The existence and boundedness of linear and multilinear Marcinkiewicz integrals on central Campanato spaces

Abstract

In this paper, we obtain the existence and boundedness of Marcinkiewicz integrals with homogeneous kernels on central Campanato spaces. Moreover, the existence and boundedness of multilinear Marcinkiewicz integrals on central Campanato spaces are also deduced. We extend various previous results to central Campanato spaces which can be seen as the local version of Campanato spaces.

1 Introduction

Let $\varOmega \in L^{1}(\mathbb{S}^{n-1})$ be homogeneous of degree zero and satisfy

$$ \int _{\mathbb{S}^{n-1}} \varOmega \bigl(x'\bigr) \,d \sigma \bigl(x'\bigr) =0, $$

(1.1)

where $x'=x/\vert x \vert$ for any $x\neq 0$. The Marcinkiewicz integral operator of higher dimension is defined by

$$ \mu _{\varOmega }(f) (x)= \biggl( \int _{0}^{\infty } \bigl\vert F_{\varOmega, t} (x) \bigr\vert ^{2} \frac{dt}{t^{3}} \biggr)^{\frac{1}{2}}, $$

where $F_{\varOmega, t} (x)= \int _{\vert x-y\vert \leq t} \frac{\varOmega (x-y)}{\vert x-y\vert ^{n-1}}f(y)\,dy$, and $\mathbb{S}^{n-1}$ is the unit sphere of $\mathbb{R}^{n}$$(n\geqslant 2)$ equipped with normalized Lebesgue measure $d \sigma =d \sigma (x')$.

Stein [1] proved that, if Ω is continuous and satisfies a $\mathrm{Lip}_{\alpha }$$(0<\alpha \leq 1)$ condition on $\mathbb{S}^{n-1}$, then $\mu _{\varOmega }$ is of type $(p,p)$$(1< p\leq 2)$ and of weak type $(1,1)$. Benedek et al. [2] proved that, if $\varOmega \in C^{1} ({\mathbb{S}^{n-1}})$, then $\mu _{\varOmega }$ is of type $(p,p)$$(1< p\leq \infty )$. Ding [3] proved that if Ω is homogeneous of degree zero satisfying a class of $L^{q}$-Dini $(1< q\leq \infty )$ conditions, and then $\mu _{\varOmega }$ is bounded on Campanato spaces.

Definition 1.1

Suppose $1\leq p <\infty$ and $-\frac{n}{p}\leq \alpha <1$. Then the Campanato space $\varepsilon ^{\alpha,p} (\mathbb{R}^{n})$ is defined as

$$ \varepsilon ^{\alpha,p}\bigl(\mathbb{R}^{n}\bigr):= \bigl\lbrace f \in L_{\mathrm{loc}}^{p} \bigl( \mathbb{R}^{n}\bigr): \Vert f \Vert _{\varepsilon ^{\alpha,p}} < \infty \bigr\rbrace , $$

where

$$ \Vert f \Vert _{\varepsilon ^{\alpha,p}} = \sup_{x \in \mathbb{R}^{n} , r>0} \frac{1}{ \vert B(x, r) \vert ^{\frac{\alpha }{n}}} \biggl( \frac{1}{ \vert B(x,r) \vert } \int _{B(x,r)} \bigl\vert f(x)- f_{B(x,r)} \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}}, $$

here $B(x,r)=\lbrace y \in \mathbb{R}^{n}: \vert x-y\vert < r \rbrace$, $\vert B(x,r)\vert$ is its Lebesgue measure and $f_{B(x,r)}= \frac{1}{\vert B(x,r)\vert }\int _{B(x,r)} f(y)\,dy$.

Remark 1.1

When $\alpha \in (0,1)$ and $p\in [1, \infty )$, $\varepsilon ^{\alpha,p} = \mathrm{Lip} _{\alpha }$, where $\mathrm{Lip} _{\alpha }$ is the Lipschitz space; when $\alpha =0$, the Campanato space is the BMO space; when $\alpha \in (-\frac{n}{p},0)$, $\varepsilon ^{\alpha,p} = L^{p, \alpha p+n}$, where $L^{p, \alpha p+n}$ is the Morrey space.

Theorem A

([3])

Let$1< p<\infty$, $1\leq q'\leq p$, Ωsatisfy (1.1) and the$L^{q}$-Dini condition. Suppose that$f\in \varepsilon ^{\alpha,p} (\mathbb{R}^{n})$for$- \infty <\alpha < 0$and there is a measurable set$E\subset \mathbb{R}^{n}$with$\vert E \vert >0$such that$\mu _{\varOmega } (f)(x)<\infty$for any$x\in E$. Then$\mu _{\varOmega } (f)(x)<\infty$a.e. on$\mathbb{R}^{n}$and$\Vert \mu _{\varOmega } (f)\Vert _{\varepsilon ^{\alpha,p}} \leq C \Vert f\Vert _{\varepsilon ^{\alpha,p}}$, where the constantCis independent off.

The multilinear theory has attracted much attention since the pioneering work of Coifman and Meyer [4, 5]. The topic was reconsidered by several authors, including Christ and Journé [6], Kenig and Stein [7], Grafakos and Torres [8, 9], and Lerner et al. [10]. Chen, Xue and Yabuta [11] proved the boundedness of a multilinear Marcinkiewicz integral on Lebesgue spaces. Xue and Yabuta [12] proved the boundedness of a multilinear Marcinkiewicz integral on Campanato spaces.

Definition 1.2

([11])

Let Ω be a function defined on $(\mathbb{R}^{n})^{m}$ with the following properties:

• (1) Ω is homogeneous of degree 0, i.e., for any $\lambda >0$ and $\vec{y}=(y_{1}, \ldots, y_{m} ) \in (\mathbb{R}^{n})^{m}$,

  $$ \varOmega (\lambda \vec{y}) = \varOmega ( \vec{y}); $$

  (1.2)

• (2) Ω is Lipschitz continuous on $(\mathbb{S}^{n-1})^{m}$, i.e. there are $0< \alpha <1$ and $C>0$ such that, for any $\vec{\xi } =(\xi _{1}, \ldots, \xi _{m})$, $\vec{\eta } =(\eta _{1}, \ldots, \eta _{m}) \in (\mathbb{R}^{n})^{m}$,

  $$ \bigl\vert \varOmega (\xi )- \varOmega (\eta ) \bigr\vert \leq C \bigl\vert \xi ' -\eta ' \bigr\vert ^{\alpha }, $$

  (1.3)

  where $(y_{1}, \ldots, y_{m} )'= \frac{(y_{1}, \ldots, y_{m} )}{\vert y_{1} \vert + \cdots + \vert y_{m} \vert }$, and we should note that $(y_{1}, \ldots, y_{m} )'$ is not an element of $(\mathbb{S}^{n-1})^{m}$;

• (3) the integration of Ω on each unit sphere vanishes,

  $$ \int _{\mathbb{S}^{n-1}} \varOmega (y_{1}, \ldots, y_{m} ) \,dy_{j} =0,\quad j=1,\ldots, m. $$

  (1.4)

For any $\vec{f}=(f_{1}, \ldots, f_{m}) \in (S(\mathbb{R}^{n}))^{m}$, we can define the operator $F_{t}$ for any $t>0$ as

$$\begin{aligned} F_{t}(\vec{f}) (x) &= \frac{\chi _{(B(0,t))^{m}} \varOmega (\vec{\cdot })}{t^{m} \vert \vec{\cdot } \vert ^{m(n-1)} } \ast (f_{1} \otimes \cdots \otimes f_{m}) (x) \\ & = \frac{1}{t^{m}} \int _{(B(0,t))^{m}} \frac{\varOmega (\vec{y})}{ \vert \vec{y} \vert ^{m(n-1)}} \prod _{i=1}^{m} f_{i}(x-y_{i}) \,d\vec{y}, \end{aligned}$$

where $\vert \vec{y}\vert = \vert y_{1}\vert + \cdots + \vert y_{m} \vert$ and $B(x,t)= \lbrace y \in \mathbb{R}^{n}: \vert y-x \vert \leq t \rbrace$. Finally, the multilinear Marcinkiewicz integral μ is defined by

$$ \mu (\vec{f}) (x)= \biggl( \int _{0}^{\infty } \bigl\vert F_{t}( \vec{f}) (x) \bigr\vert ^{2} \frac{dt}{t} \biggr)^{\frac{1}{2}}. $$

If $m=1$, it is easy to see that $\mu (\vec{f})$ coincides with $\mu _{\varOmega } (f)$.

Theorem B

([11])

Supposeμis bounded from$L^{q_{1}}\times \cdots \times L^{q_{m}}$to$L^{q}$for some$1< q_{1}, \ldots, q_{m} <\infty$with$\frac{1}{q} = \frac{1}{q_{1}} + \cdots + \frac{1}{q_{m}}$. Then, for$1< p_{1}, \ldots, p_{m} <\infty$with$\frac{1}{p} = \frac{1}{p_{1}} + \cdots + \frac{1}{p_{m}}$, there is a$C>0$such that

$$ \bigl\Vert \mu (\vec{f}) \bigr\Vert _{L^{p}} \leq C \prod _{i=1}^{m} \Vert f_{i} \Vert _{L^{p_{i}}}. $$

Yang [13] introduced the central Campanato spaces on p-adic fields and obtained the behavior of a class of p-adic singular integral operators on these spaces. In fact, the central Campanato space on Euclid spaces can be defined in a similar way. Now we are in a position to define the central Campanato space.

Definition 1.3

([13])

Suppose $1\leq p <\infty$, $q>1$, $\alpha \in \mathbb{R}$ and $n\in \mathbb{Z}$. The central Campanato space $\mathrm{CL} ^{\alpha,p}(\mathbb{R}^{n})$ is defined as

$$ \mathrm{CL} ^{\alpha,p}\bigl(\mathbb{R}^{n}\bigr):= \bigl\lbrace f \in L_{\mathrm{loc}}^{p} \bigl( \mathbb{R}^{n}\bigr): \Vert f \Vert _{\mathrm{CL} ^{\alpha,p}} < \infty \bigr\rbrace , $$

where

$$ \Vert f \Vert _{\mathrm{CL} ^{\alpha,p}} = \sup_{r>0} \frac{1}{ \vert B(0,r) \vert ^{\alpha +\frac{1}{p}}} \biggl( \int _{B(0,r)} \bigl\vert f(x)- f_{B(0,r)} \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}}. $$

Remark 1.2

When $\alpha =\frac{1}{l}-1$, the central Campanato space $\mathrm{CL}^{\alpha,p}$ is the central BMO space $\mathrm{CMO}^{p}_{l}$, where $0< l\leq 1< p<\infty$; see [14] for more about the space of $\mathrm{CMO}^{p}_{l}$.

The main results of this paper are as follows.

Theorem 1.1

Let$1< p<\infty$, $1\leq q'\leq p$, Ωsatisfy (1.1) and the$L^{q}$-Dini condition. Suppose that$f\in \mathrm{CL} ^{\alpha,p} (\mathbb{R}^{n})$for$-\infty <\alpha n <0$and there is a measurable set$E\subset \mathbb{R}^{n}$with$\vert E \vert >0$such that$\mu _{\varOmega } (f)(x)<\infty$for any$x\in E$. Then$\mu _{\varOmega } (f)(x)<\infty$a.e. on$\mathbb{R}^{n}$and$\Vert \mu _{\varOmega } (f)\Vert _{\mathrm{CL} ^{\alpha,p}} \leq C \Vert f\Vert _{\mathrm{CL} ^{\alpha,p}}$, where the constantCis independent off.

Theorem 1.2

LetΩbe a function defined on$(\mathbb{R}^{n})^{m}$, satisfying (1.2), Lipschitz continuous (1.3) with index replaced byβand (1.4). Supposeμis bounded from$L^{q_{1}}\times \cdots \times L^{q_{m}}$to$L^{q}$for some$1< q_{1}, \ldots, q_{m} <\infty$with$\frac{1}{q} = \frac{1}{q_{1}} + \cdots + \frac{1}{q_{m}}$. Suppose also that$-\infty < \alpha = \alpha _{1} + \cdots +\alpha _{m} <0$with$\alpha _{1}, \ldots, \alpha _{m} <0$and$n< p<\infty$. Then, for$1< p_{1}, \ldots, p_{m} <\infty$with$\frac{1}{p} = \frac{1}{p_{1}} + \cdots + \frac{1}{p_{m}}$, $f_{j} \in \mathrm{CL}^{\alpha _{j}, p_{j}} (\mathbb{R}^{n})$$(j=1, \ldots, m)$, $\mu (\vec{f})$is either infinite everywhere or finite almost everywhere, and in the latter case, there is a constant$C>0$such that

$$ \bigl\Vert \mu (\vec{f}) \bigr\Vert _{\mathrm{CL}^{\alpha,p}} \leq C \prod _{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}}. $$

Remark 1.3

Comparing the definition of the Campanato space and the central Campanato space, we can easily see that the selection range of the ball B is different. This leads to the difference in the range of the index about the boundedness of $\mu (\vec{f})$ on Campanato space and central Campanato space.

Throughout this paper we assume that the notation C represents a constant and its values may vary from line to line.

2 Some preliminaries and notations

Suppose that $\varOmega \in L^{q}(\mathbb{S}^{n-1})$$(q\geqslant 1)$, the integral modulus $\omega _{q}(\delta )$ of continuity of order q of Ω is defined by

$$ \omega _{q}(\delta )=\sup_{ \vert \rho \vert \leq \delta } \biggl( \int _{\mathbb{S}^{n-1}} \bigl\vert \varOmega \bigl(\rho x' \bigr)-\varOmega \bigl( x'\bigr) \bigr\vert ^{q} \,d \sigma \bigl(x'\bigr) \biggr)^{\frac{1}{q}}, $$

where ρ is a rotation on $\mathbb{S}^{n-1}$, $\vert \rho \vert =\Vert \rho -I \Vert$. We say that Ω satisfies the $L^{q}$-Dini condition, if $\omega _{q}(\delta )$ satisfies

$$ \int _{0}^{1} \frac{\omega _{q}(\delta )}{\delta } \,d\delta < \infty. $$

To prove our main results, we begin with some important lemmas.

Lemma 2.1

([3])

Suppose that$0<\lambda <n$, andΩis homogeneous of degree zero and satisfies the$L^{q}$-Dini condition for$q>1$. If there exists a constant$a_{0}>0$, such that$\vert x\vert < a_{0} R$, then we have

$$ \biggl( \int _{R< \vert y \vert < 2R} \biggl\vert \frac{\varOmega (y-x)}{ \vert y-x \vert ^{n-\lambda }} - \frac{\varOmega (y)}{ \vert y \vert ^{n-\lambda }} \biggr\vert ^{q} \,dy \biggr)^{\frac{1}{q}} \leq C R^{\frac{n}{q}-(n-\lambda )} \biggl( \frac{ \vert x \vert }{R} + \int _{\frac{ \vert x \vert }{2R} < \delta < \frac{ \vert x \vert }{R}} \frac{\omega _{q}(\delta )}{\delta } \,d\delta \biggr), $$

where the constant$C>0$is independent ofRandx.

Lemma 2.2

Suppose that$1< p <\infty$, $-\infty <\alpha <\frac{1}{2}$, and$\eta >\max \lbrace 0, \alpha pn\rbrace$. If$f\in \mathrm{CL}^{\alpha,p}$, then there exists$C>0$such that, for any ball$B_{0}$centered at origin with side lengthdand$x_{0}\in B_{0}$,

$$ \biggl( \int _{\mathbb{R}^{n}} \frac{d^{\eta } \vert f(x)-f_{B_{0}} \vert ^{p}}{d^{n+\eta }+ \vert x-x_{0} \vert ^{n+\eta }} \,dx \biggr)^{\frac{1}{p}} \leq C\,d^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{ \alpha,p}}. $$

Proof

Let $B_{k}$ be a ball with the same center of $B_{0}$ and its radius is $2^{k}$ times of $B_{0}$. Decomposing $\mathbb{R}^{n}$ into a geometrically increasing sequence of concentric balls, and using the fact that $\vert f_{B_{k}}-f_{B_{0}} \vert \leq C k \vert B_{k} \vert ^{\alpha }\Vert f \Vert _{\mathrm{CL}^{\alpha,p}}$, we have

$$\begin{aligned} & \int _{\mathbb{R}^{n}} \frac{d^{\eta } \vert f(x)-f_{B_{0}} \vert ^{p}}{d^{n+\eta }+ \vert x-x_{0} \vert ^{n+\eta }} \,dx\\ &\quad = \sum _{k=0}^{\infty } \int _{B_{k} \setminus B_{k-1}} \frac{d^{\eta } \vert f(x)-f_{B_{0}} \vert ^{p}}{d^{n+\eta }+ \vert x-x_{0} \vert ^{n+\eta }} \,dx \\ &\quad\leq C \sum_{k=0}^{\infty } \frac{d^{\eta }}{(2^{k} d)^{n+\eta }} \biggl( \int _{B_{k}} \bigl\vert f(x)-f_{B_{k}} \bigr\vert ^{p} \,dx + \vert f_{B_{k}}-f_{B_{0}} \vert ^{p} \vert B_{k} \vert \biggr) \\ & \quad\leq C \sum_{k=0}^{\infty } \frac{1}{2^{k\eta }} \bigl( \bigl(2^{k} d\bigr)^{n \alpha p} \Vert f \Vert _{\mathrm{CL}^{\alpha,p}}^{p} + k^{p} \bigl(2^{k} d \bigr)^{n \alpha p} \Vert f \Vert _{\mathrm{CL}^{\alpha,p}}^{p} \bigr) \\ &\quad \leq C d^{n\alpha p} \Vert f \Vert _{\mathrm{CL}^{\alpha,p}}^{p}. \end{aligned}$$

Then the proof is complete. □

Lemma 2.3

Suppose that$1< p <\infty$, $-\infty <\alpha <\frac{1}{2}$, $1\leq q' < p$and$\eta >\max \lbrace 0, \alpha pn\rbrace$. If$f\in \mathrm{CL}^{\alpha,p}$, then there exists$C>0$such that, for any ball$B_{0}$with side lengthdand centered at origin and$x_{0}\in B_{0}$,

$$ \biggl( \int _{\mathbb{R}^{n}} \frac{d^{\eta } \vert f(x)-f_{B_{0}} \vert ^{q'}}{d^{n+\eta }+ \vert x-x_{0} \vert ^{n+\eta }} \,dx \biggr)^{\frac{1}{q'}} \leq Cd^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{ \alpha,p}}. $$

Proof

Let $B_{k}$ be a ball with the same center of $B_{0}$ and its radius is $2^{k}$ times of $B_{0}$. Decomposing $\mathbb{R}^{n}$ into a geometrically increasing sequence of concentric balls, and using Hölder’s inequality and Lemma 2.2, we have

$$\begin{aligned} &\int _{\mathbb{R}^{n}} \frac{d^{\eta } \vert f(x)-f_{B_{0}} \vert ^{q'}}{d^{n+\eta }+ \vert x-x_{0} \vert ^{n+\eta }} \,dx \\ &\quad\leq \sum _{k=0}^{\infty } \biggl( \int _{\mathbb{R}^{n}} \frac{d^{\eta } \vert f(x)-f_{B_{0}} \vert ^{p}}{d^{n+\eta }+ \vert x-x_{0} \vert ^{n+\eta }} \,dx \biggr)^{\frac{q'}{p}} \\ &\qquad{} \times \biggl( \int _{B_{k} \setminus B_{k-1}} \frac{d^{\eta } }{d^{n+\eta }+ \vert x-x_{0} \vert ^{n+\eta }} \,dx \biggr)^{ \frac{p-q'}{p}} \\ &\quad\leq C \sum_{k=0}^{\infty } d^{n\alpha q'} \Vert f \Vert _{\mathrm{CL}^{ \alpha,p}}^{q'} \biggl( \frac{d^{\eta } (2^{k} d)^{n}}{(2^{k-1}d)^{n+\eta }} \biggr)^{ \frac{p-q'}{p}} \\ &\quad \leq C d^{n\alpha q'} \Vert f \Vert _{\mathrm{CL}^{\alpha,p}}^{q'}. \end{aligned}$$

Then the proof is complete. □

Lemma 2.4

Suppose that$1\leq p <\infty$, $\alpha <0$, $B=B(0, r)$, $x \in B$and$t>8r>0$. If$f \in \mathrm{CL} ^{\alpha,p}$, then, for$0\leq k \leq k_{0}$with$k_{0} \in \mathbb{N}$satisfying$2r \leq 2^{-k_{0}} t <4r$, we have

$$ \biggl( \frac{1}{ \vert B(x, 2^{-k}t) \vert } \int _{ B(x, 2^{-k}t)} \bigl\vert f(y)-f_{B} \bigr\vert ^{p} \,dy \biggr) ^{\frac{1}{p}} \leq C r^{n \alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha,p}}. $$

Proof

Since $x \in B$ and $0\leq k \leq k_{0}$ satisfying $2r \leq 2^{-k_{0}} t <4r$, we have $B(x, 2^{-k}t) \subset B(0, 2^{-k+1} t)$. Thus

$$\begin{aligned} & \biggl( \frac{1}{ \vert B(x, 2^{-k}t) \vert } \int _{ B(x, 2^{-k}t)} \bigl\vert f(y)-f_{B} \bigr\vert ^{p} \,dy \biggr) ^{\frac{1}{p}} \\ &\quad \leq C \biggl( \frac{1}{ \vert B(0, 2^{-k+1}t) \vert } \int _{ B(0, 2^{-k+1}t)} \bigl\vert f(y)-f_{B(0, 2^{-k+1}t)} \bigr\vert ^{p} \,dy \biggr) ^{\frac{1}{p}} \\ &\qquad{} + C \sum_{j=k}^{k_{0} } \vert f_{B(0, 2^{-j+1}t)} -f_{B(0, 2^{-j}t)} \vert + C \vert f _{B(0, 2^{-k_{0}}t)} -f_{B} \vert \\ &\quad \leq C \bigl(2^{-k+1}t\bigr)^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha,p}} + C \sum_{j=k}^{k_{0} } \vert f_{B(0, 2^{-j+1}t)} -f_{B(0, 2^{-j}t)} \vert + C \vert f _{B(0, 2^{-k_{0}}t)} -f_{B} \vert . \end{aligned}$$

For $j=k, \ldots, k_{0}$, we have

$$\begin{aligned} \vert f_{B(0, 2^{-j+1}t)} -f_{B(0, 2^{-j}t)} \vert &\leq \frac{1}{ \vert B(0,2^{-j}t) \vert } \int _{B(0,2^{-j}t)} \bigl\vert f(y)- f_{B(0, 2^{-j+1} t)} \bigr\vert \,dy \\ & \leq 2^{n} \biggl( \frac{1}{ \vert B(0, 2^{-j+1}t) \vert } \int _{B(0, 2^{-j+1} t)} \bigl\vert f(y)- f_{B(0, 2^{-j+1} t)} \bigr\vert ^{p} \,dy \biggr)^{ \frac{1}{p}} \\ & \leq 2^{n} \bigl(2^{-j+1}t\bigr)^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha,p}}. \end{aligned}$$

For the last term, since $2r \leq 2^{-k_{0}}t \leq 4r$, we have

$$\begin{aligned} \vert f _{B(0, 2^{-k_{0}}t)} -f_{B} \vert \leq 4^{n} \bigl(2^{-k_{0}} t\bigr)^{n \alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha,p}}. \end{aligned}$$

So we obtain

$$\begin{aligned} & \biggl( \frac{1}{ \vert B(x, 2^{-k}t) \vert } \int _{ B(x, 2^{-k}t)} \bigl\vert f(y)-f_{B} \bigr\vert ^{p} \,dy \biggr) ^{\frac{1}{p}} \\ &\quad \leq C \sum_{j=k}^{k_{0}} 2^{-jn\alpha } t^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{ \alpha,p}} \leq C r^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha,p}}. \end{aligned}$$

Then the proof is complete. □

Lemma 2.5

Suppose that$1\leq p <\infty$, $\alpha <0$, $B=B(0, r)$, $x \in B$and$k \in \mathbb{N}$. If$f \in \mathrm{CL} ^{\alpha,p}$, then we have

$$ \biggl( \frac{1}{ \vert B(x, 2^{k}r) \vert } \int _{ B(x, 2^{k}r)} \bigl\vert f(y)-f_{B} \bigr\vert ^{p} \,dy \biggr) ^{\frac{1}{p}} \leq C r^{n \alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha,p}}. $$

Proof

For $x \in B$ and $k \in \mathbb{N}$, we have $B(x,2^{k} r) \subset B(0, 2^{k+1}r)$. Similar to Lemma 2.4, we have

$$\begin{aligned} & \biggl( \frac{1}{ \vert B(x, 2^{k}r) \vert } \int _{ B(x, 2^{k}r)} \bigl\vert f(y)-f_{B} \bigr\vert ^{p} \,dy \biggr) ^{\frac{1}{p}} \\ &\quad \leq C \biggl( \frac{1}{ \vert B(0, 2^{k+1}r) \vert } \int _{ B(0, 2^{k+1}r)} \bigl\vert f(y)-f_{B(0, 2^{k+1}r)} \bigr\vert ^{p} \,dy \biggr) ^{\frac{1}{p}} \\ &\qquad{} + C \vert f_{B(0, 2^{k+1}r)} -f_{B(0, 2^{k}r)} \vert +\cdots + \vert f_{B(0, 2r)} -f_{B} \vert \\ &\quad \leq C \sum_{l=0}^{k+1} 2^{\ln \alpha } r^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{ \alpha,p}} \leq C r^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha,p}}. \end{aligned}$$

Then the proof is complete. □

Lemma 2.6

Suppose that$\alpha <0$, $B=B(0, r)$, $x \in B$and$t>8r>0$, $1\leq p <\infty$. If$f \in \mathrm{CL} ^{\alpha,p}$, then we have

$$ \int _{8r \leq \vert x-y \vert < t} \frac{ \vert f(y)-f_{B} \vert }{ \vert x-y \vert ^{n-1}} \,dy \leq C t r^{n \alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha,p}}. $$

Proof

Let $k_{0} \in \mathbb{N}$ satisfying $2r\leq 2^{-k_{0}}t <4r$. Using Lemma 2.4, we obtain

$$\begin{aligned} \int _{8r \leq \vert x-y \vert < t} \frac{ \vert f(y)-f_{B} \vert }{ \vert x-y \vert ^{n-1}} \,dy & \leq \sum _{k=0}^{k_{0}-1} \int _{2^{-k-1}t \leq \vert x-y \vert \leq 2^{-k}t} \frac{ \vert f(y)-f_{B} \vert }{ \vert x-y \vert ^{n-1}} \,dy \\ & \leq \sum_{k=0}^{k_{0}-1} \frac{1}{(2^{-k-1}t)^{n-1}} \int _{B(x, 2^{-k}t)} \bigl\vert f(y)-f_{B} \bigr\vert \,dy \\ & \leq C \sum_{k=0}^{k_{0}-1} 2^{-k}t \biggl( \frac{1}{ \vert B(x, 2^{-k}t) \vert } \int _{B(x, 2^{-k}t)} \bigl\vert f(y)-f_{B} \bigr\vert ^{p} \,dy \biggr)^{\frac{1}{p}} \\ & \leq Ct r^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha,p}}. \end{aligned}$$

Then the proof is complete. □

Lemma 2.7

Suppose that$m\in \mathbb{N}$, $B=B(0, r)$, $x \in B$, $1\leq p <\infty$, $0<\beta \leq 1$and$\gamma <\beta$, $\alpha =\alpha _{1} + \cdots + \alpha _{m} <0$with$\alpha _{1}, \ldots, \alpha _{m} <0$. If$f_{i} \in \mathrm{CL} ^{\alpha _{i},p_{i}}$$(i=1,\ldots, m)$, then we have

$$ \int _{(B(x,8r)^{m})^{c}} \frac{ r^{\beta }\prod_{i=1}^{m} \vert f_{i}(y_{i})-(f_{i})_{B} \vert }{(\sum_{j=1}^{m} \vert x-y_{j} \vert ) ^{mn+\beta -\gamma }} \,d \vec{y} \leq C r^{\gamma + n\alpha } \prod_{i=1}^{m} \Vert f_{i} \Vert _{\mathrm{CL}^{\alpha _{i},p_{i}}}. $$

Proof

Using Lemma 2.5 and Hölder’s inequality, we have

$$\begin{aligned} & \int _{(B(x,8r)^{m})^{c}} \frac{ r^{\beta }\prod_{i=1}^{m} \vert f_{i}(y_{i})-(f_{i})_{B} \vert }{(\sum_{j=1}^{m} \vert x-y_{j} \vert ) ^{mn+\beta -\gamma }} \,d \vec{y} \\ &\quad= \sum_{k=0}^{\infty } \int _{(B(x, 2^{k+4}r))^{m} \setminus (B(x, 2^{k+3}r))^{m}} \frac{ r^{\beta }\prod_{i=1}^{m} \vert f_{i}(y_{i})-(f_{i})_{B} \vert }{(\sum_{j=1}^{m} \vert x-y_{j} \vert ) ^{mn+\beta -\gamma }} \,d \vec{y} \\ &\quad\leq C r^{\beta }\sum_{k=0}^{\infty } \frac{1}{(2^{k+3}r)^{mn+\beta -\gamma }} \int _{(B(x, 2^{k+4}r))^{m}} \prod_{i=1}^{m} \bigl\vert f_{i}(y_{i})-(f_{i})_{B} \bigr\vert \,d \vec{y} \\ &\quad\leq C r^{\gamma }\sum_{k=0}^{\infty } \frac{1}{2^{k(\beta -\gamma )}} \prod_{i=1}^{m} \frac{1}{ \vert B(x, 2^{k+4}r) \vert } \int _{B(x, 2^{k+4}r)} \bigl\vert f_{i}(y_{i})-(f_{i})_{B} \bigr\vert \,d y_{i} \\ &\quad\leq C \sum_{k=0}^{\infty } \frac{1}{2^{k(\beta -\gamma )}} r^{ \gamma +n\alpha } \prod_{i=1}^{m} \Vert f_{i} \Vert _{\mathrm{CL}^{\alpha _{i},p_{i}}} \\ &\quad\leq C r^{\gamma +n\alpha } \prod_{i=1}^{m} \Vert f_{i} \Vert _{\mathrm{CL}^{ \alpha _{i},p_{i}}}. \end{aligned}$$

Then the proof is complete. □

3 Proof the main results

Proof of Theorem 1.1

Let $1 < p < \infty$, $1 \leq q' \leq p$ and $f \in \mathrm{CL}^{\alpha, p}$. We set B be the ball $B(0,d)$, $B^{*}=8B=B(0,8d)$, $B^{j}= 2^{j}B= B(0, 2^{j} d)$$(j\in \mathbb{Z})$. We first show that $\mu _{\varOmega } (f)(x)<\infty$, a.e. on B. Let

$$f(x)=f_{B^{*}}+\bigl(f(x)-f_{B^{*}}\bigr) \chi _{B^{*}}(x)+\bigl(f(x)-f_{B^{*}}\bigr)\bigl( 1- \chi _{B^{*}}(x)\bigr)=: f_{1}+f_{2}+f_{3}. $$

By Theorem A and (1.1), we know that $\mu _{\varOmega } (f_{1})(x)=0$ on B and

$$ \biggl( \int _{B} \bigl\vert \mu _{\varOmega } (f_{2}) (x) \bigr\vert ^{p} \,dx \biggr)^{ \frac{1}{p}} \leq \biggl( \int _{B^{*}} \bigl\vert \mu _{\varOmega } (f_{2}) (x) \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}} \leq C d^{n\alpha +\frac{n}{p}} \Vert f \Vert _{\mathrm{CL}^{\alpha, p}}. $$

Hence

$$\begin{aligned} \int _{B} \bigl\vert \mu _{\varOmega } (f_{2}) (x) \bigr\vert \,dx \leq \vert B \vert ^{ \frac{1}{p'}} \biggl( \int _{B} \bigl\vert \mu _{\varOmega } (f_{2}) (x) \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}} \leq d^{n+n\alpha } \Vert f \Vert _{\mathrm{CL}^{ \alpha, p}}. \end{aligned}$$

This shows that $\mu _{\varOmega } (f_{2})(x) <\infty$, a.e. on B. Since $\vert E\vert >0$, we have $\vert B \cap E\vert >0$. There exists an $x_{0} \in B \cap E$, such that $\mu _{\varOmega } (f)(x_{0})<\infty$ and $\mu _{\varOmega } (f_{2})(x_{0})<\infty$. Then

$$\begin{aligned} \mu _{\varOmega } (f_{3}) (x_{0}) \leq \mu _{\varOmega } (f) (x_{0}) +\mu _{ \varOmega } (f_{2}) (x_{0})< \infty. \end{aligned}$$

Fix any $x \in B$, we write

$$\begin{aligned} & \bigl\vert \mu _{\varOmega } (f_{3}) (x)- \mu _{\varOmega } (f_{3}) (x_{0}) \bigr\vert \\ &\quad\leq \biggl( \int _{0}^{\infty } \biggl( \int _{ \vert x-y \vert < t, \vert x_{0} -y \vert >t} \biggl\vert \frac{\varOmega (x-y)}{ \vert x-y \vert ^{n-1}} f_{3}(y) \biggr\vert \,dy \biggr)^{2} \frac{dt}{t^{3}} \biggr)^{\frac{1}{2}} \\ &\qquad{} + \biggl( \int _{0}^{\infty } \biggl( \int _{ \vert x-y \vert >t, \vert x_{0} -y \vert < t} \biggl\vert \frac{\varOmega (x_{0}-y)}{ \vert x_{0}-y \vert ^{n-1}} f_{3}(y) \biggr\vert \,dy \biggr)^{2} \frac{dt}{t^{3}} \biggr)^{\frac{1}{2}} \\ &\qquad{} + \biggl( \int _{0}^{\infty } \biggl( \int _{ \vert x-y \vert < t, \vert x_{0} -y \vert < t} \biggl\vert \frac{\varOmega (x-y)}{ \vert x-y \vert ^{n-1}} - \frac{\varOmega (x_{0}-y)}{ \vert x_{0}-y \vert ^{n-1}} \biggr\vert \bigl\vert f_{3}(y) \bigr\vert \,dy \biggr)^{2} \frac{dt}{t^{3}} \biggr)^{\frac{1}{2}} \\ &\quad =: I_{1}+I_{2}+I_{3}. \end{aligned}$$

We first estimate $I_{1}$. It is obvious that $\vert y-x\vert \approx \vert y-x_{0} \vert \approx \vert y \vert$, when $x\in B$, $x_{0} \in B$ and $y\in (B^{*})^{c}$. Applying Minkowski’s inequality, we get

$$\begin{aligned} I_{1} \leq {}& \int _{\mathbb{R}^{n}} \frac{ \vert \varOmega (x-y) \vert }{ \vert x-y \vert ^{n-1}} \bigl\vert f_{3}(y) \bigr\vert \biggl( \int _{ \vert x-y \vert < t, \vert x_{0} -y \vert >t} \frac{dt}{t^{3}} \biggr)^{\frac{1}{2}} \,dy \\ \leq {}& C \int _{(B^{*})^{c}} \frac{ \vert \varOmega (x-y) \vert }{ \vert x-y \vert ^{n-1}} \bigl\vert f_{3}(y) \bigr\vert \biggl\vert \frac{1}{ \vert x-y \vert ^{2}} - \frac{1}{ \vert x_{0}-y \vert ^{2}} \biggr\vert ^{\frac{1}{2}} \,dy \\ \leq {}& C \int _{(B^{*})^{c}} \frac{ \vert \varOmega (x-y) \vert }{ \vert x-y \vert ^{n-1}} \bigl\vert f_{3}(y) \bigr\vert \frac{d^{\frac{1}{2}}}{ \vert y \vert ^{\frac{3}{2}}} \,dy \\ ={}& C \int _{(B^{*})^{c}} \frac{ d^{\frac{1}{2}} \vert \varOmega (x-y) \vert }{ \vert y \vert ^{n+\frac{1}{2}}} \bigl\vert f_{3}(y) \bigr\vert \,dy. \end{aligned}$$

We take $\eta >0$, such that $\alpha n p < \eta < \frac{q'}{2}$. Then $\tau = (\frac{1}{2}-\frac{\eta }{q'})q >0$. Using Hölder’s inequality and Lemma 2.3, we have

$$\begin{aligned} I_{1} & \leq C \biggl( \int _{(B^{*})^{c}} \frac{ d^{\tau } \vert \varOmega (x-y) \vert ^{q}}{ \vert y \vert ^{n+\tau }} \,dy \biggr)^{\frac{1}{q}} \biggl( \int _{(B^{*})^{c}} \frac{ d^{\eta } \vert f_{3}(y) \vert ^{q'}}{ \vert y \vert ^{n+\eta }} \,dy \biggr)^{\frac{1}{q'}} \\ & \leq C d^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha, p}}. \end{aligned}$$

Similarly, we can get $I_{2} \leq C d^{n\alpha } \Vert f\Vert _{\mathrm{CL}^{\alpha, p}}$. For the last term $I_{3}$, we use Minkowski’s inequality to get

$$\begin{aligned} I_{3} & \leq \int _{\mathbb{R}^{n}} \biggl\vert \frac{\varOmega (x-y)}{ \vert x-y \vert ^{n-1}} - \frac{\varOmega (x_{0}-y)}{ \vert x_{0}-y \vert ^{n-1}} \biggr\vert \bigl\vert f_{3}(y) \bigr\vert \biggl( \int _{ \vert x-y \vert < t, \vert x_{0} -y \vert < t} \frac{dt}{t^{3}} \biggr)^{\frac{1}{2}} \,dy \\ & \leq C \int _{\mathbb{R}^{n}} \biggl\vert \frac{\varOmega (x-y)}{ \vert x-y \vert ^{n-1}} - \frac{\varOmega (x_{0}-y)}{ \vert x_{0}-y \vert ^{n-1}} \biggr\vert \bigl\vert f_{3}(y) \bigr\vert \biggl( \int _{ \vert x_{0} -y \vert < t}^{\infty } \frac{dt}{t^{3}} \biggr)^{\frac{1}{2}} \,dy \\ & \leq C \int _{\mathbb{R}^{n}} \biggl\vert \frac{\varOmega (x-y)}{ \vert x-y \vert ^{n-1}} - \frac{\varOmega (x_{0}-y)}{ \vert x_{0}-y \vert ^{n-1}} \biggr\vert \frac{ \vert f_{3}(y) \vert }{ \vert y \vert } \,dy \\ & \leq C \sum_{j=3}^{\infty } \int _{2^{j} d< \vert y \vert < 2^{j+1}d } \biggl\vert \frac{\varOmega (x-y)}{ \vert x-y \vert ^{n-1}} - \frac{\varOmega (x_{0}-y)}{ \vert x_{0}-y \vert ^{n-1}} \biggr\vert \frac{ \vert f_{3}(y) \vert }{ \vert y \vert } \,dy \\ & \leq C \sum_{j=3}^{\infty } \frac{ 1}{2^{j} d} \int _{2^{j} d< \vert y \vert < 2^{j+1}d } \biggl\vert \frac{\varOmega (x-y)}{ \vert x-y \vert ^{n-1}} - \frac{\varOmega (x_{0}-y)}{ \vert x_{0}-y \vert ^{n-1}} \biggr\vert \bigl\vert f_{3}(y) \bigr\vert \,dy. \end{aligned}$$

Applying Hölder’s inequality and Lemma 2.2, we have

$$\begin{aligned} I_{3} \leq{}& C \sum_{j=3}^{\infty } \frac{ 1}{2^{j} d} \biggl( \int _{2^{j} d< \vert y \vert < 2^{j+1}d } \biggl\vert \frac{\varOmega (x-y)}{ \vert x-y \vert ^{n-1}} - \frac{\varOmega (x_{0}-y)}{ \vert x_{0}-y \vert ^{n-1}} \biggr\vert ^{q} \,dy \biggr)^{\frac{1}{q}} \\ & {}\times \biggl( \int _{2^{j} d< \vert y \vert < 2^{j+1}d } \bigl\vert f_{3}(y) \bigr\vert ^{q'} \,dy \biggr)^{\frac{1}{q'}} \\ \leq{}& C \sum_{j=3}^{\infty } \frac{1}{2^{j}} \biggl( \frac{1}{(2^{j+1} d)^{n}} \int _{2^{j} d< \vert y \vert < 2^{j+1}d } \bigl\vert f_{3}(y) \bigr\vert ^{q'} \,dy \biggr)^{\frac{1}{q'}} \\ &{} + C\sum_{j=3}^{\infty } \biggl( \int _{ \frac{ \vert x-x_{0} \vert }{2^{j+1} d}} ^{ \frac{ \vert x-x_{0} \vert }{2^{j} d}} \frac{\omega _{q}(\delta )}{\delta } \,d\delta \biggr) \biggl( \frac{1}{(2^{j+1} d)^{n}} \int _{2^{j} d< \vert y \vert < 2^{j+1}d } \bigl\vert f_{3}(y) \bigr\vert ^{q'} \,dy \biggr)^{\frac{1}{q'}} \\ =:{}& I_{31}+ I_{32}. \end{aligned}$$

To estimate $I_{31}$, we take $\eta >0$. Using Lemma 2.3, it yields

$$\begin{aligned} I_{31} & \leq C \sum_{j=3}^{\infty } \frac{1}{2^{j}} \biggl( \int _{2^{j} d< \vert y \vert < 2^{j+1}d } \frac{(2^{j+1}d)^{\eta } \vert f_{3}(y) \vert ^{q'}}{ \vert y \vert ^{n+ \eta } +(2^{j+1} d)^{n+\eta }} \,dy \biggr)^{\frac{1}{q'}} \\ & \leq C \sum_{j=3}^{\infty } \frac{1}{2^{j}} \biggl( \biggl( \int _{2^{j} d< \vert y \vert < 2^{j+1}d } \frac{(2^{j+1}d)^{\eta } \vert f(y)- f_{B_{j}} \vert ^{q'}}{ \vert y \vert ^{n+ \eta } +(2^{j+1} d)^{n+\eta }} \,dy \biggr)^{\frac{1}{q'}} + \vert f_{B_{j}}- f _{B^{*}} \vert \biggr) \\ & \leq C \sum_{j=3}^{\infty } \frac{j}{2^{j}} \bigl(2^{j} d\bigr)^{n \alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha,p}} \\ & \leq C d^{n \alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha,p}}. \end{aligned}$$

As for $I_{32}$, we have

$$\begin{aligned} I_{32} ={}& C \sum_{j=3}^{\infty } \biggl( \int _{ \frac{ \vert x-x_{0} \vert }{2^{j+1} d}} ^{ \frac{ \vert x-x_{0} \vert }{2^{j} d}} \frac{\omega _{q}(\delta )}{\delta } \,d\delta \biggr) \biggl( \frac{1}{(2^{j+1} d)^{n}} \int _{2^{j} d< \vert y \vert < 2^{j+1}d } \bigl\vert f(y) - f_{B^{*}} \bigr\vert ^{q'} \,dy \biggr)^{\frac{1}{q'}} \\ \leq{}& C \sum_{j=3}^{\infty } \biggl( \int _{0} ^{1} \frac{\omega _{q}(\delta )}{\delta } \,d\delta \biggr) \biggl( \frac{1}{(2^{j+1} d)^{n}} \int _{2^{j} d< \vert y \vert < 2^{j+1}d } \bigl\vert f(y) - f_{B^{*}} \bigr\vert ^{q'} \,dy \biggr)^{\frac{1}{q'}} \\ \leq{} &C \sum_{j=3}^{\infty } \biggl( \frac{1}{(2^{j+1} d)^{n}} \int _{2^{j} d< \vert y \vert < 2^{j+1}d } \bigl\vert f(y) - f_{B_{j+1}} \bigr\vert ^{q'} \,dy \biggr)^{\frac{1}{q'}} \\ &{} + C \sum_{j=3}^{\infty } \vert f_{B_{j+1}} -f_{B^{*}} \vert \\ =:{}& I_{32}^{1} + I_{32}^{2}. \end{aligned}$$

Since $q' \leq p$ and using Hölder’s inequality, we obtain

$$\begin{aligned} I_{32}^{1} & \leq C \sum_{j=3}^{\infty } \biggl( \frac{1}{ \vert B_{j+1} \vert } \int _{B_{j+1}} \bigl\vert f(y) - f_{B_{j+1}} \bigr\vert ^{p} \,dy \biggr)^{\frac{1}{p}} \\ & \leq \sum_{j=3}^{\infty } \vert B_{j+1} \vert ^{\alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha, p}} \\ & \leq C d^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha, p}}. \end{aligned}$$

For the last term $I_{32}^{2}$, it is easy to get

$$\begin{aligned} I_{32}^{2} & \leq C \sum_{j=3}^{\infty } (j+1) \bigl(2^{j+1}d\bigr)^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha,p}} \\ & \leq C d^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha, p}}. \end{aligned}$$

Summarizing the above estimate, we conclude that

$$\begin{aligned} \bigl\vert \mu _{\varOmega } (f_{3}) (x)- \mu _{\varOmega } (f_{3}) (x_{0}) \bigr\vert \leq C d^{n\alpha } \Vert f \Vert _{\mathrm{CL}^{\alpha, p}}. \end{aligned}$$

Thus we have

$$\begin{aligned} \mu _{\varOmega } (f) (x)\leq {}& \mu _{\varOmega } (f_{1}) (x)+ \mu _{\varOmega } (f_{2}) (x) \\ &{}+\bigl\vert \mu _{\varOmega } (f_{3}) (x)- \mu _{\varOmega } (f_{3}) (x_{0}) \bigr\vert + \mu _{\varOmega } (f_{3}) (x_{0})< \infty,\quad \text{a.e. on }B. \end{aligned}$$

Because B is any ball centered at the origin, we get $\mu _{\varOmega }(f)(x)<\infty$, a.e. on $\mathbb{R}^{n}$. Finally, we show that $\Vert \mu _{\varOmega }(f) \Vert _{\mathrm{CL}^{\alpha,p}} \leq C \Vert f \Vert _{\mathrm{CL}^{ \alpha,p}}$. In fact, from the above proof we find that there exists an $x_{0} \in B$, such that $\mu _{\varOmega } (f_{3})(x_{0}) <\infty$. Repeating the above proof, we obtain

$$\begin{aligned} & \biggl( \int _{B} \bigl\vert \mu _{\varOmega } f(x)- \mu _{\varOmega } f_{3}(x_{0}) \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}} \\ & \quad\leq C \biggl( \int _{B} \bigl\vert \mu _{\varOmega } f_{2}(x) \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}} + \biggl( \int _{B} \bigl\vert \mu _{\varOmega } f_{3}(x)- \mu _{\varOmega } f_{3}(x_{0}) \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}} \\ &\quad \leq C d^{n\alpha +\frac{n}{p}} \Vert f \Vert _{\mathrm{CL}^{\alpha, p}}. \end{aligned}$$

Taking the supremum over all such B, the proof is complete. □

Now we give the proof of Theorem 1.2.

Proof of Theorem 1.2

It suffices to verify that, for any $f_{j} \in \mathrm{CL} ^{\alpha _{j}, p_{j}}(\mathbb{R}^{n})$$(j=1, \ldots, m)$, if there exists $y_{0} \in \mathbb{R}^{n}$ such that $\mu (\vec{f})(y_{0})<\infty$, then, for any ball $B=B(0,r)\subset \mathbb{R}^{n}$ with $y_{0} \in B$,

$$\begin{aligned} \biggl( \frac{1}{ \vert B \vert } \int _{B} \bigl\vert \mu (\vec{f}) (x)- \bigl( \mu ( \vec{f})\bigr)_{B} \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}} \leq C \vert B \vert ^{\alpha }\prod _{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL} ^{\alpha _{j}, p_{j}}}. \end{aligned}$$

For any $r>0$, we denote

$$ \mu ^{r} (\vec{f}) (x) = \Biggl( \int _{0}^{8r} \Biggl\vert \frac{1}{t^{m}} \int _{(B(0,t))^{m}} \frac{\varOmega (\vec{y})}{ \vert \vec{y} \vert ^{m(n-1)}} \prod _{i=1}^{m} f_{i}(x-y_{i}) \,d\vec{y} \Biggr\vert ^{2} \frac{dt}{t} \Biggr)^{ \frac{1}{2}} $$

and

$$ \mu ^{\infty } (\vec{f}) (x) = \Biggl( \int _{8r}^{\infty } \Biggl\vert \frac{1}{t^{m}} \int _{(B(0,t))^{m}} \frac{\varOmega (\vec{y})}{ \vert \vec{y} \vert ^{m(n-1)}} \prod _{i=1}^{m} f_{i}(x-y_{i}) \,d\vec{y} \Biggr\vert ^{2} \frac{dt}{t} \Biggr)^{ \frac{1}{2}}. $$

Since Ω satisfies the vanishing condition (1.4), for any $x \in B$,

$$\begin{aligned} \mu ^{r} (\vec{f}) (x) &= \mu ^{r} \bigl( \bigl(f_{1} -(f_{1})_{B}\bigr)\chi _{10B}, \ldots, \bigl(f_{m} -(f_{m})_{B} \bigr)\chi _{10B} \bigr) (x) \\ &\leq \mu \bigl(\bigl(f_{1} -(f_{1})_{B}\bigr) \chi _{10B}, \ldots, \bigl(f_{m} -(f_{m})_{B} \bigr) \chi _{10B} \bigr) (x). \end{aligned}$$

We have $\alpha _{1} + \cdots + \alpha _{m} =\alpha$ and $\frac{1}{p}= \frac{1}{p_{1}}+\cdots + \frac{1}{p_{m}}$ with $1< p_{1}, \ldots, p_{m} <\infty$. Using Theorem B, we get

$$\begin{aligned} \biggl( \int _{B} \bigl\vert \mu ^{r} (\vec{f}) (x) \bigr\vert ^{p} \,dx \biggr)^{ \frac{1}{p}} & \leq C \prod _{j=1}^{m} \biggl( \int _{10B} \bigl\vert f_{j}(y_{j})-(f_{j})_{10B} \bigr\vert ^{p_{j}}\,dy_{j} \biggr)^{\frac{1}{p_{j}}} \\ & \leq C \prod_{j=1}^{m} \vert 10B \vert ^{\alpha _{j} + \frac{1}{p_{j}}} \Vert f_{j} \Vert _{\mathrm{CL} ^{\alpha _{j}, p_{j}}} \\ & \leq C \vert B \vert ^{\alpha +\frac{1}{p}} \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL} ^{\alpha _{j}, p_{j}}}. \end{aligned}$$

We notice that

$$\begin{aligned} & \biggl( \frac{1}{ \vert B \vert } \int _{B} \bigl\vert \mu (\vec{f}) (x)- \bigl( \mu ( \vec{f})\bigr)_{B} \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}} \\ &\quad \leq C \biggl( \frac{1}{ \vert B \vert } \int _{B} \Bigl\vert \mu (\vec{f}) (x)- \inf _{y \in B}\mu (\vec{f}) (y) \Bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}} \\ &\quad\leq C \biggl( \frac{1}{ \vert B \vert } \int _{B} \bigl\vert \mu ^{r} ( \vec{f}) (x) \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}} + \biggl( \frac{1}{ \vert B \vert } \int _{B} \sup_{y \in B} \bigl\vert \mu ^{\infty } ( \vec{f}) (x)- \mu ^{\infty } (\vec{f}) (y) \bigr\vert ^{p} \,dx \biggr)^{ \frac{1}{p}} \\ &\quad \leq C \vert B \vert ^{\alpha } \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL} ^{\alpha _{j}, p_{j}}} + \biggl( \frac{1}{ \vert B \vert } \int _{B} \sup_{y \in B} \bigl\vert \mu ^{\infty } (\vec{f}) (x)- \mu ^{\infty } ( \vec{f}) (y) \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}}. \end{aligned}$$

So the proof of Theorem 1.2 reduces to proving that, for any x, $z \in B$,

$$ \bigl\vert \mu ^{\infty } (\vec{f}) (x)- \mu ^{\infty } (\vec{f}) (z) \bigr\vert \leq C \vert B \vert ^{\alpha } \prod _{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL} ^{ \alpha _{j}, p_{j}}}. $$

It is easy to see that

$$\begin{aligned} \bigl\vert \mu ^{\infty } (\vec{f}) (x)- \mu ^{\infty } (\vec{f}) (z) \bigr\vert &= \biggl\vert \biggl( \int _{8r}^{\infty } \bigl\vert F_{t}( \vec{f}) (x) \bigr\vert ^{2} \frac{dt}{t} \biggr)^{\frac{1}{2}} - \biggl( \int _{8r}^{\infty } \bigl\vert F_{t}( \vec{f}) (z) \bigr\vert ^{2} \frac{dt}{t} \biggr)^{\frac{1}{2}} \biggr\vert \\ & \leq \biggl( \int _{8r}^{\infty } \bigl\vert F_{t}( \vec{f}) (x) - F_{t}( \vec{f}) (z) \bigr\vert \bigl\vert F_{t}(\vec{f}) (x) + F_{t}(\vec{f}) (z) \bigr\vert \frac{dt}{t} \biggr)^{\frac{1}{2}}. \end{aligned}$$

For any $z \in B$, $t_{1}, \ldots, t_{m} >r$, we have $B(z, t_{i})\subset B(0, 2t_{i}) (i=1,\ldots, m)$. If $n< p<\infty$ and $- \infty < \alpha <0$, the vanishing condition of Ω and Hölder’s inequality allow us to obtain

$$\begin{aligned} & \Biggl\vert \int _{\prod _{i=1}^{m} B(z,t_{i})} \frac{\varOmega (z-y_{1}, \ldots, z-y_{m})}{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} f_{i}(y_{i}) \,d \vec{y} \Biggr\vert \\ &\quad \leq C \Biggl( \int _{\prod _{i=1}^{m} B(z,t_{i})} \Biggl\vert \prod_{j=1}^{m} \bigl( f_{j}(y_{j})- (f_{j})_{B(0, 2 t_{j})} \bigr) \Biggr\vert ^{p} \,d \vec{y} \Biggr)^{\frac{1}{p}} \\ &\qquad{} \times \biggl( \int _{\prod _{i=1}^{m} B(z,t_{i})} \frac{d \vec{y}}{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)p'}} \biggr)^{\frac{1}{p'}} \\ &\quad \leq C \Biggl( \int _{\prod _{i=1}^{m} B(0,2t_{i})} \Biggl\vert \prod_{j=1}^{m} \bigl( f_{j}(y_{j})- (f_{j})_{B(0, 2 t_{j})} \bigr) \Biggr\vert ^{p} \,d \vec{y} \Biggr)^{\frac{1}{p}} \\ & \qquad{}\times \biggl( \int _{\prod _{i=1}^{m} B(z,t_{i})} \frac{d \vec{y}}{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)p'}} \biggr)^{\frac{1}{p'}} \\ & \quad\leq C \prod_{j=1}^{m} t_{j}^{\frac{n}{p'}-n+1} \prod_{j=1}^{m} \biggl(\prod_{i\neq j} t_{i} ^{\frac{n}{p_{j}}} t_{j} ^{ \frac{n}{p_{j}}+n\alpha _{j}} \biggr) \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j} , p_{j}}} \\ &\quad =C \prod_{j=1}^{m} t_{j}^{\frac{n}{p'}-n+1} \prod_{j=1}^{m} t_{j}^{ \frac{n}{p}+n\alpha _{j}} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}} \\ & \quad= C \prod_{j=1}^{m} t_{j}^{1+n \alpha _{j}} \Vert f_{j} \Vert _{\mathrm{CL}^{ \alpha _{j}, p_{j}}}. \end{aligned}$$

For $z \in B$, $t>8r$, $B(z,t)$ can be decomposed into the following disjoint union:

$$\begin{aligned} \bigl(B(z,t)\bigr)^{m}={}& \Biggl\lbrace \bigcup _{i=1}^{m} \bigl( \bigl( B(z,t) \setminus B(z,8r) \bigr)^{i-1} \times B(z,8r)\times \bigl( B(z,t) \setminus B(z,8r) \bigr)^{m-i} \bigr) \Biggr\rbrace \\ &{} \cup \bigl( B(z,t)\setminus B(z,8r) \bigr)^{m} \cup \bigl( B(z,8r) \bigr)^{m}. \end{aligned}$$

Set $B_{i}(z,t,r):= ( B(z,t)\setminus B(z,8r) )^{i-1} \times B(z,8r)\times ( B(z,t)\setminus B(z,8r) )^{m-i}$. We write

$$\begin{aligned} t^{m} F_{t}(\vec{f}) (z)={}& \int _{ ( B(z,t) )^{m}} \frac{\varOmega (z-y_{1}, \ldots, z-y_{m})}{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} f_{i}(y_{i}) \,d \vec{y} \\ ={}& \int _{ ( B(z,t)\setminus B(z,8r) )^{m}} \frac{\varOmega (z-y_{1}, \ldots, z-y_{m})}{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} f_{i}(y_{i}) \,d \vec{y} \\ & {}+ \sum_{l=1}^{m} \int _{B_{l}(z,t,r)} \frac{\varOmega (z-y_{1}, \ldots, z-y_{m})}{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} f_{i}(y_{i}) \,d \vec{y} \\ &{} + \int _{ ( B(z,8r) )^{m}} \frac{\varOmega (z-y_{1}, \ldots, z-y_{m})}{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} f_{i}(y_{i}) \,d \vec{y}. \end{aligned}$$

According to the above estimate, we have

$$\begin{aligned} &\bigl\vert t^{m} F_{t}(\vec{f}) (z) \bigr\vert \leq C t^{m+n\alpha }\prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}}; \\ & \Biggl\vert \int _{B_{l}(z,t,r)} \frac{\varOmega (z-y_{1}, \ldots, z-y_{m})}{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} f_{i}(y_{i}) \,d \vec{y} \Biggr\vert \\ &\quad\leq C t^{m-1+n(\alpha -\alpha _{l})} r^{1+n\alpha _{l}}\prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}}; \\ &\Biggl\vert \int _{ ( B(z,8r) )^{m}} \frac{\varOmega (z-y_{1}, \ldots, z-y_{m})}{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} f_{i}(y_{i}) \,d \vec{y} \Biggr\vert \leq C r^{m+n\alpha } \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}}. \end{aligned}$$

For x, $z \in B$, $t\geqslant 8r$, we set

$$\begin{aligned} H_{t}(\vec{f}) (x,z) =:{}& \Biggl\vert \int _{ ( B(z,t)\setminus B(z,8r) )^{m}} \frac{\varOmega (z-y_{1}, \ldots, z-y_{m})}{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} f_{i}(y_{i}) \,d \vec{y} \\ &{} - \int _{ ( B(x,t)\setminus B(x,8r) )^{m}} \frac{\varOmega (x-y_{1}, \ldots, x-y_{m})}{(\sum_{j=1}^{m} \vert x-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} f_{i}(y_{i}) \,d \vec{y} \Biggr\vert . \end{aligned}$$

Consequently, we have

$$\begin{aligned} & \bigl\vert \mu ^{\infty } (\vec{f}) (x)- \mu ^{\infty } (\vec{f}) (z) \bigr\vert \\ &\quad\leq C \biggl( \int _{8r}^{\infty } \bigl\vert F_{t}( \vec{f}) (x) - F_{t}( \vec{f}) (z) \bigr\vert \frac{dt}{t^{1-n\alpha }} \biggr)^{\frac{1}{2}} \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}}^{ \frac{1}{2}} \\ & \quad\leq C \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}} \Biggl( \int _{8r}^{\infty } \Biggl(\sum _{j=1}^{m} t^{m-1+n(\alpha - \alpha _{j})} r^{1+n\alpha _{j}} + r^{m+n\alpha } \Biggr) \frac{dt}{t^{m+1-n\alpha }} \Biggr)^{\frac{1}{2}} \\ &\qquad{} + C \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}} ^{\frac{1}{2}} \biggl( \int _{8r}^{\infty } \bigl\vert H_{t}( \vec{f}) (x,z) \bigr\vert \frac{dt}{t^{m+1-n\alpha }} \biggr)^{\frac{1}{2}} \\ &\quad \leq C r^{n\alpha } \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j} , p_{j}}} +C \prod _{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}}^{ \frac{1}{2}} \biggl( \int _{8r}^{\infty } \bigl\vert H_{t}( \vec{f}) (x,z) \bigr\vert \frac{dt}{t^{m+1-n\alpha }} \biggr)^{\frac{1}{2}}. \end{aligned}$$

Fixing x, z and for $t>0$, we introduce some notations:

$$\begin{aligned} &\varXi (x,t)= \bigl\lbrace y \in \mathbb{R}^{n}: 8r \leq \vert x-y \vert < t, 8r \leq \vert z-y \vert < t \bigr\rbrace ; \\ &\varXi (z,t)= \bigl\lbrace y \in \mathbb{R}^{n}: 8r \leq \vert z-y \vert < t, 8r \leq \vert x-y \vert < t \bigr\rbrace ; \\ &\varGamma (x,t)= \bigl\lbrace y \in \mathbb{R}^{n}: 8r \leq \vert x-y \vert < t , \vert z-y \vert \geq t \bigr\rbrace ; \\ &\varGamma (z,t)= \bigl\lbrace y \in \mathbb{R}^{n}: 8r \leq \vert z-y \vert < t , \vert x-y \vert \geq t \bigr\rbrace ; \\ &\varLambda (x,t)= \bigl\lbrace y \in \mathbb{R}^{n}: 8r \leq \vert x-y \vert < t , \vert z-y \vert < 8r \bigr\rbrace ; \\ &\varLambda (z,t)= \bigl\lbrace y \in \mathbb{R}^{n}: 8r \leq \vert z-y \vert < t , \vert x-y \vert < 8r \bigr\rbrace ; \\ &\vec{\varTheta } (x,t)= \varTheta _{1}(x,t) \times \cdots \times \varTheta _{m} (x,t), \varTheta _{i}(x,t) \in \bigl\lbrace \varXi (x,t), \varGamma (x,t), \varLambda (x,t) \bigr\rbrace ; \\ &\vec{\varTheta } (z,t)= \varTheta _{1}(z,t) \times \cdots \times \varTheta _{m} (z,t), \varTheta _{i}(z,t) \in \bigl\lbrace \varXi (z,t), \varGamma (z,t), \varLambda (z,t) \bigr\rbrace ; \\ &\varXi (x,t)=\varXi (z,t) =: \varXi (t). \end{aligned}$$

For y, we denote

$$\begin{aligned} &\varXi (x,y)= \bigl\lbrace t>0: 8r \leq \vert x-y \vert < t, 8r \leq \vert z-y \vert < t \bigr\rbrace ; \\ &\varXi (z,y)= \bigl\lbrace t>0: 8r \leq \vert z-y \vert < t, 8r \leq \vert x-y \vert < t \bigr\rbrace ; \\ &\varGamma (x,y)= \bigl\lbrace t>0: 8r \leq \vert x-y \vert < t, \vert z-y \vert \geq t \bigr\rbrace ; \\ &\varGamma (z,y)= \bigl\lbrace t>0: 8r \leq \vert z-y \vert < t, \vert x-y \vert \geq t \bigr\rbrace ; \\ &\varLambda (x,y)= \bigl\lbrace t>0: 8r \leq \vert x-y \vert < t, \vert z-y \vert < 8r \bigr\rbrace ; \\ &\varLambda (z,y)= \bigl\lbrace t>0: 8r \leq \vert z-y \vert < t, \vert x-y \vert < 8r \bigr\rbrace ; \\ &\varTheta _{i}(x, y_{i}) \in \bigl\lbrace \varXi (x,y_{i}), \varGamma (x,y_{i}) \bigr\rbrace ,\quad i=1, \ldots, m; \\ &\varTheta _{i}(z, y_{i}) \in \bigl\lbrace \varXi (z,y_{i}), \varGamma (z,y_{i}) \bigr\rbrace ,\quad i=1, \ldots, m. \end{aligned}$$

It is easy to see that

$$ B(x,t)\setminus B(x,8r) = \varXi (x,t) \cup \varGamma (x,t) \cup \varLambda (x,t) $$

and

$$ B(z,t)\setminus B(z,8r) = \varXi (z,t) \cup \varGamma (z,t) \cup \varLambda (z,t). $$

We write

$$\begin{aligned} H_{t}(\vec{f}) (x,z)\leq{}& \int _{(\varXi (t))^{m}} \biggl\vert \frac{\varOmega (x-y_{1}, \ldots, x-y_{m})}{(\sum_{j=1}^{m} \vert x-y_{j} \vert )^{m(n-1)}} - \frac{\varOmega (z-y_{1}, \ldots, z-y_{m})}{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)}} \biggr\vert \\ &{} \times \prod_{i=1}^{m} \bigl\vert f_{i}(y_{i}) -(f_{i})_{B} \bigr\vert \,d \vec{y} \\ &{}+ \int _{(\varLambda (x,t))^{m}} \frac{ \vert \varOmega (x-y_{1}, \ldots, x-y_{m}) \vert }{(\sum_{j=1}^{m} \vert x-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} \bigl\vert f_{i}(y_{i}) -(f_{i})_{B} \bigr\vert \,d\vec{y} \\ &{}+ \int _{(\varLambda (z,t))^{m}} \frac{ \vert \varOmega (z-y_{1}, \ldots, z-y_{m}) \vert }{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} \bigl\vert f_{i}(y_{i}) -(f_{i})_{B} \bigr\vert \,d\vec{y} \\ &{}+ \int _{\vec{\varTheta }(x,t), \exists \varTheta _{i}(x,t)= \varGamma (x,t)} \frac{ \vert \varOmega (x-y_{1}, \ldots, x-y_{m}) \vert }{(\sum_{j=1}^{m} \vert x-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} \bigl\vert f_{i}(y_{i}) -(f_{i})_{B} \bigr\vert \,d\vec{y} \\ &{}+ \int _{\vec{\varTheta }(z,t), \exists \varTheta _{i}(z,t)= \varGamma (z,t)} \frac{ \vert \varOmega (z-y_{1}, \ldots, z-y_{m}) \vert }{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} \bigl\vert f_{i}(y_{i}) -(f_{i})_{B} \bigr\vert \,d\vec{y} \\ &{}+ \sum_{l=1}^{m-1} \int _{(\varXi (x,t))^{l}} \int _{(\varLambda (x,t))^{m-l}} \frac{ \vert \varOmega (x-y_{1}, \ldots, x-y_{m}) \vert }{(\sum_{j=1}^{m} \vert x-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} \bigl\vert f_{i}(y_{i}) -(f_{i})_{B} \bigr\vert \,d\vec{y} \\ &{}+ \sum_{l=1}^{m-1} \int _{(\varXi (z,t))^{l}} \int _{(\varLambda (z,t))^{m-l}} \frac{ \vert \varOmega (z-y_{1}, \ldots, z-y_{m}) \vert }{(\sum_{j=1}^{m} \vert z-y_{j} \vert )^{m(n-1)}} \prod _{i=1}^{m} \bigl\vert f_{i}(y_{i}) -(f_{i})_{B} \bigr\vert \,d\vec{y} \\ =:{}& \sum_{i=1}^{5} H_{t,i}( \vec{f}) (x,z) +\sum_{l=1}^{m-1} H^{l}_{t,6}( \vec{f}) (x,z)+ \sum _{l=1}^{m-1} H^{l}_{t,7} ( \vec{f}) (x,z). \end{aligned}$$

For x, $z \in B$ and Ω satisfying Lipschitz continuous condition, applying Lemma 2.7, we get

$$\begin{aligned} &\int _{8r}^{\infty } \bigl\vert H_{t,1}( \vec{f}) (x,z) \bigr\vert \frac{dt}{t^{m+1-n\alpha }} \\ &\quad \leq C \int _{((B(x,8r))^{c})^{m}} \frac{ \vert x-z \vert ^{\beta }}{(\sum_{j=1}^{m} \vert x-y_{j} \vert )^{m(n-1)+\beta }} \\ & \qquad{}\times \prod_{i=1}^{m} \bigl\vert f_{i}(y_{i}) -(f_{i})_{B} \bigr\vert \,d\vec{y} \int _{\frac{1}{m} (\sum _{j=1}^{m} \vert x-y_{j} \vert )} \frac{dt}{t^{m+1-n\alpha }} \,d \vec{y} \\ &\quad \leq C \int _{((B(x,8r))^{c})^{m}} \frac{r ^{\beta }\prod_{i=1}^{m} \vert f_{i}(y_{i}) -(f_{i})_{B} \vert }{(\sum_{j=1}^{m} \vert x-y_{j} \vert )^{mn+\beta -n\alpha }} \,d \vec{y} \\ &\quad \leq C r^{2n\alpha } \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j} , p_{j}}}. \end{aligned}$$

When x, $z \in B$, $8r\leq \vert x-y\vert < t$ and $\vert z-y\vert <8r$, we have $\vert x-y \vert < \vert x-z \vert + \vert z-y\vert <2r+8r=10r$. Then $8r < \vert x-y \vert <10r$. Applying Lemma 2.6, we get

$$\begin{aligned} \bigl\vert H_{t,2} (\vec{f}) (x,z) \bigr\vert &\leq C \int _{ ( B(x,10r) \setminus B(x,8r) )^{m}} \frac{ \prod_{i=1}^{m} \vert f_{i}(y_{i}) -(f_{i})_{B} \vert }{(\sum_{j=1}^{m} \vert x-y_{j} \vert )^{m(n-1)}} \,d \vec{y} \\ & \leq \prod_{i=1}^{m} \int _{8r \leq \vert x-y_{i} \vert \leq 10r} \frac{ \vert f_{i}(y_{i})-(f_{i})_{B} \vert }{ \vert x-y_{i} \vert ^{n-1}} \,dy_{i} \\ &\leq C r^{m+n\alpha } \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j} , p_{j}}}, \end{aligned}$$

which leads to

$$\begin{aligned} \int _{8r}^{\infty } \bigl\vert H_{t,2} ( \vec{f}) (x,z) \bigr\vert \frac{dt}{t^{m+1-n\alpha }} \leq C r^{2n\alpha } \prod _{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}}. \end{aligned}$$

Similarly,

$$\begin{aligned} \int _{8r}^{\infty } \bigl\vert H_{t,3} ( \vec{f}) (x,z) \bigr\vert \frac{dt}{t^{m+1-n\alpha }} \leq C r^{2n\alpha } \prod _{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}}. \end{aligned}$$

Now we estimate $H_{t,4} (\vec{f})$. For any x, $z \in B$,

$$\begin{aligned} \Biggl\vert \bigcap_{j=1}^{m} \varTheta _{j}(x, y_{j}) \Biggr\vert \leq \bigl\vert \varTheta _{i} (x, y_{i}) \bigr\vert = \bigl\vert \varGamma (x, y_{i}) \bigr\vert \leq \bigl\vert \vert z-y_{i} \vert - \vert x-y_{i} \vert \bigr\vert \leq \vert z-x \vert \leq 2r. \end{aligned}$$

And for any $t \in \bigcap_{j=1}^{m} \varTheta _{j}(x, y_{j})$, $t> \frac{1}{m} (\sum_{j=1}^{m} \vert x-y_{j} \vert )$. Applying Lemma 2.7, we get

$$\begin{aligned} &\int _{8r}^{\infty } \bigl\vert H_{t,4} ( \vec{f}) (x,z) \bigr\vert \frac{dt}{t^{m+1-n\alpha }} \\ &\quad\leq C \int _{((B(x,8r))^{c})^{m}} \frac{ \prod_{i=1}^{m} \vert f_{i}(y_{i}) -(f_{i})_{B} \vert }{(\sum_{j=1}^{m} \vert x-y_{j} \vert )^{m(n-1)}} \,d \vec{y} \\ &\qquad{} \times \int _{\bigcap _{j=1}^{m} \varTheta _{j}(x, y_{j})} \frac{dt}{t^{m+1-n\alpha }} \\ &\quad \leq C r \int _{((B(x,8r))^{c})^{m}} \frac{ \prod_{i=1}^{m} \vert f_{i}(y_{i}) -(f_{i})_{B} \vert }{(\sum_{j=1}^{m} \vert x-y_{j} \vert )^{mn+1-n\alpha }} \,d \vec{y} \\ &\quad \leq Cr^{2n\alpha } \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j} , p_{j}}}. \end{aligned}$$

Similarly,

$$\begin{aligned} \int _{8r}^{\infty } \bigl\vert H_{t,5} ( \vec{f}) (x,z) \bigr\vert \frac{dt}{t^{m+1-n\alpha }} \leq C r^{2n\alpha } \prod _{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}}. \end{aligned}$$

As for $H_{t,6}^{l} (\vec{f})$, suppose $y_{1}, \ldots, y_{l} \in \varXi (t)$, $y_{l+1}, \ldots, y_{m} \in \varLambda (x,t)$. Using Lemma 2.6, we have

$$\begin{aligned} H_{t,6}^{l} (\vec{f}) (x,z) \leq {}&\prod _{j=1}^{l} \int _{8r \leq \vert x-y_{j} \vert < t} \frac{ \vert f_{j}(y_{j}) -(f_{j})_{B} \vert }{ \vert x-y_{j} \vert ^{n-1}} \,dy_{j} \\ & {}\times \prod_{j=l+1}^{m} \int _{8r \leq \vert x-y_{j} \vert \leq 10r} \frac{ \vert f_{j}(y_{j}) -(f_{j})_{B} \vert }{ \vert x-y_{j} \vert ^{n-1}} \,dy_{j} \\ \leq {}&C \prod_{j=1}^{l} \bigl( \mathit{tr}^{n\alpha _{j}}\bigr) \Vert f_{j} \Vert _{\mathrm{CL}^{ \alpha _{j}, p_{j}}} \prod_{j=l+1}^{m} r^{1+n\alpha _{j}} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}} \\ = {}&C t^{l}r^{m-l+n\alpha } \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{ \alpha _{j}, p_{j}}}. \end{aligned}$$

So we obtain

$$\begin{aligned} \int _{8r}^{\infty } \bigl\vert H_{t,6}^{l} (\vec{f}) (x,z) \bigr\vert \frac{dt}{t^{m+1-n\alpha }} &\leq C r^{m-l+n\alpha } \int _{8r}^{\infty } t^{l-m-1+n\alpha } \,dt \prod _{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j} , p_{j}}} \\ & \leq C r^{2n\alpha } \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j} , p_{j}}}. \end{aligned}$$

Similarly,

$$\begin{aligned} \int _{8r}^{\infty } \bigl\vert H_{t,7}^{l} (\vec{f}) (x,z) \bigr\vert \frac{dt}{t^{m+1-n\alpha }}\leq C r^{2n\alpha } \prod _{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}}. \end{aligned}$$

Combining the above estimates, we have

$$\begin{aligned} \biggl( \int _{8r}^{\infty } \bigl\vert H_{t} ( \vec{f}) (x,z) \bigr\vert \frac{dt}{t^{m+1-n\alpha }} \biggr)^{\frac{1}{2}} \leq C r^{n\alpha } \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j}, p_{j}}}^{ \frac{1}{2}}. \end{aligned}$$

Consequently, we have

$$\begin{aligned} \bigl\vert \mu ^{\infty } (\vec{f}) (x)- \mu ^{\infty } (\vec{f}) (z) \bigr\vert \leq C r^{n\alpha } \prod_{j=1}^{m} \Vert f_{j} \Vert _{\mathrm{CL}^{\alpha _{j} , p_{j}}}. \end{aligned}$$

Then the proof is complete. □