In this section, we focus on the proof of main Theorem 2.5. First, let us begin this section with the a priori assumption that the unique weak solution \(u\in\mathcal{A}\) of the variational inequalities (1.1) satisfies
$$\begin{aligned} \bigl\Vert \vert Du\vert ^{p(x)}\bigr\Vert _{L^{(\gamma,q)}(\Omega)} < \infty. \end{aligned}$$
(4.1)
We also assume that \((A(x), \Omega)\) is \((\delta, R_{0})\)-vanishing of codimension one, where \(R_{0}\le1\) is a given number, while δ is to be determined later. Let \(R \in (0, R_{0}/(\vert \Omega \vert +1))\) and \(x_{0}\in\overline{\Omega}\) be fixed, we localize our interest in the region \(\Omega_{2R}(x_{0})\) and write
$$\begin{aligned} 2 < \gamma_{1} \le p^{-} = \inf _{\Omega_{2R}(x_{0})} p(x) \le p^{+} = \sup_{\Omega_{2R}(x_{0})} p(x) \le\gamma_{2} < \infty \end{aligned}$$
(4.2)
and
$$\begin{aligned} \lambda_{0} = \fint_{\Omega_{2R}(x_{0})} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx + \frac{1}{\delta} \biggl( \fint_{\Omega_{2R}(x_{0})} \bigl(\vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + 1\bigr)^{\eta}\,dx \biggr)^{\frac{1}{\eta}} > 1, \end{aligned}$$
(4.3)
where \(\eta>1\), and small \(\delta>0\) will be specified later. We would like to remark that the δ-flatness Reifenberg condition (2.6) for the boundary of domain is meaningful in the area of geometric measure theory only if δ is small enough, see [28]. For any \(\tau_{1}\), \(\tau_{2}\) with \(1\le\tau_{1} < \tau_{2} \le2 \), we denote an upper-level set by
$$\begin{aligned} E(\lambda) = \bigl\{ x\in\Omega_{\tau_{1}R}(x_{0}): \vert Du \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \lambda\bigr\} , \end{aligned}$$
where λ is large enough such that
$$\begin{aligned} \lambda> \biggl(\frac{16}{7} \biggr)^{d} \biggl( \frac{248}{\tau_{2}-\tau_{1}} \biggr)^{d} \lambda_{0}. \end{aligned}$$
(4.4)
We observe from the upper-level set that
$$\begin{aligned} \Omega_{r}(y) \subset\Omega_{2R}(x_{0}), \quad \forall y\in E(\lambda) \quad \mbox{and}\quad 0< r \le(\tau_{2}- \tau_{1}) R. \end{aligned}$$
Fix any point \(y\in E(\lambda)\), we consider a continuous function \(\Phi _{y}(r)\) defined by
$$\begin{aligned} \Phi_{y}(r) =& \fint_{\Omega_{r}(y)} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx \\ &{}+ \frac{1}{\delta} \biggl( \fint_{\Omega_{r}(y)}\bigl(\vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\bigr)^{\eta}\,dx \biggr)^{\frac{1}{\eta}},\quad 0< r \le(\tau_{2}-\tau_{1}) R. \end{aligned}$$
(4.5)
The Lebesgue differentiation theorem implies that
$$\begin{aligned} \lim_{r\rightarrow0} \Phi_{y}(r) > \lambda\quad \mbox{for almost every } y\in E(\lambda). \end{aligned}$$
On the other hand, if \((\tau_{2}-\tau_{1}) R/124 \le r \le(\tau _{2}-\tau_{1}) R \), then by using the fact that \(\eta>1\) and the Reifenberg flat boundary satisfies the measure density condition, see Remark 2.4, we have
$$\begin{aligned} \Phi_{y}(r) \le& \frac{\vert \Omega_{2R}(x_{0})\vert }{\vert \Omega_{r}(y)\vert } \fint_{\Omega_{2R}(x_{0})} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx \\ &{} + \biggl( \frac{\vert \Omega_{2R}(x_{0})\vert }{\vert \Omega_{r}(y)\vert } \biggr)^{\frac{1}{\eta}} \frac{1}{\delta} \biggl( \fint_{\Omega_{2R}(x_{0})} \bigl(\vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\bigr)^{\eta}\,dx \biggr)^{\frac{1}{\eta}} \\ \le& \frac{\vert \Omega_{2R}(x_{0})\vert }{\vert \Omega _{r}(y)\vert } \biggl( \fint_{\Omega_{2R}(x_{0})} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx + \frac{1}{\delta} \biggl( \fint_{\Omega_{2R}(x_{0})} \bigl(\vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\bigr)^{\eta}\,dx \biggr)^{\frac{1}{\eta}} \biggr) \\ \le& \frac{\vert B_{r}(y)\vert }{\vert \Omega\cap B_{r}(y)\vert } \frac{\vert B_{2R}(x_{0})\vert }{\vert B_{r}(y)\vert } \lambda_{0} \\ \le&\biggl( \frac{2}{1-\delta} \biggr)^{d} \biggl( \frac{248}{\tau_{2}-\tau_{1}} \biggr)^{d} \lambda_{0} \\ \le&\biggl( \frac{16}{7} \biggr)^{d} \biggl( \frac{248}{\tau_{2}-\tau_{1}} \biggr)^{d} \lambda_{0}. \end{aligned}$$
Putting the above formula into assumption (4.4) yields
$$\begin{aligned} \Phi_{y}(r) < \lambda,\quad \forall y\in E(\lambda) \mbox{ and } \forall r \in\bigl[ (\tau_{2}-\tau_{1}) R/124, (\tau_{2}- \tau_{1}) R \bigr]. \end{aligned}$$
Consequently, we conclude that for almost every \(y\in E(\lambda)\), there exists \(r_{y} = r(y) \in(0, (\tau_{2}-\tau_{1}) R/124)\) such that
$$\begin{aligned} \Phi_{y}(r_{y}) = \lambda\quad \mbox{and}\quad \Phi_{y}(r) < \lambda\quad \forall r \in\bigl(r_{y}, ( \tau_{2}-\tau_{1}) R \bigr]. \end{aligned}$$
(4.6)
Then we infer the following lemma from the well-known Vitali covering lemma due to the property of \(r_{y}\).
Lemma 4.1
Let
λ
satisfy (4.4). Then there exists a disjoint family
\(\{ \Omega_{r_{y_{i}}}(y_{i}) \}_{i=1}^{\infty}\)
with
\(y_{i}\in E(\lambda)\)
and
\(r_{y_{i}} \in(0, (\tau_{2}-\tau_{1}) R/124)\)
such that
$$\begin{aligned} \Phi_{y_{i}}(r_{y_{i}}) = \lambda\quad \textit{and}\quad \Phi_{y_{i}}(r) < \lambda\quad \textit{for all } r \in\bigl(r_{y_{i}}, ( \tau_{2}-\tau_{1}) R \bigr] \end{aligned}$$
and
$$\begin{aligned} E(\lambda) \subset\bigcup_{i=1}^{\infty} \Omega_{5r_{y_{i}}}(y_{i}). \end{aligned}$$
Lemma 4.2
Under the same hypothesis as in Lemma
4.1, we have
$$\begin{aligned} \bigl\vert \Omega_{r_{y_{i}}}(y_{i})\bigr\vert \le& C \biggl( \bigl\vert \Omega_{r_{y_{i}}}(y_{i}) \cap E(\lambda/4) \bigr\vert + \frac{1}{(\varsigma\lambda)^{\eta}} \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{r_{y_{i}}}(y_{i}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu} \\ &{} + \frac{1}{(\varsigma\lambda)^{\eta}} \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{r_{y_{i}}}(y_{i}): \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu} \biggr), \end{aligned}$$
where
\(\varsigma= \delta/6\)
and
\(C=C(d,\nu, \Lambda)\).
Proof
With Lemma 4.1 in hand, we then have that one of the following results must hold:
$$\begin{aligned}& \fint_{\Omega_{r_{y_{i}}}(y_{i})} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx \ge \frac{\lambda}{3},\qquad \fint_{\Omega_{r_{y_{i}}}(y_{i})}\vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}\eta}\,dx \ge \biggl( \frac{\delta\lambda}{3} \biggr)^{\eta}\quad \mbox{and} \\& \fint_{\Omega_{r_{y_{i}}}(y_{i})}\vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}\eta}\,dx \ge\biggl( \frac{\delta\lambda}{3} \biggr)^{\eta}. \end{aligned}$$
For the first setting, let us take any
$$\begin{aligned} 0< \epsilon< \min\biggl\{ \frac{\gamma_{1}(1+\sigma _{0})}{\gamma_{1}+\omega(2R)} - 1, \frac{\gamma p^{-}}{2} - 1 \biggr\} , \end{aligned}$$
(4.7)
where \(\sigma_{0} = \min\{\sigma_{1}, \sigma_{2}\}\) is the same as in Lemma 3.7 which is concerned with the higher integrability of Du, and \(\gamma\in[1, \infty)\). It yields the following inequality:
$$\begin{aligned} \frac{p(x)}{p^{-}} (1+ \epsilon) = \biggl(1 + \frac{p(x)-p^{-}}{p^{-}} \biggr) (1+ \epsilon) \le\biggl( 1 + \frac{\omega(2R)}{\gamma_{1}} \biggr) (1+ \epsilon) < 1 + \sigma_{0} \le\gamma_{1}. \end{aligned}$$
Then, by an additivity of the integral with respect to the domain and Hölder’s inequality, we get
$$\begin{aligned} \frac{\lambda}{3} \bigl\vert \Omega_{r_{y_{i}}}(y_{i})\bigr\vert \le& \int_{ \Omega_{r_{y_{i}}}(y_{i}) \cap E(\lambda/4) } \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx + \int_{\Omega_{r_{y_{i}}}(y_{i}) \backslash E(\lambda/4)} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx \\ \le& \bigl\vert \Omega_{r_{y_{i}}}(y_{i}) \cap E(\lambda/4) \bigr\vert ^{1-\frac{1}{1+\epsilon}} \biggl( \int_{ \Omega_{r_{y_{i}}}(y_{i}) \cap E(\lambda/4) } \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}(1+\epsilon)}\,dx \biggr)^{\frac{1}{1+\epsilon}} + \frac{\lambda}{4} \bigl\vert \Omega _{r_{y_{i}}}(y_{i}) \bigr\vert , \end{aligned}$$
which yields
$$\begin{aligned} \lambda\bigl\vert \Omega_{r_{y_{i}}}(y_{i}) \bigr\vert ^{1-\frac{1}{1+\epsilon}} \le C_{1} \bigl\vert \Omega_{r_{y_{i}}}(y_{i}) \cap E(\lambda/4) \bigr\vert ^{1-\frac{1}{1+\epsilon}} \biggl( \fint_{\Omega_{r_{y_{i}}}(y_{i})} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}(1+\epsilon)}\,dx \biggr)^{\frac{1}{1+\epsilon}}. \end{aligned}$$
(4.8)
Furthermore, on the basis of higher integrability of the gradient to weak solutions for the variational inequalities (1.1) in lines with Lemma 3.7, we obtain
$$\begin{aligned}& \biggl( \fint_{ \Omega_{r_{y_{i}}}(y_{i})} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}(1+\epsilon)}\,dx \biggr)^{\frac{1}{1+\epsilon}} \\& \quad \le C_{2} \biggl( \fint_{\Omega_{2r_{y_{i}}}(y_{i})} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx + \biggl( \fint_{\Omega_{2r_{y_{i}}}(y_{i})} \bigl(\vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\bigr)^{(1+\epsilon)}\,dx \biggr)^{\frac{1}{1+\epsilon}} \biggr). \end{aligned}$$
Now we take \(\eta= 1+ \epsilon\), using Lemma 4.1 yields
$$\begin{aligned} \biggl( \fint_{ \Omega_{r_{y_{i}}}(y_{i}) } \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}(1+\epsilon)}\,dx \biggr)^{\frac{1}{1+\epsilon}} \le C_{3} \lambda. \end{aligned}$$
Putting the above formula into (4.8), we have
$$\begin{aligned} \bigl\vert \Omega_{r_{y_{i}}}(y_{i})\bigr\vert \le C_{4} \bigl\vert \Omega_{r_{y_{i}}}(y_{i}) \cap E(\lambda/4) \bigr\vert . \end{aligned}$$
(4.9)
For the second setting, we have
$$\begin{aligned} \biggl( \frac{\delta\lambda}{3} \biggr)^{\eta} \bigl\vert \Omega_{r_{y_{i}}}(y_{i})\bigr\vert \le& \int_{\Omega_{r_{y_{i}}}(y_{i})}\vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}\eta}\,dx \\ =& \eta \int_{0}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{r_{y_{i}}}(y_{i}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu} \\ \le& ( \varsigma\lambda)^{\eta} \bigl\vert \Omega_{r_{y_{i}}}(y_{i}) \bigr\vert + \eta \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{r_{y_{i}}}(y_{i}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu}. \end{aligned}$$
Taking \(\varsigma= \delta/6\), we get
$$\begin{aligned} \bigl\vert \Omega_{r_{y_{i}}}(y_{i})\bigr\vert \le\frac{C_{5}}{(\varsigma\lambda)^{\eta}} \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{r_{y_{i}}}(y_{i}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu}. \end{aligned}$$
(4.10)
We now estimate the third setting in a similar way, just as doing it in the second setting, and conclude that
$$\begin{aligned} \bigl\vert \Omega_{r_{y_{i}}}(y_{i})\bigr\vert \le\frac{C_{6}}{(\varsigma\lambda)^{\eta}} \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{r_{y_{i}}}(y_{i}): \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu}. \end{aligned}$$
(4.11)
Finally, putting the three cases (4.9), (4.10) and (4.11) together completes the proof of Lemma 4.2. □
For any fixed point \(y_{i}\) and scale \(r_{y_{i}}\), there are now two possible cases. One is the interior case that \(B_{20r_{y_{i}}}(y_{i}) \subseteq\Omega\). The other is the boundary case that \(B_{20r_{y_{i}}}(y_{i}) \nsubseteq\Omega\). We first look at the interior case. Since \(A(x)\) is \((\delta, R_{0})\)-vanishing of codimension one, we assume that in a new coordinate system \((x_{1}, \ldots, x_{d})\), the origin is \(y_{i}\) and
$$\begin{aligned} \fint_{Q_{20r_{y_{i}}}(y_{i})} \bigl\vert A(x)- \bar {A}_{B'_{20r_{y_{i}}}(y'_{i})}( x_{1})\bigr\vert ^{2}\,dx \le\delta^{2}. \end{aligned}$$
(4.12)
For convenience, we write
$$\begin{aligned} p_{i}^{-} = \inf_{x\in Q_{20r_{y_{i}}}(y_{i})} p(x) \quad \mbox{and}\quad p_{i}^{+} = \sup_{x\in Q_{20r_{y_{i}}}(y_{i})} p(x). \end{aligned}$$
From Lemma 4.1 and the definition of \(\Phi _{y_{i}}(r_{y_{i}})\), we have
$$\begin{aligned} \fint_{Q_{20r_{y_{i}}}(y_{i})} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx \le\lambda \quad \mbox{and} \quad \biggl( \fint_{Q_{20r_{y_{i}}}(y_{i})}\bigl( \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} \bigr)^{\eta}\,dx \biggr)^{\frac{1}{\eta}} \le\delta\lambda. \end{aligned}$$
(4.13)
By the re-scaling transformation and perturbation approach based on a local comparison, we obtain
$$\begin{aligned} \fint_{Q_{20r_{y_{i}}}(y_{i})} \vert Du\vert ^{2}\,dx \le C_{0} \lambda^{\frac{p^{-}}{p_{i}^{+}}} \quad \mbox{and}\quad \fint_{Q_{20r_{y_{i}}}(y_{i})} \bigl( \vert \mathbf{f}\vert ^{2} + \vert D\psi \vert ^{2} \bigr)\,dx \le C_{0} \lambda^{\frac{p^{-}}{p_{i}^{+}}} \delta^{\frac{\gamma_{1}}{\gamma_{2}}} \end{aligned}$$
(4.14)
for some constant \(C_{0}\ge1\) independent of i, and \(\gamma_{1}\), \(\gamma_{2}\) shown as in (4.2). In fact, let \(A_{0} = \Vert \vert \mathbf{f}\vert ^{p(x)}\Vert _{L^{(\gamma, q)}(\Omega)} + \Vert \vert D\psi \vert ^{p(x)}\Vert _{L^{(\gamma, q)}(\Omega )}+1 \ge1\). A direct computation yields that
$$\begin{aligned} \biggl( \fint_{Q_{20r_{y_{i}}}(y_{i})} \vert Du\vert ^{2}\,dx \biggr)^{p_{i}^{+}-p_{i}^{-}} =& \biggl( \frac{1}{\vert Q_{20r_{y_{i}}}(y_{i})\vert } \biggr)^{p_{i}^{+}-p_{i}^{-}} \biggl( \int_{Q_{20r_{y_{i}}}(y_{i})}\vert Du\vert ^{2}\,dx \biggr)^{p_{i}^{+}-p_{i}^{-}} \\ \le& C_{1} \biggl( \frac{1}{40r_{y_{i}}} \biggr)^{d\omega(40r_{y_{i}})} \biggl( \int_{\Omega} \vert Du\vert ^{2}\,dx \biggr)^{p_{i}^{+}-p_{i}^{-}} \\ \le& C_{2} \biggl( \int_{\Omega} \vert Du\vert ^{2}\,dx \biggr)^{p_{i}^{+}-p_{i}^{-}}, \end{aligned}$$
(4.15)
where we used (2.3) based on the log-Hölder condition in the last inequality. On the other hand, by making use of the standard \(L^{2}\) estimates from Lemma 3.5 and Proposition 3.1 due to \(\frac{\gamma\gamma_{1}}{2}>1\) for \(\gamma\in[1, \infty)\) and \(q\in(0, \infty]\), we obtain
$$\begin{aligned} \int_{\Omega} \vert Du\vert ^{2}\,dx \le& C_{3} \biggl( \int_{\Omega} \vert \mathbf{f}\vert ^{2}\,dx + \int_{\Omega} \vert D\psi \vert ^{2}\,dx \biggr) \\ \le& C_{4} \biggl( \int_{\Omega} \vert \mathbf{f}\vert ^{\frac{2p(x)}{\gamma_{1}}}\,dx + \int_{\Omega} \vert D\psi \vert ^{\frac{2p(x)}{\gamma_{1}}}\,dx + \vert \Omega \vert \biggr) \\ \le& C_{5} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{\frac{2p(x)}{\gamma_{1}}} \bigr\Vert _{L^{(\frac{\gamma\gamma _{1}}{2}, \frac{q \gamma_{1}}{2})}(\Omega)} + \bigl\Vert \vert D\psi \vert ^{\frac{2p(x)}{\gamma_{1}}}\bigr\Vert _{L^{(\frac{\gamma\gamma _{1}}{2}, \frac{q \gamma_{1}}{2})}(\Omega)} + \vert \Omega \vert \bigr) \\ =& C_{5} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)} \bigr\Vert _{L^{(\gamma, q)}(\Omega)}^{\frac{2}{\gamma_{1}}} + \bigl\Vert \vert D\psi \vert ^{p(x)} \bigr\Vert _{L^{(\gamma, q)}(\Omega )}^{\frac{2}{\gamma_{1}}} + \vert \Omega \vert \bigr), \end{aligned}$$
which leads to
$$\int_{\Omega} \vert Du\vert ^{2}\,dx \le C_{6} A_{0} \bigl( 1 + \vert \Omega \vert \bigr). $$
Then we conclude
$$\begin{aligned} \biggl( \fint_{Q_{20r_{y_{i}}}(y_{i})} \vert Du\vert ^{2}\,dx \biggr)^{p_{i}^{+}-p_{i}^{-}} \le C_{2} \bigl( C_{6} A_{0} \bigl( 1 + \vert \Omega \vert \bigr) \bigr)^{p_{i}^{+}-p_{i}^{-}}. \end{aligned}$$
Recalling \(( \vert \Omega \vert +1) < \frac{R_{0}}{R} \le\frac {1}{R} \le\frac{1}{40 r_{y_{i}}}\) and (2.3) yields
$$\begin{aligned} \biggl( \fint_{Q_{20r_{y_{i}}}(y_{i})} \vert Du\vert ^{2}\,dx \biggr)^{p_{i}^{+}-p_{i}^{-}} \le C_{7} A_{0}^{p_{i}^{+}-p_{i}^{-}} \biggl( \frac{1}{40 r_{y_{i}}} \biggr)^{\omega(40r_{y_{i}})} \le C_{8} A_{0}^{p_{i}^{+}-p_{i}^{-}}, \end{aligned}$$
which follows from \(A_{0}\ge1\) that
$$ \biggl( \fint_{Q_{20r_{y_{i}}}(y_{i})} \vert Du\vert ^{2}\,dx \biggr)^{\frac{p_{i}^{+}-p_{i}^{-}}{p_{i}^{+}}} \le C_{9} A_{0}^{\frac {p_{i}^{+}-p_{i}^{-}}{p_{i}^{+}}} \le C_{9} A_{0}. $$
(4.16)
Using (4.16) and Jensen’s inequality yields
$$\begin{aligned} \fint_{Q_{20r_{y_{i}}}(y_{i})} \vert Du\vert ^{2}\,dx =& \biggl( \fint_{Q_{20r_{y_{i}}}(y_{i})} \vert Du\vert ^{2}\,dx \biggr)^{\frac{p_{i}^{+}-p_{i}^{-}}{p_{i}^{+}}} \biggl( \fint_{Q_{20r_{y_{i}}}(y_{i})} \vert Du\vert ^{2}\,dx \biggr)^{\frac{p_{i}^{-}}{p_{i}^{+}}} \\ \le& C_{9} A_{0} \biggl( \fint_{Q_{20r_{y_{i}}}(y_{i})} \vert Du\vert ^{2}\,dx \biggr)^{\frac{p_{i}^{-}}{p_{i}^{+}}} \\ \le& C_{9} A_{0} \biggl( \fint_{Q_{20r_{y_{i}}}(y_{i})} \vert Du\vert ^{\frac {2p_{i}^{-}}{p^{-}}}\,dx \biggr)^{\frac{p^{-}}{p_{i}^{+}}} \\ \le& C_{9} A_{0} \biggl( \fint_{Q_{20r_{y_{i}}}(y_{i})} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx +1 \biggr)^{\frac{p^{-}}{p_{i}^{+}}}. \end{aligned}$$
Since \(\lambda>1\), by (4.13) we get the first desired inequality in (4.14) by taking \(C_{0} = 2 C_{9} A_{0}\).
Likewise, we derive that
$$\begin{aligned} \fint_{Q_{20r_{y_{i}}}(y_{i})} \bigl( \vert \mathbf{f}\vert ^{2} + \vert D\psi \vert ^{2} \bigr)\,dx \le& C_{0} \biggl( \fint_{Q_{20r_{y_{i}}}(y_{i})} \bigl( \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} \bigr)\,dx +1 \biggr)^{\frac {p^{-}}{p_{i}^{+}}} \\ \le& C_{0} \biggl( \biggl( \fint_{Q_{20r_{y_{i}}}(y_{i})} \bigl( \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} \bigr)^{\eta}\,dx \biggr)^{\frac{1}{\eta}} + 1 \biggr)^{\frac{p^{-}}{p_{i}^{+}}} \\ \le& C_{0} ( \delta\lambda+ \delta\lambda_{0} )^{\frac{p^{-}}{p_{i}^{+}}} \le C_{0} \lambda^{\frac{p^{-}}{p_{i}^{+}}} \delta^{\frac{\gamma_{1}}{\gamma_{2}}}, \end{aligned}$$
where we also employ (4.13) in the third inequality.
Define
$$\begin{aligned}& \tilde{u}_{i}(x) = \frac{u(5r_{y_{i}}x)}{5r_{y_{i}} \sqrt{C_{0} \lambda ^{\frac{p^{-}}{p_{i}^{+}}}}},\qquad \tilde{\psi}_{i}(x) = \frac{\psi(5r_{y_{i}}x)}{5r_{y_{i}} \sqrt{C_{0} \lambda^{\frac {p^{-}}{p_{i}^{+}}}}},\\& \tilde{\mathbf{f}}_{i}(x) = \frac{\mathbf{f}(5r_{y_{i}}x)}{\sqrt{C_{0} \lambda^{\frac{p^{-}}{p_{i}^{+}}}}},\qquad \tilde{A}_{i}(x) = A(5r_{y_{i}}x). \end{aligned}$$
By using Lemma 3.6, we get that \(\tilde{u}_{i}(x)\in\tilde {\mathcal{A}}_{i} : = \{\phi_{i} \in W^{1,2}(Q_{4}): \phi_{i} \ge\tilde {\psi}_{i}, \mbox{a.e. in } Q_{4} \}\) is a weak solution of
$$\begin{aligned} \int_{Q_{4}} \tilde{A}_{i}(x) D\tilde{u}_{i} D(\tilde{\phi}_{i}-\tilde{u}_{i})\,dx \ge \int_{Q_{4}} \tilde{\mathbf{f}_{i}} D(\tilde{ \phi}_{i}-\tilde{u}_{i})\,dx \quad \mbox{for all } \tilde{ \phi}_{i}\in\tilde{\mathcal{A}}_{i}. \end{aligned}$$
Moreover, using (4.12) and (4.14), by a straightforward computation, we have
$$\begin{aligned} \fint_{Q_{4}} \bigl\vert A(x)- \bar{A}_{B'_{4}}( x_{1})\bigr\vert ^{2} \le\delta^{2},\qquad \fint_{Q_{4}} \vert D\tilde{u}\vert ^{2}\,dx \le1 \quad \mbox{and}\quad \fint_{Q_{4}} \bigl( \vert \tilde{\mathbf{f}}\vert ^{2} + \vert D\tilde{\psi} \vert ^{2}\bigr)\,dx \le \delta^{\frac{\gamma_{1}}{\gamma_{2}}}. \end{aligned}$$
Thus, by Lemmas 4.3 and 4.4 in [9], we find that for any \(\varepsilon>0 \), there exist a constant \(\delta>0\) and a function \(\tilde{v}_{i}\in W^{1,2}(Q_{2})\) to be a weak solution of
$$\begin{aligned} \operatorname{div}\bigl(\tilde{A}_{iB'_{2}}(x_{1}) D \tilde{v}_{i}\bigr)=0\quad \mbox{in } Q_{2}, \end{aligned}$$
such that
$$\begin{aligned} \fint_{Q_{2}} \bigl\vert D(\tilde{u}_{i}- \tilde{v}_{i})\bigr\vert ^{2}\,dx \le\varepsilon \quad \mbox{and}\quad \Vert D\tilde{v}_{i}\Vert ^{2}_{L^{\infty}(Q_{1})} \le N_{0} \end{aligned}$$
with a constant \(N_{0}\) being independent of i. Scaling back and denoting \(v_{i}\) by the translated function of
$$\begin{aligned} \tilde{v}_{i}(x) = \frac{v_{i}(5r_{y_{i}}x)}{5r_{y_{i}} \sqrt{C_{0} \lambda^{\frac{p^{-}}{p_{i}^{+}}}}}, \end{aligned}$$
we conclude that
$$\begin{aligned} \fint_{Q_{10r_{y_{i}}}(y_{i})} \bigl\vert D(u- v_{i})\bigr\vert ^{2}\,dx \le C_{0} \lambda^{\frac{p^{-}}{p_{i}^{+}}}\varepsilon \quad \mbox{and}\quad \Vert Dv_{i}\Vert ^{2}_{L^{\infty}(Q_{5r_{y_{i}}}(y_{i}))} \le N_{1} \lambda^{\frac{p^{-}}{p_{i}^{+}}} \end{aligned}$$
(4.17)
for some constant \(N_{1} = N_{0} C_{0} \ge1\), being independent of i.
We next consider the boundary case that \(\operatorname{dist}\{y_{i}, \partial\Omega\} = \vert y_{i}- y_{0}\vert \le20 r_{y_{i}}\) for \(y_{0}\in\partial \Omega\). Let us recall that \(r_{y_{i}} < (\tau_{2}-\tau_{1})R /124 < R_{0}/ 124\) and the geometry (2.6) of the boundary of Reifenberg flat domain, we have the following property: for any point \(y_{0}\) on the boundary of Ω, there exists a coordinate system \(\{x_{1}, \ldots , x_{d} \}\) with the origin lining somewhere in \(\Omega_{60r_{y_{i}}\delta }(y_{0})\) such that in this new coordinate system one has
$$\begin{aligned} Q^{+}_{60 r_{y_{i}}} \subset\Omega_{60 r_{y_{i}}} \subset Q_{60 r_{y_{i}}} \cap\{x_{1}> - 120 r_{y_{i}} \delta\} \end{aligned}$$
and
$$\begin{aligned} \fint_{Q_{60 r_{y_{i}}}} \bigl\vert A(x)- \bar{A}_{B'_{60r_{y_{i}}}}( x_{1})\bigr\vert ^{2}\,dx \le\delta^{2}. \end{aligned}$$
Let us now select δ so small such that \(0<\delta<\frac{1}{60}\), which yields \(\vert y_{i}\vert < (20+1)r_{y_{i}}= 21 r_{y_{i}}\) and
$$\begin{aligned} \Omega_{5r_{y_{i}}}(y_{i}) \subset \Omega_{26r_{y_{i}}} \subset\Omega_{60{r_{y_{i}}}} \subset \Omega_{96 r_{y_{i}}} (y_{i}). \end{aligned}$$
(4.18)
We write
$$\begin{aligned} p_{i}^{-} = \inf_{x\in\Omega_{96r_{y_{i}}}(y_{i})} p(x) \quad \mbox{and}\quad p_{i}^{+} = \sup_{x\in\Omega_{96r_{y_{i}}}(y_{i})} p(x). \end{aligned}$$
By Lemma 4.1, we derive that
$$\begin{aligned} \fint_{\Omega_{60r_{y_{i}}}} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\, dx \le \frac{\vert \Omega_{96 r_{y_{i}}} (y_{i})\vert }{\vert \Omega _{60{r_{y_{i}}}}\vert } \fint_{\Omega_{96 r_{y_{i}}} (y_{i})} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx \le C_{10} \lambda \end{aligned}$$
and
$$\begin{aligned}& \biggl( \fint_{\Omega_{60r_{y_{i}}}}\bigl(\vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\bigr)^{\eta}\,dx \biggr)^{\frac{1}{\eta}} \\& \quad \le\biggl( \frac{\vert \Omega_{96 r_{y_{i}}} (y_{i})\vert }{\vert \Omega_{60{r_{y_{i}}}}\vert } \biggr)^{\frac{1}{\eta}} \biggl( \fint_{\Omega_{96 r_{y_{i}}} (y_{i})}\bigl(\vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\bigr)^{\eta}\,dx \biggr)^{\frac{1}{\eta}} \le C_{10} \delta\lambda, \end{aligned}$$
where the constant \(C_{10}\) depends only on d. Once we have the above uniform bounds, one can find in the same sprit as in the interior case that
$$\begin{aligned} \fint_{\Omega_{60r_{y_{i}}}} \vert Du\vert ^{2}\,dx \le C_{11} \lambda^{\frac{p^{-}}{p_{i}^{+}}}\quad \mbox{and}\quad \fint_{\Omega_{60r_{y_{i}}}} \bigl( \vert \mathbf{f}\vert ^{2} + \vert D\psi \vert ^{2}\bigr)\,dx \le C_{11} \lambda^{\frac{p^{-}}{p_{i}^{+}}} \delta^{\frac{\gamma_{1}}{\gamma_{2}}} \end{aligned}$$
for some constant \(C_{11} = 2C_{10} A_{0} \ge1\), being independent of i. In a similar way that we have used for the interior case, also see Lemma 4.7 in [9], for any \(\varepsilon\in(0,1)\), there exist a small positive number δ and a function \(v_{i}\in W^{1,2}(Q^{+}_{52 r_{y_{i}}})\) such that
$$\begin{aligned} \fint_{\Omega_{52 r_{y_{i}}}} \bigl\vert D(u - \bar{v}_{i})\bigr\vert ^{2}\,dx \le C_{11} \lambda^{\frac{p^{-}}{p_{i}^{+}}} \varepsilon \quad \mbox{and}\quad \Vert D\bar{v}_{i}\Vert ^{2}_{L^{\infty}(\Omega_{26r_{y_{i}}})} \le N_{2} \lambda^{\frac{p^{-}}{p_{i}^{+}}} \end{aligned}$$
(4.19)
for the constant \(N_{2}\ge 1\) only dependent of \(C_{11}\). Here we have extended \(v_{i}\) by zero from \(Q^{+}_{52 r_{y_{i}}}\) to \(Q_{52 r_{y_{i}}}\) and also denote it by \(\bar{v}_{i}\). Let us write \(N = \max\{N_{1}, N_{2}\}\) being large enough, which is independent of the index i. For convenience, we also write
$$\begin{aligned} A = (4 N)^{\frac{\gamma_{2}}{\gamma_{1}}} > 1,\qquad B = \biggl( \frac{16}{7} \biggr)^{d} \biggl( \frac{248}{\tau_{2}-\tau_{1}} \biggr)^{d}. \end{aligned}$$
(4.20)
Lemma 4.3
Let
\(R_{0}>0\). For any fixed
\(0< \varepsilon<1\), we can find a small constant
\(\delta>0\)
such that if
\((A(x), \Omega)\)
is
\((\delta, R_{0})\)-vanishing codimension one and
\(u\in\mathcal{A}\)
is a weak solution of (1.1), then for any
\(1\le\tau_{1} < \tau _{2}\le2 \)
we have
$$\begin{aligned} \bigl\vert E(A\lambda)\bigr\vert \le& C \varepsilon\biggl( \bigl\vert \Omega_{\tau_{2}R}(x_{0}) \cap E(\lambda/4)\bigr\vert \\ &{} + \frac{1}{(\varsigma\lambda)^{\eta}} \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{\tau_{2}R}(x_{0}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu} \\ &{} + \frac{1}{(\varsigma\lambda)^{\eta}} \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{\tau_{2}R}(x_{0}): \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu} \biggr) \end{aligned}$$
for all
\(\lambda> B\lambda_{0}\), \(\varsigma= \delta/6\)
and
\(C=C(d,\nu ,\Lambda, \omega(\cdot), \gamma_{1}, \gamma_{2}, R_{0})\).
Proof
By Lemma 4.1 and the fact that \(E(A\lambda)\subset E(\lambda)\) for \(A>1\) as mentioned in (4.20), we obtain that \(\{\Omega _{5r_{y_{i}}}(y_{i})\}\) cover almost all \(E(A\lambda)\). Thus,
$$\begin{aligned} \bigl\vert E(A\lambda)\bigr\vert =& \bigl\vert \bigl\{ x\in \Omega_{\tau_{1}R}(x_{0}): \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > A \lambda\bigr\} \bigr\vert \\ \le& \sum_{i=1}^{\infty} \bigl\vert \bigl\{ x\in\Omega_{5r_{y_{i}}}(y_{i}): \vert Du\vert ^{2} > ( A \lambda)^{\frac{p^{-}}{p(x)}} \bigr\} \bigr\vert \\ \le& \sum_{\text{i: interior case} } \bigl\vert \bigl\{ x\in \Omega_{5r_{y_{i}}}(y_{i}): \vert Du\vert ^{2} > ( A \lambda)^{\frac{p^{-}}{p(x)}} \bigr\} \bigr\vert \\ &{} + \sum_{\text{i: boundary case} } \bigl\vert \bigl\{ x\in \Omega_{5r_{y_{i}}}(y_{i}): \vert Du\vert ^{2} > ( A \lambda)^{\frac{p^{-}}{p(x)}} \bigr\} \bigr\vert . \end{aligned}$$
(4.21)
For the interior estimate, using the fact that \(\vert Du\vert ^{2} \le2 \vert D(u-v_{i})\vert ^{2} + 2 \vert Dv_{i}\vert ^{2}\), \(\Omega_{5r_{y_{i}}}(y_{i}) = Q_{5r_{y_{i}}}(y_{i}) \) and Eq. (4.17), we find that
$$\begin{aligned}& \bigl\vert \bigl\{ x\in Q_{5r_{y_{i}}}(y_{i}): \vert Du \vert ^{2} > ( A \lambda)^{\frac{p^{-}}{p(x)}} \bigr\} \bigr\vert \\& \quad \le \bigl\vert \bigl\{ x\in Q_{5r_{y_{i}}}(y_{i}): \bigl\vert D(u-v_{i})\bigr\vert ^{2} > N_{1} \lambda^{\frac{p^{-}}{p_{i}^{+}}} \bigr\} \bigr\vert + \bigl\vert \bigl\{ x\in Q_{5r_{y_{i}}}(y_{i}): \vert Dv_{i}\vert ^{2} > N_{1} \lambda^{\frac{p^{-}}{p_{i}^{+}}} \bigr\} \bigr\vert \\& \quad \le \frac{1}{N_{1} \lambda^{\frac{p^{-}}{p_{i}^{+}}} } \int_{Q_{10r_{y_{i}}}(y_{i})} \bigl\vert D(u-v_{i})\bigr\vert ^{2}\,dx \\& \quad \le \frac{C_{0} \lambda^{\frac{p^{-}}{p_{i}^{+}}} \varepsilon \vert Q_{10r_{y_{i}}}(y_{i})\vert }{N_{0} C_{0} \lambda^{\frac{p^{-}}{p_{i}^{+}}}} \\& \quad \le C_{1} \varepsilon\bigl\vert Q_{10r_{y_{i}}}(y_{i}) \bigr\vert , \end{aligned}$$
which implies
$$\begin{aligned} \bigl\vert \bigl\{ x\in\Omega_{5r_{y_{i}}}(y_{i}): \vert Du\vert ^{2} > ( A \lambda)^{\frac{p^{-}}{p(x)}} \bigr\} \bigr\vert \le C_{2} \varepsilon\bigl\vert Q_{r_{y_{i}}}(y_{i}) \bigr\vert . \end{aligned}$$
(4.22)
For the boundary case, we carry out the same procedure as in (4.21) with Eq. (4.19) to discover that
$$\begin{aligned} \bigl\vert \bigl\{ x\in\Omega_{26r_{y_{i}}}: \vert Du\vert ^{2} > ( A \lambda)^{\frac{p^{-}}{p(x)}} \bigr\} \bigr\vert \le C_{3} \varepsilon \vert \Omega_{r_{y_{i}}}\vert . \end{aligned}$$
Then, using the measure density condition (2.8) and the geometry of Reifenberg flatness (4.18), we conclude that
$$\begin{aligned} \bigl\vert \bigl\{ x\in\Omega_{5r_{y_{i}}}(y_{i}): \vert Du\vert ^{2} > ( A \lambda)^{\frac{p^{-}}{p(x)}} \bigr\} \bigr\vert \le C_{4} \varepsilon\bigl\vert \Omega_{r_{y_{i}}}(y_{i}) \bigr\vert . \end{aligned}$$
(4.23)
Inserting (4.22) and (4.23) into (4.21), we get
$$\begin{aligned} \bigl\vert E(A\lambda)\bigr\vert \le& C_{5} \varepsilon\sum _{i=1}^{\infty} \bigl\vert \Omega_{r_{y_{i}}}(y_{i})\bigr\vert , \end{aligned}$$
where \(C_{5} = \max\{C_{2}, C_{4}\}\). Furthermore, by using Lemma 4.2, we obtain
$$\begin{aligned} \bigl\vert E(A\lambda)\bigr\vert \le& C_{5} \varepsilon\sum _{i=1}^{\infty} \biggl( \bigl\vert \Omega_{r_{y_{i}}}(y_{i}) \cap E(\lambda/4)\bigr\vert \\ &{} + \frac{1}{(\varsigma\lambda)^{\eta}} \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{r_{y_{i}}}(y_{i}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu} \\ &{} + \frac{1}{(\varsigma\lambda)^{\eta}} \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{r_{y_{i}}}(y_{i}): \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu} \biggr) \end{aligned}$$
for all \(\lambda> B\lambda_{0}\) and \(\varsigma= \delta/6\). Note that \(\{\Omega_{r_{y_{i}}}(y_{i})\}\) are non-overlapping in \(\Omega_{\tau _{2}R}(x_{0})\), then the required result follows. □
Proof of Theorem 2.5
We divide it into two steps: we first attain the estimate under the assumption of (4.1); then we prove assumption (4.1).
Step 1. Let us first establish a global estimate to the variational inequalities (1.1) under the a priori assumption (4.1). In the case \(0< q<\infty\), thanks to (3.3) in Proposition 3.1 in hand, we have
$$\begin{aligned} \bigl\Vert \vert Du\vert ^{p(x)}\bigr\Vert _{L^{(\gamma ,q)}(\Omega_{\tau_{1}R} (x_{0}))}^{q} =& \bigl\Vert \vert Du\vert ^{\frac{2 p(x)}{p^{-}}} \bigr\Vert _{L^{(\frac{\gamma p^{-}}{2}, \frac{q p^{-}}{2})}(\Omega_{\tau_{1}R} (x_{0}))}^{\frac{p^{-}}{2}q} \\ =& \frac{\gamma p^{-}}{2} \int_{0}^{\infty} \bigl( \alpha^{\frac{\gamma p^{-}}{2} } \bigl\vert \bigl\{ x\in\Omega_{\tau_{1}R} (x_{0}): \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \alpha\bigr\} \bigr\vert \bigr)^{\frac {q}{\gamma}} \frac{d \alpha}{\alpha}. \end{aligned}$$
By using the change of variables, a direct calculation shows that
$$\begin{aligned}& \bigl\Vert \vert Du\vert ^{p(x)}\bigr\Vert _{L^{(\gamma,q)}(\Omega_{\tau_{1}R} (x_{0}))}^{q} \\& \quad = \frac{\gamma p^{-}}{2} A^{\frac{q p^{-}}{2}} \int_{0}^{\infty} \lambda^{\frac{q p^{-}}{2}} \bigl\vert \bigl\{ x\in\Omega_{\tau_{1}R}(x_{0}): \vert Du\vert ^{\frac{2 p(x)}{p^{-}}}> A \lambda\bigr\} \bigr\vert ^{\frac{q}{\gamma }} \frac{d\lambda}{\lambda} \\& \quad = \frac{\gamma p^{-}}{2} A^{\frac{q p^{-}}{2}} \int_{0}^{B\lambda_{0}} \lambda^{\frac{q p^{-}}{2}} \bigl\vert E(A\lambda)\bigr\vert ^{\frac{q}{\gamma}} \frac{d\lambda}{\lambda} + \frac{\gamma p^{-}}{2} A^{\frac{q p^{-}}{2}} \int_{B\lambda_{0}}^{\infty} \lambda^{\frac{q p^{-}}{2}} \bigl\vert E(A\lambda)\bigr\vert ^{\frac{q}{\gamma}} \frac{d\lambda}{\lambda} \\& \quad := I_{1} + I_{2}. \end{aligned}$$
(4.24)
To estimate \(I_{1}\), inserting A and B into (4.20) yields
$$\begin{aligned} I_{1} \le C_{0} \bigl\vert \Omega_{2R}(x_{0}) \bigr\vert ^{\frac{q}{\gamma}} ( AB \lambda_{0} )^{\frac{q p^{-}}{2}} \le C_{1} \frac{\vert \Omega_{2R}(x_{0})\vert ^{\frac{q}{\gamma}} }{(\tau_{2}-\tau_{1})^{\frac{dq \gamma_{2} }{2}}} ( \lambda_{0} )^{\frac {q p^{-}}{2}}. \end{aligned}$$
For the estimate of \(I_{2}\), using Lemma 4.3, for any \(0<\varepsilon<1\) we have
$$\begin{aligned} I_{2} \le& C_{2} \varepsilon^{\frac{q}{\gamma}} \int_{B\lambda_{0}}^{\infty} \lambda^{\frac{q p^{-}}{2}} \biggl( \bigl\vert \Omega_{\tau_{2}R}(x_{0}) \cap E(\lambda/4)\bigr\vert \\ &{} + \frac{1}{(\varsigma\lambda)^{\eta}} \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{\tau_{2}R}(x_{0}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu} \\ &{} + \frac{1}{(\varsigma\lambda)^{\eta}} \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{\tau_{2}R}(x_{0}): \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu} \biggr)^{\frac{q}{\gamma}} \frac{d\lambda}{\lambda}. \end{aligned}$$
Note that
$$\begin{aligned} ( A_{1} + A_{2} + A_{3} )^{m} \le \max\bigl\{ 3^{m-1}, 1 \bigr\} \bigl( A_{1}^{m} + A_{2}^{m} + A_{3}^{m} \bigr) \end{aligned}$$
for any \(A_{i}>0\), \(i=1,2,3\), and \(m>0\), therefore we obtain
$$\begin{aligned} I_{2} \le& C_{3} \varepsilon^{\frac{q}{\gamma}} \biggl( \int_{B\lambda_{0}}^{\infty} \lambda^{\frac{q p^{-}}{2}} \bigl\vert \Omega_{\tau_{2}R}(x_{0}) \cap E(\lambda/4)\bigr\vert ^{\frac{q}{\gamma}} \frac{d\lambda}{\lambda} \\ &{} + \int_{B\lambda_{0}}^{\infty} \lambda^{\frac{q p^{-}}{2}} \biggl( \frac{1}{(\varsigma\lambda)^{\eta}} \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{\tau_{2}R}(x_{0}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu} \biggr)^{\frac{q}{\gamma}} \frac{d\lambda}{\lambda} \\ &{} + \int_{B\lambda_{0}}^{\infty} \lambda^{\frac{q p^{-}}{2}} \biggl( \frac{1}{(\varsigma\lambda)^{\eta}} \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{\tau_{2}R}(x_{0}): \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu} \biggr)^{\frac{q}{\gamma}} \frac{d\lambda}{\lambda} \biggr) \\ :=& C_{3} \varepsilon^{\frac{q}{\gamma}} ( I_{21} + I_{22} + I_{23} ). \end{aligned}$$
(4.25)
To estimate \(I_{21}\), a simple computation yields
$$\begin{aligned} I_{21} \le& \int_{0}^{\infty} \lambda^{\frac{q p^{-}}{2}} \bigl\vert \Omega_{\tau_{2}R}(x_{0}) \cap E(\lambda/4)\bigr\vert ^{\frac{q}{\gamma}}\frac{d\lambda}{\lambda} \\ \le& C_{4} \bigl\Vert \vert Du \vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma, q)}(\Omega_{\tau_{2}R}(x_{0}))}. \end{aligned}$$
To estimate \(I_{22}\), we examine it in two cases.
Case 1. If \(q\ge\gamma\), by using Hardy’s inequality showed in Lemma 3.2 and the change of variables, we get
$$\begin{aligned} I_{22} \le& C_{5} \int_{0}^{\infty} \lambda^{\frac{q p^{-}}{2}} \biggl( \frac{1}{(\varsigma\lambda)^{\eta}} (\varsigma\lambda)^{\eta} \bigl\vert \bigl\{ x\in \Omega_{\tau_{2}R}(x_{0}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \varsigma\lambda\bigr\} \bigr\vert \biggr)^{\frac{q}{\gamma}} \frac{d\lambda}{\lambda} \\ =& C_{5} \int_{0}^{\infty} \lambda^{\frac{q p^{-}}{2}} \bigl\vert \bigl\{ x\in\Omega_{\tau_{2}R}(x_{0}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \varsigma\lambda\bigr\} \bigr\vert ^{\frac {q}{\gamma}} \frac{d\lambda}{\lambda} \\ \le& C_{6} \bigl\Vert \vert \mathbf{f}\vert ^{p(x)} \bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{\tau_{2}R}(x_{0}))}. \end{aligned}$$
Case 2. \(0< q<\gamma\), we use the reverse Hölder inequality in Lemma 3.3 and get
$$\begin{aligned}& \biggl( \int_{\varsigma\lambda}^{\infty} \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{\tau_{2}R}(x_{0}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \frac{d\mu}{\mu} \biggr)^{\frac{q}{\gamma}} \\& \quad \le \bigl( (\varsigma\lambda)^{\eta} \bigl\vert \bigl\{ x\in \Omega_{\tau_{2}R}(x_{0}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \varsigma\lambda\bigr\} \bigr\vert \bigr)^{\frac{q}{\gamma}} \\& \qquad {} + C_{7} \int_{\varsigma\lambda}^{\infty} \bigl( \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{\tau_{2}R}(x_{0}): \vert \mathbf{f} \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \bigr)^{\frac {q}{\gamma}} \frac{d\mu}{\mu}, \end{aligned}$$
which yields
$$\begin{aligned} I_{22} \le& \int_{0}^{\infty} \lambda^{\frac{q p^{-}}{2}} \biggl( \frac{1}{(\varsigma\lambda)^{\eta}} (\varsigma\lambda) ^{\eta} \bigl\vert \bigl\{ x\in \Omega_{\tau_{2}R}(x_{0}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \varsigma\lambda\bigr\} \bigr\vert \biggr)^{\frac{q}{\gamma}} \frac{d\lambda}{\lambda} \\ &{} + C_{7} \int_{0}^{\infty} \lambda^{\frac{q p^{-}}{2}} \biggl( \frac{1}{(\varsigma\lambda)^{\eta}} \biggr)^{\frac{q}{\gamma}} \int_{\varsigma\lambda}^{\infty} \bigl( \mu^{\eta} \bigl\vert \bigl\{ x\in\Omega_{\tau_{2}R}(x_{0}): \vert \mathbf{f} \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \mu\bigr\} \bigr\vert \bigr)^{\frac {q}{\gamma}} \frac{d\mu}{\mu} \frac{d\lambda}{\lambda} \\ \le& C_{8} \int_{0}^{\infty} \lambda^{\frac{q p^{-}}{2}} \bigl\vert \bigl\{ x\in\Omega_{\tau_{2}R}(x_{0}): \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} > \varsigma\lambda\bigr\} \bigr\vert ^{\frac {q}{\gamma}} \frac{d\lambda}{\lambda} \\ \le& C_{9} \bigl\Vert \vert \mathbf{f}\vert ^{p(x)} \bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{\tau_{2}R}(x_{0}))}, \end{aligned}$$
where we also employ Hardy’s inequality of Lemma 3.2 in the second inequality. For the estimate of \(I_{23}\), using the same way as for the estimate \(I_{22}\) above, we obtain
$$\begin{aligned} I_{23} \le C_{10} \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{\tau_{2}R}(x_{0}))}. \end{aligned}$$
Putting \(I_{21}\), \(I_{22}\), \(I_{23}\) into (4.25) implies
$$\begin{aligned} I_{2} \le C_{11} \varepsilon^{\frac{q}{\gamma}} \bigl( \bigl\Vert \vert Du\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma, q)}(\Omega_{\tau_{2}R}(x_{0}))} + \bigl\Vert \vert \mathbf{f}\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{\tau_{2}R}(x_{0}))} + \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{\tau_{2}R}(x_{0}))} \bigr). \end{aligned}$$
Putting the estimates \(I_{1}\) and \(I_{2}\) into (4.24), we deduce that
$$\begin{aligned}& \bigl\Vert \vert Du\vert ^{p(x)}\bigr\Vert _{L^{(\gamma ,q)}(\Omega_{\tau_{1}R} (x_{0}))}^{q} \\& \quad \le C_{12} \varepsilon^{\frac{q}{\gamma}} \bigl\Vert \vert Du\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma, q)}(\Omega_{\tau_{2}R}(x_{0}))} + C_{12} \frac{\vert \Omega_{2R}(x_{0})\vert ^{\frac{q}{\gamma}} }{(\tau _{2}-\tau_{1})^{\frac{dq \gamma_{2} }{2}}} ( \lambda_{0} )^{\frac{q p^{-}}{2}} \\& \qquad {}+ C_{12} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{\tau_{2}R}(x_{0}))} + \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{\tau _{2}R}(x_{0}))} \bigr). \end{aligned}$$
Set \(\Phi(\tau_{i}) = \Vert \vert Du\vert ^{p(x)}\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{\tau_{i}R}(x_{0}))}\) for \(i=1,2\). Let δ be sufficiently small in Lemma 4.3, and we now select sufficiently small \(\varepsilon>0\) such that
$$\begin{aligned} 0< C_{12} \varepsilon^{\frac{q}{\gamma}} < \frac{1}{2}, \end{aligned}$$
which yields
$$\begin{aligned} \Phi(\tau_{1}) \le& \frac{1}{2} \Phi(\tau_{2}) + C_{12} \frac{\vert \Omega_{2R}(x_{0})\vert ^{\frac{q}{\gamma}} }{(\tau_{2}-\tau_{1})^{\frac{dq \gamma_{2} }{2}}} ( \lambda_{0} )^{\frac {q p^{-}}{2}} \\ &{}+ C_{12} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{\tau_{2}R}(x_{0}))} + \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{\tau _{2}R}(x_{0}))} \bigr) \end{aligned}$$
for any \(\tau_{1}\), \(\tau_{2}\) with \(1 \le\tau_{1} < \tau_{2} \le2\). We then apply the iterating Lemma 3.4 and derive
$$\begin{aligned}& \bigl\Vert \vert Du\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma ,q)}(\Omega_{R}(x_{0}))} \\& \quad \le C_{13} \bigl\vert \Omega_{2R}(x_{0}) \bigr\vert ^{\frac{q}{\gamma}} ( \lambda_{0} )^{\frac{q p^{-}}{2}} \\& \qquad {} + C_{13} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)} \bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{2R}(x_{0}))} + \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{2R}(x_{0}))} \bigr). \end{aligned}$$
Now, we recall the definition of \(\lambda_{0}\) in (4.3) and obtain
$$\begin{aligned}& \bigl\Vert \vert Du\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{R}(x_{0}))} \\& \quad \le C_{14} \bigl\vert \Omega_{2R}(x_{0})\bigr\vert ^{\frac{q}{\gamma}} \biggl( \fint_{\Omega_{2R}(x_{0})} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx \\& \qquad {}+ \frac{1}{\delta} \biggl( \fint_{\Omega_{2R}(x_{0})} \bigl(\vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + 1\bigr)^{\eta}\,dx \biggr)^{\frac{1}{\eta}} \biggr)^{\frac{q p^{-}}{2}} \\& \qquad {} + C_{14} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{2R}(x_{0}))} + \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega _{2R}(x_{0}))} \bigr). \end{aligned}$$
(4.26)
To estimate the first item on the right-hand side of (4.26), notice that
$$\begin{aligned} \frac{2 p^{+}}{p^{-}} =& 2\biggl( 1 + \frac{p^{+}-p^{-}}{p^{-}} \biggr) \\ \le&2 \biggl( 1 + \frac{\omega(2R)}{\gamma_{1}} \biggr) \\ < & 2( 1+ \sigma_{0}) \\ \le&\gamma \gamma_{1} \\ \le&\gamma p^{-}, \end{aligned}$$
(4.27)
where \(\sigma_{0} = \min\{\sigma_{1}, \sigma_{2}\}\) is the same as Lemma 3.7 for \(\gamma\in[1, \infty)\). Then it yields
$$\begin{aligned}& \biggl( \fint_{\Omega_{2R}(x_{0})} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx \biggr)^{\frac{q p^{-}}{2}} \\& \quad \le \biggl( \fint_{\Omega_{2R}(x_{0})} \vert Du\vert ^{\frac{2p^{+}}{p^{-}}}\, dx +1 \biggr)^{\frac{q p^{-}}{2}} \\& \quad \le C_{15} \biggl( \biggl( \fint_{\Omega_{4R}(x_{0})} \vert Du\vert ^{2}\,dx \biggr)^{\frac{p^{+}}{p^{-}\vphantom{\sum}}} + \fint_{\Omega_{4R}(x_{0})} \bigl( \vert \mathbf{f}\vert ^{2} + \vert D\psi \vert ^{2} \bigr)^{\frac{p^{+}}{p^{-}\vphantom{\sum}}}\,dx +1 \biggr)^{\frac{q p^{-}}{2}}, \end{aligned}$$
(4.28)
where we have employed the reverse Hölder inequality in the last inequality. By using the standard \(L^{2}\) estimate as Lemma 3.5, and the embedding inequality due to \(\frac{\gamma\gamma _{1}}{2} >1\) for \(\gamma\in[1, \infty)\) and \(q\in(0, \infty]\),we have
$$\begin{aligned}& \biggl( \biggl( \fint_{\Omega_{4R}(x_{0})} \vert Du\vert ^{2}\,dx \biggr)^{\frac{p^{+}}{p^{-}\vphantom{\sum}}} \biggr)^{\frac{q p^{-}}{2}} \\& \quad \le \biggl( \frac{1}{\vert \Omega_{4R}(x_{0})\vert } \biggr)^{\frac{q p^{+}}{2}} \biggl( \int_{\Omega} \vert Du\vert ^{2}\,dx \biggr)^{\frac{q p^{+}}{2}} \\& \quad \le C_{16} \biggl( \frac{1}{\vert \Omega_{4R}(x_{0})\vert } \biggr)^{\frac{q p^{+}}{2}} \biggl( \int_{\Omega} \vert \mathbf{f}\vert ^{2}\,dx + \int_{\Omega} \vert D\psi \vert ^{2}\,dx \biggr)^{\frac{q p^{+}}{2}} \\& \quad \le C_{17} \biggl( \frac{1}{\vert \Omega_{4R}(x_{0})\vert } \biggr)^{\frac{q p^{+}}{2}} \biggl( \int_{\Omega} \vert \mathbf{f}\vert ^{\frac{2p(x)}{\gamma_{1}}}\,dx + \int_{\Omega} \vert D\psi \vert ^{\frac{2p(x)}{\gamma_{1}}}\,dx + \vert \Omega \vert \biggr)^{\frac{q p^{+}}{2}} \\& \quad \le C_{18} \biggl( \frac{1}{\vert \Omega_{4R}(x_{0})\vert } \biggr)^{\frac{q p^{+}}{2}} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{\frac{2p(x)}{\gamma_{1}}} \bigr\Vert _{L^{(\frac{\gamma\gamma_{1}}{2}, \frac{q \gamma_{1}}{2} )}(\Omega)} + \bigl\Vert \vert D\psi \vert ^{\frac{2p(x)}{\gamma_{1}}} \bigr\Vert _{L^{(\frac{\gamma\gamma_{1}}{2}, \frac{q \gamma_{1}}{2} )}(\Omega)} + 1 \bigr)^{\frac{q p^{+}}{2}} \\& \quad = C_{18} \biggl( \frac{1}{\vert \Omega_{4R}(x_{0})\vert } \biggr)^{\frac{q p^{+}}{2}} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)} \bigr\Vert _{L^{(\gamma, q)}(\Omega)}^{\frac{2}{\gamma_{1}}} + \bigl\Vert \vert D\psi \vert ^{p(x)} \bigr\Vert _{L^{(\gamma, q)}(\Omega)}^{\frac{2}{\gamma_{1}}} + 1 \bigr)^{\frac{q \gamma_{1}}{2} \frac{ p^{+}}{\gamma_{1}}} \\& \quad \le C_{19} \biggl( \frac{1}{\vert \Omega_{4R}(x_{0})\vert } \biggr)^{\frac{q p^{+}}{2}} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)} \bigr\Vert _{L^{(\gamma, q)}(\Omega)}^{q} + \bigl\Vert \vert D\psi \vert ^{p(x)} \bigr\Vert _{L^{(\gamma, q)}(\Omega)}^{q} + 1 \bigr)^{ \frac{\gamma_{2}}{\gamma_{1}}}. \end{aligned}$$
(4.29)
In a similar way as above, we also get the following estimate. In fact, we again use (4.27) for the embedding theory and employ the fact that \(0< R<1\) to obtain
$$\begin{aligned}& \biggl( \fint_{\Omega_{4R}(x_{0})} \bigl( \vert \mathbf{f}\vert ^{2} + \vert D\psi \vert ^{2} \bigr)^{\frac{p^{+}}{p^{-}\vphantom{\sum}}}\,dx \biggr)^{\frac{q p^{-}}{2}} \\& \quad \le C_{20} \biggl( \fint_{\Omega_{4R}(x_{0})} \bigl( \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} \bigr)^{\frac{p^{+}}{p^{-}\vphantom{\sum}}}\, dx + 1 \biggr)^{\frac{q p^{-}}{2}} \\& \quad \le C_{21} \biggl( \frac{1}{\vert \Omega_{4R}(x_{0})\vert } \biggr)^{\frac{q p^{-}}{2}} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\bigr\Vert _{L^{\frac{p^{+}}{p^{-}\vphantom{\sum}}}(\Omega)} + \bigl\Vert \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\bigr\Vert _{L^{\frac{p^{+}}{p^{-}\vphantom{\sum}}}(\Omega)} + 1 \bigr)^{\frac{p^{+}}{p^{-}\vphantom{\sum}} \frac{q p^{-}}{2}} \\& \quad \le C_{22} \biggl( \frac{1}{\vert \Omega_{4R}(x_{0})\vert } \biggr)^{\frac{q p^{+}}{2}} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\bigr\Vert _{L^{( \frac{\gamma p^{-}}{2} , \frac{q p^{-}}{2})}(\Omega)} + \bigl\Vert \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\bigr\Vert _{L^{( \frac{\gamma p^{-}}{2} , \frac{q p^{-}}{2})}(\Omega)} + 1 \bigr)^{\frac{q p^{+}}{2}} \\& \quad \le C_{23} \biggl( \frac{1}{\vert \Omega_{4R}(x_{0})\vert } \biggr)^{\frac{q p^{+}}{2}} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)} \bigr\Vert _{L^{(\gamma, q)}(\Omega)}^{q} + \bigl\Vert \vert D\psi \vert ^{p(x)} \bigr\Vert _{L^{(\gamma, q)}(\Omega)}^{q} + 1 \bigr)^{ \frac{\gamma_{2}}{\gamma_{1}}}. \end{aligned}$$
(4.30)
Putting (4.29) and (4.30) into (4.28) deduces
$$\begin{aligned}& \biggl( \fint_{\Omega_{2R}(x_{0})} \vert Du\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}}\,dx \biggr)^{\frac{q p^{-}}{2}} \\& \quad \le C_{24} \biggl( \frac{1}{\vert \Omega _{4R}(x_{0})\vert } \biggr)^{\frac{qp^{+}}{2}} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma, q)}(\Omega)} + \bigl\Vert \vert D \psi \vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma, q)}(\Omega)} + 1 \bigr)^{ \frac{\gamma_{2}}{\gamma_{1}}}. \end{aligned}$$
(4.31)
Now we are in a position to estimate the second item on the right-hand side of (4.26). Since \(\eta= 1+ \epsilon\) for sufficiently small \(\epsilon>0\) as the limit of (4.7), we have \(2\eta< \gamma p^{-}\). Then, for any \(\gamma\in[1, \infty)\), by the embedding inequality it follows that
$$\begin{aligned}& \biggl( \fint_{\Omega_{2R}(x_{0})} \bigl(\vert \mathbf{f}\vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + \vert D\psi \vert ^{\frac{2p(x)}{p^{-}\vphantom{\sum}}} + 1\bigr)^{\eta}\,dx \biggr)^{\frac{q p^{-}}{2\eta}} \\& \quad \le C_{25} \biggl( \frac{1}{\vert \Omega_{2R}(x_{0})\vert } \biggr)^{\frac{q p^{-}}{2\eta}} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)}\bigr\Vert _{L^{\frac{2\eta}{p^{-}}} (\Omega_{2R}(x_{0}))} + \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert _{L^{\frac{2\eta}{p^{-}}} (\Omega_{2R}(x_{0}))} + 1 \bigr)^{q} \\& \quad \le C_{25} \biggl( \frac{1}{\vert \Omega_{2R}(x_{0})\vert } \biggr)^{\frac{q p^{+}}{2}} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)}\bigr\Vert _{L^{\frac{2\eta}{p^{-}}} (\Omega)} + \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert _{L^{\frac{2\eta}{p^{-}}} (\Omega)} + 1 \bigr)^{q} \\& \quad \le C_{26} \biggl( \frac{1}{\vert \Omega_{2R}(x_{0})\vert } \biggr)^{\frac{q p^{+}}{2}} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega)} + \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega)} + 1 \bigr). \end{aligned}$$
(4.32)
Putting (4.31) and (4.32) into (4.26), we have
$$\begin{aligned}& \bigl\Vert \vert Du\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega_{R}(x_{0}))} \\& \quad \le C_{27} \bigl( \bigl\vert \Omega_{R}(x_{0}) \bigr\vert ^{\frac{1}{\gamma}-\frac{\gamma_{2}}{2}} \bigr)^{q} \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega)} + \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma ,q)}(\Omega)} + 1 \bigr)^{\frac{\gamma_{2}}{\gamma_{1}}}. \end{aligned}$$
(4.33)
The rest of Step 1 is to use the standard finite covering argument to obtain the global estimate. In fact, since Ω̅ is compact in \(\mathbb{R}^{d}\), there exist finitely many points \(x_{0}^{k} \in\overline{\Omega}\), \(k=1,2, \ldots, N\) and the corresponding \(R_{k}\) such that \(\Omega= \bigcup_{k=1}^{N}\Omega _{R_{k}}(x^{k}_{0})\). Therefore,
$$\begin{aligned} \bigl\Vert \vert Du\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma, q)}(\Omega)} \le\sum_{k=1}^{N} \bigl\Vert \vert Du \vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma, q)}(\Omega_{R_{k}}(x^{k}_{0}))}. \end{aligned}$$
Now, thanks to estimate (4.33), it yields
$$\begin{aligned}& \bigl\Vert \vert Du\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma, q)}(\Omega)}\\& \quad \le C_{27} \sum_{k=1}^{N} \bigl( \bigl\vert \Omega_{R_{k}}\bigl(x^{k}_{0} \bigr)\bigr\vert ^{\frac{1}{\gamma}-\frac{\gamma_{2}}{2}} \bigr)^{q} \bigl(\bigl\Vert \vert \mathbf{f}\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma ,q)}(\Omega)} + \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega)} + 1 \bigr)^{\frac{\gamma_{2}}{\gamma_{1}}} \\& \quad \le C_{28} \bigl(\bigl\Vert \vert \mathbf{f}\vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega)} + \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert ^{q}_{L^{(\gamma,q)}(\Omega)} + 1 \bigr)^{\frac{\gamma_{2}}{\gamma_{1}}}, \end{aligned}$$
where \(C_{28}\) is a constant depending only on d, ν, Λ, \(\omega (\cdot)\), γ, q, δ, \(R_{0}\) and \(\vert \Omega \vert \). The proof of \(q=\infty\) is even simpler. Here for briefness we omit it, the reader may also refer to Section 5.4 in [13] for the case of \(q=\infty\).
Step 2. The remainder of our proof for Theorem 2.5 is to remove the assumption \(\vert Du\vert ^{p(x)}\in L^{(\gamma, q)}(\Omega)\) via an approximation procedure. To do this, let \(\{\vert \mathbf{f}_{k}\vert ^{p(x)}\}_{k=1}^{\infty}\) and \(\{\vert D\psi _{k}\vert ^{p(x)}\}_{k=1}^{\infty}\) be two sequences in \(C_{0}^{\infty }(\Omega)\) converging to \(\vert \mathbf{f}\vert ^{p(x)}\) and \(\vert D\psi \vert ^{p(x)}\) in \(L^{(\gamma,q)}(\Omega)\). It is clear that \(\vert \mathbf{f}_{k}\vert \) and \(\vert D\psi_{k}\vert \in L^{(\gamma\gamma_{2},q \gamma_{2})}(\Omega)\). According to the earlier work [10] and the facts that Lorentz space is an interpolation space of Lebesgue spaces, and the obstacle problems under consideration are linear, the unique weak solution
$$\begin{aligned} u_{k} \in\mathcal{A}_{k} = \bigl\{ \phi_{k} \in W^{1,2}_{0}(\Omega): \phi_{k} \ge \psi_{k} \mbox{ a.e. in } \Omega\bigr\} \end{aligned}$$
of the following variational inequalities
$$\begin{aligned} \int_{\Omega} A(x) Du_{k} D( \phi_{k} - u_{k} )\,dx \ge \int_{\Omega} \mathbf{f}_{k} D( \phi_{k} - u_{k})\,dx \quad \mbox{for all } \phi_{k} \in\mathcal{A} \end{aligned}$$
satisfies a global gradient estimate in \(L^{(\gamma\gamma_{2},q \gamma _{2})}(\Omega)\) under the assumption that \((A(x), \Omega)\) is \((\delta, R_{0})\)-vanishing of codimension one. Thus, we have
$$\begin{aligned} \vert Du_{k}\vert \in L^{(\gamma\gamma_{2},q \gamma_{2})}(\Omega) \quad \Longrightarrow\quad \vert Du_{k}\vert ^{p(x)} \in L^{(\gamma, q)}(\Omega) \end{aligned}$$
due to (3.3). As a consequence of the interpolation space, we have
$$\begin{aligned} \bigl\Vert \vert Du_{k}\vert ^{p(x)}\bigr\Vert _{L^{(\gamma, q)}(\Omega)} \le& C \bigl( \bigl\Vert \vert \mathbf{f}_{k} \vert ^{p(x)}\bigr\Vert _{L^{(\gamma, q)}(\Omega)} + \bigl\Vert \vert D \psi_{k}\vert ^{p(x)}\bigr\Vert _{L^{(\gamma, q)}(\Omega)} + 1 \bigr)^{\frac{\gamma_{2}}{\gamma_{1}}} \\ \le& C \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)}\bigr\Vert _{L^{(\gamma, q)}(\Omega)} + \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert _{L^{(\gamma, q)}(\Omega)} + 1 \bigr)^{\frac{\gamma_{2}}{\gamma_{1}}}, \end{aligned}$$
where C is independent of k. From this estimate we observe that there exists ū with \(\vert D\bar{u}\vert ^{p(x)}\in L^{(\gamma,q)}(\Omega)\) which is the weak limit of \(\{u_{k}\} _{k=1}^{\infty}\) in \(\mathcal{A}_{k}\) such that
$$\begin{aligned} \bigl\Vert \vert D\bar{u}\vert ^{p(x)}\bigr\Vert _{L^{(\gamma, q)}(\Omega)} \le& C \bigl( \bigl\Vert \vert \mathbf{f}\vert ^{p(x)}\bigr\Vert _{L^{(\gamma, q)}(\Omega)} + \bigl\Vert \vert D\psi \vert ^{p(x)}\bigr\Vert _{L^{(\gamma, q)}(\Omega)} + 1 \bigr)^{\frac{\gamma_{2}}{\gamma_{1}}}. \end{aligned}$$
Then it is easy to check that this ū is the weak solution of the original problem (1.1). So, by the uniqueness, we conclude that \(u=\bar{u}\) almost everywhere in Ω. This completes the approximation procedure. □