Harmonic Bergman Spaces on the Real Hyperbolic Ball: Atomic Decomposition, Interpolation and Inclusion Relations

Abstract

For $\alpha >-1$ and $0<p<\infty$, we study weighted Bergman spaces $\mathcal {B}^p_{\alpha }$ of harmonic functions on the real hyperbolic ball. We obtain an atomic decomposition of Bergman functions in terms of reproducing kernels. We show that an r-separated sequence $\{a_m\}$ with sufficiently large r is an interpolating sequence for $\mathcal {B}^p_{\alpha }$. Using these we determine precisely when a Bergman space $\mathcal {B}^p_{\alpha }$ is included in another Bergman space $\mathcal {B}^q_\beta$.

1 Introduction

For $x,y\in \mathbb {R}^n$, let $\langle x,y\rangle =x_1y_1+\dots +x_ny_n$ be the Euclidean inner product and $|x|=\sqrt{\langle x,x\rangle }$ be the corresponding norm. Let $\mathbb {B}=\{x\in \mathbb {R}^n:|x|<1\}$ be the unit ball and $\mathbb {S}=\partial \mathbb {B}$ be the unit sphere. The hyperbolic ball is $\mathbb {B}$ endowed with the hyperbolic metric

$$\begin{aligned} ds^2=\frac{4}{(1-|x|^2)^2}\sum _{i=1}^n dx_i^2. \end{aligned}$$

The Laplacian $\Delta _h$ and the gradient $\nabla ^h$ with respect to the hyperbolic metric are given by (see [13, Chapter 3] for more details)

$$\begin{aligned} (\Delta _h f)(a)=\Delta (f\circ \varphi _a)(0) \qquad (f\in C^2(\mathbb {B})), \end{aligned}$$

and

$$\begin{aligned} (\nabla ^h f)(a)=-\nabla (f\circ \varphi _a)(0) \qquad (f\in C^1(\mathbb {B})), \end{aligned}$$

where $\Delta =\partial ^2/\partial x_1^2+\dots +\partial ^2/\partial x_n^2$ and $\nabla =\big (\partial /\partial x_1,\dots ,\partial /\partial x_n\big )$ are the usual Euclidean Laplacian and gradient. Here $\varphi _a$ is the canonical Möbius transformation mapping $\mathbb {B}$ to $\mathbb {B}$ and exchanging a and 0 given in (9). It is easy to show that

$$\begin{aligned} \Delta _hf(a)=(1-|a|^2)^2\Delta f(a) +2(n-2)(1-|a|^2)\langle a,\nabla f(a)\rangle , \end{aligned}$$

and

$$\begin{aligned} \nabla ^hf(a)=(1-|a|^2)\nabla f(a). \end{aligned}$$

(1)

A twice continuously differentiable function $f:\mathbb {B}\rightarrow \mathbb {C}$ is called hyperbolic harmonic or $\mathcal {H}$-harmonic on $\mathbb {B}$ if $\Delta _hf(x)=0$ for every $x\in \mathbb {B}$. We denote the set of all $\mathcal {H}$-harmonic functions by $\mathcal {H}(\mathbb {B})$.

Let $\nu$ be the Lebesgue measure on $\mathbb {R}^n$ normalized so that $\nu (\mathbb {B})=1$. For $\alpha >-1$, define the weighted measure $d\nu _{\alpha }(x)$ by

$$\begin{aligned} d\nu _{\alpha }(x)=(1-|x|^2)^\alpha \,d\nu (x), \end{aligned}$$

and for $0<p<\infty$, denote the Lebesgue space with respect to $d\nu _{\alpha }$ by $L^p_{\alpha }=L^p(d\nu _{\alpha })$. The subspace of $L^p_{\alpha }$ consisting of $\mathcal {H}$-harmonic functions is called the weighted $\mathcal {H}$-harmonic Bergman space and is denoted by $\mathcal {B}^p_{\alpha }$,

$$\begin{aligned} \mathcal {B}^p_{\alpha }=\Bigl \{f\in \mathcal {H}(\mathbb {B}):\Vert f\Vert _{L^p_{\alpha }}^p =\int _{\mathbb {B}}|f(x)|^p\,d\nu _{\alpha }(x)<\infty \Bigr \}. \end{aligned}$$

These are Banach spaces when $1\le p<\infty$, and complete metric spaces with respect to the metric $d(f,g)=\Vert f-g\Vert ^p_{L^p_{\alpha }}$ when $0<p<1$.

Point evaluation functionals are bounded on all $\mathcal {B}^p_{\alpha }$ and, in particular, $\mathcal {B}^2_{\alpha }$ is a reproducing kernel Hilbert space. Therefore, for every $x\in \mathbb {B}$, there exists $\mathcal {R}_{\alpha }(x,\cdot )\in \mathcal {B}^2_{\alpha }$ such that

$$\begin{aligned} f(x)=\int _{\mathbb {B}} f(y)\overline{\mathcal {R}_{\alpha }(x,y)}\,d\nu _{\alpha }(y) \qquad (f\in \mathcal {B}^{2}_{\alpha }). \end{aligned}$$

(2)

The reproducing kernel $\mathcal {R}_{\alpha }(\cdot ,\cdot )$ is symmetric in its variables, is real valued (so conjugation in (2) can be deleted) and is $\mathcal {H}$-harmonic with respect to each variable.

For $a,b\in \mathbb {B}$, let $\rho (a,b)=|\varphi _a(b)|$ be the pseudo-hyperbolic metric, and for $0<r<1$, let $E_r(a)=\{x\in \mathbb {B}:\rho (x,a)<r\}$ be the pseudo-hyperbolic ball of radius r centered at a. For $0<r<1$, a sequence $\{a_m\}$ of points of $\mathbb {B}$ is called r-separated if $\rho (a_k,a_m)\ge r$ when $k\ne m$. An r-separated sequence $\{a_m\}$ is called an r-lattice if $\bigcup _{m=1}^\infty E_r(a_m)=\mathbb {B}$, that is, if $\{a_m\}$ is maximal.

In [6, Theorem 2], it is shown by Coifman and Rochberg that if $\{a_m\}$ is an r-lattice with r sufficiently small, then every holomorphic Bergman function $f\in A^p$ on the unit ball of $\mathbb {C}^n$ (more generally on a symmetric Siegel domain of type two) can be represented in the form $f(z)=\sum _{m=1}^\infty \lambda _m{\tilde{B}}(z,a_m)$, where $\{\lambda _m\}\in \ell ^p$ and ${\tilde{B}}(z,a_m)$ is determined by $B(\cdot ,a_m)$, the reproducing kernel at the point $a_m$. This representation is called atomic decomposition, $B(\cdot ,a_m)$ being the atoms. They further showed that a similar decomposition holds for (Euclidean) harmonic functions on the unit ball of $\mathbb {R}^n$. This last result is extended in [14, 15] to harmonic Bergman spaces on bounded symmetric domains of $\mathbb {R}^n$.

Our first aim in this work is to show that if $\{a_m\}$ is an r-lattice with small enough r, then an analogous series representation in terms of the reproducing kernels holds also for $\mathcal {H}$-harmonic Bergman spaces $\mathcal {B}^p_{\alpha }$. Atomic decomposition of $\mathcal {H}$-harmonic Hardy spaces on the real hyperbolic ball has been obtained in [7].

Theorem 1.1

Let $\alpha >-1$ and $0<p<\infty$. Pick s large enough to satisfy

$$\begin{aligned} \begin{aligned} \alpha +1&<p(s+1),\quad \hbox { if}\ p\ge 1\\ \alpha +n&<p(s+n),\quad \hbox { if}\ 0<p<1. \end{aligned} \end{aligned}$$

(3)

There is an $r_0<1/8$ depending only on $n,\alpha ,p,s$ such that if $\{a_m\}$ is an r-lattice with $r<r_0$, then for every $f\in \mathcal {B}^p_{\alpha }$, there exists $\{\lambda _m\}\in \ell ^p$ such that

$$\begin{aligned} f(x)=\sum _{m=1}^\infty \lambda _m \frac{\mathcal {R}_s(x,a_m)}{\Vert \mathcal {R}_s(\cdot ,a_m)\Vert _{\mathcal {B}^p_{\alpha }}} \qquad (x\in \mathbb {B}), \end{aligned}$$

(4)

where the series converges absolutely and uniformly on compact subsets of $\mathbb {B}$ and in $\Vert \cdot \Vert _{\mathcal {B}^p_{\alpha }}$, and the norm $\Vert \{\lambda _m\}\Vert _{\ell ^p}$ is equivalent to the norm $\Vert f\Vert _{\mathcal {B}^p_{\alpha }}$.

The decomposition above can be written in other forms. By Lemma 2.10 below, the estimate $\Vert \mathcal {R}_s(\cdot ,a_m)\Vert _{\mathcal {B}^p_{\alpha }} \sim (1-|a_m|^2)^{(\alpha +n)/p-(s+n)}$ holds and Theorem 1.1 remains true if (4) is replaced with (see Remark 3.8)

$$\begin{aligned} f(x)=\sum _{m=1}^\infty \lambda _m(1-|a_m|^2)^{s+n-(\alpha +n)/p}\, \mathcal {R}_s(x,a_m)\qquad (x\in \mathbb {B}). \end{aligned}$$

(5)

Also, $\mathcal {R}_s(a_m,a_m)\sim (1-|a_m|^2)^{-(s+n)}$ by (17), and (4) can be replaced with

$$\begin{aligned} f(x)=\sum _{m=1}^\infty \lambda _m (1-|a_m|^2)^{-(\alpha +n)/p}\, \frac{\mathcal {R}_s(x,a_m)}{\mathcal {R}_s(a_m,a_m)}\qquad (x\in \mathbb {B}). \end{aligned}$$

We next consider the interpolation problem. If $\{a_m\}$ is r-separated and $f\in \mathcal {B}^p_{\alpha }$, then the sequence (see Proposition 3.3)

$$\begin{aligned} \big \{f(a_m)(1-|a_m|^2)^{(\alpha +n)/p}\big \} \end{aligned}$$

is in $\ell ^p$. If the converse holds, that is, if for every $\{\lambda _m\}\in \ell ^p$, one can find an $f\in \mathcal {B}^p_{\alpha }$ such that $f(a_m)(1-|a_m|^2)^{(\alpha +n)/p}=\lambda _m$, then $\{a_m\}$ is called an interpolating sequence for $\mathcal {B}^p_{\alpha }$. We show that if the separation constant r is large enough, then $\{a_m\}$ is an interpolating sequence.

Theorem 1.2

Let $\alpha >-1$ and $0<p<\infty$. There is an $r_0$ with $1/2<r_0<1$ depending only on $n,\alpha ,p$ such that if $\{a_m\}$ is an r-separated sequence with $r>r_0$, then for every $\{\lambda _m\}\in \ell ^p$, there exists $f\in \mathcal {B}^p_{\alpha }$ such that

$$\begin{aligned} f(a_m)(1-|a_m|^2)^{(\alpha +n)/p}=\lambda _m, \end{aligned}$$

and the norm $\Vert f\Vert _{\mathcal {B}^p_{\alpha }}$ is equivalent to the norm $\Vert \{\lambda _m\}\Vert _{\ell ^p}$.

The holomorphic analogue of the above theorem is proved in [2] for the unit ball and polydisc, and in [11] for more general domains of $\mathbb {C}^n$. For harmonic Bergman spaces on the upper half-space of $\mathbb {R}^n$, an analogous result is proved in [5].

Finally, we determine precisely when a Bergman space $\mathcal {B}^p_{\alpha }$ is contained in an another Bergman space $\mathcal {B}^q_\beta$.

Theorem 1.3

Let $\alpha ,\beta >-1$ and $0<p,q<\infty$.

• (a) If $q\ge p$, then

  $$\begin{aligned} \mathcal {B}^p_{\alpha }\subset \mathcal {B}^q_\beta \quad \text {if and only if}\quad \frac{\alpha +n}{p}\le \frac{\beta +n}{q} \end{aligned}$$

• (b) If $q<p$, then

  $$\begin{aligned} \mathcal {B}^p_{\alpha }\subset \mathcal {B}^q_\beta \quad \text {if and only if}\quad \frac{\alpha +1}{p}<\frac{\beta +1}{q} \end{aligned}$$

In both cases the inclusion $i:\mathcal {B}^p_{\alpha }\rightarrow \mathcal {B}^q_\beta$ is continuous.

For holomorphic Bergman spaces on the unit ball of $\mathbb {C}^n$, the counterpart of this theorem has been proved in [8, Lemma 2.1]. However, this source uses gap series formed by using the so-called Ryll–Wojtaszczyk polynomials (see [12]). We do not know whether such type of $\mathcal {H}$-harmonic functions exist on the real hyperbolic ball. Our proof is based on the above atomic decomposition and interpolation theorems.

2 Preliminaries

In this section we collect some known facts about Möbius transformations and $\mathcal {H}$-harmonic Bergman spaces that will be used in the sequel.

2.1 Notation

We denote positive constants whose exact values are inessential with C. The value of C may differ from one occurrence to another. We write $X\lesssim Y$ if $X\le CY$, and $X\sim Y$ if both $X\le CY$ and $Y\le CX$.

For $x,y\in \mathbb {R}^n$, we write

$$\begin{aligned}{}[x,y]:=\sqrt{1-2\langle x,y\rangle +|x|^2|y|^2}, \end{aligned}$$

which is symmetric in the variables x, y, and the following equality holds

$$\begin{aligned}{}[x,y]^2=|x-y|^2+(1-|x|^2)(1-|y|^2). \end{aligned}$$

(6)

If either of the variables is 0, then $[x,0]=[0,y]=1$; otherwise

$$\begin{aligned}{}[x,y]=\Bigl ||y|x-\frac{y}{|y|}\Bigr |=\Bigl |\frac{x}{|x|}-|x|y\Bigr |, \end{aligned}$$

(7)

and so

$$\begin{aligned} 1-|x||y|\le [x,y] \le 1+|x||y|\qquad (x,y\in \mathbb {B}). \end{aligned}$$

(8)

2.2 Möbius Transformations

For more details about the facts listed in this subsection we refer the reader to [1] or [13].

A Möbius transformation of $\hat{\mathbb {R}}^n=\mathbb {R}^n\cup \{\infty \}$ is a finite composition of reflections (inversions) in spheres or planes. We denote the group of all Möbius transformations mapping $\mathbb {B}$ to $\mathbb {B}$ by $\mathcal {M}(\mathbb {B})$. For $a\in \mathbb {B}$, the mapping

$$\begin{aligned} \varphi _a(x)=\frac{a|x-a|^2+(1-|a|^2)(a-x)}{[x,a]^2} \qquad (x\in \mathbb {B}) \end{aligned}$$

(9)

is in $\mathcal {M}(\mathbb {B})$, exchanges a and 0, and satisfies $\varphi _a\circ \varphi _a=\text {Id}$. The group $\mathcal {M}(\mathbb {B})$ is generated by $\{\varphi _a:a\in \mathbb {B}\}$ and orthogonal transformations. A very useful identity involving $\varphi _a$ is ([13, Eqn. 2.1.7])

$$\begin{aligned} 1-|\varphi _a(x)|^2 =\frac{(1-|a|^2)(1-|x|^2)}{[x,a]^2}. \end{aligned}$$

(10)

The Jacobian $J_{\varphi _a}$ of $\varphi _a$ satisfies ([13, Theorem 3.3.1])

$$\begin{aligned} |J_{\varphi _a}(x)|=\frac{(1-|\varphi _a(x)|^2)^n}{(1-|x|^2)^n}. \end{aligned}$$

(11)

The following lemma is a special case of [10, Theorem 1.1].

Lemma 2.1

For $a,x\in \mathbb {B}$, the following equality holds

$$\begin{aligned}{}[\varphi _a(x),a]=\frac{1-|a|^2}{[x,a]}. \end{aligned}$$

Proof

Replacing x in (10) with $\varphi _a(x)$ and noting that $\varphi _a\circ \varphi _a=\text {Id}$ shows

$$\begin{aligned}{}[\varphi _a(x),a]^2 =\frac{(1-|a|^2)(1-|\varphi _a(x)|^2)}{1-|x|^2}. \end{aligned}$$

Applying (10) again, we obtain the desired result. $\square$

For $a,b\in \mathbb {B}$, the pseudo-hyperbolic metric $\rho (a,b)=|\varphi _a(b)|$ satisfies

$$\begin{aligned} \rho (a,b)=\frac{|a-b|}{[a,b]}, \end{aligned}$$

(12)

by (10) and (6). It is Möbius invariant in the sense that $\rho (\psi (a),\psi (b))=\rho (a,b)$ for every $\psi \in \mathcal {M}(\mathbb {B})$. It satisfies not only the triangle inequality, but the following strong triangle inequality (see [10, Theorem 1.2]).

Lemma 2.2

For $a,b,x\in \mathbb {B}$, the following inequalities hold

$$\begin{aligned} \frac{\big |\rho (a,x)-\rho (b,x)\big |}{1-\rho (a,x)\rho (b,x)} \le \rho (a,b) \le \frac{\rho (a,x)+\rho (b,x)}{1+\rho (a,x)\rho (b,x)}. \end{aligned}$$

Lemma 2.3

For $x,y\in \mathbb {B}$,

$$\begin{aligned} 1-\rho (x,y) \le \frac{1-|x|^2}{[x,y]} \le 1+\rho (x,y). \end{aligned}$$

Proof

The lemma clearly holds when $x=0$. Otherwise, let $x^*:=x/|x|^2$ be the inversion of x with respect to the unit sphere $\mathbb {S}$. Multiply the triangle inequality

$$\begin{aligned} |x^*-y|-|y-x|\le |x^*-x|\le |x^*-y|+|y-x|\end{aligned}$$

by $|x|$. Noting that $|x||x^*-y|=[x,y]$ by (7), and $|x||x^*-x|=1-|x|^2$, we deduce

$$\begin{aligned}{}[x,y]-|x||y-x|\le 1-|x|^2 \le [x,y]+|x||y-x|. \end{aligned}$$

The lemma follows from the facts that $|y-x|=\rho (x,y)[x,y]$ by (12), and $|x|<1$. $\square$

The following lemma is a slight modification of [3, Lemma 2.1] and immediately follows from Lemma 2.3.

Lemma 2.4

For $x,y\in \mathbb {B}$,

$$\begin{aligned} \frac{1-\rho (x,y)}{1+\rho (x,y)} \le \frac{1-|x|^2}{1-|y|^2} \le \frac{1+\rho (x,y)}{1-\rho (x,y)}. \end{aligned}$$

The next lemma is proved in [3, Lemma 2.2].

Lemma 2.5

For $a,x,y\in \mathbb {B}$,

$$\begin{aligned} \frac{1-\rho (x,y)}{1+\rho (x,y)} \le \frac{[x,a]}{[y,a]} \le \frac{1+\rho (x,y)}{1-\rho (x,y)}. \end{aligned}$$

Let $\mathbb {B}_r=\{x\in \mathbb {R}^n:|x|<r\}$. The pseudo-hyperbolic ball $E_r(a)=\{x\in \mathbb {B}:\rho (x,a)<r\}=\varphi _a(\mathbb {B}_r)$ is also a Euclidean ball with (see [13, Theorem 2.2.2])

$$\begin{aligned} \text {center}=\frac{(1-r^2)a}{1-|a|^2r^2} \quad \text {and}\quad \text {radius}=\frac{(1-|a|^2)r}{1-|a|^2r^2}. \end{aligned}$$

(13)

2.3 Separated Sequences and Lattices

There exists an r-lattice for every $0<r<1$ as explained in [6, p. 18], and every r-separated sequence can be completed to an r-lattice. The following lemma follows from an invariant volume argument.

Lemma 2.6

Let $0<r,\delta <1$ and $\{a_m\}$ be r-separated. There exists N depending only on $n,r,\delta$ such that every $x\in \mathbb {B}$ belongs to at most N of the balls $E_\delta (a_m)$.

Lemma 2.7

Let $\{a_m\}$ be an r-lattice. There exists a sequence $\{E_m\}$ of disjoint sets such that $\bigcup _{m=1}^\infty E_m=\mathbb {B}$ and

$$\begin{aligned} E_{r/2}(a_m)\subset E_m\subset E_r(a_m). \end{aligned}$$

(14)

Proof

Let $E_1=E_r(a_1)\backslash \bigcup _{m=2}^\infty E_{r/2}(a_m)$ and given $E_1,\dots ,E_{m-1}$, let

$$\begin{aligned} E_m=E_r(a_m)\backslash \biggl (\bigcup _{i=1}^{m-1} E_i\ \bigcup \bigcup _{i=m+1}^\infty E_{r/2}(a_i)\biggr ). \end{aligned}$$

$\square$

Lemma 2.8

Let $\gamma \in \mathbb {R}$ and $0<r<1$.

• (a) If $\{a_m\}$ is r-separated and $\gamma >n-1$, then $\sum _{m=1}^\infty (1-|a_m|^2)^\gamma <\infty$.

• (b) If $\{a_m\}$ is an r-lattice, then $\sum _{m=1}^\infty (1-|a_m|^2)^\gamma <\infty$ if and only if $\gamma >n-1$.

Proof

To see part (a), note that

$$\begin{aligned} \int _{E_{r/2}(a_m)}(1-|y|^2)^{\gamma -n}\,d\nu (y) \sim (1-|a_m|^2)^\gamma , \end{aligned}$$

(15)

where the implied constants depend only on the fixed parameters $n,\gamma ,r$ and are independent of $a_m$. This is true because for $y\in E_{r/2}(a_m)$, we have $(1-|y|^2)\sim (1-|a_m|^2)$ by Lemma 2.4 and $\nu (E_{r/2}(a_m))\sim (1-|a_m|^2)^n$ by (13). Thus

$$\begin{aligned} \sum _{m=1}^\infty (1-|a_m|^2)^\gamma \lesssim \int _{\mathbb {B}} (1-|y|^2)^{\gamma -n}\,d\nu (y), \end{aligned}$$

since the balls $E_{r/2}(a_m)$ are disjoint. If $\gamma >n-1$, then the above integral is finite.

For part (b), let $E_m$ be as given in Lemma 2.7. By (14), we similarly have

$$\begin{aligned} \int _{E_m}(1-|y|^2)^{\gamma -n}\,d\nu (y) \sim (1-|a_m|^2)^\gamma \end{aligned}$$

(16)

and therefore

$$\begin{aligned} \sum _{m=1}^\infty (1-|a_m|^2)^\gamma \sim \sum _{m=1}^\infty \int _{E_m}(1-|y|^2)^{\gamma -n}\,d\nu (y) =\int _{\mathbb {B}} (1-|y|^2)^{\gamma -n}\,d\nu (y). \end{aligned}$$

The last integral is finite if and only if $\gamma >n-1$. $\square$

2.4 Reproducing Kernels and Bergman Projection

The following upper estimates of the reproducing kernels $\mathcal {R}_{\alpha }$ of $\mathcal {H}$-harmonic Bergman spaces have been obtained in [16, Theorem 1.2].

Lemma 2.9

For $\alpha >-1$, there exists a constant $C>0$ such that for all $x,y\in \mathbb {B}$,

• (a) $|\mathcal {R}_{\alpha }(x,y)|\le \dfrac{C}{[x,y]^{\alpha +n}}$,

• (b) $|\nabla _1 \mathcal {R}_{\alpha }(x,y)|\le \dfrac{C}{[x,y]^{\alpha +n+1}}$.

Here $\nabla _1$ means the gradient is taken with respect to the first variable.

More is true on the diagonal $y=x$ and the two-sided estimate ([16, Lemma 6.1])

$$\begin{aligned} \mathcal {R}_{\alpha }(x,x)\sim \frac{1}{(1-|x|^2)^{\alpha +n}} \end{aligned}$$

(17)

holds. The following lemma is part of [16, Theorem 1.3].

Lemma 2.10

If $\alpha ,s>-1$, $0<p<\infty$ and $p(s+n)-(\alpha +n)>0$, then

$$\begin{aligned} \int _\mathbb {B}\bigl |\mathcal {R}_s(x,y)\bigr |^{p}\,d\nu _{\alpha }(y) \sim \dfrac{1}{(1-|x|^2)^{p(s+n)-(\alpha +n)}}. \end{aligned}$$

The implied constants depend only on $n,\alpha ,s,p$ and are independent of x.

For $s>-1$ and suitable f, we define the projection operator $P_s$ and the related operator $Q_s$ by

$$\begin{aligned} \begin{aligned} P_s f(x)&=\int _{\mathbb {B}}f(y)\mathcal {R}_s(x,y)\,d\nu _s(y),\\ Q_s f(x)&=\int _{\mathbb {B}}\frac{f(y)}{[x,y]^{s+n}}\,d\nu _s(y). \end{aligned} \end{aligned}$$

(18)

Lemma 2.11

Let $1\le p<\infty$ and $\alpha ,s>-1$. The following are equivalent:

• (a) $P_s:L^p_{\alpha }\rightarrow \mathcal {B}^p_{\alpha }$ is bounded,

• (b) $Q_s:L^p_{\alpha }\rightarrow L^p_{\alpha }$ is bounded,

• (c) $\alpha +1<p(s+1)$.

In case (c) holds, then $P_sf=f$ for every $f\in \mathcal {B}^p_{\alpha }$.

Proof

(b) $\Rightarrow$ (a) follows from Lemma 2.9 (a), (a) $\Rightarrow$ (c) is proved in [16, Theorem 1.1], and (c) $\Rightarrow$ (b) is well-known and included in the proof of [16, Theorem 1.1]. $\square$

For a proof of the following estimates, see [9, Proposition 2.2].

Lemma 2.12

Let $b>-1$ and $c\in \mathbb {R}$. For $x\in \mathbb {B}$, define

$$\begin{aligned} I_c(x):=\int _{\mathbb {S}}\frac{d\sigma (\zeta )}{|x-\zeta |^{n-1+c}} \qquad \text {and}\qquad J_{b,c}(x):=\int _{\mathbb {B}}\frac{(1-|y|^2)^b}{[x,y]^{n+b+c}}\,d\nu (y), \end{aligned}$$

where $\sigma$ is the normalized surface measure on $\mathbb {S}$. For all $x\in \mathbb {B}$,

$$\begin{aligned} I_c(x)\sim J_{b,c}(x)\sim {\left\{ \begin{array}{ll} \dfrac{1}{(1-|x|^2)^{c}},&{}\hbox { if}\ c>0;\\ 1+\log \dfrac{1}{1-|x|^2},&{}\hbox { if}\ c=0;\\ 1,&{}\hbox { if}\ c<0, \end{array}\right. } \end{aligned}$$

where the implied constants depend only on n, b, c and are independent of x.

We record the following elementary facts about the sequence spaces $\ell ^p$ for future reference.

Lemma 2.13

(i) For $0<p<q<\infty$, $\Vert \{\lambda _m\}\Vert _{\ell ^q}\le \Vert \{\lambda _m\}\Vert _{\ell ^p}$.

(ii) Let $1<p<\infty$ and $p'$ be the conjugate exponent of p, $1/p+1/p'=1$. If $\sum _{m=1}^\infty |\lambda _m\kappa _m|<\infty$ for every $\{\kappa _m\}\in \ell ^{p'}$, then $\{\lambda _m\}\in \ell ^p$.

3 Atomic Decomposition

The purpose of this section is to prove Theorem 1.1. The main problem is to show that under the assumptions of the theorem, the operator $U:\ell ^p\rightarrow \mathcal {B}^p_{\alpha }$ defined in (19) below is onto. We do this through a couple of propositions.

Proposition 3.1

For $\alpha >-1$ and $0<p<\infty$, choose s so that (3) holds. If $\{a_m\}$ is r-separated for some $0<r<1$, then the operator $U:\ell ^p\rightarrow \mathcal {B}^p_{\alpha }$ mapping $\lambda =\{\lambda _m\}$ to

$$\begin{aligned} U\lambda (x)=\sum _{m=1}^\infty \lambda _m \frac{\mathcal {R}_s(x,a_m)}{\Vert \mathcal {R}_s(\cdot ,a_m)\Vert _{\mathcal {B}^p_{\alpha }}} \qquad (x\in \mathbb {B}) \end{aligned}$$

(19)

is bounded. The above series converges absolutely and uniformly on compact subsets of $\mathbb {B}$, and also in $\Vert \cdot \Vert _{\mathcal {B}^p_{\alpha }}$.

Proof

Throughout the proof we suppress the constants that depend on the fixed parameters $n,\alpha ,p,s$ and r. Note that, by Lemma 2.10 and (3), for every $0<p<\infty$,

$$\begin{aligned} \Vert \mathcal {R}_s(\cdot ,a_m)\Vert _{\mathcal {B}^p_{\alpha }} \sim (1-|a_m|^2)^{(\alpha +n)/p-(s+n)}, \end{aligned}$$

(20)

since in (3), the inequality $p(s+n)>(\alpha +n)$ holds also in the case $p\ge 1$.

We begin with the case $0<p\le 1$. We first show that for $\lambda \in \ell ^p$, the series in (19) converges absolutely and uniformly on compact subsets of $\mathbb {B}$ which implies that $U\lambda$ is $\mathcal {H}$-harmonic on $\mathbb {B}$ since so is each $\mathcal {R}_s(\cdot ,a_m)$. If $|x|\le R<1$, then $|\mathcal {R}_s(x,a_m)|\lesssim 1$ by Lemma 2.9 (a), since $[x,a_m]\ge 1-|x|$ by (8). Thus, using also (20), the fact that $(s+n)-(\alpha +n)/p>0$ by (3), and Lemma 2.13 (i) we obtain

$$\begin{aligned} \sum _{m=1}^\infty |\lambda _m|\frac{|\mathcal {R}_s(x,a_m)|}{\Vert \mathcal {R}_s(\cdot ,a_m)\Vert _{\mathcal {B}^p_{\alpha }}} \lesssim \sum _{m=1}^\infty |\lambda _m|(1-|a_m|^2)^{(s+n)-(\alpha +n)/p} \le \sum _{m=1}^\infty |\lambda _m|\le \Vert \lambda \Vert _{\ell ^p}, \end{aligned}$$

which proves the assertion. The inequality $\Vert U\lambda \Vert _{\mathcal {B}^p_{\alpha }}\le \Vert \lambda \Vert _{\ell ^p}$ immediately follows from Lemma 2.13 (i) and shows also that the series in (19) converges in $\Vert \cdot \Vert _{\mathcal {B}^p_{\alpha }}$.

We next consider the case $1<p<\infty$. Let $p'$ be the conjugate exponent of p. The series in (19) converges absolutely and uniformly on compact subsets of $\mathbb {B}$ because we again have $|\mathcal {R}_s(x,a_m)|\lesssim 1$, and by (20) and Hölder’s inequality,

$$\begin{aligned} \sum _{m=1}^\infty |\lambda _m|\frac{|\mathcal {R}_s(x,a_m)|}{\Vert \mathcal {R}_s(\cdot ,a_m)\Vert _{\mathcal {B}^p_{\alpha }}}&\lesssim \sum _{m=1}^\infty |\lambda _m|(1-|a_m|^2)^{s+n-(\alpha +n)/p}\\&\le \Vert \lambda \Vert _{\ell ^p} \Big (\sum _{m=1}^\infty (1-|a_m|^2)^{p'(s+n-(\alpha +n)/p)}\Bigr )^{1/p'}. \end{aligned}$$

The last sum is finite by Lemma 2.8 (a) since the inequality $p'(s+n-(\alpha +n)/p)>n-1$ is equivalent to (3). Thus $U\lambda$ is $\mathcal {H}$-harmonic on $\mathbb {B}$.

To show $\Vert U\lambda \Vert _{B^p_{\alpha }}\lesssim \Vert \lambda \Vert _{\ell ^p}$, following [4, 6, 15], we use the projection theorem. Denote by $\chi _A$ the characteristic function of a set A. For $\lambda \in \ell ^p$, let

$$\begin{aligned} g(x):=\sum _{m=1}^\infty |\lambda _m|(1-|a_m|^2)^{-(\alpha +n)/p}\, \chi _{E_{r/2}(a_m)}(x) \qquad (x\in \mathbb {B}). \end{aligned}$$

We have $\Vert g\Vert _{L^p_{\alpha }}\sim \Vert \lambda \Vert _{\ell ^p}$, since the balls $E_{r/2}(a_m)$ are disjoint and

$$\begin{aligned} \Vert g\Vert _{L^p_{\alpha }}^p =\sum _{m=1}^\infty |\lambda _m|^p (1-|a_m|^2)^{-(\alpha +n)}\nu _{\alpha }(E_{r/2}(a_m)) \sim \sum _{m=1}^\infty |\lambda _m|^p, \end{aligned}$$

by (15). Next, with $Q_s$ as in (18),

$$\begin{aligned} Q_s g(x)&=\sum _{m=1}^\infty |\lambda _m|(1-|a_m|^2)^{-(\alpha +n)/p} \int _{E_{r/2}(a_m)}\frac{(1-\vert y|^2)^s}{[x,y]^{s+n}}d\nu (y)\\&\sim \sum _{m=1}^\infty \frac{|\lambda _m|(1-|a_m|^2)^{s+n-(\alpha +n)/p}}{[x,a_m]^{s+n}}, \end{aligned}$$

where in the last line we first use the fact that $[x,y]\sim [x,a_m]$ for $y\in E_{r/2}(a_m)$ by Lemma 2.5, and then use (15). This shows that $|U\lambda (x)|\lesssim Q_sg(x)$ by Lemma 2.9 (a) and (20). Since $Q_s$ is bounded by Lemma 2.11 and (3), we conclude

$$\begin{aligned} \Vert U\lambda \Vert _{\mathcal {B}^p_{\alpha }}\lesssim \Vert Q_s g\Vert _{L^p_{\alpha }} \lesssim \Vert g\Vert _{L^p_{\alpha }}\sim \Vert \lambda \Vert _{\ell ^p}. \end{aligned}$$

$\square$

To verify that the above operator $U:\ell ^p\rightarrow \mathcal {B}^p_{\alpha }$ is onto under the additional assumption that $\{a_m\}$ is an r-lattice with r small enough, we need to consider a second operator. We first recall the following sub-mean value inequality for $\mathcal {H}$-harmonic functions. For a proof see [13, Section 4.7]. Here, $d\tau (x)=(1-|x|^2)^{-n}d\nu (x)$ is the invariant measure on $\mathbb {B}$.

Lemma 3.2

Let $f\in \mathcal {H}(\mathbb {B})$ and $0<p<\infty$. For all $a\in \mathbb {B}$ and all $0<\delta <1$,

$$\begin{aligned} |f(a)|^p\le \frac{C}{\delta ^n}\int _{E_\delta (a)}|f(y)|^p\,d\tau (y), \end{aligned}$$

where $C=1$ if $p\ge 1$ and $C=2^{n/p}$ if $0<p<1$.

Proposition 3.3

Let $\alpha >-1$, $0<p<\infty$ and $\{a_m\}$ be r-separated for some $0<r<1$. Then the operator $T:\mathcal {B}^p_{\alpha }\rightarrow \ell ^p$ defined by

$$\begin{aligned} Tf=\big \{f(a_m)(1-|a_m|^2)^{(\alpha +n)/p}\big \} \end{aligned}$$

(21)

is bounded.

Proof

Applying Lemma 3.2 with $\delta =r/2$ and noting that $(1-|y|^2)\sim (1-|a_m|^2)$ for $y\in E_{r/2}(a_m)$ by Lemma 2.4, we obtain

$$\begin{aligned} |f(a_m)|^p(1-|a_m|^2)^{\alpha +n} \lesssim \int _{E_{r/2}(a_m)}|f(y)|^p\,d\nu _{\alpha }(y). \end{aligned}$$

Since the balls $E_{r/2}(a_m)$ are disjoint, we deduce

$$\begin{aligned} \Vert Tf\Vert ^p_{\ell ^p} =\sum _{m=1}^\infty |f(a_m)|^p(1-|a_m|^2)^{\alpha +n} \lesssim \sum _{m=1}^\infty \int _{E_{r/2}(a_m)}|f(y)|^p\,d\nu _{\alpha }(y) \le \Vert f\Vert _{\mathcal {B}^p_{\alpha }}^p. \end{aligned}$$

$\square$

We need a slightly modified version of the above operator T.

Proposition 3.4

For $\alpha >-1$ and $0<p<\infty$, choose s so that (3) holds. If $\{a_m\}$ is an r-lattice for some $0<r<1$ and $\{E_m\}$ is the associated sequence as given in Lemma 2.7, then the operator ${\hat{T}}:\mathcal {B}^p_{\alpha }\rightarrow \ell ^p$ defined by

$$\begin{aligned} {\hat{T}}f =\bigl \{f(a_m)\,\Vert \mathcal {R}_s(\cdot ,a_m)\Vert _{\mathcal {B}^p_{\alpha }}\, \nu _s(E_m)\bigr \} \end{aligned}$$

(22)

is bounded.

Proof

Since $\Vert \mathcal {R}_s(\cdot ,a_m)\Vert _{\mathcal {B}^p_{\alpha }}\,\nu _s(E_m) \sim (1-|a_m|^2)^{(\alpha +n)/p}$ by (16) and (20), the result follows from Proposition 3.3. $\square$

Proposition 3.5

For $\alpha >-1$ and $0<p<\infty$, choose s so that (3) holds. There exists a constant $C>0$ depending only on $n,\alpha ,p,s$ such that if $\{a_m\}$ is an r-lattice with $r<1/8$, then $\Vert I-U{\hat{T}}\Vert _{\mathcal {B}^p_{\alpha }\rightarrow \mathcal {B}^p_{\alpha }}\le Cr$.

In the proofs of the previous propositions we allowed the constants to depend on the separation constant r. This time we need to be careful that the suppressed constants are independent of r. We prove the cases $p\ge 1$ and $0<p<1$ separately.

Proof of Proposition 3.5 when $p\ge 1$

By (19) and (22),

$$\begin{aligned} U{\hat{T}}f(x)=\sum _{m=1}^\infty \int _{E_m} f(a_m)\mathcal {R}_s(x,a_m)\,d\nu _s(y), \end{aligned}$$

and by (3) and Lemma 2.11 we have $P_s f=f$ and so

$$\begin{aligned} f(x)=\sum _{m=1}^\infty \int _{E_m} f(y)\mathcal {R}_s(x,y)\,d\nu _s(y). \end{aligned}$$

Therefore

$$\begin{aligned} (I-U{\hat{T}})f(x)&=\sum _{m=1}^\infty \int _{E_m}\bigl (\mathcal {R}_s(x,y)-\mathcal {R}_s(x,a_m)\bigr )f(y)\,d\nu _s(y)\nonumber \\&\quad +\sum _{m=1}^\infty \int _{E_m}\mathcal {R}_s(x,a_m)\big (f(y)-f(a_m)\big )\,d\nu _s(y)\nonumber \\&=:h_1(x)+h_2(x). \end{aligned}$$

(23)

We first estimate $h_1$. Let $y\in E_m$. By the mean value theorem of calculus, there exists ${\tilde{y}}$ lying on the line segment joining $a_m$ and y such that

$$\begin{aligned} \mathcal {R}_s(x,y)-\mathcal {R}_s(x,a_m) =\big \langle \, y-a_m,\nabla _2\mathcal {R}_s(x,{\tilde{y}})\,\big \rangle , \end{aligned}$$

where $\nabla _2$ means the gradient is taken with respect to the second variable. Observe that because r is bounded above by 1/8, there are constants independent of r such that for $y\in E_m\subset E_r(a_m)$, we have $[y,a_m]\sim [y,y]=1-|y|^2$ by Lemma 2.5. Thus, by (12),

$$\begin{aligned} |y-a_m|=\rho (y,a_m)[y,a_m]<r[y,a_m] \lesssim r(1-|y|^2). \end{aligned}$$

Next, since $a_m$ and y are both in the ball $E_r(a_m)$, so is ${\tilde{y}}$. Hence $\rho (y,{\tilde{y}})<1/4$ and by Lemma 2.5, $[x,y]\sim [x,{\tilde{y}}]$ for every $x\in \mathbb {B}$ with the constants again not depending on r. Therefore, by Lemma 2.9 (b) and the symmetry of $\mathcal {R}_s(\cdot ,\cdot )$,

$$\begin{aligned} \big |\nabla _2\mathcal {R}_s(x,{\tilde{y}})\big |\lesssim \frac{1}{[x,{\tilde{y}}]^{s+n+1}}\sim \frac{1}{[x,y]^{s+n+1}}. \end{aligned}$$

Combining these we see that for $y\in E_m$ and $x\in \mathbb {B}$,

$$\begin{aligned} |\mathcal {R}_s(x,y)-\mathcal {R}_s(x,a_m)|\lesssim \frac{r(1-|y|^2)}{[x,y]^{s+n+1}} \lesssim \frac{r}{[x,y]^{s+n}}, \end{aligned}$$

(24)

where in the last inequality we use $[x,y]\ge 1-|y|$ by (8). Thus

$$\begin{aligned} |h_1(x)|\lesssim r\sum _{m=1}^\infty \int _{E_m}\frac{|f(y)|}{[x,y]^{s+n}}\,d\nu _s(y) =r \int _{\mathbb {B}}\frac{|f(y)|}{[x,y]^{s+n}}\,d\nu _s(y)= r Q_s(|f|)(x), \end{aligned}$$

and since $Q_s$ is bounded on $L^p_{\alpha }$ by Lemma 2.11, we obtain $\Vert h_1\Vert _{L^p_{\alpha }}\lesssim r\Vert f\Vert _{B^p_{\alpha }}$.

We now estimate $h_2$. Let $y\in E_m$. As above, we have $P_sf=f$, and so

$$\begin{aligned} f(y)-f(a_m)=\int _{\mathbb {B}}\bigl (\mathcal {R}_s(y,z)-\mathcal {R}_s(a_m,z)\bigr ) f(z)\,d\nu _s(z). \end{aligned}$$

Since $\mathcal {R}_s(\cdot ,\cdot )$ is symmetric, by (24),

$$\begin{aligned} |\mathcal {R}_s(y,z)-\mathcal {R}_s(a_m,z)|\lesssim \frac{r}{[y,z]^{s+n}}, \end{aligned}$$

for all $z\in \mathbb {B}$ with the constants not depending on r. Thus

$$\begin{aligned} |f(y)-f(a_m)|\lesssim r\int _{\mathbb {B}}\frac{|f(z)|}{[y,z]^{s+n}}\,d\nu _s(z) =rQ_s(|f|)(y), \end{aligned}$$

and so

$$\begin{aligned} |h_2(x)|\lesssim r \sum _{m=1}^\infty \int _{E_m}|\mathcal {R}_s(x,a_m)|\,Q_s(|f|)(y)\,d\nu _s(y) \lesssim r \sum _{m=1}^\infty \int _{E_m}\frac{Q_s(|f|)(y)}{[x,a_m]^{s+n}}\,d\nu _s(y), \end{aligned}$$

where in the last inequality we use Lemma 2.9 (a). By Lemma 2.4 again, we have $[x,a_m]\sim [x,y]$ for $y\in E_m\subset E_r(a_m)$ since $r<1/8$. Hence

$$\begin{aligned} |h_2(x)|\lesssim & {} r \sum _{m=1}^\infty \int _{E_m}\frac{Q_s(|f|)(y)}{[x,y]^{s+n}}d\nu _s(y) =r \int _{\mathbb {B}}\frac{Q_s(|f|)(y)}{[x,y]^{s+n}}d\nu _s(y)\\= & {} r Q_s\big (Q_s(|f|)\big )(x), \end{aligned}$$

and since $Q_s$ is bounded on $L^p_{\alpha }$ we obtain that $\Vert h_2\Vert _{L^p_{\alpha }}\lesssim r\Vert f\Vert _{B^p_{\alpha }}$.

We conclude that $\Vert (I-U{\hat{T}})f\Vert _{\mathcal {B}^p_{\alpha }}\le Cr\Vert f\Vert _{B^p_{\alpha }}$, with C depending only on $n,\alpha ,p,s$. This finishes the proof when $p\ge 1$. $\square$

In order to prove the case $0<p<1$, we need to do some preparation. The following inequality is proved in [13, Theorem 4.7.4 part (b)].

Lemma 3.6

Let $0<p<\infty$ and $0<\delta <1/2$. There exists a constant $C>0$ depending only on $n,p,\delta$ such that for all $a\in \mathbb {B}$ and $f\in \mathcal {H}(\mathbb {B})$,

$$\begin{aligned} |\nabla ^h f(a)|^p \le \frac{C}{\delta ^n}\int _{E_\delta (a)}|f(y)|^p\,d\tau (y). \end{aligned}$$

The next lemma is a special case of Theorem 1.3 part (a).

Lemma 3.7

Let $0<p<1$ and $\alpha >-1$. Then $\mathcal {B}^p_{\alpha }\subset \mathcal {B}^1_{(\alpha +n)/p-n}$ and the inclusion is continuous.

Proof

By [13, Eqn. (10.1.5)], there exists a constant $C>0$ depending only on $n,\alpha ,p$ such that

$$\begin{aligned} |f(x)|\le \frac{C}{(1-|x|^2)^{(\alpha +n)/p}} \Vert f\Vert _{\mathcal {B}^p_{\alpha }}, \end{aligned}$$

(25)

for all $x\in \mathbb {B}$ and $f\in \mathcal {B}^p_{\alpha }$. In the integral below writing $|f(x)|=|f(x)|^p|f(x)|^{1-p}$ and applying (25) to the factor $|f(x)|^{1-p}$, we deduce

$$\begin{aligned} \int _{\mathbb {B}}|f(x)|(1-|x|^2)^{(\alpha +n)/p-n}\,d\nu (x)&\le C^{1-p}\Vert f\Vert _{\mathcal {B}^p_{\alpha }}^{1-p} \int _{\mathbb {B}}|f(x)|^p(1-|x|^2)^\alpha \,d\nu (x)\\&=C^{1-p}\Vert f\Vert _{\mathcal {B}^p_{\alpha }}. \end{aligned}$$

$\square$

Proof of Proposition 3.5 when $0<p<1$

In this part of the proof we can not use the projection theorem which requires $p\ge 1$. Instead, we follow [6, p. 19] and use a suitable rearrangement of the sequence $\{a_m\}$ as described below.

Pick a 1/2-lattice $\{b_m\}$ and fix it throughout the proof. Denote the sequence of sets associated to the lattice $\{b_m\}$ as described in Lemma 2.7 by $\{D_m\}$. That is, the sets $D_m$ are disjoint with $\bigcup _{m=1}^\infty D_m=\mathbb {B}$ and

$$\begin{aligned} E_{1/4}(b_m)\subset D_m\subset E_{1/2}(b_m) \qquad (m=1,2,\dots ). \end{aligned}$$

Given an r-lattice $\{a_m\}$ with $r<1/8$, renumber $\{a_m\}$ in the following way. Call the elements of $\{a_m\}$ that are in $D_1$ as $a_{11},a_{12},\dots ,a_{1\kappa _1}$ and in general call the points of $\{a_m\}$ that are in $D_m$ as $a_{m1},a_{m2},\dots ,a_{m\kappa _m}$. Denote the sets given in Lemma 2.7 corresponding to this renumbering by $E_{mk}$. Thus, the sets $E_{mk}$ are disjoint, $\bigcup _{m=1}^\infty \bigcup _{k=1}^{\kappa _m}E_{mk}=\mathbb {B}$ and

$$\begin{aligned} E_{r/2}(a_{mk})\subset E_{mk}\subset E_r(a_{mk}) \qquad (m=1,2,\dots ,\ k=1,2,\dots ,\kappa _m). \end{aligned}$$

By the above construction, since $a_{mk}\in D_m\subset E_{1/2}(b_m)$, we have

$$\begin{aligned} \rho (a_{mk},b_m)<1/2 \qquad (m=1,2,\dots ,\ k=1,2,\dots ,\kappa _m), \end{aligned}$$

(26)

and by the triangle inequality and the fact that $r<1/8$,

$$\begin{aligned} E_{mk}\subset E_{5/8}(b_m). \end{aligned}$$

(27)

Suppose now $f\in \mathcal {B}^p_{\alpha }$. We claim that $P_s f=f$. This is true because by Lemma 3.7, f is in $\mathcal {B}^1_{(\alpha +n)/p-n}$ and for this space the required condition in Lemma 2.11 (c) is $s>(\alpha +n)/p-n$ which holds by (3). Therefore

$$\begin{aligned} f(x)=\int _{\mathbb {B}}f(y)\mathcal {R}_s(x,y)\,d\nu _s(y) =\sum _{m=1}^\infty \sum _{k=1}^{\kappa _m} \int _{E_{mk}}f(y)\mathcal {R}_s(x,y)\,d\nu _s(y). \end{aligned}$$

Next, with the above rearrangement, by (19) and (22),

$$\begin{aligned} U{\hat{T}}f(x)=\sum _{m=1}^\infty \sum _{k=1}^{\kappa _m} f(a_{mk})\nu _s(E_{mk})\mathcal {R}_s(x,a_{mk}) \end{aligned}$$

and so, similar to (23), we have

$$\begin{aligned} (I-U{\hat{T}})f(x)&=\sum _{m=1}^\infty \sum _{k=1}^{\kappa _m} \int _{E_{mk}}\bigl (\mathcal {R}_s(x,y)-\mathcal {R}_s(x,a_{mk})\bigr ) f(y)\,d\nu _s(y)\\&\quad +\sum _{m=1}^\infty \sum _{k=1}^{\kappa _m} \int _{E_{mk}}\mathcal {R}_s(x,a_{mk})\big (f(y)-f(a_{mk})\big )\,d\nu _s(y)\\&=:h_1(x)+h_2(x). \end{aligned}$$

We first estimate $h_1$. We will again be careful that in the estimates below the suppressed constants are independent of the separation constant r. Let $y\in E_{mk}$. First, as is shown in (24), for all $x\in \mathbb {B}$,

$$\begin{aligned} |\mathcal {R}_s(x,y)-\mathcal {R}_s(x,a_{mk})|\lesssim \frac{r}{[x,y]^{s+n}} \lesssim \frac{r}{[x,b_m]^{s+n}}, \end{aligned}$$

(28)

where in the last inequality we use Lemma 2.5 with (27). Next, applying Lemma 3.2 with $\delta =1/8$ and noting that $E_{1/8}(y)\subset E_{3/4}(b_m)$, we obtain

$$\begin{aligned} |f(y)|^p \lesssim \int _{E_{1/8}(y)}|f(z)|^p\, d\tau (z)&\le \int _{E_{3/4}(b_m)}|f(z)|^p\, d\tau (z)\nonumber \\&\lesssim (1-|b_m|^2)^{-(\alpha +n)} \int _{E_{3/4}(b_m)}|f(z)|^p\, d\nu _{\alpha }(z),\nonumber \\ \end{aligned}$$

(29)

where in the last inequality we use $(1-|z|^2)\sim (1-|b_m|^2)$ for $z\in E_{3/4}(b_m)$ by Lemma 2.4. Combining (28) and (29) we deduce

$$\begin{aligned} |h_1(x)|\lesssim r\sum _{m=1}^\infty \frac{(1-|b_m|^2)^{-(\alpha +n)/p}}{[x,b_m]^{s+n}} \bigg (\int _{E_{3/4}(b_m)}|f(z)|^p d\nu _{\alpha }(z)\bigg )^{\frac{1}{p}} \sum _{k=1}^{\kappa _m}\nu _s(E_{mk}).\nonumber \\ \end{aligned}$$

(30)

Since the sets $E_{mk}$ are disjoint and $E_{mk}\subset E_{5/8}(b_m)$ for every $k=1,\dots ,\kappa _m$ by (27), we have $\sum _{k=1}^{\kappa _m}\nu _s(E_{mk})\le \nu _s(E_{5/8}(b_m))$. Also $\nu _s(E_{5/8}(b_m))\sim (1-|b_m|^2)^{s+n}$ by Lemma 2.4 and (13). Using this and then Lemma 2.13 (i) yields

$$\begin{aligned} |h_1(x)|^p \lesssim r^p\sum _{m=1}^\infty \frac{(1-|b_m|^2)^{p(s+n)-(\alpha +n)}}{[x,b_m]^{p(s+n)}} \int _{E_{3/4}(b_m)}|f(z)|^p\, d\nu _{\alpha }(z). \end{aligned}$$

Integrating over $\mathbb {B}$ with respect to $d\nu _{\alpha }$, applying Fubini’s theorem, and noting that

$$\begin{aligned} (1-|b_m|^2)^{p(s+n)-(\alpha +n)} \int _{\mathbb {B}}\frac{d\nu _{\alpha }(x)}{[x,b_m]^{p(s+n)}} \lesssim 1, \end{aligned}$$

by Lemma 2.12 and (3), we obtain

$$\begin{aligned} \Vert h_1\Vert ^p_{L^p_{\alpha }} \lesssim r^p\sum _{m=1}^\infty \int _{E_{3/4}(b_m)}|f(z)|^p\, d\nu _{\alpha }(z). \end{aligned}$$

(31)

Finally, by Lemma 2.6, there exists N such that every $z\in \mathbb {B}$ belongs at most N of the balls $E_{3/4}(b_m)$, and so $\sum _{m=1}^\infty \int _{E_{3/4}(b_m)}|f(z)|^p\, d\nu _{\alpha }(z) \le N\int _{\mathbb {B}}|f(z)|^p\, d\nu _{\alpha }(z)$. We conclude that

$$\begin{aligned} \Vert h_1\Vert ^p_{L^p_{\alpha }} \lesssim r^p \Vert f\Vert ^p_{L^p_{\alpha }}. \end{aligned}$$

(32)

We next estimate $h_2$. Let $y\in E_{mk}$. By the mean-value theorem of calculus, there exists ${\tilde{y}}$ lying on the line segment joining $a_{mk}$ and y such that

$$\begin{aligned} |f(y)-f(a_{mk})|\le |y-a_{mk}|\,|\nabla f({\tilde{y}})|&=\rho (y,a_{mk})[y,a_{mk}]\, \frac{|\nabla ^h f({\tilde{y}})|}{1-|{\tilde{y}}|^2}\\&<r\,\frac{[y,a_{mk}]}{1-|{\tilde{y}}|^2}\, |\nabla ^h f({\tilde{y}})|, \end{aligned}$$

where we also use (1), (12) and the fact that $\rho (y,a_{mk})<r$ because $E_{mk}\subset E_r(a_{mk})$. Since the point ${\tilde{y}}$ is also in the ball $E_r(a_{mk})$ and $r<1/8$, we have

$$\begin{aligned} \rho ({\tilde{y}},a_{mk})<1/8, \end{aligned}$$

(33)

and therefore $(1-|{\tilde{y}}|^2)\sim (1-|a_{mk}|^2)$ by Lemma 2.4. Similarly, since $\rho (y,a_{mk})<1/8$, we have $[y,a_{mk}]\sim [a_{mk},a_{mk}]=(1-|a_{mk}|^2)$ by Lemma 2.5 and we conclude

$$\begin{aligned} |f(y)-f(a_{mk})|\lesssim r|\nabla ^h f({\tilde{y}})|. \end{aligned}$$

Next, applying Lemma 3.6 with $\delta =1/8$ and then using $E_{1/8}({\tilde{y}})\subset E_{3/4}(b_m)$ which follows from (33) and (26), we obtain

$$\begin{aligned} |\nabla ^h f({\tilde{y}})|^p \lesssim \int _{E_{1/8}({\tilde{y}})}|f(z)|^p\,d\tau (z)&\le \int _{E_{3/4}(b_m)}|f(z)|^p\,d\tau (z)\\&\lesssim (1-|b_m|^2)^{-(\alpha +n)} \int _{E_{3/4}(b_m)}|f(z)|^p\, d\nu _{\alpha }(z), \end{aligned}$$

similar to (29). Using also that

$$\begin{aligned} |\mathcal {R}_s(x,a_{mk})|\lesssim \frac{1}{[x,a_{mk}]^{s+n}} \sim \frac{1}{[x,b_m]^{s+n}}, \end{aligned}$$

which follows from Lemma 2.9 (a) and Lemma 2.5 with (26), we conclude that

$$\begin{aligned} |h_2(x)|\lesssim r\sum _{m=1}^\infty \frac{(1-|b_m|^2)^{-(\alpha +n)/p}}{[x,b_m]^{s+n}} \bigg (\int _{E_{3/4}(b_m)}|f(z)|^p d\nu _{\alpha }(z)\bigg )^{\frac{1}{p}} \sum _{k=1}^{\kappa _m}\nu _s(E_{mk}). \end{aligned}$$

This estimate is same as (30). Thus we again have $\Vert h_2\Vert ^p_{L^p_{\alpha }}\lesssim r^p\Vert f\Vert ^p_{\mathcal {B}^p_{\alpha }}$ and hence $\Vert (I-U{\hat{T}})f\Vert ^p_{\mathcal {B}^p_{\alpha }} \le \Vert h_1\Vert ^p_{L^p_{\alpha }}+\Vert h_2\Vert ^p_{L^p_{\alpha }} \lesssim r^p\Vert f\Vert ^p_{\mathcal {B}^p_{\alpha }}$. We conclude that $\Vert (I-U{\hat{T}})\Vert \le Cr$, where C depends only on $n,p,\alpha ,s$. $\square$

Proposition 3.5 immediately implies Theorem 1.1.

Proof of Theorem 1.1

By Proposition 3.5, if r is small enough, then $\Vert I-U{\hat{T}}\Vert <1$, and so $U{\hat{T}}$ has bounded inverse. Given $f\in \mathcal {B}^p_{\alpha }$, let $\lambda ={\hat{T}}(U{\hat{T}})^{-1}f$. Then $\lambda \in \ell ^p$, $U\lambda =f$, and $\Vert \lambda \Vert _{\ell ^p}\sim \Vert f\Vert _{\mathcal {B}^p_{\alpha }}$. We note that in the equivalence $\Vert \lambda \Vert _{\ell ^p}\sim \Vert f\Vert _{\mathcal {B}^p_{\alpha }}$, the suppressed constants depend also on r. $\square$

Remark 3.8

One can replace (4) in Theorem 1.1 with (5) because of Lemma 2.10. The only change needed in the above proof is to replace $U\lambda$ in (19) with

$$\begin{aligned} U\lambda (x)=\sum _{m=1}^\infty \lambda _m (1-|a_m|^2)^{s+n-(\alpha +n)/p}\,\mathcal {R}_s(x,a_m), \end{aligned}$$

and ${\hat{T}}f$ in (22) with

$$\begin{aligned} {\hat{T}}f=\{f(a_m)(1-|a_m|^2)^{(\alpha +n)/p-(s+n)}\, \nu _s(E_m)\}. \end{aligned}$$

Then $U{\hat{T}}$ remains the same and so does Proposition 3.5. In the proofs of Propositions 3.1 and 3.4 we omit the references to (20).

4 Interpolation

To prove Theorem 1.2 we again consider two operators. One is ${\hat{U}}:\ell ^p\rightarrow \mathcal {B}^p_{\alpha }$, a slightly modified version of U given in (19) and the other is $T:\mathcal {B}^p_{\alpha }\rightarrow \ell ^p$,

$$\begin{aligned} Tf=\big \{f(a_m)(1-|a_m|^2)^{(\alpha +n)/p}\big \}, \end{aligned}$$

given in (21). Our main purpose is to show that the composition $T{\hat{U}}:\ell ^p\rightarrow \ell ^p$ is invertible when the separation constant is large enough.

Proposition 4.1

For $\alpha >-1$ and $0<p<\infty$, choose s so that (3) holds. If $\{a_m\}$ is r-separated for some $0<r<1$, then the operator ${\hat{U}}:\ell ^p\rightarrow \mathcal {B}^p_{\alpha }$ mapping $\lambda =\{\lambda _m\}$ to

$$\begin{aligned} {\hat{U}}\lambda (x) =\sum _{m=1}^\infty \lambda _m(1-|a_m|^2)^{-(\alpha +n)/p}\, \frac{\mathcal {R}_s(x,a_m)}{\mathcal {R}_s(a_m,a_m)}\qquad (x\in \mathbb {B}) \end{aligned}$$

(34)

is bounded. The above series converges absolutely and uniformly on compact subsets of $\mathbb {B}$, and also in $\Vert \cdot \Vert _{\mathcal {B}^p_{\alpha }}$.

We have $\mathcal {R}_s(a_m,a_m)(1-|a_m|^2)^{(\alpha +n)/p} \sim \Vert \mathcal {R}_s(\cdot ,a_m)\Vert _{\mathcal {B}^p_{\alpha }}$ by (17) and Lemma 2.10, and this proposition can be proved in the same way as Proposition 3.1. The minor changes required are omitted.

Proposition 4.2

For $\alpha >-1$ and $0<p<\infty$, choose s so that (3) holds. There exists $1/2<r_0<1$ depending only on $n,\alpha ,p,s$ such that if $\{a_m\}$ is r-separated with $r>r_0$, then $\Vert T{\hat{U}}-I\Vert _{\ell ^p\rightarrow \ell ^p}<1$.

This proposition immediately implies Theorem 1.2, similar to the proof of Theorem 1.1 above.

To verify Proposition 4.2, let $\lambda =\{\lambda _m\}\in \ell ^p$. Then the m-th term of the sequence $(T{\hat{U}}-I)\lambda$ is given by

$$\begin{aligned} \{(T{\hat{U}}-I)\lambda \}_m =(1-|a_m|^2)^{(\alpha +n)/p} \sum _{\begin{array}{c} k=1\\ k\ne m \end{array}}^\infty \lambda _k (1-|a_k|^2)^{-(\alpha +n)/p} \frac{\mathcal {R}_s(a_m,a_k)}{\mathcal {R}_s(a_k,a_k)}, \end{aligned}$$

and by Lemma 2.9 (a) and (17), we have

$$\begin{aligned} \bigl |\{(T{\hat{U}}-I)\lambda \}_m\bigr |\le C(1-|a_m|^2)^{(\alpha +n)/p} \sum _{\begin{array}{c} k=1\\ k\ne m \end{array}}^\infty |\lambda _k|\, \frac{(1-|a_k|^2)^{s+n-(\alpha +n)/p}}{[a_m,a_k]^{s+n}}, \end{aligned}$$

(35)

where the constant C depends only on $n,\alpha ,p$ and s.

To estimate the norm $\Vert (T{\hat{U}}-I)\lambda \Vert _{\ell ^p}$, we need an estimate of the series on the right of (35) (without the $|\lambda _k|$ term) as given in Lemma 4.4 below. We first prove this lemma and complete the proof of Proposition 4.2 at the end of the section.

Observe that by Lemma 2.12, for $b>-1$ and $c>0$, there exists $C>0$ (depending only on n, b, c) such that

$$\begin{aligned} (1-|a|^2)^c \int _{\mathbb {B}}\frac{(1-|y|^2)^b}{[a,y]^{n+b+c}}\,d\nu (y) \le C, \end{aligned}$$

uniformly for all $a\in \mathbb {B}$. The next result will be needed in the proof of Lemma 4.4.

Lemma 4.3

Let $b>-1$ and $c>0$. For $\varepsilon >0$, there exists $0<r_\varepsilon <1$ such that if $r_\varepsilon<r<1$, then for all $a\in \mathbb {B}$,

$$\begin{aligned} (1-|a|^2)^c \int _{\mathbb {B}\backslash E_r(a)}\, \frac{(1-|y|^2)^b}{[a,y]^{n+b+c}}\,d\nu (y) <\varepsilon . \end{aligned}$$

Proof

Let

$$\begin{aligned} F(a,r):=(1-|a|^2)^c \int _{\mathbb {B}\backslash E_r(a)} \frac{(1-|y|^2)^b d\nu (y)}{[a,y]^{n+b+c}}, \end{aligned}$$

and in the integral make the change of variable $y=\varphi _a(z)$. Since $\varphi _a(\mathbb {B}_r)=E_r(a)$ and $|J_{\varphi _a}|$ is as given in (11), we obtain

$$\begin{aligned} F(a,r)=(1-|a|^2)^c \int _{\mathbb {B}\backslash \mathbb {B}_r} \frac{(1-|\varphi _a(z)|^2)^{b+n}\,d\nu (z)}{[a,\varphi _a(z)]^{n+b+c}(1-|z|^2)^n}. \end{aligned}$$

Applying Lemma 2.1 and (10), and simplifying shows

$$\begin{aligned} F(a,r)=\int _{\mathbb {B}\backslash \mathbb {B}_r} \frac{(1-|z|^2)^b}{[a,z]^{n+b-c}}\,d\nu (z) =n\int _r^1 t^{n-1}(1-t^2)^b\int _{\mathbb {S}} \frac{d\sigma (\zeta )}{|ta-\zeta |^{n+b-c}}\,dt, \end{aligned}$$

where in the second equality we integrate in polar coordinates and use the fact that $[a,t\zeta ]=|ta-\zeta |$ by (7). By Lemma 2.12 and the inequality $1-|a|^2t^2\ge 1-t^2$,

$$\begin{aligned} \int _{\mathbb {S}} \frac{d\sigma (\zeta )}{|ta-\zeta |^{n+b-c}} \le C g(t):= {\left\{ \begin{array}{ll} \dfrac{1}{(1-t^2)^{1+b-c}},&{}\hbox { if}\ 1+b-c>0;\\ 1+\log \dfrac{1}{1-t^2},&{}\hbox { if}\ 1+b-c=0;\\ 1,&{}\hbox { if}\ 1+b-c<0, \end{array}\right. } \end{aligned}$$

where the constant C depends only on n, b, c and do not depend on a. Thus

$$\begin{aligned} F(a,r)\le Cn\int _r^1 t^{n-1}(1-t^2)^b\,g(t)\,dt. \end{aligned}$$

In all the three cases the integral $\int _0^1 t^{n-1}(1-t^2)^b\,g(t)\,dt$ is finite because $b>-1$ and $c>0$ and hence, one can make $F(a,r)<\varepsilon$ by choosing r close to 1. $\square$

The next lemma is an analogue of Lemma 3.1 of [11].

Lemma 4.4

Let $b>n-1$ and $c>0$. For $1/2<r<1$, there exists $C(r)>0$ (depending also on n, b and c) such that for every r-separated sequence $\{a_m\}$ and for every $m=1,2,\dots$,

$$\begin{aligned} (1-|a_m|^2)^c \sum _{\begin{array}{c} k=1\\ k\ne m \end{array}}^\infty \frac{(1-|a_k|^2)^b}{[a_m,a_k]^{b+c}} \le C(r). \end{aligned}$$

Moreover, one can choose C(r) to be arbitrarily small by making r sufficiently close to 1.

Proof

By the Lemmas 2.4, 2.5 and (13), there exists $C>0$ depending only on n, b, c such that

$$\begin{aligned} \frac{(1-|a|^2)^b}{[x,a]^{b+c}} \le C\int _{E_{1/4}(a)}\frac{(1-|y|^2)^{b-n}}{[x,y]^{b+c}}\,d\nu (y), \end{aligned}$$

for all $a,x\in \mathbb {B}$. If $\{a_m\}$ is r-separated with $r>1/2$, then the balls $E_{1/4}(a_m)$ are disjoint and therefore

$$\begin{aligned} (1-|a_m|^2)^c \sum _{\begin{array}{c} k=1\\ k\ne m \end{array}}^\infty \frac{(1-|a_k|^2)^b}{[a_m,a_k]^{b+c}} \le C(1-|a_m|^2)^c \int _{\bigcup ^\infty _{\begin{array}{c} k=1\\ k\ne m \end{array}}E_{1/4}(a_k)} \frac{(1-|y|^2)^{b-n}}{[a_m,y]^{b+c}}\,d\nu (y). \end{aligned}$$

Set

$$\begin{aligned} R:=\frac{r-1/4}{1-r/4}. \end{aligned}$$

Clearly, $0<R<1$. We claim that $\bigcup ^\infty _{\begin{array}{c} k=1\\ k\ne m \end{array}}E_{1/4}(a_k) \subset \mathbb {B}\backslash E_R(a_m)$. To see this, let $z\in E_{1/4}(a_k)$ with $k\ne m$. Then, by the strong triangle inequality in Lemma 2.2,

$$\begin{aligned} \rho (z,a_m)\ge \frac{\rho (a_m,a_k)-\rho (z,a_k)}{1-\rho (a_m,a_k)\rho (z,a_k)} \ge \frac{r-\rho (z,a_k)}{1-r\rho (z,a_k)} \ge \frac{r-1/4}{1-r/4}, \end{aligned}$$

where in the second and third inequalities we use $\rho (a_m,a_k)\ge r$ and $\rho (z,a_k)<1/4$, and the elementary fact that for $0\le t_0<1$, the function $f(t)=(t-t_0)/(1-tt_0)$ is increasing on the interval $0\le t<1$ and $-f$ is decreasing. Thus

$$\begin{aligned} (1-|a_m|^2)^c \sum _{\begin{array}{c} k=1\\ k\ne m \end{array}}^\infty \frac{(1-|a_k|^2)^b}{[a_m,a_k]^{b+c}} \le C(1-|a_m|^2)^c \int _{\mathbb {B}\backslash E_R(a_m)} \frac{(1-|y|^2)^{b-n}}{[a_m,y]^{b+c}}\,d\nu (y), \end{aligned}$$

and since $R\rightarrow 1^-$ as $r\rightarrow 1^-$, the desired result follows from Lemma 4.3. $\square$

We now complete the proof of Proposition 4.2. We consider the cases $0<p\le 1$ and $p>1$ separately.

Proof of Proposition 4.2 when $0<p\le 1$

For $\lambda =\{\lambda _m\}\in \ell ^p$, by (35), Lemma 2.13 (i) and Fubini’s theorem,

$$\begin{aligned} \Vert (T{\hat{U}}-I)\lambda \Vert _{\ell ^p}^p&\le C^p\sum _{m=1}^\infty (1-|a_m|^2)^{\alpha +n} \Biggl (\sum _{\begin{array}{c} k=1\\ k\ne m \end{array}}^\infty |\lambda _k|\, \frac{(1-|a_k|^2)^{s+n-(\alpha +n)/p}}{[a_m,a_k]^{s+n}} \Biggr )^p\\&\le C^p\sum _{m=1}^\infty (1-|a_m|^2)^{\alpha +n} \sum _{\begin{array}{c} k=1\\ k\ne m \end{array}}^\infty |\lambda _k|^p\, \frac{(1-|a_k|^2)^{p(s+n)-(\alpha +n)}}{[a_m,a_k]^{p(s+n)}}\\&=C^p\sum _{k=1}^\infty |\lambda _k|^p (1-|a_k|^2)^{p(s+n)-(\alpha +n)} \sum _{\begin{array}{c} m=1\\ m\ne k \end{array}}^\infty \, \frac{(1-|a_m|^2)^{\alpha +n}}{[a_m,a_k]^{p(s+n)}}. \end{aligned}$$

By Lemma 4.4, there exists C(r) such that (note that $\alpha +n>n-1$ since $\alpha >-1$, and $p(s+n)-(\alpha +n)>0$ by (3))

$$\begin{aligned} \Vert (T{\hat{U}}-I)\lambda \Vert _{\ell ^p}^p\le C^p\,C(r)\Vert \lambda \Vert _{\ell ^p}^p. \end{aligned}$$

Since C(r) can be made arbitrarily small by making r close enough to 1, the proposition follows. $\square$

We next deal with the case $1<p<\infty$. Let $p'$ be the conjugate exponent of p. We employ Schur’s test which, for the sequence space $\ell ^p$, has the following form (see [11, Lemma 3.2]): Let $A=(A_{mk})_{1\le m,k<\infty }$ be an infinite matrix with nonnegative entries and $L_A:\ell ^p\rightarrow \ell ^p$ be the corresponding operator taking $\lambda =\{\lambda _m\}$ to

$$\begin{aligned} \{L_A\lambda \}_m=\sum _{k=1}^\infty A_{mk}\lambda _k, \qquad m=1,2,\dots . \end{aligned}$$

If there exists a constant $C>0$ and a positive sequence $\{\gamma _m\}$ such that

$$\begin{aligned} \sum _{k=1}^\infty A_{mk}\gamma _k^{p'}\le C\gamma _m^{p'},\qquad m=1,2,\dots , \end{aligned}$$

and

$$\begin{aligned} \sum _{m=1}^\infty A_{mk}\gamma _m^{p}\le C\gamma _k^{p},\qquad k=1,2,\dots , \end{aligned}$$

then the operator $L_A:\ell ^p\rightarrow \ell ^p$ is bounded and $\Vert L_A\Vert \le C$.

Proof of Proposition 4.2 when $1<p<\infty$

Without loss of generality we can assume that the r-separated sequence $\{a_m\}$ is maximal, that is $\{a_m\}$ is an r-lattice and so is an infinite sequence.

For $m,k=1,2,\dots$, let $A_{mk}=0$ if $k=m$; and if $k\ne m$, let

$$\begin{aligned} A_{mk}=(1-|a_m|^2)^{(\alpha +n)/p}\, \frac{(1-|a_k|^2)^{s+n-(\alpha +n)/p}}{[a_m,a_k]^{s+n}}. \end{aligned}$$

Let $A=(A_{mk})$ and $L_A:\ell ^p\rightarrow \ell ^p$ be the corresponding operator. Then by (35),

$$\begin{aligned} \bigl |\{(T{\hat{U}}-I)\lambda \}_m\bigr |\le C\{L_A\lambda \}_m. \end{aligned}$$

To estimate $\Vert L_A\Vert$ with the Schur’s test, we take $\{\gamma _m\}=\{(1-|a_m|^2)^{(n-1)/pp'}\}$. Then

$$\begin{aligned} \sum _{k=1}^\infty A_{mk}\gamma _k^{p'} =(1-|a_m|^2)^{(\alpha +n)/p}\, \sum _{\begin{array}{c} k=1\\ k\ne m \end{array}}^\infty \, \frac{(1-|a_k|^2)^{s+n-(\alpha +1)/p}}{[a_m,a_k]^{s+n}}, \end{aligned}$$

and by Lemma 4.4, there exists $C_1(r)$ such that (we check that $s+n-(\alpha +1)/p>n-1$ by (3), and $(\alpha +1)/p>0$)

$$\begin{aligned} \sum _{k=1}^\infty A_{mk}\gamma _k^{p'} \le (1-|a_m|^2)^{(\alpha +n)/p} \frac{C_1(r)}{(1-|a_m|^2)^{(\alpha +1)/p}}=C_1(r)\gamma _m^{p'}. \end{aligned}$$

Observe next that

$$\begin{aligned} \sum _{m=1}^\infty A_{mk}\gamma _m^{p} =(1-|a_k|^2)^{s+n-(\alpha +n)/p}\, \sum _{\begin{array}{c} m=1\\ m\ne k \end{array}}^\infty \, \frac{(1-|a_m|^2)^{(\alpha +n)/p+(n-1)/p'}}{[a_m,a_k]^{s+n}}. \end{aligned}$$

To apply Lemma 4.4 we check that $(\alpha +n)/p+(n-1)/p'=(\alpha +1)/p+n-1>n-1$, and $s+n-\bigl ((\alpha +n)/p+(n-1)/p'\bigr )=s+1-(\alpha +1)/p>0$ by (3). Thus there exists $C_2(r)$ such that

$$\begin{aligned} \sum _{m=1}^\infty A_{mk}\gamma _m^{p} \le (1-|a_k|^2)^{s+n-(\alpha +n)/p} \frac{C_2(r)}{(1-|a_k|^2)^{s+n-(\alpha +n)/p-(n-1)/p'}} =C_2(r)\gamma _k^p. \end{aligned}$$

We conclude that $L_A$ is bounded and $\Vert L_A\Vert \le \max \{C_1(r),C_2(r)\}$. Therefore $\Vert T{\hat{U}}-I\Vert \le C\max \{C_1(r),C_2(r)\}$ and since both $C_1(r)$ and $C_2(r)$ can be made arbitrarily small by making r close enough to 1, we conclude that $\Vert T{\hat{U}}-I\Vert$ can be made small. This finishes the proof of Proposition 4.2. $\square$

5 Inclusion Relations

In this section we prove Theorem 1.3.

Proof of Theorem 1.3

We first prove part (a). Suppose $\mathcal {B}^p_{\alpha }\subset \mathcal {B}^q_\beta$. Since point evaluations are bounded on $\mathcal {H}$-harmonic Bergman spaces, the inclusion $i:\mathcal {B}^p_{\alpha }\rightarrow \mathcal {B}^q_\beta$ is continuous by the closed graph theorem. For every $s>-1$ and $a\in \mathbb {B}$, the reproducing kernel $\mathcal {R}_s(a,\cdot )$ is bounded on $\mathbb {B}$ by Lemma 2.9 (a) and (8), so belongs to every Bergman space. By Lemma 2.10, for large enough s, we have

$$\begin{aligned} \frac{\Vert \mathcal {R}_s(a,\cdot )\Vert _{\mathcal {B}^q_\beta }}{\Vert \mathcal {R}_s(a,\cdot )\Vert _{\mathcal {B}^p_{\alpha }}} \sim (1-|a|^2)^{(\beta +n)/q-(\alpha +n)/p}, \end{aligned}$$

(36)

and the right-hand side is bounded as $|a|\rightarrow 1^-$ only if $(\beta +n)/q\ge (\alpha +n)/p$.

Suppose now that

$$\begin{aligned} \frac{\alpha +n}{p}\le \frac{\beta +n}{q}. \end{aligned}$$

(37)

Pick s large enough so that (3) holds both for $\alpha ,p$ and $\beta ,q$. Let $r_0$ be as asserted in the atomic decomposition theorem for $\mathcal {B}^p_{\alpha }$ and let $\{a_m\}$ be an r-lattice with $r<r_0$. Then for every $f\in \mathcal {B}^p_{\alpha }$, there exists $\{\lambda _m\}\in \ell ^p$ with $\Vert \{\lambda _m\}\Vert _{\ell ^p}\sim \Vert f\Vert _{\mathcal {B}^p_{\alpha }}$ such that

$$\begin{aligned} f(x)=\sum _{m=1}^\infty \lambda _m\frac{\mathcal {R}_s(x,a_m)}{\Vert \mathcal {R}_s(\cdot ,a_m)\Vert _{\mathcal {B}^p_{\alpha }}} =\sum _{m=1}^\infty \kappa _m \frac{\mathcal {R}_s(x,a_m)}{\Vert \mathcal {R}_s(\cdot ,a_m)\Vert _{\mathcal {B}^q_\beta }}, \end{aligned}$$

where

$$\begin{aligned} \kappa _m=\lambda _m\, \frac{\Vert \mathcal {R}_s(\cdot ,a_m)\Vert _{\mathcal {B}^q_\beta }}{\Vert \mathcal {R}_s(\cdot ,a_m)\Vert _{\mathcal {B}^p_{\alpha }}}. \end{aligned}$$

By (36) and (37), $|\kappa _m|\sim |\lambda _m|(1-|a_m|^2)^{(\beta +n)/q-(\alpha +n)/p} \le |\lambda _m|$, and so the sequence $\{\kappa _m\}$ is in $\ell ^p$. Thus, $\{\kappa _m\}\in \ell ^q$ by Lemma 2.13 (i), and it follows from Proposition 3.1 that $f\in \mathcal {B}^q_\beta$ with $\Vert f\Vert _{\mathcal {B}^q_\beta }\lesssim \Vert \{\kappa _m\}\Vert _{\ell ^q} \le \Vert \{\kappa _m\}\Vert _{\ell ^p}\lesssim \Vert \{\lambda _m\}\Vert _{\ell ^p} \lesssim \Vert f\Vert _{\mathcal {B}^p_{\alpha }}$.

We next prove part (b). Note first that in this case $p/q>1$ and the conjugate exponent of p/q is $p/(p-q)$. To see the if part, suppose

$$\begin{aligned} \frac{\alpha +1}{p}<\frac{\beta +1}{q}. \end{aligned}$$

(38)

By Hölder’s inequality,

$$\begin{aligned} \int _{\mathbb {B}}|f(x)|^q\,d\nu _\beta (x) \le \biggl (\int _{\mathbb {B}}|f(x)|^p \,d\nu _{\alpha }(x)\biggr )^{\frac{q}{p}} \biggl (\int _{\mathbb {B}} (1-|x|^2)^{(\beta -\alpha \frac{q}{p})\frac{p}{p-q}} \,d\nu (x)\biggr )^{\frac{p-q}{p}}, \end{aligned}$$

and since the exponent $(\beta -\alpha \frac{q}{p})\frac{p}{p-q}>-1$ by (38), we obtain $\Vert f\Vert _{\mathcal {B}^q_\beta }\lesssim \Vert f\Vert _{\mathcal {B}^p_{\alpha }}$.

Suppose now that $\mathcal {B}^p_{\alpha }\subset \mathcal {B}^q_\beta$. Let $r_0$ be as asserted in the interpolation theorem for $\mathcal {B}^p_{\alpha }$ and let $\{a_m\}$ be an r-lattice with $r>r_0$. Given $\{\lambda _m\}\in \ell ^{p/q}$, we have $\{|\lambda _m|^{1/q}\}\in \ell ^p$ and there exists a function $f\in \mathcal {B}^p_{\alpha }$ such that

$$\begin{aligned} f(a_m)=|\lambda _m|^{1/q}(1-|a_m|^2)^{-(\alpha +n)/p}. \end{aligned}$$

Since f is also in $\mathcal {B}^q_\beta$, the sequence $\{f(a_m)(1-|a_m|^2)^{(\beta +n)/q}\}$ is in $\ell ^q$ by Proposition 3.3, and so

$$\begin{aligned} \sum _{m=1}^\infty |\lambda _m|(1-|a_m|^2)^{(\beta +n)-(\alpha +n)q/p}<\infty . \end{aligned}$$

By Lemma 2.13 (ii), this implies that the sequence $\{(1-|a_m|^2)^{(\beta +n)-(\alpha +n)q/p}\}$ is in $\ell ^{p/(p-q)}$ and by Lemma 2.8 (b) this holds only if

$$\begin{aligned} \Bigl ((\beta +n)-(\alpha +n)\frac{q}{p}\Bigr )\frac{p}{p-q}>n-1, \end{aligned}$$

which is equivalent to (38). $\square$