考虑如下Schrödinger型算子

$H_2=(-\Delta)^2+V^2,\ \ x\in {\Bbb R}^n, $

其中$V$是非互位势且属于某一反向Höolder类$B_q,\frac n2\le q ＜n$.

当$V(x)$为非负多项式时,Zhong^[1]考虑如下Schrödinger算子 $H_1=-\Delta+V$及$H_2$. Zhong建立了关于算子$H_1$和$H_2$的基本解 的估计并证明了如下算子$V^{2-\frac2j}\nabla^{j}H_2^{-1}$ $(j=0,1,2,3,4$)和$V^kH_1^{-k}$, $V^{k-\frac12}\nabla H_1^{-k}(k\in N)$是$L^p$上的有界算子.

Shen^[2]推广了Zhong关于$H_1$的结果,此时$V$属于反向Höolder类,而此函数类包含非负多项式. 事实上Shen建立关于算子$H_1$基本解的估计且证明了 $VH_1^{-1},V^{\frac12}\nabla H_{1}^{-1}, \nabla^2 H_1^{-1}$是$L^p$上的有界算子.

Kurata和Sugano研究了Schrödinger型算子$H_2,$并建立了关于算子 $H_2$基本解的估计. 此外他们还证明了算子$V^{2-j/2}\nabla^jH_{2}^{-1}$ $ (j=0,1,2,3,4)$是$L^p$上的有界算子. 同时进一步证明 $\nabla^4H_{2}^{-1}$ 是Calderon-Zygmund算子, 详见文献^[3-6].

近来,关于与Schrödinger算子$H_1$和$H_2$相关的Riesz变换及其交换子的有界性, 涌现了许多令人感兴趣的结果 详见参考文献^{[1, 7-9]}等.

受文献^{[1-2, 8]}的启发,本文将研究将研究算子$T_1=V\nabla^{2}H_{2}^{-1},$ $T_2=V^{\frac{3}{2}}\nabla H_{2}^{-1},$ $T_1^{\ast}=\nabla^{2}H_{2}^{-1}V,$ $T_2^\ast=\nabla H_{2}^{-1}V^{\frac{3}{2}}$和它们的交换子$[b,T_1],$ $[b,T_2]$的$L^{p}$有界性,其中 $b\in BMO$.此外还将讨论算子$T_1^\ast,T_2^\ast$ 在$BMO_{L}$空间上的有界性.

为方便起见,首先给出一些定义和符号.

设$V(x)$为${\Bbb R}^n$上的非负局部$L^{q}(1＜q＜\infty)$可积函数, 如果存在常数$C>0$使得如下的反向Höolder不等式

$\begin{equation}\label{eq11}\bigg(\frac{1}{|B|}\int_{B}V(x)^{q}{\rm d}x\bigg)^{1/q}\leq\frac{C}{|B|}\int_{B}V(x){\rm d}x,\end{equation}$

(1.1)

对于所有的$B\subset {\Bbb R}^n$成立,称$V(x)$属于$B_{q}$.

对于$q_{1}\geq q_{2}>1,$由Höolder不等式有$B_{q_{1}}\subset B_{q_{2}}$. $B_{q}$类的一个重要性质:若$V\in B_{q},q>1$,则存在仅与 $n$和(1.1)中的常数$C$有关的$\epsilon>0$使得$V\in B_{q+\epsilon}$ (参见文献^[3]). 如果$V\in B_{q},q>1$,则$V(x){\rm d}x$是满足双倍性质,即对于任意的 $r>0,x\in {\Bbb R}^n$,

$\begin{equation}\label{eq12}\int_{B(x,2r)}V(y){\rm d}y\leq C_{0}\int_{B(x,r)}V(y){\rm d}y. \end{equation}$

(1.2)

由文献^[2]知,若$V\in B_{q}$,则

$\begin{equation}\label{eq21}\int_{B(x,R)}\frac{V(y)}{|x-y|^{n-2}}{\rm d}y\leq \frac{C}{{\Bbb R}^{n-2}}\int_{B(x,R)}V(y){\rm d}y,\end{equation}$

(1.3)

$\begin{equation}\label{eq22}\int_{B(x,R)}\frac{V(y)}{|x-y|^{n-1}}{\rm d}y\leq \frac{C}{{\Bbb R}^{n-1}}\int_{B(x,R)}V(y){\rm d}y,\end{equation}$

(1.4)

而且存在常数 $C>0$使得对$0＜r＜R＜\infty,$有

$\begin{equation} \frac1{r^{n-2}}\int_{B(x,r)}V(y){\rm d}y\le C\bigg(\frac Rr\bigg)^{\frac nq-2}\frac1{{\Bbb R}^{n-2}}\int_{B(x,R)}V(y){\rm d}y. \end{equation}$

(1.5)

设$f$是 ${\Bbb R}^n$上的可积函数,$B=B(x,r)$是球.又设

$$f_{B}=\frac{1}{|B|}\int_{B(x,r)}f(y){\rm d}y.$$

当$r＜\rho(x)$,则$f(B,V)=f_{B}$; 当$r>\rho(x)$,则 $f(B,V)=0$.

定义1.1 设$f$是${\Bbb R}^n$上的可积函数,如果

$\begin{equation}\|f\|_{BMO_{L}}=\mathop{\sup}\limits_{B}\frac{1}{|B|}\int_{B}|f(y)-f(B,V)|{\rm d}y＜\infty.\end{equation}$

(1.6)

称$f\in BMO_{L}({\Bbb R}^n).$

显然 $L^{\infty}({\Bbb R}^n)\subset BMO_{L}({\Bbb R}^n)\subset BMO({\Bbb R}^n)$ 且$\|f\|_{BMO}\leq 2\|f\|_{BMO_{L}}$.

注1.1 设 $1\leq p＜\infty,\,f\in BMO({\Bbb R}^n)$,则

$$\mathop{\sup}\limits_{B}\frac{1}{|B|}\bigg(\int_{B}|f(y)-f_{B,V}|^{p}{\rm d}y\bigg)^{\frac{1}{p}}\leq C\|f\|_{BMO_{L}}.$$

函数$f\in BMO_{L}({\Bbb R}^n)$当且仅当存在依赖$B=B(x,r)$且满足如下条件:当$r>\rho(x)$时$C_{B}=0$ 常数 $C_{B}$ 使得

$$\mathop{\sup}\limits_{B}\frac{1}{|B|}\bigg(\int_{B}|f(y)-C_{B}|^{p}{\rm d}y\bigg)^{\frac{1}{p}}＜\infty$$

及

$$\|f\|_{BMO_{L}}\leq C\mathop{\sup}\limits_{B}\frac{1}{|B|}\bigg(\int_{B}|f(y)-C_{B}|^{p}{\rm d}y\bigg)^{\frac{1}{p}}.$$

首先给出算子$T_1$及其交换子的$L^p$估计.

定理1.1 设 $V\in B_{q},\,q\geq \frac{n}{2}$,则对$1＜p\leq q$,有

$\begin{equation}\|T_1f\|_{L^p}\leq C\|f\|_{L^p}.\end{equation}$

(1.7)

由对偶得

推论1.2 设 $V\in B_{q},\,q\geq \frac{n}{2},$则对 $q'\leq p＜\infty$,有

$\begin{equation}\|T_1^\ast f\|_{L^p}\leq C\|f\|_{L^p}.\end{equation}$

(1.8)

由定理1.1得

推论1.3 设$V\in B_q,\,q\ge\frac n2.$ 如果 $(-\Delta)^2 u+V^2 u=f,\,x\in{\Bbb R}^n,$ 则对于$1＜p\le q$有

$$ \|V\nabla^2 u\|_{L^p}\le C_p\|f\|_{L^p}. $$

定理1.4 设$V\in B_{q},\,q\geq \frac{n}{2}.$ 设$b\in BMO$,则对$1＜p＜q$,有

$\begin{equation}\|[b,T_1]f\|_{L^p}\leq C_{p}\|b\|_{BMO}\|f\|_{L^p}.\end{equation}$

(1.9)

定理1.5 设 $V\in B_{q},\,q\geq \frac{n}{2}$,则

$\begin{equation}\|T_1^\ast f\|_{BMO_{L}({\Bbb R}^n)}\leq C\|f\|_{BMO_{L}({\Bbb R}^n)}.\end{equation}$

(1.10)

下面是关于上述算子$T_2$及其交换子的$L^p$-估计.

定理1.6 设$V\in B_{q},\,q\geq \frac{n}{2}$. 则对$1\leq p\leq p_{0}$,有

$\begin{equation}\|T_2f\|_{L^p}\leq C\|f\|_{L^p}, \end{equation}$

(1.11)

其中当$\frac{n}{2}\leq q＜n$,则$\frac{1}{p_{0}}=\frac{7}{2q}-\frac{3}{n}$;当$q\geq n,$则 $p_{0}=\frac{2q}{3}$.

由定理1.6可得

推论1.7 设 $V\in B_q,\,q\ge\frac n2.$ 假设$(-\Delta)^2 u+V^2 u=f,\,x\in{\Bbb R}^n,$ 则

$\|V^{\frac32}\nabla u\|_{L^p}\le C_p\|f\|_{L^p},\,\mbox{对于}\ \ \ 1＜p\le q,$

其中当$\frac{n}{2}\leq q＜n$,则$\frac{1}{p_{0}}=\frac{7}{2q}-\frac{3}{n}$; 当$q\geq n,$则 $p_{0}=\frac{2q}{3}$.

定理1.8 设$V\in B_{q},\,q\geq \frac{n}{2}.$ 设$b\in BMO$,则对$1＜p＜p_{0}$,

$\begin{equation}\|[b,T_2]f\|_{L^p}\leq C_{p}\|b\|_{BMO}\|f\|_{L^p}, \end{equation}$

(1.12)

其中当$\frac{n}{2}\leq q＜n$,则$\frac{1}{p_{0}}=\frac{7}{2q}-\frac{3}{n}$;当$q\geq n,$则 $p_{0}=\frac{2q}{3}$.

定理1.9 设$V\in B_{q},\,q\geq \frac{n}{2}$,则

$\begin{equation}\|T_2^\ast f\|_{BMO_{L}({\Bbb R}^n)}\leq C\|f\|_{BMO_{L}({\Bbb R}^n)}.\end{equation}$

(1.13)

本文结构安排如下：第二小节将介绍一些关键的引理;定理1.1和定理1.6的证明见第三节；第四节给出定理1.4和定理1.8的证明；第五部分我们将证明定理1.5 和定理1.9.

设$V(x)$是非负位势且属于$B_{\frac{n}{2}},H_2$是一中心Schrödinger型算子,$\Gamma_{H_{2}}(x,y)$是算子$H_{2}$的基本解.

对$x\in {\Bbb R}^n$,函数$m(x,V)$定义如下

$$\frac{1}{m(x,V)}=\sup \bigg\{r>0,\ \frac{1}{r^{n-1}}\int_{B(x,r)}V(y){\rm d}y\leq 1\bigg\}.$$

有关$m(x,v)$的信息详见文献^[2].下面引理2.1来自文献^[2].

引理2.1 设$v\in B_{q},\,q>\frac{n}{2}$,则存在常数$C>0,\ c>0,\ k_{0}>0,$使得 对任意的$ x,y \in{\Bbb R}^n$及 $0＜r＜R＜\infty,$

(i) $0＜m(x,V)＜\infty;$

(ii) 若$h=\frac{1}{m(x,V)}$,则$ \frac{1}{h^{n-2}}\int_{B(x,h)}V(y){\rm d}y=1;$

(iii) 若$|x-y|\leq \frac{C}{m(x,V)}$,则$m(x,V)\sim m(y,V);$

(iv) $m(y,V)\leq C\{1+|x-y|m(x,V)\}^{k_{0}}m(x,V);$

(v) $m(y,V)\leq Cm(x,V)\{1+|x-y|m(x,V)\}^{\frac{-k_{0}}{k_{o}+1}};$

(vi) $c\{1+|x-y|m(y,V)\}^{\frac{1}{k_{0}+1}}\leq \{1+|x-y|m(x,V)\}\leq C\{1+|x-y|m(y,V)\}^{k_{0}+1};$

(vii) $\frac{1}{r^{n-2}}\int_{B(x,r)}V(y){\rm d}y\leq C(\frac{R}{r})^{(\frac{n}{q})-2}\cdot\frac{1}{{\Bbb R}^{n-2}}\int_{B(x,R)}V(y){\rm d}y.$

而且由(ii)和(vii)得

(viii) 若 $\frac{1}{h^{n-2}}\int_{B(x,h)}V(y){\rm d}y\sim1$,则 $r\sim\frac{1}{m(x,V)}$.

下列引理参考文献^[3].

引理2.2 设$V\in B_{q},\,q\geq \frac{n}{2},$则对任意的正整数$N$,存在常数$C_{N}$使得

$\begin{equation}0\leq \Gamma_{H_{2}}(x,y)\leq \frac{C_{N}}{\{1+m(x,V)|x-y|\}^{N}}\frac{1}{|x-y|^{n-4}}.\end{equation}$

(2.1)

引理2.3 设 $j=1,2,3,$且$V\in B_{\frac{2n}{4-j}}$,则存在常数$C_{N}$使得

$\begin{equation}0\leq \nabla_{x}^{j}\Gamma_{H_{2}}(x,y)\leq \frac{C_{N}}{\{1+m(x,V)|x-y|\}^{N}}\frac{1}{|x-y|^{n-4+j}}.\end{equation}$

(2.2)

引理2.4 设$V\in B_{q},\,q\geq \frac{n}{2}$. 设$N>\log_{2}C_{0}+1$,其中$C_{0}$是(1.2)式中的常数.则对任意的 $x_{0}\in {\Bbb R}^{n}$及$R>0$,有

$$\frac{1}{\{1+m(x_{0},V)R\}^{N}}\int_{B(x_{0},R)}V(y){\rm d}y\leq C R^{n-2}.$$

下面引理2.5见文献^[4].

引理2.5 设 $j=1,2,3,\,V\in B_{q_{0}},$ $ \frac{n}{2}\leq q_{0}＜\frac{2n}{4-j}$. 设$x_{0}\in{\Bbb R}^n,$对方程$(-\Delta)^{2}u+V^{2}u=0,$ $ x\in B_{R}(x_{0})$,则存在常数$C$使得

$\begin{equation}\bigg\{\int_{B_{\frac{R}{2}}(x_{0})}|\nabla^{j}u(x)|^{t}{\rm d}x \bigg\}^{\frac{1}{t}}\leq CR^{(\frac{2n}{q_{0}})-4}\{1+Rm(x_{0},V)\}^{4}\mathop{\sup}\limits_{y\in B_{R}(x_{0})}|u(y)|,\end{equation}$

(2.3)

其中$\frac{1}{t}=\frac{2}{q_{0}}-\frac{(4-j)}{n}.$

设

$$T_1f(x)=(V\nabla^{2}H_{2}^{-1})f(x)=\int_{{\Bbb R}^n}K_1(y,x)f(y){\rm d}y, $$

其对偶算子为

$$T_1^{*}f(x)=(\nabla^{2}H_{2}^{-1}V)f(x)=\int_{{\Bbb R}^n}K_1(x,y)f(y){\rm d}y.$$

下面是核函数的估计.

引理2.6 设$V\in B_{q},\,q\geq \frac{n}{2}.$则存在常数$\delta>0$,对任意的整数$k>0$和$0＜h＜\frac{|x-y|}{16},$有

$\begin{equation}|K_1(x,y)|\leq \frac{C_{N}}{\{1+m(x,V)|x-y|\}^{k}}\frac{1}{|x-y|^{n-2}}V(y),\end{equation}$

(2.4)

$\begin{equation}|K_1(x+h,y)-K_1(x,y)|\leq \frac{C_{k}}{\{1+m(x,V)|x-y|\}^{k}}\frac{|h|^{\delta}}{|x-y|^{n-2+\delta}}V(y).\end{equation}$

(2.5)

证 我们采用文献^[8]中的证明方法.易证明$K_1(x,y)=\nabla^{2}_{x}\Gamma_{H_{2}}(x,y)V(y)$. 由引理2.3知对任意的$x,y\in {\Bbb R}^n$,有

$$|K_1(x,y)|\leq \frac{C_{k}}{\{1+m(x,V)|x-y|\}^{k}}\frac{1}{|x-y|^{n-2}}V(y).$$

对于固定的$x,y\in {\Bbb R}^n$,取 $\frac{n}{2}＜q_{0}＜\min(n,q).$ 则 $V\in B_{q_{0}}$. 设 $R=\frac{|x-y|}{8}$,$\frac{1}{t}=\frac{2}{q_{0}}-\frac{3}{n}$, 则$\delta=1-\frac{n}{t}>0.$ 对任意的 $0＜h＜\frac{R}{2}$, 由Morrey嵌入定理^[10]、引理2.3和引理2.5得

$\begin{eqnarray*} & & |K_1(x+h,y)-K_1(x,y)|\\ & = & |\nabla^{2}_{x}\Gamma_{H_2}(x+h,y)-\nabla^{2}_{x}\Gamma_{H_2}(x,y)|V(y)\\ & \leq & C|h|^{1-\frac{n}{t}}\bigg(\int_{B(x,R)}|\nabla_{z}\nabla_{x}^{2}\Gamma_{H_{2}}(z,y)|^{t}{\rm d}z\bigg)^{\frac{1}{t}}V(y)\\ & \leq & C|h|^{1-\frac{n}{t}}|R|^{\frac{2n}{q_{0}}-4}\{1+R m(x,V)\}^{4}\mathop{\sup}\limits_{z\in B_{2R}(x)}|\nabla^{2}_{x}\Gamma_{H_{2}}(z,y)|V(y)\\ & \le & C|h|^{1-\frac{n}{t}}|R|^{\frac{2n}{q_{0}}-4}\{1+R m(x,V)\}^{4}\frac{1}{\{1+R m(x,V)\}^{N}}\frac{1}{|x-y|^{n-2}}V(y)\\ & \leq & C\frac{|h|^{\delta}}{|x-y|^{n-2+\delta}}\frac{1}{\{1+m(x,V)|x-y|\}^{k}}V(y), \ \ \mbox{($k$充分大),} \end{eqnarray*}$

其中最后一个不等式我们使用了引理2.1(vi).

设

$$T_2f(x)=(V^{\frac{3}{2}}\nabla H_{2}^{-1})f(x)=\int_{{\Bbb R}^n}K_2(y,x)f(y){\rm d}y.$$

则$T_2$的对偶算子为

$$T_2^{\ast}f(x)=(\nabla H_{2}^{-1}V^{\frac{3}{2}})f(x)=\int_{{\Bbb R}^n}K_2(x,y)f(y){\rm d}y.$$

类似引理2.6,可得关于核函数$K_2(x,y)$的估计.

引理2.7 设$V\in B_{q},\,q\geq \frac{n}{2},$则存在常数$\delta>0$及对于任意整数 $k>0,\ 0＜h＜\frac{|x-y|}{16},$

$\begin{equation}|K_2(x,y)|\leq \frac{C_{N}}{\{1+m(x,V)|x-y|\}^{k}}\frac{1}{|x-y|^{n-3}}V^{\frac{3}{2}}(y),\end{equation}$

(2.6)

$\begin{equation}|K_2(x+h,y)-K_2(x,y)|\leq \frac{C_{k}}{\{1+m(x,V)|x-y|\}^{k}}\frac{|h|^{\delta}}{|x-y|^{n-3+\delta}}V^{\frac{3}{2}}(y).\end{equation}$

(2.7)

本节将分别证明定理1.1和定理1.6. 这一部分证明将采用文献^[2]的证明思想.首先定证明理1.1.

证 设$f\in L^{p}({\Bbb R}^n)$,$1＜p\leq q$且注意到

$$u(x)=(\nabla^{2}H_{2}^{-1})f(x)=\int_{{\Bbb R}^n}\nabla^{2}_{x}\Gamma_{H_{2}}(x,y)f(y){\rm d}y.$$

我们只需证明$\|Vu\|_{L^p}\leq C\|f\|_{L^p}.$ 将$u$做如下分解

$$u(x)=\int_{|y-x|＜r}\nabla^{2}_{x}\Gamma_{H_{2}}(x,y)f(y){\rm d}y+\int_{|y-x|\geq r}\nabla^{2}_{x}\Gamma_{H_{2}}(x,y)f(y){\rm d}y=u_{1}(x)+u_{2}(x),$$

其中$r=\frac{1}{m(x,v)}.$

因为 $V\in B_{q_{0}},\,q_{0}>q$并使用引理2.3,有

$\begin{eqnarray*} |u_{1}(x)| & = & \int_{|y-x|＜r}\nabla^{2}_{x}\Gamma_{H_{2}}(x,y)f(y){\rm d}y\\ & \leq & C_{N}\int_{|y-x|＜r}\frac{1}{\{1+m(x,V)|x-y|\}^{N}}\frac{1}{|x-y|^{n-2}}|f(y)|{\rm d}y\\ & \leq & C_{N}\int_{|y-x|＜r}\frac{1}{|x-y|^{n-2}}|f(y)|{\rm d}y\\ & \leq & C_{N}\bigg(\int_{|y-x|＜r}\frac{1}{|x-y|^{(n-2)q'_0}}{\rm d}y\bigg)^{\frac{1}{q'_{0}}}\bigg(\int_{|y-x|＜r}|f(y)|^{q_{0}}{\rm d}y\bigg)^{\frac{1}{q_{0}}}\\ & \leq & C_{N}r^{2-\frac{n}{q_{0}}}\bigg(\int_{|y-x|＜r}|f(y)|^{q_{0}}{\rm d}y\bigg)^{\frac{1}{q_{0}}}, \end{eqnarray*}$

其中我们使用了Höolder不等式和$q>\frac n2$. 则

$\begin{eqnarray*} \int_{{\Bbb R}^n}|V(x)u_{1}(x)|^{q_{0}}{\rm d}x & \leq & C\int_{{\Bbb R}^n}V(x)^{q_{0}}r^{2q_{0}-n}\int_{|y-x|＜r}|f(y)|^{q_{0}}{\rm d}y{\rm d}x\\ & \leq & C\int_{{\Bbb R}^n}|f(y)|^{q_{0}}\bigg\{\int_{|y-x|＜r}r^{2q_{0}-n}V(x)^{q_{0}}{\rm d}x\bigg\}{\rm d}y. \end{eqnarray*}$

设 $R=\frac{1}{m(y,V)}$,由引理2.1(iii),不等式(1.1)和引理2.4,有

$\begin{eqnarray*} \int_{|y-x|＜r}r^{2q_{0}-n}V(x)^{q_{0}}{\rm d}x & \leq & CR^{2q_{0}-n}\int_{|y-x|＜CR}V(x)^{q_{0}}{\rm d}x\\ & \leq & CR^{2q_{0}-n}R^{n}\bigg(\frac{1}{R^{n}}\int_{|y-x|＜CR}V(x)^{q_{0}}{\rm d}x\bigg)\\ & \leq & CR^{2q_{0}}\bigg(\frac{1}{R^{n}}\int_{|y-x|＜CR}V(x){\rm d}x\bigg)^{q_{0}}\\ & \leq & C. \end{eqnarray*}$

因此

$$\int_{{\Bbb R}^n}|V(x)u_{1}(x)|^{q_{0}}{\rm d}x\leq C\int_{{\Bbb R}^n}|f(y)|^{q_{0}}{\rm d}y,~\mbox{对某个} \ q_{0}>q\geq\frac{n}{2}.$$

使用引理2.3,不等式(1.3)和引理2.4,有

$\begin{eqnarray*} \int_{{\Bbb R}^n}|V(x)u_{1}(x)|{\rm d}x & \leq & C\int_{{\Bbb R}^n}|f(y)|\bigg\{\int_{|x-y|＜\frac{1}{m(x,V)}}\frac{V(x)}{|x-y|^{n-2}}{\rm d}x\bigg\}{\rm d}y\\ & \leq & C\int_{{\Bbb R}^n}|f(y)|\frac{1}{r^{n-2}}\int_{|x-y|＜\frac{1}{m(x,V)}}V(x){\rm d}x{\rm d}y\\ & \leq & C\int_{{\Bbb R}^n}|f(y)|{\rm d}y. \end{eqnarray*}$

因此由插值得

$$\|Vu_{1}\|_{L^p}\leq C\|f\|_{L^p},\ \mbox{对于} \ 1\leq p\leq q_{0}.$$

使用引理2.3和Höolder不等式得

$\begin{eqnarray*} |u_{2}(x)| & = & \bigg|\int_{|y-x|\geq r}\nabla^{2}_{x}\Gamma_{H_{2}}(x,y)f(y){\rm d}y\bigg|\\ & \leq & \int_{|y-x|\geq r}\frac{1}{\{1+m(x,V)|x-y|\}^{N}}\frac{1}{|x-y|^{n-2}}|f(y)|{\rm d}y\\ & \leq & \bigg(\int_{|y-x|\geq r}\frac{1}{\{1+m(x,V)|x-y|\}^{N}}\frac{1}{|x-y|^{n-2}}{\rm d}y\bigg)^{\frac{1}{p' }}\\ & & \times\bigg(\int_{|y-x|\geq r}\frac{|f(y)|^{p}}{\{1+m(x,V)|x-y|\}^{N}}\frac{1}{|x-y|^{n-2}}{\rm d}y\bigg)^{\frac{1}{p}}\\ & \leq & Cr^{\frac{2}{p' }}\bigg(\int_{|y-x|\geq r}\frac{|f(y)|^{p}}{\{1+m(x,V)|x-y|\}^{N}}\frac{1}{|x-y|^{n-2}}{\rm d}y\bigg)^{\frac{1}{p}}, \end{eqnarray*}$

其中$1\leq p\leq q_{0},r=\frac{1}{m(x,V)}$且注意到$N$充分大.

因此

$\begin{eqnarray*} & & \int_{{\Bbb R}^n} |V(x)u_{2}(x)|^{p}{\rm d}x\\ & \leq & C\int_{{\Bbb R}^n}|V(x)r^{\frac{2}{p' }}\bigg(\int_{|y-x|\geq r}\frac{|f(y)|^{p}}{\{1+m(x,V)|x-y|\}^{N}}\frac{1}{|x-y|^{n-2}}{\rm d}y\bigg)^{\frac{1}{p}}|^{p}{\rm d}x\\ & \leq & C\int_{{\Bbb R}^n}|f(y)|^{p}\int_{|y-x|\geq \frac{1}{m(x,V)}}\frac{V(x)^{p}}{\{1+m(x,V)|x-y|\}^{N}}\frac{1}{|x-y|^{n-2}}\frac{1}{m(x,V)^{2(p-1)}}{\rm d}x{\rm d}y. \end{eqnarray*}$

现固定$y\in {\Bbb R}^{n}$,令$R=\frac{1}{m(y,V)}$且注意到 $k_{1}=\frac{N-2(p-1)k_{o}}{k_{0}+1}$,由引理2.1得,当$k$充分大

$\begin{eqnarray*} & & \int_{|y-x|\geq \frac{1}{m(x,V)}}\frac{V(x)^{p}}{\{1+m(x,V)|x-y|\}^{N}}\frac{1}{|x-y|^{n-2}}\frac{1}{m(x,V)^{2(p-1)}}{\rm d}x\\ & \leq & C\int_{|y-x|\geq CR}\frac{V(x)^{p}}{\{1+m(y,V)|x-y|\}^{k_{1}}}\frac{1}{|x-y|^{n-2}}\frac{1}{m(x,V)^{2(p-1)}}{\rm d}x\\ & \leq & C\mathop{\sum}_{j=1}^{\infty}\int_{|y-x|\geq 2^{j}R}V(x)^{p}\frac{1}{(2^{j})^{k_{1}}(2^{j}R)^{n-2}}R^{2(p-1)}{\rm d}x\\ & \leq & C\mathop{\sum}_{j=1}^{\infty}(2^{j})^{2-k_{1}}R^{2p}\frac{1}{(2^{j}R)^{n}}\int_{|y-x|\geq 2^{j}R}V(x)^{p}{\rm d}x\\ & \leq & C\mathop{\sum}_{j=1}^{\infty}(2^{j})^{2-k_{1}}R^{2p}\bigg(\frac{1}{(2^{j}R)^{n}}\int_{|y-x|\geq 2^{j}R}V(x){\rm d}x\bigg)^{p}\\ & \leq & C\mathop{\sum}_{j=1}^{\infty}(2^{j})^{2-k_{1}}\leq C, \end{eqnarray*}$

这意味着 对于$ 1\leq p \leq q_{0}$

$$\int_{{\Bbb R}^n}|V(x)u_{2}(x)|^{p}{\rm d}x\leq C\int_{{\Bbb R}^n}|f(x)|{\rm d}x. $$

证毕.

接下来证明定理1.6.由对偶知我们只需证明 $\|T_2^{*}f(x)\|_{L^p}\leq C\|f\|_{L^p}.$

我们首先建立如下引理:

引理3.1 设 $V\in B_{q},\,q\geq \frac{n}{2}$. 则对$p_{0}' \leq p＜\infty$,

$\begin{equation}\|T_2^\ast f\|_{L^p}\leq C\|f\|_{L^p},\end{equation}$

(3.1)

其中当$\frac{n}{2}\leq q＜n$时,$\frac{1}{p'_{0}}=1-\frac{7}{2q}+\frac{3}{n};$当 $q\geq n,$则$p_{0}' =(\frac{2q}{3})'$.

证 设$V\in B_{q},\,q\geq\frac{n}{2}$.则存在$q_{1}>q$使得$V\in B_{q_{1}}$. 令$r=\frac{1}{m(x,V)}$且考虑 $\frac{n}{2}\leq q＜q_{1}＜\frac {2n}3$. 选取$t$及$p_{1}$使得 $\frac{1}{t}=\frac{2}{q_{1}}-\frac{3}{n}$, $\frac{1}{p_{1}}=1-\frac{7}{2q_{1}}+\frac{3}{n}.$ 则 $\frac{1}{t}+\frac{3}{2q_{1}}+\frac{1}{p_{1}}=1.$

由Höolder不等式得

$\begin{eqnarray*} |T_2^{*}f(x)| & \leq & \mathop{\sum}^{+\infty}_{j=-\infty}\int_{2^{j-1}r＜|x-y|\leq 2^{j}r}|\nabla_{x}\Gamma_{H_2}(x,y)|V^{\frac{3}{2}}|f(y)|{\rm d}y\\ & \leq & \mathop{\sum}^{+\infty}_{j=-\infty}\bigg(\int_{2^{j-1}r＜|x-y|\leq 2^{j}r}|\nabla_{x}\Gamma_{H_2}(x,y)|^{t}{\rm d}y\bigg)^{\frac{1}{t}}\bigg(\int_{2^{j-1}r＜|x-y|\leq 2^{j}r}V(y)^{q_{1}}{\rm d}y\bigg)^{\frac{3}{2q_{1}}}\\ & & \times \bigg(\int_{2^{j-1}r＜|x-y|\leq 2^{j}r}|f(y)|^{p_{1}}{\rm d}y\bigg)^{\frac{1}{p_{1}}}. \end{eqnarray*}$

注意到$k_{1}=N-4$并使用引理2.2和引理2.5得

$\begin{eqnarray*} & & \bigg(\int_{2^{j-1}r＜|x-y|\leq 2^{j}r}|\nabla_{x}\Gamma_{H_2}(x,y)|^{t}{\rm d}y\bigg)^{\frac{1}{t}}\\ & \leq & C(2^{j+1}r)^{\frac{2n}{q_{1}}-4}\{1+2^{j+1}r m(x,V)\}^{4}\mathop{\sup}\limits_{z\in B_{2^{j+1}R}(x)}|\Gamma_{H_2}(z,y)|\\ & \leq & C(2^{j+1}r)^{\frac{2n}{q_{1}}-4}\{1+2^{j+1}\}^{4}\frac{1}{\{1+m(x,V)|x-y|\}^{N}}\frac{1}{|x-y|^{n-4}}\\ & \leq & C(2^{j}r)^{\frac{2n}{q_{1}}-n}\frac{1}{\{1+2^{j}\}^{k_{1}}}, \end{eqnarray*}$

则

$\begin{eqnarray*} |T_2^\ast f(x)| & \leq & C\mathop{\sum}^{+\infty}_{j=-\infty}(2^{j}r)^{\frac{2n}{q_{1}}-4}\frac{1}{\{1+2^{j}\}^{k_{1}}}\bigg(\int_{|x-y|\leq 2^{j}r}V(y)^{q_{1}}{\rm d}y\bigg)^{\frac{3}{2q_{1}}}\\ & & \times \bigg(\int_{|x-y|\leq 2^{j}r}|f(y)|^{p_{1}}{\rm d}y\bigg)^{\frac{1}{p_{1}}}\\ & \leq & C\mathop{\sum}^{+\infty}_{j=-\infty}\frac{(2^{j}r)^{3}}{\{1+2^{j}\}^{k_{1}}}\bigg(\frac{1}{(2^{j}r)^{n}}\int_{|x-y|\leq 2^{j}r}V(y)^{q_{1}}{\rm d}y\bigg)^{\frac{3}{2q_{1}}}\\ & & \times \bigg(\frac{1}{(2^{j}r)^{n}}\int_{|x-y|\leq 2^{j}}|f(y)|^{p_{1}}{\rm d}y\bigg)^{\frac{1}{p_{1}}}\\ & \leq & C\{M(|f|^{p_{1}})(x)\}^{\frac{1}{p_{1}}}\mathop{\sum}^{+\infty}_{j=-\infty}\frac{(2^{j}r)^{3}}{\{1+2^{j}\}^{k_{1}}} \bigg(\frac{1}{(2^{j}r)^{n}}\int_{|x-y|\leq 2^{j}r}|V(y)|^{q_{1}}{\rm d}y\bigg)^{\frac{3}{2q_{1}}}\\ & \leq & C\{M(|f|^{p_{1}})(x)\}^{\frac{1}{p_{1}}}\mathop{\sum}^{+\infty}_{j=-\infty}\frac{1}{\{1+2^{j}\}^{k_{1}}} \bigg(\frac{1}{(2^{j}r)^{n-2}}\int_{B(x,2^{j}r)}V(y){\rm d}y\bigg)^{\frac{3}{2}}. \end{eqnarray*}$

若$j\leq 0$,由(1.5)式知

$\begin{eqnarray*} (2^{j}r)^{3}\bigg(\frac{1}{(2^{j}r)^{n}}\int_{B(x,2^{j}r)}V(y){\rm d}y\bigg)^{\frac{3}{2}} & \leq & \bigg\{(2^{-j})^{\frac{n}{q_{1}}-2}\frac{1}{r^{n-2}}\int_{B(x,r)}V(y){\rm d}y\bigg\}^{\frac{3}{2}}\\ & \le & C(2^{j})^{3-\frac{3n}{2q_{1}}}. \end{eqnarray*}$

另一方面,若$j>0$,双倍性质(1.2)式可得

$\begin{eqnarray*} (2^{j}r)^{3}\bigg(\frac{1}{(2^{j}r)^{n}}\int_{B(x,2^{j}r)}V(y){\rm d}y\bigg)^{\frac{3}{2}} & \leq & \bigg\{\frac{C_0^{j}}{(2^{j}r)^{n-2}}\int_{B(x,r)}V(y){\rm d}y\bigg\}^{\frac{3}{2}}\\ & \leq & C(2^{-j})^{\frac{3n}{2}-3}(C_0^{\frac32})^j. \end{eqnarray*}$

因此选取$k_1$充分大, $|T_2^{*}f|\leq C\{M(|f|^{p_{1}})(x)\}^{\frac{1}{p_{1}}}.$ 由此,对$p_{1}＜p＜\infty$,有 $\|T_2^{*}f\|_{L^p}\leq C\|f\|_{L^p}$. 因为$p_{0}' >p_{1}$,故(3.1)式得证.

最后考虑$q_{1}\geq n$的情形. 由$B_{q}$的性质可设 $q_{1}>n$.由引理2.2和 Höolder不等式得

$\begin{eqnarray*} |T_2^{*}f(x)| & \le & C_{N}\mathop{\sum}^{+\infty}_{j=-\infty}\frac{1}{(1+2^{j})^{N}} \frac{1}{(2^{j}r)^{n-3}}\int_{B(x,2^{j}r)}V(y)^{\frac{3}{2}}|f(y)|{\rm d}y\\ & \leq & C_{N}\mathop{\sum}^{+\infty}_{j=-\infty}\frac{(2^{j}r)^{3}}{(1+2^{j})^{N}} \bigg(\frac{1}{(2^{j}r)^{n}}\int_{B(x,2^{j}r)}V(y)^{q_{1}}{\rm d}y\bigg)^{\frac{3}{2q_{1}}} \\ & & \times\bigg(\frac{1}{(2^{j}r)^{n}}\int_{B(x,2^{j}r)}|f(y)|^{p_{1}}{\rm d}y\bigg)^{\frac{1}{p_{1}}}\\ & \leq & C_{N}\{M(|f|^{p_{1}})(x)\}^{\frac{1}{p_{1}}} \mathop{\sum}^{+\infty}_{j=-\infty}\frac{(2^{j}r)^{3}}{(1+2^{j})^{N}} \bigg(\frac{1}{(2^{j}r)^{n}}\int_{B(x,2^{j}r)}V(y)^{q_{1}}{\rm d}y\bigg)^{\frac{3}{2q_{1}}}\\ & \leq & C_{N}\{M(|f|^{p_{1}})(x)\}^{\frac{1}{p_{1}}} \mathop{\sum}^{+\infty}_{j=-\infty}\frac{1}{(1+2^{j})^{N}} \bigg(\frac{1}{(2^{j}r)^{n-2}}\int_{B(x,2^{j}r)}V(y){\rm d}y\bigg)^{\frac{3}{2}}, \end{eqnarray*}$

其中 $\frac{3}{2q_{1}}+\frac{1}{p_{1}}=1.$

因此类似第一种情形,有

$\begin{eqnarray*} |T_2^{*}f(x)| & \leq & C_{N}\{M(|f|^{p_{1}})(x)\}^{\frac{1}{p_{1}}} \mathop{\sum}^{+\infty}_{j=-\infty}\frac{1}{(1+2^{j})^{N}}\frac{1}{(2^{j}r)^{3}} \bigg(\frac{1}{(2^{j}r)^{n}}\int_{B(x,2^{j}r)}V(y){\rm d}y\bigg)^{\frac{3}{2}}\\ & \leq & C_{N}\{M(|f|^{p_{1}})(x)\}^{\frac{1}{p_{1}}}. \end{eqnarray*}$

注意到 $p_{1}=(\frac{2q_{1}}{3})'＜(\frac{2q_{0}}{3})'=p'_{0}. $ 由Hardy-Littlewood极大函数的有界性,(3.1)式得证.从而引理3.1证毕.

由对偶可证定理1.6.

本小节主要研究与Schrödinger型算子相关的Riesz变换的交换子的有界性.

引理4.1 设$V\in B_{q},\,q\geq \frac{n}{2}.$ 设$b\in BMO$,则 $\|[b,T_1^{*}]f\|_{L^p}\leq C_{p}\|b\|_{BMO}\|f\|_{L^p},$ 对$q'＜p＜+\infty$成立.

我们将采用文献^{[2, 8]}和^[11]中的思想.首先回忆sharp极大函数的定义

$$M^{\sharp}f(x)=\mathop{\sup}\limits_{B\ni x}\frac{1}{|B|}\int_{B}|f(y)-f_{B}|{\rm d}y,$$

其中$f_{B}=\frac{1}{|B|}_{B}f(y){\rm d}y,$且 上确界取遍所有 的球$B\ni x$.

定义$BMO$如下

$$BMO({\Bbb R}^{n})=\{f\in L_{loc}^{1}:\|f\|_{BMO}=\|M^{\sharp}f\|_{\infty}＜\infty\}.$$

用$2^{k}B$表示与$B$同心半径为$2^{k}$的球.

$$|f_{2^{k}B}-f_{B}|\leq C(k+1)\|f\|_{BMO},\ \mbox{对于} \ k>0.$$

下面是John-Nirenberg不等式

$\begin{equation}\|f\|_{BMO}\sim \mathop{\sup}\limits_{x\in B} \bigg\{\frac{1}{|B|}\int_{B}|f(y)-f_{B}|^{p}{\rm d}y\bigg\}^{\frac{1}{p}}, \ \mbox{对任意} \ p>1.\end{equation}$

(4.1)

首先证明引理4.2.

引理4.2 设$V\in B_{q},\,q\geq \frac{n}{2}.$ 则对于所有的$s>m'$,存在常数 $C_{s}>0$使得对于$ f\in L_{loc}^{1},b\in BMO$,

$\begin{equation}M^{\sharp}([b,T_1^{\ast}]f)(x)\leq C_{s}\|b\|_{BMO}\{M_{s}(T_1^{\ast}f)(x)+M_{s}f(x)\},\end{equation}$

(4.2)

其中 $M_{s}(f)=M(|f|^{s})^{\frac{1}{s}},$ $M$是Hardy-Littlewood 极大函数.

证 固定 $s>m',x\in {\Bbb R}^n$及球$\tilde{B}=B(x_{0},l)\ni x$. 我们需要证明

$$J=\frac{1}{|\tilde{B}|}\int_{\tilde{B}}|[b,T_1^{\ast}]f(y)-([b,T_1^{\ast}]f)_{\tilde{B}}|{\rm d}y\leq C_{s}\|b\|_{BMO}\{M_{s}(T_1^{\ast}f)(x)+M_{s}f(x)\}.$$

设 $f=f_{1}+f_{2}$,其中$f_{1}=f\chi_{32\tilde{B}}$, $f_{2}=f-f_{1}$,则

$\begin{eqnarray*} [b,T_1^{\ast}]f & = & (b-b_{\tilde{B}})T_1^{\ast}f-T_1^{\ast}(b-b_{\tilde{B}})f_{1}-T_1^{\ast}(b-b_{\tilde{B}})f_{2}\\ & := & A_{1}f+A_{2}f+A_{3}f, \end{eqnarray*}$

且有

$\begin{eqnarray*} J & \leq & \frac{1}{|\tilde{B}|}\int_{\tilde{B}}|A_{1}f(y)-(A_{1}f)_{\tilde{B}}|{\rm d}y\\ & & +\frac{1}{|\tilde{B}|}\int_{\tilde{B}}|A_{2}f(y)-(A_{2}f)_{\tilde{B}}|{\rm d}y+\frac{1}{|\tilde{B}|}\int_{\tilde{B}}|A_{3}f(y)-(A_{3}f)_{\tilde{B}}|{\rm d}y\\ & = & J_{1}+J_{2}+J_{3}. \end{eqnarray*}$

首先考虑 $J_{1}$.由Höolder不等式和$BMO$的性质,有

$\begin{eqnarray*} J_{1} & \leq & \frac{2}{|\tilde{B}|}\int_{\tilde{B}}|A_{1}f(y)|{\rm d}y=\frac{2}{|\tilde{B}|}\int_{\tilde{B}}|(b-b_{\tilde{B}})T_1^{\ast}f(y)|{\rm d}y\\ & \leq & 2\bigg(\frac{1}{|\tilde{B}|}\int_{\tilde{B}}|(b-b_{\tilde{B}})|^{s'}{\rm d}y\bigg)^{\frac{1}{s'}}\bigg(\frac{1}{|\tilde{B}|} \int_{\tilde{B}}|T_1^{\ast}f(y)|^{s}{\rm d}y\bigg)^{\frac{1}{s}}\\ & \leq & 2\|b\|_{BMO}M_{s}(T_1^{\ast}f)(x). \end{eqnarray*}$

对于$J_{2}$.固定$s_{1}$使得 $s>s_{1}>m'$,令$s_{2}=\frac{ss_{1}}{s-s_{1}}$,则

$\begin{eqnarray*} J_{2} & \leq & \frac{2}{|\tilde{B}|}\int_{\tilde{B}}|A_{2}f(y)|{\rm d}y\leq 2\bigg(\frac{1}{|\tilde{B}|}\int_{\tilde{B}}|A_{2}f(y)|^{s_{1}}{\rm d}y\bigg)^{\frac{1}{s_{1}}}\\ & \leq & 2\bigg(\frac{1}{|\tilde{B}|}\int_{32\tilde{B}}|(b-b_{\tilde{B}})f(y)|^{s_{1}}{\rm d}y\bigg)^{\frac{1}{s_{1}}}\\ & \leq & C\bigg(\frac{1}{|32\tilde{B}|}\int_{32\tilde{B}}|(b-b_{\tilde{B}})|^{s_{2}}{\rm d}y\bigg)^{s_{2}}\bigg(\frac{1}{|32\tilde{B}|}\int_{32\tilde{B}}|f(y)|^{s}{\rm d}y\bigg)^{\frac{1}{s}}\\ & \leq & C\|b\|_{BMO}M_{s}(f)(x). \end{eqnarray*}$

最后考虑$J_{3}$.设 $c_{\tilde{B}}=\int_{|z-x_{0}|>32l}K_1(x_{0},z)(b(z)-b_{\tilde{B}})f(z){\rm d}z, $ 则

$\begin{eqnarray*} J_{3} & \leq & \frac{2}{|\tilde{B}|}\int_{\tilde{B}}|A_{3}f(y)-c_{\tilde{B}}|{\rm d}y\\ & \leq & \frac{2}{|\tilde{B}|}\int_{\tilde{B}}|\int_{|z-x_{0}|\geq 32l}\{K_1(y ,z)-K_1(x_{0},z)\}(b(z)-b_{\tilde{B}})f(z){\rm d}z|{\rm d}y\\ & \leq & \frac{2}{|\tilde{B}|}\int_{\tilde{B}}\int_{|z-x_{0}|\geq 32l}|\{K_1(y ,z)-K_1(x_{0},z)\}(b(z)-b_{\tilde{B}})f(z)|{\rm d}z{\rm d}y\\ & \leq & \frac{2}{|\tilde{B}|}\int_{\tilde{B}}\mathop{\sum^{\infty}_{k=5}}\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}|\{K_1(y,z)-K_1(x_{0},z)\}(b(z)-b_{\tilde{B}})f(z)|{\rm d}z{\rm d}y. \end{eqnarray*}$

由Höolder不等式得

$\begin{eqnarray*} J_{3} & \leq & 2\frac{1}{|\tilde{B}|}\int_{\tilde{B}}\mathop{\sum^{\infty}_{k=5}}\bigg(\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}|K_1(y,z)-K_1(x_{0},z)|^{m}{\rm d}z\bigg)^{\frac{1}{m}}\\ & & \times \bigg(\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}(b(z)-b_{\tilde{B}})f(z)|^{m'}{\rm d}z\bigg)^{\frac{1}{m'}}{\rm d}y\\ & \leq & 2\frac{1}{|\tilde{B}|}\int_{\tilde{B}}\mathop{\sum^{\infty}_{k=5}}\bigg(\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}|K_1(y,z)-K_1(x_{0},z)|^{m}{\rm d}z\bigg)^{\frac{1}{m}}(2^{k}l)^{\frac{n}{m'}}k\\ & & \times\frac{1}{(2^{k+1}l)^{\frac{n}{m'}}(k+1)}\bigg(\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}(b(z)-b_{\tilde{B}})f(z)|^{m'}{\rm d}z\bigg)^{\frac{1}{m'}}{\rm d}y. \end{eqnarray*}$

设$m>\frac{n}{2}$.由引理2.6,引理2.4和(1.1)式,有

$\begin{eqnarray} & & \bigg(\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}|K_1(z,y)-K_1(x_{0},z)|^{m}{\rm d}z\bigg)^{\frac{1}{m}}\nonumber\\ & \leq & C_{N}\frac{l^{\delta}}{(2^{k}l)^{n-2+\delta}}\frac{(2^{k+1}l)^{\frac nm}}{\{1+m(x,V)2^{k}l\}^{N}} \bigg(\frac 1{(2^{k+1}l)^n}\int_{B(x_{0},2^{k+3}l)}V(y)^{m}{\rm d}y\bigg)^{\frac{1}{m}}\nonumber\\ & \leq & C_{N}\frac{l^{\delta}}{(2^{k}l)^{n-2+\delta}}\frac{1}{\{1+m(x,V)2^{k}l\}^{N}}(2^{k}l)^{-\frac{n}{m'}} \int_{B(x_{0},2^{k}l)}V(y){\rm d}y\nonumber\\ & \leq & C_{N}\frac{l^{\delta}}{(2^{k+1}l)^{n-2+\delta}}(2^{k}l)^{\frac{n}{m}-2}\leq C\frac{l^{\delta}}{(2^{k}l)^{(\frac{n}{m'})+\delta}}. \end{eqnarray}$

(4.3)

则

$\begin{equation} \mathop{\sum^{\infty}_{k=5}}\bigg(\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}|K_1(y,z)-K(x_{0},z)|^{m}{\rm d}z\bigg)^{\frac{1}{m}}(2^{k}l)^{\frac{n}{m'}}k \leq\mathop{\sum^{\infty}_{k=5}}\frac{Ck}{(2^{k})^{\delta}}\leq C. \end{equation}$

(4.4)

因此

$\begin{eqnarray*} J_{3} & \leq & C\mathop{\sup}\limits_{k\geq5}\frac{1}{(2^{k}l)^{\frac{n}{m'}}k}\bigg(\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}|(b(z)-b_{\tilde{B}})f(z)|^{m'}{\rm d}z\bigg)^{\frac{1}{m'}}\\ & \leq & C\mathop{\sup}\limits_{k\geq5}\frac{1}{k}\bigg(\frac{1}{(2^{k}l)^{n}}\int_{|z-x_{0}|\leq 2^{k+1}l}|(b(z)-b_{2^{k+1}\tilde{B}}+b_{2^{k+1}\tilde{B}}-b_{\tilde{B}})f(z)|^{m'}{\rm d}z\bigg)^{\frac{1}{m'}}\\ & \leq & C\mathop{\sup}\limits_{k\geq5}\frac{1}{k}(k+1)\|b\|_{BMO}M_{s}f(x). \end{eqnarray*}$

引理4.2证毕.

引理4.1可由引理4.2和关于sharp极大函数的Fefferman-Stein定理容易证明. 由对偶和引理4.1容易明定理1.4,故省去. 要证明定理1.8,由对偶我们只需证明如下引理.

引理4.3 设$V\in B_{q},\,q\geq \frac{n}{2}.$ 设$b\in BMO$,则 $\|[b,T_2^{*}]f\|_{L^p}\leq C_{p}\|b\|_{BMO}\|f\|_{L^p},$ 对$p'_0＜p＜+\infty$成立.

我们首先建立交换子的sharp极大函数的点态估计.

引理4.4 设$V\in B_{q},\,q\geq \frac{n}{2},$ 则对于所有的$ s>m'$,存在常数 $C_{s}>0$使得

$\begin{equation}M^{\sharp}([b,T_2^{\ast}]f)(x)\leq C_{s}\|b\|_{BMO}\{M_{s}(T_2^{\ast}f)(x)+M_{s}f(x)\},\end{equation}$

(4.5)

其中 $M_{s}(f)=M(|f|^{s})^{\frac{1}{s}},$ 和$M$是Hardy-Littlewood极大函数.

证 因为引理的证明与引理4.2非常类似,我们仅给出它们之间的主要不同之处. 由于$J_1,J_2$的估计几乎一致,故省去.对于第三项$J_3,$我们给出详细计算.

对于$J_{3}$.设 $c_{\tilde{B}}=\int_{|z-x_{0}|>32l}K_2(x_{0},z)(b(z)-b_{\tilde{B}})f(z){\rm d}z$,则

$\begin{eqnarray*} J_{3} & \leq & \frac{2}{|\tilde{B}|}\int_{\tilde{B}}|A_{3}f(y)-c_{\tilde{B}}|{\rm d}y\\ & \leq & \frac{2}{|\tilde{B}|}\int_{\tilde{B}}|\int_{|z-x_{0}|\geq 32l}\{K_2(y,z)-K_2(x_{0},z)\}(b(z)-b_{\tilde{B}})f(z){\rm d}z|{\rm d}y\\ & \leq & \frac{2}{|\tilde{B}|}\int_{\tilde{B}}\int_{|z-x_{0}|\geq 32l}|\{K_2(y,z)-K_2(x_{0},z)\}(b(z)-b_{\tilde{B}})f(z)|{\rm d}z{\rm d}y\\ & \leq & \frac{2}{|\tilde{B}|}\int_{\tilde{B}}\mathop{\sum^{\infty}_{k=5}}\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}|\{K_2(y,z)-K_2(x_{0},z)\}(b(z)-b_{\tilde{B}})f(z)|{\rm d}z{\rm d}y. \end{eqnarray*}$

由Höolder不等式得

$\begin{eqnarray*} J_{3} & \leq & 2\frac{1}{|\tilde{B}|}\int_{\tilde{B}}\mathop{\sum^{\infty}_{k=5}}\bigg(\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}|K_2(y,z)-K_2(x_{0},z)|^{m}{\rm d}z\bigg)^{\frac{1}{m}}\\ & & \times\bigg(\int_{|z-x_{0}|\leq 2^{k+1}l}(b(z)-b_{\tilde{B}})f(z)|^{m'}{\rm d}z\bigg)^{\frac{1}{m'}}{\rm d}y\\ & \leq & 2\frac{1}{|\tilde{B}|}\int_{\tilde{B}}\mathop{\sum^{\infty}_{k=5}}\bigg(\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}|K_2(y,z)-K_2(x_{0},z)|^{m}{\rm d}z\bigg)^{\frac{1}{m}}(2^{k}l)^{\frac{n}{m'}}k\\ & & \times\frac{1}{(2^{k+1}l)^{\frac{n}{m'}}(k+1)}\bigg(\int_{|z-x_{0}|\leq 2^{k+1}l}(b(z)-b_{\tilde{B}})f(z)|^{m'}{\rm d}z\bigg)^{\frac{1}{m'}}{\rm d}y. \end{eqnarray*}$

假设$m>\frac{n}{2}$.由引理2.7,(1.1)式和引理2.4得

$\begin{eqnarray} & & \bigg(\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}|K_2(y,z)-K_2(x_{0},z)|^{m}{\rm d}z\bigg)^{\frac{1}{m}}\nonumber\\ & \leq & C_{N}\frac{l^{\delta}}{(2^{k}l)^{n-3+\delta}}\frac{(2^{k+3}l)^{\frac nm}}{\{1+m(x,V)2^{k}l\}^{N}}\bigg(\frac1{(2^{k+3}l)^{n}}\int_{B(x_{0},2^{k+3}l)}V(y)^{\frac{3m}{2}}{\rm d}y \bigg)^{\frac{1}{m}}\nonumber\\ & \leq & C_{N}\frac{l^{\delta}}{(2^{k}l)^{n-3+\delta}}\frac{1}{\{1+m(x,V)2^{k}l\}^{N}} (2^{k}l)^{\frac{n}{m}-\frac{3n}{2}}\bigg(\int_{B(x_{0},2^{k+3}l)}V(y){\rm d}y\bigg)^{\frac{3}{2}}\nonumber\\ & \leq & C_{N}\frac{l^{\delta}}{(2^{k}l)^{n-3+\delta}}(2^{k+3}l)^{\frac{n}{m}-3} \leq C\frac{l^{\delta}}{(2^{k}l)^{(\frac{n}{m'})+\delta}}. \end{eqnarray}$

(4.4)

则

$\begin{equation} \mathop{\sum^{\infty}_{k=5}}\bigg(\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}|K_2(y,z)-K_2(x_{0},z)|^{m}{\rm d}z\bigg)^{\frac{1}{m}}(2^{k}l)^{\frac{n}{m'}}k \leq \mathop{\sum^{\infty}_{k=5}}\frac{C k}{(2^{k})^{\delta}}\leq C. \end{equation}$

(4.7)

因此

$\begin{eqnarray*} J_{3} & \leq & C\mathop{\sup}\limits_{k\geq5}\frac{1}{(2^{k}l)^{\frac{n}{m'}}k}\bigg(\int_{2^{k}l\leq|z-x_{0}|\leq 2^{k+1}l}|(b(z)-b_{\tilde{B}})f(z)|^{m'}{\rm d}z\bigg)^{\frac{1}{m'}}\\ & \leq & C\mathop{\sup}\limits_{k\geq5}\frac{1}{k}\bigg(\frac{1}{(2^{k}l)^{n}}\int_{|z-x_{0}|\leq 2^{k+1}l}(b(z)-b_{2^{k+1}\tilde{B}}+b_{2^{k+1}\tilde{B}}-b_{\tilde{B}})f(z)|^{m'}{\rm d}z\bigg)^{\frac{1}{m'}}\\ & \leq & C\mathop{\sup}\limits_{k\geq5}\frac{1}{k}(k+1)\|b\|_{BMO}M_{s}f(x). \end{eqnarray*}$

引理4.4证毕.

引理4.3可由引理4.4和关于sharp极大函数的Fefferman-Stein定理可得. 由对偶和引理4.3易证定理1.8.

本节将证明定理1.5和定理1.9.由于两定理证明非常相似,故我们仅证明定理1.5.

证 设$f\in BMO_{L}({\Bbb R}^{n})$ 及 $B=B(x_0,r).$ 首先设$r\geq\rho(x_{0})$.令 $B^{*}=B(x_{0},2r)$.则

$$f=f\chi_{B^{*}}+f\chi_{(B^{*})^{c}}=f_{1}+f_{2},$$

其中$\chi_{A}$ 表示集合$A$的特征函数. 因为 $\nabla^{2}H_{2}^{-1}V$在$L^{2}({\Bbb R}^{n})$上有界,由 注1.1得

$\begin{eqnarray*} \frac{1}{|B|}\int_{B}|(\nabla^{2}H_{2}^{-1}V)f_{1}(x)|{\rm d}x & \leq & \bigg(\frac{1}{|B|}\int_{B}|(\nabla^{2}H_{2}^{-1}V)f_{1}(x)|^{2}{\rm d}x\bigg)^{\frac{1}{2}}\\ & \leq & \bigg(\frac{C}{|B|}\int_{B^{*}}|f(x)|^{2}{\rm d}x\bigg)^{\frac{1}{2}}\leq C\|f\|_{BMO_{L}}. \end{eqnarray*}$

设$x\in B$.由引理 2.1并注意到$\rho(x)\leq Cr$.再由引理2.3和(1.1)式,有

$\begin{eqnarray*} & & |(\nabla^{2}H_{2}^{-1}V)f_{2}(x)|\leq \int_{(B^{*})^{c}}|\nabla_{x}^{2}\Gamma_{H_{2}}(x,y)|V(y)|f(y)|{\rm d}y\\ & \leq & \mathop{\sum}^{\infty}_{j=1}\int_{2^{j+1}B^\ast\setminus2^{j}B^\ast}|\nabla_{x}^{2}\Gamma_{H_{2}}(x,y)|V(y)|f(y)|{\rm d}y\\ & \leq & \mathop{\sum}^{\infty}_{j=1}\int_{2^{j+1}B^\ast\setminus2^{j}B^\ast}\frac{C_{N}}{\{1+m(x,V)|x-y|\}^{N}}\frac{1}{|x-y|^{n-2}}V(y)|f(y)|{\rm d}y\\ & \leq & \mathop{\sum}^{\infty}_{j=1}\rho(x_{0})^{N}\int_{2^{j+1}B^\ast\setminus2^{j}B^\ast}\frac{C_{N}}{(2^{j}r)^{N+n-2}}V(y)|f(y)|{\rm d}y\\ & \leq & C_{N}\mathop{\sum}^{\infty}_{j=1}\frac{\rho(x_{0})^{N}}{(2^{j}r)^{N+n-2}}\int_{2^{j+1}B^\ast\setminus2^{j}B^\ast}V(y)|f(y)|{\rm d}y\\ & \leq & C_{N}\mathop{\sum}^{\infty}_{j=1}\frac{\rho(x_{0})^{N}}{(2^{j}r)^{N-2}}\bigg(\frac{1}{|2^{j}B^\ast|}\int_{2^{j+1}B^\ast}V(y)^{q}{\rm d}y\bigg)^{\frac{1}{q}} \bigg(\frac{1}{|2^{j+1}B^\ast|}\int_{2^{j+1}B^\ast}|f(y)|^{q'}{\rm d}y\bigg)^{\frac{1}{q'}}\\ & \leq & C_{N}\mathop{\sum}^{\infty}_{j=1}(2^{-j})^{N-k_{0}}(\frac{\rho(x_{0})}{r})^{N-k_{0}}\|f\|_{BMO_{L}}\\ & \leq & C\|f\|_{BMO_{L}},\,\,\mbox{($N$充分大)}, \end{eqnarray*}$

则

$\begin{equation}\frac{1}{|B|}\int_{B}|(\nabla^{2}H_{2}^{-1}V)f_{2}(x)|{\rm d}x\leq C\|f\|_{BMO_{L}}.\end{equation}$

(5.1)

上述证明表明$\nabla^{2}H_{2}^{-1}V\in BMO_{L}({\Bbb R}^{n})$.

若$r＜\rho(x_{0})$. 令$B^{\sharp}=B(x_{0},2\rho(x_{0}))$,则 $f=f\chi_{B^{\sharp}}+f\chi_{(B^{\sharp})^{c}}=f_{1}^{\sharp}+f_{2}^{\sharp},$ 对于任意的$x\in B,$ $\rho(x)\sim\rho(x_{0})$,类似(5.1)式,有

$$\frac{1}{|B|}\int_{B}|(\nabla^{2}H_{2}^{-1}V)f_{2}^\sharp(x)|{\rm d}x\leq C\|f\|_{BMO_{L}}. $$

要完成定理的证明,仅需证明存在常数 $C_B$ 使得

$\begin{equation}\frac1{|B|}\int_{B}|(\nabla^{2}H_{2}^{-1}V)f_1^\sharp(x)-C_B|\,{\rm d}x\le C\|f\|_{BMO_L},\end{equation}$

(5.2)

其中 $C_B=(\nabla^{2}H_{2}^{-1}V)f(x_0).$

设$B_k^{\sharp}=B(x_0,2^{1-k}\rho(x_0)),\ k=0,1,\cdots,k_0,$ 其中 $k_0$ 满足$2^{-k_0-1}\rho(x_0)\le r\le 2^{-k_0}\rho(x_0).$

设

$\begin{eqnarray*} f_1^\sharp-f(B_0^\sharp) & = & (f_1^\sharp-f(B_{k_0}^\sharp))\chi_{B_{k_0}^\sharp} +(f_1^\sharp-f(B_{k_0}^\sharp))\chi_{B_0^\sharp\backslash B_{k_0}^\sharp}-f(B_{k_0}^\sharp)\chi_{(B_0^\sharp)^c}\\ & = & f_{11}+f_{12}+f_{13}. \end{eqnarray*}$

由算子$\nabla^{2}H_{2}^{-1}V$的$L^p$ -有界性得

$\begin{eqnarray*} \frac{1}{|B|}\int_{B}|(\nabla^{2}H_{2}^{-1}V)(f_{11})|{\rm d}x & \le & \bigg(\frac1{|B|}\int_ {B}|(\nabla^{2}H_{2}^{-1}V)(f_{11})|^p\bigg)^{\frac1p}\\ & \le & C\bigg(\frac1{|B_{k_0}^\sharp|}\int_ {B_{k_0}^\sharp}|f_1^\sharp-f(B_{k_0}^\sharp)|^p\bigg)^{\frac1p}\\ & \le & C\|f\|_{BMO}. \end{eqnarray*}$

下面考虑如下点态估计.对任意的 $x\in B$,由Höolder不等式和类似(4.4)式的证明方法,有

$\begin{eqnarray*} & & |(\nabla^{2}H_{2}^{-1}V)f_{12}(x)-(\nabla^{2}H_{2}^{-1}V)f_{12}(x_0)|\\ & \le & \sum_{k=0}^{k_0-1}\int_{B_k^\sharp\backslash B_{k+1}^\sharp}|K_1(x,y)-K_1(x_0,y)||f_1^\sharp(y)-f(B_{k_0}^\sharp)|{\rm d}y\\ & \le & \sum_{k=0}^{k_0-1}\bigg(\int_{B_k^\sharp\backslash B_{k+1}^\sharp}|K_1(x,y)-K_1(x_0,y)|^m{\rm d}y\bigg)^{1/m} \bigg(\int_{B_k^\sharp}|f_1^\sharp(y)-f(B_{k_0}^\sharp)|^{m'}{\rm d}y\bigg)^{1/m'}\\ & \le & \sum_{k=0}^{k_0-1}\frac{\rho(x_0)^\delta}{(2^{-k}\rho(x_0))^{\frac{n}{m'}+\delta}} \bigg(\int_{B_k^\sharp}|f_1^\sharp(y)-f(B_k^\sharp)|+|f(B_k^\sharp)-f(B_{k_0}^\sharp)|^{m'}{\rm d}y\bigg)^{1/m'}\\ & \le & \sum_{k=0}^{k_0-1}2^{k\delta}(k_0-k+1)\|f\|_{BMO}, \end{eqnarray*}$

这意味着

$$ \frac{1}{|B|}\int_{B}|(\nabla^{2}H_{2}^{-1}V)f_{12}(x)-(\nabla^{2}H_{2}^{-1}V)f_{12}(x_0)|{\rm d}x\le C\|f\|_{BMO_L}. $$

对最后一项.因为

$\begin{eqnarray*} & & |(\nabla^{2}H_{2}^{-1}V)f_{12}(x)-(\nabla^{2}H_{2}^{-1}V)f_{12}(x_0)|\\ & \le & |f(B_{k_0}^\sharp)|\int_{(B_{k_0}^\sharp)^c}|K_1(x,y)-K_1(x_0,y)|{\rm d}y\\ & \le & (k_0+1)\|f\|_{BMO_L}\sum_{k=1}^\infty\int_{(2^{k+1}B_{k_0}^\sharp)\backslash (2^{k+1}B_{k_0}^\sharp)} |K_1(x,y)-K_1(x_0,y)|{\rm d}y\\ & \le & (k_0+1)\|f\|_{BMO_L}\sum_{k=1}^\infty\bigg(\int_{(2^{k+1}B_{k_0}^\sharp)\backslash (2^{k+1}B_{k_0}^\sharp)} |K_1(y,x)-K_1(y,x_0)|^m{\rm d}y\bigg)^{1/m}\\ & & \times\big(2^{k+1}\rho(x_0)\big)^{n/m'}\\ & \le & C(k_0+1)\|f\|_{BMO_L}\sum_{k=1}^\infty\frac{\rho(x_0)^\delta}{(2^k\rho(x_0))^{n/m'+\delta}}\big(2^{k+1}\rho(x_0)\big)^{n/m'}\\ & \le & C\|f\|_{BMO_L}. \end{eqnarray*}$

定理证毕.