\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 51, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/51\hfil Local estimates for gradients]
{Local estimates for gradients of solutions to elliptic equations 
with variable exponents}

\author[F. Yao \hfil EJDE-2013/51\hfilneg]
{Fengping Yao}  % in alphabetical order

\address{Fengping Yao \newline
 Department of Mathematics, Shanghai University, Shanghai 200444, China}
 \email{yfp@shu.edu.cn}

\thanks{Submitted September 3, 2012. Published February 18, 2013.}
\subjclass[2000]{35J60, 35J70}
\keywords{Regularity; divergence; nonlinear; elliptic equation;
gradient; \hfill\break\indent variable exponent; $p(x)$-Laplacian}

\begin{abstract}
 In this article we  present local $L^\infty$ estimates for the  gradient  
 of  solutions to  elliptic equations with variable exponents. 
 Under proper conditions on the coefficients, we prove that
 $$
  \left| \nabla u\right|\in L^{\infty}_{loc}
  $$
 for all  weak solutions of
 $$
 \operatorname{div}  (g(|\nabla u|^2,x) \nabla u )=0\quad \text{in } \Omega.
 $$
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks


\section{Introduction}

Uhlenbeck \cite{Uhlenbeck}   obtained the interior
H\"{o}lder regularity estimates for weak solutions of
\begin{equation}
\operatorname{div}   (\rho(| \nabla u |^2)
 \nabla u )=0\quad \text{in }\Omega, \label{1.31}
\end{equation}
where $\Omega$ is an open bounded domain in $\mathbb{R}^n$ and
$\rho\in C^1( [0, \infty)) $ is a non-negative
function satisfying the   ellipticity conditions
 \begin{gather}
K^{-1}(\xi+c)^{\frac{p}{2}-1}\leq \rho( \xi)+ 2
\rho'(\xi )\xi \leq K(\xi+c)^{\frac{p}{2}-1},\label{condition1}
\\
|\rho'(\xi_1)\xi_1- \rho'(\xi_2)\xi_2|\leq K (\xi_1+\xi_2+c
)^{p/2-1-\alpha}( \xi_1-\xi_2)^{\alpha}\label{condition2}
\end{gather}
for $c \geq 0$, $\alpha>0$ and  $p \geq 2$. Especially when
$\rho(t)=t^{(p-2)/2}$, \eqref{1.31} is reduced to the  well-known
$p$-Laplace  equation.
In this paper  we discuss the  nonlinear elliptic equation  of
the  form
\begin{equation}
 \operatorname{div}  (g(| \nabla u |^2,x) \nabla u)=0\quad
\text{in }\Omega, \label{1.2}
\end{equation}
 where  $g(\xi, x)\in C^1( [0, \infty)\times \Omega) $
satisfies the  ellipticity conditions
 \begin{gather}
C_1(\xi+c)^{\frac{p(x)}{2}-1}\leq g( \xi,x)+ 2 \xi    g_{\xi}(\xi,x )
 \leq C_2(\xi+c)^{\frac{p(x)}{2}-1}, \label{condition1b}\\
 | \nabla_{x}  g(\xi,x ) | \leq C_3 (\xi+c)^{\frac{p(x)-1}{2}} |\nabla p |
| \ln (\xi+c ) | \label{condition111}
\end{gather}
for $c\geq 0$ and $C_1,C_2,C_3 >0 $. Here
$p\in W^{1,s}(\Omega)$ for some $s>n$ satisfies
\begin{equation}
 1<p_1=\inf_{\overline{\Omega}}p(x)\leq p(x)
\leq \sup_{\overline{\Omega}}p(x)= p_2< \infty. \label{pxcondition1}
\end{equation}
Especially when $g(t)=|t|^{ \frac{p(x)-2}{2} }$,
\eqref{1.2} is reduced to the $p(x)$-Laplace  elliptic equation
\begin{equation}
\operatorname{div} (|\nabla u|^{p(x)-2} \nabla
u) = 0\quad \text{in }\Omega,\label{pastequation}
 \end{equation}
whose special case is the  well-known elliptic  $p$-Laplace equation
 \[
 \operatorname{div}  ( | \nabla u |^{p-2} \nabla u )=0\quad
\text{in }\Omega, %\label{1.3}
 \]
 which  can be derived from the variational problem
  $$
\Phi(u)=  \min_{v|_{\partial
\Omega} =\varphi}\Phi(v)=:\min_{v|_{\partial \Omega}
=\varphi}\int_{\Omega}|\nabla v|^p dx.
$$

We  denote by $L^{p(x)}(\Omega)$  the variable
exponent Lebesgue-Sobolev space
 \begin{equation}
L^{p(x)}(\Omega)=\{ f:\Omega\to \mathbb{R} : f \text{ is  measurable  and } 
\int_{\Omega}|f|^{p(x)} dx < \infty \}
 \end{equation}
equipped with the Luxemburg  type norm
 \begin{equation}
\|f\|_{L^{p(x)}(\Omega)}=\inf \big\{ \lambda>0 :
 \int_{\Omega}|\frac{f}{\lambda}|^{p(x)}dx \leq 1 \big\}.
 \end{equation}
Furthermore, we define
 \begin{equation}
W^{1,p(x)}(\Omega)=\{u\in L^{p(x)}(\Omega)  : 
|\nabla u|\in L^{p(x)}(\Omega)\}
 \end{equation}
equipped with the norm 
\begin{equation} 
\|u\|_{W^{1,p(x)}(\Omega)}=\|u\|_{L^{p(x)}(\Omega)}
+\|\nabla u\|_{L^{p(x)}(\Omega)}.
 \end{equation}

By $W_{0}^{1,p(x)}(\Omega)$ we  denote 
 the closure of $C_0^{\infty}(\Omega)$  in $W^{1,p(x)}(\Omega)$.  
Actually, the $L^{p(x)}(\Omega)$, $W^{1,p(x)}(\Omega)$ and
$W_{0}^{1,p(x)}(\Omega)$ spaces are Banach spaces. 
There have been many investigations 
(see for example \cite{Diening0,Diening,Diening1,Diening2,Fan2,Fan3,Harjulehto}) 
on properties of such variable
exponent Sobolev spaces.

 As usual, the  solutions of \eqref{1.2} are taken in  a weak sense.
 We now state the definition of  weak solutions.

\begin{definition}\label{defweaksolution} \rm
 A function $u \in   W^{1, p(x)}_{\rm loc}(\Omega)$ is a local
 weak solution of  \eqref{1.2} in $\Omega$ if for
any $ \varphi \in W_0^{1,p(x)}( \Omega)$, we have 
\[
 \int_{ \Omega}g(| \nabla u |^2,x) \nabla u  \cdot \nabla \varphi dx=0.
\]
\end{definition}

 When $p(x)$ is a constant,  many authors 
\cite{Byun2,Byun4,Dibene,Duzaar,Duzaar1,Kinnumen1,Iwaniec,Mingione} 
have studied the regularity estimates for weak solutions of
quasilinear elliptic equations of $p$-Laplacian type and the general case.  
   When $p(x)$ is not a constant,   such  elliptic problems \eqref{1.2} 
appear in mathematical models of various physical phenomena, such as 
the electro-rheological fluids (see, e.g., \cite{Acerbi0,Rajagopal,Ruzicka}). 
There have been many investigations  \cite{Coscia,Fan0,Lyaghfouri} on
H\"{o}lder  estimates for the    $p(x)$-Laplacian elliptic 
equation \eqref{1.2} and the more general case. Moreover,
 Acerbi and Mingione \cite{Acerbi1}  proved that
\[
 |\mathrm{ \bf f}|^{p(x)}\in L^q_{\rm loc}(\Omega)\quad \Longrightarrow 
\quad |\nabla u|^{p(x)}\in L^q_{\rm loc}(\Omega) \quad  \text{for }   q>1 
\] 
of weak solutions  of \eqref{1.2}  under some   assumptions.
 The purpose of this paper is to extend the results in  \cite{Challal},
 where Challal and Lyaghfouri 
 obtained the local $L^{\infty}$ estimates of $| \nabla u|$ for the  
weak solutions of \eqref{pastequation}.


 We assume that $p(x)\in W^{1,s}(\Omega)$ for some $s>n$. 
Therefore, it follows  from Sobolev embedding theorem  that   
$p(x) $ is H\"{o}lder continuous with the exponent $\alpha=1-\frac{n}{s}$. 
Now let us  state the main result of this work.

\begin{theorem}\label{thm1.2} 
 Let $u \in W^{1, p(x)}_{\rm loc}(\Omega)$ be a local weak solution of
\eqref{1.2} in $\Omega$ under the assumptions  
\eqref{condition1}-\eqref{pxcondition1}.  Then
 $$
 | \nabla u| \in L^{\infty}_{\rm loc}(\Omega).
 $$
 Moreover, for each $\sigma>0$ and $\delta\in  (0,\frac{sq(1+\sigma)}{s-q}-1)$  
with a constant  $ q\in (n,s)$ there exists a positive constant  $R_0$, 
depending only on   $n,p_1,p_2,s,\sigma$  and 
$\| |\nabla u(\cdot)|^{p(\cdot)}\|_{L^1(\Omega)}$, such that, wherever 
$R \leq R_0$ and the ball $  B_{8R} \subset \Omega$,
\[
\sup_{B_{R/2}}  | \nabla u|^{p(x)}
\leq C\Big[\int\hspace{-0.38cm}- _{B_{2R}}  | \nabla u|^{p(x) } dx
  +R^{\alpha}M^{(1+\sigma)\delta}\Big( \int\hspace{-0.38cm}- _{B_{8R}} | \nabla u|^{p(x) }+ 1  dx
\Big)^{ 1+\sigma }\Big],
\]
where   $    M=\int_{B_{8R}} | \nabla u|  ^{ p(x)   } dx+1$  and 
$C$ depends on $n,p_1,p_2,s,\delta,  \|  p\|_{W^{1,s}(\Omega)}$. 
 \end{theorem}

 \section{Proof of main result}

 In this  section  we  prove Theorem  \ref{thm1.2} by  the approximation method.  
 Our approach is much influenced by \cite{Acerbi1,Challal1,Challal,Wang1}.
 We first consider the following approximation  problem
\begin{equation}
 \operatorname{div} (g(\epsilon+ | \nabla u^{\epsilon} |^2, x )
 \nabla u^{\epsilon} )=0,\quad x\in B_{R'}, \epsilon \in (0, 1], \label{2.2}
\end{equation}
where $B_{8R}\subset B_{R'}\subset  \Omega$. 
It is standard that \eqref{2.2}, with the boundary condition 
$u^{\epsilon} = u$ on $\partial B_{R'}$, has a unique solution 
$u^{\epsilon}$ for fixed $\epsilon > 0$.
Similarly to \cite{Challal}, we know 
$u^{\epsilon} \in W^{2,2}_{\rm loc}(\Omega)$.  
From \cite{Fan0} we can get $u \in C^{1, \mu }_{\rm loc}(\Omega) $ 
 for some $\mu  \in (0,1)$ and then   have 
$u^{\epsilon} \in C^{1,\nu}(\overline{B}_{R'}) $  for some 
$\nu \in (0,1)$ and  $\|u^{\epsilon}\|_{C^{1,\nu}(\overline{B}_{R'})} \le C $,
 where $C$ is a    constant independent of $\epsilon$. 
It  follows  from Ascoli-Arzel\`{a} theorem  that there exists a sequence of
$\{\epsilon_k\}$ converging to $0$
and  satisfying  $ u^{\epsilon_k}\to u $ uniformly in  
$C^{1}(\overline{B}_{R'})$.  Thus, we can get the result of
 Theorem \ref{thm1.2} by passing
to the limit as $\epsilon_k\to 0$ in \eqref{finalresult} with  $u^{\epsilon_k}$    replacing $u$. So it is sufficient to prove \eqref{finalresult}.  For simplicity, we shall drop the index $\epsilon$ on $u^{\epsilon}$ in the exposition.
Actually, from \eqref{2.2} we have
    \begin{equation}
\begin{split}
&\big[  g( \epsilon+ | \nabla u|^2, x )\delta_{ij}+ 2 g_{\xi} (
\epsilon+ | \nabla u|^2, x)u_i u_j \big] u_{ij}+ g_{x_i}(
\epsilon+ | \nabla u|^2, x)u_i \\
&=:a_{ij}u_{ij}+b_i u_i =0. 
\end{split}\label{2.3} 
\end{equation}

 \begin{lemma}\label{lemma24gaij}
If $g(\xi, x)\in C^1( [0, \infty)\times \Omega) $  satisfies  
the conditions \eqref{condition1} and \eqref{condition111}, then
\begin{gather}
C_4(\xi+c)^{\frac{p(x)}{2}-1}\le  g( \xi, x)\leq C_5(\xi+c)
^{\frac{p(x)}{2}-1},\label{condition2b}\\
| g_\xi(\xi, x )\xi|\leq
C_6(\xi+c)^{\frac{p(x)}{2}-1}\label{condition3}
\end{gather}
 for the constants $0<C_4<C_1$, $C_5>C_2>0,C_6>0 $, and
\begin{equation}
C_4\Big(c+ \epsilon+ | \nabla u|^2\Big)^{\frac{p(x)}{2}-1} | \xi|^2
\le a_{ij} \xi_i\xi_j \le C_5\Big( c+\epsilon+ | \nabla
u|^2\Big)^{\frac{p(x)}{2}-1}| \xi|^2.
\end{equation}
 \end{lemma}

 \begin{proof}
We prove only \eqref{condition2}.  First, we find that
 \[
\xi^{1/2}g( \xi, x)=\int_{0}^{\xi}(t^{1/2}g(t, x))_tdt
=\int_{0}^{\xi} \frac{1}{2}t^{-1/2} \big[  g(t, x)+ 2t g_t(t,x) \big] dt.
\]
 Moreover,
from \eqref{condition1} we deduce that 
\[ 
I_1=:\frac{C_1}{2}\int_{0}^{\xi} t^{-1/2}   ( t+c)^{\frac{p(x)}{2}-1}   dt 
\le \xi^{1/2}g(
\xi,x) \le \frac{C_2 }{2}\int_{0}^{\xi} t^{-1/2}   ( t+c)^{\frac{p(x)}{2}-1}  
 dt=:I_2.
\]
To estimate of $I_1$ and $I_2$, we consider two cases.

\textbf{Case 1: $c \leq \xi$.}  We have
\begin{align*}
 I_1 &\ge \frac{C_1}{2}  \int_{0}^{\xi}    ( t+c)^{\frac{p(x)-3}{2} }   dt\\
 &\ge  \frac{C_1}{2}  (
\frac{1}{\frac{p(x)-1}{2} })
[  ( \xi+c)^{\frac{p(x)-1}{2}}- c^{\frac{p(x)-1}{2}}]\\
 &\ge   \frac{C_1}{p(x)-1}   [  (
\xi+c)^{\frac{p(x)-1}{2}}- c^{\frac{p(x)-1}{2}}].
\end{align*}
Since $1< p_1 \le p(x) \le p_2$ and $ c \le \frac{c+\xi}{2}$, 
we obtain 
\begin{align*}
 I_1 &\ge \frac{C_1}{p_2-1}   \big[ 1- (\frac{1}{2})^{\frac{p(x)-1}{2}}\big] 
 ( \xi+c)^{\frac{p(x)-1}{2}}\\
&\ge   C_1'( \xi+c)^{\frac{p(x)-1}{2}}\\ 
&\ge C_1'( \xi+c)^{\frac{p(x)}{2}-1} \xi^{1/2}.
 \end{align*}
Moreover, we deduce that
\[
 I_2  \le   \frac{C_2 }{2}( \xi+c)^{\frac{p(x)}{2}-1}
\int_{0}^{\xi} t^{-1/2} dt
 =     C_2  ( \xi+c)^{\frac{p(x)}{2}-1} \xi^{1/2} \quad \text{for }  p(x) \ge 2
\]
and
 \begin{align*}
 I_2  &\le   \frac{C_2 }{2}  \int_{0}^{\xi}  t^{\frac{p(x)-1}{2}-1}     dt\\
 &=     \frac{C_2 }{p(x)-1} \xi^{\frac{p(x)-1}{2} } \\
 &\le   \frac{C_2 }{p_1-1} ( \xi+c)^{\frac{p(x)-1}{2} } \\
 &=  \frac{C_2 }{p_1-1} ( \xi+c)^{\frac{p(x)}{2}-1 }( \xi+c)^{1/2} 
\quad \text{for }    1< p(x)<2,
\end{align*}
which implies 
\[
I_2   \leq  \frac{C_2 }{p_1-1} ( \xi+c)^{\frac{p(x)}{2}-1 }( 2\xi)^{1/2}
 =  \frac{\sqrt{2}C_2 }{p_1-1} ( \xi+c)^{\frac{p(x)}{2}-1 } \xi^{1/2}
\]
 in view of the fact that $\xi+c \le 2\xi$.

\textbf{Case 2: $c \geq \xi$.}  Then we have
\[
 I_1 \ge \frac{C_1}{2}  \xi^{-1/2}
(\xi+c)^{-1/2}  \int_{0}^{\xi}    ( t+c)^{\frac{p(x)-1}{2} }  dt.
\]
 Furthermore,  
\[
  I_1 \ge \frac{C_1}{2}
\xi^{-1/2}   ( \xi+c)^{\frac{p(x)}{2}-1} (\frac{1}{2} )^{\frac{p(x)-1}{2}} \xi
 \ge    C_1''     ( \xi+c)^{\frac{p(x)}{2}-1}  \xi^{1/2}
\]
since
 $$
 t+c \ge c \ge\frac{1}{2} (2c ) \ge \frac{1}{2}( \xi+c).
 $$
Since the result \eqref{condition2} is trivial when $c=0$. 
Without loss of generality we may as well assume that $c>0$.   
Moreover, we first have
  \begin{align*}
   I_2 &\le \frac{C_2 }{2}
c^{-1/2}\int_{0}^{\xi} t^{-1/2}   ( t+c)^{\frac{p(x)-1}{2}}   dt \\ 
&\le \frac{C_2 }{2} c^{-1/2}( \xi+c)^{\frac{p(x)-1}{2}}   \int_{0}^{\xi}
t^{-1/2}     dt,
\end{align*}
which implies
\begin{align*}
 I_2 &\le  C_2  c^{-1/2}( \xi+c)^{\frac{p(x)-1}{2}} \xi^{1/2}  \\
  &=  C_2
(\frac{\xi+c}{c} )^{1/2}( \xi+c)^{\frac{p(x)}{2}-1} \xi^{1/2}\\
&\le \sqrt{2} C_2  ( \xi+c)^{\frac{p(x)}{2}-1} \xi^{1/2}.
\end{align*}
Thus, from Cases 1 and  2  we have
\[
g( \xi,x)  \ge \min\{ C_1', C_1''\}( \xi+c)^{\frac{p(x)}{2}-1}=:C_4 (
\xi+c)^{\frac{p(x)}{2}-1}.
\]
 and
\[
g( \xi,x)  \le \max\{\frac{\sqrt{2}C_2 }{p_1-1},  \sqrt{2} C_2\}(
\xi+c)^{\frac{p(x)}{2}-1}=:C_5 ( \xi+c)^{\frac{p(x)}{2}-1},
\]
which  completes the proof.
\end{proof}

Now we denote
\begin{equation} 
\widetilde{a_{ij}}=\frac{a_{ij}}{(c+\epsilon+ | \nabla u|^2)^{p(x)/2-1}}. 
\label{2.4} 
\end{equation}
Then, from the lemma above we have 
\begin{equation}
 C_4 | \xi|^2\le \widetilde{a_{ij}}\xi_i\xi_j \le C_5 | \xi|^2 \quad
    \text{for each }  \xi \in \mathbb{R}^n.
\end{equation}

\begin{lemma}\label{coraij}
  Let $v=(c+\epsilon+ | \nabla u|^2)^{p(x)/2}$.  Then
\begin{align*}
\operatorname{div} ( \frac{1}{p(x)}\widetilde{a_{ij}}\cdot \nabla v)
&\ge  \operatorname{div} \Big( a_{ij}  \cdot ( c+\epsilon+ | \nabla u|^2)
\cdot\ln (c+\epsilon+ | \nabla u|^2)^{1/2}     \cdot  \frac{\nabla
    p(x)}{p(x)}\Big)\\
 &=: \operatorname{div}F,
\end{align*}
where
\[
 |F|  \le  C\Big[1+(c+\epsilon+ | \nabla u|^2)^{\frac{p(x)(1+\sigma)}{2} } \big] 
\cdot |\nabla p  | \quad \text{for any }     \sigma >0.
\]
 \end{lemma}

\begin{proof}   We first find that 
\begin{align*}
&v_{x_j}\\
&=\Big((c+\epsilon+ | \nabla u|^2)\ln (c+\epsilon+ | \nabla u|^2) ^{1/2}     p_{x_j}(x)+ p(x)  u_{kj}
   u_{k}\Big)
(c+\epsilon+ | \nabla u|^2)^{\frac{p(x)}{2}-1} 
\end{align*} 
and 
\begin{align*}
&\operatorname{div} ( \frac{1}{p(x)}\widetilde{a_{ij}}\cdot \nabla v)\\
&=\operatorname{div} ( a_{ij}  \cdot(c+ \epsilon+ | \nabla u|^2)
 \cdot\ln (c+\epsilon+ | \nabla u|^2)^{1/2}  \cdot   \frac{\nabla
    p(x)}{p(x)})+  (a_{ij} u_{kj}u_{k} )_{x_i}.
\end{align*}
  Moreover, differentiating  \eqref{2.2} with respect to $x_k$, we
have \begin{equation} 
(a_{ij} u_{kj})_{x_i}+(b_k u_{i})_{x_i}=0,\label{aiju}
\end{equation}
where $b_k$ is defined in \eqref{2.3}.   Therefore, from \eqref{aiju} 
and Lemma \ref{lemma24gaij}  we have
\begin{align*}
&\operatorname{div} \Big( \frac{1}{p(x)}\widetilde{a_{ij}}\cdot \nabla v\Big)\\
&= \operatorname{div} \Big( a_{ij}\cdot  (c+ \epsilon+ | \nabla u|^2)\cdot\ln (c+\epsilon+ | \nabla u|^2)^{1/2}  \cdot   \frac{\nabla
        p(x)}{p(x)}\Big)
    +   a_{ij} u_{kj}u_{ki}-(b_k u_{i})_{x_i}  \\
&\ge  \operatorname{div} \Big( a_{ij}  \cdot(c+ \epsilon+ | \nabla u|^2)\cdot\ln (c+\epsilon+ | \nabla u|^2)^{1/2}  \cdot   \frac{\nabla
    p(x)}{p(x)}\Big)-(b_k u_{i})_{x_i}\\
&=: \operatorname{div}   F,
\end{align*}
where
  $$
  F= a_{ij}  \cdot\Big(c+ \epsilon+ | \nabla u|^2\Big)\cdot\ln 
\Big(c+\epsilon+ | \nabla u|^2\Big)^{1/2}  \cdot   \frac{\nabla
p(x)}{p(x)}- b_ku_i
  $$
 Actually, for any $\sigma>0$, we deduce that
  \begin{align*}
 |F|
&\le  C \big\{| a_{ij}| (c+\epsilon+ | \nabla u|^2)  
 |\ln (c+\epsilon+ | \nabla u|^2)^{1/2}|  |\nabla p | + | b_k||\nabla u |\big\}\\
&\le  C\big(c+\epsilon+ | \nabla u|^2\big)^{\frac{p(x)}{2}}  
 |\ln (c+\epsilon+ | \nabla u|^2)^{1/2}| |\nabla p |\\
&\le  C\big(c+\epsilon+ | \nabla u|^2\big)^{\frac{p(x)}{2}}  
 |\ln (c+\epsilon+ | \nabla u|^2)^{\frac{p(x)}{2}}| |\nabla p |\\
&\le  C \big[1+(c+\epsilon+ | \nabla u|^2)^{\frac{p(x)(1+\sigma)}{2} } \big]  
 |\nabla p |.
\end{align*}
By  \eqref{condition111},  Lemma \ref{lemma24gaij} and
 $$
 |x|| \ln x | \leq C ( 1+ |x|^{1+\sigma}) \quad \text{for any }     \sigma >0,
$$
we complete the proof. \end{proof}

Next, we shall finish the proof of the main result.
 
\begin{proof}
  Using \cite[Theorem 8.17]{GT}, we obtain
 $$
\sup_{B_{R/2}}\Big(c+\epsilon+ | \nabla
u|^2\Big)^{p(x)/2} 
\leq C\Big( \frac{1}{R^n} \int_{B_{2R}}(c+\epsilon
 + | \nabla u|^2)^{\frac{p(x)}{2}} dx+K(R)\Big),
$$ 
where
$$
K(R)=R^{1-\frac{n}{q}}\Big(\int_{B_{2R}}|F|^q dx\Big)^{1/q}
$$
and $q\in (n, s)$ is a positive constant.  Moreover, we find that
\[
 \int_{B_{2R}}|F|^q dx     
 \le  C\big\{  \int_{B_{2R}}   |\nabla p |^q dx  
 + \int_{B_{2R}} (c+\epsilon+ | \nabla
 u|^2)^{\frac{p(x)(\sigma  +1)q}{2}} |\nabla p |^q dx\big\}
\]
Furthermore, using H\"{o}lder's inequality and $p \in W^{1,s}$, we obtain
\begin{align*}
&\int_{B_{2R}}|F|^q dx\\
&\le C \Big(\int_{B_{2R}}    |\nabla p |^{s}dx\Big)^{q/s} 
\Big\{ \Big( \int_{B_{2R}}(c+\epsilon+ | \nabla
u|^2)^{\frac{p(x)sq(1+\sigma)}{2(s-q)} } dx\Big)^{ \frac{s-q}{s}}
 + R^{\frac{n(s-q)}{s}}\Big\}
\\
&\le C  \Big\{ \Big( \int_{B_{2R}}(c+\epsilon
+ | \nabla u|^2)^{\frac{p(x)sq(1+\sigma)}{2(s-q)} } dx\Big)^{ \frac{s-q}{s}}
 +R^{\frac{n(s-q)}{s}}\Big\}
\\
&\le C R^{\frac{n(s-q)}{s}}  
\Big[ \int\hspace{-0.38cm}- _{B_{2R}}(c+\epsilon+ | \nabla u|^2)^{\frac{p(x)sq(1+\sigma)}{2(s-q)} }
 +1 dx\Big]^{ \frac{s-q}{s}}.
\end{align*}
From \cite[Theorem 2]{Acerbi1},  
for any $\delta\in  (0,\frac{sq(1+\sigma)}{s-q}-1)$ there exists a 
positive constant $R_0$, depending only on $n,p_1,p_2,s,\delta,\sigma$  
and $\|\, |\nabla   u(\cdot)|^{p(\cdot)}\|_{L^1(\Omega)}$, such that,
 wherever $R \leq R_0$,
\[
\int_{B_{2R}}|F|^q dx
\le C R^{\frac{n(s-q)}{s}} M^{q(1+\sigma)\delta}
\Big(\int\hspace{-0.38cm}- _{B_{8R}}(c+\epsilon+ | \nabla u|^2)^{\frac{p(x) }{2 } }+ 1  dx
   \Big)^{q(1+\sigma)},
\]
where
$$
     M=\int_{B_{8R}} | \nabla u|  ^{ p(x)   } dx+1.
$$
Thus, 
 $$
 K(R)\le  C R^{\alpha}M^{ (1+\sigma)\delta}
\Big(\int\hspace{-0.38cm}- _{B_{8R}}(c+\epsilon+ | \nabla u|^2)^{\frac{p(x) }{2 } }+ 1  dx
   \Big)^{  1+\sigma },
 $$
where the exponent $\alpha=1-\frac{n}{s}$. Finally, we conclude that
\begin{equation}
\begin{split}
&\sup_{B_{R/2}}(c+\epsilon+ | \nabla u|^2)^{\frac{p(x)}{2}} \\
&\leq C\Big[ \int\hspace{-0.38cm}- _{B_{2R}}(c+\epsilon+ | \nabla u|^2)^{\frac{p(x)}{2}} dx
\\
&\quad +M^{(1+\sigma)\delta} R^{\alpha}
\Big(\int\hspace{-0.38cm}- _{B_{8R}}(c+\epsilon+ | \nabla u|^2)^{\frac{p(x) }{2 } }+ 1  dx
 \Big)^{ 1+\sigma }\Big],
\end{split}\label{finalresult}
\end{equation}
which completes our proof.
\end{proof}

\subsection*{Acknowledgements}
The author wishes to thank the anonymous reviewers for their valuable 
suggestions that improved this article.
This work is supported in part by the NSFC (11001165) 
and Shanghai Leading Academic Discipline Project (J50101).

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\end{document}