回顾拟微分算子的Hörmander象征类$ S^{m}_{\rho, \delta}({{\Bbb R}} ^n) $ (参见文献[8]). $ \forall m\in {{\Bbb R}} $和$ \rho, \delta\in[0, 1] $, 称$ {{\Bbb R}} ^n\times{{\Bbb R}} ^n $上的光滑函数$ a(x, \xi)\in S^{m}_{\rho, \delta}({{\Bbb R}} ^n) $, 如果对任意的重指标$ \alpha, \beta\in {\Bbb N}^{n} $, 成立

(1.1) $ \begin{eqnarray} |\partial_{x}^{\alpha}\partial_{\xi}^{\beta}a(x, \xi)|\leq C_{\alpha, \beta}(1+|\xi|)^{m-\rho|\beta|+\delta|\alpha|}, \end{eqnarray} $

其中, $ C_{\alpha, \beta} $是与$ x $和$ \xi $无关的常数. 本文约定, 象征函数$ a(x, \xi) $关于变量$ x $和$ \xi $光滑.

$ \forall f\in C^{\infty}_{0}({{\Bbb R}} ^n) $, 设$ a(x, \xi)\in S^{m}_{\rho, \delta}({{\Bbb R}} ^n) $, 则拟微分算子$ T $定义为

(1.2) $ \begin{eqnarray} Tf(x) = \int_{{{\Bbb R}} ^n}a(x, \xi)\text{e}^{2\pi {\rm i}x\cdot\xi} \hat{f}(\xi){\rm d}\xi, \end{eqnarray} $

其中$ \hat{f} $表示函数$ f $的Fourier变换, 记为$ T\in {\cal L}^{m}_{\rho, \delta} $. 由文献[18]可知, $ \forall T\in{\cal L}^{m}_{\rho, \delta} $可表示为

(1.3) $ \begin{eqnarray} Tf(x) = \int K(x, y)f(y){\rm d} y. \end{eqnarray} $

拟微分算子的形成和发展伴随着偏微分发展史上一些基本问题研究的重大进展. 对拟微分算子的研究最早可追溯到Kohn和Nirenberg^[13]以及Hörmander^[8]. 拟微分算子$ T\in{\cal L}^{m}_{\rho, \delta} $的$ L^{p} $有界性参见文献[2, 6, 8, 18].

拟微分算子$ T\in{\cal L}^{m}_{\rho, \delta} $在加权$ L^{p} $空间$ L^{p}_{\omega} $上的有界性也吸引了人们的兴趣. 早在1982年, Miller^[16]就研究了算子$ T\in{\cal L}^{0}_{1, 0} $的$ L^{p}_{\omega} $有界性问题. 随后, Chanillo和Torchinsky^[3], Alvarez和Hounie^[1]分别研究了属于$ {\cal L}^{\frac{n}{2}(\rho-1)}_{\rho, \delta} $和$ {\cal L}^{n(\rho-1)}_{\rho, \delta} $的算子$ T $的$ L^{p}_{\omega} $有界性, 其中$ 0\leq \delta \leq \rho \leq\frac{1}{2} $. 2012年, Michalowski, Rule和Staubach^[15] 改进Alvarez和Hounie文献[1]中的参数为$ \delta\in [0, 1), \rho \in (0, 1] $. 然而, 上述结果都在经典的Muckenhoupt $ A_p $权条件下获得.

在文献[19]中, Tang通过引进权$ A_{p}(\varphi) $从而改进了Miller的结果, 得到拟微分算子$ T \in {\cal L}^{0}_{1, 0} $的$ L^{p}_{\omega} $有界性估计, 其中$ p\in(1, \infty), \omega\in A_{p}(\varphi) $. 这一结果可推广到$ T\in{\cal L}^{0}_{1, \delta}, \delta\in [0, 1) $. 虽然$ A_{p}({{\Bbb R}} ^n)\subset A_{p}(\varphi) $, 然而这里的算子$ T $是Calderón-Zygmund算子^{[12, 14]}.

2018年, Wu和Wang^[20]引进了权$ A_{p}^{\theta}(\phi) $并获得了$ \phi $ -型Calderón-Zygmund算子的$ L^{p}_{\omega} $有界性, 其中$ \theta\geq 0 $, $ p\in [1, \infty) $且$ \omega\in A_{p}^{\theta}(\phi) $. 这里, $ A_{p}^{0}(\phi) = A_{p}({{\Bbb R}} ^n) $且当$ \theta>1 $时, $ A_{p}(\varphi) $包含在$ A_{p}^{\theta}(\phi) $. 拟微分算子$ T\in{\cal L}^{0}_{1, \delta} $是$ \phi $ -型Calderón-Zygmund算子. 然而, $ \phi $ -型Calderón-Zygmund算子属于Calderón-Zygmund算子的理论范畴.

受以上研究启发, 本文主要探讨拟微分算子$ T\in {\cal L}^{m}_{\rho, \delta} $及其交换子$ [b, T] $的$ L^{p}_{\omega} $有界性问题, 其中$ \omega\in A_{p}(\varphi) $. 本文改进了文献[19]中参数$ m $的取值范围. 如文献[5] 所述, 一旦在$ m\in [-(n+1), -(n+1)(1-\rho)) $时建立了拟微分算子$ T $的$ L^{p}_{\omega} $有界性, 就可以将相应的结论延拓到整个$ m\leq -(n+1)(1-\rho) $. 本文探讨了交换子$ [b, T] $的$ L^{p}_{\omega} $有界性问题, 其中$ b\in BMO({{\Bbb R}} ^n) $ (有界平均振荡函数). 类似地, 当$ \omega\in A_{p}^{\theta}(\phi) $时, 本文的结论仍成立.

设方体$ Q $的边、面平行于坐标轴; 记$ Q(x, r) $表示中心在点$ x $且其半边长为$ r>0 $的方体.$ Q = Q(x, r) $时, $ tQ $表示$ Q(x, tr), t>0 $. 设$ E $是可测函数, $ |E| $表示$ E $的Lebesgue测度; $ \chi_{E} $表示$ E $的特征函数. 设$ \omega $是权函数, 记$ \omega(E) = \int_{E}\omega(y){\rm d} y $. 设$ f\in L_{loc} $, 记$ f_{Q} = \frac{1}{|Q|}\int_{Q}|f(y)|{\rm d} y $; 当$ p\in(0, \infty) $时, 给定一个权函数$ \omega $, 满足

$ \left(\int_{{{\Bbb R}} ^n}|f(y)|^{p}{\rm d} y\right)^{\frac{1}{p}}<\infty $

的函数空间称为加权Lebesgue空间, 记为$ \|f\|_{L^{p}_{\omega}}({{\Bbb R}} ^n) $. $ C $表示一个常数, 可能会随上下文变化. $ a\lesssim b $表示$ a\leq Cb $. 若$ a\lesssim b $且$ b\lesssim a $, 则记为$ a\sim b $.

本文总是记$ \varphi(t) = (1+t)^{\alpha_{0}} $, 其中$ \alpha_{0}, t>0 $. 下面给出的$ A_{p}(\varphi) $权定义参见文献[19].

定义1.1   称权$ \omega $属于$ A_{p}(\varphi) $, 如果存在常数$ C $, 使得对任意的方体$ Q $, 有

(1.4) $ \begin{equation} \left(\frac{1}{\varphi(|Q|)|Q|}\int_{Q}\omega(y){\rm d}y\right)\left(\frac{1}{\varphi(|Q|)|Q|}\int_{Q}\omega^{-\frac{1}{p-1}}(y){\rm d}y\right)^{p-1}\leq C, 1<p<\infty. \end{equation} $

称非负函数$ \omega $满足$ A_{1}(\varphi) $条件, 如果存在常数$ C $, 使得对任意的方体$ Q $, 有

(1.5) $ \begin{equation} M_{\varphi}(\omega)(x)\leq C\omega(x), \; \; \text{a.e.}\; x\in{{\Bbb R}} ^n, \end{equation} $

其中

(1.6) $ \begin{equation} M_{\varphi}f(x) = \sup\limits_{Q\ni x}\frac{1}{\varphi(|Q|)|Q|}\int_{Q}|f(y)|{\rm d} y. \end{equation} $

Hardy-Littlewood极大函数$ M $ (H-L极大函数)定义为

(1.7) $ \begin{equation} Mf(x) = \sup\limits_{Q\ni x}\frac{1}{|Q|}\int_{Q}|f(y)|{\rm d} y. \end{equation} $

显然, 对几乎处处的$ x\in{{\Bbb R}} ^n $, $ |f(x)|\leq M_{\varphi}f(x)\leq Mf(x) $. $ \forall 1\leq p<\infty $, 易知$ A_{p}({{\Bbb R}} ^n)\subset A_{p}(\varphi) $, 其中$ A_{p}({{\Bbb R}} ^n) $表示经典的Muckenhoupt权^[7]. 众所周知, $ \omega\in A_{p}({{\Bbb R}} ^n) $意味着$ \omega(x){\rm d} x $是一个倍测度. 然而, 当$ \omega\in A_{p}(\varphi) $时, $ \omega(x){\rm d} x $只具有局部倍测度性质.

本文的主要结果叙述如下.

定理1.1   设拟微分算子$ T\in {\cal L}^{m}_{\rho, \delta} $, 其中$ \rho\in (0, 1], \delta\in [0, 1) $且

$ m\in (-(n+1), -(n+1)(1-\rho)]. $

则$ T $是$ L^{p}_{\omega}({{\Bbb R}} ^n) $上的有界算子, 即存在常数$ C>0 $, 成立不等式

(1.8) $ \begin{eqnarray} \|Tf\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\leq C\|f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}, \end{eqnarray} $

其中$ \omega\in A_{p}(\varphi) $, $ p\in(1, \infty) $. 进一步地, 参数$ m $的取值范围可延拓到$ m\leq -(n+1)(1-\rho) $.

设函数$ b\in BMO({{\Bbb R}} ^n) $ (参见文献[9]). Coifman, Rochberg和Weiss给出的交换子算子$ [b, T] $定义为^[4]

(1.9) $ \begin{equation} [b, T]f(x) = b(x)Tf(x)-T(bf)(x). \end{equation} $

类似定理1.1, 本文得到如下定理.

定理1.2   设拟微分算子$ T\in {\cal L}^{m}_{\rho, \delta} $, 其中$ \rho\in (0, 1], \delta\in [0, 1) $且

$ m\in (-(n+1), -(n+1)(1-\rho)]. $

如果函数$ b\in BMO({{\Bbb R}} ^n) $, 则交换子算子$ [b, T] $是$ L^{p}_{\omega}({{\Bbb R}} ^n) $上的有界算子, 即存在常数$ C>0 $, 成立不等式

(1.10) $ \begin{equation} \|[b, T]f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\leq C\|b\|_{BMO({{\Bbb R}} ^n)}\|f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}. \end{equation} $

其中$ \omega\in A_{p}(\varphi) $, $ p\in(1, \infty) $. 进一步地, 参数$ m $的取值范围可延拓到$ m\leq -(n+1)(1-\rho) $.

本文结构安排如下. 第2节给出了拟微分算子和极大函数的一些基本概念和辅助引理. 第3节和第4节分别证明了定理1.1和定理1.2. 证明的思路主要来自文献[17, 19]. 本文先证明函数$ f\in C_{0}^{\infty}({{\Bbb R}} ^n) $时定理1.1和定理1.2成立, 再由稠密性延拓到加权$ L^{p}({{\Bbb R}} ^n) $空间^[16]. 类似地, 本文的结果可推广到$ A_{p}^{\theta}(\phi) $权.

设$ {\cal S}({{\Bbb R}} ^n) $是$ {{\Bbb R}} ^n $中所有Schwartz函数构成的空间, $ {\cal S}^{\prime}({{\Bbb R}} ^n) $是它的对偶空间. $ {{\Bbb R}} ^n $中无穷可微($ C^{\infty} $ -函数)且具有紧支集的函数空间记为$ C_{0}^{\infty}({{\Bbb R}} ^n) $. 易知, 拟微分算子是$ {\cal S}({{\Bbb R}} ^n) $到$ {\cal S}({{\Bbb R}} ^n) $上的有界线性算子, 从而具有分布核$ K(x, y)\in {\cal S}^{\prime}({{\Bbb R}} ^n\times {{\Bbb R}} ^n) $ (参见文献[18]).

引理2.1  设拟微分算子$ T\in {\cal L}^{m}_{\rho, \delta} $, $ \rho\in (0, 1], \delta\in [0, 1) $. 则$ T $的定义为

(2.1) $ \begin{eqnarray} K(x, y) = \lim\limits_{\varepsilon\to 0}\int_{{{\Bbb R}} ^n}\text{e}^{2\pi i(x-y)\cdot\xi} a(x, \xi)\psi(\varepsilon\xi){\rm d} \xi \end{eqnarray} $

的分布核$ K(x, y) $在除对角线$ \{(x, x):x\in{{\Bbb R}} ^n\} $外光滑. 其中, 函数$ a(x, \xi)\in {\cal S}^{m}_{\rho, \delta}({{\Bbb R}} ^n) $为算子$ T $的象征; $ \psi(\xi)\in C_{0}^{\infty}({{\Bbb R}} ^n) $在$ |\xi|\leq 1 $时取值为$ 1 $; 极限在$ {\cal S}^{\prime}({{\Bbb R}} ^n) $意义下收敛且与$ \psi $的选取无关. $ K(x, y) $有估计式如下.

(1) 如果$ M\in{\Bbb N} $且$ M+m+n>0 $, 则

(2.2) $ \begin{eqnarray} \sup\limits_{|\alpha+\beta| = M}|D_{x}^{\alpha}D_{y}^{\beta}K(x, y)|\leq C_{M}\frac{1}{|x-y|^{\frac{M+m+n}{\rho}}}, x\neq y. \end{eqnarray} $

(2) 任给多重指标$ \alpha, \beta\in{\Bbb N}^{n} $和$ N\in{\Bbb N} $, 有

(2.3) $ \begin{eqnarray} \sup\limits_{|x-y|\geq \frac{1}{2}}|x-y|^{N}|D_{x}^{\alpha}D_{y}^{\beta}K(x, y)|\leq C_{\alpha, \beta, N}. \end{eqnarray} $

引理2.1首先由Alvarez和Hounie在文献[1]获得; Hounie和Kapp在文献[10, 命题3.1] 中给出其具体形式; 它在文献[11]中也有应用. 下面给出属于$ {\cal L}^{m}_{\rho, \delta} $的拟微分算子$ T $的弱$ (1, 1) $估计(参见文献[1]).

引理2.2  设$ T\in {\cal L}^{m}_{\rho, \delta} $, $ \rho\in (0, 1] $, $ \delta\in [0, 1) $, 如果$ m\leq -\frac{n}{2}(1-\rho)+\min\{0, \frac{n(\rho-\delta)}{2}\} $, 则$ T $是弱$ (1, 1) $型的.

注2.1   显然, $ m\leq -(n+1)(1-\rho) $满足引理2.2中$ m $要求的条件.

设$ f\in L_{loc} $, $ \eta\in (0, \infty) $, $ (\varphi, \eta) $ -极大算子$ M_{\varphi, \eta} $定义为

$ \begin{eqnarray*} M_{\varphi, \eta} f(x) = \sup\limits_{x\in Q}\frac{1}{\varphi(|Q|)^{\eta}|Q|}\int_{Q}|f(y)|{\rm d} y. \end{eqnarray*} $

由文献[19, 命题2.1]可知, $ (\varphi, \eta) $ -极大算子$ M_{\varphi, \eta} $是加权$ L^{p} $有界的. 即有如下引理.

引理2.3   设$ p\in(1, \infty), p^{\prime} = \frac{p}{p-1} $, 如果权$ \omega\in A_{p}(\varphi) $, 则存在常数$ C>0 $, 使得

(2.4) $ \begin{eqnarray} \|M_{\varphi, p^{\prime}}f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\leq C\|f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}. \end{eqnarray} $

注2.2  当$ \eta>p^{\prime} $时, 易知$ M_{\varphi, \eta} f(x)\leq M_{\varphi, p^{\prime}}f(x) $. 从而有

$ \begin{eqnarray*} \|M_{\varphi, \eta} f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\leq C\|M_{\varphi, p^{\prime}}f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\leq C\|f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}. \end{eqnarray*} $

设$ \eta\in(0, \infty) $, 二进$ (\varphi, \eta) $ -极大函数$ M^{\triangle}_{\varphi, \eta}f(x) $和二进Sharp$ (\varphi, \eta) $ -极大函数$ M^{\#, \triangle}_{\varphi, \eta}f(x) $分别定义为

$ M^{\triangle}_{\varphi, \eta}f(x) = \sup\limits_{Q\ni x}\frac{1}{\varphi(|Q|)^{\eta}|Q|}\int_{Q}|f(y)|{\rm d} y $

和

$ \begin{eqnarray*} M^{\#, \triangle}_{\varphi, \eta}f(x)& = &\sup\limits_{Q\ni x, r<1}\frac{1}{|Q|}\int_{Q}|f(y)-f_{Q}|{\rm d} y+\sup\limits_{Q\ni x, r\geq 1}\frac{1}{\varphi(|Q|)^{\eta}|Q|}\int_{Q}|f(y)|{\rm d} y\\ & = &\sup\limits_{Q\ni x, r<1}\inf\limits_{C}\frac{1}{|Q|}\int_{Q}|f(y)-C|{\rm d} y+\sup\limits_{Q\ni x, r\geq 1}\frac{1}{\varphi(|Q|)^{\eta}|Q|}\int_{Q}|f(y)|{\rm d} y, \end{eqnarray*} $

其中上确界是对所有的二进方体$ Q = Q(x_{0}, r) $取的.

当$ \varepsilon>0 $, 二进$ (\varepsilon, \varphi, \eta) $ - 极大函数$ M^{\triangle}_{\varepsilon, \varphi, \eta}f(x) $和二进Sharp$ (\varepsilon, \varphi, \eta) $ -极大函数$ M^{\#, \triangle}_{\varepsilon, \varphi, \eta}f(x) $分别定义为

$ M^{\triangle}_{\varepsilon, \varphi, \eta}f(x) = M^{\triangle}_{\varphi, \eta}(|f|^{\varepsilon})^{\frac{1}{\varepsilon}}(x) $

和

$ M^{\#, \triangle}_{\varepsilon, \varphi, \eta}f(x) = M^{\#, \triangle}_{\varphi, \eta}(|f|^{\varepsilon})^{\frac{1}{\varepsilon}}(x). $

由文献[19, 命题2.3]可知如下引理.

引理2.4   设$ p\in (1, \infty) $, $ \omega\in A_{p}(\varphi) $, $ \eta\in (0, \infty) $且$ \varepsilon>0 $. 如果$ f\in L^{p}_{\omega}({{\Bbb R}} ^n) $, 则

(2.5) $ \begin{eqnarray} \|f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\leq \|M^{\triangle}_{\varepsilon, \varphi, \eta}f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)} \leq C\|M^{\#, \triangle}_{\varepsilon, \varphi, \eta}f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}. \end{eqnarray} $

本节建立属于$ {\cal L}^{m}_{\rho, \delta} $的拟微分算子$ T $的$ L^{p}_{\omega}({{\Bbb R}} ^n) $估计, 其中$ \omega\in A_{p}(\varphi) $.

证   为证定理1.1, 由引理2.3和引理2.4可知, 只要证对任意的$ \eta\in(p^{\prime}, \infty) $和$ \varepsilon\in(0, 1) $有

(3.1) $ \begin{eqnarray} M^{\#, \Delta}_{\varepsilon, \varphi, \eta}(Tf)(x_{0})\leq CM_{\varphi, \eta} f(x_{0}), \; \; \text{a.e. } \; x_{0}\in{{\Bbb R}} ^n. \end{eqnarray} $

固定$ x_{0}\in{{\Bbb R}} ^n $和$ x_{0}\in Q = Q(x_{1}, r) $, 其中$ Q $是二进方体. 作$ f $的分解: $ f = f_{1}+f_{2} $, 其中, $ f_{1} = f\chi_{Q^{\ast}} $且$ Q^{\ast} = Q(x_{1}, 2\sqrt{n}r) $. 下面根据$ r $的取值分两种情形考虑$ M^{\#, \Delta}_{\varepsilon, \varphi, \eta}(Tf) $.

情形1  $ r\in(0, 1) $. 由$ \varepsilon \in(0, 1) $可知, $ \forall\alpha, \beta\in{{\Bbb R}} $有$ ||\alpha|^{\varepsilon}-|\beta|^{\varepsilon}|\leq |\alpha-\beta|^{\varepsilon} $. 从而, 选取$ C_{Q} = (Tf_{2})_{Q} $, 可得

(3.2) $ \begin{eqnarray} &&\left(\frac{1}{|Q|}\int_{Q}||Tf(x)|^{\varepsilon}-|C_{Q}|^{\varepsilon}|{\rm d} x\right)^{\frac{1}{\varepsilon}}\leq \left(\frac{1}{|Q|}\int_{Q}|Tf(x)-(Tf_{2})_{Q}|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}{}\\ &\leq &C\left(\frac{1}{|Q|}\int_{Q}|Tf_{1}(x)|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}+C\left(\frac{1}{|Q|}\int_{Q}|Tf_{2}(x)-(Tf_{2})_{Q}|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}{}\\ &\leq &C\left(\frac{1}{|Q|}\int_{Q}|Tf_{1}(x)|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}+\frac{C}{|Q|}\int_{Q}|Tf_{2}(x)-(Tf_{2})_{Q}|{\rm d} x{}\\ & = &I_{1}+I_{2}. \end{eqnarray} $

对于$ I_{1} $, 注意到算子$ T $是弱$ (1, 1) $型的(引理2.2) 且$ \varphi(|Q^{\ast}|)\leq C_{n, \alpha_{0}} $, 由Kolmogorov不等式^{[7, p485]}可得

(3.3) $ \begin{eqnarray} I_{1}\leq \frac{C}{|Q|}\|f_{1}\|_{L^{1}({{\Bbb R}} ^n)}\leq \frac{C}{|Q^{\ast}|}\int_{Q^{\ast}}|f(y)|{\rm d} y\leq CM_{\varphi, \eta} f(x_{0}). \end{eqnarray} $

对于$ I_{2} $, 记

(3.4) $ \begin{eqnarray} I_{2}& = &\frac{C}{|Q|}\int_{Q}\left|\frac{1}{|Q|}\int_{Q}\int_{{{\Bbb R}} ^n} (K(x, z)-K(y, z))f_{2}(z){\rm d} z{\rm d} y\right|{\rm d} x{}\\ &\leq&\frac{C}{|Q|}\int_{Q}\frac{1}{|Q|}\int_{Q}\int_{{{\Bbb R}} ^n\setminus Q^{\ast}} |K(x, z)-K(y, z)||f(z)|{\rm d} z{\rm d} y{\rm d} x{}\\ & = &\frac{C}{|Q|}\int_{Q}\frac{1}{|Q|}\int_{Q}\sum\limits_{l = 0}^{\infty}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}} |K(x, z)-K(y, z)||f(z)|{\rm d} z{\rm d} y{\rm d} x. \end{eqnarray} $

选取$ l_{0} = \max\{l\in{\Bbb N}:2^{l}2\sqrt{n}r\leq 1\} $, 则

(3.5) $ \begin{eqnarray} I_{2}&\leq & C\sum\limits_{l = 0}^{l_{0}}\frac{1}{|Q|}\int_{Q}\frac{1}{|Q|}\int_{Q}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}} |K(x, z)-K(y, z)||f(z)|{\rm d} z{\rm d} y{\rm d} x{}\\ &&+C\sum\limits_{l = l_{0}+1}^{\infty}\frac{1}{|Q|}\int_{Q}\frac{1}{|Q|}\int_{Q}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}} |K(x, z)-K(y, z)||f(z)|{\rm d} z{\rm d} y{\rm d} x{}\\ & = &I_{21}+I_{22}. \end{eqnarray} $

可以断言: (1) $ \forall y\in Q(x_{1}, r) $和$ \forall z\in {{\Bbb R}} ^n\setminus Q^{\ast} $, 有$ |z-x_{1}|\sim |z-y| $. 这是因为由$ |y-x_{1}|\leq \sqrt{n}r $和$ |z-x_{1}|\geq 2\sqrt{n}r $, 有

(3.6) $ \begin{eqnarray} \frac{1}{2}|z-x_{1}|\leq |z-y| \leq \frac{3}{2}|z-x_{1}|. \end{eqnarray} $

(2) $ \forall z\in 2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast} $, 有$ |z-x_{1}|\sim 2^{l}2\sqrt{n}r $. 事实上

(3.7) $ \begin{eqnarray} 2^{l}2\sqrt{n}r\leq |z-x_{1}|\leq 2\sqrt{n} 2^{l}2\sqrt{n}r\Rightarrow |z-x_{1}|\sim 2^{l}2\sqrt{n}r. \end{eqnarray} $

下面接着估计$ I_{21} $. 由中值定理和(2.2)式(选取$ M = 1 $)可得

(3.8) $ \begin{eqnarray} I_{21}&\leq &C\sum\limits_{l = 0}^{l_{0}}\frac{1}{|Q|}\int_{Q}\frac{1}{|Q|}\int_{Q}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}} \frac{|x-y|}{|z-x_{1}|^{\frac{1+n+m}{\rho}}}|f(z)|{\rm d} z{\rm d} y{\rm d} x{}\\ &\leq& C\sum\limits_{l = 0}^{l_{0}}\frac{1}{|Q|}\int_{Q}\frac{1}{|Q|}\int_{Q}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}} \frac{2\sqrt{n}r}{(2^{l}2\sqrt{n}r)^{\frac{1+n+m}{\rho}}}|f(z)|{\rm d} z{\rm d} y{\rm d} x{}\\ &\leq& C\sum\limits_{l = 0}^{l_{0}}\frac{2\sqrt{n}r}{(2^{l}2\sqrt{n}r)^{\frac{1+n+m}{\rho}}}\int_{2^{l+1}Q^{\ast}} |f(z)|{\rm d} z, \end{eqnarray} $

其中用到等价关系: $ \xi\in Q = Q(x_{1}, r) $时, $ |z-\xi|\sim|z-x_{1}| $. 注意到$ l_{0} $的取法以及$ m\leq -(1-\rho)(n+1) $, 可知(3.8)式被

(3.9) $ \begin{eqnarray} C\sum\limits_{l = 0}^{l_{0}}\frac{1}{2^{l}}\frac{1}{|2^{l+1}Q^{\ast}|}\int_{2^{l+1}Q^{\ast}} |f(z)|{\rm d} z\leq C\sum\limits_{l = 0}^{l_{0}}\frac{1}{2^{l}}M_{\varphi, \eta} f(x_{0}) \end{eqnarray} $

控制. 于是, 由(3.9)式可得

(3.10) $ \begin{eqnarray} I_{21}\leq CM_{\varphi, \eta} f(x_{0}). \end{eqnarray} $

因为$ \forall z\in 2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast} $和$ \forall y\in Q = Q(x_{1}, r) $都有

$ \begin{eqnarray*} |z-y|\geq |z-x_{1}|-|y-x_{1}|\geq 2^{l}2\sqrt{n}r-\sqrt{n}r\geq (2^{l}-1)2\sqrt{n}r\geq \frac{1}{2}. \end{eqnarray*} $

因此, 在(2.3)式中选取$ N = n+n\eta\alpha_{0}+1 $, 就有

(3.11) $ \begin{eqnarray} I_{22}&\leq &C\sum\limits_{l = l_{0}+1}^{\infty}\frac{1}{|Q|}\int_{Q}\frac{1}{|Q|}\int_{Q}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}} |K(y, z)||f(z)|{\rm d} z{\rm d} y{\rm d} x{}\\ &\leq& C\sum\limits_{l = l_{0}+1}^{\infty}\frac{1}{|Q|}\int_{Q}\frac{1}{|Q|}\int_{Q}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}} \frac{|f(z)|}{|z-y|^{n+n\eta\alpha_{0}+1}}{\rm d} z{\rm d} y{\rm d} x{}\\ &\leq &C\sum\limits_{l = l_{0}+1}^{\infty}\frac{1}{|Q|}\int_{Q}\frac{1}{|Q|}\int_{Q}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}} \frac{|f(z)|}{|z-x_{1}|^{n+n\eta\alpha_{0}+1}}{\rm d} z{\rm d} y{\rm d} x{}\\ &\leq &C\sum\limits_{l = l_{0}+1}^{\infty}\frac{1}{(2^{l}2\sqrt{n}r)^{n+n\eta\alpha_{0}+1}}\int_{2^{l+1}Q^{\ast}}|f(y)|{\rm d} y{}\\ &\leq &C\sum\limits_{l = l_{0}+1}^{\infty}\frac{1}{(2^{l}2\sqrt{n}r)}M_{\varphi, \eta} f(x_{0}){}\\ &\leq &CM_{\varphi, \eta} f(x_{0}). \end{eqnarray} $

综上所述, 有

(3.12) $ \begin{equation} I_{2}\leq CM_{\varphi, \eta} f(x_{0}). \end{equation} $

于是, 当$ r\in (0, 1) $时, (3.2)式成立.

情形2   $ r\geq 1 $. 记$ \eta_{1} = \frac{\eta}{\varepsilon} $, 则有

(3.13) $ \begin{eqnarray} &&\left(\frac{1}{\varphi(|Q|)^{\eta}|Q|}\int_{Q}|Tf(x)|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}{}\\ &\leq& \frac{C}{\varphi(|Q|)^{\eta_{1}}}\left(\frac{1}{|Q|}\int_{Q}|Tf_{1}(x)|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}+\frac{C}{\varphi(|Q|)^{\eta_{1}}}\left(\frac{1}{|Q|}\int_{Q}|Tf_{2}(x)|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}{}\\ &\leq& \frac{C}{\varphi(|Q|)^{\eta_{1}}}\left(\frac{1}{|Q|}\int_{Q}|Tf_{1}(x)|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}+\frac{C}{|Q|}\int_{Q}|Tf_{2}(x)|{\rm d} x{}\\ & = &L_{1}+L_{2}. \end{eqnarray} $

类似$ I_{1} $的估计并注意到$ \eta_{1}>\eta $, 有

(3.14) $ \begin{eqnarray} L_{1}\leq \frac{C}{\varphi(|Q|)^{\eta_{1}}}\frac{1}{|Q|}\|f_{1}\|_{L^{1}({{\Bbb R}} ^n)}\leq \frac{C}{\varphi(|Q|)^{\eta}}\frac{1}{|Q|}\int_{Q^{\ast}}|f(y)|{\rm d} y\leq CM_{\varphi, \eta} f(x_{0}). \end{eqnarray} $

另外, 由于$ \forall z\in 2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast} $和$ \forall y\in Q $, 有

$ \begin{eqnarray*} |z-y|\geq |z-x_{1}|-|y-x_{1}|\geq 2^{l}2\sqrt{n}r-\sqrt{n}r\geq \sqrt{n}r\geq 1. \end{eqnarray*} $

因此, 对$ L_{2} $采用情形1中估计$ I_{22} $的方法, 可得

(3.15) $ \begin{eqnarray} L_{2}&\leq &\frac{C}{|Q|}\int_{Q}\int_{{{\Bbb R}} ^n\setminus Q^{\ast}}|K(x, y)||f(y)|{\rm d} y{\rm d} x{}\\ & = &C\sum\limits_{l = 0}^{\infty}\frac{1}{|Q|}\int_{Q}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}}|K(x, y)||f(y)|{\rm d} y{\rm d} x{}\\ &\leq & \sum\limits_{l = 0}^{\infty}\frac{1}{|Q|}\int_{Q}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}}\frac{|f(y)|}{|y-x|^{n+n\eta\alpha_{0}+1}}{\rm d} y{\rm d} x{}\\ &\leq & C\sum\limits_{l = 0}^{\infty}\frac{1}{2^{l}}M_{\varphi, \eta} f(x_{0}){}\\ &\leq & CM_{\varphi, \eta} f(x_{0}). \end{eqnarray} $

因此, 当$ r\geq 1 $时, (3.1)式仍成立. 定理1.1得证.

本节采用文献[17, 19]的方法来证明定理1.2. 为此, 首先给出一些基本的概念和事实.

如果连续、单调不减的凸函数$ B(t):[0, \infty)\to[0, \infty) $, 它还满足$ B(0) = 0 $且$ B(t)\to \infty $ ($ t\to\infty $), 则称$ B $为Young函数. 函数$ f $在方体$ Q $上的$ B $ -平均由Luxemburg范数定义为

(4.1) $ \begin{eqnarray} \|f\|_{B, Q} = \inf\left\{\lambda>0:\frac{1}{|Q|}\int_{Q}B(\frac{|f(y)|}{\lambda}){\rm d} y\leq 1\right\}. \end{eqnarray} $

设$ \bar{B} $是Young函数$ B $的互补Young函数, 则有广义Hölder不等式

(4.2) $ \begin{eqnarray} \frac{1}{|Q|}\int_{Q}|f(y)g(y)|{\rm d} y\leq \|f\|_{B, Q}\|g\|_{\bar{B}, Q}. \end{eqnarray} $

给定Young函数$ B $, 则$ B $极大函数$ M_{B}f(x) $和$ (\varphi, \eta) $-$ B $极大函数$ M_{B, \varphi, \eta}f(x) $分别定义为

$ M_{B}f(x) = \sup\limits_{Q\ni x}\|f\|_{B, Q} $

和

$ M_{B, \varphi, \eta}f(x) = \sup\limits_{Q\ni x}\frac{1}{\varphi(|Q|)^{\eta}}\|f\|_{B, Q}, $

其中$ \eta\in(0, \infty) $.

本文选取Young函数$ B(t) = t(1+\log^{+}t) $, 相应的$ B $极大函数记为$ M_{L\log L} $. 其互补Young函数$ \bar{B}(t)\thickapprox\text{e}^{t} $, 相应的$ \bar{B} $极大函数记为$ M_{\text{exp}L} $. 下面的引理给出了$ (\varphi, \eta) $ -极大函数$ M_{\varphi, \eta} f(x) $与$ (\varphi, \eta) $-$ B $极大函数$ M_{L\log L, \varphi, \eta}f(x) $的比较^[19].

引理4.1  设$ \eta\in(0, \infty) $且$ M_{\varphi, \frac{\eta}{2}}f $局部可积, 则存在不依赖于$ f $和$ x $的常数$ C_{1}, C_{2}>0 $, 使得

(4.3) $ \begin{eqnarray} C_{2}M_{\varphi, \eta}M_{\varphi, \eta} f(x)\leq M_{L\log L, \varphi, \eta}f(x)\leq C_{1}M_{\varphi, \frac{\eta}{2}}M_{\varphi, \frac{\eta}{2}}f(x), f\in L_{loc}. \end{eqnarray} $

接下来建立二进Sharp$ (\varepsilon, \varphi, \eta) $ -极大函数$ M^{\#, \triangle}_{\varepsilon, \varphi, \eta}f(x) $的一个点态估计.

引理4.2   设$ b\in BMO $, $ 1\leq \eta<\infty $且$ 0<\varepsilon<\kappa<1 $, 则

(4.4) $ \begin{eqnarray} M^{\#, \triangle}_{\varepsilon, \varphi, \eta}([b, T]f)(x_{0})\leq C\|b\|_{BMO}(M^{\triangle}_{\kappa, \varphi, \eta}Tf(x_{0})+M_{L\log L, \varphi, \eta}f(x_{0}) \end{eqnarray} $

对几乎处处的$ x_{0}\in{{\Bbb R}} ^n $和任意的$ f\in C_{0}^{\infty}({{\Bbb R}} ^n) $成立.

证   固定$ x\in{{\Bbb R}} ^n $和$ x\in Q = Q(x_{0}, r) $, 其中$ Q $为二进方体. 作$ f $的分解: $ f = f_{1}+f_{2} $, 其中$ f_{1} = f\chi_{Q^{\ast}} $, $ Q^{\ast} = Q(x_{0}, 2\sqrt{n}r) $. 任给常数$ \lambda $, 有

$ \begin{eqnarray*} [b, T]f = (b-\lambda)Tf-T((b-\lambda)f_{1})-T((b-\lambda)f_{2}). \end{eqnarray*} $

为了证得(4.4)式, 下面根据$ r $的取值分两种情形讨论.

情形1   $ r\in(0, 1) $. 因为$ \varepsilon\in(0, 1) $, 所以有

$ \begin{eqnarray*} &&\left(\frac{1}{|Q|}\int_{Q}\left||[b, T]f(x)|^{\varepsilon}-|C_{Q}|^{\varepsilon}\right|{\rm d} x\right)^{\frac{1}{\varepsilon}}\leq \left(\frac{1}{|Q|}\int_{Q}|[b, T]f(x)-C_{Q}|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}\\ &\leq& C\left(\frac{1}{|Q|}\int_{Q}|(b(x)-\lambda)Tf(x)|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}+C\left(\frac{1}{|Q|}\int_{Q}|T((b-\lambda)f_{1})(x)|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}\\ &&+C\left(\frac{1}{|Q|}\int_{Q}|T((b-\lambda)f_{2})(x)-C_{Q}|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}\\ & = &J_{1}+J_{2}+J_{3}. \end{eqnarray*} $

为了估计$ J_{1} $, 选取$ \lambda = b_{Q^{\ast}} $($ b $在方体$ Q^{\ast} $上的平均). 然后, 对指数$ q $($ 1<q<\frac{\kappa}{\varepsilon} $)使用Hölder不等式, 再由$ BMO({{\Bbb R}} ^n) $中的John-Nirenberg不等式可得

(4.5) $ \begin{eqnarray} J_{1}&\leq & C\left(\frac{1}{|Q^{\ast}|}\int_{Q^{\ast}}|b(x)-b_{Q^{\ast}}|^{q^{\prime}\varepsilon}{\rm d} x\right)^{\frac{1}{q^{\prime}\varepsilon}}\left(\frac{1}{|Q|}\int_{Q}|Tf(x)|^{q\varepsilon}{\rm d} x\right)^{\frac{1}{q\varepsilon}}{}\\ &\leq& C\|b\|_{BMO({{\Bbb R}} ^n)}M^{\Delta}_{\kappa, \varphi, \eta}(Tf)(x_{0}), \end{eqnarray} $

其中$ q^{\prime} $是$ q $的共轭指标.

对于$ J_{2} $, 由算子$ T $的弱$ (1, 1) $有界性(引理2.2), Kolmogorov不等式以及广义Hölder不等式((4.2)式), 可得

(4.6) $ \begin{eqnarray} J_{2}&\leq & \frac{C}{|Q|}\|(b-b_{Q^{\ast}})f_{1}\|_{L^{1}({{\Bbb R}} ^n)} {}\\ &\leq & \frac{C}{|Q^{\ast}|}\int_{Q^{\ast}}|b(x)-b_{Q^{\ast}}||f(x)|{\rm d} x{}\\ &\leq & C\|b-b_{Q^{\ast}}\|_{\exp{L}, Q^{\ast}}\|f\|_{L\log{L}, Q^{\ast}} {}\\ &\leq & C\|b\|_{BMO({{\Bbb R}} ^n)}M_{L\log{L}}f(x_{0}){}\\ &\leq& C\|b\|_{BMO({{\Bbb R}} ^n)}M_{L\log{L}, \varphi, \eta}f(x_{0}), \end{eqnarray} $

其中用到了$ r\in(0, 1) $时, $ \varphi(|Q^{\ast}|)\sim C $的等价关系以及不等式$ \|b-b_{Q^{\ast}}\|_{\exp{L}, Q^{\ast}}\leq C\|b\|_{BMO({{\Bbb R}} ^n)} $ (参见文献[17, p8-9]).

为估计$ J_{3} $, 令$ C_{Q} = (T((b-b_{Q^{\ast}})f_{2}))_{Q} $. 然后, 利用Hölder不等式可得

$ \begin{eqnarray*} J_{3}&\leq & \frac{C}{|Q|}\int_{Q}|T((b-b_{Q^{\ast}})f_{2})(x)-(T((b-b_{Q^{\ast}})f_{2}))_{Q}|{\rm d} x{\nonumber}\\ &\leq&\frac{C}{|Q|^{2}}\int_{Q}\int_{Q}\int_{{{\Bbb R}} ^n\setminus Q^{\ast}} |K(x, z)-K(y, z)||b(z)-b_{Q^{\ast}}||f(z)|{\rm d} z{\rm d} y{\rm d} x{\nonumber}\\ &\leq & \frac{C}{|Q|^{2}}\int_{Q}\int_{Q}\sum\limits_{l = 0}^{\infty}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}} |K(x, z)-K(y, z)||b(z)-b_{Q^{\ast}}||f(z)|{\rm d} z{\rm d} y{\rm d} x. \end{eqnarray*} $

再取$ l_{0} = \max\{l\in{\Bbb N}:2^{l}2\sqrt{n}r\leq 1\} $, 则有

$ \begin{eqnarray*} J_{3}&\leq&\frac{C}{|Q|^{2}}\int_{Q}\int_{Q}\sum\limits_{l = 0}^{l_{0}}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}} |K(x, z)-K(y, z)||b(z)-b_{Q^{\ast}}||f(z)|{\rm d} z{\rm d} y{\rm d} x\\ &&+\frac{C}{|Q|^{2}}\int_{Q}\int_{Q}\sum\limits_{l = l_{0}+1}^{\infty}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}} |K(x, z)-K(y, z)||b(z)-b_{Q^{\ast}}||f(z)|{\rm d} z{\rm d} y{\rm d} x\\ & = &J_{31}+J_{32}. \end{eqnarray*} $

类似(3.8)式的估计, 由中值定理和(2.2)式(选取$ M = 1 $)可得

(4.7) $ \begin{eqnarray} I_{31}&\leq & \frac{C}{|Q|^{2}}\int_{Q}\int_{Q}\sum\limits_{l = 0}^{l_{0}}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}} \frac{|x-y|}{|z-x_{1}|^{\frac{1+n+m}{\rho}}}|f(z)|{\rm d} z{\rm d} y{\rm d} x{}\\ &\leq& \frac{C}{|Q|^{2}}\int_{Q}\int_{Q}\sum\limits_{l = 0}^{l_{0}}\int_{2^{l+1}Q^{\ast}\setminus 2^{l}Q^{\ast}} \frac{2\sqrt{n}r}{(2^{l}2\sqrt{n}r)^{\frac{1+n+m}{\rho}}}|b(z)-b_{Q^{\ast}}||f(z)|{\rm d} z{\rm d} y{\rm d} x{}\\ &\leq & C\sum\limits_{l = 0}^{l_{0}}\frac{2\sqrt{n}r}{(2^{l}2\sqrt{n}r)^{\frac{1+n+m}{\rho}}}\int_{2^{l+1}Q^{\ast}} |b(z)-b_{Q^{\ast}}||f(z)|{\rm d} z. \end{eqnarray} $

令$ Q^{\ast}_{l} = 2^{l+1}Q^{\ast} $, 由$ m\leq -(1-\rho)(n+1) $可知

$ \begin{eqnarray*} J_{31}&\leq & C\sum\limits_{l = 0}^{l_{0}}\frac{2^{-l}}{|Q^{\ast}_{l+1}|}\int_{Q^{\ast}_{l+1}} |b(z)-b_{Q^{\ast}}||f(z)|{\rm d} z\\ &\leq & C\sum\limits_{l = 0}^{l_{0}}\frac{2^{-l}}{|Q^{\ast}_{l+1}|}\bigg[\int_{Q^{\ast}_{l+1}} |b(z)-b_{Q^{\ast}_{l+1}}||f(z)|{\rm d} z+|b_{Q^{\ast}_{l+1}}-b_{Q^{\ast}}|\int_{Q^{\ast}_{l+1}} |f(z)|{\rm d} z\bigg]. \end{eqnarray*} $

于是, 由广义Hölder不等式((4.2)式), 不等式$ \|b-b_{Q^{\ast}}\|_{\exp{L}, Q^{\ast}}\leq C\|b\|_{BMO({{\Bbb R}} ^n)} $和$ |b_{Q^{\ast}_{l}}-b_{Q^{\ast}}|\leq Cl\|b\|_{BMO({{\Bbb R}} ^n)} $以及引理4.1, 可得

(4.8) $ \begin{eqnarray} J_{31}&\leq & C\sum\limits_{l = 0}^{l_{0}}2^{-l}\bigg[\|b\|_{BMO({{\Bbb R}} ^n)}M_{L\log{L}, \varphi, \eta}f(x_{0})+l\|b\|_{BMO({{\Bbb R}} ^n)}M_{\varphi, \eta}(f)(x_{0})\bigg]{}\\ &\leq & C\sum\limits_{l = 0}^{\infty}2^{-l}l\|b\|_{BMO({{\Bbb R}} ^n)}(M_{L\log{L}, \varphi, \eta}f(x_{0})+M_{\varphi, \eta}M_{\varphi, \eta}(f)(x_{0})){}\\ &\leq& C\|b\|_{BMO({{\Bbb R}} ^n)}M_{L\log{L}, \varphi, \eta}f(x_{0}). \end{eqnarray} $

为估计$ J_{32} $, 在(2.3)式中取$ N = n+n\eta\alpha_{0}+1 $, 从而

$ \begin{eqnarray*} J_{32}&\leq & \frac{C}{|Q|^{2}}\int_{Q}\int_{Q}\sum\limits_{l = l_{0}+1}^{\infty}\int_{Q^{\ast}_{l}\setminus Q^{\ast}_{l-1}} \frac{|x-y|}{|z-x_{0}|^{N}}|b(z)-b_{Q^{\ast}}||f(z)|{\rm d} z{\rm d} y{\rm d} x\\ &\leq & C\sum\limits_{l = l_{0}+1}^{\infty}\frac{2\sqrt{n}r}{(2^{l}2\sqrt{n}r)^{n+n\eta\alpha_{0}+1}}\int_{Q^{\ast}_{l}}|b(z)-b_{Q^{\ast}}||f(z)|{\rm d} z\\ &\leq & \sum\limits_{l = l_{0}+1}^{\infty}\frac{C2^{-l}}{\varphi(|Q^{\ast}_{l}|)^{\eta}|Q^{\ast}_{l}|}\bigg[\int_{Q^{\ast}_{l}}|b(z)-b_{Q^{\ast}_{l}}||f(z)|{\rm d} z+|b_{Q^{\ast}_{l}}-b_{Q^{\ast}}|\int_{Q^{\ast}_{l}}|f(z)|{\rm d} z\bigg]. \end{eqnarray*} $

于是

(4.9) $ \begin{eqnarray} J_{32}&\leq & C\|b\|_{BMO({{\Bbb R}} ^n)} \sum\limits_{l = l_{0}+1}^{\infty}\frac{1}{2^{l}\varphi(|Q^{\ast}_{l}|)^{\eta}|Q^{\ast}_{l}|}\bigg[\|f\|_{L\log{L}, Q^{\ast}_{l}}+l\int_{Q^{\ast}_{l}}|f(z)|{\rm d} z\bigg]{}\\ &\leq &C\|b\|_{BMO({{\Bbb R}} ^n)}\sum\limits_{l = l_{0}+1}^{\infty}\frac{1}{2^{l}}(M_{L\log{L}, \varphi, \eta}f(x_{0})+lM_{\varphi, \eta}(f)(x_{0})){}\\ &\leq &C\|b\|_{BMO({{\Bbb R}} ^n)}M_{L\log{L}, \varphi, \eta}f(x_{0})\sum\limits_{l = 1}^{\infty}\frac{l}{2^{l}}{}\\ &\leq& C\|b\|_{BMO({{\Bbb R}} ^n)}M_{L\log{L}, \varphi, \eta}f(x_{0}). \end{eqnarray} $

综上所述, 有

(4.10) $ \begin{equation} J_{3}\leq C\|b\|_{BMO({{\Bbb R}} ^n)}M_{L\log{L}, \varphi, \eta}f(x_{0}). \end{equation} $

于是, 当$ r\in(0, 1) $时, (4.4)式成立.

情形2   $ r\geq 1 $. 由于$ 0<\varepsilon<\kappa<1 $, $ \eta\geq 1 $, 记$ \eta_{2} = \frac{\eta}{\varepsilon}\geq \eta $, 则有

(4.11) $ \begin{eqnarray} &&\left(\frac{1}{\varphi(|Q|)^{\eta}|Q|}\int_{Q}|[b, T]f(x)|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}{}\\ & = & \frac{C}{\varphi(|Q|)^{\eta_{2}}}\left(\frac{1}{|Q|}\int_{Q}|(b(x)-b_{Q^{\ast}})Tf(x)-T((b-b_{Q^{\ast}})f)(x)|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}{}\\ &\leq& \frac{C}{\varphi(|Q|)^{\eta_{2}}}\left(\frac{1}{|Q|}\int_{Q}|(b(x)-b_{Q^{\ast}})Tf(x)|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}{}\\ &&+ \frac{C}{\varphi(|Q|)^{\eta_{2}}}\left(\frac{1}{|Q|}\int_{Q}|T((b-b_{Q^{\ast}})f_{1})(x)|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}{}\\ &&+\frac{C}{\varphi(|Q|)^{\eta_{2}}}\left(\frac{1}{|Q|}\int_{Q}|T((b-b_{Q^{\ast}})f_{2})(x)|^{\varepsilon}{\rm d} x\right)^{\frac{1}{\varepsilon}}{}\\ & = &S_{1}+S_{2}+S_{3}. \end{eqnarray} $

为估计$ S_{1} $, 对指数$ q $($ 1<q<\frac{\kappa}{\varepsilon} $)使用Hölder不等式, 再由$ BMO({{\Bbb R}} ^n) $中的John-Nirenberg不等式可得

(4.12) $ \begin{eqnarray} S_{1}&\leq& \left(\frac{1}{|Q^{\ast}|}\int_{Q^{\ast}}|b(x)-b_{Q^{\ast}}|^{q^{\prime}\varepsilon}{\rm d} x\right)^{\frac{1}{q^{\prime}\varepsilon}}\frac{C}{\varphi(|Q|)^{\eta_{2}}}\left(\frac{1}{|Q|}\int_{Q}|Tf(x)|^{q\varepsilon}{\rm d} x\right)^{\frac{1}{q\varepsilon}}{}\\ &\leq& C\|b\|_{BMO({{\Bbb R}} ^n)}M^{\Delta}_{q\varepsilon, \varphi, \eta}(Tf)(x_{0}){}\\ &\leq & C\|b\|_{BMO({{\Bbb R}} ^n)}M^{\Delta}_{\kappa, \varphi, \eta}(Tf)(x_{0}), \end{eqnarray} $

其中$ q^{\prime} $是$ q $的共轭指标.

对于$ S_{2} $, 由算子$ T $的弱$ (1, 1) $有界性(引理2.2), Kolmogorov不等式以及广义Hölder不等式((4.2)式), 有

(4.13) $ \begin{eqnarray} S_{2}&\leq & \frac{C}{\varphi(|Q|)^{\eta_{2}}|Q|}\|(b-b_{Q^{\ast}})f_{1}\|_{L^{1}({{\Bbb R}} ^n)}{}\\ &\leq & \frac{C}{\varphi(|Q|)^{\eta}|Q^{\ast}|}\int_{Q^{\ast}}|b(x)-b_{Q^{\ast}}||f(x)|{\rm d} x{}\\ &\leq & \frac{C}{\varphi(|Q|)^{\eta}}\|b-b_{Q^{\ast}}\|_{\exp{L}, Q^{\ast}}\|f\|_{L\log{L}, Q^{\ast}}{}\\ &\leq& C\|b\|_{BMO({{\Bbb R}} ^n)}\frac{1}{\varphi(|Q^{\ast}|)^{\eta}}\|f\|_{L\log{L}, Q^{\ast}}{}\\ &\leq &C\|b\|_{BMO({{\Bbb R}} ^n)}M_{L\log{L}, \varphi, \eta}f(x_{0}). \end{eqnarray} $

对于$ S_{3} $, 由Hölder不等式可知

$ \begin{eqnarray*} S_{3}&\leq& \frac{C}{|Q|}\int_{Q}|T((b-b_{Q^{\ast}})f_{2})(x)|{\rm d} x\\ &\leq & \frac{C}{|Q|}\int_{Q}\int_{{{\Bbb R}} ^n\setminus Q^{\ast}} |K(x, y)||b(y)-b_{Q^{\ast}}||f(y)|{\rm d} y{\rm d} x\\ &\leq & \frac{C}{|Q|}\int_{Q}\sum\limits_{l = 0}^{\infty}\int_{Q^{\ast}_{l+1}\setminus Q^{\ast}_{l}} |K(x, y)||b(y)-b_{Q^{\ast}}||f(y)|{\rm d} y{\rm d} x. \end{eqnarray*} $

于是, 在(2.3)式中取$ N = n+n\eta\alpha_{0}+1 $有

(4.14) $ \begin{eqnarray} S_{3}&\leq & \frac{C}{|Q|}\int_{Q}\sum\limits_{l = 0}^{\infty}\int_{Q^{\ast}_{l+1}\setminus Q^{\ast}_{l}} \frac{1}{|y-x|^{n+n\eta\alpha_{0}+1}}|b(y)-b_{Q^{\ast}}||f(y)|{\rm d} y{\rm d} x{}\\ &\leq & C\sum\limits_{l = 0}^{\infty}\frac{1}{2^{l}\varphi(|Q^{\ast}_{l+1}|)^{\eta}|Q^{\ast}_{l+1}|}\int_{Q^{\ast}_{l+1}} |b(y)-b_{Q^{\ast}}||f(y)|{\rm d} y{}\\ &\leq& C\|b\|_{BMO({{\Bbb R}} ^n)}M_{L\log{L}, \varphi, \eta}f(x_{0})\sum\limits_{l = 0}^{\infty}\frac{l}{2^{l}}{}\\ &\leq& C\|b\|_{BMO({{\Bbb R}} ^n)}M_{L\log{L}, \varphi, \eta}f(x_{0}). \end{eqnarray} $

因此, 当$ r\geq 1 $时, (4.4)式也成立. 综上所述, 引理4.2得证.

下面给出定理1.2的证明.

证   首先, 选取$ \eta\geq p^{\prime} $, 由引理2.4, (3.1)式和引理2.3可得

$ \begin{eqnarray*} \|M_{\kappa, \varphi, \eta}Tf\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\leq C\|M^{\#, \Delta}_{\kappa, \varphi, \eta}Tf\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\leq C\|M_{\varphi, \eta} f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\leq C\|f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}. \end{eqnarray*} $

其次, 选取$ \eta\geq 2p^{\prime} $, 由引理4.1和引理2.3可得

$ \begin{eqnarray*} \|M_{L\log{L}, \varphi, \eta}f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\leq C\|M_{\varphi, \frac{\eta}{2}}M_{\varphi, \frac{\eta}{2}}f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\leq C\|M_{\varphi, \frac{\eta}{2}}f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\leq C\|f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}. \end{eqnarray*} $

最后, 选取$ \eta\geq 2p^{\prime} $并利用引理2.4和引理4.2可得

$ \begin{eqnarray*} \|[b, T]f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}&\leq & C\|M^{\#, \Delta}_{\varepsilon, \varphi, \eta}Tf\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\\ &\leq & C\|M^{\Delta}_{\kappa, \varphi, \eta}Tf\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}+C\|M_{L\log{L}, \varphi, \eta}f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}\\ &\leq & C\|f\|_{L^{p}_{\omega}({{\Bbb R}} ^n)}. \end{eqnarray*} $

综上, 定理1.2得证.