Orthonormal Basis of Bergman Space

Question

This is a problem (#39 or #40 depending on the edition) at the end of Chapter 1 in Krantz's book Function Theory of Several Complex Variables.

Let $\Omega\subset\mathbb{C}^n$ be a smooth and bounded domain. Let $z_0\in\Omega$.

Goal: To construct an orthonormal basis of the Bergman Space $A^2(\Omega)$.

Step 1: Start the construction by choosing the unique function $\phi_0\in A^2(\Omega)$ with $\phi_0(z_0)$ real, $\|\phi_0\|=1$ and $\phi_0(z_0)$ maximal. We have an explicit description of $\phi_0$. Let $K$ be the Bergman kernel for $\Omega$. Then

$$ \phi_0(z) = \dfrac{K(z,z_0)}{\sqrt{K(z_0,z_0)}}.$$

This is because we have

$$ K(z_0,z_0) = \sup\{ |f(z_0)|^2\ :\ f\in A^2(\Omega), \|f\| =1\}.$$

The supremum is being taken over a normal family, and hence the supremum is attained for some (unique) function $\phi_0\in A^2(\Omega)$.

Step 2: Next we choose the unique function $\phi_1\in A^2(\Omega)$ with $\phi_1(z_0)=0$, $\dfrac{\partial \phi_1}{\partial z_1}(z_0)$ real, $\|\phi_1\|=1$, and $\dfrac{\partial \phi_1}{\partial z_1}(z_0)$ maximal. Again in case of

$$ \sup\left\{ \dfrac{\partial f}{\partial z_1}(z_0) :\ f\in A^2(\Omega)\ , \ f(z_0)=0, \|f\| = 1, \dfrac{\partial f}{\partial z_1}(z_0) \text{ real} \right\},$$

the suprenum is taken over a normal family and hence the supremum is attained for some unique function $\phi_1$.

Orthogonality: Note that by the reproducing property of the Bergman Kernel, we have

\begin{equation} \langle \phi_1, \phi_0\rangle = \left\langle \phi_1, \dfrac{K(\cdot,z_0)}{\sqrt{K(z_0,z_0)}}\right\rangle = \dfrac{\phi_1(z_0)}{\sqrt{K(z_0,z_0)}} = 0 \end{equation}

Further steps: Now the author says "Continue this process to create an orthogonal system on $\Omega$". Here is what I did:

Step 3: Choose the unique element $\phi_2\in A^2(\Omega)$ with $\phi_2(z_0)=\dfrac{\partial \phi_2}{\partial z_1}(z_0)=0$, $\dfrac{\partial^2 \phi_2}{\partial z_1^2}(z_0)$ real, $\|\phi_2\|=1$, and $\dfrac{\partial^2 \phi_2}{\partial z_1^2}(z_0)$ maximal. The uniqueness and existence is the same as above. Also, same as before, the reproducing property of the Bergman kernel gives us $\langle \phi_0, \phi_2\rangle = 0$.

My question: Why is $\langle \phi_1, \phi_2\rangle = 0$? Is there an explicit description of $\phi_1$ (just as we have for $\phi_0$)?

Answer 1

Theorem 2.2 in Kobayashi's paper (referenced in the problem) has the following construction:

Consider the subspace $A_0 = \{ f\in A^2(\Omega)\ :\ f(z_0) = 0$ } of $A^2(\Omega)$. Then $A_0$ is the kernel of the evaluation functional $f\mapsto f(z_0)$ and hence is a codimension $1$ subspace of $A^2(\Omega)$. Choose $\phi_0\in A^2(\Omega)$ s.t. $\|\phi_0\|=1$ and $\phi_0 \in (A_0)^\perp$. Now we claim that $A^2(\Omega) = \text{span}\{A_0, \phi_0 \}$

Proof of claim: Let $g\in A^2(\Omega)$. If $g\in A_0$, then we are done. Else $g(z_0)\ne0$. Choose a complex number $c$ such that $g(z_0)=c\phi_0(z_0)$. Then $(g-c\phi_0)\in A_0$ and hence $g\in\text{span}\{A_0, \phi_0 \}$.

The subspace $A_1^1 = \{ f\in A_0\ :\ \partial f/\partial z_1 (z_0) =0\}$ is the kernel of the linear functional $f\mapsto \partial f/ \partial z_1(z_0)$ and hence is of codimesion $1$ in $A_0$. Choose $\phi_1\in A_0$ s.t. $\|\phi_1\|=1$ and $\phi_1\in(A^1_1)^\perp$. Again we have $A^2(\Omega)=\text{span}\{A^1_1,\phi_0,\phi_1\}$.

The subspace $A_1^2 = \{ f\in A_1^1\ :\ \partial f/\partial z_2 (z_0) =0\}$ is the kernel of the linear functional $f\mapsto \partial f/ \partial z_2(z_0)$ and hence is of codimesion $1$ in $A_1^1$. Choose $\phi_2\in A_1^1$ s.t. $\|\phi_2\|=1$ and $\phi_2\in(A^2_1)^\perp$. Again we have $A^2(\Omega)=\text{span}\{A^2_1,\phi_0,\phi_1, \phi_2\}$.

Once we cover all the first order derivatives, we move on to the higher order derivatives similarly.

The subspace $A_2^{1,1} = \{ f\in A_1^n\ :\ \partial^2 f/\partial z_1^2 (z_0) =0\}$ is the kernel of the linear functional $f\mapsto \partial^2 f/\partial z_1^2(z_0)$ and hence is of codimesion $1$ in $A_1^n$. Choose $\phi_{n+1}\in A_1^n$ s.t. $\|\phi_{n+1}\|=1$ and $\phi_{n+1}\in(A^{1,1}_2)^\perp$. Again we have $A^2(\Omega)=\text{span}\{A_2^{1,1},\phi_0,\phi_1, \phi_2,\cdots,\phi_n,\phi_{n+1}\}$.

Continuing the process gives us a countable basis of orthonormal vectors in $A^2(\Omega)$. To see that this basis is a complete, note that if $f\in A^2(\Omega)$, then

$$\langle f, \phi_0\rangle =0 \implies f(z_0)=0$$ $$\langle f, \phi_1\rangle =0 \implies \partial f/ \partial z_1 (z_0)=0$$ $$\langle f, \phi_2\rangle =0 \implies \partial f/ \partial z_2 (z_0)=0$$ $$\vdots $$ $$\langle f, \phi_{n+1}\rangle =0 \implies \partial^2 f/ \partial z_2 (z_0)=0$$ $$\vdots $$

So if $\langle f, \phi_j\rangle =0$ for all $j$, then by Taylor series expansion we get that $f\equiv 0$.

I still don't know an explicit description of $\phi_1$.