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

Abstract

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 \(1< p<\infty\).

Keywords

42B35 47B47 30L99

MSC

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

[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 \(r\in(0,\infty)\),
$$ \mu\bigl(B(x,r)\bigr)\leq\lambda(x,r)\leq C_{\lambda}\lambda\biggl(x, \frac {r}{2}\biggr). $$
(1.1)
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
$$ \tilde{\lambda}(x,r)\leq C_{\tilde{\lambda}}\tilde{\lambda }(y,r), $$
(1.2)
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

[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 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
$$ K_{B,S}:=1+ \int_{2S\setminus B}\frac{1}{\lambda (c_{B},d(x,c_{B}))}\,\mathrm{d}\mu(x), $$
(1.3)
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

[1]

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
$$\frac{1}{\mu(\kappa B)} \int_{B}\bigl\vert f(y)-f_{B}\bigr\vert \, \mathrm{d}\mu(y)\leq C $$
and
$$|f_{B}-f_{R}|\leq CK_{B,R} $$
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

[14]

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,
$$\frac{1}{\mu(\sigma B)} \int_{B}\bigl[f(y)- \mathop{\operatorname{ess \,inf}}_{\tilde{B}}f\bigr]\,\mathrm{d}\mu(y)\leq C $$
and
$$\mathop{\operatorname{ess\,inf}} _{B}f- \mathop{\operatorname{ess \,inf}}_{S}f\leq CK_{B,S} $$
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 \(x\neq y\),
$$ \bigl\vert K(x,y)\bigr\vert \leq C\frac{[d(x,y)]^{1+\beta}}{\lambda(x,d(x,y))}, $$
(1.4)
and for all \(x, \tilde{x}, y\in\mathcal{X}\) with \(d(x,y)\geq 2d(x,\tilde{x})\),
$$ \bigl\vert K(x,y)-K(\tilde{x},y)\bigr\vert +\bigl\vert K(y,x)-K(y, \tilde{x})\bigr\vert \leq C\frac {[d(x,\tilde{x})]^{\delta+1+\beta}}{[d(x,y)]^{\delta}\lambda (x,d(x,y))}. $$
(1.5)
The fractional type Marcinkiewicz integral \(\mathcal{M}_{\beta,\rho ,q}(f)\) related to the above kernel \(K(x,y)\) is formally defined by
$$ \mathcal{M}_{\beta,\rho,q}(f) (x):= \biggl( \int^{\infty}_{0} \biggl\vert \frac{1}{t^{\beta+\rho}} \int_{d(x,y)< t}\frac {K(x,y)}{[d(x,y)]^{1-\rho}}f(y) \,\mathrm{d}\mu(y)\biggr\vert ^{q}\frac{\mathrm{d}t}{t} \biggr)^{\frac {1}{q}}, $$
(1.6)
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

[16]

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)

    \(\int_{\mathcal{X}}b(x)\,\mathrm{d}\mu(x)=0\);

     
  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
    $$ \|a_{i}\|_{L^{p}(\mu)}\leq\bigl[\mu(\zeta B_{i}) \bigr]^{\frac {1}{p}-1}K^{-1}_{B_{i},S}. $$
    (1.7)
     
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 \(f\in L^{1}(\mu)\),
$$ \mu\bigl(\bigl\{ x\in\mathcal{X}: \mathcal{M}_{\beta,\rho,q}(f) (x)>t\bigr\} \bigr)\leq C\frac{\|f\|_{L^{1}(\mu)}}{t}. $$
(1.8)

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. (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

[31]

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,
    $$\begin{aligned}& \frac{1}{\mu(6^{2}B_{i})} \int_{B_{i}}\bigl\vert f(x)\bigr\vert ^{p}\, \mathrm{d}\mu (x)>\frac{t^{p}}{\gamma_{0}} \quad \textit{for all }i, \\& \frac{1}{\mu(6^{2}\upsilon B_{i})} \int_{\upsilon B_{i}}\bigl\vert f(x)\bigr\vert ^{p}\, \mathrm{d}\mu(x)\leq\frac{t^{p}}{\gamma_{0}} \quad \textit {for all }i\textit{ and all } \upsilon\in(2,\infty), \end{aligned}$$
    (2.1)
    and
    $$ \bigl\vert f(x)\bigr\vert \leq t \quad \textit{for } \mu\textit{-almost every } x\in\mathcal {X}\Bigm\backslash \biggl( \bigcup_{i}6B_{i} \biggr). $$
    (2.2)
     
  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
    $$\begin{aligned}& \int_{\mathcal{X}}\varphi_{i}(x)\,\mathrm{d}\mu(x)= \int _{6B_{i}}f(x)\omega_{i}(x)\,\mathrm{d} \mu(x), \end{aligned}$$
    (2.3)
    $$\begin{aligned}& {\sum} _{i}\bigl\vert \varphi_{i}(x)\bigr\vert \leq\gamma t\quad \textit{for } \mu\textit{-almost every } x\in \mathcal{X}, \end{aligned}$$
    (2.4)
    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
    $$ \|\varphi_{i}\|_{L^{\infty}(\mu)}\mu(S_{i})\leq C \int_{\mathcal {X}}\bigl\vert f(x)\omega_{i}(x)\bigr\vert \,\mathrm{d}\mu(x), $$
    (2.5)
    and if \(p\in(1,\infty)\),
    $$ \biggl( \int_{S_{i}}\bigl\vert \varphi_{i}(x)\bigr\vert ^{p}\,\mathrm{d}\mu(x) \biggr)^{\frac{1}{p}}\bigl[ \mu(S_{i})\bigr]^{\frac{1}{p'}}\leq\frac {C}{t^{p-1}} \int_{\mathcal{X}}\bigl\vert f(x)\omega_{i}(x)\bigr\vert ^{p}\,\mathrm{d}\mu (x). $$
    (2.6)
     

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

Lemma 2.3

[14]

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 \(B\subset S\),
$$ m_{B}(f)-m_{S}(f)\leq CK_{B,S}, $$
(2.7)
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
$$ \mu\biggl(\biggl\{ x\in\mathcal{X}\Bigm\backslash \bigcup _{i}\bigl(6^{2}B_{i}\bigr): \mathcal{M}_{\beta,\rho,q}(h) (x)>t\biggr\} \biggr)\leq Ct^{-1}. $$
(3.1)
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,
$$ \frac{1}{\mu(B)} \int_{B}\mathcal{M}_{\beta,\rho,q}(f) (y)\,\mathrm{d}\mu(y) \leq C\|f\|_{L^{\infty}(\mu)}+ {\inf} _{y\in B}\mathcal{M}_{\beta,\rho,q}(f) (y). $$
(3.2)
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
$$\begin{aligned}& \frac{1}{\mu(B)} \int_{B}\mathcal{M}_{\beta,\rho ,q}(f_{1}) (y)\, \mathrm{d}\mu(y) \\& \quad \leq \frac{1}{\mu(B)} \biggl( \int_{B}\bigl[\mathcal{M}_{\beta,\rho ,q}(f_{1}) (y) \bigr]^{2}\,\mathrm{d}\mu(y) \biggr)^{\frac{1}{2}}\mu (B)^{\frac{1}{2}} \\& \quad \leq\frac{1}{[\mu(B)]^{\frac{1}{2}}}\|f_{1}\| _{L^{2}(\mu)}\leq C\|f\|_{L^{\infty}(\mu)}\frac{[\mu(6B)]^{\frac {1}{2}}}{[\mu(B)]^{\frac{1}{2}}} \\& \quad \leq C\|f\|_{L^{\infty}(\mu)}. \end{aligned}$$
(3.3)
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
$$\begin{aligned} \mathcal{M}_{\beta,\rho,q}(f_{2}) (y) \leq&\biggl( \int^{\infty}_{r_{Q}}\biggl\vert \frac{1}{t^{\beta+\rho }} \int_{d(y,z)< t}\frac{K(y,z)}{[d(y,z)]^{1-\rho}}f(z) \,\mathrm{d}\mu(z)\biggr\vert ^{q}\frac{\mathrm{d}t}{t}\biggr)^{\frac {1}{q}} \\ &{}+\biggl( \int^{\infty}_{r_{Q}}\biggl\vert \frac{1}{t^{\beta+\rho}} \int _{d(y,z)< t}\frac{K(y,z)}{[d(y,z)]^{1-\rho}}f_{1}(z) \,\mathrm{d} \mu(z)\biggr\vert ^{q}\frac{\mathrm{d}t}{t}\biggr)^{\frac {1}{q}} \\ \leq&\mathcal{M}_{\beta,\rho,q}(f) (y)+\biggl( \int^{\infty }_{r_{Q}}\biggl\vert \frac{1}{t^{\beta+\rho}} \int_{d(y,z)< 4r_{Q}}\frac {K(y,z)}{[d(y,z)]^{1-\rho}}f_{1}(z) \,\mathrm{d} \mu(z)\biggr\vert ^{q}\frac{\mathrm{d}t}{t}\biggr)^{\frac {1}{q}} \\ \leq&\mathcal{M}_{\beta,\rho,q}(f) (y)+C \int_{4Q}\frac {\vert K(y,z)\vert }{[d(y,z)]^{1-\rho}}\bigl\vert f_{1}(z) \bigr\vert \biggl( \int^{\infty}_{r_{Q}}\frac{\mathrm{d}t}{t^{q(\beta+\rho )+1}}\biggr)^{\frac{1}{q}} \,\mathrm{d}\mu(z) \\ \leq&\mathcal{M}_{\beta,\rho,q}(f) (y)+C \int_{4Q}\frac {[d(y,z)]^{1+\beta}}{\lambda(y,d(y,z))}\frac{1}{[d(y,z)]^{1-\rho}} \frac{1}{r^{\beta+\rho}_{Q}}\bigl\vert f(z)\bigr\vert \,\mathrm{d}\mu(z) \\ \leq&\mathcal{M}_{\beta,\rho,q}(f) (y)+C\|f\|_{L^{\infty}(\mu )} \int_{4Q}\frac{1}{\lambda(c_{B},d(c_{B},z))}\,\mathrm{d}\mu(z) \\ \leq&\mathcal{M}_{\beta,\rho,q}(f) (y)+C\|f\|_{L^{\infty}(\mu )}. \end{aligned}$$
(3.4)
Therefore, the estimate for (3.2) can be reduced to proving
$$ \bigl\vert \mathcal{M}_{\beta,\rho,q}(f_{2}) (x)- \mathcal{M}_{\beta,\rho ,q}(f_{2}) (y)\bigr\vert \leq C\|f \|_{L^{\infty}(\mu)}. $$
(3.5)
Write
$$\begin{aligned}& \bigl\vert \mathcal{M}_{\beta,\rho,q}(f_{2}) (x)- \mathcal{M}_{\beta,\rho ,q}(f_{2}) (y)\bigr\vert \\& \quad \leq\biggl( \int^{\infty}_{0}\biggl\vert \int_{d(x,z)< t}\frac {K(x,z)}{[d(x,z)]^{1-\rho}}f_{2}(z)\,\mathrm{d} \mu(z) \\& \qquad {}- \int_{d(y,z)< t}\frac{K(y,z)}{[d(y,z)]^{1-\rho}}f_{2}(z)\,\mathrm{d} \mu(z) \biggr\vert ^{q}\frac{\mathrm{d}t}{t^{q(\beta+\rho)+1}}\biggr)^{\frac {1}{q}} \\& \quad \leq\biggl( \int^{\infty}_{0}\biggl[ \int_{d(x,z)< t\leq d(y,z)}\frac{\vert K(x,z)\vert }{[d(x,z)]^{1-\rho}}\bigl\vert f_{2}(z) \bigr\vert \,\mathrm{d}\mu(z) \biggr]^{q}\frac{\mathrm{d}t}{t^{q(\beta+\rho)+1}} \biggr)^{\frac {1}{q}} \\& \qquad {}+\biggl( \int^{\infty}_{0}\biggl[ \int_{d(y,z)< t\leq d(x,z)}\frac{\vert K(y,z)\vert }{[d(y,z)]^{1-\rho}}\bigl\vert f_{2}(z) \bigr\vert \,\mathrm{d}\mu(z) \biggr]^{q}\frac{\mathrm{d}t}{t^{q(\beta+\rho)+1}} \biggr)^{\frac {1}{q}} \\& \qquad {}+\biggl( \int^{\infty}_{0}\biggl[ \int_{d(y,z)< t d(x,z)< t}\biggl\vert \frac{K(x,z)}{[d(x,z)]^{1-\rho}}-\frac {K(y,z)}{[d(y,z)]^{1-\rho}} \biggr\vert \bigl\vert f_{2}(z)\bigr\vert \,\mathrm{d}\mu(z) \biggr]^{q}\frac{\mathrm{d}t}{t^{q(\beta+\rho)+1}}\biggr)^{\frac {1}{q}} \\& \quad =:\mathrm{I}_{1}+\mathrm{I}_{2}+\mathrm{I}_{3}. \end{aligned}$$
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 \(B\subset S\),
$$ m_{B}\bigl(\mathcal{M}_{\beta,\rho,q}(f)\bigr)-m_{S}\bigl( \mathcal{M}_{\beta,\rho ,q}(f)\bigr)\leq CK_{B,S}. $$
(3.6)
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
$$ \mathcal{M}_{\beta,\rho,q}(f) (x)\leq\mathcal{M}_{\beta,\rho ,q}(f \chi_{2B}) (x)+\mathcal{M}_{\beta,\rho,q}(f) (y)+CK_{B,S}\|f\| _{L^{\infty}(\mu)}. $$
(3.7)
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
$$ \bigl\Vert \mathcal{M}_{\beta,\rho,q}(f)\bigr\Vert _{L^{1}(\mu)}\leq C|b|_{H^{1}(\mu )} $$
(3.8)
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. □

Declarations

Acknowledgements

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

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
College of Mathematics and Statistics, Northwest Normal University

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