设$\Omega$是欧式空间${{\mathbb{R}}^{n}}(n\ge 2)$中的有界区域.对于$T\in(0, \infty)$,记${{\Omega }_{T}}=\Omega \times (0, T)\subset {{\mathbb{R}}^{n}}\times \mathbb{R}={{\mathbb{R}}^{n+1}}$.本文研究非散度型线性抛物方程

(1.1) $\begin{equation}\label{nondiv prob} u_{t}-a_{ij}(x, t)D_{ij}u(x, t)=f(x, t), \quad (x, t)\in \Omega_{T}, \end{equation} $

其中主项系数$A(x, t)=(a_{ij}(x, t))$是$n\times n$阶对称矩阵,满足如下一致抛物型条件

A1  对任意$\xi \in {{\mathbb{R}}^{n}}$和几乎所有$(x, t)\in \Omega_{T}$,存在常数$0<\nu\leq 1\leq L<\infty$使得

(1.2) $\begin{equation}\label{uniform parabolicity} \nu|\xi|^{2}\leq \langle A(x, t)\xi, \xi\rangle \leq L|\xi|^{2}.\end{equation}$

上式及以下各处都遵照求和约定,对重复脚标$i, j$将从1至$n$求和. $Du=(D_{1}u, \cdots, D_{n}u)$, $D^{2}u=(D_{ij}u)_{i, j=1}^{n}$分别表示$u$的梯度和Hessian矩阵,其中$D_{i}u=\partial u/\partial x^{i}, D_{ij}u=\partial^{2} u/\partial x^{i}\partial x^{j}, $$ u_{t}=\partial u/\partial t$.

自Calderón和Zygmund的经典论文[8]利用奇异积分算子理论研究位势方程强解的$W^{2, p}(1<p<\infty)$估计(称之为Calderón-Zygmund估计)以来,关于具有间断系数的线性偏微分方程解的Calderón-Zygmund估计被许多著名的数学家研究. 1991年, Chiarenza, Frasca和Longo ^[6]利用Calderón-Zygmund奇异积分算子理论研究了具有VMO系数的非散度型椭圆方程解的局部$W^{2, p}$估计.之后,他们^[7]又得到了非散度型椭圆方程强解的整体正则性. Bramanti和Cerutti^[3]也将这一结论推广到非散度型抛物方程. 2003年,王立河^[16]应用修正的Vitali覆盖引理、极大值函数理论和紧性方法给出了Calderón-Zygmund估计的另外一种证明方法.利用这种方法, Byun-Lee^[4-5]分别证明了具有小的BMO系数的非散度型椭圆和抛物方程的强解在有界$C^{1, 1}$区域上具有全局$W^{2, p}$正则性.与此同时, Krylov^[12]给出了一种统一的方法研究了具有$VMO_{x}$系数的非散度型和散度型椭圆和抛物方程解的$W^{2, p}$和$W^{1, p}$正则性.这种方法的思想是基于强解(弱解)的二阶(一阶)导数的sharp极大函数的逐点估计,从而避免了使用奇异积分算子理论.该方法的另一个优点是它可以处理具有更弱正则性系数的方程,比如Kim-Krylov^[9-11]证明了具有部分小BMO系数的非散度型椭圆和抛物方程强解的$L^{p}$正则性.所谓部分小BMO系数是指,系数$a_{ij}$对某一个固定的$x^{1}$变量可测,并对其余变量$x'$或$(x', t)$的BMO半范数随着球体(或柱体)的半径趋于零而充分小,也可参见文献[17].

Lorentz空间是一类比Lebesgue空间更精细的空间,因此解的Lorentz正则性包含了其$L^{p}$正则性.近年来, Baroni^[2]、Mengesha-Phuc^[14]和Zhang-Zhou^[18]分别证明了$p$-Laplacian抛物方程组、拟线性椭圆$p$-Laplacian方程和$p(x)$-Laplacian方程的解具有Lorentz正则性.然而,这些研究都是关于散度型方程(组)且要求主项系数关于空间变量$x$满足VMO或者小BMO条件;进一步地, Zhang-Zheng^[17]对非散度椭圆方程得到了部分正则系数下的Lorentz正则性.受Kim-Krylov^{[9-10, 12]}的工作和Baroni^[2]的工作的启发,该文的目的是研究具有部分小BMO系数的非散度型线性抛物方程的强解的Lorentz估计.为此，先引入有关一些记号.设$r>0,{{x}_{0}}=(x_{0}^{1},x_{0}^{2},\cdots ,x_{0}^{n})=(x_{0}^{1},{{{x}'}_{0}})\in {{\mathbb{R}}^{n}},{{z}_{0}}=({{x}_{0}},{{t}_{0}})=(x_{0}^{1},{{{z}'}_{0}})\in {{\mathbb{R}}^{n+1}}$,定义

$\begin{eqnarray*}\label{notations} &&B_{r}(x_{0})=\{x\in \mathbb{R} ^n : |x-x_{0}|<r\}, \qquad B'_{r}(x'_{0})=\{x'\in \mathbb{R} ^{n-1} : |x'-x'_{0}|<r\}, \\ && Q_{r}(z_{0})=Q_{r}(x_{0}, t_{0})=B_{r}(x_{0})\times (t_{0}, t_{0}+r^2), \\ &&Q'_{r}(z'_{0})=Q'_{r}(x'_{0}, t_{0})=B'_{r}(x'_{0})\times (t_{0}, t_{0}+r^2), \\ && \partial_{p} Q_{r}(z_{0})=\left( \partial B_{r}(x_{0})\times (t_{0}, t_{0}+r^{2})\right)\cup \Big(B_{r}(x_{0}) \times \{t_{0}\}\Big), \end{eqnarray*}$

其中$\partial_{p} Q_{r}(z_{0})$表示柱体$Q_{r}(z_{0})$的边界.为方便,下文记$B_{r}=B_{r}(x_{0}), B'_{r}=B'_{r}(x'_{0}), Q_{r}=Q_{r}(z_{0})$等等.除了假设A1外,本文的主要假设是主项系数$A(x, t)=(a_{ij}(x, t))$满足以下部分BMO条件

A2  对任意$\delta\in (0, 1]$存在常数$ R_{0}\in (0, 1]$,使得

$\sup\limits_{z_{0}\in \mathbb{R} ^{n+1}}\sup\limits_{0<r\leq R_{0}}\int_{ Q_{r}(z_{0})} |A(x, t)-\bar{A}(x^1)|{\rm d}x{\rm d}t<\delta, $

其中

$\bar{A}(x^{1})=\int_{Q'_{r}(z'_{0})}A(x^1, x', t){\rm d}x'{\rm d}t.$

本文的主要结果如下.

定理1.1  设$(\gamma, q)\in (2, \infty)\times (0, \infty]$, $u\in W^{2, 1}_{2}(\Omega_{T})$是非散度型线性抛物方程(1.1)的强解.若存在正常数$\delta$和$R_{0}$使得其主项系数满足条件A1和A2,则当$f\in L^{\gamma, q}(\Omega_{T})$时,有$u_{t}, D^{2}u\in L^{\gamma, q}_{loc}(\Omega_{T})$;进一步地,对任意柱体$Q_{2R}:=Q_{2R}(x_{0}, t_{0})\subset\subset \Omega_{T}$, $R \in (0, R_{0}]$有估计式

(1.3) $\begin{equation}\begin{array}{rl}\label{main result-Lorentz inequality} &\|u_{t}\|_{L^{\gamma, q}(Q_{R})}+\|D^{2}u\|_{L^{\gamma, q}(Q_{R})}\\ \leq & C|Q_{2R}|^{\frac{1}{\gamma}-\frac{1}{2}}\left(\|u_{t} \|_{L^{2}(Q_{2R})}+\|D^{2}u\|_{L^{2}(Q_{2R})}\right)+ C\|f\|_{L^{\gamma, q}(Q_{2R})}, \end{array}\end{equation}$

这里常数$\delta$和$R_{0}$为依赖于$n, \nu, L, \gamma, q$, $C$为依赖于$n, \nu, L, \gamma, q, R_{0}$的常数.特别地,当$q=\infty$时, $C$仅依赖于$n, \nu, L, \gamma, R_{0}$.

下文安排如下:第2节给出一些相关概念和已知结果,如Lorentz空间嵌入性质、Hardy不等式和Hölder不等式等.第3节是主要结果的证明.

首先给出Lorentz空间和Lorentz-Sobolev空间的定义和性质.

定义2.1  设$(\gamma, q)\in [1, \infty)\times (0, \infty]$, $k$是正整数.当$q<\infty$时, $L^{\gamma, q}(\Omega_{T})$表示由所有满足

$\|g\|_{L^{\gamma, q}(\Omega_{T})}:=\left(q\int^{\infty}_{0}\Big(\lambda^{\gamma}|\{\xi \in \Omega_{T}: |g(\xi)|>\lambda\}|\Big)^{\frac{q}{\gamma}}\frac{{\rm d}\lambda}{\lambda}\right)^{\frac{1}{q}}<\infty$

的实值可测函数$g$组成的空间;当$q=\infty$时, $L^{\gamma, \infty}(\Omega_{T})$表示由所有满足

$ \|g\|_{L^{\gamma, \infty}(\Omega_{T})}:=\sup\limits_{\lambda>0} \Big(\lambda^{\gamma}|\{\xi \in \Omega_{T}: |g(\xi)|>\lambda\}|\Big)^{\frac{1}{\gamma}}<\infty$

的实值可测函数$g$组成的空间.事实上, $L^{\gamma, \infty}(\Omega_{T})$也称为$\mbox{Marcinkiewicz}$空间${\cal M}^{\gamma}(\Omega_{T})$.特别地,当$q=\gamma$时,有

$\|g\|_{L^{\gamma}(\Omega_{T})}=\left(\gamma\int^{\infty}_{0}\Big(\lambda^{\gamma}|\{\xi \in \Omega_{T}: |g(\xi)|>\lambda\}|\Big)\frac{{\rm d}\lambda}{\lambda}\right)^{\frac{1}{\gamma}}=\|g\|_{L^{\gamma, \gamma}(\Omega_{T})}.$

进一步, Lorentz-Sobolev空间$W^{2, 1}L^{\gamma, q}(\Omega_{T})$表示由所有满足$D^{r}_{t}D^{l}_{x}u\in L^{\gamma, q}(\Omega_{T})$, $0\leq 2r+l\leq 2$, $ r\in \{0, 1\}$的实值可测函数$u$组成的空间,其范数为

$\|u{{\|}_{{{W}^{2,1}}{{L}^{\gamma ,q}}({{\Omega }_{T}})}}:=\sum\limits_{j=0}^{2}{\sum\limits_{2r+l=j}{\|D_{t}^{r}D_{x}^{l}u{{\|}_{{{L}^{\gamma ,q}}({{\Omega }_{T}})}}}}.$

注2.1   $\|\cdot\|_{L^{\gamma, q}(A)}$只是一种拟范数而不是范数,因为它不满足范数定义的次可加性质.然而,映射$g\mapsto \|g\|_{L^{\gamma, q}(A)}$关于变量$g$是下半连续的,具体证明参见文献[2]的第三部分.

引理2.1  ^[2]设$U$是$\mathbb{R} ^{n+1}$中的一个有限可测集合,则下面的结论成立

(ⅰ)若$1\leq\gamma_{1}\leq \gamma_{2}<\infty$且$g\in L^{\gamma_{2}, q}(U)$, $q\in (0, \infty]$,则$L^{\gamma_{2}, q}(U)$可连续嵌入到$L^{\gamma_{1}, q}(U)$中,且

$ \|g\|_{L^{\gamma_{1}, q}(U)}\leq |U|^{\frac{1}{\gamma_{1}}-\frac{1}{\gamma_{2}}}\|g\|_{L^{\gamma_{2}, q}(U)}.$

(ⅱ)若$0<q_{1}\leq q_{2}\leq \infty$且$g\in L^{\gamma, q_{1}}(U)$, $\gamma\in [1, \infty)$,则$L^{\gamma, q_{1}}(U)$可连续嵌入到$L^{\gamma, q_{2}}(U)$中,且

$ \|g\|_{L^{\gamma, q_{2}}(U)}\leq C\|g\|_{L^{\gamma, q_{1}}(U)}.$

其中$C$为依赖于$\gamma, q_{1}, q_{2}$的常数.特别地,当$q_{2}=\infty$时, $C$仅依赖于$\gamma, q_{1}$.

(ⅲ)若$(\gamma, p, q)\in[1, \infty)\times[1, \infty)\times (0, \infty]$且$\gamma>p$,则$L^{\gamma, q}(U)$可连续嵌入到$L^{p}(U)$中.

下面介绍几个证明定理1.1所需要的重要结论.

引理2.2^[10]  设$p\in [2, \infty)$, $a_{ij}(x, t)$满足条件A1-A2.若$f\in L^{p}(Q_{2R})$,则方程

(2.1) $ \begin{equation}\label{interior eq} u_{t}-a_{ij}(x, t)D_{ij}u=f(x, t) \quad (x, t)\in Q_{2R}, \end{equation} $

有一个强解$u\in W^{2, 1}_{p}(Q_{2R})$,且存在正常数$C=C(n, \nu, L, R)$使得

(2.2) $\begin{equation}\label{interior W(2, 1)-L2-estimate inequality} \|u\|_{W^{2, 1}_{p}(Q_{R})} \leq C\left( \|f\|_{L^{p}(Q_{2R})}+\|u\|_{L^{p}(Q_{2R})}\right).\end{equation}$

引理2.3  设$a_{ij}=a_{ij}(x^{1})$, $v\in W^{2, 1}_{2}(Q_{2R})$是齐次方程

(2.3) $\begin{equation}\label{interior f=0} v_{t}-a_{ij}D_{ij}v(x, t)=0, \quad (x, t)\in Q_{2R}\end{equation}$

的强解,则$v_{t}, D^{2}v\in L^{\infty}(Q_{R})$,且有

(2.4) $\begin{equation}\label{interior f=0 inequality} \|v_{t}\|_{L^{\infty}(Q_{R})}+\|D^{2}v\|_{L^{\infty}(Q_{R})}\leq C\|v\|_{L^{2}(Q_{2R})}.\end{equation}$

证  记$v_{x'x'}$表示函数$v$对部分变量的导数,即$v_{x^{i}x^{j}}=\partial^{2} v/\partial x^{i}\partial x^{j}, i\in \{2, \cdots, n\}, $$j\in \{2, \cdots, n\}$.下文中$v_{x'}, v_{x^{1}x'}$有类似含义.根据文献[10,引理4.4]可知

(2.5) $ \begin{equation}\label{Du youjie-1} \sup\limits_{Q_{R}}|v|+\sup\limits_{Q_{R}}|v_{x^{1}}| \leq C \|v\|_{L^{2}(Q_{2R})}. \end{equation}$

由于主项系数$a_{ij}$与变量$x'$和$t$无关,容易验证$v_{t}$, $v_{x'}$和$v_{x'x'}$也是方程(2.3)的强解.因此,利用(2.5)式得

$\sup\limits_{Q_{R}}|v_{t}|\leq C \|v_{t}\|_{L^{2}(Q_{2R})}, \quad \sup\limits_{Q_{R}}|v_{x^{1}x'}| \leq C \|v_{x'}\|_{L^{2}(Q_{2R})}, \quad \sup\limits_{Q_{R}}|v_{x'x'}|\leq C \|v_{x'x'}\|_{L^{2}(Q_{2R})}.$

再利用引理2.2,有

(2.6) $ \begin{equation}\label{bufen bound} \sup\limits_{Q_{R}}|v_{t}|\leq C \|v\|_{L^{2}(Q_{2R})}, \quad \sup\limits_{Q_{R}}|v_{x^{1}x'}| \leq C \|v\|_{L^{2}(Q_{2R})}, \quad \sup\limits_{Q_{R}}|v_{x'x'}|\leq C \|v\|_{L^{2}(Q_{2R})}.\end{equation} $

注意到

$a_{11}D_{x^{1}x^{1}}v=v_{t}-\sum\limits_{j>1}a_{1j}D_{x^{1}x^{j}}v-\sum\limits_{i>1}a_{i1}D_{x^{i}x^{1}}v- \sum\limits_{i>1, j>1}a_{ij}D_{x^{i}x^{j}}v, $

于是由(2.6)式可进一步得

(2.7) $\begin{equation}\label{x1x1} \|D_{x^{1}x^{1}}v\|_{L^{\infty}(Q_{R})}\leq C\|v\|_{L^{2}(Q_{2R})}.\end{equation} $

现联合(2.6)式和(2.7)式可得(2.4)式.引理得证.

引理2.4^[10]  设$u\in W^{2, 1}_{2}(Q_{R})$是算子$L_{0}u(x, t)=u_{t}(x, t)+a_{ij}D_{ij}u(x, t), a_{ij}=a_{ij}(x^{1})$的强解,且在$\partial Q_{R}$上取值为零,则存在一个正常数$C=C(n, \nu, L)$使得

(2.8) $\begin{equation}\label{L0-estimate-1-inequality} R^{2}\int_{Q_{R}}|Du|^{2}{\rm d}x{\rm d}t+\int_{Q_{R}}|u|^{2}{\rm d}x{\rm d}t\leq CR^{4}\int_{Q_{R}}|L_{0}u|^{2}{\rm d}x{\rm d}t.\end{equation}$

引理2.5^[13]  设$p\in[1, n+2)$且$u\in W^{2, 1}_{p}(Q_{R})$, $R>0$.若$\int_{Q_{R}}u{\rm d}x{\rm d}t=0$, $\int_{Q_{R}}Du{\rm d}x{\rm d}t=0$,则有

(2.9) $ \begin{equation}\label{parabolic local bdd inequality} R^{-1}\|u\|_{L^{p^{\ast}}(Q_{R})}+\|Du\|_{L^{p^{\ast}}(Q_{R})} \leq C(n, p)\left(\|u_{t}\|_{L^{p}(Q_{R})}+\|D^{2}u\|_{L^{p}(Q_{R})}\right), \end{equation} $

其中$p^{\ast}=\frac{(n+2)p}{n+2-p}$.

引理2.6  ^[15]设$U$是$\mathbb{R} ^{n+1}$中的一个有限可测集合且$g\in {\cal M}^{\gamma}(U)$, $\gamma>2$,则$g\in L^{p}(U)$, $p\in [2, \gamma)$,且有

(2.10) $ \begin{equation}\label{Holder inequaliy in Marcin estimate} \|g\|_{L^{p}(U)}\leq \left(\frac{\gamma}{\gamma-p}\right)^{\frac{1}{p}}|U|^{\frac{1}{p}-\frac{1}{\gamma}}\|g\|_{{\cal M}^{\gamma}(U)}.\end{equation} $

引理2.7^[2] (Hardy不等式)  设$f$是一个定义在$[0, +\infty)$上的非负可测函数且满足

(2.11) $\begin{equation}\label{Hardy-1} \int_{0}^{\infty}f(\lambda){\rm d}\lambda<\infty, \end{equation}$

则对任意$\alpha\geq 1$和$r>0$,有

(2.12) $\begin{equation}\label{Hardy-2} \int^{\infty}_{0}\lambda^{r}\left(\int^{\infty}_{\lambda}f(\mu){\rm d}\mu\right)^{\alpha}\frac{{\rm d}\lambda}{\lambda}\leq \left(\frac{\alpha}{r}\right)^{\alpha}\int^{\infty}_{0}\lambda^{r}\Big(\lambda f(\lambda)\Big)^{\alpha}\frac{{\rm d}\lambda}{\lambda}.\end{equation}$

引理2.8^[2]  (Hölder不等式)  设$h$是一个定义在$[0, +\infty)$上的非负非增可测函数, $\alpha_{1}\leq \alpha_{2}\leq \infty$, $r>0$.若$\alpha_{2}<\infty$,则对任意$\varepsilon\in (0, 1]$和$\lambda\in [0, \infty)$,有

(2.13) $ \begin{equation}\label{reverse inequaliy-1} \left(\int^{\infty}_{\lambda}[\mu^{r}h(\mu)]^{\alpha_{2}}\frac{{\rm d}\mu}{\mu}\right)^{\frac{1}{\alpha_{2}}}\leq \varepsilon \lambda^{r}h(\lambda)+\frac{C}{\varepsilon^{\frac{\alpha_{2}}{\alpha_{1}}-1}}\left(\int^{\infty}_{\lambda}[\mu^{r}h(\mu)]^{\alpha_{1}}\frac{{\rm d}\mu}{\mu}\right)^{\frac{1}{\alpha_{1}}}, \end{equation}$

其中$C$为依赖于$\alpha_{1}, \alpha_{2}, r$的常数.若$\alpha_{2}=\infty$,则对任意$\lambda\in [0, \infty)$,存在常数$C=C(\alpha_{1}, r)$使得

(14) $\begin{equation}\label{reverse inequaliy-2}\sup\limits_{\mu>\lambda}[\mu^{r}h(\mu)]\leq C\lambda^{r}h(\lambda)+C\left(\int^{\infty}_{\lambda}[\mu^{r}h(\mu)]^{\alpha_{1}}\frac{{\rm d}\mu}{\mu}\right)^{\frac{1}{\alpha_{1}}}.\end{equation}$

该节主要利用Acerbi和Mingione在文献[1]中提出的"大$M$不等式原理"方法证明主要结果:定理1.1.

证  下面将分六个步骤证明定理1.1中的(1.3)式.

第一步  寻找临界点$\bar{r}$.定义变量$\lambda_{0}$如下

(3.1) $\begin{equation}\label{lambda-0 defi} \lambda_{0}=\left( \mathit{{\rlap{-} \smallint }}_{Q_{2R}}|u_{t}|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}+\left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}} +M^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{1}{\eta}}, \end{equation}$

其中$\eta=(2+\gamma)/2\in (2, \gamma)$, $M$是仅依赖于$n, \nu, L, \gamma, q$且大于1的待定常数.根据引理2.1的第三条性质可知对任意$\varepsilon>0$都有$f\in L_{loc}^{\gamma-\varepsilon}(\Omega_{T})$,也就是说,当选取$\varepsilon=\gamma-\eta$时, $f\in L^{\eta}_{loc}(\Omega_{T})$,这就保证了上述$\lambda_{0}$的定义是合理的.现取两个数$r, \lambda$满足

(3.2) $ \begin{equation}\label{r and lambda range} \frac{R}{32}\leq r\leq \frac{R}{2}, \quad \lambda>64^{n}\lambda_{0}:=B\lambda_{0}, \end{equation} $

并定义一个关于变量$r$的连续函数

$CZ(Q_{r}(\bar{z}))=\left(\mathit{{\rlap{-} \smallint }}_{Q_{r}(\bar{z})}|u_t|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}+\left(\mathit{{\rlap{-} \smallint }}_{Q_{r}(\bar{z})}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}+M^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{r}(\bar{z})}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{1}{\eta}}, $

其中$\bar{z}\in Q_{R}$.由$r\leq R/2$可知$Q_{r}(\bar{z})\subset Q_{2R}$,因此通过将积分区域$Q_{r}(\bar{z})$扩大到$Q_{2R}(\bar{z})$得

(3.3) $\begin{eqnarray}\label{CZ smaller than lambda} CZ(Q_{r}(\bar{z})) &\leq& \left(\frac{|Q_{2R}|}{|Q_{r}(\bar{z})|}\right)^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|u_{t}|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}+ \left(\frac{|Q_{2R}|}{|Q_{r}(\bar{z})|}\right)^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}\nonumber\\ &+& \left(\frac{|Q_{2R}|}{|Q_{r}(\bar{z})|}\right)^{\frac{1}{\eta}}M^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{1}{\eta}}\nonumber\\ &\leq& \left(\frac{|Q_{2R}|}{|Q_{r}(\bar{z})|}\right)^{\frac{1}{2}}\lambda_{0}\leq \left(\frac{2R}{r}\right)^{\frac{n+2}{2}}\lambda_{0}\leq \left(\frac{2R}{R/32}\right)^{n}\lambda_{0}=64^{n}\lambda_{0}<\lambda.\end{eqnarray} $

另一方面,考虑$D^{2}u$的Lebesgue点$\bar{z}$,满足$ \bar{z}\in E(\lambda, Q_{R})=\{z\in Q_{R}: |D^{2}u(z)|>\lambda\}. $应用Lebesgue微分中值定理可知对几乎每一个上述Lebesgue点$\bar{z}$都有

$\lim\limits_{r\downarrow 0}CZ(Q_{r}(\bar{z}))\geq \lim\limits_{r\downarrow 0}\left(\mathit{{\rlap{-} \smallint }}_{Q_{r}(\bar{z})}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}} =|D^{2}u(\bar{z})|>\lambda, $

也就是说,对远远小于1的正数$r$有$CZ(Q_{r}(\bar{z}))>\lambda$成立.由此可知,对于任意给定$\lambda>B\lambda_{0}$和任意点$\bar{z}\in E(\lambda, Q_{R})$,根据积分的绝对连续性,我们可以找到一个最大的数$\bar{r}=r(\bar{z})$使得

(3.4) $\begin{eqnarray}\label{CZ equi lambda} CZ(Q_{\bar{r}}(\bar{z}))&=&\left(\mathit{{\rlap{-} \smallint }}_{Q_{\bar{r}}(\bar{z})}|u_{t}|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}+\left(\mathit{{\rlap{-} \smallint }}_{Q_{\bar{r}}(\bar{z})}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}} +M^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{\bar{r}}(\bar{z})}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{1}{\eta}} \\ &=&\lambda\end{eqnarray}$

并且对于$r\in (\bar{r}, R/2]$都有$CZ(Q_{\bar{r}}(\bar{z}))<\lambda$.此外,我们还可以推导出$\bar{r}<R/32$,于是$Q_{32\bar{r}}(\bar{z})\subset Q_{2R}$且

(3.5) $\begin{eqnarray}\label{low and upper bound} \frac{\lambda}{32^{n}}&\leq& \left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|u_{t}|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}+ \left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}+ M^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{1}{\eta}} \\ &\leq&\lambda.\end{eqnarray}$

事实上,上式右端不等式可以由$\bar{r}$的定义自然得到,而对于上式左端不等式,其证明如下

$\begin{eqnarray*} &&\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|u_{t}|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}+\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}} +M^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{1}{\eta}}\nonumber\\ &\geq& \left(\frac{|Q_{\bar{r}}(\bar{z})|}{|Q_{32\bar{r}}(\bar{z})|}\right)^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{\bar{r}}(\bar{z})}|u_{t}|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}+ \left(\frac{|Q_{\bar{r}}(\bar{z})|}{|Q_{32\bar{r}}(\bar{z})|}\right)^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{\bar{r}}(\bar{z})}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}\nonumber\\ &&+\left(\frac{|Q_{\bar{r}}(\bar{z})|}{|Q_{32\bar{r}}(\bar{z})|}\right)^{\frac{1}{\eta}}M^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{\bar{r}}(\bar{z})}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{1}{\eta}}\nonumber\\ &\geq& \left(\frac{|Q_{\bar{r}}(\bar{z})|}{|Q_{32\bar{r}}(\bar{z})|}\right)^{\frac{1}{2}}\lambda\geq\left(\frac{1}{32}\right)^{\frac{n+2}{2}}\lambda\geq \frac{\lambda}{32^{n}}.\end{eqnarray*}$

根据Vitali覆盖定理,存在一族互不相交的柱体$\{Q_{i}\}, (i\in {\Bbb N})$,满足上述构造:$Q_{i}=Q_{\bar{r}_{i}}(\bar{z}_{i}) \subset Q_{2R}, \bar{z}_{i}\in E(\lambda, Q_{R}), $使得

(3.6) $\begin{equation} E(\lambda, Q_{R}) \subset \left(\bigcup\limits_{i\in {\Bbb N}}E(\lambda, 4Q_{i})\right)\cup {\cal N}_{\lambda}, \end{equation}$

其中$4Q_{i}=Q_{4\bar{r}_{i}}(\bar{z}_{i})$且${\cal N}_{\lambda}$是一个非空集合.因此,有

(3.7) $\begin{equation}\label{full level set} |E(\lambda, Q_{R})|\leq \sum\limits_{i\in {\Bbb N}} |E(\lambda, Q_{4\bar{r}_{i}}(\bar{z}_{i}))|.\end{equation}$

这说明要想估计水平集$E(\lambda, Q_{R})$的Lebesgue测度,只需估计水平集$E(\lambda, Q_{4\bar{r}_{i}}(\bar{z}_{i}))$的Lebesgue测度,具体见下文第四步.

第二步  引进两个函数$w$和$v$.设$u\in W^{2, 1}_{2}(\Omega_{T})$是方程(1.1)的强解, $Q_{\bar{r}}(\bar{z})$是第一步中定义的柱体,满足$Q_{32\bar{r}}(\bar{z})\subset Q_{2R}$和(3.5)式.首先,考虑关于函数$w\in W^{2, 1}_{2}(Q_{16\bar{r}}(\bar{z}))$的边值问题

(3.8) $\begin{equation}\label{w problem}\left\{\begin{array}{ll}w_{t}-\bar{a}_{ij}(x^{1})D_{ij}w=[a_{ij}(x, t)-\bar{a}_{ij}(x^{1})]D_{ij}u+f(x, t), &(x, t)\in Q_{16\bar{r}}(\bar{z}), \\w=0, & (x, t)\in\partial_{p} Q_{16\bar{r}}(\bar{z}).\end{array}\right.\end{equation}$

依次利用引理2.2、引理2.4、Hölder不等式和(3.5)式,有

(3.9) $\begin{eqnarray}\label{interior compare eq.-pf-3} && \mathit{{\rlap{-} \smallint }}_{Q_{8\bar{r}}(\bar{z})}|D^{2}w|^{2}{\rm d}x{\rm d}t\\\ &\leq& C\mathit{{\rlap{-} \smallint }}_{Q_{16\bar{r}}(\bar{z})}|a_{ij}(x, t)-\overline{a}_{ij}(x^{1})|^{2}|D^{2}u|^{2}{\rm d}x{\rm d}t+C\mathit{{\rlap{-} \smallint }}_{Q_{16\bar{r}}(\bar{z})}|f|^{2}{\rm d}x{\rm d}t+C\mathit{{\rlap{-} \smallint }}_{Q_{16\bar{r}}(\bar{z})}|w|^{2}{\rm d}x{\rm d}t\nonumber\\ &\leq& C\mathit{{\rlap{-} \smallint }}_{Q_{16\bar{r}}(\bar{z})}|a_{ij}(x, t)-\overline{a}_{ij}(x^{1})|^{2}|D^{2}u|^{2}{\rm d}x{\rm d}t+C\mathit{{\rlap{-} \smallint }}_{Q_{16\bar{r}}(\bar{z})}|f|^{2}{\rm d}x{\rm d}t\nonumber\\ &\leq& C\left(\mathit{{\rlap{-} \smallint }}_{Q_{16\bar{r}}(\bar{z})}|a_{ij}(x, t)-\overline{a}_{ij}(x^{1})|^{2\pi}{\rm d}x{\rm d}t\right)^{\frac{1}{\pi}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{16\bar{r}}(\bar{z})}|D^{2}u|^{2\kappa}{\rm d}x{\rm d}t\right)^{\frac{1}{\kappa}} \\ && +C\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|f|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}\cdot 2}\nonumber\\ &\leq& C\left(\mathit{{\rlap{-} \smallint }}_{Q_{16\bar{r}}(\bar{z})}|a_{ij}(x, t)-\overline{a}_{ij}(x^{1})|{\rm d}x{\rm d}t\right)^{\frac{1}{\pi}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{16\bar{r}}(\bar{z})}|D^{2}u|^{2\kappa}{\rm d}x{\rm d}t\right)^{\frac{1}{\kappa}} \\ && +C\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{1}{\eta}\cdot 2}\nonumber\\ &\leq& C\delta^{\frac{1}{\pi}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{16\bar{r}}(\bar{z})}|D^{2}u|^{2\kappa}{\rm d}x{\rm d}t\right)^{\frac{1}{\kappa}}+\frac{C\lambda^{2}}{M}, \end{eqnarray} $

其中$\pi$和$\kappa$满足

$\frac{1}{\pi}+\frac{1}{\kappa}=1, \quad 1<\kappa<\frac{q}{2}, \quad 1<\kappa<\min\left\{\frac{\eta}{2}, \frac{n+2}{2}, \frac{n+2}{n}\right\}.$

注意到函数$\hat{u}=u-(\overline{Du})_{Q_{32\bar{r}}(\bar{z})}x-(\bar{u})_{Q_{32\bar{r}}(\bar{z})}\in W^{2, 1}_{2}(Q_{32\bar{r}}(\bar{z}))$也是方程(2.1)的强解,利用引理2.2和(3.5)式,有

(3.10) $ \begin{eqnarray}\label{interior compare eq.-pf-4} \mathit{{\rlap{-} \smallint }}_{Q_{16\bar{r}}(\bar{z})}|D^{2}u|^{2\kappa}{\rm d}x{\rm d}t &=&\mathit{{\rlap{-} \smallint }}_{Q_{16\bar{r}}(\bar{z})}|D^{2}\hat{u}|^{2\kappa}{\rm d}x{\rm d}t \nonumber\\ &\leq& C\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|f|^{2\kappa}{\rm d}x{\rm d}t+C\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|\hat{u}|^{2\kappa}{\rm d}x{\rm d}t\nonumber\\ &\leq& C\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{2\kappa}{\eta}}+C|Q_{32\bar{r}}(\bar{z})|^{-1}\|\hat{u}\|^{2\kappa}_{L^{2\kappa}(Q_{32\bar{r}}(\bar{z}))}\nonumber\\ &\leq &\left(\frac{C\lambda^{2}}{M}\right)^{\kappa}+C|Q_{32\bar{r}}(\bar{z})|^{-1}\|\hat{u}\|^{2\kappa}_{L^{2\kappa}(Q_{32\bar{r}}(\bar{z}))}.\end{eqnarray} $

由引理2.5可得

(3.11) $ \begin{eqnarray}\label{interior compare eq.-pf-5} \|\hat{u}\|^{2\kappa}_{L^{2\kappa}(Q_{32\bar{r}}(\bar{z}))}&\leq& C\|\hat{u}\|^{2\kappa}_{L^{2^{\ast}}(Q_{32\bar{r}}(\bar{z}))}\nonumber\\ &\leq& C\left(\|\hat{u}_{t}\|^{2\kappa}_{L^{2}(Q_{32\bar{r}}(\bar{z}))}+\|D^{2}\hat{u}\|^{2\kappa}_{L^{2}(Q_{32\bar{r}}(\bar{z}))}\right)\nonumber\\ &=& C\left(\|u_{t}\|^{2\kappa}_{L^{2}(Q_{32\bar{r}}(\bar{z}))}+\|D^{2}u\|^{2\kappa}_{L^{2}(Q_{32\bar{r}}(\bar{z}))}\right).\end{eqnarray} $

现联合(3.9)式、(3.10)式和(3.11)式得

(3.12) $\begin{eqnarray}\label{w estimate}\mathit{{\rlap{-} \smallint }}_{Q_{8\bar{r}}(\bar{z})}|D^{2}w|^{2}{\rm d}x{\rm d}t &\leq& C\delta^{\frac{1}{\pi}}\left[\frac{\lambda^{2}}{M}+\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|u_{t}|^{2}{\rm d}x{\rm d}t+\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)\right]+\frac{C\lambda^{2}}{M}\nonumber\\ &\leq& C\left(\frac{\delta^{\frac{1}{\pi}}+1}{M}+1\right)\lambda^{2}.\end{eqnarray}$

定义另一个函数$v=\hat{u}-w$.易证函数$v\in W^{2, 1}_{2}(Q_{16\bar{r}}(\bar{z}))$是边值问题

(3.13) $\begin{equation}\label{limiting problem-1} \left\{\begin{array}{ll} v_{t}-\bar{a}_{ij}(x_{1})D_{ij}v=0, &(x, t)\in Q_{16\bar{r}}(\bar{z}), \\ v=\hat{u}, & (x, t)\in\partial_{p} Q_{16\bar{r}}(\bar{z}) \end{array}\right. \end{equation}$

的强解,再利用引理2.3得

(3.14) $ \begin{equation}\label{D2v bdd-0} \|D^{2}v\|_{L^{\infty}(Q_{4\bar{r}}(\bar{z}))}\leq C\|v\|_{L^{2}(Q_{8\bar{r}}(\bar{z}))}\leq C\|v-\hat{u}\|_{L^{2}(Q_{8\bar{r}}(\bar{z}))}+C\|\hat{u}\|_{L^{2}(Q_{8\bar{r}}(\bar{z}))}.\end{equation} $

对方程

$(v-\hat{u})_{t}-\bar{a}_{ij}(x_{1})D_{ij}(v-\hat{u})=-\hat{u}_{t}+\bar{a}_{ij}(x_{1})D_{ij}\hat{u} $

的强解$v-\hat{u}$应用引理2.4得

$ \|v-\hat{u}\|_{L^{2}(Q_{8\bar{r}}(\bar{z}))}\leq C\left(\|\hat{u}_{t}\|_{L^{2}(Q_{8\bar{r}}(\bar{z}))}+\|D^{2}\hat{u}\|_{L^{2}(Q_{8\bar{r}}(\bar{z}))}\right).$

对函数$\hat{u}$应用引理2.5有

$\|\hat{u}\|_{L^{2}(Q_{8\bar{r}}(\bar{z}))}\leq \|\hat{u}\|_{L^{2^{\ast}}(Q_{8\bar{r}}(\bar{z}))}\leq C\left(\|\hat{u}_{t}\|_{L^{2}(Q_{8\bar{r}}(\bar{z}))}+\|D^{2}\hat{u}\|_{L^{2}(Q_{8\bar{r}}(\bar{z}))}\right).$

因此,存在常数$C_{\ast}=C_{\ast}(n, \nu, L, R)$使得(3.14)式可写成

(3.15) $ \begin{equation}\label{D2v bdd} \|D^{2}v\|_{L^{\infty}(Q_{4\bar{r}}(\bar{z}))}\leq C\left[\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|u_{t}|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}} +\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}} \right]\leq C_\ast \lambda.\end{equation} $

第三步  估计柱体测度$|Q_{\bar{r}}(\bar{z})|$.记$Q\equiv Q_{\bar{r}}(\bar{z})$.由(3.5)式可知对任意给定正数$\lambda$,下面三个不等式至少有一个成立

(3.16) $\begin{equation}\label{step 2 two cases-1}\left(\frac{\lambda}{3}\right)^{2}\leq \mathit{{\rlap{-} \smallint }}_{Q}|u_{t}|^{2}{\rm d}x{\rm d}t, \end{equation}$

(3.17) $\begin{equation}\label{step 2 two cases-2}\left(\frac{\lambda}{3}\right)^{2}\leq \mathit{{\rlap{-} \smallint }}_{Q}|D^{2}u|^{2}{\rm d}x{\rm d}t, \end{equation}$

(3.18) $\begin{equation}\label{step 2 two cases-3} \left(\frac{\lambda}{3}\right)^{\eta}\leq M^{\frac{\eta}{2}}\mathit{{\rlap{-} \smallint }}_{Q}|f|^{\eta}{\rm d}x{\rm d}t.\end{equation}$

情形1  假设(3.16)式成立.根据一致抛物条件(1.2)得

(3.19) $ \begin{equation}\label{ut bdd} |u_{t}|=|f(z)+a_{ij}(z)D_{ij}u|\leq |f(z)|+L|D^{2}u|, \end{equation} $

于是

$ \begin{eqnarray*} \frac{\lambda}{3}\leq \left(\mathit{{\rlap{-} \smallint }}_{Q}|u_{t}|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}} \leq CM^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q}|f(z)|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{1}{\eta}}+C\left(\mathit{{\rlap{-} \smallint }}_{Q}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}.\end{eqnarray*} $

由此可知,只需要考虑情形2和情形3即可.

情形2  假设(3.17)式成立.利用Hölder不等式得

(3.20) $\begin{eqnarray}\label{step 2-1} \mathit{{\rlap{-} \smallint }}_{Q}|D^{2}u|^{2}{\rm d}x{\rm d}t &=&\frac{1}{|Q|}\int_{Q\backslash E(\frac{1}{6}\lambda, Q_{2R})}|D^{2}u|^{2}{\rm d}x{\rm d}t+\frac{1}{|Q|}\int_{Q\cap E(\frac{1}{6}\lambda, Q_{2R})}|D^{2}u|^{2}{\rm d}x{\rm d}t\nonumber\\ &\leq& \frac{|Q\backslash E(\frac{1}{6}\lambda, Q_{2R})|}{|Q|} \left(\frac{\lambda}{6}\right)^{2}+\left(\frac{|Q\cap E(\frac{1}{6}\lambda, Q_{2R})|}{|Q|}\right)^{\frac{\vartheta}{2+\vartheta}}\left(\mathit{{\rlap{-} \smallint }}_{Q}|D^{2}u|^{2+\vartheta}{\rm d}x{\rm d}t\right)^{\frac{2}{2+\vartheta}}\nonumber\\ &\leq& \left(\frac{\lambda}{6}\right)^{2}+\left(\frac{|Q\cap E(\frac{1}{6}\lambda, Q_{2R})|}{|Q|}\right)^{\frac{\vartheta}{2+\vartheta}}\left(\mathit{{\rlap{-} \smallint }}_{Q}|D^{2}u|^{2+\vartheta}{\rm d}x{\rm d}t\right)^{\frac{2}{2+\vartheta}}, \end{eqnarray}$

其中$\vartheta$为某一待定正常数.注意到函数$\hat{u}=u(x, t)-x_{i}(\overline{D_{i}u})_{Q_{32\bar{r}}(\bar{z})}-(\bar{u})_{Q_{32\bar{r}}(\bar{z})}$也是方程(2.1)的强解,依次应用引理2.2的第一条性质,引理2.5和(3.5)式,有

(3.21) $\begin{eqnarray}\label{step 2-2} \left(\mathit{{\rlap{-} \smallint }}_{Q}|D^{2}u|^{2+\vartheta}{\rm d}x{\rm d}t\right)^{\frac{2}{2+\vartheta}} &=& \left(\mathit{{\rlap{-} \smallint }}_{Q}|D^{2}\hat{u}|^{2+\vartheta}{\rm d}x{\rm d}t\right)^{\frac{2}{2+\vartheta}} \nonumber\\ &\leq& C\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|f|^{2+\vartheta}{\rm d}x{\rm d}t\right)^{\frac{2}{2+\vartheta}} +C\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|\hat{u}|^{2+\vartheta}{\rm d}x{\rm d}t\right)^{\frac{2}{2+\vartheta}}\nonumber\\ &\leq& C\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{2}{\eta}} \\\ && +C\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|\hat{u}_{t}|^{2}{\rm d}x{\rm d}t +\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|D^{2}\hat{u}|^{2}{\rm d}x{\rm d}t\right)\nonumber\\ &=& C\bigg(\left(\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{2}{\eta}} \\\ && +\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|u_{t}|^{2}{\rm d}x{\rm d}t+\mathit{{\rlap{-} \smallint }}_{Q_{32\bar{r}}(\bar{z})}|D^{2}u|^{2}{\rm d}x{\rm d}t\bigg)\nonumber\\ &\leq& C\lambda^{2}, \end{eqnarray}$

这里根据引理2.5的条件取$0<\vartheta<\min\{\eta-2, 4/n\}$.联合(3.17)式、(3.20)式和(3.21)式,有

$\frac{1}{12}\lambda^{2}\leq C\left(\frac{|Q\cap E(\frac{1}{6}\lambda, Q_{2R})|}{|Q|}\right)^{\frac{\vartheta}{2+\vartheta}}\lambda^{2}.$

两边同时除以$\lambda^{2}$并用$Q_{\bar{r}}(\bar{z})$替换$Q$得

(3.22) $ \begin{equation}\label{ball estimate-1} |Q_{\bar{r}}(\bar{z})|\leq C|Q_{\bar{r}}(\bar{z})\cap E(\frac{1}{6}\lambda, Q_{2R})|.\end{equation} $

情形3  假设(3.18)式成立.由Fubini定理得

$\begin{eqnarray*} \mathit{{\rlap{-} \smallint }}_{Q}|f|^{\eta}{\rm d}x{\rm d}t&=& \frac{\eta}{|Q|}\int^{\infty}_{0}\mu^{\eta}\Big|\Big\{z\in Q:|f(z)|>\mu\Big\}\Big|\frac{{\rm d}\mu}{\mu}\nonumber\\ &=& \frac{\eta}{|Q|}\int^{\tau\lambda}_{0}\mu^{\eta}\Big|\Big\{z\in Q:|f(z)|>\mu\Big\}\Big|\frac{{\rm d}\mu}{\mu} + \frac{\eta}{|Q|}\int^{\infty}_{\tau\lambda}\mu^{\eta}\Big|\Big\{z\in Q:|f(z)|>\mu\Big\}\Big|\frac{{\rm d}\mu}{\mu}\nonumber\\ &\leq& (\tau\lambda)^{\eta}+ \frac{\eta}{|Q|}\int^{\infty}_{\tau\lambda}\mu^{\eta}\Big|\Big\{z\in Q:|f(z)|>\mu\Big\}\Big|\frac{{\rm d}\mu}{\mu}.\end{eqnarray*} $

若取$\tau=\frac{1}{6M^{1/2}}$,则

$\frac{2^\eta-1}{6^\eta}\lambda^{\eta}\leq \frac{M^\frac{\eta}{2}\eta}{|Q|}\int^{\infty}_{\tau\lambda}\mu^{\eta}\Big|\Big\{z\in Q:|f(z)|>\mu\Big\}\Big|\frac{{\rm d}\mu}{\mu}.$

两边同时除以$\lambda^{\eta}$并用$Q_{\bar{r}}(\bar{z})$替换$Q$得

(3.23) $ \begin{equation}\label{ball estimate-2} |Q_{\bar{r}}(\bar{z})|\leq \frac{6^\eta M^{\frac{\eta}{2}}\eta}{(2^\eta-1)\lambda^{\eta}}\int^{\infty}_{\tau\lambda}\mu^{\eta}\Big|\Big\{z\in Q_{\bar{r}}(\bar{z}):|f(z)|>\mu\Big\}\Big|\frac{{\rm d}\mu}{\mu}.\end{equation}$

现联合(3.22)式和(3.23)式,得

(3.24) $\begin{equation}\label{estimate to special balls} |Q_{\bar{r}}(\bar{z})|\leq C|Q_{\bar{r}}(\bar{z})\cap E(\frac{1}{6}\lambda, Q_{2R})|+\frac{6^\eta M^{\frac{\eta}{2}}\eta}{(2^\eta-1)\lambda^{\eta}}\int^{\infty}_{\tau\lambda}\mu^{\eta}\Big|\Big\{z\in Q_{\bar{r}}(\bar{z}):|f(z)|>\mu\Big\}\Big|\frac{{\rm d}\mu}{\mu}.\end{equation}$

第四步  估计水平集测度$|E(\lambda, Q_{R})|$.考虑水平集合$E(A\lambda, Q_{R})$,其中$A\geq 1$是待定常数.易见$E(A\lambda, Q_{R})\subset E(\lambda, Q_{R})$,即,若$\bar{z}\in E(A\lambda, Q_{R})$,则$|D^{2}u(\bar{z})|>A\lambda$且$\bar{z}\in E(\lambda, Q_{R})$.设$Q_{\bar{r}}(\bar{z})$是第二步中定义的柱体,满足$CZ(Q_{\bar{r}}(\bar{z}))=\lambda$和(3.5)式.设$w, v$是第三步中定义在柱体$Q_{16\bar{r}}(\bar{z})$上的函数,分别满足(3.12式和(3.15)式.其中(3.15)式说明$D^{2}v$在$Q_{16\bar{r}}(\bar{z})$上是局部有界的.由此可知,若取

(3.25) $\begin{equation}\label{step 4 -1} A=A(n, \lambda, \Lambda, R), \quad A:= \max\{2C_{\ast}, N_{1}\}\geq 1, \quad N_{1}\, \mbox{是}\, \mbox{待}\, \mbox{定}\, \mbox{常}\, \mbox{数}, \end{equation}$

则$\left\{z\in Q_{4\bar{r}}(\bar{z}): |D^{2}v|>\frac{A\lambda}{2}\right\}=\emptyset$.依次利用$D^2u=D^2\hat{u}$, $u=w+v$、Hardy-Littlewood极大值函数的弱(1, 1)不等式和(3.12)式,有

(3.26) $\begin{eqnarray}\label{estimate to special level set} &&\Big|\Big\{z\in Q_{4\bar{r}}(\bar{z}): |D^{2}u|>A\lambda\Big\}\Big|\nonumber\\ &\leq& \left|\left\{z\in Q_{4\bar{r}}(\bar{z}): |D^{2}w|>2^{-1}A\lambda\right\}\right| +\left|\left\{z\in Q_{4\bar{r}}(\bar{z}): |D^{2}v|>2^{-1}A\lambda\right\}\right|\nonumber\\ &=& \left|\left\{z\in Q_{4\bar{r}}(\bar{z}): |D^{2}w|^{2}>\left(2^{-1}A\lambda\right)^{2}\right\}\right|\nonumber\\ &\leq& \frac{C}{(A\lambda)^{2}}\int_{Q_{4\bar{r}}(\bar{z})}|D^{2}w|^{2}{\rm d}x{\rm d}t\nonumber\\ &\leq& \frac{C|Q_{8\bar{r}}(\bar{z})|}{(A\lambda)^{2}}\left(\frac{\delta^{\frac{1}{\pi}}+1}{M}+1\right)\lambda^{2}\nonumber\\ &\leq& C\left(\frac{\delta^{\frac{1}{\pi}}+1}{MN_{1}^{2}}+\frac{1}{N_{1}^{2}}\right)|Q_{\bar{r}}(\bar{z})|.\end{eqnarray}$

将第三步中$|Q_{\bar{r}}(\bar{z})|$的估计(3.24)式代入到(3.26)式中得

$\begin{eqnarray*} && \Big|\Big\{z\in Q_{4\bar{r}}(\bar{z}): |D^{2}u(z)|>A \lambda\Big\}\Big|\nonumber\\ &\leq& CG(M, N_{1}, \delta)\left(\Big|Q_{\bar{r}}(\bar{z}) \cap E(\frac{1}{6}\lambda, Q_{2R})\Big|+\frac{\eta}{(\tau\lambda) ^{\eta}}\int^{\infty}_{\tau\lambda}\mu^{\eta}\Big|\Big\{z\in Q_{\bar{r}}(\bar{z}):|f(z)|>\mu\Big\}\Big|\frac{{\rm d}\mu}{\mu}\right), \end{eqnarray*}$

其中

(3.27) $\begin{equation}\label{G definition} G(M, N_{1}, \delta)=(\delta^{\frac{1}{\pi}}+1)(M N_{1}^{2})^{-1}+(N_{1}^{2})^{-1}.\end{equation}$

再结合(3.7)式有

(3.28) $\begin{eqnarray}\label{estimate to general level set}& &\Big|\Big\{z\in Q_{R}: |D^{2}u(z)|>A\lambda\Big\}\Big|\nonumber\\ & \leq& CG(M, N_{1}, \delta)\bigg(\sum\limits_{i\in {\Bbb N}}\Big| Q_{\bar{r}_{i}}(\bar{z}_{i}) \cap E(\frac{1}{6}\lambda, Q_{2R})\Big| \\ &&+\sum\limits_{i\in {\Bbb N}}\frac{\eta}{(\tau\lambda)^{\eta}}\int^{\infty} _{\tau\lambda}\mu^{\eta}\Big|\Big\{z\in Q_{\bar{r}_{i}}(\bar{z}_{i}) :|f(z)|>\mu\Big\}\Big|\frac{{\rm d}\mu}{\mu}\bigg)\nonumber\\ &\leq& CG(M, N_{1}, \delta)\left(\Big| E(\frac{1}{6}\lambda, Q_{2R}) \Big|+\frac{\eta}{(\tau\lambda)^{\eta}}\int^{\infty}_{\tau\lambda} \mu^{\eta}\Big|\Big\{z\in Q_{2R}:|f(z)|>\mu\Big\}\Big|\frac{{\rm d} \mu}{\mu}\right).\end{eqnarray}$

第五步  $q<\infty$情形下证明Lorentz估计式(1.3).由(3.19)式可知只需估计$\|D^{2}u\|_{L^{\gamma, q}(Q_{R})}$.通过简单计算可得

(3.29) $\begin{eqnarray}\label{Lorentz estimate-case 1-1} \|D^{2}u\|^{q}_{L^{\gamma, q}(Q_{R})}\leq C(B\lambda_{0})^{q}|Q_{2R}|^{\frac{q}{\gamma}}+q\int^{\infty}_{B\lambda_{0}}\left(\lambda^{\gamma}\Big|\Big\{z\in Q_{R}:|D^{2}u(z)|>\lambda\Big\}\Big|\right)^{\frac{q}{\gamma}}\frac{{\rm d}\lambda}{\lambda}.\end{eqnarray}$

将(3.28)式代入上式第二项中得

(3.30) $\begin{eqnarray}\label{Lorentz estimate J1-J2} &&q\int^{\infty}_{B\lambda_{0}}\left(\lambda^{\gamma}\Big|\Big\{z\in Q_{R}:|D^{2}u(z)|>\lambda\Big\}\Big|\right)^{\frac{q}{\gamma}}\frac{{\rm d}\lambda}{\lambda}\nonumber\\ &=& qA^{q}\int^{\infty}_{AB\lambda_{0}}\left(\lambda^{\gamma}\Big|\Big\{z\in Q_{R}:|D^{2}u(z)|>A\lambda\Big\}\Big|\right)^{\frac{q}{\gamma}}\frac{{\rm d}\lambda}{\lambda}\nonumber\\ &\leq& C[G(M, N_{1}, \delta)]^{\frac{q}{\gamma}}\int^{\infty}_{0} \left(\lambda^{\gamma}\Big|\Big\{z\in Q_{2R}:|D^{2}u(z)| >\frac{1}{6}\lambda\Big\}\Big|\right)^{\frac{q}{\gamma}}\frac{{\rm d}\lambda}{\lambda}\nonumber\\ &&+ C[G(M, N_{1}, \delta)]^{\frac{q}{\gamma}}\int^{\infty}_{0} \lambda^{q(1-\frac{\eta}{\gamma})}\left(\int^{\infty}_{ \tau\lambda}\mu^{\eta}\Big|\Big\{z\in Q_{2R}:|f(z)|>\mu\Big\} \Big|\frac{{\rm d}\mu}{\mu}\right)^{\frac{q}{\gamma}} \frac{{\rm d}\lambda}{\lambda}\nonumber\\ &:=& C[G(M, N_{1}, \delta)]^{\frac{q}{\gamma}}(J_{1}+J_{2}), \end{eqnarray}$

其中$C$仅依赖于$n, \nu, L, \gamma, q, M, R$.

对于$J_{1}$,通过变量替换得

$\begin{eqnarray*} J_{1}=6^{q}\int^{\infty}_{0}\left(\lambda^{\gamma}\Big| \Big\{z\in Q_{2R}:D^{2}u(z)>\lambda\Big\}\Big|\right) ^{\frac{q}{\gamma}}\frac{{\rm d}\lambda}{\lambda}=C(q)\|D^{2}u\|^{q} _{L^{\gamma, q}(Q_{2R})}.\end{eqnarray*}$

对于$J_{2}$,作变量替换$\tau\lambda\rightarrow \theta$,并将第三步中$\tau$的取值代入得

$J_{2}=C\int^{\infty}_{0}\theta^{q(1-\frac{\eta}{\gamma})}\left(\int^{\infty}_{\theta}\mu^{\eta}\Big|\Big\{z\in Q_{2R}:|f(z)|>\mu\Big\}\Big|\frac{{\rm d}\mu}{\mu}\right)^{\frac{q}{\gamma}}\frac{{\rm d}\theta}{\theta}.$

下面分$q\geq \gamma$和$0<q<\gamma$两种情况讨论.当$q\geq \gamma$时,令引理2.7中

$ f(\mu)=\mu^{\eta-1}\Big|\Big\{z\in Q_{2R}:|f(z)|>\mu\Big\}\Big|, \quad \alpha=\frac{q}{\gamma}\geq 1, \quad r=q(1-\frac{\eta}{\gamma})>0, $

则

$ \begin{eqnarray*} J_{2}&\leq& C(\gamma, q, M)\int^{\infty}_{0}\theta^{q(1-\frac{\eta}{\gamma})+\eta\frac{q}{\gamma}}\Big|\Big\{z\in Q_{2R}:|f(z)|>\theta\Big\}\Big|^{\frac{q}{\gamma}}\frac{{\rm d}\theta}{\theta}\nonumber\\ &=& C(\gamma, q, M)\|f\|^{q}_{L^{\gamma, q}(Q_{2R})}.\end{eqnarray*}$

当$0<q<\gamma$时,令引理2.8中

$ h(\mu)=|\{z\in Q_{2R}:|f(z)|>\mu\}|^{\frac{q}{\gamma}}\geq 0, \quad \alpha_{1}=1<\frac{\gamma}{q}=\alpha_{2}, \quad r=\frac{\eta q}{\gamma}, \quad \varepsilon=1, $

则

$\begin{eqnarray*} &&\left(\int^{\infty}_{\tau\lambda}\mu^{\eta}\Big|\Big\{z\in Q_{2R}:|f(z)|>\mu\Big\}\Big|\frac{{\rm d}\mu}{\mu}\right)^{\frac{q}{\gamma}}\nonumber\\ &\leq& (\tau\lambda)^{\frac{\eta q}{\gamma}}\Big|\Big\{z\in Q_{2R}:|f(z)|>\tau\lambda\Big\}\Big|^{\frac{q}{\gamma}}+C\int^{\infty}_{\tau\lambda}\mu^{\frac{\eta q}{\gamma}}\Big|\Big\{z\in Q_{2R}:|f(z)|>\mu\Big\}\Big|^{\frac{q}{\gamma}}\frac{{\rm d}\mu}{\mu}.\end{eqnarray*}$

结合Fubini定理,有

$\begin{eqnarray*} J_{2}&\leq& C\int^{\infty}_{0}\theta^{q(1-\frac{\eta}{\gamma})}\left(\theta^{\frac{\eta q}{\gamma}}\Big|\Big\{z\in Q_{2R}:|f(z)|>\theta\Big\}\Big|^{\frac{q}{\gamma}}\right)\frac{{\rm d}\theta}{\theta}\nonumber\\ &&+ C\int^{\infty}_{0}\theta^{q(1-\frac{\eta}{\gamma})}\left(\int^{\infty}_{\theta}\mu^{\frac{\eta q}{\gamma}}\Big|\Big\{z\in Q_{2R}:|f(z)|>\mu\Big\}\Big|^{\frac{q}{\gamma}}\frac{{\rm d}\mu}{\mu}\right)\frac{{\rm d}\theta}{\theta}\nonumber\\ &\leq& C\int^{\infty}_{0}\theta^{q}\Big|\Big\{z\in Q_{2R}:|f(z)|>\mu\Big\}\Big|^{\frac{q}{\gamma}}\frac{{\rm d}\theta}{\theta}\nonumber\\ &&+ C\int^{\infty}_{0}\mu^{\frac{\eta q}{\gamma}-1}\Big|\Big\{z\in Q_{2R}:|f(z)|>\mu\Big\}\Big|^{\frac{q}{\gamma}}\left(\int^{\mu}_{0}\theta^{q(1-\frac{\eta}{\gamma})}\frac{{\rm d}\theta}{\theta}\right){\rm d}\mu\nonumber\\ &\leq& C\|f\|^{q}_{L^{\gamma, q}(Q_{2R})}+C\int^{\infty}_{0}\mu^{q-1}\Big|\Big\{z\in Q_{2R}:|f(z)|>\mu\Big\}\Big|^{\frac{q}{\gamma}}{\rm d}\mu\nonumber\\ &\leq& C\|f\|^{q}_{L^{\gamma, q}(Q_{2R})}.\end{eqnarray*} $

综合两种情况可知,对$0<q<\infty$,总有

$ J_{2}\leq C(\gamma, q, M)\|f\|^{q}_{L^{\gamma, q}(Q_{2R})}.$

现将$J_{1}$和$J_{2}$的估计式代入(3.30)式中,并联合(3.29)式可知存在常数$C_{\ast\ast}=C_{\ast\ast}(n, \lambda, \Lambda, \gamma, $$q, M, R)$使得对任意$\gamma>2$和$0<q<\infty$有

$ \begin{eqnarray*} & &\|u_{t}\|_{L^{\gamma, q}(Q_{R})}+\|D^{2}u\|_{L^{\gamma, q}(Q_{R})}\\ &\leq & CB\lambda_{0}|Q_{2R}|^{\frac{1}{\gamma}}+C_{\ast\ast}[G(M, N_{1}, \delta)]^{\frac{1}{\gamma}}\left(\|D^{2}u\|_{L^{\gamma, q}(Q_{2R})}+\|f\|_{L^{\gamma, q}(Q_{2R})}\right), \end{eqnarray*}$

其中$G(M, N_{1}, \delta)$的定义式为(3.27)式.现在先取一个充分小的数$\delta_{0}$,再取充分大的数$M$和$N_{1}$使得对任意$\delta\in (0, \delta_{0}]$都有$C_{\ast\ast}[G(M, N_{1}, \delta)]^{\frac{1}{\gamma}}\leq \frac{1}{2}$,则

$\begin{eqnarray*} &&\|u_{t}\|_{L^{\gamma, q}(Q_{R})}+\|D^{2}u\|_{L^{\gamma, q}(Q_{R})}\\ &\leq & \frac{1}{2}\left(\|u_{t}\|_{L^{\gamma, q}(Q_{2R})}+\|D^{2}u\|_{L^{\gamma, q}(Q_{2R})}\right) +CB\lambda_{0}|Q_{2R}|^{\frac{1}{\gamma}}+C\|f\|_{L^{\gamma, q}(Q_{2R})}.\end{eqnarray*}$

一方面,当$\|u_{t}\|_{L^{\gamma, q}(Q_{2R})}+\|D^{2}u\|_{L^{\gamma, q}(Q_{2R})}<\infty$时,迭代可知

(3.31) $\begin{equation}\label{1} \|u_{t}\|_{L^{\gamma, q}(Q_{R})}+\|D^{2}u\|_{L^{\gamma, q}(Q_{R})}\leq CB\lambda_{0}|Q_{2R}|^{\frac{1}{\gamma}}+C\|f\|_{L^{\gamma, q}(Q_{2R})}.\end{equation}$

将第一步中$B$和$\lambda_{0}$的定义代入得

$\begin{eqnarray*} &&\|u_{t}\|_{L^{\gamma, q}(Q_{R})}+\|D^{2}u\|_{L^{\gamma, q}(Q_{R})}\nonumber\\ &\leq& C|Q_{2R}|^{\frac{1}{\gamma}}\left[\left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|u_{t}|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}+\left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}} +M^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{1}{\eta}}\right]\nonumber\\ &&+C\|f\|_{L^{\gamma, q}(Q_{2R})}.\end{eqnarray*} $

利用引理2.6和(2.14)式,有

(3.32) $\begin{eqnarray}\label{f estiamte} \left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{1}{\eta}} &\leq &\left[\frac{1}{|Q_{2R}|}\left(\frac{\gamma}{\gamma-\eta}|Q_{2R}|^{\frac{\gamma-\eta}{\gamma}}\|f\|^{\eta}_{{\cal M}^{\gamma}(Q_{2R})}\right)\right]^{\frac{1}{\eta}}\nonumber\\ &\leq& \frac{C(\eta, \gamma, q)}{(\gamma-\eta)^{1/\eta}}|Q_{2R}|^{-\frac{1}{\gamma}}\|f\|_{L^{\gamma, q}(Q_{2R})}, \end{eqnarray}$

因此, (3.31)式可写为

(3.33) $ \begin{eqnarray}\label{under bdd condi finial estimate} &&\|u_{t}\|_{L^{\gamma, q}(Q_{R})}+\|D^{2}u\|_{L^{\gamma, q}(Q_{R})}\\\ &\leq & C|Q_{2R}|^{\frac{1}{\gamma}-\frac{1}{2}}\left(\|u_{t}\|_{L^{2}(Q_{2R})}+\|D^{2}u\|_{L^{2}(Q_{2R})}\right)+ C\|f\|_{L^{\gamma, q}(Q_{2R})}.\end{eqnarray} $

另一方面,当$\|u_{t}\|_{L^{\gamma, q}(Q_{2R})}+\|D^{2}u\|_{L^{\gamma, q}(Q_{2R})}=\infty$时,对任意$z\in \Omega_{T}$和$k\in {\Bbb N}\cap [B\lambda_{0}, \infty)$,定义两个截断函数

$ |u_{t}(z)|_{k}:=\min\{k, u_{t}(z)\}, \quad |D^{2}u(z)|_{k}:=\min\{k, D^{2}u(z)\}.$

显然,有$\||u_{t}|_{k}\|_{L^{\gamma, q}(Q_{2R})}+\||D^{2}u|_{k}\|_{L^{\gamma, q}(Q_{2R})}<\infty$成立.同时定义水平集合

$E_{k}(\lambda, Q_{R}):=\{z\in Q_{R}: |D^{2}u|_{k}>\lambda\},$

下面证明对$k\in {\Bbb N}\cap [B\lambda_{0}, \infty)$有

(3.34) $\begin{eqnarray}\label{estimate to general level set k} &&|E_{k}(A\lambda, Q_{R})|\\ &\leq &CG(M, N_{1}, \delta)\left(| E_{k}(\frac{1}{6}\lambda, Q_{2R})|+\frac{\eta}{(\tau\lambda)^{\eta}}\int^{\infty}_{\tau\lambda}\mu^{\eta}\Big|\Big\{z\in Q_{2R}:|f(z)|>\mu\Big\}\Big|\frac{{\rm d}\mu}{\mu}\right).\end{eqnarray}$

事实上,若$k\leq A\lambda$,则由$E_{k}(A\lambda, Q_{R})=\varnothing$很容易证明(3.34)式成立.若$k> A\lambda$,则利用(3.28)式和

$E_{k}(A\lambda, Q_{R})= E(A\lambda, Q_{R})=\Big\{z\in Q_{R}:|D^{2}u(z)|>A\lambda\Big\}, $

$E_{k}(6^{-1}\lambda, Q_{2R})=E(6^{-1}\lambda, Q_{2R})$

能推出(3.34)式.因此,通过类似第一步到第五步的证明知道用$|u_{t}|_{k}$和$|D^{2}u|_{k}$分别替换$u_{t}$和$D^{2}u$后的(3.33)式仍然成立.现在取极限$k\rightarrow \infty$并应用Lorentz拟范数的下半连续性得

$\begin{eqnarray*}\|u_{t}\|_{L^{\gamma, q}(Q_{R})}+\|D^{2}u\|_{L^{\gamma, q}(Q_{R})} &\leq& \liminf\limits_{k\rightarrow \infty} \left(\||u_{t}|_{k}\|_{L^{\gamma, q}(Q_{R})}+ \||D^{2}u|_{k}\|_{L^{\gamma, q}(Q_{R})}\right)\nonumber\\ &\leq& C|Q_{2R}|^{\frac{1}{\gamma}-\frac{1}{2}}\left(\|u_{t}\|_{L^{2} (Q_{2R})}+\|D^{2}u\|_{L^{2}(Q_{2R})}\right)+ C\|f\|_{L^{\gamma, q}(Q_{2R})}, \end{eqnarray*}$

即为所证.

第六步  $q=\infty$情形下证明Lorentz估计式(1.3).设$\sigma$是待定正常数.则由

$Q=\{z\in Q: (|f(z)|+1)\leq \sigma\lambda\}\cup\{z\in Q: (|f(z)|+1)>\sigma\lambda\} $

和(3.17)式可知

(3.35) $\begin{equation}\label{case q infty-1} \left(\frac{\lambda}{3}\right)^{\eta}M^{-\frac{\eta}{2}}\leq \mathit{{\rlap{-} \smallint }}_{Q}|f|^{\eta}{\rm d}x{\rm d}t\leq (\sigma \lambda)^{\eta}+\frac{1}{|Q|}\int_{\{z\in Q:|f(z)|>\sigma\lambda\}}|f|^{\eta}{\rm d}x{\rm d}t.\end{equation} $

记$F(\sigma\lambda, Q)=\{z\in Q:|f(z)|>\sigma\lambda\}$,利用引理2.6得

(3.36) $\begin{eqnarray}\label{case q infty-2} \int_{F(\sigma\lambda, Q)}|f|^{\eta}{\rm d}x{\rm d}t &\leq& \frac{\gamma}{\gamma-\eta}|F(\sigma\lambda, Q)|^{\frac{\gamma-\eta}{\gamma}}\|f\|^{\eta}_{{\cal M}^{\gamma}(F(\sigma\lambda, Q))}\nonumber\\ &=& \frac{\gamma}{\gamma-\eta}|F(\sigma\lambda, Q)|^{\frac{\gamma-\eta}{\gamma}}\sup\limits_{\mu>0}\mu^{\eta}\Big|\Big\{z\in F(\sigma\lambda, Q):|f(z)|>\mu\Big\}\Big|^{\frac{\eta}{\gamma}}\nonumber\\ &\leq& \frac{\gamma}{\gamma-\eta}|F(\sigma\lambda, Q)| ^{\frac{\gamma-\eta}{\gamma}}\sup\limits_{\mu\leq\sigma\lambda}\mu^{\eta}\Big|\Big\{z\in F(\sigma\lambda, Q):|f(z)|>\mu\Big\}\Big|^{\frac{\eta}{\gamma}}\nonumber\\ &&+ \frac{\gamma}{\gamma-\eta}|F(\sigma\lambda, Q)|^{\frac{\gamma-\eta}{\gamma}}\sup\limits_{\mu>\sigma\lambda}\mu^{\eta}\Big|\Big\{z\in F(\sigma\lambda, Q):|f(z)|>\mu\Big\}\Big|^{\frac{\eta}{\gamma}}\nonumber\\ &\leq& \frac{\gamma}{\gamma-\eta}(\sigma\lambda)^{\eta}|F(\sigma\lambda, Q)|+ \frac{\gamma}{\gamma-\eta}|F(\sigma\lambda, Q)|^{\frac{\gamma-\eta}{\gamma}} \sup\limits_{\mu>\sigma\lambda}\mu^{\eta}| F(\mu, Q)|^{\frac{\eta}{\gamma}}.\end{eqnarray}$

联合(3.35)式和(3.36)式,有

$\begin{eqnarray*}\left(\frac{\lambda}{3}\right)^{\eta}M^{-\frac{\eta}{2}}-(\sigma\lambda)^{\eta}-\frac{\gamma}{\gamma-\eta}\frac{|F(\sigma\lambda, Q)|}{|Q|}(\sigma\lambda)^{\eta} &\leq& \frac{\gamma}{\gamma-\eta}\frac{|F(\sigma\lambda, Q)|^{\frac{\gamma-\eta}{\gamma}}}{|Q|}\sup\limits_{\mu>\sigma\lambda}\mu^{\eta}| F(\mu, Q)|^{\frac{\eta}{\gamma}}.\end{eqnarray*}$

取适当的$\sigma=\frac{C(n, \gamma)}{M^{1/2}}$使得

$ \left(\frac{\lambda}{3}\right)^{\eta}M^{-\frac{\eta}{2}}-(\sigma \lambda)^{\eta}-\frac{\gamma}{\gamma-\eta}\frac{|F(\sigma \lambda, Q)|}{|Q|}(\sigma \lambda)^{\eta}>\frac{1}{10}(\sigma\lambda)^{\eta}, $

则

$\begin{eqnarray*} |Q_{\bar{r}}(\bar{z})|=|Q|&\leq& \frac{10\gamma}{\gamma-\eta}\frac{|F(\sigma \lambda, Q)|^{\frac{\gamma-\eta}{\gamma}}}{(\sigma \lambda)^{\eta}}\sup\limits_{\mu>\sigma \lambda}\mu^{\eta}| F(\mu, Q)|^{\frac{\eta}{\gamma}}\nonumber\\ &\leq& \frac{10\gamma}{\gamma-\eta}\frac{|F(\sigma \lambda, Q)|^{\frac{\gamma-\eta}{\gamma}}}{(\sigma \lambda)^{\eta}} \sup\limits_{\mu>\sigma \lambda}\left(\mu^{\eta-\gamma}| F(\mu, Q)|^{\frac{\eta}{\gamma}-1}\right)\left(\mu^{\gamma}| F(\mu, Q)|\right)\nonumber\\ &\leq& \frac{10\gamma}{\gamma-\eta}(\sigma \lambda)^{-\gamma}\sup\limits_{\mu>\sigma \lambda}\mu^{\gamma}| F(\mu, Q)|.\end{eqnarray*}$

因此,不同于估计式(3.24),在$q=\infty$的情形下有

(3.37) $\begin{equation}\label{new estimate to special balls} |Q_{\bar{r}}(\bar{z})|\leq C|Q_{\bar{r}}(\bar{z})\cap E(\frac{1}{6}\lambda, Q_{2R})|+\frac{10\gamma}{\gamma-\eta}(\sigma \lambda)^{-\gamma}\sup\limits_{\mu>\sigma \lambda}\mu^{\gamma}\Big| \Big\{z\in Q_{\bar{r}}(\bar{z}):|f(z)|>\mu \Big\}\Big|.\end{equation}$

现在用证明(3.28)式的方法可推导出

$\begin{eqnarray*}\label{estimate to general level set for case q2} &&\Big|\Big\{z\in Q_{R}: |D^{2}u(z)|>A\lambda\Big\}\Big| \nonumber\\ &\leq& CG(M, N_{1}, \delta)\left(| E(\frac{1}{6}\lambda, Q_{2R})|+\frac{10\gamma}{\gamma-\eta}(\sigma \lambda)^{-\gamma}\sup\limits_{\mu>\sigma \lambda}\mu^{\gamma}\Big|\Big\{ z\in Q_{2R}:|f(z)|>\mu\Big\}\Big|\right), \end{eqnarray*}$

其中$G(M, N_{1}, \delta)$定义为(3.27)式.先将上式两端同时乘$(A\lambda)^{\gamma}$,再对$\lambda\in (B\lambda_{0}, \infty)$取上确界得

$\begin{eqnarray*} &&\sup\limits_{\lambda>B\lambda_{0}}(A\lambda)^{\gamma}\Big|\Big\{z\in Q_{R}: |D^{2}u(z)|>A\lambda\Big\}\Big|\nonumber\\ &\leq& CG(M, N_{1}, \delta)\left(\sup\limits_{\lambda>B\lambda_{0}}(A\lambda)^{\gamma}| E(\frac{1}{6}\lambda, Q_{2R})|+\sup\limits_{\lambda>\sigma B\lambda_{0}}\sup\limits_{\mu>\lambda}\mu^{\gamma}\Big| \Big\{z\in Q_{2R}:|f(z)|>\mu\Big\}\Big|\right)\nonumber\\ &\leq& CG(M, N_{1}, \delta)\left(\|D^{2}u\|^{\gamma}_{{\cal M}^{\gamma}(Q_{2R})}+\|f\|^{\gamma}_{{\cal M}^{\gamma}(Q_{2R})}\right).\end{eqnarray*}$

利用算子${\cal M}$和$\lambda_{0}$的定义可知

$\begin{eqnarray*}&& \|D^{2}u\|_{{\cal M}^{\gamma}(Q_{R})}\\ &\leq & C_{\ast\ast\ast}[G(M, N_{1}, \delta)]^{\frac{1}{\gamma}}\left(\|D^{2}u\|_{{\cal M}^{\gamma}(Q_{2R})}+\|f\|_{{\cal M}^{\gamma}(Q_{2R})}\right) +C|Q_{2R}|^{\frac{1}{\gamma}}B\lambda_{0}\nonumber\\ &\leq& 2^{-1}\|D^{2}u\|_{{\cal M}^{\gamma}(Q_{2R})}+C\|f\|_{{\cal M}^{\gamma}(Q_{2R})}\nonumber\\ &&+ C|Q_{2R}|^{\frac{1}{\gamma}}\left[\left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|u_{t}|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}+\left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}+M^{\frac{1}{2}}\left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{1}{\eta}}\right].\end{eqnarray*}$

这里取$\delta$充分小, $M$和$N_{1}$充分大,使得$C_{\ast\ast\ast}[G(M, N_{1}, \delta)]^{\frac{1}{\gamma}}\leq 2^{-1}$.又因为由引理2.6可得

$\left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|f|^{\eta}{\rm d}x{\rm d}t\right)^{\frac{1}{\eta}} \leq \frac{C(\eta, \gamma)}{(\gamma-\eta)^{1/\eta}}|Q_{2R}|^{-\frac{1}{\gamma}}\|f\|_{{\cal M}^{\gamma}(Q_{2R})}, $

所以

$ \begin{eqnarray*} \|D^{2}u\|_{{\cal M}^{\gamma}(Q_{R})} &\leq& 2^{-1}\|D^{2}u\|_{{\cal M}^{\gamma}(Q_{2R})}\\&&+C|Q_{2R}|^{\frac{1}{\gamma}}\Bigg[\left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|u_{t}|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}+ \left(\mathit{{\rlap{-} \smallint }}_{Q_{2R}}|D^{2}u|^{2}{\rm d}x{\rm d}t\right)^{\frac{1}{2}}\Bigg]+C\|f\|_{{\cal M}^{\gamma}(Q_{2R})}. \end{eqnarray*} $

利用第五步中截断函数方法,可以证明$q=\infty$情形下(1.3)式也成立.定理1.1得证.