sup> C | z | < r 0 ( 1 | φ a ( z ) | 2 ) q + n + 1 p | ( f φ a ) ( z ) | p d v (z)

We have $\left(1-{|{\phi }_{a}\left(z\right)|}^{2}\right)=\frac{\left(1-{|z|}^{2}\right)\left(1-{|a|}^{2}\right)}{{|1-〈z,a〉|}^{2}}$, when $|z|\le {r}_{0}$, $\frac{1-{r}_{0}^{2}}{{\left(1+{r}_{0}\right)}^{2}}\le \frac{\left(1-{|z|}^{2}\right)}{{|1-〈z,a〉|}^{2}}\le \frac{1}{{\left(1-{r}_{0}\right)}^{2}}$, and since ${|\nabla f\left(z\right)|}^{p}$ is subharmonic, that

$\begin{array}{l}{\int }_{{\mathbb{B}}_{n}}{|\stackrel{˜}{\nabla }f\left(z\right)|}^{p}{\left(1-{|z|}^{2}\right)}^{q+n+1-p}K\left(G\left(z,a\right)\right)\text{d}\lambda \left(z\right)\\ \ge C{\left(1-{|a|}^{2}\right)}^{q+n+1-p}{\int }_{|z|<{r}_{0}}{|\nabla \left(f\circ {\phi }_{a}\right)\left(z\right)|}^{p}\text{d}v\left(z\right)\\ =C{\left(1-{|a|}^{2}\right)}^{q+n+1-p}{\int }_{0}^{{r}_{0}}{r}^{2n-1}\text{d}r{\int }_{{S}_{n}}{|\nabla \left(f\circ {\phi }_{a}\right)\left(r\varsigma \right)|}^{p}\text{d}\sigma \left(\varsigma \right)\\ \ge C{\left(1-{|a|}^{2}\right)}^{q+n+1-p}{|\nabla \left(f\circ {\phi }_{a}\right)\left(z\right)|}^{p}\\ =C{\left(1-{|a|}^{2}\right)}^{q+n+1-p}{|\stackrel{˜}{\nabla }f\left(a\right)|}^{p}\end{array}$

Thus, we have $\underset{|a|\to 1}{\mathrm{lim}}{\left(1-{|a|}^{2}\right)}^{q+n+1-p}{|\stackrel{˜}{\nabla }f\left(a\right)|}^{p}=0$ when $f\in {Q}_{K,0}\left(p,q\right)$, then $f\in {\mathcal{B}}_{0}^{\frac{q+n+1}{p}}$.

The following result is the further study on the equivalence between ${Q}_{K,0}\left(p,q\right)$ and ${\mathcal{B}}_{0}^{\frac{q+n+1}{p}}$.

Theorem 2. Let $0, $\frac{p}{2}-n-1, ${Q}_{K,0}\left(p,q\right)={\mathcal{B}}_{0}^{\frac{q+n+1}{p}}$ if and only if

${\int }_{0}^{1}{\left(1-{r}^{2}\right)}^{-n-1}{r}^{2n-1}K\left(g\left(r\right)\right)\text{d}r<\infty$. (6)

Proof Sufficiency: By theorem 1, we only need to show that ${\mathcal{B}}_{0}^{\frac{q+n+1}{p}}\subset {Q}_{K,0}\left(p,q\right)$.

Since ${\int }_{0}^{1}{\left(1-{r}^{2}\right)}^{-n-1}{r}^{2n-1}K\left(g\left(r\right)\right)\text{d}r<\infty$, for given $\epsilon >0$, then there exists ${r}_{0}:0<{r}_{0}<1$, such that

${\int }_{{r}_{0}}^{1}{\left(1-{r}^{2}\right)}^{-n-1}{r}^{2n-1}K\left(g\left(r\right)\right)\text{d}r<\epsilon$.

Let $E\left(a,{r}_{0}\right)=\left\{z\in {\mathbb{B}}_{n},|{\phi }_{a}\left(z\right)|<{r}_{0}\right\}$, for any $f\in {\mathcal{B}}^{\frac{q+n+1}{p}}$, $z\in {\mathbb{B}}_{n}\E\left(a,{r}_{0}\right)$, we have

$\begin{array}{l}{\int }_{{\mathbb{B}}_{n}\E\left(a,{r}_{0}\right)}{|\stackrel{˜}{\nabla }f\left(z\right)|}^{p}{\left(1-{|z|}^{2}\right)}^{q+n+1-p}K\left(G\left(z,a\right)\right)\text{d}\lambda \left(z\right)\\ \le {‖f‖}_{{\mathcal{B}}^{\frac{q+n+1}{p}}}^{p}{\int }_{{\mathbb{B}}_{n}\E\left(a,{r}_{0}\right)}K\left(G\left(z,a\right)\right)\text{d}\lambda \left(z\right)\\ \le {‖f‖}_{{\mathcal{B}}^{\frac{q+n+1}{p}}}^{p}{\int }_{{r}_{0}<|z|<1}{\left(1-{|z|}^{2}\right)}^{-n-1}K\left(g\left(z\right)\right)\text{d}v\left(z\right)\\ \le {‖f‖}_{{\mathcal{B}}^{\frac{q+n+1}{p}}}^{p}{\int }_{{r}_{0}}^{1}{\left(1-{r}^{2}\right)}^{-n-1}{r}^{2n-1}K\left(g\left(r\right)\right)\text{d}r{\int }_{{\mathbb{S}}_{n}}\text{d}\sigma \left(\varsigma \right)\\ <\epsilon {‖f‖}_{{\mathcal{B}}^{\frac{q+n+1}{p}}}^{p}\end{array}$ (7)

And when $z\in E\left(a,{r}_{0}\right)$, we have

$\begin{array}{l}\underset{|a|\to 1}{\mathrm{lim}}{\int }_{E\left(a,{r}_{0}\right)}{|\stackrel{˜}{\nabla }f\left(z\right)|}^{p}{\left(1-{|z|}^{2}\right)}^{q+n+1-p}K\left(G\left(z,a\right)\right)\text{d}\lambda \left(z\right)\\ =\underset{|a|\to 1}{\mathrm{lim}}{\int }_{|z|<{r}_{0}}{|\stackrel{˜}{\nabla }\left(f\circ {\phi }_{a}\right)\left(z\right)|}^{p}{\left(1-{|{\phi }_{a}\left(z\right)|}^{2}\right)}^{q+n+1-p}K\left(g\left(z\right)\right)\text{d}\lambda \left(z\right)\\ \le \underset{|a|\to 1}{\mathrm{lim}}\underset{|z|<{r}_{0}}{\mathrm{sup}}{\left(1-{|{\phi }_{a}\left(z\right)|}^{2}\right)}^{q+n+1-p}{|\stackrel{˜}{\nabla }\left(f\circ {\phi }_{a}\right)\left(z\right)|}^{p}{\int }_{|z|<{r}_{0}}K\left(g\left(z\right)\right){\left(1-{|z|}^{2}\right)}^{-n-1}\text{d}V\left(z\right)\\ =\underset{|a|\to 1}{\mathrm{lim}}\underset{|z|<{r}_{0}}{\mathrm{sup}}{\left(1-{|{\phi }_{a}\left(z\right)|}^{2}\right)}^{q+n+1-p}{|\stackrel{˜}{\nabla }\left(f\circ {\phi }_{a}\right)\left(z\right)|}^{p}2n\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }×{\int }_{0}^{{r}_{0}}{\left(1-{r}^{2}\right)}^{-n-1}{r}^{2n-1}K\left(g\left(r\right)\right)\text{d}r{\int }_{{\mathbb{S}}_{n}}\text{d}\sigma \left(\varsigma \right)\\ \le C\underset{|a|\to 1}{\mathrm{lim}}\underset{|z|<{r}_{0}}{\mathrm{sup}}{\left(1-{|{\phi }_{a}\left(z\right)|}^{2}\right)}^{q+n+1-p}{|\stackrel{˜}{\nabla }\left(f\circ {\phi }_{a}\right)\left(z\right)|}^{p}\end{array}$

$\left(1-{|{\phi }_{a}\left(z\right)|}^{2}\right)=\frac{\left(1-{|z|}^{2}\right)\left(1-{|a|}^{2}\right)}{{|1-〈z,a〉|}^{2}}$, and $\frac{1-{r}_{0}^{2}}{{\left(1+{r}_{0}\right)}^{2}}\le \frac{\left(1-{|z|}^{2}\right)}{{|1-〈z,a〉|}^{2}}\le \frac{1}{{\left(1-{r}_{0}\right)}^{2}}$ when $|z|\le {r}_{0}$, so

$\underset{|a|\to 1}{\mathrm{lim}}{\left(1-{|{\phi }_{a}\left(z\right)|}^{2}\right)}^{\frac{q+n+1-p}{p}}|\stackrel{˜}{\nabla }\left(f\circ {\phi }_{a}\right)\left(z\right)|=0$,

thus

$\underset{|a|\to 1}{\mathrm{lim}}{\int }_{E\left(a,{r}_{0}\right)}{|\stackrel{˜}{\nabla }f\left(z\right)|}^{p}{\left(1-{|z|}^{2}\right)}^{q+n+1-p}K\left(G\left(z,a\right)\right)\text{d}\lambda \left(z\right)=0$,

By formula(7), then we have $\underset{|a|\to 1}{\mathrm{lim}}{\int }_{{\mathbb{B}}_{n}}{|\stackrel{˜}{\nabla }f\left(z\right)|}^{p}{\left(1-{|z|}^{2}\right)}^{q+n+1-p}K\left(G\left(z,a\right)\right)\text{d}\lambda \left(z\right)=0$, i.e. $f\in {Q}_{K,0}\left(p,q\right)$. It means ${\mathcal{B}}_{0}^{\frac{q+n+1}{p}}\subset {Q}_{K,0}\left(p,q\right)$.

Necessary: We only need to show that if ${\int }_{0}^{1}{\left(1-{r}^{2}\right)}^{-n-1}{r}^{2n-1}K\left(g\left(r\right)\right)\text{d}r=\infty$, there exists a function $f\in {\mathcal{B}}_{0}^{\frac{q+n+1}{p}}$, but $f\notin {Q}_{K,0}\left(p,q\right)$.

Let $\alpha =\left({\alpha }_{1},{\alpha }_{2},\cdots ,{\alpha }_{n}\right)$ be an n-tuple of non-negative integers, and $|\alpha |={\alpha }_{1}+{\alpha }_{2}+\cdots +{\alpha }_{n}$ satisfied $|\alpha |={2}^{N}$ where N is a integer. Let $f={|\alpha |}^{\frac{q+n+1-p}{p}}{z}^{\alpha }$, it is easy to show that $f\in {\mathcal{B}}^{\frac{q+n+1}{p}}$, and by the proof of theorem 3 in , we know that ${\int }_{{\mathbb{S}}_{n}}J{\left(r\varsigma \right)}^{\frac{p}{2}}\text{d}\sigma \left(\varsigma \right)\ge C{\left(1-r\right)}^{-\left(q+n+1\right)+\frac{p}{2}}$ when $r\in \left[\frac{3}{4},1\right)$, which

$\begin{array}{c}J\left(r\varsigma \right)={r}^{2|\alpha |-2}{|\alpha |}^{\frac{2\left(q+n+1-p\right)}{p}}\left({\alpha }_{1}^{2}{|{\varsigma }_{1}^{{\alpha }_{1}-1}{\varsigma }_{2}^{{\alpha }_{2}}\cdots {\varsigma }_{n}^{{\alpha }_{n}}|}^{2}+\cdots \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+{\alpha }_{n}^{2}{|{\varsigma }_{1}^{{\alpha }_{1}}{\varsigma }_{2}^{{\alpha }_{2}}\cdots {\varsigma }_{n}^{{\alpha }_{n}-1}|}^{2}-{r}^{2}{|\alpha |}^{2}{|{\varsigma }^{\alpha }|}^{2}\right)\end{array}$

thus

$\begin{array}{l}{\int }_{{\mathbb{B}}_{n}}{|\stackrel{˜}{\nabla }f\left(z\right)|}^{p}{\left(1-{|z|}^{2}\right)}^{q+n+1-p}K\left(G\left(z,a\right)\right)\text{d}\lambda \left(z\right)\\ \ge {\int }_{{\mathbb{B}}_{n}}{\left(1-{|z|}^{2}\right)}^{\frac{p}{2}}{\left(J\left(z\right)\right)}^{\frac{p}{2}}{\left(1-{|z|}^{2}\right)}^{q+n+1-p}K\left(g\left(z\right)\right)\text{d}\lambda \left(z\right)\\ =2n{\int }_{0}^{1}{\left(1-{r}^{2}\right)}^{q-\frac{p}{2}}{r}^{2n-1}K\left(g\left(r\right)\right)\text{d}r{\int }_{{\mathbb{S}}_{n}}J{\left(r\varsigma \right)}^{\frac{p}{2}}\text{d}\sigma \left(\varsigma \right)\\ \ge C{\int }_{\frac{3}{4}}^{1}{\left(1-{r}^{2}\right)}^{-n-1}{r}^{2n-1}K\left(g\left(r\right)\right)\text{d}r\end{array}$

Since the conclusion of theorem 1 in , we have

${\int }_{0}^{\frac{3}{4}}{\left(1-{r}^{2}\right)}^{-n-1}{r}^{2n-1}K\left(g\left(r\right)\right)\text{d}r\le C{\int }_{0}^{1}{\left(1-{r}^{2}\right)}^{2-n-1}{r}^{2n-1}K\left(g\left(r\right)\right)\text{d}r<\infty$,

Then if ${\int }_{0}^{1}{\left(1-{r}^{2}\right)}^{-n-1}{r}^{2n-1}K\left(g\left(r\right)\right)\text{d}r=\infty$, we can get

${\int }_{{\mathbb{B}}_{n}}{|\stackrel{˜}{\nabla }f\left(z\right)|}^{p}{\left(1-{|z|}^{2}\right)}^{q+n+1-p}K\left(G\left(z,a\right)\right)\text{d}\lambda \left(z\right)=\infty$,

which shows that $f\notin {Q}_{K,0}\left(p,q\right)$, the theorem is proved.

With the above conclusion, further study in this field of operator theory on ${Q}_{K,0}\left(p,q\right)$ can be conducted in the future.

Founding

Scientific Research Fund of Sichuan Provincial Education Department of China (18ZA0416).

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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