Weighted estimates for vector-valued multilinear operators with non-smooth kernels

Abstract

Let T be the multilinear Calderón-Zygmund operator with non-smooth kernel and let $T^{\ast }$ be its corresponding maximal operator. In this paper, vector-valued weighted norm inequalities for T and $T^{\ast }$ are established. As applications, weighted strong type estimates for vector-valued commutators associated with T and $T^{\ast }$ are deduced respectively.

1 Introduction and main results

Let T be a multilinear operator initially defined on the m-fold product of Schwartz spaces and taking values in the space of tempered distributions,

$$T:\mathcal{S}\left({\mathbb{R}}^n\right)\times \cdots \times \mathcal{S}\left({\mathbb{R}}^n\right)\to {\mathcal{S}}^{\mathrm{\prime }}\left({\mathbb{R}}^n\right).$$

Following [1], we say that T is a multilinear Calderón-Zygmund operator if, for some $1\le q_j<\mathrm{\infty }$, it extends to a bounded multilinear operator from $L^{q_1}\times \cdots \times L^{q_m}$ to $L^q$, where $\frac{1}{q}=\frac{1}{q_1}+\cdots +\frac{1}{q_m}$, and if there exists a function K, defined off the diagonal $x=y_1=\cdots =y_m$ in ${({\mathbb{R}}^n)}^{m+1}$, satisfying

$$T\overrightarrow{(f)}(x)=T(f_1,\dots ,f_m)(x)={\int }_{{({\mathbb{R}}^n)}^m}K(x,y_1,\dots ,y_m)f_1(y_1)\cdots f_m(y_m)d{y}_1\cdots d{y}_m,$$

(1.1)

for all $x\notin {\bigcap }_{j=1}^{m}supp{f}_j$;

$$|K(y_0,y_1,\dots ,y_m)|\le \frac{A}{{({\sum }_{k,l=0}^m|y_k-y_l|)}^{mn}};$$

(1.2)

and

$$|K(y_0,y_1,\dots ,y_j,\dots ,y_m)-K(y_0,y_1,\dots ,y_j^{\prime },\dots ,y_m)|\le \frac{A{|y_j-y_j^{\prime }|}^{ϵ}}{{({\sum }_{k,l=0}^m|y_k-y_l|)}^{mn+ϵ}}$$

(1.3)

for some $ϵ>0$ and all $0\le j\le m$, whenever $2|y_j-y_j^{\prime }|\le {max}_{0\le k\le m}|y_j-y_k|$. Such kernels are called m-linear Calderón-Zygmund kernels, and the collection of such functions is denoted by m-$\mathit{CZK}(A,ϵ)$ in [1]. As in [2], we define the maximal multilinear operator by

$$T^{\ast }(\overrightarrow{f})(x)=\underset{\delta >0}{sup}|T_{\delta }(f_1,\dots ,f_m)(x)|,$$

where $T_{\delta }$ is the smooth truncation of T given by

$$T_{\delta }(\overrightarrow{f})(x)={\int }_{{|x-y_1|}^2+\cdots +{|x-y_m|}^2>{\delta }^2}K(x,y_1,\dots ,y_m)f_1(y_1)\cdots f_m(y_m)d{y}_1\cdots d{y}_m.$$

The vector-valued multilinear Calderón-Zygmund operator $T_q$ and vector-valued maximal multilinear operator $T_q^{\ast }$ associated with T are defined and studied in [3, 4].

$$\begin{array}{c}\begin{array}{rl}{T}_q(\overrightarrow{f})(x)& ={|T(f_1,\dots ,f_m)(x)|}_q={\parallel T(f_{1\cdot },\dots ,f_{m\cdot })(x)\parallel }_{l^q}\\ ={\left(\sum _{k=1}^{\mathrm{\infty }}{|T(f_{1k},\dots ,f_{mk})(x)|}^q\right)}^{1/q},\end{array}\hfill \\ \begin{array}{rl}{T}_q^{\ast }(\overrightarrow{f})(x)& ={|T^{\ast }(f_1,\dots ,f_m)(x)|}_q={\parallel T^{\ast }(f_{1\cdot },\dots ,f_{m\cdot })(x)\parallel }_{l^q}\\ ={\left(\sum _{k=1}^{\mathrm{\infty }}{|T^{\ast }(f_{1k},\dots ,f_{mk})(x)|}^q\right)}^{1/q},\end{array}\hfill \end{array}$$

where $f_i={\{f_{ik}\}}_{k=1}^{\mathrm{\infty }}$ for $i=1,\dots ,m$.

Theorem A [3]

Assume that T is a multilinear Calderón-Zygmund operator. Let $1<p_1,\dots ,p_m<\mathrm{\infty }$, $1<q_1,\dots ,q_m<\mathrm{\infty }$ and $1/m<p,q<\mathrm{\infty }$ such that $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$, $\frac{1}{q}=\frac{1}{q_1}+\cdots +\frac{1}{q_m}$. If $({\omega }_1^{p_1},\dots ,{\omega }_m^{p_m})\in (A_{p_1},\dots ,A_{p_m})$, then there exists a constant $C>0$ such that

$${\parallel T_q(\overrightarrow{f})\parallel }_{L^p({\omega }_1^p\cdots {\omega }_m^p)}\le C\prod _{j=1}^m{\parallel {|f_j|}_{q_j}\parallel }_{L^{p_j}({\omega }_j^{p_j})}.$$

Theorem B [4]

Assume that T is a multilinear Calderón-Zygmund operator. Let $1\le p_1,\dots ,p_m<\mathrm{\infty }$, $1<q_1,\dots ,q_m<\mathrm{\infty }$ and $0<p,q<\mathrm{\infty }$ such that $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$, $\frac{1}{q}=\frac{1}{q_1}+\cdots +\frac{1}{q_m}$.

• (i) If $1<p_1,\dots ,p_m<\mathrm{\infty }$ and $\omega \in A_{p_1}\cap \cdots \cap A_{p_m}$, then there exists a constant $C>0$ such that

  $${\parallel T_q(\overrightarrow{f})\parallel }_{L^p(\omega )}\le C\prod _{j=1}^m{\parallel {|f_j|}_{q_j}\parallel }_{L^{p_j}(\omega )}.$$

• (ii) If at least one $p_j=1$ and $\omega \in A_1$, then there exists a constant $C>0$ such that

  $${\parallel T_q(\overrightarrow{f})\parallel }_{L^{p,\mathrm{\infty }}(\omega )}\le C\prod _{j=1}^m{\parallel {|f_j|}_{q_j}\parallel }_{L^{p_j}(\omega )}.$$

It is worth noting that similar results hold for $T^{\ast }$ in Theorems A and B.

We will replace (1.3) by a weaker regularity condition on the kernel K. Assume that operators $A_t$ are associated with kernels $a_t(x,y)$ in the sense that

$$A_{t}f(x)={\int }_{{\mathbb{R}}^n}{a}_t(x,y)f(y)dy$$

for every function $f\in L^p({\mathbb{R}}^n)$, $1\le p\le \mathrm{\infty }$, and $a_t(x,y)$ satisfy the following size condition:

$$|a_t(x,y)|\le h_t(x,y)=t^{-n/s}h\left(\frac{{|x-y|}^s}{t}\right),$$

(1.4)

where s is a positive fixed constant and h is a positive, bounded, decreasing function satisfying that for some $\eta >0$,

$$\underset{r\to \mathrm{\infty }}{lim}{r}^{n+\eta }h\left(r^s\right)=0.$$

(1.5)

The j th transpose $T^{\ast ,j}$ of T is defined via

$$〈T^{\ast ,j}(f_1,\dots ,f_m),g〉=〈T(f_1,\dots ,f_{j-1},g,f_{j+1},\dots ,f_m),f_j〉$$

for all $f_1,\dots ,f_m$, g in $\mathcal{S}({\mathbb{R}}^n)$. It is easy to check that the kernel $K^{\ast ,j}$ of $T^{\ast ,j}$ is related to the kernel K of T via the identity

$$K^{\ast ,j}(x,y_1,\dots ,y_{j-1},y_j,y_{j+1},\dots ,y_m)=K(y_j,y_1,\dots ,y_{j-1},x,y_{j+1},\dots ,y_m).$$

To maintain uniform notation, we may occasionally denote $T=T^{\ast ,0}$ and $K=K^{\ast ,0}$.

Assumption (H0) We always assume that there exist some $1\le q_1,\dots ,q_m<\mathrm{\infty }$ and some $0<q<\mathrm{\infty }$ with $\frac{1}{q}=\frac{1}{q_1}+\cdots +\frac{1}{q_m}$ such that both $T^{\ast }$ and T map $L^{q_1}\times \cdots \times L^{q_m}$ to $L^{q,\mathrm{\infty }}$.

Assumption (H1) Assume that there exist operators ${\{A_t^{(i)}\}}_{t>0}$ with kernels $a_t^{(i)}(x,y)$ that satisfy the conditions (1.4) and (1.5) with constants s and η for each $i=1,\dots ,m$ and that for every $j=0,1,2,\dots ,m$, there exists kernel $K_t^{\ast ,j,(i)}(x,y_1,\dots ,y_m)$ such that

$$\begin{array}{r}〈T^{\ast ,j}(f_1,\dots ,A_t^{(i)}{f}_i,\dots ,f_m),g〉\\ ={\int }_{{\mathbb{R}}^n}{\int }_{{({\mathbb{R}}^n)}^m}{K}_t^{\ast ,j,(i)}(x,y_1,\dots ,y_m)f_1(y_1)\cdots f_m(y_m)g(x)d{y}_1\cdots d{y}_{m}dx,\end{array}$$

for all $f_1,\dots ,f_m$ in $\mathcal{S}({\mathbb{R}}^n)$ with ${\bigcap }_{k=1}^{m}supp{f}_k\cap suppg=\varphi$. Also assume that there exist a non-negative function $\varphi \in C(\mathbb{R})$ with $supp\varphi \in [-1,1]$ and a constant $ϵ>0$ so that for every $j\in \{0,1,\dots ,m\}$ and every $i\in \{1,2,\dots ,m\}$, all $t>0$ and all $x,y_1,\dots ,y_n\in {\mathbb{R}}^n$, we have

$$\begin{array}{r}|K^{\ast ,j}(x,y_1,\dots ,y_m)-K_t^{\ast ,j,(i)}(x,y_1,\dots ,y_m)|\\ \le \frac{A}{{(|x-y_1|+\cdots +|x-y_m|)}^{mn}}\sum _{k=1,k\ne i}^m\varphi \left(\frac{|y_i-y_k|}{t^{1/s}}\right)\\ +\frac{A{t}^{ϵ/s}}{{(|x-y_1|+\cdots +|x-y_m|)}^{mn+ϵ}},\end{array}$$

whenever $2t^{1/s}\le |x-y_i|$.

Kernels K that satisfy the size estimate (1.2) and Assumption (H1) with parameters m, A, s, η, ε are called generalized Calderón-Zygmund kernels, and their collection is denoted by m-$\mathit{GCZK}(A,s,\eta ,\epsilon )$. We say that T is of class m-$\mathit{GCZO}(A,s,\eta ,\epsilon )$ if T has an associated kernel K in m-$\mathit{GCZK}(A,s,\eta ,\epsilon )$.

Assumption (H2) Assume that there exist operators ${\{A_t\}}_{t>0}$ with kernels $a_t(x,y)$ that satisfy conditions (1.4) and (1.5) with constants s and η, and there exist kernels $K_t^{(0)}(x,y_1,\dots ,y_m)$such that the representation is valid

$$K_t^{(0)}(x,y_1,\dots ,y_m)={\int }_{{\mathbb{R}}^n}K(z,y_1,\dots ,y_m)a_t(x,z)dz$$

and that there exist a non-negative function $\varphi \in \mathcal{C}(\mathbb{R})$ and $supp\varphi \subset [-1,1]$ and a positive constant ε such that

$$\begin{array}{r}|K(x,y_1,\dots ,y_m)-K_t^{(0)}(x,y_1,\dots ,y_m)|\\ \le \frac{A}{{({\sum }_{k=1}^m|x-y_k|)}^{mn}}\sum _{\stackrel{k=1}{k\ne j}}^m\varphi \left(\frac{|x-y_k|}{t^{1/s}}\right)+\frac{A{t}^{\epsilon /s}}{{({\sum }_{k=1}^m|x-y_k|)}^{mn+\epsilon }}\end{array}$$

(1.6)

for some $A>0$, whenever $2t^{1/s}\le {max}_{1\le j\le m}|x-y_j|$. Moreover, assume that for all $x,y_1,\dots ,y_m\in {\mathbb{R}}^n$,

$$|K_t^{(0)}(x,y_1,\dots ,y_m)|\le \frac{A}{{({\sum }_{k=1}^m|x-y_k|)}^{mn}},$$

whenever $2t^{1/s}\le {min}_{1\le j\le m}|x-y_j|$, and for all $x,x^{\mathrm{\prime }},y_1,\dots ,y_m\in {\mathbb{R}}^n$ ,

$$|K(x,y_1,\dots ,y_m)-K_t^{(0)}(x^{\mathrm{\prime }},y_1,\dots ,y_m)|\le \frac{A{t}^{\epsilon /s}}{{({\sum }_{k=1}^m|x-y_k|)}^{mn+\epsilon }},$$

whenever $2t^{1/s}\le {min}_{1\le j\le m}|x-y_j|$ and $2|x-x^{\mathrm{\prime }}|\le t^{1/s}$.

The commutators associated with T and $T^{\ast }$ are defined respectively by

$$\begin{array}{rl}{T}_{\mathrm{\Pi }\overrightarrow{b}}(\overrightarrow{f})(x)& ={[b_1,{[b_2,\dots {[b_{l-1},{[b_l,T]}_l]}_{l-1}\cdots ]}_2]}_1(\overrightarrow{f})(x)\\ ={\int }_{{({\mathbb{R}}^n)}^m}\prod _{j=1}^l(b_j(x)-b_j(y_j))K(x,y_1,\dots ,y_m)\prod _{i=1}^m{f}_i(y_i)d\overrightarrow{y},\end{array}$$

and

$$\begin{array}{rl}{T}_{\mathrm{\Pi }\overrightarrow{b}}^{\ast }(\overrightarrow{f})(x)& =\underset{\delta >0}{sup}|{[b_1,{[b_2,\dots {[b_{l-1},{[b_l,T_{\delta }]}_l]}_{l-1}\cdots ]}_2]}_1(\overrightarrow{f})(x)|\\ =\underset{\delta >0}{sup}|{\int }_{{|x-y_1|}^2+\cdots +{|x-y_m|}^2>{\delta }^2}\prod _{j=1}^l(b_j(x)-b_j(y_j))K(x,y_1,\dots ,y_m)\prod _{i=1}^m{f}_i(y_i)d\overrightarrow{y}|.\end{array}$$

Here and subsequently, we often write $\overrightarrow{y}=(y_1,\dots ,y_m)$ and $d\overrightarrow{y}=d{y}_1\cdots d{y}_m$.

For simplicity of notation, we often write $\overrightarrow{f}=(f_1,\dots ,f_m)$ with $f_j={\{f_{jk}\}}_{k=1}^{\mathrm{\infty }}$. For the sequence ${\{{\overrightarrow{f}}_k\}}_{k=1}^{\mathrm{\infty }}={\{f_{1k},\dots ,f_{mk}\}}_{k=1}^{\mathrm{\infty }}$ of vector functions, the commutators associated with vector-valued $T_q$ and $T_q^{\ast }$ can be defined by

$$\begin{array}{c}{T}_{\mathrm{\Pi }\overrightarrow{b},q}(\overrightarrow{f})(x)={|T_{\mathrm{\Pi }\overrightarrow{b}}(\overrightarrow{f})(x)|}_q={\parallel T_{\mathrm{\Pi }\overrightarrow{b}}(f_{1\cdot },\dots ,f_{m\cdot })(x)\parallel }_{l^q}={\left(\sum _{k=1}^{\mathrm{\infty }}{|T_{\mathrm{\Pi }\overrightarrow{b}}({\overrightarrow{f}}_k)(x)|}^q\right)}^{1/q},\hfill \\ T_{\mathrm{\Pi }\overrightarrow{b},q}^{\ast }(\overrightarrow{f})(x)={|T_{\mathrm{\Pi }\overrightarrow{b}}^{\ast }(\overrightarrow{f})(x)|}_q={\parallel T_{\mathrm{\Pi }\overrightarrow{b}}^{\ast }(f_{1\cdot },\dots ,f_{m\cdot })(x)\parallel }_{l^q}={\left(\sum _{k=1}^{\mathrm{\infty }}{|T_{\mathrm{\Pi }\overrightarrow{b}}^{\ast }({\overrightarrow{f}}_k)(x)|}^q\right)}^{1/q}.\hfill \end{array}$$

From now on, we always assume that T is a multilinear operator in m-$\mathit{GCZO}(A,s,\eta ,\epsilon )$ and its kernel satisfies Assumption (H2). Recently, if $l=m$, Peng et al. [5] obtained the following weighted strong type estimates for $T_{\mathrm{\Pi }\overrightarrow{b}}$ and $T_{\mathrm{\Pi }\overrightarrow{b}}^{\ast }$ with multiple weights (see Definition 2.1).

Theorem C [5]

Let $\overrightarrow{b}\in {\mathit{BMO}}^m$, $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$ with $1<p_j<\mathrm{\infty }$, $j=1,\dots ,m$. Then we have

• (i) There exists a constant C such that

  $${\parallel T_{\mathrm{\Pi }\overrightarrow{b}}^{\ast }(\overrightarrow{f})\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}\le C\prod _{i=1}^m{\parallel b_i\parallel }_{\mathit{BMO}}\prod _{i=1}^m{\parallel f_i\parallel }_{L^{p_i}(M{\omega }_i)}.$$

• (ii) If each ${\omega }_i\in A_{p_j}$, then there exists a constant C such that

  $${\parallel T_{\mathrm{\Pi }\overrightarrow{b}}^{\ast }(\overrightarrow{f})\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}\le C\prod _{i=1}^m{\parallel b_i\parallel }_{\mathit{BMO}}\prod _{i=1}^m{\parallel f_i\parallel }_{L^{p_i}({\omega }_i)},$$

where ${\nu }_{\overrightarrow{\omega }}={\prod }_{i=1}^m{\omega }_i^{p/p_j}$. Similar results still hold for $T_{\mathrm{\Pi }\overrightarrow{b}}$, which extend the results in [6]significantly.

In this work, we first pursue results parallel to Theorems A and B, then extend Theorem C to a vector-valued version. The main results can be stated as follows.

Theorem 1.1 Let $1<p_1,\dots ,p_m<\mathrm{\infty }$, $1<q_1,\dots ,q_m<\mathrm{\infty }$ and $1/m<p,q<\mathrm{\infty }$ such that $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$, $\frac{1}{q}=\frac{1}{q_1}+\cdots +\frac{1}{q_m}$. If $({\omega }_1^{p_1},\dots ,{\omega }_m^{p_m})\in (A_{p_1},\dots ,A_{p_m})$, then there exists a constant $C>0$ such that

$${\parallel T_q(\overrightarrow{f})\parallel }_{L^p({\omega }_1^p\cdots {\omega }_m^p)}\le C\prod _{j=1}^m{\parallel {|f_j|}_{q_j}\parallel }_{L^{p_j}({\omega }_j^{p_j})}.$$

Moreover, similar estimates hold for $T^{\ast }$.

Theorem 1.2 Let $1\le p_1,\dots ,p_m<\mathrm{\infty }$, $1<q_1,\dots ,q_m<\mathrm{\infty }$ and $0<p,q<\mathrm{\infty }$ such that $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$, $\frac{1}{q}=\frac{1}{q_1}+\cdots +\frac{1}{q_m}$.

• (i) If $1<p_1,\dots ,p_m<\mathrm{\infty }$ and $\omega \in A_{p_1}\cap \cdots \cap A_{p_m}$, then there exists a constant $C>0$ such that

  $${\parallel T_q(\overrightarrow{f})\parallel }_{L^p(\omega )}\le C\prod _{j=1}^m{\parallel {|f_j|}_{q_j}\parallel }_{L^{p_j}(\omega )}.$$

• (ii) If at least one $p_j=1$ and $\omega \in A_1$, then there exists a constant $C>0$ such that

  $${\parallel T_q(\overrightarrow{f})\parallel }_{L^{p,\mathrm{\infty }}(\omega )}\le C\prod _{j=1}^m{\parallel {|f_j|}_{q_j}\parallel }_{L^{p_j}(\omega )}.$$

Moreover, similar estimates hold for $T^{\ast }$.

Theorem 1.3 Let $1/m<p<\mathrm{\infty }$, $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$ with $1<p_1,\dots ,p_m<\mathrm{\infty }$, $1/m<q<\mathrm{\infty }$, and $\frac{1}{q_1}+\cdots +\frac{1}{q_m}=\frac{1}{q}$ with $1<q_1,\dots ,q_m<\mathrm{\infty }$. Suppose that $\overrightarrow{\omega }\in A_{\overrightarrow{p}}$, ${\nu }_{\overrightarrow{\omega }}={\prod }_{i=1}^m{\omega }_i^{\frac{p}{p_i}}$ and $\overrightarrow{b}\in {(\mathit{BMO})}^l$. Then we have

• (i) There then exists a constant $C>0$ such that

  $${\parallel T_{\mathrm{\Pi }\overrightarrow{b},q}(\overrightarrow{f})\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}\le \prod _{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}\prod _{j=1}^m{\parallel {|f_j|}_{q_j}\parallel }_{L^{p_j}(M{w}_j)}.$$

• (ii) If ${\omega }_j\in A_{p_j}$, then there exists a constant $C>0$ such that

  $${\parallel T_{\mathrm{\Pi }\overrightarrow{b},q}(\overrightarrow{f})\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}\le \prod _{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}\prod _{j=1}^m{\parallel {|f_j|}_{q_j}\parallel }_{L^{p_j}(w_j)}.$$

Moreover, similar estimates hold for $T^{\ast }$.

Remark 1.4 If $l=1$ and $l=m$, Theorem 1.3 can be seen as the vector-valued extension of Theorem 4.5 in [6] and Theorem C, respectively.

2 Proofs of Theorem 1.1 and Theorem 1.2

Let us begin with the definition of Hardy-Littlewood maximal operator, that is

$$Mf(x)=\underset{Q\ni x}{sup}\frac{1}{|Q|}{\int }_Q|f(y)|dy.$$

The sharp maximal function is defined by

$$M^{\mathrm{♯}}f(x)=\underset{Q\ni x}{sup}\underset{c}{inf}\frac{1}{|Q|}{\int }_Q|f(y)-c|dy\approx \underset{Q\ni x}{sup}\frac{1}{|Q|}{\int }_Q|f(y)-f_Q|dy.$$

For $\delta >0$, we also need the maximal function $M_{\delta }f=M{({|f|}^{\delta })}^{\frac{1}{\delta }}$ and $M_{\delta }^{\mathrm{♯}}f=M^{\mathrm{♯}}{({|f|}^{\delta })}^{\frac{1}{\delta }}$.

The new maximal function ℳ can be defined by

$$\mathcal{M}(\overrightarrow{f})(x)=\underset{Q\ni x}{sup}\prod _{j=1}^m\frac{1}{|Q|}{\int }_Q|f_j(y_j)|d{y}_j.$$

For $1\le l\le m$, as in [7], a modified maximal function ${\mathcal{M}}_l$ is given by

$${\mathcal{M}}_l(\overrightarrow{f})(x)=\underset{Q\ni x}{sup}\sum _{k=0}^{\mathrm{\infty }}{2}^{-knl}\left(\prod _{j=1}^l\frac{1}{|Q|}{\int }_Q|f_j(y_j)|d{y}_j\right)\left(\prod _{j=l+1}^m\frac{1}{|2^{k}Q|}{\int }_{2^{k}Q}|f_j(y_j)|d{y}_j\right).$$

For exponents $p_1,\dots ,p_m$, we will often write p for the number given by $1/p=1/p_1+\cdots +1/p_m$, and $\overrightarrow{p}$ for the vector $\overrightarrow{p}=(p_1,\dots ,p_m)$. Let us recall the definition of $A_{\overrightarrow{p}}$ weights.

Definition 2.1 Let $1\le p_1,\dots ,p_m<\mathrm{\infty }$. Given $\overrightarrow{\omega }=({\omega }_1,\dots ,{\omega }_m)$, set ${\nu }_{\overrightarrow{\omega }}={\prod }_{i=1}^m{\omega }_i^{p/p_i}$. We say that $\overrightarrow{\omega }$ satisfies the $A_{\overrightarrow{p}}$ condition if

$$\underset{Q}{sup}{\left(\frac{1}{|Q|}{\int }_Q\prod _{i=1}^m{\omega }_i^{\frac{p}{p_i}}\right)}^{\frac{1}{p}}\prod _{i=1}^m{\left(\frac{1}{|Q|}{\int }_Q{\omega }_i^{1-p_i^{\prime }}\right)}^{\frac{1}{p_i^{\prime }}}<\mathrm{\infty },$$

when $p_i=1$, ${(\frac{1}{|Q|}{\int }_Q{\omega }_i^{1-p_i^{\prime }})}^{\frac{1}{p_i^{\prime }}}$ is understood as ${({inf}_Q{\omega }_i)}^{-1}$.

We will use the following lemmas in the proof of Theorem 1.1 and Theorem 1.2.

Lemma 2.2 [3]

Let $\mathcal{T}$ be an m-linear operator, and let $1<q_1,\dots ,q_m<\mathrm{\infty }$ and $\frac{1}{m}<q<\mathrm{\infty }$ be fixed indices such that $\frac{1}{q}=\frac{1}{q_1}+\cdots +\frac{1}{q_m}$. For $({\omega }_1^{q_1},\dots ,{\omega }_m^{q_m})\in (A_{q_1},\dots ,A_{q_m})$, the following estimate holds:

$${\parallel \mathcal{T}(\overrightarrow{f})\parallel }_{L^q({\omega }_1^q\cdots {\omega }_m^q)}\le C\prod _{j=1}^m{\parallel f_j\parallel }_{L^{q_j}({\omega }_j^{q_j})}.$$

Then, for all indices, $1<p_1,\dots ,p_m<\mathrm{\infty }$ and $\frac{1}{m}<p<\mathrm{\infty }$ satisfy $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$, $1<s_1,\dots ,s_m<\mathrm{\infty }$ and $\frac{1}{m}<s<\mathrm{\infty }$ such that $\frac{1}{s}=\frac{1}{s_1}+\cdots +\frac{1}{s_m}$, and all $({\omega }_1^{p_1},\dots ,{\omega }_m^{p_m})\in (A_{p_1},\dots ,A_{p_m})$. Then the following inequality holds:

$${\parallel {\left(\sum _k\right|\mathcal{T}(f_{1k},\dots ,f_{mk}){|}^s)}^{\frac{1}{s}}\parallel }_{L^p({\omega }_1^p\cdots {\omega }_m^p)}\le C\prod _{j=1}^m{\parallel {(\sum _k|f_{jk}{|}^{s_j})}^{\frac{1}{s_j}}\parallel }_{L^{p_j}({\omega }_j^{p_j})}.$$

Lemma 2.3 [7]

Suppose that for some $1\le q_1,q_2,\dots ,q_{m-1}\le \mathrm{\infty }$, $q_m\in (1,\mathrm{\infty })$ and $q\in (0,\mathrm{\infty })$ satisfying $\frac{1}{q}=\frac{1}{q_1}+\cdots +\frac{1}{q_m}$, T maps $L^{q_1}\times \cdots \times L^{q_m}$ to $L^q$. Let $1\le p_1,\dots ,p_m<\mathrm{\infty }$, $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$, $\overrightarrow{\omega }=({\omega }_1,\dots ,{\omega }_m)\in A_{\overrightarrow{p}}$ and $\overrightarrow{p}=(p_1,\dots ,p_m)$. Then

• (i) $T^{\ast }$ can be extended to a bounded operator from $L^{p_1}({\omega }_1)\times \cdots \times L^{p_m}({\omega }_m)$ to $L^p({\nu }_{\overrightarrow{\omega }})$ if all the exponents $p_j$ are strictly greater than 1.

• (ii) $T^{\ast }$ can be extended to a bounded operator from $L^{p_1}({\omega }_1)\times \cdots \times L^{p_m}({\omega }_m)$ to $L^{p,\mathrm{\infty }}({\nu }_{\overrightarrow{\omega }})$ if some exponents $p_j$ equal 1.

Similar results hold for T.

Note that if each ${\omega }_j\in A_{p_j}$, then ${\prod }_{j=1}^m{A}_{p_j}\subset A_{\overrightarrow{p}}$ and this inclusion is strict (see [8] for details). This fact together with Lemma 2.3 yields the following weighted estimates.

Lemma 2.4 Consider an m-tuple $({\omega }_1^{p_1},\dots ,{\omega }_m^{p_m})\in (A_{p_1},\dots ,A_{p_m})$, where $1<p_1,\dots ,p_m<\mathrm{\infty }$ and $\frac{1}{m}<p<\mathrm{\infty }$ satisfy $\frac{1}{p}=\frac{1}{p_1}+\cdots +\frac{1}{p_m}$. Then there exists a constant C such that

$${\parallel T(\overrightarrow{f})\parallel }_{L^p({\omega }_1^p\cdots {\omega }_m^p)}\le C\prod _{j=1}^m{\parallel f_j\parallel }_{L^{p_j}({\omega }_j^{p_j})},$$

and

$${\parallel T^{\ast }(\overrightarrow{f})\parallel }_{L^p({\omega }_1^p\cdots {\omega }_m^p)}\le C\prod _{j=1}^m{\parallel f_j\parallel }_{L^{p_j}({\omega }_j^{p_j})}.$$

Proof of Theorem 1.1 and Theorem 1.2 As a consequence of Lemma 2.2 and Lemma 2.4, we obtain Theorem 1.1 (see the proof of Corollary 3 in [3]). From [7] we know that for all $1\le l\le m$, $\overrightarrow{f}=(f_1,\dots ,f_m)$ and $x\in {\mathbb{R}}^n$, the following two inequalities hold:

$$\begin{array}{c}\mathcal{M}(\overrightarrow{f})(x)\le {\mathcal{M}}_l(\overrightarrow{f})(x)\le 2\prod _{j=1}^{m}M(f_j)(x),\hfill \\ {\parallel T\overrightarrow{f}\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}\le C{\parallel \sum _{l=1}^m{\mathcal{M}}_l(\overrightarrow{f})\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}.\hfill \end{array}$$

So, we get ${\parallel T\overrightarrow{f}\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}\le C{\parallel {\prod }_{j=1}^{m}M(f_j)\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}$. A similar inequality still holds for $T^{\ast }$. Theorem 1.2 follows by repeating the same steps as in Corollary 3.3 in [4]. In fact, we apply Theorem 2.1 in [4] to the families

$$\mathcal{F}(T(f_1,\dots ,f_m),\prod _{j=1}^{m}M{f}_j),\mathcal{F}(T^{\ast }(f_1,\dots ,f_m),\prod _{j=1}^{m}M{f}_j).$$

Hölder’s inequality and the normal inequalities for the maximal operator yield the desired results. □

3 Proof of Theorem 1.3

We begin with some lemmas which will be used in the proof of Theorem 1.3.

Lemma 3.1 [9]

Let $0<p,\delta <\mathrm{\infty }$ and let ω be a weight in $A_{\mathrm{\infty }}$. Then there exists $C>0$ (depending upon the $A_{\mathrm{\infty }}$ condition of ω) such that

$${\int }_{{\mathbb{R}}^n}{(M_{\delta }f(x))}^p\omega (x)dx\le C{\int }_{{\mathbb{R}}^n}{(M_{\delta }^{\mathrm{♯}}f(x))}^p\omega (x)dx$$

(3.1)

for every function such that the left-hand side is finite.

Lemma 3.2 Let $0<\delta <1/m$, $1/m<q<\mathrm{\infty }$ and $1/q=1/q_1+\cdots +1/q_m$ with $1<q_1,\dots ,q_m<\mathrm{\infty }$. Then there exists a constant $C>0$ such that

$$M_{\delta }^{\mathrm{♯}}(T_q(\overrightarrow{f}))(x)\le C\prod _{j=1}^{m}M\left({|f_j|}_{q_j}\right)(x)$$

for any smooth vector function ${\{{\overrightarrow{f}}_k\}}_{k=1}^{\mathrm{\infty }}$ for any $x\in {\mathbb{R}}^n$.

Proof Fix a point $x\in {\mathbb{R}}^n$ and a cube Q centered at x. Set ${\overrightarrow{f}}_j={\overrightarrow{f}}_j^0+{\overrightarrow{f}}_j^{\mathrm{\infty }}$, where ${\overrightarrow{f}}_j^0={\overrightarrow{f}}_j{\chi }_{Q^{\ast }}$. Let ${\overrightarrow{f}}^{\alpha }=f_1^{{\alpha }_1}\cdots f_m^{{\alpha }_m}$ and $Q^{\ast }=(8\sqrt{n}+4)Q$. It is easy to see

$$|T_q(\overrightarrow{f})(z)-\mathcal{C}|\le |T_q\left({\overrightarrow{f}}^0\right)(z)|+\sum _{{\alpha }_1,\dots ,{\alpha }_m}{|T\left({\overrightarrow{f}}^{\alpha }\right)(z)-T\left({\overrightarrow{f}}^{\alpha }\right)(x)|}_q,$$

where $\mathcal{C}={\sum }_{{\alpha }_1,\dots ,{\alpha }_m}{|T(f_1^{{\alpha }_1}\cdots f_m^{{\alpha }_m})(x)|}_q$ and in the last sum each ${\alpha }_j=0$ or ∞ and in each term there is at least one ${\alpha }_j=\mathrm{\infty }$. Since $0<\delta <1/m<1$, it follows that

$$\begin{array}{r}{\left(\frac{1}{|Q|}{\int }_Q|{|T_q(\overrightarrow{f})(z)|}^{\delta }-{|\mathcal{C}|}^{\delta }|dz\right)}^{1/\delta }\\ \le C{\left(\frac{1}{|Q|}{\int }_Q{|T(\overrightarrow{f})(z)-c|}_q^{\delta }dz\right)}^{1/\delta }\\ \le C{\left(\frac{1}{|Q|}{\int }_Q{|T\left({\overrightarrow{f}}^0\right)(z)|}_q^{\delta }dz\right)}^{1/\delta }+C\sum _{{\alpha }_1,\dots ,{\alpha }_m}{\left(\frac{1}{|Q|}{\int }_Q{|T\left({\overrightarrow{f}}^{\alpha }\right)(z)-c|}_q^{\delta }dz\right)}^{1/\delta }\\ \triangleq P_1+P_2,\end{array}$$

where $\mathcal{C}={|c|}_q={({\sum }_{k\ge 1}{|c_k|}^q)}^{1/q}$.

Applying Kolmogorov’s inequality and Theorem 1.2 to $P_1$, we have

$$\begin{array}{rl}{\left(\frac{1}{|Q|}{\int }_Q{|T_q\left({\overrightarrow{f}}^0\right)(z)|}^{\delta }dz\right)}^{1/\delta }& \le C{\parallel T_q\left({\overrightarrow{f}}^0\right)\parallel }_{L^{1/m,\mathrm{\infty }}(Q,\frac{dz}{|Q|})}\\ \le C\prod _{j=1}^m\frac{1}{|Q|}{\int }_Q{|f_j(z)|}_{q_j}dz\\ \le C\prod _{j=1}^{m}M\left({|f_j|}_{q_j}\right)(x).\end{array}$$

We proceed to the estimate for $P_2$. We can take $t={[2\sqrt{n}l(Q)]}^s$. If ${\alpha }_1=\cdots ={\alpha }_m=\mathrm{\infty }$, we have

$$\begin{array}{r}{\left(\frac{1}{|Q|}{\int }_Q{|T(\overrightarrow{f})(z)-c|}_q^{\delta }dz\right)}^{1/\delta }\\ \le \frac{C}{|Q|}{\int }_Q{|T\left({\overrightarrow{f}}^{\mathrm{\infty }}\right)(z)-c|}_{q}dz\\ \le \frac{C}{|Q|}{\int }_Q{\left(\sum _{k=1}^{\mathrm{\infty }}{|T\left({\overrightarrow{f}}_k^{\mathrm{\infty }}\right)(z)-T\left({\overrightarrow{f}}_k^{\mathrm{\infty }}\right)(x)|}^q\right)}^{1/q}dz\\ \le \frac{C}{|Q|}{\int }_Q{\left(\sum _{k=1}^{\mathrm{\infty }}{|{\int }_{{({\mathbb{R}}^n\setminus Q^{\ast })}^m}|K(z,\overrightarrow{y})-K(x,\overrightarrow{y})||f_{1k}\cdots f_{mk}|d\overrightarrow{y}|}^q\right)}^{1/q}dz\\ \le \frac{C}{|Q|}{\int }_Q{\left(\sum _{k=1}^{\mathrm{\infty }}{|{\int }_{{({\mathbb{R}}^n\setminus Q^{\ast })}^m}|K(z,\overrightarrow{y})-K_t^{(0)}(z,\overrightarrow{y})||f_{1k}\cdots f_{mk}|d\overrightarrow{y}|}^q\right)}^{1/q}dz\\ +\frac{C}{|Q|}{\int }_Q{\left(\sum _{k=1}^{\mathrm{\infty }}{|{\int }_{{({\mathbb{R}}^n\setminus Q^{\ast })}^m}|K_t^{(0)}(z,\overrightarrow{y})-K(x,\overrightarrow{y})||f_{1k}\cdots f_{mk}|d\overrightarrow{y}|}^q\right)}^{1/q}dz\\ =P_{21}+P_{22}.\end{array}$$

Since $z\in Q$ and $y_j\in {\mathbb{R}}^n\setminus (8\sqrt{n}+4)Q$, we get $|y_j-z|>(4\sqrt{n}+1)l(Q)>2t^{1/s}$ for all $j=1,\dots ,m$. Applying Assumption (H2), we obtain

$$\begin{array}{rl}{P}_{21}& \le \frac{C}{|Q|}{\int }_Q{\int }_{{({\mathbb{R}}^n\setminus Q^{\ast })}^m}\frac{A{t}^{\epsilon /s}}{{(|z-y_1|+\cdots +|z-y_m|)}^{mn+\epsilon }}{|f_1|}_{q_1}\cdots {|f_m|}_{q_m}d\overrightarrow{y}dz\\ \le \sum _{k=1}^{\mathrm{\infty }}\frac{1}{2^{k\epsilon }}\prod _{j=1}^m\frac{1}{2^{(k+1)n}|Q^{\ast }|}{\int }_{2^{k+1}{Q}^{\ast }}{|f_j|}_{q_j}d{y}_j\\ \le C\prod _{j=1}^{m}M\left({|f_j|}_{q_j}\right)(x).\end{array}$$

Since $x,z\in Q$, $|z-x|\le \sqrt{n}l(Q)\le \frac{1}{2}{t}^{1/s}$. Note that $|y_j-z|>(4\sqrt{n}+1)l(Q)>2t^{1/s}$, for all $j=1,\dots ,m$, hence $\varphi (\frac{|y_j-z|}{t^{1/s}})=0$. Similarly, we get $P_{22}\le {\prod }_{j=1}^{m}M({|f_j|}_{q_j})(x)$.

Now we estimate the typical representative of $P_2$, that is, ${\alpha }_1=\cdots ={\alpha }_l=\mathrm{\infty }$ and ${\alpha }_{l+1}=\cdots ={\alpha }_m=0$.

$$\begin{array}{r}{|T(f_1^{\mathrm{\infty }},\dots ,f_l^{\mathrm{\infty }},f_{l+1}^0,\dots ,f_m^0)(z)-T(f_1^{\mathrm{\infty }},\dots ,f_l^{\mathrm{\infty }},f_{l+1}^0,\dots ,f_m^0)(x)|}_q\\ \le {\int }_{{({\mathbb{R}}^n)}^m}|K(z,\overrightarrow{y})-K_t^{(0)}(z,\overrightarrow{y})|+|K_t^{(0)}(z,\overrightarrow{y})-K(x,\overrightarrow{y})|{|f_1|}_{q_1}\cdots {|f_m|}_{q_m}d\overrightarrow{y}\\ ={\int }_{{({\mathbb{R}}^n)}^m}|K(z,\overrightarrow{y})-K_t^{(0)}(z,\overrightarrow{y})|{|f_1|}_{q_1}\cdots {|f_m|}_{q_m}d\overrightarrow{y}\\ +{\int }_{{({\mathbb{R}}^n)}^m}|K_t^{(0)}(z,\overrightarrow{y})-K(x,\overrightarrow{y})|{|f_1|}_{q_1}\cdots {|f_m|}_{q_m}d\overrightarrow{y}.\end{array}$$

Thus, we get

$$\begin{array}{r}{\left(\frac{1}{|Q|}{\int }_Q{|T(f_1^{\mathrm{\infty }},\dots ,f_l^{\mathrm{\infty }},f_{l+1}^0,\dots ,f_m^0)(z)-c|}_q^{\delta }dz\right)}^{1/\delta }\\ \le \frac{C}{|Q|}{\int }_Q{|T(f_1^{\mathrm{\infty }},\dots ,f_l^{\mathrm{\infty }},f_{l+1}^0,\dots ,f_m^0)(z)-T(f_1^{\mathrm{\infty }},\dots ,f_l^{\mathrm{\infty }},f_{l+1}^0,\dots ,f_m^0)(x)|}_{q}dz\\ \le \frac{C}{|Q|}{\int }_Q{\int }_{{({\mathbb{R}}^n)}^m}|K(z,\overrightarrow{y})-K_t^{(0)}(z,\overrightarrow{y})|{|f_1|}_{q_1}\cdots {|f_m|}_{q_m}d\overrightarrow{y}dz\\ +\frac{C}{|Q|}{\int }_Q{\int }_{{({\mathbb{R}}^n)}^m}|K_t^{(0)}(z,\overrightarrow{y})-K(x,\overrightarrow{y})|{|f_1|}_{q_1}\cdots {|f_m|}_{q_m}d\overrightarrow{y}dz\\ =P_{23}+P_{24}.\end{array}$$

For $P_{23}$, by Assumption (H2), we have

$$\begin{array}{rcl}{P}_{23}& \le & \frac{C}{|Q|}{\int }_Q({\int }_{{({\mathbb{R}}^n\setminus Q^{\ast })}^l}\frac{t^{\epsilon /s}{\prod }_{j=1}^l{|f_j(y_j)|}_{q_j}d{y}_j}{{({\sum }_{j\in \{1,2,\dots ,l\}}|z-y_j|)}^{mn+\epsilon }}\\ +{\int }_{{({\mathbb{R}}^n\setminus Q^{\ast })}^l}\frac{{\prod }_{j=1}^l{|f_j(y_j)|}_{q_j}d{y}_j}{{({\sum }_{j\in \{1,2,\dots ,l\}}|z-y_j|)}^{mn}})\prod _{j=l+1}^m{\int }_{Q^{\ast }}{|f_j(y_j)|}_{q_j}d{y}_{j}dz\\ \le & (\sum _{k=1}^{\mathrm{\infty }}\frac{{|Q^{\ast }|}^{\epsilon /n}}{{(2^k{|Q^{\ast }|}^{1/n})}^{mn+\epsilon }}{\int }_{{({\mathbb{R}}^n\setminus Q^{\ast })}^l}\prod _{j=1}^l{|f_j(y_j)|}_{q_j}d{y}_j\\ +\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{(2^k{|Q^{\ast }|}^{1/n})}^{mn}}{\int }_{{(2^{k+1}{Q}^{\ast }\setminus 2^k{Q}^{\ast })}^l}\prod _{j=1}^l{|f_j(y_j)|}_{q_j}d{y}_j)\prod _{j=l+1}^m{\int }_{Q^{\ast }}{|f_j(y_j)|}_{q_j}d{y}_j\\ \le & C\prod _{j=1}^{m}M\left({|f_j|}_{q_j}\right)(x).\end{array}$$

By a similar argument, we deduce that $P_{24}\le C{\prod }_{j=1}^{m}M({|f_j|}_{q_j})(x)$. In other case, we can also deduce the same estimate with minor modifications on the above arguments. We have thus proved Lemma 3.2. □

Lemma 3.3 Let $0<\delta <\epsilon <1/m$, $1/m<q<\mathrm{\infty }$ and $1/q=1/q_1+\cdots +1/q_m$ with $1<q_1,\dots ,q_m<\mathrm{\infty }$. Suppose that $\overrightarrow{b}\in {(\mathit{BMO})}^l$. There then exists a constant $C>0$ depending only on δ and ε such that

$$\begin{array}{rl}{M}_{\delta }^{\mathrm{♯}}(T_{\mathrm{\Pi }\overrightarrow{b},q}\overrightarrow{f})(x)\le & C\prod _{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}(\prod _{j=1}^m{M}_{L(logL)}\left({|f_j|}_{q_j}\right)(x)+M_{\epsilon }(T_q\overrightarrow{f})(x))\\ +C\sum _{j=1}^{l-1}\sum _{\sigma \in {\mathcal{C}}_j^l}\prod _{i\in \sigma }{\parallel b_i\parallel }_{\mathit{BMO}}{M}_{\epsilon }(T_{\mathrm{\Pi }{b}_{{\sigma }^{\prime }},q}\overrightarrow{f})(x)\end{array}$$

for any smooth vector function ${\{{\overrightarrow{f}}_k\}}_{k=1}^{\mathrm{\infty }}$ for any $x\in {\mathbb{R}}^n$, where ${\sigma }^{\prime }=\{1,\dots ,l\}\setminus \sigma$.

Proof For simplicity of notation, we write $F(\overrightarrow{y})$ instead of the product of m functions $f_1(y_1)\cdots f_m(y_m)$ and let ${\lambda }_j=\frac{1}{|2Q|}{\int }_{2Q}{b}_j(z)dz$, for $j=1,\dots ,l$. Let $x\in {\mathbb{R}}^n$ and Q be a cube centered at x. Then we have

$$\begin{array}{rcl}{T}_{\mathrm{\Pi }\overrightarrow{b}}(\overrightarrow{f})(x)& =& {\int }_{{({\mathbb{R}}^n)}^m}(b_1(x)-b_1(y_1))\cdots (b_l(x)-b_l(y_l))K(x,\overrightarrow{y})F(\overrightarrow{y})d\overrightarrow{y}\\ =& {\int }_{{({\mathbb{R}}^n)}^m}((b_1(x)-{\lambda }_1)-(b_1(y)-{\lambda }_1))\cdots ((b_l(x)-{\lambda }_l)-(b_l(y)-{\lambda }_l))\\ \times K(x,\overrightarrow{y})F(\overrightarrow{y})d\overrightarrow{y}\\ =& \sum _{i=0}^l\sum _{\sigma \in {\mathcal{C}}_i^l}{(-1)}^{l-j}\prod _{j\in \sigma }(b_j(x)-{\lambda }_j){\int }_{{({\mathbb{R}}^n)}^m}\prod _{j\in {\sigma }^{\prime }}(b_j(y_j)-{\lambda }_j)K(x,\overrightarrow{y})F(\overrightarrow{y})d\overrightarrow{y}\\ =& (b_1(x)-{\lambda }_1)\cdots (b_l(x)-{\lambda }_l)T(\overrightarrow{f})(x)+T((b_1({\cdot }_1)-{\lambda }_1)\cdots (b_l({\cdot }_l)-{\lambda }_l)\overrightarrow{f})(x)\\ +\sum _{i=1}^{l-1}\sum _{\sigma \in {\mathcal{C}}_i^l}{(-1)}^{l-j}\prod _{j\in \sigma }(b_j(x)-{\lambda }_j){\int }_{{({\mathbb{R}}^n)}^m}\prod _{j\in {\sigma }^{\prime }}(b_j(y_j)-{\lambda }_j)K(x,\overrightarrow{y})F(\overrightarrow{y})d\overrightarrow{y}.\end{array}$$

Noting that ${\prod }_{j\in {\sigma }^{\prime }}(b_j(y_j)-{\lambda }_j)={\prod }_{j\in {\sigma }^{\prime }}[(b_j(y_j)-b_j(x))-(b_j(x)-{\lambda }_j)]$. Then we get

$$\begin{array}{rcl}{T}_{\mathrm{\Pi }\overrightarrow{b},q}\overrightarrow{f}(x)& \le & |(b_1(x)-{\lambda }_1)\cdots (b_l(x)-{\lambda }_l)|T_q(\overrightarrow{f})(x)\\ +{|T((b_1({\cdot }_1)-{\lambda }_1)\cdots (b_l({\cdot }_l)-{\lambda }_l)\overrightarrow{f})(x)|}_q\\ +C\sum _{i=1}^{l-1}\sum _{\sigma \in {\mathcal{C}}_i^l}\prod _{j\in \sigma }|b_j(x)-{\lambda }_j|T_{\mathrm{\Pi }{b}_{{\sigma }^{\prime },q}}\overrightarrow{f}(x).\end{array}$$

Since $0<\delta <1/m<1$, it follows that

$$\begin{array}{r}{\left(\frac{1}{|Q|}{\int }_Q|{|T_{\mathrm{\Pi }\overrightarrow{b},q}(\overrightarrow{f})(z)|}^{\delta }-{|\mathcal{C}|}^{\delta }|dz\right)}^{1/\delta }\\ \le C{\left(\frac{1}{|Q|}{\int }_Q{|T_{\mathrm{\Pi }\overrightarrow{b}}(\overrightarrow{f})(z)-c|}_q^{\delta }dz\right)}^{1/\delta }\\ \le C{\left(\frac{1}{|Q|}{\int }_Q|(b_1(z)-{\lambda }_1)\cdots (b_l(z)-{\lambda }_l){|T(\overrightarrow{f})(z)|}_q^{\delta }|dz\right)}^{1/\delta }\\ +C\sum _{i=1}^{l-1}\sum _{\sigma \in {\mathcal{C}}_i^l}{\left(\frac{1}{|Q|}{\int }_Q\prod _{j\in \sigma }{(|b_j(z)-{\lambda }_j|T_{\mathrm{\Pi }{b}_{{\sigma }^{\prime },q}}\overrightarrow{f}(z))}^{\delta }dz\right)}^{1/\delta }\\ +C{\left(\frac{1}{|Q|}{\int }_Q{|T((b_1({\cdot }_1)-{\lambda }_1)\cdots (b_l({\cdot }_l)-{\lambda }_l)\overrightarrow{f})(z)-c|}_q^{\delta }dz\right)}^{1/\delta }\\ \triangleq I+\mathit{II}+\mathit{III},\end{array}$$

where $\mathcal{C}={|c|}_q={({\sum }_{k\ge 1}{|c_k|}^q)}^{1/q}$. We can choose $1<p_1,\dots ,p_l<\mathrm{\infty }$ with $\frac{1}{p_1}+\cdots +\frac{1}{p_l}+\frac{1}{\epsilon }=\frac{1}{\delta }$. Since $0<\delta <\epsilon <1/m$, Hölder’s inequality gives

$$\begin{array}{r}I\le C\prod _{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}{M}_{\epsilon }(T_q\overrightarrow{f})(x),\\ \mathit{II}\le C\sum _{i=1}^{l-1}\sum _{\sigma \in {\mathcal{C}}_i^l}\prod _{j\in \sigma }{\parallel b_j\parallel }_{\mathit{BMO}}{M}_{\epsilon }(T_{\mathrm{\Pi }{b}_{{\sigma }^{\prime }},q}\overrightarrow{f})(x).\end{array}$$

Let us estimate term III. Set ${\overrightarrow{f}}_j={\overrightarrow{f}}_j^0+{\overrightarrow{f}}_j^{\mathrm{\infty }}$, where ${\overrightarrow{f}}_j^0={\overrightarrow{f}}_j{\chi }_{Q^{\ast }}$. Let ${\overrightarrow{f}}^{\alpha }=f_1^{{\alpha }_1}\cdots f_m^{{\alpha }_m}$ and $Q^{\ast }=(8\sqrt{n}+4)Q$. Taking $\mathcal{C}={\sum }_{{\alpha }_1,\dots ,{\alpha }_m}{|T((b_1({\cdot }_1)-{\lambda }_1)\cdots (b_l({\cdot }_l)-{\lambda }_l)f_1^{{\alpha }_1}\cdots f_m^{{\alpha }_m})(x)|}_q$, we have

$$\begin{array}{r}|T_{\mathrm{\Pi }\overrightarrow{b},q}(\overrightarrow{f})(z)-\mathcal{C}|\\ \le T_q((b_1({\cdot }_1)-{\lambda }_1)\cdots (b_l({\cdot }_l)-{\lambda }_l){\overrightarrow{f}}^0)(z)\\ +C\sum _{{\alpha }_1,\dots ,{\alpha }_m}|\left(T((b_1({\cdot }_1)-{\lambda }_1)\cdots (b_l({\cdot }_l)-{\lambda }_l){\overrightarrow{f}}^{\alpha })\right)(z)\\ -\left(T((b_1({\cdot }_1)-{\lambda }_1)\cdots (b_l({\cdot }_l)-{\lambda }_l){\overrightarrow{f}}^{\alpha })\right)(x){|}_q,\end{array}$$

where in the last sum each ${\alpha }_j=0$ or ∞ and in each term there is at least one ${\alpha }_j=\mathrm{\infty }$.

If ${\alpha }_1=\cdots ={\alpha }_m=0$, applying Kolmogorov’s inequality and Theorem 1.2, we get

$$\begin{array}{r}{\left(\frac{1}{|Q|}{\int }_Q{|T_q((b_1({\cdot }_1)-{\lambda }_1)\cdots (b_l({\cdot }_l)-{\lambda }_l){\overrightarrow{f}}^0)(z)|}^{\delta }dz\right)}^{1/\delta }\\ \le C{\parallel T_q((b_1({\cdot }_1)-{\lambda }_1)\cdots (b_l({\cdot }_l)-{\lambda }_l){\overrightarrow{f}}^0)\parallel }_{L^{1/m,\mathrm{\infty }}(Q,\frac{dz}{|Q|})}\\ \le C\prod _{j=1}^l\frac{1}{|Q|}{\int }_Q|b_j(y_j)-{\lambda }_j|{|f_j(z)|}_{q_j}dz\prod _{j=l+1}^m\frac{1}{|Q|}{\int }_Q{|f_j(z)|}_{q_j}dz\\ \le C\prod _{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}{\parallel {|f_j|}_{q_j}\parallel }_{L(logL),Q}\prod _{j=l+1}^m\frac{1}{|Q|}{\int }_Q{|f_j(z)|}_{q_j}dz\\ \le C\prod _{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}\prod _{j=1}^m{M}_{L(logL)}\left({|f_j|}_{q_j}\right)(x).\end{array}$$

If ${\alpha }_1=\cdots ={\alpha }_m=\mathrm{\infty }$, we have

$$\begin{array}{r}{\left(\frac{1}{|Q|}{\int }_Q{|T_{\mathrm{\Pi }\overrightarrow{b}}\left({\overrightarrow{f}}^{\mathrm{\infty }}\right)(z)-c|}_q^{\delta }dz\right)}^{1/\delta }\\ \le \frac{C}{|Q|}{\int }_Q{|T_{\mathrm{\Pi }\overrightarrow{b}}\left({\overrightarrow{f}}^{\mathrm{\infty }}\right)(z)-c|}_{q}dz\\ \le \frac{C}{|Q|}{\int }_Q{\left(\sum _{k=1}^{\mathrm{\infty }}{|T_{\mathrm{\Pi }\overrightarrow{b}}\left({\overrightarrow{f}}_k^{\mathrm{\infty }}\right)(z)-T_{\mathrm{\Pi }\overrightarrow{b}}\left({\overrightarrow{f}}_k^{\mathrm{\infty }}\right)(x)|}^q\right)}^{1/q}dz\\ \le \frac{C}{|Q|}{\int }_Q\left(\sum _{k=1}^{\mathrm{\infty }}\right|{\int }_{{({\mathbb{R}}^n\setminus Q^{\ast })}^m}|K(z,\overrightarrow{y})-K(x,\overrightarrow{y})|\\ {\times |(b_1(y_1)-{\lambda }_1)\cdots (b_l(y_l)-{\lambda }_l)f_{1k}\cdots f_{mk}|d\overrightarrow{y}{|}^q)}^{1/q}dz\\ \le \frac{C}{|Q|}{\int }_Q\left(\sum _{k=1}^{\mathrm{\infty }}\right|{\int }_{{({\mathbb{R}}^n\setminus Q^{\ast })}^m}|K(z,\overrightarrow{y})-K_t^{(0)}(z,\overrightarrow{y})|\\ {\times |(b_1(y_1)-{\lambda }_1)\cdots (b_l(y_l)-{\lambda }_l)f_{1k}\cdots f_{mk}|d\overrightarrow{y}{|}^q)}^{1/q}dz\\ +\frac{C}{|Q|}{\int }_Q\left(\sum _{k=1}^{\mathrm{\infty }}\right|{\int }_{{({\mathbb{R}}^n\setminus Q^{\ast })}^m}|K_t^{(0)}(z,\overrightarrow{y})-K(x,\overrightarrow{y})|\\ {\times |(b_1(y_1)-{\lambda }_1)\cdots (b_l(y_l)-{\lambda }_l)f_{1k}\cdots f_{mk}|d\overrightarrow{y}{|}^q)}^{1/q}dz\\ ={\mathit{III}}_1+{\mathit{III}}_2.\end{array}$$

Consider now the term ${\mathit{III}}_1$. Taking $t={[2\sqrt{n}l(Q)]}^s$, we have by Assumption (H2)

$$\begin{array}{rl}{\mathit{III}}_1\le & \frac{C}{|Q|}{\int }_Q{\int }_{{({\mathbb{R}}^n\setminus Q^{\ast })}^m}\sum _{k=1}^{\mathrm{\infty }}\frac{{|Q^{\ast }|}^{\epsilon /n}}{{(2^k{|Q^{\ast }|}^{1/n})}^{mn+\epsilon }}\\ \times |(b_1(y_1)-{\lambda }_1)\cdots (b_l(y_l)-{\lambda }_l)|{|f_1|}_{q_1}\cdots {|f_m|}_{q_m}d\overrightarrow{y}dz\\ \le & C\prod _{j=1}^l\sum _{k=1}^{\mathrm{\infty }}\frac{1}{2^{k\epsilon }}\frac{1}{2^{(k+1)n}|Q^{\ast }|}\\ \times {\int }_{2^{k+1}{Q}^{\ast }}|b_j(y_j)-{\lambda }_j|{|f_j|}_{q_j}d{y}_j\prod _{j=l+1}^m\frac{1}{2^{(k+1)n}|Q^{\ast }|}{\int }_{2^{k+1}{Q}^{\ast }}{|f_j|}_{q_j}d{y}_j\\ \le & C\prod _{j=1}^l\sum _{k=1}^{\mathrm{\infty }}\frac{k}{2^{k\epsilon }}{\parallel b_j\parallel }_{\mathit{BMO}}{\parallel {|f_j|}_{q_j}\parallel }_{L(logL),2^{k+1}{Q}^{\ast }}\prod _{j=l+1}^m\frac{1}{2^{(k+1)n}|Q^{\ast }|}{\int }_{2^{k+1}{Q}^{\ast }}{|f_j|}_{q_j}d{y}_j\\ \le & C\prod _{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}{\parallel {|f_j|}_{q_j}\parallel }_{L(logL),2^{k+1}{Q}^{\ast }}\prod _{j=l+1}^m\frac{1}{2^{(k+1)n}|Q^{\ast }|}{\int }_{2^{k+1}{Q}^{\ast }}{|f_j|}_{q_j}d{y}_j\\ \le & C\prod _{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}\prod _{j=1}^m{M}_{L(logL)}\left({|f_j|}_{q_j}\right)(x).\end{array}$$

Similarly, we get ${\mathit{III}}_2\le {\prod }_{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}{\prod }_{j=1}^m{M}_{L(logL)}({|f_j|}_{q_j})(x)$. Now we consider only the typical representative of III. Similar to the estimates of $P_2$ in Lemma 3.2, we have

$$\begin{array}{r}{\left(\frac{1}{|Q|}{\int }_Q{|T((b_1({\cdot }_1)-{\lambda }_1)\cdots (b_l({\cdot }_l)-{\lambda }_l)f_1^{\mathrm{\infty }},\dots ,f_l^{\mathrm{\infty }},f_{l+1}^0,\dots ,f_m^0)(z)-c|}_q^{\delta }dz\right)}^{1/\delta }\\ \le \frac{C}{|Q|}{\int }_Q|T((b_1({\cdot }_1)-{\lambda }_1)\cdots (b_l({\cdot }_l)-{\lambda }_l)f_1^{\mathrm{\infty }},\dots ,f_l^{\mathrm{\infty }},f_{l+1}^0,\dots ,f_m^0)(z)\\ -T((b_1({\cdot }_1)-{\lambda }_1)\cdots (b_l({\cdot }_l)-{\lambda }_l)f_1^{\mathrm{\infty }},\dots ,f_l^{\mathrm{\infty }},f_{l+1}^0,\dots ,f_m^0)(x){|}_{q}dz\\ \le \prod _{j=1}^l({\int }_{{({\mathbb{R}}^n\setminus Q^{\ast })}^l}\frac{t^{\epsilon /s}|(b_1(y_1)-{\lambda }_1)\cdots (b_l(y_l)-{\lambda }_l)|{\prod }_{j=1}^l{|f_j(y_j)|}_{q_j}d{y}_j}{{({\sum }_{j\in \{1,2,\dots ,l\}}|z-y_j|)}^{mn+\epsilon }}\\ +{\int }_{{({\mathbb{R}}^n\setminus Q^{\ast })}^l}\frac{|(b_1(y_1)-{\lambda }_1)\cdots (b_l(y_l)-{\lambda }_l)|{\prod }_{j=1}^l{|f_j(y_j)|}_{q_j}d{y}_j}{{({\sum }_{j\in \{1,2,\dots ,l\}}|z-y_j|)}^{mn}})\prod _{j=l+1}^m{\int }_{Q^{\ast }}{|f_j(y_j)|}_{q_j}d{y}_j\\ \le C\prod _{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}{\parallel {|f_j|}_{q_j}\parallel }_{L(logL),2^{k+1}{Q}^{\ast }}\prod _{j=l+1}^m\frac{1}{|Q^{\ast }|}{\int }_{Q^{\ast }}{|f_j(z)|}_{q_j}dz\\ \le C\prod _{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}\prod _{j=1}^m{M}_{L(logL)}\left({|f_j|}_{q_j}\right)(x).\end{array}$$

Then Lemma 3.3 is proved. □

Lemma 3.4 Let $0<p<\mathrm{\infty }$, $1/m<q<\mathrm{\infty }$, and $\frac{1}{q_1}+\cdots +\frac{1}{q_m}=\frac{1}{q}$ with $1<q_1,\dots ,q_m<\mathrm{\infty }$ and let $w\in A_{\mathrm{\infty }}$. Suppose that $\overrightarrow{b}\in {(\mathit{BMO})}^l$. Then there exists a constant $C>0$ such that

$${\int }_{{\mathbb{R}}^n}{|T_{\mathrm{\Pi }\overrightarrow{b},q}\overrightarrow{f}|}^{p}w(x)dx\le C\prod _{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}^p{\int }_{{\mathbb{R}}^n}{(\prod _{j=1}^m{M}_{L(logL)}\left({|f_j|}_{q_j}\right)(x))}^{p}w(x)dx$$

(3.2)

for any smooth function $\overrightarrow{f}$ with compact support.

Proof We assume that the right-hand side of (3.2) is finite, since otherwise there is nothing to be proved. For $l=1$, by using Lemma 3.1 and Lemma 3.3, we obtain

$$\begin{array}{rl}{\parallel T_{\mathrm{\Pi }\overrightarrow{b},q}\overrightarrow{f}\parallel }_{L^p(\omega )}& \le {\parallel M_{\delta }{T}_{\mathrm{\Pi }\overrightarrow{b},q}\overrightarrow{f}\parallel }_{L^p(\omega )}\le C{\parallel M_{\delta }^{\mathrm{♯}}{T}_{\mathrm{\Pi }\overrightarrow{b},q}\overrightarrow{f}\parallel }_{L^p(\omega )}\\ \le C{\parallel b_1\parallel }_{\mathit{BMO}}[{\parallel M_{\epsilon }(T_q\overrightarrow{f})\parallel }_{L^p(\omega )}+{\parallel \prod _{j=1}^m{M}_{L(logL)}\left({|f_j|}_{q_j}\right)\parallel }_{L^p(\omega )}]\\ \le C{\parallel b_1\parallel }_{\mathit{BMO}}[{\parallel T_q\overrightarrow{f}\parallel }_{L^p(\omega )}+{\parallel \prod _{j=1}^m{M}_{L(logL)}\left({|f_j|}_{q_j}\right)\parallel }_{L^p(\omega )}]\\ \le C{\parallel b_1\parallel }_{\mathit{BMO}}[{\parallel \prod _{j=1}^{m}M\left({|f_j|}_{q_j}\right)\parallel }_{L^p(\omega )}+{\parallel \prod _{j=1}^m{M}_{L(logL)}\left({|f_j|}_{q_j}\right)\parallel }_{L^p(\omega )}]\\ \le C{\parallel b_1\parallel }_{\mathit{BMO}}{\parallel \prod _{j=1}^m{M}_{L(logL)}\left({|f_j|}_{q_j}\right)\parallel }_{L^p(\omega )}.\end{array}$$

(3.3)

For the general case $l\ge 2$, similarly to the case for $l=1$, we have

$${\parallel T_{\mathrm{\Pi }\overrightarrow{b},q}\overrightarrow{f}\parallel }_{L^p(\omega )}\le C\prod _{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}{\parallel \prod _{j=1}^m{M}_{L(logL)}\left({|f_j|}_{q_j}\right)\parallel }_{L^p(\omega )}.$$

To apply the Fefferman-Stein inequality in (3.3), one needs to verify now that ${\parallel T_{\mathrm{\Pi }\overrightarrow{b},q}\overrightarrow{f}\parallel }_{L^p(\omega )}<\mathrm{\infty }$ and ${\parallel T_q\overrightarrow{f}\parallel }_{L^p(\omega )}<\mathrm{\infty }$. We will only show the first one since the proof of another one is very similar but easier.

Suppose that the symbols $b_j$ and the weight ω are bounded functions. Since $\overrightarrow{f}$ has compact support, we may assume $supp{f}_j\subset B(0,R)$. Then we have

$$\begin{array}{rl}{\parallel T_{\mathrm{\Pi }\overrightarrow{b},q}\overrightarrow{f}\parallel }_{L^p(\omega )}& ={\int }_{|x|\le 2R}{|T_{\mathrm{\Pi }\overrightarrow{b},q}\overrightarrow{f}(x)|}^p\omega (x)dx+{\int }_{|x|>2R}{|T_{\mathrm{\Pi }\overrightarrow{b},q}\overrightarrow{f}(x)|}^p\omega (x)dx\\ =I_1+I_2.\end{array}$$

We choose $s>p$ and $s_1,\dots ,s_m>1$ such that $1/s=1/s_1+\cdots +1/s_m$. Theorem 1.1 and Hölder’s inequality imply

$$\begin{array}{rl}{I}_1& \le C{\int }_{|x|\le 2R}{|T_{\mathrm{\Pi }\overrightarrow{b},q}\overrightarrow{f}|}^{p}dx\le C{\left({\int }_{{\mathbb{R}}^n}{|T_q\overrightarrow{f}|}^{s}dx\right)}^{p/s}{R}^{n(s-p)/s}\\ \le C{\left(\prod _{j=1}^m{\parallel {|f_j|}_{q_j}\parallel }_{L^{s_j}}\right)}^p{R}^{n(s-p)/s}<\mathrm{\infty }.\end{array}$$

Consider now the term $I_2$. Since $|x|>2R$, we have by Assumption (H2)

$$\begin{array}{rcl}|T_{\mathrm{\Pi }\overrightarrow{b},q}\overrightarrow{f}(x)|& \le & C|{\int }_{{(B(0,2R))}^m}|K(x,\overrightarrow{y})||b_1(x)-b_1(y_1)|\cdots |b_l(x)-b_l(y_l)|\\ \times |f_1(y_1)|\cdots |f_m(y_m)|d\overrightarrow{y}{|}_q\\ \le & C{\int }_{{(B(0,2R))}^m}|K(x,\overrightarrow{y})-K_t^{(0)}(x,\overrightarrow{y})|\prod _{j=1}^m{|f_j(y_j)|}_{q_j}d\overrightarrow{y}\\ +{\int }_{{(B(0,2R))}^m}|K_t^{(0)}(x,\overrightarrow{y})|\prod _{j=1}^m{|f_j(y_j)|}_{q_j}d\overrightarrow{y}\\ \le & C\prod _{j=1}^m\frac{1}{{|x|}^n}{\int }_{B(0,|x|)}{|f_j(y_j)|}_{q_j}d{y}_j\le \prod _{j=1}^m{M}_{L(logL)}\left({|f_j|}_{q_j}\right)(x).\end{array}$$

Thus, $I_2\le C{\int }_{{\mathbb{R}}^n}{({\prod }_{j=1}^m{M}_{L(logL)}({|f_j|}_{q_j})(x))}^{p}w(x)dx<\mathrm{\infty }$.

For the general case, we can check the limit (see the proof of [10], Theorem 1.1, we omit the details here). Thus, (3.2) is proved. □

Proof of Theorem 1.3 (i) Since ${\nu }_{\overrightarrow{\omega }}\in A_{mp}\subset A_{\mathrm{\infty }}$, Lemma 3.4 gives

$${\int }_{R^n}{|T_{\mathrm{\Pi }\overrightarrow{b},q}(\overrightarrow{f})(x)|}^p{\nu }_{\overrightarrow{\omega }}dx\le C{\int }_{R^n}\prod _{j=1}^m{M}_{L(logL)}^p\left({|f_j|}_{q_j}\right)(x){\nu }_{\overrightarrow{\omega }}dx.$$

It follows from [8] that there exists $r>1$ such that ${\nu }_{\overrightarrow{\omega }}\in A_{(\frac{p_1}{r},\dots ,\frac{p_m}{r})}$. On the other hand, since $\mathrm{\Phi }(t)=t(1+{log}^{+}t)\le t^r$ for all $t>1$, the generalized Jensen inequality yields that

$${\parallel f_j\parallel }_{L(logL),Q}\le C{\left(\frac{1}{|Q|}{\int }_Q{|f_j(y)|}^{r}dy\right)}^{1/r}$$

for all j. One sees immediately that

$${\parallel T_{\mathrm{\Pi }\overrightarrow{b},q}(\overrightarrow{f})\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}\le C{\parallel \prod _{j=1}^m{M}_r\left({|f_j|}_{q_j}\right)\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}.$$

Since ${\nu }_{\overrightarrow{\omega }}\in A_{(\frac{p_1}{r},\dots ,\frac{p_m}{r})}$, it follows from Hölder’s inequality and the well-known inequality of Fefferman-Stein [9] that

$${\parallel \prod _{j=1}^{m}M\left({|f_j|}_{q_j}\right)\parallel }_{L^{p/r}({\nu }_{\overrightarrow{\omega }})}\le C\prod _{j=1}^m{\parallel {|f_j|}_{q_j}\parallel }_{L^{p_j/r}(M{\omega }_j)}.$$

We then have

$${\parallel T_{\mathrm{\Pi }\overrightarrow{b},q}(\overrightarrow{f})\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}\le C\prod _{j=1}^l{\parallel b_j\parallel }_{\mathit{BMO}}\prod _{j=1}^m{\parallel {|f_j|}_{q_j}\parallel }_{L^{p_j}(M{\omega }_j)}.$$

This is the desired conclusion.

• (ii) Since ${\omega }_j\in A_{p_j}$, there exists $r>0$ such that ${\omega }_j\in A_{p_j/r}$. Analysis similar to that in the proof of Theorem 1.3 (i) shows that

  $${\parallel T_{\mathrm{\Pi }\overrightarrow{b},q}(\overrightarrow{f})\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}\le C{\parallel \prod _{j=1}^m{M}_r\left({|f_j|}_{q_j}\right)\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}.$$

The Hölder inequality implies

$$\begin{array}{rl}{\parallel \prod _{j=1}^m{M}_r\left({|f_j|}_{q_j}\right)\parallel }_{L^p({\nu }_{\overrightarrow{\omega }})}& ={\left({\parallel \prod _{j=1}^{m}M\left({|f_j|}_{q_j}^r\right)\parallel }_{L^{p/r}({\nu }_{\overrightarrow{\omega }})}\right)}^{1/r}\\ \le {\left(\prod _{j=1}^m{\parallel M\left({|f_j|}_{q_j}^r\right)\parallel }_{L^{p_j/r}(\omega )}\right)}^{1/r}\\ \le \prod _{j=1}^m{\parallel {|f_j|}_{q_j}\parallel }_{L^{p_j}({\omega }_j)}.\end{array}$$

(3.4)

To prove Theorem 1.3 holds for $T^{\ast }$, it suffices to prove Lemmas 3.2-3.4 hold for $T^{\ast }$. The proof follows from similar steps in [11] and combines the argument we used in the above lemmas. The key for tackling the new complexities is a very careful deal with the supremum, we refer the reader to [11]. This concludes the proof of the theorem. □