On Berezin type operators and Toeplitz operators on Bergman spaces

Abstract

We introduce a class of integral operators called Berezin type operators. It is a generalization of the Berezin transform, and has a close relation to the Bergman–Carleson measures. The concept is partly motivated by the relationship between Hardy–Carleson measures and area operators. We mainly study the boundedness and the compactness of Berezin type operators from a Bergman space $A^{p_1}_{\alpha _1}$ to a Lebesgue space $L^{p_2}_{\alpha _2}$ with $0<p_1,p_2\le \infty$ and $\alpha _1,\alpha _2>-1$. We also show that Berezin type operators are closely related to Toeplitz operators.

1 Introduction

Let $\mathbb {B}_n=\{z\in \mathbb {C}^n:|z|<1\}$ be the unit ball in the nth dimensional complex Euclidean space $\mathbb {C}^n$. Let $\text {d}v(z)$ be the normalized volume measure on $\mathbb {B}_n$, such that $v(\mathbb {B}_n)=1$. For $0<p<\infty$ and $-1<\alpha <\infty$, let $L^{p}_{\alpha }:=L^{p}(\mathbb {B}_n,\,\text {d}v_{\alpha })$ denote the weighted Lebesgue space which contains measurable functions f on $\mathbb {B}_n$, such that

$$\begin{aligned} \Vert f\Vert _{p,\alpha }=\left( \int \limits _{\mathbb {B}_n}|f(z)|^p\,\text {d}v_{\alpha }(z)\right) ^{1/p}<\infty , \end{aligned}$$

where $\text {d}v_{\alpha }(z)=c_{\alpha }(1-|z|^2)^{\alpha }\,\text {d}v(z)$, and $c_{\alpha }$ is the normalized constant, such that $v_{\alpha }(\mathbb {B}_n)=1$. Let $L^\infty :=L^\infty (\mathbb {B}_n, \text{ d }v)$ be the space of measurable functions on $\mathbb {B}_n$, such that

$$\begin{aligned} \Vert f\Vert _\infty =\mathop {\text {ess sup}}\limits _{z\in \mathbb {B}_n}|f(z)|<\infty . \end{aligned}$$

Let $H(\mathbb {B}_n)$ be the space of all holomorphic functions on $\mathbb {B}_n$. We denote by $A^p_{\alpha }=L^{p}_{\alpha }\cap H(\mathbb {B}_n)$ the weighted Bergman space on $\mathbb {B}_n$ for $0<p<\infty$, and denote by $H^\infty :=L^\infty \cap H(\mathbb {B}_n)$ the holomorphic bounded function space.

Let $\beta >-1$ and let $\mu$ be a positive Borel measure on $\mathbb {B}_n$. For $f\in H(\mathbb {B}_n)$, we consider the following sublinear operator:

$$\begin{aligned} B_{\mu }^\beta f(z)=\int \limits _{\mathbb {B}_n}\frac{|f(w)|}{|1-\langle z, w\rangle |^{n+1+\beta }}\text {d}\mu (w). \end{aligned}$$

We call $B_{\mu }^\beta$ a Berezin type operator. Note that, if $\beta =n+1+2\alpha$ and $\text {d}\mu (w)=\text {d}v_\alpha (w)$, then $(1-|z|^2)^{n+1+\alpha }B_{\mu }^{\beta }(f)(z)$ is the $\alpha$-Berezin transform of the function |f|. If $\beta =s+\alpha$ and $f=1$, we denote

$$\begin{aligned} B_{s,\alpha }(\mu )(z): =(1-|z|^2)^s B_{\mu }^{\beta }(1)(z) =\int \limits _{\mathbb {B}_n} \frac{(1-|z|^2)^s}{|1-\langle z,w \rangle |^{n+1+\alpha +s}} \,\text {d}\mu (w). \end{aligned}$$

This is called a Berezin type transform for the measure $\mu$. We refer to [13] for more information about Berezin transforms.

Let ${\mathbb {D}}$ be the unit disk, let $\partial {\mathbb {D}}$ be the unit circle, and let $\mu$ be a nonnegative measure on ${\mathbb {D}}$. The area operator on the Hardy space $H^p$ is a sublinear operator defined by

$$\begin{aligned} A_{\mu }(f)(\zeta )=\int \limits _{\Gamma (\zeta )}\frac{|f(z)|}{1-|z|}\,\text {d}\mu (z), \qquad \forall \zeta \in \partial \mathbb {D}, \end{aligned}$$

where $\Gamma (\zeta )$ is a non-tangential approach region in $\mathbb {D}$ with vertex $\zeta \in \partial \mathbb {D}$ defined by

$$\begin{aligned} \Gamma (\zeta )=\{z\in \mathbb {D}: |\zeta -z|<2(1-|z|)\}. \end{aligned}$$

It has been proved in [1] that the $A_{\mu }$ is bounded from the Hardy space $H^p$ to $L^p(\partial \mathbb {D})$ if and only if $\mu$ is a Hardy–Carleson measure. The result has been generalized to the case $A_{\mu }:H^p\rightarrow L^q(\partial \mathbb {D})$ for possibly different p, q in [3]. Note that $A_{\mu }(1)$ is used to characterize Hardy–Carleson measures; see, for example [5, 8, 9]. It is also known that Berezin type transforms for measures are also used to characterize Bergman–Carleson measures; see [6, 11]. This observation is one of our motivations to consider the Berezin type operators on Bergman spaces.

The purpose of this paper is to study the boundedness and the compactness of the Berezin type operator from one Bergman space $A_{\alpha _1}^{p_1}$ to a Lebesgue space $L^{p_2}_{\alpha _2}$. It turns out that our characterizations are the same for Toeplitz operators.

Recall that, given $\beta >-1$ and a positive Borel measure $\mu$ on $\mathbb {B}_n$, the Toeplitz operator $T_\mu ^\beta$ is defined by

$$\begin{aligned} T_{\mu }^\beta (f)(z)=\int \limits _{\mathbb {B}_n}\frac{f(w)}{(1-\langle z, w\rangle )^{n+1+\beta }}\text {d}\mu (w), \quad \quad z\in \mathbb {B}_n. \end{aligned}$$

(1.1)

It is clear that $|T_\mu ^\beta f|\le B_\mu ^\beta f$ for $f\in H(\mathbb {B}_n)$. Therefore, boundedness of a Berezin type operator implies boundedness of the corresponding Toeplitz operator.

Our results will heavily depend on Carleson measures. For $\lambda >0$ and $\alpha >-1$, we say $\mu$ is a $(\lambda ,\alpha )$-Bergman–Carleson measure if, for any two positive numbers p and q with $q/p=\lambda$, there is a positive constant $C>0$, such that

$$\begin{aligned} \int \limits _{\mathbb {B}_n} |f(z)|^q\,\text {d}\mu (z)\le C\Vert f\Vert _{p,\alpha }^q \end{aligned}$$

for any $f\in A^p_{\alpha }$. We also denote by

$$\begin{aligned} \Vert \mu \Vert _{\lambda ,\alpha } =\sup _{f\in A^p_{\alpha }, \Vert f\Vert _{p,\alpha }\le 1}\int \limits _{\mathbb {B}_n} |f(z)|^q\,\text {d}\mu (z). \end{aligned}$$

We say a positive Borel measure $\mu$ is a vanishing $(\lambda , \alpha )$-Bergman–Carleson measure if for any two positive numbers p and q satisfying $q/p=\lambda$ and any sequence $\{f_k\}$ in $A_\alpha ^p$ with $\Vert f_k\Vert _{p, \alpha }\le 1$ and $f_k(z)\rightarrow 0$ uniformly on any compact subset of $\mathbb {B}_n$

$$\begin{aligned} \lim _{k\rightarrow \infty }\int \limits _{\mathbb {B}_n}|f_k|^q\text {d}\mu (z)=0. \end{aligned}$$

For convenience, we assume that $-1<\alpha _1,\alpha _2,\beta <\infty$ throughout the paper. We list the following conditions and notations which will be used in our main results:

$$\begin{aligned} n+1+\beta ~&>n\max \left( 1, \frac{1}{p_1}\right) +\frac{1+\alpha _1}{p_1} , \end{aligned}$$

(C-1)

$$\begin{aligned} n+1+\beta ~&>n\max \left( 1, \frac{1}{p_2}\right) +\frac{1+\alpha _2}{p_2} \end{aligned}$$

(C-2)

and

$$\begin{aligned} \lambda =1+\frac{1}{p_1}-\frac{1}{p_2}, \quad \quad \gamma =\frac{1}{\lambda }\left( \beta +\frac{\alpha _1}{p_1}-\frac{\alpha _2}{p_2}\right) . \end{aligned}$$

(1.2)

Our first two main results are for the case $0<p_1\le p_2<\infty$.

Theorem 1.1

Let $0<p_1\le p_2<\infty$, and let $-1<\alpha _1,\alpha _2,\beta <\infty$ satisfy (C-1) and (C-2). Let $\lambda , \gamma$ be given by (1.2). Then, the following statements are equivalent:

• (i) $B_\mu ^\beta$ is bounded from $A_{\alpha _1}^{p_1}$ to $L_{\alpha _2}^{p_2}$.

• (ii) $T_{\mu }^\beta$ is bounded from $A_{\alpha _1}^{p_1}$ to $A_{\alpha _2}^{p_2}$.

• (iii) The measure $\mu$ is a $(\lambda , \gamma )$-Bergman–Carleson measure.

Moreover, we have

$$\begin{aligned} \Vert B_{\mu }^\beta \Vert _{A_{\alpha _1}^{p_1}\rightarrow L_{\alpha _2}^{p_2}} \asymp \Vert T_{\mu }^\beta \Vert _{A_{\alpha _1}^{p_1}\rightarrow A_{\alpha _2}^{p_2}} \asymp \Vert \mu \Vert _{\lambda , \gamma }. \end{aligned}$$

Theorem 1.2

Let $0<p_1\le p_2<\infty$, and let $-1<\alpha _1,\alpha _2,\beta <\infty$ satisfy (C-1) and (C-2). Let $\lambda , \gamma$ be given by (1.2). Then, the following statements are equivalent:

• (i) $B_\mu ^\beta$ is compact from $A_{\alpha _1}^{p_1}$ to $L_{\alpha _2}^{p_2}$.

• (ii) $T_{\mu }^\beta$ is compact from $A_{\alpha _1}^{p_1}$ to $A_{\alpha _2}^{p_2}$.

• (iii) The measure $\mu$ is a vanishing $(\lambda , \gamma )$-Bergman–Carleson measure.

Our next main result is for the case $0<p_2<p_1<\infty$. For this result, we need a well-known result on decomposition of the unit ball $\mathbb {B}_n$.

For any $a\in \mathbb {B}_n$ with $a\ne 0$, we denote by $\varphi _a(z)$ the Möbius transformation on $\mathbb {B}_n$ that interchanges the points 0 and a. It is known that $\varphi _a$ satisfies the following properties: $\varphi _a\circ \varphi _a(z)=z$, and

$$\begin{aligned} 1-|\varphi _a(z)|^2=\frac{(1-|a|^2)(1-|z|^2)}{|1-\langle z,a\rangle |^2}, \qquad z,a\in \mathbb {B}_n. \end{aligned}$$

(1.3)

For $z,w\in \mathbb {B}_n$, the pseudo-hyperbolic distance between z and w is defined by

$$\begin{aligned} \rho (z,w)=|\varphi _z(w)|, \end{aligned}$$

and the hyperbolic distance on $\mathbb {B}_n$ between z and w induced by the Bergman metric is given by

$$\begin{aligned} \beta (z,w)=\tanh \,\rho (z,w)=\frac{1}{2}\log \frac{1+|\varphi _z(w)|}{1-|\varphi _z(w)|}. \end{aligned}$$

Throughout the paper, for $z\in \mathbb {B}_n$ and $r>0$, let D(z, r) denote the Bergman metric ball at z which is given by

$$\begin{aligned} D(z,r)=\left\{ w\in \mathbb {B}_n:\,\beta (z,w)<r \right\} . \end{aligned}$$

It is known that, for a fixed $r>0$, the weighted volume

$$\begin{aligned} v_{\alpha }(D(z,r))\asymp (1-|z|^2)^{n+1+\alpha }. \end{aligned}$$

(1.4)

We refer to [12] for the above facts.

A sequence of points $\{a_k\}$ in $\mathbb {B}_n$ is called a separated sequence (in the Bergman metric) if there exists $\delta >0$, such that $\beta (z_i,z_j)>\delta$ for any $i\ne j$.

Lemma 1.3

[12, Theorem 2.23 ] There exists a positive integer N, such that for any $0<r<1$, we can find a sequence $\{a_j\}$ in $\mathbb {B}_n$ with the following properties:

• (i) $\mathbb {B}_n=\cup _{j}D(a_j,r)$.

• (ii) The sets $D(a_j,r/4)$ are mutually disjoint.

• (iii) Each point $z\in \mathbb {B}_n$ belongs to at most N of the sets $D(a_j,4r)$.

Any sequence $\{a_j\}$ satisfying the conditions of the above lemma is called a lattice (or an r-lattice if one wants to stress the dependence on r) in the Bergman metric. Obviously, any r-lattice is separated. For convenience, we will denote by $D_j=D(a_j,r)$ and $\tilde{D}_j=D(a_j,4r)$ throughout the paper. Then, Lemma 1.3 says that $\mathbb {B}_n=\cup _{k=1}^{\infty }D_j$ and there is an positive integer N, such that every point z in $\mathbb {B}_n$ belongs to at most N of sets $\tilde{D}_j$.

Theorem 1.4

Let $0<p_2< p_1<\infty$, and let $-1<\alpha _1,\alpha _2,\beta <\infty$ satisfy (C-1) and (C-2). Let $\lambda , \gamma$ be given by (1.2). Given $0<r<1$, let $\{a_j\}$ be an r-lattice in $\mathbb {B}_n$, and let $D_j$ and $\tilde{D}_j$ be the associated Bergman metric balls given by Lemma 1.3. Then, the following statements are equivalent:

• (i) $B_{\mu }^\beta$ is compact from $A_{\alpha _1}^{p_1}$ to $L_{\alpha _2}^{p_2}$.

• (ii) $B_\mu ^\beta$ is bounded from $A_{\alpha _1}^{p_1}$ to $L_{\alpha _2}^{p_2}$.

• (iii) $T_{\mu }^\beta$ is bounded from $A_{\alpha _1}^{p_1}$ to $A_{\alpha _2}^{p_2}$.

• (iv) $T_{\mu }^\beta$ is compact from $A_{\alpha _1}^{p_1}$ to $A_{\alpha _2}^{p_2}$.

• (v)
  $$\begin{aligned}\{\mu _j\}:= \left\{ \frac{\mu (D_j)}{(1-|a_j|^2)^{(n+1+\gamma )\lambda }}\right\} \in l^{1/(1-\lambda )}. \end{aligned}$$

Moreover, we have

$$\begin{aligned} \Vert B_{\mu }^\beta \Vert _{A_{\alpha _1}^{p_1}\rightarrow L_{\alpha _2}^{p_2}} \asymp \Vert T_{\mu }^\beta \Vert _{A_{\alpha _1}^{p_1}\rightarrow A_{\alpha _2}^{p_2}} \asymp \Vert \{\mu _j\}\Vert _{l^{1/(1-\lambda )}}. \end{aligned}$$

Remark 1.5

In the above theorems, the most parts of the results on Toeplitz operators have been proved by Pau and the second author in [6]. The only exception is condition (v) in Theorem 1.4. In [6], it used the condition that “$\mu$ is a $(\lambda , \gamma )$-Bergman–Carleson measure” instead of (v) above. However, for the case $0<p_2<p_1<\infty$, it may happen that $\lambda \le 0$. In this case, the $(\lambda , \gamma )$-Bergman–Carleson measure condition does not make sense, while (v) is still valid.

Our work here is mostly built up from the work in [6]. Our main contributions are proofs of (iii)$\Rightarrow$(i) in Theorems 1.1 and 1.2 and (v)$\Rightarrow$(i) in Theorem 1.4. We would like to point out that our proofs are different from the proof of Theorem 1.2 in [6]. The key ingredients of our proofs are two technical results, Lemma 4.1 and Lemma 4.2, which allow us to treat all cases together. By comparison, in the proof of Theorem 1.2 in [6], for the case $1<p_2<\infty$, it used a new characterization of Bergman–Carleson measures discovered in that paper. For proving compactness results for $B_{\mu }^\beta$, we have to be more careful, since we are dealing with a sublinear operator. We also gave a detailed proof of a characterization of compactness of $T_{\mu }^\beta :\,A_{\alpha _1}^{p_1}\rightarrow A_{\alpha _2}^{p_2}$ for $0<p_1\le p_2<\infty$ (Proposition 3.2), which seems to be a folklore, but we could not find a proof. The proof of Proposition 3.2 for the case $0<p_1\le 1$ is actually surprisingly involved. Besides, we have also discussed the cases when $p_1=\infty$ or/and $p_2=\infty$.

The paper is organized as follows. In Sect. 2, we recall some notations and preliminary results which will be used later. In Sect. 3, we develop some tools for characterizing compactness of Berezin type operators and Toeplitz operators. We give the proofs of Theorems 1.1, 1.2 and 1.4 in Sect. 4. In Sect. 5 and Sect. 6, we study the boundedness and the compactness of $B_\mu ^\beta : A_{\alpha _1}^{p_1}\rightarrow L_{\alpha _2}^{p_2}$ and $T_\mu ^\beta : A_{\alpha _1}^{p_1}\rightarrow A_{\alpha _2}^{p_2}$ for the remaining cases when $p_1=\infty$ or/and $p_2=\infty$.

Throughout the paper, the notation $A\lesssim B$ means that there is a positive constant C, such that $A\le CB$, and the notation $A\asymp B$ means that both $A\lesssim B$ and $B\lesssim A$ are satisfied.

2 Preliminaries

2.1 Carleson measures

The following result was obtained by several authors and can be found, for example, in [11, Theorem 50], [11, p.71] and the references therein.

Theorem A

Suppose $1\le \lambda <\infty$ and $-1<\alpha <\infty$, the following statements are equivalent:

• (i) $\mu$ is a $(\lambda , \alpha )$-Bergman–Carleson measure.

• (ii) For any real number r with $0<r<1$ and any $z\in \mathbb {B}_n$

  $$\begin{aligned} \mu (D(z,r))\lesssim (1-|z|^2)^{(n+1+\alpha )\lambda }. \end{aligned}$$

• (iii) For some (every) $s>0$, the Berezin type transform of $\mu$

  $$\begin{aligned} B_{s,(n+1+\alpha )\lambda -n-1}(\mu )\in L^{\infty }(\mathbb {B}_n), \end{aligned}$$

  that is, there is a constant $C>0$

  $$\begin{aligned} \sup _{a\in \mathbb {B}_n}\int \limits _{\mathbb {B}_n} \frac{(1-|a|^2)^s}{|1-\langle z,a\rangle |^{{(n+1+\alpha )\lambda }+s}} \,d\mu (z)\le C. \end{aligned}$$

  Especially, if $\lambda =1$, we get that a positive Borel measure $\mu$ on $\mathbb {B}_n$ is a $(1,\alpha )$-Bergman–Carleson measure if and only if $B_{s,\alpha }(\mu )\in L^{\infty }(\mathbb {B}_n)$ for some (every) $s>0$.

Theorem B

Suppose $1\le \lambda <\infty$ and $-1<\alpha <\infty$, the following statements are equivalent:

• (i) $\mu$ is a vanishing $(\lambda , \alpha )$-Bergman–Carleson measure.

• (ii) For some(any) $s>0$

  $$\begin{aligned} \lim _{|a|\rightarrow 1}\int \limits _{\mathbb {B}_n}\frac{(1-|a|^2)^s}{|1-\langle z, a\rangle |^{(n+1+\alpha )\lambda +s}}d\mu (z)=0. \end{aligned}$$

• (iii) For any real number r with $0<r<1$ and any $a\in \mathbb {B}_n$

  $$\begin{aligned} \lim _{|a|\rightarrow 1}\frac{\mu (D(a,r))}{(1-|a|^2)^{(n+1+\alpha )\lambda }}=0. \end{aligned}$$

Lemma 2.1

Let $1\le \lambda <\infty$ and $-1<\gamma <\infty$. Let $\mu$ be a $(\lambda ,\gamma )$-Bergman–Carleson measure on $\mathbb {B}_n$. Then, for any $f\in H(\mathbb {B}_n)$ and any $0<p<\infty$, we have

$$\begin{aligned} \int \limits _{\mathbb {B}_n}|f(z)|^p\,d\mu (z) \lesssim \int \limits _{\mathbb {B}_n}|f(z)|^p(1-|z|^2)^{(n+1+\gamma )\lambda -(n+1)}\,dv(z). \end{aligned}$$

Proof

By [12, Lemma 2.24], we know that for $0<r<1$, we have

$$\begin{aligned} |f(z)|^p\le \frac{1}{(1-|z|^2)^{n+1}}\int \limits _{D(z,r)}|f(w)|^p\,dv(w). \end{aligned}$$

Hence, by Fubini’s theorem, the fact that $(1-|z|^2)\approx (1-|w|^2)$ for $z\in D(w,r)$, and Theorem A, we have that

$$\begin{aligned} \int \limits _{\mathbb {B}_n}|f(z)|^p\,\text {d}\mu (z)\le & {} \int \limits _{\mathbb {B}_n}\frac{1}{(1-|z|^2)^{n+1}}\int _{D(z,r)}|f(w)|^p\,\text {d}v(w)\,\text {d}\mu (z)\\= & {} \int \limits _{\mathbb {B}_n}|f(w)|^p\int \limits _{D(w,r)}\frac{\text {d}\mu (z)}{(1-|z|^2)^{n+1}}\,\text {d}v(w)\\\approx & {} \int \limits _{\mathbb {B}_n}|f(w)|^p\int \limits _{D(w,r)}\frac{\text {d}\mu (z)}{(1-|w|^2)^{n+1}}\,\text {d}v(w)\\= & {} \int \limits _{\mathbb {B}_n}|f(w)|^p\frac{\mu (D(w,r))}{(1-|w|^2)^{n+1}}\,\text {d}v(w)\\\lesssim & {} \int \limits _{\mathbb {B}_n}|f(w)|^p(1-|w|^2)^{(n+1+\gamma )\lambda -(n+1)}\,\text {d}v(w). \end{aligned}$$

The proof is complete. $\square$

2.2 Some useful estimates

The following estimate is well known, and can be found, for example, in [7, Proposition 1.4.10], [12, Theorem 1.12] and [4, Sect. 1.2].

Lemma 2.2

Suppose $z\in \mathbb {B}_n$, $t>-1$, and c is real. The integral

$$\begin{aligned} I_{c, t}(z)=\int \limits _{\mathbb {B}_n}\frac{(1-|w|^2)^t}{|1-\langle z, w\rangle |^{c}}dv(w) \end{aligned}$$

has the following asymptotic behavior as $|z|\rightarrow 1$.

• (i) If $c<n+1+t$, then $I_{c, t}(z)\asymp 1$.

• (ii) If $c=n+1+t$, then $I_{c, t}(z)\asymp \log \frac{1}{1-|z|^2}$.

• (iii) If $c>n+1+t$, then $I_{c, t}(z)\asymp (1-|z|^2)^{n+1+t-c}$.

Lemma 2.3

[6, Lemma C] Let $\{z_k\}$ be a separated sequence in $\mathbb {B}_n$, and $n<t<s$. Then

$$\begin{aligned} \sum _{k=1}^\infty \frac{(1-|z_k|^2)^t}{|1-\langle z, z_k\rangle |^s} \le C(1-|z|^2)^{t-s}, \quad \quad z\in \mathbb {B}_n. \end{aligned}$$

Lemma 2.4

Suppose $0<p<\infty ,\ \alpha >-1$. Let $\{c_j\}$ be a positive sequence, and let $\{a_j\}$ be a separated sequence in $\mathbb {B}_n$. If $s\in \mathbb {R}$, such that

$$\begin{aligned} s>n\max \left( 1, \frac{1}{p}\right) +\frac{1+\alpha }{p}, \end{aligned}$$

and f is a measurable function on $\mathbb {B}_n$, such that

$$\begin{aligned} |f(z)|\le \sum _{j=1}^\infty \frac{c_j}{|1-\langle z, a_j\rangle |^s}, \end{aligned}$$

then $f\in L_\alpha ^p$ and

$$\begin{aligned} \Vert f\Vert _{p, \alpha }^p\lesssim \sum _{j=1}^\infty \frac{c_j^p}{(1-|a_j|^2)^{sp-(n+1+\alpha )}}. \end{aligned}$$

Proof

If $0<p\le 1$, then we have that

$$\begin{aligned} |f(z)|^p\le \sum _{j=1}^\infty \frac{c_j^p}{|1-\langle z, a_j\rangle |^{sp}}. \end{aligned}$$

By Lemma 2.2, we have that

$$\begin{aligned} \int \limits _{\mathbb {B}_n}|f(z)|^p\text {d}v_\alpha (z)~&\le \sum _{j=1}^\infty c_j^p\int \limits _{\mathbb {B}_n}\frac{1}{|1-\langle z, a_j\rangle |^{sp}}\,\text {d}v_\alpha (z)\\ ~&\lesssim \sum _{j=1}^\infty \frac{c_j^p}{(1-|a_j|^2)^{sp-(n+1+\alpha )}}. \end{aligned}$$

If $p>1$, then $s>n+\frac{1+\alpha }{p}$. Let $p^\prime$ be the conjugate exponent of p, such that $1/p+1/p^\prime =1$. By Hölder’s inequality and Lemma 2.3, we have

$$\begin{aligned}&~ |f(z)|^p=\left( \sum _{j=1}^\infty \frac{c_j}{|1-\langle z, a_j\rangle |^s}\right) ^p\\ \le&~ \left( \sum _{j=1}^\infty \frac{(1-|a_j|^2)^{s-(1+\alpha )/p}}{|1-\langle z, a_j\rangle |^s}\right) ^{p-1} \left( \sum _{j=1}^\infty \frac{c_j^p(1-|a_j|^2)^{s(1-p)+(1+\alpha )/p^\prime }}{|1-\langle z, a_j\rangle |^{s}}\right) \\ \lesssim&~(1-|z|^2)^{-(1+\alpha )/p^\prime } \left( \sum _{j=1}^\infty \frac{c_j^p(1-|a_j|^2)^{s(1-p)+(1+\alpha )/p^\prime }}{|1-\langle z, a_j\rangle |^{s}}\right) . \end{aligned}$$

Hence

$$\begin{aligned} \Vert f\Vert _{p, \alpha }^p\lesssim \sum _{j=1}^\infty c_j^p(1-|a_j|^2)^{s(1-p)+(1+\alpha )/p^\prime } \int \limits _{\mathbb {B}_n}\frac{(1-|z|^2)^{-(1+\alpha )/p^\prime }}{|1-\langle z, a_j\rangle |^{s}}\text {d}v_\alpha (z). \end{aligned}$$

Since $\alpha -(1+\alpha )/p^\prime =(1+\alpha )/p-1>-1$, and

$$\begin{aligned} s-(n+1+\alpha )+(1+\alpha )/p^\prime =s-n-(1+\alpha )/p>0, \end{aligned}$$

the typical integral estimate in Lemma 2.2 gives the result. $\square$

3 Compactness of $B_{\mu }^\beta$ and $T_{\mu }^\beta$

Recall that, for a bounded linear operator T between two Banach spaces X and Y, we say that T is compact if T maps any bounded set in X to a relative compact set in Y. We also recall that a bounded linear operator $T: X\rightarrow Y$ is called completely continuous if, for every weakly convergent sequence $(x_{n})$ from X, the sequence $(Tx_{n})$ is norm-convergent in Y.

Let $-1<\beta <\infty$. Since $B_{\mu }^\beta$ is a sublinear operator, there may be different ways to define its compactness. In this paper, following the above definition, we say that $B_{\mu }^\beta :A^{p_1}_{\alpha _1}\rightarrow L^{p_2}_{\alpha _2}$ is compact if it maps any bounded set in $A^{p_1}_{\alpha _1}$ to a relative compact set in $L^{p_2}_{\alpha _2}$, where $0<p_1, p_2<\infty$, and $-1<\alpha _1, \alpha _2<\infty$. It is clear that $B_{\mu }^\beta :A^{p_1}_{\alpha _1}\rightarrow L^{p_2}_{\alpha _2}$ is compact if and only if for any bounded sequence $\{f_n\}$ in $A^{p_1}_{\alpha _1}$, the image sequence $\{B_{\mu }^\beta f_n\}$ has a convergent subsequence in $L^{p_2}_{\alpha _2}$.

We first give the following sufficient condition for the compactness of $B_{\mu }^\beta : A^{p_1}_{\alpha _1}\rightarrow L^{p_2}_{\alpha _2}$ for $0<p_1, p_2<\infty$.

Proposition 3.1

Let $0<p_1, p_2<\infty$, and $-1<\alpha _1, \alpha _2, \beta <\infty$. Assume that $B_{\mu }^\beta : A^{p_1}_{\alpha _1}\rightarrow L^{p_2}_{\alpha _2}$ is a bounded sublinear operator. Suppose that, for every bounded sequence $\{f_k\}$ in $A^{p_1}_{\alpha _1}$, such that $f_k\rightarrow 0$ uniformly on every compact subsets of $\mathbb {B}_n$ as $k\rightarrow \infty$, we have

$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert B_{\mu }^\beta f_k\Vert _{p_2,\alpha _2}=0. \end{aligned}$$

Then, $B_\mu ^\beta$ is compact from $A_{\alpha _1}^{p_1}$ to $L_{\alpha _2}^{p_2}$.

Proof

Let $\{f_k\}$ be a bounded sequence in $A^{p_1}_{\alpha _1}$. Then, there is a constant $M>0$, such that $\Vert f_k\Vert _{p_1,\alpha _1}\le M$ for all $k\ge 1$. By [12, Theorem 2.1], $\{f_k\}$ is uniformly bounded on every compact subsets of $\mathbb {B}_n$. By Montel’s Theorem, there is a subsequence of $\{f_k\}$, denoted by $\{f_{k_j}\}$, $j=1,2,3...$, such that $f_{k_j}\rightarrow f$ uniformly on every compact subsets of $\mathbb {B}_n$ for some holomorphic function f on $\mathbb {B}_n$, as $j\rightarrow \infty$. By Fatou’s Lemma

$$\begin{aligned} \int \limits _{\mathbb {B}_n}|f(z)|^{p_1}\,\text {d}v_{\alpha _1}(z)= & {} \int \limits _{\mathbb {B}_n}\lim _{j\rightarrow \infty }|f_{k_j}(z)|^{p_1}\,\text {d}v_{\alpha _1}(z)\\\le & {} \lim _{j\rightarrow \infty }\int \limits _{\mathbb {B}_n}|f_{k_j}(z)|^{p_1}\,\text {d}v_{\alpha _1}(z)\\\le & {} \lim _{j\rightarrow \infty }\Vert f_{k_j}\Vert _{p_1,\alpha _1}^{p_1}\le M. \end{aligned}$$

Thus, $f\in A^{p_1}_{\alpha _1}$. Therefore, we get that $f_{k_j}-f\rightarrow 0$ uniformly on every compact subsets of $\mathbb {B}_n$ as $j\rightarrow \infty$. By our assumption, we get that

$$\begin{aligned} \lim _{j\rightarrow \infty }\Vert B_{\mu }^\beta (f_{k_j}-f)\Vert _{p_2, \alpha _2}=0. \end{aligned}$$

We can easily check that

$$\begin{aligned} \Vert B_{\mu }^\beta f_{k_j}-B_{\mu }^\beta f\Vert _{p_2,\alpha _2} \le \Vert B_{\mu }^\beta ( f_{k_j}-f)\Vert _{p_2, \alpha _2}. \end{aligned}$$

From this inequality, we obtain that

$$\begin{aligned} \lim _{j\rightarrow \infty }\Vert B_{\mu }^\beta f_{k_j}-B_{\mu }^\beta f\Vert _{p_2,\alpha _2}=0, \end{aligned}$$

which implies that $B_{\mu }^\beta f\in L^{p_2}_{\alpha _2}$. Thus, $\{B_{\mu }^\beta f_k\}$ has a convergent subsequence in $L^{p_2}_{\alpha _2}$, and so $B_{\mu }^\beta$ is compact from $A_{\alpha _1}^{p_1}$ to $L_{\alpha _2}^{p_2}$. $\square$

The following characterization for compactness of $T_{\mu }^\beta : A_{\alpha _1}^{p_1}\rightarrow A_{\alpha _2}^{p_2}$ for $0<p_1\le p_2<\infty$ may be well known, but we cannot find a reference, so we give a proof here. The result contains the case when $0<p_1<1$ or $0<p_2<1$, in which we still define the compactness of $T_{\mu }^\beta$ in the same way as before, that is, we say that $T_{\mu }^\beta : A_{\alpha _1}^{p_1}\rightarrow A_{\alpha _2}^{p_2}$ is compact if it maps a bounded set in $A_{\alpha _1}^{p_1}$ to a relatively compact set in $A_{\alpha _2}^{p_2}$. For the case when $0<p_1\le 1$, the proof below is surprisingly involved.

Proposition 3.2

Let $0<p_1\le p_2<\infty$, and let $-1<\alpha _1,\alpha _2,\beta <\infty$. Suppose that $p_2$, $\alpha _2$ and $\beta$ satisfy (C-2), and suppose that $T_{\mu }^\beta$ is bounded from $A_{\alpha _1}^{p_1}$ to $A_{\alpha _2}^{p_2}$. Then, the following statements are equivalent:

• (i) $T_{\mu }^\beta$ is compact from $A_{\alpha _1}^{p_1}$ to $A_{\alpha _2}^{p_2}$.

• (ii) For every bounded sequence $\{f_k\}$ in $A^{p_1}_{\alpha _1}$, such that $f_k\rightarrow 0$ uniformly on every compact subsets of $\mathbb {B}_n$ as $k\rightarrow \infty$, we have

  $$\begin{aligned} \lim _{k\rightarrow \infty }\Vert T_{\mu }^{\beta }f_k\Vert _{{p_2},{\alpha _2}}=0. \end{aligned}$$

We need several lemmas to prove Proposition 3.2.

Lemma 3.3

Suppose that $1<p<\infty$. Then, $f_k\rightarrow 0$ weakly in $A^p_\alpha$ if and only if $\{f_k\}$ is bounded in $A_\alpha ^p$ and $f_k\rightarrow 0$ uniformly on every compact subsets of $\mathbb {B}_n$.

This result is well known and can be easily proved, so we omit the proof here. For the case of the unit disk, see Problem 1 of Exercise 4.7 in [13].

Lemma 3.4

Let $0<p<\infty$, and let $-1<\alpha <\infty$. Let $1\le \lambda <\infty$ and $-1<\gamma <\infty$ satisfy that

$$\begin{aligned} (n+1+\gamma )\lambda >n\max \bigg (1, \frac{1}{p}\bigg )+\frac{1+\alpha }{p}. \end{aligned}$$

(3.1)

Let $\mu$ be a $(\lambda ,\gamma )$-Bergman–Carleson measure on $\mathbb {B}_n$. If $\{f_k\}$ is a bounded sequence in $A^{p}_{\alpha }$, such that $f_k\rightarrow 0$ uniformly on every compact subsets of $\mathbb {B}_n$ as $k\rightarrow \infty$, then we also have that $B_{\mu }^{\beta }f_k\rightarrow 0$ and $T_{\mu }^{\beta }f_k\rightarrow 0$ uniformly on every compact subsets of $\mathbb {B}_n$ as $k\rightarrow \infty$.

Proof

It suffices for us to prove $B_{\mu }^{\beta }f_k\rightarrow 0$ on every compact subsets of $\mathbb {B}_n$ as $k\rightarrow \infty$, since $|T_\mu f_k|\le B_\mu f_k$. For convenience, denote $\eta =(n+1+\gamma )\lambda -(n+1)$. Since $\lambda \ge 1$ and $\gamma >-1$, we have $\eta >-1$. Let $\{f_k\}$ be a bounded sequence in $A^{p}_{\alpha }$. Then, there is a constant $M>0$, such that $\Vert f_k\Vert _{p,\alpha }\le M$ for all $k\ge 1$.

First, consider the case $0<p\le 1$. In this case, (3.1) becomes

$$\begin{aligned} \frac{n+1+\alpha }{p}<n+1+\eta , \end{aligned}$$

(3.2)

Then, there exists A, such that

$$\begin{aligned} \frac{n+1+\alpha }{p}<A<n+1+\eta . \end{aligned}$$

(3.3)

Since $0<p\le 1$, we get

$$\begin{aligned} \frac{1+\alpha }{p}\le \frac{n+1+\alpha }{p}-n<A-n<1+\eta . \end{aligned}$$

Therefore, there exists a constant B, such that $A-n<B<\min \{1+\eta , A\}$, which implies that

$$\begin{aligned} \frac{1+\alpha }{p}<B<1+\eta . \end{aligned}$$

(3.4)

Let

$$\begin{aligned} q=\frac{n}{A-B},\qquad s=\frac{nB}{A-B}-1. \end{aligned}$$

Then, since $0<A-B<n$, we see that $q>1$ and $s>-1$. Also, it is easy to check that

$$\begin{aligned} \frac{n+1+s}{q}=A, \qquad \frac{1+s}{q}=B. \end{aligned}$$

Hence, by (3.3) and (3.4), we obtain that

$$\begin{aligned} \frac{n+1+\alpha }{p}<\frac{n+1+s}{q}<n+1+\eta \end{aligned}$$

(3.5)

and

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

(3.6)

By [11, Theorem 69], (3.5) implies that $A^p_{\alpha }\subseteq A^q_s$, and so $\Vert f_k\Vert _{q,s}\lesssim \Vert f_k\Vert _{p,\alpha }\le M$. Using (3.6), we get that

$$\begin{aligned} \left( \eta -\frac{s}{q}\right) q'>-1, \end{aligned}$$

where $q'$ is the conjugate index of q, that is, it satisfies $1/q+1/q'=1$. Hence,

$$\begin{aligned} \int \limits _{\mathbb {B}_n}(1-|w|^2)^{(\eta -s/q)q'}\,\text {d}v(w)<\infty . \end{aligned}$$

Therefore, for any $\varepsilon >0$, there exists a constant $r\in (0, 1)$, such that

$$\begin{aligned} \int \limits _{\mathbb {B}_n\setminus \overline{D}_r}(1-|w|^2)^{(\eta -s/q)q'}\,\text {d}v(w)<\varepsilon ^{q'}, \end{aligned}$$

(3.7)

where $D_r=\{w\in \mathbb {B}_n:\, |w|<r\}$. By Lemma 2.1, we get that

$$\begin{aligned} |B^{\beta }_{\mu }f_k(z)|\lesssim & {} \int \limits _{\mathbb {B}_n}\frac{|f_k(w)|(1-|w|^2)^{(n+1+\gamma )\lambda -(n+1)}}{|1-\langle z,w\rangle |^{n+1+\beta }}\,\text {d}v(w)\nonumber \\= & {} \left( \int \limits _{\mathbb {B}_n\setminus \overline{D}_r}+\int \limits _{\overline{D}_r}\right) \frac{|f_k(w)|(1-|w|^2)^{\eta }}{|1-\langle z,w\rangle |^{n+1+\beta }}\,\text {d}v(w)\nonumber \\= & {} I_1(k,r)+I_2(k,r). \end{aligned}$$

(3.8)

Using Hölder’s inequality and (3.7), we get that

$$\begin{aligned} I_1(k,r)\le & {} \left( \int \limits _{\mathbb {B}_n\setminus \overline{D}_r}|f_k(w)|^q(1-|w|^2)^s\,\text {d}v(w)\right) ^{1/q}\\{} & {} \times \left( \int \limits _{\mathbb {B}_n\setminus \overline{D}_r} \frac{(1-|w|^2)^{(\eta -s/q)q'}}{|1-\langle z,w\rangle |^{(n+1+\beta )q'}}\,\text {d}v(w)\right) ^{1/q'}\\\lesssim & {} \frac{\Vert f_k\Vert _{q,s}}{(1-|z|)^{(n+1+\beta )q'}} \left( \int \limits _{\mathbb {B}_n\setminus \overline{D}_r}(1-|w|^2)^{(\eta -s/q)q'}\,\text {d}v(w)\right) ^{1/q'}\\\le & {} \frac{M\varepsilon }{(1-|z|)^{(n+1+\beta )q'}}. \end{aligned}$$

Take any compact subset K in $\mathbb {B}_n$. Then, for any $z\in K$, there is a constant $M'>0$, such that $1/(1-|z|)^{(n+1+\beta )q'}\le M'$. Thus

$$\begin{aligned} I_1(k,r)\lesssim MM'\varepsilon . \end{aligned}$$

(3.9)

Since $f_k\rightarrow 0$ uniformly on every compact subsets of $\mathbb {B}_n$ as $k\rightarrow \infty$, there exists an integer $N>0$, such that for any $k\ge N$ and any $w\in \overline{D}_r$

$$\begin{aligned} |f_k(w)|<\varepsilon . \end{aligned}$$

Remembering that $\eta >-1$, we have for any $z\in K$

$$\begin{aligned} I_2(k,r) \le \frac{\varepsilon }{(1-|z|)^{n+1+\beta }}\int \limits _{\overline{D}_r}(1-|w|^2)^{\eta }\,\text {d}v(w) \lesssim M'\varepsilon . \end{aligned}$$

(3.10)

By (3.9) and (3.10), we get that for any $k\ge N$ and any $z\in K$

$$\begin{aligned} |B^{\beta }_{\mu }f_k(z)|\le I_1(k,r)+I_2(k,r)\lesssim \varepsilon . \end{aligned}$$

Thus, $B_{\mu }^{\beta }f_k(z)\rightarrow 0$ uniformly on any compact subsets of $\mathbb {B}_n$ as $k\rightarrow \infty$ for $0<p\le 1$.

Next, consider the case $p>1$. In this case, (3.1) becomes

$$\begin{aligned} (n+1+\gamma )\lambda >n+\frac{1+\alpha }{p}, \end{aligned}$$

Recall that $\eta =(n+1+\gamma )\lambda -(n+1)$. Thus

$$\begin{aligned} \eta +1=(n+1+\gamma )\lambda -n>\frac{1+\alpha }{p}, \end{aligned}$$

which implies that

$$\begin{aligned} \left( \eta -\frac{\alpha }{p}\right) p'>-1, \end{aligned}$$

where $1/p+1/p'=1$. Hence

$$\begin{aligned} \int \limits _{\mathbb {B}_n}(1-|w|^2)^{(\eta -\alpha /p)p'}\,\text {d}v(w)<\infty . \end{aligned}$$

The rest of the proof is the same as the proof of the case $0<p\le 1$, except that we use p to replace q, and use $\alpha$ to replace s. We omit the details. $\square$

Lemma 3.5

Let $0<p_1\le p_2<\infty$, let $-1<\alpha _1,\alpha _2,\beta <\infty$, and let $\gamma , \lambda$ be given by (1.2). Suppose that $p_2$, $\alpha _2$ and $\beta$ satisfy (C-2). Then, we have

$$\begin{aligned} (n+1+\gamma )\lambda >n\max \left( 1,\frac{1}{p_1}\right) +\frac{1+\alpha _1}{p_1}. \end{aligned}$$

(3.11)

Proof

Since $0<p_1\le p_2<\infty$, it follows that $\lambda =1+1/p_1-1/p_2\ge 1$. We get the following two inequalities:

$$\begin{aligned} n+1+\beta >\frac{n+1+\alpha _2}{p_2} \end{aligned}$$

(3.12)

and

$$\begin{aligned} 1+\beta >\frac{1+\alpha _2}{p_2} \end{aligned}$$

(3.13)

by (C-2). Bearing in mind the definitions of $\lambda , \gamma$ in (1.2), then (3.13) gives that

$$\begin{aligned} (1+\gamma )\lambda =(1+\beta )+\frac{1+\alpha _1}{p_1}-\frac{1+\alpha _2}{p_2}>\frac{1+\alpha _1}{p_1}>0, \end{aligned}$$

Thus, $\gamma >-1$, and

$$\begin{aligned} (n+1+\gamma )\lambda> & {} n+\frac{n+1+\alpha _1}{p_1}-\frac{n}{p_2}\\\ge & {} n+\frac{1+\alpha _1}{p_1}, \end{aligned}$$

since $0<p_1\le p_2<\infty$. Furthermore, the inequality (3.12) implies that

$$\begin{aligned} (n+1+\gamma )\lambda =(n+1+\beta )+\frac{n+1+\alpha _1}{p_1}-\frac{n+1+\alpha _2}{p_2} >\frac{n+1+\alpha _1}{p_1}. \end{aligned}$$

The proof is complete. $\square$

Proof of Proposition 3.2

It follows from [2, Proposition 3.3 in Chapter VI] that $T_\mu : A_{\alpha _1}^{p_1}\rightarrow A_{\alpha _2}^{p_2}$ is compact if and only if $T_\mu : A_{\alpha _1}^{p_1}\rightarrow A_{\alpha _2}^{p_2}$ is completely continuous for $1<p_1<\infty$. By Lemma 3.3, we know that (ii) is equivalent to that $T_\mu : A_{\alpha _1}^{p_1}\rightarrow A_{\alpha _2}^{p_2}$ is completely continuous for $1<p_1<\infty$. Therefore, it suffices for us to prove for the case $0<p_1\le 1$.

Let $0<p_1\le 1$, and let $0<p_1\le p_2<\infty$. The proof of (ii)$\Rightarrow$(i) follows from the same discussion as in the proof of Proposition 3.1. Thus, we only need to prove that (i)$\Rightarrow$(ii). Suppose that (i) holds, i.e., that $T_{\mu }^\beta : A_{\alpha _1}^{p_1}\rightarrow A_{\alpha _2}^{p_2}$ is compact. Then, $T_{\mu }^\beta : A_{\alpha _1}^{p_1}\rightarrow A_{\alpha _2}^{p_2}$ is bounded. It follows from Theorem 1 that $\mu$ is a $(\lambda ,\gamma )$-Bergman–Carleson measure on $\mathbb {B}_n$, where $\lambda , \gamma$ are given by (1.2). Let $\{f_k\}$ be a bounded sequence in $A^{p_1}_{\alpha _1}$, such that $f_k\rightarrow 0$ uniformly on every compact subsets of $\mathbb {B}_n$ as $k\rightarrow \infty$. Suppose, on the contrary, that (ii) is not true. Then, there exist an $\varepsilon >0$ and a subsequence $\{f_{k_j}\}$ of $\{f_k\}$, such that

$$\begin{aligned} \Vert T_{\mu }^{\beta }f_{k_j}\Vert _{p_2,\alpha _2}\ge \varepsilon , \quad \text {for all }j=1,2,3.... \end{aligned}$$

(3.14)

Since $T_{\mu }^{\beta }$ is compact from $A_{\alpha _1}^{p_1}$ to $A_{\alpha _2}^{p_2}$, we can find a further subsequence $\{f_{k_{j_m}}\}$, $m=1,2,3...$, and $g\in A^{p_2}_{\alpha _2}$, such that

$$\begin{aligned} \lim _{m\rightarrow \infty }\Vert T_{\mu }^{\beta }f_{k_{j_m}}-g\Vert _{p_2,\alpha _2}=0. \end{aligned}$$

(3.15)

By [12, Theorem 2.1], we have that

$$\begin{aligned} |T_{\mu }^{\beta }f_{k_{j_m}}(z)-g(z)| \le \frac{\Vert T_{\mu }^{\beta }f_{k_{j_m}}-g\Vert _{p_2,\alpha _2}}{(1-|z|^2)^{(n+1+\alpha _2)/p_2}} \end{aligned}$$

(3.16)

for all $m\ge 1$. Hence

$$\begin{aligned} |T_{\mu }^{\beta }f_{k_{j_m}}(z)-g(z)|\rightarrow 0 \end{aligned}$$

(3.17)

uniformly on every compact subsets of $\mathbb {B}_n$, as $m\rightarrow \infty$.

By the definitions of $\lambda , \gamma$ given in (1.2), and by lemma 3.5, we have that

$$\begin{aligned} (n+1+\gamma )\lambda >\frac{n+1+\alpha _1}{p_1} \end{aligned}$$

for $0<p_1\le 1$. Since $\{f_k\}$ is a bounded sequence in $A^{p_1}_{\alpha _1}$ and $f_{k_{j_m}}(z)\rightarrow 0$ uniformly on every compact subset of $\mathbb {B}_n$ as $m\rightarrow \infty$, it follows from Lemma 3.4 that $T_{\mu }^{\beta }f_{k_{j_m}}(z)\rightarrow 0$ uniformly on compact subsets of $\mathbb {B}_n$. Thus, we must have $g=0$ by (3.17). Therefore, by (3.15), we get that

$$\begin{aligned} \lim _{m\rightarrow \infty }\Vert T_{\mu }^{\beta }f_{k_{j_m}}\Vert _{p_2,\alpha _2}=0, \end{aligned}$$

which contradicts to (3.14). Hence, (ii) must be true. The proof is complete. $\square$

4 Proofs of the main theorems

Lemma 4.1

Let $0<p_1, p_2< \infty$, let $-1<\alpha _1, \alpha _2, \beta <\infty$, and let $\lambda$ and $\gamma$ be given by (1.2). Suppose that $p_2$, $\alpha _2$ and $\beta$ satisfy (C-2). Given $0<r<1$, let $\{a_j\}$ be an r-lattice in $\mathbb {B}_n$, and let $D_j$ and $\tilde{D}_j$ be the associated Bergman metric balls given by Lemma 1.3. Then, we have

$$\begin{aligned} \Vert B_{\mu }^\beta (f)\Vert _{p_2, \alpha _2}^{p_2}\lesssim \sum _{j=1}^\infty \left( \frac{\mu (D_j)}{(1-|a_j|^2)^{(n+1+\gamma )\lambda }}\right) ^{p_2} \left( \int \limits _{\tilde{D}_j}|f(\zeta )|^{p_1}dv_{\alpha _1}(\zeta )\right) ^{p_2/p_1}. \end{aligned}$$

(4.1)

Proof

Using the fact that $|1-\langle z, w\rangle |\asymp |1-\langle z, a_j\rangle |$ for $w\in D_j$, we have that

$$\begin{aligned} |B_{\mu }^\beta (f)(z)|&~\lesssim \sum _{j=1}^\infty \int \limits _{D_j}\frac{|f(w)|}{|1-\langle z, w\rangle |^{n+1+\beta }}\text {d}\mu (w)\\&~\lesssim \sum _{j=1}^\infty \left( \sup _{w\in D_j}|f(w)|\right) \int \limits _{D_j}\frac{1}{|1-\langle z, w\rangle |^{n+1+\beta }}\, \text {d}\mu (w)\\&~\lesssim \sum _{j=1}^\infty \left( \sup _{w\in D_j}|f(w)|\right) \frac{\mu (D_j)}{|1-\langle z, a_j\rangle |^{n+1+\beta }}. \end{aligned}$$

By [12, Lemma 2.24], we have that

$$\begin{aligned} |f(w)|\lesssim \left( \frac{1}{(1-|a_j|^2)^{n+1+\alpha _1}} \int \limits _{\tilde{D}_j}|f(\zeta )|^{p_1}\text {d}v_{\alpha _1}(\zeta )\right) ^{1/p_1} \end{aligned}$$

for any $w\in D_j$. Denote

$$\begin{aligned} |\widehat{f}(a_j)|:=\left( \frac{1}{(1-|a_j|^2)^{n+1+\alpha _1}} \int \limits _{\tilde{D}_j}|f(\zeta )|^{p_1}\text {d}v_{\alpha _1}(\zeta )\right) ^{1/p_1}, \end{aligned}$$

we get that

$$\begin{aligned} |B_{\mu }^\beta (f)(z)| \lesssim \sum _{j=1}^\infty \frac{|\widehat{f}(a_j)|\mu (D_j)}{|1-\langle z, a_j\rangle |^{n+1+\beta }}. \end{aligned}$$

(4.2)

Thus, we obtain (4.1) by taking $p=p_2$, $\alpha =\alpha _2$, $s=n+1+\beta$ and $c_j=|\widehat{f}(a_j)|\mu (D_j)$ in Lemma 2.4. $\square$

Lemma 4.2

Given $0<r<1$, let $\{a_j\}$ be an r-lattice in $\mathbb {B}_n$, let $D_j=D(a_j,r)$ be the associated Bergman metric balls given by Lemma 1.3, and let $\{b_k(a_j)\}$ be a sequence depending on $a_j$, such that

$$\begin{aligned} \sup _{k}\sum \limits _{j=1}^\infty |b_k(a_j)|<\infty . \end{aligned}$$

(4.3)

Suppose, in addition, for any compact subset K of $\mathbb {B}_n$, the sequence $\{b_k(a_j)\}$ satisfies that

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\sum \limits _{j\in \Gamma }|b_k(a_j)|=0, \end{aligned}$$

(4.4)

where $\Gamma :=\{j: a_j\in K\}$. Let $\{c(a_j)\}$ be a sequence of real numbers depending on $\{a_j\}$. Then, we have the following two results.

• (i) If $\{c(a_j)\}$ is a bounded sequence, such that $\lim _{|a_j|\rightarrow 1}c(a_j)=0$, then

  $$\begin{aligned} \lim _{k\rightarrow \infty }\sum _{j=1}^\infty c(a_j)b_k(a_j)=0. \end{aligned}$$

• (ii) Let $0<t<\infty$, $0<s<1$ and $\gamma =t/(1-s)$. If $\{c(a_j)\}\in l^\gamma$, then

  $$\begin{aligned} \lim _{k\rightarrow \infty } \sum _{j=1}^\infty c(a_j)^tb_k(a_j)^{s}=0. \end{aligned}$$

Proof

(i) Since $\lim _{|a_j|\rightarrow 1}c(a_j)=0$, it follows that for any $\varepsilon >0$, there is an $r_1\in (0,1)$, such that $|c(a_j)|<\varepsilon$ for all $|a_j|>r_1$. Therefore

$$\begin{aligned}{} & {} \left| \sum _{j=1}^\infty c(a_j)b_k(a_j)\right| \nonumber \\{} & {} \qquad \le \sum _{j:\, |a_j|\le r_1}|c(a_j)b_k(a_j)|+\sum _{j:\, |a_j|> r_1}|c(a_j)b_k(a_j)|\nonumber \\{} & {} \qquad \le \sup _{j:\, |a_j|\le r_1}|c(a_j)|\sum _{j:\, |a_j|\le r_1}|b_k(a_j)|+\varepsilon \sum _{j:\, |a_j|>r_1}|b_k(a_j)|. \end{aligned}$$

(4.5)

Since the set $\{a: |a|\le r_1\}$ is a compact subset of $\mathbb {B}_n$, by (4.4), we get that

$$\begin{aligned} \lim _{k\rightarrow \infty }\sum _{j:\, |a_j|\le r_1}|b_k(a_j)|=0. \end{aligned}$$

Letting $\varepsilon \rightarrow 0$ and then letting $k\rightarrow \infty$ in (4.5), we obtain the result in (i).

(ii) Since $\{c(a_j)\}\in l^\gamma$, it follows that, for any $\varepsilon >0$, there is an $r_2\in (0,1)$, such that

$$\begin{aligned} \sum _{j:\, |a_j|>r_2} |c(a_j)|^\gamma <\varepsilon . \end{aligned}$$

Since $\{a: |a|\le r_2\}$ is a compact subset of $\mathbb {B}_n$, by (4.4), we have

$$\begin{aligned} \lim _{k\rightarrow \infty }\sum _{j:\, |a_j|\le r_2}|b_k(a_j)|=0. \end{aligned}$$

Since $1/s>1$, by Hölder inequality, we have that

$$\begin{aligned} \left| \sum _{j=1}^\infty c(a_j)^tb_k(a_j)^s\right|\le & {} \sum _{j:\, |a_j|\le r_2}|c(a_j)|^t|b_k(a_j)|^s+\sum _{j:\, |a_j|> r_2}|c(a_j)|^t|b_k(a_j)|^s\\\le & {} \left( \sum _{j:\, |a_j|\le r_2}|c(a_j)|^{\gamma }\right) ^{1-s} \left( \sum _{j:\, |a_j|\le r_2} |b_k(a_j)|\right) ^s\\{} & {} + \left( \sum _{j:\, |a_j|>r_2}|c(a_j)|^{\gamma }\right) ^{1-s} \left( \sum _{j:\, |a_j|> r_2} |b_k(a_j)|\right) ^s\\\le & {} \left( \sum _{j:\, |a_j|\le r_2}|c(a_j)|^{\gamma }\right) ^{1-s} \left( \sum _{j:\, |a_j|\le r_2} |b_k(a_j)|\right) ^s\\{} & {} +\varepsilon ^{1-s}\sum _{j:\, |a_j|> r_2}|b_k(a_j)|. \end{aligned}$$

Letting $\varepsilon \rightarrow 0$ and then letting $k\rightarrow \infty$, we obtain the result in (ii). $\square$

4.1 Proofs of Theorem 1.1 and Theorem 1.2

The implications (i)$\Rightarrow$(ii) in Theorem 1.1 and Theorem 1.2 are obvious. The implications (ii)$\Rightarrow$(iii) in Theorem 1.1 and Theorem 1.2 are given by [6, Theorem 1.2] and [6, Theorem 4.2], respectively. Thus, we only need to prove that (iii)$\Rightarrow$(i) in these two Theorems. Given $0<r<1$, let $\{a_j\}$ be an r-lattice in $\mathbb {B}_n$, and let $D_j$ and $\tilde{D}_j$ be the associated Bergman metric balls given by Lemma 1.3.

(iii)$\Rightarrow$(i) for Theorem 1.1. Suppose that $\mu$ is a $(\lambda , \gamma )$-Bergman–Carleson measure. Since the condition $0<p_1\le p_2 <\infty$ implies that $\lambda >1$ and $p_2/p_1\ge 1$, it follows from (4.1) and Theorem A. that:

$$\begin{aligned} \Vert B_{\mu }^\beta (f)\Vert _{p_2, \alpha _2}^{p_2}\lesssim & {} \Vert \mu \Vert _{\lambda ,\gamma }^{p_2} \sum _{j=1}^\infty \left( \int \limits _{\tilde{D}_j}|f(\zeta )|^{p_1}\text {d}v_{\alpha _1}(\zeta )\right) ^{p_2/p_1}\\\lesssim & {} \Vert \mu \Vert _{\lambda , \gamma }^{p_2}\left( \sum _{j=1}^\infty \int \limits _{\tilde{D}_j}|f(\zeta )|^{p_1} \text {d}v_{\alpha _1}(\zeta )\right) ^{p_2/p_1}\\\lesssim & {} \Vert \mu \Vert _{\lambda , \gamma }^{p_2}\Vert f\Vert _{p_1, \alpha _1}^{p_2}. \end{aligned}$$

Hence, $B_{\mu }^\beta$ is bounded from $A_{\alpha _1}^{p_1}$ to $L_{\alpha _2}^{p_2}$.

(iii)$\Rightarrow$(i) for Theorem 1.2. Suppose that $\mu$ is a vanishing $(\lambda , \gamma )$-Bergman–Carleson measure. It follows from Proposition 3.1 that we only need to show that $\Vert B_{\mu }^\beta f_k\Vert _{p_2, \alpha _2}\rightarrow 0$ for any bounded sequence $\{f_k\}$ in $A_{\alpha _1}^{p_1}$ converging to 0 uniformly on compact subsets of $\mathbb {B}_n$. Let

$$\begin{aligned} c(a_j)=\left( \frac{\mu (D_j)}{(1-|a_j|^2)^{(n+1+\gamma )\lambda }}\right) ^{p_2}, \end{aligned}$$

and let

$$\begin{aligned} b_k(a_j)=\left( \int \limits _{\tilde{D}_j}|f_k(\zeta )|^{p_1}\text {d}v_{\alpha _1}(\zeta )\right) ^{p_2/p_1}. \end{aligned}$$

Using (4.1) again, we get

$$\begin{aligned} \Vert B_{\mu }^\beta (f_k)\Vert _{p_2, \alpha _2}^{p_2} \lesssim \sum _{j=1}^\infty c(a_j)b_k(a_j). \end{aligned}$$

(4.6)

Since $p_2/p_1\ge 1$ and $\{f_k\}$ is a bounded sequence in $A_{\alpha _1}^{p_1}$, it follows that:

$$\begin{aligned} \sup _k\sum _{j=1}^\infty |b_k(a_j)|= & {} \sup _k \sum _{j=1}^\infty \left( \int \limits _{\tilde{D}_j}|f_k(\zeta )|^{p_1}\text {d}v_{\alpha _1}(\zeta )\right) ^{p_2/p_1}\\\lesssim & {} \sup _k\Vert f_k\Vert _{p_1, \alpha _1}^{p_2}<\infty . \end{aligned}$$

Let K be any compact subset in $\mathbb {B}_n$ and $\Gamma$ be given as in Lemma 4.2. Then, $\Gamma$ is a finite set. Since $\{f_k\}$ converges to 0 uniformly on compact subsets of $\mathbb {B}_n$, it follows that:

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\sum \limits _{j\in \Gamma }|b_k(a_j)|= & {} \lim \limits _{k\rightarrow \infty }\sum \limits _{j\in \Gamma } \left( \int \limits _{\tilde{D}_j}|f_k(\zeta )|^{p_1}\text {d}v_{\alpha _1}(\zeta )\right) ^{p_2/p_1} =0. \end{aligned}$$

By Theorem B., we have $\lim _{|a_j|\rightarrow 1}c(a_j)=0$. Therefore, by (i) of Lemma 4.2, we get that $\lim _{k\rightarrow \infty }\Vert B_{\mu }^\beta (f_k)\Vert _{p_2, \alpha _2}=0$. The proof is complete.

4.2 Proof of Theorem 1.4

We prove this theorem by showing that

(i)$\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (v) $\Rightarrow$ (i) $\Rightarrow$ (iv) $\Rightarrow$ (iii).

The implications (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) and (i) $\Rightarrow$ (iv) $\Rightarrow$ (iii) are trivial. Notice that the condition $0<p_2<p_1<\infty$ is equivalent to $-\infty<\lambda <1$, and the proof of Case 2 in (i) $\Rightarrow$ (ii) of [6, Theorem 1.2] actually works for showing our implication (iii) $\Rightarrow$ (v), with condition (v) here replacing the $(\lambda ,\gamma )$-Bergman–Carleson measure condition in [6]. Therefore, we only need to prove that (v)$\Rightarrow$(i).

(v) $\Rightarrow$ (i). Given $0<r<1$, let $\{a_j\}$ be an r-lattice in $\mathbb {B}_n$, and let $D_j$ and $\tilde{D}_j$ be the associated Bergman metric balls given by Lemma 1.3. Assume that (v) holds. It follows from Proposition 3.1 that we need only show that $\Vert B_{\mu }^\beta f_k\Vert _{p_2, \alpha _2}\rightarrow 0$ for any bounded sequence $\{f_k\}$ in $A_{\alpha _1}^{p_1}$ converging to zero uniformly on compact subsets of $\mathbb {B}_n$. We follow a similar argument as in the proof of (iii) $\Rightarrow$ (i) in Theorem 1.2. Denote by

$$\begin{aligned} c(a_j)=\frac{\mu (D_j)}{(1-|a_j|^2)^{(n+1+\gamma )\lambda }}, \ b_k(a_j)=\int \limits _{\tilde{D}_j}|f_k(\zeta )|^{p_1}\text {d}v_{\alpha _1}(\zeta ). \end{aligned}$$

As in the proof of Theorem 1.2, we know that the sequence $\{b_k(a_j)\}$ satisfies (4.3) and (4.4) in Lemma 4.2. Let $t=p_2$, and let $0<s=p_2/p_1<1$. Then

$$\begin{aligned} \gamma =\frac{t}{1-s}=\frac{p_1p_2}{p_1-p_2}=\frac{1}{1-\lambda }. \end{aligned}$$

By (v), we see that $\{c(a_j)\}\in l^{\gamma }$. Thus, by Lemma 4.2, we get that

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\Vert B_{\mu }(f_k)\Vert _{p_2, \alpha _2}=0. \end{aligned}$$

The proof is complete.

5 The case when $p_2=\infty$

In this section, we study the boundedness and compactness of $B_\mu ^\beta : A_{\alpha _1}^{p_1}\rightarrow L^\infty$ and $T_\mu ^\beta : A_{\alpha _1}^{p_1}\rightarrow H^\infty$ for $0<p_1<\infty$.

Proposition 5.1

Let $0<p_1<\infty$ and let $-1<\alpha _1, \beta <\infty$ satisfy (C-1). Let

$$\begin{aligned} \lambda =1+\frac{1}{p_1}, \quad \quad \gamma =\frac{1}{\lambda }\left( \beta +\frac{\alpha _1}{p_1}\right) . \end{aligned}$$

(5.1)

If $T_{\mu }^\beta$ is bounded from $A_{\alpha _1}^{p_1}$ to $H^\infty$, then $\mu$ is a $(\lambda , \gamma )$-Bergman–Carleson measure.

Proof

For any fixed $a\in \mathbb {B}_n$, take function

$$\begin{aligned} f_{a}(z)=\frac{(1-|a|^2)^{n+1+\beta -(n+1+\alpha _1)/p_1}}{(1-\langle z, a\rangle )^{n+1+\beta }}. \end{aligned}$$

Then, the condition (C-1) and Lemma 2.2 give that $f\in A_{\alpha _1}^{p_1}$ and $\Vert f\Vert _{p_1, \alpha _1}\asymp 1$. Since $|1-\langle w, a\rangle |\asymp 1-|a|^2$ for any $w\in D(a, r)$, it follows that:

$$\begin{aligned} T_{\mu }^\beta f_{a}(a)~&=(1-|a|^2)^{n+1+\beta -(n+1+\alpha _1)/p_1}\int \limits _{\mathbb {B}_n}\frac{\text {d}\mu (w)}{|1-\langle a, w\rangle |^{2(n+1+\beta )}}\nonumber \\ ~&\ge (1-|a|^2)^{n+1+\beta -(n+1+\alpha _1)/p_1}\int \limits _{D(a, r)}\frac{\text {d}\mu (w)}{|1-\langle a, w\rangle |^{2(n+1+\beta )}}\nonumber \\ ~&\ge C\frac{\mu (D(a, r))}{(1-|a|^2)^{(n+1+\gamma )\lambda }}. \end{aligned}$$

(5.2)

The boundedness of Toeplitz operator $T_{\mu }^\beta : A_{\alpha _1}^{p_1}\rightarrow H^\infty$ gives that

$$\begin{aligned} |T_{\mu }^\beta f_{a}(a)|\le \Vert T_{\mu }^\beta f_a\Vert _{\infty }\le \Vert T_{\mu }^\beta \Vert \Vert f_a\Vert _{p_1, \alpha _1}. \end{aligned}$$

Therefore, we get that

$$\begin{aligned} \mu (D(a, r))~&\lesssim \Vert T_{\mu }^\beta \Vert (1-|a|^2)^{(n+1+\gamma )\lambda }. \end{aligned}$$

It follows from Theorem A. that $\mu$ is a $(\lambda , \gamma )$-Bergman–Carleson measure. $\square$

Proposition 5.2

Let $0<p_1\le 1$ and let $-1<\alpha _1, \beta <\infty$ satisfy (C-1). Let $\lambda ,\gamma$ be given by (5.1). If the measure $\mu$ is a $(\lambda , \gamma )$-Bergman–Carleson measure, then $B_\mu ^\beta$ is bounded from $A_{\alpha _1}^{p_1}$ to $L^\infty$.

Proof

Given $0<r<1$, let $\{a_j\}$ be an r-lattice in $\mathbb {B}_n$, and let $D_j$ and $\tilde{D}_j$ be the associated Bergman metric balls given in Lemma 1.3. Since $0<p_1\le 1$, it follows from (4.2) that:

$$\begin{aligned} \sup _{z\in \mathbb {B}_n}|B_{\mu }^\beta (f)(z)|\lesssim & {} \sum _{j=1}^\infty \frac{\mu (D_j)}{(1-|a_j|^2)^{(n+1+\gamma )\lambda }} \left( \int \limits _{\tilde{D}_j}|f(\zeta )|^{p_1}\text {d}v_{\alpha _1}(\zeta )\right) ^{1/p_1}\nonumber \\\lesssim & {} \Vert \mu \Vert _{\lambda , \gamma }\left( \sum _{j=1}^\infty \int \limits _{\tilde{D}_j}|f(\zeta )|^{p_1}\text {d}v_{\alpha _1}(\zeta )\right) ^{1/p_1}\\\lesssim & {} \Vert \mu \Vert _{\lambda , \gamma }\Vert f\Vert _{p_1, \alpha _1}.\nonumber \end{aligned}$$

(5.3)

Thus, $B_{\mu }^\beta : A_{\alpha _1}^{p_1}\rightarrow L^\infty$ is bounded. The proof is complete. $\square$

Combining Proposition 5.1, Proposition 5.2, and the fact that $|T_\mu ^\beta f|\le B_\mu ^\beta f$, we obtain the following result.

Theorem 5.3

Let $0<p_1\le 1$ and let $-1<\alpha _1, \beta <\infty$ satisfy (C-1). Let $\lambda ,\gamma$ be given by (5.1). Then, the following statements are equivalent:

• (i) $B_\mu ^\beta$ is bounded from $A_{\alpha _1}^{p_1}$ to $L^\infty$.

• (ii) $T_{\mu }^\beta$ is bounded from $A_{\alpha _1}^{p_1}$ to $H^\infty$.

• (iii) The measure $\mu$ is a $(\lambda , \gamma )$-Bergman–Carleson measure.

Moreover, we have

$$\begin{aligned} \Vert B_{\mu }^\beta \Vert _{A_{\alpha _1}^{p_1}\rightarrow L^\infty } \asymp \Vert T_{\mu }^\beta \Vert _{A_{\alpha _1}^{p_1}\rightarrow H^\infty }\asymp \Vert \mu \Vert _{\lambda , \gamma }. \end{aligned}$$

For the case $1<p_1<\infty$, we have the following partial result.

Proposition 5.4

Let $1<p_1<\infty$ and let $-1<\alpha _1, \beta <\infty$ satisfy (C-1). Let $\lambda ,\gamma$ be given by (5.1). Then, the following statements are equivalent.

• (i) For any $(\lambda , \gamma )$-Bergman–Carleson measure $\mu$, $B_\mu ^\beta$ is bounded from $A_{\alpha _1}^{p_1}$ to $L^\infty$.

• (ii) The integral operator

  $$\begin{aligned} ({\mathcal {S}}f)(z):=\int \limits _{\mathbb {B}_n}\frac{(1-|w|^2)^{\beta +(n+1+\alpha _1)/p_1}|f(w)|}{|1-\langle z, w\rangle |^{n+1+\beta }}\text {d}v(w) \end{aligned}$$

  (5.4)

  is bounded from $A_{\alpha _1}^{p_1}$ to $L^\infty$.

Proof

(ii)$\Rightarrow$(i). Suppose that ${\mathcal {S}}: A_{\alpha _1}^{p_1} \rightarrow L^\infty$ is bounded. Let $\mu$ be an arbitrary $(\lambda , \gamma )$-Bergman–Carleson measure. It follows from Lemma 2.1 that:

$$\begin{aligned} |B_\mu ^\beta f(z)|~&=\int \limits _{\mathbb {B}_n}\frac{|f(w)|}{|1-\langle z, w\rangle |^{n+1+\beta }}\text {d}\mu (w)\\ ~&\lesssim \Vert \mu \Vert _{\lambda , \gamma }\int \limits _{\mathbb {B}_n}\frac{|f(w)|}{|1-\langle z, w\rangle |^{n+1+\beta }}(1-|w|^2)^{(n+1+\gamma )\lambda -n-1}\text {d}v(w)\\ ~&=\Vert \mu \Vert _{\lambda , \gamma }\int \limits _{\mathbb {B}_n}\frac{(1-|w|^2)^{\beta +(n+1+\alpha _1)/p_1}|f(w)|}{|1-\langle z, w\rangle |^{n+1+\beta }} \text {d}v(w)\\ ~&=\Vert \mu \Vert _{\lambda , \gamma }({\mathcal {S}}f)(w). \end{aligned}$$

Thus, the boundedness of ${\mathcal {S}}: A_{\alpha _1}^{p_1}\rightarrow L^\infty$ implies the boundedness of $B_\mu ^\beta : A_{\alpha _1}^{p_1}\rightarrow L^\infty$.

(i)$\Rightarrow$(ii). Suppose that $B_\mu ^\beta : A_{\alpha _1}^{p_1}\rightarrow L^\infty$ is bounded for any $(\lambda , \gamma )$-Bergman–Carleson measure. Consider $d\mu (z)=(1-|z|^2)^{\beta +(n+1+\alpha _1)/p_1}dv(z)$. It can be easily checked that $\mu$ is a $(\lambda , \gamma )$-Bergman–Carleson measure. Since

$$\begin{aligned} B^{\beta }_{\mu }f(z)~&=\int \limits _{\mathbb {B}_n}\frac{|f(w)|}{|1-\langle z, w\rangle |^{n+1+\beta }}\text {d}\mu (w)\\ ~&=\int \limits _{\mathbb {B}_n}\frac{(1-|w|^2)^{\beta +(n+1+\alpha _1)/p_1}|f(w)|}{|1-\langle z, w\rangle |^{n+1+\beta }}\text {d}v(w)\\ ~&=({\mathcal {S}}f)(w), \end{aligned}$$

the boundedness of $B_\mu ^\beta : A_{\alpha _1}^{p_1}\rightarrow L^\infty$ implies the boundedness of ${\mathcal {S}}: A_{\alpha _1}^{p_1}\rightarrow L^\infty$. $\square$

Remark 5.5

It follows from [10, Theorem 1.3] that for $1<p_1<\infty$, ${\mathcal {S}}: L_{\alpha _1}^{p_1} \rightarrow L^\infty$ is unbounded. However, we do not know whether ${\mathcal {S}}: A_{\alpha _1}^{p_1} \rightarrow L^\infty$ is bounded. Also, the above proposition does not fully solve the problem about when $B_\mu ^\beta$ is bounded from $A_{\alpha _1}^{p_1}$ to $L^\infty$. Therefore, we propose the following open problems.

Open Problem 1

Let $1<p<\infty$, and let $-1<\alpha <\infty$. Is the operator ${\mathcal {S}}$ bounded from the Bergman space $A_\alpha ^p$ to $L^\infty$?

Open Problem 2

Let $1<p<\infty$, and let $-1<\alpha <\infty$. How to characterize boundedness of $B_\mu ^\beta :A_{\alpha }^{p}\rightarrow L^\infty$ and $T_\mu ^\beta :A_{\alpha }^{p}\rightarrow H^\infty$?

Next let us consider compactness of $B_\mu ^\beta : A_{\alpha _1}^{p_1} \rightarrow L^\infty$. By the same discussion as in the proof of Proposition 3.1, we can obtain the following result.

Proposition 5.6

Let $0<p_1\le 1$, let $p_2=\infty$ and let $-1<\alpha _1<\infty$. Suppose that, for every bounded sequence $\{f_k\}$ in $A^{p_1}_{\alpha _1}$, such that $f_k\rightarrow 0$ uniformly on every compact subsets of $\mathbb {B}_n$ as $k\rightarrow \infty$, we have

$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert B_{\mu }^{\beta }f_k\Vert _{\infty }=0. \end{aligned}$$

Then, $B_\mu ^\beta : A_{\alpha _1}^{p_1}\rightarrow L^{\infty }$ is compact.

We can also get the the following result on $T_{\mu }^\beta$ by a similar discussion as in the proof of Proposition 3.2 combining with Proposition 5.1.

Proposition 5.7

Let $0<p_1<\infty$ and let $-1<\alpha _1, \beta <\infty$ satisfy (C-1). Suppose that $T_{\mu }^\beta$ is bounded from $A_{\alpha _1}^{p_1}$ to $H^{\infty }$. Then, the following statements are equivalent.

• (i) $T_{\mu }^\beta$ is compact from $A_{\alpha _1}^{p_1}$ to $H^{\infty }$.

• (ii) For every bounded sequence $\{f_k\}$ in $A^{p_1}_{\alpha _1}$, such that $f_k\rightarrow 0$ uniformly on every compact subsets of $\mathbb {B}_n$ as $k\rightarrow \infty$, we have

  $$\begin{aligned} \lim _{k\rightarrow \infty }\Vert T_{\mu }^{\beta }f_k\Vert _{\infty }=0. \end{aligned}$$

Proof

The implication (ii) $\Rightarrow$ (i) follows from the same discussion as in Proposition 3.1. The implication (i) $\Rightarrow$ (ii) follows from a similar discussion as in Proposition 3.2. In fact, if $T_{\mu }^\beta : A_{\alpha _1}^{p_1}\rightarrow H^{\infty }$ is compact, then $T_\mu ^\beta : A_{\alpha _1}^{p_1}\rightarrow H^{\infty }$ is bounded. It follows from Proposition 5.1 that $\mu$ is a $(\lambda , \gamma )$-Bergman–Carleson measure on $\mathbb {B}_n$, where $\lambda , \gamma$ are given by (5.1). It is easy to see that

$$\begin{aligned} (n+1+\gamma )\lambda =~&n+1+\beta +\frac{n+1+\alpha _1}{p_1} >\frac{n+1+\alpha _1}{p_1}. \end{aligned}$$

Thus, a similar discussion to Lemma 3.4 and Proposition 3.2 gives that (ii) holds. The proof is complete. $\square$

Theorem 5.8

Let $0<p_1\le 1$ and let $-1<\alpha _1, \beta <\infty$ satisfy (C-1). Let $\lambda ,\gamma$ be given by (5.1). Then, the following statements are equivalent:

• (i) $B_\mu ^\beta$ is compact from $A_{\alpha _1}^{p_1}$ to $L^\infty$.

• (ii) $T_{\mu }^\beta$ is compact from $A_{\alpha _1}^{p_1}$ to $H^\infty$.

• (iii) The measure $\mu$ is a vanishing $(\lambda , \gamma )$-Bergman–Carleson measure.

Proof

(i)$\Rightarrow$(ii). This is is trivial.

(ii)$\Rightarrow$(iii). Suppose that $T_{\mu }^\beta$ is compact from $A_{\alpha _1}^{p_1}$ to $H^\infty$. Let $\{a_k\}$ be a sequence in $\mathbb {B}_n$ with $|a_k|\rightarrow 1$. Consider the functions

$$\begin{aligned} f_k(z)=\frac{(1-|a_k|^2)^{n+1+\beta -(n+1+\alpha _1)/p_1}}{(1-\langle z, a_k\rangle )^{n+1+\beta }} \end{aligned}$$

for $k=1, 2, 3,...$. It follows from Lemma 2.2 that $\sup _k\Vert f_k\Vert _{p_1, \alpha _1}<\infty$, and it is obvious that $f_k$ converges to zero uniformly on compact subsets of $\mathbb {B}_n$. Thus, by Proposition 5.7, we have that $\Vert T_{\mu }^\beta f_k\Vert _{\infty }\rightarrow 0$. Fix any r with $0<r<1$. By the same discussion as in the proof of Proposition 5.1, we get that

$$\begin{aligned} \frac{\mu (D(a_k, r))}{(1-|a_k|^2)^{(n+1+\gamma )\lambda }} \le T_{\mu }^\beta f_k(a_k)\le {\Vert T_{\mu }^\beta f_k\Vert _\infty }\rightarrow 0 \end{aligned}$$

as $k\rightarrow \infty$. Thus, $\mu$ is a vanishing $(\lambda ,\gamma )$-Bergman–Carleson measure.

(iii)$\Rightarrow$(i). Suppose that $\mu$ is a vanishing $(\lambda , \gamma )$-Bergman–Carleson measure. By proposition 5.6, it suffices to prove that $\Vert B_{\mu }^\beta f_k\Vert _\infty \rightarrow 0$ for any bounded sequence $\{f_k\}$ in $A_{\alpha _1}^{p_1}$ converging to zero uniformly on compact subsets of $\mathbb {B}_n$. Let $\{a_j\}$ be an r-lattice in $\mathbb {B}_n$. Let

$$\begin{aligned} c(a_j)=\frac{\mu (D_j)}{(1-|a_j|^2)^{(n+1+\gamma )\lambda }} \end{aligned}$$

and

$$\begin{aligned} b_k(a_j)=\left( \int \limits _{\tilde{D}_j}|f_k(\zeta )|^{p_1}\text {d}v_{\alpha _1}(\zeta )\right) ^{1/p_1}, \end{aligned}$$

Since $0<p_1\le 1$, it follows from inequality (5.3) that:

$$\begin{aligned} \sup _{z\in \mathbb {B}_n}|B_{\mu }^\beta (f_k)(z)| \lesssim \sum _{j=1}^{\infty }c(a_j)b_k(a_j). \end{aligned}$$

By the same discussion as in the proof of Theorem 1.2, we know that the sequence $\{b_k(a_j)\}$ satisfies the condition of Lemma 4.2. Since $\mu$ is a vanishing $(\lambda ,\gamma )$-Bergman–Carleson measure, we know that $\lim _{|a_j|\rightarrow 1}c(a_j)=0$. Hence, by (i) of Lemma 4.2, we get that $\lim _{k\rightarrow \infty }\Vert B_{\mu }^\beta f_k\Vert _{\infty }=0$. The proof is complete. $\square$

Open Problem 3

Let $1<p<\infty$, and let $-1<\alpha <\infty$. How to characterize compactness of $B_\mu ^\beta :A_{\alpha }^{p}\rightarrow L^\infty$ and $T_\mu ^\beta :A_{\alpha }^{p}\rightarrow H^\infty$?

6 The case when $p_1=\infty$

In this section, we consider the case when $p_1=\infty$.

6.1 The case when $p_1=\infty , p_2=\infty$

If we follow the definition of $\lambda$ and $\gamma$ in (1.2), we get that in this case $\lambda =1$, $\gamma =\beta$. Our first result here shows that the $(1,\beta )$-Bergman–Carleson measure condition does not characterize boundedness of $B_{\mu }^\beta : H^\infty \rightarrow L^\infty$.

Lemma 6.1

Let $-1<\beta <\infty$. There exists a $(1,\beta )$-Bergman–Carleson measure $\mu$, such that $B_{\mu }^\beta$ is unbounded from $H^\infty$ to $L^\infty$.

Proof

Let $\text {d}\mu =(1-|z|^2)^{\beta }\,\text {d}v(z)$. It can be easily checked that $\mu$ is a $(1,\beta )$-Bergman–Carleson measure, and

$$\begin{aligned} B_{\mu }^\beta f(z) =\int \limits _{\mathbb {B}_n}\frac{(1-|w|^2)^\beta |f(w)|}{|1-\langle z,w\rangle |^{n+1+\beta }}\,\text {d}v(w). \end{aligned}$$

Let $f=1\in H^\infty$. Then, $B_\mu ^\beta 1\notin L^\infty$ by lemma 2.2. The proof is complete. $\square$

Open Problem 4

How to characterize boundedness and the compactness of $B_{\mu }^\beta : H^\infty \rightarrow L^\infty$ and $T_\mu ^\beta : H^\infty \rightarrow H^\infty$?

6.2 The case when $p_1=\infty , 0<\ p_2<\infty$.

We give the following sufficient condition for the boundedness of the Berezin type operator $B_\mu ^\beta : H^\infty \rightarrow L_{\alpha _2}^{p_2}$.

Proposition 6.2

Let $0<p_2<\infty$ and let $-1<\alpha _2,\beta <\infty$ satisfy (C-2). Given $0<r<1$, let $\{a_j\}$ be any r-lattice in $\mathbb {B}_n$, and let $D_j=D(a_j,r)$ be the associated Bergman metric balls given by Lemma 1.3. Suppose that

$$\begin{aligned} \{\nu _j\}=: \left\{ \frac{\mu (D_j)}{(1-|a_j|^2)^{n+1+\beta -(n+1+\alpha _2)/p_2}}\right\} \in l^{p_2}. \end{aligned}$$

(6.1)

Then, $B_\mu ^\beta$ is bounded from $H^\infty$ to $L_{\alpha _2}^{p_2}$.

Proof

Since $|1-\langle z, w\rangle |\asymp |1-\langle z, a_j\rangle |$ for $w\in D_j$, it follows that:

$$\begin{aligned} |B_{\mu }^\beta (f)(z)|&~\lesssim \Vert f\Vert _\infty \sum _{j=1}^\infty \int \limits _{D_j}\frac{1}{|1-\langle z, w\rangle |^{n+1+\beta }}\text {d}\mu (w)\\&~\lesssim \Vert f\Vert _\infty \sum _{j=1}^\infty \frac{\mu (D_j)}{|1-\langle z, a_j\rangle |^{n+1+\beta }}. \end{aligned}$$

By Lemma 2.4, we get

$$\begin{aligned} \Vert B_\mu ^\beta f\Vert _{p_2, \alpha _2}\lesssim ~&\Vert f\Vert _\infty \left( \sum _{j=1}^\infty \frac{\mu (D_j)^{p_2}}{(1-|a_j|^2)^{(n+1+\beta )p_2-(n+1+\alpha _2)}}\right) ^{1/p_2}\\ \lesssim ~&\Vert f\Vert _\infty \Vert \{\nu _j\}\Vert _{l^{p_2}}. \end{aligned}$$

Therefore, we get that $B_\mu ^\beta : H^\infty \rightarrow A_{\alpha _2}^{p_2}$ is bounded. $\square$

By a similar argument as in the proof of Proposition 3.1, we can get the following sufficient condition for the compactness of $B_{\mu }^\alpha : H^\infty \rightarrow A_{\alpha _2}^{p_2}$ for $0<p_2<\infty$.

Proposition 6.3

Let $0<p_2<\infty$, and let $-1<\alpha _2<\infty$. Assume that $B_{\mu }^{\beta }: H^\infty \rightarrow L_{\alpha _2}^{p_2}$ is bounded. Suppose that, for every bounded sequence $\{f_k\}$ in $H^\infty$, such that $f_k\rightarrow 0$ uniformly on every compact subsets of $\mathbb {B}_n$ as $k\rightarrow \infty$, we have

$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert B_{\mu }^{\beta }f_k\Vert _{p_2, \alpha _2}=0. \end{aligned}$$

Then, $B_\mu ^\beta$ is compact from $H^\infty$ to $L^{p_2}_{\alpha _2}$.

Lemma 6.4

Given $0<r<1$, let $\{a_j\}$ be an r-lattice in $\mathbb {B}_n$, let $D_j=D(a_j,r)$ be the associated Bergman metric balls given by Lemma 1.3, and let $\{b_k(a_j)\}$ be a sequence depending on $a_j$, such that

$$\begin{aligned} \sup _{k}\sup \limits _{j} |b_k(a_j)|<\infty . \end{aligned}$$

Suppose, in addition, for any compact subset K of $\mathbb {B}_n$, the sequence $\{b_k(a_j)\}$ satisfies that

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\sup \limits _{j\in \Gamma }|b_k(a_j)|\rightarrow 0, \end{aligned}$$

where $\Gamma :=\{j: a_j\in K\}$. Let $\{c(a_j)\}$ be a sequence of real numbers depending on $\{a_j\}$ satisfying that $\{c(a_j)\}\in l^1$. Then

$$\begin{aligned} \lim _{k\rightarrow \infty }\sum _{j=1}^\infty c(a_j)b_k(a_j)=0. \end{aligned}$$

Proof

Since $\{c(a_j)\}\in l^1$, then for any given $\varepsilon >0$, there is a $r_3\in (0, 1)$, such that

$$\begin{aligned} \sum _{j:\, |a_j|>r_3} |c(a_j)|<\varepsilon . \end{aligned}$$

Thus, we have

$$\begin{aligned} \left| \sum _{j=1}^\infty c(a_j)b_k(a_j)\right|\le & {} \sum _{j:\, |a_j|\le r_3} |c(a_j)b_k(a_j)|+\sum _{j:\, |a_j|>r_3}| c(a_j)b_k(a_j)|\\\le & {} \sup _{j:\, |a_j|\le r_3}|b_k(a_j)|\sum _{j:\, |a_j|<r_3} |c(a_j)|\\{} & {} +\sup _{j:\, |a_j|>r_3}|b_k(a_j)|\sum _{j:\, |a_j|>r_3} |c(a_j)|\\\le & {} \sup _{j:\, |a_j|\le r_3}|b_k(a_j)|\sum _{j:\, |a_j|<r_3} |c(a_j)|+\varepsilon \sup _k\sup _j|b_k(a_j)|. \end{aligned}$$

Letting $\varepsilon \rightarrow 0$, and then letting $k\rightarrow \infty$, we get the result, since

$$\begin{aligned} \lim _{k\rightarrow \infty }\sup _{j:\, |a_j|\le r_3}|b_k(a_j)|=0. \end{aligned}$$

$\square$

Proposition 6.5

Let $0<p_2<\infty$, and let $-1<\alpha _2,\beta <\infty$ satisfy (C-2). Let $\{\nu _j\}$ be the sequence given by (6.1) which satisfies that $\{\nu _j\} \in l^{p_2}$. Then, $B_\mu ^\beta$ is compact from $H^\infty$ to $L_{\alpha _2}^{p_2}$.

Proof

Let $\{a_j\}$ be an r-lattice. By Proposition 6.3 that it suffices for us to prove that $\lim _{k\rightarrow \infty }\Vert B_{\mu }^\beta (f_k)\Vert _{p_2, \alpha _2}=0$ for any bounded sequence $\{f_k\}\in H^\infty$, such that $f_k\rightarrow 0$ uniformly on every compact subsets of $\mathbb {B}_n$ as $k\rightarrow \infty$. First, it is easy to get that

$$\begin{aligned} |B_{\mu }^\beta (f_k)(z)| \lesssim \sum _{j=1}^\infty \frac{\Vert f_k\chi _{D_j}\Vert _\infty \mu (D_j)}{|1-\langle z, a_j\rangle |^{n+1+\beta }}. \end{aligned}$$

Let

$$\begin{aligned} c(a_j)=\left( \frac{\mu (D_j)}{(1-|a_j|^2)^{n+1+\beta -(n+1+\alpha _2)/p_2}}\right) ^{p_2} \end{aligned}$$

and

$$\begin{aligned} b_k(a_j)=\Vert f_k\chi _{D_j}\Vert _\infty ^{p_2}. \end{aligned}$$

Using Lemma 2.4, we get that

$$\begin{aligned} \Vert B_{\mu }^\beta (f_k)\Vert _{p_2, \alpha _2}^{p_2} \lesssim \sum _{j=1}^\infty c(a_j)b_k(a_j). \end{aligned}$$

(6.2)

Notice that $\{c(a_j)\}=\{\nu _j^{p_2}\}\in l^1$ and

$$\begin{aligned} \sup _k\sup _j|b_k(a_j)|=\sup _k\sup _j\Vert f_k\chi _{D_j}\Vert _\infty ^{p_2}\le \sup _k \Vert f_k\Vert _\infty ^{p_2}<\infty . \end{aligned}$$

Let K be any compact subset in $\mathbb {B}_n$ and $\Gamma =\{j:\,a_j\in K\}$. Then $\Gamma$ is a finite set. Since $f_k\rightarrow 0$ uniformly on every compact subsets of $\mathbb {B}_n$ as $k\rightarrow \infty$, it follows that for any given $\varepsilon >0$, and for all $z\in \cup _{j\in \Gamma }D_j$, we have that

$$\begin{aligned} \sup \limits _{j\in \Gamma }|b_k(a_j)|=\sup \limits _{j\in \Gamma }\Vert f_k\chi _{D_j}\Vert _\infty ^{p_2}<\varepsilon . \end{aligned}$$

Letting $\varepsilon \rightarrow 0$ and letting $k\rightarrow \infty$, we get that

$$\begin{aligned} \lim \limits _{k\rightarrow \infty }\sup \limits _{j\in \Gamma }|b_k(a_j)|\rightarrow 0. \end{aligned}$$

It follows from Lemma 6.4 that $\lim _{k\rightarrow \infty }\Vert B_{\mu }^\beta (f_k)\Vert _{p_2, \alpha _2}=0$, completing the proof. $\square$

Finally, we propose the following open problem.

Open Problem 5

How to characterize boundedness and compactness of $B_\mu ^\beta : H^\infty \rightarrow L_{\alpha _2}^{p_2}$ and $T_\mu ^\beta : H^\infty \rightarrow A_{\alpha _2}^{p_2}$?