Volterra integration operators from Hardy-type tent spaces to Hardy spaces

Abstract

In this paper, we completely characterize the boundedness and compactness of the Volterra integration operators $J_{g}$ acting from the Hardy-type tent spaces ${\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)$ to the Hardy spaces $H^t({\mathbb{B}}_n)$ in the unit ball of ${\mathbb{C}}^n$ for all $0< p,q,t<\infty$ and $\alpha >-n-1$. The duality and factorization techniques for tent spaces of sequences play an important role in the proof of the main results.

1 Introduction

Let ${\mathbb{B}}_n$ be the open unit ball in ${\mathbb{C}}^n$, and ${\mathbb{S}}_n$ the boundary of ${\mathbb{B}}_n$. Denote by $H({\mathbb{B}}_n)$ the space of all holomorphic functions on ${\mathbb{B}}_n$. A function $g\in H({\mathbb{B}}_n)$ induces an integration operator (or a Volterra operator) $J_{g}$ given by the formula:

$$J_{g}f(z)={\int }_0^{1}f(tz)Rg(tz)\frac{dt}{t},z\in {\mathbb{B}}_n,$$

where f is holomorphic on ${\mathbb{B}}_n$ and Rg is the radial derivative of g, that is,

$$Rg(z)=\sum _{k=1}^n{z}_k\frac{\partial g}{\partial z_k}(z),z=(z_1,\dots ,z_n)\in {\mathbb{B}}_n.$$

In the one-dimensional case $n=1$, the operator $J_{g}$ was first studied in the setting of the Hardy spaces by Pommerenke [22] related to the functions of bounded mean oscillation. Some important papers include the pioneering works of Aleman, Cima and Siskakis [3, 5, 6], where they described the boundedness of the operators $J_{g}$ acting on Hardy and Bergman spaces in the unit disk. Since then, much research on the Volterra operator $J_{g}$ acting on many spaces of holomorphic functions has been carried out (see [2, 4, 10, 24] for example). The higher-dimensional variant of $J_{g}$ was introduced by Hu [12]. A fundamental property of the operator $J_{g}$ is the following basic formula involving the radial derivative R and the operator $J_{g}$:

$$R(J_{g}f)(z)=f(z)Rg(z),z\in {\mathbb{B}}_n.$$

The boundedness and compactness of $J_{g}$ have been extensively studied in many spaces of holomorphic functions in the unit ball (see [20] for the corresponding study between Hardy spaces, and [9, 19] from Bergman spaces to Hardy spaces, and others [16, 23, 25] for example).

For $0< t<\infty$, the Hardy space $H^t({\mathbb{B}}_n)$ consists of those holomorphic functions f on ${\mathbb{B}}_n$ with

$${\parallel f\parallel }_{H^t({\mathbb{B}}_n)}^t=\underset{0<r<1}{sup}{\int }_{{\mathbb{S}}_n}|f(r\xi ){|}^{t}d\sigma (\xi )<\mathrm{\infty },$$

where dσ is the surface measure on the unit sphere ${\mathbb{S}}_n:=\partial {\mathbb{B}}_n$ normalized so that $\sigma ({\mathbb{S}}_n)=1$.

For $0< p,q<\infty$ and $\alpha >-n-1$, the weighted tent space ${\mathcal{T}}_{q,\alpha }^p({\mathbb{B}}_n)$ consists of all measurable functions f on ${\mathbb{B}}_n$ such that

$${\parallel f\parallel }_{{\mathcal{T}}_{q,\alpha }^p({\mathbb{B}}_n)}^p={\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|f(z){|}^q{(1-|z{|}^2)}^{\alpha }dv(z))}^{\frac{p}{q}}d\sigma (\xi )<\mathrm{\infty },$$

where dv is the volume measure on ${\mathbb{B}}_n$ normalized so that $v({\mathbb{B}}_n)=1$, and $\mathrm{\Gamma }(\xi )=\{z\in {\mathbb{B}}_n:|1-〈z,\xi 〉|<(1-{|z|}^2)\}$ is the admissible approach region. In particular, for $\alpha =0$, we write ${\mathcal{T}}_q^p({\mathbb{B}}_n)$ instead of ${\mathcal{T}}_{q,\alpha }^p({\mathbb{B}}_n)$.

Analogously, ${\mathcal{T}}_{\mathrm{\infty }}^p({\mathbb{B}}_n)$ consists of all measurable functions f on ${\mathbb{B}}_n$ such that

$${\parallel f\parallel }_{{\mathcal{T}}_{\mathrm{\infty }}^p({\mathbb{B}}_n)}^p={\int }_{{\mathbb{S}}_n}{\left(\mathrm{ess}\underset{z\in \mathrm{\Gamma }(\xi )}{sup}\right|f(z)\left|\right)}^{p}d\sigma (\xi )<\mathrm{\infty },$$

and ${\mathcal{T}}_{q,\alpha }^{\mathrm{\infty }}({\mathbb{B}}_n)$ consists of measurable functions f with

$${\parallel f\parallel }_{{\mathcal{T}}_{q,\alpha }^{\mathrm{\infty }}({\mathbb{B}}_n)}=\mathrm{ess}\underset{\xi \in {\mathbb{S}}_n}{sup}{\left(\underset{w\in \mathrm{\Gamma }(\xi )}{sup}\frac{1}{{(1-|w{|}^2)}^n}{\int }_{Q(w)}\right|f(z){|}^q{(1-|z{|}^2)}^{n+\alpha }dv(z))}^{\frac{1}{q}}<\mathrm{\infty },$$

where $Q(w)=\{z\in {\mathbb{B}}_n:|1-〈z,\frac{w}{|w|}〉|<1-{|w|}^2\}$ for $w\in {\mathbb{B}}_n\mathrm{\setminus }\{0\}$ and $Q(0)={\mathbb{B}}_n$.

For $0< p, q<\infty$ and $\alpha >-n-1$, the Hardy-type tent space ${\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)$ consists of holomorphic functions on ${\mathbb{B}}_n$ that also belong to ${\mathcal{T}}_{q,\alpha }^p({\mathbb{B}}_n)$, with the same quasinorm, and ${\mathcal{HT}}_{\mathrm{\infty }}^p({\mathbb{B}}_n)$ consists of holomorphic functions on ${\mathbb{B}}_n$ that also belong to ${\mathcal{T}}_{\mathrm{\infty }}^p({\mathbb{B}}_n)$. The space ${\mathcal{CT}}_{q,\alpha }({\mathbb{B}}_n)$ consists of those holomorphic functions that belong to ${\mathcal{T}}_{q,\alpha }^{\mathrm{\infty }}({\mathbb{B}}_n)$ that is endowed with the same norm. We refer the reader to [21] for more details on Hardy-type tent spaces.

As useful tools, tent spaces play important roles in the study of harmonic analysis and partial differential equations. By the nontangential maximal function characterization of the Hardy space, ${\mathcal{HT}}_{\mathrm{\infty }}^p({\mathbb{B}}_n)=H^p({\mathbb{B}}_n)\subseteq {\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)$, see [26], and we can consider $H^p({\mathbb{B}}_n)$ as the limit of ${\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)$ when $q\rightarrow \infty$. Hence, we describe the boundedness and compactness of $J_g:{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)\to H^t({\mathbb{B}}_n)$ for all possible ranges $0< p,q,t<\infty$ and $\alpha >-n-1$. Although only discrete characterizations are described in our theorems, continuous characterizations also can be obtained from subsequent proofs.

Our main results are as follows.

Theorem 1.1

Let $0< p,q,t<\infty$, $\alpha >-n-1$. Then, the integration operator $J_g:{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)\to H^t({\mathbb{B}}_n)$ is bounded if and only if for any $r\in (0,1)$ and an r-lattice $Z=\{a_{k}\}$ in ${\mathbb{B}}_n$, the sequence

$$ u=\{u_{k}\}= \bigl\{ \bigl\vert Rg(a_{k}) \bigr\vert \bigl(1- \vert a_{k} \vert ^{2}\bigr)^{ \frac{q-(n+1+\alpha )}{q}} \bigr\} $$

satisfies one of the following conditions:

• (a) If $p>t$ and $q>2$, then u belongs to $T_{\frac{2q}{q-2}}^{\frac{pt}{p-t}}(Z)$.

• (b) If $p>t$ and $q\leq 2$, then u belongs to $T_{\infty}^{\frac{pt}{p-t}}(Z)$.

• (c) If $p=t$ and $q>2$, then u belongs to $T_{\frac{2q}{q-2}}^{\infty}(Z)$.

• (d) If $p=t$ and $q\leq 2$ or $p< t$, then $\{u_{k}\cdot (1-|a_{k}|^{2})^{n(\frac{1}{t}-\frac{1}{p})}\}$ belongs to $l^{\infty}$.

Theorem 1.2

Let $0< p,q,t<\infty$, $\alpha >-n-1$. Then, the integration operator $J_g:{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)\to H^t({\mathbb{B}}_n)$ is compact if and only if for any $r\in (0,1)$ and an r-lattice $Z=\{a_{k}\}$ in ${\mathbb{B}}_n$, the sequence

$$ u=\{u_{k}\}= \bigl\{ \bigl\vert Rg(a_{k}) \bigr\vert \bigl(1- \vert a_{k} \vert ^{2}\bigr)^{ \frac{q-(n+1+\alpha )}{q}} \bigr\} $$

satisfies one of the following conditions:

• (a) If $p>t$ and $q>2$, then

  $${\int }_{{\mathbb{S}}_n}{\left(\underset{a_k\in \mathrm{\Gamma }(\xi )}{sup}\right|Rg(a_k){|}^{\frac{2q}{q-2}}{(1-|a_k{|}^2)}^{\frac{q-(n+1+\alpha )}{q}\cdot \frac{2q}{q-2}})}^{\frac{pt}{p-t}\cdot \frac{q-2}{2q}}d\sigma (\xi )<\mathrm{\infty }.$$

• (b) If $p>t$ and $q\leq 2$, then

  $$\underset{\rho \to 1^{-}}{lim}{\int }_{{\mathbb{S}}_n}{\left(\underset{a_k\in \mathrm{\Gamma }(\xi )\setminus \overline{D(0,\rho )}}{sup}\right|Rg(a_k)\left|{(1-|a_k{|}^2)}^{\frac{q-(n+1+\alpha )}{q}}\right)}^{\frac{pt}{p-t}}d\sigma (\xi )=0.$$

• (c) If $p=t$ and $q>2$, then

  $$ \lim_{ \vert w \vert \rightarrow 1^{-}}\frac{1}{(1- \vert w \vert ^{2})^{n}}\sum_{a_{k}\in Q(w)} \bigl( \bigl\vert Rg(a_{k}) \bigr\vert \bigl(1- \vert a_{k} \vert ^{2}\bigr)^{ \frac{q-(n+1+\alpha )}{q}} \bigr)^{\frac{2q}{q-2}}\bigl(1- \vert a_{k} \vert ^{2} \bigr)^{n}=0.$$

• (d) If $p=t$ and $q\leq 2$ or $p< t$, then

  $$ \lim_{k\rightarrow \infty} \bigl\vert Rg(a_{k}) \bigr\vert \bigl(1- \vert a_{k} \vert ^{2}\bigr)^{ \frac{q-(n+1+\alpha )}{q} + n(\frac{1}{t}-\frac{1}{p})}=0.$$

This paper is organized as follows: Sect. 2 contains some background materials and the tools used in the proofs. Theorems 1.1 and 1.2 are proved in Sect. 3 and Sect. 4, respectively.

Throughout the paper, constants are used with no attempt to calculate their exact values, and the value of a constant C may change from one occurrence to the next. We also use the notion $A\lesssim B$ to indicate that there is a constant $C>0$ with $A \leqslant CB$. The converse relation $A\gtrsim B$ is defined in an analogous manner, and if $A\lesssim B$ and $A\gtrsim B$ both hold, we write $A\asymp B$. Given $p\in [1, \infty ]$, we will denote by $p'=p/(p-1)$ its Hölder conjugate, and we agree that $1'=\infty$ and $\infty '=1$ in this paper.

2 Preliminaries

In this section, we introduce some basic results that will be used for the proofs of our main theorems.

2.1 Area methods and equivalent norms

For $\xi \in {\mathbb{S}}_n$ and $\gamma >1$, the admissible approach region $\Gamma _{\gamma}(\xi )$ is defined as

$${\mathrm{\Gamma }}_{\gamma }(\xi )=\{z\in {\mathbb{B}}_n:|1-〈z,\xi 〉|<\frac{\gamma }{2}(1-|z{|}^2)\}.$$

In this paper we agree that $\Gamma (\xi ):=\Gamma _{2}(\xi )$. It is known that for every $\delta >1$ and $\gamma >1$, there exists $\gamma '>1$ so that

$$ \bigcup_{z\in \Gamma _{\gamma}(\xi )}D(z,\delta )\subset \Gamma _{ \gamma '}(\xi ).$$

We will write $\widetilde{\Gamma}(\xi )$ to indicate this change of aperture. Given $z\in {\mathbb{B}}_n$, we can define the set $I(z)={\mathbb{S}}_n$ for $z=0$, and $I(z)=\{\xi \in {\mathbb{S}}_n:z\in \mathrm{\Gamma }(\xi )\}\subset {\mathbb{S}}_n$ for $z\neq 0$. Obviously, $\sigma (I(z))\asymp (1-|z|^{2})^{n}$, and it follows from Fubini’s theorem that, for a positive measurable function φ, and a finite positive measure ν, one has

$${\int }_{{\mathbb{B}}_n}\phi (z)d\nu (z)\asymp {\int }_{{\mathbb{S}}_n}({\int }_{\mathrm{\Gamma }(\xi )}\phi (z)\frac{d\nu (z)}{{(1-|z{|}^2)}^n})d\sigma (\xi ).$$

We will need the following well-known Calderón’s area theorem [8], which will be very important for our arguments, and the variant can be found in [1, 20].

Lemma A

Let $0< t<\infty$. If $f\in H({\mathbb{B}}_n)$ and $f(0)=0$, then

$${\parallel f\parallel }_{H^t}^t\asymp {\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|Rf(z){|}^2{(1-|z{|}^2)}^{1-n}dv(z))}^{t/2}d\sigma (\xi ).$$

Note that Lemma A shows that $f\in H({\mathbb{B}}_n)$ belongs to $H^{t}$ if and only if $Rf\in \mathcal{HT}_{2,1-n}^{t}$. This explains the special role of number 2 in Theorem 1.1 and Theorem 1.2.

2.2 Embedding theorems

We need the following embedding theorems for Hardy-type tent spaces, which are the generalizations of Lemma 15 and Lemma 23 in [21]. We prove them by a similar method.

Lemma B

Let $0< t\leq p<\infty$, $0< q\leq s<\infty$, $\alpha >-n-1$, and $\beta =\alpha +(\frac{s}{q}-1)(n+1+\alpha )$. Then,

$${\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)\subset {\mathcal{HT}}_{s,\beta }^t({\mathbb{B}}_n),$$

with bounded inclusion.

Proof

Let $\xi \in {\mathbb{S}}_n$ and $r>0$. For any $z\in \Gamma (\xi )$ and $f\in H({\mathbb{B}}_n)$, by the subharmonicity, we have

$$\begin{aligned} \bigl\vert f(z) \bigr\vert &\lesssim \frac{1}{(1- \vert z \vert ^{2})^{\frac{n+1+\alpha}{q}}} \biggl( \int _{D(z,r)} \bigl\vert f(\omega ) \bigr\vert ^{q} \bigl(1- \vert \omega \vert ^{2}\bigr)^{\alpha}\,dv(\omega ) \biggr)^{\frac{1}{q}} \\ &\lesssim \frac{1}{(1- \vert z \vert ^{2})^{\frac{n+1+\alpha}{q}}} \biggl( \int _{ \widetilde{\Gamma}(\xi )} \bigl\vert f(\omega ) \bigr\vert ^{q} \bigl(1- \vert \omega \vert ^{2}\bigr)^{\alpha}\,dv( \omega ) \biggr)^{\frac{1}{q}}. \end{aligned}$$

Writing $|f|^{s}=|f|^{q}|f|^{s-q}$ and applying this estimate to the second factor gives

$$\begin{aligned} & \int _{\Gamma (\xi )} \bigl\vert f(z) \bigr\vert ^{s} \bigl(1- \vert z \vert ^{2}\bigr)^{\beta}\,dv(z) \\ &\quad\lesssim \int _{\Gamma (\xi )} \bigl\vert f(z) \bigr\vert ^{q} \bigl(1- \vert z \vert ^{2}\bigr)^{\alpha} \biggl( \int _{\widetilde{\Gamma}(\xi )} \bigl\vert f(\omega ) \bigr\vert ^{q} \bigl(1- \vert \omega \vert ^{2}\bigr)^{ \alpha}\,dv(\omega ) \biggr)^{s/q-1}\,dv(z) \\ &\quad\lesssim \biggl( \int _{\widetilde{\Gamma}(\xi )} \bigl\vert f(z) \bigr\vert ^{q} \bigl(1- \vert z \vert ^{2}\bigr)^{ \alpha}\,dv(z) \biggr)^{\frac{s}{q}}. \end{aligned}$$

Then, for $t\leq p$, we obtain ${\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)\subset {\mathcal{HT}}_{s,\beta }^p({\mathbb{B}}_n)\subset {\mathcal{HT}}_{s,\beta }^t({\mathbb{B}}_n)$. □

Lemma C

If $0< p< t<\infty$, $0< q<\infty$ and $\alpha >-n-1$, then

$${\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)\subset A_{\eta }^t({\mathbb{B}}_n)$$

with bounded inclusion, where $A_{\eta }^t({\mathbb{B}}_n)$ is the weighted Bergman space and $\eta =(\frac{t}{p}-1)n-1+\frac{t(n+1+\alpha )}{q}$.

Proof

First, recall that if $p< t$, then $H^p({\mathbb{B}}_n)\subset A_{(\frac{t}{p}-1)n-1}^t({\mathbb{B}}_n)$ with bounded inclusion. Applying this to a fractional differential operator $\mathcal{R}^{s,\frac{n+1+\alpha}{2}}$ and according to [21, Theorem G], we have

$${\mathcal{HT}}_{2,\alpha }^p({\mathbb{B}}_n)\subset A_{(\frac{t}{p}-1)n-1+\frac{t(n+1+\alpha )}{2}}^t({\mathbb{B}}_n).$$

For any natural number k, we have $f\in {\mathcal{HT}}_{2k,\alpha }^p({\mathbb{B}}_n)$ if and only if $f^k\in {\mathcal{HT}}_{2,\alpha }^{\frac{p}{k}}({\mathbb{B}}_n)$, and then

$${\mathcal{HT}}_{2k,\alpha }^p({\mathbb{B}}_n)\subset A_{(\frac{t}{p}-1)n-1+\frac{t(n+1+\alpha )}{2k}}^t({\mathbb{B}}_n).$$

Let k be large enough such that $2k>q$. Then, by Lemma B, we have

$${\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)\subset {\mathcal{HT}}_{2k,\alpha +(\frac{2k}{q}-1)(n+1+\alpha )}^p({\mathbb{B}}_n)\subset A_{\eta }^t({\mathbb{B}}_n).$$

 □

We will also need the following Dirichlet-type embedding theorem, which can be found in [7].

Lemma D

Assume that $f\in H({\mathbb{B}}_n)$ with $f(0)=0$. If $0< p< q<\infty$, then

$${\parallel f\parallel }_{H^q({\mathbb{B}}_n)}\lesssim {\parallel Rf\parallel }_{A_{p-n-1+np/q}^p({\mathbb{B}}_n)},$$

where ${\parallel f\parallel }_{A_{\alpha }^p({\mathbb{B}}_n)}^p={\int }_{{\mathbb{B}}_n}{|f(z)|}^p{(1-{|z|}^2)}^{\alpha }dv(z)$.

2.3 Khinchine and Kahane inequalities

Let $r_{k}(u)$ be a sequence of Rademacher functions. We recall first the classical Khinchine’s inequality (see [11, Appendix A] for example).

Khinchine’s inequality: Let $0< p<\infty$. Then, for any sequence $\{c_{k}\}\in l^{2}$, we have

$$ \biggl(\sum_{k} \vert c_{k} \vert ^{2} \biggr)^{p/2}\asymp \int _{0}^{1} \biggl\vert \sum _{k}c_{k} r_{k}(u) \biggr\vert ^{p} \,dt.$$

The next result is known as Kahane’s inequality, see for instance Lemma 5 of Luecking [18].

Kahane’s inequality: Let X be a Banach space, and $0< p,q<\infty$. For any sequence $\{x_{k}\}\subset X$, one has

$$ \biggl( \int _{0}^{1} \biggl\Vert \sum _{k} r_{k}(u)x_{k} \biggr\Vert ^{q}_{X}\,dt \biggr)^{1/q}\asymp \biggl( \int _{0}^{1} \biggl\Vert \sum _{k} r_{k}(u)x_{k} \biggr\Vert ^{p}_{X}\,dt \biggr)^{1/p}.$$

2.4 Separated sequences and lattices

A sequence of points $\{z_j\}\subset {\mathbb{B}}_n$ is said to be separated if there exists $\delta >0$ such that $\beta (z_{i},z_{j})\geq \delta$ for all i and j with $i\neq j$, where $\beta (z,w)$ denotes the Bergman metric on ${\mathbb{B}}_n$. This implies that there is $\delta >0$ such that the Bergman metric balls $D_j=\{z\in {\mathbb{B}}_n:\beta (z,z_j)<\delta \}$ are pairwise disjoint.

We need a well-known result on decomposition of the unit ball ${\mathbb{B}}_n$. By Theorem 2.23 in [26], there exists a positive integer N such that for any $0< r<1$ we can find a sequence $\{a_{k}\}$ in ${\mathbb{B}}_n$ with the following properties:

• (i) ${\mathbb{B}}_n={\bigcup }_{k}D(a_k,r)$.

• (ii) The sets $D(a_{k},r/4)$ are mutually disjoint.

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

Any sequence $\{a_{k}\}$ satisfying the above conditions is called an r-lattice (in the Bergman metric). Obviously any r-lattice is a separated sequence.

2.5 Tent spaces of sequences

Let $Z=\{a_{k}\}$ be an r-lattice. We consider the complex-valued sequences enumerated by this lattice: $\lambda _{k}=f(a_{k})$. For $0 < p, q < \infty$, the tent space $T_{q}^{p}(Z)$ consists of those sequences $\lambda =\{\lambda _{k}\}$ satisfying

$${\parallel \lambda \parallel }_{T_q^p(Z)}={({\int }_{{\mathbb{S}}_n}{(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|{\lambda }_k{|}^q)}^{\frac{p}{q}}d\sigma (\xi ))}^{\frac{1}{p}}<\mathrm{\infty }.$$

Analogously, the tent space $T_{\infty}^{p}(Z)$ consists of λ with

$${\parallel \lambda \parallel }_{T_{\mathrm{\infty }}^p(Z)}={({\int }_{{\mathbb{S}}_n}\underset{a_k\in \mathrm{\Gamma }(\xi )}{sup}|{\lambda }_k{|}^{p}d\sigma (\xi ))}^{\frac{1}{p}}<\mathrm{\infty }.$$

Another tent space $T_{q}^{\infty}(Z)$ consists of λ such that

$${\parallel \lambda \parallel }_{T_q^{\mathrm{\infty }}(Z)}=\mathrm{ess}\underset{\xi \in {\mathbb{S}}_n}{sup}{(\underset{w\in \mathrm{\Gamma }(\xi )}{sup}\frac{1}{{(1-|w{|}^2)}^n}\sum _{a_k\in Q(w)}|{\lambda }_k{|}^q{(1-|a_k{|}^2)}^n)}^{\frac{1}{q}}<\mathrm{\infty }.$$

We will need the following duality results for the tent spaces of sequences. The proof can be found in [13, 14, 17].

Lemma E

Let $1\leq p<\infty$ and $Z=\{a_{k}\}$ be an r-lattice. If $1< q<\infty$, then the dual of $T_{q}^{p}(Z)$ is isomorphic to $T_{q'}^{p'}(Z)$ under the pairing

$$ \langle c,d\rangle _{T^{2}_{2}(Z)}=\sum_{k}c_{k} \overline{d_{k}}\bigl(1- \vert a_{k} \vert ^{2}\bigr)^{n}, \quad c=\{c_{k}\}\in T^{p}_{q}(Z), d=\{d_{k}\}\in T^{p'}_{q'}(Z).$$

If $0< q\leq 1$, then the dual of $T_{q}^{p}(Z)$ is isomorphic to $T_{\infty}^{p'}(Z)$ under the pairing above.

The following result originates from [20], which will be used to construct our test functions.

Lemma F

Let $0< p,q<\infty$ and $Z=\{a_{k}\}$ be an r-lattice. If $\theta >n \max (1,\frac{q}{p},\frac{1}{p},\frac{1}{q})$, then the operator

$$ S_{Z}\{\lambda _{k}\}(z)=\sum _{k=1}^{\infty}\lambda _{k} \frac{(1- \vert a_{k} \vert ^{2})^{\theta}}{(1-\langle z,a_{k}\rangle )^{\theta +\frac{n+1+\alpha}{q}}}$$

is bounded from $T_{q}^{p}(Z)$ to ${\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)$.

We will also need the following result concerning factorization of sequence tent spaces, which can be found in [19].

Theorem G

Let $0< p,q<\infty$ and $Z=\{a_{k}\}$ be a δ-lattice. If $p< p_{1},p_{2}<\infty$, $q< q_{1},q_{2}<\infty$ and satisfying

$$ \frac{1}{p_{1}}+\frac{1}{p_{2}}=\frac{1}{p} \quad {\textit{and}} \quad \frac{1}{q_{1}}+\frac{1}{q_{2}}=\frac{1}{q},$$

then

$$ T^{p}_{q}(Z)=T^{p_{1}}_{q_{1}}(Z)\cdot T^{p_{2}}_{q_{2}}(Z).$$

2.6 Discretization

We will use Khinchine’s and Kahane’s inequalities throughout the proof of our main results. These tools provide discrete version of the conditions we really need, hence, we need to obtain the continuous characterizations from the discrete ones. The following two results can be found in [19].

Lemma H

Let $0< p,q<\infty$ and $\alpha >-n-1$. There exist $r_{0}\in (0,1)$ so that if $0< r< r_{0}$ and $Z=\{a_{k}\}$ is an r-lattice, then

$$\begin{array}{rl}& {\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|f(z){|}^q{(1-|z{|}^2)}^{\alpha }dv(z))}^{p/q}d\sigma (\xi )\\ & \lesssim {\int }_{{\mathbb{S}}_n}{\left(\sum _{a_k\in \mathrm{\Gamma }(\xi )}\right|f(a_k){|}^q{(1-|a_k{|}^2)}^{n+1+\alpha })}^{p/q}d\sigma (\xi ),\end{array}$$

whenever f is holomorphic on ${\mathbb{B}}_n$ and in $\mathrm{T}_{q,\alpha}^{p}$.

Lemma I

Let $0< p<\infty$ and $\alpha \geq 0$. There exist $r_{0}\in (0,1)$ so that if $0< r< r_{0}$ and $Z=\{a_{k}\}$ is an r-lattice, then

$${\int }_{{\mathbb{S}}_n}\underset{z\in \mathrm{\Gamma }(\xi )}{sup}|f(z){|}^p{(1-|z{|}^2)}^{\alpha }d\sigma (\xi )\lesssim {\int }_{{\mathbb{S}}_n}\underset{a_k\in \mathrm{\Gamma }(\xi )}{sup}|f(a_k){|}^p{(1-|a_k{|}^2)}^{\alpha }d\sigma (\xi ),$$

whenever f is holomorphic on ${\mathbb{B}}_n$ such that the left-hand side is finite.

We also need the following similar result.

Lemma J

Let $0< p<\infty$, and $\alpha >-n-1$, $\beta >0$. There exist $r_{0}\in (0,1)$ so that if $0< r< r_{0}$ and $Z=\{a_{k}\}$ is an r-lattice, then for any $a\in {\mathbb{B}}_n$, we have

$$\begin{array}{rl}& {\int }_{{\mathbb{B}}_n}\frac{{(1-|a{|}^2)}^{\beta }}{|1-〈a,z〉{|}^{n+\beta }}|f(z){|}^p{(1-|z{|}^2)}^{\alpha }dv(z)\\ & \lesssim \sum _k\frac{{(1-|a{|}^2)}^{\beta }}{|1-〈a,a_k〉{|}^{n+\beta }}|f(a_k){|}^p{(1-|a_k{|}^2)}^{n+1+\alpha },\end{array}$$

whenever f is holomorphic on ${\mathbb{B}}_n$ such that the left-hand side is finite.

Proof

For any $a\in {\mathbb{B}}_n$ and $\beta >0$, note that

$$\begin{array}{rl}& {\int }_{{\mathbb{B}}_n}\frac{{(1-|a{|}^2)}^{\beta }}{|1-〈a,z〉{|}^{n+\beta }}|f(z){|}^p{(1-|z{|}^2)}^{\alpha }dv(z)\\ & \asymp \sum _k{\int }_{D(a_k,r)}\frac{{(1-|a{|}^2)}^{\beta }}{|1-〈a,z〉{|}^{n+\beta }}|f(z){|}^p{(1-|z{|}^2)}^{\alpha }dv(z)\\ & \lesssim \sum _k{\int }_{D(a_k,r)}\frac{{(1-|a{|}^2)}^{\beta }}{|1-〈a,z〉{|}^{n+\beta }}|f(z)-f(a_k){|}^p{(1-|z{|}^2)}^{\alpha }dv(z)\\ & +\sum _k{\int }_{D(a_k,r)}\frac{{(1-|a{|}^2)}^{\beta }}{|1-〈a,z〉{|}^{n+\beta }}|f(a_k){|}^p{(1-|z{|}^2)}^{\alpha }dv(z).\end{array}$$

By [15, Lemma 2.2], there exist $r_{0}\in (r,4r)$, such that for any $z\in D(a_{k},r)$,

$$ \bigl\vert f(z)-f(a_{k}) \bigr\vert ^{p}\lesssim \frac{r^{p}}{(1- \vert a_{k} \vert ^{2})^{n+1}} \int _{D(a_{k},r_{0})} \bigl\vert f( \omega ) \bigr\vert ^{p}\,dv(\omega ).$$

Thus, we deduce that

$$\begin{array}{rl}& {\int }_{{\mathbb{B}}_n}\frac{{(1-|a{|}^2)}^{\beta }}{|1-〈a,z〉{|}^{n+\beta }}|f(z){|}^p{(1-|z{|}^2)}^{\alpha }dv(z)\\ & \lesssim r^p\sum _k{\int }_{D(a_k,r)}\frac{{(1-|a{|}^2)}^{\beta }}{|1-〈a,z〉{|}^{n+\beta }}\frac{1}{{(1-|a_k{|}^2)}^{n+1}}{\int }_{D(a_k,r_0)}|f(\omega ){|}^{p}dv(\omega ){(1-|z{|}^2)}^{\alpha }dv(z)\\ & +\sum _k\frac{{(1-|a{|}^2)}^{\beta }}{|1-〈a,a_k〉{|}^{n+\beta }}|f(a_k){|}^p{(1-|a_k{|}^2)}^{n+1+\alpha }\\ & \lesssim r^p\sum _k{\int }_{D(a_k,4r)}\frac{{(1-|a{|}^2)}^{\beta }}{|1-〈a,\omega 〉{|}^{n+\beta }}|f(\omega ){|}^p{(1-|\omega {|}^2)}^{\alpha }dv(\omega )\\ & +\sum _k\frac{{(1-|a{|}^2)}^{\beta }}{|1-〈a,a_k〉{|}^{n+\beta }}|f(a_k){|}^p{(1-|a_k{|}^2)}^{n+1+\alpha }\\ & \lesssim r^p{\int }_{{\mathbb{B}}_n}\frac{{(1-|a{|}^2)}^{\beta }}{|1-〈a,\omega 〉{|}^{n+\beta }}|f(\omega ){|}^p{(1-|\omega {|}^2)}^{\alpha }dv(\omega )\\ & +\sum _k\frac{{(1-|a{|}^2)}^{\beta }}{|1-〈a,a_k〉{|}^{n+\beta }}|f(a_k){|}^p{(1-|a_k{|}^2)}^{n+1+\alpha }.\end{array}$$

Since the constants in “≲” do not depend on r, we can find the desired $r_{0}$, which completes the proof. □

3 Proof of Theorem 1.1

3.1 Necessity

Suppose that the integration operator $J_g:{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)\to H^t({\mathbb{B}}_n)$ is bounded. We consider first the case $p=t$, $q\leq 2$ or $p< t$. In this case, for any $a\in {\mathbb{B}}_n$ and $\theta >0$, consider the test functions

$$F_a(z)=\frac{{(1-|a{|}^2)}^{\theta }}{{(1-〈z,a〉)}^{\theta +\frac{n+1+\alpha }{q}+\frac{n}{p}}},z\in {\mathbb{B}}_n.$$

(1)

By the standard estimate for $H^t({\mathbb{B}}_n)$ functions, we have

$$|Rg(z)||F_a(z)|\lesssim \frac{{\parallel J_g(F_a)\parallel }_{H^t({\mathbb{B}}_n)}}{{(1-|z{|}^2)}^{\frac{n+t}{t}}}\lesssim \parallel J_g\parallel \cdot {\parallel F_a\parallel }_{{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)}{(1-|z{|}^2)}^{-\frac{n}{t}-1}.$$

Replacing z by a in the inequality above, we obtain

$$\begin{aligned} \bigl\vert Rg(a) \bigr\vert \bigl(1- \vert a \vert ^{2} \bigr)^{\frac{q-(n+1+\alpha )}{q}+n(\frac{1}{t}- \frac{1}{p})} \lesssim \Vert J_{g} \Vert < \infty . \end{aligned}$$

In particular, we deduce that $\sup_{k}|Rg(a_{k})|(1-|a_{k}|^{2})^{ \frac{q-(n+1+\alpha )}{q}+n(\frac{1}{t}-\frac{1}{p})} <\infty$ as desired.

Finally, it remains to deal with the other cases. Let $Z=\{a_{k}\}$ be an r-lattice and r be small enough. Consider the test functions

$$ F_{Z}(z)=\sum_{k=1}^{\infty}\lambda _{k} \frac{(1- \vert a_{k} \vert ^{2})^{\theta}r_{k}(x)}{(1-\langle z,a_{k}\rangle )^{\theta +\frac{n+1+\alpha}{q}}},$$

where $\lambda =\{\lambda _{k}\}\in T_{q}^{p}(Z)$, $r_{k}(x)$ are the Rademacher functions, and θ is large enough such that Lemma F holds. Then, by Lemma A and Lemma F, we have

$$\begin{array}{rl}{\parallel J_g(F_Z)\parallel }_{H^t}^t& \asymp {\int }_{{\mathbb{S}}_n}({\int }_{\mathrm{\Gamma }(\xi )}{|R(J_g(F_Z))|}^2{(1-|z{|}^2)}^{1-n}dv(z){)}^{t/2}d\sigma (\xi )\\ & \lesssim {\parallel J_g\parallel }^t{\parallel F_Z\parallel }_{{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)}^t\lesssim {\parallel J_g\parallel }^t{\parallel \lambda \parallel }_{T_q^p(Z)}^t,\end{array}$$

which is equivalent to

$$\begin{array}{rl}& {\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|Rg(z)\sum _{k=1}^{\mathrm{\infty }}\frac{{\lambda }_k{r}_k(x){(1-|a_k{|}^2)}^{\theta }}{{(1-〈z,a_k〉)}^{\theta +\frac{(n+1+\alpha )}{q}}}{|}^2{(1-|z{|}^2)}^{1-n}dv(z))}^{t/2}d\sigma (\xi )\\ & \lesssim {\parallel J_g\parallel }^t{\parallel \lambda \parallel }_{T_q^p(Z)}^t.\end{array}$$

Integrating with respect to x from 0 to 1, and using Fubini’s theorem, Khinchine’s inequality, and Kahane’s inequality as in the proof of Theorem 7 in [19], we obtain

$$\begin{array}{rl}& {\int }_{{\mathbb{S}}_n}{(\sum _{k=1}^{\mathrm{\infty }}|{\lambda }_k{|}^2{\int }_{\mathrm{\Gamma }(\xi )}\frac{{(1-|a_k{|}^2)}^{2\theta }}{|1-〈z,a_k〉{|}^{2\theta +\frac{2(n+1+\alpha )}{q}}}|Rg(z){|}^2{(1-|z{|}^2)}^{1-n}dv(z))}^{t/2}d\sigma (\xi )\\ & \lesssim {\parallel J_g\parallel }^t{\parallel \lambda \parallel }_{T_q^p(Z)}^t.\end{array}$$

Write $u=\{u_{k}\}$ and $u_{k}=|Rg(a_{k})|(1-|a_{k}|^{2})^{\frac{q-(n+1+\alpha )}{q}}$. Using subharmonicity and bearing in mind $\bigcup_{z\in \Gamma (\xi )}D(z,4r)\subset \widetilde{\Gamma}(\xi )$, we obtain

$$\begin{array}{rl}& {\int }_{{\mathbb{S}}_n}{(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|{\lambda }_k{|}^2|Rg(a_k){|}^2{(1-|a_k{|}^2)}^{\frac{2q-2(n+1+\alpha )}{q}})}^{t/2}d\sigma (\xi )\\ & \lesssim {\int }_{{\mathbb{S}}_n}{(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|{\lambda }_k{|}^2{\int }_{D(a_k,4r)}|Rg(z){|}^2\frac{{(1-|z{|}^2)}^{1-n}{(1-|a_k{|}^2)}^{2\theta }}{|1-〈z,a_k〉{|}^{2\theta +\frac{2(n+1+\alpha )}{q}}}dv(z))}^{t/2}d\sigma (\xi )\\ & \lesssim {\int }_{{\mathbb{S}}_n}{[{\int }_{\tilde{\mathrm{\Gamma }}(\xi )}\sum _{k=1}^{\mathrm{\infty }}|{\lambda }_k{|}^2\frac{{(1-|a_k{|}^2)}^{2\theta }}{|1-〈z,a_k〉{|}^{2\theta +\frac{2(n+1+\alpha )}{q}}}|Rg(z){|}^2{(1-|z{|}^2)}^{1-n}dv(z)]}^{t/2}d\sigma \xi )\\ & \lesssim {\parallel J_g\parallel }^t{\parallel \lambda \parallel }_{T_q^p(Z)}^t.\end{array}$$

Therefore,

$${\int }_{{\mathbb{S}}_n}{(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|{\lambda }_k{|}^2|u_k{|}^2)}^{t/2}d\sigma (\xi )\lesssim {\parallel J_g\parallel }^t{\parallel \lambda \parallel }_{T_q^p(Z)}^t.$$

(2)

(a) If $p>t$ and $q>2$, for some s large enough such that $2s>1$ and $ts>1$, we want to prove $u^{1/s}\in T_{\frac{2qs}{q-2}}^{\frac{pts}{p-t}}(Z)$, which is equivalent to $u\in T_{\frac{2q}{q-2}}^{\frac{pt}{p-t}}(Z)$. By the factorization result in Lemma G, we have

$$ T_{\frac{2qs}{q-2}}^{\frac{pts}{p-t}}(Z)= \bigl(T_{ \frac{2qs}{2qs-q+2}}^{\frac{pts}{pts-p+t}}(Z) \bigr)^{*} = \bigl(T_{ \frac{2s}{2s-1}}^{\frac{ts}{ts-1}}(Z)\cdot T_{qs}^{ps}(Z) \bigr)^{*}.$$

Take any $v=\{v_{k}\}\in T_{\frac{2qs}{2qs-q+2}}^{\frac{pts}{pts-p+t}}(Z)$ and factor it as $v_{k}=\rho _{k}\cdot \lambda _{k}^{1/s}$, where $\rho =\{\rho _{k}\}\in T_{\frac{2s}{2s-1}}^{\frac{ts}{ts-1}}(Z)$, $\lambda =\{\lambda _{k}\}\in T_{q}^{p}(Z)$. Then, by (2) and Hölder’s inequalities, we obtain

$$\begin{array}{rl}& \sum _k\left|v_k{u}_k^{1/s}\right|{(1-|a_k{|}^2)}^n\\ & \asymp {\int }_{{\mathbb{S}}_n}(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|{\rho }_k|\cdot |{\lambda }_k{|}^{1/s}\cdot |u_k{|}^{1/s})d\sigma (\xi )\\ & \lesssim {\int }_{{\mathbb{S}}_n}{(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|{\rho }_k{|}^{\frac{2s}{2s-1}})}^{\frac{2s-1}{2s}}{(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|{\lambda }_k{|}^2|u_k{|}^2)}^{\frac{1}{2s}}d\sigma (\xi )\\ & \lesssim {({\int }_{{\mathbb{S}}_n}{(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|{\rho }_k{|}^{\frac{2s}{2s-1}})}^{\frac{2s-1}{2s}\cdot \frac{ts}{ts-1}}d\sigma (\xi ))}^{\frac{ts-1}{ts}}{({\int }_{{\mathbb{S}}_n}{(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|{\lambda }_k{|}^2|u_k{|}^2)}^{\frac{t}{2}}d\sigma (\xi ))}^{\frac{1}{ts}}\\ & \lesssim {\parallel \rho \parallel }_{T_{\frac{2s}{2s-1}}^{\frac{ts}{ts-1}}(Z)}{\parallel J_g\parallel }^{1/s}{\parallel \lambda \parallel }_{T_q^p(Z)}^{1/s}\\ & \asymp {\parallel J_g\parallel }^{1/s}{\parallel v\parallel }_{T_{\frac{2qs}{2qs-q+2}}^{\frac{pts}{pts-p+t}}(Z)}.\end{array}$$

By the duality of tent spaces of sequences given in Lemma E, we have that u belongs to $T_{\frac{2q}{q-2}}^{\frac{pt}{p-t}}(Z)$.

(b) If $p>t$ and $q\leq 2$, it is sufficient to show that $u^{1/s}\in T_{\infty}^{\frac{pts}{p-t}}(Z)$ for some s large enough such that $2s>1$ and $ts>1$. By Lemma E and Lemma G, we have

$$ T_{\infty}^{\frac{pts}{p-t}}(Z)= \bigl(T_{\frac{2s}{2s-1}}^{ \frac{ts}{ts-1}}(Z)\cdot T_{qs}^{ps}(Z) \bigr)^{*}.$$

Note that if $q\leq 2$, then $\frac{2s-1}{2s}+\frac{1}{qs}=\frac{1}{\delta}$ for some $\delta \leq 1$. Thus, making some adjustments to the arguments in the proof of (a), we obtain that u belongs to $T_{\infty}^{\frac{pt}{p-t}}(Z)$.

(c) If $p=t$ and $q>2$, it suffices to prove $u^{1/s}\in T_{\frac{2qs}{q-2}}^{\infty}(Z)$ for some s large enough such that $2s>1$ and $ts>1$. An appeal to Lemma G gives that

$$ T_{\frac{2qs}{q-2}}^{\infty}(Z)= \bigl(T_{\frac{2qs}{2qs-q+2}}^{1}(Z) \bigr)^{*}= \bigl(T_{\frac{2s}{2s-1}}^{\frac{ps}{ps-1}}(Z)\cdot T_{qs}^{ps}(Z) \bigr)^{*}.$$

Proceeding with the argument as above again, we have that u belongs to $T_{\frac{2q}{q-2}}^{\infty}(Z)$, which finishes the proof of necessity.

3.2 Sufficiency

To prove the sufficiency of Theorem 1.1, we split it into four cases.

(a) If $p>t$, $q>2$ and $u\in T_{\frac{2q}{q-2}}^{\frac{pt}{p-t}}(Z)$, let $\eta =(1-n-\frac{2\alpha}{q})\frac{q}{q-2}$. By considering the dilated functions $Rg_{\rho}(z)=Rg(\rho z)$ ($0<\rho <1$), an approximation argument (see [21, Lemma 7]) shows that according to Lemma H, we have

$$\begin{array}{rl}& {\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|Rg(z){|}^{\frac{2q}{q-2}}{(1-|z{|}^2)}^{\eta }dv(z))}^{\frac{q-2}{2q}\frac{pt}{p-t}}d\sigma (\xi )\\ & \lesssim {\int }_{{\mathbb{S}}_n}{\left(\sum _{a_k\in \mathrm{\Gamma }(\xi )}\right|Rg(a_k){|}^{\frac{2q}{q-2}}{(1-|a_k{|}^2)}^{n+1+\eta })}^{\frac{q-2}{2q}\frac{pt}{p-t}}d\sigma (\xi )\\ & ={\parallel u\parallel }_{T_{\frac{2q}{q-2}}^{\frac{pt}{p-t}}(Z)}^{\frac{pt}{p-t}}<\mathrm{\infty },\end{array}$$

which means $Rg\in {\mathcal{HT}}_{\frac{2q}{q-2},\eta }^{\frac{pt}{p-t}}({\mathbb{B}}_n)$. Then, by Lemma A and Holder’s inequalities, we have

$$\begin{array}{rl}& {\parallel J_{g}f\parallel }_{H^t({\mathbb{B}}_n)}^t\\ & \asymp {\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|f(z){|}^2|Rg(z){|}^2{(1-|z{|}^2)}^{1-n}dv(z))}^{\frac{t}{2}}d\sigma (\xi )\\ & \lesssim {\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|f(z){|}^q{(1-|z{|}^2)}^{\alpha }dv(z))}^{\frac{t}{q}}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|Rg(z){|}^{\frac{2q}{q-2}}{(1-|z{|}^2)}^{\eta }dv(z))}^{\frac{t(q-2)}{2q}}d\sigma (\xi )\\ & \lesssim {({\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|f(z){|}^q{(1-|z{|}^2)}^{\alpha }dv(z))}^{\frac{p}{q}}d\sigma (\xi ))}^{\frac{t}{p}}\\ & \cdot {({\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|Rg(z){|}^{\frac{2q}{q-2}}{(1-|z{|}^2)}^{\eta }dv(z))}^{\frac{t(q-2)}{2q}\frac{p}{p-t}}d\sigma (\xi ))}^{\frac{p-t}{p}}\\ & \lesssim {\parallel f\parallel }_{{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)}^t\cdot {\parallel Rg\parallel }_{{\mathcal{HT}}_{\frac{2q}{q-2},\eta }^{\frac{pt}{p-t}}({\mathbb{B}}_n)}^t.\end{array}$$

(b) If $p>t$ and $q\leq 2$ and $u\in T_{\infty}^{\frac{pt}{p-t}}(Z)$, define

$$U_g(\xi )=\underset{z\in \mathrm{\Gamma }(\xi )}{sup}|Rg(z)|{(1-|z{|}^2)}^{\frac{q-(n+1+\alpha )}{q}},\xi \in {\mathbb{S}}_n.$$

Using the approximation argument with Lemma I, we obtain

$${\int }_{{\mathbb{S}}_n}|U_g(\xi ){|}^{\frac{pt}{p-t}}d\sigma (\xi )\lesssim {\int }_{{\mathbb{S}}_n}\underset{a_k\in \mathrm{\Gamma }(\xi )}{sup}|u_k{|}^{\frac{pt}{p-t}}d\sigma (\xi )={\parallel u\parallel }_{T_{\mathrm{\infty }}^{\frac{pt}{p-t}}(Z)}^{\frac{pt}{p-t}}<\mathrm{\infty },$$

which means $U_{g}$ belongs to $L^{\frac{pt}{p-t}}({\mathbb{S}}_n)$. Let $\beta =\alpha +(\frac{2}{q}-1)(n+1+\alpha )$. Then, applying Lemma A, Hölder’s inequality, and Lemma B, we have

$$\begin{array}{rl}& {\parallel J_{g}f\parallel }_{H^t({\mathbb{B}}_n)}^t\\ & \asymp {\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|f(z){|}^2|Rg(z){|}^2{(1-|z{|}^2)}^{1-n}dv(z))}^{t/2}d\sigma (\xi )\\ & \lesssim {\int }_{{\mathbb{S}}_n}\underset{z\in \mathrm{\Gamma }(\xi )}{sup}|Rg(z){|}^t{(1-|z{|}^2)}^{\frac{(1-n-\beta )t}{2}}\cdot {\left({\int }_{\mathrm{\Gamma }(\xi )}\right|f(z){|}^2{(1-|z{|}^2)}^{\beta }dv(z))}^{t/2}d\sigma (\xi )\\ & \lesssim {({\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|f(z){|}^2{(1-|z{|}^2)}^{\beta }dv(z))}^{p/2}d\sigma (\xi ))}^{t/p}\\ & \cdot {\left({\int }_{{\mathbb{S}}_n}\underset{z\in \mathrm{\Gamma }(\xi )}{sup}\right|Rg(z){|}^{\frac{pt}{p-t}}{(1-|z{|}^2)}^{\frac{q-(n+1+\alpha )}{q}\frac{pt}{p-t}}d\sigma (\xi ))}^{\frac{p-t}{p}}\\ & ={\parallel f\parallel }_{{\mathcal{HT}}_{2,\beta }^p({\mathbb{B}}_n)}^t\cdot {\parallel U_g\parallel }_{L^{\frac{pt}{p-t}}({\mathbb{S}}_n)}^t\lesssim {\parallel f\parallel }_{{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)}^t\cdot {\parallel U_g\parallel }_{L^{\frac{pt}{p-t}}({\mathbb{S}}_n)}^t.\end{array}$$

(c) If $p=t$, $q>2$ and $u\in T_{\frac{2q}{q-2}}^{\infty}(Z)$, by Lemma J, we can obtain

$$\underset{w\in {\mathbb{B}}_n}{sup}\frac{1}{{(1-|w{|}^2)}^n}{\int }_{Q(w)}|Rg(z){|}^{\frac{2q}{q-2}}{(1-|z{|}^2)}^{\frac{q-(n+1+\alpha )}{q}\frac{2q}{q-2}-1}dv(z)\lesssim {\parallel u\parallel }_{T_{\frac{2q}{q-2}}^{\mathrm{\infty }}(Z)}^{\frac{2q}{q-2}}<\mathrm{\infty },$$

which means $Rg\in {\mathcal{CT}}_{\frac{2q}{q-2},\eta }({\mathbb{B}}_n)$, where $\eta =(1-n-\frac{2\alpha}{q})\frac{q}{q-2}$. Applying the embedding theorem for Hardy spaces, we obtain that for any $\xi \in {\mathbb{S}}_n$,

$$\begin{array}{rl}{\int }_{\mathrm{\Gamma }(\xi )}|Rg(z){|}^{\frac{2q}{q-2}}{(1-|z{|}^2)}^{\eta }dv(z)& \lesssim {\int }_{{\mathbb{B}}_n}\frac{{\chi }_{\mathrm{\Gamma }(\xi )}(z)}{|1-〈z,\xi 〉{|}^n}|Rg(z){|}^{\frac{2q}{q-2}}{(1-|z{|}^2)}^{n+\eta }dv(z)\\ & \lesssim {\parallel Rg\parallel }_{{\mathcal{CT}}_{\frac{2q}{q-2},\eta }({\mathbb{B}}_n)}^{\frac{2q}{q-2}}\underset{0<\rho <1}{sup}{\parallel \frac{{\chi }_{\mathrm{\Gamma }(\xi )}(\cdot )}{{(1-〈\cdot ,\xi 〉)}^n}\parallel }_{L^1(\rho {\mathbb{S}}_n)}\\ & \lesssim {\parallel Rg\parallel }_{{\mathcal{CT}}_{\frac{2q}{q-2},\eta }({\mathbb{B}}_n)}^{\frac{2q}{q-2}},\end{array}$$

where $\chi _{\Gamma (\xi )}$ is the characteristic function of $\Gamma (\xi )$. Then, Lemma A and Hölder’s inequality give that

$$\begin{array}{rl}& {\parallel J_{g}f\parallel }_{H^t({\mathbb{B}}_n)}^t\\ & \asymp {\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|f(z){|}^2|Rg(z){|}^2{(1-|z{|}^2)}^{1-n}dv(z))}^{\frac{t}{2}}d\sigma (\xi )\\ & \lesssim {\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|f(z){|}^q{(1-|z{|}^2)}^{\alpha }dv(z))}^{\frac{t}{q}}{\left({\int }_{\mathrm{\Gamma }(\xi )}\right|Rg(z){|}^{\frac{2q}{q-2}}{(1-|z{|}^2)}^{\eta }dv(z))}^{\frac{t(q-2)}{2q}}d\sigma (\xi )\\ & \lesssim \underset{\xi \in {\mathbb{S}}_n}{sup}{\int }_{\mathrm{\Gamma }(\xi )}|Rg(z){|}^{\frac{2q}{q-2}}{(1-|z{|}^2)}^{\eta }dv(z)\cdot {\parallel f\parallel }_{{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)}^t\\ & \lesssim {\parallel Rg\parallel }_{{\mathcal{CT}}_{\frac{2q}{q-2},\eta }({\mathbb{B}}_n)}^{\frac{2q}{q-2}}\cdot {\parallel f\parallel }_{{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)}^t.\end{array}$$

(d) First, the case for $p=t$, $q\leqslant 2$ is particularly simple. Indeed, in this case, Lemma B implies that ${\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)\subset {\mathcal{HT}}_{2,\beta }^p({\mathbb{B}}_n)$, where $\beta =\alpha +(\frac{2}{q}-1)(n+1+\alpha )$. Since $u_{k}(1-|a_{k}|^{2})^{n(\frac{1}{t}-\frac{1}{p})}\in l^{\infty}$, we can obtain

$$\underset{z\in {\mathbb{B}}_n}{sup}|Rg(z)|{(1-|z{|}^2)}^{\frac{q-(n+1+\alpha )}{q}+n(\frac{1}{t}-\frac{1}{p})}<\mathrm{\infty }.$$

Then, we have

$${\parallel J_{g}f\parallel }_{H^t({\mathbb{B}}_n)}^t\lesssim {\parallel f\parallel }_{{\mathcal{HT}}_{2,\beta }^t({\mathbb{B}}_n)}^t\cdot \underset{z\in {\mathbb{B}}_n}{sup}|Rg(z){|}^t{(1-|z{|}^2)}^{\frac{q-(n+1+\alpha )}{q}\cdot t}\lesssim {\parallel f\parallel }_{{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)}^t.$$

Next, for the remaining case $p< t$, there exists some r such that $p< r< t$ and denote that $\eta =(\frac{r}{p}-1)n-1+\frac{r(n+1+\alpha )}{q}$. Then, according to Lemma D and Lemma C, we have

$$\begin{array}{rl}{\parallel J_{g}f\parallel }_{H^t({\mathbb{B}}_n)}^t& \lesssim {\parallel R(J_{g}f)\parallel }_{A_{r-n-1+\frac{nr}{t}}^r({\mathbb{B}}_n)}^t\\ & ={\left({\int }_{{\mathbb{B}}_n}\right|f(z){|}^r|Rg(z){|}^r{(1-|z{|}^2)}^{r-n-1+\frac{nr}{t}}dv(z))}^{t/r}\\ & \lesssim {\left({\int }_{{\mathbb{B}}_n}\right|f(z){|}^r{(1-|z{|}^2)}^{\eta }dv(z))}^{t/r}\cdot \underset{z\in {\mathbb{B}}_n}{sup}|Rg(z){|}^t{(1-|z{|}^2)}^{\frac{q-(n+1+\alpha )}{q}\cdot t+nt(\frac{1}{t}-\frac{1}{p})}\\ & \lesssim {\parallel f\parallel }_{A_{\eta }^r({\mathbb{B}}_n)}^t\lesssim {\parallel f\parallel }_{{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)}^t.\end{array}$$

Theorem 1.1 is now proven.

4 Proof of Theorem 1.2

4.1 Necessity

Suppose $J_g:{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)\to H^t({\mathbb{B}}_n)$ is compact. It is obvious that (a) holds by Theorem 1.1, so we only need to prove (b), (c), and (d). Denote

$$ E= \bigl\{ \lambda =\{\lambda _{k}\}\in T^{p}_{q}(Z): \Vert \lambda \Vert _{T^{p}_{q}(Z)} = 1 \bigr\} $$

to be the unit sphere of $T^{p}_{q}(Z)$, and let

$$S_Z(\lambda )(z)=\sum _{k=1}^{\mathrm{\infty }}{\lambda }_k\frac{{(1-|a_k{|}^2)}^{\theta }}{{(1-〈z,a_k〉)}^{\theta +\frac{n+1+\alpha }{q}}},z\in {\mathbb{B}}_n$$

be the bounded operator defined in Lemma F, where $Z=\{a_{k}\}$ is an r-lattice and r is small enough. Since $S_{Z}(E)$ is a bounded set and $J_{g}$ is compact, the set $J_{g}\circ S_{Z}(E)$ is relatively compact in $H^t({\mathbb{B}}_n)$. It is well known that a relatively compact set must be a totally bounded set, and then for any $\varepsilon >0$, there exist a finite number of functions $h_{1},\ldots,h_{N}$, such that $J_{g}\circ S_{Z}(E)\subset \bigcup_{i=1}^{N} B(h_{i}, \frac{\varepsilon}{2})$, where $B(h,\frac{\epsilon }{2}):=\{f\in J_g\circ S_Z(E):{\parallel f-h\parallel }_{H^t({\mathbb{B}}_n)}<\frac{\epsilon }{2}\}$. Observing that ${sup}_{i=1,\dots ,N}{\parallel h_i\parallel }_{H^t({\mathbb{B}}_n)}<\mathrm{\infty }$, for the above $\varepsilon >0$, there exists $\rho _{0}\in (0,1)$ such that

$$\underset{i=1,\dots ,N}{sup}{({\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )\setminus \overline{D(0,\rho )}}\right|R{h}_i(z){|}^2{(1-|z{|}^2)}^{1-n}dv(z))}^{t/2}d\sigma (\xi ))}^{1/t}<\frac{\epsilon }{2}$$

whenever $\rho >\rho _{0}$. Thus, for any $\lambda \in E$, there exists some $i_{0}\in \{1,\ldots,N\}$ such that $J_{g}\circ S_{Z}(\lambda )\in B(h_{i_{0}},\frac{\varepsilon}{2})$, and we can deduce that

$$\begin{array}{rl}& {({\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )\setminus \overline{D(0,\rho )}}\right|Rg(z)S_Z(\lambda )(z){|}^2{(1-|z{|}^2)}^{1-n}dv(z))}^{t/2}d\sigma (\xi ))}^{1/t}\\ & \lesssim {({\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )\setminus \overline{D(0,\rho )}}\right|Rg(z)S_Z(\lambda )(z)-R{h}_{i_0}(z){|}^2{(1-|z{|}^2)}^{1-n}dv(z))}^{t/2}d\sigma (\xi ))}^{1/t}\\ & +{({\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )\setminus \overline{D(0,\rho )}}\right|R{h}_{i_0}(z){|}^2{(1-|z{|}^2)}^{1-n}dv(z))}^{t/2}d\sigma (\xi ))}^{1/t}\\ & \lesssim {\parallel J_g\circ S_Z(\lambda )-h_{i_0}\parallel }_{H^t({\mathbb{B}}_n)}+\frac{\epsilon }{2}<\epsilon \end{array}$$

whenever $\rho >\rho _{0}$, which is the same as

$$\begin{array}{rl}& {\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )\setminus \overline{D(0,\rho )}}\right|\sum _{k=1}^{\mathrm{\infty }}{\lambda }_k\frac{{(1-|a_k{|}^2)}^{\theta }}{{(1-〈z,a_k〉)}^{\theta +\frac{n+1+\alpha }{q}}}{|}^2|Rg(z){|}^2{(1-|z{|}^2)}^{1-n}dv(z))}^{t/2}d\sigma (\xi )\\ & \lesssim {\epsilon }^t{\parallel \lambda \parallel }_{T_q^p(Z)}^t\end{array}$$

for any $\lambda \in T_{q}^{p}(Z)$ and $\rho >\rho _{0}$. Let $r_{k}(x)$ be the Rademacher functions. Replacing $\lambda _{k}$ by $\lambda _{k}r_{k}(x)$, and utilizing the same method as in the proof of the corresponding case in Theorem 1.1, we obtain that

$${\int }_{{\mathbb{S}}_n}{(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|{\lambda }_k{|}^2|Rg(a_k){|}^2{(1-|a_k{|}^2)}^{\frac{2q-2(n+1+\alpha )}{q}}\cdot {\chi }_{\{|z|\ge \rho \}}(a_k))}^{t/2}d\sigma (\xi )\lesssim {\epsilon }^t{\parallel \lambda \parallel }_{T_q^p(Z)}^t$$

for $\rho >\rho _{0}':=\inf \{|a_{k}|:D(a_{k},\delta )\subset \{|z|\geq \rho _{0}\}\}$, where $\chi _{\{|z|\geq \rho \}}$ is the characteristic function. Denote

$$ u_{\rho}=\{u_{\rho ,k}\}= \bigl\{ \bigl\vert Rg(a_{k}) \bigr\vert \bigl(1- \vert a_{k} \vert ^{2}\bigr)^{ \frac{q-(n+1+\alpha )}{q}}\cdot \chi _{\{ \vert z \vert \geq \rho \}}(a_{k}) \bigr\} .$$

Then, we have

$${\int }_{{\mathbb{S}}_n}{(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|{\lambda }_k{|}^2|u_{\rho ,k}{|}^2)}^{t/2}d\sigma (\xi )\lesssim {\epsilon }^t{\parallel \lambda \parallel }_{T_q^p(Z)}^t\text{for any}\rho >{\rho }_0^{\prime }.$$

(3)

(b) If $p>t$ and $q\leq 2$, applying the duality and factorization of sequence tent spaces as in the proof of Theorem 1.1, we can obtain the desired result. To this end, it is sufficient to prove that for some s large enough such that $2s>1$ and $ts>1$, $\|u_{\rho}^{1/s}\|_{T^{\frac{pts}{p-t}}_{\infty}(Z)}\lesssim \varepsilon ^{t}$ whenever $\rho >\rho _{0}'$, i.e.,

$$\underset{\rho >{\rho }_0^{\prime }}{sup}{\left({\int }_{{\mathbb{S}}_n}\underset{a_k\in \mathrm{\Gamma }(\xi )\setminus \overline{D(0,\rho )}}{sup}\right|Rg(a_k){|}^{\frac{pt}{p-t}}{(1-|a_k{|}^2)}^{\frac{q-(n+1+\alpha )}{q}\frac{pt}{p-t}}d\sigma (\xi ))}^{\frac{p-t}{pts}}\lesssim {\epsilon }^t.$$

By Lemma E and Lemma G, we have

$$ T_{\infty}^{\frac{pts}{p-t}}(Z)= \bigl(T_{\delta}^{ \frac{pts}{pts-p+t}}(Z) \bigr)^{*}= \bigl(T_{\frac{2s}{2s-1}}^{ \frac{ts}{ts-1}}(Z)\cdot T_{qs}^{ps}(Z) \bigr)^{*}.$$

Note that if $q\leq 2$, then $\frac{2s-1}{2s}+\frac{1}{qs}=\frac{1}{\delta}$ for some $\delta \leq 1$. Take $v=\{v_{k}\}\in T_{\delta}^{\frac{pts}{pts-p+t}}(Z)$ and factor it as $v_{k}=l_{k}\cdot \lambda _{k}^{1/s}$, where $l=\{l_{k}\}\in T_{\frac{2s}{2s-1}}^{\frac{ts}{ts-1}}(Z)$, $\lambda =\{\lambda _{k}\}\in T_{q}^{p}(Z)$. Then, using Hölder’s inequalities, we obtain

$$\begin{array}{rl}\left|\sum _k{v}_k{u}_{\rho ,k}^{1/s}{(1-|a_k{|}^2)}^n\right|& \lesssim {\int }_{{\mathbb{S}}_n}(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|l_k|\cdot |{\lambda }_k{|}^{1/s}\cdot |u_{\rho ,k}{|}^{1/s})d\sigma (\xi )\\ & \lesssim {\int }_{{\mathbb{S}}_n}{(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|l_k{|}^{\frac{2s}{2s-1}})}^{\frac{2s-1}{2s}}{(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|{\lambda }_k{|}^2|u_{\rho ,k}{|}^2)}^{\frac{1}{2s}}d\sigma (\xi )\\ & \lesssim {\parallel l\parallel }_{T_{\frac{2s}{2s-1}}^{\frac{ts}{ts-1}}(Z)}{({\int }_{{\mathbb{S}}_n}{(\sum _{a_k\in \mathrm{\Gamma }(\xi )}|{\lambda }_k{|}^2|u_{\rho ,k}{|}^2)}^{\frac{t}{2}}d\sigma (\xi ))}^{\frac{1}{ts}}.\end{array}$$

Combining this with (3), we establish that

$$ \biggl\vert \sum_{k}v_{k}u_{\rho ,k}^{1/s} \bigl(1- \vert a_{k} \vert ^{2}\bigr)^{n} \biggr\vert \lesssim \Vert l \Vert _{T_{\frac{2s}{2s-1}}^{\frac{ts}{ts-1}}(Z)} \varepsilon ^{1/s} \Vert \lambda \Vert _{T_{q}^{p}(Z)}^{1/s}$$

whenever $\rho >\rho _{0}'$. Considering all possible factorizations yields

$$ \biggl\vert \sum_{k}v_{k}u_{\rho ,k}^{1/s} \bigl(1- \vert a_{k} \vert ^{2}\bigr)^{n} \biggr\vert \lesssim \varepsilon ^{1/s} \Vert v \Vert _{T_{\delta}^{\frac{pts}{pts-p+t}}(Z)}$$

whenever $\rho >\rho _{0}'$. By the duality of tent spaces of sequences given in Lemma E, we have $\|u_{\rho}^{1/s}\|_{T^{\frac{pts}{p-t}}_{\infty}(Z)}\lesssim \varepsilon ^{t}$ whenever $\rho >\rho _{0}'$.

(c) If $p=t$ and $q>2$, observing that

$$ \lim_{ \vert w \vert \rightarrow 1^{-}}\frac{1}{(1- \vert w \vert ^{2})^{n}}\sum_{a_{k}\in Q(w)} \bigl( \bigl\vert Rg(a_{k}) \bigr\vert \bigl(1- \vert a_{k} \vert ^{2}\bigr)^{ \frac{q-(n+1+\alpha )}{q}} \bigr)^{\frac{2q}{q-2}}\bigl(1- \vert a_{k} \vert ^{2} \bigr)^{n}=0,$$

is equivalent to

$$\underset{\rho \to 1^{-}}{lim}\underset{w\in {\mathbb{B}}_n}{sup}\frac{1}{{(1-|w{|}^2)}^n}\sum _{a_k\in Q(w)}|u_{\rho ,k}{|}^{2q/(q-2)}{(1-|a_k{|}^2)}^n=0,$$

it suffices to prove for some s large enough such that $2s>1$ and $ts>1$, $\|u_{\rho}^{1/s}\|_{T^{\infty}_{2q/(q-2)}(Z)}\lesssim \varepsilon ^{t}$ whenever $\rho >\rho _{0}'$. An appeal to Lemma G gives that

$$ T_{\frac{2qs}{q-2}}^{\infty}(Z)= \bigl(T_{\frac{2qs}{2qs-q+2}}^{1}(Z) \bigr)^{*}= \bigl(T_{\frac{2s}{2s-1}}^{\frac{ps}{ps-1}}(Z)\cdot T_{qs}^{ps}(Z) \bigr)^{*}.$$

Proceeding with the similar argument as above, we can obtain the desired result.

(d) If $p=t$, $q\leq 2$ or $p< t$, note that $|F_{a}(z)|\rightarrow 0$ uniformly on any compact subsets of ${\mathbb{B}}_n$, as $|a|\rightarrow 1^{-}$, where $F_{a}$ are defined in (1). The compactness of $J_{g}$ implies that

$$ \lim_{ \vert a \vert \rightarrow 1^{-}} \Vert J_{g}F_{a} \Vert _{H^{t}}=0.$$

By the standard pointwise estimate for the derivative of $H^t({\mathbb{B}}_n)$ functions, and replacing z by a, we obtain

$$ \lim_{ \vert a \vert \rightarrow 1^{-}} \bigl\vert Rg(a) \bigr\vert \bigl(1- \vert a \vert ^{2}\bigr)^{ \frac{q-(n+1+\alpha )}{q}+ n(\frac{1}{t}-\frac{1}{p})}=0,$$

which is the same as

$$ \lim_{k\rightarrow \infty} \bigl\vert Rg(a_{k}) \bigr\vert \bigl(1- \vert a_{k} \vert ^{2}\bigr)^{ \frac{q-(n+1+\alpha )}{q}+ n(\frac{1}{t}-\frac{1}{p})}=0.$$

Then, the proof of necessity is complete.

4.2 Sufficiency

To prove the compactness of $J_g:{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)\to H^t({\mathbb{B}}_n)$, let ${\{f_k\}}_{k=1}^{\mathrm{\infty }}\subset {\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)$ and satisfy ${sup}_k{\parallel f_k\parallel }_{{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)}<\mathrm{\infty }$. Then, $\{f_{k}\}$ is uniformly bounded on compact subsets of ${\mathbb{B}}_n$, and hence $\{f_{k}\}$ forms a normal family by Montel’s theorem. Therefore, we can extract a subsequence $\{f_{n_{k}}\}_{k=1}^{\infty}$ that converges uniformly on compact subsets of ${\mathbb{B}}_n$ to a holomorphic function f. Fatou’s Lemma shows that $f\in {\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)$. Denote $h_{k}=f_{n_{k}}-f$, then $h_k\in {\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)$. We just need to prove that $\lim_{k\rightarrow \infty}\|J_{g}h_{k}\|_{H^{t}}=0$, which can yield that $J_g:{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)\to H^t({\mathbb{B}}_n)$ is compact.

(a) If $p>t$ and $q>2$ with

$${\int }_{{\mathbb{S}}_n}{\left(\underset{a_k\in \mathrm{\Gamma }(\xi )}{sup}\right|Rg(a_k){|}^{\frac{2q}{q-2}}{(1-|a_k{|}^2)}^{\frac{q-(n+1+\alpha )}{q}\cdot \frac{2q}{q-2}})}^{\frac{pt}{p-t}\cdot \frac{q-2}{2q}}d\sigma (\xi )<\mathrm{\infty },$$

according to the proof of Theorem 1.1, we have $Rg\in {\mathcal{HT}}_{\frac{2q}{q-2},\eta }^{\frac{pt}{p-t}}({\mathbb{B}}_n)$, where $\eta =(1-n-\frac{2\alpha}{q})\frac{q}{q-2}$. Thus, by the dominated convergence theorem, for any $\varepsilon > 0$, there exists $\rho _{0}\in (0,1)$ such that

$$\underset{\rho \ge {\rho }_0}{sup}{({\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )\setminus \overline{D(0,{\rho }_0)}}\right|Rg(z){|}^{\frac{2q}{q-2}}{(1-|z{|}^2)}^{\eta }dv(z))}^{\frac{pt}{p-t}\frac{q-2}{2q}}d\sigma (\xi ))}^{\frac{p-t}{pt}}<\epsilon .$$

Observing that $|h_{k}(z)|\rightarrow 0$ uniformly on any compact subsets of ${\mathbb{B}}_n$, we can choose $k_{0}$ large enough such that $|h_{k}(z)|<\varepsilon$ for any $k\geq k_{0}$ and $|z|\leq \rho _{0}$, and then we have

$$\begin{array}{rl}& {\parallel J_g{h}_k\parallel }_{H^t({\mathbb{B}}_n)}^t\\ & \asymp {\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )\cap \{|z|\le {\rho }_0\}}\right|h_k(z){|}^2|Rg(z){|}^2{(1-|z{|}^2)}^{1-n}dv(z))}^{t/2}d\sigma (\xi )\\ & +{\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )\setminus \overline{D(0,{\rho }_0)}}\right|h_k(z){|}^2|Rg(z){|}^2{(1-|z{|}^2)}^{1-n}dv(z))}^{t/2}d\sigma (\xi )\\ & \lesssim {\epsilon }^t+{\parallel h_k\parallel }_{{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)}^t{({\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )\setminus \overline{D(0,{\rho }_0)}}\right|Rg(z){|}^{\frac{2q}{q-2}}{(1-|z{|}^2)}^{\eta }dv(z))}^{\frac{pt}{p-t}\frac{q-2}{2q}}d\sigma (\xi ))}^{\frac{p-t}{p}}\\ & \lesssim {\epsilon }^t.\end{array}$$

(b) If $p>t$ and $q\leq 2$, the assumption

$$\underset{\rho \to 1^{-}}{lim}{\int }_{{\mathbb{S}}_n}{\left(\underset{a_k\in \mathrm{\Gamma }(\xi )\setminus \overline{D(0,\rho )}}{sup}\right|Rg(a_k)\left|{(1-|a_k{|}^2)}^{\frac{q-(n+1+\alpha )}{q}}\right)}^{\frac{pt}{p-t}}d\sigma (\xi )=0$$

implies that for any $\varepsilon > 0$, there exists $\rho _{0}\in (0,1)$ such that

$$\underset{\rho \ge {\rho }_0}{sup}{({\int }_{{\mathbb{S}}_n}{\left({\int }_{\mathrm{\Gamma }(\xi )\setminus \overline{D(0,{\rho }_0)}}\right|Rg(z)|{(1-|z{|}^2)}^{\frac{q-(n+1+\alpha )}{q}}dv(z))}^{\frac{pt}{p-t}}d\sigma (\xi ))}^{\frac{p-t}{pt}}<\epsilon .$$

Choose $k_{0}$ such that $\sup_{k\geq k_{0},|z|\leq \rho _{0}}|h_{k}(z)|<\varepsilon$. By a similar argument as the previous case, we have

$$\begin{array}{rl}& {\parallel J_g{h}_k\parallel }_{H^t({\mathbb{B}}_n)}^t\\ & \lesssim {\epsilon }^t+{\parallel h_k\parallel }_{{\mathcal{HT}}_{q,\alpha }^p({\mathbb{B}}_n)}^t{({\int }_{{\mathbb{S}}_n}{\left(\underset{z\in \mathrm{\Gamma }(\xi )\setminus \overline{D(0,{\rho }_0)}}{sup}\right|Rg(z)\left|{(1-|z{|}^2)}^{\frac{q-(n+1+\alpha )}{q}}\right)}^{\frac{pt}{p-t}}d\sigma (\xi ))}^{\frac{p-t}{p}}\\ & \lesssim {\epsilon }^t.\end{array}$$

(c) If $p=t$ and $q>2$, the assumption

$$ \lim_{ \vert w \vert \rightarrow 1^{-}}\frac{1}{(1- \vert w \vert ^{2})^{n}}\sum_{a_{k}\in Q(w)} \bigl( \bigl\vert Rg(a_{k}) \bigr\vert \bigl(1- \vert a_{k} \vert ^{2}\bigr)^{ \frac{q-(n+1+\alpha )}{q}} \bigr)^{\frac{2q}{q-2}}\bigl(1- \vert a_{k} \vert ^{2} \bigr)^{n}=0$$

implies that

$$\underset{\rho \to 1^{-}}{lim}\underset{w\in {\mathbb{B}}_n}{sup}\frac{1}{{(1-|w{|}^2)}^n}{\int }_{Q(w)\setminus \overline{D(0,\rho )}}|Rg(z){|}^{\frac{2q}{q-2}}{(1-|z{|}^2)}^{n+\eta }dv(z)=0,$$

where $\eta =(1-n-\frac{2\alpha}{q})\frac{q}{q-2}$. Thus, for any $\varepsilon >0$, there exists $\rho _{0}\in (0,1)$ such that

$$\underset{w\in {\mathbb{B}}_n,\rho \ge {\rho }_0}{sup}\frac{1}{{(1-|w{|}^2)}^n}{\int }_{Q(w)\setminus \overline{D(0,\rho )}}|Rg(z){|}^{\frac{2q}{q-2}}{(1-|z{|}^2)}^{n+\eta }dv(z)<\epsilon .$$

Then, we can obtain $\|J_{g}h_{k}\|_{H^{t}}^{t}\lesssim \varepsilon$ by a similar technique as the proof of Theorem 1.1.

(d) If $p=t$ and $q\leq 2$ or $p< t$, the assumption implies that

$$ \lim_{ \vert z \vert \rightarrow 1^{-}} \bigl\vert Rg(z) \bigr\vert \bigl(1- \vert z \vert ^{2}\bigr)^{ \frac{q-(n+1+\alpha )}{q} + n(\frac{1}{t}-\frac{1}{p})}=0.$$

Then, we can complete the proof of Theorem 1.2 by following the standard modifying arguments as in the proof of Theorem 1.1.