多线性傅里叶乘子的研究源于Coifman和Meyer的重要工作^[2-3].设$\sigma\in L^{\infty}(\mathbb{R} ^{mn})$,多线性傅里叶乘子算子$T_{\sigma}$定义为

(1.1)$\begin{equation}T_{\sigma}(f_1, \cdots, f_m)(x)=\int_{\mathbb{R} ^{mn}}\exp(2\pi{\rm i}x(\xi_1+\cdots+\xi_m))\sigma(\xi_1, \cdots, \xi_m)\hat{f_1}(\xi_1)\cdots\hat{f_m}(\xi_m){\rm d}\vec{\xi}, \end{equation}$

其中$f_1, \cdots, f_m\in {\cal S}(\mathbb{R} ^{n}), {\rm d}\vec{\xi}={\rm d}\xi_1\cdots {\rm d}\xi_m$.

Coifman和Meyer在文献[3]中证明了对于所有的$|\alpha_1|+\cdots+|\alpha_m|\leq s, s\geq 2mn+1$,如果$\sigma\in C^{s}(\mathbb{R} ^{mn}\backslash\{0\})$满足条件

(1.2)$\begin{equation}|\partial^{\alpha_1}_{\xi_1}\cdots\partial^{\alpha_m}_{\xi_m}\sigma(\xi_1, \cdots, \xi_m)|\leqC_{\alpha_1, \cdots, \alpha_m}(|\xi_1|+\cdots+|\xi_m|)^{-(|\alpha_1|+\cdots+|\alpha_m|)}, \end{equation}$

则对于所有的$1<p_1, \cdots, p_m, p<\infty$,其中$1/p=\sum\limits_{1\leq k\leq m}1/p_k$, $T_{\sigma}$从$L^{p_1}(\mathbb{R} ^{n})\times\cdots\times L^{p_m}(\mathbb{R} ^{n})$到$L^{p}(\mathbb{R} ^{n})$是有界的.对于$s\geq mn+1$的情形, Grafakos和Torres^[4]利用多线性Calderón-Zygmund算子理论,把Coifman和Meyer的多线性结果推广到指数$1/m\leq p\leq 1$.

关于这一主题, Tomita给出了一个非常重要的进展.设$\Phi\in {\cal S}(\mathbb{R} ^{mn})$,使得$\rm{supp}\Phi\subset \{(\xi_1, \cdots, \xi_m):1/2\leq|\xi_1|+\cdots+|\xi_m|\leq2\}$,且对于所有的$(\xi_1, \cdots, \xi_m)\in \mathbb{R} ^{mn}\backslash\{0\}$,有

$\sum\limits_{l\in{\Bbb Z}}\Phi(2^{-l}\xi_1, \cdots, 2^{-l}\xi_m)=1.$

令

$\sigma_l(\xi_1, \cdots, \xi_m)=\Phi(\xi_1, \cdots, \xi_m)\sigma(2^{l}\xi_1, \cdots, 2^{l}\xi_m), $

且

$\|\sigma_l\|_{W^{s}(\mathbb{R} ^{mn})}=\left(\int_{\mathbb{R} ^{mn}}(1+|\xi_1|^2+\cdots+|\xi_m|^2)^s|{\cal F}\sigma_l(\xi_1, \cdots, \xi_m)| ^2{\rm d}\vec{\xi}\right)^{1/2}, $

其中${\cal F}\sigma_l$是$\sigma_l$的Fourier变换. Tomita^[5]证明了如果对于某个$s\in (mn/2, mn]$,有

(1.3)$ \begin{equation} \sup\limits_{l\in {\Bbb Z}}\|\sigma_l\|_{W^{s}(\mathbb{R} ^{mn})}<\infty , \end{equation}$

那么$T_{\sigma}$是从$L^{p_1}(\mathbb{R} ^{n})\times\cdots\times L^{p_m}(\mathbb{R} ^{n})$到$L^{p}(\mathbb{R} ^{n})$有界的,其中$p_1, \cdots, p_m, p\in(1, \infty)$且$1/p=\sum\limits_{1\leq k\leq m}1/p_k$. Grafakos和Si^[6]考虑了当$p\leq 1$时$T_{\sigma}$从$L^{p_1}(\mathbb{R} ^{n})\times\cdots\times L^{p_m}(\mathbb{R} ^{n})$到$L^{p}(\mathbb{R} ^{n})$的映射性质.特别地,文献[6,定理1.1]的证明表明,如果对于某个$s>n$, $\sigma$满足(1.3)式,那么$T_{\sigma}$是从$L^{p_1}(\mathbb{R} ^{n})\times\cdots\times L^{p_m}(\mathbb{R} ^{n})$到$L^{p}(\mathbb{R} ^{n})$有界的,其中$p_1, \cdots, p_m\in(mn/s, \infty)$且$1/p=\sum\limits_{1\leq k\leq m}1/p_k$.

焦在文献[1]中考虑了多线性傅里叶乘子算子$T_{\sigma}$的加权估计.令$\sigma$满足(1.3)式, $s\in(mn/2, mn], $$t_1, \cdots, t_m\in [1, 2)$,使得$1/t_1+\cdots+1/t_m=s/n$.如果$ p_k\in(t_k, \infty)$, $k=1, \cdots, m$,且权函数$\omega_1, \cdots, \omega_m$满足$\vec{\omega}\in A_{\vec{p}/\vec{t}}(\mathbb{R} ^{mn})$ ($A_{\vec{p}/\vec{t}}(\mathbb{R} ^{mn})$的定义见下面的定义1.1),那么

(1.4)$\begin{equation}\|T_{\sigma}(f_1, \cdots, f_m)\|_{L^{p}(\mathbb{R} ^{n}, \nu_{\vec{\omega}})}\lesssim\prod\limits_{k=1}^{m}\|f_k\|_{L^{p_k}(\mathbb{R} ^{n}, \omega_k)}.\end{equation}$

另一方面,用稀疏算子控制Calderón-Zygmund算子始于文献[7], $A_2$理论(Hytönen^[8])是其简单的推论.后来,这一结果同时分别被Conde-Alonso和Rey^[9], Lerner和Nazarov^[10]改进到点态有界.以上所有涉及到的结果均要求核函数满足log-Dini条件.最终, Lacey^[11]把log-Dini条件减弱到Dini条件, Roncal和Tapiola^[12]把证明中的界提高到更精确的常数.最近, Lerner^[13]也给出了一种新的证明,且对于某些更一般的算子也适用. Li^[14]将Dini条件减弱到$L^r$-Höremander条件,给出了多线性奇异积分算子的稀疏控制定理.

受以上工作的启发,本文将给出带有多重权的双线性傅里叶乘子算子$T_{\sigma}$的量化的估计.更确切地说,通过用稀疏算子控制双线性奇异积分算子,我们得到了文献[1,定理1.1]的量化的加权估计.为了介绍我们的结果,我们首先介绍一类权函数.

定义1.1  设$m\geq 1$为整数, $\omega_1, \cdots, \omega_m$是权函数, $p_1, \cdots, p_m, p\in (0, \infty)$且$1/p=\sum\limits_{k=1}^m 1/p_k, r_k\in(0, p_k]\ (1\leq k\leq m)$.令$\vec{\omega}=(\omega_1, \cdots, \omega_m), \vec{p}=(p_1, \cdots, p_m)$且$\nu_{\vec{\omega}}=\prod\limits_{k=1}^m \omega_k^{p/p_k}$.我们称$\vec{\omega}\in A_{\vec{p}/\vec{r}}(\mathbb{R} ^{mn})$,如果

$\sup\limits_{Q\subset \mathbb{R} ^{n}}\left(\frac{1}{|Q|}\int_Q\nu_{\vec{\omega}}(x){\rm d}x\right)^{1/p}\prod\limits_{k=1}^m\bigg(\frac{1}{|Q|}\int_Q\omega_k^{-\frac{1}{\frac{p_k}{r_k}-1}}(x){\rm d}x\bigg)^{1/r_k-1/p_k}<\infty, $

当$p_k=r_k, \Big(\frac{1}{|Q|}\int_Q \omega_k^{-\frac{1}{\frac{p_k}{r_k}-1}}(x){\rm d}x\Big)^{1/r_k-1/p_k}$可理解为$(\inf\limits_{x\in Q}\omega_k)^{-1/p_k}$.

注1.1  当$r_1=\cdots=r_m=1, A_{\vec{p}/\vec{r}}(\mathbb{R} ^{mn})$记作$A_{\vec{p}}(\mathbb{R} ^{mn})$ (见文献[15]),当$r_1=\cdots=r_m>1, A_{\vec{p}/\vec{r}}(\mathbb{R} ^{mn})$在文献[16]中给出了定义.

本文我们的主要结果如下.

定理1.1  设$\sigma$是一个多重权且满足(1.3)式,对于某个$s\in(n, 2n], t_1, t_2\in [1, 2)$,使得$1/t_1+1/t_2=s/n$.假设权函数$\omega_1, \omega_2$满足$\vec{\omega}\in A_{\vec{p}/\vec{t}}(\mathbb{R} ^{2n})$,那么存在$\sigma_1, \sigma_2>1$,使得对于$r_1, r_2$ ($t_1<r_1<t_1\sigma_1, t_2<r_2<t_2\sigma_2$),且$p_k\in(t_k, \infty)$ ($k=1, 2$),有

(1.5)$\begin{equation}\|T_{\sigma}(f_1, f_2)\|_{L^{p}(\mathbb{R} ^{n}, \nu_{\vec{\omega}})}\lesssim\frac{1}{s-\frac{n}{r_1}-\frac{n}{r_2}}[\vec{\omega}]_{A_{\vec{p}/\vec{t}}}^{\max\left(1, (\frac{p_1}{r_1})', (\frac{p_2}{r_2})'\right)}\prod\limits_{k=1}^{2}\|f_k\|_{L^{p_k}(\mathbb{R} ^{n}, \omega_k)}.\end{equation}$

注1.2  虽然我们仅仅讨论的是双线性傅里叶乘子算子,但定理1.1的证明对于多线性傅里叶乘子算子仍适用.

文中, $C$表示与主要参数无关的正的常数，但在不同地方取值可能不同.我们用符号$A\lesssim B$表示存在正的常数$C$使得$A\leq CB$.对于任意集合$E\subset \mathbb{R} ^{n}, \chi _E$表示它的特征函数.对于方体$Q\subset \mathbb{R} ^{n}$及$\lambda\in (0, \infty)$,我们用$l(Q)$ (diam$Q$)表示$Q$的边长(直径),用$ \lambda Q$表示中心相同,边长是$Q$的$\lambda$倍的方体.对于$x\in \mathbb{R} ^{n}$且$r>0, B(x, r)$表示以$x$为中心, $r$为半径的球.

在这一部分,我们首先介绍几个在定理1.1的证明中需要用到的引理.我们先回顾一些定义.

$\mathbb{R} ^{n}$中的标准二进制网格包含了所有形如$2^{-k}([0, 1)^n+j), k\in{\Bbb Z}, j\in{{\Bbb Z}}^n$的方体,记作${\cal D}$.对于固定的方体$Q$,用${\cal D}(Q)$表示与$Q$有关的二进制方体构成的集合.

我们称$\mathbb{R} ^n$中的方体族${\cal S}$为$\eta$ -稀疏, $0<\eta<1$,如果对每一个$Q\in {\cal S}$,存在一个子集$E_Q\subset Q$使得$|E_Q|\geq\eta|Q|$,且集合$\{E_Q\}_{Q\in {\cal S}}$两两互不相交.通常$\eta$仅依赖于维数,当参数不重要时我们可以忽略不计.

给定一个稀疏族${\cal S}$, $ r_1, r_2\in [1, \infty)$,定义稀疏算子${\cal A}_{{\cal S}, r_1, r_2}$为

${\cal A}_{{\mathcalS}, r_1, r_2}(f_1, f_2)(x)=\sum\limits_{Q\in{\mathcalS}}\prod\limits_{j=1}^2\left(\frac{1}{|Q|}\int_Q|f_j(y)|^{r_j}{\rm d}y\right)^{\frac{1}{r_j}}.$

首先,受文献[17]的启发,我们建立了带有多重权的稀疏算子${\cal A}_{{\cal S}, r_1, r_2}$的加权常数估计.

设$\sigma\in A_\infty=\bigcup\limits_{p\geq1}A_p$.关于$\sigma$的二进制极大算子定义为

$M_{\sigma}^{{\cal D}}(f)(x)=\sup\limits_{{x\inQ}\atop {Q\in{\cal D}}}\frac{1}{\sigma(Q)}\int_Q|f|\sigma.$

Moen^[18]证明了如下结果,它在我们的证明过程中起到了非常重要的作用.

(2.1)$\begin{equation}\|M_{\sigma}^{{\cal D}}f\|_{L^p(\sigma)}\leqp'\|f\|_{L^p(\sigma)}, 1<p<\infty.\end{equation}$

现在我们给出如下的引理,得到了稀疏算子${\cal A}_{{\cal S}, r_1, r_2}$的量化的加权估计.

引理2.1  设$\omega_1, \omega_2$为权函数, $p_1, p_2, p\in(0, \infty)$其中$1/p=1/p_1+1/p_2, r_k\in(0, p_k]$$ (k=1, 2), 1/r=1/r_1+1/r_2.$令$\vec{\omega}=(\omega_1, \omega_2), \vec{p}=(p_1, p_2), \vec{r}=(r_1, r_2)$且$\nu_{\vec{\omega}}=\prod\limits_{k=1}^2\omega_k^{p/p_k}$.假设$\vec{\omega}\in A_{\vec{p}/\vec{r}}$,则

(2.2)$\begin{equation}\|{\cal A}_{{\mathcalS}, r_1, r_2}(f_1, f_2)\|_{L^p(\nu_{\vec{\omega}})}\lesssim[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\max\left(1, (\frac{p_1}{r_1})', (\frac{p_2}{r_2})'\right)}\prod\limits_{k=1}^2\|f_k\|_{L^{p_k}(\omega_k)}.\end{equation}$

证  令$a_k=p_k/r_k, k=1, 2$.设$\sigma_k=\omega_k^{1-a_k'}$且$f_k\geq 0$,我们有$\sigma_k, \nu_{\vec{\omega}}\in A_{\infty}$ (见文献[15,定理3.6]).我们只需证明

(2.3)$\begin{equation}\|{\cal A}_{{\mathcalS}, r_1, r_2}(f_1\sigma_1^{\frac{1}{r_1}}, f_2\sigma_2^{\frac{1}{r_2}})\|_{L^p(\nu_{\vec{\omega}})}\lesssim[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\max\left(1, (\frac{p_1}{r_1})', (\frac{p_2}{r_2})'\right)}\prod\limits_{k=1}^2\|f_k\|_{L^{p_k}(\omega_k)}.\end{equation}$

根据定义,对任意方体$Q\subset \mathbb{R} ^{n}$,我们有

$[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}\geq\left(\frac{1}{|Q|}\int_Q\nu_{\vec{\omega}}\right)^{\frac{1}{p}}\prod\limits_{k=1}^2\left(\frac{1}{|Q|}\int_Q\omega_k^{1-a_k'}\right)^{\frac{1}{r_k}-\frac{1}{p_k}}.$

记$\beta=\max\left(1, a_1', a_2'\right)$,且假设$0\leq g\leq L^{p'}(\nu_{\vec{\omega}})$.我们有

$\int_{\mathbb{R} ^{n}}{\cal A}_{{\mathcalS}, r_1, r_2}(f_1\sigma_1^{\frac{1}{r_1}}, f_2\sigma_2^{\frac{1}{r_2}})g\nu_{\vec{\omega}}=\sum\limits_{Q\in{\mathcalS}}\int_Q g\nu_{\vec{\omega}}\times\prod\limits_{k=1}^2\bigg(\frac{1}{|Q|}\int_Q|f_k|^{r_k}\sigma_k\bigg)^{\frac{1}{r_k}}.$

由此以及$[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}$的定义,我们可以得到

$\begin{eqnarray*} &&\sum\limits_{Q\in{\cal S}}\int_Q g \nu_{\vec{\omega}}\times\left(\prod\limits_{k=1}^2\frac{1}{|Q|}\int_Q|f_k|^{r_k}\sigma_k\right)^{\frac{1}{r_k}}\\ &\leq &[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\beta}\sum\limits_{Q\in{\mathcal S}}\frac{|Q|^{\frac{\beta}{r}-\frac{1}{r}}}{\nu_{\vec{\omega}}(Q)^{\frac{\beta}{p}-1}\prod\limits_{k=1}^2\sigma_k(Q)^{\beta(\frac{1}{r_k}-\frac{1}{p_k})-\frac{1}{r_k}}} \times\left(\frac{1}{\nu_{\vec{\omega}}(Q)}\int_Q g \nu_{\vec{\omega}}\right)\\ &&\times\left(\prod\limits_{k=1}^2\frac{1}{\sigma_k(Q)}\int_Q|f_k|^{r_k}\sigma_k\right)^{\frac{1}{r_k}}\\ &\leq &2^{\frac{\beta}{r}-\frac{1}{r}}[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\beta}\sum\limits_{Q\in{\mathcal S}}\frac{|E_Q|^{\frac{\beta}{r}-\frac{1}{r}}}{\nu_{\vec{\omega}}(Q)^{\frac{\beta}{p}-1}\prod\limits_{k=1}^2\sigma_k(Q)^{\beta(\frac{1}{r_k}-\frac{1}{p_k})-\frac{1}{r_k}}} \times\left(\frac{1}{\nu_{\vec{\omega}}(Q)}\int_Q g \nu_{\vec{\omega}}\right)\\ &&\times\left(\prod\limits_{k=1}^2\frac{1}{\sigma_k(Q)}\int_Q|f_k|^{r_k}\sigma_k\right)^{\frac{1}{r_k}}\\ &\leq &2^{\frac{\beta}{r}-\frac{1}{r}}[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\beta}\sum\limits_{Q\in{\mathcal S}}\frac{|E_Q|^{\frac{\beta}{r}-\frac{1}{r}}}{\nu_{\vec{\omega}}(E_Q)^{\frac{\beta}{p}-1}\prod\limits_{k=1}^2\sigma_k(E_Q)^{\beta(\frac{1}{r_k}-\frac{1}{p_k})-\frac{1}{r_k}}} \times\left(\frac{1}{\nu_{\vec{\omega}}(Q)}\int_Q g \nu_{\vec{\omega}}\right)\\ &&\times\left(\prod\limits_{k=1}^2\frac{1}{\sigma_k(Q)}\int_Q|f_k|^{r_k}\sigma_k\right)^{\frac{1}{r_k}}, \end{eqnarray*} $

其中最后一个不等式我们用到了$\nu_{\vec{\omega}}(Q)\geq \nu_{\vec{\omega}}(E_Q), \sigma_k(Q)\geq\sigma_k(E_Q)$,且指数为正.另一方面,由Hölder's不等式,我们有

$|E_Q|=\int_{E_Q}\nu_{\vec{\omega}}^{~\frac{r}{p}}\prod\limits_{k=1}^2\sigma_k^{r(\frac{1}{r_k}-\frac{1}{p_k})}\leq\nu_{\vec{\omega}}(E_Q)^{\frac{r}{p}} \prod\limits_{k=1}^2\sigma_k(E_Q)^{r(\frac{1}{r_k}-\frac{1}{p_k})}.$

由此我们可以得到

$ \begin{eqnarray*}&&\sum\limits_{Q\in{\cal S}}\int_Q g \nu_{\vec{\omega}}\times\left(\prod\limits_{k=1}^2\frac{1}{|Q|}\int_Q|f_k|^{r_k}\sigma_k\right)^{\frac{1}{r_k}}\\ &\leq &2^{\frac{\beta}{r}-\frac{1}{r}}[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\beta}\sum\limits_{Q\in{\mathcal S}}\frac{\nu_{\vec{\omega}}(E_Q)^{\frac{\beta-1}{p}}\prod\limits_{k=1}^2\sigma_k(E_Q)^{(\beta-1)(\frac{1}{r_k}-\frac{1}{p_k})}} {\nu_{\vec{\omega}}(E_Q)^{\frac{\beta}{p}-1}\prod\limits_{k=1}^2\sigma_k(E_Q)^{\beta(\frac{1}{r_k}-\frac{1}{p_k})-\frac{1}{r_k}}} \\&&\times\left(\frac{1}{\nu_{\vec{\omega}}(Q)}\int_Q g \nu_{\vec{\omega}}\right)\left(\prod\limits_{k=1}^2\frac{1}{\sigma_k(Q)}\int_Q|f_k|^{r_k}\sigma_k\right)^{\frac{1}{r_k}}\\ &\leq &2^{\frac{\beta-1}{r}}[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\beta}\sum\limits_{Q\in{\mathcal S}}\nu_{\vec{\omega}}(E_Q)^{1-\frac{1}{p}}\prod\limits_{k=1}^2\sigma_k(E_Q)^{\frac{1}{p_k}} \left(\frac{1}{\nu_{\vec{\omega}}(Q)}\int_Q g \nu_{\vec{\omega}}\right) \left(\prod\limits_{k=1}^2\frac{1}{\sigma_k(Q)}\int_Q|f_k|^{r_k}\sigma_k\right)^{\frac{1}{r_k}} \\&=&2^{\frac{\beta-1}{r}}[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\beta}\sum\limits_{Q\in{\mathcalS}}\nu_{\vec{\omega}}(E_Q)^{\frac{1}{p'}}\prod\limits_{k=1}^2\sigma_k(E_Q)^{\frac{1}{p_k}}\left(\frac{1}{\nu_{\vec{\omega}}(Q)}\int_Q g\nu_{\vec{\omega}}\right)\left(\prod\limits_{k=1}^2\frac{1}{\sigma_k(Q)}\int_Q|f_k|^{r_k}\sigma_k\right)^{\frac{1}{r_k}}\\&=&2^{\frac{\beta-1}{r}}[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\beta}\sum\limits_{Q\in{\mathcalS}}\left[\left(\frac{1}{\nu_{\vec{\omega}}(Q)}\int_Q g\nu_{\vec{\omega}}\right)\nu_{\vec{\omega}}(E_Q)^{\frac{1}{p'}}\right]\left[\prod\limits_{k=1}^2\left(\frac{1}{\sigma_k(Q)}\int_Q|f_k|^{r_k}\sigma_k\right)\sigma_k(E_Q)^{\frac{r_k}{p_k}}\right]^{\frac{1}{r_k}}.\end{eqnarray*} $

再由Hölder's不等式及$\{E_Q\}_{Q\in {\cal S}}$的两两互不相交,有

$\begin{eqnarray*}&&\sum\limits_{Q\in{\cal S}}\int_Q g\nu_{\vec{\omega}}\times\left(\prod\limits_{k=1}^2\frac{1}{|Q|}\int_Q|f_k|^{r_k}\sigma_k\right)^{\frac{1}{r_k}}\\&\leq&2^{\frac{\beta-1}{r}}[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\beta}\left[\sum\limits_{Q\in{\mathcalS}}\left(\frac{1}{\nu_{\vec{\omega}}(Q)}\int_Q g\nu_{\vec{\omega}}\right)^{p'}\nu_{\vec{\omega}}(E_Q)\right]^{\frac{1}{p'}}\\&&\times\prod\limits_{k=1}^2\left[\sum\limits_{Q\in{\mathcalS}}\left(\frac{1}{\sigma_k(Q)}\int_Q|f_k|^{r_k}\sigma_k\right)^{\frac{p_k}{r_k}}\sigma_k(E_Q)\right]^{\frac{1}{p_k}}\\&\leq&2^{\frac{\beta-1}{r}}[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\beta}\|M_{\nu_{\vec{\omega}}}^{\mathcal{D}}(g)\|_{L^{p'}(\nu_{\vec{\omega}})}\times\prod\limits_{k=1}^2\|M_{\sigma_i}^{\mathcal{D}}(|f_k|^{r_k})\|_{L^{p_k/r_k}(\sigma_k)}^{1/r_k}\\&\lesssim&2^{\frac{\beta-1}{r}}[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\beta}\|g\|_{L^{p'}(\nu_{\vec{\omega}})}\times\prod\limits_{k=1}^2\|f_k^{r_k}\|_{L^{p_k/r_k}(\sigma_k)}^{1/r_k}\\&=&2^{\frac{\beta-1}{r}}[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\beta}\|g\|_{L^{p'}(\nu_{\vec{\omega}})}\times\prod\limits_{k=1}^2\|f_k\|_{L^{p_k}(\sigma_k)}, \end{eqnarray*}$

其中最后一个不等式是由(2.1)式得到.这样我们就证明了(2.3)式.

设$r_1, \cdots, r_m\in (0, \infty)$,令$\vec{r}=(r_1, \cdots, r_m)$.极大算子${\mathcal M}_{\vec{r}}$定义为

${\mathcalM}_{\vec{r}}(f_1, \cdots, f_m)(x)=\sup\limits_{B\nix}\prod\limits_{k=1}^m\left(\frac{1}{|B|}\int_B|f_k(x)|^{r_k}{\rm d}x\right)^{1/r_k}, $

其中上确界取遍包含$x$的所有球.

对于一个给定的算子$T$,主极大截断算子${\cal M}_T$定义为

${\cal M}_T f(x)=\sup\limits_{Q\ni x}{\rm ess}\sup\limits_{\xi\in Q}|T(f\chi_{\mathbb{R} ^{n}\backslash 3Q})(\xi)|, $

其中上确界取遍$\mathbb{R} ^{n}$中包含$x$的所有方体.

给定一个方体$Q_0$, $x\in Q_0$,局部极大算子定义为

${\cal M}_{T, Q_0} f(x)=\sup\limits_{Q\ni x, Q\subset Q_0} {\rm ess}\sup\limits_{\xi\in Q} |T(f\chi_{3Q_0\backslash3Q})(\xi)|.$

为了叙述我们的最主要结果,我们需要给出多线性主极大截断算子的定义.给定一个多线性算子$T$,定义

${\cal M}_T (f_1, \cdots, f_m)(x)=\sup\limits_{Q\ni x} {\rm ess}\sup\limits_{\xi\in Q} |T(f_1, \cdots, f_m)(\xi)-T(f_1 \chi _{3Q}, \cdots, f_m \chi _{3Q})(\xi)|.$

给定方体$Q_0$, $x\in Q_0$我们定义局部极大算子为

$\begin{eqnarray*}&&{\cal M}_{T, Q_0} (f_1, \cdots, f_m)(x)\\&=&\sup\limits_{Q\ni x, Q\subset Q_0} {\rm ess}\sup\limits_{\xi\in Q} |T(f_1 \chi _{3Q_0}, \cdots, f_m \chi_{3Q_0})(\xi)-T(f_1\chi _{3Q}, \cdots, f_m \chi _{3Q})(\xi)|.\end{eqnarray*}$

下面,我们证明如下引理.

引理2.2  假设$T$从$L^{r_1}\times L^{r_2}$到$ L^{r, \infty}$有界, $r_1, r_2\in[1, \infty)$,且$r\in (\frac{1}{2}, \infty)$, $1/r=1/r_1+1/r_2.$那么对于a.e. $x\in Q_0$,有

(2.4)$\begin{equation}|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)|\leq C_n\|T\|_{L^{r_1}\timesL^{r_2}\rightarrow L^{r, \infty}}|f_1(x) f_2(x)|+{\mathcalM}_{T, Q_0}(f_1, f_2)(x).\end{equation}$

证  假设$x\in \rm{int} Q_0$,且$x$是$T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})$的连续逼近点.那么对任意$\varepsilon>0$,集合

$E_s(x):=\{y\in B(x, s):|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(y)-T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)|<\varepsilon\}$

满足$\lim\limits_{s\rightarrow 0}\frac{|E_s(x)|}{|B(x, s)|}=1$,其中$B(x, s)$是以$x$为中心, $s$为半径的开球.

以$x$为中心,包含$B(x, s)$的最小方体记为$Q(x, s)$.则对于a.e. $y\in E_s(x)$,有

$\begin{eqnarray*}|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)|&\leq& |T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(y)|+\varepsilon \\&\leq&|T(f_1\chi_{3Q(x, s}, f_2\chi_{3Q(x, s)})(y)|+{\cal M}_{T, Q_0}(f_1, f_2)(x)+\varepsilon, \end{eqnarray*}$

因此

$\begin{eqnarray*}&&|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)|\\&\leq&{\rm ess}\inf\limits_{y\inE_s(x)}|T(f_1\chi_{3Q(x, s}, f_2\chi_{3Q(x, s)})(y)|+{\mathcalM}_{T, Q_0} (f_1, f_2)(x)+\varepsilon\\&\leq&|E_s(x)|^{-r}\|T(f_1\chi_{3Q(x, s}, f_2\chi_{3Q(x, s)})\|_{L^{r, \infty}}+{\mathcalM}_{T, Q_0} (f_1, f_2)(x)+\varepsilon\\&\leq& \|T\|_{L^{r_1}\times L^{r_2}\rightarrowL^{r, \infty}}\frac{1}{|E_s(x)|^r}\prod\limits_{i=1}^2\bigg(\int_{3Q(x, s)}|f_i(y)|^{r_i}{\rm d}y\bigg)^{\frac{1}{r_i}}+{\mathcalM}_{T, Q_0} (f_1, f_2)(x)+\varepsilon.\end{eqnarray*}$

另外假设$x$是$|f_1|^{r_1}$和$|f_2|^{r_2}$的Lebesgue点,令$s\rightarrow 0$, $\varepsilon\rightarrow 0$,引理得证.

接下来,我们给出一类双线性奇异积分算子的稀疏控制公式.它的证明与Lerner^[13]和Li^[14]的方法类似.

引理2.3  假设$T$是双线性奇异积分算子, $r_1, r_2\in[1, \infty)$,且$r\in (\frac{1}{2}, \infty)$.如果$T$及其相应的主极大截断算子${\cal M}_T$从$L^{r_1}(\mathbb{R} ^{n})\times L^{r_2}(\mathbb{R} ^{n})$到$L^{r, \infty}(\mathbb{R} ^{n})$都是有界的,那么对于具有紧支集的函数$f_i\in L^{r_i}(\mathbb{R} ^{n}), i=1, 2, $存在一个$\frac{1}{2}\frac{1}{3^n}$ -稀疏族${\cal S}$,使得对于a.e. $x\in \mathbb{R} ^{n}$,有

$|T(f_1, f_2)(x)|\lesssim {\cal A}_{{\cal S}, r_1, r_2}(f_1, f_2)(x).$

证  固定一个方体$Q_0\subset \mathbb{R} ^{n}$,我们证明存在一个$\frac{1}{2}$ -稀疏族${\cal F}\subset {\cal D}(Q_0)$,使得对于a.e. $x\in Q_0$,有

$\begin{eqnarray*}|T(f_1\chi_{3Q}, f_2\chi_{3Q})(x)|&\leq &c_n\left(\|T\|_{L^{r_1}\times L^{r_2}\rightarrow L^{r, \infty}}+\|{\cal M}T\|_{L^{r_1}\times L^{r_2}\rightarrowL^{r, \infty}}\right)\\&&\times \sum\limits_{Q\in{\mathcalS}}\prod\limits_{j=1}^2\left(\frac{1}{|Q|}\int_Q|f_j(y)|^{r_j}{\rm d}y\right)^{\frac{1}{r_j}}.\end{eqnarray*}$

我们只需证明如下的结果

(2.5)$\begin{equation}|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)|\chi_{Q_0}\leqc_{n, T}\langle|f_1|^{r_1}\rangle_{3Q_0}^{\frac{1}{r_1}}\langle|f_2|^{r_2}\rangle_{3Q_0}^{\frac{1}{r_2}}+\sum\limits_j |T(f_1\chi_{3P_j}, f_2\chi_{3P_j})(x)|\chi_{P_j}, \end{equation}$

其中$P_j$是$Q_0$的互不相交的二进制子方体,即$P_j\in {\cal D}(Q_0)$且$\sum\limits_j|P_j|\leq\frac{1}{2}|Q_0|$.考虑到两两互不相交的方体$P_j\in \mathcal {D}(Q_0)$,我们有

$\begin{eqnarray*}&&|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)|\chi_{Q_0}\\&=&|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)|\chi_{Q_0\backslash\cup_jP_j}+\sum\limits_j|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)|\chi_{P_j}\\&\leq &|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)|\chi_{Q_0\backslash\cup_j P_j}+\sum\limits_j|T(f_1\chi_{3P_j}, f_2\chi_{3P_j})(x)|\chi_{P_j}\\&&+\sum\limits_j|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)-T(f_1\chi_{3P_j}, f_2\chi_{3P_j})(x)|\chi_{P_j}, \end{eqnarray*}$

因此,为了证明(2.5)式,只需证明存在两两互不相交的方体$P_j\in \mathcal {D}(Q_0)$且$\sum\limits_j|P_j|\leq\frac{1}{2}|Q_0|$,使得对于a.e. $x\in Q_0$,有

$\begin{eqnarray*}&&\sum\limits_j|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)-T(f_1\chi_{3P_j}, f_2\chi_{3P_j})(x)|\chi_{P_j}\\&&+|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)|\chi_{Q_0\backslash\cup_jP_j}\leqc_{n, T}\langle|f_1|^{r_1}\rangle_{3Q_0}^{\frac{1}{r_1}}\langle|f_2|^{r_2}\rangle_{3Q_0}^{\frac{1}{r_2}}.\end{eqnarray*}$

根据假设, ${\cal M}_T$是从$L^{r_1}\times L^{r_2}$到$L^{r, \infty}$有界的.因此,存在某个足够大的$c_n$使得集合

$\begin{eqnarray*}E:&=&\{x\inQ_0:|f_1(x)f_2(x)|>c_n\langle|f_1|^{r_1}\rangle_{3Q_0}^{\frac{1}{r_1}}\langle|f_2|^{r_2}\rangle_{3Q_0}^{\frac{1}{r_2}}\}\\&&\cup\{x\in Q_0:{\cal M}_{T, Q_0}(f_1, f_2)(x)>c_n\|{\mathcalM}_T\|_{L^{r_1}\times L^{r_2}\rightarrowL^{r, \infty}}\langle|f_1|^{r_1}\rangle_{3Q_0}^{\frac{1}{r_1}}\langle|f_2|^{r_2}\rangle_{3Q_0}^{\frac{1}{r_2}}\}\end{eqnarray*}$

满足$|E|\leq \frac{1}{2^{n+1}}|Q_0|$.对于函数$\chi_E$应用Calder$\acute{0}$n-Zygmund分解,则存在两两互不相交的方体$P_j\in {\cal D}(Q_0)$,使得

$\frac{1}{2^{n+1}}|P_j|\leq|P_j\cap E|<\frac{1}{2}|P_j|$

且$|E\backslash \cup_jP_j |=0$.因此$\sum\limits_j|P_j|\leq\frac{1}{2}|Q_0|$且$P_j\cap E^c \neq\varnothing$.故

$\begin{eqnarray*}&&{\rm ess}\sup\limits_{\xi\in P_j}|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)-T(f_1\chi_{3P_j}, f_2\chi_{3P_j})(x)|\\&\leq& c_n\|{\cal M}_T\|_{L^{r_1}\times L^{r_2}\rightarrowL^{r, \infty}}\langle|f_1|^{r_1}\rangle_{3Q_0}^{\frac{1}{r_1}}\langle|f_2|^{r_2}\rangle_{3Q_0}^{\frac{1}{r_2}}.\end{eqnarray*}$

另一方面,由引理2.2,对于a.e. $x\in Q_0\backslash \cup_jP_j $,我们有

$|T(f_1\chi_{3Q_0}, f_2\chi_{3Q_0})(x)|\leq c_n (\|T\|_{L^{r_1}\times L^{r_2}\rightarrow L^{r, \infty}}+\|{\mathcalM}_T\|_{L^{r_1}\times L^{r_2}\rightarrowL^{r, \infty}})\langle|f_1|^{r_1}\rangle_{3Q_0}^{\frac{1}{r_1}}\langle|f_2|^{r_2}\rangle_{3Q_0}^{\frac{1}{r_2}}.$

因此,由以上估计我们得到(2.5)式,其中

$c_{n, T}\overline{\sim} \|T\|_{L^{r_1}\times L^{r_2}\rightarrow L^{r, \infty}}+\|{\mathcalM}_T\|_{L^{r_1}\times L^{r_2}\rightarrow L^{r, \infty}}.$

下面的证明与文献[13]中的一样,这里就不再赘叙.因此我们完成了引理的证明.

接下来,我们证明双线性奇异积分算子$T$的主极大截断算子${\mathcal M}_T$从$L^{r_1}(\mathbb{R} ^{n})\times L^{r_2}(\mathbb{R} ^{n})$到$L^{r, \infty}(\mathbb{R} ^{n})$是有界的.

设$K(x;y_1, y_2)$是定义在$(\mathbb{R} ^{n})^3\backslash \{x=y_1=y_2\}$上的局部可积函数.算子$T$被称为具有核函数$K$的双线性奇异积分算子,如果$T$是双线性的,且满足

(2.6)$\begin{equation}T(f_1, f_2)(x)=\int_{(\mathbb{R} ^{n})^2}K(x;y_1, y_2)f(y_1)f(y_2){\rm d}y_1{\rm d}y_2, \end{equation}$

其中$f_1, f_2$是具有紧支集的有界函数,且a.e. $x\in \mathbb{R} ^{n}\backslash\bigcap\limits _{j=1}^2 {\rm supp} f_j$.

引理2.4  设$T$是具有核函数$K$的双线性奇异积分算子,满足(2.6)式.对于$x, x', y_1, $$y_2\in \mathbb{R} ^{n}$,令

$V(x, x';y_1, y_2)=|K(x;y_1, y_2)-K(x';y_1, y_2)|.$

假设对于某固定的$r_1, r_2\in (1, \infty)$,有

(ⅰ) $T$从$L^{r_1}(\mathbb{R} ^{n})\times L^{r_2}(\mathbb{R} ^{n})$到$L^{r, \infty}(\mathbb{R} ^{n})$有界,其中$1/r=1/r_1+1/r_2$;

(ⅱ)存在常数$s\in(\frac{n}{r_1}+\frac{n}{r_2}, \frac{n}{r_1}+\frac{n}{r_2}+1)$,使得对于任意半径为$R$的球$B$, $x, x'\in \frac{1}{4}B$,以及非负整数$j_1, j_2$,且$j^{*}=\max\limits_{1\leq k\leq2}j_k>2, $

$\bigg(\int_{S_{j_1}(B)}\bigg(\int_{S_{j_2}(B)}|V(x, x';y_1, y_2)|^{r'_2}{\rm d}y_2\bigg)^{\frac{r'_1}{r'_2}}{\rm d}y_1\bigg)^{\frac{1}{r'_1}}\lesssim\frac{R^{s-n/r_1-n/r_2}}{|2^{j^{*}}B|^{s/n}}, $

则${\cal M}_T$从$L^{r_1}(\mathbb{R} ^{n})\times L^{r_2}(\mathbb{R} ^{n})$到$L^{r, \infty}$是有界的.

证  设$x, x', \xi\in Q\subset \frac{1}{2}\cdot3Q$.以$x$为中心, 2diam$Q$为半径的闭球记为$B$.则$3Q\subset B $,且我们有

$\begin{eqnarray*}|T(f_1\chi_{\mathbb{R} ^{n}\backslash3Q}, f_2\chi_{\mathbb{R} ^{n}\backslash 3Q})(\xi)|&\leq&|T(f_1\chi_{\mathbb{R} ^{n}\backslashB}, f_2\chi_{\mathbb{R} ^{n}\backslashB})(\xi)-T(f_1\chi_{\mathbb{R} ^{n}\backslashB}, f_2\chi_{\mathbb{R} ^{n}\backslash B})(x')|\\&&+|T(f_1\chi_{B\backslash 3Q}, f_2\chi_{B\backslash3Q})(\xi)|+|T(f_1\chi_{\mathbb{R} ^{n}\backslashB}, f_2\chi_{\mathbb{R} ^{n}\backslash B})(x')|.\end{eqnarray*}$

根据光滑性条件,

$\begin{eqnarray*}&&|T(f_1\chi_{\mathbb{R} ^{n}\backslashB}, f_2\chi_{\mathbb{R} ^{n}\backslashB})(\xi)-T(f_1\chi_{\mathbb{R} ^{n}\backslashB}, f_2\chi_{\mathbb{R} ^{n}\backslash B})(x')| \\&\leq& \int_{\mathbb{R} ^{n}\backslashB}\int_{\mathbb{R} ^{n}\backslash B}|k(\xi;y_1, y_2)-k(x';y_1, y_2)||f_1(y_1)| |f_2(y_2)|{\rm d}y_1{\rm d}y_2\\&=&\sum\limits_{j_1=1}^{\infty}\sum\limits_{j_2=1}^{\infty}\int_{{2^{j_1}B\backslash}2^{j_1-1}B}\int_{{2^{j_2}B\backslash}2^{j_2-1}B}|V(\xi, x';y_1, y_2)| |f_1(y_1)| |f_2(y_2)|{\rm d}y_1{\rm d}y_2\\&\leq&\sum\limits_{j_1=1}^{\infty}\sum\limits_{j_2=1}^{\infty}\left(\int_{2^{j_1}B}|f_1(y_1)|^{r_1}{\rm d}y_1\right)^{\frac{1}{r_1}}\left(\int_{2^{j_2}B}|f_2(y_2)|^{r_2}{\rm d}y_2\right)^{\frac{1}{r_2}}\\&&\times\bigg(\int_{S_{j_1}(B)}\bigg(\int_{S_{j_2}(B)}|V(\xi, x';y_1, y_2)|^{r'_2}{\rm d}y_2\bigg)^{\frac{r'_1}{r'_2}}{\rm d}y_1\bigg)^{\frac{1}{r'_1}}\\&\lesssim&\sum\limits_{j_1=1}^{\infty}\sum\limits_{j_2=1}^{\infty}\bigg(\frac{1}{|2^{j_1}B|}\int_{2^{j_1}B}|f_1(y_1)|^{r_1}{\rm d}y_1\bigg)^{\frac{1}{r_1}}\bigg(\frac{1}{|2^{j_2}B|}\int_{2^{j_2}B}|f_2(y_2)|^{r_2}{\rm d}y_2\bigg)^{\frac{1}{r_2}}\\&&\times|2^{j_1}B|^{\frac{1}{r_1}}|2^{j_2}B|^{\frac{1}{r_2}}\frac{|B|^{s/n-1/r_1-1/r_2}}{|2^{j^{*}}B|^{s/n}}\\&\lesssim&\sum\limits_{j^*=1}^{\infty}\frac{|B|^{s/n-1/r_1-1/r_2}}{|2^{j^{*}}B|^{s/n}}|2^{j^{*}}B|^{1/r_1+1/r_2}{\mathcalM}_{\vec{r}}(f_1, f_2)(x)\\&\lesssim&\sum\limits_{j^*=1}^{\infty}\frac{1}{2^{j^{*}(\frac{s}{n}-\frac{1}{r_1}-\frac{1}{r_2})}}{\mathcalM}_{\vec{r}}(f_1, f_2)(x)\\ &\lesssim& \frac{1}{\frac{s}{n}-\frac{1}{r_1}-\frac{1}{r_2}} {\mathcalM}_{\vec{r}}(f_1, f_2)(x)\\&\lesssim &\frac{1}{s-\frac{n}{r_1}-\frac{n}{r_2}}{\mathcalM}_{\vec{r}}(f_1, f_2)(x).\end{eqnarray*}$

然后取$x'\in Q$的$L^{r/2}$平均,我们得到

$\begin{eqnarray*}|T(f_1\chi_{B\backslash 3Q}, f_2\chi_{B\backslash 3Q})(\xi)|&\leq&\left(\frac{1}{|Q|}\int_Q|T(f_1\chi_{B\backslash3Q}, f_2\chi_{B\backslash3Q})(x')|^{\frac{r}{2}}{\rm d}x'\right)^{\frac{2}{r}}\\&\lesssim& \|T(f_1\chi_{B\backslash 3Q}, f_2\chi_{B\backslash3Q})\|_{L^{r, \infty}(Q, \frac{{\rm d}x'}{|Q|})}\\&\lesssim &\|T\|_{L^{r_1}\times L^{r_2}\rightarrowL^{r, \infty}}\left(\frac{1}{|B|}\int_B|f_1|^{r_1}\right)^{\frac{1}{r_1}}\left(\frac{1}{|B|}\int_B|f_2|^{r_2}\right)^{\frac{1}{r_2}}\\&\lesssim &{\cal M}_{\vec{r}}(f_1, f_2)(x), \end{eqnarray*}$

并且

$\begin{eqnarray*}|T(f_1\chi_{\mathbb{R} ^{n}\backslashB}, f_2\chi_{\mathbb{R} ^{n}\backslash B})(x')|&\leq&\left(\frac{1}{|Q|}\int_Q|T(f_1\chi_{\mathbb{R} ^{n}\backslashB}, f_2\chi_{\mathbb{R} ^{n}\backslashB})(x')|^{\frac{r}{2}}{\rm d}x'\right)^{\frac{2}{r}}\\&\leq &M_{\frac{r}{2}}(T(f_1, f_2))(x).\end{eqnarray*}$

因此我们可以得到

${\cal M}_{T}(f_1, f_2)(x)\lesssim \frac{1}{s-\frac{n}{r_1}-\frac{n}{r_2}}({\cal M}_{\vec{r}}(f_1, f_2)(x)+M_{\frac{r}{2}}(T(f_1, f_2))(x)).$

显然${\mathcal M}_{\vec{r}}(f_1, f_2)(x)$从$L^{r_1}\times L^{r_2}$到$L^{r, \infty}$是有界的.因此我们仅需证明

$\|M_{\frac{r}{2}}(T(f_1, f_2))\|_{L^{r, \infty}}\lesssim\|f_1\|_{L^{r_1}}\|f_2\|_{L^{r_2}}.$

事实上,利用文献[19]的(11)式可以得到这一结果.这样,我们就证明了${\cal M}_T$从$L^{r_1}\times L^{r_2}$到$ L^{r, \infty}$是有界的.

最后,我们证明$A_{\vec{p}}$的性质.

引理2.5  假设$\vec{\omega}\in A_{\vec{p}/\vec{t}}(\mathbb{R} ^{2n})$ ($\vec{p}, \vec{t}$见定义1.1).那么存在$\sigma_1, \sigma_2>1$,使得$\vec{\omega}\in A_{\vec{p}/\vec{t\sigma}}(\mathbb{R} ^{2n})$,其中$\vec{t\sigma}=(t_1\sigma_1, t_2\sigma_2)$,且$[\vec{\omega}]_{A_{\vec{p}/\vec{t\sigma}}}\lesssim[\vec{\omega}]_{A_{\vec{p}/\vec{t}}}.$

证  假设$\vec{\omega}\in A_{\vec{p}/\vec{t}}(\mathbb{R} ^{2n})$,则

$\omega_k^{-1/(\frac{p_k}{t_k}-1)}\inA_{\frac{p_kt_k}{t(p_k-t_k)}}(\mathbb{R} ^n), \ \ k=1, 2$

(见文献[1,定理2.1]).由反向Hölder's不等式,存在$\varepsilon_k>0, k=1, 2, $使得

$\frac{1}{|Q|}\int_Q\omega_k^{-\frac{1+\varepsilon_k}{\frac{p_k}{t_k}-1}}{\rm d}x\leq\left(\frac{C}{|Q|}\int_Q\omega_k^{-\frac{1}{\frac{p_k}{t_k}-1}}{\rm d}x\right)^{1+\varepsilon_k}.$

令$\frac{1+\varepsilon_k}{\frac{p_k}{t_k}-1}=\frac{1}{\frac{p_k}{t_k\sigma_k}-1}$,则$\sigma_k>1\ (k=1, 2), $且有

$\begin{eqnarray*}[\vec{\omega}]_{A_{\vec{p}/\vec{t\sigma}}} &=&\sup\limits_{Q\subset\mathbb{R} ^{n}}\left(\frac{1}{|Q|}\int_Q\nu_{\vec{\omega}}(x){\rm d}x\right)^{1/p}\prod\limits_{k=1}^2\left(\frac{1}{|Q|}\int_Q\omega_k^{-\frac{1}{\frac{p_k}{t_k\sigma_k}-1}}(x){\rm d}x\right)^{1/t_k\sigma_k-1/p_k}\\&\leq&\sup\limits_{Q\subset \mathbb{R} ^{n}}\left(\frac{1}{|Q|}\int_Q\nu_{\vec{\omega}}(x){\rm d}x\right)^{1/p}\prod\limits_{k=1}^2\left(\frac{C}{|Q|}\int_Q\omega_k^{-\frac{1}{\frac{p_k}{t_k}-1}}(x){\rm d}x\right)^{(1+\varepsilon_k)(1/t_k\sigma_k-1/p_k)}\\&\lesssim&\sup\limits_{Q\subset\mathbb{R} ^{n}}\left(\frac{1}{|Q|}\int_Q\nu_{\vec{\omega}}(x){\rm d}x\right)^{1/p}\prod\limits_{k=1}^2\left(\frac{1}{|Q|}\int_Q\omega_k^{-\frac{1}{\frac{p_k}{t_k}-1}}(x){\rm d}x\right)^{1/t_k-1/p_k}\\&=&[\vec{\omega}]_{A_{\vec{p}/\vec{t}}}.\end{eqnarray*}$

证毕.

定理1.1的证明  在文献[1]中,焦已经证明了双线性傅里叶乘子算子$T_\sigma$满足引理2.4的条件.因此双线性傅里叶乘子算子$T_\sigma$和相应的主极大截断算子${\cal M}_T$都是从$L^{r_1}(\mathbb{R} ^{n})\times L^{r_2}(\mathbb{R} ^{n})$到$L^{r, \infty}(\mathbb{R} ^{n})$有界的.

由引理2.2,我们可以得到

$|T_\sigma(f_1, f_2)(x)|\lesssim \frac{1}{s-\frac{n}{r_1}-\frac{n}{r_2}}{\cal A}_{{\mathcalS}, r_1, r_2}(f_1, f_2)(x).$

因此若$\vec{\omega}\in A_{\vec{p}/\vec{r}}(\mathbb{R} ^{2n})$,由引理2.1,我们可以得到

$\|T_{\sigma}(f_1, f_2)\|_{L^{p}(\mathbb{R} ^{n}, \nu_{\vec{\omega}})}\lesssim\frac{1}{s-\frac{n}{r_1}-\frac{n}{r_2}}[\vec{\omega}]_{A_{\vec{p}/\vec{r}}}^{\max\left(1, (\frac{p_1}{r_1})', (\frac{p_2}{r_2})'\right)}\prod\limits_{k=1}^{2}\|f_k\|_{L^{p_k}(\mathbb{R} ^{n}, \omega_k)}.$

根据引理2.5,如果$\vec{\omega}\in A_{\vec{p}/\vec{t}}(\mathbb{R} ^{2n})$,那么存在$\sigma_1, \sigma_2>1$,使得$\vec{\omega}\in A_{\vec{p}/\vec{t\sigma}}(\mathbb{R} ^{2n})$,其中$\vec{t\sigma}=(t_1\sigma_1, t_2\sigma_2)$,且

$[\vec{\omega}]_{A_{\vec{p}/\vec{t\sigma}}}\lesssim[\vec{\omega}]_{A_{\vec{p}/\vec{t}}}.$

选取$r_1, r_2$满足$t_1<r_1<t_1\sigma_1, t_2<r_2<t_2\sigma_2$,则$\vec{\omega}\in A_{\vec{p}/\vec{r}}(\mathbb{R} ^{2n})$,且

$[\vec{\omega}]_{A_{\vec{p}/\vec{r}}} \lesssim[\vec{\omega}]_{A_{\vec{p}/\vec{t}}}.$

因此我们完成了定理1.1的证明.