Assume that the function \(u\in W_{X}^{1,p-\delta}(\Omega)(\delta <\frac{1}{2})\) is a very weak solution to the Dirichlet problem (1.2). Choose a ball \(B_{0}\) such that \(\overline{\Omega}\subset\frac {1}{2}B_{0}\) and let B be a ball of radius R with \(3B\subset B_{0}\) for fixed \(0< R<1\). There are two cases: (i) \(3B\subset\Omega\) or (ii) \(3B\backslash\Omega\neq\emptyset\). In the case (i), the following estimate has been proved in [7]:
(3.1)
where θ small enough, \(b>1\), \(\max \{ 1,(p-\delta)_{*} \} < t< p-\delta\).
When \(3B\backslash\Omega\neq\emptyset\), a similar inequality (see (3.31) below) will be achieved.
Step 1. Let η be a smooth cut-off function on 2B, i.e. \(\eta\in C_{0}^{\infty}(2B)\) such that
$$ 0\leq\eta\leq1, \qquad\eta=1\quad \mbox{on } B \quad\mbox{and} \quad \vert X \eta \vert \leq c/R. $$
Define \(\hat{u}=\eta(u-u_{0})\) and
$$ E_{\mu}=\bigl\{ x\in\mathbf{R}^{n}:M\vert X \hat{u} \vert (x)\leq\mu\bigr\} , \quad\operatorname{for } \mu>0. $$
(3.2)
We conclude from Lemma 2.5 and the assumption \((H_{1})\) that û is Lipschitz continuous on \(E_{\mu}\cup(\mathbf {R}^{n}\setminus\Omega)\).
Indeed, if \(x,y\in E_{\mu}\cap\Omega\), then Lemma 2.5 implies \(\vert \hat{u}(x)-\hat{u}(y) \vert \leq c\mu d(x,y)\); if \(x,y\in\mathbf{R}^{n}\setminus\Omega\), then \(\hat{u}(x)=\hat {u}(y)=0\). We set \(B_{\rho_{x}}=B(x,\rho_{x})\) with \(\rho _{x}=2\operatorname{dist}(x,\mathbf{R}^{n}\setminus\Omega)\) for the case \(x\in E_{\mu}\cap\Omega\) and \(y\in\mathbf{R}^{n}\setminus\Omega\). Since û is zero on \(\mathbf{R}^{n}\setminus\Omega\), it follows that
$$\begin{aligned} \int_{{B_{{\rho_{x}}}} \cap({\mathbf{{R}}^{n}}\setminus\Omega )}\vert \hat{u} - {\hat{u}_{{B_{{\rho_{x}}}}}} \vert \,dz & =\int_{{B_{{\rho_{x}}}} \cap({\mathbf{{R}}^{n}}\setminus\Omega)} \vert {\hat{u}_{{B_{{\rho_{x}}}}}} \vert \,dz \\ &=\bigl\vert {B_{{\rho_{x}}}} \cap\bigl({\mathbf{{R}}^{n}} \setminus\Omega\bigr) \bigr\vert \vert {\hat{u}_{{B_{{\rho _{x}}}}}} \vert \end{aligned}$$
and then, from assumption \((H_{1})\) and Lemma 2.2,
Therefore, we have by (2.2) and (3.2)
$$\begin{aligned} \bigl\vert \hat{u}(x)-\hat{u}(y) \bigr\vert &=\bigl\vert \hat{u}(x) \bigr\vert \\ &\leq\bigl\vert \hat{u}(x)-\hat{u}_{B_{\rho_{x}}} \bigr\vert +\vert \hat{u}_{B_{\rho_{x}}} \vert \\ &\leq c{\rho_{x}}M\vert X\hat{u} \vert (x)+c C_{1}\mu {\rho_{x}} \\ &\leq c C_{1}\mu{\rho_{x}} \\ &\leq cC_{1}\mu d(x,y). \end{aligned}$$
It follows that û is a Lipschitz function on \(E_{\mu}\cup (\mathbf{R}^{n}\setminus\Omega)\) with the Lipschitz constant \(cC_{1}\mu\).
As in [7], we can use the Kirszbraun theorem (see e.g. [17]) to extend û to a Lipschitz function \(v_{\mu}\) defined on \(\mathbf{R}^{n}\) with the same Lipschitz constant. Moreover, there exists \(\mu_{0}\) such that, for every \(\mu\geq\mu_{0}\), \(\operatorname{supp}{v_{\mu}}\subset3B\cap\Omega\).
In fact, let \(D=2B\cap\Omega\) and \(x\in\mathbf{R}^{n}\backslash (3B\cap\Omega)\), we have by Lemma 2.1 that
where \(\vert B' \vert >\vert B \vert \), \(C_{d}\) is the doubling constant. Setting
$$ \mu_{0}=\frac{C_{d}}{\vert 2B \vert } \int_{D}\vert X\hat{u} \vert (y)\,dy, $$
then \(M\vert X\hat{u} \vert (x)\leq\mu, \mu\geq\mu_{0}\), which implies \(v_{\mu}(x)=\hat{u}(x)=0\) for \(x\in\mathbf {R}^{n}\backslash(3B\cap\Omega)\). So we can take the function \(v_{\mu}\) as a test function in (1.6).
Let \(\mu\geq\mu_{0}\) and take \(v_{\mu}\) as a test function in (1.6) to have
$$\begin{aligned} \int_{3B \cap\Omega}A(x,u,Xu) \cdot X{v_{\mu}}\,dx + \int_{3B \cap\Omega} B(x,u,Xu){v_{\mu}}\,dx=0. \end{aligned}$$
Noting that \({v_{\mu}}=\hat{u}\) on \((3B\cap\Omega)\cap{E_{\mu}}\) and that \(\operatorname{supp}\hat{u}\subset D\), we have by the structure conditions on \(A(x,u,\xi)\) and \(B(x,u,\xi)\)
$$\begin{aligned} & \int_{D\cap E_{\mu}} A(x,u,Xu)\cdot X\hat{u}\,dx+ \int_{D\cap E_{\mu}} B(x,u,Xu)\hat{u}\,dx \\ &\quad\leq \int_{(3B\cap\Omega)\backslash E_{\mu}} \bigl\vert A(x,u,Xu) \bigr\vert \vert Xv_{\mu} \vert \,dx+ \int_{(3B\cap\Omega)\backslash E_{\mu}} \bigl\vert B(x,u,Xu) \bigr\vert \vert v_{\mu} \vert \,dx \\ &\quad\leq c\mu \int_{(3B\cap\Omega)\backslash E_{\mu}} \bigl( \vert u \vert ^{p-1}+\vert Xu \vert ^{p-1} \bigr)\,dx, \end{aligned}$$
(3.3)
where in the last inequality we use the fact that \(\vert Xv_{\mu} \vert \leq c\mu\), \(\vert v_{\mu} \vert \leq cR\mu\) (see [7]).
Multiplying both sides of (3.3) by \(\mu^{-(1+\delta)}\) and integrating over \((\mu_{0},\infty)\), we get
$$\begin{aligned} L&:= \int_{\mu_{0}}^{\infty} \int_{D\cap E_{\mu}}{\mu}^{-(1+\delta)} \bigl( A(x,u,Xu)\cdot X \hat{u}+B(x,u,Xu)\hat{u} \bigr) \,dx\,d\mu \\ &\leq c \int_{\mu_{0}}^{\infty} \int_{(3B\cap\Omega)\backslash E_{\mu}}\mu^{-\delta} \bigl( \vert u \vert ^{p-1}+\vert Xu \vert ^{p-1} \bigr) \,dx\,d\mu:=P. \end{aligned}$$
(3.4)
Interchanging the order of integration and applying (3.2), we have
$$\begin{aligned} P &=c \int_{3B} \int_{{\mu_{0}}}^{M\vert X\hat{u} \vert } {{\mu^{ - \delta}} \bigl( { \vert u{ \vert ^{p - 1}} + \vert Xu{ \vert ^{p - 1}}} \bigr)\, d\mu\,dx} \\ &\leq\frac{c}{1-\delta} \int_{(3B\cap\Omega)\backslash E_{\mu_{0}}}\bigl(M\vert X\hat{u} \vert \bigr)^{1-\delta}\bigl(\vert u \vert ^{p-1}+\vert Xu \vert ^{p-1}\bigr)\,dx \\ &\leq c \int_{3B\cap\Omega} \bigl( \vert u \vert ^{p-\delta}+\vert Xu \vert ^{p-\delta} \bigr)\,dx+c \int_{3B\cap\Omega}\bigl(M\vert X\hat{u} \vert \bigr)^{p-\delta}\,dx. \end{aligned}$$
(3.5)
Using Lemma 2.4 and Lemma 2.8, we have
$$\begin{aligned} & c \int_{3B\cap\Omega}\bigl(M\vert X\hat{u} \vert \bigr)^{p-\delta}\,dx \\ &\quad \leq c \int_{D}\vert X\hat{u} \vert ^{p-\delta}\,dx \\ &\quad \leq c \int_{D}\vert Xu-Xu_{0} \vert ^{p-\delta}\,dx+\frac{c}{R^{p-\delta}} \int_{2B}\vert u-u_{0} \vert ^{p-\delta}\,dx \\ &\quad \leq c \int_{D}\vert Xu-Xu_{0} \vert ^{p-\delta}\,dx +\frac{c\vert 2B \vert }{R^{p-\delta}} \biggl( \frac{1}{\operatorname{cap}_{p-\delta}(N(u-u_{0}),2B)} \int_{2B}\vert Xu-Xu_{0} \vert ^{p-\delta}\,dx \biggr) , \end{aligned}$$
where \(N(u-u_{0})= \{ x\in\bar{B}:u(x)=u_{0}(x) \} \). Since \(u-u_{0}\) vanishes outside Ω, we have \(\mathbf{R}^{n}\setminus\Omega \subset\{u-u_{0}=0\}\). On the other hand, by Lemma 2.7 and assumption \((H_{2})\), there exists \(\delta_{0}\) such that if \(0<\delta<\delta _{0}\), \(\mathbf{R}^{n}\setminus\Omega\) is uniformly \((X,p-\delta )\)-fat, and hence
$$\begin{aligned} {\operatorname{cap}}_{p-\delta}\bigl(N(u-u_{0}),2B \bigr)&\geq{\operatorname{cap}}_{p-\delta}\bigl(\bar{B}\cap\bigl( \mathbf{R}^{n}\setminus\Omega\bigr),2B\bigr) \\ &\geq c \operatorname{cap}_{p-\delta}(\bar{B},2B)\geq c\vert B \vert R^{-(p-\delta)}. \end{aligned}$$
(3.6)
From (3.6) and the doubling condition, we derive
$$\begin{aligned} c \int_{3B\cap\Omega}\bigl(M\vert X\hat{u} \vert \bigr)^{p-\delta}\,dx&\leq c \int_{D}\vert X\hat{u} \vert ^{p-\delta}\,dx \\ &\leq c \int_{D}\vert Xu \vert ^{p-\delta}\,dx+c \int_{D}\vert Xu_{0} \vert ^{p-\delta}\,dx, \end{aligned}$$
(3.7)
and then (3.5) becomes
$$ P\leq c \int_{3B\cap\Omega} \vert u \vert ^{p-\delta}\,dx+c \int_{3B\cap\Omega} \vert Xu_{0} \vert ^{p-\delta}\,dx+c \int_{3B\cap\Omega} \vert Xu \vert ^{p-\delta}\,dx. $$
(3.8)
As regards the estimation of L, by changing the order of integration, we have
$$\begin{aligned} L&= \int_{\mu_{0}}^{\infty} \int_{D}\mu^{-(1 +\delta)}\bigl(A(x,u,Xu)\cdot X\hat{u} + B(x,u,Xu)\hat{u}\bigr)\chi_{\{M\vert X\hat{u} \vert (x)\leq\mu\} }\,dx\,d\mu \\ &= \int_{D\backslash{E_{\mu_{0}}}} \int_{M\vert X\hat{u} \vert }^{\infty}{\mu^{-(1 + \delta )}}\bigl(A(x,u,Xu) \cdot X\hat{u} + B(x,u,Xu)\hat{u}\bigr)\,dx\,d\mu \\ & \quad {}+ \int_{D \cap{E_{{\mu_{0}}}}} \int_{{\mu_{0}}}^{\infty}{\mu^{-(1 +\delta)}}\bigl(A(x,u,Xu) \cdot X\hat{u} + B(x,u,Xu)\hat{u}\bigr)\,dx\,d\mu \\ &=\frac{1}{\delta} \int_{D\backslash E_{\mu_{0}}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}\bigl(A(x,u,Xu)\cdot X\hat{u}+B(x,u,Xu)\hat{u}\bigr)\,dx \\ &\quad{}+\frac{1}{\delta} \int_{D\cap E_{\mu_{0}}}\mu_{0}^{-\delta}\bigl(A(x,u,Xu) \cdot X\hat{u}+B(x,u,Xu)\hat{u}\bigr)\,dx. \end{aligned}$$
Since \(D\setminus E_{\mu_{0}}=D\setminus(D\cap E_{\mu_{0}})\), (1.3) and (1.4) imply
$$\begin{aligned} L&=\frac{1}{\delta} \int_{D}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}A(x,u,Xu)\cdot X\hat{u}\,dx-\frac{1}{\delta} \int_{D\cap E_{\mu_{0}}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}A(x,u,Xu)\cdot X\hat{u}\,dx \\ &\quad{}+\frac{1}{\delta} \int_{D\backslash E_{\mu_{0}}} \bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}B(x,u,Xu)\hat{u}\,dx \\ &\quad{}+\frac{1}{\delta} \int_{D\cap E_{\mu_{0}}}\mu_{0}^{-\delta}\bigl(A(x,u,Xu) \cdot X\hat{u}+B(x,u,Xu)\hat{u}\bigr)\,dx \\ &\geq\frac{1}{\delta} \int_{D}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}A(x,u,Xu)\cdot X\hat{u}\,dx \\ &\quad{}-\frac{2\alpha}{\delta} \int_{D\cap E_{\mu_{0}}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}\bigl(\vert u \vert ^{p-1}+\vert Xu \vert ^{p-1}\bigr)\vert X\hat{u} \vert \,dx \\ &\quad{}-\frac{\alpha}{\delta} \int_{D}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}\bigl(\vert u \vert ^{p-1}+\vert Xu \vert ^{p-1}\bigr)\vert \hat{u} \vert \,dx \\ &:=\frac{1}{\delta}(I_{1}-2\alpha I_{2}-\alpha I_{3}). \end{aligned}$$
(3.9)
Step 2. Next, we will estimate \(I_{i}\) (\(i=1,2,3\)) one by one.
Now for estimation of \(I_{1}\). To this end, define the sets
$$\begin{aligned}& {D_{1}} = \bigl\{ x \in D\setminus B:M\vert X\hat{u} \vert \le \delta\bigl({M_{D}}\vert Xu-Xu_{0} \vert \bigr)\bigr\} , \\& {D_{2}}= \bigl\{ x \in D\setminus B:M\vert X\hat{u} \vert > \delta \bigl({M_{D}}\vert Xu-Xu_{0} \vert \bigr)\bigr\} \end{aligned}$$
and \(B_{\Omega}=B\cap\Omega\). Thus
$$\begin{aligned} I_{1} &= \int_{B_{\Omega}\cup D_{2}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}\bigl(A(x,u,Xu)-A(x,u_{0},Xu_{0})\bigr) \cdot\eta X(u-u_{0})\,dx \\ &\quad{} + \int_{B_{\Omega}\cup D_{2}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}A(x,u_{0},Xu_{0})\cdot\eta (Xu-Xu_{0})\,dx \\ &\quad{} + \int_{D_{2}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}A(x,u,Xu)\cdot X\eta(u-u_{0})\,dx \\ &\quad{} + \int_{D_{1}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}A(x,u,Xu)\cdot X\hat{u}\,dx. \end{aligned}$$
Since \(( M\vert X\hat{u} \vert ) ^{-\delta}\le \vert X\hat{u} \vert ^{-\delta}\) a.e., it follows from (1.5) and (1.3) that
$$\begin{aligned} I_{1}&\geq\beta \int_{B_{\Omega}}\bigl( M\vert X\hat{u} \vert \bigr)^{-\delta} \vert Xu-Xu_{0} \vert ^{p}\,dx \\ &\quad{} -\alpha\biggl( \int_{B_{\Omega}} \vert X\hat{u} \vert ^{-\delta} \bigl( \vert u_{0} \vert ^{p-1}+ \vert Xu_{0} \vert ^{p-1} \bigr) \\ &\quad{} \times \vert Xu-Xu_{0} \vert \,dx + \int_{D_{2}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}\bigl(\vert u_{0} \vert ^{p-1}+\vert Xu_{0} \vert ^{p-1}\bigr)\vert Xu-Xu_{0} \vert \,dx \\ &\quad{} + \int_{D_{2}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}\bigl(\vert u \vert ^{p-1}+\vert Xu \vert ^{p-1}\bigr)\bigl\vert X\eta(u-u_{0}) \bigr\vert \,dx \\ &\quad{} + \int_{D_{1}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}\bigl(\vert u \vert ^{p-1}+\vert Xu \vert ^{p-1}\bigr)\vert X\hat{u} \vert \,dx \biggr) \\ &:=I_{11}-\alpha(I_{12}+I_{13}+I_{14}+I_{15}). \end{aligned}$$
(3.10)
Since the function \((M\vert X\hat{u} \vert )^{-\delta}\) is an \(A_{p}\)-weight, we obtain from Lemma 2.6 that
$$ I_{11}\geq c\beta \int_{B_{\Omega}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}\bigl(M_{B_{\Omega}} \vert Xu-Xu_{0} \vert \bigr)^{p}\,dx. $$
By the doubling condition and Lemma 2.8 we see that, for \(x\in\frac{B}{2}\cap\Omega\),
(3.11)
where \(\max \{ 1,(p-\delta)_{*} \} < s'< p-\delta\) is such that \(\mathbf{R}^{n}\setminus\Omega\) is uniformly \((X,s')\)-fat and the last inequality comes from an argument similar to (3.6).
To continue, we define
$$ G= \biggl\{ x\in\frac{B}{2}\cap\Omega:M_{B_{\Omega}}\bigl\vert X(u-u_{0}) \bigr\vert \geq c \biggl( \frac{1}{\vert 2B \vert } \int_{D}\vert Xu-Xu_{0} \vert ^{s'}\,dx \biggr) ^{\frac{1}{s'}} \biggr\} . $$
So from (3.11) we see that \(M\vert X\hat{u} \vert \leq cM_{B_{\Omega}} \vert X(u-u_{0}) \vert \) on G, and then
$$\begin{aligned} I_{11}&\geq c \int_{G}\bigl(M_{B_{\Omega}} \vert Xu-Xu_{0} \vert \bigr)^{-\delta}\bigl(M_{B_{\Omega}} \vert Xu-Xu_{0} \vert \bigr)^{p}\,dx \\ &\geq c \int_{\frac{B}{2}\cap\Omega} \vert Xu-Xu_{0} \vert ^{p-\delta}\,dx-c\vert B \vert \biggl( \frac{1}{\vert 2B \vert } \int_{D}\vert Xu-Xu_{0} \vert ^{s'}\,dx \biggr) ^{\frac{p-\delta}{s'}} \\ &\geq c \int_{\frac{B}{2}\cap\Omega} \vert Xu \vert ^{p-\delta}\,dx-c \int_{D}\vert Xu_{0} \vert ^{p-\delta}\,dx-c\vert B \vert \biggl( \frac{1}{\vert 2B \vert } \int_{D}\vert Xu \vert ^{s'}\,dx \biggr) ^{\frac{p-\delta}{s'}}. \end{aligned}$$
(3.12)
Using the fact \(X\hat{u}=X(u-u_{0})\) on B and Young’s inequality, we have
$$\begin{aligned} I_{12} &\leq c \int_{D} \bigl( \vert u_{0} \vert ^{p-\delta}+\vert Xu_{0} \vert ^{p-\delta} \bigr)\,dx +c \varepsilon \int_{D}\vert Xu-Xu_{0} \vert ^{p-\delta}\,dx \\ &\leq c \int_{D}\vert u_{0} \vert ^{p-\delta}\,dx+c \int_{D}\vert Xu_{0} \vert ^{p-\delta}\,dx+c\varepsilon \int_{D}\vert Xu \vert ^{p-\delta}\,dx. \end{aligned}$$
(3.13)
Next from the definition of \(D_{2}\) and Lemma 2.4, we see
$$\begin{aligned} I_{13} &\leq c\delta^{-\delta} \int_{D}\bigl(M_{D}\vert Xu-Xu_{0} \vert \bigr)^{1-\delta}\bigl(\vert u_{0} \vert ^{p-1}+\vert Xu_{0} \vert ^{p-1}\bigr)\,dx \\ &\leq c \int_{D} \bigl( \vert u_{0} \vert ^{p-\delta}+\vert Xu_{0} \vert ^{p-\delta} \bigr)\,dx +c \varepsilon \int_{D}\vert Xu-Xu_{0} \vert ^{p-\delta}\,dx \\ &\leq c \int_{D}\vert u_{0} \vert ^{p-\delta}\,dx+c \int_{D}\vert Xu_{0} \vert ^{p-\delta}\,dx+c\varepsilon \int_{D}\vert Xu \vert ^{p-\delta}\,dx. \end{aligned}$$
(3.14)
For \(I_{14}\), we have by using \(\vert X\eta(u-u_{0}) \vert \leq \vert X\hat{u} \vert +\vert Xu-Xu_{0} \vert \)
$$\begin{aligned} I_{14} &\leq c \int_{D_{2}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}\bigl(\vert u \vert ^{p-1}+\vert Xu-Xu_{0} \vert ^{p-1}+\vert Xu_{0} \vert ^{p-1}\bigr)\bigl\vert X\eta(u-u_{0}) \bigr\vert \,dx \\ &\leq c \int_{D_{2}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}\bigl(\vert u \vert ^{p-1}+\vert Xu_{0} \vert ^{p-1}\bigr) \bigl(\vert X\hat{u} \vert + \vert Xu-Xu_{0} \vert \bigr)\,dx \\ & \quad {} +c \int_{D_{2}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta} \vert Xu-Xu_{0} \vert ^{p-1}\vert X \eta \vert \vert u-u_{0} \vert \,dx \\ &\leq c \int_{D_{2}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}\bigl(\vert u \vert ^{p-1}+\vert Xu_{0} \vert ^{p-1}\bigr)\vert X\hat{u} \vert \,dx \\ & \quad {} +c \int_{D_{2}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta}\bigl(\vert u \vert ^{p-1}+\vert Xu_{0} \vert ^{p-1}\bigr)\vert Xu-Xu_{0} \vert \,dx \\ & \quad {} +\frac{c}{R} \int_{D_{2}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta} \vert Xu-Xu_{0} \vert ^{p-1}\vert u-u_{0} \vert \,dx \\ &:=K_{1}+K_{2}+K_{3}. \end{aligned}$$
Using Young’s inequality and (3.7), we get
$$\begin{aligned} K_{1}&\leq c \int_{D}\bigl(\vert u \vert ^{p-1}+\vert Xu_{0} \vert ^{p-1}\bigr)\vert X\hat{u} \vert ^{1-\delta}\,dx \\ &\leq c \int_{D}\vert u \vert ^{p-\delta}\,dx+c \int_{D}\vert Xu_{0} \vert ^{p-\delta}\,dx+c\varepsilon \int_{D}\vert X{u} \vert ^{p-\delta}\,dx. \end{aligned}$$
(3.15)
By the definition of \(D_{2}\) and noting that \(\vert X(u-u_{0}) \vert \le M_{D}\vert X(u-u_{0}) \vert \) a.e. D,
$$\begin{aligned} K_{2} &\leq c\delta^{-\delta} \int_{D}\vert Xu-Xu_{0} \vert ^{1-\delta} \bigl(\vert u \vert ^{p-1}+\vert Xu_{0} \vert ^{p-1}\bigr)\,dx \\ &\leq c \int_{D}\vert u \vert ^{p-\delta}\,dx+c \int_{D}\vert Xu_{0} \vert ^{p-\delta}\,dx+c\varepsilon \int_{D}\vert Xu \vert ^{p-\delta}\,dx. \end{aligned}$$
(3.16)
Finally, by Young’s inequality,
$$\begin{aligned} K_{3} &\leq\frac{c\delta^{-\delta}}{R} \int_{D_{2}}\bigl(M_{D}\vert Xu-Xu_{0} \vert \bigr)^{-\delta} \vert Xu-Xu_{0} \vert ^{p-1} \vert u-u_{0} \vert \,dx \\ &\leq\frac{c\delta^{-\delta}}{R} \int_{D}\vert Xu-Xu_{0} \vert ^{p-1-\delta} \vert u-u_{0} \vert \,dx \\ &\leq c\varepsilon \int_{{D}}\vert Xu-Xu_{0} \vert ^{p -\delta}\,dx + c \int_{{D}}{\biggl\vert {\frac{{u-{u_{0}}}}{R}} \biggr\vert ^{p-\delta}}\,dx. \end{aligned}$$
(3.17)
In order to estimate the second component of the right-hand side, we let \(s''=(p-\delta)(1-\vartheta)\), where \(0<\vartheta<\frac{p-\delta }{p-\delta+Q}\) if \(p-\delta\leq Q\) and \(0<\vartheta<\min \{ \frac {p-\delta-Q}{p-\delta},\frac{1}{2} \} \) if \(p-\delta> Q\). Denote
$$ \kappa= \textstyle\begin{cases} \frac{Q}{Q-s''},&s''< Q,\\ 2,&s''>Q, \end{cases} $$
then \(\kappa s''\geq p-\delta\). Using Lemma 2.7 and Lemma 2.8, we derive
where the proof of the last inequality is similar to (3.6). Therefore,
$$ c \int_{2B}\biggl\vert \frac{u-u_{0}}{R}\biggr\vert ^{p-\delta}\,dx \leq c\vert 2B \vert \biggl( \frac{1}{\vert 2B \vert } \int_{D}\vert Xu-Xu_{0} \vert ^{s''}\,dx \biggr) ^{\frac{{p-\delta}}{s''}}. $$
(3.18)
Inserting (3.18) into (3.17), we have
$$\begin{aligned} K_{3}&\leq c\varepsilon \int_{D} \vert Xu \vert ^{p -\delta}\,dx + c \int_{D} \vert Xu_{0} \vert ^{p - \delta}\,dx \\ &\quad {}+ c\vert 2B \vert \biggl( \frac{1}{\vert 2B \vert } \int_{D} \vert Xu \vert ^{s''}\,dx \biggr) ^{\frac{{p-\delta}}{s''}}. \end{aligned}$$
(3.19)
A combination of (3.15), (3.16) and (3.19) implies
$$\begin{aligned} I_{14}&\leq c \int_{D}\bigl(\vert u \vert ^{p-\delta}+\vert Xu_{0} \vert ^{p-\delta}\bigr)\,dx \\ &\quad{} +c\varepsilon \int_{D}\vert Xu \vert ^{p-\delta}\,dx + c\vert 2B \vert \biggl( \frac{1}{\vert 2B \vert } \int_{D} \vert Xu \vert ^{s''}\,dx \biggr) ^{\frac{{p-\delta}}{s''}}. \end{aligned}$$
(3.20)
The definition of \(D_{1}\) and Lemma 2.4 give
$$\begin{aligned} I_{15} &\leq c \int_{D_{1}}\bigl(M\vert X\hat{u} \vert \bigr)^{1-\delta}\bigl(\vert u \vert ^{p-1}+\vert Xu \vert ^{p-1}\bigr)\,dx \\ &\leq c\delta^{1-\delta} \int_{D}\bigl(M_{D}\vert Xu-Xu_{0} \vert \bigr)^{1-\delta}\bigl(\vert u \vert ^{p-1}+\vert Xu \vert ^{p-1}\bigr)\,dx \\ &\leq c\delta^{1-\delta} \biggl[ \int_{D}\vert Xu-Xu_{0} \vert ^{p-\delta}\,dx + \int_{D} \bigl( \vert u \vert ^{p-\delta}+\vert Xu \vert ^{p-\delta} \bigr)\,dx \biggr] \\ &\leq c \int_{D} \bigl( \vert u \vert ^{p-\delta}+\vert Xu_{0} \vert ^{p-\delta} \bigr)\,dx+c\delta \int_{D}\vert Xu \vert ^{p-\delta}\,dx. \end{aligned}$$
(3.21)
The previous estimates show that
$$\begin{aligned} I_{1}&\geq c \int_{\frac{B}{2}\cap\Omega} \vert Xu \vert ^{p-\delta}\,dx -c \int_{D} \bigl( \vert u \vert ^{p-\delta}+\vert u_{0} \vert ^{p-\delta}+\vert Xu_{0} \vert ^{p-\delta} \bigr)\,dx \\ & \quad{} - c(\varepsilon+\delta) \int_{D}\vert Xu \vert ^{p-\delta}\,dx-c\vert 2B \vert \biggl( \frac{1}{\vert 2B \vert } \int_{D}\vert Xu \vert ^{t}\,dx \biggr) ^{\frac{p-\delta}{t}}, \end{aligned}$$
(3.22)
where \(t=\max\{s',s''\}< p-\delta\).
Now we address the estimation of \(I_{2}\). Using (3.7), we have
$$\begin{aligned} I_{2}&\leq \int_{D}\vert u \vert ^{p-1}\vert X\hat{u} \vert ^{1-\delta}\,dx + \int_{D\cap{E_{\mu_{0}}}}\bigl(M\vert X\hat{u} \vert \bigr)^{-\delta} \vert Xu \vert ^{p-1}\vert X\hat{u} \vert \,dx \\ &\leq c \int_{D}\vert u \vert ^{p-\delta}\,dx+c\varepsilon \int_{D}\vert X\hat{u} \vert ^{p-\delta}\,dx + \int_{D\cap E_{\mu_{0}}}\vert Xu \vert ^{p-1}\bigl(M\vert X \hat{u} \vert \bigr)^{1-\delta}\,dx \\ &\leq c \int_{D} \bigl( \vert u \vert ^{p-\delta}+\vert Xu_{0} \vert ^{p-\delta} \bigr)\,dx \\ &\quad{} + c\varepsilon \int_{D}\vert Xu \vert ^{p-\delta}\,dx + \int_{D\cap E_{\mu_{0}}}\vert Xu \vert ^{p-1}\bigl(M\vert X \hat{u} \vert \bigr)^{1-\delta}\,dx. \end{aligned}$$
(3.23)
To estimate the last integral in (3.23), let \(0<\tau<\frac {1}{2}\) and \(x\in D\cap E_{\mu_{0}}\). If \(\vert Xu \vert \geq \tau^{-1}\mu_{0}\), then \(M\vert X\hat{u} \vert \leq\mu_{0}\leq \tau \vert Xu \vert \) and
$$ \vert Xu{ \vert ^{p - 1}} {\bigl(M\vert X\hat{u} \vert \bigr)^{1 - \delta}} \le \vert Xu{ \vert ^{p - 1}} {\bigl( \tau \vert Xu \vert \bigr)^{1 - \delta}} = {\tau ^{1 - \delta}} \vert Xu{ \vert ^{p - \delta}}; $$
(3.24)
if \(\vert Xu \vert <\tau^{-1}\mu_{0}\), then
$$ \vert Xu{ \vert ^{p - 1}} {\bigl(M\vert X\hat{u} \vert \bigr)^{1 - \delta}} \le{ \bigl( {{\tau^{ - 1}} {\mu _{0}}} \bigr) ^{p - 1}}\mu_{0}^{1 - \delta} \le{ \tau^{1 - p}}\mu_{0}^{p - \delta}. $$
(3.25)
By (3.24) and (3.25), we deduce that, for any \(x\in D\cap E_{\mu_{0}}\),
$$ \vert Xu \vert ^{p-1}\bigl(M\vert X\hat{u} \vert \bigr)^{1-\delta}\leq c \bigl( \tau^{1-\delta} \vert Xu \vert ^{p-\delta}+\tau^{1-p}\mu_{0}^{p-\delta} \bigr) . $$
(3.26)
For the second term in (3.26), we first observe from the proof of (3.11) that
Noticing \(\mu_{0}=\frac{c}{\vert 2B \vert }\int_{D}\vert X\hat {u} \vert \,dx\), we have from Hölder’s inequality
(3.27)
By (3.26) and (3.27), it follows that
$$\begin{aligned} & \int_{D\cap E_{\mu_{0}}}\vert Xu \vert ^{p-1}\bigl(M\vert X \hat{u} \vert \bigr)^{1-\delta}\,dx \\ &\quad\leq c\tau^{1-\delta} \int_{D}\vert Xu \vert ^{p-\delta}\,dx+c \int_{D}\vert Xu_{0} \vert ^{p-\delta}\,dx +c\tau^{1-p}\vert 2B \vert \biggl( \frac{1}{\vert 2B \vert } \int_{D}\vert Xu \vert ^{s'}\,dx \biggr) ^{\frac{p-\delta}{s'}}. \end{aligned}$$
(3.28)
Taking (3.28) into (3.23), we have
$$\begin{aligned} I_{2}&\leq c \int_{D}\vert u \vert ^{p-\delta}\,dx+ c \int_{D}\vert Xu_{0} \vert ^{p-\delta}\,dx \\ &\quad{} +c \bigl( \varepsilon+\tau^{1-\delta} \bigr) \int_{D}\vert Xu \vert ^{p-\delta}\,dx+c\tau ^{1-p}\vert 2B \vert \biggl( \frac{1}{\vert 2B \vert } \int_{D}\vert Xu \vert ^{s'}\,dx \biggr) ^{\frac{p-\delta}{s'}}. \end{aligned}$$
(3.29)
For the estimation of \(I_{3}\): From (2.3), Lemma 2.8 and a similar process to the proof of (3.18), we have
$$\begin{aligned} I_{3}&\leq c \int_{D}\bigl(\vert u \vert ^{p-1}+\vert Xu \vert ^{p-1}\bigr)\vert \hat{u} \vert ^{1-\delta}\,dx \\ &\leq \int_{D} \vert u{ \vert ^{p - \delta}}+ c\varepsilon \int_{D} \vert Xu{ \vert ^{p - \delta}}\,dx + c \int_{D}\vert u - u_{0} \vert ^{p -\delta}\,dx \\ &\leq c \int_{D}\bigl(\vert u \vert ^{p - \delta}+ \vert X{u_{0}} { \vert ^{p - \delta}}\bigr)\,dx \\ &\quad{} + c\varepsilon \int_{D}\vert Xu \vert ^{p -\delta}\,dx + c\vert 2B \vert \biggl( \frac{1}{\vert 2B \vert } \int_{D}\vert Xu \vert ^{t}\,dx \biggr) ^{\frac{{p -\delta} }{{t}}}, \end{aligned}$$
(3.30)
where \(t=\max\{s',s''\}< p-\delta\).
Step 3. Taking into account (3.4), (3.8), substituting (3.22), (3.29) and (3.30) into (3.9), and letting \(\varepsilon=\tau^{1-\delta}\), it follows that
$$\begin{aligned} &\int_{\frac{B}{2}\cap\Omega} \vert Xu \vert ^{p-\delta}\, dx \\ & \quad \leq c \int_{3B \cap\Omega} \bigl( \vert u \vert ^{p - \delta} +\vert {u_{0}} \vert ^{p - \delta} +\vert X{u_{0}} \vert ^{p - \delta} \bigr)\,dx \\ &\quad \quad{}+c \bigl( \delta+\tau^{1-\delta} \bigr) \int_{3B\cap\Omega} \vert Xu \vert ^{p-\delta}\,dx +c\tau ^{1-p}\vert 2B \vert \biggl( \frac{1}{\vert 2B \vert } \int_{3B\cap\Omega} \vert Xu \vert ^{t}\,dx \biggr) ^{\frac{p-\delta}{t}}. \end{aligned}$$
(3.31)
To sum up the cases \(3B\subset\Omega\) and \(3B\backslash\Omega\neq \emptyset\), we let
$$ g(x)= \textstyle\begin{cases} \vert Xu \vert ^{t}, &x\in\Omega,\\ 0, & x\in\mathbf{R}^{n}\backslash\Omega, \end{cases} $$
and
$$ f(x)= \textstyle\begin{cases} ( \vert u-u_{0} \vert +\vert u_{0} \vert +\vert Xu_{0} \vert ) ^{t}, & x\in\Omega,\\ 0, & x\in\mathbf{R}^{n}\backslash\Omega. \end{cases} $$
Thus we have from (3.1) and (3.31)
where \(q=\frac{p-\delta}{t}\), \(\theta=c ( \delta+\tau^{1-\delta} ) \) and \(b = c{\tau^{1 - p}}\). Choosing τ, δ small enough, we see by Lemma 2.3 that there exists \(t_{1}=p-\delta+\varepsilon_{0}\), for some \(\varepsilon_{0}>0\), such that \(\vert Xu \vert \in L^{t_{1}}(\Omega)\).
Furthermore, we will show that there exists \({t_{2}}>r=p-\delta\) such that \(u \in L^{t_{2}}(\Omega)\). Since \(u-u_{0}\in W_{X,0}^{1,r}(\Omega)\), we obtain from Lemma 2.2 that, for \(r< Q\), \(r^{*}=Qr/(Q-r)\),
$$ \biggl( \int_{\Omega} \vert u-u_{0} \vert ^{r^{*}}\,dx \biggr) ^{\frac{1}{r^{*}}}\leq C(\Omega) \biggl( \int_{\Omega}\bigl\vert X(u-u_{0}) \bigr\vert ^{r}\,dx \biggr) ^{\frac{1}{r}}< \infty. $$
Taking \(t_{2}=\min\{s,r^{*}\}>r\), we have
$$\begin{aligned} \biggl( \int_{\Omega} \vert u \vert ^{t_{2}}\,dx \biggr) ^{\frac{1}{t_{2}}} &\leq\biggl( \int_{\Omega} \vert u-u_{0} \vert ^{t_{2}}\,dx \biggr) ^{\frac{1}{t_{2}}}+ \biggl( \int_{\Omega} \vert u_{0} \vert ^{t_{2}}\,dx \biggr) ^{\frac{1}{t_{2}}} \\ &\leq C \biggl( \int_{\Omega} \vert u-u_{0} \vert ^{r^{*}}\,dx \biggr) ^{\frac{1}{r^{*}}}+ \biggl( \int_{\Omega} \vert u_{0} \vert ^{t_{2}}\,dx \biggr) ^{\frac{1}{t_{2}}} \end{aligned}$$
and then \(u\in L^{t_{2}}(\Omega)\) by \(u_{0}\in L^{s}(\Omega)\). If \(r\geq Q\) then we can apply the above reasoning for any \(r^{*}<\infty\) to obtain \(u\in L^{t_{2}}(\Omega)\).
We set \(\tilde{p}=\min\{t_{1},t_{2}\}>p-\delta\) and \(u\in W_{X}^{1,\tilde{p}}(\Omega)\). Repeating the preceding reasoning, we know that there exists \(\tilde{\delta}>0\) such that \(u\in W_{X}^{1,p+\tilde{\delta}}(\Omega)\) and the proof is complete.