).