\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2012 (2012), No. 25, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2012 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2012/25\hfil $p(x)$-Laplacian problems with critical exponents] {Existence of solutions for discontinuous $p(x)$-Laplacian problems with critical exponents} \author[X. Shang, Z. Wang \hfil EJDE-2012/25\hfilneg] {Xudong Shang, Zhigang Wang} % in alphabetical order \address{ School of Mathematics, Nanjing Normal University Taizhou College, 225300, Jiangsu, China} \email[Xudong Shang ]{xudong-shang@163.com} \email[Zhigang Wang]{wzg19.scut@163.com} \thanks{Submitted November 7, 2011. Published February 7, 2012.} \subjclass[2000]{35J92, 35J70, 35R70} \keywords{$p(x)$-Laplacian problem; critical Sobolev exponents; \hfill\break\indent discontinuous nonlinearities} \begin{abstract} In this article, we study the existence of solutions to the problem \begin{gather*} -\operatorname{div}(|\nabla u|^{p(x)-2}\nabla u) =\lambda |u|^{p^*(x)-2}u + f(u)\quad x \in \Omega ,\\ u = 0 \quad x \in \partial\Omega, \end{gather*} where $\Omega$ is a smooth bounded domain in ${\mathbb{R}}^{N}$, $p(x)$ is a continuous function with $1
0$.
The function $f(u)$ can have discontinuities, so that functionals
associated with \eqref{e1.1} may not be differentiable, and
standard variational techniques can not be applied.
There are many publications for the case when $p(x)$ is a constant function;
see for example \cite{AB,ABG,Ba,BT,S}.
For the existence of solutions for $p(x)$-Laplacian problems
we refer the reader to \cite{Bo,CF,D,FH,FZ,Mi}.
The existence of solutions for $p(x)$-Laplacian problems with critical
growth is relatively new. In 2012, Bonder and Silva \cite{BS} extended the
concentration-compactness principle of Lions to the variable exponent spaces and
proved the existence of solutions to the problem
\begin{gather*}
-\Delta_{p(x)}u = |u|^{q(x)-2}u + \lambda(x)|u|^{r(x)-2}u\quad x \in \Omega ,\\
u = 0 \quad x \in \partial\Omega.
\end{gather*}
Where $\Omega$ is a smooth bounded domain in ${\mathbb{R}}^{N}$, with
$q(x)\leq p^*(x)$ and the set $\{q(x)= p^*(x)\}\neq\emptyset$, we
can find a similar result in \cite{F}.
Fu \cite{FZh} studied the existence of solutions for
$p(x)$-Laplacian equationd involving the critical exponent and obtained a
sequence of radially symmetric solutions.
In the present paper, we study the discontinuous $p(x)$-Laplacian problems
with critical growth for \eqref{e1.1}. To handle the gaps at the
discontinuity points, our approach uses nonsmooth critical point theory
for locally Lipschitz functionals, we obtain some general results
for the simple case when $f$ has only one point of discontinuity.
Because $f$ is discontinuous, we say that a function $u \in W_0^{1,p(x)}(\Omega)$
is a solution of the multivalued problem associated to \eqref{e1.1} if $u$ satisfies
\[
-\Delta_{p(x)}u - \lambda |u|^{p^*(x)-2}u \in \widehat{f}(u) \quad \text{a.e. in } \Omega ,
\]
where $\widehat{f}(u)$ is the multivalued function
$\widehat{f}(u) = [\underline{f}(u),\overline{f}(u)]$ with
\[
\underline{f}(t) =\liminf_{s\to t}f(s),\quad \overline{f}(t) = \limsup_{s\to t}f(s).
\]
In this article, we assume $f: {{\mathbb{R}}} \to {\mathbb{R}}$ is a
measurable function satisfying:
\begin{itemize}
\item[(F1)] $f(t) = 0$ if $t\leq0$ and for all $t \in {\mathbb{R}}$,
there exist the limits:
\[
f(t+0)=\lim_{\delta\to 0^{+}}f(t+\delta);\quad
f(t-0)=\lim_{\delta\to 0^{+}}f(t-\delta).
\]
\item[(F2)] there exist $C_1,C_2> 0$ such that
$|f(t)|\leq C_1 + C_2|t|^{q(x)-1}$, where $q(x)\in C(\overline{\Omega})$
such that $p(x) < q(x) < p^*(x)$.
\item[(F3)] $f(t) = o({|t|^{p(x)-1}})$ as $t\to 0$.
\item[(F4)] $\underline{f}(t)t \geq q_{-}F(t)> 0$, for all
$t\in {\mathbb{R}}\backslash \{0\}$,
where $F(t)=\int_0^tf(s)ds$.
\end{itemize}
Note that by hypothesis (F1),
\[
\overline{f}(u) = \max\{f(u-0), f(u+0) \},\quad
\underline{f}(u) = \min\{ f(u-0), f(u+0)\}.
\]
\begin{theorem} \label{thm1.1}
Suppose $f$ satisfies {\rm (F1)-(F4)}. Then there exists
$\lambda_0 > 0$ such that \eqref{e1.1} has a nontrivial solution
for every $\lambda\in(0,\lambda_0)$.
\end{theorem}
One of the main motivations is to consider the particular case associated
with \eqref{e1.1},
\begin{equation} \label{e1.2}
\begin{gathered}
-\Delta_{p(x)}u= \lambda |u|^{p^*(x)-2}u + bh(u-a)|u|^{q(x)-2}u\quad x \in \Omega ,
\\ u = 0 \quad x \in \partial\Omega,
\end{gathered}
\end{equation}
where $h(t)=0$ if $t \leq 0$ and $h(t)=1$ if $t > 0$, $a$ and $b$ are
positive real parameters, $p(x) 0$, there exists
$\lambda_0 > 0$ such that for every $\lambda\in(0,\lambda_0)$, Equation
\eqref{e1.2} has a nontrivial solution satisfying
$\operatorname{meas}\{x\in\Omega: u(x)>a\}>0$.
\end{theorem}
The rest of this article is organized as follows:
In section 2 we introduce some necessary preliminary knowledge;
in section 3 contains the proof of our main results.
\section{Preliminaries}
First, we recall some definitions and properties of generalized gradient of
locally Lipschitz functionals, which will be used later.
Let $X$ be a Banach space, $X^*$ be its topological dual and
$\langle\cdot,\cdot\rangle$ be the duality.
A functional $I: X\to {\mathbb{R}}$ is said to be locally Lipschitz
if for every $u\in X$ there exists a neighborhood $ U$ of $u$ and a constant
$K>0$ depending on $ U$ such that
\[
|I(u)-I(v)|\leq K\|u-v\|,\quad \forall u,v\in U.
\]
For a locally Lipschitz functional $I$, we define the generalized directional
derivative at $u\in X$ in the direction $v\in X$ by
\[
I^0(u;v)=\limsup_{h\to 0,\, \delta\downarrow 0}
\frac{I(u+h+\delta v)-I(u+h)}{\delta}.
\]
It is easy to show that $I^0(u;v)$ is subadditive and positively homogeneus.
The generalized gradient of $I$ at $u$ is the set
\[
\partial I(u)=\{w\in X^*:I^0(u;v) \geq \langle w,v\rangle, \forall v\in X \}.
\]
Then, for each $v \in X$,
$I^0(u;v) = \max \{ \langle \omega, v \rangle: \omega \in \partial I(u) \}$.
A point $u \in X$ is a critical point of $I$ if $0 \in \partial I(u)$.
It is easy to see that if $u \in X$ is a local minimum or maximum, then
$0 \in \partial I(u)$.
Next, we recall some definitions and basic properties of the
generalized Lebesgue-Sobolev spaces $L^{p(x)}(\Omega)$ and $W_0^{1,p(x)}(\Omega)$,
where $\Omega \subset {\mathbb{R}}^{N}$ is an arbitrary domain with smooth boundary.
Set
\begin{gather*}
C_{+}(\overline{\Omega})= \{p(x)\in C(\overline{\Omega}):
p(x)> 1, \forall x\in\overline{\Omega}\},\\
p_{+} = \max_{x\in\Omega}p(x),\quad
p_{-} = \min_{x\in\Omega}p(x).
\end{gather*}
For any $p(x)\in C_{+}(\overline{\Omega})$, we define the variable exponent
Lebesgue space
\[
L^{p(x)}(\Omega) = \{u\in M(\Omega):\int_{\Omega}|u(x)|^{p(x)}dx< \infty\},
\]
with the norm
\[
|u|_{p(x)}=\inf\{\mu > 0 : \int_{\Omega}|\frac{u}{\mu}|^{p(x)}dx\leq 1\},
\]
where $M(\Omega)$ is the set of all measurable real functions defined on $\Omega$.
Define the space
\[
W_0^{1,p(x)}(\Omega)=\{u\in L^{p(x)}(\Omega): |\nabla u|\in L^{p(x)}(\Omega)\},
\]
with the norm
\[
\|u\|= |u|_{p(x)}+|\nabla u|_{p(x)}.
\]
\begin{proposition}[\cite{FZH,KR}] \label{prop2.1}
There is a constant $C > 0$ such that for all $u\in W_0^{1,p(x)}(\Omega)$,
\[
|u|_{p(x)}\leq C|\nabla u|_{p(x)}.
\]
So $|\nabla u|_{p(x)}$ and $\|u\|$ are equivalent norms in $W_0^{1,p(x)}(\Omega)$.
Hence we will use the norm
$\|u\| = |\nabla u|_{p(x)}$ for all $u\in W_0^{1,p(x)}(\Omega)$.
\end{proposition}
\begin{proposition}[\cite{FZH,KR}] \label{prop2.2}
Set $ \rho(u)=\int_{\Omega}|u|^{p(x)}dx$. For $u,u_n\in L^{p(x)}(\Omega)$, we have:
\begin{itemize}
\item[(1)] $|u|_{p(x)}<1$ $(=1;>1)$ $\Leftrightarrow \rho(u)<1$ $(=1;>1)$.
\item[(2)] If $|u|_{p(x)} > 1$, then $|u|^{p_{-}}_{p(x)}\leq \rho(u)
\leq |u|^{p_{+}}_{p(x)}$.
\item[(3)] If $|u|_{p(x)}< 1$, then $|u|^{p_{+}}_{p(x)}\leq \rho(u)
\leq |u|^{p_{-}}_{p(x)}$.
\item[(4)] $\lim_{n\to\infty} u_n=u \Leftrightarrow
\lim_{n\to\infty} \rho (u_n-u) = 0$.
\item[(5)] $\lim_{n\to\infty} |u_n|_{p(x)}
=\infty \Leftrightarrow \lim_{n\to\infty} \rho (u_n) = \infty$.
\end{itemize}
\end{proposition}
\begin{proposition}[\cite{FZH}] \label{prop2.3}
If $q(x)\in C_{+}(\Omega)$ and $q(x) < p^*(x)$ for any $x\in\Omega$, the imbedding
$W^{1,p(x)}(\Omega)\to L^{q(x)}(\Omega) $ is compact.
\end{proposition}
\begin{proposition}[\cite{KR}] \label{prop2.4}
The conjugate space of $L^{p(x)}(\Omega) $ is $L^{q(x)}(\Omega) $, where
$\frac{1}{p(x)}+\frac{1}{q(x)} =1$.
For any $u \in L^{p(x)}(\Omega)$ and $v \in L^{q(x)}(\Omega)$,
\[
\int_{\Omega}|uv|dx\leq (\frac{1}{p_{-}}+ \frac{1}{q_{-}})|u|_{p(x)}|v|_{q(x)}.
\]
\end{proposition}
\begin{proposition}[\cite{FH}] \label{prop2.5}
If $|u|^{q(x)}\in L^{\frac{s(x)}{q(x)}}(\Omega)$, where
$q(x),s(x) \in L^{\infty}_{+}(\Omega)$, $q(x)\leq s(x)$, then $u\in L^{s(x)}(\Omega)$
and there is a number $\overline{q}\in [q_{-},q_{+}]$ such
that $||u|^{q(x)}|_{\frac{s(x)}{q(x)}} = (|u|_{s(x)})^{\overline{q}}$.
\end{proposition}
Let $I_{\lambda}(u): W_0^{1,p(x)}(\Omega) \to {\mathbb{R}}$ be the energy
functional defined as
\begin{equation}
I_{\lambda}(u) = \int_{\Omega}\frac{1}{p(x)}|\nabla u|^{p(x)}dx
- \lambda\int_{\Omega}\frac{1}{p^*(x)}|u|^{p^*(x)}dx - \int_{\Omega}F(u)dx,
\end{equation}
denote $\Phi(u)= \int_{\Omega}F(u)dx$.
We say that $I_{\lambda}(u)$ satisfies the nonsmooth $(PS)_{c}$ condition,
if any sequence $\{u_n\}\subseteq X$ such that $I_{\lambda}(u_n)\to c$
and $m(u_n) = \min \{ \|w\|_{X^*}: w\in\partial I_{\lambda}(u_n)\} \to 0$,
as $n\to\infty$, possesses a convergent subsequence.
To prove our main results, we use the generalizations of the mountain pass
theorem \cite{Ch}.
\begin{theorem} \label{thm2.1}
Let $X$ be a reflexive Banach space, $I: X\to {\mathbb{R}}$ is locally Lipschitz
functional which satisfies the nonsmooth $(PS)_{c}$ condition,
$I(0)=0$ and there are $\rho, r > 0$ and $e\in X$ with
$\|e\| > r$, such that
\[
I(u)\geq \beta \quad \text{if } \|u\| = r \quad \text{and} \quad I(e)\leq 0.
\]
Then there exists $u \in X$ such that
$0 \in \partial I(u)$ and $I(u)=c$. Where
\begin{gather*}
c = \inf_{\gamma\in \Gamma}\max_{0\leq t\leq 1}I(\gamma(t)),\\
\Gamma = \{\gamma \in C([0,1], X)|\gamma(0)=0, \gamma(1) = e \}.
\end{gather*}
\end{theorem}
Recall the concentration-compactness principle for variable exponent spaces.
This will be the keystone that enable us to verify that $I_{\lambda}$ satisfies the
nonsmooth $(PS)_{c}$ condition.
\begin{proposition}[\cite{BS}] \label{prop2.6}
Let $\{u_n\} $ converge weakly to $u$ in $W_0^{1,p(x)}(\Omega)$ such that
$|u_n|^{p^*(x)}$ and $|\nabla u_n|^{p(x)}$ converge weakly to nonnegative
measures $\nu$ and $\mu$ on ${\mathbb{R}}^{N}$ respectively. Then, for some
countable set $J$, we have:
\begin{itemize}
\item[(i)] $\nu = |u|^{p^*(x)} + \sum_{j\in J}\nu_{j}\delta_{x_{j}}$,
\item[(ii)] $\mu \geq |\nabla u|^{p(x)} + \sum_{j\in J}\mu_{j}\delta_{x_{j}}$,
\item[(iii)] $S\nu_{j}^{\frac{1}{p^*(x_{j})}} \leq \mu_{j}^{\frac{1}{p(x_{j})}}$,
\end{itemize}
where $x_{j}\in \Omega$, $\delta_{x_{j}}$ is the Dirac measure at $x_{j}$,
$\nu_{j}$ and $\mu_{j}$ are constants and $S$ is the best constant in the
Gagliardo-Nirenberg-Sobolev inequality for variable exponents, namely
$$
S = \inf \big\{\frac{\|u\|_{1,p(x)}}{|u|_{p^*(x)}}: u\in W_0^{1,p(x)}(\Omega),
u\not=0\big\}.
$$
\end{proposition}
\section{Proof of main results}
In this section, we denote by $u_n \rightharpoonup u$ the weak convergence of
a sequence $u_n$ to $u$ in $W_0^{1,p(x)}(\Omega)$, and $o(1)$ denote a real
vanishing sequence, $C$ and $C_{i}, i= 1, 2, \dots$ are positive constants,
$|A|$ is the Lebesgue measure of $A$ and $p'(x)$ as the conjugate function
of $p(x)$. $u \in W_0^{1,p(x)}(\Omega)$ is called a solution of \eqref{e1.1} if $u$
is a critical point of $I_{\lambda}(u)$ and satisfies
\[
-\operatorname{div}(|\nabla u|^{p(x)-2}\nabla u)
- \lambda |u|^{p^*(x)-2}u \in [\underline{f}(u),\overline{f}(u)]\quad
\text{a.e. } {x\in\Omega }.
\]
\begin{lemma} \label{lem3.1}
The function $\Phi(u)$ is locally Lipschitz on $W_0^{1,p(x)}(\Omega)$.
\end{lemma}
\begin{proof} By (F2), Proposition \ref{prop2.4} and \ref{prop2.5}, for all
$u,v \in W_0^{1,p(x)}(\Omega)$,
\begin{align*}
|\Phi(u)- \Phi(v)|& \leq \int_{\Omega}\big|\int^{v}_u|f(t)|dt\big|dx \\
&\leq \int_{\Omega}\big|\int^{v}_u|C_1 + C_2|t|^{q(x)-1}|dt\big|dx \\
&\leq (|C_1|_{\frac{q(x)}{q(x)-1}} + C_3|u|^{\overline{q}-1}_{q(x)}
+ C_3|v|^{\overline{q}-1}_{q(x)})|u-v|_{q(x)}.
\end{align*}
From Proposition \ref{prop2.3}, we obtain that there is a neighborhood
$U \subset W_0^{1,p(x)}(\Omega)$ of $u$ such that
\[
|\Phi(u)- \Phi(v)| \leq K\|u-v\|,
\]
where $K>0$ depends on $\max \{ \|u\|, \|v\|\}$.
So, $\Phi(u)$ is locally Lipschitz in $W_0^{1,p(x)}(\Omega)$.
The proof is complete.
\end{proof}
From Lemma \ref{lem3.1}, by Chang's results we have that $I_{\lambda}(u)$ is locally Lipschitz
and $ \omega \in \partial I_{\lambda}(u)$ if and only if there is
$\overline{\omega}\in W^{-1,p^{'}(x)}(\Omega)$ such that for all
$\varphi\in W_0^{1,p(x)}(\Omega)$,
\begin{equation} \label{e3.1}
\langle \omega, \varphi\rangle
= \int_{\Omega}|\nabla u|^{p(x)-2}\nabla u\nabla \varphi dx
-\lambda\int_{\Omega}|u|^{p^*(x)-2}u\varphi dx
- \int_{\Omega}\overline{\omega}\varphi dx,
\end{equation}
and
\begin{equation}
\overline{\omega}(x)\in [\underline{f}(u(x)),\overline{f}(u(x))]\quad
\text{a.e. } {x\in\Omega }.
\end{equation}
\begin{lemma} \label{lem3.2}
Assume {\rm (F1), (F2)}. Let $\{u_n\}$ be a bounded sequence
in $W_0^{1,p(x)}(\Omega)$ such that
$I_{\lambda}(u_n) \to c$ and $m(u_n)\to 0$. Then there exists a
subsequence (denoted again by $u_n$) and some $u \in W_0^{1,p(x)}(\Omega)$,
such that
\[
|\nabla u_n|^{p(x)-2}\nabla u_n \rightharpoonup |\nabla u|^{p(x)-2}\nabla u
\quad \text{weakly in } [L^{\frac{p(x)}{p(x)-1}}(\Omega)]^{N}.
\]
\end{lemma}
\begin{proof}
The proof is similar to that of \cite[Theorem 1]{Zh}.
Because $\{u_n\}$ is bounded in $W_0^{1,p(x)}(\Omega)$, there exist
a subsequence and $u \in W_0^{1,p(x)}(\Omega)$ such that
$u_n \rightharpoonup u$ in $W_0^{1,p(x)}(\Omega)$ and
$u_n \to u$ in $L^{p(x)}(\Omega)$ as $n\to \infty$.
We claim that the set $J$ given by Proposition \ref{prop2.6} is finite.
Choose a function $\varphi(x)\in C_0^{\infty}({\mathbb{R}}^{N})$
such that $0 \leq \varphi(x)\leq 1$, $\varphi(x) \equiv 1$ on $B(0,1)$
and $\varphi(x) \equiv 0$ on ${\mathbb{R}}^{N}\setminus B(0,2) $.
Let $\varphi_{j,\varepsilon}(x)= \varphi (\frac{x-x_{j}}{\varepsilon})$,
for any $x\in {\mathbb{R}}^{N}$, $\epsilon > 0$ and $j\in J$.
It is clear that $\{\varphi_{j,\varepsilon}u_n\} \subset W_0^{1,p(x)}(\Omega)$
for any $j\in J$, and is bounded in $ W_0^{1,p(x)}(\Omega)$.
Take $\varphi = \varphi_{j,\varepsilon}u_n$ in $\langle \omega_n, \varphi\rangle$,
we obtain
\begin{equation} \label{e3.3}
\begin{split}
&\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\cdot u_n\nabla \varphi_{j,\varepsilon}dx
+ \int_{\Omega}|\nabla u_n|^{p(x)}\varphi_{j,\varepsilon}dx \\
& -\lambda\int_{\Omega}| u_n|^{p^*(x)}\varphi_{j,\varepsilon}dx
-\int_{\Omega}\overline{\omega}_n\varphi_{j,\varepsilon}u_ndx = o(1).
\end{split}
\end{equation}
Taking into account that $\omega_n \in \partial I_{\lambda}(u_n)$, by (F2)
and $u_n \rightharpoonup u$ in $W_0^{1,p(x)}(\Omega)$, we infer that
$\overline{\omega}_n$ is bounded in $W^{-1,p^{'}(x)}(\Omega)$, and
so there exists $\overline{\omega}_0 \in W^{-1,p^{'}(x)}(\Omega)$ such that
\begin{equation} \label{e3.4}
\overline{\omega}_n\rightharpoonup \overline{\omega}_0 \quad \text{in }
W^{-1,p^{'}(x)}(\Omega)\quad \text{and}\quad
\overline{\omega}_0 \in [\underline{f}(u),\overline{f}(u)].
\end{equation}
Let $n\to \infty$ in \eqref{e3.3}, by Proposition \ref{prop2.6}, we have
\begin{equation} \label{e3.5}
\begin{split}
&\lim_{n\to \infty} \int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n
\cdot u_n\nabla \varphi_{j,\varepsilon}dx \\
&=\lambda\int_{\Omega}\varphi_{j,\varepsilon}d\nu
- \int_{\Omega}\varphi_{j,\varepsilon}d\mu
+ \int_{\Omega}\overline{\omega}_0\varphi_{j,\varepsilon}u\,dx.
\end{split}
\end{equation}
By H\"older inequality it is easy to check that
\begin{align*}
0&\leq |\lim_{n\to\infty}\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\nabla
\varphi_{j,\varepsilon}\cdot u_n\,dx| \\
&\leq \Big(\int_{\Omega}|\nabla u_n|^{p_{+}}dx\Big)^{\frac{p_{+}-1}{p_{+}}}
\Big(\int_{\Omega}| u_n|^{p_{+}}|\nabla\varphi_{j,\varepsilon}|^{p_{+}}dx
\Big)^{\frac{1}{p_{+}}}\\
&\quad + \Big(\int_{\Omega}|\nabla u_n|^{p_{-}}dx\Big)^{\frac{p_{-}-1}{p_{-}}}
\Big(\int_{\Omega}| u_n|^{p_{-}}|\nabla\varphi_{j,\varepsilon}|^{p_{-}}dx
\Big)^{\frac{1}{p_{-}}}\\
&\leq C_4(\int_{\Omega}|\nabla u_n|^{p_{+}}dx)^{\frac{p_{+}-1}{p_{+}}}
\Big(\int_{B(x_{j},2\varepsilon)}| u_n|^{(p_{+})^*}dx\Big)^{\frac{1}{(p_{+})^*}}\\
&\quad + C_5\Big(\int_{\Omega}|\nabla u_n|^{p_{-}}dx\Big)^{\frac{p_{-}-1}{p_{-}}}
\Big(\int_{B(x_{j},2\varepsilon)}| u_n|^{(p_{-})^*}dx\Big)^{\frac{1}{(p_{-})^*}}
&\to 0, \quad \text{as }\epsilon \to 0.
\end{align*}
From \eqref{e3.5}, as $\epsilon \to 0$, we obtain
$\lambda \nu_{j} = \mu_{j}$.
From Proposition \ref{prop2.6}, we conclude that
\begin{equation} \label{e3.7}
\nu_{j} = 0 \quad \text{or}\quad \nu_{j} \geq S^{N}
\max\{\lambda^{-\frac{N}{p_{+}}},\lambda^{-\frac{N}{p_{-}}}\}.
\end{equation}
It implies that $J$ is a finite set.
Without loss of generality, let $J = \{1,2,\dots,m\}$. For any $\delta > 0$,
we denote $\Omega_{\delta}= \{x\in \Omega|\text{dist}(x,x_{j})> \delta\}$.
Choose $R$ large enough such that
$\overline{\Omega} \subset \{x\in {\mathbb{R}}^{N}| |x|< R \}$,
$\psi (x) \in C^{\infty}({\mathbb{R}}^{N})$, $0 \leq \psi (x) \leq 1$,
$\psi (x)\equiv 0$ on $B(0,2R)$ and $\psi (x)\equiv 1$ on
${\mathbb{R}}^{N}\setminus B(0,3R) $. Take $\epsilon > 0$ small enough such that
$B(x_{i},\epsilon) \cap B(x_{j},\epsilon) =\emptyset$, $\forall i,j \in J$,
$i \neq j$ and $\cup_{j=1}^{m}B(x_{j},\epsilon) \subset B(0,2R)$. We take
\[
\psi_{\epsilon}(x) = 1 - \sum_{j=1}^{m}\varphi_{j,\varepsilon} - \psi (x), \quad
x\in {\mathbb{R}}^{N}.
\]
Then $\psi_{\epsilon}(x) \in C^{\infty}({\mathbb{R}}^{N})$,
$\operatorname{supp}\psi_{\epsilon} \subset B(0,3R)$ with
$\psi_{\epsilon}(x) = 0 $ on $ \cup_{j=1}^{m}B(x_{j},\epsilon/2)$
and $\psi_{\epsilon}(x) = 1 $ on
$ ({\mathbb{R}}^{N}\setminus B(x_{j},\epsilon))\cap B(0,2R)$.
As $\{\psi_{\epsilon}u_n\}$ is bounded in $W_0^{1,p(x)}(\Omega)$,
let $\varphi = \psi_{\epsilon}u_n$ in $\langle \omega_n, \varphi\rangle$, we obtain
\begin{align*} %3.8
&\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\cdot u_n\nabla \psi_{\epsilon}dx
+ \int_{\Omega}|\nabla u_n|^{p(x)}\psi_{\epsilon}dx \\
&-\lambda\int_{\Omega}| u_n|^{p^*(x)}\psi_{\epsilon}dx
-\int_{\Omega}\overline{\omega}_n\psi_{\epsilon} u_ndx = o (1).
\end{align*}
By \eqref{e3.4} and $u_n \rightharpoonup u$ in $W_0^{1,p(x)}(\Omega)$, we can easily
obtain
\[
\lim_{n\to \infty}\int_{\Omega}\overline{\omega}_n\psi_{\epsilon} u_ndx
= \int_{\Omega}\overline{\omega}_0\psi_{\epsilon} u dx.
\]
Since $\psi_{\epsilon}(x) = 0 $ on $ \cup_{j=1}^{m}B(x_{j},\frac{\epsilon}{2})$ and
$\nu = |u|^{p^*(x)} + \sum_{j\in J}\nu_{j}\delta_{x_{j}}$, we obtain
\[
\lim_{n\to \infty}\int_{\Omega}| u_n|^{p^*(x)}\psi_{\epsilon}dx
= \int_{\Omega}\psi_{\epsilon}d\nu = \int_{\Omega}| u|^{p^*(x)}\psi_{\epsilon}dx.
\]
Hence
\begin{equation} \label{e3.9}
\begin{split}
\lim_{n\to \infty}\int_{\Omega}|\nabla u_n|^{p(x)}\psi_{\epsilon}dx
& = \lim_{n\to \infty}(-\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n
\cdot u_n\nabla \psi_{\epsilon}dx) \\
&\quad +\lambda\int_{\Omega}| u|^{p^*(x)}\psi_{\epsilon}dx
+\int_{\Omega}\overline{\omega}_0\psi_{\epsilon} udx.
\end{split}
\end{equation}
In the same way, taking $\varphi = \psi_{\epsilon}u$ in $\langle \omega_n, \varphi\rangle$,
we obtain
\begin{align*}
&\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\cdot u\nabla \psi_{\epsilon}dx
+ \int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\nabla u\cdot\psi_{\epsilon}dx\\
& -\lambda\int_{\Omega}| u_n|^{p^*(x)-2}u_nu\psi_{\epsilon}dx
-\int_{\Omega}\overline{\omega}_n\psi_{\epsilon} udx = o(1).
\end{align*}
Thus
\begin{equation} \label{e3.10}
\begin{split}
&\lim_{n\to \infty}\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\nabla u
\cdot\psi_{\epsilon}dx \\
&= \lim_{n\to \infty}(-\int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\cdot
u\nabla \psi_{\epsilon}dx)
+\lambda\int_{\Omega}| u|^{p^*(x)}\psi_{\epsilon}dx
+\int_{\Omega}\overline{\omega}_0\psi_{\epsilon} udx.
\end{split}
\end{equation}
So, from \eqref{e3.9} and \eqref{e3.10}, as $n\to \infty$, we have
\begin{align*}
0&\leq \int_{\Omega_{\delta}}(|\nabla u_n|^{p(x)-2}\nabla u_n
- |\nabla u|^{p(x)-2}\nabla u)(\nabla u_n- \nabla u)dx \\
& \leq \int_{\Omega}\psi_{\epsilon}(|\nabla u_n|^{p(x)-2}\nabla u_n
- |\nabla u|^{p(x)-2}\nabla u)(\nabla u_n- \nabla u)dx\\
&= \int_{\Omega}|\nabla u_n|^{p(x)-2}\nabla u_n\nabla \psi_{\epsilon}\cdot(u-u_n)dx\\
&\quad + \int_{\Omega}\psi_{\epsilon}|\nabla u|^{p(x)-2}\nabla u\cdot(\nabla u- \nabla u_n)dx
+ o(1) \\
&\leq \|\nabla\psi_{\epsilon}\|_{\infty}\cdot||\nabla u_n|^{p(x)-1}|_{p^{'}(x)}
\cdot|u-u_n|_{p(x)} \\
&\quad + \int_{\Omega}\psi_{\epsilon}|\nabla u|^{p(x)-2}\nabla u\cdot(\nabla u_n- \nabla u)dx
+ o(1).
\end{align*}
Thus we have
\begin{equation} \label{e3.11}
\lim_{n\to \infty}\int_{\Omega_{\delta}}(|\nabla u_n|^{p(x)-2}\nabla u_n
- |\nabla u|^{p(x)-2}\nabla u)(\nabla u_n- \nabla u)dx = 0.
\end{equation}
Denote
\[
g_n(x)=(|\nabla u_n|^{p(x)-2}\nabla u_n- |\nabla u|^{p(x)-2}\nabla u)(\nabla u_n- \nabla u),
\]
then $g_n(x) \geq 0$, and by \eqref{e3.11}, $g_n(x) \to 0$ a.e. on $\Omega_{\delta}$.
Let $E$ be a compact subset of $\Omega_{\delta}$, suppose
$g_n(x) \to 0$ a.e. on $E$. If $\nabla u_n$ were not convergence to $\nabla u$
everywhere on $E$, there would at least exist $x_0\in E$ such that
\[
\lim_{n\to \infty} \nabla u_n(x_0) \neq \nabla u(x_0).
\]
Then
\begin{align*}
|\nabla u_n(x_0)|^{p(x_0)}
&= |\nabla u_n(x_0)|^{p(x_0)-2}\nabla u_n(x_0)\nabla u(x_0) \\
& + |\nabla u(x_0)|^{p(x_0)-2}\nabla u_n(x_0)\nabla u(x_0)
- |\nabla u(x_0)|^{p(x_0)} + g_n(x_0).
\end{align*}
By the interpolation inequality,
\begin{align*}
||\nabla u_n(x_0)|^{p(x_0)-2}\nabla u_n(x_0)\nabla u(x_0)|
& \leq |\nabla u_n(x_0)|^{p(x_0)-1}\cdot|\nabla u(x_0)| \\
& \leq \epsilon_1|\nabla u_n(x_0)|^{p(x_0)}
+ C_{\epsilon_1}|\nabla u(x_0)|^{p(x_0)},
\end{align*}
and
\begin{align*}
||\nabla u(x_0)|^{p(x_0)-2}\nabla u_n(x_0)\nabla u(x_0)|
& \leq |\nabla u(x_0)|^{p(x_0)-1}\cdot|\nabla u_n(x_0)| \\
& \leq \epsilon_2|\nabla u(x_0)|^{p(x_0)} + C_{\epsilon_2}|\nabla u_n(x_0)|^{p(x_0)}.
\end{align*}
We choose $\epsilon_1,\epsilon_2$ properly, because $g_n(x_0)$ is bounded,
then $|\nabla u(x_0)|\leq C$.
Let $\nabla u(x_0) = \eta$, so we can assume
$\nabla u_n(x_0)\to \overline{\eta} \neq \eta$. Thus
\[
g_n(x_0) \to (|\overline{\eta}|^{p(x_0)-2}\overline{\eta}
- |\eta|^{p(x_0)-2}\eta)(\overline{\eta} - \eta) > 0.
\]
This contradicts $g_n(x_0) \to 0$. Hence, $\nabla u_n(x_0)\to \nabla u(x_0)$
everywhere on $E$. So $\nabla u_n(x_0)\to \nabla u(x_0)$ a.e. on $\Omega_{\delta}$.
Since $\delta$ is arbitrary, we obtain $\nabla u_n(x_0)\to \nabla u(x_0)$
a.e. on $\Omega$.
Since $\{|\nabla u_n|^{p(x)-2}\nabla u_n\}$ is integrable in $L^{1}(\Omega)$,
we obtain that as $n\to \infty$,
\[
|\nabla u_n|^{p(x)-2}\nabla u_n \rightharpoonup |\nabla u|^{p(x)-2}\nabla u
\quad \text{weakly in } [L^{\frac{p(x)}{p(x)-1}}(\Omega)]^{N}.
\]
The proof is complete
\end{proof}
\begin{lemma} \label{lem3.3}
Suppose $f$ satisfies {\rm (F2), (F4)}.
Then $I_{\lambda}$ satisfies the nonsmooth $(PS)_{c}$ condition provided
$c< (\frac{1}{p_{+}}-\frac{1}{q_{-}})S^{N}\max\{\lambda^{1-\frac{N}{p_{+}}},
\lambda^{1-\frac{N}{p_{-}}}\}$.
\end{lemma}
\begin{proof}
Let $\{u_n\}\subset W_0^{1,p(x)}(\Omega)$ be such that $I_{\lambda}(u_n) \to c$
and $m(u_n)\to 0$ as $n\to \infty$. We must show the existence of a subsequence of
$\{u_n\}$ which converges strongly in $W_0^{1,p(x)}(\Omega)$. First, we show
that $\{u_n\}$ is bounded.
We know that
\begin{equation} \label{e3.12}
\begin{split}
I_{\lambda}(u_n)
&= \int_{\Omega}\frac{1}{p(x)}|\nabla u_n|^{p(x)}dx
- \lambda\int_{\Omega}\frac{1}{p^*(x)}|u_n|^{p^*(x)}dx - \int_{\Omega}F(u_n)dx \\
& \geq \frac{1}{p_{+}}\int_{\Omega}|\nabla u_n|^{p(x)}dx
- \frac{\lambda}{p^*_{-}}\int_{\Omega}|u_n|^{p^*(x)}dx - \int_{\Omega}F(u_n)dx.
\end{split}
\end{equation}
Let $\omega_n\in \partial I_{\lambda}(u_n)$ such that
$\|\omega_n \| = m(u_n)= o (1)$. From \eqref{e3.1} we have
\begin{equation} \label{e3.13}
\langle \omega_n, u_n\rangle = \int_{\Omega}|\nabla u_n|^{p(x)}dx
-\lambda\int_{\Omega}|u_n|^{p^*(x)}dx - \int_{\Omega}\overline{\omega}_nu_ndx,
\end{equation}
where
$\overline{\omega}_n(x)\in [\underline{f}(u_n),\overline{f}(u_n)]$ for a.e.
$x\in\Omega$.
By (F4) we obtain
\begin{equation} \label{e3.14}
\frac{1}{q_{-}}\overline{\omega}_nu_n
\geq \frac{1}{q_{-}}\underline{f}(u_n)u_n \geq F(u_n).
\end{equation}
From \eqref{e3.12}, \eqref{e3.13} and \eqref{e3.14}, we obtain
\begin{equation} \label{e3.15}
C_{6}(1 + \|u_n\|) \geq I_{\lambda}(u_n)
- \frac{1}{q_{-}}\langle \omega_n, u_n\rangle \geq (\frac{1}{p_{+}}
-\frac{1}{q_{-}})\int_{\Omega}|\nabla u_n|^{p(x)}dx.
\end{equation}
If $\|u_n\| > 1$, by Proposition \ref{prop2.2}, we obtain
\[
(\frac{1}{p_{+}}-\frac{1}{q_{-}})\|u_n\|^{p_{-}} \leq C_{6}(1 + \|u_n\|).
\]
Thus $\{ u_n\}$ is bounded in $W_0^{1,p(x)}(\Omega)$. Then there exist a
subsequence and $u \in W_0^{1,p(x)}(\Omega)$ such that $u_n \rightharpoonup u$
in $W_0^{1,p(x)}(\Omega)$, so we know that $\{ |u_n|^{p^*(x)-2}u_n\varphi\} $
is uniformly integrable in $L^{1}(\Omega)$. By this fact,
Lemma \ref{lem3.2} and $m(u_n)\to 0$, taking $n\to \infty$ in $\langle \omega_n, \varphi\rangle$,
we have
\[
0 = \int_{\Omega}|\nabla u|^{p(x)-2}\nabla u\nabla \varphi dx
-\lambda\int_{\Omega}|u|^{p^*(x)-2} u\varphi dx
- \int_{\Omega}\overline{\omega}\varphi dx,
\hskip 0.3cm \forall\hskip0.1cm \varphi\in C_0^{\infty}(R^{N}).
\]
So we derive that
\begin{equation} \label{e3.16}
-\Delta_{p(x)}u - \lambda |u|^{p^*(x)-2}u \in [\underline{f}(u),\overline{f}(u)].
\end{equation}
Now we applying Proposition \ref{prop2.6} to prove that $\nu_{j} =0$ in \eqref{e3.7}.
Assume $\nu_{j}\neq 0$ for some $j\in J$.
From \eqref{e3.15}, we have
\[
I_{\lambda}(u_n)- \frac{1}{q_{-}}\langle \omega_n, u_n\rangle
\geq (\frac{1}{p_{+}}-\frac{1}{q_{-}})\int_{\Omega}|\nabla u_n|^{p(x)}dx.
\]
Since $I_{\lambda}(u_n) \to c$ and $m(u_n)\to 0$, using Proposition \ref{prop2.6},
taking $n\to \infty$, we obtain
\begin{align*}
c &\geq (\frac{1}{p_{+}}-\frac{1}{q_{-}})\int_{\Omega}|\nabla u|^{p(x)}dx
+ (\frac{1}{p_{+}}-\frac{1}{q_{-}})\sum_{j\in J}\mu_{j} \\
&\geq(\frac{1}{p_{+}}-\frac{1}{q_{-}})S^{N}
\max\{\lambda^{1-\frac{N}{p_{+}}},\lambda^{1-\frac{N}{p_{-}}}\}.
\end{align*}
Since $ c< (\frac{1}{p_{+}}-\frac{1}{q_{-}})S^{N}
\max\{\lambda^{1-\frac{N}{p_{+}}},\lambda^{1-\frac{N}{p_{-}}}\}$,
then $ \nu_{j} =0$ for all $j\in J$.
Hence we have
\begin{equation} \label{e3.17}
\int_{\Omega}| u_n|^{p^*(x)}dx \to \int_{\Omega}| u|^{p^*(x)}dx.
\end{equation}
So we can use \cite[Lemma 2.1]{F}. Set $v_n= u_n - u$ and we have
\begin{gather}
\int_{\Omega}| u_n|^{p^*(x)}dx = \int_{\Omega}| v_n|^{p^*(x)}dx
+ \int_{\Omega}| u|^{p^*(x)}dx + o(1), \label{e3.18} \\
\int_{\Omega}|\nabla u_n|^{p(x)}dx
= \int_{\Omega}|\nabla v_n|^{p(x)}dx + \int_{\Omega}|\nabla u|^{p(x)}dx + o(1).
\label{e3.19}
\end{gather}
Thus, by \eqref{e3.17} and \eqref{e3.18},
$u_n\to u$ strongly in $ {L^{p^*(x)} (\Omega)}$.
From \eqref{e3.13}, using \eqref{e3.16} and \eqref{e3.19}, we obtain
\[
\langle \omega_n, u_n\rangle = \int_{\Omega}|\nabla v_n|^{p(x)}dx
+ \int_{\Omega}|\nabla u|^{p(x)}dx
-\lambda\int_{\Omega}|u_n|^{p^*(x)}dx
- \int_{\Omega}\overline{\omega}_nu_ndx + o(1).
\]
By \eqref{e3.4} and \eqref{e3.17}, letting $n \to \infty$, we conclude that
\[
\int_{\Omega}|\nabla v_n|^{p(x)}dx \to 0.
\]
This fact and Proposition \ref{prop2.2} imply that
$u_n\to u$ strongly in $W_0^{1,p(x)}(\Omega)$.
The proof is complete.
\end{proof}
\begin{lemma} \label{lem3.4}
Suppose $f$ satisfies {\rm (F2), (F3)}. Then, for every $\lambda > 0$, there
are $\alpha, \rho > 0$, such that $I_{\lambda}(u) \geq \alpha$, $\|u\|=\rho$.
\end{lemma}
\begin{proof} By (F2) and (F3), we have
\[
|f(t)|\leq \epsilon |t|^{p(x)-1}+ C|t|^{q(x)}\leq \epsilon |t|^{p(x)-1}
+ C(\epsilon)|t|^{p^*(x)-1}.
\]
Therefore,
\begin{equation} \label{e3.20}
\begin{split}
I_{\lambda}(u)&= \int_{\Omega}\frac{1}{p(x)}|\nabla u|^{p(x)}dx
- \lambda\int_{\Omega}\frac{1}{p^*(x)}|u|^{p^*(x)}dx - \int_{\Omega}F(u)dx \\
&\geq \frac{1}{p_{+}}\int_{\Omega}|\nabla u|^{p(x)}dx
- \frac{\epsilon}{p_{-}}\int_{\Omega}| u|^{p(x)}dx
- \frac{\lambda + C(\epsilon)}{p^*_{-}}\int_{\Omega}|u|^{p^*(x)}dx.
\end{split}
\end{equation}
we can take $\|u\| < 1$ sufficiently small such that $|u|_{p(x)} < 1$ and
$|u|_{p^*(x)} < 1$. From \eqref{e3.20}, Propositions \ref{prop2.1} and \ref{prop2.4},
and the definition of $S$, using the usual arguments, we obtain
\begin{align*}
I_{\lambda}(u) &\geq \frac{1}{p_{+}}\|u\|^{p_{+}}
- \frac{\epsilon}{p_{-}}|u|^{p_{-}}_{p(x)}
-\frac{\lambda + C(\epsilon)}{p^*_{-}}|u|^{p^*_{-}}_{p^*(x)} \\
&\geq \frac{1}{2p_{+}}\|u\|^{p_{+}} - \frac{\lambda
+ C(\epsilon)}{p^*_{-}}S^{-p^*_{-}}\|u\|^{p^*_{-}} \\
&= (\frac{1}{2p_{+}} - \frac{\lambda + C(\epsilon)}{p^*_{-}}S^{-p^*_{-}}\|u\|^{p^*_{-}
-p_{+}})\|u\|^{p_{+}}.
\end{align*}
Considering
\[
g(t) = \frac{1}{2p_{+}} - \frac{\lambda
+ C(\epsilon)}{p^*_{-}}S^{-p^*_{-}}\|t\|^{p^*_{-}-p_{+}},
\]
since $p_{+}< p^*_{-}$, we have $g(t) \to \frac{1}{2p_{+}}$ as $t \to 0$.
Hence, there exists $ \rho > 0$ such that
$g(\rho) > 0$.
So, we obtain $\alpha$ and $\rho > 0$, such that
\[
I_{\lambda}(u) \geq \alpha, \quad \|u\|=\rho.
\]
The proof is complete.
\end{proof}
Next, we choose $\varphi(x) \in W_0^{1,p(x)}(\Omega)$, such that $\|\varphi\| = 1$.
\begin{lemma} \label{lem3.5}
Suppose $f$ satisfies {\rm (F4)}. Then, there exists $\lambda_0 > 0$, $t_0 > 0$
such that $I_{\lambda}(t_0\varphi) < 0$, and
for all $\lambda \in (0, \lambda_0)$,
\[
\sup_{t \geq 0}I_{\lambda}(t\varphi) < (\frac{1}{p_{+}}-\frac{1}{q_{-}})S^{N}
\max\{\lambda^{1-\frac{N}{p_{+}}},\lambda^{1-\frac{N}{p_{-}}}\}.
\]
\end{lemma}
\begin{proof} By (F4), we have
\[
f(u)u \geq\underline{f}(u)u\geq q_{-}F(u), \quad \forall u\neq0.
\]
This implies
$F(t u) \geq t^{q_{-}}F(u)$, for all $t\geq 1$.
Then, for any $t > 1$,
\[ % \label{e3.21}
I_{\lambda}(t\varphi) \leq \frac{t^{p_{+}}}{p_{-}}
- \frac{\lambda t^{p^*_{+}}}{p^*_{+}}\int_{\Omega}|\varphi|^{p^*(x)}dx
- \int_{\Omega}F(t\varphi)dx
\leq \frac{t^{p}_{+}}{p_{-}} - t^{q_{-}}\int_{\Omega}F(\varphi)dx = J_1(t\varphi).
\]
Since $q_{-} > p_{+}$ and $ F(\varphi) > 0$, there exists $t_0 > 0$
sufficiently large such that $I_{\lambda}(t_0\varphi) < 0$ and
$ \|t_0\varphi\| > \rho$ with $\rho$ given by Lemma \ref{lem3.4}.
If $ 0\leq t < 1$, then
\[
I_{\lambda}(t\varphi) \leq \frac{t^{p_{-}}}{p_{-}}
- \int_{\Omega}F(t\varphi)dx = J_2(t\varphi).
\]
Let $J(t\varphi) = \max \{ J_1(t\varphi), J_2(t\varphi)\}$, so we have
\[
\sup_{t \geq 0}I_{\lambda}(t\varphi) \leq \sup_{t \geq 0}J(t\varphi).
\]
Hence, we can find $\lambda_0 > 0$ such that
\[
\sup_{t \geq 0}J_{\lambda}(t\varphi) < (\frac{1}{p_{+}}
-\frac{1}{q_{-}})S^{N}\max\{\lambda^{1-\frac{N}{p_{+}}},\lambda^{1-\frac{N}{p_{-}}}\}.
\]
So, for all $\lambda\in (0, \lambda_0)$, we have
\[
\sup_{t \geq 0}I_{\lambda}(t\varphi) <(\frac{1}{p_{+}}-\frac{1}{q_{-}})S^{N}
\max\{\lambda^{1-\frac{N}{p_{+}}},\lambda^{1-\frac{N}{p_{-}}}\}.
\]
The proof is complete.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm1.1}]
It is obvious that $I_{\lambda}(0) = 0$. By Lemmas \ref{lem3.1},
\ref{lem3.3}--\ref{lem3.5},
according to Theorem \ref{thm2.1}, there exist $\lambda_0 > 0$, and
for all $\lambda \in (0, \lambda_0)$, we can find an $u \in W_0^{1,p(x)}(\Omega)$
such that $I_{\lambda}(u) > 0$ and $0 \in \partial I_{\lambda}(u)$.
Hence, $u$ is a nontrivial solution of \eqref{e1.1}.
The proof is complete.
\end{proof}
\begin{proof}[Proof of Theorem \ref{thm1.2}]
In \eqref{e1.2}, $f(u)= bh(u-a)|u|^{q(x)-2}u $ has only one discontinuity point $a$,
so by the consequence of Theorem \ref{thm1.1}, we obtain that an $u\in W_0^{1,p(x)}(\Omega)$
is a nontrivial nonnegative solution of \eqref{e1.2}. That is,
\begin{equation} \label{e3.22}
-\Delta_{p(x)}u - \lambda |u|^{p^*(x)-2}u \in \widehat{f}(u) \quad \text{a.e. in }
\Omega
\end{equation}
where $\widehat{f}(u)$ is the multivalued function given by
\begin{equation} \label{e3.23}
\widehat{f}(s) = \begin{cases}
\{f(s)\} & s\neq a,\\
[0, b h(x)u^{q(x)-1}] & s = a.
\end{cases}
\end{equation}
If $V = \{ x\in \Omega: u(x)=a\}$ exists, by \eqref{e3.22} and \eqref{e3.23}, we have
\[
-\Delta_{p(x)}u - \lambda |u|^{p^*(x)-2}u \in [0, b h(x)u^{q(x)-1}]\quad
\text{a.e. in } V.
\]
Using the Morrey-Stampacchia's theorem \cite{Mo}, we have $- \Delta_{p(x)}u =0$
a.e. $x\in V$. So
\[
-\lambda a^{p^*(x)-1} \geq 0 \quad \text{a.e. in } V.
\]
This is a contradiction. Thus $|V|= 0$.
The proof is complete.
\end{proof}
\subsection*{Acknowledgments}
The author wants to thank the anonymous referees for their carefully
reading this paper and their useful comments.
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\end{document}