Iterated Commutators for Multilinear Singular Integral Operators on Morrey Space with Non-Doubling Measures

Tie Li, Yinsheng Jiang, Yaoming Niu

Baotou Teacher’s College of Inner Mongolia University of Science and Technology, Baotou, China.

DOI: 10.4236/jamp.2020.81005                

Abstract

Let μ be a non-negative Radon measure on R^d which only satisfies the following growth condition that there exists a positive constant C such that μ(B(x,r)) ≤ Crⁿ for all x∈ R^d, r > 0 and some fixed n ∈ (0,d]. This paper is interested in the properties of the iterated commutators of multilinear singular integral operators on Morrey spaces .Precisely speaking, we show that the iterated commutators generated by multilinear singular integrals operators are bounded from to where (Regular Bounded Mean Oscillation space) and 1 < q_j ≤ p_j <∞ with 1/p = 1/p₁ + ... + 1/p_m and 1/q = 1/q₁+ ... + 1/q_m.

Keywords

Non-Doubling Measures, Morrey Space, Multilinear Singular Integral Operators, RBMO, Commutator

1. Introduction

Let $\mu$ be a positive Radon measures on ${ℝ}^d$ satisfying only the growth condition, that is, there exists a constant $C>0$ and $n\in \left(\mathrm{0,}d\right]$ such that

$\mu \left(Q\right)\le Cl{\left(Q\right)}^n$ (1)

for any cube $Q\subset {ℝ}^d$ with sides parallel to the coordinate axes. $Q\left(x\mathrm{,}l\left(Q\right)\right)$ will be the cube centered at x with side length $l\left(Q\right)$. For $r>0$, rQ will denote the cube with the same center as Q and with $l\left(rQ\right)=rl\left(Q\right)$. The set of all cubes $Q\subset {ℝ}^d$, satisfying $\mu \left(Q\right)>0$ is denoted by $\mathfrak{L}\left(\mu \right)$. In this note, we do not assume that $\mu$ is doubling.

Nazarov, Treil and Volberg developed the theory of the singular integrals for the measures with growth condition to investigate the analytic capacity on the complex plane [1] [2]. Tolsa showed that the analytic capacity is subadditive and that it is bi-Lipschitz invariant [3] [4] and defined for the growth measures RBMO (regular bounded mean oscillation) space, the Hardy space $H_1\left(\mu \right)$ and the Littlewood-Paley decomposition [5] [6]. He also gave his $H_1\left(\mu \right)$ space in terms of the grand maximal operator [7]. Recently many people paid attention to the measure with growth condition because of recovering the Calderón-Zygmund theory and solving the long-standing open Painlevé problem.

The boundedness of fractional integral operators on the classical Morrey spaces was studied by Adams [8], Chiarenza et al. [9]. In [9], by establishing a pointwise estimate of fractional integrals in terms of the Hardy-Littlewood maximal function, they showed the boundedness of fractional integral operators on the Morrey spaces. In 2005, Sawano and Tanaka [10] gave a natural definition of Morrey spaces for Radon measures which might be non-doubling but satisfied the growth condition, and they investigated the boundedness in these spaces of some classical operators in harmonic analysis. Later on, Sawano [11] defined the generalized Morrey spaces on ${ℝ}^n$ for non-doubling measure and showed the properties of maximal operators, fractional integral operators and singular operators in this space.

A classical result of commutator is due to Coifman, Rochberg and Weiss [12], if $b\in BMO$ and T is a Calderón-Zygmund operator, then the commutator $\left[b\mathrm{,}T\right]$ is bounded on $L^p$ spaces for $1<p<\infty$. The same result for the multilinear commutator was obtained by Pérez and Trujillo-Gonzalez [13]. Tolsa [5] developed the theory of Calderón-Zygmund operators and their commutators with RBMO functions in the setting of non-doubling measures. Hu, Meng and Yang [14] considered the multilinear commutator on Lebesgue spaces with non-doubling measures. Chen and Sawyer [15] modified the definition of RBMO to investigate the commutators of the potential operators and RBMO functions.

In the last decade, multilinear singular integrals of Calderón-Zygmund type have attracted great attentions. Some interesting results refer to [16] [17] [18] [19] [20] in the text of Lebesgue measures. It points out that Perez and Pradolini [21] introduced a said iterated commutators generated by the multilinear singular integral operators with Calderón-Zygmund type and vector function $\stackrel{\to }{b}\in RBM{O}^m$ and obtained the boundedness from $L^{p_1}\times \cdots \times L^{p_m}$ to $L^p$ with $1/p=1/p_1+\cdots +1/p_m$ for $1<p_1,\cdots ,p_m<\infty$ (in fact, they considered the weighted case). Xu [22] extended the result to the case of the non-doubling measures. Very recently, Tao and He [23] obtained the boundedness of the multilinear Calderón-Zygmund operators on the generalized Morrey spaces over the quasi-metric space of non-homogeneoustype. The aim of this paper is to study the iterated commutators of multilinear singular integral operators on Morrey spaces with non-doubling measures.

Before stating our result, we recall some definitions and notation. Given ${\beta }_d\ge 2^{d+1}$ large enough but depending only on the dimension d, we say that a cube $Q\subset {ℝ}^d$ is doubling if $\mu \left(2Q\right)\le {\beta }_d\mu \left(Q\right)$. For any fixed cube $Q\subset {ℝ}^d$, let N be the smallest nonnegative integer such that $2^{N}Q$ is doubling. We denote this cube by $\tilde{Q}$.

For two cubes $Q\subset R$ in ${ℝ}^d$, we suppose

$K_{Q,R}=1+{\displaystyle \underset{k=1}{\overset{N_{Q,R}}{\sum }}}\frac{\mu \left(2^{k}Q\right)}{l{\left(2^{k}Q\right)}^n},$ (2)

where $N_{Q\mathrm{,}R}$ is the first positive integer k such that $l\left(2^{k}Q\right)\ge l\left(R\right)$. This was introduced by Tolsa in [5].

Let $m_{Q}f$ be the mean value of f on Q, namely, $m_{Q}f=\frac{1}{\mu \left(Q\right)}{\displaystyle {\int }_Q}f\left(x\right)\text{d}\mu$. The regularity bounded mean oscillations function spaces were introduced by Tolsa [5].

Definition 1.1. Let $\eta >1$ be a fixed constant. We say that $f\in L_{loc}^1\left(\mu \right)$ is in RBMO if there exists a constant $\mathcal{A}$ such that

$\frac{1}{\mu \left(\eta Q\right)}{\displaystyle {\int }_Q}\left|f\left(y\right)-m_{\tilde{Q}}f\right|\text{d}\mu \left(y\right)\le \mathcal{A}\mathrm{,}$ (3)

for any cube Q, and

$\left|m_{Q}f-m_{R}f\right|\le \mathcal{A}{K}_{Q\mathrm{,}R}\mathrm{,}$ (4)

for any two doubling cubes $Q\subset R$. The mininal constant $\mathcal{A}$ is the $RBMO\left(\mu \right)$ norm of f, and it will be denoted by ${\Vert f\Vert }_{\ast }$.

The definition of the Morrey space with non-doubling measure is given in the following [10].

Definition 1.2. Let $k>1$ and $1\le q\le p<\infty$, the Morrey space $M_q^p\left(k\mathrm{,}\mu \right)$ is defined as

$M_q^p\left(k,\mu \right):=\left\{f\in L_{loc}^q\left(\mu \right):{\Vert f\Vert }_{M_q^p\left(k,\mu \right)}<\infty \right\},$

where

${\Vert f\Vert }_{M_q^p\left(k,\mu \right)}:=\underset{Q\in \mathfrak{L}\left(\mu \right)}{\sup }\mu {\left(kQ\right)}^{\frac{1}{p}-\frac{1}{q}}{\left({\displaystyle {\int }_Q}{\left|f\right|}^q\text{d}\mu \right)}^{\frac{1}{q}}.$ (5)

It is easy to observe that $L^p\left(\mu \right)=M_p^p\left(k\mathrm{,}\mu \right)$, and Hölder’s inequality tells us ${\Vert f\Vert }_{M_{q_2}^p\left(k,\mu \right)}\le {\Vert f\Vert }_{M_{q_1}^p\left(k,\mu \right)}$ for all $1\le q_2\le q_1\le p$, then we have $L^p\left(\mu \right)=M_p^p\left(k,\mu \right)\subset M_{q_1}^p\left(k,\mu \right)\subset M_{q_2}^p\left(k,\mu \right)$. The space $M_q^p\left(k,\mu \right)$ is a Banach space with its norm ${\Vert f\Vert }_{M_q^p\left(k,\mu \right)}$ and the parameter $k>1$ appearing in the definition does not affect it. The Morrey space norm reflects local regularity of f more precisely than the Lebesgue space norm. See [10] [11] [24] for details. We will denote $M_q^p\left(\mathrm{2,}\mu \right)$ by $M_q^p\left(\mu \right)$.

Denoting by $\stackrel{\to }{f}=\left(f_1,f_2,\cdots ,f_m\right)$, we consider the multilinear singular integral operator $T_m$ as follows,

$\begin{array}{l}{T}_m\left(\stackrel{\to }{f}\right)\left(x\right)\\ ={\displaystyle {\int }_{{\left({ℝ}^d\right)}^m}}K\left(x\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_m\right)f_1\left(y_1\right)\cdots f_m\left(y_m\right)\text{d}\mu \left(y_1\right)\cdots \text{d}\mu \left(y_m\right)\text{for}x\in {ℝ}^d\mathrm{,}\end{array}$ (6)

whenever $f_1\mathrm{,}\cdots \mathrm{,}{f}_m$ are $C^{\infty }$ -functions with compact support and $x\notin {\displaystyle {\cap }_{j=1}^m}\text{supp}{f}_j$. Moreover,

$\left|K\left(y_0\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_m\right)\right|\le A{\left({\displaystyle \underset{k\mathrm{,}l=0}{\overset{m}{\sum }}}\left|y_k-y_l\right|\right)}^{-mn}$ (7)

and, for some $ϵ>0$,

$\left|K\left(y_0\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_j\mathrm{,}\cdots \mathrm{,}{y}_m\right)-K\left(y_0\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y^{\prime }}_j\mathrm{,}\cdots \mathrm{,}{y}_m\right)\right|\le \frac{A{\left|y_j-{y^{\prime }}_j\right|}^{ϵ}}{{\left({\displaystyle {\sum }_{k\mathrm{,}l=0}^m\left|y_k-y_l\right|}\right)}^{mn+ϵ}}$ (8)

provided that $0\le j\le m$ and $\left|y_j-{y^{\prime }}_j\right|\le \frac{1}{2}{\max }_{0\le k\le m}\left|y_j-y_k\right|$.

Let $b_i\in RBMO\left(\mu \right)$ for $i=1,2,\cdots ,m$ and let $\stackrel{\to }{b}=\left(b_1\mathrm{,}{b}_2\mathrm{,}\cdots \mathrm{,}{b}_m\right)$, then the iterated commutator ${\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}$ is formally defined as

$\begin{array}{c}{\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\left(x\right)={\displaystyle {\int }_{{\left({ℝ}^d\right)}^m}}{\displaystyle \underset{j=1}{\overset{m}{\prod }}}\left(b_j\left(x\right)-b_j\left(y_j\right)\right)K\left(x\mathrm{,}{y}_1\mathrm{,}\cdots \mathrm{,}{y}_m\right)\\ \times f_1\left(y_1\right)\cdots f_m\left(y_m\right)\text{d}\mu \left(y_1\right)\cdots \text{d}\mu \left(y_m\right)\mathrm{.}\end{array}$ (9)

Suppose $\Vert \mu \Vert =\infty$ the main result in this paper can be stated as follow.

Theorem 1.1. Let ${\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}$ as in (9) and satisfying conditions (7) and (8). Let $1<q_j\le p_j<\infty$ with $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$ and $\frac{1}{q}=\frac{1}{q_1}+\cdots +\frac{1}{q_m}$. Suppose $b_i\in RBMO\left(\mu \right)$ for $i=1,2,\cdots ,m$. If $T_m$ maps $L^1\left(\mu \right)\times \cdots \times L^1\left(\mu \right)$ to $L^{1/m\mathrm{,}\infty }\left(\mu \right)$, then the commutators ${\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}$ are bounded from $M_{q_1}^{p_1}\left(\mu \right)\times \cdots \times M_{q_m}^{p_m}\left(\mu \right)$ to $M_q^p\left(\mu \right)$, that is,

${\Vert {\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\left(x\right)\Vert }_{M_q^p\left(\mu \right)}\le C{\displaystyle \underset{i\mathrm{<mo>\; =\; </mo>1}}{\overset{m}{\prod }}}{\Vert b_i\Vert }_{\ast }{\Vert f_i\Vert }_{M_{q_i}^{p_i}\left(\mu \right)}\mathrm{.}$ (10)

More generally, denote by $C_i^m$ the family of all subsets $\sigma =\left\{{\sigma }_1\mathrm{,}{\sigma }_2\mathrm{,}\cdots \mathrm{,}{\sigma }_i\right\}$ of i different elements of $\left\{\mathrm{1,2,}\cdots \mathrm{,}m\right\}$, and let ${\sigma }^{\prime }=\left\{1,2,\cdots ,m\right\}\backslash \sigma$ and ${\stackrel{\to }{b}}_{\sigma }=\left\{b_{{\sigma }_1}\mathrm{,}{b}_{{\sigma }_2}\mathrm{,}\cdots \mathrm{,}{b}_{{\sigma }_i}\right\}$. For any $\sigma \in C_i^m$, we define

$\begin{array}{c}{\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}\left(\stackrel{\to }{f}\right)\left(x\right)={\displaystyle {\int }_{{\left({ℝ}^d\right)}^m}}{\displaystyle \underset{{\sigma }_j\in \sigma }{\prod }}\left(b_{{\sigma }_j}\left(x\right)-b_{{\sigma }_j}\left(y_{{\sigma }_j}\right)\right)K\left(x,y_1,\cdots ,y_m\right)\\ \times f_1\left(y_1\right)\cdots f_m\left(y_m\right)\text{d}\mu \left(y_1\right)\cdots \text{d}\mu \left(y_m\right).\end{array}$ (11)

In case $\sigma =\left\{1,2,\cdots ,m\right\}$, one sees that ${\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}$ is just the commutator ${\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}$. So we have a more generalization version of the theorem as following.

Theorem 1.2. Let ${\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}$ as in (11) and satisfying conditions (7) and (8). Let $1<q_j\le p_j<\infty$ with $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$ and $\frac{1}{q}=\frac{1}{q_1}+\cdots +\frac{1}{q_m}$. Suppose $b_i\in RBMO\left(\mu \right)$ for $i=1,2,\cdots ,m$. If $T_m$ maps $L^1\left(\mu \right)\times \cdots \times L^1\left(\mu \right)$ to $L^{1/m\mathrm{,}\infty }\left(\mu \right)$, then for all $\sigma \subset \left\{\mathrm{1,2,}\cdots \mathrm{,}m\right\}$, the commutators ${\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}$ are bounded from $M_{q_1}^{p_1}\left(\mu \right)\times \cdots \times M_{q_m}^{p_m}\left(\mu \right)$ to $M_q^p\left(\mu \right)$, that is,

${\Vert {\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}\left(\stackrel{\to }{f}\right)\left(x\right)\Vert }_{M_q^p\left(\mu \right)}\le C{\displaystyle \underset{j\in \sigma }{\prod }}{\Vert b_j\Vert }_{\ast }{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\Vert f_i\Vert }_{M_{q_i}^{p_i}\left(\mu \right)}\mathrm{.}$ (12)

2. Proof of Main Results

Before proving our theorem, we recall the following maximal operator,

$M_{k}f\left(x\right)=\underset{x\in Q\subset \mathfrak{L}\left(\mu \right)}{\sup }\frac{1}{\mu \left(kQ\right)}{\displaystyle {\int }_Q}\left|f\left(y\right)\right|\text{d}\mu \left(y\right)\mathrm{.}$ (13)

we will use the sharp maximal estimates. Let f be a function in $L_{loc}^1\left(\mu \right)$, the sharp maximal function of f is defined by

$M^{\#}f\left(x\right)=\underset{Q\ni x}{\sup }\frac{1}{\mu \left(\frac{3}{2}Q\right)}{\displaystyle {\int }_Q}\left|f\left(y\right)-m_{\tilde{Q}}f\right|\text{d}\mu \left(y\right)+\underset{\begin{array}{c}R\supset Q\ni x\\ Q\mathrm{,}R\text{doubling}\end{array}}{\sup }\frac{\left|m_{Q}f-m_{R}f\right|}{K_{Q\mathrm{,}R}}$ (14)

The non-centered doubling maximal operator is defined by

$Nf\left(x\right)=\underset{Q\ni x\mathrm{,}Q\text{doubling}}{\sup }\frac{1}{\mu \left(Q\right)}{\displaystyle {\int }_Q}\left|f\left(y\right)\right|\text{d}\mu \left(y\right)\mathrm{.}$ (15)

By the Lebesgue differential theorem, it is easy to see that $\left|f\left(x\right)\right|\le Nf\left(x\right)$ for any $f\in L_{loc}^1\left(\mu \right)$ and $\mu -a\mathrm{.}e\mathrm{.}x\in {ℝ}^d$. Define the non-centered maximal operator,

$M_{\tau \mathrm{,}\eta }f\left(x\right)=\underset{Q\ni x}{\sup }{\left\{\frac{1}{\mu \left(\eta Q\right)}{\displaystyle {\int }_Q}{\left|f\left(y\right)\right|}^{\tau }\text{d}\mu \left(y\right)\right\}}^{\frac{1}{\tau }}\mathrm{.}$ (16)

for $\eta >1$ and $\tau >1$, where the supremum is taking over all the cubes Q containing the point x.

To prove Theorem 1.2 is reduced to the following lemmas.

Lemma 2.1. Let $\tau >1,s_i>1,b_i\in RBMO\left(\mu \right)$ and $f_i\in L^{q_i}\left(\mu \right)\mathrm{,}i=\mathrm{1,2,}\cdots \mathrm{,}m$ and $\sigma \subset \left\{\mathrm{1,2,}\cdots \mathrm{,}m\right\}$. If $T_m$ maps $L^1\left(\mu \right)\times \cdots \times L^1\left(\mu \right)$ to $L^{1/m\mathrm{,}\infty }\left(\mu \right)$ and satisfying conditions (7) and (8). Then we have

$\begin{array}{l}{M}^{\#}\left({\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}\left(\stackrel{\to }{f}\right)\right)\left(x\right)\\ \le C[{\displaystyle \underset{j\in \sigma }{\prod }}{\Vert b_j\Vert }_{\ast }{M}_{\tau \mathrm{,}\frac{3}{2}}\left(T_m\left(\stackrel{\to }{f}\right)\right)\left(x\right)\\ +{\displaystyle \underset{\begin{array}{c}{\sigma }_1\cup {\sigma }_2=\sigma \\ {\sigma }_1\ne \varnothing \mathrm{,}{\sigma }_2\ne \varnothing \end{array}}{\sum }}{\displaystyle \underset{j\notin {\sigma }_1}{\prod }}{\Vert b_j\Vert }_{\ast }{M}_{\tau \mathrm{,}\frac{3}{2}}{\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }_2}}}\left(\stackrel{\to }{f}\right)\left(x\right)\\ +{\displaystyle \underset{j\in \sigma }{\prod }}{\Vert b_j\Vert }_{\ast }{\displaystyle \underset{j=1}{\overset{m}{\prod }}}{M}_{s_j\mathrm{,}\frac{9}{8}}{f}_j\left(x\right)]\mathrm{.}\end{array}$ (17)

We postpone the proof of Lemma 2.1 after of Theorem 1.2.

Lemma 2.2. [10] Let $q>\tau >1,\eta >1$, and $1<q\le p<\infty$, then the operator $M_{\tau \mathrm{,}\eta }$ is bounded on $M_q^p\left(\mu \right)$ and

${\Vert M_{\tau \mathrm{,}\eta }\left(f\right)\Vert }_{M_q^p\left(\mu \right)}\le C{\Vert f\Vert }_{M_q^p\left(\mu \right)}\mathrm{.}$

with the constant C independent of f.

Lemma 2.3. [24] Suppose that $1<q\le p<\infty$, and there exists an increasing sequence of concentric doubling cubes, $I_0\subset I_1\subset \cdots \subset I_k\subset \cdots$, such that

$\underset{k\to \infty }{\lim }{m}_{I_k}\left(f\right)=0\text{and}{\displaystyle \underset{k=0}{\overset{\infty }{\cup }}{I}_k}={ℝ}^d\mathrm{.}$

Then there exist a constant $C>0$ independent on f such that

${\Vert Nf\Vert }_{M_q^p\left(\mu \right)}\le C{\Vert M^{\#}\Vert }_{M_q^p\left(\mu \right)}\mathrm{.}$

Lemma 2.4. [25] Let $1<p_i<\infty$ and $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$. If $T_m$ maps $L^1\left(\mu \right)\times \cdots \times L^1\left(\mu \right)$ to $L^{1/m\mathrm{,}\infty }\left(\mu \right)$ and satisfying conditions (7) and (8). Then there exists a constant C independent of $f_i$ such that

${\Vert T_m\left(\stackrel{\to }{f}\right)\Vert }_{L^p\left(\mu \right)}\le C{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\Vert f_i\Vert }_{L^{p_i}\left(\mu \right)}\mathrm{.}$

Remark 2.1. If $T_m$ maps $L^1\left(\mu \right)\times \cdots \times L^1\left(\mu \right)$ to $L^{1/m\mathrm{,}\infty }\left(\mu \right)$ and satisfying conditions (7) and (8), with $1<q_i\le p_i<\infty$, $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$ and $\frac{1}{q}=\frac{1}{q_1}+\cdots +\frac{1}{q_m}$. From Corollary 1.8 in [23], we can easily get that $T_m$ is bounded from $M_{q_1}^{p_1}\left(\mu \right)\times \cdots \times M_{q_m}^{p_m}\left(\mu \right)$ to $M_q^p\left(\mu \right)$.

Proof of Theorem 1.2. Using Lemma 2.1, Lemma 2.2, Lemma 2.3 and Lemma 2.4, we get that

$\begin{array}{l}{\Vert {\left(T_m\right)}_{b_i}\left(\stackrel{\to }{f}\right)\Vert }_{M_q^p\left(\mu \right)}\\ \le C{\Vert N\left({\left(T_m\right)}_{b_i}\left(\stackrel{\to }{f}\right)\right)\Vert }_{M_q^p\left(\mu \right)}\le C{\Vert M^{\#}\left({\left(T_m\right)}_{b_i}\left(\stackrel{\to }{f}\right)\right)\Vert }_{M_q^p\left(\mu \right)}\\ \le C{\Vert b_i\Vert }_{\ast }{\Vert M_{\tau ,\frac{3}{2}}\left(T_m\left(\stackrel{\to }{f}\right)\right)\left(x\right)+{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{M}_{s_i,\frac{9}{8}}{f}_i\Vert }_{M_q^p\left(\mu \right)}\\ \le C{\Vert b_i\Vert }_{\ast }{\Vert M_{\tau ,\frac{3}{2}}\left(T_m\left(\stackrel{\to }{f}\right)\right)\Vert }_{M_q^p\left(\mu \right)}+C{\Vert b_i\Vert }_{\ast }{\Vert {\displaystyle \underset{i=1}{\overset{m}{\prod }}}{M}_{s_i,\frac{9}{8}}{f}_i\Vert }_{M_q^p\left(\mu \right)}\\ \le C{\Vert b_i\Vert }_{\ast }{\Vert T_m\left(\stackrel{\to }{f}\right)\Vert }_{M_q^p\left(\mu \right)}+C{\Vert b_i\Vert }_{\ast }{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\Vert f_i\Vert }_{M_{q_j}^{p_j}\left(\mu \right)}\\ \le C{\Vert b_i\Vert }_{\ast }{\displaystyle \underset{j=1}{\overset{m}{\prod }}}{\Vert f_i\Vert }_{M_{q_j}^{p_j}\left(\mu \right)}.\end{array}$

Applying the inequality (17) in Lemma 2.1, for $\sigma \subset \left\{\mathrm{1,2,}\cdots \mathrm{,}m\right\}$, we have

$\begin{array}{l}{\Vert {\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}\left(\stackrel{\to }{f}\right)\Vert }_{M_q^p\left(\mu \right)}\\ \le C{\Vert N\left({\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}\left(\stackrel{\to }{f}\right)\right)\Vert }_{M_q^p\left(\mu \right)}\le C{\Vert M^{\#}\left({\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}\left(\stackrel{\to }{f}\right)\right)\Vert }_{M_q^p\left(\mu \right)}\\ \le C\Vert {\displaystyle \underset{j\in \sigma }{\prod }}{\Vert b_j\Vert }_{\ast }{M}_{\tau ,\frac{3}{2}}\left(T_m\left(\stackrel{\to }{f}\right)\right)+{\displaystyle \underset{\begin{array}{c}{\sigma }_1\cup {\sigma }_2=\sigma \\ {\sigma }_1\ne \varnothing {\sigma }_2\ne \varnothing \end{array}}{\sum }}{\displaystyle \underset{j\in {\sigma }_1}{\prod }}{\Vert b_j\Vert }_{\ast }{M}_{\tau ,\frac{3}{2}}\left({\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }_2}}}\left(\stackrel{\to }{f}\right)\right)\\ +{{\displaystyle \underset{j\in \sigma }{\prod }}{\Vert b_j\Vert }_{\ast }{\displaystyle \underset{j=1}{\overset{m}{\prod }}}{M}_{s_j,\frac{9}{8}}{f}_j\Vert }_{M_q^p\left(\mu \right)}\\ \le C{\displaystyle \underset{j\in \sigma }{\prod }}{\Vert b_j\Vert }_{\ast }{\displaystyle \underset{j=1}{\overset{m}{\prod }}}{\Vert f_j\Vert }_{M_{q_j}^{p_j}\left(\mu \right)}+C{\displaystyle \underset{\begin{array}{c}{\sigma }_1\cup {\sigma }_2=\sigma \\ {\sigma }_1\ne \varnothing {\sigma }_2\ne \varnothing \end{array}}{\sum }}{\displaystyle \underset{j\in {\sigma }_1}{\prod }}{\Vert b_j\Vert }_{\ast }{\Vert M_{\tau ,\frac{3}{2}}\left({\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }_2}}}\left(\stackrel{\to }{f}\right)\right)\Vert }_{M_q^p\left(\mu \right)}\\ \le C{\displaystyle \underset{j\in \sigma }{\prod }}{\Vert b_j\Vert }_{\ast }{\displaystyle \underset{j=1}{\overset{m}{\prod }}}{\Vert f_j\Vert }_{M_{q_j}^{p_j}\left(\mu \right)}+C{\displaystyle \underset{\begin{array}{c}{\sigma }_1\cup {\sigma }_2=\sigma \\ {\sigma }_1\ne \varnothing {\sigma }_2\ne \varnothing \end{array}}{\sum }}{\displaystyle \underset{j\in {\sigma }_1}{\prod }}{\Vert b_j\Vert }_{\ast }{\Vert {\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }_2}}}\left(\stackrel{\to }{f}\right)\Vert }_{M_q^p(\mu )}\end{array}$

where ${\sigma }_1$ and ${\sigma }_2$ are two nonempty subsets of $\sigma$ and ${\sigma }_1\cap {\sigma }_2=\varnothing$. Hence, we can make use of induction on $\sigma \subset \left\{\mathrm{1,2,}\cdots \mathrm{,}m\right\}$ to get that

${\Vert {\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}\left(\stackrel{\to }{f}\right)\Vert }_{M_q^p\left(\mu \right)}\le C{\displaystyle \underset{j\in \sigma }{\prod }}{\Vert b_j\Vert }_{\ast }{\displaystyle \underset{i\mathrm{<mo>\; =\; </mo>1}}{\overset{m}{\prod }}}{\Vert f_i\Vert }_{M_{q_i}^{p_i}\left(\mu \right)}\mathrm{.}$ (18)

This completes the proof of Theorem 1.2.☐

For simplicity of the notation, we only show the special case $\sigma =\left\{1,\cdots ,m\right\}$ of Lemma 2.1. The similar process with minor modification will be to able prove Lemma 2.1 for the general case.

Lemma 2.5. Let $\tau >1,s_i>1,b_i\in RBMO\left(\mu \right)$ and $f_i\in L^{q_i}\left(\mu \right),i=1,2,\cdots ,m$. If $T_m$ maps $L^1\left(\mu \right)\times \cdots \times L^1\left(\mu \right)$ to $L^{1/m\mathrm{,}\infty }\left(\mu \right)$ and satisfying conditions (7) and (8), then there exists a constant $C>0$ independent of $b_i$ and $f_i$ such that

$\begin{array}{l}{M}^{\#}\left({\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\right)\left(x\right)\\ \le C[{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\Vert b_i\Vert }_{\ast }{M}_{\tau ,\frac{3}{2}}\left(T_m\left(\stackrel{\to }{f}\right)\right)\left(x\right)\\ +{\displaystyle \underset{i=1}{\overset{m-1}{\sum }}}{\displaystyle \underset{\sigma \in C_i^m}{\sum }}{\displaystyle \underset{j\in \sigma }{\prod }}{\Vert b_j\Vert }_{\ast }{M}_{\eta ,\frac{3}{2}}\left({\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }^{\prime }}}}\left(\stackrel{\to }{f}\right)\right)\left(x\right)\\ +{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\Vert b_i\Vert }_{\ast }{M}_{s_i,\frac{9}{8}}{f}_i\left(x\right)]\end{array}$ (19)

In order to prove Lemma 2.5, we have the following decomposition for the commutators ${\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}$. For any $\stackrel{\to }{\lambda }=\left({\lambda }_1\mathrm{,}{\lambda }_2\mathrm{,}\cdots \mathrm{,}{\lambda }_m\right)\in {ℝ}^m$, writing ${\stackrel{\to }{b}}_{\sigma }\left(x\right)-{\stackrel{\to }{b}}_{\sigma }\left(y\right)=\left({\stackrel{\to }{b}}_{\sigma }\left(x\right)-{\stackrel{\to }{\lambda }}_{\sigma }\right)-\left({\stackrel{\to }{b}}_{\sigma }\left(y\right)-{\stackrel{\to }{\lambda }}_{\sigma }\right)$, thus it is clear that ${\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}\left(\stackrel{\to }{f}\right)={\left(T_m\right)}_{{\displaystyle \prod \left({\stackrel{\to }{b}}_{\sigma }-{\stackrel{\to }{\lambda }}_{\sigma }\right)}}\left(\stackrel{\to }{f}\right)$. Moreover,

$\begin{array}{l}{\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\left(x\right)\\ ={\displaystyle \underset{i=0}{\overset{m}{\sum }}}{\displaystyle \underset{\sigma \in C_i^m}{\sum }}{\left(-1\right)}^{m-i}{\displaystyle \underset{{\sigma }_j\in \sigma }{\prod }}\left(b_{{\sigma }_j}\left(x\right)-{\lambda }_{{\sigma }_j}\right)\times {\displaystyle {\int }_{{\left({ℝ}^d\right)}^m}}{\displaystyle \underset{{\sigma }_k\in {\sigma }^{\prime }}{\prod }}\left(b_{{\sigma }_k}\left(y\right)-{\lambda }_{{\sigma }_k}\right)\\ \times K\left(x\mathrm{,}{y}_1,\cdots ,y_m\right)f_1\left(y_1\right)\cdots f_m\left(y_m\right)\text{d}\mu \left(y_1\right)\cdots \text{d}\mu \left(y_m\right)\mathrm{.}\end{array}$ (20)

By expanding $b_{{\sigma }_k}\left(y\right)-{\lambda }_{{\sigma }_k}=\left(b_{{\sigma }_k}\left(y\right)-b_{{\sigma }_k}\left(x\right)\right)-\left(b_{{\sigma }_k}\left(x\right)-{\lambda }_{{\sigma }_k}\right)$ and

$\begin{array}{l}{\displaystyle \underset{{\sigma }_k\in {\sigma }^{\prime }}{\prod }}\left(b_{{\sigma }_k}\left(y\right)-{\lambda }_{{\sigma }_k}\right)\\ ={\displaystyle \underset{\begin{array}{l}{\sigma }^{\left(1\right)}\cup {\sigma }^{\left(2\right)}={\sigma }^{\prime }\\ {\sigma }^{\left(1\right)}\cap {\sigma }^{\left(2\right)}=\varnothing \end{array}}{\sum }}{\displaystyle \underset{{\sigma }_u\in {\sigma }^{\left(1\right)}}{\prod }}\left(b_{{\sigma }_u}\left(y\right)-b_{{\sigma }_u}\left(x\right)\right){\displaystyle \underset{{\sigma }_v\in {\sigma }^{\left(2\right)}}{\prod }}\left(b_{{\sigma }_v}\left(x\right)-{\lambda }_{{\sigma }_v}\right)\mathrm{,}\end{array}$

hence we can obtain from the equality (20) that

$\begin{array}{l}{\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\left(x\right)\\ ={\displaystyle \underset{j=1}{\overset{m}{\prod }}}\left(b_j\left(x\right)-{\lambda }_j\right)T_m\left(\stackrel{\to }{f}\right)\left(x\right)+{\left(-1\right)}^m{T}_m\left(\left(\stackrel{\to }{b}-\stackrel{\to }{\lambda }\right)\stackrel{\to }{f}\right)\left(x\right)\\ +{\displaystyle \underset{i=1}{\overset{m-1}{\sum }}}{\displaystyle \underset{\sigma \in C_i^m}{\sum }}{C}_{m,i}{\displaystyle \underset{{\sigma }_j\in \sigma }{\prod }}\left(b_{{\sigma }_j}\left(x\right)-{\lambda }_{{\sigma }_j}\right)\times {\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }^{\prime }}}}\left(\stackrel{\to }{f}\right)\left(x\right)\end{array}$ (21)

where $C_{m\mathrm{,}i}$ are constants depending only on m and i.

Proof of Lemma 2.5. For simplicity, we denote by $\mathcal{R}\left(\stackrel{\to }{b}\mathrm{,}\stackrel{\to }{f}\right)\left(x\right)$ the quantities on the right hand side of the inequality (19). Recall the definition of the sharp maximal operator $M^{\#}$, and use the standard technique, see [15] for example, we only need to prove that

$\frac{1}{\mu \left(\frac{3}{2}Q\right)}{\displaystyle {\int }_Q}\left|{\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\left(z\right)-h_Q\right|\text{d}\mu \left(z\right)\le C\mathcal{R}\left(\stackrel{\to }{b}\mathrm{,}\stackrel{\to }{f}\right)\left(x\right)$ (22)

and

$\left|h_Q-h_R\right|\le C{\left(K_{Q\mathrm{,}R}\right)}^{2m}\mathcal{R}\left(\stackrel{\to }{b}\mathrm{,}\stackrel{\to }{f}\right)\left(x\right)$ (23)

with the absolute constant C independent of $\stackrel{\to }{b}\mathrm{,}\stackrel{\to }{f}\mathrm{,}Q$ and R, where R is any doubling cube with $Q\subset R$. In fact, we take

$h_Q={\left(-1\right)}^m{m}_Q\left(T_m\left(\left(m_{\tilde{Q}}\left(b_1\right)-b_1\right)f_1{\chi }_{{ℝ}^d\backslash \frac{4}{3}Q}\mathrm{,}\cdots \mathrm{,}\left(m_{\tilde{Q}}\left(b_m\right)-b_m\right)f_m{\chi }_{{ℝ}^d\backslash \frac{4}{3}Q}\right)\right)$ (24)

and clearly

$h_R={\left(-1\right)}^m{m}_R\left(T_m\left(\left(m_{\tilde{R}}\left(b_1\right)-b_1\right)f_1{\chi }_{{ℝ}^d\backslash \frac{4}{3}R}\mathrm{,}\cdots \mathrm{,}\left(m_{\tilde{R}}\left(b_m\right)-b_m\right)f_m{\chi }_{{ℝ}^d\backslash \frac{4}{3}R}\right)\right)$ (25)

Recall the equality (21), for any $z\in Q$, we have that

$\begin{array}{l}\left|{\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\left(z\right)-h_Q\right|\\ \le \left|{\displaystyle \underset{j=1}{\overset{m}{\prod }}}\left(b_j\left(x\right)-m_{\tilde{Q}}\left(b_j\right)\right)T_m\left(\stackrel{\to }{f}\right)\left(z\right)\right|\\ +\left|{\displaystyle \underset{i=1}{\overset{m-1}{\sum }}}{\displaystyle \underset{\sigma \in C_i^m}{\sum }}{C}_{i\mathrm{,}m}{\displaystyle \underset{{\sigma }_j\in \sigma }{\prod }}\left(b_{{\sigma }_j}\left(z\right)-m_{\tilde{Q}}\left(b_{{\sigma }_j}\right)\right){\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }^{\prime }}}}\left(\stackrel{\to }{f}\right)\left(z\right)\right|\\ +\left|{\left(-1\right)}^m{T}_m\left(\left(\stackrel{\to }{b}-m_{\tilde{Q}}\left(\stackrel{\to }{b}\right)\right)\stackrel{\to }{f}\right)\left(z\right)-h_Q\right|\\ =\mathrm{<mo>\; :\; </mo>}I\left(z\right)+II\left(z\right)+III\left(z\right)\mathrm{.}\end{array}$ (26)

In order to show the inequality (22), we will calculate the integrals for the three functions above, respectively. Firstly, for $\tau >1$, by the Hölder inequality one sees that

$\begin{array}{l}\frac{1}{\mu \left(\frac{3}{2}Q\right)}{\displaystyle {\int }_Q}\left|I\left(z\right)\right|\text{d}\mu \left(z\right)\\ \le {\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\left(\frac{1}{\mu \left(\frac{3}{2}Q\right)}{\displaystyle {\int }_Q}{\left|b_i\left(z\right)-m_{\tilde{Q}}\left(b_i\right)\right|}^{{\tau }_i}\text{d}\mu \left(z\right)\right)}^{\frac{1}{{\tau }_i}}{\left(\frac{1}{\mu \left(\frac{3}{2}Q\right)}{\displaystyle {\int }_Q}{\left|T_m\left(\stackrel{\to }{f}\right)\right|}^{\tau }\text{d}\mu \left(z\right)\right)}^{\frac{1}{\tau }}\\ \le C{\Vert b_1\Vert }_{\ast }{\Vert b_2\Vert }_{\ast }\cdots {\Vert b_m\Vert }_{\ast }{M}_{\tau \mathrm{,}\frac{3}{2}}\left(T_m\left(\stackrel{\to }{f}\right)\right)\left(x\right)\end{array}$ (27)

where we have choose ${\tau }_i>1$ such that $\frac{1}{{\tau }_1}+\frac{1}{{\tau }_2}+\cdots +\frac{1}{{\tau }_m}+\frac{1}{\tau }=1$.

Similarly, for $\tau >1$, by the Hölder inequality, we also deduce that

$\begin{array}{l}\frac{1}{\mu \left(\frac{3}{2}Q\right)}{\displaystyle {\int }_Q}\left|II\left(z\right)\right|\text{d}\mu \left(z\right)\\ \le C{\displaystyle \underset{i=1}{\overset{m-1}{\sum }}}{\displaystyle \underset{\sigma \in C_i^m}{\sum }}{\displaystyle \underset{{\sigma }_j\in \sigma }{\prod }}{\Vert b_{{\sigma }_j}\Vert }_{\ast }{M}_{\tau \mathrm{,}\frac{3}{2}}\left({\left(T_m\right)}_{b_{{\sigma }^{\prime }}}\left(\stackrel{\to }{f}\right)\right)\left(x\right)\mathrm{.}\end{array}$ (28)

To estimate the integral related to the function $III\left(z\right)$, we split $f_i$ as $f_i=f_i^0+f_i^{\infty }$, where $f_i^0=f{\chi }_{\frac{4}{3}Q}$ and $f_i^{\infty }=f_i-f_i^0$, this yields

$\begin{array}{c}{\displaystyle \underset{j=1}{\overset{m}{\prod }}{f}_j\left(y_j\right)}={\displaystyle \underset{{\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_m\in \left\{0,\infty \right\}}{\sum }}{f}_1^{{\alpha }_1}\left(y_1\right)\cdots f_m^{{\alpha }_m}\left(y_m\right)\\ ={\displaystyle \underset{j=1}{\overset{m}{\prod }}{f}_j^0\left(y_j\right)}+{\displaystyle \underset{\ast }{\sum }{f}_1^{{\alpha }_1}\left(y_1\right)\cdots f_m^{{\alpha }_m}\left(y_m\right)}+{\displaystyle \underset{j=1}{\overset{m}{\prod }}{f}_j^{\infty }\left(y_j\right)},\end{array}$ (29)

where each term in $\underset{\ast }{{\displaystyle \sum }}$ satisfies that ${\alpha }_{j_1}={\alpha }_{j_2}=\cdots ={\alpha }_{j_{\lambda }}=0$, for some $1\le \lambda <m$ and some $\left\{j_1\mathrm{,}{j}_2\mathrm{,}\cdots \mathrm{,}{j}_{\lambda }\right\}\subset \left\{\mathrm{1,2,}\cdots \mathrm{,}m\right\}$. So we can decompose the function $III\left(z\right)$ further into three parts as follows

$\begin{array}{c}III\left(z\right)\le \left|T_m\left(\left(b_1-m_{\tilde{Q}}\left(b_1\right)\right)f_1^0\mathrm{,}\cdots \mathrm{,}\left(b_m-m_{\tilde{Q}}\left(b_m\right)\right)f_m^0\right)\left(z\right)\right|\\ +{\displaystyle \underset{\ast }{\sum }}\left|T_m\left(\left(b_1-m_{\tilde{Q}}\left(b_1\right)\right)f_1^{{\alpha }_1}\mathrm{,}\cdots \mathrm{,}\left(b_m-m_{\tilde{Q}}\left(b_m\right)\right)f_1^{{\alpha }_m}\right)\left(z\right)\right|\\ +\left|{\left(-1\right)}^m{T}_m\left(\left(b_1-m_{\tilde{Q}}\left(b_1\right)\right)f_1^{\infty }\mathrm{,}\cdots \mathrm{,}\left(b_m-m_{\tilde{Q}}\left(b_m\right)\right)f_1^{\infty }\right)\left(z\right)-h_Q\right|\\ =\mathrm{<mo>\; :\; </mo>}II{I}_1\left(z\right)+II{I}_2\left(z\right)+II{I}_3\left(z\right)\mathrm{.}\end{array}$ (30)

For $s_i>1$,we can take $1<{\mu }_i<s_i$ such that $\frac{1}{v}=\frac{1}{u_1}+\frac{1}{u_2}+\cdots +\frac{1}{u_m}$ and $v>1$. Let $\frac{1}{u_i}=\frac{1}{s_i}+\frac{1}{t_i}$ for each $i=1,2,\cdots ,m$, then $1<t_i<\infty$. Using Lemma 2.4, we know that the $T_m$ is bounded from $L^{u_1}\left(\mu \right)\times L^{u_2}\left(\mu \right)\times \cdots \times L^{u_m}\left(\mu \right)$ to $L^v\left(\mu \right)$. Hence, by this boundedness and Hölder inequality, we have

$\begin{array}{l}\frac{1}{\mu \left(\frac{3}{2}Q\right)}{\displaystyle {\int }_Q}\left|II{I}_1\left(z\right)\right|\text{d}\mu \left(z\right)\\ \le \frac{\mu {\left(Q\right)}^{1-\frac{1}{v}}}{\mu \left(\frac{3}{2}Q\right)}{\Vert T_m\left(\left(b_1-m_{\tilde{Q}}\left(b_1\right)\right)f_1^0,\cdots ,\left(b_m-m_{\tilde{Q}}\left(b_m\right)\right)f_m^0\right)\left(z\right)\Vert }_{L^v\left(\mu \right)}\\ \le C\mu {\left(\frac{3}{2}Q\right)}^{-\frac{1}{v}}{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\Vert \left(b_i-m_{\tilde{Q}}\left(b_i\right)\right)f_i^0\Vert }_{L^{u_i}\left(\mu \right)}\\ \le C{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\Vert b_i\Vert }_{\ast }{M}_{s_i,\frac{9}{8}}{f}_i\left(x\right).\end{array}$ (31)

In order to estimate the integral of terms $II{I}_2\left(z\right)$ and $II{I}_3\left(z\right)$ over Q, we will give their point-wise estimates. In fact, for $z\in Q$, since $1\le \lambda \le m-1$ we observe that

$\begin{array}{c}II{I}_2\left(z\right)={\displaystyle \underset{\ast }{\sum }}\left|T_m\left(\left(b_1-m_{\tilde{Q}}\left(b_1\right)\right)f_1^{{\alpha }_1},\cdots ,\left(b_m-m_{\tilde{Q}}\left(b_m\right)\right)f_1^{{\alpha }_m}\right)\left(z\right)\right|\\ \le C{\displaystyle \underset{\ast }{\sum }}{\displaystyle \underset{j\in \left\{j_1,\cdots ,j_{\lambda }\right\}}{\prod }}{\displaystyle {\int }_{\frac{4}{3}Q}}\left|b_j\left(y_j\right)-m_{\tilde{Q}}\left(b_j\right)\right|\left|f_j\left(y_j\right)\right|\text{d}\mu \left(y_j\right)\\ \times {\displaystyle {\int }_{{\left({ℝ}^d\backslash \frac{4}{3}Q\right)}^{m-\lambda }}}\frac{{\displaystyle {\sum }_{j\notin \left\{j_1,\cdots ,j_{\lambda }\right\}}\left|b_j\left(y_j\right)-m_{\tilde{Q}}\left(b_j\right)\right|\left|f_j\left(y_j\right)\right|\text{d}\mu \left(y_j\right)}}{{\left({\displaystyle {\sum }_{j\notin \left\{j_1,\cdots ,j_{\lambda }\right\}}\left|z-y_j\right|}\right)}^{mn}}\\ \le C{\displaystyle \underset{\ast }{\sum }}{\displaystyle \underset{j\in \left\{j_1,j_2,\cdots ,j_{\lambda }\right\}}{\prod }}{\Vert b_j\Vert }_{\ast }{M}_{s_j,\frac{9}{8}}{f}_j\left(x\right)\\ \times {\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{2}^{-k\lambda n}{\displaystyle \underset{j\notin \left\{j_1,j_2,\cdots ,j_{\lambda }\right\}}{\prod }}{\Vert b_j\Vert }_{\ast }\left(k+1\right)M_{s_j,\frac{9}{8}}{f}_j\left(x\right)\\ \le C{\displaystyle \underset{j=1}{\overset{\infty }{\prod }}}{\Vert b_j\Vert }_{\ast }{M}_{s_j,\frac{9}{8}}{f}_j\left(x\right)\end{array}$ (32)

where we have used the fact (see [5] ) that, there is an absolute constant C such that, for any $b\in RBMO$, integer $k\ge 0$ and cubes Q,

$\begin{array}{c}\left|m_{\widetilde{2^k\frac{4}{3}Q}}\left(b\right)-m_{\tilde{Q}}\left(b\right)\right|\le C{\Vert b\Vert }_{\ast }{K}_{\tilde{Q}\mathrm{,}\widetilde{2^k\frac{4}{3}Q}}\\ \le C{\Vert b\Vert }_{\ast }{K}_{Q{\mathrm{,2}}^k\frac{4}{3}Q}\\ \le Ck{\Vert b\Vert }_{\ast }\mathrm{.}\end{array}$ (33)

On the other hand, for $II{I}_3\left(z\right)$, we note for any $z\mathrm{,}y\in Q$ that

$\begin{array}{l}|T_m\left(\left(b_1-m_{\tilde{Q}}\left(b_1\right)\right)f_1^{\infty },\cdots ,\left(b_m-m_{\tilde{Q}}\left(b_m\right)\right)f_m^{\infty }\right)\left(z\right)\\ -T_m\left(\left(b_1-m_{\tilde{Q}}\left(b_1\right)\right)f_1^{\infty },\cdots ,\left(b_m-m_{\tilde{Q}}\left(b_m\right)\right)f_m^{\infty }\right)\left(y\right)|\\ \le {\displaystyle {\int }_{{\left({ℝ}^d\backslash \frac{4}{3}Q\right)}^m}}\left|K\left(z,y_1,\cdots ,y_m\right)-K\left(y,y_1,\cdots ,y_m\right)\right|\\ \times \left|{\displaystyle \underset{i=1}{\overset{m}{\prod }}}\left(b_i\left(y_i\right)-m_{\tilde{Q}}\left(b_i\right)\right)f_i^{\infty }\right|\text{d}\mu \left(y_1\right)\text{d}\mu \left(y_2\right)\cdots \text{d}\mu (ym)\end{array}$

$\begin{array}{l}\le {\displaystyle {\int }_{{\left({ℝ}^d\backslash \frac{4}{3}Q\right)}^m}}\frac{{\left|z-y\right|}^{ϵ}{\displaystyle {\prod }_{i=1}^m\left|\left(b_i\left(y_i\right)-m_{\tilde{Q}}\left(b_i\right)\right)f_i^{\infty }\right|}}{{\left({\displaystyle {\sum }_{k.l=0}^m\left|y_k-y_l\right|}\right)}^{mn+ϵ}}\text{d}\mu \left(y_1\right)\cdots \text{d}\mu \left(y_m\right)\\ \le C{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{\displaystyle {\int }_{2^k\frac{4}{3}Q\backslash 2^{k-1}\frac{4}{3}Q}}\frac{l{\left(Q\right)}^{\frac{ϵ}{m}}}{l{\left(2^k\frac{3}{2}Q\right)}^{n+\frac{ϵ}{m}}}\left|\left(b_i\left(y_i\right)-m_{\tilde{Q}}\left(b_i\right)\right)f_i\right|\text{d}\mu \left(y_i\right)\\ \le C{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{2}^{-\frac{k}{m}}\left(k+1\right){\Vert b_i\Vert }_{\ast }{M}_{s_i,\frac{9}{8}}{f}_i\left(x\right)\\ \le C{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\Vert b_i\Vert }_{\ast }{M}_{s_i,\frac{9}{8}}{f}_i\left(x\right),\end{array}$ (34)

where we have use the inequation (33) again.

Taking the mean over $y\in Q$, we can obtain that

$\frac{1}{\mu \left(\frac{3}{2}Q\right)}{\displaystyle {\int }_Q}\left(\left|II{I}_2\left(z\right)\right|+\left|II{I}_3\left(z\right)\right|\right)\text{d}\mu \left(z\right)\le C{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\Vert b_i\Vert }_{\ast }{M}_{s_i\mathrm{,}\frac{9}{8}}{f}_i\left(x\right)\mathrm{.}$ (35)

Combing the inequalities (26) (27), (28), (30), (31) and (33), we see from the estimates of $I\mathrm{,}II\mathrm{,}II{I}_1\mathrm{,}II{I}_2$ and $II{I}_3$ that the desired inequality (22) holds.

Next we turn to estimate the inequality (23). For any cubes $Q\subset R$ with $x\in Q$, where R is doubling. We denote $N_{Q\mathrm{,}R}+2$ by N, then $2^{N}Q\supset 2Q$ and $2^{N}Q\supset 2R$. We recall the equality (29) and let $f_i^0=f_i{\chi }_{2^{N}Q\backslash \frac{4}{3}Q}$ and $f_i^R=f_i{\chi }_{2^{N}Q\backslash \frac{4}{3}R}$, and let $f_i^{\infty }=f_i{\chi }_{{ℝ}^d\backslash 2^{N}Q}$. Then we can write

$\begin{array}{c}\left|h_Q-h_R\right|=|m_Q\left[T_m\left(\left(m_{\tilde{Q}}\left(b_1\right)-b_1\right)f_1{\chi }_{{ℝ}^d\backslash \frac{4}{3}Q},\cdots ,\left(m_{\tilde{Q}}\left(b_m\right)-b_m\right)f_m{\chi }_{{ℝ}^d\backslash \frac{4}{3}Q}\right)\right]\\ -m_R\left[T_m\left(\left(m_{\tilde{R}}\left(b_1\right)-b_1\right)f_1{\chi }_{{ℝ}^d\backslash \frac{4}{3}R},\cdots ,\left(m_{\tilde{R}}\left(b_m\right)-b_m\right)f_m{\chi }_{{ℝ}^d\backslash \frac{4}{3}R}\right)\right]|\\ \le |m_Q\left[T_m\left(\left(m_{\tilde{Q}}\left(b_1\right)-b_1\right)f_1^{\infty },\cdots ,\left(m_{\tilde{Q}}\left(b_m\right)-b_m\right)f_m^{\infty }\right)\right]\\ -m_R\left[T_m\left(\left(m_{\tilde{Q}}\left(b_1\right)-b_1\right)f_1^{\infty },\cdots ,\left(m_{\tilde{Q}}\left(b_m\right)-b_m\right)f_m^{\infty }\right)\right]|\end{array}$

$\begin{array}{l}+|m_R\left[T_m\left(\left(m_{\tilde{Q}}\left(b_1\right)-b_1\right)f_1^{\infty },\cdots ,\left(m_{\tilde{Q}}\left(b_m\right)-b_m\right)f_m^{\infty }\right)\right]\\ -m_R\left[T_m\left(\left(m_R\left(b_1\right)-b_1\right)f_1^{\infty },\cdots ,\left(m_R\left(b_m\right)-b_m\right)f_m^{\infty }\right)\right]|\\ +{\displaystyle \underset{\begin{array}{c}{\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_m\in \left\{0,\infty \right\}\\ \text{at}\text{least}\text{one}{\alpha }_i\ne \infty \end{array}}{\sum }}\left|m_Q\left[T_m\left(\left(m_{\tilde{Q}}\left(b_1\right)-b_1\right)f_1^{{\alpha }_1},\cdots ,\left(m_{\tilde{Q}}\left(b_m\right)-b_m\right)f_m^{{\alpha }_m}\right)\right]\right|\\ +{\displaystyle \underset{\begin{array}{c}{\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_m\in \left\{R,\infty \right\}\\ \text{at}\text{least}\text{one}{\alpha }_i\ne \infty \end{array}}{\sum }}\left|m_R\left[T_m\left(\left(m_R\left(b_1\right)-b_1\right)f_1^{{\alpha }_1},\cdots ,\left(m_R\left(b_m\right)-b_m\right)f_m^{{\alpha }_m}\right)\right]\right|\\ =:A_1+A_2+A_3+A_4.\end{array}$ (36)

For the term $A_1$, noting

$\left|m_R\left(b_i\right)-m_{\tilde{Q}}\left(b_i\right)\right|\le C{\Vert b_i\Vert }_{\ast }{K}_{Q\mathrm{,}R}$ (37)

and the similar argument as that for the estimate of $II{I}_3$, we can obtain that

$A_1\le C{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{K}_{Q,R}{\Vert b_i\Vert }_{\ast }{M}_{s_i,\frac{9}{8}}{f}_i\left(x\right)\le C{\left(K_{Q,R}\right)}^m{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\Vert b_i\Vert }_{\ast }{M}_{s_i,\frac{9}{8}}{f}_i\left(x\right).$ (38)

To estimate $A_2$, we recall the notations and note that, for any sequences ${\xi }_j$ and ${\zeta }_j$,

${\displaystyle \underset{j=1}{\overset{m}{\prod }}}\left({\xi }_j+{\zeta }_j\right)={\displaystyle \underset{i=0}{\overset{m}{\sum }}}{\displaystyle \underset{\sigma \in C_i^m}{\sum }}{\displaystyle \underset{j\in \sigma }{\prod }}{\xi }_j{\displaystyle \underset{j^{\prime }\in {\sigma }^{\prime }}{\prod }}{\zeta }_{j^{\prime }}.$ (39)

Using this equality and expanding $m_{\tilde{Q}}\left(b_j\right)-b_j\left(y\right)=\left[m_{\tilde{Q}}\left(b_j\right)-b_j\left(z\right)\right]+\left[b_j\left(z\right)-b_j\left(y\right)\right]$, we observe that

$\begin{array}{l}{T}_m\left(\left(m_{\tilde{Q}}\left(b_1\right)-b_1\right)f_1^{\infty },\cdots ,\left(m_{\tilde{Q}}\left(b_m\right)-b_m\right)f_m^{\infty }\right)\left(z\right)\\ ={\displaystyle \underset{i=0}{\overset{m}{\sum }}}{\displaystyle \underset{\sigma \in C_i^m}{\sum }}{\displaystyle \underset{j\in \sigma }{\prod }}\left(m_{\tilde{Q}}\left(b_j\right)-b_j\left(z\right)\right){\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }^{\prime }}}}\left(\stackrel{\to }{f}{\chi }_{{ℝ}^d\backslash 2^{N}Q}\right)\left(z\right)\end{array}$ (40)

Similarly,

$\begin{array}{l}{T}_m\left(\left(m_R\left(b_1\right)-b_1\right)f_1^{\infty },\cdots ,\left(m_R\left(b_m\right)-b_m\right)f_m^{\infty }\right)\left(z\right)\\ ={\displaystyle \underset{i=0}{\overset{m}{\sum }}}{\displaystyle \underset{\sigma \in C_i^m}{\sum }}{\displaystyle \underset{j\in \sigma }{\prod }}\left(m_R\left(b_j\right)-b_j\left(z\right)\right){\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }^{\prime }}}}\left(\stackrel{\to }{f}{\chi }_{{ℝ}^d\backslash 2^{N}Q}\right)\left(z\right).\end{array}$ (41)

Thus

$\begin{array}{l}|T_m\left(\left(m_{\tilde{Q}}\left(b_1\right)-b_1\right)f_1^{\infty },\cdots ,\left(m_{\tilde{Q}}\left(b_m\right)-b_m\right)f_m^{\infty }\right)\left(z\right)\\ -T_m\left(\left(m_R\left(b_1\right)-b_1\right)f_1^{\infty },\cdots ,\left(m_R\left(b_m\right)-b_m\right)f_m^{\infty }\right)\left(z\right)|\\ \le \left|{\displaystyle \underset{i=1}{\overset{m}{\sum }}}{\displaystyle \underset{\sigma \in C_i^m}{\sum }}{\displaystyle \underset{j\in \sigma }{\prod }}\left(m_{\tilde{Q}}\left(b_j\right)-b_j\left(z\right)\right){\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }^{\prime }}}}\left(\stackrel{\to }{f}{\chi }_{{ℝ}^d\backslash 2^{N}Q}\right)\left(z\right)\right|\\ +\left|{\displaystyle \underset{i=1}{\overset{m}{\sum }}}{\displaystyle \underset{\sigma \in C_i^m}{\sum }}{\displaystyle \underset{j\in \sigma }{\prod }}\left(m_R\left(b_j\right)-b_j\left(z\right)\right){\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }^{\prime }}}}\left(\stackrel{\to }{f}{\chi }_{{ℝ}^d\backslash 2^{N}Q}\right)\left(z\right)\right|\\ =:B_1\left(z\right)+B_2\left(z\right).\end{array}$ (42)

To estimate the integrals above, we recall that $f_i^{\infty }=f_i{\chi }_{{ℝ}^d\backslash 2^{N}Q}$ and let $f_j^Q=f_j{\chi }_{2^{N}Q}$, then we can write that $f_j^{\infty }=f_j-f_j^Q$ and $f_j=f_j^{\infty }+f_j^Q$, and thus we have

$\begin{array}{l}{\displaystyle \underset{j=1}{\overset{m}{\prod }}{f}_j^{\infty }\left(y_j\right)}={\displaystyle \underset{j=1}{\overset{m}{\prod }}{f}_j\left(y_j\right)}+{\displaystyle \underset{i=1}{\overset{m}{\sum }}}{\left(-1\right)}^i{\displaystyle \underset{\rho \in C_i^m}{\sum }}{\displaystyle \underset{j\in \rho }{\prod }}{f}_j^Q\left(y_j\right){\displaystyle \underset{j^{\prime }\in {\rho }^{\prime }}{\prod }}{f}_{j^{\prime }}\left(y_{j^{\prime }}\right)\\ ={\displaystyle \underset{j=1}{\overset{m}{\prod }}{f}_j\left(y_j\right)}+{\displaystyle \underset{\lambda =1}{\overset{m}{\sum }}}{\displaystyle \underset{\left\{j_1,j_2,\cdots ,j_{\lambda }\right\}\subset \left\{1,2,\cdots ,m\right\}}{\sum }}\\ \times C_{j_1,j_2,\cdots ,j_m}{f}_{j_1}^Q\left(y_{j_1}\right),\cdots ,f_{j_{\lambda }}^Q\left(y_{j_{\lambda }}\right)f_{j_{\lambda +1}}^{\infty }\left(y_{j_{\lambda +1}}\right),\cdots ,f_{j_m}^{\infty }\left(y_{j_m}\right)\end{array}$ (43)

where $C_{j_1\mathrm{,}{j}_2\mathrm{,}\cdots \mathrm{,}{j}_m}$ are constant independent of $\stackrel{\to }{f}$ and Q. From the equality (43), we can deduce that

$\begin{array}{l}{\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }^{\prime }}}}\left(\stackrel{\to }{f}{\chi }_{{ℝ}^d\backslash 2^{N}Q}\right)\left(z\right)\\ ={\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }^{\prime }}}}\stackrel{\to }{f}\left(z\right)+{\displaystyle \underset{\lambda =1}{\overset{m}{\sum }}}{\displaystyle \underset{\left\{j_1,j_2,\cdots ,j_m\right\}\subset \left\{1,2,\cdots ,m\right\}}{\sum }}{C}_{j_1,j_2,\cdots ,j_m}{T}_m\left(g_1{F}_1,g_2{F}_2,\cdots ,g_m{F}_m\right)\left(z\right)\end{array}$ (44)

where

$g_j\left(y\right)=\{\begin{array}{ll}{b}_j\left(z\right)-b_j\left(y\right),\hfill & \text{if}j\in {\sigma }^{\prime }\hfill \\ 1,\hfill & \text{if}j\in \sigma \hfill \end{array}$

and

$F_j\left(y\right)=\{\begin{array}{ll}{f}_j^Q\left(y\right)\mathrm{,}\hfill & \text{if}j\in \left\{j_1\mathrm{,}\cdots \mathrm{,}{j}_{\lambda }\right\}\hfill \\ f_j^{\infty }\left(y\right)\mathrm{,}\hfill & \text{if}j\notin \left\{j_1\mathrm{,}\cdots \mathrm{,}{j}_{\lambda }\right\}\hfill \end{array}$

Along the same lines as that of the pointwise estimates of $II{I}_2\left(z\right)$,we can obtain that, for $x\mathrm{,}z\in R\subset 2^{N-1}Q$ and if $1\le \lambda \le m$,

$\begin{array}{l}\left|T_m\left(g_1{F}_1,g_2{F}_2,\cdots ,g_m{F}_m\right)\left(z\right)\right|\\ \le C{\displaystyle \underset{j\in \left\{j_1,j_2,\cdots ,j_{\lambda }\right\}}{\prod }}\frac{1}{l{\left(2^{N}Q\right)}^n}{\displaystyle {\int }_{2^{N}Q}}\left|g_j\left(y_j\right)f_j\left(y_j\right)\right|\text{d}\mu \left(y_j\right)\\ \times {\displaystyle \underset{k=1}{\overset{\infty }{\sum }}}{2}^{-k\lambda n}{\displaystyle \underset{\begin{array}{c}j\notin \left\{j_1,j_2,\cdots ,j_{\lambda }\right\}\\ 1\le j\le m\end{array}}{\prod }}\frac{1}{l{\left(2^k{2}^{N}Q\right)}^n}{\displaystyle {\int }_{2^k{2}^{N}Q}}\left|g_j\left(y_j\right)f_j\left(y_j\right)\right|\text{d}\mu \left(y_j\right).\end{array}$ (45)

Let ${\left(g_j\right)}_{k,{s^{\prime }}_j}^{*}=1$ if $j\in \sigma$ ; and

${\left(g_j\right)}_{k,{s^{\prime }}_j}^{*}=\{\begin{array}{ll}{\left(\frac{1}{l{\left(2^{N}Q\right)}^n}{\displaystyle {\int }_{2^{N}Q}}{\left|b_j\left(z\right)-b_j\left(y\right)\right|}^{{s^{\prime }}_j}\text{d}\mu \left(y\right)\right)}^{\frac{1}{{s^{\prime }}_j}},\hfill & \text{if}j\in {\sigma }^{\prime }\cap \left\{j_1,j_2,\cdots ,j_{\lambda }\right\}\hfill \\ {\left(\frac{1}{l{\left(2^{k+N}Q\right)}^n}{\displaystyle {\int }_{2^{k+N}Q}}{\left|b_j\left(z\right)-b_j\left(y\right)\right|}^{{s^{\prime }}_j}\text{d}\mu \left(y\right)\right)}^{\frac{1}{{s^{\prime }}_j}},\hfill & \text{if}j\in {\sigma }^{\prime }\backslash \left\{j_1,j_2,\cdots ,j_{\lambda }\right\}\hfill \end{array}$

thus for $j\in {\sigma }^{\prime }$,

${\left(g_j\right)}_{k\mathrm{,}{s^{\prime }}_j}^{\mathrm{<mo>\; *\; </mo>}}\le C{\Vert b_j\Vert }_{\ast }+C\left|m_{2^{N}Q}\left(b_j\right)-b_j\left(z\right)\right|+C\left|m_{2^{k+N}}Q\left(b_j\right)-b_j\left(z\right)\right|\mathrm{.}$

Hence we get from (45) that, for $\tau >1$,

$\begin{array}{l}{\left(\frac{1}{\mu \left(R\right)}{\displaystyle {\int }_R}{\left|T_m\left(g_1{F}_1,g_2{F}_2,\cdots ,g_m{F}_m\right)\left(z\right)\right|}^{\tau }\text{d}\mu \left(z\right)\right)}^{\frac{1}{\tau }}\\ \le C\left({\displaystyle \underset{j=1}{\overset{m}{\prod }}}{M}_{s_j,\frac{9}{8}}{f}_j\left(x\right)\right){\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}{2}^{-k\lambda n}\times {\left(\frac{1}{\mu \left(R\right)}{\displaystyle {\int }_R}{\left|{\displaystyle \underset{j\in {\sigma }^{\prime }}{\prod }}{\left(g_j\right)}_{k,{s^{\prime }}_j}^{*}\right|}^{\tau }\text{d}\mu \left(z\right)\right)}^{\frac{1}{\tau }}\\ \le C\left({\displaystyle \underset{j=1}{\overset{m}{\prod }}}{M}_{s_j,\frac{9}{8}}{f}_j\left(x\right)\right){\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}{2}^{-k\lambda n}{\displaystyle \underset{j\in {\sigma }^{\prime }}{\prod }}{\Vert b_j\Vert }_{\ast }\left(1+K_{R,2^{N}Q}+K_{R,2^{k+N}Q}\right)\\ \le C\left({\displaystyle \underset{j=1}{\overset{m}{\prod }}}{M}_{s_j,\frac{9}{8}}{f}_j\left(x\right)\right){\displaystyle \underset{j\in {\sigma }^{\prime }}{\prod }}{\Vert b_j\Vert }_{\ast },\end{array}$ (46)

where we have used the fact that the cubes R and $2^{N}Q$ are comparable, which implies $K_{R{\mathrm{,2}}^{N}Q}\le C$ and $K_{R{\mathrm{,2}}^{k+N}Q}\le C\left(1+k\right)$. Using the inequality (46) above and the identity (44), we obtain that, for $\tau >1$,

$\begin{array}{c}{A}_2\le \frac{1}{\mu \left(R\right)}{\displaystyle {\int }_R}\left(\left|B_1\left(z\right)\right|+\left|B_2\left(z\right)\right|\right)\text{d}\mu \left(z\right)\\ \le C{\displaystyle \underset{i=1}{\overset{m}{\sum }}}{\displaystyle \underset{\sigma \in C_i^m}{\sum }}\left({\displaystyle \underset{j\in \sigma }{\prod }}{K}_{Q,R}{\Vert b_j\Vert }_{\ast }\right)\\ \times {\left(\frac{1}{\mu \left(R\right)}{\displaystyle {\int }_R}{\left|{\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }^{\prime }}}}\left(\stackrel{\to }{f}{\chi }_{{ℝ}^d\backslash 2^{N}Q}\right)\right|}^{\tau }\text{d}\mu \left(z\right)\right)}^{\frac{1}{\tau }}\\ \le C{\displaystyle \underset{i=1}{\overset{m}{\sum }}}{\displaystyle \underset{\sigma \in C_i^m}{\sum }}\left({\displaystyle \underset{j\in \sigma }{\prod }}{K}_{Q,R}{\Vert b_j\Vert }_{\ast }\right)M_{\tau ,\frac{3}{2}}\left({\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{{\sigma }^{\prime }}}}\left(\stackrel{\to }{f}\right)\right)\left(x\right)\\ +C{\displaystyle \underset{j=1}{\overset{m}{\prod }}}{K}_{Q,R}{\Vert b_j\Vert }_{\ast }{M}_{s_j,\frac{9}{8}}{f}_j\left(x\right).\end{array}$

The estimates of $A_3$ and $A_4$ is very similar to the one used in the estimate of $A_2$. In fact, repeating the similar procedures used in (45) and (46) for $\tau >1$, and noting that $K_{Q{\mathrm{,2}}^{k+N}Q}\le K_{Q\mathrm{,}R}+K_{R{\mathrm{,2}}^{k+N}Q}\le C\left(1+k\right)+K_{Q\mathrm{,}R}$ since $2^{N-3}Q\subset 2R\subset 2^{N}Q$ by the definition of N, we can deduce that

$\begin{array}{l}{\displaystyle \underset{\begin{array}{c}{\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_m\in \left\{0,\infty \right\}\\ \text{at}\text{least}\text{one}{\alpha }_i\ne \infty \\ \text{and}\text{one}{\alpha }_j\ne 0\end{array}}{\sum }}\left|m_Q\left[T_m\left(\left(m_{\tilde{Q}}\left(b_1\right)-b_1\right)f_1^{{\alpha }_1},\cdots ,\left(m_{\tilde{Q}}\left(b_m\right)-b_m\right)f_m^{{\alpha }_m}\right)\right]\right|\\ +{\displaystyle \underset{\begin{array}{c}{\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_m\in \left\{R,\infty \right\}\\ \text{at}\text{least}\text{one}{\alpha }_i\ne \infty \end{array}}{\sum }}\left|m_R\left[T_m\left(\left(m_R\left(b_1\right)-b_1\right)f_1^{{\alpha }_1},\cdots ,\left(m_R\left(b_m\right)-b_m\right)f_m^{{\alpha }_m}\right)\right]\right|\\ \le C{\displaystyle \underset{j=1}{\overset{m}{\prod }}}{K}_{Q,R}{\Vert b_j\Vert }_{\ast }{M}_{s_j,\frac{9}{8}}{f}_j\left(x\right).\end{array}$ (47)

It is left to estimate the term in $A_3$ of the case ${\alpha }_1={\alpha }_2=\cdots ={\alpha }_m=0$. A small modification is needed to estimate this term. For $z\in Q$ and $x\in Q$, one sees

$\begin{array}{l}\left|T_m\left(\left(m_{\tilde{Q}}\left(b_1\right)-b_1\right)f_1^0,\cdots ,\left(m_{\tilde{Q}}\left(b_j\right)-b_j\right)f_j^0,\cdots ,\left(m_{\tilde{Q}}\left(b_m\right)-b_m\right)f_m^0\right)\left(z\right)\right|\\ \le C{\displaystyle \underset{k=1}{\overset{N}{\sum }}}{\displaystyle \underset{j=1}{\overset{m}{\prod }}}\frac{1}{l{\left(2^k\frac{4}{3}Q\right)}^n}{\displaystyle {\int }_{2^k\frac{4}{3}Q}}\left|\left(m_{\tilde{Q}}\left(b_j\right)-b_j\left(y_j\right)\right)f_j\left(y_j\right)\right|\text{d}\mu \left(y_j\right)\\ \le C{\displaystyle \underset{k=1}{\overset{N}{\sum }}}{\displaystyle \underset{j=1}{\overset{m}{\prod }}}\frac{\mu \left(2^k\frac{3}{2}Q\right)}{l{\left(2^k\frac{4}{3}Q\right)}^n}\left(1+K_{Q,2^k\frac{4}{3}Q}\right){\Vert b_j\Vert }_{\ast }{M}_{s_j,\frac{9}{8}}{f}_j\left(x\right)\\ \le C{\left(K_{Q,R}\right)}^{2m}{\displaystyle \underset{j=1}{\overset{m}{\prod }}}{\Vert b_j\Vert }_{\ast }{M}_{s_j,\frac{9}{8}}{f}_j\left(x\right).\end{array}$

This and the inequality (47) follows

$A_3+A_4\le C{\left(K_{Q,R}\right)}^{2m}{\displaystyle \underset{j=1}{\overset{m}{\prod }}}{\Vert b_j\Vert }_{\ast }{M}_{s_j,\frac{9}{8}}{f}_j\left(x\right).$

Moreover, combing the estimates of $A_1\mathrm{,}{A}_2\mathrm{,}{A}_3$ and $A_4$, we obtain the desired inequality (23).

Finally, let us show how to acquire the inequality (19) from the two inequalities (22) and (23). Fix the point x and let Q be any cube that $x\in Q$. notice $K_{Q\mathrm{,}\tilde{Q}}\le C$, hence we see from the inequalities (22) and (23) that

$\begin{array}{l}\frac{1}{\mu \left(\frac{3}{2}Q\right)}{\displaystyle {\int }_Q}\left|{\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\left(z\right)-m_{\tilde{Q}}\left({\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\right)\right|\text{d}\mu \left(z\right)\\ \le \frac{1}{\mu \left(\frac{3}{2}Q\right)}{\displaystyle {\int }_Q}\left|{\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\left(z\right)-h_Q\right|\text{d}\mu \left(z\right)+\frac{1}{\mu \left(\frac{3}{2}Q\right)}{\displaystyle {\int }_Q}\left|h_Q-h_{\tilde{Q}}\right|\text{d}\mu \left(z\right)\\ +\frac{1}{\mu \left(\frac{3}{2}Q\right)}{\displaystyle {\int }_Q}\left|m_{\tilde{Q}}\left({\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\right)-h_{\tilde{Q}}\right|\text{d}\mu \left(z\right)\\ \le C\mathcal{R}\left(\stackrel{\to }{b},\stackrel{\to }{f}\right)\left(x\right).\end{array}$ (48)

On the other hand for all doubling cubes $Q\subset R$ with $x\in Q$ such that $K_{Q\mathrm{,}R}\le P_0$, where $P_0$ is the constant in Lemma 6 in [15], using (23), we have

$\left|h_Q-h_R\right|\le C{K}_{Q\mathrm{,}R}{P}_0{}^{2m-1}\mathcal{R}\left(\stackrel{\to }{b}\mathrm{,}\stackrel{\to }{f}\right)\left(x\right)$ (49)

and moreover the inequality (49) holds for any doubling cubes Q, R with $Q\subset R$. Therefore,

$\begin{array}{l}{m}_Q\left({\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\right)-m_R\left({\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\right)\mathrm{<mo>\; |\; </mo>}\\ \le \left|m_Q\left({\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\right)-h_Q\right|+\left|h_R-m_R\left({\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\right)\right|+\left|h_Q-h_R\right|\\ \le C{K}_{Q\mathrm{,}R}\mathcal{R}\left(\stackrel{\to }{b}\mathrm{,}\stackrel{\to }{f}\right)\left(x\right)\mathrm{.}\end{array}$ (50)

According to the estimates (48) (50) and the definition of the sharp maximal function, we deduce the inequality (19) and so finish the proof of the Lemma 2.5.☐

3. Conclusions

The proof of Lemma 2.5 can be slightly modified to prove the conclusion of Lemma 2.1. Therefore we show that the iterated commutators generated by multilinear singular integrals operators ${\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}$ are bounded from

$M_{q_1}^{p_1}\left(\mu \right)\times \cdots \times M_{q_m}^{p_m}\left(\mu \right)$ to $M_q^p\left(\mu \right)$. Suppose $\Vert \mu \Vert =\infty$, the detailed conclusion can be described as follows: Let ${\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}$ as in (9) and satisfying conditions (7)

and (8). Let $1<q_j\le p_j<\infty$ with $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$ and $\frac{1}{q}=\frac{1}{q_1}+\cdots +\frac{1}{q_m}$. Suppose $b_i\in RBMO\left(\mu \right)$ for $i=1,2,\cdots ,m$. If $T_m$ maps $L^1\left(\mu \right)\times \cdots \times L^1\left(\mu \right)$ to $L^{1/m\mathrm{,}\infty }\left(\mu \right)$, then the commutators ${\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}$ are bounded from $M_{q_1}^{p_1}\left(\mu \right)\times \cdots \times M_{q_m}^{p_m}\left(\mu \right)$ to $M_q^p\left(\mu \right)$, that is,

${\Vert {\left(T_m\right)}_{{\displaystyle \prod \stackrel{\to }{b}}}\left(\stackrel{\to }{f}\right)\left(x\right)\Vert }_{M_q^p\left(\mu \right)}\le C{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\Vert b_i\Vert }_{\ast }{\Vert f_i\Vert }_{M_{q_i}^{p_i}\left(\mu \right)}\mathrm{.}$

More generally, let ${\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}$ as in (11) and satisfying conditions (7) and (8). Let $1<q_j\le p_j<\infty$ with $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$ and $\frac{1}{q}=\frac{1}{q_1}+\cdots +\frac{1}{q_m}$. Suppose $b_i\in RBMO\left(\mu \right)$ for $i=1,2,\cdots ,m$. If $T_m$ maps $L^1\left(\mu \right)\times \cdots \times L^1\left(\mu \right)$ to $L^{1/m\mathrm{,}\infty }\left(\mu \right)$, then for all $\sigma \subset \left\{\mathrm{1,2,}\cdots \mathrm{,}m\right\}$, the commutators ${\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}$ are bounded from $M_{q_1}^{p_1}\left(\mu \right)\times \cdots \times M_{q_m}^{p_m}\left(\mu \right)$ to $M_q^p\left(\mu \right)$, that is,

${\Vert {\left(T_m\right)}_{{\displaystyle \prod {\stackrel{\to }{b}}_{\sigma }}}\left(\stackrel{\to }{f}\right)\left(x\right)\Vert }_{M_q^p\left(\mu \right)}\le C{\displaystyle \underset{j\in \sigma }{\prod }}{\Vert b_j\Vert }_{\ast }{\displaystyle \underset{i=1}{\overset{m}{\prod }}}{\Vert f_i\Vert }_{M_{q_i}^{p_i}\left(\mu \right)}.$

Funding

Supported by the National Natural Science Foundation of China (Grant No. 11161044 and Grant No. 11661061) and Natural science foundation of Inner Mongolia (No. 2019MS01003).

Conflicts of Interest

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

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