Fractional type Marcinkiewicz integrals over non-homogeneous metric measure spaces

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 $1< p<\infty$.

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, [3–9]). 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 [10–16]. 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 [17–20] 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) $(\mathcal{X},d)$ is geometrically doubling.

• (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) 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) 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 B̃.

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 [22–24]. 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) 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) 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) 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 [26–30].

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

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

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

• (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) 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) 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) There exists a positive constant c̃ 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) 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) 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. □