双调和抛物方程解的局部Lorentz估计

Local Lorentz Estimates for the Bi-Harmonic Parabolic Equation

摘要: Lp估计是偏微分方程中基本的正则性估计，本文我们主要研究一类四阶双调和抛物方程的解的新的正则性估计——Lorentz估计。

Abstract: Lp estimates are the basic regularity estimates in the partial differential equations. In this paper, we mainly study a new class of regularity estimates—Lorentz estimates for the Bi-harmonic parabolic equation.

1. 引言

微分方程解的存在唯一性及其正则性研究是偏微分方程中的经典问题，其中 $L^p$ 估计在正则性理论的研究中具有非常重要的作用。2003年，Wang [1] 在Caffarelli和Peral [2] 基础上利用紧方法、Vitali覆盖引理、极大函数等技巧给出了Possion方程和热传导方程的内估计的几何化证明方法。之后，Byun和Wang利用类似技巧在 [3] [4] [5] [6] [7] 中得到了各类二阶散度型椭圆方程与抛物方程在不同区域中的估计。2013年，Baroni在 [8] 中证明了退化的和非退化的具有VMO系数的散度型椭圆和抛物方程组在Lorentz空间中的正则性估计。在那篇文章中，Baroni引入了Calderón-Zygmund算子和相应的水平估计得到了文章的结论。另外，Yao和Zhou在 [9] [10] 中又给出了一类四阶双调和抛物方程解的 $L^p$ 和Schauder估计。2015年，Yu在 [11] 中证明了这类四阶双调和抛物方程的加权 $L^p$ 估计。本文主要研究下述四阶双调和抛物方程解的局部Lorentz估计

$u_t+{\Delta }^{2}u=f,z\in Q_2\subset {ℝ}^{n+1}$ (1)

这里我们定义 $Q_r=B_r\times \left(-r^4,r^4\right)$ ， $Q_r\left(z_0\right)=Q_r+z_0$ ，其中 $z_0=\left(x_0,t_0\right)$ 。另外我们记

$\left|D^{4}u\right|=\left|D_x^{4}u\left(x,t\right)\right|={\left({\displaystyle \underset{i,j,k,l=1,\cdots ,n}{\sum }{\left|D_{x_i{x}_j{x}_k{x}_l}u\left(x,t\right)\right|}^2}\right)}^{\frac{1}{2}},$

${\Vert D^{4}u\Vert }_{L^p\left({ℝ}^{n+1}\right)}={\displaystyle \underset{i,j,k,l=1,\cdots ,n}{\sum }{\Vert D_{x_i{x}_j{x}_k{x}_l}u\left(x,t\right)\Vert }_{L^p\left({ℝ}^{n+1}\right)}}.$

本文的主要目的是得到问题(1)的解u的如下估计

${\Vert D^{4}u\Vert }_{L\left(\gamma ,q\right)\left(Q_1\right)}\le c\left({\Vert D^{4}u\Vert }_{L^2\left(Q_2\right)}+{\Vert f\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}\right).$

其中 $c>0$ 与u和f无关。

下面给出Lorentz空间 [8] 的定义和嵌入性质。对于开集 $Q\subset {ℝ}^{n+1}$ 和参数 $1\le \gamma <\infty$ ， $0<q<\infty$ ，Lorentz空间 $L\left(\gamma ,q\right)\left(Q\right)$ 是包含所有满足下式的可测函数 $g:Q\to ℝ$ 组成的空间

${\Vert g\Vert }_{L\left(\gamma ,q\right)\left(Q\right)}^q:=q{\displaystyle {\int }_0^{\infty }{\lambda }^{q-1}{\left|\left\{\xi \in Q:\left|g\left(\xi \right)\right|>\lambda \right\}\right|}^{\frac{q}{\gamma }}\text{d}\lambda }<+\infty .$

事实上，当 $q=\infty$ ，Marcinkiewicz空间 ${\mathcal{M}}^{\gamma }\left(Q\right)=L\left(\gamma ,\infty \right)\left(Q\right)$ ，且有

${\Vert g\Vert }_{L\left(\gamma ,\infty \right)\left(Q\right)}={\Vert g\Vert }_{{\mathcal{M}}^{\gamma }\left(Q\right)}:=\underset{\lambda >0}{\sup }{\left({\lambda }^{\gamma }\left|\left\{\xi \in Q:\left|g\left(\xi \right)\right|>\lambda \right\}\right|\right)}^{\frac{1}{\gamma }}<+\infty .$

性质1 [8] ：对于 $1\le {\gamma }_1\le {\gamma }_2<\infty$ 和 $0<q\le \infty$ ，有以下连续嵌入成立

$$ (2)

且有 ${\Vert g\Vert }_{L\left({\gamma }_1,q\right)\left(Q\right)}\le {\left|A\right|}^{\frac{1}{{\gamma }_1}-\frac{1}{{\gamma }_2}}{\Vert g\Vert }_{L\left({\gamma }_2,q\right)\left(Q\right)}$ 。

性质2 [8] ：对于 $0<q_1\le q_2\le \infty$ 和 $1<\gamma <\infty$ ，有以下连续嵌入成立

$$ (3)

且有 ${\Vert g\Vert }_{L\left(\gamma ,q_2\right)\left(Q\right)}\le c\left(\gamma ,q_1,q_2\right){\Vert g\Vert }_{L\left(\gamma ,q_1\right)\left(Q\right)}$ 。

下面我们给出本文所要证明的主要结论。

定理3：对于 $\gamma >2$ ， $0<q\le \infty$ 。假设 $f\in L\left(\gamma ,q\right)\left(Q_2\right)$ ， $f\in L^2\left(Q_2\right)$ 。如果u满足双调和抛物方程(1)且 $\left|D^{4}u\right|\in L^2\left(Q_2\right)$ ，那么就有 $\left|D^{4}u\right|\in L\left(\gamma ,q\right)\left(Q_1\right)$ ，且有以下估计式成立

${\Vert D^{4}u\Vert }_{L\left(\gamma ,q\right)\left(Q_1\right)}\le c\left({\Vert D^{4}u\Vert }_{L^2\left(Q_2\right)}+{\Vert f\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}\right),$ (4)

其中常数 $c>0$ 与u和f无关。

2. 引理和准备工作

引理4 [12] ：Hardy不等式：可测函数 $F:\left[0,+\infty \right)\to \left[0,+\infty \right)$ 使得

${\displaystyle {\int }_0^{\infty }F\left(\lambda \right)\text{d}\lambda }<\infty .$

那么对于任意 $\alpha \ge 1$ 和 $r>0$ 都有

${\displaystyle {\int }_0^{\infty }{\lambda }^r{\left({\displaystyle {\int }_{\lambda }^{\infty }F\left(\mu \right)\text{d}\mu }\right)}^{\alpha }\frac{\text{d}\lambda }{\lambda }}\le {\left(\frac{\alpha }{r}\right)}^{\alpha }{\displaystyle {\int }_0^{\infty }{\lambda }^r{\left[\lambda F\left(\lambda \right)\right]}^{\alpha }\frac{\text{d}\lambda }{\lambda }}.$

引理5 [13] ：反Hölder不等式： $h:\left[0,+\infty \right)\to \left[0,+\infty \right)$ 是一个不增可测函数，其中 ${\alpha }_1\le {\alpha }_2\le \infty$ ， $r>0$ 。那么当 ${\alpha }_2<\infty$ 时

${\left[{\displaystyle {\int }_{\lambda }^{\infty }{\left[{\mu }^{r}h\left(\mu \right)\right]}^{{\alpha }_2}\frac{\text{d}\mu }{\mu }}\right]}^{\frac{1}{{\alpha }_2}}\le \epsilon {\lambda }^{r}h\left(\lambda \right)+\frac{c}{{\epsilon }^{\frac{{\alpha }_2}{{\alpha }_1}-1}}{\left[{\displaystyle {\int }_{\lambda }^{\infty }{\left[{\mu }^{r}h\left(\mu \right)\right]}^{{\alpha }_1}\frac{\text{d}\mu }{\mu }}\right]}^{\frac{1}{{\alpha }_1}},$

对每个 $\epsilon \in \left(0,1\right]$ 和任意 $\lambda \ge 0$ ；如果 ${\alpha }_2=\infty$ 那么

$\underset{\mu >\lambda }{\sup }\left[{\mu }^{r}h\left(\mu \right)\right]\le c{\lambda }^{r}h\left(\lambda \right)+c{\left({\displaystyle {\int }_{\lambda }^{\infty }{\left[{\mu }^{r}h\left(\mu \right)\right]}^{{\alpha }_1}\frac{\text{d}\mu }{\mu }}\right)}^{\frac{1}{{\alpha }_1}},$

常数c依赖于 ${\alpha }_1$ ， ${\alpha }_2$ ，r。当 ${\alpha }_2=\infty$ 时，常数 $c\equiv c\left({\alpha }_1,r\right)$ 。

引理6 [14] ：迭代引理：设 $0\le T_0<T_1$ ， $\phi \left(t\right)$ 是 $\left[T_0,T_1\right]$ 上的有界非负函数，且对于任意的 $T_0\le t<s\le T_1$ ， $\phi$ 满足

$\phi \left(t\right)\le \theta \phi \left(s\right)+\frac{A}{{\left(s-t\right)}^{\alpha }}+B,$

其中 $\theta$ ，A，B和 $\alpha$ 是非负常数，且 $\theta <1$ ，则

$\phi \left(\rho \right)\le c\left(\frac{A}{{\left(R-\rho \right)}^{\alpha }}+B\right),$

对任意 $T_0\le \rho <R\le T_1$ ，其中常数 $c=c\left(\alpha ,\theta \right)$ 。

引理7 [15] ：Hölder不等式：如果 $f\in {\mathcal{M}}^t\left(Q\right)\left(t>1\right)$ ，那么 $$ ， $1\le q<t$ 且有以下不等式成立

${\Vert f\Vert }_{L^q\left(Q\right)}\le {\left(\frac{t}{t-q}\right)}^{\frac{1}{q}}{\left|Q\right|}^{\frac{1}{q}-\frac{1}{t}}{\Vert f\Vert }_{{\mathcal{M}}^t\left(Q\right)}.$

下面我们给出证明定理3之前所需的准备工作。

令

${\lambda }_0={\left(\frac{1}{\left|Q_2\right|}{\displaystyle {\int }_{Q_2}{\left|D^{4}u\right|}^2\text{d}z}\right)}^{\frac{1}{2}}+{\left(M^{\frac{\eta }{2}}\frac{1}{\left|Q_2\right|}{\displaystyle {\int }_{Q_2}{\left|f\right|}^{\eta }\text{d}z}\right)}^{\frac{1}{\eta }}.$

其中 $M>1$ ， $2<\eta =\frac{2+\gamma }{2}<\gamma$ 。

定义Calderón-Zygmund算子为

$CZ\left(Q_r\left(z_i\right)\right):={\left(\frac{1}{\left|Q_r\left(z_i\right)\right|}{\displaystyle {\int }_{Q_r\left(z_i\right)}{\left|D^{4}u\right|}^2\text{d}z}\right)}^{\frac{1}{2}}+{\left(M^{\frac{\eta }{2}}\frac{1}{\left|Q_r\left(z_i\right)\right|}{\displaystyle {\int }_{Q_r\left(z_i\right)}{\left|f\right|}^{\eta }\text{d}z}\right)}^{\frac{1}{\eta }},$

其中 $Q_r\left(z_i\right)\subset Q_2$ 且 $z_i\in Q_1$ 。令 $B={40}^{\frac{4+n}{2}}$ ，则对于固定的 $\lambda >B{\lambda }_0$ ， $1/20\le r\le 1/2$ ，我们将积分区域从 $Q_r\left(z_i\right)$ 扩大到 $Q_2$ ，就有

$\begin{array}{c}CZ\left(Q_r\left(z_i\right)\right)={\left(\frac{1}{\left|Q_r\left(z_i\right)\right|}{\displaystyle {\int }_{Q_r\left(z_i\right)}{\left|D^{4}u\right|}^2\text{d}z}\right)}^{\frac{1}{2}}+{\left(M^{\frac{\eta }{2}}\frac{1}{\left|Q_r\left(z_i\right)\right|}{\displaystyle {\int }_{Q_r\left(z_i\right)}{\left|f\right|}^{\eta }\text{d}z}\right)}^{\frac{1}{\eta }}\\ \le {\left(\frac{\left|Q_2\right|}{\left|Q_r\left(z_i\right)\right|}\right)}^{\frac{1}{2}}\left[{\left(\frac{1}{\left|Q_2\right|}{\displaystyle {\int }_{Q_2}{\left|D^{4}u\right|}^2\text{d}z}\right)}^{\frac{1}{2}}+{\left(M^{\frac{\eta }{2}}\frac{1}{\left|Q_2\right|}{\displaystyle {\int }_{Q_2}{\left|f\right|}^{\eta }\text{d}z}\right)}^{\frac{1}{\eta }}\right]\\ \le {40}^{\frac{4+n}{2}}{\lambda }_0<\lambda .\end{array}$

现在对于点 $z_i$ 的水平集 $E\left(\lambda ,Q_1\right)=\left\{z\in Q_1:\left|D^{4}u\right|>\lambda \right\}$ ，由Lebesgue定理可得 $CZ\left(Q_r\left(z_i\right)\right)>\lambda$ ，当 $0<r\ll 1$ 。因此，一旦固定 $\lambda >B{\lambda }_0$ ，由积分的绝对连续性可知存在最大的 $r_i$ 使得

$CZ\left(Q_{r_i}\left(z_i\right)\right)={\left(\frac{1}{\left|Q_{r_i}\left(z_i\right)\right|}{\displaystyle {\int }_{Q_{r_i}\left(z_i\right)}{\left|D^{4}u\right|}^2\text{d}z}\right)}^{\frac{1}{2}}+{\left(M^{\frac{\eta }{2}}\frac{1}{\left|Q_{r_i}\left(z_i\right)\right|}{\displaystyle {\int }_{Q_{r_i}\left(z_i\right)}{\left|f\right|}^{\eta }\text{d}z}\right)}^{\frac{1}{\eta }}=\lambda ,$ (5)

其中 $r_i<1/20$ 且对任意的 $r\in \left(r_i,1/2\right]$ ，有

$CZ\left(Q_r\left(z_i\right)\right)={\left(\frac{1}{\left|Q_r\left(z_i\right)\right|}{\displaystyle {\int }_{Q_r\left(z_i\right)}{\left|D^{4}u\right|}^2\text{d}z}\right)}^{\frac{1}{2}}+{\left(M^{\frac{\eta }{2}}\frac{1}{\left|Q_r\left(z_i\right)\right|}{\displaystyle {\int }_{Q_r\left(z_i\right)}{\left|f\right|}^{\eta }\text{d}z}\right)}^{\frac{1}{\eta }}<\lambda .$ (6)

现令 $Q\equiv Q_{r_i}\left(z_i\right)$ ，那么对于固定的 $\lambda$ ，那么必有以下式子中的其中一个成立

${\left(\frac{\lambda }{2}\right)}^2\le \frac{1}{\left|Q\right|}{\displaystyle {\int }_Q{\left|D^{4}u\right|}^2\text{d}z}$ 或 ${\left(\frac{\lambda }{2}\right)}^{\eta }\le M^{\frac{\eta }{2}}\frac{1}{\left|Q\right|}{\displaystyle {\int }_Q{\left|f\right|}^{\eta }\text{d}z}.$ (7)

假设第一种情况成立，那么我们得到

$\begin{array}{c}{\left(\frac{\lambda }{2}\right)}^2\le \frac{1}{\left|Q\right|}{\displaystyle {\int }_Q{\left|D^{4}u\right|}^2\text{d}z}=2{\displaystyle {\int }_0^{\infty }\mu }\left|\left\{z\in Q:\left|D^{4}u\right|>\mu \right\}\right|\text{d}\mu \\ \le 2{\displaystyle {\int }_0^{\frac{\lambda }{4}}\mu }\left|\left\{z\in Q:\left|D^{4}u\right|>\mu \right\}\right|\text{d}\mu +2{\displaystyle {\int }_{\frac{\lambda }{4}}^{\infty }\mu }\left|\left\{z\in Q:\left|D^{4}u\right|>\mu \right\}\right|\text{d}\mu \\ \le \frac{{\lambda }^2}{16}\left|Q\right|+2{\displaystyle {\int }_{\frac{\lambda }{4}}^{\infty }\mu }\left|\left\{z\in Q:\left|D^{4}u\right|>\mu \right\}\right|\text{d}\mu \end{array}$

吸收右端第一项就有

$\left|Q\right|\le \frac{c}{{\lambda }^2}{\displaystyle {\int }_{\frac{\lambda }{4}}^{\infty }\mu }\left|\left\{z\in Q:\left|D^{4}u\right|>\mu \right\}\right|\text{d}\mu .$ (8)

另一方面，如果第二种情况成立，那么选取 $\varsigma =\frac{1}{4M^{1/2}}$ ，分离积分得到

$\begin{array}{c}{\left(\frac{\lambda }{2}\right)}^{\eta }\frac{1}{M^{\eta /2}}\le \frac{1}{\left|Q\right|}{\displaystyle {\int }_Q{\left|f\right|}^{\eta }\text{d}z}=\frac{\eta }{\left|Q\right|}{\displaystyle {\int }_0^{\infty }{\mu }^{\eta -1}}\left|\left\{z\in Q:\left|f\right|>\mu \right\}\right|\text{d}\mu \\ \le {\left(\zeta \lambda \right)}^{\eta }+\frac{\eta }{\left|Q\right|}{\displaystyle {\int }_{\zeta \lambda }^{\infty }{\mu }^{\eta -1}}\left|\left\{z\in Q:\left|f\right|>\mu \right\}\right|\text{d}\mu \end{array}$

吸收右端第一项可得

$\left|Q\right|\le \frac{c}{{\lambda }^{\eta }}{\displaystyle {\int }_{\zeta \lambda }^{\infty }{\mu }^{\eta -1}}\left|\left\{z\in Q:\left|f\right|>\mu \right\}\right|\text{d}\mu .$ (9)

结合式(8)和(9)，我们就有

$\left|Q\right|\le \frac{c}{{\lambda }^2}{\displaystyle {\int }_{\frac{\lambda }{4}}^{\infty }\mu }\left|\left\{z\in Q:\left|D^{4}u\right|>\mu \right\}\right|\text{d}\mu +\frac{c}{{\lambda }^{\eta }}{\displaystyle {\int }_{\zeta \lambda }^{\infty }{\mu }^{\eta -1}}\left|\left\{z\in Q:\left|f\right|>\mu \right\}\right|\text{d}\mu .$ (10)

令 $u_{\lambda }=u/\lambda$ ， $f_{\lambda }=f/\lambda$ 。事实上，根据式(6)可得如下两个不等式

$\frac{1}{\left|Q_{10r_i}\left(z_i\right)\right|}{\displaystyle {\int }_{Q_{10r_i}\left(z_i\right)}{\left|D^4{u}_{\lambda }\right|}^2\text{d}z}\le 1,$ (11)

$\frac{1}{\left|Q_{10r_i}\left(z_i\right)\right|}{\displaystyle {\int }_{Q_{10r_i}\left(z_i\right)}{\left|f_{\lambda }\right|}^2\text{d}z}\le c{\left(\frac{1}{\left|Q_{10r_i}\left(z_i\right)\right|}{\displaystyle {\int }_{Q_{10r_i}\left(z_i\right)}{\left|f_{\lambda }\right|}^{\eta }\text{d}z}\right)}^{\frac{2}{\eta }}\le \frac{c}{M}.$ (12)

对于固定的 $i\ge 1$ ，我们首先对 $f_{\lambda }$ 作从 $Q_{10r_i}\left(z_i\right)$ 到 ${ℝ}^n\times \left(0,T\right]$ 上的零延拓，并记为 $\overline{f_{\lambda }}$ 。则 $\overline{f_{\lambda }}\in L^2\left({ℝ}^n\times \left(0,T\right]\right)$ ，现在我们引入下述参照方程

由 $L^p$ 估计 [9] ，我么可得： ${\Vert D^{4}w\Vert }_{L^2\left({ℝ}^n\times \left(0,T\right]\right)}\le c{\Vert \overline{f_{\lambda }}\Vert }_{L^2\left({ℝ}^n\times \left(0,T\right]\right)}$ 。从而可得

${\displaystyle {\int }_{Q_{10r_i}\left(z_i\right)}{\left|D^{4}w\right|}^2\text{d}z}\le {\displaystyle {\int }_{{ℝ}^n\times \left(0,T\right]}{\left|D^{4}w\right|}^2\text{d}z}\le c{\displaystyle {\int }_{{ℝ}^n\times \left(0,T\right]}{\left|\overline{f_{\lambda }}\right|}^2\text{d}z}=c{\displaystyle {\int }_{Q_{10r_i}\left(z_i\right)}{\left|\overline{f_{\lambda }}\right|}^2\text{d}z},$

由上式及式(12)得

$\begin{array}{l}\frac{1}{\left|Q_{10r_i}\left(z_i\right)\right|}{\displaystyle {\int }_{Q_{10r_i}\left(z_i\right)}{\left|D^{4}w\right|}^2\text{d}z}\\ \le c\frac{1}{\left|Q_{10r_i}\left(z_i\right)\right|}{\displaystyle {\int }_{Q_{10r_i}\left(z_i\right)}{\left|f_{\lambda }\right|}^2\text{d}z}\\ \le c{\left(\frac{1}{\left|Q_{10r_i}\left(z_i\right)\right|}{\displaystyle {\int }_{Q_{10r_i}\left(z_i\right)}{\left|f_{\lambda }\right|}^{\eta }\text{d}z}\right)}^{\frac{2}{\eta }}\le \frac{c}{M}\le c\end{array}$ (13)

令 $v=u_{\lambda }-w$ ，可知， $z\in Q_{10r_i}\left(z_i\right)$ 。

由式(11)和(13)得

$\begin{array}{l}\frac{1}{\left|Q_{10r_i}\left(z_i\right)\right|}{\displaystyle {\int }_{Q_{10r_i}\left(z_i\right)}{\left|D^{4}v\right|}^2\text{d}z}\\ \le 2\left[\frac{1}{\left|Q_{10r_i}\left(z_i\right)\right|}{\displaystyle {\int }_{Q_{10r_i}\left(z_i\right)}{\left|D^{4}w\right|}^2\text{d}z}+\frac{1}{\left|Q_{10r_i}\left(z_i\right)\right|}{\displaystyle {\int }_{Q_{10r_i}\left(z_i\right)}{\left|D^4{u}_{\lambda }\right|}^2\text{d}z}\right]\le c\end{array}$

故由基本的 $W_{loc}^{4,\infty }$ 内部正则性理论，可得： $\underset{Q_{5r_i}\left(z_i\right)}{\sup }\left|D^{4}v\right|\le \frac{A}{2}$ ，这里取 $A\ge 1$ 。

对任意的 $z_i\in E\left(A\lambda ,Q_1\right)=\left\{z\in Q_1:\left|D^{4}u\right|>A\lambda \right\}$ ，因此 $\left|D^{4}u\right|>A\lambda$ ，故有 $\left|D^{4}u\right|>\lambda$ 。再对于任意的

$z_i\in Q_{5r_i}\left(z_i\right)\cap E\left(A\lambda ,Q_1\right)$ ，结合式(12)得到

$\begin{array}{l}\left|\left\{z\in Q_{5r_i}\left(z_i\right):\left|D^{4}u\right|>A\lambda \right\}\right|=\left|\left\{z\in Q_{5r_i}\left(z_i\right):\left|D^4{u}_{\lambda }\right|>A\right\}\right|\\ \le \left|\left\{z\in Q_{5r_i}\left(z_i\right):\left|D^{4}w\right|>\frac{A}{2}\right\}\right|+\left|\left\{z\in Q_{5r_i}\left(z_i\right):\left|D^{4}v\right|>\frac{A}{2}\right\}\right|\\ =\left|\left\{z\in Q_{5r_i}\left(z_i\right):\left|D^{4}w\right|>\frac{A}{2}\right\}\right|\le \frac{c}{A^2}{\displaystyle {\int }_{Q_{5r_i}\left(z_i\right)}{\left|D^{4}w\right|}^2\text{d}z}\\ \le {\displaystyle {\int }_{Q_{10r_i}\left(z_i\right)}{\left|f_{\lambda }\right|}^2\text{d}z}\le \frac{c}{M}\left|Q_{10r_i}\left(z_i\right)\right|\le \frac{c}{M}\left|Q_{10r_i}\left(z_i\right)\right|\end{array}$ (14)

下面考虑式(5)，根据Vitali覆盖引理，可知存在一族互不相交的 ${\left\{Q_{r_i}\left(z_i\right)\right\}}_{i\in \mathbb{N}}$ ， $r_i=r\left(z_i\right)<1/20$ 且

$E\left(A\lambda ,Q_1\right)\subset {\displaystyle \underset{i\in \mathbb{N}}{\cup }{Q}_{5r_i}\left(z_i\right)}\cup \left\{零测集\right\}$ 。再由式(10)和(14)得到

$\begin{array}{l}\left|\left\{z\in Q_1:\left|D^{4}u\right|>A\lambda \right\}\right|\\ \le {\displaystyle \underset{i=1}{\overset{\infty }{\sum }}\left|\left\{z\in Q_{5r_i}\left(z_i\right):\left|D^{4}u\right|>A\lambda \right\}\right|}\le {\displaystyle \underset{i=1}{\overset{\infty }{\sum }}\frac{c}{M}\left|Q_{r_i}\left(z_i\right)\right|}\\ \le \frac{c}{M}{\displaystyle \underset{i=1}{\overset{\infty }{\sum }}\left[\frac{1}{{\lambda }^2}{\displaystyle {\int }_{\frac{\lambda }{4}}^{\infty }\mu }\left|\left\{z\in Q_{r_i}\left(z_i\right):\left|D^{4}u\right|>\mu \right\}\right|\text{d}\mu +\frac{1}{{\lambda }^{\eta }}{\displaystyle {\int }_{\zeta \lambda }^{\infty }{\mu }^{\eta -1}\left|\left\{z\in Q_{r_i}\left(z_i\right):\left|f\right|>\mu \right\}\right|\text{d}\mu }\right]}\\ \le \frac{c}{M}\left[\frac{1}{{\lambda }^2}{\displaystyle {\int }_{\frac{\lambda }{4}}^{\infty }\mu }\left|\left\{z\in Q_2:\left|D^{4}u\right|>\mu \right\}\right|\text{d}\mu +\frac{1}{{\lambda }^{\eta }}{\displaystyle {\int }_{\zeta \lambda }^{\infty }{\mu }^{\eta -1}\left|\left\{z\in Q_2:\left|f\right|>\mu \right\}\right|\text{d}\mu }\right]\end{array}$ . (15)

3. 定理3的证明

下面证明当 $q<\infty$ 时的结论。

对于 $\gamma >2$ 和 $q<\infty$ ，首先根据Lorentz范数的定义得到

$\begin{array}{c}{\Vert D^{4}u\Vert }_{L\left(\gamma ,q\right)\left(Q_1\right)}^q=q{A}^q{\displaystyle {\int }_0^{\infty }{\lambda }^{q-1}}{\left|\left\{z\in Q_1:\left|D^{4}u\right|>A\lambda \right\}\right|}^{\frac{q}{\gamma }}\text{d}\lambda \\ =q{A}^q{\displaystyle {\int }_0^{B{\lambda }_0}{\lambda }^{q-1}{\left|\left\{z\in Q_1:\left|D^{4}u\right|>A\lambda \right\}\right|}^{\frac{q}{\gamma }}\text{d}\lambda }\\ +q{A}^q{\displaystyle {\int }_{B{\lambda }_0}^{\infty }{\lambda }^{q-1}{\left|\left\{z\in Q_1:\left|D^{4}u\right|>A\lambda \right\}\right|}^{\frac{q}{\gamma }}\text{d}\lambda }\\ :=S_1+S_2\end{array}$

对于 $S_1$ ，考虑 ${\lambda }_0$ 的定义和不等式 ${\Vert g\Vert }_{L^t\left(Q_2\right)}\le c{\Vert g\Vert }_{{\mathcal{M}}^{\gamma }\left(Q_2\right)}$ (见引理7)和 ${\Vert g\Vert }_{{\mathcal{M}}^{\gamma }\left(Q_2\right)}\le c{\Vert g\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}$ (见式3)，我们有

$\begin{array}{c}{S}_1=q{A}^q{\displaystyle {\int }_0^{B{\lambda }_0}{\lambda }^{q-1}}{\left|\left\{z\in Q_1:\left|D^{4}u\right|>A\lambda \right\}\right|}^{\frac{q}{\gamma }}\text{d}\lambda \\ \le c{\left[B{\lambda }_0\right]}^q\le c{\left[{\left(\frac{1}{\left|Q_2\right|}{\displaystyle {\int }_{Q_2}{\left|D^{4}u\right|}^2\text{d}z}\right)}^{\frac{1}{2}}+M^{\frac{1}{2}}{\left(\frac{1}{\left|Q_2\right|}{\displaystyle {\int }_{Q_2}{\left|f\right|}^{\eta }\text{d}z}\right)}^{\frac{1}{\eta }}\right]}^q\\ \le c\left[{\Vert D^{4}u\Vert }_{L^2\left(Q_2\right)}^q+{\Vert f\Vert }_{{\mathcal{M}}^{\gamma }\left(Q_2\right)}^q\right]\\ \le c\left[{\Vert D^{4}u\Vert }_{L^2\left(Q_2\right)}^q+{\Vert f\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}^q\right]\end{array}$ (16)

对于 $S_2$ ，结合式(15)有

$\begin{array}{c}{S}_2\le c{\left(\frac{1}{M}\right)}^{\frac{q}{\gamma }}[{\displaystyle {\int }_{B{\lambda }_0}^{\infty }{\lambda }^{q-1-\frac{2q}{\gamma }}}{\left({\displaystyle {\int }_{\frac{\lambda }{4}}^{\infty }\mu }\left|\left\{z\in Q_2:\left|D^{4}u\right|>\mu \right\}\right|\text{d}\mu \right)}^{\frac{q}{\gamma }}\text{d}\lambda \\ +{\displaystyle {\int }_{B{\lambda }_0}^{\infty }{\lambda }^{q-1-\frac{\eta q}{\gamma }}}{\left({\displaystyle {\int }_{\zeta \lambda }^{\infty }{\mu }^{\eta -1}}\left|\left\{z\in Q_2:\left|f\right|>\mu \right\}\right|\text{d}\mu \right)}^{\frac{q}{\gamma }}\text{d}\lambda ]\end{array}$ (17)

进一步，当时，利用引理4得到

$\begin{array}{c}{S}_2\le c{\left(\frac{1}{M}\right)}^{\frac{q}{\gamma }}[{\displaystyle {\int }_0^{\infty }{\lambda }^{q-1-\frac{2q}{\gamma }}}{\lambda }^{\frac{q}{\gamma }}{\left(\lambda \left|\left\{z\in Q_2:\left|D^{4}u\right|>\lambda \right\}\right|\right)}^{\frac{q}{\gamma }}\text{d}\lambda +{\displaystyle {\int }_0^{\infty }{\lambda }^{q-1-\frac{\eta q}{\gamma }}}{\lambda }^{\frac{q}{\gamma }}{\left({\lambda }^{\eta -1}\left|\left\{z\in Q_2:\left|f\right|>\lambda \right\}\right|\right)}^{\frac{q}{\gamma }}\text{d}\lambda ]\\ \le c{\left(\frac{1}{M}\right)}^{\frac{q}{\gamma }}\left[{\Vert D^{4}u\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}^q+{\Vert f\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}^q\right]\end{array}$ (18)

当 $0<q<\gamma$ ，利用引理5 $\left({\alpha }_1=1,{\alpha }_2=\frac{\gamma }{q},r=\frac{2q}{\gamma },\epsilon =1\right)$ 得到

$\begin{array}{l}{\left({\displaystyle {\int }_{\frac{\lambda }{4}}^{\infty }\mu }\left|\left\{z\in Q_2:\left|D^{4}u\right|>\mu \right\}\right|\text{d}\mu \right)}^{\frac{q}{\gamma }}\\ \le c{\lambda }^{\frac{2q}{\gamma }}{\left|\left\{z\in Q_2:\left|D^{4}u\right|>\frac{\lambda }{4}\right\}\right|}^{\frac{q}{\gamma }}+c{\displaystyle {\int }_{\frac{\lambda }{4}}^{\infty }{\mu }^{\frac{2q}{\gamma }-1}}{\left|\left\{z\in Q_2:\left|D^{4}u\right|>\mu \right\}\right|}^{\frac{q}{\gamma }}\text{d}\mu \end{array}$

因此，再利用Fubini定理我们有

$\begin{array}{l}{\displaystyle {\int }_{B{\lambda }_0}^{\infty }{\lambda }^{q-1-\frac{2q}{\gamma }}}{\left({\displaystyle {\int }_{\frac{\lambda }{4}}^{\infty }\mu }\left|\left\{z\in Q_2:\left|D^{4}u\right|>\mu \right\}\right|\text{d}\mu \right)}^{\frac{q}{\gamma }}\text{d}\lambda \\ \le c{\displaystyle {\int }_0^{\infty }{\lambda }^{q-1}}{\left|\left\{z\in Q_2:\left|D^{4}u\right|>\frac{\lambda }{4}\right\}\right|}^{\frac{q}{\gamma }}\text{d}\lambda +c{\displaystyle {\int }_0^{\infty }{\lambda }^{q-1-\frac{2q}{\gamma }}}{\displaystyle {\int }_{\frac{\lambda }{4}}^{\infty }{\mu }^{\frac{2q}{\gamma }-1}}{\left|\left\{z\in Q_2:\left|D^{4}u\right|>\mu \right\}\right|}^{\frac{q}{\gamma }}\text{d}\mu \text{d}\lambda \\ \le c{\Vert D^{4}u\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}^q+c{\displaystyle {\int }_0^{\infty }{\mu }^{\frac{2q}{\gamma }-1}}{\left|\left\{z\in Q_2:\left|D^{4}u\right|>\mu \right\}\right|}^{\frac{q}{\gamma }}{\displaystyle {\int }_0^{4\mu }{\lambda }^{q-1-\frac{2q}{\gamma }}}\text{d}\lambda \text{d}\mu \\ \le c{\Vert D^{4}u\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}^q\end{array}$ (19)

同理可得

${\displaystyle {\int }_{B{\lambda }_0}^{\infty }{\lambda }^{q-1-\frac{\eta q}{\gamma }}}{\left({\displaystyle {\int }_{\zeta \lambda }^{\infty }{\mu }^{\eta -1}\left|\left\{z\in Q_2:\left|f\right|>\mu \right\}\right|\text{d}\mu }\right)}^{\frac{q}{\gamma }}\text{d}\lambda \le c{\Vert f\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}^q.$ (20)

因此，由式(18)，(19)和(20)我们有

$S_2\le c{\left(\frac{1}{M}\right)}^{\frac{q}{\gamma }}\left[{\Vert D^{4}u\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}^q+{\Vert f\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}^q\right].$

最后结合估计式 $S_1$ 和 $S_2$ ，可以得到

${\Vert D^{4}u\Vert }_{L\left(\gamma ,q\right)\left(Q_1\right)}^q\le c{\left(\frac{1}{M}\right)}^{\frac{q}{\gamma }}{\Vert D^{4}u\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}^q+c{\Vert D^{4}u\Vert }_{L^2\left(Q_2\right)}^q+c{\Vert f\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}^q.$

不妨假设 ${\Vert D^{4}u\Vert }_{L\left(\gamma ,q\right)\left(Q_2\right)}^q<+\infty$ 。事实上，考虑截断梯度 ${\left|D^{4}u\right|}_k=\min \left\{\left|D^{4}u\right|,k\right\}$ 。然后选择M充分大，

再利用迭代引理就可以得到当 $q<\infty$ 时的结论。

下面证明当 $q=\infty$ 时的结论。

考虑式(7)中的第二种情形， $\tau$ 为一个待定常数，有

${\left(\frac{\lambda }{2}\right)}^{\eta }\le \frac{M^{\eta /2}}{\left|Q\right|}{\displaystyle {\int }_Q{\left|f\right|}^{\eta }\text{d}z}\le M^{\eta /2}{\left(\tau \lambda \right)}^{\eta }+\frac{M^{\eta /2}}{\left|Q\right|}{\displaystyle \underset{\left\{z\in Q:\left|f\right|>\tau \lambda \right\}}{\int }{\left|f\right|}^{\eta }\text{d}z}$

接下来，利用引理7，定义 $F_1\left(\tau \lambda ,Q\right)=\left\{z\in Q:\left|f\right|>\tau \lambda \right\}$ ，那么

$\begin{array}{l}{\left(\frac{\lambda }{2}\right)}^{\eta }-M^{\eta /2}{\left(\tau \lambda \right)}^{\eta }\le \frac{M^{\eta /2}}{\left|Q\right|}{\displaystyle \underset{F_1\left(\tau \lambda ,Q\right)}{\int }{\left|f\right|}^{\eta }\text{d}z}\\ \le c{M}^{\eta /2}\frac{{\left|F_1\left(\tau \lambda ,Q\right)\right|}^{1-\frac{\eta }{\gamma }}}{\left|Q\right|}\underset{\mu >0}{\sup }{\mu }^{\eta }{\left|\left\{z\in F_1\left(\tau \lambda ,Q\right):\left|f\right|>\mu \right\}\right|}^{\frac{\eta }{\gamma }}\\ \le c{M}^{\eta /2}\frac{{\left|F_1\left(\tau \lambda ,Q\right)\right|}^{1-\frac{\eta }{\gamma }}}{\left|Q\right|}\left[{\left(\tau \lambda \right)}^{\eta }{\left|F_1\left(\tau \lambda ,Q\right)\right|}^{\frac{\eta }{\gamma }}+\underset{\mu >\tau \lambda }{\sup }{\mu }^{\eta }{\left|F_1\left(\mu ,Q\right)\right|}^{\frac{\eta }{\gamma }}\right]\\ \le c{M}^{\eta /2}\left[{\left(\tau \lambda \right)}^{\eta }+\frac{{\left|F_1\left(\tau \lambda ,Q\right)\right|}^{1-\frac{\eta }{\gamma }}}{\left|Q\right|}\underset{\mu >\tau \lambda }{\sup }{\mu }^{\eta }{\left|F_1\left(\tau \lambda ,Q\right)\right|}^{\frac{\eta }{\gamma }}\right]\end{array}$

选择适当的 $\tau$ 使得 $\frac{1}{2^{\eta }}-M^{\eta /2}{\tau }^{\eta }-c{M}^{\eta /2}{\tau }^{\eta }\ge \frac{1}{4^{\eta }}$ ，事实上 $\tau =\frac{c}{M^{1/2}}$ ，那么

$\begin{array}{c}\left|Q\right|\le c\frac{{\left|F_1\left(\tau \lambda ,Q\right)\right|}^{1-\frac{\eta }{\gamma }}}{{\left(\tau \lambda \right)}^{\eta }}{\left[\underset{\mu >\tau \lambda }{\sup }{\mu }^{\gamma }\left|F_1\left(\tau \lambda ,Q\right)\right|\right]}^{\frac{\eta }{\gamma }}\\ \le c{\left(\tau \lambda \right)}^{-\gamma }{\left[{\left(\tau \lambda \right)}^{\gamma }\left|F_1\left(\tau \lambda ,Q\right)\right|\right]}^{1-\frac{\eta }{\gamma }}{\left[\underset{\mu >\tau \lambda }{\sup }{\mu }^{\gamma }\left|F_1\left(\mu ,Q\right)\right|\right]}^{\frac{\eta }{\gamma }}\\ \le c{\lambda }^{-\gamma }{\left[\underset{\mu \ge \tau \lambda }{\sup }{\mu }^{\gamma }\left|F_1\left(\mu ,Q\right)\right|\right]}^{1-\frac{\eta }{\gamma }}{\left[\underset{\mu >\tau \lambda }{\sup }{\mu }^{\gamma }\left|F_1\left(\mu ,Q\right)\right|\right]}^{\frac{\eta }{\gamma }}\\ \le c{\lambda }^{-\gamma }\underset{\mu >\tau \lambda }{\sup }{\mu }^{\gamma }\left|F_1\left(\mu ,Q\right)\right|\end{array}$ (21)

同样的，根据(7)式中的第一种情形我们可以得到

$\left|Q\right|\le c{\lambda }^{-\gamma }\underset{\mu >\tau \lambda }{\sup }{\mu }^{\gamma }\left|F_2\left(\mu ,Q\right)\right|.$ (22)

其中 $F_2\left(\mu ,Q\right)=\left\{z\in Q:\left|D^{4}u\right|>\mu \right\}$ 。因此，结合以上两式得到

$\left|Q\right|\le c\left[{\lambda }^{-\gamma }\underset{\mu >\tau \lambda }{\sup }{\mu }^{\gamma }\left|F_2\left(\mu ,Q\right)\right|+{\lambda }^{-\gamma }\underset{\mu >\tau \lambda }{\sup }{\mu }^{\gamma }\left|F_1\left(\mu ,Q\right)\right|\right].$

类似于在式(15)中的做法，由Vitali覆盖引理我们得到

$\left|\left\{z\in Q_1:\left|D^{4}u\right|>A\lambda \right\}\right|\le \frac{c}{M}\left[{\lambda }^{-\gamma }\underset{\mu >\tau \lambda }{\sup }{\mu }^{\gamma }\left|F_2\left(\mu ,Q\right)\right|+{\lambda }^{-\gamma }\underset{\mu >\tau \lambda }{\sup }{\mu }^{\gamma }\left|F_1\left(\mu ,Q\right)\right|\right]$ (23)

下面根据范数定义，我们有

$\begin{array}{c}{\Vert D^{4}u\Vert }_{{\mathcal{M}}^{\gamma }\left(Q_1\right)}=\underset{\lambda >0}{\sup }A\lambda {\left|\left\{z\in Q_1:\left|D^{4}u\right|>A\lambda \right\}\right|}^{\frac{1}{\gamma }}\\ =\underset{0<\lambda \le B{\lambda }_0}{\sup }A\lambda {\left|\left\{z\in Q_1:\left|D^{4}u\right|>A\lambda \right\}\right|}^{\frac{1}{\gamma }}+\underset{\lambda >B{\lambda }_0}{\sup }A\lambda {\left|\left\{z\in Q_1:\left|D^{4}u\right|>A\lambda \right\}\right|}^{\frac{1}{\gamma }}\\ :=J_1+J_2\end{array}$

先估计 $J_1$ ，考虑 ${\lambda }_0$ 的定义和不等式 ${\Vert g\Vert }_{L^t\left(Q_2\right)}\le c{\Vert g\Vert }_{{\mathcal{M}}^{\gamma }\left(Q_2\right)}$ (见引理7)，得到

$\begin{array}{c}{J}_1=\underset{0<\lambda \le B{\lambda }_0}{\sup }A\lambda {\left|\left\{z\in Q_1:\left|D^{4}u\right|>A\lambda \right\}\right|}^{\frac{1}{\gamma }}\\ \le cB{\lambda }_0\le c\left[{\left(\frac{1}{\left|Q\right|}{\displaystyle {\int }_{\left|Q\right|}{\left|D^{4}u\right|}^2\text{d}z}\right)}^{\frac{1}{2}}+M^{\frac{1}{2}}{\left(\frac{1}{\left|Q\right|}{\displaystyle {\int }_{\left|Q\right|}{\left|f\right|}^{\eta }\text{d}z}\right)}^{\frac{1}{\eta }}\right]\\ \le c\left[{\Vert D^{4}u\Vert }_{L^2\left(Q_2\right)}+{\Vert f\Vert }_{{\mathcal{M}}^{\gamma }\left(Q_1\right)}\right]\end{array}$ (24)

对于，考虑式(23)，进行变量替换之后得到

$\begin{array}{c}{J}_2=\underset{\lambda >B{\lambda }_0}{\sup }A\lambda {\left|\left\{z\in Q_1:\left|D^{4}u\right|>A\lambda \right\}\right|}^{\frac{1}{\gamma }}\\ \le c{\left(\frac{1}{M}\right)}^{\frac{1}{\gamma }}\left[\underset{\lambda >B\tau {\lambda }_0}{\sup }\underset{\mu >\lambda }{\sup }\mu {\left|F_2\left(\mu ,Q_2\right)\right|}^{\frac{1}{\gamma }}+\underset{\lambda >B\tau {\lambda }_0}{\sup }\underset{\mu >\lambda }{\sup }\mu {\left|F_1\left(\mu ,Q_2\right)\right|}^{\frac{1}{\gamma }}\right]\end{array}$ (25)

因为 $\underset{\lambda >B\tau {\lambda }_0}{\sup }\underset{\mu >\lambda }{\sup }\mu {\left|F_2\left(\mu ,Q_2\right)\right|}^{\frac{1}{\gamma }}\le {\Vert D^{4}u\Vert }_{{\mathcal{M}}^{\gamma }\left(Q_2\right)}$ ， $\underset{\lambda >B\tau {\lambda }_0}{\sup }\underset{\mu >\lambda }{\sup }\mu {\left|F_1\left(\mu ,Q_2\right)\right|}^{\frac{1}{\gamma }}\le {\Vert f\Vert }_{{\mathcal{M}}^{\gamma }\left(Q_2\right)}$ 。最后，结合估计

式 $J_1$ 和 $J_2$ 我们得到

${\Vert D^{4}u\Vert }_{{\mathcal{M}}^{\gamma }\left(Q_1\right)}\le c{\left(\frac{1}{M}\right)}^{\frac{1}{\gamma }}{\Vert D^{4}u\Vert }_{{\mathcal{M}}^{\gamma }\left(Q_1\right)}+c{\Vert D^{4}u\Vert }_{L^2\left(Q_2\right)}+c{\Vert f\Vert }_{{\mathcal{M}}^{\gamma }\left(Q_1\right)}.$

因此，类似于在 $q<\infty$ 中的情形，利用迭代引理我们就得到了当 $q=\infty$ 时的结论。

综上所述，定理3得证。

致谢

首先我要感谢我的导师姚锋平老师，在老师一直的鼓励和无私的指导之下我才能够完成本篇论文的写作。其次，我要感谢我同师门的师姐，她们在我的论文写作过程中给出了宝贵的建议。最后，我要感谢我的父母对我学业上的支持。