§10.3 The Bergman Kernel

10.3 The Bergman Kernel

Fix $Ω$ be an open region and let $𝑧_{0} \in Ω$ be fixed. Then let $𝐴^{2} (Ω)$ be the Hilbert space of square integrable holomorphic functions on $Ω$ . Recall Example 10.1.15 $Example 10.1.15 Example 10.1.15$ :

Example 10.1.15 (Bergman Space).

Let $Ω \subseteq ℂ$ be an open region. Let

𝐴^{2} (Ω) = {𝑓 holomorphic on Ω : \iint_{Ω} {| 𝑓 (𝑧) |}^{2} d 𝑥 d 𝑦 < \infty} .

Then $𝐴^{2} (Ω)$ is a vector space over $ℂ$ with the standard definition of addition and scalar multiplication. Then the inner product defined by

⟨ 𝑓, 𝑔 ⟩ = \iint_{Ω} 𝑓 (𝑧) \overline{𝑔 (𝑧)} d 𝑥 d 𝑦 where 𝑧 = 𝑥 + i 𝑦 (10.3 . 1)

for any two

𝑓, 𝑔 \in 𝐴^{2} (Ω)

defines a Hilbert space with

𝐴^{2} (Ω)

For each $𝑓 \in Ω$ , Proposition 10.1.16 $Proposition 10.1.16 Proposition 10.1.16$ gives:

| 𝑓 (𝑧_{0}) | \leq \frac{1}{𝑟 \sqrt{π}} ‖ 𝑓 ‖ .

Moreover, the mapping $𝜙$ between $𝐴^{2} (Ω)$ and $ℂ$ given by

𝜙 : 𝑓 (\in 𝐴^{2} (Ω)) \mapsto 𝑓 (𝑧_{0}) (\in ℂ)

is a bounded linear functional on $𝐴^{2} (Ω)$ : linearity is trivial and boundedness follows from Proposition 10.1.16 $Proposition 10.1.16 Proposition 10.1.16$ . Then by the Riesz Representation Theorem (Theorem 10.1.36 $Theorem 10.1.36 Theorem 10.1.36$ ), there as a unique $𝑘_{𝑧_{0}} \in 𝐴^{2} (Ω)$ , dependent on $𝑧_{0}$ but not $𝑓 \in 𝐴^{2} (Ω)$ , such that

𝑓 (𝑧_{0}) = 𝜙 (𝑓) = ⟨ 𝑓, 𝑘_{𝑧_{0}} ⟩ . (for all 𝑓 \in 𝐴^{2} (Ω))

Then

𝑓 (𝑧_{0}) = \iint_{Ω} 𝑓 (𝜁) \overline{𝑘_{𝑧_{0}} (𝜁)} d 𝜉 d 𝜂 . (where 𝜁 = 𝜉 + i 𝜂)

Definition 10.3.1 (Bergman Kernel).

Let

Ω

be an open region, define the Bergman kernel to be a complex function

𝐾 (𝑧, 𝑤)

of two variables in

𝐴^{2} (Ω)

, given by

𝐾 (𝑧, 𝑤) = \overline{𝑘_{𝑧} (𝑤)}

, where

𝑘_{𝑧} (𝑤)

is the unique Riesz representer for the linear functional

𝑓 \mapsto 𝑓 (𝑧)

Then the previous equation is simply

𝑓 (𝑧_{0}) = \iint_{Ω} 𝑓 (𝜁) 𝐾 (𝑧_{0}, 𝜁) d 𝜉 d 𝜂 . (*)

Theorem 10.3.2 (Conjugate-symmetry of the Bergman Kernel).

For any $𝑧, 𝑤 \in Ω$ ,

𝐾 (𝑧, 𝑤) = \overline{𝐾 (𝑤, 𝑧)} .

Proof.

By the integral representation in $(∗) Part 2 Part 2$ , $\forall 𝑧, 𝑤 \in Ω$ , letting $𝑓 = 𝑧 \mapsto \overline{𝐾 (𝑤, 𝑧)} = 𝑘_{𝑤} (𝑧)$ ,

\begin{aligned} \overline{𝐾 (𝑤, 𝑧)} & = 𝑘_{𝑤} (𝑧) = \iint_{Ω} 𝑘_{𝑤} (𝜁) \overline{𝑘_{𝑧} (𝜁)} d 𝜉 d 𝜂 \\ = \overline{\iint_{Ω} 𝑘_{𝑧} (𝜁) \overline{𝑘_{𝑤} (𝜁)} d 𝜉 d 𝜂} \\ = \overline{𝑘_{𝑧} (𝑤)} = \overline{\overline{𝐾 (𝑧, 𝑤)}} = 𝐾 (𝑧, 𝑤), (by (∗) Part 2 Part 2) \end{aligned}

proving the assertion. ⁠

□

Corollary 10.3.2.1.

The Bergman Kernel

𝐾 (𝑧, 𝑤)

is holomorphic in the first argument:

\frac{𝜕}{𝜕 \overline{𝑧}} 𝐾 (𝑧, 𝑤) = 0

for all

𝑧 \in Ω

Proof.

Since

\overline{𝐾 (𝑤, 𝑧)} = 𝑘_{𝑤} (𝑧)

is holomorphic in

𝑧

, it follows from Theorem 10.3.2

Theorem 10.3.2 Theorem 10.3.2

that

𝐾 (𝑧, 𝑤)

is holomorphic in

𝑧

. ⁠

□

Moreover, certain properties of the Bergman kernel that we have discussed previously uniquely determine the Bergman kernel:

Theorem 10.3.3 (Reproducing Property of $𝐾 (\cdot, \cdot)$ ).

If $\tilde{𝐾} : Ω^{2} \to ℂ$ satisfies $𝑤 \mapsto \overline{\tilde{𝐾} (𝑧, 𝑤)} \in 𝐴^{2} (Ω)$ for each $𝑧 \in Ω$ , and if for all $𝑓 \in 𝐴^{2} (Ω)$ ,

𝑓 (𝑧) = \iint_{Ω} 𝑓 (𝜁) \tilde{𝐾} (𝑧, 𝜁) d 𝜉 d 𝜂,

then

\tilde{𝐾} \equiv 𝐾

Ω^{2}

Proof.

Let $𝑓$ be the Bergman kernel $𝜁 \mapsto 𝐾 (𝜁, 𝑤)$ (holomorphic in the left argument by Corollary 10.3.2.1 $Corollary 10.3.2.1 Corollary 10.3.2.1$ ). Then $\forall 𝑧, 𝑤 \in Ω$ ,

\begin{aligned} 𝐾 (𝑧, 𝑤) & = \iint_{Ω} 𝐾 (𝜁, 𝑤) \tilde{𝐾} (𝑧, 𝜁) d 𝜉 d 𝜂 \\ = \overline{\iint_{Ω} \overline{\tilde{𝐾} (𝑧, 𝜁)} 𝐾 (𝑤, 𝜁) d 𝜉 d 𝜂} (by Theorem 10.3.2 Theorem 10.3.2 Theorem 10.3.2) \\ = \overline{\overline{\tilde{𝐾} (𝑧, 𝑤)}}, \end{aligned}

where the last step uses

(∗) Part 2 Part 2

and

𝜁 \mapsto \overline{\tilde{𝐾} (𝑧, 𝜁)} \in 𝐴^{2} (Ω)

. Therefore,

𝐾 (𝑧, 𝑤) = \tilde{𝐾} (𝑧, 𝑤) ⟹ 𝐾 \equiv \tilde{𝐾}

Ω^{2}

. ⁠

□

Therefore, in the search of a Bergman kernel for a specified domain, if one can construct a function satisfying the properties outlined in Theorem 10.3.3 $Theorem 10.3.3 Theorem 10.3.3$ , then one can ascertain that it is the Bergman kernel.

Theorem 10.3.4.

Let

𝐴^{2} (Ω)

be the Bergman space of square-integrable functions on any open region

Ω \subseteq ℂ

. Then

𝐴^{2} (Ω)

has a complete countable orthonormal system

{𝑒_{𝑗}}

(There are several approaches to prove this theorem.)

Proof.

Enumerate

Ω \cap (ℚ + i ℚ) = {𝑧_{𝑛}}_{𝑛 \in ℕ} \subseteq Ω,

such that $\overline{{𝑧_{𝑛}}_{𝑛}} = \overline{Ω}$ , or in other words ${𝑧_{𝑛}}_{𝑛}$ is countable dense in $Ω$ . For each fixed $𝑛 \in ℕ$ , $𝑓 \mapsto 𝑓 (𝑧_{𝑛})$ is a bounded linear functional by Proposition 10.1.16 $Proposition 10.1.16 Proposition 10.1.16$ . Then let $𝑘_{𝑧_{𝑛}} \in 𝐴^{2} (Ω)$ be its unique Riesz representer, so that

𝑓 (𝑧_{𝑛}) = ⟨ 𝑓, 𝑘_{𝑧_{𝑛}} ⟩

for every $𝑓 \in 𝐴^{2} (Ω)$ . Define $𝑆 = \overline{{span {𝑘_{𝑧_{𝑛}}}}_{𝑛}}$ , which is Hilbert space in $𝐴^{2} (Ω)$ . Suppose $𝑓 \in 𝑆^{⟂}$ . Then for any $𝑛 \in ℕ$ ,

𝑓 (𝑧_{𝑛}) = ⟨ 𝑓, 𝑘_{𝑧_{𝑛}} ⟩ = 0 .

Because ${𝑧_{𝑛}}$ has an accumulation point in $Ω$ and $𝑓 (𝑧_{𝑛}) = 0$ for each $𝑛$ , by the Identity Theorem (Theorem 3.3.3 $Theorem 3.3.3 Theorem 3.3.3$ ), $𝑓 \equiv 0$ . Then $𝑆^{⟂} = {0}$ . Since $𝑆$ is closed, any $𝑓 \in 𝐴^{2} (Ω)$ can be decomposed, by Theorem 10.1.28 $Theorem 10.1.28 Theorem 10.1.28$ , into $𝑓_{1} + 𝑓_{2}$ , where $𝑓_{1} \in 𝑆$ and $𝑓_{2} \in 𝑆^{⟂} = {0}$ . Then $𝑓 \in 𝑆$ , implying $𝐴^{2} (Ω) = 𝑆$ ; then $𝑘_{𝑧_{𝑛}}$ densely spans $𝐴^{2} (Ω)$ .

For any $𝑓 \in 𝐴^{2} (Ω)$ for any $𝜀 > 0$ , choose

\tilde{𝑓} = \sum_{𝑗 = 1}^{𝑙} 𝑐_{𝑗} 𝑘_{𝑧_{𝑛_{𝑗}}}

such that $‖ \tilde{𝑓} - 𝑓 ‖ < \frac{𝜀}{2}$ . Letting $\tilde{𝑐_{𝑗}} \in ℚ + i ℚ$ be chosen such that

| 𝑐_{𝑗} - \tilde{𝑐_{𝑗}} | < \frac{𝜀}{2 𝑙 max {1, {max}_{𝑗} ‖ 𝑘_{𝑧_{𝑛_{𝑗}}} ‖}}

and letting

\tilde{\tilde{𝑓}} = \sum_{𝑗 = 1}^{𝑙} \tilde{𝑐_{𝑗}} 𝑘_{𝑧_{𝑛_{𝑗}}},

we see that

‖ \tilde{𝑓} - \tilde{\tilde{𝑓}} ‖ = ‖ \sum_{𝑗 = 1}^{𝑙} (𝑐_{𝑗} - \tilde{𝑐_{𝑗}}) 𝑘_{𝑧_{𝑛_{𝑗}}} ‖ \leq \sum_{𝑗 = 1}^{𝑙} \frac{𝜀}{2 𝑙 max {1, {max}_{𝑗} ‖ 𝑘_{𝑧_{𝑛_{𝑗}}} ‖}} ‖ 𝑘_{𝑧_{𝑛_{𝑗}}} ‖ \leq \frac{𝜀}{2},

whence

‖ 𝑓 - \tilde{\tilde{𝑓}} ‖ \leq ‖ \tilde{𝑓} - 𝑓 ‖ + ‖ \tilde{𝑓} - \tilde{\tilde{𝑓}} ‖ < 𝜀 .

Therefore, any $𝑓 \in 𝐴^{2} (Ω)$ can be arbitrarily approximated by an element in

{\sum_{𝑗 = 1}^{𝑙} \tilde{𝑐_{𝑗}} 𝑘_{𝑧_{𝑛_{𝑗}}} : 𝑙 \in ℕ, (\forall 𝑗 \in ℕ_{\leq 𝑙}) 𝑛_{𝑗} \in ℕ, 𝑐_{𝑗} \in ℚ + i ℚ},

which is countable (see equivalent justification in the proof of Proposition 10.2.8 $Proposition 10.2.8 Proposition 10.2.8$ ; essentially the set $⋃_{𝑙 = 1}^{\infty} ℕ^{𝑙} \times {(ℚ + i ℚ)}^{𝑙}$ is countable). Therefore Proposition 10.2.8 $Proposition 10.2.8 Proposition 10.2.8$ implies, from separability, the existence of a complete countable orthonormal system.

⁠ $□$

Proof (Alternative).

A reader who has learned measure theory could note that

𝐿^{2} (Ω)

is a separable Hilbert space, and by Theorem 10.2.9

Theorem 10.2.9 Theorem 10.2.9

𝐴^{2} (Ω)

is separable and has a complete countable orthonormal system by Proposition 10.2.8

Proposition 10.2.8 Proposition 10.2.8

. ⁠

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