Open Access

Fractional type Marcinkiewicz integrals over non-homogeneous metric measure spaces

Journal of Inequalities and Applications20162016:259

DOI: 10.1186/s13660-016-1203-0

Received: 8 March 2016

Accepted: 12 October 2016

Published: 21 October 2016


The main goal of the paper is to establish the boundedness of the fractional type Marcinkiewicz integral \(\mathcal{M}_{\beta,\rho,q}\) on non-homogeneous metric measure space which includes the upper doubling and the geometrically doubling conditions. Under the assumption that the kernel satisfies a certain Hörmander-type condition, the authors prove that \(\mathcal{M}_{\beta,\rho,q}\) is bounded from Lebesgue space \(L^{1}(\mu)\) into the weak Lebesgue space \(L^{1,\infty}(\mu)\), from the Lebesgue space \(L^{\infty}(\mu)\) into the space \(\operatorname{RBLO}(\mu)\), and from the atomic Hardy space \(H^{1}(\mu)\) into the Lebesgue space \(L^{1}(\mu)\). Moreover, the authors also get a corollary, that is, \(\mathcal{M}_{\beta,\rho,q}\) is bounded on \(L^{p}(\mu)\) with \(1< p<\infty\).


42B35 47B47 30L99


non-homogeneous metric measure space fractional type Marcinkiewicz integral Lebesgue space Hardy space \(\operatorname{RBLO}(\mu)\)

1 Introduction

In 2010, Hytönen in [1] first introduced a new class of metric measure spaces which satisfy the so-called upper doubling and the geometrically doubling conditions (see also Definitions 1.1 and 1.2 below, respectively), for convenience, the new spaces are called non-homogeneous metric measure spaces. As special cases, the new spaces not only contain the homogeneous type spaces (see [2]), but also they include metric spaces endowed with measures satisfying the polynomial growth condition (see, for example, [39]). Further, it is meaningful to pay much attention to a study of the properties of some classical operators, commutators, and function spaces on non-homogeneous metric measure spaces; see [1016]. In addition, we know that the harmonic analysis has important applications in many fields including geometrical analysis, functional analysis, partial differential equations, and fuzzy fractional differential equations, we refer the reader to [1720] and the references therein.

In the present paper, let \((\mathcal{X}, d, \mu)\) be a non-homogeneous metric measure space in the sense of Hytönen [1]. In 2007, Hu et al. [5] obtained the boundedness of the Marcinkiewicz with non-doubling measure. Besides, Lin and Yang [13] established some equivalent boundedness of Marcinkiewicz integral on \((\mathcal{X}, d, \mu)\). Inspired by this, we will mainly consider the boundedness of the fractional type Marcinkiewicz integrals introduced in [21] on \((\mathcal{X}, d, \mu)\).

To state the main consequences of this article, we first of all recall some necessary notions and notation. Hytönen [1] originally introduced the following notions of the upper doubling condition and the geometrically doubling condition.

Definition 1.1


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, for all \(x\in \mathcal{X}\) and \(r\in(0,\infty)\),
$$ \mu\bigl(B(x,r)\bigr)\leq\lambda(x,r)\leq C_{\lambda}\lambda\biggl(x, \frac {r}{2}\biggr). $$
Hytönen et al. [16] have proved that there is another dominating function λ̃ such that \(\tilde{\lambda }\leq\lambda\), \(C_{\tilde{\lambda}}\leq C_{\lambda}\), and
$$ \tilde{\lambda}(x,r)\leq C_{\tilde{\lambda}}\tilde{\lambda }(y,r), $$
where \(x,y\in\mathcal{X}\) and \(d(x,y)\leq r\). Based on this, we also assume the dominating function λ that in (1.1) satisfies (1.2) in this paper.

Definition 1.2


A metric space \((\mathcal{X},d)\) is said to be geometrically doubling, if there exists some \(N_{0}\in\mathbb{N}\) such that, for any ball \(B(x,r)\subset\mathcal{X}\), 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 1.3

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

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

  2. (2)

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

  3. (3)

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

  4. (4)

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

Now we recall the definition of coefficient \(K_{B,S}\) introduced by Hytönen in [1], which is analogous to the quantity \(K_{Q,R}\) introduced in [4], that is, for any two balls \(B\subset S\) in \(\mathcal {X}\), define
$$ K_{B,S}:=1+ \int_{2S\setminus B}\frac{1}{\lambda (c_{B},d(x,c_{B}))}\,\mathrm{d}\mu(x), $$
where \(c_{B}\) is the center of the ball B.

Though the measure doubling condition is not assumed uniformly for all balls on \((\mathcal{X},d, \mu)\), it was proved in [1] that there still exist many balls satisfying the property of the \((\alpha,\eta )\)-doubling, namely, we say that a ball \(B\subset\mathcal{X}\) is \((\alpha,\eta)\)-doubling if \(\mu(\alpha B)\leq\eta\mu(B)\), for \(\alpha,\eta>1\). In the rest of this paper, unless α and \(\eta_{\alpha}\) are specified, otherwise, by an \((\alpha,\eta _{\alpha})\)-doubling ball we mean a \((6,\beta_{6})\)-doubling ball with a fixed number \(\eta_{6}>\max\{C^{3\log_{2}6}_{\lambda}, 6^{n}\}\), where \(n:=\log _{2}N_{0}\) is viewed as a geometric dimension of the space. Moreover, the smallest \((6,\eta_{6})\)-doubling ball of the from \(6^{j}B\) with \(j\in\mathbb{N}\) is denoted by \(\tilde{B}^{6}\), and \(\tilde{B}^{6}\) is simply denoted by .

Next, we recall the following definition of \(\operatorname{RBMO}(\mu)\) from [1].

Definition 1.4


Let \(\kappa>1\) be a fixed constant. 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, a number \(f_{B}\) such that
$$\frac{1}{\mu(\kappa B)} \int_{B}\bigl\vert f(y)-f_{B}\bigr\vert \, \mathrm{d}\mu(y)\leq C $$
$$|f_{B}-f_{R}|\leq CK_{B,R} $$
for any two balls B and R such that \(B\subset R\). Moreover, the \(\operatorname{RBMO}(\mu)\) norm of f is defined to be the minimal constant C as above and denoted by \(\|f\|_{\operatorname{RBMO}(\mu)}\).

From [1], Hytönen showed that the space \(\operatorname{RBMO}(\mu)\) is not dependent on the choice of κ. Lin and Yang [14] introduced the following definition of the space \(\operatorname{RBLO}(\mu)\) and proved that \(\operatorname{RBLO}(\mu)\subset\operatorname{RBMO}(\mu)\).

Definition 1.5


A function \(f\in L^{1}_{\mathrm{loc}}(\mu)\) is said to belong to the space \(\operatorname{RBLO}(\mu)\) if there exists a positive constant C such that. for any \((6,\beta_{6})\)-doubling ball B,
$$\frac{1}{\mu(\sigma B)} \int_{B}\bigl[f(y)- \mathop{\operatorname{ess \,inf}}_{\tilde{B}}f\bigr]\,\mathrm{d}\mu(y)\leq C $$
$$\mathop{\operatorname{ess\,inf}} _{B}f- \mathop{\operatorname{ess \,inf}}_{S}f\leq CK_{B,S} $$
for any two \((6,\beta_{6})\)-doubling balls \(B\subset S\). The minimal constant C above is defined to be the norm of f in \(\operatorname{RBLO}(\mu)\) and denoted by \(\|f\|_{\operatorname{RBLO}(\mu)}\).

Now we give the notion of the fractional type Marcinkiewicz integral slightly changed from [21].

Definition 1.6

Let \(\Delta=\{(x,x): x\in\mathcal{X}\}\). A stand kernel is a mapping \(K: \mathcal{X}\times\mathcal{X}\setminus\Delta \rightarrow\mathbb{C}\) for which there exist positive constants \(\delta\in(0,1]\), \(\beta\geq0\), and C such that, for \(x, y\in \mathcal{X}\) with \(x\neq y\),
$$ \bigl\vert K(x,y)\bigr\vert \leq C\frac{[d(x,y)]^{1+\beta}}{\lambda(x,d(x,y))}, $$
and for all \(x, \tilde{x}, y\in\mathcal{X}\) with \(d(x,y)\geq 2d(x,\tilde{x})\),
$$ \bigl\vert K(x,y)-K(\tilde{x},y)\bigr\vert +\bigl\vert K(y,x)-K(y, \tilde{x})\bigr\vert \leq C\frac {[d(x,\tilde{x})]^{\delta+1+\beta}}{[d(x,y)]^{\delta}\lambda (x,d(x,y))}. $$
The fractional type Marcinkiewicz integral \(\mathcal{M}_{\beta,\rho ,q}(f)\) related to the above kernel \(K(x,y)\) is formally defined by
$$ \mathcal{M}_{\beta,\rho,q}(f) (x):= \biggl( \int^{\infty}_{0} \biggl\vert \frac{1}{t^{\beta+\rho}} \int_{d(x,y)< t}\frac {K(x,y)}{[d(x,y)]^{1-\rho}}f(y) \,\mathrm{d}\mu(y)\biggr\vert ^{q}\frac{\mathrm{d}t}{t} \biggr)^{\frac {1}{q}}, $$
where \(x\in\mathcal{X}\), \(\rho>0\), \(\beta\geq0\), and \(q>1\).

Recently, many authors have studied the properties of the fractional type Marcinkiewicz integrals; see [2224]. To the fractional type Marcinkiewicz integral operator \(\mathcal {M}_{\beta,\rho,q}\) as in (1.6), one can return to the Marcinkiewicz integrals on different function spaces when the indices are replaced by some fixed numbers; see the following remark.

Remark 1.7

  1. (1)

    When \(\rho=1\), \(\beta=0\), and \(q=2\), the operator \(\mathcal {M}_{\beta,\rho,q}(f)\) as in (1.6) is just the Marcinkiewicz integral on \((\mathcal{X},d,\mu)\) in [13].

  2. (2)

    If we take \((\mathcal{X},d,\mu)=(\mathbb{R}^{n},|\cdot|,\mu)\), \(\rho=1\), \(\beta=0\), and \(q=2\), the operator \(\mathcal{M}_{\beta ,\rho,q}(f)\) as in (1.6) is just the Marcinkiewicz integral with non-doubling measures (see [5]).

  3. (3)
    If we take \((\mathcal{X},d,\mu)=(\mathbb{R}^{n},|\cdot|,\mathrm {d}x)\), \(K(x,y)=\frac{\Omega(x-y)}{|x-y|^{n-1}}\), \(\rho=1\), \(\beta =0\), and \(q=2\), then the operator \(\mathcal{M}_{\beta,\rho,q}(f)\) as in (1.6) is just the classical Marcinkiewicz integral introduced in [25] and its form is as follows:
    $$\mathcal{M}_{\Omega}(f) (x):= \biggl( \int^{\infty}_{0}\biggl\vert \int _{|x-y|\leq t}\frac{\Omega(x-y)}{|x-y|^{n-1}}f(y)\,\mathrm{d}y \biggr\vert ^{2}\frac{\mathrm{d}t}{t^{3}} \biggr)^{\frac{1}{2}}, \quad x\in \mathbb{R}^{n}; $$
    for more about behaviors of the \(\mathcal{M}_{\Omega}\), see [2630].

Further, we recall the notion of the atomic Hardy spaces given in [16].

Definition 1.8


Let \(\zeta\in(1,\infty)\) and \(p>1\). A function \(b\in L^{1}_{\mathrm{loc}}(\mu)\) is called a \((p,1)_{\tau}\)-atomic block if
  1. (1)

    there exists a ball S such that \(\operatorname{supp} b\subset S\);

  2. (2)


  3. (3)
    for any \(i\in\{1,2\}\), there exists a function \(a_{i}\) supported on a ball \(B_{i}\subset S\) and \(\tau_{i}\in\mathbb{C}\) such that \(b=\tau_{1}a_{1}+\tau_{2}a_{2}\) and
    $$ \|a_{i}\|_{L^{p}(\mu)}\leq\bigl[\mu(\zeta B_{i}) \bigr]^{\frac {1}{p}-1}K^{-1}_{B_{i},S}. $$
Moreover, let
$$|b|_{H^{1,p}_{\mathrm{atb}}(\mu)}:=|\tau_{1}|+|\tau_{2}|. $$

We say that a function \(f\in L^{1}(\mu)\) belongs to the atomic Hardy space \(H^{1,p}_{\mathrm{atb}}(\mu)\), if there exist \((p,1)_{\tau }\)-atomic blocks \(\{b_{i}\}^{\infty}_{i=1}\) such that \(f= {\sum } ^{\infty}_{i=1}b_{i}\) in \(L^{1}(\mu)\) and \({\sum } ^{\infty}_{i=1}|b_{i}|_{H^{1,p}_{\mathrm{atb}}(\mu)}<\infty \). The norm of f in \(H^{1,p}_{\mathrm{atb}}(\mu)\) is defined by \(\|f\|_{H^{1,p}_{\mathrm{atb}}(\mu)}:=\inf\{ {\sum} _{i}|b_{i}|_{H^{1,p}_{\mathrm{\mathrm{atb}}}(\mu)}\}\), where the infimum is taken over all the possible decompositions of f as above.

Also, in [16], Hytönen et al. proved that, for each \(p\in (1,\infty]\), the atomic Hardy space \(H^{1,p}_{\mathrm{atb}}(\mu)\) is independent of the choice of ζ and that the spaces \(H^{1,p}_{\mathrm{atb}}(\mu)\) and \(H^{1,\infty}_{\mathrm{atb}}(\mu )\) have the same norms for all \(p\in(1,\infty]\). Thus, we always denote \(H^{1,p}_{\mathrm{atb}}(\mu)\) simply by \(H^{1}(\mu)\).

Finally, we state the main results of this article.

Theorem 1.9

Let \(K(x,y)\) satisfy (1.4) and (1.5), and \(\mathcal{M}_{\beta,\rho,q}\) be as in (1.6), where \(\rho >0\), \(\beta\geq0\), and \(q>1\). If \(\mathcal{M}_{\beta,\rho,q}\) is bounded on \(L^{2}(\mu)\), then it is also bounded from \(L^{1}(\mu)\) into \(L^{1,\infty}(\mu)\), that is, there exists a positive constant C such that, for all \(t>0\) and \(f\in L^{1}(\mu)\),
$$ \mu\bigl(\bigl\{ x\in\mathcal{X}: \mathcal{M}_{\beta,\rho,q}(f) (x)>t\bigr\} \bigr)\leq C\frac{\|f\|_{L^{1}(\mu)}}{t}. $$

Theorem 1.10

Let \(K(x,y)\) satisfy (1.4) and (1.5), \(\rho >0\), \(\beta\geq0\), and \(q>1\). Suppose that \(\mathcal{M}_{\beta,\rho ,q}\) is as in (1.6) and bounded on \(L^{2}(\mu)\). Then for \(f\in L^{\infty}(\mu)\), \(\mathcal{M}_{\beta,\rho,q}\) is either infinite everywhere or finite μ-finite almost everywhere; more precisely, if \(\mathcal{M}_{\beta,\rho,q}\) is finite at some point \(x_{0}\in\mathcal{X}\), then \(\mathcal{M}_{\beta,\rho,q}\) is μ-almost everywhere and
$$\bigl\Vert \mathcal{M}_{\beta,\rho,q}(f)\bigr\Vert _{\operatorname{RBLO}(\mu)}\leq C \Vert f\Vert _{L^{\infty}(\mu)}, $$
where the positive constant C is not dependent on f.

By Theorem 1.9, Theorem 1.10, and Theorem 1.1 in [15], it is easy to obtain the following corollary.

Corollary 1.11

Under the assumption of Theorem 1.9, then \(\mathcal {M}_{\beta,\rho,q}\) is bounded on \(L^{p}(\mu)\) for any \(p\in (1,\infty)\).

Theorem 1.12

Let \(K(x,y)\) satisfy (1.4) and (1.5), and \(\mathcal{M}_{\beta,\rho,q}\) be as in (1.6). If \(\mathcal{M}_{\beta,\rho,q}\) is bounded on \(L^{2}(\mu)\), then it is also bounded from \(H^{1}(\mu)\) into \(L^{1}(\mu)\).

Throughout the paper, C represents for a positive constant which is independent of the main parameters involved, but it may be different from line to line. For a μ-measurable set E, \(\chi_{E}\) denotes its characteristic function. For any \(p\in[1,\infty]\), we denote by \(p'\) its conjugate index, that is, \(\frac{1}{p}+\frac{1}{p'}=1\).

2 Preliminaries

In this section, in order to prove our main theorems, we need some lemmas. First, we recall some useful properties of \(K_{B,S}\) as in (1.3) (see [1]).

Lemma 2.1


  1. (1)

    For all balls \(B\subset R\subset S\), it holds true that \(K_{B,R}\leq K_{B,S}\).

  2. (2)

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

  3. (3)

    For any \(\varrho\in(1,\infty)\), there exists a positive constant \(C_{\varrho}\), depending on ϱ, such that, for all balls B, \(K_{B,\tilde{B}^{\varrho}}\leq C_{\varrho}\).

  4. (4)

    There is a positive constant c such that, for all balls \(B\subset R\subset S\), \(K_{B,S}\leq K_{B,R}+cK_{R,S}\). In particular, if B and R are concentric, then \(c=1\).

  5. (5)

    There exists a positive constant such that, for all balls \(B\subset R\subset S\), \(K_{B,R}\leq\tilde {c}K_{B,S}\); moreover, if B and R are concentric, then \(K_{R,S}\leq K_{B,S}\).


Next, we recall the Calderón-Zygmund decomposition theorem from [31] as follows. Let \(\gamma_{0}\) be a fixed non-negative constant and satisfy \(\gamma_{0}>\max\{C^{3\log_{2}6}_{\lambda},6^{3n}\}\), where \(C_{\lambda}\) is as in (1.1) and n as in Remark 1.3.

Lemma 2.2


Let \(p\in[1,\infty)\), \(f\in L^{p}(\mu)\), and \(t\in(0,\infty )\) (\(t>\frac{\gamma_{0}\|f\|_{L^{p}(\mu)}}{\mu(\mathcal{X})}\) when \(\mu(\mathcal{X})<\infty\)). Then:
  1. (1)
    There exists a family of finite overlapping balls \(\{ 6B_{i}\}_{i}\) such that \(\{B_{i}\}_{i}\) is pairwise disjoint,
    $$\begin{aligned}& \frac{1}{\mu(6^{2}B_{i})} \int_{B_{i}}\bigl\vert f(x)\bigr\vert ^{p}\, \mathrm{d}\mu (x)>\frac{t^{p}}{\gamma_{0}} \quad \textit{for all }i, \\& \frac{1}{\mu(6^{2}\upsilon B_{i})} \int_{\upsilon B_{i}}\bigl\vert f(x)\bigr\vert ^{p}\, \mathrm{d}\mu(x)\leq\frac{t^{p}}{\gamma_{0}} \quad \textit {for all }i\textit{ and all } \upsilon\in(2,\infty), \end{aligned}$$
    $$ \bigl\vert f(x)\bigr\vert \leq t \quad \textit{for } \mu\textit{-almost every } x\in\mathcal {X}\Bigm\backslash \biggl( \bigcup_{i}6B_{i} \biggr). $$
  2. (2)
    For each i, let \(S_{i}\) be the smallest \((3\times 6^{2}, C^{\log_{2}(3\times6^{2})+1}_{\lambda})\)-doubling ball of the family \(\{(3\times6^{2})^{k}B_{i}\}_{k\in\mathbb{N}}\), and \(\omega _{i}=\chi_{6B_{i}}/( {\sum} _{k}\chi_{6B_{k}})\). Then there exist a family \(\{\varphi_{i}\}_{i}\) of functions that, for each i, \(\operatorname{supp}(\varphi_{i})\subset S_{i}\), \(\varphi_{i}\) has a constant sign on \(S_{i}\) and
    $$\begin{aligned}& \int_{\mathcal{X}}\varphi_{i}(x)\,\mathrm{d}\mu(x)= \int _{6B_{i}}f(x)\omega_{i}(x)\,\mathrm{d} \mu(x), \end{aligned}$$
    $$\begin{aligned}& {\sum} _{i}\bigl\vert \varphi_{i}(x)\bigr\vert \leq\gamma t\quad \textit{for } \mu\textit{-almost every } x\in \mathcal{X}, \end{aligned}$$
    where γ is some positive constant depending only on \((\mathcal {X},\mu)\), and there exists a positive constant C, independent of f, t, and i, such that, if \(p=1\), then
    $$ \|\varphi_{i}\|_{L^{\infty}(\mu)}\mu(S_{i})\leq C \int_{\mathcal {X}}\bigl\vert f(x)\omega_{i}(x)\bigr\vert \,\mathrm{d}\mu(x), $$
    and if \(p\in(1,\infty)\),
    $$ \biggl( \int_{S_{i}}\bigl\vert \varphi_{i}(x)\bigr\vert ^{p}\,\mathrm{d}\mu(x) \biggr)^{\frac{1}{p}}\bigl[ \mu(S_{i})\bigr]^{\frac{1}{p'}}\leq\frac {C}{t^{p-1}} \int_{\mathcal{X}}\bigl\vert f(x)\omega_{i}(x)\bigr\vert ^{p}\,\mathrm{d}\mu (x). $$

Finally, we recall the following characterizations of \(\operatorname{RBLO}(\mu)\) given in [14].

Lemma 2.3


If \(f\in L^{1}_{\mathrm{loc}}(\mu)\) is said to be in the space \(\operatorname{RBLO}(\mu)\), then there exists a non-negative constant C satisfying that, for all \((6,\beta_{6})\)-doubling balls B,
$$\frac{1}{\mu(B)} \int_{B}\bigl[f(y)- \mathop{\operatorname{ess\, inf}} _{B}f\bigr]\,\mathrm{d}\mu(y)\leq C $$
and, for all \((6,\beta_{6})\)-doubling balls \(B\subset S\),
$$ m_{B}(f)-m_{S}(f)\leq CK_{B,S}, $$
in this paper, \(m_{B}(f)\) represents the mean of f over B, that is,
$$m_{B}(f):=\frac{1}{\mu(B)} \int_{B}f(y)\,\mathrm{d}\mu(y). $$
Moreover, the minimal constant C above equals \(\|f\|_{\operatorname{RBLO}(\mu)}\).

3 Proofs of the main theorems

Proof of Theorem 1.9

Without loss of generality, by homogeneity, we can assume that \(\|f\| _{L^{1}(\mu)}=1\). It is easy to see that the conclusion of Theorem 1.9 automatically holds true if \(t\leq\eta_{6}\|f\|_{L^{1}(\mu)}/\mu (\mathcal{X})\) when \(\mu(\mathcal{X})<\infty\). Therefore, we only need consider the case \(t>\eta_{6}\|f\|_{L^{1}(\mu)}/\mu(\mathcal {X})\). For any given \(f\in L^{1}(\mu)\) and \(t>\eta_{6}\|f\| _{L^{1}(\mu)}/\mu(\mathcal{X})\), applying Lemma 2.2 to f and t, and letting \(S_{i}\) be as in Lemma 2.2, we may write \(f=g+h\), where \(g:=f\chi_{\mathcal{X}\setminus \bigcup _{i}6B_{i}}+ {\sum} _{i}\varphi_{i}\) and \(h:= {\sum} _{i}(\omega_{i}f-\varphi_{i})=: {\sum} _{i}h_{i}\). By applying (2.2), (2.4), and the assumption \(\|f\| _{L^{1}(\mu)}=1\), we easily obtain \(\|g\|_{L^{\infty}(\mu)}\leq Ct\) and \(\|g\|_{L^{1}(\mu)}\leq C\). Thus, by the \(L^{2}(\mu)\)-boundedness of \(\mathcal{M}_{\beta,\rho,q}\), we conclude that
$$\mu\bigl(\bigl\{ x\in\mathcal{X}: \mathcal{M}_{\beta,\rho,q}(g) (x)>t\bigr\} \bigr)\leq Ct^{-2}\bigl\Vert \mathcal{M}_{\beta,\rho,q}(g)\bigr\Vert _{L^{2}(\mu)}\leq Ct^{-2}\Vert g\Vert _{L^{2}(\mu)}\leq Ct^{-1}. $$
On the other hand, by (2.1) with \(p=1\), and the fact that \(\{B_{i}\} _{i}\) is a sequence of pairwise disjoint balls, we have
$$\mu\biggl( \bigcup_{i}6^{2}B_{i} \biggr)\leq {\sum} _{i}\mu\bigl(6^{2}B_{i} \bigr)\leq Ct^{-1} {\sum} _{i} \int _{B_{i}}\bigl\vert f(x)\bigr\vert \,\mathrm{d}\mu(x)\leq Ct^{-1}, $$
and therefore, the proof of Theorem 1.9 can be reduced to proving
$$ \mu\biggl(\biggl\{ x\in\mathcal{X}\Bigm\backslash \bigcup _{i}\bigl(6^{2}B_{i}\bigr): \mathcal{M}_{\beta,\rho,q}(h) (x)>t\biggr\} \biggr)\leq Ct^{-1}. $$
Notice that
$$\begin{aligned}& \mu\biggl(\biggl\{ x\in\mathcal{X}\Bigm\backslash \bigcup _{i}\bigl(6^{2}B_{i}\bigr): \mathcal{M}_{\beta,\rho,q}(h) (x)>t\biggr\} \biggr) \\ & \quad \leq t^{-1} {\sum} _{i} \int_{\mathcal {X}\setminus6S_{i}}\mathcal{M}_{\beta,\rho,q}(h_{i}) (x)\, \mathrm{d}\mu(x) +t^{-1} {\sum} _{i} \int_{6S_{i}\setminus 6^{2}B_{i}}\mathcal{M}_{\beta,\rho,q}(h_{i}) (x)\, \mathrm{d}\mu(x) \\ & \quad =:\mathrm{E}_{1}+\mathrm{E}_{2}. \end{aligned}$$
For \(\mathrm{E}_{1}\). Let \(S_{i}\) be as in Lemma 2.2. Denote its center and radius by \(c_{S_{i}}\) and \(r_{S_{i}}\), respectively. Write
$$\begin{aligned}& \int_{\mathcal{X}\setminus6S_{i}}\mathcal{M}_{\beta,\rho ,q}(h_{i}) (x)\, \mathrm{d}\mu(x) \\ & \quad \leq \int_{\mathcal{X}\setminus6S_{i}}\biggl( \int ^{d(x,c_{S_{i}})+r_{S_{i}}}_{0}\biggl\vert \frac{1}{t^{\beta+\rho}} \int_{d(x,y)< t}\frac{K(x,y)}{[d(x,y)]^{1-\rho}}h_{i}(y) \,\mathrm{d} \mu(y)\biggr\vert ^{q}\frac{\mathrm{d}t}{t}\biggr)^{\frac {1}{q}}\, \mathrm{d}\mu(x) \\& \qquad {}+ \int_{\mathcal{X}\setminus6S_{i}}\biggl( \int^{\infty }_{d(x,c_{S_{i}})+r_{S_{i}}}\biggl\vert \frac{1}{t^{\beta+\rho}} \int_{d(x,y)< t}\frac{K(x,y)}{[d(x,y)]^{1-\rho}}h_{i}(y) \,\mathrm{d} \mu(y)\biggr\vert ^{q}\frac{\mathrm{d}t}{t}\biggr)^{\frac {1}{q}}\, \mathrm{d}\mu(x) \\& \quad =:\mathrm{F}_{1}+\mathrm{F}_{2}. \end{aligned}$$
Applying the Minkowski inequality and (1.4), we have
$$\begin{aligned} \mathrm{F}_{1} \leq& \int_{\mathcal{X}\setminus6S_{i}} \int _{\mathcal{X}}\frac{\vert h_{i}(y)\vert [d(x,y)]^{1+\beta}}{\lambda (x,d(x,y))[d(x,y)]^{1-\rho}} \biggl( \int^{d(x,c_{S_{i}})+r_{S_{i}}}_{d(x,y)}\frac{\mathrm {d}t}{t^{q(\beta+\rho)+1}} \biggr)^{\frac{1}{q}}\,\mathrm{d}\mu (y)\,\mathrm{d}\mu(x) \\ \leq&C \int_{\mathcal{X}\setminus6S_{i}} \int_{\mathcal{X}}\frac {\vert h_{i}(y)\vert [d(x,y)]^{\rho+\beta}}{\lambda(x,d(x,y))} \\ &{}\times\biggl(\frac{1}{[d(x,y)]^{q(\beta+\rho)}}- \frac {1}{[d(x,c_{S_{i}})+r_{S_{i}}]^{q(\beta+\rho)}} \biggr)^{\frac {1}{q}}\,\mathrm{d}\mu(y)\,\mathrm{d}\mu(x) \\ \leq&C \int_{\mathcal{X}\setminus6S_{i}} \int_{\mathcal{X}}\frac {\vert h_{i}(y)\vert [d(x,y)]^{\rho+\beta}}{\lambda(x,d(x,y))} \frac{(r_{S_{i}})^{\frac{1}{q}(\beta+\rho)}}{[d(x,y)]^{(\beta+\rho )}[d(x,c_{S_{i}})+r_{S_{i}}]^{\frac{1}{q}(\beta+\rho)}}\,\mathrm{d} \mu(y)\,\mathrm{d}\mu(x) \\ \leq&C(r_{S_{i}})^{\frac{1}{q}(\beta+\rho)} \int_{\mathcal {X}}\bigl\vert h_{i}(y)\bigr\vert \, \mathrm{d}\mu(y) \int_{\mathcal{X}\setminus 6S_{i}}\frac{\mathrm{d}\mu(x)}{\lambda (x,d(x,c_{S_{i}}))[d(x,c_{S_{i}})+r_{S_{i}}]^{\frac{1}{q}(\beta+\rho )}} \\ \leq&C \int_{\mathcal{X}}\bigl\vert h_{i}(y)\bigr\vert \, \mathrm{d}\mu(y) {\sum } ^{\infty}_{k=1}6^{-k\frac{1}{q}(\beta+\rho)} \int _{6^{k+1}S_{i}\setminus6^{k}S_{i}} \frac{\mathrm{d}\mu(x)}{\lambda(x,d(x,c_{S_{i}}))} \\ \leq&C\|h_{i}\|_{L^{1}(\mu)}. \end{aligned}$$
For \(x\in\mathcal{X}\setminus6S_{i}\) and \(y\in S_{i}\), it holds true that \(d(x,y)< d(x,c_{S_{i}})+r_{S_{i}}\). Thus, by the vanishing moment of \(h_{i}\) and (1.5), we can conclude
$$\begin{aligned} \mathrm{F}_{2} =& \int_{\mathcal{X}\setminus6S_{i}}\biggl( \int^{\infty }_{d(x,c_{S_{i}})+r_{S_{i}}}\biggl\vert \frac{1}{t^{\beta+\rho}} \int_{\mathcal{X}}\biggl[\frac{K(x,y)}{[d(x,y)]^{1-\rho}}-\frac {K(x,c_{S_{i}})}{[d(x,c_{S_{i}})]^{1-\rho}}\biggr] \\ &{}\times h_{i}(y) \,\mathrm{d}\mu(y)\biggr\vert ^{q} \frac{\mathrm{d}t}{t}\biggr)^{\frac {1}{q}}\,\mathrm{d}\mu(x) \\ \leq& \int_{\mathcal{X}\setminus6S_{i}}\biggl( \int^{\infty }_{d(x,c_{S_{i}})+r_{S_{i}}}\biggl\vert \frac{1}{t^{\beta+\rho}} \int_{\mathcal{X}}\biggl[\frac{K(x,y)}{[d(x,y)]^{1-\rho}}-\frac {K(x,y)}{[d(x,c_{S_{i}})]^{1-\rho}}\biggr] \\ &{}\times h_{i}(y) \,\mathrm{d}\mu(y)\biggr\vert ^{q} \frac{\mathrm{d}t}{t}\biggr)^{\frac {1}{q}}\,\mathrm{d}\mu(x) \\ &{}+ \int_{\mathcal{X}\setminus6S_{i}}\biggl( \int^{\infty }_{d(x,c_{S_{i}})+r_{S_{i}}}\biggl\vert \frac{1}{t^{\beta+\rho}} \int_{\mathcal{X}}\biggl[\frac{K(x,y)}{[d(x,y)]^{1-\rho}}-\frac {K(x,c_{S_{i}})}{[d(x,y)]^{1-\rho}}\biggr] \\ &{}\times h_{i}(y) \,\mathrm{d}\mu(y)\biggr\vert ^{q} \frac{\mathrm{d}t}{t}\biggr)^{\frac {1}{q}}\,\mathrm{d}\mu(x) \\ =&:\mathrm{F}_{21}+\mathrm{F}_{22}. \end{aligned}$$
For \(\mathrm{F}_{21}\). By applying the Minkowski inequality, (1.2), and (1.4), we have
$$\begin{aligned} \mathrm{F}_{21} \leq& \int_{\mathcal{X}\setminus6S_{i}} \int _{\mathcal{X}} \bigl\vert h_{i}(y)\bigr\vert \biggl\vert \frac{K(x,y)}{[d(x,y)]^{1-\rho}}-\frac {K(x,y)}{[d(x,c_{S_{i}})]^{1-\rho}}\biggr\vert \\ &{}\times\biggl( \int^{\infty}_{d(x,c_{S_{i}})+r_{S_{i}}}\frac{\mathrm {d}t}{t^{q(\beta+\rho)+1}}\biggr)^{\frac{1}{q}} \,\mathrm{d}\mu(y)\,\mathrm{d}\mu(x) \\ \leq&C \int_{\mathcal{X}\setminus6S_{i}} \int_{\mathcal{X}} \bigl\vert h_{i}(y)\bigr\vert \biggl\vert \frac{K(x,y)}{[d(x,y)]^{1-\rho}}-\frac {K(x,y)}{[d(x,c_{S_{i}})]^{1-\rho}}\biggr\vert \frac{1}{[d(x,c_{S_{i}})+r_{S_{i}}]^{\beta+\rho}} \, \mathrm{d}\mu(y)\,\mathrm{d}\mu(x) \\ \leq&Cr_{S_{i}} \int_{\mathcal{X}\setminus6S_{i}} \int_{\mathcal{X}} \frac{\vert h_{i}(y)\vert [d(x,y)]^{\rho+\beta-1}}{\lambda(x,d(x,y))} \frac{1}{[d(x,c_{S_{i}})+r_{S_{i}}]^{\beta+\rho}} \,\mathrm{d} \mu(y)\,\mathrm{d}\mu(x) \\ \leq&Cr_{S_{i}} \int_{\mathcal{X}\setminus6S_{i}} \int_{\mathcal{X}} \frac{\vert h_{i}(y)\vert }{\lambda(x,d(x,c_{S_{i}}))} \frac{\mathrm{d}\mu(y)\,\mathrm{d}\mu(x)}{d(x,c_{S_{i}})} \\ \leq&Cr_{S_{i}} \int_{\mathcal{X}}\bigl\vert h_{i}(y)\bigr\vert \, \mathrm{d}\mu (y) {\sum} ^{\infty}_{k=1} \int_{6^{k+1}S_{i}\setminus 6^{k}S_{i}}\frac{\mathrm{d}\mu(x)}{\lambda (x,d(x,c_{S_{i}}))d(x,c_{S_{i}})} \\ \leq&C \int_{\mathcal{X}}\bigl\vert h_{i}(y)\bigr\vert \, \mathrm{d}\mu(y) {\sum } ^{\infty}_{k=1}6^{-k} \int_{6^{k+1}S_{i}\setminus 6^{k}S_{i}}\frac{\mathrm{d}\mu(x)}{\lambda (c_{S_{i}},d(x,c_{S_{i}}))} \\ \leq&C\|h_{i}\|_{L^{1}(\mu)}. \end{aligned}$$
Next we estimate \(\mathrm{F}_{22}\). By the Minkowski inequality and (1.5), we deduce that
$$\begin{aligned} \mathrm{F}_{22} \leq&C \int_{\mathcal{X}\setminus6S_{i}} \int_{\mathcal {X}}\bigl\vert h_{i}(y)\bigr\vert \biggl\vert \frac{K(x,y)}{[d(x,y)]^{1-\rho}}-\frac {K(x,c_{S_{i}})}{[d(x,y)]^{1-\rho}}\biggr\vert \\ &{}\times\biggl( \int^{\infty}_{d(x,c_{S_{i}})+r_{S_{i}}}\frac{\mathrm {d}t}{t^{q(\beta+\rho)+1}}\biggr)^{\frac{1}{q}} \,\mathrm{d}\mu (y)\,\mathrm{d}\mu(x) \\ \leq&C \int_{\mathcal{X}\setminus6S_{i}} \int_{\mathcal {X}}\bigl\vert h_{i}(y)\bigr\vert \bigl\vert K(x,y)-K(x,c_{S_{i}})\bigr\vert \frac{1}{[d(x,y)]^{\beta+1}}\,\mathrm{d} \mu(y)\,\mathrm{d}\mu(x) \\ \leq&C \int_{\mathcal{X}\setminus6S_{i}} \int_{\mathcal {X}}\bigl\vert h_{i}(y)\bigr\vert \frac{[d(c_{S_{i}},y)]^{\delta+\beta +1}}{[d(x,y)]^{\delta}\lambda(x,d(x,y))} \frac{1}{[d(x,y)]^{\beta+1}}\,\mathrm{d}\mu(y)\,\mathrm{d}\mu(x) \\ \leq&C \int_{\mathcal{X}\setminus6S_{i}} \int_{\mathcal {X}}\bigl\vert h_{i}(y)\bigr\vert \frac{r^{\delta+\beta+1}_{S_{i}}}{[d(x,y)]^{\delta +\beta+1}\lambda(x,d(x,c_{S_{i}}))} \,\mathrm{d}\mu(y)\,\mathrm{d}\mu(x) \\ \leq&C \int_{\mathcal{X}}\bigl\vert h_{i}(y)\bigr\vert \, \mathrm{d}\mu(y) \biggl( {\sum} ^{\infty}_{k=1}6^{-k(\delta+\beta +1)} \int_{6^{k+1}S_{i}\setminus6^{k}S_{i}}\frac{\mathrm{d}\mu (x)}{\lambda(x,d(x,c_{S_{i}}))} \biggr) \\ \leq&C\|h_{i}\|_{L^{1}(\mu)}. \end{aligned}$$
Combining the estimates for \(\mathrm{F}_{21}\), \(\mathrm{F}_{22}\), \(\mathrm{F}_{1}\), and the fact that
$$\|h_{i}\|_{L^{1}(\mu)}\leq \int_{\mathcal{X}}\bigl\vert f(y)\omega _{i}(y)\bigr\vert \,\mathrm{d}\mu(y), $$
we have \(\mathrm{E}_{1}\leq Ct^{-1}\).
Now we turn to an estimate of \(\mathrm{E}_{2}\). Let \(N_{1}\) be the positive integer satisfying \(S_{i}=(3\times6^{2})^{N_{1}}B_{i}\). By \(h_{i}:=\omega_{i}f-\varphi_{i}\), (1.4), the Minkowski inequality, the Hölder inequality, and (2.5) together with the \(L^{2}(\mu )\)-boundedness of \(\mathcal{M}_{\beta,\rho,q}\), we get
$$\begin{aligned} \mathrm{E}_{2} \leq&t^{-1} {\sum} _{i} \int _{6S_{i}\setminus6^{2}B_{i}}\bigl\vert \mathcal{M}_{\beta,\rho,q}(\omega _{i}f) (x)\bigr\vert \,\mathrm{d}\mu(x) +t^{-1} {\sum } _{i} \int_{6S_{i}\setminus 6^{2}B_{i}}\bigl\vert \mathcal{M}_{\beta,\rho,q}( \varphi_{i}) (x)\bigr\vert \,\mathrm{d}\mu(x) \\ \leq&Ct^{-1} {\sum} _{i} \int_{6S_{i}\setminus 6^{2}B_{i}} \int_{\mathcal{X}}\frac{[d(x,y)]^{\rho+\beta}}{\lambda (x,d(x,y))}\bigl\vert f(y) \omega_{i}(y)\bigr\vert \biggl( \int^{\infty}_{d(x,y)}\frac{\mathrm{d}t}{t^{q(\beta+\rho )+1}}\biggr)^{\frac{1}{q}} \,\mathrm{d}\mu(y)\,\mathrm{d}\mu(x) \\ &{}+Ct^{-1} {\sum} _{i}\biggl( \int_{6S_{i}}\bigl\vert \mathcal{M}_{\beta,\rho,q}(\varphi _{i}) (x)\bigr\vert ^{2}\,\mathrm{d}\mu(x) \biggr)^{\frac{1}{2}}\mu (6S_{i})^{\frac{1}{2}} \\ \leq&Ct^{-1} {\sum} _{i} \int_{6S_{i}\setminus 6^{2}B_{i}} \int_{\mathcal{X}}\frac{\vert f(y)\omega_{i}(y)\vert }{\lambda(x,d(x,y))} \,\mathrm{d}\mu(y)\,\mathrm{d} \mu(x)+Ct^{-1} {\sum} _{i}\|\varphi_{i} \|_{L^{2}(\mu)}\mu(6S_{i})^{\frac{1}{2}} \\ \leq&Ct^{-1} {\sum} _{i} \int_{\mathcal {X}}\bigl\vert f(y)\omega_{i}(y)\bigr\vert \,\mathrm{d}\mu(y) {\sum} ^{N_{1}+1}_{k=1} \frac{\mu((3\times 6^{2})^{k}B_{i})}{\lambda(c_{B_{i}},(3\times6^{2})^{k}r_{B_{i}})} \\ &{}+Ct^{-1} {\sum} _{i}\| \varphi_{i}\|_{L^{\infty}(\mu )}\mu(S_{i}) \\ \leq&Ct^{-1} \int_{\mathcal{X}}\bigl\vert f(y)\bigr\vert \,\mathrm{d}\mu(y)\leq Ct^{-1}. \end{aligned}$$

Combining the estimates for \(\mathrm{E}_{1}\) and \(\mathrm{E}_{2}\), we obtain (3.1) and hence the proof of Theorem 1.9 is finished. □

Proof of Theorem 1.10

We first claim that there exists a positive constant C such that, for any \(f\in L^{\infty}(\mu)\) and \((6,\beta_{6})\)-doubling ball B,
$$ \frac{1}{\mu(B)} \int_{B}\mathcal{M}_{\beta,\rho,q}(f) (y)\,\mathrm{d}\mu(y) \leq C\|f\|_{L^{\infty}(\mu)}+ {\inf} _{y\in B}\mathcal{M}_{\beta,\rho,q}(f) (y). $$
In order to prove (3.2), for each fixed ball B, we assume that Q is the smallest ball which includes B and has the same center as B, so \(2Q\subset6B\). Decompose f as
$$f(x)=f\chi_{2Q}(x)+f\chi_{\mathcal{X}\setminus2Q}(x)=:f_{1}(x)+f_{2}(x). $$
By applying Hölder inequality and the \(L^{2}(\mu)\)-boundedness of \(\mathcal{M}_{\beta,\rho,q}\), we have
$$\begin{aligned}& \frac{1}{\mu(B)} \int_{B}\mathcal{M}_{\beta,\rho ,q}(f_{1}) (y)\, \mathrm{d}\mu(y) \\& \quad \leq \frac{1}{\mu(B)} \biggl( \int_{B}\bigl[\mathcal{M}_{\beta,\rho ,q}(f_{1}) (y) \bigr]^{2}\,\mathrm{d}\mu(y) \biggr)^{\frac{1}{2}}\mu (B)^{\frac{1}{2}} \\& \quad \leq\frac{1}{[\mu(B)]^{\frac{1}{2}}}\|f_{1}\| _{L^{2}(\mu)}\leq C\|f\|_{L^{\infty}(\mu)}\frac{[\mu(6B)]^{\frac {1}{2}}}{[\mu(B)]^{\frac{1}{2}}} \\& \quad \leq C\|f\|_{L^{\infty}(\mu)}. \end{aligned}$$
Let \(r_{Q}\) be the radius of the ball Q. Noticing that \(d(y,z)\geq r_{Q}\) for any \(y\in B\), and \(z\in\mathcal{X}\setminus2Q\), by the Minkowski inequality, (1.2), and (1.4), we can deduce
$$\begin{aligned} \mathcal{M}_{\beta,\rho,q}(f_{2}) (y) \leq&\biggl( \int^{\infty}_{r_{Q}}\biggl\vert \frac{1}{t^{\beta+\rho }} \int_{d(y,z)< t}\frac{K(y,z)}{[d(y,z)]^{1-\rho}}f(z) \,\mathrm{d}\mu(z)\biggr\vert ^{q}\frac{\mathrm{d}t}{t}\biggr)^{\frac {1}{q}} \\ &{}+\biggl( \int^{\infty}_{r_{Q}}\biggl\vert \frac{1}{t^{\beta+\rho}} \int _{d(y,z)< t}\frac{K(y,z)}{[d(y,z)]^{1-\rho}}f_{1}(z) \,\mathrm{d} \mu(z)\biggr\vert ^{q}\frac{\mathrm{d}t}{t}\biggr)^{\frac {1}{q}} \\ \leq&\mathcal{M}_{\beta,\rho,q}(f) (y)+\biggl( \int^{\infty }_{r_{Q}}\biggl\vert \frac{1}{t^{\beta+\rho}} \int_{d(y,z)< 4r_{Q}}\frac {K(y,z)}{[d(y,z)]^{1-\rho}}f_{1}(z) \,\mathrm{d} \mu(z)\biggr\vert ^{q}\frac{\mathrm{d}t}{t}\biggr)^{\frac {1}{q}} \\ \leq&\mathcal{M}_{\beta,\rho,q}(f) (y)+C \int_{4Q}\frac {\vert K(y,z)\vert }{[d(y,z)]^{1-\rho}}\bigl\vert f_{1}(z) \bigr\vert \biggl( \int^{\infty}_{r_{Q}}\frac{\mathrm{d}t}{t^{q(\beta+\rho )+1}}\biggr)^{\frac{1}{q}} \,\mathrm{d}\mu(z) \\ \leq&\mathcal{M}_{\beta,\rho,q}(f) (y)+C \int_{4Q}\frac {[d(y,z)]^{1+\beta}}{\lambda(y,d(y,z))}\frac{1}{[d(y,z)]^{1-\rho}} \frac{1}{r^{\beta+\rho}_{Q}}\bigl\vert f(z)\bigr\vert \,\mathrm{d}\mu(z) \\ \leq&\mathcal{M}_{\beta,\rho,q}(f) (y)+C\|f\|_{L^{\infty}(\mu )} \int_{4Q}\frac{1}{\lambda(c_{B},d(c_{B},z))}\,\mathrm{d}\mu(z) \\ \leq&\mathcal{M}_{\beta,\rho,q}(f) (y)+C\|f\|_{L^{\infty}(\mu )}. \end{aligned}$$
Therefore, the estimate for (3.2) can be reduced to proving
$$ \bigl\vert \mathcal{M}_{\beta,\rho,q}(f_{2}) (x)- \mathcal{M}_{\beta,\rho ,q}(f_{2}) (y)\bigr\vert \leq C\|f \|_{L^{\infty}(\mu)}. $$
$$\begin{aligned}& \bigl\vert \mathcal{M}_{\beta,\rho,q}(f_{2}) (x)- \mathcal{M}_{\beta,\rho ,q}(f_{2}) (y)\bigr\vert \\& \quad \leq\biggl( \int^{\infty}_{0}\biggl\vert \int_{d(x,z)< t}\frac {K(x,z)}{[d(x,z)]^{1-\rho}}f_{2}(z)\,\mathrm{d} \mu(z) \\& \qquad {}- \int_{d(y,z)< t}\frac{K(y,z)}{[d(y,z)]^{1-\rho}}f_{2}(z)\,\mathrm{d} \mu(z) \biggr\vert ^{q}\frac{\mathrm{d}t}{t^{q(\beta+\rho)+1}}\biggr)^{\frac {1}{q}} \\& \quad \leq\biggl( \int^{\infty}_{0}\biggl[ \int_{d(x,z)< t\leq d(y,z)}\frac{\vert K(x,z)\vert }{[d(x,z)]^{1-\rho}}\bigl\vert f_{2}(z) \bigr\vert \,\mathrm{d}\mu(z) \biggr]^{q}\frac{\mathrm{d}t}{t^{q(\beta+\rho)+1}} \biggr)^{\frac {1}{q}} \\& \qquad {}+\biggl( \int^{\infty}_{0}\biggl[ \int_{d(y,z)< t\leq d(x,z)}\frac{\vert K(y,z)\vert }{[d(y,z)]^{1-\rho}}\bigl\vert f_{2}(z) \bigr\vert \,\mathrm{d}\mu(z) \biggr]^{q}\frac{\mathrm{d}t}{t^{q(\beta+\rho)+1}} \biggr)^{\frac {1}{q}} \\& \qquad {}+\biggl( \int^{\infty}_{0}\biggl[ \int_{d(y,z)< t d(x,z)< t}\biggl\vert \frac{K(x,z)}{[d(x,z)]^{1-\rho}}-\frac {K(y,z)}{[d(y,z)]^{1-\rho}} \biggr\vert \bigl\vert f_{2}(z)\bigr\vert \,\mathrm{d}\mu(z) \biggr]^{q}\frac{\mathrm{d}t}{t^{q(\beta+\rho)+1}}\biggr)^{\frac {1}{q}} \\& \quad =:\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}. \end{aligned}$$
For any \(x, y\in B\), applying the Minkowski inequality, (1.2), and (1.4), we have
$$\begin{aligned} \mathrm{I}_{1} \leq&C \int_{\mathcal{X}}\frac{[d(y,z)]^{\beta+\rho }}{\lambda(y,d(y,z))}\bigl\vert f_{2}(z) \bigr\vert \biggl( \int^{d(y,z)}_{d(x,z)} \frac{\mathrm{d}t}{t^{q(\beta+\rho)+1}} \biggr)^{\frac {1}{q}}\,\mathrm{d}\mu(z) \\ \leq&C \int_{\mathcal{X}}\frac{[d(y,z)]^{\beta+\rho}}{\lambda (y,d(y,z))}\bigl\vert f_{2}(z) \bigr\vert \frac{[d(x,y)]^{\beta+\rho}}{[d(x,z)]^{\beta +\rho} [d(y,z)]^{\beta+\rho}}\,\mathrm{d}\mu(z) \\ \leq&Cr^{\beta+\rho}_{B} \int_{\mathcal{X}\setminus2Q}\frac {\vert f(z)\vert }{\lambda(c_{B},d(c_{B},z))}\frac{1}{[d(c_{B},z)]^{\beta+\rho }}\,\mathrm{d} \mu(z) \\ \leq&Cr^{\beta+\rho}_{B} {\sum} ^{\infty}_{i=1} \int _{2^{i+1}Q\setminus2^{i}Q} \frac{\vert f(z)\vert }{\lambda(c_{B},d(c_{B},z))}\frac{1}{[d(c_{B},z)]^{\beta +\rho}}\,\mathrm{d} \mu(z) \\ \leq&C\|f\|_{L^{\infty}(\mu)} {\sum} ^{\infty }_{i=1}2^{-i(\beta+\rho)} \frac{\mu(2^{i+1}Q)}{\lambda(c_{B},2^{i}r_{Q})} \\ \leq&C\|f\|_{L^{\infty}(\mu)}. \end{aligned}$$
With a similar argument to that used in the proof of \(\mathrm{I}_{1}\), it is not difficult to obtain
$$\mathrm{I}_{2}\leq C\|f\|_{L^{\infty}(\mu)}. $$
Now we turn to an estimate of \(\mathrm{I}_{3}\). Write
$$\begin{aligned} \mathrm{I}_{3} \leq& \int_{\mathcal{X}}\biggl\vert \frac {K(x,z)}{[d(x,z)]^{1-\rho}}-\frac{K(y,z)}{[d(y,z)]^{1-\rho}} \biggr\vert \bigl\vert f_{2}(z)\bigr\vert \biggl( \int^{\infty}_{d(y,z)}\frac{\mathrm{d}t}{t^{q(\beta+\rho )+1}}\biggr)^{\frac{1}{q}} \,\mathrm{d}\mu(z) \\ \leq& \int_{\mathcal{X}}\biggl\vert \frac{K(x,z)}{[d(x,z)]^{1-\rho }}-\frac{K(y,z)}{[d(x,z)]^{1-\rho}} \biggr\vert \bigl\vert f_{2}(z)\bigr\vert \biggl( \int^{\infty}_{d(y,z)}\frac{\mathrm{d}t}{t^{q(\beta+\rho )+1}}\biggr)^{\frac{1}{q}} \,\mathrm{d}\mu(z) \\ &{}+ \int_{\mathcal{X}}\biggl\vert \frac{K(y,z)}{[d(x,z)]^{1-\rho}}-\frac {K(y,z)}{[d(y,z)]^{1-\rho}} \biggr\vert \bigl\vert f_{2}(z)\bigr\vert \biggl( \int^{\infty}_{d(y,z)}\frac{\mathrm{d}t}{t^{q(\beta+\rho )+1}}\biggr)^{\frac{1}{q}} \,\mathrm{d}\mu(z) \\ \leq&C \int_{\mathcal{X}}\bigl\vert K(x,z)-K(y,z)\bigr\vert \bigl\vert f_{2}(z)\bigr\vert \frac {1}{[d(y,z)]^{\beta+1}}\,\mathrm{d}\mu(z) \\ &{}+C \int_{\mathcal{X}}\biggl\vert \frac{1}{[d(x,z)]^{1-\rho}}-\frac {1}{[d(y,z)]^{1-\rho}} \biggr\vert \frac{\vert f_{2}(z)\vert }{[d(y,z)]^{\rho -1}\lambda(y,d(y,z))}\,\mathrm{d}\mu(z) \\ =:&\mathrm{I}_{31}+\mathrm{I}_{32}. \end{aligned}$$
By applying (1.5), we have
$$\begin{aligned} \mathrm{I}_{31} \leq&C\|f\|_{L^{\infty}(\mu)} \int_{\mathcal {X}\setminus2Q}\frac{r^{\delta+1+\beta}_{B}}{\lambda(c_{B},d(c_{B},z))} \frac{1}{[d(c_{B},z)]^{\delta+\beta+1}}\,\mathrm{d} \mu(z) \\ \leq&C\|f\|_{L^{\infty}(\mu)} {\sum} ^{\infty }_{k=1} \int_{2^{k+1}Q\setminus2^{k}Q}\frac{r^{\delta+1+\beta }_{B}}{\lambda(c_{B},d(c_{B},z))} \frac{1}{[d(c_{B},z)]^{\delta+\beta+1}}\,\mathrm{d} \mu(z) \\ \leq&C\|f\|_{L^{\infty}(\mu)} {\sum} ^{\infty }_{k=1}2^{-k(1+\beta+\delta)} \frac{\mu(2^{k+1}Q)}{\lambda (c_{B},2^{k}Q)} \\ \leq&C\|f\|_{L^{\infty}(\mu)}. \end{aligned}$$

Now we turn to an estimate of \(\mathrm{I}_{32}\) by two steps: \(0<\rho <1\) and \(\rho\geq1\).

As \(0<\rho<1\), we get
$$\begin{aligned} \mathrm{I}_{32} \leq&C\|f\|_{L^{\infty}(\mu)} \int_{\mathcal {X}\setminus2Q}\frac{r_{B}[d(x,z)]^{-\rho}}{[d(x,z)]^{1-\rho}} \frac{1}{\lambda(c_{B},d(c_{B},z))}\,\mathrm{d} \mu(z) \\ \leq&C\|f\|_{L^{\infty}(\mu)} {\sum} ^{\infty }_{k=1} \int_{2^{k+1}Q\setminus2^{k}Q}\frac{r_{B}}{d(x,z)} \frac{1}{\lambda(c_{B},d(c_{B},z))}\,\mathrm{d} \mu(z) \\ \leq&C\|f\|_{L^{\infty}(\mu)} {\sum} ^{\infty }_{k=1}2^{-k} \frac{\mu(2^{k+1}Q)}{\lambda(c_{B},2^{k}r_{Q})} \\ \leq&C\|f\|_{L^{\infty}(\mu)}. \end{aligned}$$
As \(\rho\geq1\), we deduce that
$$\begin{aligned} \mathrm{I}_{32} \leq&C\|f\|_{L^{\infty}(\mu)} \int_{\mathcal {X}\setminus6B}\frac{r_{B}[d(y,z)]^{\rho-2}}{[d(y,z)]^{\rho-1}} \frac{1}{\lambda(c_{B},d(c_{B},z))}\,\mathrm{d} \mu(z) \\ \leq&C\|f\|_{L^{\infty}(\mu)} {\sum} ^{\infty }_{k=1} \int_{2^{k+1}Q\setminus2^{k}Q}\frac{r_{B}}{d(c_{B},z)} \frac{1}{\lambda(c_{B},d(c_{B},z))}\,\mathrm{d} \mu(z) \\ \leq&C\|f\|_{L^{\infty}(\mu)} {\sum} ^{\infty }_{k=1}2^{-k} \frac{\mu(2^{k+1}Q)}{\lambda(c_{B},2^{k}r_{Q})}\leq C\| f\|_{L^{\infty}(\mu)}. \end{aligned}$$
Thus, we have
$$\mathrm{I}_{3}\leq C\|f\|_{L^{\infty}(\mu)}, $$
which, together with \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\), implies (3.5). Therefore, the estimate of (3.2) is completed.
By (3.2), it follows that, for \(f\in L^{\infty}(\mu)\), if \(\mathcal {M}_{\beta,\rho,q}(f)(x_{0})<\infty\) with some point \(x_{0}\in \mathcal{X}\), we can get \(\mathcal{M}_{\beta,\rho,q}(f)\) is μ-finite a.e., and in this case, for any \((6,\beta_{6})\)-doubling ball B, we have
$$\frac{1}{\mu(B)} \int_{B}\bigl[\mathcal{M}_{\beta,\rho,q}(f) (x)- \mathop{\operatorname{ess \,inf} }_{B}\mathcal{M}_{\beta,\rho ,q}(f)\bigr]\,\mathrm{d}\mu(x)\leq C \|f\|_{L^{\infty}(\mu)}. $$
To prove that \(\mathcal{M}_{\beta,\rho,q}(f)\in\operatorname{RBLO}(\mu )\), applying Lemma 2.3, it only suffices to prove that \(\mathcal {M}_{\beta,\rho,q}(f)\) satisfies (2.7), namely, for any two \((6,\beta _{6})\)-doubling balls B and S with \(B\subset S\),
$$ m_{B}\bigl(\mathcal{M}_{\beta,\rho,q}(f)\bigr)-m_{S}\bigl( \mathcal{M}_{\beta,\rho ,q}(f)\bigr)\leq CK_{B,S}. $$
For any \(x\in B\) and \(y\in S\), write
$$\begin{aligned} \mathcal{M}_{\beta,\rho,q}(f) (x) \leq&\mathcal{M}_{\beta,\rho ,q}(f \chi_{2B}) (x)+\mathcal{M}_{\beta,\rho,q}(f\chi_{2S\setminus 2B}) (x) \\ &{}+\bigl[\mathcal{M}_{\beta,\rho,q}(f\chi_{\mathcal{X}\setminus 2S}) (x)- \mathcal{M}_{\beta,\rho,q}(f\chi_{\mathcal{X}\setminus2S}) (y)\bigr] +\mathcal{M}_{\beta,\rho,q}(f \chi_{\mathcal{X}\setminus2S}) (y). \end{aligned}$$
With an argument similar to that used in the proof of \(\mathcal {M}_{\beta,\rho,q}(f_{2})(y)\) in (3.4), we have
$$\mathcal{M}_{\beta,\rho,q}(f\chi_{\mathcal{X}\setminus2S}) (y)\leq \mathcal{M}_{\beta,\rho,q}(f) (y)+C\|f\|_{L^{\infty}(\mu)}. $$
By (3.5), for any \(x, y\in S\), it is not difficult to get
$$\bigl\vert \mathcal{M}_{\beta,\rho,q}(f\chi_{\mathcal{X}\setminus 2S}) (x)- \mathcal{M}_{\beta,\rho,q}(f\chi_{\mathcal{X}\setminus 2S}) (y)\bigr\vert \leq C\|f \|_{L^{\infty}(\mu)}. $$
For any \(x\in B\), by the Minkowski inequality and (1.4), we obtain
$$\begin{aligned}& \mathcal{M}_{\beta,\rho,q}(f\chi_{2S\setminus2B}) (x) \\& \quad \leq C \int _{\mathcal{X}}\frac{|f\chi_{2S\setminus2B}(y)|[d(x,y)]^{\rho+\beta }}{\lambda(x,d(x,y))} \biggl( \int^{\infty}_{d(x,y)}\frac{\mathrm{d}t}{t^{q(\beta+\rho )+1}} \biggr)^{\frac{1}{q}}\,\mathrm{d}\mu(y) \\& \quad \leq C \int_{2S\setminus2B}\frac{|f(y)|}{\lambda(x,d(x,y))}\,\mathrm{d}\mu(y) \\& \quad \leq C\|f\|_{L^{\infty}(\mu)} \int_{2S\setminus B}\frac{1}{\lambda (c_{B},d(c_{B},y))}\,\mathrm{d}\mu(y) \\& \quad \leq CK_{B,S}\|f\|_{L^{\infty}(\mu)}. \end{aligned}$$
Thus, for any \(x\in B\) and \(y\in S\), we have
$$ \mathcal{M}_{\beta,\rho,q}(f) (x)\leq\mathcal{M}_{\beta,\rho ,q}(f \chi_{2B}) (x)+\mathcal{M}_{\beta,\rho,q}(f) (y)+CK_{B,S}\|f\| _{L^{\infty}(\mu)}. $$
For (3.7), taking the mean value over B for x and over S for y, it follows that
$$\begin{aligned} \begin{aligned} m_{B}\bigl(\mathcal{M}_{\beta,\rho,q}(f)\bigr)-m_{S}\bigl( \mathcal{M}_{\beta,\rho ,q}(f)\bigr)&\leq C\bigl[m_{B}\bigl( \mathcal{M}_{\beta,\rho,q}(f\chi _{2B})\bigr)+K_{B,S}\|f \|_{L^{\infty}(\mu)}\bigr] \\ &\leq CK_{B,S}\|f\|_{L^{\infty}(\mu)}, \end{aligned} \end{aligned}$$
which, together with (3.3), we finish the proof of Theorem 1.10. □

Proof of Theorem 1.12

Because the definition of \(H^{1}(\mu)\) is independent of the choice of ς, thus, for convenience, we assume \(\zeta=2\) as in (1.7). By a standard argument, it suffices to prove that
$$ \bigl\Vert \mathcal{M}_{\beta,\rho,q}(f)\bigr\Vert _{L^{1}(\mu)}\leq C|b|_{H^{1}(\mu )} $$
for any atomic block b with \(\operatorname{supp} b\subset S\). Write
$$\begin{aligned} \int_{\mathcal{X}}\bigl\vert \mathcal{M}_{\beta,\rho,q}(b) (x)\bigr\vert \,\mathrm{d}\mu (x) =& \int_{\mathcal{X}\setminus2S}\bigl\vert \mathcal{M}_{\beta,\rho ,q}(b) (x)\bigr\vert \,\mathrm{d}\mu(x)+ \int_{2S}\bigl\vert \mathcal{M}_{\beta,\rho,q}(b) (x)\bigr\vert \,\mathrm{d}\mu (x) \\ =:&\mathrm{J}_{1}+\mathrm{J}_{2}. \end{aligned}$$
In a way similar to that used in the proof of \(\mathrm{E}_{1}\) in Theorem 1.9, we can obtain
$$\mathrm{J}_{1}\leq C\|b\|_{L^{1}(\mu)}\leq C|b|_{H^{1}(\mu)}. $$
Let \(b= {\sum} _{i}\tau_{i}a_{i}\) be as in Definition 1.8 and we have
$$\begin{aligned} \mathrm{J}_{2} \leq& {\sum} _{i}\vert \tau_{i}\vert \int _{2B_{i}}\bigl\vert \mathcal{M}_{\beta,\rho,q}(b_{i}) (x)\bigr\vert \,\mathrm{d}\mu(x)+ {\sum} _{i}\vert \tau_{i}\vert \int_{2S\setminus 2B_{i}}\bigl\vert \mathcal{M}_{\beta,\rho,q}(b_{i}) (x)\bigr\vert \,\mathrm{d}\mu(x) \\ =:&\mathrm{J}_{21}+\mathrm{J}_{22}. \end{aligned}$$
By applying the Hölder inequality, the \(L^{2}(\mu)\)-boundedness of \(\mathcal{M}_{\beta,\rho,q}\), and (1.7), we have
$$\begin{aligned} \mathrm{J}_{21} \leq& {\sum} _{i}| \tau_{i}| \biggl( \int_{2B_{i}}\bigl\vert \mathcal{M}_{\beta,\rho,q}(a_{i}) (x)\bigr\vert ^{2}\,\mathrm{d}\mu(x) \biggr)^{\frac{1}{2}} \mu(2B_{i})^{\frac{1}{2}} \\ \leq&C {\sum} _{i}|\tau_{i}|\|a_{i} \|_{L^{2}(\mu )}\mu(2B_{i})^{\frac{1}{2}} \\ \leq&C {\sum} _{i}|\tau_{i}|\|a_{i} \|_{L^{\infty }(\mu)}\mu(2B_{i})\leq C {\sum} _{i}| \tau_{i}|. \end{aligned}$$
Now we turn to an estimate of \(\mathrm{J}_{22}\). Also, in a way similar to \(\mathrm{E}_{2}\) in the proof of Theorem 1.9, we have
$$\begin{aligned} \mathrm{J}_{22} \leq&C {\sum} _{i}| \tau_{i}| \int _{2S\setminus2B_{i}}\frac{1}{\lambda(c_{B_{i}},d(x,c_{B_{i}}))}\| a_{i} \|_{L^{1}(\mu)} \\ \leq&C {\sum} _{i}|\tau_{i}|K_{B_{i},S} \|a_{i}\| _{L^{\infty}(\mu)}\mu(2B_{i})\leq C {\sum} _{i}|\tau_{i}|. \end{aligned}$$
Combining with the above estimates, this implies (3.8) and hence the proof of Theorem 1.12. □



This paper is supported by National Natural Foundation of China (Grant No. 11561062).

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Authors’ Affiliations

College of Mathematics and Statistics, Northwest Normal University


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