Multilinear fractional integral operators on non-homogeneous metric measure spaces

Abstract

In this paper, the boundedness in Lebesgue spaces for multilinear fractional integral operators and commutators generated by multilinear fractional integrals with an $\operatorname{RBMO}(\mu)$ function on non-homogeneous metric measure spaces is obtained.

1 Introduction and preliminaries

A measure μ is called a doubling measure, if there exists a positive constant C such that $\mu(B(x,2l))\leq C\mu(B(x,l))$, for all $x\in\operatorname{supp} \mu$ and all $l>0$, which is the main condition in homogeneous spaces. Also μ is a non-doubling measure, if there exists an integer $k\in(0, n]$ and a positive constant $C_{0}$, such that

$$ \mu\bigl(B(x, l)\bigr) \leq C_{0}l^{k}. $$

(1.1)

This innovation caused the tremendous development of harmonic analysis (see [1–8]). It is worthy to mention that this theory solves the Painlevé’s problem and Vitushkin’s conjectures (see [7, 9]). Hytönen [10] introduced the non-homogeneous metric measure spaces $(X,d,\mu)$, which contains the homogeneous spaces and non-doubling measure spaces. Many researchers obtained the boundedness of operators on the non-homogeneous metric measure spaces; see, e.g., [10–25].

For multilinear integral operators, the bilinear theory for Calderón-Zygmund operators was studied by Coifman-Meyers [26], then, the boundedness on Lebesgue spaces or Hardy spaces for multilinear singular integrals was proved by Gorafakos-Torres [27, 28]. In non-doubling measure spaces, Xu [29, 30] and Lian-Wu [31] obtained the boundedness of multilinear singular integrals or multilinear fractional integrals and commutators respectively. In non-homogeneous metric measure spaces, Hu et al. [32] established the weighted norm inequalities for multilinear Calderón-Zygmund operators. The authors of [23] proved the boundedness on Lebesgue spaces for commutators of multilinear singular integrals.

In this paper, we introduce multilinear fractional integrals and its commutators on non-homogeneous metric spaces, then we study the boundedness in Lebesgue spaces for these operators, provided that fractional integral is bounded from $L^{r}(\mu)$ to $L^{s}(\mu)$, for some $r\in(1, 1/\beta)$ and $1/s=1/r-\beta$ with $0<\beta<1$. Our results include both the results for the homogeneous spaces and the non-doubling measure spaces.

Throughout this paper, $L_{c}^{\infty}(\mu)$ denotes $L^{\infty}(\mu)$ with compact support. C always denotes a positive constant independent of the main parameters involved, but it may be different in different currents. And $p'$ is the conjugate index of p, namely, $1/p+1/p'=1$. Next let us give some definitions and notations.

Definition 1.1

[10]

A metric space $(X,d)$ is geometrically doubling, if there is a positive integer $N_{0}$ such that, for all ball $B(x,r)\subset X$, one can find a finite ball covering $\{B(x_{j},r/2)\}_{j=1}^{N_{0}}$.

Definition 1.2

[10]

For a metric measure space $(X,d,\mu)$, if μ is a Borel measure on X, and there is a function $\lambda: X\times(0,+\infty) \rightarrow(0,+\infty)$ and a positive constant $C_{\lambda}$, such that for all $x\in X$, the function $l \longmapsto\lambda(x,l)$ is non-decreasing, and for all $x\in X$, $l >0$, the following holds:

$$ \mu\bigl(B(x, l)\bigr)\leq\lambda(x, l)\leq C_{\lambda}\lambda(x, l/2), $$

(1.2)

then $(X,d,\mu)$ is called upper doubling.

Remark 1.3

• (i) If $\lambda(x, l)$ equals to $\mu(B(x,l))$, then the homogeneous spaces is upper doubling spaces. Also, if $\lambda(x, l)$ equals $Cl^{k}$, then a metric space $(X,d,\mu)$ satisfying (1.1) is upper doubling.

• (ii) By [18], we know that there exists another function $\tilde{\lambda}\leq\lambda$, $\forall x, y \in X$ with $d(x, y)\leq l$, and the following holds:

  $$ \tilde{\lambda}(x, l)\leq \widetilde {C}\tilde{\lambda}(y, l). $$

  (1.3)

  Thus one always assumes that λ satisfies (1.3) throughout this paper. Because the singularity of the commutators is stronger than that of the fractional integral, we need to assume $\lambda(x, al) \geq a^{m}\lambda(x, l)$, for all $x \in X$ and $a,l > 0$, in the proof of boundedness of commutators.

• (iii) The upper doubling condition is equivalent to the weak growth condition introduced by Tan-Li in [33].

A measure μ is $(\alpha, \beta)$-doubling, if $\mu(\alpha B)\leq\beta\mu(B)$, for $\alpha ,\beta\in(1,+\infty)$ and all ball $B\subset X$. Bui-Duong [11] pointed out that there exist many doubling balls. One always means that $(\alpha,\beta)$-doubling ball is a $(6,\beta_{0})$-doubling ball throughout the paper, for some fixed number $\beta_{0} >\max\{C_{\lambda}^{3\log_{2}6}, 6^{n}\}$, where $n=\log_{2}N_{0}$ is viewed as a geometric dimension of the space, except α and β are designated.

Definition 1.4

[15]

For $0\leq\gamma<1$, B and R be two arbitrary balls with $B\subset R$ and $N_{B,R}$ be the smallest integer satisfying $6^{N_{B,R}}l_{B}\geq l_{R}$. One defines

$$ K^{(\gamma)}_{B,R} = 1+\sum_{j=1}^{N_{B,R}} \biggl[\frac{\mu(6^{j}B)}{\lambda(x_{B},6^{j}l_{B})} \biggr]^{(1-\gamma)}. $$

(1.4)

For $\gamma=0$, one simply writes $K^{(0)}_{B,R}=K_{B,R}$.

Definition 1.5

Let $\alpha\in(0,m)$. We call K is an m-linear fractional integral kernel, if

$$ K(\cdot,\ldots,\cdot)\in L_{\mathrm{loc}}^{1} \bigl((X)^{m+1} \backslash\bigl\{ (x,y_{1}\cdots,y_{i},\ldots,y_{m}):x=y_{i}, 1\leq i\leq m\bigr\} \bigr), $$

and the following two items hold:

(i):

$$ \bigl\vert K(x,y_{1},\ldots,y_{i},\ldots, y_{m})\bigr\vert \leq \frac{C}{ [\sum_{i=1}^{m}\lambda(x,d(x,y_{i})) ]^{m-\alpha}}, $$

(1.5)

$\forall(x,y_{1},\ldots,y_{i},\ldots,y_{m})\in(X)^{m+1}$, with $x\neq y_{i}$ for some i;

(ii):

there is a constant $0<\delta\leq1$,

$$\begin{aligned}& \bigl\vert K(x,y_{1},\ldots,y_{i}, \ldots,y_{m})-K\bigl(x',y_{1}, \ldots,y_{i},\ldots ,y_{m}\bigr)\bigr\vert \\& \quad \leq \frac{Cd(x,x')^{\delta}}{ [\sum_{i=1}^{m}d(x,y_{i}) ]^{\delta} [\sum_{i=1}^{m}\lambda(x,d(x,y_{i})) ]^{m-\alpha}}, \end{aligned}$$

(1.6)

if $Cd(x,x')\leq\max_{1\leq i\leq m}d(x,y_{i})$, and for every i,

$$\begin{aligned}& \bigl\vert K(x,y_{1},\ldots,y_{i}, \ldots,y_{m})-K\bigl(x,y_{1},\ldots,y'_{i}, \ldots ,y_{m}\bigr)\bigr\vert \\& \quad \leq \frac{Cd(y_{i},y'_{i})^{\delta}}{ [\sum_{i=1}^{m}d(x,y_{i}) ]^{\delta} [\sum_{i=1}^{m}\lambda(x,d(x,y_{i})) ]^{m-\alpha}}, \end{aligned}$$

(1.7)

if $Cd(y_{i},y'_{i})\leq\max_{1\leq i\leq m}d(x,y_{i})$.

For any m compactly supported bounded functions $f_{1},\ldots,f_{m}$, and any point $x\notin \bigcap_{i=1}^{m}\operatorname{supp} f_{i}$, the multilinear fractional integral operators $I_{\alpha,m}$ is defined by

$$\begin{aligned}& I_{\alpha,m}(f_{1},\ldots,f_{m}) (x) \\& \quad = \int_{X^{m}}K(x,y_{1},\ldots,y_{m})f_{1}(y_{1}) \cdots f_{m}(y_{m})\, d\mu(y_{1})\cdots \, d \mu(y_{m}). \end{aligned}$$

(1.8)

Remark 1.6

As $\max_{1\leq i\leq m}d(x,y_{i})\leq\sum_{i=1}^{m}d(x,y_{i})\leq m\max_{1\leq i\leq m}d(x,y_{i})$, (ii) in Definition 1.5 is equivalent to the following:

(ii′):

There is a constant $0<\delta\leq1$,

$$\begin{aligned}& \bigl\vert K(x,y_{1},\ldots,y_{i},\ldots,y_{m})-K \bigl(x',y_{1},\ldots,y_{i},\ldots ,y_{m}\bigr)\bigr\vert \\& \quad \leq \frac{Cd(x,x')^{\delta}}{ [\max_{1\leq i\leq m}d(x,y_{i}) ]^{\delta} [\sum_{i=1}^{m}\lambda (x,d(x,y_{i})) ]^{m-\alpha}}, \end{aligned}$$

if $Cd(x,x')\leq\max_{1\leq i\leq m}d(x,y_{i})$, and for every i,

$$\begin{aligned}& \bigl\vert K(x,y_{1},\ldots,y_{i},\ldots,y_{m})-K \bigl(x,y_{1},\ldots,y'_{i},\ldots ,y_{m}\bigr)\bigr\vert \\& \quad \leq \frac{Cd(y_{j},y'_{i})^{\delta}}{ [\max_{1\leq i\leq m}d(x,y_{i}) ]^{\delta} [\sum_{i=1}^{m}\lambda(x,d(x,y_{i})) ]^{m-\alpha}}, \end{aligned}$$

if $Cd(y_{i},y'_{i})\leq\max_{1\leq i\leq m}d(x,y_{i})$.

Definition 1.7

[11]

Given $\rho>1$, $b\in L_{\mathrm{loc}}^{1}(\mu)$ is an $\operatorname{RBMO}(\mu)$ function, if there is a positive constant C, for all B, we have

$$ \frac{1}{\mu(\rho B)} \int_{B}\bigl\vert b(x)-m_{\widetilde {B}}b\bigr\vert \, d \mu(x)\leq C, $$

(1.9)

and for all two doubling balls $B, R$ with $B\subset R$,

$$ \bigl\vert m_{B}(b)-m_{R}(b)\bigr\vert \leq CK_{B,R}, $$

(1.10)

where B̃ is the smallest $(\alpha,\beta)$-doubling ball with the form $6^{k}B$, $k\in{\mathbf{N}}\cup\{0\}$, and

$$m_{\widetilde{B}}(b)=\frac{1}{\mu(\widetilde{B})} \int_{\widetilde {B}}b(x)\,d\mu(x). $$

The $\operatorname{RBMO}(\mu)$ norm of b, denoted by $\|b\|_{\ast}$, is the minimal constant C in (1.9) and (1.10).

For $1\leq j \leq m$, let $C_{j}^{m}$ be the family of subsets $\sigma =\{\sigma(1),\sigma(2),\ldots,\sigma(j)\}$ of $\{1,2,\ldots,m\}$ with j different elements. For each $\sigma\in C_{j}^{m}$, $\sigma'=\{1,2,\ldots,m\}\backslash \sigma$. For $b_{j}\in \operatorname{RBMO}(\mu)$, $j=1,\ldots,m$, set $\vec{b}=(b_{1},b_{2},\ldots,b_{m})$, $\vec{b}_{\sigma}=(b_{\sigma(1)},\ldots ,b_{\sigma(j)})$, $b_{\sigma}(x)=b_{\sigma(1)}(x)\cdot\cdot\cdot b_{\sigma(j)}(x)$. Denote $\vec{f}=(f_{1},\ldots,f_{m})$, $\vec{f}_{\sigma}=(f_{\sigma(1)},\ldots,f_{\sigma(j)})$, and $\vec{b}_{\sigma'}\vec{f}_{\sigma'}=(b_{\sigma'(j+1)}f_{\sigma '(j+1)},\ldots, b_{\sigma'(m)}f_{\sigma'(m)})$.

Definition 1.8

For $b_{j}\in \operatorname{RBMO}(\mu)$, $j=1,\ldots, m$, and multilinear fractional integral operators $I_{\alpha,m}$, we define the commutators $[\vec{b},I_{\alpha,m}]$ by

$$ [\vec{b},I_{\alpha,m}](\vec{f}) (x)=\sum_{j=0}^{m} \sum_{\sigma\in C_{j}^{m}}(-1)^{m-j}b_{\sigma}(x)I_{\alpha,m}( \vec{f}_{\sigma},\vec {b}_{\sigma'}\vec{f}_{\sigma'}) (x). $$

For $m=2$,

$$\begin{aligned}{} [b_{1},b_{2},I_{\alpha,2}](f_{1},f_{2}) (x) =&b_{1}(x)b_{2}(x)I_{\alpha ,2}(f_{1},f_{2}) (x)-b_{1}(x)I_{\alpha,2}(f_{1},b_{2}f_{2}) (x) \\ &{}-b_{2}(x)I_{\alpha,2}(b_{1}f_{1},f_{2}) (x)+I_{\alpha ,2}(b_{1}f_{1},b_{2}f_{2}) (x). \end{aligned}$$

(1.11)

$[b_{1},I_{\alpha,2}]$ and $[b_{2},I_{\alpha,2}]$ are defined thus:

$$\begin{aligned}& [b_{1},I_{\alpha,2}](f_{1},f_{2}) (x)=b_{1}(x)I_{\alpha ,2}(f_{1},f_{2}) (x)-I_{\alpha,2}(b_{1}f_{1},f_{2}) (x), \\& [b_{2},I_{\alpha,2}](f_{1},f_{2}) (x)=b_{2}(x)I_{\alpha ,2}(f_{1},f_{2}) (x)-I_{\alpha,2}(f_{1},b_{2}f_{2}) (x). \end{aligned}$$

In this paper, one only considers the case of $m=2$ for simplicity.

Theorem 1.9

Let $0<\alpha<2$, $1< p_{1}$, $p_{2}<+\infty$, $0<\frac{1}{q}=\frac {1}{p_{1}}+\frac{1}{p_{2}}-\alpha<1$, $g_{1}\in L^{p_{1}}(\mu)$ and $g_{2}\in L^{p_{2}}(\mu)$. If $I_{\beta}$ is bounded from $L^{r}(\mu)$ into $L^{s}(\mu)$, for some $r\in(1, 1/\beta)$ and $1/s=1/r-\beta$, with $0<\beta<1$, then there is a positive constant C,

$$ \bigl\Vert I_{\alpha,2}(g_{1},g_{2})\bigr\Vert _{L^{q}(\mu)}\leq C \|g_{1}\|_{L^{p_{1}}(\mu)}\|g_{2} \|_{L^{p_{2}}(\mu)}, $$

where $I_{\beta}$ is defined by

$$ I_{\beta} f(x):= \int_{X} \frac{f(y)}{[\lambda(y,d(x,y))]^{1-\beta}} \,d\mu(y). $$

Theorem 1.10

Set $\|\mu\|=\infty$, $0<\alpha<2$, $1< p_{1}$, $p_{2}<+\infty$, $0<\frac{1}{q}=\frac {1}{p_{1}}+\frac{1}{p_{2}}-\alpha<1$, $g_{1}\in L^{p_{1}}(\mu)$, $g_{2}\in L^{p_{2}}(\mu)$, $b_{1},b_{2}\in \operatorname{RBMO}(\mu)$ and if $I_{\beta}$ is bounded from $L^{r}(\mu)$ into $L^{s}(\mu )$ for some $r\in(1, 1/\beta)$, $1/s=1/r-\beta$ with $0<\beta<1$, then there is a positive constant C,

$$ \bigl\Vert [b_{1},b_{2},I_{\alpha,2}](g_{1},g_{2}) \bigr\Vert _{L^{q}(\mu)}\leq C \|g_{1}\|_{L^{p_{1}}(\mu)} \|g_{2}\|_{L^{p_{2}}(\mu)}. $$

Remark 1.11

For the case that $\|\mu\|<\infty$, by Lemma 2.1 in Section 2 below, Theorem 1.10 also holds, if we assume that

$$\begin{aligned}& \int_{X}I_{\alpha,2}(g_{1},g_{2}) (x)\,d\mu(x)=0,\qquad \int_{X}[b_{1},I_{\alpha ,2}](g_{1},g_{2}) (x)\,d\mu(x)=0, \\& \int_{X}[b_{2},I_{\alpha,2}](g_{1},g_{2}) (x)\,d\mu(x)=0\quad \text{and}\quad \int _{X}[b_{1},b_{2},I_{\alpha,2}](g_{1},g_{2}) (x)\,d\mu(x)=0. \end{aligned}$$

This paper is organized as follows. Theorem 1.9 and Theorem 1.10 are proved in Section 2. In Section 3, some applications are stated.

2 Proof of main results

Proof of Theorem 1.9

Let $\alpha=\alpha _{1}+\alpha_{2}$, $0<\alpha_{i}<1/p_{i}<1$, $i=1,2$. It is easy to check that

$$ \prod_{i=1}^{2}\bigl[\lambda \bigl(x,d(x,y_{i})\bigr)\bigr]^{1-\alpha_{i}}\leq \Biggl[\sum _{i=1}^{2}\lambda\bigl(x,d(x,y_{i})\bigr) \Biggr]^{2-\alpha}. $$

Thus

$$\begin{aligned} \bigl\vert I_{\alpha,2}(g_{1},g_{2}) (x)\bigr\vert \leq& C \int_{X^{2}}\frac {\vert g_{1}(y_{1})g_{2}(y_{2})\vert }{ [\sum_{i=1}^{2}\lambda(x,d(x,y_{i})) ]^{2-\alpha}}\,d\mu(y_{1})\,d \mu(y_{2}) \\ \leq& \prod_{i=1}^{2} \int_{X}\frac{\vert g_{i}(y_{i})\vert }{[\lambda (x,d(x,y_{i}))]^{1-\alpha_{i}}}\,d\mu(y_{i}) \\ =& \prod_{i=1}^{2}I_{\alpha_{i}} \bigl(\vert g_{i}\vert \bigr) (x). \end{aligned}$$

Let $1/q_{i}=1/p_{i}-\alpha_{i}$ and $1/q_{1}+1/q_{2}=1/q$, $1< q_{i}<\infty$. It follows from the Hölder’s inequality and the $L^{p_{i}}(\mu )-L^{q_{i}}(\mu)$ boundedness of $I_{\alpha_{i}}$, $i=1,2$, that

$$\begin{aligned}& \bigl\Vert I_{\alpha,2}(g_{1},g_{2})\bigr\Vert _{L^{q}(\mu)} \\& \quad \leq \Biggl\Vert \prod_{i=1}^{2}I_{\alpha_{i}} \bigl(\vert g_{i}\vert \bigr) \Biggr\Vert _{L^{q}(\mu)} \\& \quad \leq \bigl\Vert I_{\alpha_{1}}\bigl(\vert g_{1}\vert \bigr)\bigr\Vert _{L^{q_{1}}(\mu)}\bigl\Vert I_{\alpha _{2}}\bigl(\vert g_{2}\vert \bigr) (x)\bigr\Vert _{L^{q_{2}}(\mu)} \\& \quad \leq \Vert g_{1}\Vert _{L^{p_{1}}(\mu)}\Vert g_{2}\Vert _{L^{p_{2}}(\mu)}. \end{aligned}$$

Thus the proof of Theorem 1.9 is completed. □

In order to prove Theorem 1.10, we need some lemmas.

For $f \in L_{\mathrm{loc}}^{1}(\mu)$ and $0<\beta<1$, one defines the sharp maximal operator

$$ M^{\sharp,(\beta)}f(x)=\sup_{B\ni x}\frac{1}{\mu(6B)} \int_{B}\bigl\vert f(y)-m_{\widetilde{B}}(f)\bigr\vert \,d\mu(y) +\sup_{(B,R)\in\Delta_{x}}\frac{\vert m_{B}(f)-m_{R}(f)\vert }{K^{(\beta)}_{B,R}}, $$

here $\Delta_{x}:=\{(B,R):x\in B\subset R \text{ and } B, R \text{ are doubling balls}\}$.

One defines the non-centered doubling maximal operator

$$Nf(x)=\sup_{B\ni x, B\text{ doubling}}\frac{1}{\mu(B)} \int_{B}\bigl\vert f(y)\bigr\vert \,d\mu(y). $$

It is easy to see

$$ \bigl\vert f(x)\bigr\vert \leq Nf(x), $$

for every $f\in L^{1}_{\mathrm{loc}}(\mu)$ and μ-a.e. $x \in X$.

For $\rho>1$, $\alpha\in(0,1)$ and $t\in(1,\infty)$, one defines the non-centered maximal operator $M^{(\alpha)}_{t,(\rho)}f$ as follows:

$$ M^{(\alpha)}_{t,(\rho)}f(x)=\sup_{B\ni x} \biggl\{ \frac{1}{[\mu(\rho B)]^{1-\alpha t}} \int_{B}\bigl\vert f(y)\bigr\vert ^{t}\,d \mu(y) \biggr\} ^{1/t}. $$

For simplicity, write $M^{(0)}_{1,(\rho)}f(x)$ as $M_{(\rho)}f$. If $\rho\geq5$ and for every $p>1$, then $\|M_{(\rho)}f\|_{L^{p}(\mu)}$ $\leq C\|f\|_{L^{p}(\mu)}$ and for $p\in(t,1/\alpha)$ and $1/q=1/p-\alpha$, $\|M^{(\alpha)}_{t,(\rho)}\|_{L^{q}(\mu)}\leq C\|f\| _{L^{p}(\mu)}$ (see [15]).

Lemma 2.1

[15]

For $f \in L^{1}_{\mathrm{loc}}(\mu)$, $\int_{X}f(x)\,d\mu(x)=0$ if $\|\mu\|<\infty$. Assume $0<\beta<1$ and $\inf(1,Nf)\in L^{p}(\mu)$, $1< p<\infty$, then

$$ \bigl\Vert N(f)\bigr\Vert _{L^{p}(\mu)}\leq C\bigl\Vert M^{\sharp,(\beta)}(f)\bigr\Vert _{L^{p}(\mu)}. $$

Lemma 2.2

[11, 15]

For $1<\rho<\infty$ and $1\leq p<\infty$, if $b\in \operatorname{RBMO}(\mu)$, then for all balls $B\in X$,

$$ \biggl\{ \frac{1}{\mu(\rho B)} \int_{B}\bigl\vert b_{B}-m_{\widetilde {B}}(b) \bigr\vert ^{p}\,d\mu(X) \biggr\} ^{1/p}\leq C\|b \|_{\ast}. $$

(2.1)

Lemma 2.3

[3]

For $b\in \operatorname{RBMO}(\mu)$,

$$ \bigl\vert m_{\widetilde{6^{i}\frac{6}{5}B}}(b)-m_{\widetilde {B}}(b)\bigr\vert \leq Ci\|b \|_{\ast}. $$

Lemma 2.4

For $0<\alpha<2$, $1< p_{1}, p_{2}, q<\infty$, $1< r< q$ and $b_{1}, b_{2}\in \operatorname{RBMO}(\mu)$. If $I_{\beta}$ is bounded from $L^{r}(\mu)$ to $L^{s}(\mu)$, for some $r\in(1, 1/\beta)$ and $1/s=1/r-\beta$, with $0<\beta<1$, then, for every $x\in X$, $g_{1}\in L^{p_{1}}(\mu)$, and $g_{2}\in L^{p_{2}}(\mu)$,

$$\begin{aligned}& M^{\sharp,(\alpha/2)}[b_{1},b_{2},I_{\alpha,2}](g_{1},g_{2}) (x) \\& \quad \leq C \bigl\{ \|b_{1}\|_{\ast}\|b_{2} \|_{\ast}M_{r,(6)}\bigl(I_{\alpha ,2}(g_{1},g_{2}) \bigr) (x) +\|b_{1}\|_{\ast}M_{r,(6)} \bigl([b_{2},I_{\alpha ,2}](g_{1},g_{2}) \bigr) (x) \\& \qquad {}+\|b_{2}\|_{\ast}M_{r,(6)} \bigl([b_{1},I_{\alpha,2}](g_{1},g_{2}) \bigr) (x) +\|b_{1}\|_{\ast}\|b_{2} \|_{\ast}M^{(\alpha /2)}_{p_{1},(5)}g_{1}(x)M_{p_{2},(5)}g_{2}(x) \bigr\} , \end{aligned}$$

(2.2)

$$\begin{aligned}& M^{\sharp,(\alpha/2)}[b_{1},I_{\alpha,2}](g_{1},g_{2}) (x) \\& \quad \leq C \bigl\{ \|b_{1}\|_{\ast}M_{r,(6)} \bigl(I_{\alpha,2}(g_{1},g_{2})\bigr) (x) + \|b_{1}\|_{\ast}M^{(\alpha/2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha /2)}_{p_{2},(5)}g_{2}(x) \bigr\} , \end{aligned}$$

(2.3)

$$\begin{aligned}& M^{\sharp,(\alpha/2)}[b_{2},I_{\alpha,2}](g_{1},g_{2}) (x) \\& \quad \leq C \bigl\{ \|b_{2}\|_{\ast}M_{r,(6)} \bigl(I_{\alpha,2}(g_{1},g_{2})\bigr) (x) + \|b_{2}\|_{\ast}M^{(\alpha/2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha /2)}_{p_{2},(5)}g_{2}(x) \bigr\} . \end{aligned}$$

(2.4)

Proof

Choose $b_{1}, b_{2}\in L^{\infty}(\mu)$ according to Lemma 3.11 in [14]. As $L_{c}^{\infty}(\mu)$ is dense in $L^{p}(\mu)$ for $1< p<\infty$, by standard density arguments, we only need to consider the case that $g_{1}, g_{2}\in L_{c}^{\infty}(\mu)$.

Similar to Theorem 9.1 in [6], in order to obtain (2.2), we only need to prove that, for every $x\in B$,

$$\begin{aligned}& \frac{1}{\mu(6B)} \int_{B}\bigl\vert [b_{1},b_{2}, I_{\alpha,2}](g_{1},g_{2}) (z)-H_{B}\bigr\vert \,d\mu(z) \\& \quad \leq C \bigl\{ \|b_{1}\|_{\ast}\|b_{2} \|_{\ast}M_{r,(6)}\bigl(I_{\alpha ,2}(g_{1},g_{2}) \bigr) (x) +\|b_{1}\|_{\ast}M_{r,(6)} \bigl([b_{2},I_{\alpha ,2}](g_{1},g_{2}) \bigr) (x) \\& \qquad {}+\|b_{2}\|_{\ast}M_{r,(6)} \bigl([b_{1},I_{\alpha,2}](g_{1},g_{2}) \bigr) (x) +C\|b_{1}\|_{\ast}\|b_{2} \|_{\ast}M^{(\alpha/2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha /2)}_{p_{2},(5)}g_{2}(x) \bigr\} , \end{aligned}$$

(2.5)

and, for every ball $B\subset R$, with $x\in B$, R is a doubling ball,

$$\begin{aligned} |H_{B}-H_{R}| \leq& CK_{B,R}^{2}K_{B,R}^{(\alpha/2)} \bigl[\|b_{1}\| _{\ast}\|b_{2} \|_{\ast}M_{r,(6)}\bigl(I_{\alpha,2}(g_{1},g_{2}) \bigr) (x) \\ &{}+\|b_{1}\|_{\ast}\|b_{2}\|_{\ast}M^{(\alpha /2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha/2)}_{p_{2},(5)}g_{2}(x) \\ &{}+\|b_{1}\|_{\ast}M_{r,(6)}\bigl([b_{2},I_{\alpha ,2}](g_{1},g_{2}) \bigr) (x) \\ &{}+\|b_{2}\|_{\ast}M_{r,(6)}\bigl([b_{1},I_{\alpha,2}](g_{1},g_{2}) \bigr) (x) \bigr]. \end{aligned}$$

(2.6)

For every ball B, let

$$\begin{aligned}& H_{B}:= m_{B}\bigl(I_{\alpha,2}\bigl( \bigl(b_{1}- m_{\widetilde {B}}(b_{1})\bigr)g_{1} \chi_{X\backslash\frac {6}{5}B},\bigl(b_{2}-m_{\widetilde {B}}(b_{2}) \bigr)g_{2}\chi_{X\backslash\frac{6}{5}B}\bigr)\bigr), \\& H_{R}:= m_{R}\bigl(I_{\alpha,2}\bigl( \bigl(b_{1}- m_{R}(b_{1})\bigr)g_{1} \chi_{X\backslash\frac {6}{5}R},\bigl(b_{2}-m_{R}(b_{2}) \bigr)g_{2}\chi_{X\backslash\frac{6}{5}R}\bigr)\bigr). \end{aligned}$$

It is easy to see that

$$[b_{1},b_{2},I_{\alpha,2}]=I_{\alpha ,2}\bigl( \bigl(b_{1}-b_{1}(z)\bigr)g_{1}, \bigl(b_{2}-b_{2}(z)\bigr)g_{2}\bigr) $$

and

$$\begin{aligned}& I_{\alpha,2}\bigl(\bigl(b_{1}-m_{\widetilde {B}}(b_{1}) \bigr)g_{1},\bigl(b_{2}-m_{\widetilde {B}}(b_{2}) \bigr)g_{2}\bigr) \\& \quad = I_{\alpha,2}\bigl(\bigl(b_{1}-b_{1}(z)+b_{1}(z)-m_{\widetilde {B}}(b_{1}) \bigr)g_{1},\bigl(b_{2}-b_{2}(z)+b_{2}(z)-m_{\widetilde {B}}(b_{2}) \bigr)g_{2}\bigr) \\& \quad = \bigl(b_{1}(z)-m_{\widetilde {B}}(b_{1})\bigr) \bigl(b_{2}(z)-m_{\widetilde {B}}(b_{2})\bigr)I_{\alpha,2}(g_{1},g_{2}) \\& \qquad {}-\bigl(b_{1}(z)-m_{\widetilde {B}}(b_{1}) \bigr)I_{\alpha ,2}\bigl(g_{1},\bigl(b_{2}-b_{2}(z) \bigr)g_{2}\bigr) \\& \qquad {}-\bigl(b_{2}(z)-m_{\widetilde {B}}(b_{2}) \bigr)I_{\alpha ,2}\bigl(\bigl(b_{1}-b_{1}(z) \bigr)g_{1},g_{2}\bigr) \\& \qquad {}+I_{\alpha,2}\bigl(\bigl(b_{1}-b_{1}(z) \bigr)g_{1},\bigl(b_{2}-b_{2}(z) \bigr)g_{2}\bigr). \end{aligned}$$

Thus

$$\begin{aligned}& \frac{1}{\mu(6B)} \int_{B}\bigl\vert [b_{1},b_{2},I_{\alpha ,2}](g_{1},g_{2}) (z)-H_{B}\bigr\vert \,d\mu(z) \\& \quad \leq C \biggl(\frac{1}{\mu(6B)} \int_{B}\bigl\vert \bigl(b_{1}(z)-m_{\widetilde {B}}(b_{1}) \bigr) \bigl(b_{2}(z)-m_{\widetilde {B}}(b_{2}) \bigr)I_{\alpha ,2}(g_{1},g_{2}) (z)\bigr\vert \,d \mu(z) \biggr) \\& \qquad {}+C \biggl(\frac{1}{\mu(6B)} \int_{B}\bigl\vert \bigl(b_{1}(z)-m_{\widetilde {B}}(b_{1}) \bigr)I_{\alpha,2}\bigl(g_{1},\bigl(b_{2}-b_{2}(z) \bigr)g_{2}\bigr) (z)\bigr\vert \,d\mu(z) \biggr) \\& \qquad {}+C \biggl(\frac{1}{\mu(6B)} \int_{B}\bigl\vert \bigl(b_{2}(z)-m_{\widetilde {B}}(b_{2}) \bigr)I_{\alpha,2}\bigl(\bigl(b_{1}-b_{1}(z) \bigr)g_{1},g_{2}\bigr) (z)\bigr\vert \,d\mu(z) \biggr) \\& \qquad {}+C \biggl(\frac{1}{\mu(6B)} \int_{B}\bigl\vert I_{\alpha,2}\bigl( \bigl(b_{1}-m_{\widetilde {B}}(b_{1})\bigr)g_{1}, \bigl(b_{2}-m_{\widetilde {B}}(b_{2})\bigr)g_{2} \bigr) (z)-H_{B}\bigr\vert \,d\mu (z) \biggr) \\& \quad =: F_{1}+F_{2}+F_{3}+F_{4}. \end{aligned}$$

(2.7)

For $F_{1}$, choose $r_{1}, r_{2}>1$, such that $\frac{1}{r}+\frac{1}{r_{1}}+\frac{1}{r_{2}}=1$. It follows from Hölder’s inequality that

$$\begin{aligned} F_{1} \leq& C \biggl(\frac{1}{\mu(6B)} \int_{B}\bigl\vert b_{1}(z)-m_{\widetilde {B}}b_{1} \bigr\vert ^{r_{1}}\,d\mu(z) \biggr)^{1/r_{1}} \\ &{}\times \biggl(\frac{1}{\mu(6B)} \int_{B}\bigl\vert b_{2}(z)-m_{\widetilde {B}}b_{2} \bigr\vert ^{r_{2}}\,d\mu(z) \biggr)^{1/r_{2}} \\ &{}\times \biggl(\frac{1}{\mu(6B)} \int_{B}\bigl\vert I_{\alpha ,2}(g_{1},g_{2}) \bigr\vert ^{r}\,d\mu(z) \biggr)^{1/r} \\ \leq& C\|b_{1}\|_{\ast}\|b_{2} \|_{\ast}M_{r,(6)}\bigl(I_{\alpha,2}(g_{1},g_{2}) \bigr) (x). \end{aligned}$$

For $F_{2}$, choose $s>1$ such that $\frac{1}{s}+\frac{1}{r}=1$, it follows that

$$\begin{aligned} F_{2} \leq& C \biggl(\frac{1}{\mu(6B)} \int_{B}\bigl\vert b_{1}(z)-m_{\widetilde {B}}b_{1} \bigr\vert ^{s}\,d\mu(z) \biggr)^{1/s} \\ &{} \times \biggl(\frac{1}{\mu(6B)} \int_{B}\bigl\vert [b_{2},I_{\alpha ,2}](g_{1},g_{2}) \bigr\vert ^{r}\,d\mu(z) \biggr)^{1/r} \\ \leq& C\|b_{1}\|_{\ast}M_{r,(6)} \bigl([b_{2},I_{\alpha,2}](g_{1},g_{2}) \bigr) (x). \end{aligned}$$

For $F_{3}$, in the same way, one obtains

$$ F_{3}\leq C\|b_{2}\|_{\ast}M_{r,(6)} \bigl([b_{1},I_{\alpha,2}](g_{1},g_{2}) \bigr) (x). $$

For $F_{4}$, let $g_{k}^{1}=g_{k}\chi_{\frac{6}{5}B}$ and $g_{k}^{2}=g_{k}-g_{k}^{1}$ for $k=1,2$. Therefore,

$$\begin{aligned} F_{4} \leq& C \biggl(\frac{1}{\mu(6B)} \int_{B}\bigl\vert I_{\alpha,2}\bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{1}^{1}(z),(b_{2}-m_{\widetilde {B}}b_{2})g_{2}^{1} \bigr) (z)\bigr\vert \,d\mu (z) \biggr) \\ &{}+ C \biggl(\frac{1}{\mu(6B)} \int_{B}\bigl\vert I_{\alpha,2}\bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{1}^{1}(z),(b_{2}-m_{\widetilde {B}}b_{2})g_{2}^{2} \bigr) (z)\bigr\vert \,d\mu (z) \biggr) \\ &{}+ C \biggl(\frac{1}{\mu(6B)} \int_{B}\bigl\vert I_{\alpha,2}\bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{1}^{2}(z),(b_{2}-m_{\widetilde {B}}b_{2})g_{2}^{1} \bigr) (z)\bigr\vert \,d\mu (z) \biggr) \\ &{}+ C \biggl(\frac{1}{\mu(6B)} \int_{B}\bigl\vert I_{\alpha,2}\bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{1}^{2}(z),(b_{2}-m_{\widetilde {B}}b_{2})g_{2}^{2} \bigr) (z)-H_{B}\bigr\vert \,d\mu(z) \biggr) \\ =:&F_{41}+F_{42}+F_{43}+F_{44}. \end{aligned}$$

For $1< p_{i}<\infty$, $i=1,2$, choose $s_{1}=\sqrt{p_{1}}$, $s_{2}=\sqrt {p_{2}}$, $\frac{1}{v}=\frac{1}{s_{1}}+\frac{1}{s_{2}}-\alpha$, $\frac {1}{s_{1}}=\frac{1}{p_{1}}+\frac{1}{v_{1}}$ and $\frac{1}{s_{2}}=\frac{1}{p_{2}}+\frac{1}{v_{2}}$. It follows from Hölder’s inequality and Theorem 1.9 that

$$\begin{aligned} F_{41} \leq& C\frac{\mu(B)^{1-1/v}}{\mu(6B)}\bigl\Vert I_{\alpha,2} \bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{1}^{1},(b_{2}-m_{\widetilde {B}}b_{2})g_{2}^{1} \bigr)\bigr\Vert _{L^{v}(\mu )} \\ \leq& C\frac{1}{\mu(6B)^{1/v}}\bigl\Vert (b_{1}-m_{\widetilde {B}}b_{1})g_{1}^{1} \bigr\Vert _{L^{s_{1}}(\mu)}\bigl\Vert (b_{2}-m_{\widetilde {B}}b_{2})g_{2}^{1} \bigr\Vert _{L^{s_{2}}(\mu)} \\ \leq& C\frac{1}{\mu(6B)^{1/v}} \biggl( \int_{\frac{6}{5}B}\vert b_{1}-m_{\widetilde {B}}b_{1} \vert ^{v_{1}}\,d\mu(z) \biggr)^{1/v_{1}} \biggl( \int_{\frac{6}{5}B}\bigl\vert g_{1}(z)\bigr\vert ^{p_{1}}\,d\mu(z) \biggr)^{1/p_{1}} \\ &{} \times \biggl( \int_{\frac{6}{5}B}\vert b_{2}-m_{\widetilde {B}}b_{2} \vert ^{v_{2}}\,d\mu(z)\biggr)^{1/v_{2}} \biggl( \int_{\frac {6}{5}B}\bigl\vert g_{2}(z)\bigr\vert ^{p_{2}}\,d\mu(z) \biggr)^{1/p_{2}} \\ \leq& C\prod_{i=1}^{2} \biggl( \frac{\int_{\frac{6}{5}B}\vert b_{i}-m_{\widetilde {B}}b_{i}\vert ^{v_{i}}\,d\mu(z)}{\mu(6B)} \biggr)^{1/v_{i}} \biggl(\frac{\int_{\frac{6}{5}B}\vert g_{i}(z)\vert ^{p_{i}}\,d\mu(z)}{\mu (6B)^{1-\alpha p_{i}/2}} \biggr)^{1/p_{i}} \\ \leq& C\Vert b_{1}\Vert _{\ast} \Vert b_{2} \Vert _{\ast}M^{(\alpha /2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha/2)}_{p_{2},(5)}g_{2}(x). \end{aligned}$$

For $F_{42}$, it follows from (i) of Definition 1.5, Lemmas 2.2-2.3, the condition of $\lambda(x,al) \geq a^{m}\lambda(x,l)$, and Hölder’s inequality that

$$\begin{aligned} F_{42} \leq& C\frac{1}{\mu(6B)} \int_{B} \int_{X} \int_{X} \frac{\vert b_{1}(y_{1})-m_{\widetilde {B}}b_{1}\vert \vert g_{1}^{1}(y_{1})\vert }{ [\lambda(z,d(z,y_{1}))+\lambda(z,d(z,y_{2}))]^{2-\alpha}} \\ &{} \times\bigl\vert b_{2}(y_{2})-m_{\widetilde {B}}b_{2} \bigr\vert \bigl\vert g_{2}^{2}(y_{2})\bigr\vert \,d\mu (y_{1})\,d\mu(y_{2})\,d\mu(z) \\ \leq& C\frac{1}{\mu(6B)} \int_{B} \int_{\frac{6}{5}B} \bigl\vert b_{1}(y_{1})-m_{\widetilde {B}}b_{1} \bigr\vert \bigl\vert g_{1}(y_{1})\bigr\vert \,d \mu(y_{1}) \\ &{}\times \int_{X\backslash\frac{6}{5}B} \frac{\vert b_{2}(y_{2})-m_{\widetilde {B}}b_{2}\vert \vert g_{2}(y_{2})\vert \,d\mu (y_{2})}{[\lambda(z,d(z,y_{2}))]^{2-\alpha}}\,d\mu(z) \\ \leq& C \biggl(\frac{\int_{\frac{6}{5}B}\vert b_{1}(y_{1})-m_{\widetilde {B}}b_{1}\vert ^{p'_{1}}\,d\mu(y_{1})}{\mu(6B)} \biggr)^{1/p'_{1}} \biggl( \frac{\int_{\frac{6}{5}B}\vert g_{1}(y_{1})\vert ^{p_{1}}\,d\mu(y_{1})}{\mu (6B)^{1-\alpha p_{1}/2}} \biggr)^{1/p_{1}} \\ &{}\times \mu(6B)^{-\alpha/2}\mu(B)\sum_{i=1}^{\infty} \int_{6^{i}\frac {6}{5}B\backslash6^{i-1}\frac{6}{5}B}\frac{\vert b_{2}(y_{2})-m_{\widetilde {B}} b_{2}\vert \vert g_{2}(y_{2})\vert }{[\lambda(x,6^{i-1}\frac{6}{5}l_{B})]^{2-\alpha }}\,d\mu(y_{2}) \\ \leq& C\|b_{1}\|_{\ast}M^{(\alpha/2)}_{p_{1},(5)}g_{1}(x) \sum_{i=1}^{\infty }6^{-km(1-\alpha/2)} \biggl[ \frac{\mu(B)}{\mu(\frac{6}{5}B)} \biggr]^{1-\alpha/2} \biggl[\frac {\mu(\frac{6}{5}B)}{\lambda(x,\frac{6}{5}l_{B})} \biggr]^{1-\alpha/2} \\ &{}\times\frac{1}{[\lambda(x,5\times6^{i}\frac{6}{5}l_{B})]^{1-\alpha/2}} \int_{6^{i}\frac{6}{5}B}\bigl\vert b_{2}(y_{2})-m_{\widetilde {B}} b_{2}\bigr\vert \bigl\vert g_{2}(y_{2})\bigr\vert \,d\mu(y_{2}) \\ \leq& C\|b_{1}\|_{\ast}M^{(\alpha/2)}_{p_{1},(5)}g_{1}(x) \sum_{i=1}^{\infty }6^{-im(1-\alpha/2)} \frac{1}{[\mu(5\times6^{i}\frac{6}{5}B)]^{1-\alpha /2}} \\ &{}\times \int_{6^{i}\frac{6}{5}B} \bigl\vert b_{2}(y_{2})-m_{\widetilde{6^{i}\frac{6}{5}B}}(b_{2})+m_{\widetilde {6^{i}\frac{6}{5}B}}(b_{2})-m_{\widetilde {B}}b_{2} \bigr\vert \bigl\vert g_{2}(y_{2})\bigr\vert \,d\mu (y_{2}) \\ \leq& C\|b_{1}\|_{\ast}M^{(\alpha/2)}_{p_{1},(5)}g_{1}(x) \sum_{i=1}^{\infty }6^{-im(1-\alpha/2)} \\ &{} \times \biggl[ \biggl(\frac{\int_{6^{i}\frac{6}{5}B}\vert b_{2}(y_{2}) -m_{\widetilde{6^{i}\frac{6}{5}B}}(b_{2})\vert ^{p'_{2}}\,d\mu(y_{2})}{\mu (5\times6^{i}\frac{6}{5}B)} \biggr)^{1/p'_{2}} \biggl( \frac{\int_{6^{i}\frac{6}{5}B}\vert g_{2}(y_{2})\vert ^{p_{2}}\,d\mu (y_{2})}{\mu(5\times6^{i}\frac{6}{5}B)^{1-\alpha p_{2}/2}} \biggr)^{1/p_{2}} \\ &{} +C k\|b_{2}\|_{\ast} \biggl(\frac{\int_{6^{k}\frac {6}{5}B}\vert g_{2}(y_{2})\vert ^{p_{2}}\,d\mu(y_{2})}{\mu(5\times6^{i}\frac {6}{5}B)^{1-\alpha p_{2}/2}} \biggr)^{1/p_{2}} \biggl(\frac{\int _{6^{i}\frac{6}{5}B}\,d\mu(y_{2})}{\mu(5\times6^{i}\frac{6}{5}B)} \biggr)^{1/p'_{2}} \biggr] \\ \leq& C\|b_{1}\|_{\ast}\|b_{2} \|_{\ast}M^{(\alpha /2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha/2)}_{p_{2},(5)}g_{2}(x). \end{aligned}$$

In the same way, one obtains

$$ F_{43}\leq C\|b_{1}\|_{\ast}\|b_{2} \|_{\ast}M^{(\alpha /2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha/2)}_{p_{2},(5)}g_{2}(x). $$

For $F_{44}$, let $z, z_{0}\in B$, it follows from (ii) of Definition 1.5, Lemmas 2.2-2.3, the condition of λ, and Hölder’s inequality that

$$\begin{aligned}& \bigl\vert I_{\alpha,2}\bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{2}^{2},(b_{2}-m_{\widetilde {B}}b_{2})g_{2}^{2} \bigr) (z) \\& \qquad {}-I_{\alpha,2}\bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{2}^{2},(b_{2}-m_{\widetilde {B}}b_{2})g_{2}^{2} \bigr) (z_{0})\bigr\vert \\& \quad \leq C \int_{X\backslash \frac{6}{5}B} \int_{X\backslash\frac {6}{5}B}\bigl\vert K(z,y_{1},y_{2})-K(z_{0},y_{1},y_{2}) \bigr\vert \\& \qquad {}\times\prod_{j=1}^{2}\bigl\vert \bigl(b_{j}(y_{j})-m_{\widetilde {B}}b_{j} \bigr)g_{j}(y_{j})\bigr\vert \,d\mu(y_{j}) \\& \quad \leq C \int_{X\backslash \frac{6}{5}B} \int_{X\backslash\frac{6}{5}B}\frac{d(z,z_{0})^{\delta} \prod_{j=1}^{2}\vert (b_{j}(y_{j})-m_{\widetilde {B}}b_{j})g_{j}(y_{j})\vert \,d\mu(y_{j})}{ (d(z,y_{1})+d(z,y_{2}))^{\delta}[\sum_{j=1}^{2}\lambda (x,d(x,y_{j}))]^{2-\alpha}} \\& \quad \leq C\prod_{j=1}^{2} \int_{X\backslash \frac{6}{5}B}\frac{d(z,z_{0})^{\delta_{i}} \vert b_{j}(y_{j})-m_{\widetilde {B}}b_{j}\vert \vert g_{j}(y_{j})\vert \,d\mu (y_{j})}{d(z,y_{j})^{\delta_{j}}[\lambda(z,d(z,y_{j}))]^{1-\alpha/2}} \\& \quad \leq C\prod_{j=1}^{2}\sum _{k=1}^{\infty} \int_{6^{k}\frac {6}{5}B\backslash6^{k-1}\frac{6}{5}B}6^{-k\delta_{j}} \biggl[\frac{\mu (5\times6^{k}\frac{6}{5}B)}{ \lambda(z,5\times6^{k}\frac{6}{5}l_{B})} \biggr]^{1-\alpha/2} \\& \qquad {} \times\frac{\vert b_{j}(y_{j})-m_{\widetilde {B}}b_{j}\vert \vert g_{j}(y_{j})\vert \,d\mu(y_{j})}{[\mu(5\times6^{k}\frac{6}{5}B)]^{1-\alpha /2}} \\& \quad \leq C\prod_{j=1}^{2}\sum _{k=1}^{\infty}6^{-k\delta_{j}} \biggl(\frac {\int_{6^{k}\frac{6}{5}B}\vert b_{j}(y_{j})-m_{\widetilde {B}}b_{j}\vert ^{p'_{j}}\,d\mu(y_{j})}{ [\mu(5\times6^{k}\frac{6}{5}B)]^{1-\alpha p_{j}/2}} \biggr)^{1/p'_{j}} \\& \qquad {} \times \biggl(\frac{\int_{6^{k}\frac{6}{5}B}\vert g_{j}(y_{j})\vert ^{p_{j}}\,d\mu (y_{j})}{\mu(5\times6^{k}\frac{6}{5}B)} \biggr)^{1/p_{i}} \\& \quad \leq C\prod_{j=1}^{2}\sum _{k=1}^{\infty}6^{-k\delta_{j}}M_{p_{j},(6)}g_{j}(x) \biggl(\frac{1}{[\mu(5\times6^{k}\frac{6}{5}B)]^{1-\alpha p_{j}/2}} \\& \qquad {} \times \int_{6^{k}\frac{6}{5}B}\bigl\vert b_{j}(y_{j})-m_{\widetilde{6^{k}\frac{6}{5}B}} +m_{\widetilde{6^{k}\frac{6}{5}B}}-m_{\widetilde {B}}b_{j}\bigr\vert ^{p'_{j}} \,d\mu (y_{j}) \biggr)^{1/p'_{j}} \\& \quad \leq C\prod_{j=1}^{2}\sum _{k=1}^{\infty}6^{-k\delta_{j}}k\|b_{j} \|_{\ast }M_{p_{j},(6)}g_{j}(x) \\ & \quad \leq C\|b_{1}\|_{\ast}\|b_{2} \|_{\ast}M^{(\alpha /2)}_{p_{1},(6)}g_{1}(x)M^{(\alpha/2)}_{p_{2},(6)}g_{2}(x), \end{aligned}$$

where $\delta=\delta_{1}+\delta_{2}$, $\delta_{1},\delta_{2}>0$.

It follows from taking the mean over $z_{0}\in B$ that

$$ F_{44}\leq C\|b_{1}\|_{\ast} \|b_{2}\|_{\ast}M^{(\alpha /2)}_{p_{1},(6)}g_{1}(x)M^{(\alpha/2)}_{p_{2},(6)}g_{2}(x). $$

(2.8)

Thus (2.5) is obtained from (2.7) to (2.8).

Now we turn to the proof of (2.6). Set $N=N_{B,R}+1$. For two balls $B\subset R$ with $x\in B$, here R is a doubling ball and B is an every ball,

$$\begin{aligned}& \bigl\vert \bigl\vert m_{B}\bigl[I_{\alpha,2} \bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{1}^{2},(b_{2}-m_{\widetilde {B}}b_{2})g_{2}^{2} \bigr)\bigr]\bigr\vert -\bigl\vert m_{R}\bigl[I_{\alpha ,2} \bigl((b_{1}-m_{R}b_{1})g_{1}^{2},(b_{2}-m_{R}b_{2})g_{2}^{2} \bigr)\bigr]\bigr\vert \bigr\vert \\ & \quad \leq \bigl\vert m_{B}\bigl[I_{\alpha,2} \bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{1}\chi _{X\backslash6^{N}B},(b_{2}-m_{\widetilde {B}}b_{2})g_{2} \chi_{X\backslash 6^{N}B}\bigr)\bigr] \\& \qquad {}-m_{R}\bigl[I_{\alpha,2}\bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{1} \chi_{X\backslash 6^{N}B},(b_{2}-m_{\widetilde {B}}b_{2})g_{2} \chi_{X\backslash6^{N}B}\bigr)\bigr] \bigr\vert \\& \qquad {}+\bigl\vert m_{R}\bigl[I_{\alpha,2} \bigl((b_{1}-m_{R}b_{1})g_{1} \chi_{X\backslash 6^{N}B},(b_{2}-m_{R}b_{2})g_{2} \chi_{X\backslash6^{N}B}\bigr)\bigr] \\& \qquad {}-m_{R}\bigl[I_{\alpha,2}\bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{1} \chi_{X\backslash 6^{N}B},(b_{2}-m_{\widetilde {B}}b_{2})g_{2} \chi_{X\backslash6^{N}B}\bigr)\bigr] \bigr\vert \\& \qquad {}+\bigl\vert m_{B}\bigl[I_{\alpha,2} \bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{1}\chi _{6^{N}B\backslash\frac{6}{5}B},(b_{2}-m_{\widetilde {B}}b_{2})g_{2} \chi _{X\backslash\frac{6}{5}B}\bigr)\bigr]\bigr\vert \\& \qquad {}+\bigl\vert m_{B}\bigl[I_{\alpha,2} \bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{1}\chi _{X\backslash\frac{6}{5}B},(b_{2}-m_{\widetilde {B}}b_{2})g_{2} \chi _{6^{N}B\backslash\frac{6}{5}B}\bigr)\bigr]\bigr\vert \\& \qquad {}+\bigl\vert m_{R}\bigl[I_{\alpha,2} \bigl((b_{1}-m_{R}b_{1})g_{1} \chi_{6^{N}B\backslash \frac{6}{5}R},(b_{2}-m_{R}b_{2})g_{2} \chi_{X\backslash6^{N}B}\bigr)\bigr] \bigr\vert \\& \qquad {}+\bigl\vert m_{R}\bigl[I_{\alpha,2} \bigl((b_{1}-m_{R}b_{1})g_{1} \chi_{X\backslash\frac {6}{5}R},(b_{2}-m_{R}b_{2})g_{2} \chi_{6^{N}B\backslash\frac {6}{5}R}\bigr)\bigr]\bigr\vert \\ & \quad =: G_{1}+G_{2}+G_{3}+G_{4}+G_{5}+G_{6}. \end{aligned}$$

(2.9)

Similar to the estimate of $F_{44}$,

$$ G_{1}\leq C\|b_{1}\|_{\ast}\|b_{2} \|_{\ast}M^{(\alpha /2)}_{p_{1},(6)}g_{1}(x)M^{(\alpha/2)}_{p_{2},(6)}g_{2}(x). $$

For $G_{2}$, it is easy to see that

$$\begin{aligned}& I_{\alpha,2}\bigl((b_{1}-m_{R}b_{1})g_{1} \chi_{X\backslash 6^{N}B},(b_{2}-m_{R}b_{2})g_{2} \chi_{X\backslash6^{N}B}\bigr) (z) \\ & \qquad {}-I_{\alpha,2}\bigl((b_{1}-m_{\widetilde {B}}b_{1})g_{1} \chi_{X\backslash 6^{N}B},(b_{2}-m_{\widetilde {B}}b_{2})g_{2} \chi_{X\backslash6^{N}B}\bigr) (z) \\& \quad = (m_{R}b_{2}-m_{\widetilde {B}}b_{2})I_{\alpha ,2} \bigl((b_{1}-m_{R}b_{1})g_{1} \chi_{X\backslash6^{N}B},g_{2}\chi _{X\backslash6^{N}B}\bigr) (z) \\& \qquad {}+(m_{R}b_{1}-m_{\widetilde {B}}b_{1})I_{\alpha,2} \bigl(g_{1}\chi_{X\backslash 6^{N}B},(b_{2}-m_{R}b_{2})g_{2} \chi_{X\backslash6^{N}B}\bigr) (z) \\& \qquad {}+(m_{R}b_{1}-m_{\widetilde {B}}b_{1}) (m_{R}b_{2}-m_{\widetilde {B}}b_{2})I_{\alpha,2}(g_{1} \chi_{X\backslash6^{N}B},g_{2}\chi _{X\backslash6^{N}B}) (z). \end{aligned}$$

Thus

$$\begin{aligned} G_{2} \leq&\biggl\vert (m_{R}b_{2}-m_{\widetilde {B}}b_{2}) \frac{1}{\mu(R)} \int_{R}I_{\alpha,2}\bigl((b_{1}-m_{R}b_{1})g_{1} \chi _{X\backslash6^{N}B},g_{2}\chi_{X\backslash6^{N}B}\bigr) (z)\,d\mu(z) \biggr\vert \\ &{}+\biggl\vert (m_{R}b_{1}-m_{\widetilde {B}}b_{1}) \frac{1}{\mu(R)} \int_{R}I_{\alpha,2}\bigl(g_{1} \chi_{X\backslash 6^{N}B},(b_{2}-m_{R}b_{2})g_{2} \chi_{X\backslash6^{N}B}\bigr) (z)\,d\mu(z) \biggr\vert \\ &{}+\biggl\vert (m_{R}b_{1}-m_{\widetilde {B}}b_{1}) (m_{R}b_{2}-m_{\widetilde {B}}b_{2}) \frac{1}{\mu(R)} \int_{R}I_{\alpha,2}(g_{1} \chi_{X\backslash 6^{N}B},g_{2}\chi_{X\backslash6^{N}B}) (z)\, d\mu(z)\biggr\vert \\ =:&G_{21}+G_{22}+G_{23}. \end{aligned}$$

For $G_{21}$,

$$\begin{aligned}& I_{\alpha,2}\bigl((b_{1}-m_{R}b_{1})g_{1} \chi_{X\backslash6^{N}R},g_{2}\chi _{X\backslash6^{N}R}\bigr) (z) \\& \quad = I_{\alpha ,2}\bigl((b_{1}-m_{R}b_{1})g_{1},g_{2} \bigr) (z)-T\bigl((b_{1}-m_{R}b_{1})g_{1} \chi _{6^{N}B}\chi_{\frac{6}{5}R},g_{2}\chi_{\frac{6}{5}R} \bigr) (z) \\& \qquad {}-I_{\alpha,2}\bigl((b_{1}-m_{R}b_{1})g_{1} \chi_{\frac{6}{5}R},g_{2}\chi _{6^{N}B}\chi_{\frac{6}{5}R} \bigr) (z) \\& \qquad {}+I_{\alpha,2}\bigl((b_{1}-m_{R}b_{1})g_{1} \chi_{6^{N}B}\chi_{\frac {6}{5}R},g_{2}\chi_{6^{N}B} \chi_{\frac{6}{5}R}\bigr) (z) \\& \qquad {}-I_{\alpha,2}\bigl((b_{1}-m_{R}b_{1})g_{1} \chi_{X\backslash\frac {6}{5}R},g_{2}\chi_{6^{N}B}\bigr) (z) \\& \qquad {}-I_{\alpha,2}\bigl((b_{1}-m_{R}b_{1})g_{1} \chi_{6^{N}B},g_{2}\chi _{X\backslash\frac{6}{5}R}\bigr) (z) \\& \qquad {}+I_{\alpha,2}\bigl((b_{1}-m_{R}b_{1})g_{1} \chi_{6^{N}B\backslash\frac {6}{5}R},g_{2}\chi_{6^{N}B\backslash\frac{6}{5}R}\bigr) (z) \\& \quad =: E_{1}(z)+E_{2}(z)+E_{3}(z)+E_{4}(z)+E_{5}(z)+E_{6}(z)+E_{7}(z). \end{aligned}$$

For $E_{1}(z)$, it is easy to see that

$$\frac{1}{\mu(R)} \int_{R}\bigl\vert I_{\alpha,2}\bigl(b_{1}-b_{1}(z)g_{1},g_{2} \bigr) (z)\bigr\vert \,d\mu (z)\leq CM_{r,(6)}\bigl([b_{1},I_{\alpha,2}]g_{1},g_{2} \bigr) (x). $$

It follows from Hölder’s inequality that

$$\frac{1}{\mu(R)} \int_{R}\bigl\vert \bigl(b_{1}(z)-m_{R}(b_{1}) \bigr)I_{\alpha ,2}(g_{1},g_{2}) (z)\bigr\vert \,d \mu(z)\leq C\|b_{1}\|_{\ast}M_{r,(6)} \bigl(I_{\alpha ,2}(g_{1},g_{2})\bigr) (x). $$

Therefore

$$\begin{aligned} \bigl\vert m_{R}(E_{1})\bigr\vert \leq& \bigl\vert m_{R}\bigl(I_{\alpha ,2}\bigl(b_{1}-b_{1}(z)g_{1},g_{2} \bigr)\bigr)\bigr\vert +\bigl\vert m_{R}\bigl( \bigl(b_{1}(z)-m_{R}(b_{1})\bigr)I_{\alpha ,2}(g_{1},g_{2}) \bigr)\bigr\vert \\ \leq& C \bigl\{ M_{r,(6)}\bigl([b_{1},I_{\alpha,2}]g_{1},g_{2} \bigr) (x)+\|b_{1}\|_{\ast }M_{r,(6)} \bigl(I_{\alpha,2}(g_{1},g_{2})\bigr) (x) \bigr\} . \end{aligned}$$

For $E_{2}(z)$, denote $s_{1}=\sqrt{p_{1}}$, $s_{2}=p_{2}$, $\frac{1}{v}=\frac {1}{s_{1}}+\frac{1}{s_{2}}-\alpha$ and $\frac{1}{s_{1}}=\frac{1}{p_{1}}+\frac {1}{v_{1}}$. Noting that R is a doubling ball, by Theorem 1.9, one obtains

$$\begin{aligned} \bigl\vert m_{R}(E_{2})\bigr\vert \leq& C \frac{\mu(R)^{1-1/v}}{\mu(6R)}\bigl\Vert I_{\alpha ,2}\bigl((b_{1}-m_{R}b_{1})g_{1} \chi_{6^{N}B}\chi_{6/5R},g_{2}\chi_{6/5R} \bigr)\bigr\Vert _{L^{v}(\mu)} \\ \leq& C\mu(6R)^{-1/v}\bigl\Vert (b_{1}-m_{R}b_{1})g_{1} \chi_{6^{N}B}\chi_{6/5R}\bigr\Vert _{L^{s_{1}}(\mu)}\Vert g_{2}\chi_{6/5R}\Vert _{L^{s_{2}}(\mu)} \\ \leq& C\frac{1}{\mu(6R)^{1/v}} \biggl( \int_{\frac {6}{5}R}\vert b_{1}-m_{R}b_{1} \vert ^{v_{1}}\,d\mu(z) \biggr)^{1/v_{1}} \biggl( \int _{\frac{6}{5}R}\bigl\vert g_{1}(z)\bigr\vert ^{p_{1}}\,d\mu(z) \biggr)^{1/p_{1}} \\ &{}\times \biggl( \int_{\frac{6}{5}R}\bigl\vert g_{2}(z)\bigr\vert ^{p_{2}}\,d\mu(z) \biggr)^{1/p_{2}} \\ \leq& C \biggl(\frac{1}{\mu(6R)} \int_{\frac {6}{5}R}\vert b_{1}-m_{R}b_{1} \vert ^{v_{1}}\,d\mu(z) \biggr)^{1/v_{1}} \\ &{}\times\prod_{j=1}^{2} \biggl( \frac{1}{\mu(6R)^{1-\alpha p_{j}/2}} \int _{\frac{6}{5}R}\bigl\vert g_{j}(z)\bigr\vert ^{p_{j}}\,d\mu(z) \biggr)^{1/p_{j}} \\ \leq& C\Vert b_{1}\Vert _{\ast}M^{(\alpha/2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha /2)}_{p_{2},(5)}g_{2}(x). \end{aligned}$$

Also one deduces

$$ \bigl\vert m_{R}(E_{3})\bigr\vert +\bigl\vert m_{R}(E_{4})\bigr\vert \leq C\|b_{1} \|_{\ast}M^{(\alpha/2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha /2)}_{p_{2},(5)}g_{2}(x). $$

For $E_{5}$, as $z\in R$, noting that R is a doubling ball, it follows from (i) of Definition 1.5, Lemmas 2.2-2.3, and the conditions of λ that

$$\begin{aligned} \bigl\vert E_{5}(z)\bigr\vert \leq& C \int_{6^{N}B} \int_{X\backslash\frac{6}{5}R}\frac {\vert b_{1}(y_{1})-m_{R}b_{1}\vert \vert g_{1}(y_{1})\vert \vert g_{2}(y_{2})\vert \,d\mu(y_{1})\,d\mu(y_{2})}{ [\sum_{i=1}^{2}\lambda(x,d(x,y_{i}))]^{2-\alpha}} \\ \leq& C \int_{6^{N}B}\bigl\vert g_{2}(y_{2})\bigr\vert \,d\mu(y_{2})\sum_{j=1}^{\infty} \int _{6^{j}\frac{6}{5}R\backslash6^{j-1}\frac{6}{5}R} \frac{\vert b_{1}(y_{1})-m_{R}b_{1}\vert \vert g_{1}(y_{1})\vert }{(\lambda(z,5\times 6^{j}\frac{6}{5}l_{R}))^{2-\alpha}}\,d\mu(y_{1}) \\ \leq& C \int_{6^{N}B}\bigl\vert g_{2}(y_{2})\bigr\vert \,d\mu(y_{2})\sum_{j=1}^{\infty }6^{-jm(1-\alpha/2)} \\ &{}\times \int_{6^{j}\frac{6}{5}R} \frac{1}{[\lambda(z,5\times\frac{6}{5}l_{R})]^{1-\alpha/2}}\frac {\vert b_{1}(y_{1})-m_{R}b_{1}\vert \vert g_{1}(y_{1})\vert \,d\mu(y_{1})}{ [\lambda(z,5\times6^{j}\frac{6}{5}l_{R})]^{1-\alpha/2}} \\ \leq& C\frac{1}{[\lambda(z,6l_{R})]^{1-\alpha/2}} \int _{6^{N}B}\bigl\vert g_{2}(y_{2})\bigr\vert \,d\mu(y_{2})\sum_{j=1}^{\infty}6^{-jm(1-\alpha /2)} \\ &{} \times\frac{1}{[\lambda(z,5\times6^{j}\frac{6}{5}l_{R})]^{1-\alpha /2}}\times \biggl[ \int_{6^{j}\frac{6}{5}R}\bigl\vert b_{1}(y_{1})-m_{6^{j}\frac {6}{5}R}(b_{1}) \bigr\vert \bigl\vert g_{1}(y_{1})\bigr\vert \,d \mu(y_{1}) \\ &{} + \int_{6^{j}\frac{6}{5}R}\bigl\vert m_{6^{j}\frac {6}{5}R}(b_{1})-m_{R}b_{1} \bigr\vert \bigl\vert g_{1}(y_{1})\bigr\vert \,d \mu(y_{1}) \biggr] \\ \leq& C\frac{1}{[\lambda(z,6l_{R})]^{1-\alpha/2}} \int _{6^{N}B}\bigl\vert g_{2}(y_{2})\bigr\vert \,d\mu(y_{2})\sum_{j=1}^{\infty}6^{-jm(1-\alpha /2)} \\ &{} \times \biggl[ \biggl(\frac{1}{\lambda(z,5\times6^{j}\frac {6}{5}l_{R})} \int_{6^{j}\frac{6}{5}R} \bigl\vert b_{1}(y_{1})-m_{6^{j}\frac{6}{5}R}(b_{1}) \bigr\vert ^{p'_{1}}\,d\mu(y_{1}) \biggr)^{1/p'_{1}} \\ &{}\times \biggl(\frac{1}{[\lambda(z,6^{j+1}\frac{6}{5}l_{R})]^{1-\alpha p_{1}/2}} \int_{6^{j}\frac{6}{5}R}\bigl\vert g_{1}(y_{1})\bigr\vert ^{p_{1}}\,d\mu(y_{1}) \biggr)^{1/p_{1}} \\ &{} +j\|b_{1}\|_{\ast}\frac{1}{[\lambda(z,5\times6^{j}\frac {6}{5}l_{R})]^{1-\alpha/2}} \int_{6^{j}\frac{6}{5}R}\bigl\vert g_{1}(y_{1})\bigr\vert \,d\mu (y_{1}) \biggr] \\ \leq& C\|b_{1}\|_{\ast}M^{(\alpha/2)}_{p_{1},(5)}g_{1}(x) \sum_{j=1}^{N_{B,R}}\frac{1}{[\lambda(z,6l_{R})]^{1-\alpha/2}} \\ &{} \times \biggl[ \int_{6^{j+1}B\backslash6^{j}B}\bigl\vert g_{2}(y_{2})\bigr\vert \,d\mu (y_{2})+ \int_{6B}\bigl\vert g_{2}(y_{2})\bigr\vert \,d\mu(y_{2}) \biggr] \\ \leq& C\|b_{1}\|_{\ast}M^{(\alpha/2)}_{p_{1},(5)}g_{1}(x) \sum_{j=1}^{N_{B,R}}\frac{\int_{6^{j+1}B}\vert g_{2}(y_{2})\vert \,d\mu(y_{2})}{[\mu (5\times6^{j+1}B)]^{1-\alpha/2}} \\ &{}\times \biggl[\frac{\mu(5\times6^{j+1}B)}{\lambda(z,5\times 6^{j+1}l_{B})} \biggr]^{1-\alpha/2} \biggl[ \frac{\lambda(z,5\times6^{j}l_{B})}{\lambda(z,6l_{R})} \biggr]^{1-\alpha/2} \\ &{}+C\|b_{1}\|_{\ast}M^{(\alpha/2)}_{p_{1},(5)}g_{1}(x) \frac{1}{[\lambda (z,6l_{R})]^{1-\alpha/2}} \int_{6B}\bigl\vert g_{2}(y_{2})\bigr\vert \,d\mu(y_{2}) \\ \leq& CK^{(\alpha/2)}_{B,R}\|b_{1}\|_{\ast}M^{(\alpha /2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha/2)}_{p_{2},(5)}g_{2}(x). \end{aligned}$$

Therefore

$$ |m_{R}(E_{5})|\leq CK^{(\alpha/2)}_{B,R} \|b_{1}\|_{\ast}M^{(\alpha /2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha/2)}_{p_{2},(5)}g_{2}(x). $$

Similar to the estimate of $m_{R}(E_{5})$,

$$ \bigl\vert m_{R}(E_{6})\bigr\vert +\bigl\vert m_{R}(E_{7})\bigr\vert \leq CK^{(\alpha/2)}_{B,R} \|b_{1}\|_{\ast}M^{(\alpha /2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha/2)}_{p_{2},(5)}g_{2}(x). $$

By (1.10), it yields

$$\begin{aligned} G_{21} \leq& C \bigl[\|b_{1}\|_{\ast} \|b_{2}\|_{\ast}M_{r,(6)}\bigl(I_{\alpha ,2}(g_{1},g_{2}) \bigr) (x) +\|b_{1}\|_{\ast}M_{r,(6)} \bigl([b_{2},I_{\alpha,2}](g_{1},g_{2}) \bigr) (x) \\ &{}+\|b_{2}\|_{\ast}M_{r,(6)}\bigl([b_{1},I_{\alpha,2}](g_{1},g_{2}) \bigr) (x) +\|b_{1}\|_{\ast}\|b_{2} \|_{\ast}M^{(\alpha /2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha/2)}_{p_{2},(5)}g_{2}(x) \bigr]. \end{aligned}$$

For $G_{22}$ and $G_{23}$, they are similar to $G_{21}$, thus

$$\begin{aligned} G_{2} \leq& C \bigl[\|b_{1}\|_{\ast} \|b_{2}\|_{\ast}M_{r,(6)}\bigl(I_{\alpha ,2}(g_{1},g_{2}) \bigr) (x) +\|b_{1}\|_{\ast}M_{r,(6)} \bigl([b_{2},I_{\alpha,2}](g_{1},g_{2}) \bigr) (x) \\ &{}+\|b_{2}\|_{\ast}M_{r,(6)}\bigl([b_{1},I_{\alpha,2}](g_{1},g_{2}) \bigr) (x) +\|b_{1}\|_{\ast}\|b_{2} \|_{\ast}M^{(\alpha /2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha/2)}_{p_{2},(5)}g_{2}(x) \bigr]. \end{aligned}$$

For $G_{3}$ to $G_{6}$, in the same way as $E_{5}(z)$, it yields

$$ G_{3}+G_{4}+G_{5}+G_{6} \leq C\|b_{1}\|_{\ast}\|b_{2}\|_{\ast}M^{(\alpha /2)}_{p_{1},(5)}g_{1}(x)M^{(\alpha/2)}_{p_{2},(5)}g_{2}(x). $$

(2.10)

Thus (2.6) is obtained from (2.9) to (2.10) and (2.2) is proved. Also, one obtains (2.3) and (2.4) in a similar way to (2.2). Here the details is omitted. Thus Lemma 2.4 is proved. □

Proof of Theorem 1.10

Set $0<\alpha<2$, $1< p_{1}, p_{2} <\infty$, $0<\frac{1}{q}=\frac{1}{p_{1}}+\frac{1}{p_{2}}-\alpha<1$, $1< r< q$, $f_{1}\in L^{p_{1}}(\mu)$, $g_{2}\in L^{p_{2}}(\mu)$, $b_{1}, b_{2}\in \operatorname{RBMO}(\mu)$. Noticing that $|g(x)|\leq Ng(x)$, recalling the boundedness of $M^{(\alpha/2)}_{r,(\rho)}$ and $M_{r,(\rho)}$, for $\rho\geq5$ and $r< q$, and using Hölder’s inequality, it follows from Lemmas 2.1-2.4 and Theorem 1.9 that

$$\begin{aligned}& \bigl\Vert [b_{1},b_{2},I_{\alpha,2}](g_{1},g_{2}) \bigr\Vert _{L^{q}(\mu)} \\& \quad \leq \bigl\Vert N\bigl([b_{1},b_{2},I_{\alpha,2}](g_{1},g_{2}) \bigr)\bigr\Vert _{L^{q}(\mu)} \\& \quad \leq C\bigl\Vert M^{\sharp,(\alpha/2)}\bigl([b_{1},b_{2},I_{\alpha ,2}](g_{1},g_{2}) \bigr)\bigr\Vert _{L^{q}(\mu)} \\& \quad \leq C\Vert b_{1}\Vert _{\ast} \Vert b_{2}\Vert _{\ast}\bigl\Vert M_{r,(6)} \bigl(I_{\alpha ,2}(g_{1},g_{2})\bigr)\bigr\Vert _{L^{q}(\mu)} \\& \qquad {}+C\Vert b_{1}\Vert _{\ast}\bigl\Vert M_{r,(6)}\bigl([b_{2},I_{\alpha,2}](g_{1},g_{2}) \bigr)\bigr\Vert _{L^{q}(\mu)} \\& \qquad {}+C\Vert b_{2}\Vert _{\ast}\bigl\Vert M_{r,(6)}\bigl([b_{1},I_{\alpha,2}](g_{1},g_{2}) \bigr)\bigr\Vert _{L^{q}(\mu)} \\& \qquad {}+C\Vert b_{1}\Vert _{\ast} \Vert b_{2}\Vert _{\ast}\bigl\Vert M^{(\alpha /2)}_{p_{1},(5)}g_{1}M^{(\alpha/2)}_{p_{2},(5)}g_{2} \bigr\Vert _{L^{q}(\mu)} \\& \quad \leq C\Vert b_{1}\Vert _{\ast} \Vert b_{2}\Vert _{\ast} \Vert g_{1}\Vert _{L^{p_{1}}(\mu)}\Vert g_{2}\Vert _{L^{p_{2}}(\mu)} \\& \qquad {}+C\Vert b_{1}\Vert _{\ast}\bigl\Vert [b_{2},I_{\alpha,2}](g_{1},g_{2})\bigr\Vert _{L^{q}(\mu )} \\& \qquad {}+C\Vert b_{2}\Vert _{\ast}\bigl\Vert [b_{1},I_{\alpha,2}](g_{1},g_{2})\bigr\Vert _{L^{q}(\mu)} \\& \quad \leq C\Vert b_{1}\Vert _{\ast} \Vert b_{2}\Vert _{\ast} \Vert g_{1}\Vert _{L^{p_{1}}(\mu)}\Vert g_{2}\Vert _{L^{p_{2}}(\mu)} \\& \qquad {}+C\Vert b_{1}\Vert _{\ast}\bigl\Vert M^{\sharp,(\alpha/2)}\bigl([b_{2},I_{\alpha ,2}](g_{1},g_{2}) \bigr)\bigr\Vert _{L^{q}(\mu)} \\& \qquad {}+C\Vert b_{2}\Vert _{\ast}\bigl\Vert M^{\sharp,(\alpha/2)}\bigl([b_{1},I_{\alpha ,2}](g_{1},g_{2}) \bigr)\bigr\Vert _{L^{q}(\mu)} \\& \quad \leq \Vert b_{1}\Vert _{\ast} \Vert b_{2}\Vert _{\ast} \Vert g_{1}\Vert _{L^{p_{1}}(\mu)}\Vert g_{2}\Vert _{L^{p_{2}}(\mu)} \\& \qquad {}+C\Vert b_{1}\Vert _{\ast}\bigl\Vert M_{r,(6)}\bigl(I_{\alpha,2}(g_{1},g_{2})\bigr) \bigr\Vert _{L^{q}(\mu )} \\& \qquad {}+C\Vert b_{1}\Vert _{\ast}\bigl\Vert M^{(\alpha/2)}_{p_{1},(5)}g_{1}M^{(\alpha /2)}_{p_{2},(5)}g_{2} \bigr\Vert _{L^{q}(\mu)} \\& \qquad {}+C\Vert b_{2}\Vert _{\ast}\bigl\Vert M_{r,(6)}\bigl(I_{\alpha,2}(g_{1},g_{2})\bigr) \bigr\Vert _{L^{q}(\mu )} \\& \qquad {}+C\Vert b_{2}\Vert _{\ast}\bigl\Vert M^{(\alpha/2)}_{p_{1},(5)}g_{1}M^{(\alpha /2)}_{p_{2},(5)}g_{2} \bigr\Vert _{L^{q}(\mu)} \\& \quad \leq C\Vert b_{1}\Vert _{\ast} \Vert b_{2}\Vert _{\ast} \Vert g_{1}\Vert _{L^{p_{1}}(\mu)}\Vert g_{2}\Vert _{L^{p_{2}}(\mu)}. \end{aligned}$$

This proves Theorem 1.10. □

3 Applications

In this section, we apply Theorem 1.9 and Theorem 1.10 to the study of a fractional integral operator.

Lemma 3.1

[15]

Suppose $\operatorname{diam}(X)=\infty$, $\alpha\in(0,1)$, $p\in(1,1/\alpha)$, and $1/q=1/p-\alpha$. If λ satisfies the ϵ-weak reverse doubling condition, for some $\epsilon\in(0,\min\{\alpha,1-\alpha,1/q\})$, then

$$\|T_{\alpha}f\|_{L^{q}(\mu)}\leq C \|f\|_{L^{p}(\mu)}, $$

where $T_{\alpha}$ is defined by

$$ T_{\alpha} f(x):= \int_{X} \frac{f(y)}{[\lambda(y,d(x,y))]^{1-\alpha}} \,d\mu(y). $$

Theorem 3.2

Under the same conditions as that in Lemma  3.1, the results of Theorem  1.9 and Theorem  1.10 hold true, on replacing $I_{\beta}$ there by $T_{\alpha}$.

4 Conclusion

In this paper, we prove that multilinear fractional integral operators and commutators, generated by multilinear fractional integrals, with an $\operatorname{RBMO}(\mu)$ function on non-homogeneous metric measure spaces, are bounded in Lebesgue spaces. The results are established for both the homogeneous spaces and the non-doubling measure spaces.