Boundedness for Multilinear Operators of Pseudo-Differential Operators

Jiasheng Zeng^1,2, Rongping Chen^1,2, Jiangang Liu^1,2, Xiaojing Wang^1,2, Chunsheng Liu^1,2

¹School of Mathematics and Statistics, Hunan University of Technology and Business, Changsha, China.
²Key Laboratory of Hunan Province for Statistical Learning and Intelligent Computation, Hunan University of Technology and Business, Changsha, China.

DOI: 10.4236/apm.2023.137029                

Abstract

In this paper, we establish a sharp function estimate for the multilinear integral operators associated to the pseudo-differential operators. As the application, we obtain the L^p (1 < p < ∞) norm inequalities for the multilinear operators.

Keywords

Multilinear Operator, Pseudo-Differential Operator, Sharp Function Estimate, BMO

1. Introduction and Results

Let b be a locally integrable function on $R^n$ and T be an integral operator. For a suitable function f, the commutator generated by b and T is defined by $\left[b\mathrm{,}T\right]f=bT\left(f\right)-T\left(bf\right)$ . The investigation of the commutator begins with Coifman-Rochberg-Weiss pioneering study and classical result (see [1] ). The major reason for considering the problem of commutators is that the boundedness of commutator can produce some characterizations of function spaces (see [1] [2] ). Now, with the development of the Calderón-Zygmund singular integral operators, their commutators and multilinear operators have been well studied (see [1] [3] - [7] ). In [8] , Hu and Yang proved a variant sharp function estimate for the multilinear singular integral operators. In [9] [10] [11] [12] , C. Pérez, G. Pradolini and R. Trujillo-Gonzalez obtained a sharp weighted estimate for the singular integral operators and their commutators. The boundedness of the pseudo-differential operators was studied by many authors (see [13] - [21] ). In [15] , the boundedness of the commutators associated to the pseudo-differential operators is obtained. The main purpose of this paper is to study the multilinear pseudo-differential operators as follows.

We say a symbol $\sigma \left(x\mathrm{,}\xi \right)$ is in the class $S_{\rho \mathrm{,}\delta }^m$ or $\sigma \in S_{\rho \mathrm{,}\delta }^m$ , if for $x\mathrm{,}\xi \in R^n$ ,

$\left|\frac{{\partial }^{\alpha }}{\partial x^{\alpha }}\frac{{\partial }^{\beta }}{\partial {\xi }^{\beta }}\sigma \left(x\mathrm{,}\xi \right)\right|\le C_{\alpha \mathrm{,}\beta }{\left(1+\left|\xi \right|\right)}^{m-\rho \left|\beta \right|+\delta \left|\alpha \right|}\mathrm{.}$

A pseudo-differential operator with symbol $\sigma \left(x\mathrm{,}\xi \right)\in S_{\rho \mathrm{,}\delta }^m$ is defined by

$T\left(f\right)\left(x\right)={\displaystyle {\int }_{R^n}}{\text{e}}^{2\pi ix\cdot \xi }\sigma \left(x\mathrm{,}\xi \right)\hat{f}\left(\xi \right)\text{d}\xi \mathrm{,}$

where f is a Schwartz function and $\hat{f}$ denotes the Fourier transform of f. We know there exists a kernel $K\left(x\mathrm{,}y\right)$ such that

$T\left(f\right)\left(x\right)={\displaystyle {\int }_{R^n}}K\left(x,x-y\right)f\left(y\right)\text{d}y,$

where, formally,

$K\left(x\mathrm{,}y\right)={\displaystyle {\int }_{R^n}}{\text{e}}^{2\pi i\left(x-y\right)\cdot \xi }\sigma \left(x\mathrm{,}\xi \right)\text{d}\xi \mathrm{.}$

In [14] , the boundedness of the pseudo-differential operators with symbol $\sigma \in S_{1-\theta ,\delta }^{-\beta }\left(\beta <n\theta /2,0\le \delta <1-\theta \right)$ is obtained. In [14] , the boundedness of the pseudo-differential operators with symbol of order 0 and $-\infty$ is obtained. In [17] , the sharp function estimate of the pseudo-differential operators with symbol $\sigma \in S_{1-\theta ,\delta }^{-n\theta /2}\left(0<\theta <1,0\le \delta <1-\theta \right)$ is obtained. In [15] , the boundedness of the pseudo-differential operators and their commutators with symbol $\sigma \in S_{1-\theta ,\delta }^{-n\theta /2}\left(0<\theta <1,0\le \delta <1-\theta \right)$ is obtained. Our results are motivated by these papers.

Suppose T is a pseudo-differential operator with symbol $\sigma \left(x\mathrm{,}\xi \right)\in S_{\rho \mathrm{,}\delta }^m$ . Let $m_j$ be the positive integers ( $j=1,\cdots ,l$ ), $m_1+\cdots +m_l=L$ and $b_j$ be the functions on $R^n$ ( $j=1,\cdots ,l$ ). Set, for $1\le j\le m$ ,

$R_{m_j+1}\left(b_j;x,y\right)=b_j\left(x\right)-{\displaystyle \underset{\left|\alpha \right|\le m_j}{\sum }}\frac{1}{\alpha !}{D}^{\alpha }{b}_j\left(y\right){\left(x-y\right)}^{\alpha }.$

The multilinear operator associated to T is defined by

$T_b\left(f\right)\left(x\right)={\displaystyle {\int }_{R^n}}\frac{{\displaystyle {\prod }_{j=1}^l{R}_{m_j+1}}\left(b_j;x,y\right)}{{\left|x-y\right|}^L}K\left(x,x-y\right)f\left(y\right)\text{d}y.$

Note that when $L=0$ , $T_b$ is just the multilinear commutator of T and $b_j$ (see [11] ). While when $L>0$ , $T_b$ is non-trivial generalizations of the commutator. It is well known that multilinear operators are of great interest in harmonic analysis and have been widely studied by many authors. Hu and Yang (see [8] ) proved a variant sharp estimate for the multilinear singular integral operators. In [11] , Pérez and Trujillo-Gonzalez prove a sharp estimate for the multilinear commutator when $b_j\in Os{c}_{exp{L}^{r_j}}\left(R^n\right)$ . The main purpose of this paper is to prove a sharp function inequality for the multilinear operators associated to the pseudo-differential operators with symbol $\sigma \in S_{1-\theta ,\delta }^{-n\theta /2}\left(0<\theta <1,0\le \delta <1-\theta \right)$ when $D^{\alpha }{b}_j\in BMO\left(R^n\right)$ for all $\alpha$ with $\left|\alpha \right|=m_j$ . As the application, we obtain the $L^p\left(p>1\right)$ norm inequality for the multilinear operators.

First, let us introduce some notations. Throughout this paper, $Q=Q\left(x\mathrm{,}d\right)$ will denote a cube of $R^n$ with sides parallel to the axes, whose center is x and side length is d. For a locally integrable function b, the sharp function of b is defined by

$b^{\mathrm{<mo>\; \#\; </mo>}}\left(x\right)=\underset{Q\ni x}{\sup }\frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}\left|b\left(y\right)-b_Q\right|\text{d}y\mathrm{,}$

where, and in what follows, $b_Q={\left|Q\right|}^{-1}{\displaystyle {\int }_Q}b\left(x\right)\text{d}x$ . It is well-known that (see [22] [23] [24] )

$b^{\mathrm{<mo>\; \#\; </mo>}}\left(x\right)\approx \underset{Q\ni x}{\sup }\underset{c\in C}{\inf }\frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}\left|b\left(y\right)-c\right|\text{d}y$

and

${\Vert b-b_{2^{k}Q}\Vert }_{BMO}\le Ck{\Vert b\Vert }_{BMO}\text{for}k\ge 1.$

We say that b belongs to $BMO\left(R^n\right)$ if $b^{\mathrm{<mo>\; \#\; </mo>}}$ belongs to $L^{\infty }\left(R^n\right)$ and ${\Vert b\Vert }_{BMO}={\Vert b^{\mathrm{<mo>\; \#\; </mo>}}\Vert }_{L^{\infty }}$ . Let M be the Hardy-Littlewood maximal operator defined by

$M\left(f\right)\left(x\right)=\underset{x\in Q}{\sup }\frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}\left|f\left(y\right)\right|\text{d}y\mathrm{,}$

we write that $M_p\left(f\right)={\left(M\left(f^p\right)\right)}^{1/p}$ for $0<p<\infty$ .

We shall prove the following theorems.

Theorem 1. Suppose T is the pseudo-differential operator with symbol $\sigma \in S_{1-\theta ,\delta }^{-n\theta /2}\left(0<\theta <1,0\le \delta <1-\theta \right)$ . Let $D^{\alpha }{b}_j\in BMO\left(R^n\right)$ for all $\alpha$ with $\left|\alpha \right|=m_j$ and $j=1,\cdots ,l$ . Then there exists a constant $C>0$ such that for every $f\in C_0^{\infty }\left(R^n\right)$ , $2<r<\infty$ and $\tilde{x}\in R^n$ ,

${\left(T_b\left(f\right)\right)}^{\#}\left(\tilde{x}\right)\le C{\displaystyle \underset{j=1}{\overset{l}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).$

Theorem 2. Suppose T is the pseudo-differential operator with symbol $\sigma \in S_{1-\theta ,\delta }^{-n\theta /2}\left(0<\theta <1,0\le \delta <1-\theta \right)$ . Let $D^{\alpha }{b}_j\in BMO\left(R^n\right)$ for all $\alpha$ with $\left|\alpha \right|=m_j$ and $j=1,\cdots ,l$ .

1) If $w\in A_{\infty }$ and $2<p<\infty$ , then

${\Vert T_b\left(f\right)\Vert }_{L^p\left(w\right)}\le C{\displaystyle \underset{j=1}{\overset{l}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right){\Vert M_r\left(f\right)\Vert }_{L^p\left(w\right)}.$

2) If $w\in A_1$ and $2<p<\infty$ , then

${\Vert T_b\left(f\right)\Vert }_{L^p\left(w\right)}\le C{\displaystyle \underset{j=1}{\overset{l}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right){\Vert f\Vert }_{L^p\left(w\right)}.$

2. Proofs of Theorems

To prove the theorems, we need the following lemmas.

Lemma 1. (see [4] ) Let b be a function on $R^n$ and $D^{\alpha }b\in L^q\left(R^n\right)$ for all $\alpha$ with $\left|\alpha \right|=L$ and some $q>n$ . Then

$\left|R_L\left(b;x,y\right)\right|\le C{\left|x-y\right|}^L{\displaystyle \underset{\left|\alpha \right|=L}{\sum }}{\left(\frac{1}{\left|\tilde{Q}\left(x,y\right)\right|}{\displaystyle {\int }_{\tilde{Q}\left(x,y\right)}}{\left|D^{\alpha }b\left(z\right)\right|}^q\text{d}z\right)}^{1/q},$

where $\tilde{Q}$ is the cube centered at x and having side length $5\sqrt{n}\left|x-y\right|$ .

Lemma 2. (see [13] ) Let T be the pseudo-differential operator with symbol $\sigma \in S_{1-\theta ,\delta }^{-n\theta /2}\left(0<\theta <1,0\le \delta <1-\theta \right)$ . Then, for every $f\in L^p\left(R^n\right),1<p<\infty$ ,

${\Vert T\left(f\right)\Vert }_{L^p}\le C{\Vert f\Vert }_{L^p}\mathrm{.}$

Lemma 3. (see [13] ) Let $\sigma \in S_{1-\theta ,\delta }^{-n\theta /2}\left(0<\theta <1,0\le \delta <1-\theta \right)$ and K be the kernel of the pseudo-differential operator T with symbol $\sigma \in S_{1-\theta \mathrm{,}\delta }^{-n\theta /2}$ . Then, for $\left|x_0-x\right|\le d<1$ and $k\ge 1$ ,

${\left({\displaystyle {\int }_{{\left(2^{k}d\right)}^{1-\theta }\le \left|y-x_0\right|<{\left(2^{k+1}d\right)}^{1-\theta }}}{\left|K\left(x,x-y\right)-K\left(x_0,x_0-y\right)\right|}^2\text{d}y\right)}^{1/2}\le C\frac{{\left|x_0-x\right|}^{\left(1-\theta \right)\left(m-n/2\right)}}{{\left(2^{k}d\right)}^{m\left(1-\theta \right)}},$

provided m is an integer such that $n/2<m<n/2+1/\left(1-\theta \right)$ .

Lemma 4. (see [13] ) Let $\sigma \in S_{\rho ,\delta }^0\left(0<\rho <1\right)$ and

$K\left(x\mathrm{,}w\right)={\displaystyle {\int }_{R^n}}{\text{e}}^{2\pi iw\cdot \xi }\sigma \left(x\mathrm{,}\xi \right)\text{d}\xi \mathrm{.}$

Then, for $\left|w\right|\ge 1/4$ and any integer $N\ge 1$ ,

$\left|K\left(x\mathrm{,}w\right)\right|\le C_N{\left|w\right|}^{-2N}\mathrm{.}$

Proof of Theorem 1. It suffices to prove for $f\in C_0^{\infty }\left(R^n\right)$ and some constant $C_0$ , the following inequality holds:

$\frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}\left|T_b\left(f\right)\left(x\right)-C_0\right|\text{d}x\le C{\displaystyle \underset{j=1}{\overset{l}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(x\right).$

Without loss of generality, we may assume $l=2$ . Fix a cube $Q=Q\left(x_0\mathrm{,}d\right)$ and $\tilde{x}\in Q$ . We consider the following two cases:

Case 1. $d\le 1$ . In this case, let $Q^{\ast }$ be the cube concentric with Q of side length $d^{1-\theta }$ . Let $\tilde{Q}=5\sqrt{n}{Q}^{\ast }$ and ${\tilde{b}}_j\left(x\right)=b_j\left(x\right)-{\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}\frac{1}{\alpha !}{\left(D^{\alpha }{b}_j\right)}_{\tilde{Q}}{x}^{\alpha }$ , then $R_{m_j}\left(b_j;x,y\right)=R_{m_j}\left({\tilde{b}}_j;x,y\right)$ and $D^{\alpha }{\tilde{b}}_j=D^{\alpha }{b}_j-{\left(D^{\alpha }{b}_j\right)}_{\tilde{Q}}$ for $\left|\alpha \right|=m_j$ . We write, for $f=f{\chi }_{\tilde{Q}}+f{\chi }_{R^n\backslash \tilde{Q}}=f_1+f_2$ ,

$\begin{array}{l}{T}_b\left(f\right)\left(x\right)={\displaystyle {\int }_{R^n}}\frac{{\displaystyle {\prod }_{j=1}^2{R}_{m_j+1}\left({\tilde{b}}_j;x,y\right)}}{{\left|x-y\right|}^L}K\left(x,x-y\right)f\left(y\right)\text{d}y\\ ={\displaystyle {\int }_{R^n}}\frac{{\displaystyle {\prod }_{j=1}^2{R}_{m_j}\left({\tilde{b}}_j;x,y\right)}}{{\left|x-y\right|}^L}K\left(x,x-y\right)f_1\left(y\right)\text{d}y\\ -{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}\frac{1}{{\alpha }_1!}{\displaystyle {\int }_{R^n}}\frac{R_{m_2}\left({\tilde{b}}_2;x,y\right){\left(x-y\right)}^{{\alpha }_1}{D}^{{\alpha }_1}{\tilde{b}}_1\left(y\right)}{{\left|x-y\right|}^L}K\left(x,x-y\right)f_1\left(y\right)\text{d}y\\ -{\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}\frac{1}{{\alpha }_2!}{\displaystyle {\int }_{R^n}}\frac{R_{m_1}\left({\tilde{b}}_1;x,y\right){\left(x-y\right)}^{{\alpha }_2}{D}^{{\alpha }_2}{\tilde{b}}_2\left(y\right)}{{\left|x-y\right|}^L}K\left(x,x-y\right)f_1\left(y\right)\text{d}y\\ +{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}\frac{1}{{\alpha }_1!{\alpha }_2!}{\displaystyle {\int }_{R^n}}\frac{{\left(x-y\right)}^{{\alpha }_1+{\alpha }_2}{D}^{{\alpha }_1}{\tilde{b}}_1\left(y\right)D^{{\alpha }_2}{\tilde{b}}_2\left(y\right)}{{\left|x-y\right|}^L}K\left(x,x-y\right)f_1\left(y\right)\text{d}y\\ +{\displaystyle {\int }_{R^n}}\frac{{\displaystyle {\prod }_{j=1}^2{R}_{m_j+1}\left({\tilde{b}}_j;x,y\right)}}{{\left|x-y\right|}^L}K\left(x,x-y\right)f_2\left(y\right)\text{d}y,\end{array}$

then

$\begin{array}{l}\frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}\left|T_b\left(f\right)\left(x\right)-T_{\tilde{b}}\left(f_2\right)\left(x_0\right)\right|\text{d}x\\ \le \frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}\left|{\displaystyle {\int }_{R^n}}\frac{{\displaystyle {\prod }_{j=1}^2{R}_{m_j}\left({\tilde{b}}_j;x,y\right)}}{{\left|x-y\right|}^L}K\left(x,x-y\right)f_1\left(y\right)\text{d}y\right|\text{d}x\\ +\frac{C}{\left|Q\right|}{\displaystyle {\int }_Q}\left|{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\displaystyle {\int }_{R^n}}\frac{R_{m_2}\left({\tilde{b}}_2;x,y\right){\left(x-y\right)}^{{\alpha }_1}}{{\left|x-y\right|}^L}{D}^{{\alpha }_1}{\tilde{b}}_1\left(y\right)K\left(x,x-y\right)f_1\left(y\right)\text{d}y\right|\text{d}x\\ +\frac{C}{\left|Q\right|}{\displaystyle {\int }_Q}\left|{\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}{\displaystyle {\int }_{R^n}}\frac{R_{m_1}\left({\tilde{b}}_1;x,y\right){\left(x-y\right)}^{{\alpha }_2}}{{\left|x-y\right|}^L}{D}^{{\alpha }_2}{\tilde{b}}_2\left(y\right)K\left(x,x-y\right)f_1\left(y\right)\text{d}y\right|\text{d}x\end{array}$

$\begin{array}{l}+\frac{C}{\left|Q\right|}{\displaystyle {\int }_Q}\left|{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\displaystyle {\int }_{R^n}}\frac{{\left(x-y\right)}^{{\alpha }_1+{\alpha }_2}{D}^{{\alpha }_1}{\tilde{b}}_1\left(y\right)D^{{\alpha }_2}{\tilde{b}}_2\left(y\right)}{{\left|x-y\right|}^L}K\left(x,x-y\right)f_1\left(y\right)\text{d}y\right|\text{d}x\\ +\frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}\left|T_{\tilde{b}}\left(f_2\right)\left(x\right)-T_{\tilde{b}}\left(f_2\right)\left(x_0\right)\right|\text{d}x\\ :=I_1+I_2+I_3+I_4+I_5.\end{array}$

Now, let us estimate $I_1$ , $I_2$ , $I_3$ , $I_4$ and $I_5$ , respectively. First, for $x\in Q$ and $y\in \tilde{Q}$ , by Lemma 1, we get

$R_L\left({\tilde{b}}_j;x,y\right)\le C{\left|x-y\right|}^L{\displaystyle \underset{\left|{\alpha }_j\right|=L}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}.$

Now, let $\sigma \left(x,\xi \right)=\sigma \left(x,\xi \right){\left|\xi \right|}^{n\theta /2}{\left|\xi \right|}^{-n\theta /2}=q\left(x,\xi \right){\left|\xi \right|}^{-n\theta /2}$ , we have $q\left(x\mathrm{,}\xi \right)\in S_{1-\theta \mathrm{,}\delta }^0$ , set S be the pseudo-differential operator with symbol $q\left(x\mathrm{,}\xi \right)$ , by the Hardy-Littlewood-Soboleve fractional integration theorem and the $L^2$ -boundedness of S (see [13] ), we obtain, for $1/p=1/2-\theta /2$ ,

$\begin{array}{c}{I}_1\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right)\frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}\left|T\left(f_1\right)\left(x\right)\right|\text{d}x\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right){\left(\frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}{\left|T\left(f_1\right)\left(x\right)\right|}^p\text{d}x\right)}^{1/p}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right){\left|Q\right|}^{-1/p}{\left({\displaystyle {\int }_{R^n}}{\left|S\left(f_1\right)\left(x\right)\right|}^2\text{d}x\right)}^{1/2}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right){\left|Q\right|}^{-1/p}{\left({\displaystyle {\int }_{R^n}}{\left|f_1\left(x\right)\right|}^2\text{d}x\right)}^{1/2}\end{array}$

$\begin{array}{l}\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right)\frac{{\left|\tilde{Q}\right|}^{1/2}}{{\left|Q\right|}^{1/p}}{\left(\frac{1}{\left|\tilde{Q}\right|}{\displaystyle {\int }_{\tilde{Q}}}{\left|f\left(x\right)\right|}^2\text{d}x\right)}^{1/2}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right){\left(\frac{1}{\left|\tilde{Q}\right|}{\displaystyle {\int }_{\tilde{Q}}}{\left|f\left(x\right)\right|}^r\text{d}x\right)}^{1/r}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).\end{array}$

For $I_2$ , by Lemma 1 and Hölder’s inequality, we get, for $1/r+1/r^{\prime }=1/2$ ,

$\begin{array}{c}{I}_2\le C{\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}{\Vert D^{{\alpha }_2}{b}_2\Vert }_{BMO}{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\left(\frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}{\left|T\left(D^{{\alpha }_1}{\tilde{b}}_1{f}_1\right)\left(x\right)\right|}^p\text{d}x\right)}^{1/p}\\ \le C{\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}{\Vert D^{{\alpha }_2}{b}_2\Vert }_{BMO}{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\left|Q\right|}^{-1/p}{\left({\displaystyle {\int }_{R^n}}{\left|S\left(D^{{\alpha }_1}{\tilde{b}}_1{f}_1\right)\left(x\right)\right|}^2\text{d}x\right)}^{1/2}\\ \le C{\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}{\Vert D^{{\alpha }_2}{b}_2\Vert }_{BMO}{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\left|Q\right|}^{-1/p}{\left({\displaystyle {\int }_{R^n}}{\left|D^{{\alpha }_1}{\tilde{b}}_1\left(x\right)f_1\left(x\right)\right|}^2\text{d}x\right)}^{1/2}\\ \le C{\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}{\Vert D^{{\alpha }_2}{b}_2\Vert }_{BMO}\frac{{\left|\tilde{Q}\right|}^{1/2}}{{\left|Q\right|}^{1/p}}{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\left(\frac{1}{\left|\tilde{Q}\right|}{\displaystyle {\int }_{\tilde{Q}}}{\left|D^{{\alpha }_1}{b}_1\left(x\right)-{\left(D^{\alpha }{b}_j\right)}_{\tilde{Q}}\right|}^{r^{\prime }}\text{d}x\right)}^{1/r^{\prime }}\\ \times {\left(\frac{1}{\left|\tilde{Q}\right|}{\displaystyle {\int }_{\tilde{Q}}}{\left|f\left(x\right)\right|}^r\text{d}x\right)}^{1/r}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).\end{array}$

For $I_3$ , similar to the proof of $I_2$ , we get

$I_3\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).$

Similarly, for $I_4$ , taking $r_1,r_2>1$ such that $1/r+1/r_1+1/r_2=1/2$ , we obtain

$\begin{array}{c}{I}_4\le C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\left(\frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}{\left|T\left(D^{{\alpha }_1}{\tilde{b}}_1{D}^{{\alpha }_2}{\tilde{b}}_2{f}_1\right)\left(x\right)\right|}^p\text{d}x\right)}^{1/p}\\ \le C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\left|Q\right|}^{-1/p}{\left({\displaystyle {\int }_{R^n}}{\left|S\left(D^{{\alpha }_1}{\tilde{b}}_1{D}^{{\alpha }_2}{\tilde{b}}_2{f}_1\right)\left(x\right)\right|}^2\text{d}x\right)}^{1/2}\\ \le C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\left|Q\right|}^{-1/p}{\left({\displaystyle {\int }_{R^n}}{\left|D^{{\alpha }_1}{\tilde{b}}_1\left(x\right)D^{{\alpha }_2}{\tilde{b}}_2\left(x\right)f_1\left(x\right)\right|}^2\text{d}x\right)}^{1/2}\\ \le C\frac{{\left|\tilde{Q}\right|}^{1/2}}{{\left|Q\right|}^{1/p}}{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\displaystyle \underset{j=1}{\overset{2}{\prod }}}{\left(\frac{1}{\left|\tilde{Q}\right|}{\displaystyle {\int }_{\tilde{Q}}}{\left|D^{{\alpha }_j}{\tilde{b}}_j\left(x\right)\right|}^{r_j}\text{d}x\right)}^{1/r_j}{\left(\frac{1}{\left|\tilde{Q}\right|}{\displaystyle {\int }_{\tilde{Q}}}{\left|f\left(x\right)\right|}^r\text{d}x\right)}^{1/r}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).\end{array}$

For $I_5$ , we write

$\begin{array}{l}{T}_{\tilde{b}}\left(f_2\right)\left(x\right)-T_{\tilde{b}}\left(f_2\right)\left(x_0\right)\\ ={\displaystyle {\int }_{R^n}}\left(\frac{K\left(x,x-y\right)}{{\left|x-y\right|}^L}-\frac{K\left(x_0,x_0-y\right)}{{\left|x_0-y\right|}^L}\right){\displaystyle \underset{j=1}{\overset{2}{\prod }}}{R}_{m_j}\left({\tilde{b}}_j;x,y\right)f_2\left(y\right)\text{d}y\\ +{\displaystyle {\int }_{R^n}}\left(R_{m_1}\left({\tilde{b}}_1;x,y\right)-R_{m_1}\left({\tilde{b}}_1;x_0,y\right)\right)\frac{R_{m_2}\left({\tilde{b}}_2;x,y\right)}{|x_0-y{|}^L}K\left(x_0,x_0-y\right)f_2\left(y\right)\text{d}y\\ +{\displaystyle {\int }_{R^n}}\left(R_{m_2}\left({\tilde{b}}_2;x,y\right)-R_{m_2}\left({\tilde{b}}_2;x_0,y\right)\right)\frac{R_{m_1}\left({\tilde{b}}_1;x_0,y\right)}{{\left|x_0-y\right|}^L}K\left(x_0,x_0-y\right)f_2\left(y\right)\text{d}y\end{array}$

$\begin{array}{l}-{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}\frac{1}{{\alpha }_1!}{\displaystyle {\int }_{R^n}}[\frac{R_{m_2}\left({\tilde{b}}_2;x,y\right){\left(x-y\right)}^{{\alpha }_1}}{{\left|x-y\right|}^L}K\left(x,x-y\right)\\ -\frac{R_{m_2}\left({\tilde{b}}_2;x_0,y\right){\left(x_0-y\right)}^{{\alpha }_1}}{{\left|x_0-y\right|}^L}K\left(x_0,x_0-y\right)]D^{{\alpha }_1}{\tilde{b}}_1\left(y\right)f_2\left(y\right)\text{d}y\\ -{\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}\frac{1}{{\alpha }_2!}{\displaystyle {\int }_{R^n}}[\frac{R_{m_1}\left({\tilde{b}}_1;x,y\right){\left(x-y\right)}^{{\alpha }_2}}{{\left|x-y\right|}^L}K\left(x,x-y\right)\\ -\frac{R_{m_1}\left({\tilde{b}}_1;x_0,y\right){\left(x_0-y\right)}^{{\alpha }_2}}{{\left|x_0-y\right|}^L}K\left(x_0,x_0-y\right)]D^{{\alpha }_2}{\tilde{b}}_2\left(y\right)f_2\left(y\right)\text{d}y\end{array}$

$\begin{array}{l}+{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}\frac{1}{{\alpha }_1!{\alpha }_2!}{\displaystyle {\int }_{R^n}}[\frac{{\left(x-y\right)}^{{\alpha }_1+{\alpha }_2}}{{\left|x-y\right|}^L}K\left(x,x-y\right)\\ -\frac{{\left(x_0-y\right)}^{{\alpha }_1+{\alpha }_2}}{{\left|x_0-y\right|}^L}K\left(x_0,x_0-y\right)]D^{{\alpha }_1}{\tilde{b}}_1\left(y\right)D^{{\alpha }_2}{\tilde{b}}_2\left(y\right)f_2\left(y\right)\text{d}y\\ =I_5^{\left(1\right)}+I_5^{\left(2\right)}+I_5^{\left(3\right)}+I_5^{\left(4\right)}+I_5^{\left(5\right)}+I_5^{\left(6\right)}.\end{array}$

By Lemma 1 and the following inequality (see [24] )

$\left|b_{Q_1}-b_{Q_2}\right|\le C\log \left(\left|Q_2\right|/\left|Q_1\right|\right){\Vert b\Vert }_{BMO}\text{for}{Q}_1\subset Q_2\mathrm{,}$

we know that, for $x\in Q$ and $y\in Q\left(x_0\mathrm{,}{\left(2^{k+1}d\right)}^{1-\theta }\right)\backslash Q\left(x_0\mathrm{,}{\left(2^{k}d\right)}^{1-\theta }\right)$ ,

$\begin{array}{c}\left|R_L\left(\tilde{b};x,y\right)\right|\le C{\left|x-y\right|}^L{\displaystyle \underset{\left|\alpha \right|=L}{\sum }}\left({\Vert D^{\alpha }b\Vert }_{BMO}+\left|{\left(D^{\alpha }b\right)}_{\tilde{Q}\left(x,y\right)}-{\left(D^{\alpha }b\right)}_{\tilde{Q}}\right|\right)\\ \le Ck{\left|x-y\right|}^L{\displaystyle \underset{\left|\alpha \right|=L}{\sum }}{\Vert D^{\alpha }b\Vert }_{BMO}.\end{array}$

Note that $\left|x-y\right|~\left|x_0-y\right|$ for $x\in \tilde{Q}$ and $y\in R^n\backslash \tilde{Q}$ , we obtain

$\begin{array}{c}\left|I_5^{\left(1\right)}\right|\le {\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}{k}^2{\displaystyle {\int }_{{\left(2^{k}d\right)}^{1-\theta }\le \left|y-x_0\right|<{\left(2^{k+1}d\right)}^{1-\theta }}}\left|K\left(x,x-y\right)-K\left(x_0,x_0-y\right)\right|\\ \times \frac{1}{{\left|x-y\right|}^m}{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left|R_{m_j}\left({\tilde{b}}_j;x,y\right)\right|\left|f\left(y\right)\right|\text{d}y\\ +{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}{k}^2{\displaystyle {\int }_{{\left(2^{k}d\right)}^{1-\theta }\le \left|y-x_0\right|<{\left(2^{k+1}d\right)}^{1-\theta }}}\left|\frac{1}{{\left|x-y\right|}^L}-\frac{1}{{\left|x_0-y\right|}^L}\right|\\ \times \left|K\left(x_0,x_0-y\right)\right|{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left|R_{m_j}\left({\tilde{b}}_j;x,y\right)\right|\left|f\left(y\right)\right|\text{d}y\end{array}$

$\begin{array}{c}\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right){\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}{k}^2{\left({\displaystyle {\int }_{\left|y-x_0\right|<{\left(2^{k+1}d\right)}^{1-\theta }}}{\left|f\left(y\right)\right|}^2\text{d}y\right)}^{1/2}\\ \times {\left({\displaystyle {\int }_{{\left(2^{k}d\right)}^{1-\theta }\le \left|y-x_0\right|<{\left(2^{k+1}d\right)}^{1-\theta }}}{\left|K\left(x,x-y\right)-K\left(x_0,x_0-y\right)\right|}^2\text{d}y\right)}^{1/2}\\ +C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right){\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}{k}^2{\left({\displaystyle {\int }_{\left|y-x_0\right|<{\left(2^{k+1}d\right)}^{1-\theta }}}{\left|f\left(y\right)\right|}^2\text{d}y\right)}^{1/2}\\ \times {\left({\displaystyle {\int }_{{\left(2^{k}d\right)}^{1-\theta }\le \left|y-x_0\right|<{\left(2^{k+1}d\right)}^{1-\theta }}}\frac{{\left|x_0-x\right|}^2}{{\left|x_0-y\right|}^2}{\left|K\left(x_0,x_0-y\right)\right|}^2\text{d}y\right)}^{1/2},\end{array}$

for the second term above, similar to the proof of Lemma 2.1 in [13] , we have

$\begin{array}{l}{\left({\displaystyle {\int }_{{\left(2^{k}d\right)}^{1-\theta }\le \left|y-x_0\right|<{\left(2^{k+1}d\right)}^{1-\theta }}}\frac{{\left|x_0-x\right|}^2}{{\left|x_0-y\right|}^2}{\left|K\left(x_0,x_0-y\right)\right|}^2\text{d}y\right)}^{1/2}\\ \le C\frac{{\left|x_0-x\right|}^{\left(1-\theta \right)\left(m-n/2\right)}}{{\left(2^{k}d\right)}^{m\left(1-\theta \right)}}\mathrm{,}\end{array}$

thus, by Lemma 3 and recall that $n/2<m$ ,

$\begin{array}{c}\left|I_5^{\left(1\right)}\right|\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right){\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}{k}^2\frac{d^{\left(1-\theta \right)\left(m-n/2\right)}}{{\left(2^{k}d\right)}^{m\left(1-\theta \right)}}{\left({\displaystyle {\int }_{\left|y-x_0\right|<{\left(2^{k+1}d\right)}^{1-\theta }}}{\left|f\left(y\right)\right|}^2\text{d}y\right)}^{1/2}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right){\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{k}^2{2}^{k\left(1-\theta \right)\left(n/2-m\right)}{\left(\frac{1}{\left|Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)\right|}{\displaystyle {\int }_{Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)}}{\left|f\left(y\right)\right|}^r\text{d}y\right)}^{1/r}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right){\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{k}^2{2}^{k\left(1-\theta \right)\left(n/2-m\right)}{M}_r\left(f\right)\left(\tilde{x}\right)\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).\end{array}$

For $I_5^{\left(2\right)}$ , by the formula (see [4] ):

$R_L\left(\tilde{b};x,y\right)-R_L\left(\tilde{b};x_0,y\right)={\displaystyle \underset{\left|\beta \right|<L}{\sum }}\frac{1}{\beta !}{R}_{L-\left|\beta \right|}\left(D^{\beta }\tilde{b};x,x_0\right){\left(x-y\right)}^{\beta }$

and Lemma 1, we have

$\left|R_L\left(\tilde{b};x,y\right)-R_L\left(\tilde{b};x_0,y\right)\right|\le C{\displaystyle \underset{\left|\beta \right|<L}{\sum }}{\displaystyle \underset{\left|\alpha \right|=L}{\sum }}{\left|x-x_0\right|}^{L-\left|\beta \right|}{\left|x-y\right|}^{\left|\beta \right|}{\Vert D^{\alpha }b\Vert }_{BMO},$

thus

$\begin{array}{c}\left|I_5^{\left(2\right)}\right|\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right){\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}{\displaystyle {\int }_{{\left(2^{k}d\right)}^{1-\theta }\le \left|y-x_0\right|<{\left(2^{k+1}d\right)}^{1-\theta }}}k\frac{\left|x-x_0\right|}{\left|x_0-y\right|}\left|K\left(x_0,x_0-y\right)\right|\left|f\left(y\right)\right|\text{d}y\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right){\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}k{2}^{k\left(1-\theta \right)\left(n/2-m\right)}{\left(\frac{1}{\left|Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)\right|}{\displaystyle {\int }_{Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)}}{\left|f\left(y\right)\right|}^r\text{d}y\right)}^{1/r}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).\end{array}$

Similarly,

$\left|I_5^{\left(3\right)}\right|\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).$

For $I_5^{\left(4\right)}$ , recall that $\left|b_{2^{k}Q}-b_{2Q}\right|\le Ck{\Vert b\Vert }_{BMO}$ , similar to the proofs of $I_5^{\left(1\right)}$ and $I_5^{\left(2\right)}$ , we get, for $1/r+1/r^{\prime }=1/2$

$\begin{array}{c}\left|I_5^{\left(4\right)}\right|\le C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\displaystyle {\int }_{R^n}}\left|\frac{{\left(x-y\right)}^{{\alpha }_1}}{{\left|x-y\right|}^L}-\frac{{\left(x_0-y\right)}^{{\alpha }_1}}{{\left|x_0-y\right|}^L}\right|\left|R_{m_2}\left({\tilde{b}}_2;x,y\right)\right|\left|K\left(x,x-y\right)\right|\left|D^{{\alpha }_1}{\tilde{b}}_1\left(y\right)\right|\left|f_2\left(y\right)\right|\text{d}y\\ +C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\displaystyle {\int }_{R^n}}\left|R_{m_2}\left({\tilde{b}}_2;x,y\right)-R_{m_2}\left({\tilde{b}}_2;x_0,y\right)\right|\frac{\left|{\left(x_0-y\right)}^{{\alpha }_1}\right|}{{\left|x_0-y\right|}^L}\left|K\left(x,x-y\right)\right|\left|D^{{\alpha }_1}{\tilde{b}}_1\left(y\right)\right|\left|f_2\left(y\right)\right|\text{d}y\\ +C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\displaystyle {\int }_{R^n}}\left|K\left(x,x-y\right)-K\left(x_0,x_0-y\right)\right|\frac{\left|{\left(x_0-y\right)}^{{\alpha }_1}\right|}{{\left|x_0-y\right|}^L}\left|R_{m_2}\left({\tilde{b}}_2;x_0,y\right)\right|\left|D^{{\alpha }_1}{\tilde{b}}_1\left(y\right)\right|\left|f_2\left(y\right)\right|\text{d}y\\ \le C{\displaystyle \underset{\left|\alpha \right|=m_2}{\sum }}{\Vert D^{\alpha }{b}_2\Vert }_{BMO}{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}k{2}^{k\left(1-\theta \right)\left(n/2-m\right)}\\ \times {\left(\frac{1}{\left|Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)\right|}{\displaystyle {\int }_{Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)}}{\left|f\left(y\right)D^{{\alpha }_1}{\tilde{b}}_1\left(y\right)\right|}^2\text{d}y\right)}^{1/2}\end{array}$

$\begin{array}{c}\le C{\displaystyle \underset{\left|\alpha \right|=m_2}{\sum }}{\Vert D^{\alpha }{b}_2\Vert }_{BMO}{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}k{2}^{k\left(1-\theta \right)\left(n/2-m\right)}{\left(\frac{1}{\left|Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)\right|}{\displaystyle {\int }_{Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)}}{\left|f\left(y\right)\right|}^r\text{d}y\right)}^{1/r}\\ \times {\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\left(\frac{1}{\left|Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)\right|}{\displaystyle {\int }_{Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)}}{\left|D^{{\alpha }_1}{b}_1\left(y\right)-{\left(D^{{\alpha }_1}{b}_1\right)}_{\tilde{Q}}\right|}^{r^{\prime }}\text{d}y\right)}^{1/r^{\prime }}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right){\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{k}^2{2}^{k\left(1-\theta \right)\left(n/2-m\right)}{M}_r\left(f\right)\left(\tilde{x}\right)\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).\end{array}$

Similarly,

$\left|I_5^{\left(5\right)}\right|\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).$

For $I_5^{\left(6\right)}$ , similar to the proof of $I_5^{\left(1\right)}$ , we get, for $1/r+1/r_1+1/r_2=1/2$ ,

$\begin{array}{c}\left|I_5^{\left(6\right)}\right|\le C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\displaystyle {\int }_{R^n}}\left|\frac{{\left(x-y\right)}^{{\alpha }_1+{\alpha }_2}}{{\left|x-y\right|}^L}-\frac{{\left(x_0-y\right)}^{{\alpha }_1+{\alpha }_2}}{{\left|x_0-y\right|}^L}\right|\\ \times \left|K\left(x,x-y\right)\right|\left|D^{{\alpha }_1}{\tilde{b}}_1\left(y\right)\right|\left|D^{{\alpha }_2}{\tilde{b}}_2\left(y\right)\right|\left|f_2\left(y\right)\right|\text{d}y\\ +C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\displaystyle {\int }_{R^n}}\left|K\left(x,x-y\right)-K\left(x_0,x_0-y\right)\right|\frac{\left|{\left(x_0-y\right)}^{{\alpha }_1+{\alpha }_2}\right|}{{\left|x_0-y\right|}^L}\\ \times \left|D^{{\alpha }_1}{\tilde{b}}_1\left(y\right)\right|\left|D^{{\alpha }_2}{\tilde{b}}_2\left(y\right)\right|\left|f_2\left(y\right)\right|\text{d}y\end{array}$

$\begin{array}{c}\le C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{2}^{k\left(1-\theta \right)\left(n/2-m\right)}\\ \times {\left(\frac{1}{\left|Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)\right|}{\displaystyle {\int }_{Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)}}{\left|f\left(y\right)D^{{\alpha }_1}{\tilde{b}}_1\left(y\right)D^{{\alpha }_2}{\tilde{b}}_2\left(y\right)\right|}^2\text{d}y\right)}^{1/2}\end{array}$

$\begin{array}{c}\le C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{2}^{k\left(1-\theta \right)\left(n/2-m\right)}{\left(\frac{1}{\left|Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)\right|}{\displaystyle {\int }_{Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)}}{\left|f\left(y\right)\right|}^r\text{d}y\right)}^{1/r}\\ \times {\displaystyle \underset{j=1}{\overset{2}{\prod }}}{\left(\frac{1}{\left|Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)\right|}{\displaystyle {\int }_{Q\left(x_0,{\left(2^{k}d\right)}^{1-\theta }\right)}}{\left|D^{{\alpha }_j}{b}_j\left(y\right)-{\left(D^{{\alpha }_j}{b}_j\right)}_{\tilde{Q}}\right|}^{r_j}\text{d}y\right)}^{1/r_j}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).\end{array}$

Thus

$\left|I_5\right|\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).$

Case 2. $d>1$ . In this case, let $\tilde{Q}=5\sqrt{n}Q$ and ${\tilde{b}}_j\left(x\right)=b_j\left(x\right)-{\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}\frac{1}{\alpha !}{\left(D^{\alpha }{b}_j\right)}_{\tilde{Q}}{x}^{\alpha }$ , then $R_{m_j}\left(b_j;x,y\right)=R_{m_j}\left({\tilde{b}}_j;x,y\right)$ and $D^{\alpha }{\tilde{b}}_j=D^{\alpha }{b}_j-{\left(D^{\alpha }{b}_j\right)}_{\tilde{Q}}$ for $\left|\alpha \right|=m_j$ . Write, for $f=f{\chi }_{\tilde{Q}}+f{\chi }_{R^n\backslash \tilde{Q}}=f_1+f_2$ ,

$\begin{array}{l}\frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}\left|T_b\left(f\right)\left(x\right)\right|\text{d}x\\ \le \frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}\left|{\displaystyle {\int }_{R^n}}\frac{{\displaystyle {\prod }_{j=1}^2{R}_{m_j}}\left({\tilde{b}}_j;x,y\right)}{{\left|x-y\right|}^L}K\left(x,x-y\right)f_1\left(y\right)\text{d}y\right|\text{d}x\\ +\frac{C}{\left|Q\right|}{\displaystyle {\int }_Q}\left|{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\displaystyle {\int }_{R^n}}\frac{R_{m_2}\left({\tilde{b}}_2;x,y\right){\left(x-y\right)}^{{\alpha }_1}}{{\left|x-y\right|}^L}{D}^{{\alpha }_1}{\tilde{b}}_1\left(y\right)K\left(x,x-y\right)f_1\left(y\right)\text{d}y\right|\text{d}x\\ +\frac{C}{\left|Q\right|}{\displaystyle {\int }_Q}\left|{\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}{\displaystyle {\int }_{R^n}}\frac{R_{m_1}\left({\tilde{b}}_1;x,y\right){\left(x-y\right)}^{{\alpha }_2}}{{\left|x-y\right|}^L}{D}^{{\alpha }_2}{\tilde{b}}_2\left(y\right)K\left(x,x-y\right)f_1\left(y\right)\text{d}y\right|\text{d}x\end{array}$

$\begin{array}{l}+\frac{C}{\left|Q\right|}{\displaystyle {\int }_Q}\left|{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\displaystyle {\int }_{R^n}}\frac{{\left(x-y\right)}^{{\alpha }_1+{\alpha }_2}{D}^{{\alpha }_1}{\tilde{b}}_1\left(y\right)D^{{\alpha }_2}{\tilde{b}}_2\left(y\right)}{{\left|x-y\right|}^L}K\left(x,x-y\right)f_1\left(y\right)\text{d}y\right|\text{d}x\\ +\frac{1}{\left|Q\right|}{\displaystyle {\int }_Q}\left|T_{\tilde{b}}\left(f_2\right)\left(x\right)\right|\text{d}x\\ :=J_1+J_2+J_3+J_4+J_5.\end{array}$

Similar to the proof of $I_1$ , $I_2$ , $I_3$ and $I_4$ , we get, by the $L^p\left(1<p<\infty \right)$ -boundedness of T (see Lemma 2),

$\begin{array}{c}{J}_1\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right){\left(\frac{1}{\left|Q\right|}{\displaystyle {\int }_{R^n}}{\left|T\left(f_1\right)\left(x\right)\right|}^r\text{d}x\right)}^{1/r}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right){\left|Q\right|}^{-1/r}{\left({\displaystyle {\int }_{R^n}}{\left|f_1\left(x\right)\right|}^r\text{d}x\right)}^{1/r}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right){\left(\frac{1}{\left|\tilde{Q}\right|}{\displaystyle {\int }_{\tilde{Q}}}{\left|f\left(x\right)\right|}^r\text{d}x\right)}^{1/r}\end{array}$

$\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|{\alpha }_j\right|=m_j}{\sum }}{\Vert D^{{\alpha }_j}{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right);$

$\begin{array}{c}{J}_2\le C{\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}{\Vert D^{{\alpha }_2}{b}_2\Vert }_{BMO}{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\left(\frac{1}{\left|Q\right|}{\displaystyle {\int }_{R^n}}{\left|T\left(D^{{\alpha }_1}{\tilde{b}}_1{f}_1\right)\left(x\right)\right|}^2\text{d}x\right)}^{1/2}\\ \le C{\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}{\Vert D^{{\alpha }_2}{b}_2\Vert }_{BMO}{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\left|Q\right|}^{-1/2}{\left({\displaystyle {\int }_{R^n}}{\left|D^{{\alpha }_1}{\tilde{b}}_1\left(x\right)f_1\left(x\right)\right|}^2\text{d}x\right)}^{1/2}\\ \le C{\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}{\Vert D^{{\alpha }_2}{b}_2\Vert }_{BMO}{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\left(\frac{1}{\left|\tilde{Q}\right|}{\displaystyle {\int }_{\tilde{Q}}}{\left|D^{{\alpha }_1}{b}_1\left(x\right)-{\left(D^{\alpha }{b}_1\right)}_{\tilde{Q}}\right|}^{r^{\prime }}\text{d}x\right)}^{1/r^{\prime }}\\ \times {\left(\frac{1}{\left|\tilde{Q}\right|}{\displaystyle {\int }_{\tilde{Q}}}{\left|f\left(x\right)\right|}^r\text{d}x\right)}^{1/r}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right);\end{array}$

$J_3\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right);$

$\begin{array}{c}{J}_4\le C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\left|Q\right|}^{-1/2}{\left({\displaystyle {\int }_{R^n}}{\left|T\left(D^{{\alpha }_1}{\tilde{b}}_1{D}^{{\alpha }_2}{\tilde{b}}_2{f}_1\right)\left(x\right)\right|}^2\text{d}x\right)}^{1/2}\\ \le C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\left|Q\right|}^{-1/2}{\left({\displaystyle {\int }_{R^n}}{\left|D^{{\alpha }_1}{\tilde{b}}_1\left(x\right)D^{{\alpha }_2}{\tilde{b}}_2\left(x\right)f_1\left(x\right)\right|}^2\text{d}x\right)}^{1/2}\\ \le C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\displaystyle \underset{j=1}{\overset{2}{\prod }}}{\left(\frac{1}{\left|\tilde{Q}\right|}{\displaystyle {\int }_{\tilde{Q}}}{\left|D^{{\alpha }_j}{\tilde{b}}_j\left(x\right)\right|}^{r_j}\text{d}x\right)}^{1/r_j}{\left(\frac{1}{\left|\tilde{Q}\right|}{\displaystyle {\int }_{\tilde{Q}}}{\left|f\left(x\right)\right|}^r\text{d}x\right)}^{1/r}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).\end{array}$

For $J_5$ , we write

$\begin{array}{l}{T}_{\tilde{b}}\left(f_2\right)\left(x\right)={\displaystyle {\int }_{R^n}}\frac{{\displaystyle {\prod }_{j=1}^2{R}_{m_j}}\left({\tilde{b}}_j;x,y\right)}{{\left|x-y\right|}^L}K\left(x,x-y\right)f_2\left(y\right)\text{d}y\\ -{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}\frac{1}{{\alpha }_1!}{\displaystyle {\int }_{R^n}}\frac{R_{m_2}\left({\tilde{b}}_2;x,y\right){\left(x-y\right)}^{{\alpha }_1}}{{\left|x-y\right|}^L}K\left(x,x-y\right)D^{{\alpha }_1}{\tilde{b}}_1\left(y\right)f_2\left(y\right)\text{d}y\\ -{\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}\frac{1}{{\alpha }_2!}{\displaystyle {\int }_{R^n}}\frac{R_{m_1}\left({\tilde{b}}_1;x,y\right){\left(x-y\right)}^{{\alpha }_2}}{{\left|x-y\right|}^L}K\left(x,x-y\right)D^{{\alpha }_2}{\tilde{b}}_2\left(y\right)f_2\left(y\right)\text{d}y\\ +{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}\frac{1}{{\alpha }_1!{\alpha }_2!}{\displaystyle {\int }_{R^n}}\frac{{\left(x-y\right)}^{{\alpha }_1+{\alpha }_2}}{{\left|x-y\right|}^L}K\left(x,x-y\right)D^{{\alpha }_1}{\tilde{b}}_1\left(y\right)D^{{\alpha }_2}{\tilde{b}}_2\left(y\right)f_2\left(y\right)\text{d}y,\end{array}$

similar to the proof of $I_5$ and by using lemma 4, we get

$\begin{array}{c}\left|T_{\tilde{b}}\left(f_2\right)\left(x\right)\right|\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right){\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}{k}^2{\displaystyle {\int }_{2^{k+1}\tilde{Q}\backslash 2^k\tilde{Q}}}{\left|x-y\right|}^{-2n}\left|f\left(y\right)\right|\text{d}y\\ +C{\displaystyle \underset{\left|\alpha \right|=m_2}{\sum }}{\Vert D^{\alpha }{b}_2\Vert }_{BMO}{\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}k{\displaystyle {\int }_{2^{k+1}\tilde{Q}\backslash 2^k\tilde{Q}}}{\left|x-y\right|}^{-2n}\left|D^{{\alpha }_1}{\tilde{b}}_1\left(y\right)\right|\left|f\left(y\right)\right|\text{d}y\end{array}$

$\begin{array}{c}+C{\displaystyle \underset{\left|\alpha \right|=m_1}{\sum }}{\Vert D^{\alpha }{b}_1\Vert }_{BMO}{\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}k{\displaystyle {\int }_{2^{k+1}\tilde{Q}\backslash 2^k\tilde{Q}}}{\left|x-y\right|}^{-2n}\left|D^{{\alpha }_2}{\tilde{b}}_2\left(y\right)\right|\left|f\left(y\right)\right|\text{d}y\\ +C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{\displaystyle \underset{k=0}{\overset{\infty }{\sum }}}{\displaystyle {\int }_{2^{k+1}\tilde{Q}\backslash 2^k\tilde{Q}}}{\left|x-y\right|}^{-2n}\left|D^{{\alpha }_1}{\tilde{b}}_1\left(y\right)\right|\left|D^{{\alpha }_2}{\tilde{b}}_2\left(y\right)\right|\left|f\left(y\right)\right|\text{d}y\end{array}$

$\begin{array}{c}\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)d^{-n}{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{k}^2{2}^{-kn}{\left(\frac{1}{\left|2^k\tilde{Q}\right|}{\displaystyle {\int }_{2^k\tilde{Q}}}{\left|f\left(y\right)\right|}^r\text{d}y\right)}^{1/r}\\ +C{\displaystyle \underset{\left|\alpha \right|=m_2}{\sum }}{\Vert D^{\alpha }{b}_2\Vert }_{BMO}{d}^{-n}{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}k{2}^{-kn}{\left(\frac{1}{\left|2^k\tilde{Q}\right|}{\displaystyle {\int }_{2^k\tilde{Q}}}{\left|f\left(y\right)\right|}^r\text{d}y\right)}^{1/r}\\ \times {\displaystyle \underset{\left|{\alpha }_1\right|=m_1}{\sum }}{\left(\frac{1}{\left|2^k\tilde{Q}\right|}{\displaystyle {\int }_{2^k\tilde{Q}}}{\left|D^{{\alpha }_1}{b}_1\left(y\right)-{\left(D^{{\alpha }_1}{b}_1\right)}_{\tilde{Q}}\right|}^{r^{\prime }}\text{d}y\right)}^{1/r^{\prime }}\\ +C{\displaystyle \underset{\left|\alpha \right|=m_1}{\sum }}{\Vert D^{\alpha }{b}_1\Vert }_{BMO}{d}^{-n}{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}k{2}^{-kn}{\left(\frac{1}{\left|2^k\tilde{Q}\right|}{\displaystyle {\int }_{2^k\tilde{Q}}}{\left|f\left(y\right)\right|}^r\text{d}y\right)}^{1/r}\\ \times {\displaystyle \underset{\left|{\alpha }_2\right|=m_2}{\sum }}{\left(\frac{1}{\left|2^k\tilde{Q}\right|}{\displaystyle {\int }_{2^k\tilde{Q}}}{\left|D^{{\alpha }_2}{b}_2\left(y\right)-{\left(D^{{\alpha }_1}{b}_2\right)}_{\tilde{Q}}\right|}^{r^{\prime }}\text{d}y\right)}^{1/r^{\prime }}\end{array}$

$\begin{array}{c}+C{\displaystyle \underset{\left|{\alpha }_1\right|=m_1,\left|{\alpha }_2\right|=m_2}{\sum }}{d}^{-n}{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{2}^{-kn}{\left(\frac{1}{\left|2^k\tilde{Q}\right|}{\displaystyle {\int }_{2^k\tilde{Q}}}{\left|f\left(y\right)\right|}^r\text{d}y\right)}^{1/r}\\ \times {\displaystyle \underset{j=1}{\overset{2}{\prod }}}{\left(\frac{1}{\left|2^k\tilde{Q}\right|}{\displaystyle {\int }_{2^k\tilde{Q}}}{\left|D^{{\alpha }_j}{b}_j\left(y\right)-{\left(D^{{\alpha }_j}{b}_j\right)}_{\tilde{Q}}\right|}^{r_j}\text{d}y\right)}^{1/r_j}\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right){\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{k}^2{2}^{-kn}{M}_r\left(f\right)\left(\tilde{x}\right)\\ \le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right),\end{array}$

thus

$\left|J_5\right|\le C{\displaystyle \underset{j=1}{\overset{2}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right)M_r\left(f\right)\left(\tilde{x}\right).$

This completes the proof of Theorem 1.

Proof of Theorem 2. (a) follows from Theorem 1. For (b), Choose $1<r<p$ in Theorem 1, we get

$\begin{array}{c}{\Vert T_b\left(f\right)\Vert }_{L^p}\le {\Vert M\left(T_b\left(f\right)\right)\Vert }_{L^p}\le C{\Vert {\left(T_b\left(f\right)\right)}^{\#}\Vert }_{L^p}\\ \le C{\displaystyle \underset{j=1}{\overset{l}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right){\Vert M_r\left(f\right)\Vert }_{L^p}\\ \le C{\displaystyle \underset{j=1}{\overset{l}{\prod }}}\left({\displaystyle \underset{\left|\alpha \right|=m_j}{\sum }}{\Vert D^{\alpha }{b}_j\Vert }_{BMO}\right){\Vert f\Vert }_{L^p}.\end{array}$

This finishes the proof.

Acknowledgements

The authors would like to express their gratitude to the referee for his/her valuable comments and suggestions.

Funding

Project supported by Scientific Research Fund of Hunan Provincial Education Departments (19C1037).

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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