Bergman Kernel for $L_p$ space (p $\neq$ 2)

Question

For a given space $\mathcal{X}$, consider the Banach space $L_p(\mathcal{X},\mu)$ for the measure space $(\mathcal{X},\mu)$. I'm trying to understand the Bergman kernel for $L_p$ when $p \neq 2$.

For $p = 2$, for the set of holomorphic functions $A^2 \subset L_2$, we know the Bergman kernel. Given that the space $A^2$ is a Hilbert space under the usual inner product $\langle f,g \rangle = \int_{\mathcal{X}} f(x)\bar{g}(x)d\mu(x)$, this turns out be an RKHS (reproducing kernel Hilbert space).

I'm trying to construct an RKBS (reproducing kernel Banach space) where both the dual map and kernel are known, and that's why considering $L_p$ where $p \neq 2$. Note the reproducing kernel for a Banach space is slightly different. More specifically, the notion of reproducing kernel$K$, which I'm considering has the following properties: for a Banach space $B$ and its dual $B'$, it should satisfy

1. $K(x,⋅) \in B$, $K(⋅,x) \in B'$ for all $x\in \mathcal{X}$
2. $f(x)=(f,K(⋅,x))_B$, $f^∗(x)=(K(x,⋅),f^∗)_B$ for $f \in B$, $f^∗ \in B'$ where $f^*$ is the dual of $f$.

Now, when considering the case of $L_p$, one of the semi-inner products is $(f,g)_{L_p} = \frac{\int_{\mathcal{X}} f\bar{g}|g|^{p-2}}{||g||^{p-2}_{L_p}}$ for $f,g \in L_p$. It is clear that the dual map in this case has the following form: $f^* = \frac{\bar{f}|f|^{p-2}}{||f||^{p-2}_{L_p}}$. But it is not clear to me what could be valid reproducing kernels. A potential one could be an $L_p$ version of the Bergman kernel. One idea could be to consider the set of holomorphic functions $A^p \subset L_p$, but the issue is we have a different semi-inner product compared to the $L_2$ case and thus we can't use the same Bergman kernel.

Could we find an explicit form of the Bergman kernel for the Bergman space $A^p$ of $L_p$ under any smoothness assumption on the space $\mathcal{X}$ ,e.g. polydisc for $L_2$ gives an explicit form? There is some useful discussion in some prior work but I haven't been to get a concrete answer to this.

Some prior work: https://arxiv.org/pdf/1201.4148.pdf

Answer 1

NOT A COMPLETE ANSWER. IT REMAINS TO SOLVE (1) BELOW.

This link is surely useful:

https://math.libretexts.org/Bookshelves/Analysis/Tasty_Bits_of_Several_Complex_Variables_(Lebl)/05%3A_Integral_Kernels/5.02%3A_The_Bergman_Kernel

This is the chapter on the Bergman kernel of the book on several complex variables of Jiří Lebl (Jiří is active here and I am a huge fan of his contributions, by the way).

The task is to redo as much as possible of that in the space $A^p(U)$ rather than $A^2(U)$. Since the OP is seeking explicit formulas, $U$ must be the unit disk $\mathbb D$.

Now Lemma (5.2.4) in Jiří's chapter easily generalizes to $A^p$; just use Hölder rather than Cauchy-Schwarz. Thus, the evaluation at $z\in \mathbb D$, that is the linear functional $f\mapsto f(z)$, is continuous in $A^p$. Standard abstract nonsense immediately implies that there is a kernel $K_p$ such that $$ f(z)=(f, K_p( z, \cdot))_{L^p}, $$ for all $f\in A^p$. In the case $p=2$ this kernel must reduce to $$ K_2(z, w)= \frac{1}{\pi} \frac{1}{{(1-z\bar{w})}^2},$$ as I read in the displayed formula immediately after (5.2.14) of Jiri's chapter.

(MESSAGE TO JIRI. By the way, if you are reading this, that display formula does not compile properly. You seem to be missing an \end{align}).

Recall the definition of $(\cdot, \cdot)_{L^p}$, that is, $$ (f, K_p(z, \cdot))_{L^p}=\frac{\int_{\mathbb D} f(w)\overline{K_p(z, w)}\lvert K_p(z, w)\rvert^{p-2}\, dV(w)}{\left( \int_{\mathbb D} \lvert K_p(z,w)\rvert^p\, dV(w)\right)^{\frac{p-2}{p}}}.$$ We must have $$ (f, K_p(z, \cdot))_{L^p}=(f, K_2(z, \cdot))_{L^2}, $$ so, since $f$ is arbitrary here, $K_p$ is obtained from $K_2$ by solving the equation $$\tag{1} \overline{K_2(z,w)}=\frac{\overline{K_p(z, w)}\lvert K_p(z, w)\rvert^{p-2}}{\left( \int_{\mathbb D} \lvert K_p(z,w')\rvert^p\, dV(w')\right)^{\frac{p-2}{p}}},$$ where $K_2$ is explicit and it has been given previously. Note that here $dV$ denotes the "volume element" in $\mathbb D$, that is $$%dV(w)=dx\wedge dy=\frac{i}{2}dw\wedge d\overline w,\qquad w=x+iy. dV(w)=dx dy=\frac{1}{2}dwd\overline w,\qquad w=x+iy.$$(This should be called "area element" and denoted $dA$, but the letter $A$ is already used here).