数学物理学报, 2026, 46(6): 2225-2238

非齐次椭圆障碍问题弱解梯度的局部 Besov 估计

要佳慧, 佟玉霞,*

华北理工大学理学院 河北唐山 063210

Local Besov Estimates for the Gradient of Weak Solutions to Nonhomogeneous Elliptic Obstacle Problems

Yao Jiahui, Tong Yuxia,*

College of Science, North China University of Science and Technology, Hebei Tangshan 063210

通讯作者: 佟玉霞, E-mail: tongyuxia@126.com

收稿日期: 2025-05-13   修回日期: 2025-10-17  

基金资助: 华北理工大学研究生创新项目(2026S28)

Received: 2025-05-13   Revised: 2025-10-17  

Fund supported: Graduate Student Innovation Fund of North China University of Science and Technology(2026S28)

摘要

该文研究了与非齐次椭圆方程相对应的障碍问题, 通过建立适当的容许函数, 并利用变分不等式、$N$-函数的 Young 不等式和有限差分法等技巧, 建立了椭圆障碍问题弱解梯度在有界开区域上的局部 Besov 估计.

关键词: 障碍问题; 弱解; Besov 估计; 有限差分

Abstract

This paper studies the obstacle problem corresponding to non-homogeneous elliptic equations. By establishing appropriate admissible functions and using techniques such as variational inequalities, Young's inequality for $N$-functions, and the finite difference method, a local Besov estimate for the gradient of the weak solution of the elliptic obstacle problem in a bounded open domain is established.

Keywords: obstacle problem; weak solution; Besov estimate; finite difference

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本文引用格式

要佳慧, 佟玉霞. 非齐次椭圆障碍问题弱解梯度的局部 Besov 估计[J]. 数学物理学报, 2026, 46(6): 2225-2238

Yao Jiahui, Tong Yuxia. Local Besov Estimates for the Gradient of Weak Solutions to Nonhomogeneous Elliptic Obstacle Problems[J]. Acta Mathematica Scientia, 2026, 46(6): 2225-2238

1 引言

$\Omega$$\mathbb{R}^{n}$ 中的有界开区域, $n\geq 2.$$\Omega$ 上考虑方程

$\begin{equation} \textrm{div}~\big({a}(|\nabla{u}|)\nabla{u}\big)=f(x), \end{equation}$

其中函数 $a:(0,\infty)\rightarrow(0,\infty)\in C^{1}(0,\infty)$ 满足

$\begin{equation} 0\leq i_{a}=:\inf_{t>0}\frac{ta^{\prime}(t)}{a(t)}\leq\sup_{t>0}\frac{ta^{\prime}(t)}{a(t)}=:s_{a}<\infty. \end{equation}$

2011 年, Cianchi 和 Maz'ya[1] 证明了方程 (1.1) 在满足条件 (1.2) 情况下的 Dirichlet 和 Neumann 边值问题弱解的全局 Lipschitz 正则性. 2014 年, Cianchi 和 Maz'ya[2] 研究了方程 (1.1), 在满足条件 (1.2) 的情况下, 获得了方程的 Dirichlet 和 Neumann 边值问题弱解的梯度估计. 2017 年, Yao 和 Zhou[3] 在有界区域 $\Omega$ 上研究了拟线性椭圆方程

$\textrm{div}~\big({a}(|\nabla{u}|)\nabla{u}\big)=\textrm{div}~\big({a}(|\textbf{f}|)\textbf{f}\big)$

的局部 Calderón-Zygmund 估计. 2020 年, Yao, Zhang 和 Zhou[4] 在全空间 $\mathbb{R}^{n}$ 上研究了椭圆方程

$\begin{equation} \textrm{div}~\big({a}(|\nabla{u}|)\nabla{u}\big)=\textrm{div}~\textbf{f}, \end{equation}$

使用 Hardy-Littlewood 极大函数获得了该方程弱解的全局 {BMO} 估计. 2021 年, Ma 和 Yao[5] 在有界区域 $\Omega$ 上研究了椭圆方程 (1.3), 获得了该方程弱解的局部 Besov 估计. 关于椭圆方程的更多研究参见文献[6,7,8].

本文在有界区域 $\Omega$ 上研究与椭圆方程 (1.3) 相对应的障碍问题, 其中 $\textbf{f}=(f^1,f^2,\ldots,f^n)$ 是给定的向量函数, 函数 $a:(0,\infty)\rightarrow(0,\infty)\in C^{1}(0,\infty)$ 满足条件 (1.2), 且在 $(0,\infty)$ 上单调递增. 特别地, 当 $a(t)=t^{p-2},$$p\geq2$ 时, 椭圆方程 (1.3) 退化为 $p$-Laplace 方程

$\textrm{div}~\big(|\nabla{u}|^{p-2}\nabla{u}\big)=\textrm{div}~\textbf{f}.$

$\forall\xi\in\mathbb{R}^n$, 作辅助函数$\textit{V}$

$\begin{equation} V(\xi)=\sqrt{a(|\xi|)}\xi. \end{equation}$

定义

$\begin{equation} b(t)=ta(t) \end{equation}$

$\begin{equation} \phi(t)=\int_{0}^{t}\tau a(\tau)\textrm{d}\tau=\int_{0}^{t}b(\tau)\textrm{d}\tau, ~~~\forall\tau\geq0. \end{equation}$

由 (1.2) 式可知 $b(t)$$[0,+\infty)$ 上连续且严格单调递增, $\phi(t)$$[0,+\infty)$ 上单调递增.

$\psi\in W^{1,\phi}(\Omega)$ 是定义在 $\Omega$ 上取值为 $\mathbb{R}\cup\{-\infty,\infty\}$ 的任意函数, 定义容许集

$\mathcal {K}_{\psi}^{\phi}(\Omega):=\left\{v\in W^{1,\phi}(\Omega):v\geq\psi~~a.e.~~ x\in\Omega\right\},$

其中函数 $\psi$ 为障碍, ${W}^{1,\phi}(\Omega)$ 为 Orlicz-Sobolev 空间, 其介绍见下节. 显然, 集合 $\mathcal {K}_{\psi}^{\phi}(\Omega)$ 非空. 下面给出本文障碍问题弱解的定义.

定义 1.1 假设 $\textbf{f}\in L_{loc}^{\tilde{\phi}}(\Omega).$ 若对于任意的 $\textit{v}$$\in\mathcal {K}_{\psi}^{\phi}(\Omega)$, 都有

$\begin{equation} \int_{\Omega}\left\langle a(|\nabla u|)\nabla u,\nabla (v-u)\right\rangle \textrm{d}x\geq\int_{\Omega}\left\langle \textbf{f},\nabla (v-u)\right\rangle \textrm{d}x, \end{equation}$

则称 $u\in\mathcal {K}_{\psi}^{\phi}(\Omega)$ 为方程 (1.3) 对应的 $\mathcal {K}_{\psi}^{\phi}(\Omega)$-障碍问题的弱解.

上述定义中的 $L_{loc}^{\tilde{\phi}}$ 为 Orlicz 空间, $\tilde{\phi}$$N$-函数 $\phi$ 的共轭函数, 其介绍见第 2 节. 障碍问题是数学研究中的一个重要领域, 其目的是在一定条件下寻找给定障碍约束的解. 它在多孔介质中的流体过滤、受约束加热、弹塑性、最优化和金融数学等领域具有广泛应用[9]. 近年来, 障碍问题受到了众多学者的关注, 关于障碍问题的研究参见文献[10,11,12,13,14,15,16,17,18,19], 其中椭圆型障碍问题参考文献[10,11,12,13,14,15], 抛物型障碍问题参考文献[16,17,18,19]. 障碍问题包括单障碍问题与双侧障碍问题, 本文主要考虑单障碍问题.

本文主要受到 Ma 和 Yao 在文献[5] 以及 Byun 和 Namkyeong 在文献[14] 中关于 Besov 估计以及障碍问题的处理思想的启发, 运用有限差分理论, 借助 Hölder 不等式和 Young 不等式, 获得了非线性椭圆障碍问题弱解梯度在有界区域 $\Omega$ 上的局部 Besov 正则性. 下面是本文的主要结论.

定理 1.1$0<\alpha<\frac{s_{a}+2}{2(s_{a}+1)}<1,$$a(t)$ 满足 (1.6) 式, $\phi(t)$ 满足 (1.6)式. 对任意 $1\leq q<\infty$$\frac{2\alpha(s_a+1)}{s_{a}+2}<\beta<1,$$\textbf{f}\in B_{2,\frac{q(i_{a}+2)}{2(i_{a}+1)},loc}^{\beta}(\Omega)\cap L_{loc}^{\tilde{\phi}}(\Omega),$$u\in\mathcal {K}_{\psi}^{\phi}(\Omega)$ 是方程 (1.3) 对应的 $\mathcal {K}_{\psi}^{\phi}(\Omega)$-障碍问题的弱解且 $\nabla\psi\in B^{\alpha}_{\phi,q,loc}(\Omega)\cap L^{\phi}_{loc}(\Omega)$, 则 $V(\nabla{u})\in B_{2,q,loc}^{\alpha}(\Omega).$

$a(t)=t^{p-2},$$p\geq2$ 时, 方程 (1.3) 退化成 $p$-Laplace 方程, 该结论仍然成立.

2 预备知识

本文中与常数 $C$ 有关的参数将使用括号强调, 例如: $C=C(n,p,q,\ldots).$ 对于任意给定的 $x_0\in \Omega,$ 常数 $R>0,$$B_R(x_0):=B(x_0,R)=\{x:|x-x_0|<R\}$ 表示以 $x_0$ 为球心, $R$ 为半径的开球, 在不致混淆的情况下, 省略球心 $x_0.$$x\in\Omega,h\in\mathbb{R}^{n},$$|h|$ 足够小, 有 $x+h\in\Omega.$ 对任意函数 $F:\Omega\subset\mathbb{R}^n\rightarrow\mathbb{R},$ 有限差分算子定义为

$\Delta_hF(x):=F(x+h)-F(x).$

首先给出 Orlicz-Sobolev 空间中的一些定义.

定义 2.1[5] 若函数 $\phi$ 是凸函数, 并且满足

$\phi(0)=0,~~\lim\limits_{t\rightarrow+\infty}\frac{\phi(t)}{t}=\lim\limits_{t\rightarrow0^+}\frac{t}{\phi(t)}=+\infty,$

则函数 $\phi:[0,\infty)\rightarrow[0,\infty)$ 称为 $\textit{N}$-函数. 当 $t\geq0$ 时, $N$-函数 $\phi$ 的共轭函数 $\tilde{\phi}$ 定义为

$\tilde{\phi}(t)=\sup_{s\geq0}\{st-\phi(s)\}.$

定义 2.2[3] 若存在正常数 $K$, 使得对任意 $t>0$, 有 $\phi(2t)\leq K\phi(t),$ 则称 $N$-函数 $\phi$ 满足全局 $\Delta_{2}$ 条件, 表示为 $\phi\in\Delta_2$. 若存在正常数 $\theta>1$, 使得对任意 $t>0$, 有

$\phi(t)\leq\frac{\phi(\theta t)}{2\theta},$

则称 $N$-函数 $\phi$ 满足全局 $\nabla_{2}$ 条件, 表示为 $\phi\in\nabla_2$.

定义 2.3[4] 给定 $N$-函数 $\phi$, Orlicz 类 $K^\phi(\Omega)$ 是所有满足 $ \int_\Omega \phi\left(|g|\right)\textrm{d}x<\infty$ 的可测函数 $g:\Omega\rightarrow\mathbb{R}$ 的集合. Orlicz 空间 $L^\phi(\Omega)$$K^\phi(\Omega)$ 的线性闭包且具有 Luxemburg 范数

$\begin{eqnarray*} {\left\| g\right\|}_{L^\phi(\Omega)}:=\textrm{inf}\left\{\lambda>0:\int_{\Omega} \phi\left(\frac{|g(x)|}{\lambda}\right) \textrm{d}x\leq1\right\}. \end{eqnarray*}$

此外, Orlicz-Sobolev 空间 $W^{1,\phi}(\Omega):=\left\{g\in L^{\phi}(\Omega)|\nabla g\in L^\phi(\Omega)\right\}$, 具有范数 ${\left\| g\right\|}_{W^{1,\phi}(\Omega)}:={\left\| g\right\|}_{L^\phi(\Omega)}+{\left\| \nabla g\right\|}_{L^\phi(\Omega)}$.

通常情况下, 有 $K^\phi(\Omega)\subset L^\phi(\Omega)$ (见参考文献[20], 第 8 章). 若 $\phi\in\Delta_2$, 则 $K^\phi(\Omega)= L^\phi(\Omega).$

下面给出 Besov-Orlicz 空间的定义, 这在证明定理 1.1 中具有重要作用.

定义 2.4[14,21]$0<\alpha<1,$$1\leq q <\infty$.$v\in L^{\phi}(\Omega)$ 且满足

${\left\| v\right\|}_{B_{\phi,q}^{\alpha}(\Omega)}={\left\| v\right\|}_{L^\phi(\Omega)}+[v]_{B_{\phi,q}^{\alpha}(\Omega)}<+\infty,$

这里

$[v]_{{B}_{\phi,q}^\alpha(\Omega)}:=\left(\int_{\Omega}\frac{\left\|\Delta_h v\right\|_{L^\phi(\Omega)}^q}{|h|^{\alpha q}}\frac{\textrm{d}h}{|h|^n}\right)^{\frac{1}{q}},$

则称 $v$ 属于 Besov-Orlicz 空间 $B_{\phi,q}^{\alpha}(\Omega)$.特别地, 当 $\phi(t)=t^{p},$${\left\| v\right\|}_{B_{\phi,q}^{\alpha}(\Omega)}={\left\| v\right\|}_{B_{p,q}^{\alpha}(\Omega)}.$

由于

$\left(\int_{|h|\geq\delta}\frac{\left\|\Delta_h v\right\|_{L^\phi(\Omega)}^q}{|h|^{\alpha q}}\frac{\textrm{d}h}{|h|^n}\right)^{\frac{1}{q}}\leq c(\delta,n,\alpha,q)\left\|v\right\|_{L^{\phi}(\Omega)},$

所以对任意足够小的 $\delta>0,$$B^{\alpha}_{\phi,q}(\Omega)$ 的范数也可以表示为

$\left\| v\right\|_{B_{\phi,q}^{\alpha}(\Omega)}:={\left\| v\right\|}_{L^\phi(\Omega)}+\left(\int_{|h|\geq\delta}\frac{\left\|\Delta_h v\right\|_{L^\phi(\Omega)}^q}{|h|^{\alpha q}}\frac{\textrm{d}h}{|h|^n}\right)^{\frac{1}{q}}.$

对任意 ${\Omega}\subset\mathbb{R}^n,$ 若截断函数 $\eta\in C_0^\infty(\Omega)$$\eta v\in L^\phi(\Omega),$ 满足

${\left\| \eta v\right\|}_{B_{\phi,q}^{\alpha}(\Omega)}={\left\|\eta v\right\|}_{L^\phi(\Omega)}+[\eta v]_{B_{\phi,q}^{\alpha}(\Omega)}<\infty,$

则称 $v\in B_{\phi,q,loc}^\alpha(\Omega).$

为完成证明, 需要给出下面几个引理.

引理 2.1[5] 设函数 $F$$G$ 满足 $F\in W^{1,\phi}(B_{R}),$$G\in L^{\tilde{\phi}}(B_{R})$, 则

(1) 对任意 $0<|h|<R$, 有 $\Delta_hF\in W^{1,\phi}(B_{R-|h|}),$$\Delta_h{(FG)(x)}=F(x+h)\Delta_hG(x)+G(x)\Delta_hF(x)$$D_i(\Delta_hF)=\Delta_h(D_iF).$

(2) 若函数 $\textit{F}$$\textit{G}$ 至少有一个支集包含在 $B_{R-|h|}$ 中, 则有

$\int_{B_R}F\Delta_hG\textrm{d}x=\int_{B_R}G\Delta_{-h}F\textrm{d}x.$

引理 2.2[3]$\phi$ 是一个 $N$-函数, 则有 $st\leq \widetilde{\phi}(s)+\phi(t),\ \mbox{对任意 } s,t\geq0.$$\phi\in\Delta_2\cap\nabla_2$, 则有 Young 不等式, $st\leq\varepsilon\tilde{\phi}(s)+C(\varepsilon)\phi(t),\ \mbox{对任意 } s,t\geq0 \mbox{ 和 } \varepsilon>0.$

引理 2.3[3,5] 假设 $a(t)$ 满足 (1.2) 式, $\phi(t)$ 满足 (1.6) 式, 则

(1) $\phi$ 是严格凸的 $\textit{N}$-函数且满足 $\tilde{\phi}\left( b(t)\right)\leq C_0\phi\left(t\right),$ 这里 $C_{0}>0,$$t\geq0$$\phi$ 的共轭函数 $\tilde{\phi}$ 满足定义 2.1.

(2) $\phi(t)\in\Delta_2\cap\nabla_2.$

(3) 对任意 $t>0,$ 下面的不等式成立

$ \begin{align*} a(t)\theta^{i_a}\leq a(\theta t)\leq a(t)\theta^{s_a},\quad \mbox{当 } \theta\geq1\mbox{ 时};\quad a(t)\theta^{s_a}\leq a(\theta t)\leq a(t)\theta^{i_a},\quad \mbox{当 } 0<\theta<1\mbox{ 时}. \end{align*} $

引理 2.4[5] 假设 $a(t)$ 满足 (2.1) 式, $V(z)$ 满足 (1.4) 式. 对任意 $\xi,\eta\in \mathbb{R}^{n}$$x\in \Omega,$ 存在正常数 $C,L,\mu$, 使得

(1) $\left\langle a(|\xi|)\xi-a(|\eta|)\eta,(\xi-\eta)\right\rangle\geq\mu a(|\xi|+|\eta|)|\xi-\eta|^2, $

(2) $|a(|\xi|)\xi-a(|\eta|)\eta|\leq L a(|\xi|+|\eta|)|\xi-\eta|,$

(3) $|V(\xi)-V(\eta)|^2\leq Ca(|\xi|+|\eta|)|\xi-\eta|^2.$

引理 2.5[21]$B_s\subset B_t\subset\Omega$ 是同心球, $\phi$$N$-函数. 对任意 $f\in W^{1,\phi}(B_t)$$|h|\leq t-s, |h|\in\mathbb{R}^{+},$

$\int_{B_s}\phi\left(\frac{|\Delta_hf(x)|}{|h|}\right)\textrm{d}x\leq\int_{B_t}\phi\left(|\nabla f(x)|\right)\textrm{d}x.$

3 主要结论的证明

定理 1.1 的证明$u$ 是障碍问题 (1.7) 的弱解. 对任意 $ B_{2R}\Subset\Omega$, 取截断函数 $\eta\in C_{0}^{\infty}(\Omega)$, 满足

$ \begin{align*} \textrm{在 }B_{\frac{R}{2}}\textrm{ 内 }\eta \equiv 1, \ \textrm{在 }\Omega\setminus B_{R} \textrm{ 内 }\eta \equiv 0, \ 0\leq \eta \leq 1, \ \left | \nabla\eta \right |\leq \frac{C}{R}. \end{align*} $

$v=u-{\frac{1}{2}}\Delta_{-h}\big(\eta^2\Delta_h(u-\psi)\big)$, 取 $\left|h\right|$ 足够小, 使得 $x+h\in\Omega$, 从而 $v\in\mathcal {K}_{\psi}^{\phi}(\Omega),$ 这是因为由有限差分算子定义可得

$\begin{eqnarray*} v-{\psi} &=& \displaystyle u-{\frac{1}{2}}\Delta_{-h}\big(\eta^2\Delta_{h}(u-\psi)\big)-\psi \nonumber\\ &=& \displaystyle u-\psi+{\frac{1}{2}}\eta^2(x)(u(x+h)-\psi(x+h))-{\frac{1}{2}}\eta^2(x)(u(x)-\psi(x)) \nonumber\\ && \displaystyle -{\frac{1}{2}}\eta^2(x-h)(u(x)-\psi(x))+{\frac{1}{2}}\eta^2(x-h)(u(x-h)-\psi(x-h)) \nonumber\\ &=& \displaystyle {\frac{1}{2}}\eta^2(x)(u(x+h)-\psi(x+h))+{\frac{1}{2}}\eta^2(x-h)(u(x-h)-\psi(x-h)) \nonumber\\ && \displaystyle +({\frac{1}{2}}-{\frac{1}{2}}\eta^2(x))(u(x)-\psi(x))+({\frac{1}{2}}-{\frac{1}{2}}\eta^2(x-h))(u(x)-\psi(x)) \nonumber\\ &\geq& 0, \end{eqnarray*}$

这里最后一个不等式用到了不等式 $u\geq\psi$$0\leq\eta\leq1$. 于是可用 $\textit{v}(x)$ 作为变分不等式 (1.7) 的容许函数, 得到

$\int_{\Omega}\left\langle a(|\nabla u|)\nabla u,\nabla\big(-{\frac{1}{2}}\Delta_{-h}(\eta^2\Delta_{h}(u-\psi))\big)\right\rangle \textrm{d}x\geq\int_{\Omega}\left\langle \textbf{f},\nabla\big(-{\frac{1}{2}}\Delta_{-h}(\eta^2\Delta_{h}(u-\psi))\big)\right\rangle \textrm{d}x.$

对上式应用引理 2.1 (1) 和 (2), 得

$\int_{B_{R}}\left\langle\Delta_{h}\big(a(|\nabla u|)\nabla u\big),\nabla\big(\eta^2{\Delta_{h}(u-\psi)}\big)\right\rangle \textrm{d}x\leq\int_{B_{R}}\left\langle \Delta_{h}\textbf{f},\nabla\big(\eta^2{\Delta_{h}(u-\psi)}\big)\right\rangle \textrm{d}x.$

将上述不等式整理可得

$\begin{eqnarray*} &&\int_{B_{R}}\eta^2\left\langle \Delta_{h}[a(|\nabla u|)\nabla u],\Delta_{h}\nabla u\right\rangle \textrm{d}x \nonumber\\ &\leq& \displaystyle -2\int_{B_{R}}\eta \left\langle\Delta_{h}[a(|\nabla u|)\nabla u],\nabla\eta \Delta_{h}(u-\psi)\right\rangle \textrm{d}x \nonumber\\ && \displaystyle +\int_{B_{R}}\left\langle \Delta_{h}[a(|\nabla u|)\nabla u],\eta^{2}\Delta_{h}\nabla \psi\right\rangle \textrm{d}x \nonumber\\ && \displaystyle +\int_{B_{R}}\left\langle \Delta_{h}\textbf{f},\eta^2(\Delta_h\nabla u-\Delta_h\nabla \psi)\right\rangle \textrm{d}x \nonumber\\ && \displaystyle +\int_{B_{R}}\left\langle \Delta_{h}\textbf{f},2\eta \nabla\eta\Delta_{h}(u-\psi)\right\rangle \textrm{d}x. \nonumber \end{eqnarray*}$

上式可以表示为

$\begin{equation} I_1\leq I_2+I_3+I_4+I_5. \end{equation}$

下面分别估计 $I_i(i=1,2,3,4,5).$ 估计 $I_1.$ 根据引理 2.4 (1), 有

$\begin{eqnarray*} I_1 &=& \displaystyle \int_{B_{R}}\eta^{2}\left\langle a\left(|\nabla u(x+h)|\right)\nabla u(x+h)-a\left(|\nabla u(x)|\right)\nabla u(x),\nabla(\Delta_{h}u)\right\rangle \textrm{d}x \nonumber\\ &\geq& \displaystyle \mu\int_{B_{R}}\eta^{2} a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\Delta_{h}(\nabla u(x))\right|^2 \textrm{d}x. \end{eqnarray*}$

估计 $I_2.$ 根据引理 2.4(2) 和 Young 不等式, 有

$\begin{eqnarray*} \left|I_2\right| &\leq& \displaystyle 2\int_{B_{R}}\left| a\left(|\nabla u(x+h)|\right)\nabla u(x+h)-a\left(|\nabla u(x)|\right)\nabla u(x)\right|\left|\eta\right|\left|\nabla\eta\right| \left|\Delta_{h}(u-\psi)\right|\textrm{d}x \nonumber\\ &\leq& \displaystyle 2L \int_{B_R}a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\Delta_{h}\nabla u(x)\right|\left|\eta\right|\left|\nabla\eta\right|\left(\left|\Delta_{h} u\right|+\left|\Delta_{h} \psi\right|\right) \textrm{d}x \nonumber\\ &\leq& \displaystyle \varepsilon_1\int_{B_R}a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\Delta_{h}\nabla u(x)\right|^2\eta^2 \textrm{d}x \nonumber\\ && \displaystyle +C(\varepsilon_1,L)\int_{B_R}a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\nabla\eta\right|^2 \left|\Delta_{h} u(x)\right|^2\textrm{d}x \nonumber\\ && \displaystyle +C(\varepsilon_1,L)\int_{B_R}a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\nabla\eta\right|^2 \left|\Delta_{h} \psi(x)\right|^2\textrm{d}x. \end{eqnarray*}$

估计 $I_3.$ 根据引理 2.4(2) 和 Young 不等式, 有

$\begin{eqnarray*} \left|I_3\right| &\leq& \displaystyle \int_{B_{R}}\left| a\left(|\nabla u(x+h)|\right)\nabla u(x+h)-a\left(|\nabla u(x)|\right)\nabla u(x)\right|\eta^{2}\left|\Delta_{h}\nabla\psi\right| \textrm{d}x \nonumber\\ &\leq& \displaystyle L\int_{B_R}a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\Delta_{h}\nabla u(x)\right|\eta^{2}\left|\Delta_{h}\nabla\psi\right| \textrm{d}x \nonumber\\ &\leq& \displaystyle \varepsilon_2\int_{B_R}a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\Delta_{h}\nabla u(x)\right|^2\eta^2 \textrm{d}x \nonumber\\ && \displaystyle +C(\varepsilon_2,L)\int_{B_R}a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\Delta_{h}\nabla \psi(x)\right|^2\eta^2 \textrm{d}x. \end{eqnarray*}$

估计 $I_4.$ 由三角不等式, 得

$\begin{eqnarray*} I_{4} \leq \displaystyle \int_{B_{R}}|\Delta_h\textbf{f}||\Delta_h\nabla u|\eta^2 \textrm{d}x+\int_{B_{R}}|\Delta_h\textbf{f}||\Delta_h\nabla \psi|\eta^2 \textrm{d}x =: \displaystyle I_{41}+I_{42}. \end{eqnarray*}$

下面分别估计 $I_{41}$$I_{42}.$ 首先估计 $I_{41}.$ 因为 $0\leq i_a\leq s_a<\infty,$ 所以 $\frac{2(i_a+1)}{i_a+2}\geq 1,\frac{2(s_a+1)}{s_a+2}\geq 1.$ 利用 Young 不等式、Hölder 不等式和引理 2.3(3), 有

$\begin{eqnarray*} I_{41} &\leq& \displaystyle \int_{\{x\in B_R:|\Delta_h\nabla u|\geq1\}}|\Delta_h\textbf{f}||\Delta_h\nabla u|\eta^2 \textrm{d}x+\int_{\{x\in B_R:|\Delta_h\nabla u|<1\}}|\Delta_h\textbf{f}||\Delta_h\nabla u|\eta^2 \textrm{d}x \nonumber\\ &\leq& \displaystyle C(\varepsilon_3)\int_{\{x\in B_{R}:|\Delta_{h}\nabla u|\geq1\}}|\Delta_{h}\mathbf{f}|^{\frac{i_{a}+2}{i_{a}+1}}\mathrm{d}x+\varepsilon_3\int_{\{x\in B_{R}:|\Delta_{h}\nabla u|\geq1\}}\eta^{2}|\Delta_{h}\nabla u|^{i_{a}+2}\mathrm{d}x \nonumber\\ && \displaystyle +C(\varepsilon_3)\int_{\{x\in B_R:|\Delta_h\nabla u|<1\}}|\Delta_h\mathbf{f}|^{\frac{s_a+2}{s_a+1}}\mathrm{d}x+\varepsilon_3\int_{\{x\in B_R:|\Delta_h\nabla u|<1\}}\eta^2|\Delta_h\nabla u|^{s_a+2}\mathrm{d}x \nonumber\\ &=& \displaystyle C(\varepsilon_3)\int_{\{x\in B_{R}:|\Delta_{h}\nabla u|\geq1\}}|\Delta_{h}\mathbf{f}|^{\frac{i_{a}+2}{i_{a}+1}}\mathrm{d}x+\varepsilon_3\int_{\{x\in B_{R}:|\Delta_{h}\nabla u|\geq1\}}\eta^{2}|\Delta_{h}\nabla u|^{2}|\Delta_{h}\nabla u|^{i_{a}}\mathrm{d}x \nonumber\\ && \displaystyle +C(\varepsilon_3)\int_{\{x\in B_R:|\Delta_h\nabla u|<1\}}|\Delta_h\mathbf{f}|^{\frac{s_a+2}{s_a+1}}\mathrm{d}x+\varepsilon_3\int_{\{x\in B_R:|\Delta_h\nabla u|<1\}}\eta^2|\Delta_h\nabla u|^{2}|\Delta_{h}\nabla u|^{s_{a}}\mathrm{d}x \nonumber\\ &\leq& \displaystyle C(\varepsilon_3)\int_{\{x\in B_{R}:|\Delta_{h}\nabla u|\geq1\}}|\Delta_{h}\mathbf{f}|^{\frac{i_{a}+2}{i_{a}+1}}\mathrm{d}x+\varepsilon_3\int_{\{x\in B_{R}:|\Delta_{h}\nabla u|\geq1\}}\eta^{2}|\Delta_{h}\nabla u|^{2}\frac{1}{a(1)}a(|\Delta_h\nabla u|)\mathrm{d}x \nonumber\\ && \displaystyle +C(\varepsilon_3)\int_{\{x\in B_R:|\Delta_h\nabla u|<1\}}|\Delta_h\mathbf{f}|^{\frac{s_a+2}{s_a+1}}\mathrm{d}x+\varepsilon_3\int_{\{x\in B_R:|\Delta_h\nabla u|<1\}}\eta^2|\Delta_h\nabla u|^{2}\frac{1}{a(1)}a(|\Delta_h\nabla u|)\mathrm{d}x \nonumber\\ &\leq& \displaystyle C(\varepsilon_3)\int_{B_R}|\Delta_h\mathbf{f}|^{\frac{i_a+2}{i_a+1}}+|\Delta_h\mathbf{f}|^{\frac{s_a+2}{s_a+1}}\mathrm{d}x+C\varepsilon_3\int_{B_R}a(|\Delta_h\nabla u|)|\Delta_h\nabla u|^2\eta^2\mathrm{d}x \nonumber\\ &\leq& \displaystyle C(\varepsilon_3)|h|^{\frac{\beta(i_a+2)}{i_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{i_a+2}{2(i_a+1)}} +C(\varepsilon_3)|h|^{\frac{\beta(s_a+2)}{s_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{s_a+2}{2(s_a+1)}} \nonumber\\ && \displaystyle +C\varepsilon_3\int_{B_R}a(|\nabla u(x+h)|+|\nabla u(x)|)|\Delta_h\nabla u|^2\eta^2\mathrm{d}x. \end{eqnarray*}$

上述推导过程使用了不等式 $0\leq\eta\leq1$. 接着估计 $I_{42}.$ 类似 $I_{41}$ 的推导, 有

$\begin{eqnarray*} I_{42} &\leq& \displaystyle C(\varepsilon_3)|h|^{\frac{\beta(i_a+2)}{i_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{i_a+2}{2(i_a+1)}} +C(\varepsilon_3)|h|^{\frac{\beta(s_a+2)}{s_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{s_a+2}{2(s_a+1)}} \nonumber\\ && \displaystyle +C\varepsilon_3\int_{B_R}a(|\nabla \psi(x+h)|+|\nabla \psi(x)|)|\Delta_h\nabla \psi|^2\eta^2\mathrm{d}x. \end{eqnarray*}$

将 (3.6) 式和 (3.7) 式代入 (3.5) 式, 得

$\begin{matrix} I_{4} &\leq C(\varepsilon_3)|h|^{\frac{\beta(i_a+2)}{i_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{i_a+2}{2(i_a+1)}} +C(\varepsilon_3)|h|^{\frac{\beta(s_a+2)}{s_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{s_a+2}{2(s_a+1)}} \nonumber\\ &~~~+C\varepsilon_3\int_{B_R}a(|\nabla u(x+h)|+|\nabla u(x)|)|\Delta_h\nabla u|^2\eta^2\mathrm{d}x \nonumber\\ &~~~+C\varepsilon_3\int_{B_R}a(|\nabla \psi(x+h)|+|\nabla \psi(x)|)|\Delta_h\nabla \psi|^2\eta^2\mathrm{d}x. \end{matrix}$

估计 $I_5.$ 根据三角不等式和 $0\leq\eta\leq1$, 有

$\begin{eqnarray*} I_5 &\leq& \displaystyle 2\int_{B_R}|\Delta_h\textbf{f}||\nabla\eta|(|\Delta_h u|+|\Delta_h\psi|)\eta \textrm{d}x \nonumber\\ &\leq& \displaystyle 2\int_{B_R}|\Delta_h\textbf{f}||\nabla\eta||\Delta_h u|\textrm{d}x+2\int_{B_R}|\Delta_h\textbf{f}||\nabla\eta||\Delta_h \psi|\textrm{d}x \nonumber\\ &\leq& \displaystyle 2\int_{\{x\in B_R:|\Delta_h u|\geq1\}}|\Delta_h\textbf{f}||\nabla\eta||\Delta_h u|\textrm{d}x+2\int_{\{x\in B_R:|\Delta_h u|<1\}}|\Delta_h\textbf{f}||\nabla\eta||\Delta_h u|\textrm{d}x \nonumber\\ && \displaystyle +2\int_{\{x\in B_R:|\Delta_h \psi|\geq1\}}|\Delta_h\textbf{f}||\nabla\eta||\Delta_h \psi|\textrm{d}x+2\int_{\{x\in B_R:|\Delta_h \psi|<1\}}|\Delta_h\textbf{f}||\nabla\eta||\Delta_h \psi|\textrm{d}x \nonumber\\ &=:& \displaystyle I_{51}+I_{52}+I_{53}+I_{54}. \end{eqnarray*}$

下面估计 $I_{51},I_{52},I_{53},I_{54}.$ 由 Young 不等式、Hölder 不等式、引理 $2.3(3)$$\eta$ 的定义式, 有

$\begin{eqnarray*} I_{51} &\leq& \displaystyle C(\varepsilon_4)\int_{B_R}|\Delta_{h}\mathbf{f}|^{\frac{i_{a}+2}{i_{a}+1}}\mathrm{d}x+\varepsilon_4\int_{\{x\in B_{R}:|\Delta_{h} u|\geq1\}}|\Delta_{h} u|^{i_{a}+2}|\nabla\eta|^{i_{a}+2}\mathrm{d}x \nonumber\\ &\leq& \displaystyle C(\varepsilon_4)\int_{B_R}|\Delta_{h}\mathbf{f}|^{\frac{i_{a}+2}{i_{a}+1}}\mathrm{d}x+\varepsilon_4\left(\frac{C}{R}\right)^{i_a+2}\int_{\{x\in B_{R}:|\Delta_{h} u|\geq1\}}|\Delta_{h} u|^{i_{a}}|\Delta_{h} u|^{2}\mathrm{d}x \nonumber\\ &\leq& \displaystyle C(\varepsilon_4)|h|^{\frac{\beta(i_a+2)}{i_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{i_a+2}{2(i_a+1)}}+C(\varepsilon_4,R,i_a) \int_{B_R}a(|\Delta_{h}u|)|\Delta_{h}u|^{2}\textrm{d}x. \nonumber\\ I_{52} &\leq& \displaystyle C(\varepsilon_4)\int_{B_R}|\Delta_{h}\mathbf{f}|^{\frac{s_{a}+2}{s_{a}+1}}\mathrm{d}x+\varepsilon_4\int_{\{x\in B_{R}:|\Delta_{h} u|<1\}}|\Delta_{h} u|^{s_{a}+2}|\nabla\eta|^{s_{a}+2}\mathrm{d}x \nonumber\\ &\leq& \displaystyle C(\varepsilon_4)\int_{B_R}|\Delta_{h}\mathbf{f}|^{\frac{s_{a}+2}{s_{a}+1}}\mathrm{d}x+\varepsilon_4\left(\frac{C}{R}\right)^{s_a+2}\int_{\{x\in B_{R}:|\Delta_{h} u|<1\}}|\Delta_{h} u|^{s_{a}}|\Delta_{h} u|^{2}\mathrm{d}x \nonumber\\ &\leq& \displaystyle C(\varepsilon_4)|h|^{\frac{\beta(s_a+2)}{s_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{s_a+2}{2(s_a+1)}}+C(\varepsilon_4,R,s_a) \int_{B_R}a(|\Delta_{h}u|)|\Delta_{h}u|^{2}\textrm{d}x. \end{eqnarray*}$

同理可得:

$\begin{eqnarray*} I_{53} &\leq& \displaystyle C(\varepsilon_4)|h|^{\frac{\beta(i_a+2)}{i_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{i_a+2}{2(i_a+1)}}+C(\varepsilon_4,R,i_a) \int_{B_R}a(|\Delta_{h}\psi|)|\Delta_{h}\psi|^{2}\textrm{d}x. \nonumber\\ I_{54} &\leq& \displaystyle C(\varepsilon_4)|h|^{\frac{\beta(s_a+2)}{s_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{s_a+2}{2(s_a+1)}}+C(\varepsilon_4,R,s_a) \int_{B_R}a(|\Delta_{h}\psi|)|\Delta_{h}\psi|^{2}\textrm{d}x. \end{eqnarray*}$

根据 $I_{51}, I_{52}, I_{53}$$I_{54}$ 的估计, 于是有

$\begin{eqnarray*} I_5&\leq& \displaystyle C(\varepsilon_4)|h|^{\frac{\beta(i_a+2)}{i_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{i_a+2}{2(i_a+1)}} +C(\varepsilon_4)|h|^{\frac{\beta(s_a+2)}{s_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{s_a+2}{2(s_a+1)}} \nonumber\\ && \displaystyle +C(\varepsilon_4,R,i_a,s_a)\int_{B_R}a(|\Delta_{h}u|)|\Delta_{h}u|^{2}\textrm{d}x+C(\varepsilon_4,R,i_a,s_a)\int_{B_R}a(|\Delta_{h}\psi|)|\Delta_{h}\psi|^{2}\textrm{d}x. \end{eqnarray*}$

结合 (3.1)-(3.4), (3.8) 和 (3.9), 并取 $\varepsilon_1, \varepsilon_2, \varepsilon_3$ 足够小, 使得 $\left(\varepsilon_1+\varepsilon_2+C\varepsilon_3\right)\leq \mu$, 从而有

$\begin{eqnarray*} && \displaystyle \int_{B_{R}}\eta^{2} a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\Delta_{h}(\nabla u(x))\right|^2 \textrm{d}x \nonumber\\ &\leq& \displaystyle C(L,\mu)\int_{B_R}a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\nabla\eta\right|^2 \left|\Delta_{h} u(x)\right|^2\textrm{d}x \nonumber\\ && \displaystyle +C(L,\mu)\int_{B_R}a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\nabla\eta\right|^2 \left|\Delta_{h} \psi(x)\right|^2\textrm{d}x \nonumber\\ && \displaystyle +C(\varepsilon_4,\mu,R,i_a,s_a)\int_{B_R}a(|\Delta_{h}u|)|\Delta_{h}u|^{2}\textrm{d}x +C(\varepsilon_4,\mu,R,i_a,s_a)\int_{B_R}a(|\Delta_{h}\psi|)|\Delta_{h}\psi|^{2}\textrm{d}x \nonumber\\ && \displaystyle +C(L,\mu)\int_{B_R}a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\Delta_{h}\nabla \psi(x)\right|^2\eta^2 \textrm{d}x \nonumber\\ && \displaystyle +C(\mu)\int_{B_{R}}\eta^{2} a\left(|\nabla \psi(x+h)|+|\nabla\psi(x)|\right)\left|\Delta_{h}(\nabla \psi(x))\right|^2 \textrm{d}x \nonumber\\ && \displaystyle +C(\varepsilon_4,\mu)|h|^{\frac{\beta(i_a+2)}{i_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{i_a+2}{2(i_a+1)}} +C(\varepsilon_4,\mu)|h|^{\frac{\beta(s_a+2)}{s_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{s_a+2}{2(s_a+1)}}_{.} \end{eqnarray*}$

由参考文献[1] 中的定理 $2.1$ 可知 $C^{\infty}(\overline{{B}_{R+|h|}})$$W^{1,B}(B_{R+|h|})$ 中稠密, 于是根据逼近讨论, 可假设 $u\in C^{\infty}(\overline{{B}_{R+|h|}}),\psi\in C^{\infty}(\overline{{B}_{R+|h|}})$. 对 (3.10)式右端第一个积分应用拉格朗日中值定理, 再根据引理 2.2、引理 2.3、(1.5) 式和 $\eta$ 的定义, 有

$\begin{eqnarray*} && \displaystyle \int_{B_R}a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\nabla\eta\right|^2 \left|\Delta_{h} u(x)\right|^2\textrm{d}x \nonumber\\ &=& \displaystyle C(L,\mu)\left|h\right|^2\int_{B_{R}} a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\nabla\eta\right|^2\left|\nabla u(x+\theta h)\right|^2 \textrm{d}x \nonumber\\ &\leq& \displaystyle C(L,\mu)\frac{|h|^2}{R^{2}}\int_{\{x\in B_{R}:|\nabla u(x+h)|+|\nabla u(x)|\geq|\nabla u(x+\theta h)|\}} a(\left|\nabla u(x+h)\right|+\left|\nabla u(x)\right|)\left|\nabla u(x+\theta h)\right|^2 \textrm{d}x \nonumber\\ && \displaystyle +C(L,\mu)\frac{|h|^2}{R^2}\int_{\{x\in B_{R}:|\nabla u(x+h)|+|\nabla u(x)|<|\nabla u(x+\theta h)|\}} a(\left|\nabla u(x+h)\right|+\left|\nabla u(x)\right|)\left|\nabla u(x+\theta h)\right|^2 \textrm{d}x \nonumber\\ &\leq& \displaystyle C(L,\mu)\frac{|h|^2}{R^{2}}\int_{B_{R}}a(\left|\nabla u(x+h)\right|+\left|\nabla u(x)\right|)(\left|\nabla u(x+h)\right|+\left|\nabla u(x)\right|)^{2}\textrm{d}x \nonumber\\ && \displaystyle +C(L,\mu)\frac{|h|^2}{R^{2}}\int_{B_{R}}a(\left|\nabla u(x+\theta h)\right|)(\left|\nabla u(x+\theta h)\right|)^{2}\textrm{d}x \nonumber\\ &\leq& \displaystyle C(L,\mu)\frac{|h|^2}{R^{2}}\int_{B_{R}}\tilde{\phi}\left[a(\left|\nabla u(x+h)\right|+\left|\nabla u(x)\right|)(\left|\nabla u(x+h)\right|+\left|\nabla u(x)\right|)\right]\textrm{d}x \nonumber\\ && \displaystyle +C(L,\mu)\frac{|h|^2}{R^{2}}\int_{B_{R}}\phi(\left|\nabla u(x+h)\right|+\left|\nabla u(x)\right|)\textrm{d}x \nonumber\\ && \displaystyle +C(L,\mu)\frac{|h|^2}{R^{2}}\int_{B_{R}}\tilde{\phi}\left[a(\left|\nabla u(x+\theta h)\right|)\left|\nabla u(x+\theta h)\right|\right]\textrm{d}x+C(L,\mu)\frac{|h|^2}{R^2}\int_{B_{R}}\phi(\left|\nabla u(x+\theta h)\right|)\textrm{d}x \nonumber\\ &\leq& \displaystyle C(L,\mu,C_0,K)\frac{|h|^2}{R^{2}}\int_{B_{R+|h|}}\phi(\left|\nabla u\right|) \textrm{d}x, \end{eqnarray*}$

这里 $\theta\in (0,1),$ 最后一个不等式使用了定义 $2.2$. 同理可得 (3.10) 式右端第二个积分

$\begin{eqnarray*} && \displaystyle \int_{B_{R}}a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\nabla\eta\right|^2\left|\Delta_{h} \psi(x)\right|^2 \textrm{d}x \nonumber\\ &\leq& \displaystyle C(L,\mu,C_0,K)\frac{|h|^2}{R^2}\int_{B_{R+|h|}}\phi\big(|\nabla u|\big)\textrm{d}x+C(L,\mu,C_0,K)\frac{|h|^2}{R^2}\int_{B_{R+|h|}} \phi\big(|\nabla \psi|\big)\textrm{d}x. \end{eqnarray*}$

现在估计 (3.10) 式右端第三、四个积分. 若 $|h|$ 足够小, 则 $|\Delta_h u|\leq\frac{|\Delta_h u|}{|h|},$ 因为 $a(t)$$(0,+\infty)$ 上的增函数, 故 $a\left(|\Delta_h u|\right)\leq a\left(\frac{|\Delta_h u|}{|h|}\right).$ 因此, 由引理 2.2、引理 2.3 和引理 2.5, 得

$\begin{matrix} \int_{B_{R}} a\left(|\Delta_{h} u|\right)|\Delta_{h} u|^2\textrm{d}x &\leq C|h|^2\int_{B_{R}}a\left(\frac{|\Delta_h u|}{|h|}\right)\frac{|\Delta_{h} u|^{2}}{|h|^{2}}\textrm{d}x \nonumber\\ &\leq C|h|^2\int_{B_{R}}\tilde{\phi}\left[a\left(\frac{|\Delta_h u|}{|h|}\right)\frac{|\Delta_h u|}{|h|}\right]\textrm{d}x \nonumber\\ &~~~+C|h|^2\int_{B_{R}}\phi\left(\frac{|\Delta_h u|}{|h|}\right)\textrm{d}x \nonumber\\ &\leq C|h|^2\int_{B_{R}}\phi\left(\frac{|\Delta_h u|}{|h|}\right)\textrm{d}x \nonumber\\ &\leq C|h|^2\int_{B_{R+|h|}}\phi\big(|\nabla u|\big)\textrm{d}x. \end{matrix}$

同理可得

$\begin{eqnarray*} \int_{B_{R}} a\left(|\Delta_{h}\psi|\right)|\Delta_{h} \psi|^2\textrm{d}x\leq C|h|^2\int_{B_{R+|h|}}\phi\big(|\nabla \psi|\big)\textrm{d}x. \end{eqnarray*}$

将 (3.13) 式与 (3.14) 式相加, 得到

$\begin{eqnarray*} && \displaystyle \int_{B_R}a(|\Delta_{h}u|)|\Delta_{h}u|^{2}\textrm{d}x+\int_{B_R}a(|\Delta_{h}\psi|)|\Delta_{h}\psi|^{2}\textrm{d}x \nonumber\\ &\leq& \displaystyle C|h|^2\int_{B_{R+|h|}}\phi\big(|\nabla u|\big)\textrm{d}x +C|h|^2\int_{B_{R+|h|}}\phi\big(|\nabla \psi|\big)\textrm{d}x, \end{eqnarray*}$

这里正常数 $C=C(\varepsilon_4,\mu,R,C_0,i_a,s_a).$ 下面估计 (3.10) 式右端第五个积分, 应用引理 2.2、引理 2.3 和 (1.5) 式, 有

$\begin{eqnarray*} && \displaystyle \int_{B_R}a(\left|\nabla u(x+h)\right|+\left|\nabla u(x)\right|)\left|\Delta_{h}\nabla \psi(x)\right|^{2}\eta^{2} \textrm{d}x \nonumber\\ &=& \displaystyle C(L,\mu)\lambda^{2}\int_{B_R}a\big(|\nabla u(x+h)|+|\nabla u(x)|\big)\frac{\left|\Delta_{h}\nabla \psi(x)\right|^{2}}{\lambda^{2}}\eta^{2} \textrm{d}x \nonumber\\ &\leq& \displaystyle C(L,\mu)\lambda^{2}\int_{\{x\in B_{R}:|\nabla u(x+h)|+|\nabla u(x)|\geq\frac{\left|\Delta_{h}\nabla\psi(x)\right|}{\lambda}\}} a\big(|\nabla u(x+h)|+|\nabla u(x)|\big)\frac{|\Delta_{h}\nabla \psi(x)|^{2}}{\lambda^{2}} \textrm{d}x \nonumber\\ && \displaystyle +C(L,\mu)\lambda^{2}\int_{\{x\in B_{R}:|\nabla u(x+h)|+|\nabla u(x)|<\frac{|\Delta_{h}\nabla\psi(x)|}{\lambda}\}} a\big(|\nabla u(x+h)|+|\nabla u(x)|\big)\frac{|\Delta_{h}\nabla \psi(x)|^{2}}{\lambda^{2}} \textrm{d}x \nonumber\\ &\leq& \displaystyle C(L,\mu)\lambda^{2}\int_{B_R} a\big(|\nabla u(x+h)|+|\nabla u(x)|\big)\big(|\nabla u(x+h)|+|\nabla u(x)|\big)^{2} \textrm{d}x \nonumber\\ && \displaystyle +C(L,\mu)\lambda^{2}\int_{B_R}a\big(\frac{|\Delta_{h}\nabla\psi(x)|}{\lambda}\big)\big(\frac{|\Delta_{h}\nabla\psi(x)|}{\lambda}\big)^{2}\textrm{d}x \nonumber\\ &\leq& \displaystyle C(L,\mu)\lambda^{2}\int_{B_R}\tilde{\phi}[a\big(|\nabla u(x+h)|+|\nabla u(x)|\big)\big(|\nabla u(x+h)|+|\nabla u(x)|\big)]\textrm{d}x \nonumber\\ && \displaystyle +C(L,\mu)\lambda^{2}\int_{B_R}\phi\big(|\nabla u(x+h)|+|\nabla u(x)|\big)\textrm{d}x \nonumber\\ && \displaystyle +C(L,\mu)\lambda^{2}\int_{B_R}\tilde{\phi}[a\big(\frac{|\Delta_{h}\nabla\psi(x)|}{\lambda}\big)\frac{|\Delta_{h}\nabla\psi(x)|}{\lambda}]\textrm{d}x +C(L,\mu)\lambda^{2}\int_{B_R}\phi\big(\frac{|\Delta_{h}\nabla\psi(x)|}{\lambda}\big)\textrm{d}x \nonumber\\ &\leq& \displaystyle C(L,\mu,C_0)\lambda^{2}\int_{B_R}\phi\big(|\nabla u(x+h)|+|\nabla u(x)|\big)\textrm{d}x+C(L,\mu,C_0)\lambda^{2}\int_{B_R}\phi\big(\frac{|\Delta_{h}\nabla\psi(x)|}{\lambda}\big)\textrm{d}x \nonumber\\ &\leq& \displaystyle C(L,\mu,C_0,K)\lambda^{2}\int_{B_{R+|h|}}\phi\big(|\nabla u|\big)\textrm{d}x+C(L,\mu,C_0)\lambda^{2}\int_{B_{R}}\phi\big(\frac{|\Delta_{h}\nabla\psi(x)|}{\lambda}\big)\textrm{d}x, \end{eqnarray*}$

这里使用了不等式 $0\leq\eta\leq1$ 和定义 2.2. 取 $\lambda=\left\|\Delta_{h}\nabla\psi\right\|_{L^{\phi}{(B_R)}}>0,$ 根据 (2.1) 式, 得

$\begin{eqnarray*} && \displaystyle \int_{B_R}a\big(|\nabla u(x+h)|+|\nabla u(x)|\big)\left|\Delta_{h}\nabla \psi(x)\right|^{2}\eta^{2} \textrm{d}x \nonumber\\ &\leq& \displaystyle C(L,\mu,C_0,K)\left\|\Delta_{h}\nabla\psi\right\|^{2}_{L^{\phi}{(B_{R})}}\int_{B_{R+|h|}}\phi\big(|\nabla u|\big)\textrm{d}x+ C(L,\mu,C_0)\left\|\Delta_{h}\nabla\psi\right\|^{2}_{L^{\phi}{(B_{R})}}. \end{eqnarray*}$

下面估计 (3.10) 式右端第六个积分. 同理可得:

$\begin{eqnarray*} && \displaystyle \int_{B_R}\eta^2 a\left(|\nabla \psi(x+h)|+|\nabla \psi(x)|\right)\left|\Delta_{h}\nabla\psi\right|^2\textrm{d}x \nonumber\\ &\leq& \displaystyle C(\mu,C_0,K)\left\|\Delta_{h}\nabla\psi\right\|^{2}_{L^{\phi}{(B_{R})}}\int_{B_{R+|h|}}\phi(\left|\nabla \psi\right|)\textrm{d}x+C(\mu,C_0)\left\|\Delta_{h}\nabla\psi\right\|^{2}_{L^{\phi}{(B_{R})}}. \end{eqnarray*}$

将 (3.11) 式、(3.12) 式、(3.15)-(3.17) 式代入 (3.10) 式, 有

$\begin{eqnarray*} && \displaystyle \int_{B_{R}}\eta^{2} a\left(|\nabla u(x+h)|+|\nabla u(x)|\right)\left|\Delta_{h}(\nabla u(x))\right|^2 \textrm{d}x \nonumber\\ &\leq& \displaystyle C\left(\frac{|h|^2}{R^2}+\left\|\Delta_{h}\nabla\psi\right\|^{2}_{L^{\phi}{(B_{R})}}+|h|^2\right)\int_{B_{R+|h|}}\phi(\left|\nabla u\right|)\textrm{d}x \nonumber\\ && \displaystyle +C\left(\frac{|h|^2}{R^2}+\left\|\Delta_{h}\nabla\psi\right\|^{2}_{L^{\phi}{(B_{R})}}+|h|^2\right)\int_{B_{R+|h|}}\phi(\left|\nabla \psi\right|)\textrm{d}x +C\left\|\Delta_{h}\nabla\psi\right\|^{2}_{L^{\phi}{(B_{R})}} \nonumber\\ && \displaystyle +C|h|^{\frac{\beta(i_a+2)}{i_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{i_a+2}{2(i_a+1)}} +C|h|^{\frac{\beta(s_a+2)}{s_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{s_a+2}{2(s_a+1)}}_{,} \end{eqnarray*}$

其中正常数 $C=C(\varepsilon_4,L,\mu,C_0,K,i_a,s_a).$

因为由引理 2.4(3), 有

$\begin{eqnarray*} && \displaystyle \left|\Delta_{h}V(\nabla u)\right|^2=\left|V(\nabla u(x+h))-V(\nabla u(x))\right|^2\leq Ca(|\nabla u(x+h)|+|\nabla u(x)|)|\Delta_{h}\nabla u(x)|^2, \nonumber\\ \end{eqnarray*}$

所以根据截断函数 $\eta$ 的定义式、 (3.18) 式和 (3.19) 式, 有

$\begin{eqnarray*} && \displaystyle \nonumber \int_{B_{R/2}}\left|\Delta_{h}V(\nabla u)\right|^{2}\textrm{d}x\\ \nonumber\\ &\leq& \displaystyle \int_{B_R}\left|\Delta_{h}V(\nabla u)\right|^{2}\eta^{2}\textrm{d}x \nonumber\\ &\leq& \displaystyle C\int_{B_R}\eta^{2}a(\left|\nabla u(x+h)\right|+\left|\nabla u(x)\right|)\left|\Delta_{h}(\nabla u(x))\right|^{2}\textrm{d}x \nonumber\\ &\leq& \displaystyle C\left(\frac{|h|^2}{R^2}+\left\|\Delta_{h}\nabla\psi\right\|^{2}_{L^{\phi}{(B_{R})}}+|h|^2\right)\int_{B_{R+|h|}}\phi(\left|\nabla u\right|)\textrm{d}x \nonumber\\ && \displaystyle +C\left(\frac{|h|^2}{R^2}+\left\|\Delta_{h}\nabla\psi\right\|^{2}_{L^{\phi}{(B_{R})}}+|h|^2\right)\int_{B_{R+|h|}}\phi(\left|\nabla \psi\right|)\textrm{d}x +C\left\|\Delta_{h}\nabla\psi\right\|^{2}_{L^{\phi}{(B_{R})}} \nonumber\\ && \displaystyle +C|h|^{\frac{\beta(i_a+2)}{i_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{i_a+2}{2(i_a+1)}} +C|h|^{\frac{\beta(s_a+2)}{s_a+1}}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{s_a+2}{2(s_a+1)}}_{.} \end{eqnarray*}$

将 (3.20) 式两端同除 $\left|h\right|^{2\alpha},$ 得到

$\begin{eqnarray*} && \displaystyle \left(\int_{B_{R/2}}\left|\frac{\Delta_{h}V(\nabla u)}{\left|h\right|^{\alpha}}\right|^{2}\textrm{d}x\right)^{\frac{1}{2}} \nonumber\\ &\leq& \displaystyle C\left(|h|^{1-\alpha}+\frac{\left\|\Delta_{h}\nabla\psi\right\|_{L^{\phi}{(B_{R})}}}{|h|^{\alpha}}\right)\left(\int_{B_{R+|h|}}\phi(\left|\nabla u\right|)+\phi(\left|\nabla \psi\right|) \textrm{d}x\right)^{\frac{1}{2}} +C\frac{\left\|\Delta_{h}\nabla\psi\right\|_{L^{\phi}{(B_{R})}}}{|h|^{\alpha}} \nonumber\\ && \displaystyle +C|h|^{\frac{\beta(i_a+2)}{2(i_a+1)}-\alpha}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{i_a+2}{4(i_a+1)}} +C|h|^{\frac{\beta(s_a+2)}{2(s_a+1)}-\alpha}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{s_a+2}{4(s_a+1)}}_{,} \end{eqnarray*}$

这里 $C=C(\varepsilon_4,L,\mu,C_0,K,R,i_a,s_a).$ 不失一般性, 假设 $s_a>i_a.$ 因为 $0<\alpha<\frac{s_{a}+2}{2(s_{a}+1)}<1$$\frac{2\alpha(s_a+1)}{s_{a}+2}<\beta<1,$ 所以

$\begin{eqnarray*} \alpha<\frac{\beta(s_{a}+2)}{2(s_{a}+1)}\leq\frac{\beta(i_{a}+2)}{2(i_{a}+1)}. \end{eqnarray*}$

对 (3.21) 式取 $L^{q}$ 范数, 并将测度 $\frac{\textrm{d}h}{|h|^n}$ 限制在球 $B_{\delta}(|h|<\delta<R)$ 上, 再根据 (3.22) 式, 有

$\begin{matrix} &~~~\left(\int_{B_{\delta}}\left(\int_{B_{R/2}}\left|\frac{\Delta_h V(\nabla u)}{|h|^{\alpha}}\right|^{2}\textrm{d}x\right)^{\frac{q}{2}}\frac{\textrm{d}h}{|h|^{n}}\right)^{\frac{1}{q}} \nonumber\\ &\leq C\left[\int_{B_{\delta}}\left(|h|^{(1-\alpha)q}+\frac{\left\|\Delta_{h}\nabla\psi\right\|^{q}_{L^{\phi}{(B_{R})}}}{|h|^{q\alpha}} \right)\left(\int_{B_{2R}}\phi(\left|\nabla u\right|)\textrm{d}x\right)^{\frac{q}{2}}\frac{\textrm{d}h}{|h|^{n}}\right]^{\frac{1}{q}} \nonumber\\ &~~~+C\left[\int_{B_{\delta}}\left(|h|^{(1-\alpha)q}+\frac{\left\|\Delta_{h}\nabla\psi\right\|^{q}_{L^{\phi}{(B_{R})}}}{|h|^{q\alpha}}\right)\left(\int_{B_{2R}}\phi(\left|\nabla \psi\right|) \textrm{d}x\right)^{\frac{q}{2}}\frac{\textrm{d}h}{|h|^{n}}\right]^{\frac{1}{q}} \nonumber\\ &~~~+C\left(\int_{B_{\delta}}\frac{\left\|\Delta_{h}\nabla\psi\right\|^{q}_{L^{\phi}{(B_{R})}}}{|h|^{q\alpha}}\frac{\textrm{d}h}{|h|^{n}}\right)^{\frac{1}{q}} \nonumber\\ &~~~+C\left[\int_{B_{\delta}}|h|^{q\left(\frac{\beta(i_a+2)}{2(i_a+1)}-\alpha\right)}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{(i_a+2)q}{4(i_a+1)}} \frac{\textrm{d}h}{|h|^{n}}\right]^{\frac{1}{q}} \nonumber\\ &~~~+C\left[\int_{B_{\delta}}|h|^{q\left(\frac{\beta(s_a+2)}{2(s_a+1)}-\alpha\right)}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{(s_a+2)q}{4(s_a+1)}} \frac{\textrm{d}h}{|h|^{n}}\right]^{\frac{1}{q}} \nonumber\\ &\leq C\left(\int_{B_{2R}}\phi(\left|\nabla u\right|)\textrm{d}x\right)^{\frac{1}{2}}\left[\left(\int_{0}^{\delta}\rho^{(1-\alpha)q-1}\textrm{d}\rho\right)^{\frac{1}{q}} +\left(\int_{B_{\delta}}\frac{\left\|\Delta_{h}\nabla\psi\right\|^{q}_{L^{\phi}{(B_{R})}}}{|h|^{q\alpha}}\frac{\textrm{d}h}{|h|^{n}}\right)^{\frac{1}{q}}\right] \nonumber\\ &~~~+C\left(\int_{B_{2R}}\phi(\left|\nabla\psi\right|)\textrm{d}x\right)^{\frac{1}{2}}\left[\left(\int_{0}^{\delta}\rho^{(1-\alpha)q-1}\textrm{d}\rho\right)^{\frac{1}{q}} +\left(\int_{B_{\delta}}\frac{\left\|\Delta_{h}\nabla\psi\right\|^{q}_{L^{\phi}{(B_{R})}}}{|h|^{q\alpha}}\frac{\textrm{d}h}{|h|^{n}}\right)^{\frac{1}{q}}\right] \nonumber\\ &~~~+C\left(\int_{B_{\delta}}\frac{\left\|\Delta_{h}\nabla\psi\right\|^{q}_{L^{\phi}{(B_{R})}}}{|h|^{q\alpha}}\frac{\textrm{d}h}{|h|^{n}}\right)^{\frac{1}{q}} \nonumber\\ &~~~+C\left[\int_{B_{\delta}}|h|^{q\left(\frac{\beta(s_a+2)}{2(s_a+1)}-\alpha\right)}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{(s_a+2)q}{4(s_a+1)}} \frac{\textrm{d}h}{|h|^{n}}\right]^{\frac{1}{q}} \nonumber\\ &~~~+C\left[\int_{B_{\delta}}|h|^{q\left(\frac{\beta(i_a+2)}{2(i_a+1)}-\alpha\right)}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{(i_a+2)q}{4(i_a+1)}} \frac{\textrm{d}h}{|h|^{n}}\right]^{\frac{1}{q}}. \end{matrix}$

因为由 Hölder 不等式可得

$\begin{matrix} &\quad\left[\int_{B_\delta}|h|^{q\left(\frac{\beta(s_a+2)}{2(s_a+1)}-\alpha\right)}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right) ^{\frac{(s_a+2)q}{4(s_a+1)}}\frac{\mathrm{d}h}{|h|^n}\right]^{\frac{1}{q}}\nonumber\\ &=\left[\int_{B_\delta}|h|^{q\left(\frac{\beta(s_a+2)}{2(s_a+1)}-\alpha\right)}|h|^{-n\left[\frac{(i_a+1)(s_a+2)}{(i_a+2)(s_a+1)}+\frac{s_a-i_a}{(i_a+2)(s_a+1)}\right]} \left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{(s_a+2)q}{4(s_a+1)}}\mathrm{d}h\right]^{\frac{1}{q}}\nonumber\\ &\leq\left[\int_{B_{\delta}}|h|^{q\left(\frac{\beta(s_{a}+2)}{2(s_{a}+1)}-\alpha\right)\cdot\frac{(i_{a}+2)(s_{a}+1)}{s_{a}-i_{a}}}|h|^{-n}\mathrm{d}h\right] ^{\frac{s_{a}-i_{a}}{q(i_{a}+2)(s_{a}+1)}}\nonumber\\ &~~~\times\left[\int_{B_\delta}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^{\frac{(i_a+2)q} {4(i_a+1)}}\frac{\mathrm{d}h}{|h|^n}\right]^{\frac{(i_a+1)(s_a+2)}{q(i_a+2)(s_a+1)}}\nonumber\\ &\leq C(\mu,q,\alpha,\beta,i_a,s_a)\left\|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right\|_{L^{\frac{q(i_a+2)}{2(i_a+1)}}\left(\frac{\mathrm{d}h}{|h|^n};L^2(B_{2R})\right)}^{\frac{s_a+2}{2(s_a+1)}}\nonumber \end{matrix}$

$\begin{eqnarray*} &&\left[\int_{B_\delta}|h|^{q\left(\frac{\beta(i_a+2)}{2(i_a+1)}-\alpha\right)}\left(\int_{B_R}\left|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right|^2\mathrm{d}x\right)^ {\frac{(i_a+2)q}{4(i_a+1)}}\frac{\textrm{d}h}{|h|^n}\right]^{\frac{1}{q}}\\ &\leq& C(\mu,q,\alpha,\beta,i_a,s_a)\left\|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right\|_{L^{\frac{q(i_a+2)}{2(i_a+1)}}\left(\frac{\mathrm{d}h}{|h|^n};L^2(B_{2R})\right)}^{\frac{i_a+2}{2(i_a+1)}}, \end{eqnarray*}$

所以 (3.23) 式进一步简化为

$\begin{eqnarray*} && \displaystyle \left(\int_{B_{\delta}}\left(\int_{B_{R/2}}\left|\frac{\Delta_h V(\nabla u)}{|h|^{\alpha}}\right|^{2}\textrm{d}x\right)^{\frac{q}{2}}\frac{\textrm{d}h}{|h|^{n}}\right)^{\frac{1}{q}} \nonumber\\ &\leq& \displaystyle C\left(\int_{B_{2R}}\phi(\left|\nabla u\right|)\textrm{d}x\right)^{\frac{1}{2}}\left[\left(\int_{0}^{\delta}\rho^{(1-\alpha)q-1}\textrm{d}\rho\right)^{\frac{1}{q}} +\left(\int_{B_{\delta}}\frac{\left\|\Delta_{h}\nabla\psi\right\|^{q}_{L^{\phi}{(B_{R})}}}{|h|^{q\alpha}}\frac{\textrm{d}h}{|h|^{n}}\right)^{\frac{1}{q}}\right] \nonumber\\ && \displaystyle +C\left(\int_{B_{2R}}\phi(\left|\nabla\psi\right|)\textrm{d}x\right)^{\frac{1}{2}}\left[\left(\int_{0}^{\delta}\rho^{(1-\alpha)q-1}\textrm{d}\rho\right)^{\frac{1}{q}} +\left(\int_{B_{\delta}}\frac{\left\|\Delta_{h}\nabla\psi\right\|^{q}_{L^{\phi}{(B_{R})}}}{|h|^{q\alpha}}\frac{\textrm{d}h}{|h|^{n}}\right)^{\frac{1}{q}}\right] \nonumber\\ && \displaystyle +C\left(\int_{B_{\delta}}\frac{\left\|\Delta_{h}\nabla\psi\right\|^{q}_{L^{\phi}{(B_{R})}}}{|h|^{q\alpha}}\frac{\textrm{d}h}{|h|^{n}}\right)^{\frac{1}{q}} \nonumber\\ && \displaystyle +C\left\|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right\|_{L^{\frac{q(i_a+2)}{2(i_a+1)}}\left(\frac{\mathrm{d}h}{|h|^n};L^2(B_{2R})\right)}^{\frac{s_a+2}{2(s_a+1)}} +C\left\|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right\|_{L^{\frac{q(i_a+2)}{2(i_a+1)}}\left(\frac{\mathrm{d}h}{|h|^n};L^2(B_{2R})\right)}^{\frac{i_a+2}{2(i_a+1)}} \nonumber\\ &\leq& \displaystyle C\left(\int_{B_{2R}}\phi(\left|\nabla u\right|)\textrm{d}x\right)^{\frac{1}{2}} \left[1+\left(\int_{B_{\delta}}\frac{\left\|\Delta_{h}\nabla\psi\right\|^{q}_{L^{\phi}{(B_{R})}}}{|h|^{q\alpha}}\frac{\textrm{d}h}{|h|^{n}}\right)^{\frac{1}{q}}\right] \nonumber\\ && \displaystyle +C\left(\int_{B_{2R}}\phi(\left|\nabla \psi\right|)\textrm{d}x\right)^{\frac{1}{2}} \left[1+\left(\int_{B_{\delta}}\frac{\left\|\Delta_{h}\nabla\psi\right\|^{q}_{L^{\phi}{(B_{R})}}}{|h|^{q\alpha}}\frac{\textrm{d}h}{|h|^{n}}\right)^{\frac{1}{q}}\right] \nonumber\\ && \displaystyle +C\left(\int_{B_{\delta}}\frac{\left\|\Delta_{h}\nabla\psi\right\|^{q}_{L^{\phi}{(B_{R})}}}{|h|^{q\alpha}}\frac{\textrm{d}h}{|h|^{n}}\right)^{\frac{1}{q}} \nonumber\\ && \displaystyle +C\left\|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right\|_{L^{\frac{q(i_a+2)}{2(i_a+1)}}\left(\frac{\mathrm{d}h}{|h|^n};L^2(B_{2R})\right)}^{\frac{s_a+2}{2(s_a+1)}} +C\left\|\frac{\Delta_h\mathbf{f}}{|h|^\beta}\right\|_{L^{\frac{q(i_a+2)}{2(i_a+1)}}\left(\frac{\mathrm{d}h}{|h|^n};L^2(B_{2R})\right)}^{\frac{i_a+2}{2(i_a+1)}} \nonumber\\ &<& \displaystyle \infty, \end{eqnarray*}$

这里 $C=C(\varepsilon_4,L,\mu,C_0,K,R,q,\alpha,\beta,i_a,s_a,\delta)$. 因为 $\nabla\psi\in B^{\alpha}_{\phi,q,loc}(\Omega)\cap L^{\phi}_{loc}(\Omega),u\in W_{loc}^{1,\phi}(\Omega)$$\mathbf{f}\in B_{2,\frac{q(i_a+2)}{2(i_a+1)},loc}^\beta(\Omega)\cap L^{\tilde{\phi}}_{loc}(\Omega),$ 故对任意 $h$$0<\alpha<\frac{s_a+2}{2(s_a+1)}<1,$ 上式都有界, 因此 $V(\nabla u)\in B_{2,q,loc}^{\alpha}(\Omega).$ 定理 1.1 得证.

在定理 1.1 中取 $a(t)=t^{p-2}(p\geq 2)$, 此时 $\phi(t)=t^p/p$, $\widetilde{\phi}(t)=\frac{p-1}{p}t^{\frac{p}{p-1}}$, $V(\xi)=\sqrt{|\xi|^{p-2}}\xi$, 结论仍然成立.

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