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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.

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 [18]). 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., [1025].

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

  1. (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.

  2. (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.

  3. (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 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.

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. □

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 \(T_{\alpha}\).

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.

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Acknowledgements

The authors would like to thank the referees for their careful reading and valuable suggestions. Huajun Gong is partially supported by Posted-doctor Foundation of China (No. 2015M580728) and Shenzhen University (No. 2014-62). Rulong Xie is partially supported by NSF of Anhui Province (No. 1608085QA12) and NSF of Education Committee of Anhui Province (No. KJ2016A506). Chen Xu is partially supported by NSFC (No. 61472257), Guangdong Provincial Science and Technology Plan Project (No. 2013B040403005) and The HD Video R and D Platform for Intelligent Analysis and Processing in Guangdong Engineering Technology Research Center of Colleges and Universities (No. GCZX-A1409).

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Correspondence to Rulong Xie.

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The authors declare that they have no competing interests.

Authors’ contributions

HG and RX proposed this problem and finished the proof of Theorem 1.9 and Theorem 1.10 together. CX finished the proof of Lemma 2.4. All authors read and approved the final manuscript.

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Gong, H., Xie, R. & Xu, C. Multilinear fractional integral operators on non-homogeneous metric measure spaces. J Inequal Appl 2016, 275 (2016). https://doi.org/10.1186/s13660-016-1218-6

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  • DOI: https://doi.org/10.1186/s13660-016-1218-6

MSC

  • 42B25
  • 47B47

Keywords

  • multilinear fractional integrals
  • commutators
  • non-homogeneous metric measure spaces