The Patterson–Sullivan Reconstruction of Pluriharmonic Functions for Determinantal Point Processes on Complex Hyperbolic Spaces

Abstract

The Patterson–Sullivan reconstruction is proved almost surely to recover a Bergman function from its values on a random discrete subset sampled with the determinantal point process induced by the Bergman kernel on the unit ball ${\mathbb {B}}_d$ in ${\mathbb {C}}^d$. For supercritical weighted Bergman spaces, the reconstruction is uniform when the functions range over the unit ball of the weighted Bergman space. We obtain a necessary and sufficient condition for reconstruction of a fixed pluriharmonic function in the complex hyperbolic space of arbitrary dimension; prove simultaneous uniform reconstruction for weighted Bergman spaces as well as strong simultaneous uniform reconstruction for weighted harmonic Hardy spaces; and establish the impossibility of the uniform simultaneous reconstruction for the Bergman space $A^2({\mathbb {B}}_d)$ on ${\mathbb {B}}_d$.

Appendix: The Proof of Proposition 6.2

Lemma 9.1

There exist two constants $c, C>0$ depending only on d, such that

$$\begin{aligned}&\frac{c}{(1 - |z|^2)^d} \log \left( \frac{2}{1 - |z|^2}\right) \le K_{W_{\mathrm {cr}}}(z, z) \\&\quad \le \frac{C}{(1 - |z|^2)^d} \log \left( \frac{2}{1 - |z|^2}\right) \quad \hbox { for all}\ z\in {\mathbb {B}}_d. \end{aligned}$$

Lemma 9.2

For any integer $d\ge 1$, there exist constants $c_d, C_d> 0$ such that

$$\begin{aligned}&\frac{c_d}{(1 - t)^d} \log \left( \frac{2}{1 - t}\right) \le \sum _{k =0}^\infty (k+1)^{d-1} \log (k + 2) t^k \\&\quad \le \frac{C_d}{(1 - t)^d} \log \left( \frac{2}{1 - t}\right) \quad \hbox { for all}\ t\in (0,1). \end{aligned}$$

Proof

For any integer $m\ge 1$ and any $t\in (0, 1)$, we have

$$\begin{aligned} (1 - t) \sum _{k=0}^\infty (k+1)^m \log (k+2) t^k = \sum _{k=0}^\infty \left[ \underbrace{(k+1)^m \log (k+2) - k^m \log (k +1)}_{\hbox { denoted}\ a_{k,m}}\right] t^k. \end{aligned}$$

Clearly, there exist constants $c_m, C_m>0$ depending only on m such that

$$\begin{aligned} c_m \cdot (k+1)^{m-1} \log (k+2) \le a_{k,m} \le C_m \cdot (k + 1)^{m-1} \log (k +2) \quad \hbox { for all}\ k\ge 0. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \frac{c_m}{1-t} \le \frac{\sum _{k=0}^\infty (k+1)^m \log (k+2) t^k}{\sum _{k=0}^\infty (k+1)^{m-1} \log (k+2) t^k} \le \frac{C_m}{1-t} \quad \hbox { for all}\ t\in (0,1). \end{aligned}$$

It follows that, there exist constants $c_d', C_d' >0$ such that

$$\begin{aligned} \frac{c_d'}{(1-t)^{d-1}} \le \frac{\sum _{k=0}^\infty (k+1)^{d-1}\log (k+2) t^k}{\sum _{k=0}^\infty \log (k+2) t^k} \le \frac{C_d'}{(1-t)^{d-1}} \quad \hbox { for all}\ t\in (0,1). \end{aligned}$$

Finally, note that for any $t\in (0, 1)$,

$$\begin{aligned} (1 - t) \sum _{k=0}^\infty \log (k + 2) t^k = \sum _{k=0}^\infty \log \left( 1 + \frac{1}{k+1}\right) t^k \le \log 2 + \sum _{k=1}^\infty \frac{t^k}{k} = \log \left( \frac{2}{ 1 - t}\right) \end{aligned}$$

and there exists $c''>0$ such that for any $t\in (0, 1)$,

$$\begin{aligned}&(1 - t) \sum _{k=0}^\infty \log (k + 2) t^k = \sum _{k=0}^\infty \log \left( 1 + \frac{1}{k+1}\right) t^k\\&\quad \ge c'' \left( \log 2 + \sum _{k=1}^\infty \frac{t^k}{k} \right) = c'' \log \left( \frac{2}{ 1 - t}\right) . \end{aligned}$$

Combining the above inequalities, we complete the proof of the lemma. $\square$

Proof of Lemma 9.1

Since $W_{\mathrm {cr}}$ is radial, we have

$$\begin{aligned} K_{W_{\mathrm {cr}}}(z, w) = \sum _{n\in {\mathbb {N}}_0^d} a_n(W_{\mathrm {cr}}) z^n {\bar{w}}^n \quad \hbox {with} z^{n} : = z_1^{n_1} \cdots z_d^{n_d} \hbox {,} a_n(W_{\mathrm {cr}})= \Vert z^n\Vert _{A^2({\mathbb {B}}_d, W_{\mathrm {cr}})}^{-2}. \end{aligned}$$

(9.1)

By the formula of integration in polar coordinates and [ZHU05, Lem. 1.11],

$$\begin{aligned} a_n(W_{\mathrm {cr}})^{-1}&= \int _{{\mathbb {B}}_d} | z^n|^2 W_{\mathrm {cr}} (z) dv_d(z) = 2d \int _0^1 r^{2d + 2|n|-1} W_{\mathrm {cr}}(r)dr \int _{{\mathbb {S}}_d} |\zeta ^n|^2 d\sigma _{{\mathbb {S}}_d}(\zeta ) \\&= d \cdot \frac{n_1! \cdots n_d!}{ (d-1 + |n|)!} \int _0^1 t^{|n| + d -1} ( 1 - t)^{-1} \log ^{- 2} \left( \frac{4}{ 1 - t}\right) dt \\&= d \cdot \frac{n_1! \cdots n_d!}{ (d-1 + |n|)!} \cdot \int _{\log 4}^\infty \frac{(1 - 4 e^{-x})^{|n| + d -1}}{x^2} dx. \end{aligned}$$

Claim: There exist two constants $c_1, c_2 > 0$ such that

$$\begin{aligned} \frac{c_1}{\log (4k +4)} \le \int _{\log 4}^\infty \frac{(1 - 4 e^{-x})^k}{x^2} dx \le \frac{c_2}{\log (4k +4)} \quad \hbox { for all integer}\ k\ge 0. \end{aligned}$$

Indeed, for any integer $k \ge 0$, we have the lower estimate of the integral:

$$\begin{aligned} \int _{\log 4}^\infty (1 - 4 e^{-x})^k \frac{dx}{x^2}\ge & {} \int _{\log (4k+4)}^\infty (1 - 4 e^{-x})^k \frac{dx}{x^2} \\\ge & {} \int _{\log (4k+4)}^\infty \left( 1 - \frac{1}{k+1}\right) ^k \frac{dx}{x^2} \ge \frac{c_1}{\log (4k +4)}. \end{aligned}$$

Now set $H_k(x): = (1 - 4 e^{-x})^k/x^2$ for $x \in [\log 4, \infty )$. For $k\ge 1$, we can show, by studying the derivative $H_k'(x)$, that $H_k$ is increasing on the interval $[\log 4, \log (4k +4)]$ and hence

$$\begin{aligned} H_k(x) \le H_k(\log (4k+4)) \le \frac{1}{\log ^2 (4k +4)} \quad \hbox { for all}\ x\in [\log 4, \log (4k +4)]. \end{aligned}$$

Therefore, for any $k\ge 1$, we have

$$\begin{aligned} \int _{\log 4}^\infty H_k(x) dx \!\le \! \sup _{\log 4 \le x \le \log (4k +4)} H_k(x) \!\cdot \! \int _{\log 4}^{\log (4k +4)} dx \!+\! \int _{\log (4k + 4)}^\infty \frac{ 1}{x^2}dx \!\le \! \frac{2}{\log (4k \!+\!4)}. \end{aligned}$$

Consequently, there exist constants $c_3, c_4> 0$ such that for any $n\in {\mathbb {N}}_0^d$, we have

$$\begin{aligned} c_3 \le \frac{ a_n(W_{\mathrm {cr}})^{-1}}{ \displaystyle d \cdot \frac{n_1! \cdots n_d!}{ (d-1 + |n|)!} \frac{1}{\log (4(|n|+d))}} \le c_4. \end{aligned}$$

(9.2)

By using the elementary identity (see e.g. Zhu [ZHU05, formula (1.1)])

$$\begin{aligned} \sum _{n \in {\mathbb {N}}_0^d: |n| = k } \frac{r_1^{n_1} \cdots r_d^{n_d}}{n_1!\cdots n_d!} = \frac{(r_1 + \cdots + r_d)^k}{k!}, \end{aligned}$$

we obtain, for any $z\in {\mathbb {B}}_d$, that

$$\begin{aligned} \Phi (z) = : \sum _{n \in {\mathbb {N}}_0^d} \frac{(d -1 + |n|)!}{d} \frac{\log (4(|n|+d)) }{n_1!\cdots n_d!} |z^n|^{2}= \sum _{k=0}^\infty \frac{(d -1 +k)!}{ d \cdot k!} \log (4(k+d)) |z|^{2k}. \end{aligned}$$

Therefore, by the limit equalities

$$\begin{aligned} \lim _{k\rightarrow \infty }\frac{(d-1+k)!}{(k+1)^{d-1} \cdot k!} = 1 \text { and }\lim _{k\rightarrow \infty }\frac{\log (4 (k + d))}{\log (k + 2)} = 1 \end{aligned}$$

and Lemma 9.2, there exist constants $c_5, c_6> 0$, such that

$$\begin{aligned} \frac{c_5}{(1 - |z|^2)^d} \log \left( \frac{2}{1 - |z|^2}\right) \le \Phi (z)\le \frac{c_6}{(1 - |z|^2)^d} \log \left( \frac{2}{1 - |z|^2}\right) . \end{aligned}$$

Finally, comparing (9.1) with the definition of $\Phi (z)$ and using (9.2), we obtain $c_3 \Phi (z) \le K_{W_{\mathrm {cr}}}(z,z) \le c_4 \Phi (z)$ for all $z\in {\mathbb {B}}_d$. This completes the whole proof. $\square$

Let $\Phi : [0, 1) \rightarrow {\mathbb {R}}_{+}$ be a function in $L^1([0,1])$ such that $\int _{\delta }^1 \Phi (r)dr > 0$ for any $\delta \in (0, 1)$. Let $B: [0, 1)\rightarrow {\mathbb {R}}_{+}$ be a function such that $B\Phi \in L^1([0,1])$ and $\lim _{r\rightarrow 1^{-}} B(r) = \infty .$ Define two radial Bergman-admissible weights on ${\mathbb {B}}_d$ by $W_\Phi (z) = \Phi (|z|)$ and $W_{B\Phi }(z) = B(|z|) \Phi (|z|)$. We shall compare the reproducing kernels $K_{W_\Phi }$ and $K_{W_{B\Phi }}$. Clearly,

$$\begin{aligned} \lim _{k\rightarrow \infty } \frac{ \int _0^1 r^k \Phi (r) dr}{ \int _0^1 r^k B(r) \Phi (r)dr} = 0. \end{aligned}$$

(9.3)

Lemma 9.3

Under the above assumptions on $\Phi$ and B, we have

$$\begin{aligned} \lim _{|z|\rightarrow 1^{-}} \frac{K_{W_{B\Phi }}(z,z)}{K_{W_\Phi }(z,z)} = 0. \end{aligned}$$

Proof

Since $W_\Phi : {\mathbb {B}}_d\rightarrow {\mathbb {R}}_{+}$ is radial, we have

$$\begin{aligned} K_{W_\Phi }(z, w)= & {} \sum _{n\in {\mathbb {N}}_0^d} a_n(\Phi ) z^n {\bar{w}}^n \quad \hbox { with}\ a_n(\Phi )= \Vert z^n\Vert _{A^2({\mathbb {B}}_d, W_\Phi )}^{-2} \\= & {} \Vert z_1^{n_1} \cdots z_d^{n_d}\Vert _{A^2({\mathbb {B}}_d, W_\Phi )}^{-2}. \end{aligned}$$

For any $n\in {\mathbb {N}}_0^d$, by the formula of integration in polar coordinates,

$$\begin{aligned} a_n(\Phi )^{-1} = \int _{{\mathbb {B}}_d} | z^n|^2 \Phi (|z|) dv_d(z) = 2d \int _0^1 r^{2d + 2|n|-1} \Phi (r)dr \int _{{\mathbb {S}}_d} |\zeta ^n|^2 d\sigma _{{\mathbb {S}}_d}(\zeta ). \end{aligned}$$

Replacing $\Phi$ by $B\Phi$, we obtain the corresponding formulas for $K_{W_{B\Phi }}$ and for $a_n(B\Phi )$. In particular, we see that the ratio $\frac{a_n(B\Phi )}{a_n(\Phi )}$ depends only on |n|:

$$\begin{aligned} R_{|n|}: = \frac{a_n(B\Phi )}{a_n(\Phi )} = \frac{ \int _0^1 r^{2d + 2|n|-1} W(r)dr}{ \int _0^1 r^{2d + 2|n|-1} B(r) W(r)dr}. \end{aligned}$$

Then for any $z\in {\mathbb {B}}_d$, by writing $z = r\zeta$ with $r = |z|$ and $\zeta = z/r\in {\mathbb {S}}_d$, we have

$$\begin{aligned} K_{W_\Phi }(z,z)&= \sum _{k=0}^\infty \left( \sum _{n\in {\mathbb {N}}_0^d: |n| = k} a_n(\Phi ) |z^n|^2\right) = \sum _{k=0}^\infty \left( \sum _{n\in {\mathbb {N}}_0^d: |n| = k} a_n(\Phi ) |\zeta ^n|^2\right) r^{2k},\\ K_{W_{B\Phi }}(z,z)&= \sum _{k=0}^\infty \left( \sum _{n\in {\mathbb {N}}_0^d: |n| = k} a_n(B\Phi ) |z^n|^2\right) = \sum _{k=0}^\infty R_k \cdot \left( \sum _{n\in {\mathbb {N}}_0^d: |n| = k} a_n(\Phi ) |\zeta ^n|^2\right) r^{2k}. \end{aligned}$$

Since $W_\Phi$ is radial, the transformation $f(\cdot ) \mapsto f (U\cdot )$ is unitary on $A^2({\mathbb {B}}_d, W_\Phi )$ for any $d\times d$ unitary matrix U. Hence by (3.9), the function $z \mapsto K_{W_\Phi }(z,z)$ is radial and thus

$$\begin{aligned} A_k(\Phi ): = \sum _{n\in {\mathbb {N}}_0^d: |n| = k} a_n(\Phi ) |\zeta ^n|^2 \quad \hbox { does not depend on}\ \zeta \in {\mathbb {S}}_d. \end{aligned}$$

Now by (9.3), we have $\lim _{k\rightarrow \infty } R_{k} = 0$. It follows that

$$\begin{aligned} \limsup _{|z|\rightarrow 1^{-}} \frac{K_{W_{B\Phi }}(z,z)}{K_{W_\Phi }(z,z)} = \limsup _{r\rightarrow 1^{-}} \frac{ \sum _{k=0}^\infty R_k A_k(\Phi ) r^{2k} }{ \sum _{k=0}^\infty A_k(\Phi ) r^{2k} } = 0. \end{aligned}$$

In the last step, we use the following elementary fact: for any two sequences $(a_k)_{k\in {\mathbb {N}}_0}, (b_k)_{k\in {\mathbb {N}}_0}$ in ${\mathbb {R}}_{+}$ with $\lim _{k\rightarrow \infty } a_k/b_k = 0$, the following limit

$$\begin{aligned} \lim _{t \rightarrow 1^{-}} \frac{ \sum _{k\in {\mathbb {N}}_0} a_k t^k }{\sum _{k\in {\mathbb {N}}_0} b_k t^k} = 0 \end{aligned}$$

holds provided that the two series $\sum _{k\in {\mathbb {N}}_0} a_k t^k, \sum _{k\in {\mathbb {N}}_0} b_k t^k$ converge for all $t\in (0, 1)$ and $\sum _{k\in {\mathbb {N}}_0} b_k = \infty$. $\square$

Proof of Proposition 6.2

We can write $W_{\mathrm {cr}} = W_{\Phi }$ for a function $\Phi : [0, 1) \rightarrow {\mathbb {R}}_{+}$. Let W be a supercritical weight on ${\mathbb {B}}_d$. Set $B(r) : = \inf _{\zeta \in {\mathbb {S}}_d} W(r \zeta )/\Phi (r)$ for any $r\in [0,1)$. By (6.1), $\lim _{r\rightarrow 1^{-}} B(r)= \infty .$ Hence by Lemma 9.3, we have

$$\begin{aligned} \lim _{|z|\rightarrow 1^{-}} \frac{K_{W_{B\Phi }} (z, z) }{K_{W_{\mathrm {cr}}} (z, z)} = \lim _{|z|\rightarrow 1^{-}} \frac{K_{W_{B\Phi }} (z, z) }{K_{W_{\Phi }} (z, z)} = 0. \end{aligned}$$

Since $B(|z|) \Phi (|z|) \le W(z)$, by (3.9), we have $K_W(z,z) \le K_{W_{B\Phi }}(z,z).$ The desired estimate of $K_W(z,z)$ now follows from Lemma 9.1. $\square$