与薛定谔算子相关的Marcinkiewicz积分在混合Morrey空间上的加权估计

Weighted Estimates of Marcinkiewicz Integrals Associated with Schr?dinger Operator on Mixed Morrey Space

摘要: 利用A_p的估计以及函数分解方法，借助L^p空间上的加权估计，证明了与薛定谔算子相关的Marcinkiewicz积分在混合Morrey空间上的加权有界性，并给出与薛定谔算子相关的Marcinkiewicz积分的BMO交换子的加权有界性。

Abstract: By using the weight estimation of A_p and the function decomposition method, and owing to the weighted estimation on the L^p spaces, we proved the weighted boundedness of Marcinkiewicz in-tegrals associated with Schrödinger operator on mixed Morrey spaces, and the weighted boundedness of BMO commutators of Marcinkiewicz integrals associated with Schrödinger oper-ator.

1. 引言及主要结果

我们考虑薛定谔算子

$L=-\Delta +V\left(x\right)\mathrm{,\; <mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>}{ℝ}^d\mathrm{,\; <mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>\; <mtext>\; \hspace{0.17em}\; </mtext>}d\ge 3$

其中 $V\left(x\right)$ 是非负位势，属于反向Hölder类 $R{H}_s\mathrm{,}s\ge \frac{d}{2}$，因此，存在一个常数C，使得对任意球 $B\subset {ℝ}^n$

${\left(\frac{1}{\left|B\right|}{\displaystyle {\int }_B}V{\left(y\right)}^s\text{d}y\right)}^{\frac{1}{s}}\le \frac{C}{\left|B\right|}{\displaystyle {\int }_B}V\left(y\right)\text{d}y\mathrm{.}$ (1)

对任意 $V\in R{H}_s\mathrm{,}s\ge \frac{d}{2}$，临界半径函数 $\rho \left(x\right)=\rho \left(x\mathrm{,}V\right)$，给定

$\rho \left(x\right):=\sup \left\{r>0:\frac{1}{r^{d-2}}{\displaystyle {\int }_{B\left(x,r\right)}}V\left(y\right)\text{d}y\le 1\right\}.$ (2)

其中 $B\left(x,r\right)$ 是以x为中心，以r为半径的球。对任意 $x\in {ℝ}^d$ 这个辅助函数满足 $0<\rho \left(x\right)<\infty$。

与薛定谔算子L相关的Marcinkiewicz积分 ${\mu }_{j\mathrm{,}\Omega }^L$ 定义为

${\mu }_{j\mathrm{,}\Omega }^L\left(f\right)\left(x\mathrm{,}t\right)={\left({\displaystyle {\int }_0^{\infty }}{\left|{\displaystyle {\int }_{\left|x-y\right|\le h}}\left|\Omega \left(x-y\right)\right|K_j^L\left(x\mathrm{,}y\right)f\left(y\mathrm{,}t\right)\text{d}y\right|}^2\frac{\text{d}h}{h^3}\right)}^{\frac{1}{2}}$ (3)

其中 $K_j^L\left(x\mathrm{,}y\right)=\widetilde{K_j^L}\left(x\mathrm{,}y\right)\left|x-y\right|$ 并且 $\widetilde{K_j^L}\left(x\mathrm{,}y\right)$ 是 $R_j=\frac{\partial }{\partial x_j}{L}^{-\frac{1}{2}},j=1,\cdots ,n$ 的核。当 $V=0$ 时， $K_j^{\Delta }\left(x,y\right)=\widetilde{K_j^{\Delta }}\left(x\mathrm{,}y\right)\left|x-y\right|=\frac{\frac{x_j-y_j}{\left|x-y\right|}}{{\left|x-y\right|}^{n-1}}$ 并且 $\widetilde{K_j^{\Delta }}\left(x\mathrm{,}y\right)$ 是 $R_j=\frac{\partial }{\partial x_j}{\Delta }^{-\frac{1}{2}},j=1,\cdots ,n$ 的核。

下面我们定义交换子 $\left[b\mathrm{,}{\mu }_{j\mathrm{,}\Omega }^L\right]$

$\left[b,{\mu }_{j,\Omega }^L\right]\left(f\right)\left(x,t\right)={\left({\displaystyle {\int }_0^{\infty }}{\left|{\displaystyle {\int }_{\left|x-y\right|\le h}}\left|\Omega \left(x-y\right)\right|K_j^L\left(x,y\right)\left[b\left(x\right)-b\left(y\right)\right]f\left(x,t\right)\text{d}y\right|}^2\frac{\text{d}h}{h^3}\right)}^{\frac{1}{2}}$ (4)

当 $f\left(y\mathrm{,}t\right)\equiv f\left(y\right)$ 时，上面定义的与薛定谔算子L相关的Marcinkiewicz积分 ${\mu }_{j\mathrm{,}\Omega }^L$ 是一般的与薛定谔算子L相关的Marcinkiewicz积分 ${\mu }_{j\mathrm{,}\Omega }^L$。2020年，Ferit GÜRBÜZ在文献 [1] 中得到了 ${\mu }_{j\mathrm{,}\Omega }^L$ 与 ${\mu }_{j\mathrm{,}\Omega \mathrm{,}b}^L$ 在加权Lebesgue空间上的有界性，进一步关于与薛定谔算子L相关的Marcinkiewicz积分 ${\mu }_{j\mathrm{,}\Omega }^L$ 的结果可参见文献 [2] [3] [4] [5]。

Morrey空间作为Lebesgue空间的一个重要推广，在调和分析及其偏微分方程等领域有非常重要的应用 [6] [7] [8] [9]。2020年，Wang在文献 [10] 中定义了一类加权Morrey空间 $L_{\rho \mathrm{,}\theta }^{p\mathrm{,}k}\left(\mu \mathrm{,}\nu \right)$。2017年，Ragusa-Scapellato在文献 [11] 中定义了一类时空混合范Morry空间 $L^{q\mathrm{,}\mu }\left(\mathrm{0,}T\mathrm{,}{L}^{p\mathrm{,}\lambda }\left({ℝ}^n\right)\right)$，它的优点之一是允许我们将时间和空间分开对待，这一特点在研究进化算子(例如Kolmogorov算子)和抛物型偏微分方程有重要应用。文献 [11] [12] [13] 得到了Riesz位势，Marcinkiewicz积分和带Gaussian核算子在时空混合范Morry空间 $L^{q\mathrm{,}\mu }\left(\mathrm{0,}T\mathrm{,}{L}^{p\mathrm{,}\lambda }\left({ℝ}^n\right)\right)$ 上的有界性。本文将主要研究与薛定谔算子L相关的Marcinkiewicz积分 ${\mu }_{j\mathrm{,}\Omega }^L$ 在时空混合范Morrey空间上的加权有界性，并给出Marcinkiewicz积分交换子的加权有界性。为此，我们首先引入下面定义。

定义1 [10] 设 $1\le p<\infty$， $0<\lambda <1$ 并且对于 ${ℝ}^d$ 中两个权函数u和v，给定 $0<\theta <\infty$，加权Morrey空间 $L_{\rho \mathrm{,}\theta }^{p\mathrm{,}k}\left(\mu \mathrm{,}\nu \right)$ 定义为 ${ℝ}^d$ 上所有p-局部可积函数f的集合，使

${\left(\frac{1}{\upsilon {\left(B\right)}^{\lambda }}{\displaystyle {\int }_B}{\left|f\left(x\right)\right|}^p\mu \left(x\right)\text{d}x\right)}^{\frac{1}{p}}\le C\cdot {\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{\theta }$ (5)

对 ${ℝ}^d$ 上任意的球 $B=B\left(x_0\mathrm{,}r\right)$，则

${\Vert f\Vert }_{L_{\rho \mathrm{,}\theta }^{p\mathrm{,}\lambda }\left(\mu \mathrm{,}\nu \right)}\mathrm{<mo>\; :\; </mo>}=\underset{B}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\theta }\cdot {\left(\frac{1}{\nu {\left(B\right)}^{\lambda }}{\displaystyle {\int }_B}{\left|f\left(x\right)\right|}^p\mu \left(x\right)\text{d}x\right)}^{\frac{1}{p}}<\infty$ (6)

定义2 设 $T>0$， $1\le p,q<\infty$， $0<\lambda ,\mu <1$。u，v为 ${ℝ}^d$ 上的非负可测函数，时空混合Morrey空间定义为

$L^{q,\mu }\left(0,T,L_{\rho ,\theta }^{p,\lambda }\left(u,v\right)\right):=\left\{f\left(x,t\right):{\Vert f\Vert }_{L^{q,\mu }\left(0,T,L_{\rho ,\theta }^{p,\lambda }\left(u,v\right)\right)}<\infty \right\},$

其中，

${\Vert f\Vert }_{L^{q,\mu }\left(0,T,L_{\rho ,\theta }^{p,\lambda }\left(u,v\right)\right)}:={\left(\underset{t_0\in \left[0,T\right],\rho >0}{\sup }\frac{1}{{\rho }^{\mu }}{\displaystyle {\int }_{\left(0,T\right)\cap \left(t_0-\rho ,t_0+\rho \right)}}\underset{B}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\theta q}{\left(\frac{1}{v{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_B}{\left|f\left(x,t\right)\right|}^{p}u\left(x\right)\text{d}x\right)}^{\frac{q}{p}}\text{d}t\right)}^{\frac{1}{q}}.$ (7)

这里， $B_{\rho }\left(x\right)=\left\{y\in {ℝ}^d:\left|y-x\right|<\rho \right\}$。当 $\mu =\nu$ 时，简记为 $L^{q\mathrm{,}\mu }\left(\mathrm{0,}T\mathrm{,}{L}_{\rho \mathrm{,}\theta }^{p\mathrm{,}\lambda }\left(u\right)\right)$。

容易看出，时空混合范加权Morrey空间 $L^{q,\mu }\left(0,T,L_{\rho ,\theta }^{p,\lambda }\left(u,v\right)\right)$ 是文献中所定义的加权Morrey空间 $L_{\rho \mathrm{,}\theta }^{p\mathrm{,}\lambda }\left(u\mathrm{,}v\right)$ 的一种自然推广。叙述本文主要结果之前，先回顾 $A_p^{\rho \mathrm{,}\infty }$ 和 $A_{p\mathrm{,}q}^{\rho \mathrm{,}\infty }$ 权的定义 [14]。

设 $1<p<\infty$， $0<\theta <\infty$，称非负可测函数 $w\in A_p^{\rho \mathrm{,}\infty }$，如果对任意的球 $B\subseteq {ℝ}^d$，存在与B无关的常数 $C>0$，使得

${\left(\frac{1}{\left|B\right|}{\displaystyle {\int }_B}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}{\left(\frac{1}{\left|B\right|}{\displaystyle {\int }_B}w{\left(x\right)}^{-\frac{p^{\prime }}{p}}\text{d}x\right)}^{\frac{1}{p^{\prime }}}\le C\cdot {\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{\theta }\mathrm{.}$ (8)

设 $1<p<q<\infty$， $0<\theta <\infty$，称非负可测函数 $w\in A_{p\mathrm{,}q}^{\rho \mathrm{,}\infty }$，如果对任意的球 $B\subseteq {ℝ}^d$，存在与B无关的常数 $C>0$，使得

${\left(\frac{1}{\left|B\right|}{\displaystyle {\int }_B}w{\left(x\right)}^q\text{d}x\right)}^{\frac{1}{q}}{\left(\frac{1}{\left|B\right|}{\displaystyle {\int }_B}w{\left(x\right)}^{-p^{\prime }}\text{d}x\right)}^{\frac{1}{p^{\prime }}}\le C\cdot {\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{\theta }\mathrm{,}$ (9)

其中， $p^{\prime }=\frac{p}{p-1}$ 为p的对偶指标。

设 $1<r<\infty$， $0<\theta <\infty$，如果对任意的球 $B\subseteq {ℝ}^d$，存在与B无关的常数 $C>0$，使得下面的反向Hölder不等式成立

${\left(\frac{1}{\left|B\right|}{\displaystyle {\int }_B}w{\left(x\right)}^r\text{d}x\right)}^{\frac{1}{r}}\le C\left(\frac{1}{\left|B\right|}{\displaystyle {\int }_B}w\left(x\right)\text{d}x\right){\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{\theta }\mathrm{,}$ (10)

记 $w\in R{H}_r^{\rho \mathrm{,}\theta }$。

在文献 [15] 中引入了一类新的 ${\text{BMO}}_{\rho \mathrm{,}\infty }\left({ℝ}^n\right)$ 空间的定义，即 ${\text{BMO}}_{\rho \mathrm{,}\infty }\left({ℝ}^n\right)\mathrm{<mo>\; :\; </mo>}={\displaystyle {\cup }_{\theta >0}}{\text{BMO}}_{\rho \mathrm{,}\infty }\left({ℝ}^n\right)$ 其中 $0<\theta <\infty$，被 ${\text{BMO}}_{\rho \mathrm{,}\infty }\left({ℝ}^n\right)$ 空间定义的局部可积函数b满足

$\frac{1}{\left|B\left(x_0\mathrm{,}r\right)\right|}{\displaystyle {\int }_{B\left(x_0\mathrm{,}r\right)}}\left|b\left(x\right)-b_B\left(x_0\mathrm{,}r\right)\right|\le C\cdot {\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{\theta }$ (11)

对所有的球 $B\left(x_0\mathrm{,}r\right)$，其中 $x_0\in {ℝ}^d$， $r>0$，并且 $b_B\left(x_0\mathrm{,}r\right)$ 被定义为b在 $B\left(x_0\mathrm{,}r\right)$，即 $b_{B\left(x_0\mathrm{,}r\right)}\mathrm{<mo>\; :\; </mo>}=\frac{1}{\left|B\left(x_0\mathrm{,}r\right)\right|}{\displaystyle {\int }_{B\left(x_0\mathrm{,}r\right)}}b\left(y\right)\text{d}y$。 $b\in {\text{BMO}}_{\rho \mathrm{,}\infty }\left({ℝ}^n\right)$ 的范数用 ${\Vert b\Vert }_{{\text{BMO}}_{\rho \mathrm{,}\theta }}$，即

${\Vert b\Vert }_{{\text{BMO}}_{\rho \mathrm{,}\theta }}\mathrm{<mo>\; :\; </mo>}=\underset{B\left(x_0\mathrm{,}r\right)}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\theta }\cdot \left(\frac{1}{\left|B\left(x_0\mathrm{,}r\right)\right|}{\displaystyle {\int }_{B\left(x_0\mathrm{,}r\right)}}\left|b\left(x\right)-b_{B\left(x_0\mathrm{,}r\right)}\right|\text{d}x\right)\mathrm{,}$ (12)

其中上确界取所有的球 $B\left(x_0\mathrm{,}r\right)$。

本文的主要结果如下。

定理1 设 $\Omega \in L^s\left(S^{d-1}\right)$， $1<s\le \infty$。则对任意 $1<p,q<\infty$， $0<\lambda <1$， $0<\mu <1$，且 $w\in A_p^{\rho \mathrm{,}\infty }$，存在与f无关的常数 $C>0$，使得

${\Vert {\mu }_{j\mathrm{,}\Omega }^L\left(f\right)\Vert }_{L^{q\mathrm{,}\mu }\left(\mathrm{0,}T\mathrm{,}{L}_{\rho \mathrm{,}\infty }^{p\mathrm{,}\lambda }\left(w\right)\right)}\le C{\Vert f\Vert }_{L^{q\mathrm{,}\mu }\left(\mathrm{0,}T\mathrm{,}{L}_{\rho \mathrm{,}\infty }^{p\mathrm{,}\lambda }\left(w\right)\right)}\mathrm{.}$

定理2 设 $\Omega \in L^s\left(S^{d-1}\right)$， $1<s\le \infty$，且 $b\in \text{BMO}\left({ℝ}^d\times \left[\mathrm{0,}T\right]\right)$。则对任意 $1<p,q<\infty$， $0<\lambda <1$， $0<\mu <1$，且 $w\in A_p^{\rho \mathrm{,}\infty }$，存在与f无关的常数 $C>0$，使得

${\Vert \left[b\mathrm{,}{\mu }_{j\mathrm{,}\Omega }^L\right]\left(f\right)\Vert }_{L^{q\mathrm{,}\mu }\left(\mathrm{0,}T\mathrm{,}{L}_{\rho \mathrm{,}\infty }^{p\mathrm{,}\lambda }\left(w\right)\right)}\le C{\Vert f\Vert }_{L^{q\mathrm{,}\mu }\left(\mathrm{0,}T\mathrm{,}{L}_{\rho \mathrm{,}\infty }^{p\mathrm{,}\lambda }\left(w\right)\right)}\mathrm{.}$

定理3 设 $\Omega \in L^s\left(S^{d-1}\right)$， $1<s\le \infty$，且 $b\in \text{BMO}\left({ℝ}^d\times \left[\mathrm{0,}T\right]\right)$。则对任意 $1<p,q<\infty$， $0<\lambda <1$， $0<\mu <1$，且 $w\in A_p^{\rho \mathrm{,}\infty }$，存在与f无关的常数 $C>0$，使得

${\left(\underset{t_0\in \left(0,T\right),\rho >0}{\sup }\frac{1}{{\rho }^{\mu }}{\displaystyle {\int }_{\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)}}{\Vert {\mu }_{j\mathrm{,}\Omega }^L\left(f\right)\Vert }_{BM{O}_{\rho \mathrm{,}\infty }}^q\text{d}t\right)}^{\frac{1}{q}}\le C{\Vert f\Vert }_{L^{q\mathrm{,}\mu }\left(\mathrm{0,}T\mathrm{,}{L}_{\rho \mathrm{,}\infty }^{p\mathrm{,}\lambda }\left(w\right)\right)}\mathrm{.}$

2. 定理的证明

本节介绍定理证明中需要用到的结论和引理。

引理1 [1] 设零阶齐次函数 $\Omega \in L^s\left(S^{d-1}\right)\left(1<s\le \infty \right)$，对任意 $\mu >0$， $\Omega \left(\mu x\right)=\Omega \left(x\right)$， $x\in R^n\backslash 0$ 并且 $V\in R{H}_n$。则对任意 $q^{\prime }<p<\infty$ 且 $w\in A_{\frac{p}{q^{\prime }}}$，下列不等式成立

${\Vert {\mu }_{j\mathrm{,}\Omega }^L\left(f\right)\Vert }_{L^p\left(w\right)}\le C{\Vert f\Vert }_{L^p\left(w\right)}\mathrm{.}$

引理2 [1] 设零阶齐次函数 $\Omega \in L^s\left(S^{d-1}\right)\left(1<s\le \infty \right)$，对任意 $\mu >0$， $\Omega \left(\mu x\right)=\Omega \left(x\right)$， $x\in R^n\backslash 0$， $V\in R{H}_n$ 并且 $b\in \text{BMO}\left(R^n\right)$。则对任意的 $q^{\prime }<p<\infty$ 且 $w\in A_{\frac{p}{q^{\prime }}}$，下列不等式成立

${\Vert \left[b\mathrm{,}{\mu }_{j\mathrm{,}\Omega }^L\right]\left(f\right)\Vert }_{L^p\left(w\right)}\le C{\Vert f\Vert }_{L^p\left(w\right)}\mathrm{.}$

引理3 [16] 设 $V\in R{H}_d$，则，

(a) 对任意N，存在一个常数C，使得

$\left|K_j^L\left(x\mathrm{,}z\right)\right|\le C{\left(1+\frac{\left|x-z\right|}{\rho \left(x\right)}\right)}^{-N}\cdot \frac{1}{{\left|x-z\right|}^{d-1}}\mathrm{<mo>\; ;\; </mo>}$

(b) 对任意N和 $0<\delta <\min 1,1-\frac{d}{q_0}$，存在一个常数C，使得

$\left|K_j^L\left(x\mathrm{,}z\right)-K_j^L\left(y\mathrm{,}z\right)\right|\le C{\left(1+\frac{\left|x-z\right|}{\rho \left(x\right)}\right)}^{-N}\frac{{\left|x-y\right|}^{\delta }}{{\left|x-z\right|}^{d-1+\delta }}\mathrm{,}$

其中 $\left|x-y\right|<\frac{2}{3}\left|x-z\right|$。

引理4 [16] 设 $V\in R{H}_s$ 且 $s\ge \frac{d}{2}$，则存在两个正常数 $C_0\ge 1$ 和 $N_0>0$，使得对任意的 $x\mathrm{,}y\in {ℝ}^d$，都有

$\frac{1}{C_0}{\left(1+\frac{\left|x-y\right|}{\rho \left(x\right)}\right)}^{-N_0}\le \frac{\rho \left(y\right)}{\rho \left(x\right)}\le C_0{\left(1+\frac{\left|x-y\right|}{\rho \left(x\right)}\right)}^{\frac{N_0}{N_0+1}}\mathrm{.}$

作为上式的一个直接结果，我们可以得到，对任意的整数 $k\ge 1$，有下面的估计

$1+\frac{2^{k}r}{\rho \left(y\right)}\ge \frac{1}{C_0}{\left(1+\frac{r}{\rho \left(x\right)}\right)}^{-\frac{N_0}{N_0+1}}\left(1+\frac{2^{k}r}{\rho \left(x\right)}\right)\mathrm{,}$

对任意 $y\in B\left(x\mathrm{,}r\right)$，其中 $x\in {ℝ}^n$ 且 $r>0$， $C_0$ 是(3)式中所定义的。

引理5 [10] 设 $w\in A_p^{\rho \mathrm{,}\theta }$，其中 $0<\theta <\infty$， $1\le p<\infty$。则存在两个数 $\delta ,\eta >0$ 和常数 $C>0$，对球B的任意可测集E，使得

$\frac{\omega \left(E\right)}{\omega \left(B\right)}\le C{\left(\frac{\left|E\right|}{\left|B\right|}\right)}^{\delta }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{\eta }\mathrm{.}$

引理6 [10] 设 $b\in {\text{BMO}}_{\rho \mathrm{,}\infty }\left({ℝ}^d\right)$ 且 $w\in A_p^{\rho \mathrm{,}\infty }$，其中 $1\le p<\infty$，则存在常数 $C>0$ 和 $\mu >0$，使得对 ${ℝ}^d$ 上任意球 $B=B\left(x_0,r\right)$，我们有

${\left({\displaystyle {\int }_B}{\left|b\left(x\right)-b_B\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}\le C\cdot w{\left(B\right)}^{\frac{1}{p}}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{\mu }\mathrm{.}$

引理7 [10] 设 $b\in {\text{BMO}}_{\rho \mathrm{,}\theta }\left({ℝ}^d\right)$，其中 $0<\theta <\infty$，则对任意整数 $k>0$，存在一个常数 $C>0$，使得对 ${ℝ}^d$ 上任意球 $B=B\left(x_0\mathrm{,}r\right)$，我们有

$\left|b_{2^{k+1}B}-b_B\right|\le C\cdot \left(k+1\right){\left(1+\frac{2^{k+1}r}{\rho \left(x_0\right)}\right)}^{\theta }\mathrm{.}$

定理1的证明 设 $B=B\left(x_0\mathrm{,}r\right)$ 是 ${ℝ}^d$ 中的一个以 $x_0$ 为中心，以r为半径的球，记 $f=f_1+f_2$，其中 $f_1=f{\chi }_{2B}$， ${\chi }_B$ 表示B的特征函数。则有

$\begin{array}{l}\frac{1}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left({\displaystyle {\int }_B}{\left|{\mu }_{j\mathrm{,}\Omega }^{L}f\left(x\mathrm{,}t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}\\ \le \frac{1}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left({\displaystyle {\int }_B}{\left|{\mu }_{j\mathrm{,}\Omega }^L{f}_1\left(x\mathrm{,}t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{P}}+\frac{1}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left({\displaystyle {\int }_B}{\left|{\mu }_{j\mathrm{,}\Omega }^L{f}_2\left(x\mathrm{,}t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}\\ \mathrm{<mo>\; :\; </mo>}=I_1\left(t\right)+I_2\left(t\right)\mathrm{.}\end{array}$

由引理1，可得

$\begin{array}{c}{I}_1\left(t\right)\mathrm{<mo>\; =\; </mo>}C\frac{1}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left({\displaystyle {\int }_B}{\left|{\mu }_{j\mathrm{,}\Omega }^L{f}_1\left(x\mathrm{,}t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}\\ \le C\frac{w{\left(2B\right)}^{\frac{\lambda }{p}}}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left(\frac{1}{w{\left(2B\right)}^{\lambda }}{\displaystyle {\int }_{2B}}{\left|f\left(x\mathrm{,}t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}\\ \le C{\Vert f\Vert }_{L_{\rho \mathrm{,}\theta }^{p\mathrm{,}\lambda }\left(w\right)}\cdot \frac{w{\left(2B\right)}^{\frac{\lambda }{p}}}{w{\left(B\right)}^{\frac{\lambda }{p}}}\cdot {\left(1+\frac{2r}{\rho \left(x_0\right)}\right)}^{\theta }\mathrm{.}\end{array}$

对任意权 $v\in A_p^{\rho \mathrm{,}{\theta }^{\prime }}$ 及 ${ℝ}^n$ 中任意球B，则存在与v和B无关的常数 $C>0$，使得

$v\left(2B\left(x_0\mathrm{,}r\right)\right)\le C\cdot {\left(1+\frac{2r}{\rho \left(x_0\right)}\right)}^{p{\theta }^{\prime }}v\left(B\left(x_0\mathrm{,}r\right)\right)\mathrm{,}$

事实上，对 $1<p<\infty$，应用Hölder不等式，可得

$\begin{array}{l}\frac{1}{2B}{\displaystyle {\int }_{2B}}\left|\hslash \left(x\right)\right|\text{d}x=\frac{1}{2B}{\displaystyle {\int }_{2B}}\left|\hslash \left(x\right)\right|v{\left(x\right)}^{\frac{1}{p}}v{\left(x\right)}^{-\frac{1}{p}}\text{d}x\\ \le \frac{1}{2B}{\left({\displaystyle {\int }_{2B}}{\left|\hslash \left(x\right)\right|}^{p}v\left(x\right)\text{d}x\right)}^{\frac{1}{p}}{\left({\displaystyle {\int }_{2B}}v{\left(x\right)}^{-\frac{p^{\prime }}{p}}\text{d}x\right)}^{\frac{1}{p^{\prime }}}\\ \le \frac{C}{v{\left(2B\right)}^{\frac{1}{p}}}{\left({\displaystyle {\int }_{2B}}{\left|\hslash \left(x\right)\right|}^{p}v\left(x\right)\text{d}x\right)}^{\frac{1}{p}}{\left(1+\frac{2r}{\rho \left(x_0\right)}\right)}^{{\theta }^{\prime }}\mathrm{.}\end{array}$

如果我们令 $\hslash \left(x\right)={\chi }_B\left(x\right)$，则以上的表达式就为

$\frac{\left|B\right|}{\left|2B\right|}\le C\cdot \frac{v{\left(B\right)}^{\frac{1}{t}}}{v{\left(2B\right)}^{\frac{1}{t}}}{\left(1+\frac{2r}{\rho \left(x_0\right)}\right)}^{p{\theta }^{\prime }}\mathrm{.}$

即我们就得到

$\begin{array}{c}{I}_1\left(t\right)\le C{\Vert f\Vert }_{L_{\rho \mathrm{,}\theta }^{p\mathrm{,}\lambda }\left(w\right)}\cdot {\left(1+\frac{2r}{\rho \left(x_0\right)}\right)}^{p{\theta }^{\prime }}\cdot {\left(1+\frac{2r}{\rho \left(x_0\right)}\right)}^{\theta }\\ \le C{\Vert f\Vert }_{L_{\rho \mathrm{,}\theta }^{p\mathrm{,}\lambda }\left(w\right)}\cdot {\left(1+\frac{2r}{\rho \left(x_0\right)}\right)}^{{\vartheta }^{\prime }}\mathrm{.}\end{array}$

其中 ${\vartheta }^{\prime }:=p{\theta }^{\prime }+\theta$ 下面我们估计 $I_2\left(t\right)$，我们注意到，如果 $x\in B$， $y\in 2^{j+1}B\backslash 2^{j}B$， $j\ge 1$ 则应用引理3(a)，可得

$\begin{array}{c}{\mu }_{j,\Omega }^L\left(f_2\right)\left(x,t\right)\le {\left({\displaystyle {\int }_0^{\infty }}{\left|{\displaystyle {\int }_{{\left(2B\right)}^c}}\left|\Omega \left(x-y\right)\right|K_j^L\left(x,y\right)f\left(y,t\right)\text{d}y\right|}^2\frac{\text{d}t}{t^3}\right)}^{\frac{1}{2}}\\ \le {\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\left({\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|K_j^L\left(x,y\right)\right|\left|f\left(y,t\right)\right|\text{d}y\right)\cdot {\left({\displaystyle {\int }_{2^{j-1}r}^{\infty }}\frac{\text{d}h}{h^3}\right)}^{\frac{1}{2}}\\ \le {\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\frac{1}{{\left|2^{j+1}B\right|}^{\frac{1}{n}}}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\frac{1}{{\left(1+\frac{\left|x-y\right|}{\rho \left(x\right)}\right)}^N}\cdot \frac{1}{{\left|x-y\right|}^{n-1}}\left|f\left(y,t\right)\right|\text{d}y.\end{array}$

我们注意到，如果 $x\in B$， $y\in {\left(2B\right)}^c$，则 $\left|y-x\right|~\left|y-x_0\right|$，应用Hölder不等式和引理4。可得

$\begin{array}{l}{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\frac{1}{{\left|2^{j+1}B\right|}^{\frac{1}{n}}}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\frac{1}{{\left(1+\frac{\left|x-y\right|}{\rho \left(x\right)}\right)}^N}\cdot \frac{1}{{\left|x-y\right|}^{n-1}}\left|f\left(y,t\right)\right|\text{d}y\\ \le C{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\frac{1}{\left|2^{j+1}B\right|}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|{\left(1+\frac{2^{j}r}{\rho \left(x\right)}\right)}^{-N}\left|f\left(y,t\right)\right|\text{d}y\\ \le C{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\frac{1}{\left|2^{j+1}B\right|}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\cdot \frac{N_0}{N_0+1}}\cdot {\left(1+\frac{2^{j}r}{\rho \left(x_0\right)}\right)}^{-N}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|f\left(y,t\right)\right|\text{d}y\\ \le C{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\frac{1}{\left|2^{j+1}B\right|}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\cdot \frac{N_0}{N_0+1}}{\left(1+\frac{2^{j}r}{\rho \left(x_0\right)}\right)}^{-N}{\left({\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}{\left|\Omega \left(x-y\right)\right|}^s\text{d}y\right)}^{\frac{1}{s}}\cdot {\left({\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}{\left|f\left(y,t\right)\right|}^{s^{\prime }}\text{d}y\right)}^{\frac{1}{s^{\prime }}}.\end{array}$

下面利用球坐标变换，我们估计

$\begin{array}{c}{\left({\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}{\left|\Omega \left(x-y\right)\right|}^s\text{d}y\right)}^{\frac{1}{s}}={\left({\displaystyle {\int }_{2^{j-1}r\le \left|z\right|<2^{j+2}r}}{\left|\Omega \left(z\right)\right|}^s\text{d}y\right)}^{\frac{1}{s}}\\ ={\left({\displaystyle {\int }_{2^{j-1}r}^{2^{j+2}r}}{\displaystyle {\int }_{S^{n-1}}}{\left|\Omega \left(z\right)\right|}^s{\rho }^{n-1}\text{d}\sigma \left(z\right)\text{d}\rho \right)}^{\frac{1}{s}}\\ \le C{\left({\displaystyle {\int }_{S^{n-1}}}{\left|\Omega \left(z\right)\right|}^s\text{d}\sigma \left(z\right)\right)}^{\frac{1}{s}}\cdot {\left({\displaystyle {\int }_{2^{j-1}r}^{2^{j+2}r}}{\rho }^{n-1}\text{d}\rho \right)}^{\frac{1}{s}}\\ \le C{\Vert \Omega \Vert }_{L^s\left(S^{n-1}\right)}{\left|2^{j+1}B\right|}^{\frac{1}{s}}.\end{array}$

下面应用Hölder不等式，且 $w\in A_p^{\rho \mathrm{,}{\theta }^{\prime }}$，则

$\begin{array}{c}{\left({\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}{\left|f\left(y,t\right)\right|}^{s^{\prime }}\text{d}y\right)}^{\frac{1}{s^{\prime }}}\le {\left({\displaystyle {\int }_{2^{j+1}B}}{\left|f\left(y,t\right)\right|}^{s^{\prime }}w{\left(y\right)}^{\frac{1}{p_1}-\frac{1}{p_1}}\text{d}y\right)}^{\frac{1}{s^{\prime }}}\\ \le {\left({\displaystyle {\int }_{2^{j+1}B}}{\left|f\left(y,t\right)\right|}^{p_1{s}^{\prime }}w\left(y\right)\text{d}y\right)}^{\frac{1}{p_1{s}^{\prime }}}\cdot {\left({\displaystyle {\int }_{2^{j+1}B}}w{\left(y\right)}^{-\frac{{p^{\prime }}_1}{p_1}}\text{d}y\right)}^{\frac{1}{{p^{\prime }}_1{s}^{\prime }}}\\ \le C{\left({\displaystyle {\int }_{2^{j+1}B}}{\left|f\left(y,t\right)\right|}^{p}w\left(y\right)\text{d}y\right)}^{\frac{1}{p}}\cdot {\left({\displaystyle {\int }_{2^{j+1}B}}w{\left(y\right)}^{-\frac{1}{p_1-1}}\text{d}y\right)}^{\frac{p_1-1}{p}}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}w{\left(2^{j+1}B\right)}^{\frac{\lambda -1}{p}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{\theta +{\theta }^{\prime }}{\left|2^{j+1}B\right|}^{\frac{1}{p_1{s}^{\prime }}+\frac{1}{{p^{\prime }}_1{s}^{\prime }}}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}w{\left(2^{j+1}B\right)}^{\frac{\lambda -1}{p}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{\theta +{\theta }^{\prime }}{\left|2^{j+1}B\right|}^{\frac{1}{s^{\prime }}}.\end{array}$

因此，由上面的估计我们可以得到

$\left|{\mu }_{j,\Omega }^L{f}_2\left(x,t\right)\right|\le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}w{\left(2^{j+1}B\right)}^{\frac{\lambda -1}{p}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{-N+\theta +{\theta }^{\prime }}\cdot {\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\left(\frac{N_0}{N_0+1}\right)}.$

因此，可以得到

$\begin{array}{c}{I}_2\left(t\right)\le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\left(\frac{N_0}{N_0+1}\right)}\cdot {\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\frac{w{\left(B\right)}^{\frac{1-\lambda }{p}}}{w{\left(2^{j+1}B\right)}^{\frac{1-\lambda }{p}}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{-N+\theta +{\theta }^{\prime }}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\left(\frac{N_0}{N_0+1}\right)}\cdot {\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}{\left(\frac{B}{2^{j+1}B}\right)}^{\delta \left(\frac{1-\lambda }{p}\right)}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{-N+\theta +{\theta }^{\prime }+\eta \left(\frac{1-\lambda }{p}\right)}.\end{array}$

因此，取N足够大，使得 $N>\theta +{\theta }^{\prime }+\eta \left(\frac{1-\lambda }{p}\right)$ 我们可得到

$\begin{array}{c}{I}_2\left(t\right)\le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\left(\frac{N_0}{N_0+1}\right)}\cdot {\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}{\left(\frac{B}{2^{j+1}B}\right)}^{\delta \left(\frac{1-\lambda }{p}\right)}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\left(\frac{N_0}{N_0+1}\right)}.\end{array}$

结合 $I_1\left(t\right)$ 和 $I_2\left(t\right)$ 的估计，我们令 $\vartheta =\max \left\{{\vartheta }^{\prime },N\left(\frac{N_0}{N_0+1}\right)\right\}$，可以得到下面式子

$\begin{array}{l}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\vartheta }{\left(\frac{1}{w{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|{\mu }_{j,\Omega }^{L}f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}\\ \le C\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\theta }{\left(\frac{1}{w{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}.\end{array}$

在上式两边q次方，再在 $\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)$ 上积分，得到

$\begin{array}{l}{\displaystyle {\int }_{\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)}}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\vartheta q}{\left(\frac{1}{w{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|{\mu }_{j,\Omega }^{L}f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{q}{p}}\text{d}t\\ \le C{\displaystyle {\int }_{\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)}}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\theta q}{\left(\frac{1}{w{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{q}{p}}\text{d}t.\end{array}$

不等号两边乘 $\frac{1}{{\rho }^{\mu }}$，再取上确界，两边再 $\frac{1}{q}$ 次方，就可得到

$\begin{array}{l}{\left(\underset{t_0\in \left(0,T\right),\rho >0}{\sup }\frac{1}{{\rho }^{\mu }}{\displaystyle {\int }_{\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)}}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\vartheta q}{\left(\frac{1}{w{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|{\mu }_{j,\Omega }^{L}f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}y\right)}^{\frac{q}{p}}\text{d}t\right)}^{\frac{1}{q}}\\ \le {\left(\underset{t_0\in \left(0,T\right),\rho >0}{\sup }\frac{1}{{\rho }^{\mu }}{\displaystyle {\int }_{\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)}}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\theta q}{\left(\frac{1}{w{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{q}{p}}\text{d}t\right)}^{\frac{1}{q}}.\end{array}$

定理2的证明 设 $B=B\left(x_0,r\right)$ 是 ${ℝ}^d$ 中的一个以 $x_0$ 为中心，以r为半径的球，记 $f=f_1+f_2$，其中 $f_1=f{\chi }_{2B}$， ${\chi }_B$ 表示B的特征函数。则有

$\begin{array}{l}\frac{1}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left({\displaystyle {\int }_B}{\left|\left[b,{\mu }_{j,\Omega }^L\right]f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}\\ \le \frac{1}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left({\displaystyle {\int }_B}{\left|\left[b,{\mu }_{j,\Omega }^L\right]f_1\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}+\frac{1}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left({\displaystyle {\int }_B}{\left|\left[b,{\mu }_{j,\Omega }^L\right]f_2\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}\\ :=J_1\left(t\right)+J_2\left(t\right).\end{array}$

由引理2可得

$\begin{array}{c}{J}_1\left(t\right)=\frac{1}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left({\displaystyle {\int }_B}{\left|\left[b,{\mu }_{j,\Omega }^L\right]f_1\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}\\ \le C\frac{1}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left({\displaystyle {\int }_{2B}}{\left|f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}\\ \le C\frac{w{\left(2B\right)}^{\frac{\lambda }{p}}}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left(\frac{1}{w{\left(2B\right)}^{\lambda }}{\displaystyle {\int }_{2B}}{\left|f\left(x,t\right)\right|}^{p}w{\left(x\right)}^p\text{d}x\right)}^{\frac{1}{p}}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}\cdot \frac{w{\left(2B\right)}^{\frac{\lambda }{p}}}{w{\left(B\right)}^{\frac{\lambda }{p}}}\cdot {\left(1+\frac{2r}{\rho \left(x_0\right)}\right)}^{\theta }.\end{array}$

下面我们估计 $J_2\left(t\right)$，对任意的 $x\in B$，我们就要

$\begin{array}{c}\left|\left[b,{\mu }_{j,\Omega }^L\right]f_2\left(x,t\right)\right|\le \left|b\left(x\right)-b_B\right|{\left({\displaystyle {\int }_0^{\infty }}{\left|{\displaystyle {\int }_{{\left(2B\right)}^c\cap \left\{y:\left|x-y\right|\le h\right\}}}\left|\Omega \left(x-y\right)\right|K_j^L\left(x,y\right)f\left(y,t\right)\text{d}y\right|}^2\frac{\text{d}h}{h^3}\right)}^{\frac{1}{2}}\\ +{\left({\displaystyle {\int }_0^{\infty }}{\left|{\displaystyle {\int }_{{\left(2B\right)}^c\cap \left\{y:\left|x-y\right|\le h\right\}}}\left|\Omega \left(x-y\right)\right|\left|b\left(y\right)-b_B\right|K_j^L\left(x,y\right)f\left(y,t\right)\text{d}y\right|}^2\frac{\text{d}h}{h^3}\right)}^{\frac{1}{2}}\\ :=J_{21}+J_{22}.\end{array}$

由定理1的证明可知

$\left|{\mu }_{j,\Omega }^L{f}_2\left(x,t\right)\right|\le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}w{\left(2^{j+1}B\right)}^{\frac{\lambda -1}{p}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{-N+\theta +{\theta }^{\prime }}\cdot {\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\left(\frac{N_0}{N_0+1}\right)}$

则

$J_{21}\le C\left|b\left(x\right)-b_B\right|{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}w{\left(2^{j+1}B\right)}^{\frac{\lambda -1}{p}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{-N+\theta +{\theta }^{\prime }}\cdot {\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\left(\frac{N_0}{N_0+1}\right)}$

因此，应用Hölder不等式和引理5，可得

$\begin{array}{l}\frac{1}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left({\displaystyle {\int }_B}{J}_{21}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\frac{w{\left(B\right)}^{\frac{1-\lambda }{p}}}{w{\left(2^{j+1}B\right)}^{\frac{1-\lambda }{p}}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{-N+\theta +{\theta }^{\prime }}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\left(\frac{N_0}{N_0+1}\right)}{\left(\frac{1}{w\left(B\right)}{\displaystyle {\int }_B}{\left|b\left(x\right)-b_B\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\left(\frac{N_0}{N_0+1}\right)+\mu }\cdot {\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\frac{w{\left(B\right)}^{\frac{1-\lambda }{p}}}{w{\left(2^{j+1}B\right)}^{\frac{1-\lambda }{p}}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{-N+\theta +{\theta }^{\prime }}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\left(\frac{N_0}{N_0+1}\right)+\mu }\cdot {\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}{\left(\frac{\left|B\right|}{\left|2^{j+1}B\right|}\right)}^{\delta \left(\frac{1-\lambda }{p}\right)}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{-N+\theta +{\theta }^{\prime }+\eta \left(\frac{1-\lambda }{p}\right)}.\end{array}$

下面我们估计 $J_{22}$，我们注意到，如果 $x\in B$， $y\in {\left(2B\right)}^c$，则 $\left|y-x\right|~\left|y-x_0\right|$。因此应用引理3，可得

$\begin{array}{c}{J}_{22}\le C{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\frac{1}{{\left|2^{j+1}B\right|}^{\frac{1}{n}}}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|b\left(y\right)-b_B\right|\left|K_j^L\left(x,y\right)f\left(y,t\right)\right|\text{d}y\\ \le C{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\frac{1}{{\left|2^{j+1}B\right|}^{\frac{1}{n}}}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|b\left(y\right)-b_B\right|{\left(1+\frac{\left|x-y\right|}{\rho (x)}\right)}^{-N}\frac{1}{{\left|x-y\right|}^{n-1}}\left|f\left(y,t\right)\right|\text{d}y\\ \le C{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\frac{1}{\left|2^{j+1}B\right|}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|b\left(y\right)-b_B\right|{\left(1+\frac{\left|x_0-y\right|}{\rho \left(x\right)}\right)}^{-N}\left|f\left(y,t\right)\right|\text{d}y\\ \le C{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\frac{1}{\left|2^{j+1}B\right|}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|b\left(y\right)-b_B\right|{\left(1+\frac{\left|2^{j}r\right|}{\rho \left(x\right)}\right)}^{-N}\left|f\left(y,t\right)\right|\text{d}y\\ \le C{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\frac{1}{\left|2^{j+1}B\right|}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|b\left(y\right)-b_B\right|{\left(1+\frac{\left|r\right|}{\rho \left(x\right)}\right)}^{-N\left(\frac{N_0}{N_0+1}\right)}{\left(1+\frac{\left|2^{j+1}r\right|}{\rho \left(x\right)}\right)}^{-N}\left|f\left(y,t\right)\right|\text{d}y\\ \le C{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}{\left(1+\frac{\left|r\right|}{\rho \left(x\right)}\right)}^{-N\left(\frac{N_0}{N_0+1}\right)}{\left(1+\frac{\left|2^{j+1}r\right|}{\rho \left(x\right)}\right)}^{-N}\frac{1}{\left|2^{j+1}B\right|}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|b\left(y\right)-b_B\right|\left|f\left(y,t\right)\right|\text{d}y.\end{array}$

下面对任意的整数 $j\ge 1$，我们有

$\begin{array}{l}\frac{1}{\left|2^{j+1}B\right|}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|b\left(y\right)-b_B\right|\left|f\left(y,t\right)\right|\text{d}y\\ \le \frac{1}{\left|2^{j+1}B\right|}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|b\left(y\right)-b_{2^{j+1}B}\right|\left|f\left(y,t\right)\right|\text{d}y\\ +\frac{1}{\left|2^{j+1}B\right|}\left|b_{2^{j+1}B}-b_B\right|{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|f\left(y,t\right)\right|\text{d}y\\ =\mathrm{<mo>\; :\; </mo>}{I}_{21}+I_{22}\mathrm{.}\end{array}$

下面应用Hölder不等式和引理6可得

$\begin{array}{c}{I}_{21}\le \frac{1}{\left|2^{j+1}B\right|}{\left({\displaystyle {\int }_{2^{j+1}B}}{\left|b\left(y\right)-b_{2^{j+1}B}\right|}^{s^{\prime }}{\left|f\left(y,t\right)\right|}^{s^{\prime }}\text{d}y\right)}^{\frac{1}{s^{\prime }}}{\left({\displaystyle {\int }_{2^{j+1}B}}{\left|\Omega \left(x-y\right)\right|}^s\right)}^{\frac{1}{s}}\\ \le C\frac{1}{{\left|2^{j+1}B\right|}^{1-\frac{1}{s}}}{\left({\displaystyle {\int }_{2^{j+1}B}}{\left|f\left(y,t\right)\right|}^{p_1{s}^{\prime }}w\left(y\right)\text{d}y\right)}^{\frac{1}{p_1{s}^{\prime }}}{\left({\displaystyle {\int }_{2^{j+1}B}}{\left|b\left(y\right)-b_{2^{j+1}B}\right|}^{{p^{\prime }}_1{s}^{\prime }}w{\left(y\right)}^{-\frac{{p^{\prime }}_1}{p_1}}\text{d}y\right)}^{\frac{1}{{p^{\prime }}_1{s}^{\prime }}}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}\frac{w{\left(2^{j+1}B\right)}^{\frac{\lambda }{p}}}{\left|2^{j+1}B\right|}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{\theta }{w}^{-\frac{p^{\prime }}{p}}{\left(2^{j+1}B\right)}^{\frac{1}{p^{\prime }}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{\mu }\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}\frac{w{\left(2^{j+1}B\right)}^{\frac{\lambda }{p}}}{\left|2^{j+1}B\right|}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{\theta +\mu }{\left|2^{j+1}B\right|}^{\frac{1}{p^{\prime }}}\frac{{\left|2^{j+1}B\right|}^{\frac{1}{p}}}{w{\left(2^{j+1}B\right)}^{\frac{1}{p}}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{{\theta }^{\prime }}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}w{\left(2^{j+1}B\right)}^{\frac{\lambda -1}{p}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{\theta +\mu +{\theta }^{\prime }}.\end{array}$

由定理1的证明可得

$\begin{array}{l}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|f\left(y,t\right)\right|\text{d}y\\ \le C{\Vert \Omega \Vert }_{L^s\left(S^{n-1}\right)}{\left|2^{j+1}B\right|}^{\frac{1}{s}}{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }}\left(w\right)w{\left(2^{j+1}B\right)}^{\frac{\lambda -1}{p}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{\theta +{\theta }^{\prime }}{\left|2^{j+1}B\right|}^{\frac{1}{s^{\prime }}}.\end{array}$

因此，应用引理7，可得

$\begin{array}{c}{I}_{22}=\frac{1}{\left|2^{j+1}B\right|}\left|b_{2^{j+1}B}-b_B\right|{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|f\left(y,t\right)\right|\text{d}y\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}\left(j+1\right)w{\left(2^{j+1}B\right)}^{\frac{\lambda -1}{p}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{{{\theta }^{\prime }}^{\prime }+\theta +{\theta }^{\prime }}.\end{array}$

因此，结合 $I_{21}$ 和 $I_{22}$ 的估计可得

$\begin{array}{l}\frac{1}{\left|2^{j+1}B\right|}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-y\right)\right|\left|b\left(y\right)-b_B\right|\left|f\left(y,t\right)\right|\text{d}y\\ \le C{\Vert f\Vert }_{L_{\rho \mathrm{,}\theta }^{p\mathrm{,}\lambda }\left(w\right)}\left(j+1\right)w{\left(2^{j+1}B\right)}^{\frac{\lambda -1}{p}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{{\theta }^{″}+\theta +{\theta }^{\prime }+\mu }\end{array}$

因此，应用引理5可得

$\begin{array}{l}\frac{1}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left({\displaystyle {\int }_B}{J}_{22}^p\text{d}x\right)}^{\frac{1}{p}}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\left(\frac{N_0}{N_0+1}\right)}{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\left(j+1\right)\frac{w{\left(B\right)}^{\frac{1-\lambda }{p}}}{w{\left(2^{j+1}B\right)}^{\frac{1-\lambda }{p}}}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{{\theta }^{″}+\theta +{\theta }^{\prime }+\mu -N}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{N\left(\frac{N_0}{N_0+1}\right)}{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\left(j+1\right){\left(\frac{\left|B\right|}{\left|2^{j+1}B\right|}\right)}^{\delta \left(\frac{1-\lambda }{p}\right)}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{{\theta }^{″}+\theta +{\theta }^{\prime }+\mu -N+\eta \left(\frac{1-\lambda }{p}\right)}.\\ \end{array}$

结合以上的估计，我们可得

$\begin{array}{c}{J}_2\left(t\right)=\frac{1}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left({\displaystyle {\int }_B}{J}_{21}^p\text{d}x\right)}^{\frac{1}{p}}+\frac{1}{w{\left(B\right)}^{\frac{\lambda }{p}}}{\left({\displaystyle {\int }_B}{J}_{22}^p\text{d}x\right)}^{\frac{1}{p}}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{\mu +N\left(\frac{N_0}{N_0+1}\right)}{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\left(j+1\right){\left(\frac{\left|B\right|}{\left|2^{j+1}B\right|}\right)}^{\delta \left(\frac{1-\lambda }{p}\right)}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{{\theta }^{″}+\theta +{\theta }^{\prime }+\mu -N+\eta \left(\frac{1-\lambda }{p}\right)}.\end{array}$

取N足够大，使得 $N>{\theta }^{″}+\theta +{\theta }^{\prime }+\mu +\eta \left(\frac{1-\lambda }{p}\right)$，则

$\begin{array}{c}{J}_2\left(t\right)\le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{\mu +N\left(\frac{N_0}{N_0+1}\right)}{\displaystyle \underset{j=1}{\overset{\infty }{\sum }}}\left(j+1\right){\left(\frac{\left|B\right|}{\left|2^{j+1}B\right|}\right)}^{\delta \left(\frac{1-\lambda }{p}\right)}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w\right)}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{\mu +N\left(\frac{N_0}{N_0+1}\right)}.\end{array}$

结合 $J_1\left(t\right)$ 和 $J_2\left(t\right)$ 的估计，我们令 $\vartheta =\max \left\{{\vartheta }^{\prime },\mu +N\left(\frac{N_0}{N_0+1}\right)\right\}$，则可以得到

$\begin{array}{l}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\vartheta }{\left(\frac{1}{w{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|\left[b,{\mu }_{j,\Omega }^L\right]f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}\\ \le C\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\theta }{\left(\frac{1}{w{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{1}{p}}.\end{array}$

在上式两边q次方，再在 $\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)$ 上积分，得到

$\begin{array}{l}{\displaystyle {\int }_{\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)}}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\vartheta q}{\left(\frac{1}{w{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|\left[b,{\mu }_{j,\Omega }^L\right]f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{q}{p}}\text{d}t\\ \le C{\displaystyle {\int }_{\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)}}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\theta q}{\left(\frac{1}{w{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{q}{p}}\text{d}t\mathrm{.}\end{array}$

不等号两边乘 $\frac{1}{{\rho }^{\mu }}$，再取上确界，两边再 $\frac{1}{q}$ 次方，就可得到

$\begin{array}{l}{\left(\underset{t_0\in \left(0,T\right),\rho >0}{\sup }\frac{1}{{\rho }^{\mu }}{\displaystyle {\int }_{\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)}}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\vartheta q}{\left(\frac{1}{w{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|\left[b,{\mu }_{j,\Omega }^L\right]f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{q}{p}}\text{d}t\right)}^{\frac{1}{q}}\\ \le {\left(\underset{t_0\in \left(0,T\right),\rho >0}{\sup }\frac{1}{{\rho }^{\mu }}{\displaystyle {\int }_{\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)}}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\theta q}{\left(\frac{1}{w{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|f\left(x,t\right)\right|}^{p}w\left(x\right)\text{d}x\right)}^{\frac{q}{p}}\text{d}t\right)}^{\frac{1}{q}}\mathrm{.}\end{array}$

定理3的证明 设 $B=B\left(x_0,r\right)$ 是 ${ℝ}^d$ 中的一个以 $x_0$ 为中心，以r为半径的球，记 $f=f_1+f_2$，其中 $f_1=f{\chi }_{4B}$， ${\chi }_B$ 表示B的特征函数。则有

$\begin{array}{l}\frac{1}{\left|B\right|}{\displaystyle {\int }_B}\left|{\mu }_{j,\Omega }^{L}f\left(x,t\right)-{\left({\mu }_{j,\Omega }^{L}f\right)}_B\right|\text{d}x\\ \le \frac{1}{\left|B\right|}{\displaystyle {\int }_B}\left|{\mu }_{j,\Omega }^L{f}_1\left(x,t\right)-{\left({\mu }_{j,\Omega }^L{f}_1\right)}_B\right|\text{d}x+\frac{1}{\left|B\right|}{\displaystyle {\int }_B}\left|{\mu }_{j,\Omega }^L{f}_2\left(x,t\right)-{\left({\mu }_{j,\Omega }^L{f}_2\right)}_B\right|\text{d}x\\ =:K_1\left(t\right)+K_2\left(t\right).\end{array}$

下面我们估计 $K_1\left(t\right)$，令 $\lambda =\frac{p}{q}$ 由Hölder不等式和引理1可得

$\begin{array}{c}{K}_1\left(t\right)\le \frac{2}{\left|B\right|}{\displaystyle {\int }_B}\left|{\mu }_{j,\Omega }^L{f}_1\left(x,t\right)\right|\text{d}x\\ \le \frac{2}{\left|B\right|}{\left({\displaystyle {\int }_B}{\left|{\mu }_{j,\Omega }^L{f}_1\left(x,t\right)\right|}^{q}w{\left(x\right)}^q\text{d}x\right)}^{\frac{1}{q}}{\left({\displaystyle {\int }_B}w{\left(x\right)}^{-q^{\prime }}\text{d}x\right)}^{\frac{1}{q^{\prime }}}\\ \le \frac{C}{\left|B\right|}{\left({\displaystyle {\int }_{4B}}{\left|f\left(x,t\right)\right|}^{p}w{\left(x\right)}^p\text{d}x\right)}^{\frac{1}{p}}{\left({\displaystyle {\int }_B}w{\left(x\right)}^{-q^{\prime }}\text{d}x\right)}^{\frac{1}{q^{\prime }}}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w^p,w^q\right)}\cdot \frac{w^q{\left(4B\right)}^{\frac{\lambda }{p}}}{\left|B\right|}{\left({\displaystyle {\int }_B}w{\left(x\right)}^{-q^{\prime }}\text{d}x\right)}^{\frac{1}{q^{\prime }}}{\left(1+\frac{4r}{\rho \left(x_0\right)}\right)}^{\theta }\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w^p,w^q\right)}\cdot \frac{w^q{\left(4B\right)}^{\frac{1}{q}}}{w^q{\left(B\right)}^{\frac{1}{q}}}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{{\theta }^{\prime }}{\left(1+\frac{4r}{\rho \left(x_0\right)}\right)}^{\theta }\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w^p,w^q\right)}{\left(1+\frac{4r}{\rho \left(x_0\right)}\right)}^{2{\theta }^{\prime }}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{{\theta }^{\prime }}{\left(1+\frac{4r}{\rho \left(x_0\right)}\right)}^{\theta }.\end{array}$

因此，我们令 ${\vartheta }^{\prime }:=3{\theta }^{\prime }+\theta$，则

$K_1\left(t\right)\le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w^p,w^q\right)}{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{{\vartheta }^{\prime }}$

下面我们估计 $K_2\left(t\right)$，对任意 $x\in B\left(x_0\mathrm{,}r\right)$，应用引理3(b)，则

$\begin{array}{l}\left|{\mu }_{j,\Omega }^L{f}_2\left(x,t\right)-{\left({\mu }_{j,\Omega }^L{f}_2\right)}_B\right|\\ =\left|\frac{1}{\left|B\right|}{\displaystyle {\int }_B}\left[{\mu }_{j,\Omega }^L{f}_2\left(x,t\right)-{\mu }_{j,\Omega }^L{f}_2\left(y,t\right)\right]\text{d}y\right|\\ =\left|\frac{1}{\left|B\right|}{\displaystyle {\int }_B}{\left({\displaystyle {\int }_0^{\infty }}\left|{\displaystyle {\int }_{{\left(4B\right)}^c}}\left|\Omega \left(x-z\right)\right|K_j^L\left(x,z\right)f\left(z,t\right)\text{d}z\right|\frac{\text{d}h}{h^3}\right)}^{\frac{1}{2}}\text{d}y\right|\\ -\left|\frac{1}{\left|B\right|}{\displaystyle {\int }_B}{\left({\displaystyle {\int }_0^{\infty }}\left|{\displaystyle {\int }_{{\left(4B\right)}^c}}\left|\Omega \left(y-z\right)\right|K_j^L\left(y,z\right)f\left(z,t\right)\text{d}z\right|\frac{\text{d}h}{h^3}\right)}^{\frac{1}{2}}\text{d}y\right|\\ \le C\frac{1}{\left|B\right|}{\displaystyle \underset{j=2}{\overset{\infty }{\sum }}}\frac{1}{{\left|2^{j+1}B\right|}^{\frac{1}{n}}}{\displaystyle {\int }_B}\left({\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-z\right)-\Omega \left(y-z\right)\right|\left|K_j^L\left(x,z\right)-K_j^L\left(y,z\right)\right|\cdot \left|f\left(z,t\right)\right|\text{d}z\right)\text{d}y\\ \le C\frac{1}{\left|B\right|}{\displaystyle \underset{j=2}{\overset{\infty }{\sum }}}\frac{1}{{\left|2^{j+1}B\right|}^{\frac{1}{n}}}{\displaystyle {\int }_B}\left({\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-z\right)-\Omega \left(y-z\right)\right|{\left(1+\frac{\left|x-z\right|}{\rho \left(x\right)}\right)}^{-N}\frac{{\left|x-y\right|}^{\delta }}{{\left|x-z\right|}^{n-1+\delta }}\cdot \left|f\left(z,t\right)\right|\text{d}z\right)\text{d}y.\end{array}$

我们注意到 $\left|x-y\right|\le \frac{\left|x-z\right|}{2}$，若 $x\mathrm{,}y\in B$， $z\in {\left(4B\right)}^c$，则 $\left|x-z\right|~\left|x_0-z\right|$，因此，可得

$\begin{array}{l}\left|{\mu }_{j,\Omega }^L{f}_2\left(x,t\right)-{\left({\mu }_{j,\Omega }^L{f}_2\right)}_B\right|\\ \le C\frac{1}{\left|B\right|}{\displaystyle \underset{j=2}{\overset{\infty }{\sum }}}\frac{1}{{\left|2^{j+1}B\right|}^{\frac{1}{n}}}{\displaystyle {\int }_B}\left({\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-z\right)-\Omega \left(y-z\right)\right|{\left(1+\frac{\left|x_0-z\right|}{\rho \left(x\right)}\right)}^{-N}\frac{r^{\delta }}{{\left|x_0-z\right|}^{n-1+\delta }}\cdot \left|f\left(z,t\right)\right|\text{d}z\right)\text{d}y\\ \le C{\displaystyle \underset{j=2}{\overset{\infty }{\sum }}}\frac{1}{2^{j\delta }}\cdot \frac{1}{\left|2^{j+1}B\right|}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-z\right)-\Omega \left(y-z\right)\right|{\left(1+\frac{2^{j}r}{\rho \left(x\right)}\right)}^{-N}\left|f\left(z,t\right)\right|\text{d}z\\ \le C{\displaystyle \underset{j=2}{\overset{\infty }{\sum }}}\frac{1}{2^{j\delta }}\cdot {\left(1+\frac{r}{\rho \left(x\right)}\right)}^{-N\left(\frac{N_0}{N_0+1}\right)}{\left(1+\frac{2^{j+1}r}{\rho \left(x\right)}\right)}^{-N}\frac{1}{\left|2^{j+1}B\right|}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|\Omega \left(x-z\right)-\Omega \left(y-z\right)\right|\left|f\left(z,t\right)\right|\text{d}z\\ \le C{\displaystyle \underset{j=2}{\overset{\infty }{\sum }}}\frac{1}{2^{j\delta }}\cdot {\left(1+\frac{r}{\rho \left(x\right)}\right)}^{-N\left(\frac{N_0}{N_0+1}\right)}{\left(1+\frac{2^{j+1}r}{\rho \left(x\right)}\right)}^{-N}{\omega }_{\infty }\left(2\frac{\left|x-y\right|}{\left|x-z\right|}\right)\cdot \frac{1}{\left|2^{j+1}B\right|}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|f\left(z,t\right)\right|\text{d}z.\end{array}$

下面令 $k=\frac{p}{q}$ 且应用Hölder不等式和引理5可得

$\begin{array}{l}\frac{1}{\left|2^{j+1}B\right|}{\displaystyle {\int }_{2^{j+1}B\backslash 2^{j}B}}\left|f\left(z,t\right)\right|\text{d}z\\ \le \frac{1}{\left|2^{j+1}B\right|}{\left({\displaystyle {\int }_{2^{j+1}B}}{\left|f\left(z,t\right)\right|}^{p}w{\left(z\right)}^p\text{d}z\right)}^{\frac{1}{p}}{\left({\displaystyle {\int }_{2^{j+1}B}}w{\left(z\right)}^{-p^{\prime }}\text{d}z\right)}^{\frac{1}{p^{\prime }}}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w^p,w^q\right)}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{\theta }\cdot \frac{w^q{\left(2^{j+1}B\right)}^{\frac{\lambda }{p}}}{w^q{\left(2^{j+1}B\right)}^{\frac{1}{q}}}\cdot {\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{{\theta }^{\prime }}\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w^p,w^q\right)}{\left(1+\frac{2^{j+1}r}{\rho \left(x_0\right)}\right)}^{\theta +{\theta }^{\prime }}.\end{array}$

因此，取N足够大，使得 $N>\theta +{\theta }^{\prime }$

$\begin{array}{l}\left|{\mu }_{j,\Omega }^L{f}_2\left(x,t\right)-{\left({\mu }_{j,\Omega }^L{f}_2\right)}_B\right|\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w^p,w^q\right)}{\displaystyle \underset{j=2}{\overset{\infty }{\sum }}}\frac{1}{2^{j\delta }}\cdot {\left(1+\frac{r}{\rho \left(x\right)}\right)}^{-N\left(\frac{N_0}{N_0+1}\right)}{\left(1+\frac{2^{j+1}r}{\rho \left(x\right)}\right)}^{-N+\theta +{\theta }^{\prime }}\cdot {\omega }_{\infty }\left(2\frac{\left|x-y\right|}{\left|x-z\right|}\right)\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w^p,w^q\right)}{\left(1+\frac{r}{\rho \left(x\right)}\right)}^{-N\left(\frac{N_0}{N_0+1}\right)}{\displaystyle \underset{j=2}{\overset{\infty }{\sum }}}\frac{1}{2^{j\delta }}\cdot {\omega }_{\infty }\left(2\frac{\left|x-y\right|}{\left|x-z\right|}\right)\\ \le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w^p,w^q\right)}{\left(1+\frac{r}{\rho \left(x\right)}\right)}^{-N\left(\frac{N_0}{N_0+1}\right)}\cdot {\displaystyle \underset{j=2}{\overset{\infty }{\sum }}}{\omega }_{\infty }\left(\frac{1}{2^{k-1}}\right).\end{array}$

由于

${\displaystyle \underset{j=2}{\overset{\infty }{\sum }}}{\omega }_{\infty }\left(\frac{1}{2^{k-1}}\right)=\frac{1}{\log 2}{\displaystyle \underset{j=2}{\overset{\infty }{\sum }}}{\omega }_{\infty }\left(\frac{1}{2^{k-1}}\right){\displaystyle {\int }_{\frac{1}{2^{k-1}}}^{\frac{1}{2^{k-2}}}}\frac{\text{d}\delta }{\delta }\le \frac{1}{\log 2}{\displaystyle \underset{j=2}{\overset{\infty }{\sum }}}{\displaystyle {\int }_{\frac{1}{2^{k-1}}}^{\frac{1}{2^{k-2}}}}{\omega }_{\infty }\left(\delta \right)\frac{\text{d}\delta }{\delta }\le \frac{1}{\log 2}{\displaystyle {\int }_0^1}{\omega }_{\infty }\left(\delta \right)\frac{\text{d}\delta }{\delta }<\infty .$

因此

$\left|{\mu }_{j,\Omega }^L{f}_2\left(x,t\right)-{\left({\mu }_{j,\Omega }^L{f}_2\right)}_B\right|\le C{\Vert f\Vert }_{L_{\rho ,\theta }^{p,\lambda }\left(w^p,w^q\right)}{\left(1+\frac{r}{\rho \left(x\right)}\right)}^{-N\left(\frac{N_0}{N_0+1}\right)}.$

我们令 $\vartheta =\max \left\{{\vartheta }^{\prime },N\left(\frac{N_0}{N_0+1}\right)\right\}$，则可以得到下面不等式

$\begin{array}{l}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\vartheta }\cdot \frac{1}{\left|B\right|}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}\left|{\mu }_{j,\Omega }^{L}f\left(x,t\right)-{\left({\mu }_{j,\Omega }^{L}f\right)}_B\right|\text{d}x\\ \le C\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\theta }{\left(\frac{1}{w^q{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|f\left(x,t\right)\right|}^{p}w{\left(x\right)}^p\text{d}x\right)}^{\frac{1}{p}}.\end{array}$

在上式两边q次方，再在 $\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)$ 上积分，得到

$\begin{array}{l}{\displaystyle {\int }_{\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)}}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\vartheta q}\cdot {\left(\frac{1}{\left|B\right|}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}\left|{\mu }_{j,\Omega }^{L}f\left(x,t\right)-{\left({\mu }_{j,\Omega }^{L}f\right)}_B\right|\text{d}x\right)}^q\text{d}t\\ \le C{\displaystyle {\int }_{\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)}}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\theta q}{\left(\frac{1}{w^q{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|f\left(x,t\right)\right|}^{p}w{\left(x\right)}^p\text{d}x\right)}^{\frac{q}{p}}\text{d}t.\end{array}$

不等号两边乘 $\frac{1}{{\rho }^{\mu }}$，再取上确界，两边再 $\frac{1}{q}$ 次方，就可得到

$\begin{array}{l}{\left(\underset{t_0\in \left(\mathrm{0,}T\right)\mathrm{,}\rho \mathrm{<mo>\; >\; </mo>0}}{\sup }\frac{1}{{\rho }^{\mu }}{\displaystyle {\int }_{\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)}}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\vartheta q}\cdot {\left(\frac{1}{\left|B\right|}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}\left|{\mu }_{j,\Omega }^{L}f\left(x,t\right)-{\left({\mu }_{j,\Omega }^{L}f\right)}_B\right|\text{d}x\right)}^q\text{d}t\right)}^{\frac{1}{q}}\\ \le {\left(\underset{t_0\in \left(\mathrm{0,}T\right)\mathrm{,}\rho \mathrm{<mo>\; >\; </mo>0}}{\sup }\frac{1}{{\rho }^{\mu }}{\displaystyle {\int }_{\left(\mathrm{0,}T\right)\cap \left(t_0-\rho \mathrm{,}{t}_0+\rho \right)}}\underset{x\in R^n,\rho >0}{\sup }{\left(1+\frac{r}{\rho \left(x_0\right)}\right)}^{-\theta q}{\left(\frac{1}{w^q{\left(B_{\rho }\left(x\right)\right)}^{\lambda }}{\displaystyle {\int }_{B_{\rho }\left(x\right)}}{\left|f\left(x,t\right)\right|}^{p}w{\left(x\right)}^p\text{d}x\right)}^{\frac{q}{p}}\text{d}t\right)}^{\frac{1}{q}}\mathrm{.}\end{array}$