# Commutators of multilinear strongly singular integrals on non-homogeneous
metric measure spaces

Hailian Wang and Rulong Xie∗ Hailian Wang, Department of Mathematics, Chaohu
University, Hefei 238000, China<EMAIL_ADDRESS>Rulong Xie ($\ast$
Corresponding author), 1. Department of Mathematics, Chaohu University, Hefei
238000, China; 2. School of Mathematical Sciences, University of Science and
Technology of China, Hefei 230026, China<EMAIL_ADDRESS>
###### Abstract.

Let $(X,d,\mu)$ denotes non-homogeneous metric measure space satisfying
geometrically doubling and the upper doubling measure condition. In this
paper, the boundedness in Lebesgue spaces for two kinds of commutators, which
are iterated commutators and commutators in summation form, generated by
multilinear strongly singular integral operators with RBMO$(\mu)$ function on
non-homogeneous metric measure spaces $(X,d,\mu)$ is obtained.

###### Key words and phrases:

multilinear strongly singular integral; commutators; non-homogeneous metric
measure spaces; RBMO$(\mu)$.

###### 1991 Mathematics Subject Classification:

42B20, 42B30

##  1\. Introduction

It is well known that Hytönen [17] introduced non-homogeneous metric measure
spaces $(X,d,\mu)$, satisfying geometrically doubling and the upper doubling
measure condition (see Definitions 1.1 and 1.2). The non-homogeneous metric
measure space includes both the homogeneous spaces and non-doubling measure
spaces as special cases. We recall that the homogeneous space is a space
equipped with a non-negative doubling measure $\mu$, where $\mu$ 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 $x\in\rm{supp}\mu$ and $r>0$. The non-
doubling measure space is a space equipped with a non-negative measure $\mu$,
where $\mu$ only needs to satisfy the polynomial growth condition, i.e., for
all $x\in R^{n}$ and $r>0$, there exist a constant $C_{0}>0$ and $k\in(0,n]$
such that

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

where $B(x,r)=\\{y\in R^{n}:|y-x|<r\\}$. There are many important results in
non-doubling measure spaces (see [14], [15], [31], [34], [35], [38], [39] and
the references therein). And the analysis on non-doubling measures has
important applications in solving the long-standing open Painlev$\acute{e}$’s
problem (see [34]).

Next let us recall some results on non-homogeneous metric measure spaces.
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
RBMO$(\mu)$. In [3], the authors also proved that Calderón-Zygmund operator
and commutators generated by Calderón-Zygmund operators and RBMO function are
bounded in $L^{p}(\mu)(1<p<+\infty)$. Recently, some equivalent
characterizations are established by Liu et al. [30] for the boundedness of
Calderón-Zygmund operators on $L^{p}(\mu)(1<p<+\infty)$. In [9], Fu et al.
established boundedness of multilinear commutators of Calderón-Zygmund
operators on Orlicz spaces on non-homogeneous spaces. The more results on non-
homogeneous metric measure spaces have been obtained (see [4], [10], [11],
[18]-[24], [29], [41] and the references therein).

The theory on multilinear singular integral operators has been considered by
some researchers. In [6], Coifman and Meyer firstly established the theory of
bilinear Calderón-Zygmund operators. Later, Grafakos and Torres [12, 13]
established 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 measure spaces $(R^{n},\mu)$ was
established by Xu [39, 40]. Weighted norm inequalities for multilinear
Calderón-Zygmund operators on non-homogeneous metric measure spaces were also
constructed in [16]. Boundedness for commutators of multilinear Calderón-
Zygmund operators and multilinear fractional integral operators on non-
homogeneous metric measure spaces was also established in [37, 11]. Zheng and
Tao [42] established the boundedness for iterated commutators of multilinear
singular integrals of Dini’s type on non-homogeneous metric measure spaces.

The introduction of 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 [7] enlarged the multiplier operators onto a
class of convolution operators. Coifman [5] 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 strongly singular integral and
its commutators on Lebesgue spaces, Morrey spaces and Hardy spaces. Recently,
we [36] established the boundedness for multilinear strongly singular integral
operators on non-homogeneous metric measure spaces.

In this paper, two kinds of commutators generated by multilinear strongly
singular integral operators and RBMO$(\mu)$ function on non-homogeneous metric
measure spaces are introduced. And we will prove that they are bounded in
$m$-multiple Lebesgue spaces, provided that multilinear strongly singular
integral is bounded from $m$-multiple $L^{1}(\mu)\times\ldots\times
L^{1}(\mu)$ to $L^{1/m,\infty}(\mu)$, where $L^{p}(\mu)$ and
$L^{p,\infty}(\mu)$ denote the Lebesgue space and weak Lebesgue space
respectively.

Before stating the main results, let us 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 $\mu$
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 $x\in X,r\longmapsto\lambda(x,r)$ is non-
decreasing, and for all $x\in X,r>0$,

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

######  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 $\mu$
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 $\lambda$ be a dominating
function on $X\times(0,+\infty)$ as in Definition 1.2. In [20], it was shown
that there exists another dominating function $\tilde{\lambda}$ such that for
all $x,y\in X$ with $d(x,y)\leq r$,

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

Thus, we assume that $\lambda$ always satisfies (1.3) in this paper.

######  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 in Lemma 2.3 of [3], there exist plenty of doubling balls with
small radii and with large radii. In the rest of this paper, unless $\alpha$
and $\beta$ 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.

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

(i)

(1.4) $|K(x,y_{1},\cdots,y_{j},\cdots,y_{m})|\leq
C\biggl{[}\sum_{j=1}^{m}\lambda(x,d(x,y_{j}))\biggr{]}^{-m}$

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

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

(1.5)
$\begin{split}&|K(x,y_{1},\cdots,y_{j},\cdots,y_{m})-K(x^{\prime},y_{1},\cdots,y_{j},\cdots,y_{m})|\\\
\leq&\frac{Cd(x,x^{\prime})^{\delta}}{\biggl{[}\sum\limits_{j=1}^{m}d(x,y_{j})\biggr{]}^{\delta/\alpha}\biggl{[}\sum\limits_{j=1}^{m}\lambda(x,d(x,y_{j}))\biggr{]}^{m}},\end{split}$

provided that $Cd(x,x^{\prime})^{\alpha}\leq\max\limits_{1\leq j\leq
m}d(x,y_{j})$ and for each $j$,

(1.6)
$\begin{split}&|K(x,y_{1},\cdots,y_{j},\cdots,y_{m})-K(x,y_{1},\cdots,y^{\prime}_{j},\cdots,y_{m})|\\\
&\leq\frac{Cd(y_{j},y^{\prime}_{j})^{\delta}}{\biggl{[}\sum\limits_{j=1}^{m}d(x,y_{j})\biggr{]}^{\delta/\alpha}\biggl{[}\sum\limits_{j=1}^{m}\lambda(x,d(x,y_{j}))\biggr{]}^{m}},\end{split}$

provided that $Cd(y_{j},y^{\prime}_{j})^{\alpha}\leq\max\limits_{1\leq j\leq
m}d(x,y_{j})$, where $C$ is a positive constant.

A multilinear operator $T$ is called a multilinear strongly singular integral
operator with the above kernel $K$ satisfying $(\ref{1.4})$, $(\ref{1.5})$ and
$(\ref{1.6})$ if, for $f_{1},\cdots f_{m}$ are $L^{\infty}$ functions with
bounded support and $x\notin\bigcap_{j=1}^{m}\text{supp}f_{j}$,

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

######  Definition 1.7.

[3] For any two balls $B\subset Q$, let $N_{B,Q}$ be the smallest integer
satisfying $6^{N_{B,Q}}r_{B}\geq r_{Q}$, denote

(1.8)
$K_{B,Q}=1+\sum_{k=1}^{N_{B,Q}}\frac{\mu(6^{k}B)}{\lambda(x_{B},6^{k}r_{B})},$

where $x_{B}$ and $r_{B}$ denote center and radius of ball $B$ respectively.

######  Definition 1.8.

[3] Let $\rho>1$ be some fixed constant. A function $b\in L_{loc}^{1}(\mu)$ is
said to belong to RBMO$(\mu)$ if there exists a constant $C>0$ such that for
any ball $B$,

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

and for any two doubling balls $B\subset Q$,

(1.10) $|m_{B}(b)-m_{Q}(b)|\leq CK_{B,Q},$

where $\widetilde{B}$ is the smallest $(\alpha,\beta)$-doubling ball of the
form $6^{k}B$ with $k\in{\mathbf{N}}\bigcup\\{0\\}$, and
$m_{\widetilde{B}}(b)$ is the mean value of $b$ on $\widetilde{B}$, namely,

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

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

Let us introduce the definitions of two kinds of commutators, one can refer to
the references [33, 32].

######  Definition 1.9.

Iterated commutators generated by multilinear strongly singular integral $T$
with $\vec{b}=(b_{1},\cdots,b_{m})\in\text{RBMO}(\mu)^{m}$ is defined by

(1.11)
$T_{\prod\vec{b}}(f_{1},\cdots,f_{m})=[b_{1},[b_{2},...[b_{m-1},[b_{m},T]_{m}]_{m-1}\cdots]_{2}]_{1}(f_{1},\cdots,f_{m}).$

In particular, when $m=2$, we obtain

(1.12)
$\begin{split}[b_{1},b_{2},T](f_{1},f_{2})(x)=&b_{1}(x)b_{2}(x)T(f_{1},f_{2})(x)-b_{1}(x)T(f_{1},b_{2}f_{2})(x)\\\
&-b_{2}(x)T(b_{1}f_{1},f_{2})(x)+T(b_{1}f_{1},b_{2}f_{2})(x).\end{split}$

Also, we define $[b_{1},T]$ and $[b_{2},T]$ as follows respectively.

(1.13)
$[b_{1},T](f_{1},f_{2})(x)=b_{1}(x)T(f_{1},f_{2})(x)-T(b_{1}f_{1},f_{2})(x),$
(1.14)
$[b_{2},T](f_{1},f_{2})(x)=b_{2}(x)T(f_{1},f_{2})(x)-T(f_{1},b_{2}f_{2})(x).$

######  Definition 1.10.

The commutators in the summation form generated by multilinear strongly
singular integral $T$ with
$\vec{b}=(b_{1},\cdots,b_{m})\in\text{RBMO}(\mu)^{m}$ is defined by

(1.15)
$T_{\sum\vec{b}}(f_{1},\cdots,f_{m})=\sum_{j=1}^{m}T_{b_{j}}^{j}(f_{1},\cdots,f_{m}),$

where

$T_{b_{j}}^{j}(f_{1},\cdots,f_{m})=b_{j}T(f_{1},\cdots,f_{j},\cdots,f_{m})-T(f_{1},\cdots,b_{j}f_{j},\cdots,f_{m}).$

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 results as follows.

######  Theorem 1.11.

Suppose that $\mu$ is a Radon measure with $||\mu||=\infty$. Let
$[b_{1},b_{2},T]$ be defined by (1.12). Let $1<p_{1},p_{2},q<+\infty$,
$f_{1}\in L^{p_{1}}(\mu)$, $f_{2}\in L^{p_{2}}(\mu)$,
$b_{1}\in\text{RBMO}(\mu)$ and $b_{2}\in\text{RBMO}(\mu)$. 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

(1.16) $||[b_{1},b_{2},T](f_{1},f_{2})||_{L^{q}(\mu)}\leq
C||f_{1}||_{L^{p_{1}}(\mu)}||f_{2}||_{L^{p_{2}}(\mu)},$

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

######  Theorem 1.12.

Suppose that $\mu$ is a Radon measure with $||\mu||=\infty$. Let
$T_{\sum\vec{b}}$ be defined by (1.15) with $m=2$. Let
$1<p_{1},p_{2},q<+\infty$, $f_{1}\in L^{p_{1}}(\mu)$, $f_{2}\in
L^{p_{2}}(\mu)$, $b_{1}\in RBMO(\mu)$ and $b_{2}\in RBMO(\mu)$. 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

(1.17) $||T_{\sum\vec{b}}(f_{1},f_{2})||_{L^{q}(\mu)}\leq
C||f_{1}||_{L^{p_{1}}(\mu)}||f_{2}||_{L^{p_{2}}(\mu)},$

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

######  Remark 1.13.

For $||\mu||<\infty$, by Lemma 2.1 in Section 2, Theorem 1.11 also holds if
one assumes that $\int_{X}G(f_{1},f_{2})(x)d\mu(x)=0$ with the operator $G$ be
replaced by $T$, $[b_{1},T]$, $[b_{2},T]$ and $[b_{1},b_{2},T]$. Also Theorem
1.12 holds if one assumes that $\int_{X}G(f_{1},f_{2})(x)d\mu(x)=0$ with the
operator $G$ be replaced by $T$, $[b_{1},T]$ and $[b_{2},T]$.

######  Remark 1.14.

Since the proof of Theorem 1.12 is similar with that of Theorem 1.11, we only
give the proof of Theorem 1.11 in this paper.

Throughout the paper, $\chi_{E}$ denotes the characteristic function of set
$E$. And $p^{\prime}$ is the conjugate index of $p$, namely,
$\frac{1}{p}+\frac{1}{p^{\prime}}=1$. $C$ denotes a positive constant
independent of the main parameters involved, but it may be different in
different places.

##  2\. Proof of Main Result

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

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

$M^{\sharp}f(x)=\sup_{B\ni
x}\frac{1}{\mu(6B)}\int_{B}|f(y)-m_{\widetilde{B}}(f)|d\mu(y)+\sup_{(B,Q)\in\Delta_{x}}\frac{|m_{B}(f)-m_{Q}(f)|}{K_{B,Q}},$

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 defined
by

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

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

$M^{\sharp}_{\delta}f(x)=\\{M^{\sharp}(|f|^{\delta})(x)\\}^{1/\delta}$

and

$N_{\delta}f(x)=\\{N(|f|^{\delta})(x)\\}^{1/\delta}.$

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

$|f(x)|\leq N_{\delta}f(x)$

for $\mu-a.e.\ x\in X$.

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

$M_{r,(\rho)}f(x)=\sup_{B\ni x}\biggl{\\{}\frac{1}{\mu(\rho
B)}\int_{B}|f(y)|^{r}d\mu(y)\biggr{\\}}^{1/r}.$

When $r=1$, we simply write $M_{1,(\rho)}f(x)$ as $M_{(\rho)}f$. If $\rho\geq
5$, then the operator $M_{(\rho)}f$ is bounded on $L^{p}(\mu)$ for $p>1$ and
$M_{r,(\rho)}$ is bounded on $L^{p}(\mu)$ for $p>r$ (see [3]).

######  Lemma 2.1.

[40] Let $f\in L^{1}_{loc}(\mu)$ 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

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

######  Lemma 2.2.

[9] Let $1\leq p<+\infty$ and $1<\rho<+\infty$. Then $b\in RBMO(\mu)$ if and
only if for any ball $B\subset X$,

$\biggl{\\{}\frac{1}{\mu(\rho
B)}\int_{B}|b_{B}-m_{\widetilde{B}}(b)|^{p}d\mu(x)\biggr{\\}}^{1/p}\leq C,$

and for any two doubling balls $B\subset Q$,

$|m_{B}(b)-m_{Q}(b)|\leq CK_{B,Q}.$

######  Lemma 2.3.

[9]

$|m_{\widetilde{6^{j}\frac{6}{5}B}}(b)-m_{\widetilde{B}}(b)|\leq
Cj||b||_{\ast}.$

######  Lemma 2.4.

[9] $K(B,Q)$ has the following properties:

(i) For all balls $B\subset R\subset Q$, $K(B,R)\leq 2K(B,Q)$.

(ii) For any $\rho\in[1,+\infty)$, there exists a positive constant $C(\rho)$,
depending on $\rho$, such that, for all balls $B\subset Q$ with $r_{Q}\leq\rho
r_{B}$, $K(B,Q)\leq C(\rho)$.

(iii) There exists a positive constant $C$ such that, for all balls $B$,
$K(B,\tilde{B})\leq C$.

(iv) There exists a positive constant $C$ such that, for all balls $B\subset
R\subset Q$, $K(B,Q)\leq K(B,R)+CK(R,Q)$.

(v) There exists a positive constant $C$ such that, for all balls $B\subset
R\subset Q$, $K(R,Q)\leq CK(B,Q)$.

######  Lemma 2.5.

[36] Suppose that $\mu$ is a Radon measure with $||\mu||=\infty$. Let $T$ be
defined by (1.7). Let $1<p_{1},p_{2},q<+\infty$, $f_{1}\in L^{p_{1}}(\mu)$ and
$f_{2}\in L^{p_{2}}(\mu)$. 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

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

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

######  Lemma 2.6.

Suppose that $[b_{1},b_{2},T]$ is defined by (1.12), $0<\delta<1/2$,
$1<p_{1},p_{2},q<+\infty$, $1<r<q$ and $b_{1},b_{2}\in RBMO(\mu)$. 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 for any $x\in X$, any ball $B$ with
$B\ni x$, $f_{1}\in L^{p_{1}}(\mu)$ and $f_{2}\in L^{p_{2}}(\mu)$, we have the
following two cases:

(i)When $l=l(B):=\sup\limits_{x,y\in B}d(x,y)\geq 1$, assume that
$\widetilde{B}$ is the smallest $(\alpha,\beta)$-doubling ball of the form
$6^{k}B$ with $k\in{\mathbf{N}}\bigcup\\{0\\}$, it follows

(2.1) $\begin{split}&M_{\delta}^{\sharp}[b_{1},b_{2},T](f_{1},f_{2})(x)\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x)\\\ &\ \
+C||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1},f_{2}))(x)+C||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1},f_{2}))(x)\\\
&\ \
+C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x)+C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))(x)\\\
&\ \
+C||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))(x)+C||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))(x),\\\
\end{split}$

(2.2) $\begin{split}M_{\delta}^{\sharp}[b_{1},T](f_{1},f_{2})(x)&\leq
C||b_{1}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x)+C||b_{1}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x)\\\
&\ \
+C||b_{1}||_{\ast}M_{r,(6)}(T(f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))(x)\end{split}$

and

(2.3) $\begin{split}M_{\delta}^{\sharp}[b_{2},T](f_{1},f_{2})(x)&\leq
C||b_{2}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x)+C||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x)\\\
&\ \
+C||b_{2}||_{\ast}M_{r,(6)}(T(f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))(x).\end{split}$

(ii)When $0<l<1$, assume that $B_{0}$, $Q_{0}$ and $\tilde{B}_{0}$ are
concentric with $B$, $Q$ and $\tilde{B}$ respectively and
$l(B_{0})=l(B)^{\alpha}$, $l(Q_{0})=l(Q)^{\alpha}$,
$l(\tilde{B}_{0})=l(\tilde{B})^{\alpha}$, it follows

(2.4) $\begin{split}&M_{\delta}^{\sharp}[b_{1},b_{2},T](f_{1},f_{2})(x)\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x)\\\ &\ \
+C||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1},f_{2}))(x)+C||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1},f_{2}))(x)\\\
&\ \
+C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x)+C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}))(x)\\\
&\ \
+C||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}))(x)+C||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}))(x),\\\
\end{split}$

(2.5) $\begin{split}M_{\delta}^{\sharp}[b_{1},T](f_{1},f_{2})(x)&\leq
C||b_{1}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x)+C||b_{1}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x)\\\
&\ \
+C||b_{1}||_{\ast}M_{r,(6)}(T(f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}))(x)\end{split}$

and

(2.6) $\begin{split}M_{\delta}^{\sharp}[b_{2},T](f_{1},f_{2})(x)&\leq
C||b_{2}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x)+C||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x)\\\
&\ \
+C||b_{2}||_{\ast}M_{r,(6)}(T(f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}))(x).\end{split}$

###### Proof.

Because $L^{\infty}(\mu)$ with bounded 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 bounded support. Also, by Corollary 3.11 in [35],
without loss of generality, we can assume that $b_{1},b_{2}\in
L^{\infty}(\mu)$. Next we divide two cases for proving the result.

Case 1: $l(B)=l\geq 1$. As in the proof of Theorem 9.1 in [35], to obtain
(2.1), it suffices to show that

(2.7)
$\begin{split}&\biggl{(}\frac{1}{\mu(6B)}\int_{B}||[b_{1},b_{2},T](f_{1},f_{2})(z)|^{\delta}-|h_{B}|^{\delta}|d\mu(z)\biggr{)}^{1/\delta}\\\
\leq&C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x)+C||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1},f_{2}))(x)\\\
&+C||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1},f_{2}))(x)+C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x),\end{split}$

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

(2.8)
$\begin{split}|h_{B}-h_{Q}|\leq&CK_{B,Q}^{4}\biggr{[}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x)+||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1},f_{2}))(x)\\\
&+||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1},f_{2}))(x)+||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x)\\\
&+||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))(x)+||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))(x)\\\
&+||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))(x)\biggr{]}\end{split}$

for all balls $B\subset Q$ with $x\in B$, where $B$ is an arbitrary ball, $Q$
is a doubling ball, $\widetilde{B}$ is the smallest $(\alpha,\beta)$-doubling
ball of the form $6^{k}B$ with $k\in{\mathbf{N}}\bigcup\\{0\\}$. We denote

$\begin{split}h_{B}:=&m_{B}\biggl{[}T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{\frac{6}{5}B},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}B}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}B},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{\frac{6}{5}B}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}B},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}B}\biggr{)}\biggr{]}\end{split}$

and

$\begin{split}h_{Q}:=&m_{Q}\biggl{[}T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}Q},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}Q}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}Q},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}Q}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}Q},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}Q}\biggr{)}\biggr{]}.\end{split}$

It is easy to see that

$[b_{1},b_{2},T]=T((b_{1}-b_{1}(z))f_{1},(b_{2}-b_{2}(z))f_{2})$

and

$\begin{split}&T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2})\\\
=&T((b_{1}-b_{1}(z)+b_{1}(z)-m_{\tilde{B}}(b_{1}))f_{1},(b_{2}-b_{2}(z)+b_{2}(z)-m_{\tilde{B}}(b_{2}))f_{2})\\\
=&(b_{1}(z)-m_{\tilde{B}}(b_{1}))(b_{2}(z)-m_{\tilde{B}}(b_{2}))T(f_{1},f_{2})\\\
&\ +(b_{1}(z)-m_{\tilde{B}}(b_{1}))T(f_{1},(b_{2}-b_{2}(z))f_{2})\\\ &\
+(b_{2}(z)-m_{\tilde{B}}(b_{2}))T((b_{1}-b_{1}(z))f_{1},f_{2})+T((b_{1}-b_{1}(z))f_{1},(b_{2}-b_{2}(z))f_{2}).\end{split}$

Then

$\begin{split}&\biggl{(}\frac{1}{\mu(6B)}\int_{B}||[b_{1},b_{2},T](f_{1},f_{2})(z)|^{\delta}-|h_{B}|^{\delta}|d\mu(z)\biggr{)}^{1/\delta}\\\
\leq&C\biggl{(}\frac{1}{\mu(6B)}\int_{B}|[b_{1},b_{2},T](f_{1},f_{2})(z)-h_{B}|^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
\leq&C\biggl{(}\frac{1}{\mu(6B)}\int_{B}|(b_{1}(z)-m_{\tilde{B}}(b_{1}))(b_{2}(z)-m_{\tilde{B}}(b_{2}))T(f_{1},f_{2})(z)|^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
&\ \
+C\biggl{(}\frac{1}{\mu(6B)}\int_{B}|(b_{1}(z)-m_{\tilde{B}}(b_{1}))T(f_{1},(b_{2}-b_{2}(z))f_{2})(z)|^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
&\ \
+C\biggl{(}\frac{1}{\mu(6B)}\int_{B}|(b_{2}(z)-m_{\tilde{B}}(b_{2}))T((b_{1}-b_{1}(z))f_{1},f_{2})(z)|^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
&\ \
+C\biggl{(}\frac{1}{\mu(6B)}\int_{B}|T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2})(z)-h_{B}|^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
=&:E_{1}+E_{2}+E_{3}+E_{4}.\end{split}$

We first estimate $E_{1}$. Choosing $r_{1},r_{2}>1$ such that
$\dfrac{1}{r}+\dfrac{1}{r_{1}}+\dfrac{1}{r_{2}}=\dfrac{1}{\delta}$. By
Hölder’s inequality and Lemma 2.2, it follows

(2.9) $\begin{split}E_{1}&\leq
C\biggl{(}\frac{1}{\mu(6B)}\int_{B}|b_{1}(z)-m_{\tilde{B}}(b_{1})|^{r_{1}}d\mu(z)\biggr{)}^{1/r_{1}}\\\
&\ \
\times\biggl{(}\frac{1}{\mu(6B)}\int_{B}|b_{2}(z)-m_{\tilde{B}}(b_{2})|^{r_{2}}d\mu(z)\biggr{)}^{1/r_{2}}\\\
&\ \
\times\biggl{(}\frac{1}{\mu(6B)}\int_{B}|T(f_{1},f_{2})|^{r}d\mu(z)\biggr{)}^{1/r}\\\
&\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x).\end{split}$

For $E_{2}$, choosing $s>1$ such that
$\dfrac{1}{s}+\dfrac{1}{r}=\dfrac{1}{\delta}$, by Hölder’s inequality and
Lemma 2.2, we obtain

(2.10) $\begin{split}E_{2}&\leq
C\biggl{(}\frac{1}{\mu(6B)}\int_{B}|b_{1}(z)-m_{\tilde{B}}(b_{1})|^{s}d\mu(z)\biggr{)}^{1/s}\\\
&\ \
\times\biggl{(}\frac{1}{\mu(6B)}\int_{B}|[b_{2},T](f_{1},f_{2})|^{r}d\mu(z)\biggr{)}^{1/r}\\\
&\leq C||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1},f_{2}))(x).\end{split}$

Similar to estimate $E_{2}$, we immediately have

(2.11) $E_{3}\leq C||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1},f_{2}))(x).$

Let us turn to estimate $E_{4}$. Denote $f_{j}^{1}=f_{j}\chi_{\frac{6}{5}B}$
and $f_{j}^{2}=f_{j}-f_{j}^{1}$ for $j=1,2$, we have

$\begin{split}&|T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2})(z)-h_{B}|\\\
\leq&|T((b_{1}-m_{\tilde{B}}(b_{1}))f^{1}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{1}_{2})(z)|\\\
&\ \
+|T((b_{1}-m_{\tilde{B}}(b_{1}))f^{1}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})(z)-m_{B}[T((b_{1}-m_{\tilde{B}}(b_{1}))f^{1}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})]|\\\
&\ \
+|T((b_{1}-m_{\tilde{B}}(b_{1}))f^{2}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{1}_{2})(z)-m_{B}[T((b_{1}-m_{\tilde{B}}(b_{1}))f^{2}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{1}_{2})]|\\\
&\ \
+|T((b_{1}-m_{\tilde{B}}(b_{1}))f^{2}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})(z)-m_{B}[T((b_{1}-m_{\tilde{B}}(b_{1}))f^{2}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})]|\\\
=&:E_{41}(z)+E_{42}(z)+E_{43}(z)+E_{44}(z).\end{split}$

Then

$\begin{split}E_{4}&\leq
C\biggl{(}\frac{1}{\mu(6B)}\int_{B}E_{41}(z)^{\delta}d\mu(z)\biggr{)}^{1/\delta}+C\biggl{(}\frac{1}{\mu(6B)}\int_{B}E_{42}(z)^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
&\ \
+C\biggl{(}\frac{1}{\mu(6B)}\int_{B}E_{43}(z)^{\delta}d\mu(z)\biggr{)}^{1/\delta}+C\biggl{(}\frac{1}{\mu(6B)}\int_{B}E_{44}(z)^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
&=:E_{41}+E_{42}+E_{43}+E_{44}.\end{split}$

To estimate $E_{41}$, we need the classical Kolmogorov’s theorem: Let
$(X,\mu)$ be a probability measure space and let $0<p<q<\infty$, then there
exists a constant $C>0$, such that $||f||_{L^{p}(\mu)}\leq
C||f||_{L^{q,\infty}(\mu)}$ for any measurable function $f$. Let $p=\delta$
and $q=1/2$ such that $0<\delta<1/2$. Using Kolmogorov’s theorem, Lemma 2.2
and Hölder’s inequality, we obtain

$\begin{split}E_{41}\leq&C||T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1}^{1},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}^{1})||_{L^{1/2,\infty}(\frac{6}{5}B,\frac{d\mu(z)}{\mu(6B)})}\\\
\leq&C\frac{1}{\mu(6B)}\int_{\frac{6}{5}B}|(b_{1}(z)-m_{\tilde{B}}(b_{1}))f_{1}(z)|d\mu(z)\\\
&\ \
\times\frac{1}{\mu(6B)}\int_{\frac{6}{5}B}|(b_{2}(z)-m_{\tilde{B}}(b_{2}))f_{2}(z)|d\mu(z)\\\
\leq&C(\frac{1}{\mu(6B)}\int_{\frac{6}{5}B}|b_{1}(z)-m_{\tilde{B}}(b_{1})|^{p^{\prime}_{1}}d\mu(z))^{1/p^{\prime}_{1}}\\\
&\ \
\times(\frac{1}{\mu(6B)}\int_{\frac{6}{5}B}|f_{1}(z)|^{p_{1}}d\mu(z))^{1/p_{1}}\\\
&\ \
\times(\frac{1}{\mu(6B)}\int_{\frac{6}{5}B}|b_{2}(z)-m_{\tilde{B}}(b_{2})|^{p^{\prime}_{2}}d\mu(z))^{1/p^{\prime}_{2}}\\\
&\ \
\times(\frac{1}{\mu(6B)}\int_{\frac{6}{5}B}|f_{2}(z)|^{p_{2}}d\mu(z))^{1/p_{2}}\\\
\leq&C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).\end{split}$

To compute $E_{42}$, let $z,y\in B$, $y_{1}\in\frac{6}{5}B$ and $y_{2}\in
X\backslash\frac{6}{5}B$, then $\max\limits_{1\leq i\leq 2}d(z,y_{i})\geq
d(z,y_{2})\geq Cl(B)\geq Cl(B)^{\alpha}\geq Cd(z,y)^{\alpha}$. By Definition
1.6, Lemma 2.2, Lemma 2.3, Hölder’s inequality and the properties of
$\lambda$, it follows

$\begin{split}&|T((b_{1}-m_{\tilde{B}}(b_{1}))f^{1}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})(z)-T((b_{1}-m_{\tilde{B}}(b_{1}))f^{1}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})(y)|\\\
\leq&C\int_{X\backslash\frac{6}{5}B}\int_{\frac{6}{5}B}|K(z,y_{1},y_{2})-K(y,y_{1},y_{2})||\prod_{i=1}^{2}(b_{i}(y_{i})-m_{\tilde{B}}(b_{i}))f_{i}(y_{i})|d\mu(y_{1})d\mu(y_{2})\\\
\leq&C\int_{X\backslash\frac{6}{5}B}\int_{\frac{6}{5}B}\frac{d(z,y)^{\delta}}{\biggl{[}\sum\limits_{i=1}^{2}d(z,y_{i})\biggr{]}^{\delta/\alpha}\biggl{[}\sum\limits_{i=1}^{2}\lambda(z,d(z,y_{i}))\biggr{]}^{2}}\\\
&\ \ \
\times|\prod_{i=1}^{2}(b_{i}(y_{i})-m_{\tilde{B}}(b_{i}))f_{i}(y_{i})|d\mu(y_{1})d\mu(y_{2})\\\
\leq&C\int_{\frac{6}{5}B}\frac{|b_{1}(y_{1})-m_{\tilde{B}}(b_{1})||f_{1}(y_{1})|}{\lambda(z,d(z,y_{1}))}d\mu(y_{1})\\\
&\ \ \
\times\int_{X\backslash\frac{6}{5}B}\frac{d(z,y)^{\delta}}{d(z,y_{2})^{\delta/\alpha}}\frac{|b_{2}(y_{2})-m_{\tilde{B}}(b_{2})||f_{2}(y_{2})|}{\lambda(z,d(z,y_{2}))}d\mu(y_{2})\\\
\leq&C\frac{\mu(6B)}{\lambda(x_{B},\frac{6}{5}r_{B})}\frac{1}{\mu(6B)}\int_{\frac{6}{5}B}|b_{1}(y_{1})-m_{\tilde{B}}(b_{1})||f_{1}(y_{1})|d\mu(y_{1})\\\
&\ \ \ \times\sum_{k=1}^{\infty}\int_{6^{k}\frac{6}{5}B\backslash
6^{k-1}\frac{6}{5}B}\frac{d(z,y)^{\delta}}{d(z,y_{2})^{\delta/\alpha}}\frac{|b_{2}(y_{2})-m_{\tilde{B}}(b_{2})||f_{2}(y_{2})|}{\lambda(z,d(z,y_{2}))}d\mu(y_{2})\\\
\leq&C(\frac{1}{\mu(6B)}\int_{\frac{6}{5}B}|b_{1}(y_{1})-m_{\tilde{B}}(b_{1})|^{p^{\prime}_{1}}d\mu(y_{1}))^{1/p^{\prime}_{1}}\\\
&\ \ \
\times(\frac{1}{\mu(6B)}\int_{\frac{6}{5}B}|f_{1}(y_{1})|^{p_{1}}d\mu(y_{1}))^{1/p_{1}}\\\
&\ \ \
\times\sum_{k=1}^{\infty}\frac{1}{\lambda(x_{B},6^{k-1}\frac{6}{5}r_{B})}6^{-k\delta/\alpha}l^{\delta(1-1/\alpha)}\int_{6^{k}\frac{6}{5}B}|b_{2}(y_{2})-m_{\tilde{B}}(b_{2})||f_{2}(y_{2})|d\mu(y_{2})\\\
\leq&C||b_{1}||_{\ast}M_{p_{1},(5)}f_{1}(x)\sum_{k=1}^{\infty}6^{-k\delta/\alpha}\frac{\mu(5\times
6^{k}\frac{6}{5}B)}{\lambda(x_{B},6^{k-1}\frac{6}{5}r_{B})}\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B)}\\\ &\ \ \
\times\int_{6^{k}\frac{6}{5}B}|b_{2}(y_{2})-m_{\widetilde{6^{k}\frac{6}{5}B}}(b_{2})+m_{\widetilde{6^{k}\frac{6}{5}B}}(b_{2})-m_{\tilde{B}}(b_{2})||f_{2}(y_{2})|d\mu(y_{2})\\\
\leq&C||b_{1}||_{\ast}M_{p_{1},(5)}f_{1}(x)\sum_{k=1}^{\infty}6^{-k\delta/\alpha}\\\
&\ \ \ \times\biggl{[}\biggl{(}\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B)}\int_{6^{k}\frac{6}{5}B}|b_{2}(y_{2})-m_{\widetilde{6^{k}\frac{6}{5}B}}(b_{2})|^{p^{\prime}_{2}}d\mu(y_{2})\biggr{)}^{1/p^{\prime}_{2}}\\\
&\ \ \ \times\biggl{(}\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B)}\int_{6^{k}\frac{6}{5}B}|f_{2}(y_{2})|^{p_{2}}d\mu(y_{2})\biggr{)}^{1/p_{2}}\\\
&\ \ \ +Ck||b_{2}||_{\ast}\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B)}\int_{6^{k}\frac{6}{5}B}|f_{2}(y_{2})|d\mu(y_{2})\biggr{]}\\\
\leq&C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).\\\
\end{split}$

Taking the mean over $y\in B$, we have

$E_{42}(z)\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Therefore

$E_{42}\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Similarly, we get

$E_{43}\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

For $E_{44}$, by Definition 1.6, Lemma 2.2, Hölder’s inequality and the
properties of $\lambda$, we obtain

$\begin{split}&|T((b_{1}-m_{\tilde{B}}(b_{1}))f^{2}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})(z)-T((b_{1}-m_{\tilde{B}}(b_{1}))f^{2}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})(y)|\\\
\leq&C\int_{X\backslash\frac{6}{5}B}\int_{X\backslash\frac{6}{5}B}|K(z,y_{1},y_{2})-K(y,y_{1},y_{2})|\\\
&\ \ \
\times|\prod_{i=1}^{2}(b_{i}(y_{i})-m_{\tilde{B}}(b_{i}))f_{i}(y_{i})|d\mu(y_{1})d\mu(y_{2})\\\
\leq&C\int_{X\backslash\frac{6}{5}B}\int_{X\backslash\frac{6}{5}B}\frac{d(z,y)^{\delta}|\prod_{i=1}^{2}(b_{i}(y_{i})-m_{\tilde{B}}(b_{i})f_{i}(y_{i})|d\mu(y_{1})d\mu(y_{2})}{(d(z,y_{1})+d(z,y_{2}))^{\delta/\alpha}[\sum\limits_{j=1}^{2}\lambda(z,d(z,y_{j}))]^{2}}\\\
\leq&C\prod_{i=1}^{2}\int_{X\backslash\frac{6}{5}B}\frac{d(z,y)^{\delta_{i}}|b_{i}(y_{i})-m_{\tilde{B}}(b_{i})||f_{i}(y_{i})|d\mu(y_{i})}{d(z,y_{i})^{\delta_{i}/\alpha}\lambda(z,d(z,y_{i}))}\\\
\leq&C\prod_{i=1}^{2}\sum\limits_{k=1}^{\infty}6^{-k\delta_{i}/\alpha}l^{\delta(1-1/\alpha)}\frac{\mu(5\times
6^{k}\frac{6}{5}B)}{\lambda(x_{B},6^{k-1}\frac{6}{5}r_{B})}\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B)}\\\ &\ \
\times\int_{6^{k}\frac{6}{5}B}|b_{i}(y_{i})-m_{\tilde{B}}(b_{i})||f_{i}(y_{i})|d\mu(y_{i})\\\
\leq&C\prod_{i=1}^{2}\sum_{k=1}^{\infty}6^{-k\delta_{i}/\alpha}(\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B)}\int_{6^{k}\frac{6}{5}B}|b_{i}(y_{i})-m_{\tilde{B}}(b_{i})|^{p^{\prime}_{i}}d\mu(y_{i}))^{1/p^{\prime}_{i}}\\\
&\ \ \ \times(\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B)}\int_{6^{k}\frac{6}{5}B}|f_{i}(y_{i})|^{p_{i}})^{1/p_{i}}\\\
\leq&C\prod_{i=1}^{2}\sum_{k=1}^{\infty}6^{-k\delta_{i}/\alpha}M_{p_{i},(5)}f_{i}(x)(\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B)}\int_{6^{k}\frac{6}{5}B}|b_{i}(y_{i})-m_{\widetilde{6^{k}\frac{6}{5}B}}(b_{i})\\\
&\ \ \
+m_{\widetilde{6^{k}\frac{6}{5}B}}(b_{i})-m_{\tilde{B}}(b_{i})|^{p^{\prime}_{i}}d\mu(y_{i}))^{1/p^{\prime}_{i}}\\\
\leq&C\prod_{i=1}^{2}\sum_{k=1}^{\infty}k6^{-k\delta_{i}/\alpha}||b_{i}||_{\ast}M_{p_{i},(5)}f_{i}(x)\\\
\leq&C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).\end{split}$

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

Taking the mean over $y\in B$, then

$E_{44}(z)\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Thus

$E_{44}\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

So (2.7) can be obtained.

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. Recall that $\widetilde{B}$
is the smallest $(\alpha,\beta)$-doubling ball of the form $6^{k}B$ with
$k\in{\mathbf{N}}\bigcup\\{0\\}$. Now we have the two cases: $B\subset
Q\subset\tilde{B}$ or $B\subset\tilde{B}\subset Q$. By Lemma 2.4, we can
obtain the following facts. If $B\subset Q\subset\tilde{B}$, we have
$K_{Q,\tilde{B}}\leq CK_{B,\tilde{B}}\leq C$. If $B\subset\tilde{B}\subset Q$,
we have $K_{\tilde{B},Q}\leq CK_{B,Q}$. Without loss of generality, we only
consider $B\subset\tilde{B}\subset Q$ in this paper. The another case can be
considered by the same method.

Let $N=N_{\tilde{B},Q}+1$. It is easy to see that

$|h_{B}-h_{Q}|\leq|h_{B}-h_{\tilde{B}}|+|h_{\tilde{B}}-h_{Q}|.$

Now we first consider $|h_{\tilde{B}}-h_{Q}|$. We write

$\begin{split}|h_{\tilde{B}}-h_{Q}|\leq&\biggl{|}m_{\tilde{B}}\biggl{[}T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}}\biggr{)}\biggr{]}\\\
&\ \
-m_{\tilde{B}}\biggl{[}T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}}\biggr{)}\biggr{]}\biggr{|}\\\
&\ \
+\biggl{|}m_{\tilde{B}}\biggl{[}T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}}\biggr{)}\biggr{]}\\\
&\ \
-m_{Q}\biggl{[}T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}Q},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}Q}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}Q},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}Q}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}Q},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}Q}\biggr{)}\biggr{]}\biggr{|}\\\
=:&F_{1}+F_{2}.\end{split}$

To estimate $F_{1}$, recall the fact that

$T(f_{1},f_{2})=T(f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}})+T(f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}})+T(f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}})+T(f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}},f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}}),$

then

$\begin{split}F_{1}&\leq\biggl{|}m_{\tilde{B}}\biggl{[}T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2})-T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}})\biggr{]}\\\
&\ \
-m_{\tilde{B}}\biggl{[}T((b_{1}-m_{Q}(b_{1}))f_{1},(b_{2}-m_{Q}(b_{2}))f_{2})-T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}})\biggr{]}\biggr{|}\\\
&\leq\biggl{|}m_{\tilde{B}}\biggl{[}T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1},(b_{2}-m_{\tilde{\tilde{B}}}(b_{2}))f_{2})-T((b_{1}-m_{Q}(b_{1}))f_{1},(b_{2}-m_{Q}(b_{2}))f_{2})\biggr{]}\biggr{|}\\\
&\ \
+\biggl{|}m_{\tilde{B}}\biggl{[}T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}})\\\
&\ \
-T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}})\biggr{]}\biggr{|}\\\
&=:F_{11}+F_{12}.\end{split}$

For $F_{11}$, recall the fact that

$\begin{split}&T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2})\\\
=&T((b_{1}-b_{1}(z)+b_{1}(z)-m_{\tilde{B}}(b_{1}))f_{1},(b_{2}-b_{2}(z)+b_{2}(z)-m_{\tilde{B}}(b_{2}))f_{2})\\\
=&(b_{1}(z)-m_{\tilde{B}}(b_{1}))(b_{2}(z)-m_{\tilde{B}}(b_{2}))T(f_{1},f_{2})\\\
&\ +(b_{1}(z)-m_{\tilde{B}}(b_{1}))T(f_{1},(b_{2}-b_{2}(z))f_{2})\\\ &\
+(b_{2}(z)-m_{\tilde{B}}(b_{2}))T((b_{1}-b_{1}(z))f_{1},f_{2})+T((b_{1}-b_{1}(z))f_{1},(b_{2}-b_{2}(z))f_{2})\\\
=&(b_{1}(z)-m_{\tilde{B}}(b_{1}))(b_{2}(z)-m_{\tilde{B}}(b_{2}))T(f_{1},f_{2})-(b_{1}(z)-m_{\tilde{B}}(b_{1}))[b_{2},T](f_{1},f_{2})\\\
&\
-(b_{2}(z)-m_{\tilde{B}}(b_{2}))[b_{1},T](f_{1},f_{2})+[b_{1},b_{2},T](f_{1},f_{2}).\end{split}$

Then

$\begin{split}F_{11}\leq&\biggl{|}m_{\tilde{B}}[(b_{1}(z)-m_{\tilde{B}}(b_{1}))(b_{2}(z)-m_{\tilde{B}}(b_{2}))T(f_{1},f_{2})]\biggr{|}\\\
&\
+\biggl{|}m_{\tilde{B}}[(b_{1}(z)-m_{\tilde{B}}(b_{1}))[b_{2},T](f_{1},f_{2})]\biggr{|}\\\
&\
+\biggl{|}m_{\tilde{B}}[(b_{2}(z)-m_{\tilde{B}}(b_{2}))[b_{1},T](f_{1},f_{2})]\biggr{|}\\\
&\
+\biggl{|}m_{\tilde{B}}[(b_{1}(z)-m_{Q}(b_{1}))(b_{2}(z)-m_{Q}(b_{2}))T(f_{1},f_{2})]\biggr{|}\\\
&\
+\biggl{|}m_{\tilde{B}}[(b_{1}(z)-m_{Q}(b_{1}))[b_{2},T](f_{1},f_{2})]\biggr{|}\\\
&\
+\biggl{|}m_{\tilde{B}}[(b_{2}(z)-m_{Q}(b_{2}))[b_{1},T](f_{1},f_{2})]\biggr{|}\\\
=&:J_{1}+J_{2}+J_{3}+J_{4}+J_{5}+J_{6}.\end{split}$

Recall that $K_{\tilde{B},Q}\leq CK_{B,Q}$ with $B\subset\tilde{B}\subset Q$.
Note that $\tilde{B}$ is a doubling ball, similar to estimate for (2.9),
(2.10) and (2.11) respcetively, we immediately obtain

$\begin{split}J_{1}+J_{4}&\leq
CK^{2}_{B,Q}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x),\\\
J_{2}+J_{5}&\leq
CK_{B,Q}||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1},f_{2}))(x),\\\
J_{3}+J_{6}&\leq
CK_{B,Q}||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1},f_{2}))(x).\\\ \end{split}$

For $F_{12}$, with the same method to estimate $F_{11}$, we have

$\begin{split}F_{12}&\leq
CK^{2}_{B,Q}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))(x)\\\
&\ \
+CK_{B,Q}||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))(x)\\\
&\ \
+CK_{B,Q}||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))(x).\\\
\end{split}$

Combining the estimates of $F_{11}$ and $F_{12}$, we complete the estimate for
$F_{1}$.

Now we turn to estimate $F_{2}$. By decomposing the region of the integral, we
have

$\begin{split}F_{2}\leq&|m_{\tilde{B}}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}\backslash\frac{6}{5}\tilde{B}})\\}|\\\
&+|m_{\tilde{B}}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{6^{N}\tilde{B}\backslash\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}})\\}|\\\
&+|m_{\tilde{B}}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{6^{N}\tilde{B}\backslash\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}\backslash\frac{6}{5}\tilde{B}})\\}|\\\
&+|m_{\tilde{B}}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{6^{N}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash
6^{N}\tilde{B}})\\}\\\
&-m_{Q}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{6^{N}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash
6^{N}\tilde{B}})\\}|\\\
&+|m_{\tilde{B}}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash
6^{N}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}})\\}\\\
&-m_{Q}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash
6^{N}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}})\\}|\\\
&+|m_{\tilde{B}}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash
6^{N}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash
6^{N}\tilde{B}})\\}\\\ &-m_{Q}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash
6^{N}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash
6^{N}\tilde{B}})\\}|\\\
&+|m_{Q}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}Q},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}\backslash\frac{6}{5}Q})\\}|\\\
&+|m_{Q}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{6^{N}\tilde{B}\backslash\frac{6}{5}Q},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}Q})\\}|\\\
&+|m_{Q}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{6^{N}\tilde{B}\backslash\frac{6}{5}Q},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}\backslash\frac{6}{5}Q})\\}|\\\
=:&\sum_{i=1}^{9}F_{2i}.\end{split}$

For $F_{21}$, we first compute
$T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}\backslash\frac{6}{5}\tilde{B}})(z)$.

$\begin{split}&|T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}\backslash\frac{6}{5}\tilde{B}})(z)|\\\
\leq&C\int_{6^{N}\tilde{B}\backslash\frac{6}{5}\tilde{B}}\int_{\frac{6}{5}\tilde{B}}\frac{|(b_{1}(y_{1})-m_{Q}(b_{1}))f_{1}(y_{1})(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|}{[\sum\limits_{i=1}^{2}\lambda(z,d(z,y_{i}))]^{2}}d\mu(y_{1})d\mu(y_{2})\\\
\leq&C\int_{\frac{6}{5}\tilde{B}}\frac{|(b_{1}(y_{1})-m_{Q}(b_{1}))f_{1}(y_{1})|}{\lambda(z,d(z,y_{1}))}d\mu(y_{1})\biggl{\\{}\sum_{k=1}^{N-1}\int_{6^{k+1}\tilde{B}\backslash
6^{k}\tilde{B}}\frac{|(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|}{\lambda(z,d(z,y_{2}))}d\mu(y_{2})\\\
&\ \
+\int_{6\tilde{B}\backslash\frac{6}{5}\tilde{B}}\frac{|(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|}{\lambda(z,d(z,y_{2}))}d\mu(y_{2})\biggr{\\}}\\\
\leq&C\frac{\mu(6\tilde{B})}{\lambda(x_{\tilde{B}},\frac{6}{5}r_{\tilde{B}})}\frac{1}{\mu(6\tilde{B})}\int_{\frac{6}{5}\tilde{B}}|(b_{1}(y_{1})-m_{Q}(b_{1}))f_{1}(y_{1})|d\mu(y_{1})\\\
&\ \ \times\biggl{\\{}\sum_{k=1}^{N-1}\frac{\mu(5\times
6^{k+1}\tilde{B})}{\lambda(x_{\tilde{B}},6^{k}r_{\tilde{B}})}\frac{1}{\mu(5\times
6^{k+1}\tilde{B})}\int_{6^{k+1}\tilde{B}}|(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|d\mu(y_{2})\\\
&\ \ +\frac{\mu(5\times
6\tilde{B})}{\lambda(x_{\tilde{B}},6r_{\tilde{B}})}\frac{1}{\mu(5\times
6\tilde{B})}\int_{6\tilde{B}}|(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|d\mu(y_{2})\biggr{\\}}\\\
\leq&CK^{3}_{\tilde{B},Q}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x)\\\
\leq&CK^{3}_{B,Q}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x),\end{split}$

here we have used the property $K_{\tilde{B},Q}\leq CK_{B,Q}$ with
$B\subset\tilde{B}\subset Q$.

Taking the mean over $z\in\tilde{B}$, we obtain

$F_{21}\leq
CK^{3}_{B,Q}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Similar with the method to estimate $F_{21}$, we immediately have

$F_{22}+F_{27}+F_{28}\leq
CK^{3}_{B,Q}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Now we estimate $F_{23}$.

$\begin{split}&|T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{6^{N}\tilde{B}\backslash\frac{6}{5}\tilde{B}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}\backslash\frac{6}{5}\tilde{B}})(z)\\\
\leq&C\int_{6^{N}\tilde{B}\backslash\frac{6}{5}\tilde{B}}\int_{6^{N}\tilde{B}\backslash\frac{6}{5}\tilde{B}}\frac{|(b_{1}(y_{1})-m_{Q}(b_{1}))f_{1}(y_{1})(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|}{[\sum\limits_{i=1}^{2}\lambda(z,d(z,y_{i}))]^{2}}d\mu(y_{1})d\mu(y_{2})\\\
\leq&C\sum_{j=1}^{N-1}\sum_{k=1}^{N-1}\int_{6^{j+1}\tilde{B}\backslash
6^{j}\tilde{B}}\int_{6^{k+1}\tilde{B}\backslash
6^{k}\tilde{B}}\frac{|(b_{1}(y_{1})-m_{Q}(b_{1}))f_{1}(y_{1})(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|}{[\sum\limits_{i=1}^{2}\lambda(z,d(z,y_{i}))]^{2}}d\mu(y_{1})d\mu(y_{2})\\\
&\ \
+\int_{6\tilde{B}\backslash\frac{6}{5}\tilde{B}}\int_{6\tilde{B}\backslash\frac{6}{5}\tilde{B}}\frac{|(b_{1}(y_{1})-m_{Q}(b_{1}))f_{1}(y_{1})(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|}{[\sum\limits_{i=1}^{2}\lambda(z,d(z,y_{i}))]^{2}}d\mu(y_{1})d\mu(y_{2})\\\
\leq&C\sum_{k=1}^{N-1}\frac{\mu(5\times
6^{k+1}\tilde{B})}{\lambda(x_{\tilde{B}},6^{k}r_{\tilde{B}})}\frac{1}{\mu(5\times
6^{k+1}\tilde{B})}\int_{6^{k+1}\tilde{B}}|(b_{1}(y_{1})-m_{Q}(b_{1}))f_{1}(y_{1})|d\mu(y_{1})\\\
&\ \ \times\sum_{j=1}^{N-1}\frac{\mu(5\times
6^{j+1}\tilde{B})}{\lambda(x_{\tilde{B}},6^{j}r_{\tilde{B}})}\frac{1}{\mu(5\times
6^{j+1}\tilde{B})}\int_{6^{j+1}\tilde{B}}|(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|d\mu(y_{2})\\\
&\ \
+\biggl{[}\frac{\mu(30\tilde{B})}{\lambda(x_{\tilde{B}},6r_{\tilde{B}})}\biggr{]}^{2}\prod_{i=1}^{2}\frac{1}{\mu(30\tilde{B})}\int_{6\tilde{B}}|(b_{i}(y_{i})-m_{Q}(b_{i}))f_{i}(y_{i})|d\mu(y_{i})\\\
\leq&CK_{\tilde{B},Q}^{4}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).\\\
\leq&CK_{B,Q}^{4}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).\\\
\end{split}$

Taking the mean over $z\in\tilde{B}$, we obtain

$F_{23}\leq
CK_{B,Q}^{4}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

For $F_{29}$, with the similar method to estimate $F_{23}$, we obtian

$F_{29}\leq
CK_{B,Q}^{4}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Using the similar method to estimate $E_{42}(z)$, it follows

$F_{24}+F_{25}\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Using the similar method to estimate $E_{44}(z)$, we have

$F_{26}\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Next we consider $|h_{B}-h_{\tilde{B}}|$. With the similar method to estimate
$F_{2}$, we easily obtain that

$\begin{split}|h_{B}-h_{\tilde{B}}|&=\biggl{|}m_{B}\biggl{[}T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{\frac{6}{5}B},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}B}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}B},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{\frac{6}{5}B}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}B},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}B}\biggr{)}\biggr{]}\\\
&\ \
-m_{\tilde{B}}\biggl{[}T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}}\biggr{)}\biggr{]}\biggr{|}\\\
&\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).\end{split}$

Thus (2.8) holds and hence (2.1) is proved. With the same method to prove
(2.1), we can obtain that (2.2) and (2.3) are also hold. Here we omit the
details. Thus Lemma 2.6 in this case is proved.

Case 2: $0<l(B)=l<1$. Recall that $\widetilde{B}$ is the smallest
$(\alpha,\beta)$-doubling ball of the form $6^{k}B$ with
$k\in{\mathbf{N}}\bigcup\\{0\\}$. Assume that $B_{0}$, $Q_{0}$ and
$\tilde{B}_{0}$ are concentric with $B$, $Q$ and $\tilde{B}$ respectively and
$l(B_{0})=l(B)^{\alpha}$, $l(Q_{0})=l(Q)^{\alpha}$,
$l(\tilde{B}_{0})=l(\tilde{B})^{\alpha}$, then $B\subset B_{0}$, $Q\subset
Q_{0}$ and $\tilde{B}\subset\tilde{B}_{0}$. Let $\rho$ be a number such that
$6B_{0}=\rho B$. As in the proof of Theorem 9.1 in [35], to obtain (2.4), it
suffices to show that

(2.12)
$\begin{split}&\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}||[b_{1},b_{2},T](f_{1},f_{2})(z)|^{\delta}-|h_{B}|^{\delta}|d\mu(z)\biggr{)}^{1/\delta}\\\
\leq&C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x)+C||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1},f_{2}))(x)\\\
&+C||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1},f_{2}))(x)+C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x),\end{split}$

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

(2.13) $\begin{split}|h_{B}-h_{Q}|&\leq
CK_{B,Q}^{4}\biggr{[}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x)+||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1},f_{2}))(x)\\\
+&||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1},f_{2}))(x)+||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x)\\\
+&||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}))(x)+||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}))(x)\\\
+&||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}))(x)\biggr{]}.\end{split}$

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

$\begin{split}h_{B}:=&m_{B}\biggl{[}T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{\frac{6}{5}B_{0}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}B_{0}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}B_{0}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{\frac{6}{5}B_{0}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}B_{0}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}B_{0}}\biggr{)}\biggr{]}\end{split}$

and

$\begin{split}h_{Q}:=&m_{Q}\biggl{[}T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}Q_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}Q_{0}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}Q_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}Q_{0}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}Q_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}Q_{0}}\biggr{)}\biggr{]}.\end{split}$

Then

$\begin{split}&\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}||[b_{1},b_{2},T](f_{1},f_{2})(z)|^{\delta}-|h_{B}|^{\delta}|d\mu(z)\biggr{)}^{1/\delta}\\\
\leq&C\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}|[b_{1},b_{2},T](f_{1},f_{2})(z)-h_{B}|^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
\leq&C\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}|(b_{1}(z)-m_{\tilde{B}}(b_{1}))(b_{2}(z)-m_{\tilde{B}}(b_{2}))T(f_{1},f_{2})(z)|^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
&\ \
+C\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}|(b_{1}(z)-m_{\tilde{B}}(b_{1}))T(f_{1},(b_{2}-b_{2}(z))f_{2})(z)|^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
&\ \
+C\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}|(b_{2}(z)-m_{\tilde{B}}(b_{2}))T((b_{1}-b_{1}(z))f_{1},f_{2})(z)|^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
&\ \
+C\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}|T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2})(z)-h_{B}|^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
=&:G_{1}+G_{2}+G_{3}+G_{4}.\end{split}$

We first estimate $G_{1}$. Choosing $r_{1},r_{2}>1$ such that
$\dfrac{1}{r}+\dfrac{1}{r_{1}}+\dfrac{1}{r_{2}}=\dfrac{1}{\delta}$. By
Hölder’s inequality, Lemma 2.2 and the fact $B\subset B_{0}$, we have

$\begin{split}G_{1}&\leq
C\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}|b_{1}(z)-m_{\tilde{B}}b_{1}|^{r_{1}}d\mu(z)\biggr{)}^{1/r_{1}}\\\
&\ \
\times\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}|b_{2}(z)-m_{\tilde{B}}b_{2}|^{r_{2}}d\mu(z)\biggr{)}^{1/r_{2}}\\\
&\ \
\times\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B_{0}}|T(f_{1},f_{2})|^{r}d\mu(z)\biggr{)}^{1/r}\\\
&\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x).\end{split}$

For $G_{2}$, let $s>1$ such that
$\dfrac{1}{s}+\dfrac{1}{r}=\dfrac{1}{\delta}$, by Hölder’s inequality, Lemma
2.2 and the fact $B\subset B_{0}$, we deduce

$\begin{split}G_{2}&\leq
C\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}|b_{1}(z)-m_{\tilde{B}}b_{1}|^{s}d\mu(z)\biggr{)}^{1/s}\\\
&\ \
\times\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B_{0}}|[b_{2},T](f_{1},f_{2})|^{r}d\mu(z)\biggr{)}^{1/r}\\\
&\leq C||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1},f_{2}))(x).\end{split}$

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

$G_{3}\leq C||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1},f_{2}))(x).$

Let us turn to estimate $G_{4}$. Denote
$f_{j}^{1}=f_{j}\chi_{\frac{6}{5}B_{0}}$ and $f_{j}^{2}=f_{j}-f_{j}^{1}$ for
$j=1,2$, we have

$\begin{split}&|T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2})(z)-h_{B}|\\\
\leq&|T((b_{1}-m_{\tilde{B}}(b_{1}))f^{1}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{1}_{2})(z)|\\\
&\ \
+|T((b_{1}-m_{\tilde{B}}(b_{1}))f^{1}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})(z)-m_{B}[T((b_{1}-m_{\tilde{B}}(b_{1}))f^{1}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})]|\\\
&\ \
+|T((b_{1}-m_{\tilde{B}}(b_{1}))f^{2}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{1}_{2})(z)-m_{B}[T((b_{1}-m_{\tilde{B}}(b_{1}))f^{2}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{1}_{2})]|\\\
&\ \
+|T((b_{1}-m_{\tilde{B}}(b_{1}))f^{2}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})(z)-m_{B}[T((b_{1}-m_{\tilde{B}}(b_{1}))f^{2}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})]|\\\
=&:G_{41}(z)+G_{42}(z)+G_{43}(z)+G_{44}(z).\end{split}$

Then

$\begin{split}G_{4}&\leq
C\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}E_{41}(z)^{\delta}d\mu(z)\biggr{)}^{1/\delta}+C\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}E_{42}(z)^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
&\ \
+C\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}E_{43}(z)^{\delta}d\mu(z)\biggr{)}^{1/\delta}+C\biggl{(}\frac{1}{\mu(6B_{0})}\int_{B}E_{44}(z)^{\delta}d\mu(z)\biggr{)}^{1/\delta}\\\
&=:G_{41}+G_{42}+G_{43}+G_{44}.\end{split}$

To estimate $G_{41}$, using Kolmogorov’s theorem, Lemma 2.2 and Hölder’s
inequality, we obtain

$\begin{split}G_{41}\leq&C||T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1}^{1},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}^{1})||_{L^{1/2,\infty}(\frac{6}{5}B_{0},\frac{d\mu(z)}{\mu(6B_{0})})}\\\
\leq&C\frac{1}{\mu(6B_{0})}\int_{\frac{6}{5}B_{0}}|(b_{1}(z)-m_{\tilde{B}}(b_{1}))f_{1}(z)|d\mu(z)\\\
&\ \
\times\frac{1}{\mu(6B_{0})}\int_{\frac{6}{5}B_{0}}|(b_{2}(z)-m_{\tilde{B}}(b_{2}))f_{2}(z)|d\mu(z)\\\
\leq&C(\frac{1}{\mu(6B_{0})}\int_{\frac{6}{5}B_{0}}|b_{1}(z)-m_{\tilde{B}}(b_{1})|^{p^{\prime}_{1}}d\mu(z))^{1/p^{\prime}_{1}}\\\
&\ \
\times(\frac{1}{\mu(6B_{0})}\int_{\frac{6}{5}B_{0}}|f_{1}(z)|^{p_{1}}d\mu(z))^{1/p_{1}}\\\
&\ \
\times(\frac{1}{\mu(6B_{0})}\int_{\frac{6}{5}B_{0}}|b_{2}(z)-m_{\tilde{B}}(b_{2})|^{p^{\prime}_{2}}d\mu(z))^{1/p^{\prime}_{2}}\\\
&\ \
\times(\frac{1}{\mu(6B_{0})}\int_{\frac{6}{5}B_{0}}|f_{2}(z)|^{p_{2}}d\mu(z))^{1/p_{2}}\\\
\leq&C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).\end{split}$

To compute $G_{42}$, let $z,y\in B$, $y_{1}\in\frac{6}{5}B_{0}$ and $y_{2}\in
X\backslash\frac{6}{5}B_{0}$, then $\max\limits_{1\leq i\leq 2}d(z,y_{i})\geq
d(z,y_{2})\geq Cl(B_{0})=Cl(B)^{\alpha}\geq Cd(z,y)^{\alpha}$. Using
Definition 1.6, Lemma 2.2, Lemma 2.3, Hölder’s inequality and the properties
of $\lambda$, we know

$\begin{split}&|T((b_{1}-m_{\tilde{B}}(b_{1}))f^{1}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})(z)-T((b_{1}-m_{\tilde{B}}(b_{1}))f^{1}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})(y)|\\\
\leq&C\int_{X\backslash\frac{6}{5}B_{0}}\int_{\frac{6}{5}B_{0}}|K(z,y_{1},y_{2})-K(y,y_{1},y_{2})||\prod_{i=1}^{2}(b_{i}(y_{i})-m_{\tilde{B}}(b_{i}))f_{i}(y_{i})|d\mu(y_{1})d\mu(y_{2})\\\
\leq&C\int_{X\backslash\frac{6}{5}B_{0}}\int_{\frac{6}{5}B_{0}}\frac{d(z,y)^{\delta}}{\biggl{[}\sum\limits_{i=1}^{2}d(z,y_{i})\biggr{]}^{\delta/\alpha}\biggl{[}\sum\limits_{i=1}^{2}\lambda(z,d(z,y_{i}))\biggr{]}^{2}}\\\
&\ \ \
\times|\prod_{i=1}^{2}(b_{i}(y_{i})-m_{\tilde{B}}(b_{i}))f_{i}(y_{i})|d\mu(y_{1})d\mu(y_{2})\\\
\leq&C\int_{\frac{6}{5}B_{0}}\frac{|b_{1}(y_{1})-m_{\tilde{B}}(b_{1})||f_{1}(y_{1})|}{\lambda(z,d(z,y_{1}))}d\mu(y_{1})\\\
&\ \ \
\times\int_{X\backslash\frac{6}{5}B_{0}}\frac{d(z,y)^{\delta}}{d(z,y_{2})^{\delta/\alpha}}\frac{|b_{2}(y_{2})-m_{\tilde{B}}(b_{2})||f_{2}(y_{2})|}{\lambda(z,d(z,y_{2}))}d\mu(y_{2})\\\
\leq&C\frac{\mu(6B_{0})}{\lambda(x_{B},\frac{6}{5}r_{B_{0}})}\frac{1}{\mu(6B_{0})}\int_{\frac{6}{5}B_{0}}|b_{1}(y_{1})-m_{\tilde{B}}(b_{1})||f_{1}(y_{1})|d\mu(y_{1})\\\
&\ \ \ \times\sum_{k=1}^{\infty}\int_{6^{k}\frac{6}{5}B_{0}\backslash
6^{k-1}\frac{6}{5}B_{0}}\frac{d(z,y)^{\delta}}{d(z,y_{2})^{\delta/\alpha}}\frac{|b_{2}(y_{2})-m_{\tilde{B}}(b_{2})||f_{2}(y_{2})|}{\lambda(z,d(z,y_{2}))}d\mu(y_{2})\\\
\end{split}$

$\begin{split}&\leq
C(\frac{1}{\mu(6B_{0})}\int_{\frac{6}{5}B_{0}}|b_{1}(y_{1})-m_{\tilde{B}}(b_{1})|^{p^{\prime}_{1}}d\mu(y_{1}))^{1/p^{\prime}_{1}}\\\
&\ \ \
\times(\frac{1}{\mu(6B_{0})}\int_{\frac{6}{5}B_{0}}|f_{1}(y_{1})|^{p_{1}}d\mu(y_{1}))^{1/p_{1}}\\\
&\ \ \
\times\sum_{k=1}^{\infty}\frac{1}{\lambda(x_{B},6^{k-1}\frac{6}{5}r_{B_{0}})}6^{-k\delta/\alpha}\int_{6^{k}\frac{6}{5}B_{0}}|b_{2}(y_{2})-m_{\tilde{B}}(b_{2})||f_{2}(y_{2})|d\mu(y_{2})\\\
&\leq
C||b_{1}||_{\ast}M_{p_{1},(5)}f_{1}(x)\sum_{k=1}^{\infty}6^{-k\delta/\alpha}\biggl{[}\biggl{(}\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B_{0})}\\\ &\ \ \
\times\int_{6^{k}\frac{6}{5}B_{0}}|b_{2}(y_{2})-m_{\widetilde{6^{k}\frac{6}{5}B_{0}}}(b_{2})|^{p^{\prime}_{2}}d\mu(y_{2})\biggr{)}^{1/p^{\prime}_{2}}\\\
&\ \ \ \times\biggl{(}\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B_{0})}\int_{6^{k}\frac{6}{5}B_{0}}|f_{2}(y_{2})|^{p_{2}}d\mu(y_{2})\biggr{)}^{1/p_{2}}\\\
&\ \ \ +Ck||b_{2}||_{\ast}\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B_{0})}\int_{6^{k}\frac{6}{5}B_{0}}|f_{2}(y_{2})|d\mu(y_{2})\biggr{]}\\\
&\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).\\\
\end{split}$

Taking the mean over $y\in B$, we have

$G_{42}(z)\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Thus

$G_{42}\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Similarly, we get

$G_{43}\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

For $G_{44}$, by Definition 1.6, Lemma 2.2, Lemma 2.3, Hölder’s inequality and
the properties of $\lambda$, we obtain

$\begin{split}&|T((b_{1}-m_{\tilde{B}}(b_{1}))f^{2}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})(z)-T((b_{1}-m_{\tilde{B}}(b_{1}))f^{2}_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f^{2}_{2})(y)|\\\
\leq&C\int_{X\backslash\frac{6}{5}B_{0}}\int_{X\backslash\frac{6}{5}B_{0}}|K(z,y_{1},y_{2})-K(y,y_{1},y_{2})|\\\
&\ \ \
\times|\prod_{i=1}^{2}(b_{i}(y_{i})-m_{\tilde{B}}b_{i})f_{i}(y_{i})|d\mu(y_{1})d\mu(y_{2})\\\
\leq&C\int_{X\backslash\frac{6}{5}B_{0}}\int_{X\backslash\frac{6}{5}B_{0}}\frac{d(z,y)^{\delta}|\prod_{i=1}^{2}(b_{i}(y_{i})-m_{\tilde{B}}b_{i})f_{i}(y_{i})|d\mu(y_{1})d\mu(y_{2})}{(d(z,y_{1})+d(z,y_{2}))^{\delta/\alpha}[\sum_{j=1}^{2}\lambda(z,d(z,y_{j}))]^{2}}\\\
\leq&C\prod_{i=1}^{2}\int_{X\backslash\frac{6}{5}B_{0}}\frac{d(z,y)^{\delta_{i}}|b_{i}(y_{i})-m_{\tilde{B}}b_{i}||f_{i}(y_{i})|d\mu(y_{i})}{d(z,y_{i})^{\delta_{i}/\alpha}\lambda(x_{B},d(z,y_{i}))}\\\
\leq&C\prod_{i=1}^{2}\sum_{k=1}^{\infty}\int_{6^{k}\frac{6}{5}B_{0}}6^{-k\delta_{i}/\alpha}\frac{\mu(5\times
6^{k}\frac{6}{5}B_{0})}{\lambda(x_{B},5\times 6^{k}\frac{6}{5}r_{B_{0}})}\\\
&\ \ \times\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B_{0})}|b_{i}(y_{i})-m_{\tilde{B}}(b_{i})||f_{i}(y_{i})|d\mu(y_{i})\\\
\leq&C\prod_{i=1}^{2}\sum_{k=1}^{\infty}6^{-k\delta_{i}/\alpha}(\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B_{0})}\int_{6^{k}\frac{6}{5}B_{0}}|b_{i}(y_{i})-m_{\tilde{B}}(b_{i})|^{p^{\prime}_{i}}d\mu(y_{i}))^{1/p^{\prime}_{i}}\\\
&\ \ \ \times(\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B_{0})}\int_{6^{k}\frac{6}{5}B_{0}}|f_{i}(y_{i})|^{p_{i}})^{1/p_{i}}\\\
\leq&C\prod_{i=1}^{2}\sum_{k=1}^{\infty}6^{-k\delta_{i}/\alpha}M_{p_{i},(5)}f_{i}(x)(\frac{1}{\mu(5\times
6^{k}\frac{6}{5}B_{0})}\int_{6^{k}\frac{6}{5}B_{0}}|b_{i}(y_{i})-m_{\widetilde{6^{k}\frac{6}{5}B_{0}}}(b_{i})\\\
&\ \ \
+m_{\widetilde{6^{k}\frac{6}{5}B_{0}}}(b_{i})-m_{\tilde{B}}(b_{i})|^{p^{\prime}_{i}}d\mu(y_{i}))^{1/p^{\prime}_{i}}\\\
\leq&C\prod_{i=1}^{2}\sum_{k=1}^{\infty}6^{-k\delta_{i}/\alpha}k||b_{i}||_{\ast}M_{p_{i},(5)}f_{i}(x)\\\
\leq&C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).\end{split}$

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

Taking the mean over $y\in B$, then

$G_{44}(z)\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Therefore

$G_{44}\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

So (2.12) can be obtained.

Next we prove (2.13). Consider two balls $B\subset Q$ with $x\in B$, where $B$
is an arbitrary ball and $Q$ is a doubling ball. Assume that $B_{0}$, $Q_{0}$
and $\tilde{B}_{0}$ are concentric with $B$, $Q$ and $\tilde{B}$ respectively
and $l(B_{0})=l(B)^{\alpha}$, $l(Q_{0})=l(Q)^{\alpha}$,
$l(\tilde{B}_{0})=l(\tilde{B})^{\alpha}$, then $B\subset B_{0}$, $Q\subset
Q_{0}$ and $\tilde{B}\subset\tilde{B}_{0}$. Thus we have the four cases:
$B\subset Q\subset Q_{0}\subset\tilde{B}\subset\tilde{B}_{0}$, $B\subset
Q\subset\tilde{B}\subset Q_{0}\subset\tilde{B}_{0}$,
$B\subset\tilde{B}\subset\tilde{B}_{0}\subset Q\subset Q_{0}$,
$B\subset\tilde{B}\subset Q\subset\tilde{B}_{0}\subset Q_{0}$. Without loss of
generality, we only consider $B\subset\tilde{B}\subset\tilde{B}_{0}\subset
Q\subset Q_{0}$ in this paper. The other cases can be considered by the same
method.

Let $N=N_{\tilde{B}_{0},Q_{0}}+1$. It is easy to see that

$|h_{B}-h_{Q}|\leq|h_{B}-h_{\tilde{B}}|+|h_{\tilde{B}}-h_{Q}|.$

Now we first consider $|h_{\tilde{B}}-h_{Q}|$. Write

$\begin{split}|h_{\tilde{B}}-h_{Q}|\leq&\biggl{|}m_{\tilde{B}}\biggl{[}T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}_{0}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}_{0}}\biggr{)}\biggr{]}\\\
&\ \
-m_{\tilde{B}}\biggl{[}T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}_{0}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}_{0}}\biggr{)}\biggr{]}\biggr{|}\\\
&\ \
+\biggl{|}m_{\tilde{B}}\biggl{[}T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}_{0}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}\tilde{B}_{0}}\biggr{)}\biggr{]}\\\
&\ \
-m_{Q}\biggl{[}T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}Q_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}Q_{0}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}Q_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}Q_{0}}\biggr{)}\\\
&\ \
+T\biggl{(}(b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash\frac{6}{5}Q_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash\frac{6}{5}Q_{0}}\biggr{)}\biggr{]}\biggr{|}\\\
=:&H_{1}+H_{2}.\end{split}$

To estimate $H_{1}$, write

$\begin{split}H_{1}&\leq\biggl{|}m_{\tilde{B}}\biggl{[}T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2})-T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}})\biggr{]}\\\
&\ \
-m_{\tilde{B}}\biggl{[}T((b_{1}-m_{Q}(b_{1}))f_{1},(b_{2}-m_{Q}(b_{2}))f_{2})-T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}})\biggr{]}\biggr{|}\\\
&\leq\biggl{|}m_{\tilde{B}}\biggl{[}T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2})-T((b_{1}-m_{Q}(b_{1}))f_{1},(b_{2}-m_{Q}(b_{2}))f_{2})\biggr{]}\biggr{|}\\\
&\ \
+\biggl{|}m_{\tilde{B}}\biggl{[}T((b_{1}-m_{\tilde{B}}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{\tilde{B}}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}})\\\
&\ \
-T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}})\biggr{]}\biggr{|}\\\
&=:H_{11}+H_{12}.\end{split}$

For $H_{11}$, since it equals to $F_{11}$, we immediately have

$\begin{split}H_{11}&\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1},f_{2}))(x)+C||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1},f_{2}))(x)\\\
&\ \ +C||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1},f_{2}))(x).\\\ \end{split}$

For $H_{12}$, with the same method to estimate $F_{11}$, we have

$\begin{split}H_{12}&\leq
CK^{2}_{B,Q}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{r,(6)}(T(f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}))(x)\\\
&\ \
+CK_{B,Q}||b_{1}||_{\ast}M_{r,(6)}([b_{2},T](f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}))(x)\\\
&\ \
+CK_{B,Q}||b_{2}||_{\ast}M_{r,(6)}([b_{1},T](f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}}))(x).\\\
\end{split}$

Combining the estimates of $H_{11}$ and $H_{12}$, we complete the estimate for
$H_{1}$.

Now we turn to estimate $H_{2}$. By decomposing the region of the integral, we
have

$\begin{split}H_{2}\leq&|m_{\tilde{B}}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}_{0}\backslash\frac{6}{5}\tilde{B}_{0}})\\}|\\\
&+|m_{\tilde{B}}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{6^{N}\tilde{B}_{0}\backslash\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}\tilde{B}_{0}})\\}|\\\
&+|m_{\tilde{B}}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{6^{N}\tilde{B}_{0}\backslash\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}_{0}\backslash\frac{6}{5}\tilde{B}_{0}})\\}|\\\
&+|m_{\tilde{B}}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{6^{N}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash
6^{N}\tilde{B}_{0}})\\}\\\
&-m_{Q}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{6^{N}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash
6^{N}\tilde{B}_{0}})\\}|\\\
&+|m_{\tilde{B}}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash
6^{N}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}_{0}})\\}\\\
&-m_{Q}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash
6^{N}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}_{0}})\\}|\\\
&+|m_{\tilde{B}}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash
6^{N}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash
6^{N}\tilde{B}_{0}})\\}\\\
&-m_{Q}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{X\backslash
6^{N}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{X\backslash
6^{N}\tilde{B}_{0}})\\}|\\\
&+|m_{Q}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}Q_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}B_{0}\backslash\frac{6}{5}Q_{0}})\\}|\\\
&+|m_{Q}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{6^{N}\tilde{B}_{0}\backslash\frac{6}{5}Q_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{\frac{6}{5}Q_{0}})\\}|\\\
&+|m_{Q}\\{T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{6^{N}\tilde{B}_{0}\backslash\frac{6}{5}Q_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}_{0}\backslash\frac{6}{5}Q_{0}})\\}|\\\
=:&\sum_{i=1}^{9}H_{2i}.\end{split}$

Similar to estimate $F_{2i},1\leq i\leq 9$, we obtain that

$\begin{split}H_{21}+H_{22}+H_{27}+H_{28}&\leq
CK^{3}_{B,Q}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x),\\\
H_{23}+H_{29}&\leq
CK_{B,Q}^{4}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x),\\\
H_{24}+H_{25}+H_{26}&\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).\\\
\end{split}$

For simplicity, we only computer $H_{21}$. We write

$\begin{split}&|T((b_{1}-m_{Q}(b_{1}))f_{1}\chi_{\frac{6}{5}\tilde{B}_{0}},(b_{2}-m_{Q}(b_{2}))f_{2}\chi_{6^{N}\tilde{B}_{0}\backslash\frac{6}{5}\tilde{B}_{0}})(z)|\\\
\leq&C\int_{6^{N}\tilde{B}_{0}\backslash\frac{6}{5}\tilde{B}_{0}}\int_{\frac{6}{5}\tilde{B}_{0}}\frac{|(b_{1}(y_{1})-m_{Q}(b_{1}))f_{1}(y_{1})(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|}{[\sum\limits_{i=1}^{2}\lambda(z,d(z,y_{i}))]^{2}}d\mu(y_{1})d\mu(y_{2})\\\
\leq&C\int_{\frac{6}{5}\tilde{B}_{0}}\frac{|(b_{1}(y_{1})-m_{Q}(b_{1}))f_{1}(y_{1})|}{\lambda(z,d(z,y_{1}))}d\mu(y_{1})\biggl{\\{}\sum_{k=1}^{N-1}\int_{6^{k+1}\tilde{B}_{0}\backslash
6^{k}\tilde{B}_{0}}\frac{|(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|}{\lambda(z,d(z,y_{2}))}d\mu(y_{2})\\\
&\ \
+\int_{6\tilde{B}_{0}\backslash\frac{6}{5}\tilde{B}_{0}}\frac{|(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|}{\lambda(z,d(z,y_{2}))}d\mu(y_{2})\biggr{\\}}\\\
\leq&C\frac{\mu(6\tilde{B}_{0})}{\lambda(x_{\tilde{B}_{0}},\frac{6}{5}r_{\tilde{B}_{0}})}\frac{1}{\mu(6\tilde{B}_{0})}\int_{\frac{6}{5}\tilde{B}_{0}}|(b_{1}(y_{1})-m_{Q}(b_{1}))f_{1}(y_{1})|d\mu(y_{1})\\\
&\ \ \times\biggl{\\{}\sum_{k=1}^{N-1}\frac{\mu(5\times
6^{k+1}\tilde{B}_{0})}{\lambda(x_{\tilde{B}_{0}},6^{k}r_{\tilde{B}_{0}})}\frac{1}{\mu(5\times
6^{k+1}\tilde{B}_{0})}\int_{6^{k+1}\tilde{B}_{0}}|(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|d\mu(y_{2})\\\
&\ \ +\frac{\mu(5\times
6\tilde{B}_{0})}{\lambda(x_{\tilde{B}_{0}},6r_{\tilde{B}_{0}})}\frac{1}{\mu(5\times
6\tilde{B}_{0})}\int_{6\tilde{B}_{0}}|(b_{2}(y_{2})-m_{Q}(b_{2}))f_{2}(y_{2})|d\mu(y_{2})\biggr{\\}}\\\
\leq&CK^{2}_{\tilde{B}_{0},Q}K_{\tilde{B}_{0},Q_{0}}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x)\\\
\leq&CK^{3}_{B,Q}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x),\end{split}$

here we have used the conclusions $K_{\tilde{B}_{0},Q}\leq CK_{B,Q}$ and
$K_{\tilde{B}_{0},Q_{0}}\leq CK_{B,Q}$, which can be obtained by Lemma 2.4.

Taking the mean over $z\in\tilde{B}$, we obtain

$H_{21}\leq
CK^{3}_{B,Q}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Combining the estimate for $H_{2i},1\leq i\leq 9$, then

$H_{2}\leq
CK_{B,Q}^{4}||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Next we consider $|h_{B}-h_{\tilde{B}}|$. With the similar method to estimate
$H_{2}$, we easily obtain that

$|h_{B}-h_{\tilde{B}}|\leq
C||b_{1}||_{\ast}||b_{2}||_{\ast}M_{p_{1},(5)}f_{1}(x)M_{p_{2},(5)}f_{2}(x).$

Thus (2.13) holds and hence (2.4) is proved. With the same method to prove
(2.4), we can obtain that (2.5) and (2.6) also hold. Here we omit the details.
Thus Lemma 2.6 has been proved. ∎

###### Proof of Theorem 1.11.

Let $0<\delta<1/2$, $1<p_{1},p_{2},q<\infty$,
$\dfrac{1}{q}=\dfrac{1}{p_{1}}+\dfrac{1}{p_{2}}$, $1<r<q$, $f_{1}\in
L^{p_{1}}(\mu)$, $f_{2}\in L^{p_{2}}(\mu)$, $b_{1}\in\text{RBMO}(\mu)$ and
$b_{2}\in\text{RBMO}(\mu)$.

When $l(B)\geq 1$, by $|f(x)|\leq N_{\delta}f(x)$, Lemma 2.1, Lemma 2.5, Lemma
2.6, Hölder’s inequality and the boundedness of $M_{(\rho)}$ and
$M_{r,(\rho)}$ for $\rho\geq 5$ and $q>r$, it follows

$\begin{split}&||[b_{1},b_{2},T](f_{1},f_{2})||_{L^{q}(\mu)}\leq||N_{\delta}([b_{1},b_{2},T](f_{1},f_{2}))||_{L^{q}(\mu)}\leq
C||M^{\sharp}_{\delta}([b_{1},b_{2},T](f_{1},f_{2}))||_{L^{q}(\mu)}\\\
\leq&C||b_{1}||_{\ast}||b_{2}||_{\ast}||M_{r,(6)}(T(f_{1},f_{2}))||_{L^{q}(\mu)}+C||b_{1}||_{\ast}||M_{r,(6)}([b_{2},T](f_{1},f_{2}))||_{L^{q}(\mu)}\\\
&+C||b_{2}||_{\ast}||M_{r,(6)}([b_{1},T](f_{1},f_{2}))||_{L^{q}(\mu)}+C||b_{1}||_{\ast}||b_{2}||_{\ast}||M_{p_{1},(5)}f_{1}M_{p_{2},(5)}f_{2}||_{L^{q}(\mu)}\\\
&+C||b_{1}||_{\ast}||b_{2}||_{\ast}||M_{r,(6)}(T(f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))||_{L^{q}(\mu)}\\\
&+C||b_{1}||_{\ast}||M_{r,(6)}([b_{2},T](f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))||_{L^{q}(\mu)}\\\
&+C||b_{2}||_{\ast}||M_{r,(6)}([b_{1},T](f_{1}\chi_{\frac{6}{5}\tilde{B}},f_{2}\chi_{\frac{6}{5}\tilde{B}}))||_{L^{q}(\mu)}\\\
\leq&C||b_{1}||_{\ast}||b_{2}||_{\ast}||f_{1}||_{L^{p_{1}}(\mu)}||f_{2}||_{L^{p_{2}}(\mu)}+C||b_{1}||_{\ast}||([b_{2},T](f_{1},f_{2}))||_{L^{q}(\mu)}\\\
&+C||b_{2}||_{\ast}||([b_{1},T](f_{1},f_{2}))||_{L^{q}(\mu)}\\\
\leq&C||b_{1}||_{\ast}||b_{2}||_{\ast}||f_{1}||_{L^{p_{1}}(\mu)}||f_{2}||_{L^{p_{2}}(\mu)}+C||b_{1}||_{\ast}||M^{\sharp}_{\delta}([b_{2},T](f_{1},f_{2}))||_{L^{q}(\mu)}\\\
&+C||b_{2}||_{\ast}||M^{\sharp}_{\delta}([b_{1},T](f_{1},f_{2}))||_{L^{q}(\mu)}\\\
\leq&C||b_{1}||_{\ast}||b_{2}||_{\ast}||f_{1}||_{L^{p_{1}}(\mu)}||f_{2}||_{L^{p_{2}}(\mu)}+C||b_{1}||_{\ast}||M_{r,(6)}(T(f_{1},f_{2}))||_{L^{q}(\mu)}\\\
&+C||b_{1}||_{\ast}||M_{p_{1},(5)}f_{1}M_{p_{2},(5)}f_{2}||_{L^{q}(\mu)}+C||b_{2}||_{\ast}||M_{r,(6)}(T(f_{1},f_{2}))||_{L^{q}(\mu)}\\\
&+C||b_{2}||_{\ast}||M_{p_{1},(5)}f_{1}M_{p_{2},(5)}f_{2}||_{L^{q}(\mu)}\\\
\leq&C||b_{1}||_{\ast}||b_{2}||_{\ast}||f_{1}||_{L^{p_{1}}(\mu)}||f_{2}||_{L^{p_{2}}(\mu)}+C||M_{p_{1},(5)}f_{1}||_{L^{p_{1}}(\mu)}||M_{p_{2},(5)}f_{2}||_{L^{p_{2}}(\mu)}\\\
\leq&C||b_{1}||_{\ast}||b_{2}||_{\ast}||f_{1}||_{L^{p_{1}}(\mu)}||f_{2}||_{L^{p_{2}}(\mu)}.\end{split}$

When $0<l(B)<1$, using the above method, we also obtain the results of Theorem
1.11. Thus the proof of Theorem 1.11 has been completed. ∎

Acknowledgements. The authors would like to thank the referees for their very
carefully reading and many valuable suggestions. The work was supported by
Natural Science Foundation of China (No.11971026), Natural Science Foundation
of Education Committee of Anhui Province (Nos.KJ2016A506, KJ2017A454) and
Excellent Young Talents Foundation of Anhui Province (Nos.GXYQ2017070,
GXYQ2020049).

## References

  * [1] J. Alvarez, M. Milman. $H^{p}$ continuity properties of Calderón-Zygmund-type operators. J. Math. Anal. Appl., 1986, 118: 63-79.
  * [2] J. Alvarez, M. Milman. Vector valued inequalities for strongly singular Calderón-Zygmund operators. Rev. Mat. Iberoam., 1986, 2: 405-426.
  * [3] T. Bui, X. Duong. Hardy Spaces, regularized BMO spaces and the boundedness of Calderón-Zygmund operators on non-homogeneous spaces. J. Geom. Anal., 2013, 23: 895-932.
  * [4] J. Chen, H. Lin. Maximal multilinear commutators on non-homogeneous metric measure spaces. Taiwanese J. Math., 2017, 21: 1133-1160.
  * [5] R. Coifman. A real variable characterization of $H^{p}$. Studia Math., 1974, 51: 269-274.
  * [6] R. Coifman, Y. Meyer. On commutators of singular integrals and bilinear singular integrals. Trans. Amer. Math. Soc., 1975, 212: 315-331.
  * [7] C. Fefferman, E. Stein. $H^{p}$ spaces of several variables. Acta Math., 1972, 129: 137-193.
  * [8] X. Fu, H. Lin, Da. Yang, Do. Yang. Hardy spaces $\mathcal{H}^{p}$ over non-homogeneous metric measure spaces and their applications. Sci. China Math., 2015, 58: 309-388.
  * [9] X. Fu, D. Yang, W. Yuan. Boundedness of multilinear commutators of Calderón-Zygmund operators on Orlicz spaces over non-homogeneous spaces. Taiwanese J. Math., 2012, 16: 2203-2238.
  * [10] X. Fu, D. Yang, W. Yuan. Generalized fractional integrals and their commutators over non-homogeneous metric measure spaces. Taiwanese J. Math., 2014, 18: 509-557.
  * [11] H. Gong, R. Xie, C. Xu. Multilinear fractional integral operators on non-homogeneous metric measure spaces. J. Ineq. Appl., 2016, 275: 1-17.
  * [12] L. Grafakos, R. Torres. Multilinear Calderón-Zygmund theory. Adv. Math., 2002, 165: 124-164.
  * [13] L. Grafakos, R. Torres. On multilinear singular integrals of Calderón-Zygmund type. Publ. Mat., 2002, 57-91. (extra)
  * [14] G. Hu, Y. Meng, D. Yang. Multilinear commutators of singular integrals with non-doubling measures. Integral Equ. Oper. Theory, 2005, 51: 235-255.
  * [15] G. Hu, Y. Meng, D. Yang. New atomic characterization of $\mathcal{H}^{1}$ space with non-doubling measures and its applications. Math. Proc. Cambridge, 2005, 138: 151-171.
  * [16] G. Hu, Y. Meng, D. Yang. Weighted norm inequalities for multilinear Calderón-Zygmund operators on non-homogeneous metric measure spaces. Forum Math., 2014, 26: 1289-1322.
  * [17] T. Hytönen. A framework for non-homogeneous analysis on metric spaces, and the RBMO space of Tolsa. Publ. Mat., 2010, 54: 485-504.
  * [18] T. Hytönen, H. Martikainen. Non-homogeneous $Tb$ theorem and random dyadic cubes on metric measure spaces. J. Geom. Anal., 2012, 22: 1071-1107.
  * [19] T. Hytönen, S. Liu, Da. Yang, Do. Yang. Boundedness of Calderón-Zygmund operators on non-homogeneous metric measure spaces. Canad. J. Math., 2012, 64(4): 892-923.
  * [20] T. Hytönen, Da. Yang, Do. Yang. The Hardy space $\mathcal{H}^{1}$ on non-homogeneous metric spaces. Math Proc. Cambridge, 2012, 153(01): 9-31.
  * [21] H. Lin, S. Wu, D. Yang. Boundedness of certain commutators over non-homogeneous metric measure spaces. Anal. Math. Phys., 2017, 7: 187-218.
  * [22] H. Lin, D. Yang. Spaces of type BLO on non-homogeneous metric measure spaces. Front Math. China, 2011, 6: 271-292.
  * [23] H. Lin, D. Yang. An interpolation theorem for sublinear operators on non-homogeneous metric measure spaces. Banach J. Math. Anal, 2012, 6: 168-179.
  * [24] H. Lin, D. Yang. Equivalent boundedness of Marcinkiewicz integrals on non-homogeneous metric measure spaces. Sci. China Math., 2014, 57: 123-144.
  * [25] Y. Lin. Strongly singular Calderón-Zygmund operator and commutator on Morrey type spaces. Acta Math. Sinica, English Series, 2007, 23: 2097-2110.
  * [26] Y. Lin, S. Lu. Toeplitz operators related to strongly singular Calderón-Zygmund operators. Science in China, Ser. A, 2006, 49: 1048-1064.
  * [27] Y. Lin, S. Lu. Boundedness of commutators on Hardy-type spaces. Integral Equ. Oper. Theory, 2007, 57: 381-396.
  * [28] Y. Lin, S. Lu. Strongly singular Calderón-Zygmund operators and their commutators. Jordan J. Math. Stat., 2008, 1: 31-49.
  * [29] S. Liu, Y. Meng, D. Yang. Boundedness of maximal Calderón-Zygmund operators on non-homogeneous metric measure spaces. Proc. Roy. Soc. Edinburgh Sect. A., 2014, 144: 567-589.
  * [30] S. Liu, Da. Yang, Do. Yang. Boundedness of Calderón-Zygmund operators on non-homogeneous metric measure spaces: equivalent characterizations. J. Math. Anal. Appl., 2012, 386: 258-272.
  * [31] F. Nazarov, S. Treil, A. Volberg. The $Tb$-theorem on non-homogeneous spaces. Acta Math., 2003, 190: 151-239.
  * [32] C. P$\acute{e}$rez, R. Torres. Sharp maximal function estimates for multilinear singular integrals. Contemp. Math., 2003, 320: 323-331.
  * [33] C. P$\acute{e}$rez, G. Pradolini, R. Torres, R. Trujillo-Gonz$\acute{a}$lez. End-point estimates for iterated commutators of multilinear singular integrals. Bull Lond. Math. Soc., 2014, 46: 26-42.
  * [34] X. Tolsa. Painlevé’s problem and the semiadditivity of analytic capacity. Acta Math., 2003, 190: 105-149.
  * [35] X. Tolsa. BMO, $\mathcal{H}^{1}$ and Calderón-Zygmund operators for non-doubling measures. Math. Ann., 2001, 319: 89-149.
  * [36] H. Wang, R. Xie. Multilinear strongly singular integral operators on non-homogeneous metric measure spaces. Submitted.
  * [37] R. Xie, H. Gong, X. Zhou. Commutators of multilinear singular integral operators on non-homogeneous metric measure spaces. Taiwanese J. Math., 2015, 19: 703-723.
  * [38] R. Xie, L. Shu. $\theta$-type Calderón-Zygmund operators with non-doubling measures. Acta Math. Appl. Sinica, English Series, 2013, 29: 263-280.
  * [39] J. Xu. Boundedness of multilinear singular integrals for non-doubling measures. J. Math. Anal. Appl., 2007, 327: 471-480.
  * [40] J. Xu. Boundedness in Lebesgue spaces for commutators of multilinear singular integrals and RBMO functions with non-doubling measures. Science in China. Series A, 2007, 50: 361-376.
  * [41] Da. Yang, Do. Yang, G. Hu. The Hardy Space $H^{1}$ with non-doubling measures and their applications. Lecture Notes in Math. 2084, Springer-Verlag, Berlin, 2013.
  * [42] T. Zheng, X. Tao. Boundedness for iterated commutators of multilinear singular integrals of Dini’s type on non-homogeneous metric measure spaces (in Chinese). Sci. Sin. Math., 2017, 47: 1029-1046.