Open Access

The boundedness of Marcinkiewicz integrals commutators on non-homogeneous metric measure spaces

Journal of Inequalities and Applications20152015:259

Received: 3 March 2015

Accepted: 13 August 2015

Published: 27 August 2015


Let \((\mathcal{X},d,\mu)\) be a metric measure space satisfying the upper doubling condition and geometrically doubling condition in the sense of Hytönen. In this paper, the authors establish the boundedness of the commutator generated by the \(\operatorname {RBMO}(\mu)\) function and the Marcinkiewicz integral with kernel satisfying a Hörmander-type condition, respectively, from \(L^{p}(\mu)\) with \(1< p<\infty\) to itself.


non-homogeneous metric measure spaces Marcinkiewicz integral commutator Lebesgue space \(\operatorname {RBMO}(\mu)\)


42B20 42B35 47B47 30L99

1 Introduction

In 1938, Marcinkiewicz [1] introduced the integral on one-dimensional Euclidean space \(\mathbb{R}\), which is today called the Marcinkiewicz integral, and conjectured that it is bounded on \(L^{p}([0,2\pi])\), \(1< p<\infty\). Zygmund in [2] proved the Marcinkiewicz conjecture. In 1958, Stein [3] generalized the above Marcinkiewicz integral to the higher-dimensional case. Let Ω be homogeneous of degree zero in \(\mathbb{R}^{n}\), \(n\geq2\), integrable and have mean value zero on the unit sphere \(\mathcal{S}^{n-1}\). The higher-dimensional Marcinkiewicz integral is then defined by
$$\mathcal{M}_{\Omega}(f) (x)= \biggl\{ \int_{0}^{\infty} \biggl\vert \int_{|x-y|\leq t}\frac {\Omega(x-y)}{|x-y|^{n-1}}f(y)\,dy\biggr\vert ^{2}\frac{dt}{t^{3}}\biggr\} ^{\frac{1}{2}},\quad x\in \mathbb{R}^{n}. $$
Stein [3] proved that if \(\Omega\in \operatorname {Lip}_{\alpha}(\mathcal {S}^{n-1})\) for some \(\alpha\in(0,1]\), then \(\mathcal{M}_{\Omega}\) is bounded on \(L^{p}(\mathbb{R}^{n})\) for \(p\in(1,2]\) and also bounded from \(L^{1}(\mathbb{R}^{n})\) to \(L^{1,\infty}(\mathbb{R}^{n})\). Since then, many papers focused on the boundedness of this operator on various function spaces. We refer the reader to [412] for its developments and applications.

The main purpose of this paper is to establish the bound of the commutator generated by the Marcinkiewicz integral and the \(\operatorname {RBMO}(\mu)\) function on the non-homogeneous metric measure spaces.

During the past 10 to 15 years, considerable attention has been paid to the study of the classical theory of harmonic analysis on Euclidean spaces with non-doubling measures only satisfying the polynomial growth condition (see [1321]). To be precise, let μ be a positive Radon measure on \(\mathbb{R}^{d}\) with satisfies the polynomial growth condition that, for all \(x\in\mathbb{R}^{d}\) and \(r>0\),
$$ \mu\bigl(B(x,r)\bigr)\leq c_{0}r^{n}, $$
where \(c_{0}\) is a positive constants and \(0< n\leq d\), and \(B(x,r)\) is the open ball centered at x and having radius r. The analysis associated with such non-doubling measure μ has proved to play a striking role in solving the long-standing open Painlevé’s problem and Vitushkin’s conjecture by Tolsa [19]. The non-doubling measure μ may not satisfy the well-known doubling condition, which is a key assumption in harmonic analysis on spaces of homogeneous type in the sense of Coifman and Weiss [22, 23]. To unify both spaces of homogeneous type and the metric spaces endowed with measures satisfying the polynomial growth condition, Hytönen [24] introduced a new class of metric measure spaces satisfying both the so-called geometrically doubling and the upper doubling condition, which are called non-homogeneous metric measure spaces (see Definition 1.3 below). Many classical results have been proved still valid if the underlying spaces are non-homogeneous metric measure spaces (see [2532]). From now on, we always assume that \((\mathcal{X},d,\mu)\) is a non-homogeneous metric measure spaces in the sense of Hytönen [24]. In this setting, Hytönen [24] introduced the space \(\operatorname {RBMO}(\mu)\), and Hytönen and Martikainen [27] established a version of the Tb theorem. About Marcinkiewicz integral, Lin and Yang [31] have proved that the \(L^{p}(\mu)\)-boundedness with \(p\in (1,\infty)\) is equivalent to either of its boundedness from \(L^{1}(\mu)\) into \(L^{1,\infty}(\mu)\) or from the atomic Hardy space \(H^{1}(\mu)\) (see [28]) to \(L^{1}(\mu)\). They also showed that if the Marcinkiewicz integral is bounded from \(H^{1}(\mu)\) to \(L^{1}(\mu)\), then it is bounded from \(L^{\infty}(\mu)\) to RBLO(μ) (see [33]), which is a proper subset of \(\operatorname {RBMO}(\mu)\). These results essentially improve the existing results in [34].

Now we recall some necessary notions and notation.

The following notion of the geometrically doubling is well known in analysis on metric spaces, which was originally introduced by Coifman and Weiss in [22, 23] and is also known as metrically doubling.

Definition 1.1

A metric space \((\mathcal{X},d)\) is said to be geometrically doubling if there exists some \(N_{0}\in\mathbb{N}\) such that, for all balls \(B(x,r)\subset\mathcal{X}\), there exists a finite ball covering \(\{B(x_{i},\frac{r}{2})\}_{i}\) of \(B(x,r)\) such that the cardinality of this covering is at most \(N_{0}\).

Remark 1.2

Let \((\mathcal{X},d)\) be a metric space. In [24], Hytönen showed that the following statements are mutually equivalent:
  1. (1)

    \((\mathcal{X},d)\) is geometrically doubling.

  2. (2)

    For any \(\varepsilon\in(0,1)\) and any ball \(B(x,r)\subset \mathcal{X}\), there exists a finite ball covering \(\{B(x_{i},\varepsilon r)\}_{i}\) of \(B(x,r)\) such that the cardinality of this covering is at most \(N_{0}\varepsilon^{-n}\), where \(n=\log_{2}{N_{0}}\).

  3. (3)

    For any \(\varepsilon\in(0,1)\) and any ball \(B(x,r)\subset \mathcal{X}\) contains at most \(N_{0}\varepsilon^{-n}\) centers of disjoint balls \(\{B(x_{i},\varepsilon r)\}_{i}\).

  4. (4)

    There exists \(M\in\mathbb{N}\) such that any ball \(B(x,r)\subset \mathcal{X}\) contains at most M centers \(\{x_{i}\}_{i}\) of disjoint balls \(\{B(x_{i},r/4)\}_{i=1}^{M}\).


Definition 1.3

A metric measure space \((\mathcal{X},d,\mu)\) is said to be upper doubling if μ is a Borel measure on \(\mathcal{X}\) and there exist a dominating function \(\lambda: \mathcal{X}\times(0,\infty) \rightarrow(0,\infty) \) and a positive constant \(c_{\lambda}\) such that, for each \(x\in\mathcal{X}\), \(r\rightarrow\lambda(x,r)\) is non-decreasing and
$$ \mu\bigl(B(x,r)\bigr)\leq\lambda(x,r)\leq c_{\lambda}\lambda(x,r/2),\quad \mbox{for all } x\in \mathcal{X},r>0. $$

It was proved in [28] that there exists a dominating function λ̃ related to λ satisfying the property that there exists a positive constant \(c_{\tilde{\lambda}}\) such that \(\tilde{\lambda}\leq\lambda\), \(c_{\tilde{\lambda}}\leq c_{\lambda}\) and, for all \(x,y\in\mathcal{X}\), \(r>0\) with \(d(x,y)\leq r\), \(\tilde{\lambda}(x,r)\leq c_{\tilde{\lambda}}\tilde{\lambda}(y,r)\). Based on this, in this paper, we always assume that the dominating function λ also satisfies it.

The following coefficients \(\delta(B,S)\) for all ball B and S were introduced in [24] as analogs of Tolsa’s number \(K_{B,S}\) in [18].

Definition 1.4

For all balls \(B\subset S\), let
$$ \delta(B,S)=1+\int_{(2S-B)}\frac{d\mu(x)}{\lambda (c_{B},d(x,c_{B}) )}, $$
where above and in that follows, for a ball \(B=B(c_{B},r_{B})\) and \(\rho>0\), \(\rho B=B(c_{B},\rho r_{B})\).

Remark 1.5

The following discrete version \(K_{B,S}\) of \(\delta (B,S)\) was first introduced by Bui and Duong [25] in non-homogeneous metric measure spaces, which is more close to the quantity \(K_{B,S}\) introduced by Tolsa [18] in the setting of non-doubling measures. For all balls \(B\subset S\), let \(K_{B,S}\) be defined by
$$ K_{B,S}=1+\sum_{i=1}^{N_{B,S}} \frac{\mu(6^{i}B)}{\lambda (c_{B},6^{i}r_{B})}, $$
where \(N_{B,S}\) denote the smallest integer satisfying \(6^{N_{B,S}}r_{B}\geq r_{s}\). Obviously \(\delta(B,S)\lesssim K_{B,S}\). As was pointed out by Bui and Duong [25], it is not true that \(\delta(B,S)\thicksim K_{B,S}\).

Definition 1.6

Let \(\alpha,\beta\in(0,\infty)\). A ball \(B\subset\mathcal{X}\) is called \((\alpha,\beta)\)-doubling if \(\mu(\alpha B)\leq\beta\mu(B)\).

It was proved in [24] that if a metric measure space \((\mathcal{X},d,\mu)\) is upper doubling and \(\alpha,\beta\in(0,\infty)\) satisfying \(\beta>c_{\lambda}^{\log_{2}{\alpha}}=\alpha^{v}\), then, for any ball B, there exists some \(j\in\mathbb{N}\cup\{0\}\) such that \(\alpha^{j}B\) is \((\alpha,\beta)\)-doubling. Moreover, let \((\mathcal{X},d,\mu)\) be geometrically doubling, \(\beta>\alpha^{n}\) with \(n=\log_{2}N_{0}\) and μ a Borel measure on \(\mathcal{X}\) which is finite on bounded sets. Hytönen [24] also showed that for μ-almost every \(x\in\mathcal{X}\), there exist arbitrary small \((\alpha,\beta)\)-doubling balls centered at x. Furthermore, the radii of these balls may be chosen to be of the form \(\alpha^{-j}B\) for \(j\in\mathbb{N}\) and any preassigned number \(r>0\). Throughout this paper, for any \(\alpha\in(1,\infty)\) and ball B, the smallest \((\alpha,\beta_{\alpha})\)-doubling ball of the form \(\alpha^{j}B\) with \(j\in\mathbb{N}\) is denoted by \(\tilde{B}^{\alpha}\), where
$$\beta_{\alpha}=\max\bigl\{ \alpha^{3n}, \alpha^{3v}\bigr\} +30^{n}+30^{v}. $$

In what follows, by a doubling ball we mean a \((6,\beta_{6})\)-doubling ball and \(\tilde{B}^{6}\) is simply denoted by .

Now we recall the definition of \(\operatorname {RBMO}(\mu)\) from [24].

Definition 1.7

Let \(\rho\in(1,\infty)\). A function \(f\in L^{1}_{\mathrm{loc}}(\mu)\) is said to be in the space \(\operatorname {RBMO}(\mu)\) if there exist a positive constant c and, for any ball \(B\subset\mathcal{X}\), a number \(f_{B}\) such that
$$\frac{1}{\mu(\rho B)}\int_{B}\bigl\vert f(x)-f_{B} \bigr\vert \,d\mu(x)\leq c $$
and, for any two balls \(B\subset S\),
$$\vert f_{B}-f_{S}\vert \leq c\delta_{B,S}. $$
The infimum of the positive constant c is defined to be the \(\operatorname {RBMO}(\mu )\) norm of f and denote by \(\Vert f\Vert _{\mathrm{RBM}(\mu)}\).

In [24], it follows that the definition of \(\operatorname {RBMO}(\mu)\) is independent of the choice of \(\rho\in(1,\infty)\).

The following equivalent characterization of \(\operatorname {RBMO}(\mu)\) was established in [28].

Lemma 1.8

Let \(\rho\in(1,\infty)\) and \(f\in L^{1}_{\mathrm{loc}}(\mu)\). Then the following statements are equivalent:
  1. (1)

    \(f\in \operatorname {RBMO}(\mu)\);

  2. (2)
    there exist a positive constant c and, for any ball \(B\subset\mathcal{X}\), such that
    $$\frac{1}{\mu(\rho B)}\int_{B}\bigl\vert f(x)-m_{\tilde{B}}f \bigr\vert \,d\mu(x)\leq c $$
    and, for any doubling balls \(B\subset S\),
    $$\vert m_{B}-m_{S}\vert \leq c\delta_{B,S}. $$
Moreover, let \(\Vert f\Vert _{*}\) be the infimum of the positive constant c in (2). Then there exists a constant such that \(\frac{\Vert f\Vert _{*}}{\tilde{c}}\leq \Vert f\Vert _{\operatorname {RBMO}(\mu)}\leq\tilde{c} \Vert f\Vert _{*}\).

Now we give the definition of Marcinkiewicz integral (see [31]).

Definition 1.9

Let K be a locally integrable function on \((\mathcal{X}\times\mathcal{X})\setminus\{(x,x):x\in\mathcal{X}\}\). Assume that there exists a positive constant c such that, for all \(x,y\in\mathcal{X}\) with \(x\neq y\),
$$ \bigl\vert K(x,y)\bigr\vert \leq c\frac{d(x,y)}{\lambda(x,d(x,y))} $$
and, for all \(y, y'\in\mathcal{X}\),
$$ \int_{d(x,y)\geq2d(y,y')}\bigl[\bigl\vert K(x,y)-K\bigl(x,y' \bigr)\bigr\vert +\bigl\vert K(y,x)-K\bigl(y',x\bigr)\bigr\vert \bigr]\frac {1}{d(x,y)}\,d\mu(x)\leq c. $$
The Marcinkiewicz integral \(\mathcal{M}f\) associated to the above kernel K is defined by setting, for all \(x\in\mathcal{X}\),
$$ \mathcal{M}(f) (x)=\biggl[\int_{0}^{\infty}\biggl\vert \int_{d(x,y)< t}K(x,y)f(y)\,d\mu (y)\biggr\vert ^{2}\frac{dt}{t^{3}}\biggr]^{\frac{1}{2}}. $$
We show that the commutator \(\mathcal{M}_{b}\), associating with \(b\in \operatorname {RBMO}(\mu)\) and \(\mathcal{M}\), which is defined by
$$ \mathcal{M}_{b}(f) (x)=[b,\mathcal{M}](f) (x)=b(x)\mathcal{M}(f) (x)- \mathcal {M}(bf) (x),\quad x\in\mathcal{X}. $$

In [31], the authors have proven that the Marcinkiewicz integral \(\mathcal{M}\) is bounded from \(L^{p}(\mu)\) to \(L^{p}(\mu)\), \(1< p<\infty\). Our main result is as follows.

Theorem 1.10

Let K satisfy (1.5) and the following Hörmander-type condition:
$$\begin{aligned} &\sup_{r>0, d(y,y')\leq r}\sum_{i=1}^{\infty}i \int_{6^{i}r< d(x,y)\leq 6^{i+1}r}\bigl[\bigl\vert K(x,y)-K\bigl(x,y' \bigr)\bigr\vert \\ &\quad {}+\bigl\vert K(y,x)-K\bigl(y',x\bigr)\bigr\vert \bigr]\frac{1}{d(x,y)}\,d\mu(x)\leq c. \end{aligned}$$
If \(\mathcal{M}\) is bounded on \(L^{2}(\mu)\), then, for any \(b\in \operatorname {RBMO}(\mu)\), the commutator \(\mathcal{M}_{b}\) is bounded on \(L^{p}(\mu)\) with the bound no more than \(c_{p}\Vert b\Vert _{\operatorname {RBMO}(\mu)}\), where \(1< p<\infty\).

Remark 1.11

The Hörmander-type condition (1.9) is slightly stronger than (1.6).

The organization of this paper is as follows. In Section 2, we introduce the sharp maximal operator \(M^{\#}\), associated with \(K_{B,S}\) and prove Lemma 2.6. This technical lemma is of independent interest. Section 3 is devoted to the proof of Theorem 1.10.

Throughout this paper, we denote c a positive constant which is independent of the main parameters involved, but may vary from line to line. For any ball \(B\subset\mathcal{X}\), we denote its center and radius by \(c_{B}\) and \(r_{B}\). \(m_{B}f\) means that \(\frac{1}{\mu(B)}\int_{B}f(y)\,d\mu(y)\).

2 The sharp maximal function

For a locally integrable function f, let \(M^{\#}f\) be the sharp maximal function of f, namely, for \(x\in\mathcal{X}\),
$$ M^{\#}f(x)=\sup_{x\in B}\frac{1}{\mu(6B)}\int _{B}\bigl\vert f(y)-m_{\tilde{B}}f\bigr\vert \,d\mu (y)+ \sup_{(B,S)\in\Delta_{x}}\frac{\vert m_{B}f-m_{S}f\vert }{K_{B,S}}, $$
where \(\Delta_{x}=\{(B,S):x\in B\subset S \mbox{ and } B, S \mbox{ are doubling balls}\}\).
For \(0< r<\infty\), let \(M_{r}^{\#}f(x)=[M^{\#}(\vert f\vert ^{r})(x)]^{\frac{1}{r}}\) for \(x\in\mathcal{X}\). A simple computation proves that if \(0< r<1\),
$$ M_{r}^{\#}f(x)\leq c_{r}M^{\#}f(x), $$
where \(c_{r}>0\) is independent of f and x.

We recall some results in [32].

Lemma 2.1

  1. (1)
    Let \(p\in(1,\infty)\), \(r\in(1,p)\) and \(\rho\in [5,\infty)\). The following maximal operators defined, respectively, by setting, for all \(f\in L^{1}_{\mathrm{loc}}(\mu)\) and \(x\in\mathcal{X}\):
    $$\begin{aligned}& M_{r,\rho}f(x)=\sup_{x\in B}\biggl\{ \frac{1}{\mu(\rho B)}\int _{B}\bigl\vert f(y)\bigr\vert ^{r}\,d\mu (y) \biggr\} ^{\frac{1}{r}}, \\& Nf(x)=\sup_{x\in B: \textrm{doubling}}\frac{1}{\mu(B)}\int_{B} \bigl\vert f(y)\bigr\vert \,d\mu(y), \end{aligned}$$
    $$M_{\rho}f(x)=\sup_{x\in B}\frac{1}{\mu(\rho B)}\int _{B}\bigl\vert f(y)\bigr\vert \,d\mu(y) $$
    are bounded on \(L^{p}(\mu)\) and also bounded from \(L^{1}(\mu)\) to \(L^{1,\infty}(\mu)\).
  2. (2)

    For all \(f\in L^{1}_{\mathrm{loc}}(\mu)\) it holds true that \(\vert f(x)\vert \leq Nf(x)\) for μ-almost every \(x\in\mathcal{X}\).


In Lemma 2.1, if \(0< r<1\), using the Hölder inequality, we have \(M_{r,\rho}f(x)< M_{\rho} f(x)\). So Lemma 2.1 is right when \(0< r<1\).

We also need the following Calderón-Zygmund decomposition theorem obtained by Bui and Duong [25]. Let γ be a fixed positive constant satisfying that \(\gamma>\max\{c_{\lambda}^{3\log _{2}6},6^{3n}\}\) where \(c_{\lambda}\) is as in Definition 1.3 and n as in Remark 1.2.

Lemma 2.2

Let \(p\in[1,\infty)\), \(f\in L^{p}(\mu)\) and \(t\in (0,\infty)\) (\(t>\frac{\gamma^{\frac{1}{p}}\Vert f\Vert _{L^{p}(\mu)}}{[\mu(\mathcal {X})]^{1/p}} \) when \(\mu(\mathcal{X)}<\infty\)). Then
  1. (1)
    there exists a family of finite overlapping balls \(\{B_{j}\}_{j}\), pairwise disjoint,
    $$\begin{aligned}& \frac{1}{\mu(6^{2}B_{j})}\int_{B_{j}}\bigl\vert f(x)\bigr\vert ^{p}\,d\mu(x)>\frac{t^{p}}{\gamma}\quad \textit{for all } j, \\& \frac{1}{\mu(6^{2}\eta B_{j})}\int_{\eta B_{j}}\bigl\vert f(x)\bigr\vert ^{p}\,d\mu(x)\leq\frac {t^{p}}{\gamma} \quad\textit{ for all } j \textit{ and all } \eta \in(2,\infty), \end{aligned}$$
    $$\bigl\vert f(x)\bigr\vert \leq t \quad \textit{for } \mu\textit{-almost every } x\in \mathcal {X}\setminus\biggl(\bigcup_{j}6B_{j}\biggr); $$
  2. (2)
    for each j, let \(S_{j}\) be a \((3\times6^{2}, c_{\lambda}^{\log _{2}{3\times6^{2}}+1})\)-doubling ball of the family \(\{(3\times6^{2})^{k}B_{j}\} _{k\in\mathbb{N}}\), and \(\omega_{j}=\frac{\chi_{6B_{j}}}{\sum_{k}\chi _{6B_{k}}}\). Then there exists a family \(\{\varphi_{j}\}_{j}\) of functions such that, for each j, supp \((\varphi_{j})\subset S_{j}\), \(\varphi_{j}\) has a constant sign on \(S_{j}\),
    $$\begin{aligned}& \int_{\mathcal{X}}\varphi_{j}(x)\,d\mu(x)=\int _{6B_{j}}f(x)\omega_{j}(x)\,d\mu(x), \\& \sum_{j}\bigl\vert \varphi_{j}(x)\bigr\vert \leq\gamma_{0} t \quad \textit{for } \mu\textit{-almost every } x\in \mathcal{X}, \end{aligned}$$
    where \(\gamma_{0}\) is some positive constant depending only on \((\mathcal {X},\mu)\), and there exists a positive constant c, independent of f, t and j such that, when \(p=1\), it holds true that
    $$\Vert \varphi_{j}\Vert _{L^{\infty}(\mu)}\mu(S_{j})\leq c \int_{\mathcal {X}}\bigl\vert f(x)\omega_{j}(x)\bigr\vert \,d\mu(x) $$
    and, if \(p\in(1,\infty)\), it holds true that
    $$\biggl\{ \int_{S_{j}}\bigl\vert \varphi_{j}(x)\bigr\vert ^{p}\,d\mu(x)\biggr\} ^{1/p}\bigl[\mu(S_{j}) \bigr]^{1/p'}\leq\frac {c}{t^{p-1}}\int_{\mathcal{X}}\bigl\vert f(x)\omega_{j}(x)\bigr\vert ^{p}\,d\mu(x). $$

The following John-Nirenberg inequality was established by Hytönen in [24].

Lemma 2.3

let \((\mathcal{X}, d,\mu)\) be geometrically doubling and upper doubling. For every \(\rho>1\), there is a constant c so that, for every \(f\in \operatorname {RBMO}(\mu)\) and every ball B,
$$\mu\bigl(x\in B:\bigl\vert f(x)-f_{B}\bigr\vert >t\bigr)\leq2\mu( \rho B) \exp\bigl(-ct/\Vert f\Vert _{\operatorname {RBMO}(\mu)}\bigr), $$
where \(f_{B}\) can be seen in definition of \(\operatorname {RBMO}(\mu)\).
From Lemma 2.3, it is easy to prove that there are two positive \(c_{1}\), \(c_{2}\) such that, for any ball B and \(b\in \operatorname {RBMO}(\mu)\),
$$ \frac{1}{\mu(\rho B)}\int_{B}\exp\biggl(\frac{\vert b(x)-m_{\widetilde{B}}(b)\vert }{c_{1}\Vert b\Vert _{\operatorname {RBMO}(\mu)}} \biggr)\,d\mu(x)\leq c_{2}. $$

Lemma 2.4

There is a constant c such that, for any \(a>0\) and \(t_{1}, t_{2}>0\),
$$t_{1}t_{2}\leq c\bigl[t_{1}\log(2+at_{1})+a^{-1} \exp t_{2}\bigr]. $$

This lemma had been established in [34].

We also need some useful properties of \(K_{B,S}\), which were proved in [25, 32].

Lemma 2.5

  1. (1)

    For all balls \(B\subset R\subset S\), \(K_{B,R}\leq 2K_{B,S}\).

  2. (2)

    For any \(\rho\in[1,\infty)\), there exists a positive constant \(c_{\rho}\), depending only on ρ, such that, for all balls \(B\subset S\) with \(r_{S}\leq\rho r_{B}\), \(K_{B,S}\leq c_{\rho}\).

  3. (3)

    There exists a positive constant c, such that, for all balls B, \(K_{B,\tilde{B}}\leq c\).

  4. (4)

    There exists a positive constant c, depending on \(c_{\lambda}\), such that, for all balls \(B\subset R\subset S\), \(K_{B,S}\leq K_{B,R}+cK_{R,S}\).

  5. (5)

    There exists a positive constant c, depending on \(c_{\lambda}\), such that, for all balls \(B\subset R\subset S\), \(K_{R,S}\leq cK_{B,S}\).


Now we give and prove the main result about the sharp maximal function \(M^{\#}\).

Lemma 2.6

Let K satisfy (1.5) and the Hörmander-type condition (1.9). We have \(s\in(1,\infty)\), \(p_{0}\in(1,\infty)\) and \(b\in L^{\infty}(\mu)\). If \(\mathcal{M}\) is bounded on \(L^{2}(\mu)\), then there is a positive constant c such that, for all \(f\in L^{\infty}(\mu)\cap L^{p_{0}}(\mu)\) and for all \(x\in\mathcal{X}\),
$$M^{\#}\bigl[\mathcal{M}_{b}(f)\bigr](x)\leq c\bigl[\Vert b\Vert _{\operatorname {RBMO}(\mu )}M_{s,6}\bigl[\mathcal{M}(f)\bigr](x)+\Vert b \Vert _{\operatorname {RBMO}(\mu)}\Vert f\Vert _{L^{\infty }(\mu)}\bigr]. $$


Without loss of generality, we may assume \(\Vert b\Vert _{\operatorname {RBMO}(\mu )}=1\). To prove Lemma 2.6, it suffices to prove that
$$ \frac{1}{\mu(6B)}\int_{B}\bigl\vert \mathcal{M}_{b}(f) (y)-h_{B}\bigr\vert \,d\mu(y)\leq cM_{s,6}\bigl[\mathcal{M}(f)\bigr](x)+\Vert f\Vert _{L^{\infty}(\mu)} $$
for all \(x\in B\) and
$$ \vert h_{B}-h_{S}\vert \leq c (K_{B,S})^{2} \bigl[M_{s,6}\bigl[\mathcal{M}(f)\bigr](x)+\Vert f\Vert _{L^{\infty }(\mu)} \bigr] $$
for all balls \(B\subset S\) with \(x\in B\), where B is an arbitrary ball and S is a doubling ball,
$$h_{B}=m_{B}\bigl[\mathcal{M}\bigl(\bigl(b-m_{\widetilde{B}}(b) \bigr)f\chi_{\mathcal{X}\setminus \frac{6}{5}B}\bigr)\bigr] $$
$$h_{S}=m_{S}\bigl[\mathcal{M}\bigl(\bigl(b-m_{S}(b) \bigr)f\chi_{\mathcal{X}\setminus\frac{6}{5}S}\bigr)\bigr]. $$
To prove (2.4), for a fixed ball B, \(x\in B\) and \(f\in L^{\infty}(\mu )\), we write
$$f(y)= f(y)\chi_{ \frac{6}{5}B}(y)+f(y)\chi_{\mathcal{X}\setminus\frac {6}{5}B}(y)=f_{1}(y)+f_{2}(y) $$
$$\mathcal{M}_{b}(f) (y)=\bigl(b(y)-m_{\tilde{B}}(b)\bigr) \mathcal{M}(f) (y)-\mathcal {M}\bigl(\bigl(b(y)-m_{\tilde{B}}(b) \bigr)f_{1}\bigr) (y)- \mathcal{M}\bigl(\bigl(b(y)-m_{\tilde{B}}(b) \bigr)f_{2}\bigr) (y). $$
So we can write
$$\begin{aligned} &\frac{1}{\mu(6B)}\int_{B}\bigl\vert \mathcal{M}_{b}(f) (y)-h_{B}\bigr\vert \,d\mu(y) \\ &\quad \leq\frac{1}{\mu(6B)}\int_{B}\bigl\vert b(y)-m_{\tilde{B}}(b)\bigr\vert \mathcal {M}(f) (y)\,d\mu(y) \\ &\qquad {}+\frac{1}{\mu(6B)}\int_{B}\mathcal{M}\bigl( \bigl(b(y)-m_{\tilde{B}}(b)\bigr)f_{1}\bigr) (y)\,d\mu (y) \\ &\qquad {}+\frac{1}{\mu(6B)}\int_{B}\bigl\vert \mathcal{M}\bigl( \bigl(b(y)-m_{\tilde {B}}(b)\bigr)f_{2}\bigr) (y)-h_{B} \bigr\vert \,d\mu(y) \\ &\quad =A_{1}+A_{2}+A_{3}. \end{aligned}$$
By the Hölder inequality and Corollary 2.3 in [32], we see that
$$\begin{aligned} A_{1} \leq&\frac{1}{\mu(6B)^{\frac{1}{s}+\frac{1}{s'}}}\biggl[\int_{B}\bigl\vert b(y)-m_{\tilde{B}}(b)\bigr\vert ^{s'}\,d\mu(y) \biggr]^{1/s'}\biggl[\int_{B}\bigl(\mathcal {M}(f) \bigr)^{s}(y)\biggr]^{1/s} \\ \leq& M_{s,6}\bigl[\mathcal{M}(f)\bigr](x). \end{aligned}$$
To estimate \(A_{2}\), from the Hölder inequality, the \(L^{2}(\mu )\)-boundedness of \(\mathcal{M}\) and Corollary 2.3 in [32], it follows that
$$\begin{aligned} A_{2} \leq&\frac{\mu(B)^{1/2}}{\mu(6B)}\biggl[\int_{B}\bigl\vert \mathcal{M}\bigl[\bigl(b(y)-m_{\tilde {B}}(b)\bigr)f_{1} \bigr](y)\bigr\vert ^{2}\,d\mu(y)\biggr]^{1/2} \\ \leq&\biggl[\frac{1}{\mu(6B)}\int_{B}\bigl\vert \bigl(b(y)-m_{\tilde{B}}(b)\bigr)f_{1}(y)\bigr\vert ^{2}\,d\mu (y)\biggr]^{1/2} \\ \leq&\mu(6B)^{-1/2}\bigl\Vert \bigl(b(y)-m_{\widetilde{\frac{6}{5}B}}(b) \bigr)f_{1}(y)\bigr\Vert _{L^{2}(\mu)} +\mu(6B)^{-1/2}\bigl\Vert \bigl(m_{\widetilde{\frac{6}{5}B}}(b)-m_{\tilde {B}}(b)\bigr)f_{1}(y) \bigr\Vert _{L^{2}(\mu)} \\ \leq& \Vert f\Vert _{L^{\infty}(\mu)}\biggl[\frac{1}{\mu(6B)}\int _{\frac {6}{5}B}\bigl\vert b(y)-m_{\widetilde{\frac{6}{5}B}}(b)\bigr\vert ^{2}\,d\mu(y)\biggr]^{1/2}\\ &{}+ c\Vert f\Vert _{L^{\infty}(\mu)} \biggl[\frac{\mu(\frac{6}{5}B)}{\mu(6B)}\biggr]^{1/2} \\ \leq& c\Vert f\Vert _{L^{\infty}(\mu)}, \end{aligned}$$
where we use the fact that \(\vert m_{\widetilde{\frac{6}{5}B}}(b)-m_{\tilde {B}}(b)\vert \leq c(K_{B,\widetilde{B}}+K_{\frac{6}{5}B,\widetilde{\frac {6}{5}B}}+K_{B,\frac{6}{5}B})\leq c\).
To obtain (2.4), we still need to estimate \(A_{3}\). Set
$$\begin{aligned}& M_{1}(x,y)=\biggl(\int_{0}^{\infty}\biggl[ \int_{d(y,z)\leq t\leq d(x,z)}\bigl\vert K(y,z) \bigl(b(z)-m_{\widetilde{B}}(b) \bigr)f_{2}(z)\bigr\vert \,d\mu(z)\biggr]^{2}\frac{dt}{t^{3}} \biggr)^{1/2}, \\& M_{2}(x,y)=\biggl(\int_{0}^{\infty}\biggl[ \int_{d(x,z)\leq t\leq d(y,z)}\bigl\vert K(y,z) \bigl(b(z)-m_{\widetilde{B}}(b) \bigr)f_{2}(z)\bigr\vert \,d\mu(z)\biggr]^{2}\frac{dt}{t^{3}} \biggr)^{1/2}, \end{aligned}$$
$$M_{3}(x,y)=\biggl(\int_{0}^{\infty}\biggl[ \int_{\max\{d(y,z), d(x,z)\} }\bigl\vert \bigl(K(y,z)-K(x,z)\bigr) \bigl(b(z)-m_{\widetilde{B}}(b)\bigr)f_{2}(z)\bigr\vert \,d\mu(z) \biggr]^{2}\frac {dt}{t^{3}}\biggr)^{1/2}. $$
For any \(x,y\in\mathcal{X}\), we have (see also [31], p.134)
$$\bigl\vert \mathcal{M}\bigl[\bigl(b-m_{\widetilde{B}}(b)\bigr)f_{2} \bigr](y)-\mathcal {M}\bigl[\bigl(b-m_{\widetilde{B}}(b)\bigr)f_{2} \bigr](x)\bigr\vert \leq\sum_{i=1}^{3}M_{i}(x,y). $$
Applying the Minkowski inequality and (1.9) we conclude that, for all \(x, y\in B\),
$$\begin{aligned} M_{1}(x,y) \leq& \int_{d(y,z)< d(x,z)}\bigl\vert K(y,z)f_{2}(z) \bigl(b(z)-m_{\widetilde {B}}(b)\bigr)\bigr\vert \biggl[\int_{d(y,z)\leq t < d(x,z)}\frac{dt}{t^{3}}\biggr]^{1/2}\,d \mu(z) \\ \leq&c \int_{\mathcal{X}\setminus5B}\frac{r_{B}^{1/2}}{d(z,c_{B})^{1/2}} \frac{\vert b(z)-m_{\widetilde{B}}(b)\vert }{\lambda(c_{B}, d(z,c_{B}))}f(z)\,d \mu(z) \\ \leq&c \sum_{i=1}^{\infty}\int _{6^{i}5B\setminus6^{i-1}5B}\frac {r_{B}^{1/2}}{d(z,c_{B})^{1/2}} \frac{\vert m_{\widetilde{6^{i}5B}}(b)-m_{\widetilde{B}}(b)\vert }{\lambda(c_{B}, d(z,c_{B}))}f(z)\,d\mu(z) \\ &{}+\sum_{i=1}^{\infty}\int_{6^{i}5B\setminus6^{i-1}5B} \frac {r_{B}^{1/2}}{d(z,c_{B})^{1/2}} \frac{\vert m_{\widetilde{6^{i}5B}}(b)-b(z)\vert }{\lambda(c_{B}, d(z,c_{B}))}f(z)\,d\mu (z) \\ \leq&c \sum_{i=1}^{\infty}i6^{(-i/2)} \frac{1}{\lambda (c_{B},6^{i-1}5r_{B})}\int_{6^{i}5B}\bigl\vert f(z)\bigr\vert d \mu(z) \\ &{}+\sum_{i=1}^{\infty}6^{(-i/2)} \frac{1}{\lambda(c_{B},6^{i-1}5r_{B})}\int_{6^{i}5B} \bigl\vert m_{\widetilde{6^{i}5B}}(b)-b(z) \bigr\vert \bigl\vert f(z)\bigr\vert \,d\mu(z) \\ \leq&c\Vert f\Vert _{L^{\infty}(\mu)}\sum_{i=1}^{\infty}i6^{(-i/2)} \frac{\mu (6^{i+1}5B)}{\lambda(c_{B},6^{i-1}5r_{B})} \\ \leq&c\Vert f\Vert _{L^{\infty}(\mu)}, \end{aligned}$$
where we use the doubling condition of λ, \(\vert m_{\widetilde {6^{i}5B}}(b)-m_{\widetilde{B}}(b)\vert \leq ci\) and \(\lambda(c_{B},d(x,c_{B}))\backsim\lambda(x,d(x,c_{B}))\backsim\lambda (x,d(x,y)) \) for \(y\in B\) and \(x\in\mathcal{X}\setminus kB\) (\(k>1\)).
Similarly, \(M_{2}(x,y)\leq c\Vert f\Vert _{L^{\infty}(\mu)}\). Now for all \(x,y\in B\), by the Minkowski inequality we have
$$\begin{aligned} &M_{3}(x,y)\\ &\quad \leq \int_{\mathcal {X}}\bigl\vert \bigl(K(y,z)-K(x,z)\bigr)f_{2}(z) \bigl(b(z)-m_{\widetilde{B}}(b)\bigr) \bigr\vert \biggl[\int_{\max\{ d(y,z),d(x,z)\}\leq t} \frac{dt}{t^{3}} \biggr]^{1/2}\,d\mu(z) \\ &\quad \leq c \int_{\mathcal{X}}\bigl\vert \bigl(K(y,z)-K(x,z) \bigr)f_{2}(z) \bigl(b(z)-m_{\widetilde {B}}(b)\bigr)\bigr\vert \frac{1}{d(y,z)}\,d\mu(z) \\ &\quad \leq c \Vert f\Vert _{L^{\infty}(\mu)}\sum_{i=1}^{\infty} \int_{6^{i}5B\setminus 6^{i-1}5B}\bigl\vert \bigl(K(y,z)-K(x,z)\bigr) \bigl(b(z)-m_{\widetilde{6^{i}5B}}(b)\bigr)\bigr\vert \frac {1}{d(y,z)}\,d\mu(z) \\ &\qquad {}+c \Vert f\Vert _{L^{\infty}(\mu)}\sum_{i=1}^{\infty} \int_{6^{i}5B\setminus 6^{i-1}5B}\bigl\vert \bigl(K(y,z)-K(x,z)\bigr) \bigl(m_{\widetilde{6^{i}5B}}(b)\bigr)-m_{\widetilde {B}}(b)\bigr\vert \frac{1}{d(y,z)}\,d\mu(z) \\ &\quad =M_{31}+M_{32}. \end{aligned}$$
In Lemma 2.4, we write \(a=6^{i}\mu(6^{i+1}5B)\), \(t_{1}=\frac {\vert K(y,z)-K(x,z)\vert }{d(y,z)}\), and \(t_{2}=\frac{\vert b(z)-m_{\widetilde {6^{i}5B}}\vert }{c_{2}}\). From this we have
$$\begin{aligned} M_{31}(x,y) \leq&c \Vert f\Vert _{L^{\infty}(\mu)}\sum _{i=1}^{\infty}\int_{6^{i}5B\setminus6^{i-1}5B} \frac {\vert K(y,z)-K(x,z)\vert }{d(y,z)}\bigl\vert b(z)-m_{\widetilde{6^{i}5B}}(b)\bigr\vert \,d\mu(z) \\ \leq&c \Vert f\Vert _{L^{\infty}(\mu)}\sum_{i=1}^{\infty} \int_{6^{i}5B\setminus 6^{i-1}5B} \biggl[\frac{\vert K(y,z)-K(x,z)\vert }{d(y,z)}\log\biggl[2+6^{i} \mu\bigl(6^{i+1}5B\bigr) \\ &{}\times\frac{\vert K(y,z)-K(x,z)\vert }{d(y,z)}\biggr]+\frac{1}{6^{i}\mu(6^{i+1}5B)}\exp\biggl(\frac {\vert b(z)-m_{\widetilde{6^{i}5B}}\vert }{c_{2}} \biggr)\biggr]\,d\mu(z) \\ \leq&c \Vert f\Vert _{L^{\infty}(\mu)}\sum_{i=1}^{\infty}i \int_{6^{i}5B\setminus 6^{i-1}5B} \frac{\vert K(y,z)-K(x,z)\vert }{d(y,z)}\log\biggl(2+\frac{\mu(6^{i+1}5B)}{\lambda(c_{B},d(y,z))} \biggr)\,d\mu(z) \\ &{}+c \Vert f\Vert _{L^{\infty}(\mu)}\sum_{i=1}^{\infty} \int_{6^{i}5B\setminus 6^{i-1}5B}\frac{1}{6^{i}\mu(6^{i+1}5B)}\exp\biggl(\frac{\vert b(z)-m_{\widetilde {6^{i}5B}}\vert }{c_{2}} \biggr)\,d\mu(z) \\ \leq&c \Vert f\Vert _{L^{\infty}(\mu)}\sum_{i=1}^{\infty}i \int_{6^{i}5B\setminus6^{i-1}5B}\frac{\vert K(y,z)-K(x,z)\vert }{d(y,z)}\,d\mu(z) \\ &{}+c\Vert f\Vert _{L^{\infty}(\mu)}\sum_{i=1}^{\infty} \frac{1}{6^{i}}\frac{1}{\mu(6^{i+1}5B)}\int_{6^{i}5B}\exp\biggl( \frac {\vert b(z)-m_{\widetilde{6^{i}5B}}\vert }{c_{2}}\biggr)\,d\mu(z) \\ \leq&c \Vert f\Vert _{L^{\infty}(\mu)}, \end{aligned}$$
where we use (2.3). For \(M_{32}\) we estimate
$$\begin{aligned} M_{32} \leq&c \Vert f\Vert _{L^{\infty}(\mu)}\sum _{i=1}^{\infty}\bigl\vert m_{\widetilde {6^{i}5B}}(b))-m_{\widetilde{Q}}(b) \bigr\vert \int_{6^{i}5B\setminus6^{i-1}5B}\bigl\vert \bigl(K(y,z)-K(x,z)\bigr) \bigr\vert \frac{1}{d(y,z)}\,d\mu (z) \\ \leq&c \Vert f\Vert _{L^{\infty}(\mu)}\sum_{i=1}^{\infty}i \int_{6^{i}5B\setminus6^{i-1}5B}\bigl\vert \bigl(K(y,z)-K(x,z)\bigr)\bigr\vert \frac{1}{d(y,z)}\,d\mu (z) \\ \leq&c \Vert f\Vert _{L^{\infty}(\mu)}. \end{aligned}$$
Combining these estimates above, we get
$$\begin{aligned} A_{3} \leq&\frac{1}{\mu(6B)}\int_{B}\bigl\vert \mathcal{M}\bigl(\bigl(b(y)-m_{\tilde {B}}(b)\bigr)f_{2}\bigr) (y)-h_{B}\bigr\vert \,d\mu(y) \\ \leq& c\frac{1}{\mu(6B)}\frac{1}{\mu(B)}\int_{B}\int _{B}\sum_{i=1}^{3}M_{i}(x,y)\,d \mu(x)\,d\mu(y) \\ \leq&c \Vert f\Vert _{L^{\infty}(\mu)}. \end{aligned}$$
So the estimate (2.4) is proved.
Now we prove (2.5). Consider two balls \(B\subset S\) with \(x\in B\) and let \(N=N_{B,S}+1\), where S is a doubling ball. Write \(\vert h_{B}-h_{S}\vert \) as
$$\begin{aligned} &\vert h_{B}-h_{S}\vert \\ &\quad \leq\bigl\vert m_{S}\bigl[\mathcal{M}\bigl(\bigl(b-m_{\widetilde{B}}(b) \bigr)f\chi_{\mathcal {X}\setminus6^{N}B}\bigr)\bigr] - m_{B}\bigl[\mathcal{M}\bigl( \bigl(b-m_{\widetilde{B}}(b)\bigr)f\chi_{\mathcal{X}\setminus 6^{N}B}\bigr)\bigr]\bigr\vert \\ &\qquad {}+\bigl\vert m_{S}\bigl[\mathcal{M}\bigl(\bigl(b-m_{S}(b) \bigr)f\chi_{\mathcal{X}\setminus6^{N}B}\bigr)\bigr] - m_{S}\bigl[\mathcal{M}\bigl( \bigl(b-m_{\widetilde{B}}(b)\bigr)f\chi_{\mathcal{X}\setminus 6^{N}B}\bigr)\bigr]\bigr\vert \\ &\qquad {}+ \bigl\vert m_{B}\bigl[\mathcal{M}\bigl(\bigl(b-m_{\widetilde{B}}(b) \bigr)f\chi_{6^{N}B\setminus\frac {6}{5}B}\bigr)\bigr]\bigr\vert + \bigl\vert m_{S}\bigl[\mathcal{M}\bigl(\bigl(b-m_{S}(b)\bigr)f \chi_{6^{N}B\setminus\frac{6}{5}S}\bigr)\bigr]\bigr\vert \\ &\quad = B_{1}+B_{2}+B_{3}+B_{4}. \end{aligned}$$
As in the estimate for the \(A_{3}\), we have \(B_{1}\leq c\Vert f\Vert _{L^{\infty }(\mu)}\). To estimate \(B_{2}\), for \(y\in\mathcal{X}\), we get
$$\begin{aligned} B_{2} \leq&\bigl\vert m_{S}\bigl[\mathcal{M}\bigl( \bigl(b-m_{S}(b)\bigr)f\chi_{\mathcal{X}\setminus6^{N}B}\bigr)\bigr] - m_{S}\bigl[\mathcal{M}\bigl(\bigl(b-m_{\widetilde{B}}(b)\bigr)f \chi_{\mathcal{X}\setminus 6^{N}B}\bigr)\bigr]\bigr\vert \\ \leq& c m_{S}\bigl\vert \bigl(m_{S}(b)-m_{\widetilde{B}}(b) \bigr)\mathcal{M}(f\chi _{\mathcal{X}\setminus6^{N}B})\bigr\vert \\ \leq& c\frac{K_{B,S}+K_{B,\widetilde{B}}}{\mu(S)}\int_{S}\mathcal {M}(f \chi_{\mathcal{X}\setminus6^{N}B}) (y)\,d\mu(y) \\ \leq& c \frac{K_{B,S}}{\mu(S)}\mu(S)^{1/s'}\biggl(\int_{S} \mathcal{M}^{s}(f\chi _{\mathcal{X}\setminus6^{N}B}) (y)\,d\mu(y)\biggr)^{1/s} \\ \leq& cK_{B,S}M_{s,6}\bigl[\mathcal{M}(f)\bigr]. \end{aligned}$$
For \(y\in R\), we have
$$\begin{aligned} B_{4} \leq&\bigl\vert m_{S}\bigl[\mathcal{M}\bigl( \bigl(b-m_{S}(b)\bigr)f\chi_{6^{N}B\setminus\frac {6}{5}S}\bigr)\bigr]\bigr\vert \\ \leq& \int_{\mathcal{X}}\bigl\vert K(y,z)\bigr\vert \bigl\vert b(z)-m_{S}(b)\bigr\vert \bigl\vert f(z)\chi_{6^{N}B\setminus \frac{6}{5}S} \bigr\vert \biggl(\int_{d(y,z)< t}\frac{dt}{t^{3}} \biggr)^{1/2}\,d\mu(z) \\ \leq& c \int_{\mathcal{X}}\frac{\vert b(z)-m_{S}(b)\vert }{\lambda (y,d(y,z))}\bigl\vert f(z) \chi_{6^{N}B\setminus\frac{6}{5}S}\bigr\vert \,d\mu(z) \\ \leq& c \Vert f\Vert _{L^{\infty}(\mu)}\int_{6^{N}B} \frac{\vert b(z)-m_{S}(b)\vert }{\lambda (c_{B},6^{N}r_{B})}\,d\mu \\ \leq& c \Vert f\Vert _{L^{\infty}(\mu)}\frac{1}{\lambda(c_{B},6^{N}r_{B})}\int _{6^{N}B}\bigl\vert m_{S}(b)-m_{\widetilde{6S}}(b) \bigr\vert +\bigl\vert b(z)-m_{\widetilde{6S}}\bigr\vert \,d\mu (z) \\ \leq& c\Vert f\Vert _{L^{\infty}(\mu)}\frac{1}{\mu(6^{2}S)}\int _{6S}\bigl\vert b(z)-m_{\widetilde{6S}}\bigr\vert \,d\mu(z) +c\Vert f\Vert _{L^{\infty}(\mu)}\bigl\vert m_{S}(b) \\ &{}-m_{\widetilde{6S}}(b) \bigr\vert \frac{\mu (6^{N}B)}{\lambda(c_{B},6^{N}r_{B})} \\ \leq& c\Vert f\Vert _{L^{\infty}(\mu)}, \end{aligned}$$
where we have used \(\vert m_{S}(b)-m_{\widetilde{6S}}(b)\vert \leq c(K_{S,6S}+K_{6S,\widetilde{6S}})\leq c\).
In order to estimate \(B_{3}\), for \(y\in B\), we get
$$\begin{aligned} &\bigl\vert m_{B}\bigl[\mathcal{M}\bigl(\bigl(b-m_{\widetilde{B}}(b) \bigr)f\chi_{6^{N}B\setminus\frac {6}{5}B}\bigr)\bigr](y)\bigr\vert \\ &\quad \leq \bigl\vert m_{B}\bigl[\mathcal{M}\bigl(\bigl(b-m_{\widetilde{B}}(b) \bigr)f\chi_{6^{N}B\setminus6B}\bigr)\bigr](y) -m_{B}\bigl[\mathcal{M}\bigl( \bigl(b-m_{\widetilde{B}}(b)\bigr)f\chi_{6B\setminus\frac {6}{5}B}\bigr)\bigr](y)\bigr\vert \\ &\quad \leq\int_{6B\setminus\frac{6}{5}B}\bigl\vert K(y,z)\bigr\vert \bigl\vert b(z)-m_{\widetilde {B}}(b)\bigr\vert \bigl\vert f(z)\bigr\vert \biggl(\int_{d(y,z)< t}\frac{dt}{t^{3}}\biggr)^{1/2}\,d \mu(z) \\ &\qquad {}+\int_{6^{N}B\setminus6B}\bigl\vert K(y,z)\bigr\vert \bigl\vert b(z)-m_{\widetilde{B}}(b)\bigr\vert \bigl\vert f(z)\bigr\vert \biggl(\int _{d(y,z)< t}\frac{dt}{t^{3}}\biggr)^{1/2}\,d\mu(z) \\ &\quad \leq c\Vert f\Vert _{L^{\infty}(\mu)}\int_{6B\setminus\frac{6}{5}B} \frac {\vert b(z)-m_{\widetilde{B}}(b)\vert }{\lambda(y,d(y,z))}\,d\mu(z) +c\Vert f\Vert _{L^{\infty}(\mu)}\int _{6^{N}B\setminus6B}\frac {\vert b(z)-m_{\widetilde{B}}(b)\vert }{\lambda(y,d(y,z))}\,d\mu(z) \\ &\quad \leq c\Vert f\Vert _{L^{\infty}(\mu)}\frac{\mu(6^{2}B)}{\lambda(c_{B},6r_{B})}\frac {1}{\mu(6^{2}B)} \int_{6B}\bigl\vert b(z)-m_{\widetilde{B}}(b)\bigr\vert \,d \mu(z) \\ &\qquad {}+c\Vert f\Vert _{L^{\infty}(\mu)}\sum_{k=1}^{N-1} \int_{6^{k+1}B\setminus 6^{k}B}\frac{\vert b(z)-m_{\widetilde{B}}(b)\vert }{\lambda(y,d(y,z))}\,d\mu(z) \\ &\quad \leq c\Vert f\Vert _{L^{\infty}(\mu)}+c\Vert f\Vert _{L^{\infty}(\mu)}\sum _{k=1}^{N-1} \frac{\mu(6^{k+2}B)}{\lambda(c_{B},6^{k+1}r_{B})} \frac{1}{\mu(6^{k+2}B)} \int_{6^{k+1}B}\bigl\vert b(z)-m_{\widetilde{6^{k+1}B}}(b) \bigr\vert \,d\mu(z) \\ &\qquad {}+c\Vert f\Vert _{L^{\infty}(\mu)}\sum_{k=1}^{N-1} \frac{\mu(6^{k+2}B)}{\lambda(c_{B},6^{k+1}r_{B})}\frac{1}{\mu(6^{k+2}B)} \int_{6^{k+1}B}\bigl\vert m_{\widetilde{6^{k+1}B}}(b)-m_{\widetilde{B}}(b)\bigr\vert \,d\mu (z) \\ &\quad \leq c\Vert f\Vert _{L^{\infty}(\mu)}+c\Vert f\Vert _{L^{\infty}(\mu)}\sum _{k=1}^{N-1} \frac{\mu(6^{k+2}B)}{\lambda(c_{B},6^{k+1}r_{B})}(cK_{B,6^{k+1}B}+1) \\ &\quad \leq cK^{2}_{B,S}\Vert f\Vert _{L^{\infty}(\mu)}. \end{aligned}$$
That is to say, \(B_{3}\leq cK^{2}_{B,S}\Vert f\Vert _{L^{\infty}(\mu)}\).

Combining the estimates through \(B_{1}\) to \(B_{4}\) establishes (2.5), which completes the proof of Lemma 2.6. □

3 Proof of Theorem 1.10

In this section, we prove Theorem 1.10. Let \(0< r<1\), we prove that, for any \(p\in(1, \infty)\), \(b\in L^{\infty}(\mu)\), and all bounded functions f with compact support,
$$ \mu\bigl(\bigl\{ x\in\mathcal{X}:M_{r}^{\#}\bigl[ \mathcal{M}_{b}(f)\bigr](x)>\lambda\bigr\} \bigr)\leq c \lambda^{-p}\Vert b\Vert ^{p}_{\operatorname {RBMO}(\mu)}\Vert f \Vert ^{p}_{L^{p}(\mu)}. $$
Once (3.1) is established, it follows from the Marcinkiewicz interpolation theorem that
$$ \bigl\Vert M_{r}^{\#}\bigl[\mathcal{M}_{b}(f) \bigr]\bigr\Vert _{L^{p}(\mu)}\leq c\Vert b\Vert _{\operatorname {RBMO}(\mu )}\Vert f\Vert _{L^{p}(\mu)}. $$
This via Theorem 4.2 in [25] states that, for any \(p\in(1, \infty)\), \(b\in L^{\infty}(\mu)\), and all bounded functions f with compact support and integral zero,
$$ \bigl\Vert \mathcal{M}_{b}(f)]\bigr\Vert _{L^{p}(\mu)}\leq c \Vert b\Vert _{\operatorname {RBMO}(\mu)}\Vert f\Vert _{L^{p}(\mu)}. $$
In [25], Theorem 6.4, the authors show the density in \(L^{p}(\mu)\) of bounded functions with compact support and integral zero. Similar to [18], Lemma 3.3, using the truncation argument, a routine argument leads to (3.3) for all \(b\in \operatorname {RBMO}(\mu)\) and \(f\in L^{p}(\mu)\).
Now we prove (3.1). Without loss of generality, we assume that \(\rho=6\) in Lemma 2.1 and \(\Vert b\Vert _{\operatorname {RBMO}(\mu)}=1\). For each fixed \(t>0\) and bounded function f with compact support, applying the Calderón-Zygmund decomposition to \(\vert f\vert ^{p}\) at level \(t^{p}\) as Lemma 2.2, we decompose \(f(x)= g(x)+h(x)\), where
$$g(x)=f(x)\chi_{\mathcal{X}\setminus\bigcup_{i}6B_{i}}(x)+\sum_{i} \varphi_{i}(x), h(x)=\sum_{i}\bigl[ \omega_{i}(x)f(x)-\varphi_{i}(x)\bigr]=\sum _{i} h_{i}(x). $$
It is obvious that \(\Vert g\Vert _{L^{\infty}(\mu)}\leq ct\). Using Lemma 2.1(2) we have
$$\begin{aligned} \biggl\Vert \sum_{i}\varphi_{i}\biggr\Vert ^{p}_{L^{p}(\mu)} \leq& \biggl\Vert \sum _{i}\vert \varphi_{i}\vert \biggr\Vert ^{p-1}_{L^{\infty}(\mu)}\biggl\Vert \sum_{i} \varphi_{i}\biggr\Vert _{L^{1}(\mu)} \\ \leq& ct^{p-1} \sum_{i}\biggl(\int _{R_{i}}\bigl\vert \varphi_{i}(x)\bigr\vert ^{p}\,d\mu(x)\biggr)^{1/p}\mu (S_{i})^{1/p'} \\ \leq& c\sum_{i}\int_{6B_{i}}\bigl\Vert f(x)\bigr\Vert ^{p}\,d\mu(x)\leq c\Vert f\Vert _{p}^{p}. \end{aligned}$$
That is to say, \(\Vert g\Vert _{L^{p}(\mu)}\leq c\Vert f\Vert _{L^{p}(\mu)}\). Using (3.2) and Lemma 2.6 we have
$$\begin{aligned} &\mu\bigl(\bigl\{ x\in\mathcal{X}:M^{\#}_{r}\bigl( \mathcal{M}_{b}(g)\bigr) (x)>2ct\bigr\} \bigr) \\ &\quad \leq c \mu\bigl(\bigl\{ x\in\mathcal{X}:M_{s,6}\bigl(\mathcal{M}(g) \bigr) (x)>t\bigr\} \bigr) \\ &\quad \leq c t^{-p}\bigl\Vert M_{s,6}\bigl(\mathcal{M}(g) \bigr)\bigr\Vert ^{p}_{L^{p}(\mu)}\leq ct^{-p}\Vert f \Vert ^{p}_{L^{p}(\mu)}, \end{aligned}$$
where \(1< s< p\).
Similar to [25], Section 4.1, we have, for any f,
$$M^{\#}_{r}f(x)\leq M_{r,6}(f) (x)+3N_{r}(f) (x)\leq cM_{r,6}(f) (x). $$
From this we write
$$\begin{aligned} &\mu\bigl(\bigl\{ x\in\mathcal{X}:M^{\#}_{r}\bigl[ \mathcal{M}_{b}(h)\bigr](x)>t\bigr\} \bigr) \\ &\quad \leq\mu\biggl(\biggl\{ x\in \mathcal{X}:M_{r,6}\biggl[\mathcal{M}\biggl(\sum _{i}\bigl(b-m_{\widetilde {6B_{i}}}(b)\bigr)h_{i}\biggr) \biggr](x)>ct\biggr\} \biggr) \\ &\qquad {}+\mu\biggl(\biggl\{ x\in\mathcal{X}:M_{r,6}\biggl[\sum _{i}\bigl\vert b-m_{\widetilde {6B_{i}}}(b)\bigr\vert \mathcal{M}(h_{i})\biggr](x)>t/2\biggr\} \biggr) \\ &\quad = D_{1}+D_{2}. \end{aligned}$$
According to the weak type 1-1 estimate for \(M_{\rho}\), we have, for any \(\lambda>0\),
$$\lambda\mu\bigl(\bigl\{ x\in\mathcal{X}:M_{r,6}(g) (x)>\lambda\bigr\} \bigr)\leq c\sup_{\delta>c\lambda}\delta\mu\bigl(\bigl\{ x\in\mathcal{X}: \bigl\vert g(x)\bigr\vert >c\delta\bigr\} \bigr). $$
Taking \(1< p_{1}< p\), it follows that
$$\begin{aligned} D_{1} \leq& t^{-1}\sup_{\delta>ct}\delta\mu \biggl(\biggl\{ x\in\mathcal{X}:\mathcal {M}\biggl(\sum _{i}\bigl(b-m_{\widetilde{B}}(b)\bigr)h_{i}\biggr) (x)>c\delta\biggr\} \biggr) \\ \leq&ct^{-p_{1}}\biggl\Vert \sum_{i} \bigl(b-m_{\widetilde{6B_{i}}}(b)\bigr)h_{i}\biggr\Vert ^{p_{1}}_{L^{p_{1}}(\mu)} \\ \leq&ct^{-p_{1}}\biggl\Vert \sum_{i} \bigl(b-m_{\widetilde{6B_{i}}}(b)\bigr)f\omega_{i}\biggr\Vert ^{p_{1}}_{L^{p_{1}}(\mu)} +ct^{-p_{1}}\biggl\Vert \sum _{i}\bigl(b-m_{\widetilde{6B_{i}}}(b)\bigr)\varphi_{i} \biggr\Vert ^{p_{1}}_{L^{p_{1}}(\mu)} \\ \leq& D_{11}+D_{12}. \end{aligned}$$
For \(D_{11}\), it follows that
$$\begin{aligned} D_{11} \leq& ct^{-p_{1}}\sum_{i} \biggl[\int_{6B_{i}}\bigl\vert f(x)\bigr\vert ^{p}d \mu(x)\biggr]^{p_{1}/p} \biggl[\int_{6B_{i}}\bigl\vert b(x)-m_{\widetilde{6B_{i}}}(b)\bigr\vert ^{p_{1}(p/p_{1})'}\,d\mu (x) \biggr]^{1-p_{1}/p} \\ \leq&ct^{-p_{1}}\sum_{i}\biggl[\int _{6B_{i}}\bigl\vert f(x)\bigr\vert ^{p}\,d\mu(x) \biggr]^{p_{1}/p}\mu \bigl(6^{2}B_{i} \bigr)^{1-p_{1}/p} \\ \leq&ct^{-p_{1}}\sum_{i}\int _{6B_{i}}\bigl\vert f(x)\bigr\vert ^{p}\,d\mu(x) \biggl[\int_{6B_{i}}\bigl\vert f(x)\bigr\vert ^{p}d \mu (x)\biggr]^{p_{1}/p-1}\mu\bigl(6^{3}B_{i} \bigr)^{1-p_{1}/p} \\ \leq&ct^{-p}\Vert f\Vert ^{p}_{L^{p}(\mu)}, \end{aligned}$$
where we use Lemma 2.2(1). To estimate \(D_{12}\), by the fact \(\sum_{i}\varphi_{i}\leq ct\), we have
$$\begin{aligned} D_{12} \leq& c\biggl\Vert \sum_{i} \bigl(b-m_{\widetilde{6B_{i}}}(b)\bigr)\varphi_{i}t^{-1}\biggr\Vert ^{p_{1}}_{L^{p_{1}}(\mu)} \\ \leq&c \biggl\Vert \biggl[\sum_{i}t^{-1} \vert \varphi_{i}\vert \bigl\vert b-m_{\widetilde {6B_{i}}}(b)\bigr\vert ^{p_{1}}\biggr]^{1/p_{1}}\biggl[\sum _{i}\bigl\vert t^{-1}\varphi_{i}\bigr\vert \biggr]^{1/p_{1}'}\biggr\Vert ^{p_{1}}_{L^{p_{1}}(\mu)} \\ \leq&c \biggl\Vert \biggl[\sum_{i}t^{-1} \vert \varphi_{i}\vert \bigl\vert b-m_{\widetilde {6B_{i}}}(b)\bigr\vert ^{p_{1}}\biggr]^{1/p_{1}}\biggr\Vert ^{p_{1}}_{L^{p_{1}}(\mu)} \\ \leq&ct^{-1}\sum_{i}\int _{R_{i}}\bigl\vert \varphi_{i}(x)\bigr\vert \bigl\vert b(x)-m_{\widetilde {6B_{i}}}(b)\bigr\vert ^{p_{1}}\,d\mu(x) \\ \leq&ct^{-1}\sum_{i}\biggl(\int _{R_{i}}\bigl\vert \varphi_{i}(x)\bigr\vert ^{p}\,d\mu(x)\biggr)^{1/p}\biggl(\int_{R_{i}} \bigl\vert b(x)-m_{\widetilde{R_{i}}}(b)\bigr\vert ^{p_{1}p'}\,d\mu(x) \biggr)^{1/p'} \\ &{}+ct^{-1}\sum_{i}\int _{R_{i}}\bigl\vert \varphi_{i}(x)\bigr\vert \bigl\vert m_{\widetilde {R_{i}}}(b)-m_{\widetilde{6B_{i}}}(b)\bigr\vert ^{p_{1}}\,d \mu(x) \\ \leq&ct^{-1}\sum_{i}\biggl(\int _{R_{i}}\bigl\vert \varphi_{i}(x)\bigr\vert ^{p}\,d\mu(x)\biggr)^{1/p}\mu(R_{i})^{1/p'}+ ct^{-1}\sum_{i}\int_{R_{i}} \bigl\vert \varphi_{i}(x)\bigr\vert \,d\mu(x) \\ \leq&ct^{-p}\sum_{i}\int _{6B_{i}}\bigl\vert f(x)\bigr\vert ^{p}\,d \mu(x)+ct^{-1}\sum_{i}\biggl(\int _{R_{i}}\bigl\vert \varphi_{i}(x)\bigr\vert ^{p}\,d\mu(x)\biggr)^{1/p}\mu(S_{i})^{1/p'} \\ \leq&ct^{-p} \Vert f\Vert ^{p}_{L^{p}(\mu)}. \end{aligned}$$
In order to estimate \(D_{2}\), we write
$$\begin{aligned} D_{2} \leq& ct^{-1}\sum_{i}\int _{\mathcal{X}\setminus 6S_{i}}\bigl\vert b(x)-m_{\widetilde{6B_{i}}}(b)\bigr\vert \mathcal{M}(h_{i}) (x)\,d\mu(x) \\ &{}+ct^{-1}\sum_{i}\int _{ 6S_{i}}\bigl\vert b(x)-m_{\widetilde{6B_{i}}}(b)\bigr\vert \mathcal {M}(\varphi_{i}) (x)\,d\mu(x) \\ &{}+ct^{-1}\sum_{i}\int _{\frac{6}{5}6B_{i}}\bigl\vert b(x)-m_{\widetilde {6B_{i}}}(b)\bigr\vert \mathcal{M}(\omega_{i}f) (x)\,d\mu(x) \\ &{}+ct^{-1}\sum_{i}\int _{6S_{i}\setminus\frac{6}{5}6B_{i}}\bigl\vert b(x)-m_{\widetilde {6B_{i}}}(b)\bigr\vert \mathcal{M}(\omega_{i}f) (x)\,d\mu(x) \\ =& D_{21}+D_{22}+D_{23}+D_{24}. \end{aligned}$$
For each i, we have
$$\begin{aligned} &\int_{\mathcal{X}\setminus6S_{i}}\bigl\vert b(x)-m_{\widetilde{6B_{i}}}(b)\bigr\vert \mathcal {M}(h_{i}) (x)\,d\mu(x) \\ &\quad \leq\int_{\mathcal{X}\setminus6S_{i}}\bigl\vert b(x)-m_{\widetilde {6B_{i}}}(b)\bigr\vert \biggl[\int_{0}^{d(x,c_{S_{i}})+r_{6S_{i}}}\biggl\vert \int _{d(x,y)< t}K(x,y)h_{i}(y)\,d\mu(y)\biggr\vert ^{2}\frac{dt}{t^{3}}\biggr]^{1/2}\,d\mu(x) \\ &\qquad {}+\int_{\mathcal{X}\setminus6S_{i}}\bigl\vert b(x)-m_{\widetilde{6B_{i}}}(b)\bigr\vert \biggl[\int^{\infty}_{d(x,c_{S_{i}})+r_{6S_{i}}}\biggl\vert \int _{d(x,y)< t}K(x,y)h_{i}(y)\,d\mu (y)\biggr\vert ^{2}\frac{dt}{t^{3}}\biggr]^{1/2}\,d\mu(x) \\ &\quad =D_{21}^{1}+D_{21}^{2}. \end{aligned}$$
$$\bigl\vert m_{\widetilde{6^{k+1}6S_{i}}}(b)-m_{\widetilde{6B_{i}}}(x)\bigr\vert \leq c(K_{6B_{i},\widetilde {6B_{i}}}+K_{6B_{i},6S_{i}}+K_{6S_{i},6^{k+1}6S_{i}}+K_{6^{k+1}6S_{i},\widetilde {6^{k+1}6S_{i}}})\leq ck, $$
we get
$$\begin{aligned} D_{21}^{1} \leq& \int_{\mathcal{X}\setminus6S_{i}}\bigl\vert b(x)-m_{\widetilde{6B_{i}}}(b)\bigr\vert \int_{\mathcal{X}}\bigl\vert K(x,y)h_{i}(y)\bigr\vert \biggl(\int_{d(x,y)}^{d(x,c_{S_{i}})+r_{6S_{i}}} \frac{dt}{t^{3}}\biggr)^{1/2}\,d\mu(y)\,d\mu(x) \\ \leq&c\int_{\mathcal{X}\setminus6S_{i}}\bigl\vert b(x)-m_{\widetilde{6B_{i}}}(b)\bigr\vert \int_{\mathcal{X}}\bigl\vert h_{i}(y)\bigr\vert \frac {r_{6S_{i}}^{1/2}}{d(x,c_{S_{i}})^{1/2}\lambda(x,d(x,c_{S_{i}}))}\,d\mu(y)\,d\mu (x) \\ =&c\Vert h_{i}\Vert _{L^{1}(\mu)}\sum _{j}\int_{6^{j+1}6S_{i}\setminus 6^{j}6S_{i}}\bigl\vert b(x)-m_{\widetilde{6B_{i}}}(b)\bigr\vert \frac{r_{6S_{i}}^{1/2}}{d(x,c_{S_{i}})^{1/2}\lambda(x,d(x,c_{S_{i}}))}\,d\mu (x) \\ \leq&c\Vert h_{i}\Vert _{L^{1}(\mu)}\sum _{j}\int_{6^{j+1}6S_{i}\setminus 6^{j}6S_{i}}\bigl\vert b(x)-m_{\widetilde{6^{j+1}6S_{i}}}(b)\bigr\vert \frac{r_{6R_{i}}^{1/2}}{d(x,c_{S_{i}})^{1/2}\lambda(x,d(x,c_{S_{i}}))}\,d\mu (x) \\ &{}+c\Vert h_{i}\Vert _{L^{1}(\mu)}\sum _{j}\bigl\vert m_{\widetilde {6^{j+1}6S_{i}}}(b)-m_{\widetilde{6B_{i}}}(b) \bigr\vert \\ &{}\times \int_{6^{j+1}6S_{i}\setminus6^{j}6S_{i}} \frac{r_{6S_{i}}^{1/2}}{d(x,c_{S_{i}})^{1/2}\lambda(x,d(x,c_{S_{i}}))}\,d\mu (x) \\ \leq&c\Vert h_{i}\Vert _{L^{1}(\mu)}\sum _{j}(j+1)\frac{r^{1/2}_{6S_{i}}\mu (6^{j+2}6S_{i})}{(6^{j}r_{6S_{i}})^{1/2}\lambda(c_{S_{i}},6^{j}r_{6S_{i}})} \\ \leq&c\Vert h_{i}\Vert _{L^{1}(\mu)}. \end{aligned}$$
By the vanishing moment of \(h_{i}\), it follows that
$$\begin{aligned} D_{21}^{2} \leq& \int_{\mathcal{X}\setminus6S_{i}}\bigl\vert b(x)-m_{\widetilde{6B_{i}}}(b)\bigr\vert \int_{\mathcal{X}}\bigl\vert \bigl[K(x,y)-K(x,c_{R_{i}})\bigr]h_{i}(y)\bigr\vert \\ &{}\times\biggl( \int^{\infty }_{d(x,c_{S_{i}})+r_{6S_{i}}}\frac{dt}{t^{3}} \biggr)^{1/2}\,d\mu(y)\,d\mu(x) \\ \leq& c\int_{\mathcal{X}\setminus6S_{i}}\bigl\vert b(x)-m_{\widetilde{6B_{i}}}(b) \bigr\vert \int_{\mathcal{X}}\bigl\vert \bigl[K(x,y)-K(x,c_{R_{i}}) \bigr]h_{i}(y)\bigr\vert \frac {1}{d(x,c_{S_{i}})}\,d\mu(y)\,d\mu(x) \\ \leq& \int_{\mathcal{X}}h_{i}(y)\biggl[\sum _{j}\int_{6^{j+1}6S_{i}\setminus 6^{j}6S_{i}}\frac{\vert [K(x,y)-K(x,c_{S_{i}})\vert }{d(x,c_{S_{i}})} \\ &{}\times\bigl(\bigl\vert b(x)-m_{6^{j+1}6S_{i}}\bigr\vert +\vert m_{6^{j+1}6S_{i}}-m_{\widetilde{6B_{i}}}\vert \bigr)\,d\mu (x)\biggr]\,d\mu(y) \\ \leq&c\Vert h_{i}\Vert _{L^{1}(\mu)}. \end{aligned}$$
$$\begin{aligned} \Vert h_{i}\Vert _{L^{1}(\mu)} \leq& c\biggl(\int _{6B_{i}}\bigl\vert f(x)\bigr\vert ^{p}\,d\mu(x) \biggr)^{1/p}\mu (6B_{i})^{1/p'}+\biggl(\int _{6S_{i}}\bigl\vert \varphi(x)\bigr\vert ^{p}d \mu(x)\biggr)^{1/p}\mu(6S_{i})^{1/p'} \\ \leq&ct^{1-p}\Vert f\Vert ^{p}_{L^{p}(\mu)}, \end{aligned}$$
so \(D_{21}\leq ct^{-1}t^{1-p}\Vert f\Vert ^{p}_{L^{p}(\mu)}\leq ct^{-p}\Vert f\Vert ^{p}_{L^{p}(\mu)}\).
For \(D_{22}\), it follows from the \(L^{p}(\mu)\) boundedness of \(\mathcal {M}\) that
$$\begin{aligned} D_{22} \leq& ct^{-1}\sum_{i} \int_{ 6S_{i}}\bigl\vert b(x)-m_{\widetilde {6B_{i}}}(b)\bigr\vert \mathcal{M}(\varphi_{i}) (x)\,d\mu(x) \\ &{}+ct^{-1}\sum_{i}\bigl\vert m_{\widetilde{6B_{i}}}(b)-m_{\widetilde{6B_{i}}}(b)\bigr\vert \int_{ 6S_{i}} \mathcal{M}(\varphi_{i}) (x)\,d\mu(x) \\ \leq& ct^{-1}\sum_{i}\biggl(\int _{ 6S_{i}}\bigl\vert b(x)-m_{\widetilde {6B_{i}}}(b)\bigr\vert ^{p'}\,d\mu(x)\biggr)^{1/p'}\bigl\Vert \mathcal{M}( \varphi_{i})\bigr\Vert _{L^{p}(\mu)} \\ &{}+ct^{-1}\sum_{i}\bigl\Vert \mathcal{M}(\varphi_{i})\bigr\Vert _{L^{p}(\mu)} \mu(6S_{i})^{1/p'} \\ \leq& ct^{-1}\sum_{i}\biggl(\int _{S_{i}}\varphi_{i}(x)\,d\mu(x)\biggr)^{1/p}\mu \bigl(6^{2}S_{i}\bigr)^{1/p'} \\ \leq& ct^{-1}t^{1-p}\sum_{i} \int_{6B_{i}}\bigl\vert f(x)\bigr\vert ^{p}\,d\mu(x) \leq ct^{-p}\Vert f\Vert ^{p}_{L^{p}(\mu)}. \end{aligned}$$
Similar to \(D_{22}\), we have
$$\begin{aligned} D_{23} \leq& ct^{-1}\sum_{i} \biggl(\int_{ \frac{6}{5}6B_{i}}\bigl\vert b(x)-m_{\widetilde {6B_{i}}}(b)\bigr\vert ^{p'}\,d\mu(x)\biggr)^{1/p'}\bigl\Vert \mathcal{M}( \omega_{i}f)\bigr\Vert _{L^{p}(\mu)} \\ \leq& ct^{-1}\sum_{i} \mu(S_{i})^{1/p'}\Vert \omega_{i}f\Vert _{L^{p}(\mu)} \\ \leq& ct^{-1}\sum_{i}t^{1-p} \Vert \omega_{i}f\Vert _{L^{p}(\mu)}\int_{6B_{i}} \bigl\vert f(x)\bigr\vert ^{p}\,d\mu(x) \biggl(\int _{S_{i}}\bigl\vert \varphi_{i}(x)^{p}\bigr\vert \,d\mu(x)\biggr)^{-1} \\ \leq& ct^{-p}\Vert f\Vert _{L^{p}(\mu)}. \end{aligned}$$
Next we estimate \(D_{24}\). If \(\operatorname {supp}f\subset B\) for some ball then, for any \(\rho>1\) and \(x\in\mathcal{X}\setminus\rho B\), we have
$$\begin{aligned} \mathcal{M}(f) (x) \leq& c\int_{B}\bigl\vert K(x,y)f(y) \bigr\vert \biggl(\int_{d(x,y)}^{\infty} \frac {dt}{t^{3}}\biggr)^{1/2}\,d\mu(y) \\ \leq& c\int_{B}\frac{\vert f(y)\vert }{\lambda(x,d(x,y))}\leq\frac{c}{\lambda (c_{B},d(x,c_{B}))} \int_{B}\bigl\vert f(y)\bigr\vert \,d\mu(y). \end{aligned}$$
For any i we write \(S_{i}=(3\times6^{2})^{k_{i}}B_{i}\). It follows that
$$\begin{aligned} D_{24} \leq& t^{-1}\sum_{i} \int_{6S_{i}\setminus\frac{6}{5}6B_{i}} \frac{\vert b(x)-m_{\widetilde{6B_{i}}}(b)\vert }{\lambda(c_{B_{i}},d(x,c_{B_{i}}))}\int_{6B_{i}} \bigl\vert f(y)\omega_{i}(y)\bigr\vert \,d\mu(y)\,d\mu(x) \\ \leq& ct^{-1}\sum_{i}\biggl(\int _{6B_{i}}\bigl\vert f(y)\bigr\vert \,d\mu(y)\biggr)\int _{6S_{i}\setminus S_{i}} \frac{\vert b(x)-m_{\widetilde{6B_{i}}}(b)\vert }{\lambda(c_{B_{i}},d(x,c_{B_{i}}))}\,d\mu (x) \\ &{}+ ct^{-1}\sum_{i}\biggl(\int _{6B_{i}}\bigl\vert f(y)\bigr\vert \,d\mu(y)\biggr)\int _{S_{i}\setminus\frac{6}{5}6B_{i}} \frac{\vert b(x)-m_{\widetilde{6B_{i}}}(b)\vert }{\lambda(c_{B_{i}},d(x,c_{B_{i}}))}\,d\mu (x) \\ \leq& ct^{-1}\sum_{i}\biggl(\int _{6B_{i}}\bigl\vert f(y)\bigr\vert \,d\mu(y)\biggr)\int _{6S_{i}\setminus S_{i}} \frac{\vert b(x)-m_{\widetilde{6S_{i}}}(b)\vert +\vert m_{\widetilde {6S_{i}}}(b)-m_{\widetilde{6B_{i}}}(b)\vert }{\lambda(c_{B_{i}},d(x,c_{B_{i}}))}\,d\mu (x) \\ &{}+ ct^{-1}\sum_{i}\biggl(\int _{6B_{i}}\bigl\vert f(y)\bigr\vert \,d\mu(y)\biggr)\\ &{}\times\sum _{j=0}^{k_{i}-1}\int_{6^{j+1}6B_{i}\setminus6^{j}6B_{i}} \frac{\vert b(x)-m_{\widetilde{6S_{i}}}(b)\vert +\vert m_{\widetilde {6S_{i}}}(b)-m_{\widetilde{6B_{i}}}(b)\vert }{\lambda(c_{B_{i}},d(x,c_{B_{i}}))}\,d\mu (x) \\ \leq& ct^{-1}\sum_{i}\biggl(\int _{6B_{i}}\bigl\vert f(y)\bigr\vert \,d\mu(y)\biggr) \biggl(1+ \frac{\mu (6S_{i})}{\lambda(c_{B_{i}},r_{S_{i}})}\biggr) \\ &{}+ ct^{-1}\sum_{i}\biggl(\int _{6B_{i}}\bigl\vert f(y)\bigr\vert \,d\mu(y)\biggr)\Biggl[1+ \sum_{j=0}^{k_{i}-1}\frac {\mu(6^{j+2}6B_{i})}{\lambda(c_{B_{i}},6^{j+1}r_{B_{i}})}K_{B_{i},S_{i}} \Biggr] \\ \leq& ct^{-1}\sum_{i}\biggl(\int _{6B_{i}}\bigl\vert f(y)\bigr\vert \,d\mu(y)\biggr)\leq ct^{-1}\sum_{i}\biggl(\int _{6B_{i}}\bigl\vert f(y)\bigr\vert ^{p}\,d\mu(y) \biggr)^{1/p}\mu(6B_{i})^{1/p'} \\ \leq& ct^{-p}\Vert f\Vert _{L^{p}(\mu)}^{p}. \end{aligned}$$
Combining these estimates for the term \(D_{21}\), \(D_{22}\), \(D_{23}\), and \(D_{24}\) yields the desired estimate for \(D_{2}\). So we complete the proof of Theorem 1.10.



Project (Grant No. 11261055, 11161044) supported by NSFC. The authors cordially thank the referees for their careful reading and helpful comments.

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Authors’ Affiliations

College of Mathematics and System Science, Xinjiang University


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© Yonghui and Jiang 2015