Estimates for fractional type Marcinkiewicz integrals with non-doubling measures

Abstract

Under the assumption that μ is a non-doubling measure on ${\mathbb{R}}^d$ satisfying the growth condition, the authors prove that the fractional type Marcinkiewicz integral ℳ is bounded from the Hardy space $H_{\mathrm{fin}}^{1,\mathrm{\infty },0}(\mu )$ to the Lebesgue space $L^q(\mu )$ for $\frac{1}{q}=1-\frac{\alpha }{n}$ with kernel satisfying a certain Hörmander-type condition. In addition, the authors show that for $p=\frac{n}{\alpha }$, ℳ is bounded from the Morrey space $M_q^p(\mu )$ to the space $RBMO(\mu )$ and from the Lebesgue space $L^{\frac{n}{\alpha }}(\mu )$ to the space $RBMO(\mu )$.

MSC:46A20, 42B25, 42B35.

1 Introduction

Let μ be a nonnegative Radon measure on ${\mathbb{R}}^d$ which satisfies the following growth condition: for all $x\in {\mathbb{R}}^d$ and all $r>0$,

$$\mu (B(x,r))\le C_0{r}^n,$$

(1.1)

where $C_0$ and n are positive constants and $n\in (0,d]$, $B(x,r)$ is the open ball centered at x and having radius r. So μ is claimed to be non-doubling measure. If there exists a positive constant C such that for any $x\in supp(\mu )$ and $r>0$, $\mu (B(x,2r))\le C\mu (B(x,r))$, the μ is said to be doubling measure. It is well known that the doubling condition on underlying measures is a key assumption in the classical theory of harmonic analysis. Especially, in recent years, many classical results concerning the theory of Calderón-Zygmund operators and function spaces have been proved still valid if the underlying measure is a nonnegative Radon measure on ${\mathbb{R}}^d$ which only satisfies (1.1) (see [1–8]). The motivation for developing the analysis with non-doubling measures and some examples of non-doubling measures can be found in [9]. We only point out that the analysis with non-doubling measures played a striking role in solving the long-standing open Painlevé’s problem by Tolsa in [10].

Let $K(x,y)$ be a μ-locally integrable function on ${\mathbb{R}}^d\times {\mathbb{R}}^d\setminus \{(x,y):x=y\}$. Assume that there exists a positive constant C such that for any $x,y\in {\mathbb{R}}^d$ with $x\ne y$,

$$|K(x,y)|\le C{|x-y|}^{-(n-1)},$$

(1.2)

and for any $x,y,y^{\prime }\in {\mathbb{R}}^d$,

$${\int }_{|x-y|\ge 2|y-y^{\prime }|}\left[\right|K(x,y)-K(x,y^{\prime })|+|K(y,x)-K(y^{\prime },x)\left|\right]\frac{1}{|x-y|}d\mu (x)\le C.$$

(1.3)

The fractional type Marcinkiewicz integral ℳ associated to the above kernel $K(x,y)$ and the measure μ as in (1.1) is defined by

$$\mathcal{M}(f)(x)={\left({\int }_0^{\mathrm{\infty }}\right|{\int }_{|x-y|\le t}\frac{K(x,y)}{{|x-y|}^{-\alpha }}f(y)d\mu (y){|}^2\frac{dt}{t^3})}^{\frac{1}{2}},x\in {\mathbb{R}}^d,0<\alpha <n.$$

(1.4)

If μ is the d-dimensional Lebesgue measure in ${\mathbb{R}}^d$, and

$$K(x,y)=\frac{\mathrm{\Omega }(x-y)}{{|x-y|}^{n-1}},$$

(1.5)

with Ω homogeneous of degree zero and $\mathrm{\Omega }\in {Lip}_{\gamma }(S^{d-1})$ for some $\gamma \in (0,1]$, then K satisfies (1.2) and (1.3). Under these conditions, ℳ in (1.4) is introduced by Si et al. in [11]. As a special case, by letting $\alpha =0$, we recapture the classical Marcinkiewicz integral operators that Stein introduced in 1958 (see [12]). Since then, many works have appeared about Marcinkiewicz type integral operators. A nice survey has been given by Lu in [13].

In 2007, the Hörmander-type condition was introduced by Hu et al. in [14], which was slightly stronger than (1.3) and was defined as follows:

$$\begin{array}{c}\underset{\begin{array}{c}\ell >0,y,y^{\prime }\in {\mathbb{R}}^d\\ |y-y^{\prime }|\le \ell \end{array}}{sup}\sum _{k=1}^{\mathrm{\infty }}k{\int }_{2^k\ell <|x-y|\le 2^{k+1}\ell }\left[\right|K(x,y)-K(x,y^{\prime })|\hfill \\ +|K(y,x)-K(y^{\prime },x)|]\frac{1}{|x-y|}d\mu (x)\le C.\hfill \end{array}$$

(1.6)

However, in this paper, we discover that the kernel should satisfy some other kind of smoothness condition to replace (1.6).

Definition 1.1 Let $1\le s<\mathrm{\infty }$, $0<\epsilon <1$. The kernel K is said to satisfy a Hörmander-type condition if there exist $c_s>1$ and $C_s>0$ such that for any $x\in {\mathbb{R}}^d$ and $\ell >c_s|x|$,

$$\begin{array}{c}\underset{\begin{array}{c}\ell >0,y,y^{\prime }\in {\mathbb{R}}^d\\ |y-y^{\prime }|\le \ell \end{array}}{sup}\sum _{k=1}^{\mathrm{\infty }}{2}^{k\epsilon }{\left(2^k\ell \right)}^n\left(\frac{1}{{(2^k\ell )}^n}{\int }_{2^k\ell <|x-y|\le 2^{k+1}\ell }\right[\left(\right|K(x,y)-K(x,y^{\prime })|\hfill \\ {{+|K(y,x)-K(y^{\prime },x)|\left)\frac{1}{|x-y|}\right]}^{s}d\mu (x))}^{\frac{1}{s}}\le C_s.\hfill \end{array}$$

(1.7)

We denote by ${\mathcal{H}}^s$ the class of kernels satisfying this condition. It is clear that these classes are nested,

$${\mathcal{H}}^{s_2}\subset {\mathcal{H}}^{s_1}\subset {\mathcal{H}}^1,1<s_1<s_2<\mathrm{\infty }.$$

We should point out that ${\mathcal{H}}^1$ is not condition (1.6).

The purpose of this paper is to get some estimates for the fractional type Marcinkiewicz integral ℳ with kernel K satisfying (1.2) and (1.7) on the Hardy-type space and the $RBMO(\mu )$ space. To be precise, we establish the boundedness of ℳ in $H_{\mathrm{fin}}^{1,\mathrm{\infty },0}(\mu )$ for $\frac{1}{q}=1-\frac{\alpha }{n}$ in Section 2. In Section 3, we prove that ℳ is bounded from the space $RBMO(\mu )$ to the Morrey space $M_q^p(\mu )$, from the space $RBMO(\mu )$ to the Lebesgue space $L^{\frac{n}{\alpha }}(\mu )$ for $p=\frac{n}{\alpha }$.

Before stating our results, we need to recall some necessary notation and definitions. For a cube $Q\subset {\mathbb{R}}^d$, we mean a closed cube whose sides are parallel to the coordinate axes. We denote its center and its side length by $x_Q$ and $\ell (Q)$, respectively. Let $\eta >1$, ηQ denote the cube with the same center as Q and $\ell (\eta Q)=\eta \ell (Q)$. Given two cubes $Q\subset R$ in ${\mathbb{R}}^d$, set

$$S_{Q,R}=1+\sum _{k=1}^{N_{Q,R}}\frac{\mu (2^{k}Q)}{{[\ell (2^{k}Q)]}^n},$$

where $N_{Q,R}$ is the smallest positive integer k such that $\ell (2^{k}Q)\ge \ell (R)$. The concept $S_{Q,R}$ was introduced in [15], where some useful properties of $S_{Q,R}$ can be found.

Lemma 1.2 For a function $b\in L_{\mathrm{loc}}^1(\mu )$, $0<\beta \le 1$, conditions (i) and (ii) below are equivalent.

• (i) There exist some constant $C_2$ and a collection of numbers $b_Q$ such that these two properties hold: for any cube Q,

  $$\frac{1}{\mu (2Q)}{\int }_Q|b(x)-b(y)|d\mu (x)\le C_2\ell {(Q)}^{\beta },$$

  (1.8)

and for any cube R such that $Q\subset R$ and $\ell (R)\le 2\ell (Q)$,

$$|b_Q-b_R|\le C_2\ell {(Q)}^{\beta }.$$

(1.9)

• (ii) For any given p, $1\le p\le \mathrm{\infty }$, there is a constant $C(p)\ge 0$ such that for every cube Q, then

  $${\left[\frac{1}{\mu (Q)}{\int }_Q\right|b(x)-m_Q(b){|}^{p}d\mu (x)]}^{\frac{1}{p}}\le C(p)\ell {(Q)}^{\beta },$$

  (1.10)

where

$$m_Q(b)=\frac{1}{\mu (Q)}{\int }_{Q}b(y)d\mu (y),$$

and also for any cube R such that $Q\subset R$ and $\ell (R)\le 2\ell (Q)$,

$$|m_Q(b)-m_R(b)|\le C(p)\ell {(Q)}^{\beta }.$$

Remark 1.3 Lemma 1.2 is a slight variant of Theorem 2.3 in [16]. To be precise, if we replace all balls in Theorem 2.3 of [16] by cubes, we then obtain Lemma 1.2.

Remark 1.4 For $0<\beta \le 1$, (1.9) is equivalent to

$$|b_Q-b_R|\le C{S}_{Q,R}\ell {(R)}^{\beta }$$

(1.11)

for any two cubes $Q\subset R$ with $\ell (R)\le 2\ell (Q)$ (see Remark 2.7 in [16]).

Lemma 1.5 Let $0<\alpha <n$, $1<p<\frac{n}{\alpha }$, $\frac{1}{r}=\frac{1}{p}-\frac{\alpha }{n}$ and $q\ge \frac{n}{n-\alpha }$. Then the fractional integral operator $I_{\alpha }$ defined by

$$I_{\alpha }f(x)={\int }_{{\mathbb{R}}^d}\frac{f(y)}{{|x-y|}^{n-\alpha }}dy$$

is bounded from $L^p(\mu )$ to $L^r(\mu )$ (see [17]).

Lemma 1.6 Let $0<\alpha <n$, $1<p<\frac{n}{\alpha }$, $\frac{1}{q}=\frac{1}{p}-\frac{\alpha }{n}$. Suppose that $K(x,y)$ satisfies (1.2) and (1.3) and ℳ is as in (1.4). Then there exists a positive constant $C>0$ such that for all bounded functions f with compact support,

$${\parallel \mathcal{M}(f)\parallel }_{L^q(\mu )}\le C{\parallel f\parallel }_{L^p(\mu )}.$$

Proof of Lemma 1.6 By Minkowski’s inequality, we have

$$\begin{array}{rcl}\mathcal{M}(f)(x)& =& {\left({\int }_0^{\mathrm{\infty }}\right|{\int }_{|x-y|\le t}\frac{K(x,y)}{{|x-y|}^{-\alpha }}f(y)d\mu (y){|}^2\frac{dt}{t^3})}^{1/2}\\ \le & {\int }_{{\mathbb{R}}^d}\frac{|K(x,y)|}{{|x-y|}^{-\alpha }}|f(y)|{\left({\int }_{|x-y|}^{\mathrm{\infty }}\frac{dt}{t^3}\right)}^{\frac{1}{2}}d\mu (y)\\ \le & C{\int }_{{\mathbb{R}}^d}\frac{1}{{|x-y|}^{n-\alpha -1}}|f(y)|\frac{1}{|x-y|}d\mu (y)\\ \le & C{\int }_{{\mathbb{R}}^d}\frac{|f(y)|}{{|x-y|}^{n-\alpha }}d\mu (y)\\ \le & C{I}_{\alpha }(|f|)(x).\end{array}$$

By Lemma 1.5 then

$${\parallel \mathcal{M}(f)\parallel }_{L^q(\mu )}\le C{\parallel f\parallel }_{L^p(\mu )}.$$

 □

Throughout this paper, we use the constant C with subscripts to indicate its dependence on the parameters. For a μ-measurable set E, ${\chi }_E$ denotes its characteristic function. For any $p\in [1,\mathrm{\infty }]$, we denote by $p^{\prime }$ its conjugate index, namely $\frac{1}{p}+\frac{1}{p^{\prime }}=1$.

2 Boundedness of ℳ in Hardy spaces

This section is devoted to the behavior of ℳ in Hardy spaces. In order to define the Hardy space $H^1(\mu )$, Tolsa introduced the grand maximal operator $M_{\varphi }$ in [18].

Definition 2.1 Given $f\in L_{\mathrm{loc}}^1(\mu )$, $M_{\varphi }f$ is defined as

$$M_{\varphi }f(x)=\underset{\phi \sim x}{sup}\left|{\int }_{{\mathbb{R}}^d}f\phi d\mu \right|,$$

where the notation $\phi \sim x$ means that $\phi \in L^1(\mu )\cap C^1({\mathbb{R}}^d)$ and satisfies

• (1) ${\parallel \phi \parallel }_{L^1(\mu )}\le 1$,

• (2) $0\le \phi (y)\le \frac{1}{{|x-y|}^n}$ for all $y\in {\mathbb{R}}^d$,

• (3) $|{\phi }^{\prime }(y)|\le \frac{1}{{|x-y|}^{n+1}}$ for all $y\in {\mathbb{R}}^d$.

Based on Theorem 1.2 in [18], we can define the Hardy space $H^1(\mu )$ as follows (see [15]).

Definition 2.2 The Hardy space $H^1(\mu )$ is the set of all functions $f\in L^1(\mu )$ satisfying that ${\int }_{{\mathbb{R}}^d}fd\mu =0$ and $M_{\varphi }f\in L^1(\mu )$. Moreover, the norm of $f\in H^1(\mu )$ is defined by

$${\parallel f\parallel }_{H^1(\mu )}={\parallel f\parallel }_{L^1(\mu )}+{\parallel M_{\varphi }f\parallel }_{L^1(\mu )}.$$

We recall the atomic Hardy space $H_{\mathrm{atb}}^{1,\mathrm{\infty },0}(\mu )$ as follows.

Definition 2.3 Let $\rho >1$. A function $h\in L_{\mathrm{loc}}^1(\mu )$ is called an atomic block if

• (1) there exists some cube R such that $supph\subset R$,

• (2) ${\int }_{{\mathbb{R}}^d}h(x)d\mu (x)=0$,

• (3) for $i=1,2$, there are functions $a_i$ supported on cubes $Q_i\subset R$ and numbers ${\lambda }_i\in \mathbb{R}$ such that $h={\lambda }_1{a}_1+{\lambda }_2{a}_2$, and

  $${\parallel a_i\parallel }_{L^{\mathrm{\infty }}(\mu )}\le {[\mu (\rho Q_i)S_{Q_i,R}]}^{-1}.$$

Then define

$${|h|}_{H_{\mathrm{atb}}^{1,\mathrm{\infty },0}(\mu )}=|{\lambda }_1|+|{\lambda }_2|.$$

Define $H_{\mathrm{atb}}^{1,\mathrm{\infty },0}(\mu )$ and $H_{\mathrm{fin}}^{1,\mathrm{\infty },0}(\mu )$ as follows:

$${\parallel f\parallel }_{H_{\mathrm{atb}}^{1,\mathrm{\infty },0}(\mu )}=inf\{\sum _j^{\mathrm{\infty }}{|h_j|}_{H_{\mathrm{atb}}^{1,\mathrm{\infty },0}(\mu )}:f=\sum _{j=1}^{\mathrm{\infty }}{h}_j,{\{h_j\}}_{j\in \mathbb{N}}\text{ are }(1,\mathrm{\infty },0)\text{-atoms}\}$$

and

$${\parallel f\parallel }_{H_{\mathrm{fin}}^{1,\mathrm{\infty },0}(\mu )}=inf\{\sum _j^k{|h_j|}_{H_{\mathrm{atb}}^{1,\mathrm{\infty },0}(\mu )}:f=\sum _{j=1}^k{h}_j,{\{h_j\}}_{j=1}^k\text{ are }(1,\mathrm{\infty },0)\text{-atoms}\},$$

where the infimum is taken over all possible decompositions of f in atomic blocks, $H_{\mathrm{fin}}^{1,\mathrm{\infty },0}(\mu )$ is the set of all finite linear combinations of $(1,\mathrm{\infty },0)$-atoms.

Remark 2.4 It was proved in [15] that for each $\rho >1$, the atomic Hardy space $H_{\mathrm{atb}}^{1,\mathrm{\infty },0}(\mu )$ is independent of the choice of ρ.

Applying the theory of Meda et al. in [19], we easily get the result as follows.

Theorem 2.5 Let $0<\alpha <n$, $\frac{1}{q}=1-\frac{\alpha }{n}$. Suppose that K satisfies (1.2) and the ${\mathcal{H}}^q$ condition and $f\in H_{\mathrm{fin}}^{1,\mathrm{\infty },0}(\mu )$. Then ℳ is bounded from the Hardy space into the Lebesgue space, namely there exists a positive constant C such that

$${\parallel \mathcal{M}(f)\parallel }_{L^q(\mu )}\le C{\parallel f\parallel }_{H_{\mathrm{fin}}^{1,\mathrm{\infty },0}(\mu )}.$$

Proof of Theorem 2.5 Without loss of generality, we may assume that $\rho =4$ and $f=\sum h$ as a finite of atomic blocks defined in Definition 2.3. It is easy to see that we only need to prove the theorem for one atomic block h. Let R be a cube such that $supph\subset R$, ${\int }_{{\mathbb{R}}^d}h(x)d\mu (x)=0$, and

$$h(x)={\lambda }_1{a}_1(x)+\lambda a_2(x),$$

(2.1)

where ${\lambda }_i$ for $i=1,2$ is a real number, ${|h_i|}_{H_{\mathrm{atb}}^{1,\mathrm{\infty },0}(\mu )}={\lambda }_1+{\lambda }_2$, $a_i$ for $i=1,2$ is a bounded function supported on some cubes $Q_i\subset R$ and it satisfies

$${\parallel a_i\parallel }_{L^{\mathrm{\infty }}(\mu )}\le {[\mu (4Q_i)S_{Q_i,R}]}^{-1}.$$

(2.2)

Write

$$\begin{array}{rcl}{\parallel \mathcal{M}(h)\parallel }_{L^q(\mu )}& \le & {\left({\int }_{2R}\right|\mathcal{M}(h)(x){|}^{q}d\mu (x))}^{\frac{1}{q}}+{\left({\int }_{{\mathbb{R}}^d\setminus 2R}\right|\mathcal{M}(h)(x){|}^{q}d\mu (x))}^{\frac{1}{q}}\\ \le & {\left({\int }_{2R}\right|\mathcal{M}(h)(x){|}^{q}d\mu (x))}^{\frac{1}{q}}\\ +{\{{\int }_{{\mathbb{R}}^d\setminus 2R}{\left({\int }_0^{|x-x_R|+2\ell (R)}\right|{\int }_{|x-y|\le t}\frac{K(x,y)}{{|x-y|}^{-\alpha }}h(y)d\mu (y){|}^2\frac{dt}{t^3})}^{\frac{q}{2}}d\mu (x)\}}^{\frac{1}{q}}\\ +{\{{\int }_{{\mathbb{R}}^d\setminus 2R}{\left({\int }_{|x-x_R|+2\ell (R)}^{\mathrm{\infty }}\right|{\int }_{|x-y|\le t}\frac{K(x,y)}{{|x-y|}^{-\alpha }}h(y)d\mu (y){|}^2\frac{dt}{t^3})}^{\frac{q}{2}}d\mu (x)\}}^{\frac{1}{q}}\\ =& \mathrm{I}+\mathrm{II}+\mathrm{III}.\end{array}$$

By (2.1), we have

$$\begin{array}{rcl}\mathrm{I}& =& {\left({\int }_{2R}\right|\mathcal{M}(h)(x){|}^{q}d\mu (x))}^{\frac{1}{q}}\\ \le & |{\lambda }_1|{\left({\int }_{2R}\right|\mathcal{M}(a_1)(x){|}^{q}d\mu (x))}^{\frac{1}{q}}+|{\lambda }_2|{\left({\int }_{2R}\right|\mathcal{M}(a_2)(x){|}^{q}d\mu (x))}^{\frac{1}{q}}\\ =& {\mathrm{I}}_1+{\mathrm{I}}_2.\end{array}$$

To estimate ${\mathrm{I}}_1$, we write

$$\begin{array}{rcl}{\mathrm{I}}_1& \le & |{\lambda }_1|{\left({\int }_{2Q_1}\right|\mathcal{M}(a_1)(x){|}^{q}d\mu (x))}^{\frac{1}{q}}+|{\lambda }_1|{\left({\int }_{2R\setminus 2Q_1}\right|\mathcal{M}(a_1)(x){|}^{q}d\mu (x))}^{\frac{1}{q}}\\ =& {\mathrm{I}}_{11}+{\mathrm{I}}_{12}.\end{array}$$

Choose $p_1$ and $q_1$ such that $1<p_1<\frac{n}{\alpha }$, $1<q<q_1$ and $\frac{1}{q_1}=\frac{1}{p_1}-\frac{n}{\alpha }$. By the Hölder inequality, the fact that $S_{Q_1,R}\ge 1$ and the $(L^{p_1}(\mu ),L^{q_1}(\mu ))$-boundedness of ℳ (see Lemma 1.6), we have that

$$\begin{array}{rcl}{\mathrm{I}}_{11}& \le & |{\lambda }_1|{\left[{\int }_{2Q_1}\right|\mathcal{M}(a_1)(x){|}^{q_1}d\mu (x)]}^{\frac{1}{q_1}}\mu {(2Q_1)}^{\frac{1}{q}-\frac{1}{q_1}}\\ \le & C|{\lambda }_1|{\parallel a_1\parallel }_{L^{p_1}(\mu )}\mu {(2Q_1)}^{\frac{1}{q}-\frac{1}{q_1}}\\ \le & C|{\lambda }_1|{\parallel a_1\parallel }_{L^{\mathrm{\infty }}(\mu )}\mu {(2Q_1)}^{\frac{1}{p_1}+\frac{1}{q}-\frac{1}{q_1}}\\ \le & C|{\lambda }_1|.\end{array}$$

Denote $N_{2Q_1,2R}$ simply by $N_1$. Invoking the fact that ${\parallel a_1\parallel }_{L^{\mathrm{\infty }}(\mu )}\le {[\mu (4Q_i)S_{Q_i,R}]}^{-1}$, we thus get

$$\begin{array}{rcl}{\mathrm{I}}_{12}& \le & C|{\lambda }_1|{\{\sum _{k=1}^{N_1+1}{\int }_{2^{k+1}{Q}_1\setminus 2^k{Q}_1}{\left[{\int }_0^{\mathrm{\infty }}\right|{\int }_{|x-y|\le t}\frac{a_1(y)}{{|x-y|}^{n-\alpha -1}}d\mu (y){|}^2\frac{dt}{t^3}]}^{\frac{q}{2}}d\mu (x)\}}^{\frac{1}{q}}\\ \le & C|{\lambda }_1|\{\sum _{k=1}^{N_1+1}\ell {\left(2^k{Q}_1\right)}^{q(\alpha -n)}\\ {\times {\int }_{2^{k+1}{Q}_1\setminus 2^k{Q}_1}{[{\int }_{Q_1}\frac{|a_1(y)|}{{|x-y|}^{n-1-\alpha }}{\left({\int }_{|x-y|}^{\mathrm{\infty }}\frac{dt}{t^3}\right)}^{\frac{1}{2}}d\mu (y)]}^{q}d\mu (x)\}}^{\frac{1}{q}}\\ \le & C|{\lambda }_1|{\{\sum _{k=1}^{N_1+1}\ell {\left(2^k{Q}_1\right)}^{q(\alpha -n)}{\int }_{2^{k+1}{Q}_1\setminus 2^k{Q}_1}{\left[{\int }_{Q_1}\right|a_1(y)|d\mu (y)]}^{q}d\mu (x)\}}^{\frac{1}{q}}\\ \le & C|{\lambda }_1|{\left\{\sum _{k=1}^{N_1+1}\ell {\left(2^k{Q}_1\right)}^{q(\alpha -n)}\mu \left(2^{k+1}{Q}_1\right){\parallel a_1\parallel }_{L^{\mathrm{\infty }}(\mu )}^q\mu {(Q_1)}^q\right\}}^{\frac{1}{q}}\\ \le & C|{\lambda }_1|{\left\{\sum _{k=1}^{N_1+1}\ell {\left(2^k{Q}_1\right)}^{q(\alpha -n)}\mu {(4Q_1)}^{-q}{S}_{Q_1,R}^{-q}\mu \left(2^{k+1}{Q}_1\right){\parallel a_1\parallel }_{L^{\mathrm{\infty }}(\mu )}^q\mu {(Q_1)}^q\right\}}^{\frac{1}{q}}\\ \le & C|{\lambda }_1|{\left(S_{Q_1,R}^{-q}\sum _{k=2}^{N_1+1}\frac{\mu (2^k{Q}_1)}{\ell {(2^k{Q}_1)}^n}\right)}^{\frac{1}{q}}\\ \le & C|{\lambda }_1|.\end{array}$$

Here we have used the fact that

$$\sum _{k=2}^{N_1+1}\frac{\mu (2^{k}Q)}{\ell {(2^{k}Q)}^n}\le C{S}_{Q,R},$$

see [16] for details.

The estimates for ${\mathrm{I}}_{11}$ and ${\mathrm{I}}_{12}$ give the desired estimate for ${\mathrm{I}}_1$. With a similar argument, we have

$${\mathrm{I}}_2\le C|{\lambda }_2|.$$

Combining the estimates for ${\mathrm{I}}_1$ and ${\mathrm{I}}_2$ yields the estimate for I.

For $i=1,2$, $y\in Q_i\subset R$, $x\in {\mathbb{R}}^d\setminus (2R)$, we have $|x-y|\sim |x-x_R|\sim |x-x_R|+2\ell (R)$, by Minkowski’s inequality, we get

$$\begin{array}{rcl}\mathrm{II}& \le & {\{{\int }_{{\mathbb{R}}^d\setminus (2R)}{\left[{\int }_R\frac{h(y)}{{|x-y|}^{n-1-\alpha }}{\left({\int }_{|x-y|}^{|x-x_R|+2\ell (R)}\frac{dt}{t^3}\right)}^{\frac{1}{2}}\right]}^{q}d\mu (x)\}}^{\frac{1}{q}}\\ \le & C{\int }_R{\{{\int }_{{\mathbb{R}}^d\setminus (2R)}{\left[\right|\frac{1}{{(|x-x_R|+2\ell (R))}^2}-\frac{1}{{|x-y|}^2}{|}^{\frac{1}{2}}\frac{|h(y)|}{{|x-y|}^{n-1-\alpha }}]}^{q}d\mu (x)\}}^{\frac{1}{q}}d\mu (y)\\ \le & C{\int }_R{\{{\int }_{{\mathbb{R}}^d\setminus (2R)}{(\frac{\ell {(R)}^{\frac{1}{2}}}{{|x-y|}^{\frac{3}{2}}}\cdot \frac{|h(y)|}{{|x-y|}^{n-1-\alpha }})}^{q}d\mu (x)\}}^{\frac{1}{q}}d\mu (y)\\ \le & C{\int }_R{\{\sum _{k=1}^{\mathrm{\infty }}{\int }_{2^{k+1}R\setminus (2^{k}R)}{\left(\frac{\ell {(R)}^{\frac{1}{2}}}{{|x-y|}^{n-\alpha +\frac{1}{2}}}\right)}^{q}d\mu (x)\}}^{\frac{1}{q}}|h(y)|d\mu (y)\\ \le & C(\sum _{j=1}^2|{\lambda }_j|{\parallel a_j\parallel }_{L^1(\mu )})\left\{\sum _{k=1}^{\mathrm{\infty }}\ell {(R)}^{\frac{1}{2}}\ell {\left(2^{k}R\right)}^{-n+\alpha -\frac{1}{2}}\mu {\left(2^{k+1}R\right)}^{\frac{1}{q}}\right\}\\ \le & C(\sum _{j=1}^2|{\lambda }_j|).\end{array}$$

For any $y\in R$, we have $|x-y|\le |x-x_R|+|y-x_R|\le |x-x_R|+2\ell (R)\le t$. It follows that

$$\begin{array}{rcl}\mathrm{III}& \le & {\{{\int }_{{\mathbb{R}}^d\setminus 2R}{\left({\int }_{|x-x_R|+2\ell (R)}^{\mathrm{\infty }}\right|{\int }_{|x-y|\le t}[\frac{K(x,y)}{{|x-y|}^{-\alpha }}-\frac{K(x,x_R)}{{|x-x_R|}^{-\alpha }}]h(y)d\mu (y){|}^2\frac{dt}{t^3})}^{\frac{q}{2}}d\mu (x)\}}^{\frac{1}{q}}\\ \le & {\{{\int }_{{\mathbb{R}}^d\setminus 2R}{\left[{\int }_R|\frac{K(x,y)}{{|x-y|}^{-\alpha }}-\frac{K(x,x_R)}{{|x-x_R|}^{-\alpha }}|{\left({\int }_{|x-x_R|+2\ell (R)}^{\mathrm{\infty }}\frac{dt}{t^3}\right)}^{\frac{1}{2}}\right|h(y)|d\mu (y)]}^{q}d\mu (x)\}}^{\frac{1}{q}}\\ \le & C{\int }_R\sum _{k=1}^{\mathrm{\infty }}{\{{\int }_{2^{k+1}R\setminus 2^{k}R}{\left[\right|\frac{K(x,y)}{{|x-y|}^{-\alpha }}-\frac{K(x,x_R)}{{|x-x_R|}^{-\alpha }}|\cdot \frac{1}{|x-y|}]}^{q}d\mu (x)\}}^{\frac{1}{q}}|h(y)|d\mu (y)\\ \le & C{\int }_R\sum _{k=1}^{\mathrm{\infty }}\left\{{\int }_{2^{k+1}R\setminus 2^{k}R}\right[|\frac{K(x,y)}{{|x-y|}^{-\alpha }}-\frac{K(x,y)}{{|x-x_R|}^{-\alpha }}\\ {{+\frac{K(x,y)}{{|x-x_R|}^{-\alpha }}-\frac{K(x,x_R)}{{|x-x_R|}^{-\alpha }}|\cdot \frac{1}{|x-y|}]}^{q}d\mu (x)\}}^{\frac{1}{q}}|h(y)|d\mu (y)\\ \le & C{\int }_R\sum _{k=1}^{\mathrm{\infty }}{\{{\int }_{2^{k+1}R\setminus 2^{k}R}{\left[\right|\frac{K(x,y)}{{|x-y|}^{-\alpha }}-\frac{K(x,y)}{{|x-x_R|}^{-\alpha }}|\cdot \frac{1}{|x-y|}]}^{q}d\mu (x)\}}^{\frac{1}{q}}|h(y)|d\mu (y)\\ +C{\int }_R\sum _{k=1}^{\mathrm{\infty }}{\{{\int }_{2^{k+1}R\setminus 2^{k}R}{\left[\right|\frac{K(x,y)}{{|x-x_R|}^{-\alpha }}-\frac{K(x,x_R)}{{|x-x_R|}^{-\alpha }}|\cdot \frac{1}{|x-y|}]}^{q}d\mu (x)\}}^{\frac{1}{q}}|h(y)|d\mu (y)\\ \le & C{\int }_R\sum _{k=1}^{\mathrm{\infty }}\ell (R){\{{\int }_{2^{k+1}R\setminus 2^{k}R}\frac{1}{{|x-y|}^{q(n-\alpha +1)}}d\mu (x)\}}^{\frac{1}{q}}|h(y)|d\mu (y)\\ +{\int }_R\sum _{k=1}^{\mathrm{\infty }}{({\int }_{2^{k+1}R\setminus 2^{k}R}{\left[\ell {\left(2^{k}R\right)}^{\alpha }\frac{|K(x,y)-K(x,x_R)|}{|x-y|}\right]}^{q}d\mu (x))}^{\frac{1}{q}}|h(y)|d\mu (y)\\ \le & C(\sum _{j=1}^2|{\lambda }_j|).\end{array}$$

Here we have used the fact that $\frac{1}{q}=1-\frac{\alpha }{n}$.

Combining the estimates for I, II and III yields that

$${\parallel \mathcal{M}(h)\parallel }_{L^q(\mu )}\le C{|h|}_{H_{\mathrm{atb}}^{1,\mathrm{\infty },0}(\mu )},$$

and this is the result of Theorem 2.5. □

3 Boundedness of ℳ in $RBMO(\mu )$ spaces

In this section, we discuss the boundedness for ℳ as in (1.4) in the space $RBMO(\mu )$ for $f\in M_p^q(\mu )$ and $f\in L^{\frac{n}{\alpha }}(\mu )$, respectively.

Firstly, we need to recall the definition of Morrey space with non-doubling measure denoted by $M_q^p(\mu )$, which was introduced by Sawano and Tanaka in [20].

Definition 3.1 Let $\nu >1$ and $1\le q\le p<\mathrm{\infty }$. The Morrey space $M_q^p(\mu )$ is defined by

$$M_q^p(\mu )=\{f\in L_{\mathrm{loc}}^q(\mu ):{\parallel f\parallel }_{M_q^p(\mu )}<\mathrm{\infty }\},$$

where the norm ${\parallel f\parallel }_{M_q^p(\mu )}$ is given by

$${\parallel f\parallel }_{M_q^p(\mu )}=\underset{Q}{sup}\mu {(\nu Q)}^{\frac{1}{p}-\frac{1}{q}}{\left({\int }_Q\right|f(x){|}^{q}d\mu (x))}^{\frac{1}{q}}.$$

We should note that the parameter $\nu >1$ appearing in the definition does not affect the definition of the space $M_q^p(\mu )$, and $M_q^p(\mu )$ is a Banach space with its norms (see [20]). By using the Hölder inequality to (1.4), it is easy to see that for all $1\le q_2\le q_1\le p$, then

$$L^p(\mu )=M_p^p(\mu )\subset M_{q_1}^p(\mu )\subset M_{q_2}^p(\mu ).$$

Theorem 3.2 Let $0<\alpha <n$, $1\le q<p=\frac{n}{\alpha }$. Suppose that $K(x,y)$ satisfies (1.2) and the ${\mathcal{H}}^{p^{\prime }}$ condition, ℳ is defined as in (1.4). Then there exists a positive constant C such that for all $f\in M_q^p(\mu )$,

$${\parallel \mathcal{M}(f)\parallel }_{RBMO(\mu )}\le C{\parallel f\parallel }_{M_q^p(\mu )}.$$

Theorem 3.3 Let $0<\alpha <n$ and $p=\frac{n}{\alpha }$. Suppose that $K(x,y)$ satisfies (1.2) and the ${\mathcal{H}}^{\frac{n}{n-\alpha }}$ condition, ℳ is defined as in (1.4). Then there exists a positive constant C such that for all bounded functions f with compact support,

$${\parallel \mathcal{M}(f)\parallel }_{RBMO(\mu )}\le C{\parallel f\parallel }_{L^{\frac{n}{\alpha }}(\mu )}.$$

Remark 3.4 As a special condition, we take $p=q=\frac{n}{\alpha }$, Theorem 3.3 can be deduced with a similar method of Theorem 3.2.

Proof of Theorem 3.2 For any cubes Q and R in ${\mathbb{R}}^d$ such that $Q\subset R$ satisfies $\ell (R)\le 2\ell (Q)$, let

$$a_Q=m_Q[\mathcal{M}(f{\chi }_{{\mathbb{R}}^d\setminus \frac{3}{2}Q})]$$

and

$$a_R=m_R[\mathcal{M}(f{\chi }_{{\mathbb{R}}^d\setminus \frac{3}{2}R})].$$

It is easy to see that $a_Q$ and $a_R$ are real numbers. By Lemma 1.2, we need to show that for some fixed $r>q$, there exists a constant $C>0$ such that

$${\left(\frac{1}{\mu (2Q)}{\int }_Q\right|\mathcal{M}(f)(x)-a_Q{|}^{r}d\mu (x))}^{\frac{1}{r}}\le C{\parallel f\parallel }_{M_q^p(\mu )}$$

(3.1)

and

$$|a_Q-a_R|\le C{\parallel f\parallel }_{M_q^p(\mu )}.$$

(3.2)

Let us first prove estimate (3.1). For a fixed cube Q and $x\in Q$, decompose $f=f_1+f_2$, where $f_1=f_{{\chi }_{\frac{3}{2}Q}}$ and $f_2=f-f_1$. Write that

$$\begin{array}{c}\frac{1}{\mu (2Q)}{\int }_Q|\mathcal{M}(f)(x)-a_Q{|}^{r}d\mu (x)\hfill \\ =\frac{1}{\mu (2Q)}{\int }_Q|\mathcal{M}(f_1+f_2)(x)-a_Q{|}^{r}d\mu (x)\hfill \\ \le \frac{1}{\mu (2Q)}{\int }_Q|\mathcal{M}(f_1)(x){|}^{r}d\mu (x)+\frac{1}{\mu (2Q)}{\int }_Q|\mathcal{M}(f_2)(x)-a_Q{|}^{r}d\mu (x)\hfill \\ ={\mathrm{I}}_1+{\mathrm{I}}_2.\hfill \end{array}$$

For $\frac{1}{r}=\frac{1}{q}-\frac{\alpha }{n}$ and $p=\frac{\alpha }{n}$, it follows that

$$\begin{array}{rcl}{\mathrm{I}}_1& =& \frac{1}{\mu (2Q)}{\int }_Q|\mathcal{M}(f_1)(x){|}^{r}d\mu (x)\\ \le & C\frac{1}{\mu (2Q)}{\left({\int }_{\frac{3}{2}Q}\right|f(x){|}^{q}d\mu (x))}^{\frac{r}{q}}\\ \le & C\frac{1}{\mu (2Q)}{\left(\mu {(2Q)}^{\frac{1}{p}-\frac{1}{q}}{\int }_{\frac{3}{2}Q}\right|f(x){|}^{q}d\mu (x))}^{\frac{r}{q}}\mu {(2Q)}^{r(\frac{1}{q}-\frac{1}{p})}\\ \le & C{\parallel f\parallel }_{M_q^p(\mu )}^r\mu {(2Q)}^{r(\frac{1}{q}-\frac{1}{p})-1}\\ \le & C{\parallel f\parallel }_{M_q^p(\mu )}^r.\end{array}$$

Now let us estimate the term ${\mathrm{I}}_2$,

$$\begin{array}{rl}{\mathrm{I}}_2& =\frac{1}{\mu (2Q)}{\int }_Q|\mathcal{M}(f_2)(x)-a_Q{|}^{r}d\mu (x)\\ =\frac{1}{\mu (2Q)}{\int }_Q|\mathcal{M}(f_2)(x)-\frac{1}{\mu (Q)}{\int }_Q\mathcal{M}(f{\chi }_{{\mathbb{R}}^d\setminus \frac{3}{2}Q})(y)d\mu (y){|}^{r}d\mu (x)\\ =\frac{1}{\mu (2Q)}{\int }_Q|\frac{1}{\mu (Q)}{\int }_Q\mathcal{M}(f_2)(x)d\mu (y)-\frac{1}{\mu (Q)}{\int }_Q\mathcal{M}(f{\chi }_{{\mathbb{R}}^d\setminus \frac{3}{2}Q})(y)d\mu (y){|}^{r}d\mu (x)\\ \le \frac{1}{\mu (2Q)}\frac{1}{\mu (Q)}{\int }_Q{\int }_Q|\mathcal{M}(f_2)(x)-\mathcal{M}(f_2)(y){|}^{r}d\mu (x)d\mu (y).\end{array}$$

In order to estimate $|\mathcal{M}(f_2)(x)-\mathcal{M}(f_2)(y)|$, we write

$$\begin{array}{c}{D}_1(x,y)={\left({\int }_0^{\mathrm{\infty }}{[{\int }_{|x-z|\le t<|y-z|}\frac{|K(x,z)|}{{|x-z|}^{-\alpha }}{f}_2(z)d\mu (z)]}^2\frac{dt}{t^3}\right)}^{\frac{1}{2}},\hfill \\ D_2(x,y)={\left({\int }_0^{\mathrm{\infty }}{[{\int }_{|y-z|\le t<|x-z|}\frac{|K(y,z)|}{{|y-z|}^{-\alpha }}{f}_2(z)d\mu (z)]}^2\frac{dt}{t^3}\right)}^{\frac{1}{2}}\hfill \end{array}$$

and

$$D_3(x,y)={\left({\int }_0^{\mathrm{\infty }}{\left[{\int }_{\begin{array}{c}|x-z|\le t\\ |y-z|\le t\end{array}}\right|\frac{K(x,z)}{{|x-z|}^{-\alpha }}-\frac{K(y,z)}{{|y-z|}^{-\alpha }}\left|\right|f_2(z)|d\mu (z)]}^2\frac{dt}{t^3}\right)}^{\frac{1}{2}}.$$

It is easy to get that for any $x,y\in Q$,

$$\begin{array}{c}|\mathcal{M}(f_2)(x)-\mathcal{M}(f_2)(y)|\hfill \\ =|{\left({\int }_0^{\mathrm{\infty }}\right|{\int }_{|x-z|\le t}\frac{K(x,z)}{{|x-z|}^{\alpha }}d\mu (z){|}^2\frac{dt}{t^3})}^{\frac{1}{2}}-{\left({\int }_0^{\mathrm{\infty }}\right|{\int }_{|y-z|\le t}\frac{K(y,z)}{{|y-z|}^{\alpha }}d\mu (z){|}^2\frac{dt}{t^3})}^{\frac{1}{2}}|\hfill \\ \le {\left({\int }_0^{\mathrm{\infty }}\right|{\int }_{|x-z|\le t}\frac{K(x,z)}{{|x-z|}^{-\alpha }}{f}_2(z)d\mu (z)-{\int }_{|y-z|\le t}\frac{K(y,z)}{{|y-z|}^{-\alpha }}{f}_2(z)d\mu (z){|}^2\frac{dt}{t^3})}^{\frac{1}{2}}\hfill \\ \le \left({\int }_0^{\mathrm{\infty }}\right|{\int }_{|x-z|\le t<|y-z|}\frac{K(x,z)}{{|x-z|}^{-\alpha }}{f}_2(z)d\mu (z)+{\int }_{|y-z|\le t}\frac{K(x,z)}{{|x-z|}^{-\alpha }}{f}_2(z)d\mu (z)\hfill \\ {-{\int }_{|y-z|\le t<|x-z|}\frac{K(y,z)}{{|y-z|}^{-\alpha }}{f}_2(z)d\mu (z)-{\int }_{|x-z|\le t}\frac{K(y,z)}{{|y-z|}^{-\alpha }}{f}_2(z)d\mu (z){|}^2\frac{dt}{t^3})}^{\frac{1}{2}}\hfill \\ \le {\left({\int }_0^{\mathrm{\infty }}\right|{\int }_{|x-z|\le t<|y-z|}\frac{K(x,z)}{{|x-z|}^{-\alpha }}{f}_2(z)d\mu (z){|}^2\frac{dt}{t^3})}^{\frac{1}{2}}\hfill \\ +{\left({\int }_0^{\mathrm{\infty }}\right|{\int }_{|y-z|\le t<|x-z|}\frac{K(y,z)}{{|y-z|}^{-\alpha }}{f}_2(z)d\mu (z){|}^2\frac{dt}{t^3})}^{\frac{1}{2}}\hfill \\ +{\left\{{\int }_0^{\mathrm{\infty }}{[{\int }_{\begin{array}{c}|x-z|\le t\\ |y-z|\le t\end{array}}(\frac{K(x,z)}{{|x-z|}^{-\alpha }}-\frac{K(y,z)}{{|y-z|}^{-\alpha }})f_2(z)d\mu (z)]}^2\frac{dt}{t^3}\right\}}^{\frac{1}{2}}\hfill \\ \le \sum _{j=1}^3{D}_j(x,y).\hfill \end{array}$$

For $D_1(x,y)$, since $x,y\in Q$, $z\in \frac{3}{2}Q$, thus we get

$$\begin{array}{rcl}{D}_1(x,y)& \le & C{\left({\int }_0^{\mathrm{\infty }}{[{\int }_{|x-z|\le t<|y-z|}\frac{|f_2(z)|}{{|x-z|}^{n-\alpha -1}}d\mu (z)]}^2\frac{dt}{t^3}\right)}^{\frac{1}{2}}\\ \le & C{\int }_{|x-z|<|y-z|}\frac{|f_2(z)|}{{|x-z|}^{n-\alpha -1}}{\left({\int }_{|x-z|}^{|y-z|}\frac{dt}{t^3}\right)}^{\frac{1}{2}}d\mu (z)\\ \le & C\ell {(Q)}^{\frac{1}{2}}{\int }_{|x-z|<|y-z|}\frac{|f_2(z)|}{{|x-z|}^{n-\alpha +\frac{1}{2}}}d\mu (z)\\ \le & C\ell {(Q)}^{\frac{1}{2}}{\int }_{{\mathbb{R}}^d\setminus \frac{3}{2}Q}\frac{|f_2(z)|}{{|x-z|}^{n-\alpha +\frac{1}{2}}}d\mu (z)\\ \le & C\ell {(Q)}^{\frac{1}{2}}\sum _{k=1}^{\mathrm{\infty }}{\int }_{2^{k+1}Q\setminus 2^{k}Q}\frac{|f_2(z)|}{{|x-z|}^{n-\alpha +\frac{1}{2}}}d\mu (z)\\ \le & C\ell {(Q)}^{\frac{1}{2}}\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\ell {(\frac{3}{2}{2}^{k}Q)}^{n-\alpha +\frac{1}{2}}}{\int }_{2^{k+1}Q}|f_2(z)|d\mu (z)\\ \le & C\sum _{k=1}^{\mathrm{\infty }}{2}^{-\frac{k}{2}}\frac{1}{\ell {(\frac{3}{2}{2}^{k}Q)}^{n-\alpha }}{({\int }_{2^{k+1}Q}{|f_2(z)|}^{q}d\mu (z))}^{\frac{1}{q}}\mu {\left(\frac{3}{2}{2}^{k}Q\right)}^{1-\frac{1}{q}}\\ \le & C{\parallel f\parallel }_{M_q^p(\mu )}\sum _{k=1}^{\mathrm{\infty }}{2}^{-\frac{k}{2}}\\ \le & C{\parallel f\parallel }_{M_q^p(\mu )}.\end{array}$$

By a similar argument, it follows that

$$D_2(x,y)\le C{\parallel f\parallel }_{M_q^p(\mu )}.$$

Finally, by the condition ${\mathcal{H}}^{P^{\prime }}$, which the kernel $K(x,y)$ conditions, applying Minkowski’s inequality, and the fact that $\alpha =\frac{n}{p}$, we have

$$\begin{array}{rcl}{D}_3(x,y)& =& {\left({\int }_0^{\mathrm{\infty }}{\left[{\int }_{\begin{array}{c}|x-z|\le t\\ |y-z|\le t\end{array}}\right|\frac{K(x,z)}{{|x-z|}^{-\alpha }}-\frac{K(y,z)}{{|y-z|}^{-\alpha }}\left|\right|f_2(z)|d\mu (z)]}^2\frac{dt}{t^3}\right)}^{\frac{1}{2}}\\ \le & C{\int }_{{\mathbb{R}}^d\setminus \frac{3}{2}Q}|\frac{K(x,z)}{{|x-z|}^{-\alpha }}-\frac{K(y,z)}{{|y-z|}^{-\alpha }}||f(z)|{\left({\int }_{\begin{array}{c}|x-z|\le t\\ |y-z|\le t\end{array}}\frac{dt}{t^3}\right)}^{\frac{1}{2}}d\mu (z)\\ \le & C\sum _{k=1}^{\mathrm{\infty }}{\int }_{\frac{3}{2}{2}^{k+1}Q\setminus \frac{3}{2}{2}^{k}Q}|\frac{K(x,z)}{{|x-z|}^{-\alpha }}-\frac{K(y,z)}{{|y-z|}^{-\alpha }}|\frac{|f(z)|}{|y-z|}d\mu (z)\\ \le & C{\parallel f\parallel }_{M_q^p(\mu )}\sum _{k=1}^{\mathrm{\infty }}\mu {\left(2^{k}Q\right)}^{\frac{1}{q}-\frac{1}{p}}\\ \times {\{{\int }_{\frac{3}{2}{2}^{k+1}Q\setminus \frac{3}{2}{2}^{k}Q}{\left[\frac{1}{|y-z|}\right|\frac{K(x,z)}{{|x-z|}^{-\alpha }}-\frac{K(y,z)}{{|y-z|}^{-\alpha }}\left|\right]}^{q^{\prime }}d\mu (z)\}}^{\frac{1}{q^{\prime }}}\\ \le & C{\parallel f\parallel }_{M_q^p(\mu )}\sum _{k=1}^{\mathrm{\infty }}\ell {\left(\frac{3}{2}{2}^{k}Q\right)}^{\frac{n}{q}-\frac{n}{p}}\\ \times \left\{{\int }_{\frac{3}{2}{2}^{k+1}Q\setminus \frac{3}{2}{2}^{k}Q}\right[\frac{1}{|y-z|}|\frac{K(x,z)}{{|x-z|}^{-\alpha }}-\frac{K(x,z)}{{|y-z|}^{-\alpha }}\\ {{+\frac{K(x,z)}{{|y-z|}^{-\alpha }}-\frac{K(y,z)}{{|y-z|}^{-\alpha }}\left|\right]}^{q^{\prime }}d\mu (z)\}}^{\frac{1}{q^{\prime }}}\\ \le & C{\parallel f\parallel }_{M_q^p(\mu )}\sum _{k=1}^{\mathrm{\infty }}\ell {\left(\frac{3}{2}{2}^{k}Q\right)}^{\alpha -\frac{n}{p}}\ell {\left(\frac{3}{2}{2}^{k}Q\right)}^n\\ \times {\{\frac{1}{\ell {(\frac{3}{2}{2}^{k}Q)}^n}{\int }_{\frac{3}{2}{2}^{k+1}Q\setminus \frac{3}{2}{2}^{k}Q}{\left[\right|K(x,z)-K(y,z)\left|\frac{1}{|y-z|}\right]}^{q^{\prime }}d\mu (z)\}}^{\frac{1}{q^{\prime }}}\\ +C{\parallel f\parallel }_{M_q^p(\mu )}\sum _{k=1}^{\mathrm{\infty }}\ell {\left(\frac{3}{2}{2}^{k}Q\right)}^{\frac{n}{q}-\frac{n}{p}}\ell {(Q)}^{\alpha }{({\int }_{\frac{3}{2}{2}^{k+1}Q\setminus \frac{3}{2}{2}^{k}Q}\frac{1}{{|y-z|}^{n{q}^{\prime }}}d\mu (z))}^{\frac{1}{q^{\prime }}}\\ \le & C{\parallel f\parallel }_{M_q^p(\mu )}.\end{array}$$

Combining these estimates, we conclude that

$${\mathrm{I}}_2\le C{\parallel f\parallel }_{M_q^p(\mu )},$$

and so estimate (3.1) is proved.

We proceed to show (3.2). For any cubes $Q\subset R$ with $x\in Q$, denote $N_{Q,R+1}$ simply by N. Write

$$\begin{array}{rcl}|a_Q-a_R|& \le & |m_R[\mathcal{M}(f{\chi }_{{\mathbb{R}}^d\setminus 2^{N}Q})]-m_Q[\mathcal{M}(f{\chi }_{{\mathbb{R}}^d\setminus 2^{N}R})]|\\ +\left|m_Q[\mathcal{M}(f{\chi }_{2^{N}Q\setminus \frac{3}{2}Q})]\right|+\left|m_R[\mathcal{M}(f{\chi }_{2^{N}Q\setminus \frac{3}{2}R})]\right|\\ =& E_1+E_2+E_3.\end{array}$$

As in the estimate for the term ${\mathrm{I}}_2$, then

$$E_2\le C{\parallel f\parallel }_{M_q^p(\mu )}.$$

We conclude from $y\in R$, $z\in 2^{N}Q\setminus \frac{3}{2}Q$ that

$$\begin{array}{rcl}\mathcal{M}(f{\chi }_{2^{N}Q\setminus \frac{3}{2}R})(y)& \le & C{\int }_{2^{N}Q\setminus \frac{3}{2}R}\left|\frac{K(y,z)}{{|y-z|}^{-\alpha }}\right|{\left({\int }_{|y-z|}^{\mathrm{\infty }}\frac{dt}{t^3}\right)}^{\frac{1}{2}}d\mu (z)\\ \le & C{\int }_{2^{N}Q\setminus \frac{3}{2}R}\frac{|f(z)|}{{|y-z|}^{n-\alpha }}d\mu (z)\\ \le & C\ell {(R)}^{\alpha -n}{\int }_{2^{N}Q\setminus \frac{3}{2}R}|f(z)|d\mu (z)\\ \le & C\ell {(R)}^{\alpha -n}{\left({\int }_{2^{N}Q\setminus \frac{3}{2}R}\right|f(z){|}^{q}d\mu (z))}^{\frac{1}{q}}\mu {\left(2^{N}Q\right)}^{1-\frac{1}{q}}\\ \le & C\ell {(R)}^{\alpha -n}\mu {\left(2^{N}Q\right)}^{\frac{1}{p}-\frac{1}{q}}{\left({\int }_{2^{N}Q}\right|f(z){|}^{q}d\mu (z))}^{\frac{1}{q}}\mu {\left(2^{N}Q\right)}^{1-\frac{1}{p}}\\ \le & C{\parallel f\parallel }_{M_q^p(\mu )}\ell {\left(2^{N}Q\right)}^{\alpha -\frac{n}{p}}\\ \le & C{\parallel f\parallel }_{M_q^p(\mu )}.\end{array}$$

Taking mean over $y\in R$, we obtain

$$E_3\le C{\parallel f\parallel }_{M_q^p(\mu )}.$$

Analysis similar to that in the estimates for $E_3$ shows that

$$E_2\le C{\parallel f\parallel }_{M_q^p(\mu )}.$$

Finally, we get (3.2) and this is precisely the assertion of Theorem 3.2. □