Weighted Trudinger inequality associated with rough multilinear fractional type operators

Abstract

Let $I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}$ be the multilinear fractional type operator defined by $I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)={\int }_{{\mathbb{R}}^n}\mathrm{\Omega }(y){\prod }_{j=1}^m{f}_j(x-{\theta }_{j}y){|y|}^{(\alpha -n)}dy$. In this paper, we study the weighted estimates for the Trudinger inequality associated to $I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}$ with rough homogeneous kernels, which improve some known results significantly. A similar Trudinger inequality holds for another type of fractional integral defined by ${\overline{I}}_{\mathrm{\Omega },\alpha }(\overrightarrow{f})(x)={\int }_{{({\mathbb{R}}^n)}^m}\frac{{\prod }_{j=1}^m|f_j(y_j)||{\mathrm{\Omega }}_j(x-y_j)|}{{|(x-y_1,x-y_2,\dots ,x-y_m)|}^{mn-\alpha }}d\overrightarrow{y}$, where $d\overrightarrow{y}=d{y}_1\cdots d{y}_m$.

1 Introduction

The Trudinger inequality (also sometimes called the Moser-Trudinger inequality) is named after N. Trudinger who first put forward this inequality in [22]. Later, J. Moser [14] gave a sharp form of this Trudinger inequality. It provides an inequality between a certain Sobolev space norm and an Orlicz space norm of a function. In [14], J. Moser gave the largest positive number ${\beta }_0$, such that if $u\in C^1({\mathbb{R}}^n)$, normalized and supported in a domain D with finite measure in ${\mathbb{R}}^n$, such that ${\int }_D{|\mathrm{\nabla }u(x)|}^{n}dx\le 1$, then there is a constant $c_0$ depending only on n such that for all $\beta \le {\beta }_0=n{w}_{n-1}^{1/(n-1)}$, where $w_{n-1}$ is the area of the surface of the unit n-ball. The following inequality holds:

$${\int }_{D}exp\left(\beta {|u(x)|}^{n/(n-1)}\right)dx\le c_0|D|.$$

(1.1)

In 1971, D. Adams [1] considered the similar inequality of J. Moser for higher order derivatives. The key, for him, was to write the function u as a potential $I_{\alpha }$ (see the definition below) and prove the analogue of (1.1) as follows:

$${\int }_{D}exp\left(\frac{n}{w_{n-1}}{\left|\frac{I_{\alpha }f(x)}{{\parallel f\parallel }_p}\right|}^{n/(n-\alpha )}\right)dx\le c_0|D|,\text{for }\alpha =n/p,f\in L^p(1<p<\mathrm{\infty }).$$

(1.2)

Variant forms of the Trudinger inequality as a generalization of the classical results, especially in the literature associated with multilinear Riesz potential or multilinear fractional integral, have been studied in recently years (see, for example, [2, 3, 6, 7, 10, 14, 16–18, 20, 21]). This kind of inequality plays an important role in Harmonic analysis and other fields, such as PDE.

We begin by introducing a class of multilinear maximal function and multilinear fractional integral operators. Suppose that $n\ge 2$, $0<\alpha <n$, Ω is homogeneous of degree zero, and $\mathrm{\Omega }\in L^s(S^{n-1})$ ($s>1$), where $S^{n-1}$ denotes the unit sphere of ${\mathbb{R}}^n$. The multilinear maximal function and multilinear fractional integral is defined by

$$I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)={\int }_{{\mathbb{R}}^n}\mathrm{\Omega }(y)\prod _{j=1}^m{f}_j(x-{\theta }_{j}y){|y|}^{(\alpha -n)}dy$$

(1.3)

and the fractional maximal operator $M_{\mathrm{\Omega },\alpha }$ defined by

$$M_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)=\underset{r>0}{sup}\frac{1}{r^{n-\alpha }}{\int }_{|y|<r}|\mathrm{\Omega }(y)|\prod _{j=1}^m|f_j(x-{\theta }_{j}y)|dy.$$

(1.4)

Multilinear fractional integral $I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}$ can be looked at as a natural generalization of the classical fractional integral, which has a very profound background of partial differential equations and is a very important operator in Harmonic analysis. In fact, if we take $K=1$, ${\theta }_j=1$, and $\mathrm{\Omega }=1$, then $I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}$ is just the well-known classical fractional integral operator studied by Muckenhoupt and Wheeden in [15]. We denote it by $I_{\alpha }$. If $\mathrm{\Omega }\equiv 1$, we simply denote $I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}=I_{\alpha }^{\mathrm{\Theta }}$. In recent years, the study of the Trudinger inequality associated to multilinear type operators has received increasing attention. Among them, it is well known that Grafakos considered the boundedness of a family of related fractional integrals in [7]. After that, in [6], Y. Ding and S. Lu gave the following Trudinger inequality with rough kernels.

Theorem A ([6])

Let $0<\alpha <n$, $s=\frac{n}{\alpha }$, $\frac{1}{s}=\frac{1}{p_1}+\frac{1}{p_2}+\cdots +\frac{1}{p_m}$, $p_j>1$, $j=1,2,\dots ,m$, $m\ge 2$. Denote B as a ball with a radius R in ${\mathbb{R}}^n$. If $f_j\in L^{p_j}(B)$, $supp(f_j)\subset B$, and $\mathrm{\Omega }\in L^{n/(n-\alpha )}(S^{n-1})$, then for any $\gamma <1$, there is a constant C, independent of n, α, ${\theta }_j$, γ, such that

$${\int }_{B}exp\left(n\gamma {\left(\frac{L{I}_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)}{{\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}}{\prod }_{j=1}^m{\parallel f_j\parallel }_{L^{p_j}}}\right)}^{n/(n-\alpha )}\right)dx\le C{R}^n,$$

where $L={\prod }_{j=1}^m{|{\theta }_j|}^{n/p_j}$, $\mathrm{\Theta }=({\theta }_1,{\theta }_2,\dots ,{\theta }_m)$, $\overrightarrow{f}=(f_1,f_2,\dots ,f_m)$ and

$${\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}}={({\int }_{S^{n-1}}{|\mathrm{\Omega }(x)|}^{n/(n-\alpha )}d\sigma (x))}^{(n-\alpha )/n}.$$

The definition of multiple weights $A_{\overrightarrow{p},q}$ was given in [5] and [13] independently, including some weighted estimates for a class of multilinear fractional type operators. These results together with [12] answered an open problem in [8], namely the existence of the multiple weights.

In 2010, W. Li, Q. Xue, and K. Yabuta [16] obtained the weighted estimates for the Trudinger inequality associated to $I_{\alpha }^{\mathrm{\Theta }}$ as follows.

Theorem B ([16])

Let $0<\alpha <n$, $s=\frac{n}{\alpha }$, $\frac{1}{s}=\frac{1}{p_1}+\frac{1}{p_2}+\cdots +\frac{1}{p_m}$, $p_j>1$, ${\omega }_j(x)\in A_{p_j}$, and ${\omega }_j\ge 1$, $j=1,2,\dots ,m$, $m\ge 2$, ${\nu }_{\overrightarrow{\omega }}={\prod }_{j=1}^m{\omega }_j^{s/p_j}$. Denote B as a ball with the radius R in ${\mathbb{R}}^n$, if $f_j\in L_{{\omega }_j}^{p_j}(B)$, $supp(f_j)\subset B$, $j=1,2,\dots ,m$, then for any $\gamma <1$, there is a constant C, independent of n, α, ${\theta }_j$, γ, such that

$${\int }_{B}exp\left(\frac{n}{{\omega }_{n-1}}\gamma {\left(\frac{L{I}_{\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)}{{\prod }_{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{n/(n-\alpha )}\right){\nu }_{\overrightarrow{\omega }}dx\le C\prod _{j=1}^m{\omega }_j(B),$$

where $L={\prod }_{j=1}^m{|{\theta }_j|}^{n/p_j}$, $\mathrm{\Theta }=({\theta }_1,{\theta }_2,\dots ,{\theta }_m)$, $\overrightarrow{f}=(f_1,f_2,\dots ,f_m)$.

On the other hand, in 1999, Kenig and Stein [11] considered another more general type of multilinear fractional integral which was defined by

$$I_{\alpha ,A}(\overrightarrow{f})(x)={\int }_{{({\mathbb{R}}^n)}^m}\frac{1}{{|(y_1,\dots ,y_m)|}^{mn-\alpha }}\prod _{i=1}^m{f}_i({\ell }_i(y_1,\dots ,y_m,x))d{y}_i,$$

where ${\ell }_i$ is a linear combination of $y_j$s and x depending on the matrix A. They showed that $I_{\alpha ,A}$ was of strong type $(L^{p_1}\times \cdots \times L^{p_m},L^q)$ and weak type $(L^{p_1}\times \cdots \times L^{p_m},L^{q,\mathrm{\infty }})$. When ${\ell }_i(y_1,\dots ,y_m,x)=x-y_i$, we denote this multilinear fractional type operator by ${\overline{I}}_{\alpha }$. In 2008, L. Tang [20] obtained the estimation of the exponential integrability of the above operator ${\overline{I}}_{\alpha }$, which is quite similar to Theorem B.

Thus, it is natural to ask whether Theorem B is true or not for $I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}$ with rough kernels. Moreover, one may ask if Theorem B still holds or not for the operator with rough kernels defined by

$${\overline{I}}_{\mathrm{\Omega },\alpha }(\overrightarrow{f})(x)={\int }_{{({\mathbb{R}}^n)}^m}\frac{{\prod }_{j=1}^m|f_j(y_j)||{\mathrm{\Omega }}_j(x-y_j)|}{{|(x-y_1,x-y_2,\dots ,x-y_m)|}^{mn-\alpha }}d\overrightarrow{y}.$$

Inspired by the works above, in this paper, we study the Trudinger inequality associated to multilinear fractional integral operators $I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}$ and ${\overline{I}}_{\mathrm{\Omega },\alpha }$ with rough homogeneous kernels. Precisely, we obtain the following theorems, which give a positive answer to the above questions.

Theorem 1.1 Let $0<\alpha <n$, $s=\frac{n}{\alpha }$, $\frac{1}{s}=\frac{1}{p_1}+\frac{1}{p_2}+\cdots +\frac{1}{p_m}$, $p_j>1$, $j=1,2,\dots ,m$, $m\ge 2$. Denote B as a ball with radius R in ${\mathbb{R}}^n$; if $f_j\in L_{{\omega }_j}^{p_j}(B)$, $supp(f_j)\subset B$ ($j=1,2,\dots ,m$), $\mathrm{\Omega }\in L^{n/(n-\alpha )}(S^{n-1})$, and ${\nu }_{\overrightarrow{\omega }}={\prod }_{j=1}^m{\omega }_j^{\frac{s}{p_j}}$, where ${\omega }_j\in A_s$, ${\omega }_j\ge 1$. Then for any $\gamma <1$, there is a constant C, independent of n, α, ${\theta }_j$, γ, such that

$${\int }_{B}exp\left(n\gamma {\left(\frac{L{I}_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)}{{\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}}{\prod }_{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{n/(n-\alpha )}\right){\nu }_{\overrightarrow{\omega }}dx\le C\prod _{j=1}^m{\omega }_j(B),$$

where $L={\prod }_{j=1}^m{|{\theta }_j|}^{n/p_j}$, $\mathrm{\Theta }=({\theta }_1,{\theta }_2,\dots ,{\theta }_m)$, $\overrightarrow{f}=(f_1,f_2,\dots ,f_K)$.

Remark 1.1 If we take $\mathrm{\Omega }=1$, then Theorem 1.1 coincides with Theorem B. If $w_j\equiv 1$ for $j=1,\dots ,K$, then Theorem 1.1 is just Theorem A that appeared in [6]. We give an example of ${\nu }_{\overrightarrow{\omega }}$ as follows: Let ${\omega }_j(x)={(1+|x|)}^{{\alpha }_j}$ (${\alpha }_j\ge 0$ for each j), then ${\nu }_{\omega }(x)$ satisfy the conditions of the above Theorem 1.1.

Remark 1.2 Assume $m=1$, ${\omega }_j=1$. If $\alpha =1$, Trudinger [20] proved exponential integrability of $I_{\alpha }(\overrightarrow{f})$, and Strichartz [19] for other α. In 1972, Hedberg [9] gave a simpler proof for all α. In 1970, Hempel-Morris-Trudinger [10] showed that if $\gamma >1$, for $\alpha =1$ the inequality in Theorem 1.1 cannot hold, and later Adams [1] obtained the same conclusion for all α; meanwhile, in the endpoint case $\gamma =1$, it is true. In 1985, Chang and Marshall [4] proved a similar sharp exponential inequality concerning the Dirichlet integral. Assume $m\ge 2$, $w_j=1$, then the result was obtained by Grafakos [7] as we have already mentioned above.

Corollary 1.2 Let B, $f_j$, $p_j$, s, and ${\nu }_{\overrightarrow{\omega }}$ be the same as in Theorem  1.1, then $I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})$ is in $L^q({\nu }_{\overrightarrow{\omega }}(B))$ for every $q>0$, that is,

$${\parallel I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})\parallel }_{L^q({\nu }_{\overrightarrow{\omega }}(B))}\le C{\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}(S^{n-1})}\prod _{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}$$

for some constant C depending only on q on n on α and on the ${\theta }_j$ ’s.

Theorem 1.3 Let $m\ge 2$, $0<\alpha <mn$, $1/p=1/p_1+1/p_2+\cdots +1/p_m=\alpha /n$ with $1<p_i<\mathrm{\infty }$ for $i=1,2,\dots ,m$. Let B be a ball with radius R in ${\mathbb{R}}^n$ and let $f_j\in L^{p_j}(B)$ be supported in B, and if ${\mathrm{\Omega }}_j$ is homogeneous of degree zero, and ${\mathrm{\Omega }}_j\in L^{p_j^{\prime }}(S^{n-1})$, where $S^{n-1}$ denotes the sphere of ${\mathbb{R}}^n$, and ${\nu }_{\overrightarrow{\omega }}(\overrightarrow{y})={\prod }_{j=1}^m{\omega }_j^{1/p_j}(y_j)$, where $\overrightarrow{y}=(y_1,y_2,\dots ,y_m)$ and ${\omega }_j\in A_s$, ${\omega }_j\ge 1$. Then there exist constants $k_1$, $k_2$ depending only on n, m, α, p, and the $p_j$ such that

$${\int }_{B}exp\left(k_1{\left(\frac{|{\overline{I}}_{\mathrm{\Omega },\alpha }(\overrightarrow{f})(x)|}{{\prod }_{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{n/(mn-\alpha )}\right){\nu }_{\overrightarrow{\omega }}(x)dx\le k_2\prod _{j=1}^m{\omega }_j(B).$$

Remark 1.3 If we take $\mathrm{\Omega }=1$, $w_j\equiv 1$ for $j=1,\dots ,m$, then Theorem 1.3 is just as Theorem 1.3 appeared in [20]. But there is something that needs to be changed in the proof of Theorem 1.3 in [20]. In the case $r_1=r_2=\cdots =r_{m-1}=0$, one cannot obtain the conclusion that $F_2\le C_2{[log\frac{2\sqrt{m}R}{\delta }]}^{(mn-\alpha )/n}$. Thus, our proof gives an alternative correction of Theorem 1.3 in [20].

Corollary 1.4 Let B, $f_j$, $p_j$, s, and ${\nu }_{\overrightarrow{\omega }}$ be the same as in Theorem  1.3. Then ${\overline{I}}_{\mathrm{\Omega },\alpha }(\overrightarrow{f})$ is in $L^q({\nu }_{\overrightarrow{\omega }}(B))$ for every $q>0$, that is,

$${\parallel {\overline{I}}_{\mathrm{\Omega },\alpha }(\overrightarrow{f})\parallel }_{L^q({\nu }_{\overrightarrow{\omega }}(B))}\le C\prod _{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}$$

for some constant C depending only on q on n on α.

Corollary 1.2 and Corollary 1.4 follow since exponential integrability of ${\overline{I}}_{\mathrm{\Omega },\alpha }(\overrightarrow{f})$ implies integrability to any power q.

On the other hand, we shall study the boundedness of the multilinear fractional maximal operator with a weighted norm. It follows the following theorem.

Theorem 1.5 If $1<p_j<\mathrm{\infty }$, $\frac{1}{s}={\sum }_{j=1}^m\frac{1}{p_j}$, $\frac{1}{r}=\frac{1}{s}-\frac{\alpha }{n}$, ${\omega }_j^{\frac{p_j}{s}}\in A(s,\frac{s{r}_j}{p_j})$, $1/r_j=1/p_j(1-\alpha s/n)$, $j=1,2,\dots ,m$, ${\nu }_{\overrightarrow{\omega }}={\prod }_{j=1}^m{\omega }_j$, then there is a constant C, independent $f_j$, such that

$${\left({\int }_{{\mathbb{R}}^n}{(M_{1,\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x){\nu }_{\overrightarrow{\omega }}(x))}^{r}dx\right)}^{\frac{1}{r}}\le C\prod _{j=1}^m{\left({\int }_{{\mathbb{R}}^n}{|f_j(x){\omega }_j(x)|}^{p_j}dx\right)}^{\frac{1}{p_j}},$$

where $\overrightarrow{f}=(f_1,f_2,\dots ,f_m)$, $f_j\in L_{{\omega }_j}^{p_j}({\mathbb{R}}^n)$.

2 The proof of Theorem 1.1

In this section, we will prove Theorem 1.1.

Proof For any $\delta >0$,

$$|I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)|\le C{\delta }^{\alpha }{M}_{\mathrm{\Omega }}(\overrightarrow{f})(x)+{\int }_{|y|\ge \delta }\frac{|\mathrm{\Omega }(y)|}{{|y|}^{n-\alpha }}\prod _{j=1}^m{f}_j(x-{\theta }_{j}y)dy.$$

Set $P=2min\{\frac{1}{{\theta }_j}:j=1,2,\dots ,K\}$. For any $R>0$, denote $B(R)$ as a ball with radius R in ${\mathbb{R}}^n$, then for any $x\in B(R)$, when $|x-{\theta }_{j}y|<R$, $|{\theta }_{j}y|<2R$ for $j=1,\dots ,m$. Therefore, $|y|<RP$. So,

$${\int }_{|y|\ge \delta }\prod _{j=1}^m{f}_j(x-{\theta }_{j}y){|y|}^{\alpha -n}dy={\int }_{\delta \le |y|<PR}\prod _{j=1}^m{f}_j(x-{\theta }_{j}y){|y|}^{\alpha -n}dy.$$

According to the relationship between s and $p_j$: $\frac{1}{p_1}+\frac{1}{p_2}+\cdots +\frac{1}{p_m}+\frac{1}{n/(n-\alpha )}=1$, from the Hölder’s inequality and ${\nu }_{\overrightarrow{\omega }}\ge 1$, it follows that

$$\begin{array}{c}{\int }_{\delta \le |y|<PR}\mathrm{\Omega }(y)\prod _{j=1}^m{f}_j(x-{\theta }_{j}y){|y|}^{\alpha -n}dy\hfill \\ \le {\left({\int }_{\delta \le |y|\le PR}{(\prod _{j=1}^m{f}_j(x-{\theta }_{j}y))}^{s}dy\right)}^{1/s}{\left({\int }_{\delta \le |y|\le PR}{\left(\frac{|\mathrm{\Omega }(y)|}{{|y|}^{n-\alpha }}\right)}^{s^{\prime }}dy\right)}^{1/s^{\prime }}\hfill \\ \le {({\int }_{\delta \le |y|\le PR}\prod _{j=1}^m{f}_j{(x-{\theta }_{j}y)}^s{\nu }_{\overrightarrow{\omega }}(x-{\theta }_{j}y)dy)}^{1/s}{\parallel \mathrm{\Omega }\parallel }_{L^{s^{\prime }}}{(ln\frac{PR}{\delta })}^{\frac{n-\alpha }{n}}\hfill \\ \le \prod _{j=1}^m{({\int }_{\delta \le |y|\le PR}{|f_j(x-{\theta }_{j}y)|}^{p_j}{\omega }_j(x-{\theta }_{j}y)dy)}^{\frac{1}{p_j}}{\parallel \mathrm{\Omega }\parallel }_{L^{s^{\prime }}}{(\frac{1}{n}ln{\left(\frac{PR}{\delta }\right)}^n)}^{\frac{n-\alpha }{n}}\hfill \\ \le L^{-1}\prod _{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}{\parallel \mathrm{\Omega }\parallel }_{L^{s^{\prime }}}{(\frac{1}{n}ln{\left(\frac{PR}{\delta }\right)}^n)}^{\frac{n-\alpha }{n}}.\hfill \end{array}$$

Hence, we obtain that

$$|I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)|\le C{\delta }^{\alpha }{M}_{\mathrm{\Omega }}\overrightarrow{f}(x)+L^{-1}\prod _{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}{\parallel \mathrm{\Omega }\parallel }_{L^{s^{\prime }}}{(\frac{1}{n}ln{\left(\frac{PR}{\delta }\right)}^n)}^{\frac{n-\alpha }{n}}.$$

Set $\delta =\epsilon {(|I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)|/C{M}_{\mathrm{\Omega }}(\overrightarrow{f})(x))}^{1/\alpha }$, then

$$exp\left\{n\gamma {\left(\frac{L{I}_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)}{{\parallel \mathrm{\Omega }\parallel }_{L^{s^{\prime }}}{\prod }_{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{\frac{n}{n-\alpha }}\right\}\le lnC{R}^n{\left(\frac{M_{\mathrm{\Omega }}(\overrightarrow{f})(x)}{I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)}\right)}^{n/\alpha }.$$

Now we put $B_1=\{x\in B:\frac{I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)}{{\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}}{\prod }_{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\ge 1\}$, $B_2=B-B_1$, thus

$$\begin{array}{c}{\int }_{B_1}exp\left(n\gamma {\left(\frac{L{I}_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)}{{\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}}{\prod }_{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{n/(n-\alpha )}\right){\nu }_{\overrightarrow{\omega }}(x)dx\hfill \\ \le C{R}^n{\int }_{B_1}{\left(\frac{M_{\mathrm{\Omega }}(\overrightarrow{f})(x)}{I_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)}\right)}^{n/\alpha }{\nu }_{\overrightarrow{\omega }}(x)dx\hfill \\ \le C{R}^n{\int }_{B_1}{\left(\frac{M_{\mathrm{\Omega }}(\overrightarrow{f})(x)}{{\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}}{\prod }_{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{n/\alpha }{\nu }_{\overrightarrow{\omega }}(x)dx.\hfill \end{array}$$

By the fact that

$$\begin{array}{c}{M}_{\mathrm{\Omega }}(\overrightarrow{f})(x)=\underset{r>0}{sup}{\int }_{|y|<r}{|\mathrm{\Omega }(y)|}^{{\sum }_{j=1}^m\frac{s}{p_j}}\prod _{j=1}^m{f}_j(x-{\theta }_{j}y)dy\hfill \\ \le \underset{r>0}{sup}\prod _{j=1}^m{(\frac{1}{r^n}{\int }_{|y|<r}|\mathrm{\Omega }(y)|f_j^{\frac{p_j}{s}}(x-{\theta }_{j}y)dy)}^{\frac{s}{p_j}}\hfill \\ \le \prod _{j=1}^m{(M_{\mathrm{\Omega }}\left(f^{\frac{p_j}{s}}\right)(x))}^{\frac{s}{p_j}}.\hfill \end{array}$$

Therefore, we get

$$\begin{array}{c}{\int }_{B_1}exp\left(n\gamma {\left(\frac{L{I}_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)}{{\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}}{\prod }_{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{n/(n-\alpha )}\right){\nu }_{\overrightarrow{\omega }}(x)dx\hfill \\ \le \frac{C{R}^n}{{\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}}{\prod }_{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}^s}{\int }_{B_1}\prod _{j=1}^m{\left(M_{\mathrm{\Omega }}(f_j^{\frac{p_j}{s}}(x))\right)}^{\frac{s^2}{p_j}}{\nu }_{\overrightarrow{\omega }}(x)dx\hfill \\ \le \frac{C{R}^n}{{\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}}{\prod }_{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p-j}}^s}\prod _{j=1}^m{({\int }_{B_1}{\left(M_{\mathrm{\Omega }}(f_j^{\frac{p_j}{s}}(x))\right)}^s{\omega }_j(x)dx)}^{\frac{1}{s}\frac{s^2}{p_j}}\hfill \\ \le \frac{C{R}^n}{{\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}}{\prod }_{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}^s}\prod _{j=1}^m{\parallel f_j^{\frac{p_j}{s}}\parallel }_{L_{{\omega }_j}^s}^{\frac{s^2}{p_j}}\hfill \\ \le C{R}^n.\hfill \end{array}$$

Here, in the above third inequality, we have used the well-known weighted result of Hardy-Littlewood maximal function.

From ${\omega }_j\ge 1$ ($j=1,2,\dots ,m$), we get

$$R^n=c{\int }_{B}dx\le c{\int }_B{\omega }_j(x)dx=c{\omega }_j(B).$$

Hence,

$${\int }_{B_1}exp\left(n\gamma {\left(\frac{L{I}_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)}{{\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}}{\prod }_{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{n/(n-\alpha )}\right){\nu }_{\overrightarrow{\omega }}(x)dx\le C^{\prime }\prod _{j=1}^m{\omega }_j(B).$$

On the other hand,

$$\begin{array}{c}{\int }_{B_2}exp\left(n\gamma {\left(\frac{L{I}_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)}{{\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}}{\prod }_{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{n/(n-\alpha )}\right){\nu }_{\overrightarrow{\omega }}(x)dx\hfill \\ \le exp(n\gamma ){\left(\frac{L}{{\parallel \mathrm{\Omega }\parallel }_{L^{s^{\prime }}}}\right)}^{\frac{n}{n-\alpha }}{\int }_{B_2}{\nu }_{\overrightarrow{\omega }}(x)dx\hfill \\ \le C\prod _{j=1}^m{\omega }_j(B).\hfill \end{array}$$

From the above all, we obtain that

$${\int }_{B}exp\left(n\gamma {\left(\frac{L{I}_{\mathrm{\Omega },\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)}{{\parallel \mathrm{\Omega }\parallel }_{L^{n/(n-\alpha )}}{\prod }_{j=1}^m{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{n/(n-\alpha )}\right){\nu }_{\overrightarrow{\omega }}(x)dx\le C\prod _{j=1}^m{\omega }_j(B).$$

 □

3 The proof of Theorem 1.5

In this section, we will prove Theorem 1.5.

Proof By the well-known Hölder’s inequality, we get

$$\begin{array}{rcl}{M}_{1,\alpha }(\overrightarrow{f})(x)& =& \underset{r>0}{sup}\frac{1}{{|r|}^{n-\alpha }}{\int }_{|y|<r}\prod _{j=1}^m{f}_j(x-y)dy\\ \le & \underset{r>0}{sup}\frac{1}{{|r|}^{n-\alpha }}\prod _{j=1}^m{({\int }_{|y|<r}{f}_j^{\frac{p_j}{s}}(x-y)dy)}^{\frac{s}{p_j}}\\ \le & \prod _{j=1}^m{(\underset{r>0}{sup}\frac{1}{{|r|}^{n-\alpha }}{\int }_{|y|<r}{f}_j^{\frac{p_j}{s}}(x-y)dy)}^{\frac{s}{p_j}}\\ =& \prod _{j=1}^m{(M_{1,\alpha }\left(f^{p_j/s}\right)(x))}^{\frac{s}{p_j}}.\end{array}$$

Hence,

$$\begin{array}{rcl}{\left({\int }_{{\mathbb{R}}^n}{(M_{1,\alpha }(\overrightarrow{f})(x){\nu }_{\overrightarrow{\omega }}(x))}^{r}dx\right)}^{1/r}& \le & {\left[{\int }_{{\mathbb{R}}^n}{\left(\prod _{j=1}^m{[M_{1,\alpha }\left(f^{p_j/s}\right)(x){\omega }_j^{p_j/s}(s)]}^{\frac{s}{p_j}}\right)}^{r}dx\right]}^{1/r}\\ \le & \prod _{j=1}^m{\left[{\int }_{{\mathbb{R}}^n}{(M_{1,\alpha }\left(f_j^{p_j/s}\right)(x){\omega }^{p_j/s}(x))}^{s{r}_j/p_j}dx\right]}^{\frac{p_j}{s{r}_j}\frac{s}{p_j}}.\end{array}$$

In addition, from the condition ${\omega }_j^{p_j/s}(x)\in A(s,\frac{s{r}_j}{p_j})$, it follows that

$${\left[{\int }_{{\mathbb{R}}^n}{(M_{1,\alpha }\left(f_j^{p_j/s}\right)(x){\omega }^{p_j/s}(x))}^{s{r}_j/p_j}dx\right]}^{\frac{p_j}{s{r}_j}\frac{s}{p_j}}\le C_j{\left[{\int }_{{\mathbb{R}}^n}{(f_j^{p_j/s}(x){\omega }_j^{p_j/s}(x))}^{s}dx\right]}^{1/p_j}.$$

According to the above, we obtain that

$${\left({\int }_{R^n}{(M_{1,\alpha }(\overrightarrow{f})(x){\nu }_{\overrightarrow{\omega }}(x))}^{r}dx\right)}^{1/r}=C\prod _{j=1}^m{\left({\int }_{{\mathbb{R}}^n}{(f_j(x){\omega }_j(x))}^{p_j}dx\right)}^{1/p_j}.$$

It is easy to see that

$$M_{1,\alpha }^{\mathrm{\Theta }}(\overrightarrow{f})(x)=\underset{r>0}{sup}\frac{1}{r^{n-\alpha }}{\int }_{|y|<r}\prod _{j=1}^m|f_j(x-{\theta }_{j}y)|dy,$$

where $\mathrm{\Theta }=({\theta }_1,{\theta }_2,\dots ,{\theta }_m)$, ${\theta }_j\in \mathbb{R}$ holds, also. □

4 The proof of Theorem 1.3

In this section, we will prove Theorem 1.3.

Proof For any $\delta >0$ and $x\in B$,

$$\begin{array}{c}|{\overline{I}}_{\mathrm{\Omega },\alpha }(f_1,f_2,\dots ,f_m)(x)|\hfill \\ \le {\int }_{|(x-y_1,x-y_2,\dots ,x-y_m)|<\delta }\frac{{\prod }_{j=1}^m|{\mathrm{\Omega }}_j(y_j)f_j(y_j)|}{{|(x-y_1,x-y_2,\dots ,x-y_m)|}^{mn-\alpha }}d\overrightarrow{y}\hfill \\ +{\int }_{|(x-y_1,x-y_2,\dots ,x-y_m)|\ge \delta }\frac{{\prod }_{j=1}^m|{\mathrm{\Omega }}_j(y_j)f_j(y_j)|}{{|(x-y_1,x-y_2,\dots ,x-y_m)|}^{mn-\alpha }}d\overrightarrow{y}\hfill \\ :=F_1+F_2.\hfill \end{array}$$

For $F_1$, let $\alpha ={\sum }_{j=1}^m{\alpha }_j$ with ${\alpha }_j=n/p_j$ for $j=1,2,\dots ,m$. Then

$$\begin{array}{rcl}{F}_1& \le & {\int }_{|(x-y_1,x-y_2,\dots ,x-y_m)|<\delta }\frac{|{\mathrm{\Omega }}_j(y_j)f_j(y_j)|}{{\prod }_{j=1}^m{|x-y_j|}^{n-{\alpha }_j}}d\overrightarrow{y}\\ \le & \prod _{j=1}^m{\int }_{|x-y_j|<\delta }\frac{|{\mathrm{\Omega }}_j(y_j)f_j(y_j)|}{{|x-y_j|}^{n-{\alpha }_j}}d{y}_j\\ \le & C\prod _{j=1}^m{\delta }^{{\alpha }_j}{M}_{{\mathrm{\Omega }}_j}(f_j)(x)\\ :=& C_1{\delta }^{\alpha }\prod _{j=1}^m{M}_{{\mathrm{\Omega }}_j}(f_j)(x),\end{array}$$

where $M_{\mathrm{\Omega }}$ denotes as $M_{\mathrm{\Omega }}(f)(x)={sup}_{r>0}\frac{1}{r^n}{\int }_{|x-y|<r}|\mathrm{\Omega }(y)f(y)|dy$.

For $F_2$, if $(y_1,y_2,\dots ,y_m)$ satisfies $|(x-y_1,x-y_2,\dots ,x-y_m)|\ge \delta$, then for some $j\in 1,2,\dots ,m$, $|x-y_j|\le \frac{\delta }{\sqrt{m}}$. Without losing the generalization, we set $j=m$.

Thus,

$$F_2\le {\int }_{\delta /\sqrt{m}\le |x-y_m|\le 2R}{\int }_{{({\mathbb{R}}^n)}^{m-1}}\frac{{\prod }_{j=1}^m|{\mathrm{\Omega }}_j(y_j)f_j(y_j)|}{{|(x-y_1,x-y_2,\dots ,x-y_m)|}^{mn-\alpha }}d\overrightarrow{y}.$$

Define that $f_j^0=f_j{\chi }_{B(x,\delta /\sqrt{m})}$ and $f_j^{\mathrm{\infty }}=f-f_j^0$ for $j=1,2,\dots ,m$. By the condition of ${\nu }_{\overrightarrow{\omega }}$, we have

$$\begin{array}{rcl}{F}_2& \le & \sum _{\overrightarrow{r}\in {\{0,\mathrm{\infty }\}}^m}{\int }_{\delta /\sqrt{m}\le |x-y_m|\le 2R}{\int }_{{({\mathbb{R}}^n)}^{m-1}}\frac{{\prod }_{j=1}^{m-1}|{\mathrm{\Omega }}_j(y_j)f_j^{r_j}(y_j)||{\mathrm{\Omega }}_m(y_m)f_m(y_m)|}{{|(x-y_1,x-y_2,\dots ,x-y_m)|}^{mn-\alpha }}d\overrightarrow{y}\\ \le & \sum _{\overrightarrow{r}\in {\{0,\mathrm{\infty }\}}^m}{\int }_{\delta /\sqrt{m}\le |x-y_m|\le 2R}{\int }_{{({\mathbb{R}}^n)}^{m-1}}\frac{{\prod }_{j=1}^{m-1}|{\mathrm{\Omega }}_j(y_j)f_j^{r_j}(y_j)||{\mathrm{\Omega }}_m(y_m)f_m(y_m)|}{{|(x-y_1,x-y_2,\dots ,x-y_m)|}^{mn-\alpha }}{\nu }_{\overrightarrow{\omega }}(\overrightarrow{y})d\overrightarrow{y},\end{array}$$

where $\overrightarrow{r}=(r_1,r_2,\dots ,r_m)$. In the case that $r_1=r_2=\cdots =r_{m-1}=0$, by the fact that

$$\begin{array}{rcl}{|(x-y_1,x-y_2,\dots ,x-y_m)|}^{mn-\alpha }& \ge & {|x-y_m|}^{mn-\alpha }\\ =& {|x-y_m|}^{n-{\alpha }_m}{|x-y_m|}^{{\sum }_{j=1}^{m-1}n/p_j^{\prime }}\\ \ge & {|x-y_m|}^{n-{\alpha }_m}{\left(\frac{\delta }{\sqrt{m}}\right)}^{{\sum }_{j=1}^{m-1}n/p_j^{\prime }},\end{array}$$

we have

$$\begin{array}{c}{\int }_{\delta /\sqrt{m}\le |x-y_m|\le 2R}{\int }_{{({\mathbb{R}}^n)}^{m-1}}\frac{{\prod }_{j=1}^{m-1}|{\mathrm{\Omega }}_j(y_j)f_j^0(y_j)||\mathrm{\Omega }(y_m)f_m(y_m)|}{{|(x-y_1,x-y_2,\dots ,x-y_m)|}^{mn-\alpha }}{\nu }_{\overrightarrow{\omega }}(\overrightarrow{y})d\overrightarrow{y}\hfill \\ \le \prod _{j=1}^{m-1}{\delta }^{-\frac{n}{p_j^{\prime }}}{\int }_{\frac{\delta }{\sqrt{m}}\le |x-y_m|\le 2R}\frac{|{\mathrm{\Omega }}_m(y_m)f_m(y_m)|}{{|x-y_m|}^{n-{\alpha }_m}}{\omega }_m^{1/p_m}(y_m)d{y}_m\hfill \\ \times \prod _{j=1}^{m-1}{\int }_{|x-y_j|<\delta /\sqrt{m}}|{\mathrm{\Omega }}_j(y_j)f_j(y_j)|{\omega }_j^{1/p_j}(y_j)d{y}_j\hfill \\ \le C\prod _{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}{(log\frac{2R\sqrt{m}}{\delta })}^{1/p_m^{\prime }}\hfill \\ \le C\prod _{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}{(log\frac{2R\sqrt{m}}{\delta })}^{(mn-\alpha )/n}.\hfill \end{array}$$

Consider the case where exactly l of the $r_j$ are ∞ for some $1\le l\le m$. Without losing the generalization, we only give the argument for $r_j=\mathrm{\infty }$, $j=1,2,\dots ,l$, then

$$\begin{array}{c}{\int }_{\delta /\sqrt{m}\le |x-y_m|\le 2R}{\int }_{{({\mathbb{R}}^n)}^{m-1}}\frac{{\prod }_{j=1}^m{\mathrm{\Omega }}_j(y_j){\prod }_{j=1}^l|f_j^{\mathrm{\infty }}(y_j){\prod }_{k=l+1}^{m-1}{f}_k^0(y_k)f_m(y_m)|}{{|(x-y_1,x-y_2,\dots ,x-y_m)|}^{mn-\alpha }}{\nu }_{\overrightarrow{\omega }}d\overrightarrow{y}\hfill \\ \le \prod _{k=l+1}^{m-1}{\int }_{|x-y_k|<\delta /\sqrt{m}}|{\mathrm{\Omega }}_k(y_k)f_k(y_k)|{\omega }_k^{1/p_m}(y_k)d{y}_k\hfill \\ \times \prod _{j=1}^l{\int }_{\delta /\sqrt{m}\le |x-y_j|\le 2R}\frac{|{\mathrm{\Omega }}_j(y_j)f_j(y_j)|}{{|x-y_j|}^{n-{\alpha }_j}}{\omega }_j^{1/p_j}(y_j)d{y}_j\hfill \\ \times {\int }_{\delta /\sqrt{m}\le |x-y_m|\le 2R}\frac{|{\mathrm{\Omega }}_m(y_m)f_m(y_m)|}{{|x-y_m|}^{(m-l)n-{\sum }_{k=l+1}^m{\alpha }_k}}{\omega }_m^{1/p_m}(y_m)d{y}_m\hfill \\ \le C{[log\frac{2\sqrt{m}R}{\delta }]}^{{\sum }_{k=1}^l\frac{1}{p_m^{\prime }}}\prod _{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}\hfill \\ \le C\prod _{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}{[log\frac{2\sqrt{m}R}{\delta }]}^{(mn-\alpha )/n}.\hfill \end{array}$$

Combining the above cases, we obtain

$$F_2\le C_2\prod _{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}{[log\frac{2\sqrt{m}R}{\delta }]}^{(mn-\alpha )/n}.$$

Thus, by the estimates for $F_1$, $F_2$, we have

$$\begin{array}{rcl}{\overline{I}}_{\mathrm{\Omega },\alpha }(f_1,f_2,\dots ,f_m)(x)& \le & C_1{\delta }^{\alpha }\prod _{j=1}^m{M}_{{\mathrm{\Omega }}_j}(f_j)(x)\\ +C_2\prod _{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}{[log\frac{2\sqrt{m}R}{\delta }]}^{(mn-\alpha )/n}.\end{array}$$

In particular, we chose $\delta =2\sqrt{m}R$ for all $x\in B$, then

$${\overline{I}}_{\mathrm{\Omega },\alpha }(f_1,f_2,\dots ,f_m)(x)\le C_1{\delta }^{\alpha }\prod _{j=1}^m{M}_{{\mathrm{\Omega }}_j}(f_j)(x).$$

Now, we set

$$\delta =\delta (x)=\epsilon {[|{\overline{I}}_{\mathrm{\Omega },\alpha }(f_1,f_2,\dots ,f_m)(x)|/C_1\prod _{j=1}^m{M}_{{\mathrm{\Omega }}_j}(f_j)(x)]}^{1/\alpha },$$

where $\epsilon <1$.

Then

$$\begin{array}{c}|{\overline{I}}_{\mathrm{\Omega },\alpha }(f_1,f_2,\dots ,f_m)(x)|\hfill \\ \le {\epsilon }^{\alpha }|{\overline{I}}_{\mathrm{\Omega },\alpha }(f_1,f_2,\dots ,f_m)(x)|\hfill \\ +C_2\prod _{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}{[\frac{1}{n}log\left(\frac{{(2\sqrt{m}R)}^n{[C_1{\prod }_{j=1}^m{M}_{{\mathrm{\Omega }}_j}(f_j)(x)]}^{n/\alpha }}{{\epsilon }^n{|{\overline{I}}_{\mathrm{\Omega },\alpha }(f_1,f_2,\dots ,f_m)(x)|}^{n/\alpha }}\right)]}^{(mn-\alpha )/n}.\hfill \end{array}$$

Hence,

$$exp\left(k_1{\left(\frac{|{\overline{I}}_{\mathrm{\Omega },\alpha }(f_1,f_2,\dots ,f_m)(x)|}{{\prod }_{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{n/(mn-\alpha )}\right)\le \frac{C{[{\prod }_{j=1}^m{M}_{{\mathrm{\Omega }}_j}(f_j)(x)]}^{n/\alpha }}{{|{\overline{I}}_{\mathrm{\Omega },\alpha }(f_1,f_2,\dots ,f_m)(x)|}^{n/\alpha }}.$$

Let $B_1=\{x\in B:\frac{|{\overline{I}}_{\mathrm{\Omega },\alpha }(f_1,f_2,\dots ,f_m)(x)|}{{\prod }_{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L^{p_j}}}\ge 1\}$ and $B_2=B-B_1$, then

$$\begin{array}{c}{\int }_{B_1}exp\left(k_1{\left(\frac{|{\overline{I}}_{\mathrm{\Omega },\alpha }(f_1,f_2,\dots ,f_m)(x)|}{{\prod }_{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{{p_j}^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{n/(mn-\alpha )}\right){\nu }_{\overrightarrow{\omega }}dx\hfill \\ \le C{R}^n{\int }_{B_1}{\left(\frac{{\prod }_{j=1}^m{M}_{{\mathrm{\Omega }}_j}(f_j)(x)}{{\prod }_{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{n/\alpha }{\nu }_{\overrightarrow{\omega }}dx\hfill \\ \le C{R}^n{\left(\prod _{j=1}^m\frac{{\parallel M_{{\mathrm{\Omega }}_j}(f_j)\parallel }_{L_{{\omega }_j}^{p_j}}}{{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L_{{\omega }_j}^{p_j}}}\right)}^{n/\alpha }\hfill \\ \le C{R}^n\hfill \\ \le C\prod _j^m{\omega }_j(B).\hfill \end{array}$$

On the other hand,

$$\begin{array}{c}{\int }_{B_2}exp\left(k_1{\left(\frac{|{\overline{I}}_{\mathrm{\Omega },\alpha }(f_1,f_2,\dots ,f_m)(x)|}{{\prod }_{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L^{p_j}}}\right)}^{n/(mn-\alpha )}\right){\nu }_{\overrightarrow{\omega }}(x)dx\hfill \\ \le exp(k_1)\prod _{j=1}^m{\int }_{B_2}{\omega }_j(x)dx\hfill \\ \le C\prod _{j=1}^m{\omega }_j(B).\hfill \end{array}$$

Combining the above results, we obtain

$${\int }_{B}exp\left(k_1{\left(\frac{|{\overline{I}}_{\mathrm{\Omega },\alpha }(f_1,f_2,\dots ,f_m)(x)|}{{\prod }_{j=1}^m{\parallel {\mathrm{\Omega }}_j\parallel }_{L^{p_j^{\prime }}(S^{n-1})}{\parallel f_j\parallel }_{L^{p_j}}}\right)}^{n/(mn-\alpha )}\right){\nu }_{\overrightarrow{\omega }}(x)dx\le k_2\prod _{j=1}^m{\omega }_j(B),$$

where $k_1$, $k_2$ are constants depending only on n, m, α, p, and the $p_j$. □

Authors’ information

• H. Feng’s current address: Department of Mathematical and Statistical Sciences, University of Alberta, Canada.