Multilinear strongly singular integral operators on non-homogeneous metric measure spaces

Abstract

Let $(X,d,\mu )$ be a non-homogeneous metric measure space satisfying the geometrically and upper doubling measure conditions. In this paper, the boundedness in Lebesgue spaces for multilinear strongly singular integral operators on non-homogeneous metric measure spaces is proved. As an application, the boundedness in Morrey spaces for multilinear strongly singular integral operators is also obtained.

1 Introduction and main results

It is well known that a metric measure space $(X,d,\mu )$ equipped with a non-negative doubling measure μ is called a space of homogeneous type. μ is said to satisfy the doubling condition if there exists a constant $C>0$ such that $\mu (B(x,2r))\leq C\mu (B(x,r))$ for all $x\in \operatorname{supp} \mu$ and $r>0$. In the case of non-doubling measures, a non-negative measure μ only should satisfy the polynomial growth condition, i.e., for all $x \in R^{n}$ and $r>0$, there exists a constant $C_{0}>0$ and $k\in (0, n]$ such that

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

(1.1)

where $B(x, r)= \{y \in R^{n}: |y-x| < r\}$. This breakthrough brings rapid development in harmonic analysis (see [14, 15, 31, 34, 35, 37, 38] and their therein). And the analysis of non-doubling measures has important applications in solving the long-standing open Painlevé problem (see [35]).

Hytönen [17] stated that the measure satisfying (1.1) does not include the doubling measure as a special case. He introduced non-homogeneous metric measure spaces $(X,d,\mu )$, satisfying the geometrically and upper doubling measure conditions (see Definition 1.1 and 1.2). The highlight of this kind of spaces is that it includes both the homogeneous and metric spaces with polynomial growth measures as special cases. From then on, some results on non-homogeneous metric measure spaces were obtained. Hytönen et al. [20] and Bui and Duong [3] independently introduced the atomic Hardy space $H^{1}(\mu )$ and proved that the dual space of $H^{1}(\mu )$ is $\operatorname{RBMO}(\mu )$. In [3], the authors also proved that the Calderón-Zygmund operator and commutators of the Calderón-Zygmund operators and RBMO functions are bounded in $L^{p}(\mu )$ for $1< p<+\infty$. Recently, some equivalent characterizations have been established by Liu et al. [29] for the boundedness of Carderón-Zygmund operators on $L^{p}(\mu )$ for $1< p<+\infty$. In [9], Fu et al. established boundedness of multilinear commutators of the Calderón-Zygmund operators on the Orlicz spaces on non-homogeneous spaces. More results on non-homogeneous metric measure spaces have also been obtained in [4, 5, 10, 11, 18–24] and the references therein.

Some researchers considered the theory of multilinear singular integral operators; for example, in [7], Coifman and Meyers firstly established the theory of bilinear Calderón-Zygmund operators. Later, Grafakos and Torres [12, 13] demonstrated the boundedness of multilinear singular integral on the product Lebesgue spaces and Hardy spaces. The boundedness of multilinear singular integrals and commutators on non-doubling measures spaces $(R^{n},\mu )$ was established by Xu [38, 39]. Weighted norm inequalities for multilinear Calderón-Zygmund operators on non-homogeneous metric measure spaces were also constructed in [16]. The boundedness for commutators of multilinear Calderón-Zygmund operators and multilinear fractional integral operators on non-homogeneous metric measure spaces was also established in [11, 36].

The introduction of the strongly singular integral operator is motivated by a class of multiplier operators whose symbol is given by $e^{i|\xi |\alpha } /|\xi |\beta$ away from the origin, where $0<\alpha <1$ and $\beta >0$. Fefferman and Stein [8] enlarged the multiplier operators onto a class of convolution operators. Coifman [6] also considered a related class of operators for $n = 1$. The strongly singular non-convolution operators were introduced and researched by Alvarez and Milman [1, 2], whose properties are similar to those of Calderón-Zygmund operators, but the kernel is more singular near the diagonal than those of the standard case. Furthermore, Lin and Lu [25–28] obtained the boundedness for the strongly singular integral and its commutators on Lebesgue spaces, Morrey spaces, and Hardy spaces.

In this paper, we first introduce the multilinear strongly singular integral operators on non-homogeneous metric spaces. Then we will also prove that it is bounded in m-multiple Lebesgue spaces, provided that multilinear strongly singular integrals are bounded from m-multiple $L^{1}(\mu )\times \cdots \times L^{1}(\mu )$ to $L^{1/m,\infty }(\mu )$, where $L^{p}(\mu )$ and $L^{p,\infty }(\mu )$ denote the Lebesgue space and weak Lebesgue space, respectively. As an application, the boundedness in Morrey spaces for multilinear strongly singular integral on non-homogeneous metric spaces is obtained. A variant of sharp maximal operator $M^{\sharp }$, Kolmogorov’s theorem and some good properties of the dominating function λ (see Definition 1.2) are the main tools for proving the results in this paper.

Before stating the main results of this paper, we first recall some notations and definitions.

Definition 1.1

([17])

A metric space $(X,d)$ is called geometrically doubling if there exists some $N_{0}\in \mathbf{N}$ such that, for any ball $B(x,r)\subset X$, there exists a finite ball covering $\{B(x_{i},r/2)\}_{i}$ of $B(x,r)$ such that the cardinality of this covering is at most $N_{0}$.

Definition 1.2

([17])

A metric measure space $(X,d,\mu )$ is said to be upper doubling if μ is a Borel measure on X and there exists a dominating function $\lambda : X \times (0,+\infty ) \rightarrow (0,+\infty )$, and a constant $C_{\lambda } >0$ such that for each $x\in X, r \longmapsto (x,r)$ is non-decreasing, and for all $x\in X$, $r >0$,

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

(1.2)

Remark 1.3

(i) A space of homogeneous type is a special case of upper doubling spaces, where one can take the dominating function $\lambda (x, r)= \mu (B(x,r))$. On the other hand, a metric space $(X,d,\mu )$ satisfying the polynomial growth condition (1.1) (in particular, $(X,d,\mu )=(R^{n}, |\cdot |, \mu )$ with μ satisfying (1.1) for some $k\in (0, n]$)) is also an upper doubling measure space if we take $\lambda (x, r)=Cr^{k}$.

(ii) Let $(X,d,\mu )$ be an upper doubling space and λ be a dominating function on $X \times (0,+\infty )$ as in Definition 1.2. In [20], it was showed that there exists another dominating function λ̃ such that for all $x, y \in X$ with $d(x, y)\leq r$,

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

(1.3)

Thus, in this paper, we suppose that λ always satisfies (1.3).

Definition 1.4

([3])

Let $1<\alpha ,\beta <+\infty$. A ball $B\subset X$ is called $(\alpha , \beta )$-doubling if $\mu (\alpha B)\leq \beta \mu (B)$.

Remark 1.5

As pointed out in Lemma 2.3 in [3], there exist plenty of doubling balls with small radii and with large radii. For the rest of this paper, unless α and β are specified otherwise, by an $(\alpha ,\beta )$ doubling ball, we mean a $(6,\beta _{0})$-doubling with a 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.

Definition 1.6

([3])

Let $N_{B,Q}$ be the smallest integer satisfying $6^{N_{B,Q}}r_{B}\geq r_{Q}$ denote

$$ K_{B,Q} = 1+\sum_{k=1}^{N_{B,Q}} \frac{\mu (6^{k}B)}{\lambda (x_{B},6^{k}r_{B})}, $$

(1.4)

where $x_{B}$ and $r_{B}$ are center and radius of B, respectively.

Let us first introduce m-linear strongly singular integral kernel.

Definition 1.7

A kernel $K(\cdot ,\ldots ,\cdot )\in L_{\mathrm{loc}}^{1}((X)^{m+1}\backslash \{(x,y_{1} \cdots ,y_{j},\ldots ,y_{m}):x=y_{1}=\cdots =y_{j}=\cdots =y_{m}\})$ is called an m-linear strongly singular integral kernel if it satisfies:

(i)

$$ \bigl\vert K(x,y_{1},\ldots ,y_{j}, \ldots , y_{m}) \bigr\vert \leq C \Biggl[\sum _{j=1}^{m} \lambda \bigl(x,d(x,y_{j}) \bigr) \Biggr]^{-m} $$

(1.5)

for all $(x,y_{1}\cdots ,y_{j},\ldots ,y_{m})\in (X)^{m+1}$ with $x\neq y_{j}$ for some j.

(ii) There exist $0<\alpha <1$ and $0<\delta \leq 1$ such that

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

(1.6)

provided that $\operatorname{Cd}(x,x')^{\alpha }\leq \max_{1\leq j\leq m}d(x,y_{j})$ and for each j,

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

(1.7)

provided that $\operatorname{Cd}(y_{j},y'_{j})^{\alpha }\leq \max_{1\leq j\leq m}d(x,y_{j})$.

A multilinear operator T is called a multilinear strongly singular integral operator with the above kernel K, satisfying (1.5), (1.6), and (1.7), if for $f_{1},\ldots f_{m}$ are $L^{\infty }$ functions with compact support and $x\notin \bigcap_{j=1}^{m}\text{supp}f_{j}$,

$$ T(f_{1},\ldots f_{m}) (x)= \int _{X^{m}}K(x,y_{1},\ldots y_{m})f_{1}(y_{1}) \cdots f_{m}(y_{m})\,d\mu (y_{1})\cdots \,d\mu (y_{m}). $$

(1.8)

Definition 1.8

([4])

Let $k>1$ and $1\leq q\leq p<+\infty$. The Morrey spaces are defined by

$$ M^{p}_{q}(k,\mu ):=\bigl\{ f\in L^{q}_{\mathrm{loc}}( \mu ): \Vert f \Vert _{M^{p}_{q}(k,\mu )}< + \infty \bigr\} ,$$

where

$$ \Vert f \Vert _{M^{p}_{q}(k,\mu )}=\sup_{B\in X}\mu (kB)^{\frac{1}{p}- \frac{1}{q}} \biggl( \int _{B} \vert f \vert ^{q}\,d\mu \biggr)^{\frac{1}{q}}.$$

Remark 1.9

The definition of the Morrey spaces are independent of the constant $k>1$, and the norms are equivalent for different $k>1$, see [4, 30, 32, 33, 40].

Definition 1.10

([10])

Let $\epsilon \in (0,\infty )$. A dominating function λ is said to satisfy the ϵ-weak reverse doubling condition if, for all $r\in (0, 2 \operatorname{diam}(X))$ and $a\in (1, 2 \operatorname{diam}(X)/r)$, there exists a number $C(a)\in [1,\infty )$, depending only on a and X, such that for all $x\in X$,

$$ \lambda (x, ar)\geq C(a)\lambda (x, r) $$

(1.9)

and

$$ \sum_{k=1}^{\infty }\frac{1}{[ C(a)]^{k}}< \infty . $$

(1.10)

For the sake of simplicity and without loss of generality, we only consider the case of $m=2$ in this paper. Let us state the main result as follows.

Theorem 1.11

Let T be defined by (1.8). Assume $1< p_{1},p_{2},q<+\infty$ and $f_{1}\in L^{p_{1}}(\mu )$, $f_{2}\in L^{p_{2}}(\mu )$ with $\int _{X}T(f_{1},f_{2})(x)\,d\mu (x)=0$ if $\|\mu \|:=\mu (X)<\infty$. If T is bounded from $L^{1}(\mu )\times L^{1}(\mu )$ to $L^{1/2,\infty }(\mu )$, then there exists a constant $C>0$ such that

$$ \bigl\Vert T(f_{1},f_{2}) \bigr\Vert _{L^{q}(\mu )}\leq C \Vert f_{1} \Vert _{L^{p_{1}}(\mu )} \Vert f_{2} \Vert _{L^{p_{2}}( \mu )}, $$

(1.11)

where $\frac{1}{q}=\frac{1}{p_{1}}+\frac{1}{p_{2}}$.

As an application of the main result in this paper, we obtain the following result.

Theorem 1.12

Let T be defined by (1.8). Assume that $1< q_{1}\leq p_{1}<+\infty$, $1< q_{2}\leq p_{2}<+\infty$, and $f_{1}\in L^{p_{1}}(\mu )$, $f_{2}\in L^{p_{2}}(\mu )$ with $\int _{X}T(f_{1},f_{2})(x)\,d\mu (x)=0$ if $\|\mu \|<\infty$. Assume that λ satisfies the ϵ-weak reverse doubling condition. If T is bounded from $L^{1}(\mu )\times L^{1}(\mu )$ to $L^{1/2,\infty }(\mu )$, then there exists a constant $C>0$ such that

$$ \bigl\Vert T(f_{1},f_{2}) \bigr\Vert _{M^{p}_{q}(\mu )}\leq C \Vert f_{1} \Vert _{M^{p_{1}}_{q_{1}}( \mu )} \Vert f_{2} \Vert _{M^{p_{2}}_{q_{2}}(\mu )}, $$

(1.12)

where $\frac{1}{q}=\frac{1}{q_{1}}+\frac{1}{q_{2}}$, $\frac{1}{p}=\frac{1}{p_{1}}+\frac{1}{p_{2}}$.

Throughout the paper, C denotes a positive constant independent of the main parameters involved, but it may be different in different places.

2 Proof of main results

To prove Theorem 1.11, we first give some notations and lemmas.

Let $f \in L_{\mathrm{loc}}^{1}(\mu )$, the sharp maximal operator is defined as follows

$$ M^{\sharp }f(x)=\sup_{B\ni x}\frac{1}{\mu (6B)} \int _{B} \bigl\vert f(y)-m_{ \widetilde{B}}(f) \bigr\vert \,d\mu (y) +\sup_{(B,Q)\in \Delta _{x}} \frac{ \vert m_{B}(f)-m_{Q}(f) \vert }{K_{B,Q}}, $$

(2.1)

where $\Delta _{x}:=\{(B,Q):x\in B\subset Q\ \text{and}\ B,\ Q\ \text{are doubling balls}\}$. And the non-centered doubling maximal operator is denoted as follows

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

For any $0<\delta < 1$, we define that

$$ M^{\sharp }_{\delta }f(x)=\bigl\{ M^{\sharp }\bigl( \vert f \vert ^{\delta }\bigr) (x)\bigr\} ^{1/\delta },\qquad N_{ \delta }f(x)= \bigl\{ N\bigl( \vert f \vert ^{\delta }\bigr) (x)\bigr\} ^{1/\delta }. $$

We can obtain that for any $f\in L^{1}_{\mathrm{loc}}(\mu )$,

$$ \bigl\vert f(x) \bigr\vert \leq N_{\delta }f(x) $$

(2.2)

for $\mu -a.e. \ x \in X$ (see [36]).

Let $\rho >1$ and $1< p<\infty$, the non-centered maximal operator $M_{(\rho )}f$ is defined as follows

$$ M_{(\rho )}f(x)=\sup_{B\ni x} \biggl\{ \frac{1}{\mu (\rho B)} \int _{B} \bigl\vert f(y) \bigr\vert \,d\mu (y) \biggr\} . $$

(2.3)

The operator $M_{(\rho )}f$ is bounded on $L^{p}(\mu )$ for $\rho \geq 5$ and $p>1$ (see [3]).

Lemma 2.1

([3, 36])

Let $f \in L^{1}_{\mathrm{loc}}(\mu )$ with $\int _{X}f(x)\,d\mu (x)=0$ if $\|\mu \|<\infty$. For $1< p<\infty$ and $0<\delta <1$, if $\inf (1,N_{\delta }f)\in L^{p}(\mu )$, then there exists a constant $C>0$ such that

$$ \bigl\Vert N_{\delta }(f) \bigr\Vert _{L^{p}(\mu )}\leq C \bigl\Vert M_{\delta }^{\sharp }(f) \bigr\Vert _{L^{p}( \mu )}. $$

(2.4)

Lemma 2.2

([38])

Let $(X,d,\mu )$ be probability measure spaces and let $0< s< t<+\infty$, then there exists a constant C such that for any measurable function f,

$$ \Vert f \Vert _{L^{s}(\mu )}\leq C \Vert f \Vert _{L^{t,\infty }(\mu )}. $$

(2.5)

Lemma 2.3

Suppose that T is defined by (1.8), $0<\delta <1/2$, $1< p_{1},p_{2},q<\infty$. If T is bounded from $L^{1}(\mu )\times L^{1}(\mu )$ to $L^{1/2,\infty }(\mu )$, then there exists a constant $C>0$ such that

$$ M_{\delta }^{\sharp }T(f_{1},f_{2}) (x)\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

(2.6)

Proof of Theorem 1.11

Because $L^{\infty }(\mu )$ with compact support is dense in $L^{p}(\mu )$ for $1< p<\infty$, we only consider the situation of $f_{1},f_{2}\in L^{\infty }(\mu )$ with compact support. Let $l(B):=\sup_{x,y\in B}d(x,y)$. Next, we consider two cases for proving the result.

Case 1: $\boldsymbol{l(B)=l\geq 1}$. As in the proof of Theorem 9.1 in [34], to obtain (2.6), it suffices to show that

$$ \biggl(\frac{1}{\mu (6B)} \int _{B} \bigl\vert { \bigl\vert T(f_{1},f_{2}) (z) \bigr\vert }^{\delta }- \vert h_{B} \vert ^{ \delta } \bigr\vert \,d\mu (z) \biggr)^{1/\delta }\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x) $$

(2.7)

holds for any x and ball B with $x\in B$, and

$$ \vert h_{B}-h_{Q} \vert \leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x) $$

(2.8)

for all balls $B\subset Q$ with $x\in B$, where B is an arbitrary ball, and Q is a doubling ball. For any ball B, we denote

$$ h_{B}:= m_{B}\bigl(T\bigl(f_{1}^{0},f_{2}^{\infty } \bigr)+T\bigl(f_{1}^{\infty },f_{2}^{0} \bigr)+T\bigl(f_{1}^{ \infty },f_{2}^{\infty } \bigr)\bigr),$$

and

$$ h_{Q}:= m_{Q}\bigl(T\bigl(f_{1}^{0},f_{2}^{\infty } \bigr)+T\bigl(f_{1}^{\infty },f_{2}^{0} \bigr)+T\bigl(f_{1}^{ \infty },f_{2}^{\infty } \bigr)\bigr),$$

where we split each $f_{i}$ as $f_{i}=f_{i}^{0}+f_{i}^{\infty }$, $f_{i}^{0}=f_{i}\chi _{\frac{6}{5}B}$ and $f_{i}^{\infty }=f_{i}-f_{i}^{0}$, $i=1,2$. Since

$$ \begin{aligned} \bigl\vert T(f_{1},f_{2}) (y) \bigr\vert &\leq \bigl\vert T\bigl(f_{1}^{0},f_{2}^{0} \bigr) (y) \bigr\vert + \bigl\vert T\bigl(f_{1}^{0},f_{2}^{ \infty } \bigr) (y) \bigr\vert + \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{0} \bigr) (y) \bigr\vert + \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{ \infty } \bigr) (y) \bigr\vert , \end{aligned} $$

then

$$\begin{aligned}& \biggl(\frac{1}{\mu (6B)} \int _{B} \bigl\vert { \bigl\vert T(f_{1},f_{2}) (z) \bigr\vert }^{ \delta }- \vert h_{B} \vert ^{\delta } \bigr\vert \,d\mu (z) \biggr)^{1/\delta } \\& \quad \leq C \biggl(\frac{1}{\mu (6B)} \int _{B} \bigl\vert T(f_{1},f_{2}) (z)-h_{B} \bigr\vert ^{ \delta }\,d\mu (z) \biggr)^{1/\delta } \\& \quad \leq C \biggl(\frac{1}{\mu (6B)} \int _{B} \bigl\vert T\bigl(f_{1}^{0},f_{2}^{0} \bigr) (z) \bigr\vert ^{ \delta }\,d\mu (z) \biggr)^{1/\delta } \\& \qquad {}+C \biggl(\frac{1}{\mu (6B)} \int _{B} \bigl\vert T\bigl(f_{1}^{0},f_{2}^{\infty } \bigr) (z)-T\bigl(f_{1}^{0},f_{2}^{ \infty } \bigr) (y) \bigr\vert ^{\delta }\,d\mu (z) \biggr)^{1/\delta } \\& \qquad {} +C \biggl(\frac{1}{\mu (6B)} \int _{B} \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{0} \bigr) (z)-T\bigl(f_{1}^{ \infty },f_{2}^{0} \bigr) (y) \bigr\vert ^{\delta }\,d\mu (z) \biggr)^{1/\delta } \\& \qquad {} +C \biggl(\frac{1}{\mu (6B)} \int _{B} \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{\infty } \bigr) (z)-T\bigl(f_{1}^{ \infty },f_{2}^{\infty } \bigr) (y) \bigr\vert ^{\delta }\,d\mu (z) \biggr)^{1/\delta } \\& \quad =:I_{1}+I_{2}+I_{3}+I_{4}. \end{aligned}$$

First, we estimate $I_{1}$. Applying Lemma 2.2 with $s=\delta$ ($0<\delta <1/2$) and $t=1/2$, we get

$$\begin{aligned} I_{1}&\leq C \biggl(\frac{1}{\mu (6B)} \int _{B} \bigl\vert T\bigl(f_{1}^{0},f_{2}^{0} \bigr) (z) \bigr\vert ^{ \delta }\,d\mu (z) \biggr)^{1/\delta } \\ &\leq C \biggl( \frac{\mu (B)}{\mu (6B)} \biggr)^{1/\delta } \bigl\Vert T(f_{1} \chi _{\frac{6}{5}B},f_{2}\chi _{\frac{6}{5}B}) \bigr\Vert _{L^{1/2,\infty }(B, \frac{\,d\mu }{\mu (B)})} \\ &\leq C \biggl( \frac{\mu (B)}{\mu (6B)} \biggr)^{1/\delta }\prod _{i=1}^{2} \frac{1}{\mu (B)} \int _{\frac{6}{5}B} \bigl\vert f_{i}(z_{i}) \bigr\vert \,d\mu (z_{i}) \\ &\leq C \biggl( \frac{\mu (B)}{\mu (6B)} \biggr)^{1/\delta }\prod _{i=1}^{2} \biggl[\frac{\mu (5\times \frac{6}{5}B)}{\mu (B)} \frac{1}{\mu (5\times \frac{6}{5}B)} \int _{\frac{6}{5}B} \bigl\vert f_{i}(z_{i}) \bigr\vert \,d\mu (z_{i}) \biggr] \\ &\leq C \biggl( \frac{\mu (B)}{\mu (6B)} \biggr)^{1/\delta -2} M_{(5)}f_{1}(x)M_{(5)}f_{2}(x) \\ &\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned}$$

For $I_{2}$, let $z,y\in B$, $z_{1}\in \frac{6}{5}B$ and $z_{2}\in X\backslash \frac{6}{5}B$, then $\max_{1\leq i\leq 2}d(z,z_{i})\geq d(z,z_{2})\geq Cl(B) \geq Cl(B)^{\alpha }\geq \operatorname{Cd}(z,y)^{\alpha }$. By Definition 1.7 and the properties of λ, we deduce

$$\begin{aligned}& \bigl\vert T\bigl(f_{1}^{0},f_{2}^{\infty } \bigr) (z)-T\bigl(f_{1}^{0},f_{2}^{ \infty } \bigr) (y) \bigr\vert \\& \quad \leq C \int _{X\backslash \frac{6}{5}B} \int _{ \frac{6}{5}B} \frac{1}{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}} \frac{d(z,y)^{\delta }}{[\sum_{i=1}^{2}d(z,z_{i})]^{\delta /\alpha }} \bigl\vert f_{1}(z_{1}) \bigr\vert \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad \leq C \int _{X\backslash \frac{6}{5}B} \int _{ \frac{6}{5}B} \frac{1}{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}} \frac{d(z,y)^{\delta }}{[d(z,z_{2})]^{\delta /\alpha }} \bigl\vert f_{1}(z_{1}) \bigr\vert \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad \leq C \int _{\frac{6}{5}B} \frac{ \vert f_{1}(z_{1}) \vert }{\lambda (z,d(z,z_{1}))}\,d\mu (z_{1})\sum _{k=1}^{ \infty } \int _{6^{k+1}\frac{6}{5}B\backslash 6^{k}\frac{6}{5}B} \frac{1}{\lambda (z,d(z,z_{2}))} \frac{d(z,y)^{\delta }}{[d(z,z_{2})]^{\delta /\alpha }} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\& \quad \leq C \frac{\mu (5\times \frac{6}{5}B)}{\lambda (x_{B},\frac{6}{5}r_{B})} \frac{1}{\mu (5\times \frac{6}{5}B)} \int _{\frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\& \qquad {} \times \sum_{k=1}^{\infty }6^{-k\delta /\alpha }l^{\delta (1-1/ \alpha )} \frac{\mu (5\times 6^{k+1}\frac{6}{5}B)}{\lambda (x_{B},6^{k}\frac{6}{5}r_{B})} \frac{1}{\mu (5\times 6^{k+1}\frac{6}{5}B)} \int _{6^{k+1}\frac{6}{5}B} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\& \quad \leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned}$$

Therefore,

$$ I_{2}\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x).$$

Similar to estimate $I_{2}$, we immediately obtain

$$ I_{3}\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x).$$

Let us move on to $I_{4}$ estimate. By Definition 1.7, we have

$$\begin{aligned}& \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{\infty } \bigr) (z)-T\bigl(f_{1}^{\infty },f_{2}^{ \infty } \bigr) (y) \bigr\vert \\& \quad \leq C \int _{X\backslash \frac{6}{5}B} \int _{X\backslash \frac{6}{5}B} \frac{ \vert f_{1}(z_{1}) \vert \vert f_{2}(z_{2}) \vert }{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}} \frac{d(z,y)^{\delta }}{[\sum_{i=1}^{2}d(z,z_{i})]^{\delta /\alpha }}\,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad \leq C\sum_{k=1}^{\infty } \int _{6^{k+1}\frac{6}{5}B\backslash 6^{k} \frac{6}{5}B}\sum_{j=1}^{\infty } \int _{6^{j+1}\frac{6}{5}B \backslash 6^{j}\frac{6}{5}B} \frac{ \vert f_{1}(z_{1}) \vert \vert f_{2}(z_{2}) \vert }{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}} \\& \qquad {}\times\frac{d(z,y)^{\delta }}{[\sum_{i=1}^{2}d(z,z_{i})]^{\delta /\alpha }}\,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad \leq C\sum_{k=1}^{\infty }\sum _{j=k}^{\infty } \int _{6^{k+1} \frac{6}{5}B\backslash 6^{k}\frac{6}{5}B} \bigl\vert f_{2}(z_{2}) \bigr\vert \int _{6^{j+1} \frac{6}{5}B\backslash 6^{j}\frac{6}{5}B} \frac{ \vert f_{1}(z_{1}) \vert }{[\lambda (z,d(z,z_{1}))]^{2}}\\& \qquad {}\times \frac{d(z,y)^{\delta }}{[d(z,z_{1})]^{\delta /\alpha }}\,d\mu (z_{1})\,d\mu (z_{2}) \\& \qquad {} +C\sum_{k=1}^{\infty }\sum _{j=1}^{k-1} \int _{6^{k+1}\frac{6}{5}B \backslash 6^{k}\frac{6}{5}B} \frac{ \vert f_{2}(z_{2}) \vert }{[\lambda (z,d(z,z_{2}))]^{2}} \frac{d(z,y)^{\delta }}{[d(z,z_{2})]^{\delta /\alpha }} \\& \qquad {}\times\int _{6^{j+1} \frac{6}{5}B\backslash 6^{j}\frac{6}{5}B}f_{1}(z_{1})\,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad =:I_{41}+I_{42}. \end{aligned}$$

To estimate $I_{41}$, we take into account the properties of λ and the fact of $r\geq 1$ and $0<\alpha <1$, so we have

$$\begin{aligned} I_{41} \leq {}& C\sum _{j=1}^{\infty } \biggl( \frac{1}{[\lambda (x_{B},6^{j}\frac{6}{5}r_{B})]^{2}} \int _{6^{j+1} \frac{6}{5}B\backslash 6^{j}\frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1})6^{-j \delta /\alpha }l^{\delta (1-1/\alpha )} \biggr) \\ &{} \times \sum_{k=1}^{j} \int _{6^{k+1}\frac{6}{5}B\backslash 6^{k} \frac{6}{5}B} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\ \leq {}& C\sum_{j=1}^{\infty } \biggl( \frac{1}{[\lambda (x_{B},5\times 6^{j}\frac{6}{5}r_{B})]^{2}} \int _{6^{j+1} \frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1})6^{-j\delta /\alpha }l^{\delta (1-1/ \alpha )} \biggr) \\ &{} \times \int _{6^{j+1}\frac{6}{5}B} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\ \leq{} & C\sum_{j=1}^{\infty }6^{-j\delta /\alpha }l^{\delta (1-1/ \alpha )} \frac{1}{\mu (5\times 6^{j+1}\frac{6}{5}B)} \int _{6^{j+1} \frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\ &{} \times \frac{1}{\mu (5\times 6^{j+1}\frac{6}{5}B)} \int _{6^{j+1} \frac{6}{5}B} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\ \leq {}& CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned}$$

For $I_{42}$, by the properties of λ, we have

$$\begin{aligned} I_{42} \leq{} & C\sum _{k=1}^{\infty } \biggl( \frac{1}{[\lambda (x_{B},6^{k}\frac{6}{5}r_{B})]^{2}} \int _{6^{k+1} \frac{6}{5}B\backslash 6^{k}\frac{6}{5}B} \bigl\vert f_{1}(z_{2}) \bigr\vert \,d\mu (z_{2})6^{-k \delta /\alpha }l^{\delta (1-1/\alpha )} \biggr) \\ &{} \times \sum_{j=1}^{k-1} \int _{6^{j+1}\frac{6}{5}B\backslash 6^{j} \frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\ \leq{} & C\sum_{k=1}^{\infty } \biggl( \frac{1}{[\lambda (x_{B},5\times 6^{k}\frac{6}{5}r_{B})]^{2}} \int _{6^{k+1} \frac{6}{5}B} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2})6^{-k\delta /\alpha }l^{\delta (1-1/ \alpha )} \biggr) \\ &{} \times \int _{6^{k}\frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\ \leq{} & C\sum_{k=1}^{\infty }6^{-k\delta /\alpha }l^{\delta (1-1/ \alpha )} \frac{1}{\mu (5\times 6^{k+1}\frac{6}{5}B)} \int _{6^{k+1} \frac{6}{5}B} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\ &{} \times \frac{1}{\mu (5\times 6^{k}\frac{6}{5}B)} \int _{6^{k} \frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\ \leq{} & CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned}$$

Thus,

$$ \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{\infty } \bigr) (z)-T\bigl(f_{1}^{\infty },f_{2}^{\infty } \bigr) (y) \bigr\vert \leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

By the above estimate, we obtain (2.7).

Next, we prove (2.8). Consider two balls $B\subset Q$ with $x\in B$, where B is an arbitrary ball, and Q is a doubling ball. Let $N=N_{B,Q}+1$, then we obtain

$$\begin{aligned} \vert h_{B}-h_{Q} \vert \leq{} & \bigl\vert m_{B}\bigl(T(f_{1}\chi _{\frac{6}{5}B},f_{2}\chi _{6^{N}\frac{6}{5}B \backslash \frac{6}{5}B})+T(f_{1} \chi _{6^{N}\frac{6}{5}B\backslash \frac{6}{5}B},f_{2}\chi _{\frac{6}{5}B})\bigr) \bigr\vert \\ &{} + \bigl\vert m_{B}\bigl(T(f_{1}\chi _{6^{N}\frac{6}{5}B\backslash \frac{6}{5}B},f_{2} \chi _{6^{N}\frac{6}{5}B\backslash \frac{6}{5}B})\bigr) \bigr\vert \\ &{} + \bigl\vert m_{B}\bigl(T(f_{1}\chi _{6^{N}\frac{6}{5}B},f_{2}\chi _{6^{N} \frac{6}{5}B}) \bigr)-m_{Q}\bigl(T(f_{1}\chi _{6^{N}\frac{6}{5}B},f_{2} \chi _{6^{N} \frac{6}{5}B})\bigr) \bigr\vert \\ &{} + \bigl\vert m_{B}\bigl(T(f_{1}\chi _{X\backslash 6^{N}\frac{6}{5}B},f_{2}\chi _{6^{N} \frac{6}{5}B}) \bigr)-m_{Q}\bigl(T(f_{1}\chi _{X\backslash 6^{N}\frac{6}{5}B},f_{2} \chi _{6^{N}\frac{6}{5}B})\bigr) \bigr\vert \\ &{} + \bigl\vert m_{B}\bigl(T(f_{1}\chi _{6^{N}\frac{6}{5}B},f_{2}\chi _{X \backslash 6^{N}\frac{6}{5}B}) \bigr)-m_{Q}\bigl(T(f_{1}\chi _{6^{N}\frac{6}{5}B},f_{2} \chi _{X\backslash 6^{N}\frac{6}{5}B})\bigr) \bigr\vert \\ &{} + \bigl\vert m_{Q}\bigl(T(f_{1}\chi _{\frac{6}{5}Q},f_{2}\chi _{6^{N}\frac{6}{5}B \backslash \frac{6}{5}Q})+T(f_{1} \chi _{6^{N}\frac{6}{5}B\backslash \frac{6}{5}Q},f_{2}\chi _{6^{N}\frac{6}{5}B})\bigr) \bigr\vert \\ &{} + \bigl\vert m_{Q}\bigl(T(f_{1}\chi _{6^{N}\frac{6}{5}B\backslash \frac{6}{5}Q},f_{2} \chi _{6^{N}\frac{6}{5}B\backslash \frac{6}{5}Q})\bigr) \bigr\vert \\ =:{}&J_{1}+J_{2}+J_{3}+J_{4}+J_{5}+J_{6}+J_{7}. \end{aligned}$$

For the estimate $J_{1}$, by the condition (1.5) in Definition 1.7 and the properties of λ, we obtain

$$\begin{aligned}& \bigl\vert T(f_{1}\chi _{\frac{6}{5}B},f_{2}\chi _{6^{N} \frac{6}{5}B\backslash \frac{6}{5}B}) (z) \bigr\vert \\& \quad \leq C \int _{6^{N}\frac{6}{5}B\backslash \frac{6}{5}B} \int _{ \frac{6}{5}B} \frac{ \vert f_{1}(z_{1}) \vert \vert f_{2}(z_{2}) \vert }{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}}\,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad \leq C \int _{\frac{6}{5}B} \frac{ \vert f_{1}(z_{1}) \vert }{\lambda (z,d(z,z_{1}))}\,d\mu (z_{1}) \int _{6^{N} \frac{6}{5}B\backslash \frac{6}{5}B} \frac{ \vert f_{2}(z_{2}) \vert }{\lambda (z,d(z,z_{2}))}\,d\mu (z_{2}) \\& \quad \leq C\frac{1}{\mu (5\times \frac{6}{5}B)} \int _{\frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1})\sum_{k=1}^{N_{B,Q}} \int _{6^{k+1}\frac{6}{5}B\backslash 6^{k} \frac{6}{5}B}\frac{ \vert f_{2}(z_{2}) \vert }{\lambda (z,d(z,z_{2}))}\,d\mu (z_{2}) \\& \quad \leq CM_{(5)}f_{1}(x)\sum_{k=1}^{N_{B,Q}} \frac{\mu (5\times 6^{k+1}\frac{6}{5}B)}{\lambda (x_{B},5\times 6^{k+1}\frac{6}{5}r_{B})} \frac{1}{\mu (5\times 6^{k+1}\frac{6}{5}B)} \int _{6^{k+1}\frac{6}{5}B} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\& \quad \leq CK_{B,Q}M_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned}$$

Also,

$$ \bigl\vert T(f_{1}\chi _{6^{N}\frac{6}{5}B\backslash \frac{6}{5}B},f_{2}\chi _{ \frac{6}{5}B}) (z) \bigr\vert \leq CK_{B,Q}M_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

Then,

$$ J_{1}\leq CK_{B,Q}M_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

Let us estimate $J_{2}$. By the condition (1.5) in Definition 1.7 and the properties of λ, we have

$$\begin{aligned}& \bigl\vert T(f_{1}\chi _{6^{N}\frac{6}{5}B\backslash \frac{6}{5}B},f_{2} \chi _{6^{N}\frac{6}{5}B\backslash \frac{6}{5}B}) \bigr\vert \\& \quad \leq C \int _{6^{N}\frac{6}{5}B\backslash \frac{6}{5}B} \int _{6^{N} \frac{6}{5}B\backslash \frac{6}{5}B} \frac{ \vert f_{1}(z_{1}) \vert \vert f_{2}(z_{2}) \vert }{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}}\,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad \leq C\sum_{k=1}^{N-1}\sum _{j=1}^{N-1} \int _{6^{k+1}\frac{6}{5}B \backslash 6^{k}\frac{6}{5}B} \int _{6^{j+1}\frac{6}{5}B\backslash 6^{j} \frac{6}{5}B} \frac{ \vert f_{1}(z_{1}) \vert \vert f_{2}(z_{2}) \vert }{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}}\,d\mu (z_{1})\,d\mu (z_{2}) \\& \qquad {} +C \int _{6\times \frac{6}{5}B\backslash \frac{6}{5}B} \int _{6 \times \frac{6}{5}B\backslash \frac{6}{5}B} \frac{ \vert f_{1}(z_{1}) \vert \vert f_{2}(z_{2}) \vert }{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}}\,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad =:J_{21}+J_{22}. \end{aligned}$$

For $J_{22}$, since $z\in B$ and $z_{i}\in 6\times \frac{6}{5}B\backslash \frac{6}{5}B$, then $\frac{1}{5}r_{B}\leq d(z,z_{i})\leq \frac{41}{5}r_{B}$ for $i=1,2$. Therefore,

$$\begin{aligned} J_{22}&\leq \frac{C}{[\lambda (z,\frac{1}{5}r_{B})]^{2}} \int _{6\times \frac{6}{5}B\backslash \frac{6}{5}B} \int _{6\times \frac{6}{5}B \backslash \frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \bigl\vert f_{2}(z_{2}) \bigr\vert \mu (z_{1})\,d\mu (z_{2}) \\ & \leq \frac{C}{[\lambda (x_{B},\frac{1}{5}r_{B})]^{2}} \int _{6 \times \frac{6}{5}B\backslash \frac{6}{5}B} \int _{6\times \frac{6}{5}B \backslash \frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \bigl\vert f_{2}(z_{2}) \bigr\vert \mu (z_{1})\mu (z_{2}) \\ &\leq C\prod_{i=1}^{2} \frac{\mu (30\times \frac{6}{5}B)}{\lambda (x_{B},\frac{1}{5}r_{B})} \frac{1}{\mu (30\times \frac{6}{5}B)} \int _{6\times \frac{6}{5}B} \bigl\vert f_{i}(z_{i}) \bigr\vert \,d\mu (z_{i}) \\ &\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned}$$

For $J_{21}$, by the properties of λ, we have

$$\begin{aligned} J_{21}\leq {}&C\sum _{k=1}^{N-1}\sum_{j=k}^{N-1} \int _{6^{k+1} \frac{6}{5}B\backslash 6^{k}\frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \int _{6^{j+1} \frac{6}{5}B\backslash 6^{j}\frac{6}{5}B} \frac{ \vert f_{2}(z_{2}) \vert }{[\lambda (z,d(z,z_{2}))]^{2}}\,d\mu (z_{2})\,d\mu (z_{1}) \\ &{} +C\sum_{k=1}^{N-1}\sum _{j=1}^{k-1} \int _{6^{k+1}\frac{6}{5}B \backslash 6^{k}\frac{6}{5}B} \frac{ \vert f_{1}(z_{1}) \vert }{[\lambda (z,d(z,z_{1}))]^{2}} \int _{6^{j+1} \frac{6}{5}B\backslash 6^{j}\frac{6}{5}B} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2})\,d\mu (z_{1}) \\ \leq{}& C\sum_{j=1}^{N-1} \frac{1}{[\lambda (x_{B},6^{j}\frac{6}{5}r_{B})]^{2}} \int _{6^{j+1} \frac{6}{5}B\backslash 6^{j}\frac{6}{5}B} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \sum_{k=1}^{j} \int _{6^{j+1}\frac{6}{5}B\backslash 6^{j}\frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\ &{} +C\sum_{k=1}^{N-1} \int _{6^{k+1}\frac{6}{5}B\backslash 6^{k} \frac{6}{5}B}\frac{ \vert f_{1}(z_{1}) \vert }{[\lambda (z,d(z,z_{1}))]^{2}} \int _{6^{k} \frac{6}{5}B} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2})\,d\mu (z_{1}) \\ \leq {}&C\sum_{j=1}^{N-1} \frac{1}{[\lambda (x_{B},6^{j}\frac{6}{5}r_{B})]^{2}} \int _{6^{j+1} \frac{6}{5}B} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \int _{6^{j+1}\frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\ &{} +C\sum_{j=1}^{N-1} \frac{1}{[\lambda (x_{B},6^{j}\frac{6}{5}r_{B})]^{2}} \int _{6^{j+1} \frac{6}{5}B} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \int _{6^{k}B} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\ \leq{}& CK_{B,Q}M_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned}$$

For $J_{3}$, similar to estimate $I_{1}$, we have

$$ J_{3}\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

For $J_{4}$ and $J_{5}$, similar to estimate $I_{2}$, we have

$$ J_{4}+J_{5}\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

By a similar method to estimate $J_{1}$, we can obtain that

$$ J_{6}\leq CK_{B,Q}M_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

By a similar method to estimate $J_{2}$, we also obtain that

$$ J_{7}\leq CK_{B,Q}M_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

Hence, (2.8) is proved. Thus, Lemma 2.3, in this case, is proved.

Case 2: $\boldsymbol{0< l(B)=l<1}$. Assume that $B_{0}$ and $Q_{0}$ are concentric with B and Q, respectively, and $l(B_{0})=l(B)^{\alpha }$, $l(Q_{0})=l(Q)^{\alpha }$. As in the proof of Theorem 9.1 in [34], to obtain (2.6), it suffices to show that

$$ \biggl(\frac{1}{\mu (6B_{0})} \int _{B} \bigl\vert { \bigl\vert T(f_{1},f_{2}) (z) \bigr\vert }^{\delta }- \vert \widetilde{h}_{B} \vert ^{\delta } \bigr\vert \,d\mu (z) \biggr)^{1/\delta }\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x) $$

(2.9)

holds for any x and ball B with $x\in B$, and

$$ \vert \widetilde{h}_{B}-\widetilde{h}_{Q} \vert \leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x) $$

(2.10)

for all balls $B\subset Q$ with $x\in B$, where B is an arbitrary ball, Q is a doubling ball. For any ball B, we denote

$$ \widetilde{h}_{B}:= m_{B}\bigl(T\bigl(f_{1}^{0},f_{2}^{\infty } \bigr)+T\bigl(f_{1}^{\infty },f_{2}^{0} \bigr)+T\bigl(f_{1}^{ \infty },f_{2}^{\infty } \bigr)\bigr),$$

and

$$ \widetilde{h}_{Q}:= m_{Q}\bigl(T\bigl(f_{1}^{0},f_{2}^{\infty } \bigr)+T\bigl(f_{1}^{\infty },f_{2}^{0} \bigr)+T\bigl(f_{1}^{ \infty },f_{2}^{\infty } \bigr)\bigr),$$

where we split each $f_{i}$ as $f_{i}=f_{i}^{0}+f_{i}^{\infty }$, $f_{i}^{0}=f_{i}\chi _{\frac{6}{5}B_{0}}$ and $f_{i}^{\infty }=f_{i}-f_{i}^{0}$, $i=1,2$. Since

$$ \begin{aligned} \bigl\vert T(f_{1},f_{2}) (y) \bigr\vert &\leq \bigl\vert T\bigl(f_{1}^{0},f_{2}^{0} \bigr) (y) \bigr\vert + \bigl\vert T\bigl(f_{1}^{0},f_{2}^{ \infty } \bigr) (y) \bigr\vert + \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{0} \bigr) (y) \bigr\vert + \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{ \infty } \bigr) (y) \bigr\vert , \end{aligned} $$

then

$$ \begin{aligned} & \biggl(\frac{1}{\mu (6B_{0})} \int _{B} \bigl\vert { \bigl\vert T(f_{1},f_{2}) (z) \bigr\vert }^{ \delta }- \vert \widetilde{h}_{B} \vert ^{\delta } \bigr\vert \,d\mu (z) \biggr)^{1/\delta } \\ & \quad \leq C \biggl(\frac{1}{\mu (6B_{0})} \int _{B} \bigl\vert T(f_{1},f_{2}) (z)- \widetilde{h}_{B} \bigr\vert ^{\delta }\,d\mu (z) \biggr)^{1/\delta } \\ & \quad \leq C \biggl(\frac{1}{\mu (6B_{0})} \int _{B} \bigl\vert T\bigl(f_{1}^{0},f_{2}^{0} \bigr) (z) \bigr\vert ^{ \delta }\,d\mu (z) \biggr)^{1/\delta } \\ &\qquad {} +C \biggl(\frac{1}{\mu (6B_{0})} \int _{B} \bigl\vert T\bigl(f_{1}^{0},f_{2}^{ \infty } \bigr) (z)-T\bigl(f_{1}^{0},f_{2}^{\infty } \bigr) (y) \bigr\vert ^{\delta }\,d\mu (z) \biggr)^{1/ \delta } \\ &\qquad {} +C \biggl(\frac{1}{\mu (6B_{0})} \int _{B} \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{0} \bigr) (z)-T\bigl(f_{1}^{ \infty },f_{2}^{0} \bigr) (y) \bigr\vert ^{\delta }\,d\mu (z) \biggr)^{1/\delta } \\ &\qquad {} +C \biggl(\frac{1}{\mu (6B_{0})} \int _{B} \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{ \infty } \bigr) (z)-T\bigl(f_{1}^{\infty },f_{2}^{\infty } \bigr) (y) \bigr\vert ^{\delta }\,d\mu (z) \biggr)^{1/\delta } \\ &\quad =:L_{1}+L_{2}+L_{3}+L_{4}. \end{aligned} $$

We first estimate $L_{1}$. Applying Lemma 2.2 with $s=\delta$ ($0<\delta <1/2$) and $t=1/2$, we get

$$ \begin{aligned} L_{1}&\leq C \biggl(\frac{1}{\mu (6B_{0})} \int _{B} \bigl\vert T\bigl(f_{1}^{0},f_{2}^{0} \bigr) (z) \bigr\vert ^{ \delta }\,d\mu (z) \biggr)^{1/\delta } \\ &\leq C \biggl( \frac{\mu (B)}{\mu (6B_{0})} \biggr)^{1/\delta } \bigl\Vert T(f_{1} \chi _{\frac{6}{5}B_{0}},f_{2}\chi _{\frac{6}{5}B_{0}}) \bigr\Vert _{L^{1/2, \infty }(B,\frac{\,d\mu }{\mu (B)})} \\ &\leq C \biggl( \frac{\mu (B)}{\mu (6B_{0})} \biggr)^{1/\delta }\prod _{i=1}^{2} \biggl[\frac{\mu (5\times \frac{6}{5}B_{0})}{\mu (B)} \frac{1}{\mu (5\times \frac{6}{5}B_{0})} \int _{\frac{6}{5}B_{0}} \bigl\vert f_{i}(z_{i}) \bigr\vert \,d\mu (z_{i}) \biggr] \\ &\leq C \biggl( \frac{\mu (B)}{\mu (6B_{0})} \biggr)^{1/\delta -2} M_{(5)}f_{1}(x)M_{(5)}f_{2}(x) \\ &\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned} $$

For $L_{2}$, let $z,y\in B$, $z_{1}\in \frac{6}{5}B_{0}$ and $z_{2}\in X\backslash \frac{6}{5}B_{0}$, then $\max_{1\leq i\leq 2}d(z,z_{i})\geq d(z,z_{2})\geq Cl(B_{0})= Cl(B)^{\alpha }\geq \operatorname{Cd}(z,y)^{\alpha }$. By Definition 1.7 and the properties of λ, we obtain

$$\begin{aligned}& \bigl\vert T\bigl(f_{1}^{0},f_{2}^{\infty } \bigr) (z)-T\bigl(f_{1}^{0},f_{2}^{ \infty } \bigr) (y) \bigr\vert \\& \quad \leq C \int _{X\backslash \frac{6}{5}B_{0}} \int _{ \frac{6}{5}B_{0}} \frac{1}{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}} \frac{d(z,y)^{\delta }}{[\sum_{i=1}^{2}d(z,z_{i})]^{\delta /\alpha }} \bigl\vert f_{1}(z_{1}) \bigr\vert \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad \leq C \int _{X\backslash \frac{6}{5}B_{0}} \int _{ \frac{6}{5}B_{0}} \frac{1}{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}} \frac{d(z,y)^{\delta }}{[d(z,z_{2})]^{\delta /\alpha }} \bigl\vert f_{1}(z_{1}) \bigr\vert \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad \leq C \int _{\frac{6}{5}B_{0}} \frac{ \vert f_{1}(z_{1}) \vert }{\lambda (z,d(z,z_{1}))}\,d\mu (z_{1})\\& \qquad {} \times\sum _{k=1}^{ \infty } \int _{6^{k+1}\frac{6}{5}B_{0}\backslash 6^{k}\frac{6}{5}B_{0}} \frac{1}{\lambda (z,d(z,z_{2}))} \frac{d(z,y)^{\delta }}{[d(z,z_{2})]^{\delta /\alpha }} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\& \quad \leq C \frac{\mu (5\times \frac{6}{5}B_{0})}{\lambda (x_{B},\frac{6}{5}r_{B_{0}})} \frac{1}{\mu (5\times \frac{6}{5}B_{0})} \int _{\frac{6}{5}B_{0}} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\& \qquad {} \times \sum_{k=1}^{\infty }6^{-k\delta /\alpha } \frac{\mu (5\times 6^{k+1}\frac{6}{5}B_{0})}{\lambda (x_{B},6^{k}\frac{6}{5}r_{B_{0}})} \frac{1}{\mu (5\times 6^{k+1}\frac{6}{5}B_{0})} \int _{6^{k+1} \frac{6}{5}B_{0}} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\& \quad \leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned}$$

Therefore,

$$ L_{2}\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x).$$

Similar to estimate $L_{2}$, we also obtain that

$$ L_{3}\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x).$$

Let us turn to estimate $L_{4}$. Write

$$\begin{aligned}& \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{\infty } \bigr) (z)-T\bigl(f_{1}^{\infty },f_{2}^{ \infty } \bigr) (y) \bigr\vert \\& \quad \leq C \int _{X\backslash \frac{6}{5}B_{0}} \int _{X\backslash \frac{6}{5}B_{0}} \frac{ \vert f_{1}(z_{1}) \vert \vert f_{2}(z_{2}) \vert }{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}} \frac{d(z,y)^{\delta }}{[\sum_{i=1}^{2}d(z,z_{i})]^{\delta /\alpha }}\,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad \leq C\sum_{k=1}^{\infty } \int _{6^{k+1}\frac{6}{5}B_{0}\backslash 6^{k} \frac{6}{5}B_{0}}\sum_{j=1}^{\infty } \int _{6^{j+1}\frac{6}{5}B_{0} \backslash 6^{j}\frac{6}{5}B_{0}} \frac{ \vert f_{1}(z_{1}) \vert \vert f_{2}(z_{2}) \vert }{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}}\\& \qquad {} \times \frac{d(z,y)^{\delta }}{[\sum_{i=1}^{2}d(z,z_{i})]^{\delta /\alpha }}\,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad \leq C\sum_{k=1}^{\infty }\sum _{j=k}^{\infty } \int _{6^{k+1} \frac{6}{5}B_{0}\backslash 6^{k}\frac{6}{5}B_{0}} \bigl\vert f_{2}(z_{2}) \bigr\vert \\& \qquad {} \times \int _{6^{j+1} \frac{6}{5}B_{0}\backslash 6^{j}\frac{6}{5}B_{0}} \frac{ \vert f_{1}(z_{1}) \vert }{[\lambda (z,d(z,z_{1}))]^{2}} \frac{d(z,y)^{\delta }}{[d(z,z_{1})]^{\delta /\alpha }}\,d\mu (z_{1})\,d\mu (z_{2}) \\& \qquad {} +C\sum_{k=1}^{\infty }\sum _{j=1}^{k-1} \int _{6^{k+1}\frac{6}{5}B_{0} \backslash 6^{k}\frac{6}{5}B_{0}} \frac{ \vert f_{2}(z_{2}) \vert }{[\lambda (z,d(z,z_{2}))]^{2}} \frac{d(z,y)^{\delta }}{[d(z,z_{2})]^{\delta /\alpha }} \\& \qquad {} \times\int _{6^{j+1} \frac{6}{5}B_{0}\backslash 6^{j}\frac{6}{5}B_{0}}f_{1}(z_{1})\,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad =:L_{41}+L_{42}. \end{aligned}$$

To estimate $L_{41}$, by the properties of λ, we obtain

$$\begin{aligned} L_{41} \leq & C\sum _{j=1}^{\infty } \biggl( \frac{1}{[\lambda (x_{B},6^{j}\frac{6}{5}r_{B_{0}})]^{2}} \int _{6^{j+1} \frac{6}{5}B_{0}\backslash 6^{j}\frac{6}{5}B_{0}} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1})6^{-j \delta /\alpha } \biggr) \\ &{} \times \sum_{k=1}^{j} \int _{6^{k+1}\frac{6}{5}B_{0}\backslash 6^{k} \frac{6}{5}B_{0}} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\ \leq & C\sum_{j=1}^{\infty } \biggl( \frac{1}{[\lambda (x_{B},5\times 6^{j}\frac{6}{5}r_{B_{0}})]^{2}} \int _{6^{j+1}\frac{6}{5}B_{0}} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1})6^{-j\delta / \alpha } \biggr) \\ &{} \times \int _{6^{j+1}\frac{6}{5}B_{0}} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\ \leq & C\sum_{j=1}^{\infty }6^{-j\delta /\alpha } \frac{1}{\mu (5\times 6^{j+1}\frac{6}{5}B_{0})} \int _{6^{j+1} \frac{6}{5}B_{0}} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\ &{} \times \frac{1}{\mu (5\times 6^{j+1}\frac{6}{5}B_{0})} \int _{6^{j+1} \frac{6}{5}B_{0}} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\ \leq & CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned}$$

For $L_{42}$, by the properties of λ, we have

$$\begin{aligned} L_{42} \leq & C\sum _{k=1}^{\infty } \biggl( \frac{1}{[\lambda (x_{B},6^{k}\frac{6}{5}r_{B_{0}})]^{2}} \int _{6^{k+1} \frac{6}{5}B_{0}\backslash 6^{k}\frac{6}{5}B_{0}} \bigl\vert f_{1}(z_{2}) \bigr\vert \,d\mu (z_{2})6^{-k \delta /\alpha } \biggr) \\ &{} \times \sum_{j=1}^{k-1} \int _{6^{j+1}\frac{6}{5}B_{0} \backslash 6^{j}\frac{6}{5}B_{0}} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\ \leq & C\sum_{k=1}^{\infty } \biggl( \frac{1}{[\lambda (x_{B},5\times 6^{k}\frac{6}{5}r_{B_{0}})]^{2}} \int _{6^{k+1}\frac{6}{5}B_{0}} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2})6^{-k\delta / \alpha } \biggr) \\ &{} \times \int _{6^{k}\frac{6}{5}B_{0}} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\ \leq & C\sum_{k=1}^{\infty }6^{-k\delta /\alpha } \frac{1}{\mu (5\times 6^{k+1}\frac{6}{5}B_{0})} \int _{6^{k+1} \frac{6}{5}B_{0}} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\ &{} \times \frac{1}{\mu (5\times 6^{k}\frac{6}{5}B_{0})} \int _{6^{k} \frac{6}{5}B_{0}} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\ \leq & CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned}$$

Thus,

$$ \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{\infty } \bigr) (z)-T\bigl(f_{1}^{\infty },f_{2}^{\infty } \bigr) (y) \bigr\vert \leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

By the above estimate, we obtain (2.9).

Next, we prove (2.10). Consider two balls $B\subset Q$ with $x\in B$, where B is an arbitrary ball, and Q is a doubling ball. Denote $N=N_{B,Q}+1$. Recall that $B_{0}$ and $Q_{0}$ are concentric with B and Q, respectively, and $l(B_{0})=l(B)^{\alpha }$, $l(Q_{0})=l(Q)^{\alpha }$. Then,

$$ \begin{aligned} \vert \widetilde{h}_{B}- \widetilde{h}_{Q} \vert \leq {}& \bigl\vert m_{B}\bigl(T(f_{1}\chi _{\frac{6}{5}B_{0}},f_{2}\chi _{6^{N} \frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}})+T(f_{1} \chi _{6^{N} \frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}},f_{2}\chi _{\frac{6}{5}B_{0}})\bigr) \bigr\vert \\ &{} + \bigl\vert m_{B}\bigl(T(f_{1}\chi _{6^{N}\frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}},f_{2}\chi _{6^{N}\frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}})\bigr) \bigr\vert \\ &{} + \bigl\vert m_{B}\bigl(T(f_{1}\chi _{6^{N}\frac{6}{5}B_{0}},f_{2}\chi _{6^{N} \frac{6}{5}B_{0}}) \bigr)-m_{Q}\bigl(T(f_{1}\chi _{6^{N}\frac{6}{5}B_{0}},f_{2} \chi _{6^{N}\frac{6}{5}B_{0}})\bigr) \bigr\vert \\ &{} + \bigl\vert m_{B}\bigl(T(f_{1}\chi _{X\backslash 6^{N}\frac{6}{5}B_{0}},f_{2} \chi _{6^{N}\frac{6}{5}B_{0}}) \bigr)-m_{Q}\bigl(T(f_{1}\chi _{X\backslash 6^{N} \frac{6}{5}B_{0}},f_{2} \chi _{6^{N}\frac{6}{5}B_{0}})\bigr) \bigr\vert \\ &{} + \bigl\vert m_{B}\bigl(T(f_{1}\chi _{6^{N}\frac{6}{5}B_{0}},f_{2}\chi _{X \backslash 6^{N}\frac{6}{5}B_{0}}) \bigr)-m_{Q}\bigl(T(f_{1}\chi _{6^{N} \frac{6}{5}B_{0}},f_{2} \chi _{X\backslash 6^{N}\frac{6}{5}B_{0}})\bigr) \bigr\vert \\ &{} + \bigl\vert m_{Q}\bigl(T(f_{1}\chi _{\frac{6}{5}Q_{0}},f_{2}\chi _{6^{N} \frac{6}{5}B_{0}\backslash \frac{6}{5}Q_{0}})+T(f_{1} \chi _{6^{N} \frac{6}{5}B_{0}\backslash \frac{6}{5}Q_{0}},f_{2}\chi _{6^{N} \frac{6}{5}B_{0}})\bigr) \bigr\vert \\ &{} + \bigl\vert m_{Q}\bigl(T(f_{1}\chi _{6^{N}\frac{6}{5}B_{0}\backslash \frac{6}{5}Q_{0}},f_{2}\chi _{6^{N}\frac{6}{5}B_{0}\backslash \frac{6}{5}Q_{0}})\bigr) \bigr\vert \\ =:{}&M_{1}+M_{2}+M_{3}+M_{4}+M_{5}+M_{6}+M_{7}. \end{aligned} $$

To estimate $M_{1}$, by the condition (1.5) in Definition 1.7 and the properties of λ, we obtain

$$ \begin{aligned} & \bigl\vert T(f_{1}\chi _{\frac{6}{5}B_{0}},f_{2}\chi _{6^{N} \frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}}) (z) \bigr\vert \\ &\quad\leq C \int _{6^{N}\frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}} \int _{ \frac{6}{5}B_{0}} \frac{ \vert f_{1}(z_{1}) \vert \vert f_{2}(z_{2}) \vert }{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}}\,d\mu (z_{1})\,d\mu (z_{2}) \\ &\quad\leq C \int _{\frac{6}{5}B_{0}} \frac{ \vert f_{1}(z_{1}) \vert }{\lambda (z,d(z,z_{1}))}\,d\mu (z_{1}) \int _{6^{N} \frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}} \frac{ \vert f_{2}(z_{2}) \vert }{\lambda (z,d(z,z_{2}))}\,d\mu (z_{2}) \\ &\quad\leq C\frac{1}{\mu (5\times \frac{6}{5}B_{0})} \int _{\frac{6}{5}B_{0}} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1})\sum_{k=1}^{N_{B,Q}} \int _{6^{k+1}\frac{6}{5}B_{0} \backslash 6^{k}\frac{6}{5}B_{0}} \frac{ \vert f_{2}(z_{2}) \vert }{\lambda (z,d(z,z_{2}))}\,d\mu (z_{2}) \\ &\quad\leq CM_{(5)}f_{1}(x)\sum_{k=1}^{N_{B,Q}} \frac{\mu (5\times 6^{k+1}\frac{6}{5}B_{0})}{\lambda (x_{B},5\times 6^{k+1}\frac{6}{5}r_{B_{0}})} \frac{1}{\mu (5\times 6^{k+1}\frac{6}{5}B_{0})} \int _{6^{k+1} \frac{6}{5}B_{0}} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\ &\quad\leq CK_{B,Q}M_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned} $$

Also,

$$ \bigl\vert T(f_{1}\chi _{6^{N}\frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}},f_{2} \chi _{\frac{6}{5}B_{0}}) (z) \bigr\vert \leq CK_{B,Q}M_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

Then,

$$ M_{1}\leq CK_{B,Q}M_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

Let us estimate $M_{2}$. By the condition (1.5) in Definition 1.7 and the properties of λ, we obtain

$$\begin{aligned}& \bigl\vert T(f_{1}\chi _{6^{N}\frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}},f_{2}\chi _{6^{N}\frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}}) \bigr\vert \\& \quad \leq C \int _{6^{N}\frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}} \int _{6^{N} \frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}} \frac{ \vert f_{1}(z_{1})\|f_{2}(z_{2}) \vert }{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}}\,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad \leq C\sum_{k=1}^{N-1}\sum _{j=1}^{N-1} \int _{6^{k+1}\frac{6}{5}B_{0} \backslash 6^{k}\frac{6}{5}B_{0}} \int _{6^{j+1}\frac{6}{5}B_{0} \backslash 6^{j}\frac{6}{5}B_{0}} \frac{ \vert f_{1}(z_{1}) \vert \vert f_{2}(z_{2}) \vert }{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}}\,d\mu (z_{1})\,d\mu (z_{2}) \\& \qquad {} +C \int _{6\times \frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}} \int _{6\times \frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}} \frac{ \vert f_{1}(z_{1}) \vert \vert f_{2}(z_{2}) \vert }{[\sum_{i=1}^{2}\lambda (z,d(z,z_{i}))]^{2}}\,d\mu (z_{1})\,d\mu (z_{2}) \\& \quad =:M_{21}+M_{22}. \end{aligned}$$

For $M_{22}$, since $z\in B_{0}$ and $z_{i}\in 6\times \frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}$, then $\frac{1}{5}r_{B_{0}}\leq d(z,z_{i})\leq \frac{41}{5}r_{B_{0}}$ for $i=1,2$. Therefore,

$$ \begin{aligned} M_{22}&\leq \frac{C}{[\lambda (z,\frac{1}{5}r_{B_{0}})]^{2}} \int _{6\times \frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}} \int _{6\times \frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}} \frac{ \vert f_{1}(z_{1}) \vert \vert f_{2}(z_{2}) \vert }{d}\mu (z_{1})\,d\mu (z_{2}) \\ & \leq \frac{C}{[\lambda (x_{B},\frac{1}{5}r_{B_{0}})]^{2}} \int _{6 \times \frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}} \int _{6\times \frac{6}{5}B_{0}\backslash \frac{6}{5}B_{0}} \frac{ \vert f_{1}(z_{1}) \vert \vert f_{2}(z_{2}) \vert }{d}\mu (z_{1})\,d\mu (z_{2}) \\ &\leq C\prod_{i=1}^{2} \frac{\mu (30\times \frac{6}{5}B_{0})}{\lambda (x_{B},\frac{1}{5}r_{B_{0}})} \frac{1}{\mu (30\times \frac{6}{5}B_{0})} \int _{6\times \frac{6}{5}B_{0}} \bigl\vert f_{i}(z_{i}) \bigr\vert \,d\mu (z_{i}) \\ &\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned} $$

For $M_{21}$, by the properties of λ, we obtain

$$ \begin{aligned} M_{21}\leq {}&C\sum _{k=1}^{N-1}\sum_{j=k}^{N-1} \int _{6^{k+1} \frac{6}{5}B_{0}\backslash 6^{k}\frac{6}{5}B_{0}} \bigl\vert f_{1}(z_{1}) \bigr\vert \int _{6^{j+1} \frac{6}{5}B_{0}\backslash 6^{j}\frac{6}{5}B_{0}} \frac{ \vert f_{2}(z_{2}) \vert }{[\lambda (z,d(z,z_{2}))]^{2}}\,d\mu (z_{2})\,d\mu (z_{1}) \\ &{} +C\sum_{k=1}^{N-1}\sum _{j=1}^{k-1} \int _{6^{k+1}\frac{6}{5}B_{0} \backslash 6^{k}\frac{6}{5}B_{0}} \frac{ \vert f_{1}(z_{1}) \vert }{[\lambda (z,d(z,z_{1}))]^{2}} \int _{6^{j+1} \frac{6}{5}B_{0}\backslash 6^{j}\frac{6}{5}B_{0}} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2})\,d\mu (z_{1}) \\ \leq{} &C\sum_{j=1}^{N-1} \frac{1}{[\lambda (x_{B},6^{j}\frac{6}{5}r_{B_{0}})]^{2}} \int _{6^{j+1} \frac{6}{5}B_{0}\backslash 6^{j}\frac{6}{5}B_{0}} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\ &{}\times\sum_{k=1}^{j} \int _{6^{j+1}\frac{6}{5}B_{0}\backslash 6^{j} \frac{6}{5}B_{0}} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\ &{} +C\sum_{k=1}^{N-1} \int _{6^{k+1}\frac{6}{5}B_{0}\backslash 6^{k} \frac{6}{5}B_{0}}\frac{ \vert f_{1}(z_{1}) \vert }{[\lambda (z,d(z,z_{1}))]^{2}} \int _{6^{k}\frac{6}{5}B_{0}} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2})\,d\mu (z_{1}) \\ \leq{} &C\sum_{j=1}^{N-1} \frac{1}{[\lambda (x_{B},6^{j}\frac{6}{5}r_{B_{0}})]^{2}} \int _{6^{j+1} \frac{6}{5}B_{0}} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \int _{6^{j+1}\frac{6}{5}B_{0}} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \\ &{} +C\sum_{j=1}^{N-1} \frac{1}{[\lambda (x_{B},6^{j}\frac{6}{5}r_{B_{0}})]^{2}} \int _{6^{j+1} \frac{6}{5}B_{0}} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \int _{6^{k}B_{0}} \bigl\vert f_{2}(z_{2}) \bigr\vert \,d\mu (z_{2}) \\ \leq{}& CK_{B,Q}M_{(5)}f_{1}(x)M_{(5)}f_{2}(x). \end{aligned} $$

For $M_{3}$, similar to estimate $L_{1}$, we have

$$ M_{3}\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

For $M_{4}$ and $M_{5}$, similar to estimate $L_{2}$, we have

$$ M_{4}+M_{5}\leq CM_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

By a similar method to estimate $M_{1}$, we also obtain that

$$ M_{6}\leq CK_{B,Q}M_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

By a similar method to estimate $M_{2}$, we also obtain that

$$ M_{7}\leq CK_{B,Q}M_{(5)}f_{1}(x)M_{(5)}f_{2}(x). $$

Hence, (2.10) is proved. Thus, the proof of Lemma 2.3 is completed. □

Proof of Theorem 1.11

Let $0<\delta <1/2$, $1< p_{1},p_{2},q<\infty$, $\frac{1}{q}=\frac{1}{p_{1}}+\frac{1}{p_{2}}$, $f_{1}\in L^{p_{1}}(\mu )$ and $f_{2}\in L^{p_{2}}(\mu )$. By $|f(x)|\leq N_{\delta }f(x)$, Lemma 2.1, Lemma 2.3, Hörder’s inequality and the boundedness of $M_{(\rho )}$ for $\rho \geq 5$, we obtain

$$ \begin{aligned} \bigl\Vert T(f_{1},f_{2}) \bigr\Vert _{L^{q}(\mu )}& \leq \bigl\Vert N_{\delta } \bigl(T(f_{1},f_{2})\bigr) \bigr\Vert _{L^{q}( \mu )} \\ &\leq C \bigl\Vert M^{\sharp }_{\delta }\bigl(T(f_{1},f_{2}) \bigr) \bigr\Vert _{L^{q}(\mu )}\leq C \bigl\Vert M_{(5)}(f_{1})M_{(5)}(f_{2}) \bigr\Vert _{L^{q}( \mu )} \\ &\leq C \bigl\Vert M_{(5)}(f_{1}) \bigr\Vert _{L^{p_{1}}(\mu )} \bigl\Vert M_{(5)}(f_{1}) \bigr\Vert _{L^{p_{2}}( \mu )}\leq C \Vert f_{1} \Vert _{L^{p_{1}}(\mu )} \Vert f_{2} \Vert _{L^{p_{2}}(\mu )}. \end{aligned} $$

Thus, the proof of Theorem 1.11 is finished. □

Next, let us prove Theorem 1.12. We first prove the following lemma.

Lemma 2.4

Assume that λ satisfies the ϵ-weak reverse doubling condition. Let $1< q\leq p<\infty$. If $f\in M_{q}^{p}(\mu )$, then for $x\in B(x_{B},r_{B})$, we have

$$ \int _{X\backslash 2B}\frac{ \vert f(y) \vert }{\lambda (x,d(x,y))}\,d\mu (y)\leq C\bigl( \lambda (x_{B},r_{B})\bigr)^{-\frac{1}{p}} \Vert f \Vert _{M_{q}^{p}(\mu )}.$$

Proof

By Hölder’s inequality, we have

$$\begin{aligned}& \int _{X\backslash 2B}\frac{ \vert f(y) \vert }{\lambda (x,d(x,y))}\,d\mu (y) \\& \quad =\sum_{j=1}^{\infty } \int _{2^{j+1}B\backslash 2^{j}B} \frac{ \vert f(y) \vert }{\lambda (x,d(x,y))}\,d\mu (y) \\& \quad \leq C\sum_{j=1}^{\infty } \frac{1}{\lambda (x_{B},2^{j}r_{B})} \int _{2^{j+1}B \backslash 2^{j}B} \bigl\vert f(y) \bigr\vert \,d\mu (y) \\& \quad \leq C\sum_{j=1}^{\infty } \frac{1}{\lambda (x_{B},2^{j}r_{B})}\biggl[ \int _{2^{j+1}B} \bigl\vert f(y) \bigr\vert ^{q}\,d\mu (y)\biggr]^{1/q}\mu \bigl(2^{j+1}B\bigr)^{1/q'} \\& \quad \leq C\sum_{j=1}^{\infty } \frac{1}{\lambda (x_{B},2^{j}r_{B})} \Vert f \Vert _{L^{q}( \mu ,2^{j+1}B)}\bigl[\mu \bigl(2^{j+2}B\bigr))\bigr]^{1/p-1/q}\bigl[\mu \bigl(2^{j+2}B\bigr))\bigr]^{1/q-1/p} \mu \bigl(2^{j+1}B \bigr)^{1/q'} \\& \quad \leq C\sum_{j=1}^{\infty } \frac{1}{\lambda (x_{B},2^{j}r_{B})} \Vert f \Vert _{M_{q}^{p}( \mu )}\bigl[\mu \bigl(2^{j+2}B\bigr))\bigr]^{1-1/p} \\& \quad \leq C \Vert f \Vert _{M_{q}^{p}(\mu )}\sum_{j=1}^{\infty } \bigl[\lambda \bigl(x_{B},2^{j}r_{B}\bigr) \bigr]^{-1/p} \\& \quad \leq C \Vert f \Vert _{M_{q}^{p}(\mu )}\sum_{j=1}^{\infty } \frac{1}{[C(2^{j})]^{1/p}}\bigl[\lambda (x_{B},r_{B}) \bigr]^{-1/p} \\& \quad \leq C \Vert f \Vert _{M_{q}^{p}(\mu )}\bigl[\lambda (x_{B},r_{B}) \bigr]^{-1/p}. \end{aligned}$$

Thus, the proof of Lemma 2.4 has been completed. □

Now we give the proof of Theorem 1.12.

Proof of Theorem 1.12

Fixing a ball $B\in X$, we have

$$ \begin{aligned} \bigl\vert T(f_{1},f_{2}) (z) \bigr\vert &\leq \bigl\vert T\bigl(f_{1}^{0},f_{2}^{0} \bigr) (z) \bigr\vert + \bigl\vert T\bigl(f_{1}^{0},f_{2}^{ \infty } \bigr) (z) \bigr\vert + \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{0} \bigr) (z) \bigr\vert + \bigl\vert T\bigl(f_{1}^{\infty },f_{2}^{0} \bigr) (z) \bigr\vert \\ &:=G_{1}+G_{2}+G_{3}+G_{4}, \end{aligned} $$

where $f^{0}_{i}=f_{i}\chi _{2B}$ and $f^{\infty }_{i}=f_{i}-f^{0}_{i}$ for $i=1,2$.

For $G_{1}$, by the result of Theorem 1.11, we have

$$ \begin{aligned} \bigl\Vert T\bigl(f^{0}_{1},f^{0}_{2} \bigr) \bigr\Vert _{M^{p}_{q}}(\mu )& \leq \sup_{B \in X} \mu (6B)^{\frac{1}{p}-\frac{1}{q}}\biggl( \int _{B} \bigl\vert T\bigl(f^{0}_{1},f^{0}_{2} \bigr) (z) \bigr\vert ^{q}\,d\mu (y)\biggr)^{\frac{1}{q}} \\ & \leq \sup_{B\in X}\mu (6B)^{\frac{1}{p}-\frac{1}{q}} \bigl\Vert T \bigl(f^{0}_{1},f^{0}_{2}\bigr) \bigr\Vert _{L^{q}( \mu )} \\ & \leq C\sup_{B\in X}\mu (6B)^{\frac{1}{p}-\frac{1}{q}}\prod _{i=1}^{2} \bigl\Vert f^{0}_{i} \bigr\Vert _{L^{q_{i}}( \mu )} \\ & \leq C\sup_{B\in X}\mu (6B)^{\frac{1}{p}-\frac{1}{q}}\prod _{i=1}^{2}\biggl( \int _{2B} \vert f_{i} \vert ^{q_{i}}\,d\mu (y)\biggr)^{\frac{1}{q_{i}}} \\ & \leq C\sup_{B\in X}\prod_{i=1}^{2} \mu (6B)^{\frac{1}{p_{i}}- \frac{1}{q_{i}}}\biggl( \int _{2B} \vert f_{i} \vert ^{q_{i}}\,d\mu (y)\biggr)^{\frac{1}{q_{i}}} \\ & \leq C\prod_{i=1}^{2} \Vert f_{i} \Vert _{M^{p_{i}}_{q_{i}}}(\mu ). \end{aligned} $$

For $G_{2}$, by condition (1.5) in Definition 1.7 and Lemma 2.4, we have

$$\begin{aligned} \bigl\vert T\bigl(f^{0}_{1},f^{\infty }_{2} \bigr) (z) \bigr\vert & \leq C \int _{2B} \bigl\vert f_{1}(z_{1}) \bigr\vert \int _{X\backslash 2B} \frac{ \vert f_{2}(z_{2}) \vert }{[\lambda (x,d(x,z_{1}))+\lambda (x,d(x,z_{2}))]^{2}}\,d\mu (z_{2})\,d\mu (z_{1}) \\ & \leq C \int _{2B}\frac{ \vert f_{1}(z_{1}) \vert }{\lambda (x,d(x,z_{1}))}\,d\mu (z_{1}) \int _{X\backslash 2B}\frac{ \vert f_{2}(z_{2}) \vert }{\lambda (x,d(x,z_{2}))}\,d\mu (z_{2}) \\ & \leq C\frac{1}{\mu (2B)} \int _{2B} \bigl\vert f_{1}(z_{1}) \bigr\vert \,d\mu (z_{1}) \int _{X \backslash 2B}\frac{ \vert f_{2}(z_{2}) \vert }{\lambda (x,d(x,z_{2}))}\,d\mu (z_{2}) \\ & \leq C\frac{1}{\mu (2B)}\bigl(\mu (2B)\bigr)^{1-1/q_{1}} \Vert f_{1} \Vert _{L^{q_{1}}( \mu ,2B)}\bigl(\lambda (x_{B},r_{B}) \bigr)^{-1/p_{2}} \Vert f_{2} \Vert _{M^{p_{2}}_{q_{2}}( \mu )} \\ & \leq C\mu (2B)^{-1/p_{1}}\bigl(\lambda (x_{B},r_{B}) \bigr)^{-1/p_{2}}\prod_{i=1}^{2} \Vert f_{i} \Vert _{M^{p_{i}}_{q_{i}}}( \mu ). \end{aligned}$$

Therefore,

$$\begin{aligned} \bigl\Vert T\bigl(\bigl(f^{0}_{1},f^{\infty }_{2} \bigr)\bigr) \bigr\Vert _{M^{p}_{q}(\mu )}& \leq \sup_{B\in X} \mu (6B)^{1/p-1/q} \biggl( \int _{B} \bigl\vert T\bigl(\bigl(f^{0}_{1},f^{ \infty }_{2} \bigr)\bigr) (z) \bigr\vert ^{q} \,d\mu (z) \biggr)^{1/q} \\ &\leq C\mu (6B)^{1/p-1/q}\mu (2B)^{-1/p_{1}}\bigl(\lambda (x,r) \bigr)^{-1/p_{2}} \mu (B)^{1/q}\prod _{i=1}^{2} \Vert f_{i} \Vert _{M^{p_{i}}_{q_{i}}}(\mu ) \\ &\leq C\mu (2B)^{1/p-1/q}\mu (2B)^{-1/p_{1}}\bigl(\mu (2B) \bigr)^{-1/p_{2}}\bigl( \mu (B)\bigr)^{1/q}\prod _{i=1}^{2} \Vert f_{i} \Vert _{M^{p_{i}}_{q_{i}}}(\mu ) \\ &\leq C\biggl[\frac{\mu (B)}{\mu (2B)}\biggr]^{1/q}\prod _{i=1}^{2} \Vert f_{i} \Vert _{M^{p_{i}}_{q_{i}}}( \mu ) \\ &\leq C\prod_{i=1}^{2} \Vert f_{i} \Vert _{M^{p_{i}}_{q_{i}}}(\mu ). \end{aligned}$$

For $G_{3}$, similar to $G_{2}$, we have

$$ \bigl\Vert T\bigl(\bigl(f^{\infty }_{1},f^{0}_{2} \bigr)\bigr) \bigr\Vert _{M^{p}_{q}(\mu )}\leq \prod _{i=1}^{2} \Vert f_{i} \Vert _{M^{p_{i}}_{q_{i}}}( \mu ). $$

Let us move on to estimate $G_{4}$. For $y\in B$, by Lemma 2.4, we have

$$\begin{aligned} \bigl\vert T\bigl(f^{\infty }_{1},f^{\infty }_{2} \bigr) (z) \bigr\vert & \leq C \int _{X \backslash 2B} \int _{X\backslash 2B} \frac{ \vert f_{1}(z_{1})f_{2}(z_{2}) \vert }{[\lambda (x,d(x,z_{1}))+\lambda (x,d(x,z_{2}))]^{2}}\,d\mu (z_{2})\,d\mu (z_{1}) \\ & \leq C\prod_{i=1}^{2} \int _{X\backslash 2B} \frac{ \vert f_{i}(z_{i}) \vert }{\lambda (x,d(x,z_{i}))}\,d\mu (z_{i}) \\ & \leq C\bigl(\lambda (x_{B},r_{B}) \bigr)^{-\frac{1}{p}} \Vert f \Vert _{M_{q_{i}}^{p_{i}}( \mu )}. \end{aligned}$$

Thus,

$$\begin{aligned} \bigl\Vert T\bigl(f^{\infty }_{1},f^{\infty }_{2} \bigr) \bigr\Vert _{M_{q}^{p}(\mu )} & \leq \sup_{B\in X}\mu (6B)^{1/p-1/q} \biggl( \int _{B} \bigl\vert T\bigl(\bigl(f^{\infty }_{1},f^{ \infty }_{2} \bigr)\bigr) (z) \bigr\vert ^{q} \,d\mu (z) \biggr)^{1/q} \\ &\leq C\mu (6B)^{1/p-1/q}\bigl(\lambda (x,r)\bigr)^{-1/p}\mu (B)^{1/q}\prod_{i=1}^{2} \Vert f_{i} \Vert _{M^{p_{i}}_{q_{i}}}( \mu ) \\ &\leq C\mu (2B)^{1/p-1/q}\bigl(\mu (2B)\bigr)^{-1/p}\bigl(\mu (B)\bigr)^{1/q}\prod_{i=1}^{2} \Vert f_{i} \Vert _{M^{p_{i}}_{q_{i}}}( \mu ) \\ &\leq C\biggl[\frac{\mu (B)}{\mu (2B)}\biggr]^{1/q}\prod _{i=1}^{2} \Vert f_{i} \Vert _{M^{p_{i}}_{q_{i}}}( \mu ) \\ &\leq C\prod_{i=1}^{2} \Vert f_{i} \Vert _{M^{p_{i}}_{q_{i}}}(\mu ). \end{aligned}$$

The proof of Theorem 1.12 is finished. □