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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 241, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/241\hfil Spectrum for anisotropic equations]
{Spectrum for anisotropic equations involving weights and variable exponents}

\author[I.-L. St\u{a}ncu\c{t} \hfil EJDE-2014/241\hfilneg]
{Ionela-Loredana St\u{a}ncu\c{t}}  % in alphabetical order

\address{Ionela-Loredana St\u{a}ncu\c{t}\newline
Department of Mathematics, University of Craiova, 200585, Romania}
\email{stancutloredana@yahoo.com}

\thanks{Submitted June 26, 2014. Published November 18, 2014.}
\subjclass[2000]{35D30, 35J60, 58E05}
\keywords{$\vec{p}(\cdot)$-Laplace operator;
 anisotropic variable exponent Sobolev space;  \hfill\break\indent
 critical point; weak solution; eigenvalue}

\begin{abstract}
 We study the problem
 $$
 -\sum_{i=1}^{N}\Big[\partial_{x_{i}}\Big(|\partial_{x_{i}}u|^{p_{i}(x)-2}
 \partial_{x_{i}}u\Big)
 +|u|^{p_{i}(x)-2}u\Big]+|u|^{q(x)-2}u
 =\lambda g(x)|u|^{r(x)-2}u
 $$
 in $\Omega$, $u=0$ on $\partial\Omega$, where $\Omega$ is a bounded domain
 in $\mathbb{R}^{N}$ ($N\geq3$), with smooth boundary, $\lambda$ is a
 positive real number, the functions $p_{i}, q, r:\overline\Omega\to[2,\infty)$
 are Lipschitz continuous, $g:\overline\Omega\to[0,\infty)$ is measurable and
 these fulfill certain conditions. The main result of this paper establish
 the existence of two positive constants $\lambda_0$ and $\lambda_{1}$
 with $0<\lambda_0\leq\lambda_{1}$ such that any $\lambda\in[\lambda_{1},\infty)$
 is an eigenvalue, while any $\lambda\in(0,\lambda_0)$ is not an eigenvalue of
 our problem.
\end{abstract}

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

\section{Introduction} 

The purpose of this paper is to study the eigenvalue problem
\begin{equation}
\begin{gathered}
-\sum_{i=1}^{N}\big[\partial_{x_{i}}(|\partial_{x_{i}}u|^{p_{i}(x)-2}
\partial_{x_{i}}u)
+|u|^{p_{i}(x)-2}u\big]+|u|^{q(x)-2}u
=\lambda g(x)|u|^{r(x)-2}u \quad \text{in } \Omega,\\
u=0 \quad \text{on } \partial\Omega,
\end{gathered} \label{pb} 
\end{equation}
where $\Omega$ is a bounded domain with smooth boundary in
$\mathbb{R}^{N}$ $(N\geq3)$. The functions
$p_{i}, q, r:\overline{\Omega}\to[2,\infty)$ are Lipschitz continuous,
while $g:\overline{\Omega}\to[0,\infty)$ is a measurable function
for which there exists an open subset $\Omega_0\subset\Omega$ such that
$g(x)>0$ for any $x\in\Omega_0$, and $\lambda\geq0$ is a real number.

A motivation for the study of problem \eqref{pb} is given in
\cite{m2,m3}. In \cite{m2} the problem  studied involves the Laplace
operator and $\Omega\subset\mathbb{R}^{N}$ is a bounded domain with
smooth boundary, while in \cite{m3} the authors deal with a problem
involving the $p(\cdot)$-Laplace operator and $\Omega\subset\mathbb{R}^{N}$
($N\geq3$) is a smooth exterior domain.

We emphasize the presence of $\vec{p}(\cdot)$-Laplace operator in problem \eqref{pb}.
 This is a natural extension of the $p(\cdot)$-Laplace operator.
Both $p(\cdot)$-Laplace operator and $\vec{p}(\cdot)$-Laplace operator are
nonhomogeneous, unlike the $p$-Laplace operator, where $p$ is a positive constant.
The study of nonlinear elliptic equations involving quasilinear homogeneous
type operators like the $p$-Laplace operator is based on the theory of
standard Sobolev spaces to find weak solutions, while
in the case of operators $p(\cdot)$-Laplace and $\vec{p}(\cdot)$-Laplace the
natural setting is the use of the isotropic variable exponent Sobolev spaces and
anisotropic variable exponent Sobolev spaces respectively (for our approach).

Thanks to the applicability to diverse fields of variable Sobolev spaces,
in the past decades appeared many papers which involve such spaces.
These are used to model various phenomena in image restoration (see \cite{c1}),
in elastic mechanics (see \cite{z1}) and for the modelling of electrorheological
fluids (or smart fluids). The first major discovery on electrorheological fluids
was in 1949 due to  Winslow  \cite{w1}. These fluids have the
interesting property that their viscosity can undergoes a significant
change (namely can raise by up to five orders of magnitude) which depends on
the electric field in the fluid. This phenomenon is known as the Winslow
effect. Electrorheological fluids have been used in robotics and space technology.
The experimental research has been done mainly in the USA, for instance in
NASA laboratories.

\section{Abstract framework}

First, we introduce briefly a variable exponent Lebesgue-Sobolev setting.
For more information on properties of variable exponent  Lebesgue-Sobolev
spaces we refer to
\cite{e1,e2,e3,k1,m4,s1}.

Throughout this paper, for any Lipschitz continuous function
$p:\overline{\Omega}\to(1,\infty)$ we define
$$
p^{+}=\operatorname{ess\,sup}_{x\in\Omega}p(x)\quad\text{and}\quad
p^{-}=\operatorname{ess\,inf}_{x\in\Omega}p(x).
$$
We define the \emph{variable exponent Lebesgue space}
$$
L^{p(\cdot)}(\Omega)=\Big\{u; u \text{is a measurable real-valued function and }
\int_{\Omega}|u|^{p(x)}dx<\infty\Big\},
$$
endowed with the so-called \emph{Luxemburg norm}
$$
|u|_{p(\cdot)}=\inf\Big\{\mu >0; \int_{\Omega}\Big|\frac{u(x)}{\mu}\Big|^{p(x)}dx
\leq1\Big\},
$$
which is a separable and reflexive Banach space. If $0<|\Omega|<\infty$
and $p_{1},\, p_{2}$ are variable exponents such that $p_{1}(x)\leq p_{2}(x)$
almost everywhere in $\Omega$, then the embedding
$L^{p_{2}(\cdot)}(\Omega)\hookrightarrow L^{p_{1}(\cdot)}(\Omega)$ is continuous.

We denote by $L^{p'(\cdot)}(\Omega)$ the conjugate space of $L^{p(\cdot)}(\Omega)$,
where $\frac{1}{p(x)}+\frac{1}{p'(x)}=1$. For any $u\in L^{p(\cdot)}(\Omega)$
and $v\in L^{p'(\cdot)}(\Omega)$ the following H\"{o}lder type inequality
\begin{equation}
\label{Holder}
\Big|\int_{\Omega}uv\, dx\Big|\leq\Big(\frac{1}{p^{-}}
+\frac{1}{p'^{-}}\Big)|u|_{p(\cdot)}|v|_{p'(\cdot)}
\leq2|u|_{p(\cdot)}|v|_{p'(\cdot)}
\end{equation}
holds.

An important role in handling the generalized Lebesgue spaces is played by
the $p(\cdot)$-\emph{modular} of $L^{p(\cdot)}(\Omega)$ space, which
is the mapping $\rho_{p(\cdot)}:L^{p(\cdot)}(\Omega)\to\mathbb{R}$ defined by
$$
\rho_{(\cdot)}(u)=\int_{\Omega}|u|^{p(x)}dx.
$$
If $(u_{n}), u\in L^{p(\cdot)}(\Omega)$, then the following relations hold:
\begin{gather}
\label{|u|>1}
|u|_{p(\cdot)}>1 \Rightarrow |u|_{p(\cdot)}^{p^{-}}
\leq\rho_{p(\cdot)}(u)\leq|u|_{p(\cdot)}^{p^{+}}, \\
\label{|u|<1}
|u|_{p(\cdot)}<1 \Rightarrow |u|_{p(\cdot)}^{p^{+}}
\leq\rho_{p(\cdot)}(u)\leq|u|_{p(\cdot)}^{p^{-}}, \\
\label{conv 0}
|u_{n}-u|_{p(\cdot)}\to0\Leftrightarrow \rho_{p(\cdot)}(u_{n}-u)\to0.
\end{gather}
We denote by $W_0^{1,p(\cdot)}$ the \emph{variable exponent Sobolev space}
defined by
$$
W^{1,p(\cdot)}_0(\Omega)=\big\{u;  u_{\mid\partial\Omega}=0, \;
 u\in L^{p(\cdot)}(\Omega)\text{ and }|\nabla u|\in L^{p(\cdot)}(\Omega)\big\},
$$
endowed with the equivalent norms
$$
\|u\|_{p(\cdot)}=|u|_{p(\cdot)}+|\nabla u|_{p(\cdot)}
$$
and
$$
\|u\|=\inf\Big\{\mu>0; \ \int_{\Omega}\Big(\Big|
\frac{\nabla u(x)}{\mu}\Big|^{p(x)}
+\Big|\frac{u(x)}{\mu}\Big|^{p(x)}\Big)dx\leq1\Big\},
$$
where, in the definition of $\|u\|_{p(\cdot)}$, $|\nabla u|_{p(\cdot)}$ is
the Luxemburg norm of $|\nabla u|$.
We remember that $W_0^{1,p(\cdot)}(\Omega)$ is a separable and reflexive
Banach space. Also, we note that if $p, s:\overline{\Omega}\to(1,\infty)$
are Lipschitz continuous functions with $p^{+}<N$ and $p(x)\leq s(x)\leq p^{*}(x)$
for all $x\in\overline{\Omega}$, then the there exists the continuous embedding
$W_0^{1,p(\cdot)}(\Omega)\hookrightarrow L^{s(\cdot)}(\Omega)$, where
$p^{*}(x)=\frac{Np(x)}{N-p(x)}$.

Next, we present  the \emph{anisotropic variable exponent Sobolev space}
$W_0^{1,\vec{p}(\cdot)}(\Omega)$, where $\vec{p}:\overline{\Omega}\to\mathbb{R}^{N}$
is the vectorial function $\vec{p}(\cdot)=(p_{1}(\cdot),\dots, p_{N}(\cdot))$
and the components $p_{i}$, $i\in\{1,\dots, N\}$, are logarithmic H\"{o}lder
continuous, that is, there exists $M>0$ such that
$|p_{i}(x)-p_{i}(y)|\leq-M/\log(|x-y|)$ for any $x,\, y\in\Omega$ with
$|x-y|\leq1/2$ and $i\in\{1,\dots,N\}$. Also, we define
$W_0^{1,\vec{p}(\cdot)}(\Omega)$ as the closure of $C_0^{\infty}(\Omega)$
under the norm
$$
\|u\|_{\vec{p}(\cdot)}=\sum_{i=1}^{N}
\big(|\partial_{x_{i}}u|_{p_{i}(\cdot)}+|u|_{p_{i}(\cdot)}\big),
$$
and is a reflexive Banach space (see \cite{m1}).

Now, we introduce $\vec{P}_{+}, \vec{P}_{-}\in\mathbb{R}^{N}$ as
$$
\vec{P}_{+}=(p_{1}^{+},\dots,p_{N}^{+}), \quad
\vec{P}_{-}=(p_{1}^{-},\dots,p_{N}^{-}),
$$
and $P_{+}^{+}, P_{-}^{+}, P_{-}^{-}\in\mathbb{R}^{+}$ as
$$
P_{+}^{+}=\max\{p_{1}^{+},\dots,p_{N}^{+}\}, \quad
P_{-}^{+}=\max\{p_{1}^{-},\dots,p_{N}^{-}\}, \quad
 P_{-}^{-}=\min\{p_{1}^{-},\dots,p_{N}^{-}\}.
$$
We also always assume that
$$
\sum_{i=1}^{N}\frac{1}{p_{i}^{-}}>1,
$$
and define $P_{-}^{*}, \ P_{-,\infty}\in\mathbb{R}^{+}$ by
$$
P_{-}^{*}=\frac{N}{\sum_{i=1}^{N}1/p_{i}^{-}-1}, \quad
P_{-,\infty}=\max\{P_{-}^{+}, P_{-}^{*}\}.
$$


\section{Main result}

We study the problem \eqref{pb} assuming that the functions $p_{i}$, $q$
and $r$ satisfy the hypotheses
\begin{gather}\label{p}
2\leq P_{-}^{-}\leq P_{+}^{+}<N, \\
\label{p,r,q}
P_{-}^{+}\leq P_{+}^{+}<r^{-}\leq r^{+}<q^{-}\leq q^{+}<P_{-}^{*}\leq p_{i}^{*}(x)
\quad \forall x\in\overline{\Omega},\; \forall i\in\{1,\dots,N\}.
\end{gather}
Furthermore, we assume that the weight function $g(x)$ satisfies the hypotheses
\begin{gather} \label{g infty}
\int_{\Omega}(\lambda g(x))^{\frac{q(x)}{q(x)-r(x)}}dx<\infty,\\
\label{g}
g\in L^{\infty}(\Omega)\cap L^{{p}_{i}^{0}(\cdot)}(\Omega),
\end{gather}
where $p_{i}^{0}(x)=\frac{p_{i}^{*}(x)}{p_{i}^{*}(x)-r^{-}}$
for any $x\in\overline{\Omega}$ and any $i\in\{1,\dots,N\}$.

We look for weak solutions for problem \eqref{pb} in the space
$W_0^{1,\vec{p}(\cdot)}(\Omega)$.
We say that $\lambda\in\mathbb{R}$ is an \emph{eigenvalue} of problem
\eqref{pb} if there exists a $u\in W_0^{1,\vec{p}(\cdot)}(\Omega)\setminus\{0\}$
such that
\begin{align*}
&\int_{\Omega}\Big[\sum_{i=1}^{N}\big(|\partial_{x_{i}}u|^{p_{i}(x)
-2}\partial_{x_{i}}u\partial_{x_{i}}v
+|u|^{p_{i}(x)-2}uv\big)+|u|^{q(x)-2}uv\Big]dx\\
&-\lambda\int_{\Omega}g(x)|u|^{r(x)-2}uv\,dx=0,
\end{align*}
for all $v\in W_0^{1,\vec{p}(\cdot)}(\Omega)$. We point out that
if $\lambda$ is an eigenvalue of problem \eqref{pb} then the corresponding
$u\in W_0^{1,\vec{p}(\cdot)}(\Omega)\setminus\{0\}$ is a \emph{weak solution}
of problem \eqref{pb}.

Define
\begin{gather*}
\lambda_{1}:=\inf_{u\in W_0^{1,\vec{p}(\cdot)}(\Omega)\setminus\{0\}}
\frac{\int_{\Omega}\sum_{i=1}^{N}\Big(\frac{|\partial_{x_{i}}u|^{p_{i}(x)}}{p_{i}(x)}
+\frac{|u|^{p_{i}(x)}}{p_{i}(x)}\Big)dx+\int_{\Omega}\frac{|u|^{q(x)}}{q(x)}dx}
{\int_{\Omega}\frac{g(x)}{r(x)}|u|^{r(x)}dx},
\\
\lambda_0:=\inf_{u\in W_0^{1,\vec{p}(\cdot)}(\Omega)\setminus\{0\}}
\frac{\int_{\Omega}\sum_{i=1}^{N}\big(|\partial_{x_{i}}u|^{p_{i}(x)}
+|u|^{p_{i}(x)}\big)dx+\int_{\Omega}|u|^{q(x)}dx}
{\int_{\Omega}g(x)|u|^{r(x)}dx}.
\end{gather*}
Our main result is given by the following theorem.

\begin{theorem} \label{thm1}
Assume that conditions \eqref{p}--\eqref{g} are satisfied. Then
$$
0<\lambda_0\leq\lambda_{1}.
$$
In addition, any $\lambda\in(0,\lambda_0)$ is not an eigenvalue of
problem \eqref{pb}, while each $\lambda\in[\lambda_{1},\infty)$
is an eigenvalue of our problem.
\end{theorem}

\section{Proof of the main result}

In what follows we denote by $E$ the generalized Sobolev space
$W_0^{1,\vec{p}(\cdot)}(\Omega)$.
We need to define the functionals $J_{1},I_{1},J_0, I_0:E\to\mathbb{R}$ by
\begin{gather*}
J_{1}(u)=\int_{\Omega}\sum_{i=1}^{N}
\Big(\frac{|\partial_{x_{i}}u|^{p_{i}(x)}}{p_{i}(x)}
+\frac{|u|^{p_{i}(x)}}{p_{i}(x)}\Big)dx+\int_{\Omega}\frac{|u|^{q(x)}}{q(x)}dx,
\\
I_{1}(u)=\int_{\Omega}\frac{g(x)}{r(x)}|u|^{r(x)}dx,
\\
J_0(u)=\int_{\Omega}\sum_{i=1}^{N}\big(|\partial_{x_{i}}u|^{p_{i}(x)}
+|u|^{p_{i}(x)}\big)dx+\int_{\Omega} |u|^{q(x)}dx,
\\
I_0(u)=\int_{\Omega}g(x)|u|^{r(x)}dx.
\end{gather*}
\cite[Theorem 1]{m1} assures that $J_{1},I_{1}\in C^{1}(E,\mathbb{R})$
and the Fr\'{e}chet derivatives are given by
\begin{gather*}
\langle J_{1}'(u),v\rangle=\int_{\Omega}\Big[\sum_{i=1}^{N}
\big(|\partial_{x_{i}}u|^{p_{i}(x)-2}\partial_{x_{i}}u
\partial_{x_{i}}v+|u|^{p_{i}(x)-2}uv\big)+|u|^{q(x)-2}uv\Big]dx,
\\
\langle I_{1}'(u),v\rangle=\int_{\Omega}g(x)|u|^{r(x)-2}uv\,dx.
\end{gather*}
Also, we define for any $\lambda>0$ the functional
$$
T_{\lambda}^{1}(u)=J_{1}(u)-\lambda\cdot I_{1}(u)\quad \forall u\in E.
$$
We point out that $\lambda$ is an eigenvalue of problem \eqref{pb}
if and only if there is an element $u_{\lambda}\in E\setminus\{0\}$,
which is a critical point of the functional $T_{\lambda}^{1}$.

To give a clear view of what needs to be proved, we divide the proof
of the theorem in four steps.
\smallskip

\noindent\textbf{Step 1.} We show that $\lambda_0, \lambda_{1}>0$.
It should be noticed that from the condition \eqref{p,r,q}, we have
$p_{i}(x)<r(x)<q(x)$ for any $x\in\overline{\Omega}$ and any
$i\in\{1,\dots,N\}$, and therefore
$$
|u(x)|^{r(x)}\leq|u(x)|^{p_{i}(x)}+|u(x)|^{q(x)}\quad  \forall u\in E,\;
 \forall x\in\overline{\Omega},\;\forall i\in\{1,\dots,N\}.
$$
Thus,
\begin{equation}\label{|g|}
\int_{\Omega}\big(|u|^{p_{i}(x)}+|u|^{q(x)}\big)dx
\geq\frac{1}{|g|_{\infty}}\cdot\int_{\Omega}g(x)|u|^{r(x)}dx
\quad  \forall u\in E,\; \forall i\in\{1,\dots,N\}.
\end{equation}
It is obvious that
$$
\int_{\Omega}\sum_{i=1}^{N}\big(|\partial_{x_{i}}u|^{p_{i}(x)}
+|u|^{p_{i}(x)}\big)dx
+\int_{\Omega}|u|^{q(x)}dx\geq\int_{\Omega}\big(|u|^{p_{i}(x)}+|u|^{q(x)}\big)dx,
$$
which together with relation \eqref{|g|} we can deduce that
$$
\frac{\int_{\Omega}\sum_{i=1}^{N}\big(|\partial_{x_{i}}u|^{p_{i}(x)}+|u|^{p_{i}(x)}
\big)dx +\int_{\Omega}|u|^{q(x)}dx}
{\int_{\Omega}g(x)|u|^{r(x)}dx}\geq\frac{1}{|g|_{\infty}}>0.
$$
Hence we obtain that $\lambda_0>0$.

Next, using \eqref{|g|}, by a simple computation we arrive at
\[
\int_{\Omega}\frac{|u|^{p_{i}(x)}}{p_{i}(x)}dx+\int_{\Omega}
\frac{|u|^{q(x)}}{q(x)}dx
\geq\frac{r^{-}}{q^{+}\cdot|g|_{\infty}}\int_{\Omega}
\frac{g(x)}{r(x)}|u|^{r(x)}dx.
\]
It is clear that
\[
\int_{\Omega}\sum_{i=1}^{N}\Big(\frac{|\partial_{x_{i}}u|^{p_{i}(x)}}{p_{i}(x)}
+\frac{|u|^{p_{i}(x)}}{p_{i}(x)}\Big)dx
+\int_{\Omega}\frac{|u|^{q(x)}}{q(x)}dx
\geq\int_{\Omega}\frac{|u|^{p_{i}(x)}}{p_{i}(x)}dx
+\int_{\Omega}\frac{|u|^{q(x)}}{q(x)}dx,
\]
and considering the previous inequality we derive that
\[
\frac{\int_{\Omega}\sum_{i=1}^{N}\Big(\frac{|\partial_{x_{i}}u|^{p_{i}(x)}}{p_{i}(x)}
+\frac{|u|^{p_{i}(x)}}{p_{i}(x)}\Big)dx
+\int_{\Omega}\frac{|u|^{q(x)}}{q(x)}dx}{\int_{\Omega}\frac{g(x)}{r(x)}|u|^{r(x)}dx}
\geq\frac{r^{-}}{q^{+}\cdot|g|_{\infty}}>0,
\]
wherefrom $\lambda_{1}>0$. Step 1 is verified.
\smallskip

\noindent\textbf{Step 2.}
We prove that any $\lambda\in(0,\lambda_0)$ is not an eigenvalue of
 problem \eqref{pb}.

We argue indirectly. So, suppose that there is $\lambda\in(0,\lambda_0)$,
an eigenvalue of problem \eqref{pb}. Thereby we can deduce the existence of
an element $u_{\lambda}\in E\setminus\{0\}$ such that
\begin{align*}
&\int_{\Omega}\Big[\sum_{i=1}^{N}\big(|\partial_{x_{i}}u_{\lambda}|^{p_{i}(x)-2}
\partial_{x_{i}}u_{\lambda}\partial_{x_{i}}v
 +|u_{\lambda}|^{p_{i}(x)-2}u_{\lambda}v\big)
+|u_{\lambda}|^{q(x)-2}u_{\lambda}v\Big]dx\\
&=\lambda\int_{\Omega}g(x)|u_{\lambda}|^{r(x)-2}u_{\lambda}v\,dx\quad
 \forall v\in E.
\end{align*}
Taking $v=u_{\lambda}$ in the above equality we obtain
\begin{equation} \label{u lambda}
J_0(u_{\lambda})=\lambda\cdot I_0(u_{\lambda}).
\end{equation}
By  $u_{\lambda}\in E\setminus\{0\}$ we have $J_0(u_{\lambda})>0$ and
$I_0(u_{\lambda})>0$.
On the other hand
$$
\frac{J_0(u_{\lambda})}{I_0(u_{\lambda})}=\frac{\int_{\Omega}\sum_{i=1}^{N}
\big(|\partial_{x_{i}}u_{\lambda}|^{p_{i}(x)}+|u_{\lambda}|^{p_{i}(x)}\big)dx
+\int_{\Omega}
|u_{\lambda}|^{q(x)}dx}{\int_{\Omega}g(x)|u_{\lambda}|^{r(x)}dx}\geq\lambda_0.
$$
This, together with \eqref{u lambda} yield
$$
J_0(u_{\lambda})\geq\lambda_0\cdot I_0(u_{\lambda})
>\lambda\cdot I_0(u_{\lambda})=J_0(u_{\lambda}),
$$
which is a contradiction. This proves the Step 2.
\smallskip

\noindent\textbf{Step 3.}
We verify that each $\lambda\in(\lambda_{1}, \infty)$ is an eigenvalue for
problem \eqref{pb}.
With an eye to show what we proposed in this step, we start by proving
the following three lemmas.

\begin{lemma}\label{lemr+<s<P-*}
Assume that conditions \eqref{p}--\eqref{g} are fulfilled and $s$ is a
real number such that
$r^{+}<s<P_{-}^{*}$.
Then $g\in L^{\frac{s}{s-r^{-}}}(\Omega)\cap L^{\frac{s}{s-r^{+}}}(\Omega)$ and
\begin{equation} \label{g--s}
\int_{\Omega}g(x)|u|^{r(x)}dx\leq|g|_{\frac{s}{s-r^{-}}}|u|_{s}^{r^{-}}
+|g|_{\frac{s}{s-r^{+}}}|u|_{s}^{r^{+}}\quad \forall u\in E.
\end{equation}
\end{lemma}

\begin{proof}
In the first instance we highlight the inequalities
\begin{equation} \label{s-r}
\frac{s}{s-r^{+}}\geq\frac{s}{s-r^{-}}>\frac{P_{-}^{*}}{P_{-}^{*}-r^{-}}
\geq\frac{p_{i}^{*}(x)}{p_{i}^{*}(x)-r^{-}}=p_{i}^{0}(x)
\end{equation}
for all $x\in\overline\Omega$ and all $i\in\{1,\dots,N\}$.
Also, we have
\[
(p_{i}^{0})^{-}\leq p_{i}^{0}(x)\leq\frac{P_{-}^{*}}{P_{-}^{*}-r^{-}}\quad
 \forall x\in\overline{\Omega},\; \forall i\in\{1,\dots,N\}.
\]
So we arrive at
\begin{equation}\label{g-infty}
|g|_{\infty}^{\frac{s}{s-r^{-}}-(p_{i}^{0})^{-}}+|g|_{\infty}^{\frac{s}{s-r^{-}}
-\frac{P_{-}^{*}}{P_{-}^{*}-r^{-}}}
\geq|g|_{\infty}^{\frac{s}{s-r^{-}}-p_{i}^{0}(x)}\quad
\forall x\in\overline{\Omega},\; \forall i\in\{1,\dots,N\}.
\end{equation}
By \eqref{g}, \eqref{s-r} and \eqref{g-infty} we can easily see that
\begin{align*}
\int_{\Omega}[g(x)]^{\frac{s}{s-r^{-}}}dx
&=\int_{\Omega}[g(x)]^{p_{i}^{0}(x)}
\cdot[g(x)]^{\frac{s}{s-r^{-}}-p_{i}^{0}(x)}dx\\
&\leq\int_{\Omega}[g(x)]^{p_{i}^{0}(x)}\cdot|g|_{\infty}^{\frac{s}{s-r^{-}}
 -p_{i}^{0}(x)}dx\\
&\leq\Big(|g|_{\infty}^{\frac{s}{s-r^{-}}-(p_{i}^{0})^{-}}
 +|g|_{\infty}^{\frac{s}{s-r^{-}}
-\frac{P_{-}^{*}}{P_{-}^{*}-r^{-}}}\Big)
\int_{\Omega}[g(x)]^{p_{i}^{0}(x)}dx<\infty,
\end{align*}
that is $g\in L^{\frac{s}{s-r^{-}}}(\Omega)$. In a similar fashion, we can
show that $g\in L^{\frac{s}{s-r^{+}}}(\Omega)$.

From
\begin{equation}
\label{r+r-}
|u(x)|^{r^{-}}+|u(x)|^{r^{+}}\geq|u(x)|^{r(x)}\quad \forall u\in E,\;
\forall x\in\overline{\Omega},
\end{equation}
and H\"{o}lder type inequality \eqref{Holder}, we deduce
\begin{align*}
\int_{\Omega}g(x)|u|^{r(x)}dx
&\leq\int_{\Omega}g(x)|u|^{r^{-}}dx+\int_{\Omega}g(x)|u|^{r^{+}}dx\\
&\leq|g|_{\frac{s}{s-r^{-}}}|u|_{s}^{r^{-}}+|g|_{\frac{s}{s-r^{+}}}|u|_{s}^{r^{+}}
\quad \forall u\in E\,.
\end{align*}
The proof of Lemma \ref{lemr+<s<P-*} is complete.
\end{proof}

\begin{lemma} \label{lim}
For each $\lambda>0$ we have
$$
\lim_{\|u\|_{\vec{p}(\cdot)}\to\infty}T_{\lambda}^{1}(u)=\infty.
$$
\end{lemma}

\begin{proof}
Let $s\in\mathbb{R}$ be such that
\begin{equation}
\label{s1}
r^{+}<s<q^{-}<P_{-}^{*}.
\end{equation}
Without loss of generality we assume that $\|u\|_{\vec{p}(\cdot)}>1$
for each $u\in E$. By \eqref{p,r,q} and \eqref{s1} we have
\[
|u(x)|^{p_{i}(x)}+|u(x)|^{q(x)}\geq|u(x)|^{s}\quad \forall u\in E,\;
\forall x\in\overline{\Omega},\;\forall i\in\{1,\dots,N\}\,.
\]
This implies
\begin{equation}\label{pi,q,s}
\int_{\Omega}\Big(\sum_{i=1}^{N}|u|^{p_{i}(x)}+|u|^{q(x)}\Big)dx
\geq\int_{\Omega}|u|^{s}dx.
\end{equation}
Now, using \eqref{pi,q,s} and Lemma \ref{lemr+<s<P-*} we have
\begin{align*}
T_{\lambda}^{1}(u)
&=\int_{\Omega}\sum_{i=1}^{N}\Big(\frac{|\partial_{x_{i}}u|^{p_{i}(x)}}{p_{i}(x)}
+\frac{|u|^{p_{i}(x)}}{p_{i}(x)}\Big)dx+\int_{\Omega}\frac{|u|^{q(x)}}{q(x)}dx
-\lambda\int_{\Omega}\frac{g(x)}{r(x)}|u|^{r(x)}dx\\
&\geq\Big(\frac{1}{P_{+}^{+}}\int_{\Omega}\sum_{i=1}^{N}|\partial_{x_{i}}u|^{p_{i}(x)}dx
+\frac{1}{2P_{+}^{+}}\int_{\Omega}\sum_{i=1}^{N}|u|^{p_{i}(x)}dx\Big)\\
&\quad +\Big(\frac{1}{2P_{+}^{+}}\int_{\Omega}\sum_{i=1}^{N}|u|^{p_{i}(x)}dx
+\frac{1}{q^{+}}\int_{\Omega}|u|^{q(x)}dx\Big)
-\frac{\lambda}{r^{-}}\int_{\Omega}g(x)|u|^{r(x)}dx\\
&\geq\frac{1}{2P_{+}^{+}}\int_{\Omega}\sum_{i=1}^{N}\big(|\partial_{x_{i}}u|^{p_{i}(x)}+|u|^{p_{i}(x)}\big)dx +\frac{1}{\max\{2P_{+}^{+},q^{+}\}}\int_{\Omega}|u|^{s}dx\\
&\quad -C_{1}|u|_{s}^{r^{-}}-C_{2}|u|_{s}^{r^{+}},
\end{align*}
where $C_{1}=\frac{\lambda}{r^{-}}|g|_{\frac{s}{s-r^{-}}}$ and
$C_{2}=\frac{\lambda}{r^{-}}|g|_{\frac{s}{s-r^{+}}}$.
To go further we need to define for each $i\in\{1,\dots,N\}$ the following:
\[
\alpha_{i}=\begin{cases}
P_{+}^{+} & \text{for } |\partial_{x_{i}}u|_{p_{i}(\cdot)}<1 \\
P_{-}^{-} & \text{for } |\partial_{x_{i}}u|_{p_{i}(\cdot)}>1,
\end{cases} \quad
\beta_{i}=\begin{cases}
P_{+}^{+} & \text{for } |u|_{p_{i}(\cdot)}<1 \\
P_{-}^{-} & \text{for } |u|_{p_{i}(\cdot)}>1.
\end{cases}
\]
From that fact and applying the Jensen's inequality to the convex function
$a:\mathbb{R}^{+}\to\mathbb{R}^{+}$, $a(t)=t^{P_{-}^{-}}$, $P_{-}^{-}\geq2$,
 we can write
\begin{equation} \label{T lambda}
\begin{aligned}
T_{\lambda}^{1}(u)
&\geq\frac{1}{2P_{+}^{+}}\sum_{i=1}^{N}\big(|\partial_{x_{i}}u|_{p_{i}
(\cdot)}^{\alpha_{i}}+|u|_{p_{i}(\cdot)}^{\beta_{i}}\big)\\
&\quad +\frac{1}{\max\{2P_{+}^{+},q^{+}\}}\int_{\Omega}|u|^{s}dx-
C_{1}|u|_{s}^{r^{-}}-C_{2}|u|_{s}^{r^{+}} \\
&\geq\frac{1}{2P_{+}^{+}}\sum_{i=1}^{N}|\partial_{x_{i}}u|_{p_{i}(\cdot)}^{P_{-}^{-}}
-\frac{1}{2P_{+}^{+}}\sum_{\{i;\, \alpha_{i}=P_{+}^{+}\}}\big(|\partial_{x_{i}}u|_{p_{i}(\cdot)}^{P_{-}^{-}}
-|\partial_{x_{i}}u|_{p_{i}(\cdot)}^{P_{+}^{+}}\big)\\
&\quad +\frac{1}{2P_{+}^{+}}\sum_{i=1}^{N}|u|_{p_{i}(\cdot)}^{P_{-}^{-}}
-\frac{1}{2P_{+}^{+}}\sum_{\{i;\, \beta_{i}=P_{+}^{+}\}}\big(|u|_{p_{i}(\cdot)}^{P_{-}^{-}}
-|u|_{p_{i}(\cdot)}^{P_{+}^{+}}\big) \\
&\quad +\frac{1}{\max\{2P_{+}^{+},q^{+}\}}\int_{\Omega}|u|^{s}dx
-C_{1}|u|_{s}^{r^{-}}-C_{2}|u|_{s}^{r^{+}} \\
&\geq\frac{\|u\|_{\vec{p}(\cdot)}^{P_{-}^{-}}}{2P_{+}^{+}(2N)^{P_{-}^{-}-1}}
-\frac{N}{P_{+}^{+}}
+(C_{3}|u|_{s}^{s}-C_{1}|u|_{s}^{r^{-}})+(C_{3}|u|_{s}^{s}
-C_{2}|u|_{s}^{r^{+}}),
\end{aligned}
\end{equation}
where $C_{3}=\frac{1}{2\max\{2P_{+}^{+},\, q^{+}\}}$.
We are going to show that for each $u\in E$ there are two positive
constants $L_{1}=L_{1}(r^{-},s,C_{1},C_{3})$ and
$L_{2}=L_{2}(r^{+},s,C_{2},C_{3})$ such that
\begin{gather}\label{L1'}
C_{3}|u|_{s}^{s}-C_{1}|u|_{s}^{r^{-}}\geq-L_{1},\\
\label{L2'}
C_{3}|u|_{s}^{s}-C_{2}|u|_{s}^{r^{+}}\geq-L_{2}.
\end{gather}
For this purpose, we define the functional $\Upsilon:(0,\infty)\to\mathbb{R}$ as
$$
\Upsilon(t)=\alpha t^{a}-\beta t^{b},
$$
where $\alpha,\, \beta,\, a,\, b$ are positive constants with $a>b$.
By a usual computation we find that $\Upsilon$ achieves its negative
global minimum
$$
\Upsilon(t_0)=-(a-b)\Big(\frac{b^{b}}{a^{a}}\Big)^{\frac{1}{a-b}}
\alpha^{\frac{b}{b-a}}\cdot\beta^{\frac{a}{a-b}},
$$
where $t_0=\big(\frac{\beta b}{\alpha a}\big)^{\frac{1}{a-b}}>0$. Consequently,
\begin{equation}\label{C(a,b)}
\alpha t^{a}-\beta t^{b}\geq-(a-b)\Big(\frac{b^{b}}{a^{a}}
\Big)^{\frac{1}{a-b}}\alpha^{\frac{b}{b-a}}\cdot\beta^{\frac{a}{a-b}}
\quad \forall t>0.
\end{equation}
Taking in \eqref{C(a,b)} $a=s$, $b=r^{-}$, $\alpha=C_{3}$ and
$\beta=C_{1}$ we find
\[
L_{1}=C(s,\, r^{-})\alpha^{\frac{r^{-}}{r^{-}-s}}
\beta^{\frac{s}{s-r^{-}}}.
\]
 In a similar manner, taking in \eqref{C(a,b)} $a=s$, $b=r^{+}$,
 $\alpha=C_{3}$ and $\beta=C_{2}$ we deduce that \eqref{L2'} holds  for
\[
L_{2}=C(s,\, r^{+})\alpha^{\frac{r^{+}}{r^{+}-s}}\beta^{\frac{s}{s-r^{+}}}.
\]
 Finally, putting together \eqref{T lambda}--\eqref{L2'} we conclude Lemma
\ref{lim}.
\end{proof}

\begin{lemma} \label{semicontinuous}
For any $\lambda>0$, the functional $T_{\lambda}^{1}$ is weakly lower 
semicontinuous on $E$.
\end{lemma}

\begin{proof}
Let $(u_{n})\subset E$ be such that $u_{n}\rightharpoonup u_0$ in $E$.
We define
\begin{gather*}
F(x,u)=\frac{1}{q(x)}|u|^{q(x)}-\frac{\lambda g(x)}{r(x)}|u|^{r(x)},\\
f(x,u)=F_{u}(x,u)=|u|^{q(x)-2}u-\lambda g(x)|u|^{r(x)-2}u.
\end{gather*}
Using ordinary rule of the derivation we find
\begin{equation}\label{f-u}
f_{u}(x,u)=(q(x)-1)|u|^{q(x)-2}-\lambda g(x)(r(x)-1)|u|^{r(x)-2}.
\end{equation}
We shall employ in what follows the following inequality: 
for any $k_{1},\, k_{2}>0$ and $0<q<r$ we have
$$
k_{1}|t|^{q}-k_{2}|t|^{r}\leq C k_{1}
\Big(\frac{k_{1}}{k_{2}}\Big)^{\frac{q}{r-q}}\quad \forall t\in\mathbb{R},
$$
where $C=C(q,\, r)>0$ is a constant depending on $q$ and $r$.

If we make the substitutions $k_{1}=q(x)-1$, $k_{2}=\lambda g(x)(r(x)-1)$, 
$q=q(x)-2$ and $r=r(x)-2$, then \eqref{f-u} becomes
\[
f_{u}(x,u)\leq C(q(x)-1)\Big(\frac{q(x)-1}{r(x)-1}\Big)^{\frac{q(x)-2}{r(x)-q(x)}}
(\lambda g(x))^{\frac{q(x)-2}{q(x)-r(x)}}.
\]
As a result of the fact that 
$C(q(x)-1)\big(\frac{q(x)-1}{r(x)-1}\big)^{\frac{q(x)-2}{r(x)-q(x)}}$ 
is a bounded expression, we arrive at
\begin{equation}\label{f-u C1}
f_{u}(x,u)\leq C_{1}(\lambda g(x))^{\frac{q(x)-2}{q(x)-r(x)}},
\end{equation}
where $C_{1}$ is a positive constant. Also, the equalities
\begin{align*}
\int_0^{s}f_{u}(x,u_0+t(u_{n}-u_0))dt
&=\frac{f(x,u_0+s(u_{n}-u_0))-f(x,u_0)}{u_{n}-u_0}\\
&=\frac{F_{u}(x,u_0+s(u_{n}-u_0))-F_{u}(x,u_0)}{u_{n}-u_0}
\end{align*}
hold. Integrating over $[0,1]$ it results that
\begin{align*}
&\int_0^{1}\int_0^{s}f_{u}(x,u_0+t(u_{n}-u_0))dt\, ds\\
&=\frac{\int_0^{1}\big[F_{u}(x,u_0+s(u_{n}-u_0))-F_{u}(x,u_0)\big]ds}{u_{n}-u_0}\\
&=\frac{F(x,u_{n})-F(x,u_0)}{(u_{n}-u_0)^{2}}-\frac{f(x,u_0)}{u_{n}-u_0},
\end{align*}
which can be also written in the equivalent form
\begin{equation}\label{F}
\begin{aligned}
F(x,u_{n})-F(x,u_0)
&=(u_{n}-u_0)^{2}\int_0^{1}\int_0^{s}f_{u}(x,u_0+t(u_{n}-u_0))dt\, ds\\
&\quad +(u_{n}-u_0)f(x,u_0).
\end{aligned}
\end{equation}
Taking into account \eqref{f-u C1}, \eqref{F} and using the definition of 
$T_{\lambda}^{1}$ it follows that
\begin{equation} \label{T lambda 1}
\begin{aligned}
&T_{\lambda}^{1}(u_0)-T_{\lambda}^{1}(u_{n})\\
&=\int_{\Omega}\sum_{i=1}^{N}\Big(\frac{|\partial_{x_{i}}u_0|^{p_{i}(x)}}{p_{i}(x)}
-\frac{|u_0|^{p_{i}(x)}}{p_{i}(x)}\Big)dx
-\int_{\Omega}\sum_{i=1}^{N}\Big(\frac{|\partial_{x_{i}}u_{n}|^{p_{i}(x)}}{p_{i}(x)}
-\frac{|u_{n}|^{p_{i}(x)}}{p_{i}(x)}\Big)dx\\
&\quad +\int_{\Omega}[F(x,u_{n})-F(x,u_0)]dx\\
&\leq\int_{\Omega}\sum_{i=1}^{N}\Big(\frac{|\partial_{x_{i}}u_0|^{p_{i}(x)}}{p_{i}(x)}
-\frac{|u_0|^{p_{i}(x)}}{p_{i}(x)}\Big)dx
-\int_{\Omega}\sum_{i=1}^{N}\Big(\frac{|\partial_{x_{i}}u_{n}|^{p_{i}(x)}}{p_{i}(x)}
-\frac{|u_{n}|^{p_{i}(x)}}{p_{i}(x)}\Big)dx\\
&\quad +\int_{\Omega}(u_{n}-u_0)^{2}\int_0^{1}
 \int_0^{s}f_{u}(x,u_0+t(u_{n}-u_0))dt\, ds\, dx\\
&\quad +\int_{\Omega}(u_{n}-u_0)f(x,u_0)dx\\
&\leq\int_{\Omega}\sum_{i=1}^{N}
 \Big(\frac{|\partial_{x_{i}}u_0|^{p_{i}(x)}}{p_{i}(x)}
 -\frac{|u_0|^{p_{i}(x)}}{p_{i}(x)}\Big)dx
 -\int_{\Omega}\sum_{i=1}^{N}\Big(\frac{|\partial_{x_{i}}u_{n}|^{p_{i}(x)}}{p_{i}(x)}
 -\frac{|u_{n}|^{p_{i}(x)}}{p_{i}(x)}\Big)dx\\
&\quad +C_{2}\int_{\Omega}(u_{n}-u_0)^{2}(\lambda g(x))^{\frac{q(x)-2}{q(x)-r(x)}}dx
+\int_{\Omega}(u_{n}-u_0)f(x,u_0)dx,
\end{aligned}
\end{equation}
where $C_{2}$ is a positive constant. We intend to prove that the last two
integrals converge to $0$ as $n\to\infty$.

Relying on \cite[Theorem 1]{m1} we find that $E$ is compactly embedded in
 $L^{q(\cdot)}(\Omega)$, and since $u_{n}\rightharpoonup u_0$ in $E$ 
we obtain $u_{n}\to u_0$ in $L^{q(\cdot)}(\Omega)$. This implies 
\[
\int_{\Omega}|u_{n}-u_0|^{q(x)}dx\to 0,
\]
yielding
$(u_{n}-u_0)^{2}\in L^{\frac{q(\cdot)}{2}}(\Omega)$.
Based on H\"{o}lder type inequality \eqref{Holder} and the hypothesis 
\eqref{g infty} we derive that
\[
\int_{\Omega}(u_{n}-u_0)^{2}\cdot(\lambda g(x))^{\frac{q(x)-2}{q(x)-r(x)}}dx
\leq2\Big|(\lambda g(x))^{\frac{q(x)-2}{q(x)-r(x)}}
\Big|_{\frac{q(\cdot)}{q(\cdot)-2}}\big|(u_{n}-u_0)^{2}\big|_{\frac{q(\cdot)}{2}}.
\]
On the other hand, 
\[
\rho_{\frac{q(\cdot)}{2}}\big((u_{n}-u_0)^{2}\big)
=\int_{\Omega}\big|(u_{n}-u_0)^{2}\big|^{\frac{q(x)}{2}}dx=\int_{\Omega}|u_{n}-u_0|^{q(x)}dx\to0.
\]
Thereupon, relation \eqref{conv 0} implies
$\big|(u_{n}-u_0)^{2}\big|_{\frac{q(\cdot)}{2}}\to 0$,
and for this reason we obtain
\begin{equation}\label{lim1}
\int_{\Omega}(u_{n}-u_0)^{2}\cdot(\lambda g(x))^{\frac{q(x)-2}{q(x)-r(x)}}dx\to0.
\end{equation}

Next, we define $\Theta:E\to\mathbb{R}$ by
$$
\Theta(v)=\int_{\Omega}f(x,u_0)v\, dx.
$$
In the first instance, it is clear that $\Theta$ is linear. On the other hand,
\begin{equation} \label{Theta}
\begin{aligned}
|\Theta(v)|&\leq\int_{\Omega}|f(x,u_0)v|dx\\
&=\int_{\Omega}\big||u_0|^{q(x)-2}u_0 -\lambda g(x)|u_0|^{r(x)-2}u_0\big|\, |v|dx\\
&\leq\int_{\Omega}|u_0|^{q(x)-1}|v|dx+\lambda\int_{\Omega}g(x)|u_0|^{r(x)-1}|v|dx.
\end{aligned}
\end{equation}
In accordance with the H\"{o}lder type inequality \eqref{Holder} we obtain
\[
\int_{\Omega}|u_0|^{q(x)-1}|v|dx\leq2\big||u_0|^{q(x)-1}
\big|_{\frac{q(\cdot)}{q(\cdot)-1}}|v|_{q(\cdot)}.
\]
We know that the embedding $E\hookrightarrow L^{q(\cdot)}(\Omega)$ is continuous;
that is, there is a positive constant $C$ such that
\[
|v|_{q(\cdot)}\leq C\|v\|_{\vec{p}(\cdot)}\ \ \ \forall v\in E.
\]
The last two inequalities lead us to
\[
\int_{\Omega}|u_0|^{q(x)-1}|v|dx\leq C_{1}\|v\|_{\vec{p}(\cdot)},
\]
where $C_{1}>0$ is a constant. Also, reasoning as above we have
\begin{align*}
\int_{\Omega}g(x)|u_0|^{r(x)-1}|v|dx
&\leq|g|_{\infty}\int_{\Omega}|u_0|^{r(x)-1}|v|dx\\
&\leq 2|g|_{\infty}\big||u_0|^{r(x)-1}\big|_{\frac{r(\cdot)}{r(\cdot)-1}}|v|_{r(\cdot)}
\leq C_{2}\|v\|_{\vec{p}(\cdot)},
\end{align*}
where $C_{2}>0$ is a constant.

In light of the above, \eqref{Theta} becomes
\[
|\Theta(v)|\leq\overline{C}\|v\|_{\vec{p}(\cdot)}\quad \forall v\in E
\]
(where $\overline{C}>0$ is a constant); that is to say, $\Theta$ is continuous.
 Accordingly, we conclude that
$\Theta(u_{n})\to\Theta(u_0)$,
and therefore
\begin{equation}\label{lim2}
\int_{\Omega}f(x,u_0)(u_{n}-u_0)dx\to0.
\end{equation}

To complete the proof of lemma, we must prove that the functional 
$\Xi_{1}:E\to\mathbb{R}$,
\[
\Xi_{1}(u)=\int_{\Omega}\sum_{i=1}^{N}\frac{|u|^{p_{i}(x)}}{p_{i}(x)}dx
\]
is convex. Considering that the function
$[0,\infty)\ni t\mapsto t^{\gamma}$
is convex for each $\gamma>1$,  for any $x\in\Omega$ fixed we can say that
\begin{equation}\label{conv}
\Big|\frac{\alpha+\beta}{2}\Big|^{p_{i}(x)}\leq\Big|\frac{|\alpha|
+|\beta|}{2}\Big|^{p_{i}(x)}
\leq\frac{1}{2}|\alpha|^{p_{i}(x)}+\frac{1}{2}|\beta|^{p_{i}(x)}
\end{equation}
for all $\alpha, \beta\in\mathbb{R}$ and all $i\in\{1,\dots,N\}$.
If we take $\alpha=u$ and $\beta=v$ in \eqref{conv}, multiply by 
$1/p_{i}(x)$, sum from $1$ to $N$ and intergate over $\Omega$, we obtain 
$$
\Xi_{1}\Big(\frac{u+v}{2}\Big)\leq\frac{1}{2}\Xi_{1}(u)+\frac{1}{2}\Xi_{1}(v)
\quad \forall u,v\in E.
$$

In the same manner we can prove that the functional $\Xi_{2}:E\to\mathbb{R}$ 
defined by
$$
\Xi_{2}(u)=\int_{\Omega}\sum_{i=1}^{N}
\frac{|\partial_{x_{i}}u|^{p_{i}(x)}}{p_{i}(x)}dx
$$
is convex. Thereby $\Xi_{1}+\Xi_{2}$ is convex on $E$. Next, we propose 
to show that the functional $\Xi_{1}+\Xi_{2}$ is weakly lower semicontinuous 
on $E$. Making use of Corollary III.8 in \cite{b1} we ascertain that is
enough to demonstrate the lower semicontinuity of  $\Xi_{1}+\Xi_{2}$. 
Therefor, we fix $u\in E$ and $\varepsilon>0$. Let $v\in E$ be arbitrary. 
By convexity of $\Xi_{1}+\Xi_{2}$ and H\"{o}lder type inequality \eqref{Holder} 
we have
\begin{align*}
&\Xi_{1}(v)+\Xi_{2}(v)\\
&\geq\Xi_{1}(u)+\Xi_{2}(u)+\langle\Xi_{1}'(u)+\Xi_{2}'(u),v-u\rangle \\
&=\Xi_{1}(u)+\Xi_{2}(u)
 +\int_{\Omega}\sum_{i=1}^{N}|\partial_{x_{i}}u|^{p_{i}(x)-2}
 \partial_{x_{i}}u\partial_{x_{i}}(v-u)dx \\
&\quad +\int_{\Omega}\sum_{i=1}^{N}|u|^{p_{i}(x)-2}u(v-u)dx \\
&\geq\Xi_{1}(u)+\Xi_{2}(u)-\int_{\Omega}\sum_{i=1}^{N}|
 \partial_{x_{i}}u|^{p_{i}(x)-1}|\partial_{x_{i}}(v-u)|dx
 -\int_{\Omega}\sum_{i=1}^{N}|u|^{p_{i}(x)-1}|v-u|dx \\
&\geq\Xi_{1}(u)+\Xi_{2}(u)
 -2\Big(\sum_{i=1}^{N}\big||\partial_{x_{i}}u|^{p_{i}(x)-1}\big|
 _{\frac{p_{i}(\cdot)}{p_{i}(\cdot)-1}}|\partial_{x_{i}}(v-u)|_{p_{i}(\cdot)}\\
&\quad +\sum_{i=1}^{N} \big||u|^{p_{i}(x)-1}\big|_{\frac{p_{i}(\cdot)}{p_{i}(\cdot)-1}}
 |v-u|_{p_{i}(\cdot)}\Big) \\
&=\Xi_{1}(u)+\Xi_{2}(u)
 -2\sum_{i=1}^{N}\Big(\big||\partial_{x_{i}}u|^{p_{i}(x)-1}\big|
_{\frac{p_{i}(\cdot)}{p_{i}(\cdot)-1}}|\partial_{x_{i}}(v-u)|_{p_{i}(\cdot)} \\
&\quad +\big||u|^{p_{i}(x)-1}\big|_{\frac{p_{i}(\cdot)}{p_{i}(\cdot)-1}}|v-u|_{p_{i}(\cdot)}
 \Big) \\
&\geq\Xi_{1}(u)+\Xi_{2}(u)
-2\sum_{i=1}^{N}\Big(\big||\partial_{x_{i}}u|^{p_{i}(x)-1}
 \big|_{\frac{p_{i}(\cdot)}{p_{i}(\cdot)-1}}\\
&\quad +\big||u|^{p_{i}(x)-1}\big|_{\frac{p_{i}(\cdot)}{p_{i}(\cdot)-1}}\Big)
\big(|\partial_{x_{i}}(v-u)|_{p_{i}(\cdot)}+|v-u|_{p_{i}(\cdot)}\big) \\
&\geq\Xi_{1}(u)+\Xi_{2}(u)-C\sum_{i=1}^{N}\big(|\partial_{x_{i}}
 (v-u)|_{p_{i}(\cdot)}+|v-u|_{p_{i}(\cdot)}\big) \\
&=\Xi_{1}(u)+\Xi_{2}(u)-C\|v-u\|_{\vec{p}(\cdot)}
\end{align*}
for all $v\in E$ with $\|v-u\|_{\vec{p}(\cdot)}\leq\varepsilon/C$, where $C>0$ 
is a constant, whence we obtain
the weakly lower semicontinuity of $\Xi_{1}+\Xi_{2}$ on $E$; that is,
\begin{equation}\label{xi}
\liminf_{n\to\infty}(\Xi_{1}+\Xi_{2})(u_{n})\geq(\Xi_{1}+\Xi_{2})(u_0).
\end{equation}
Passing to the limit in \eqref{T lambda 1} and making use of 
\eqref{lim1}, \eqref{lim2} and \eqref{xi} it follows that
$$
\liminf_{n\to\infty}T_{\lambda}^{1}(u_{n})\geq T_{\lambda}^{1}(u_0)
$$
meaning that Lemma \ref{semicontinuous} holds.
\end{proof}

Then on the basis of these three lemmas above mentioned, we are going to
 show what we have proposed to Step 3. We fix $\lambda\in(\lambda_{1},\infty)$. 
In the light of coercivity and weakly lower semicontinuity of $T_{\lambda}^{1}$ 
we can use \cite[Theorem 1.2]{s3} to obtain the existence of a global minimum
point of $T_{\lambda}^{1}$, $u_{\lambda}\in E$. Ultimately, to complete 
Step 3 we have to show only that $u_{\lambda}$ is not trivial. In truth, we 
have $\lambda_{1}=\inf_{u\in E\setminus\{0\}}\frac{J_{1}(u)}{I_{1}(u)}$
and $\lambda_{1}<\lambda$ whence we obtain that there is a $v_{\lambda}\in E$ 
so that $T_{\lambda}^{1}(v_{\lambda})<0$. Thus
\[
\inf_{E}T_{\lambda}^{1}<0,
\]
and so we can conclude that $u_{\lambda}$ is a nontrivial critical point 
of $T_{\lambda}^{1}$ or, in other words, $\lambda$ is an eigenvalue of 
problem \eqref{pb} leading to Step 3 is verified.
\smallskip

\noindent\textbf{Step 4.} 
In this last step we show that $\lambda_{1}$ is an eigenvalue of problem \eqref{pb}.
First of all we prove two lemmas.

\begin{lemma}\label{lemlem}
We have that
$$
\lim_{\|u\|_{\vec{p}(\cdot)}\to0}\frac{J_0(u)}{I_0(u)}=+\infty.
$$
\end{lemma}

\begin{proof}
We fix $s\in\mathbb{R}$ such that
$$
r^{+}<s<q^{-}<P_{-}^{*}.
$$
It should be noticed that from the condition \eqref{p,r,q}, we have 
$P_{-,\infty}=P_{-}^{*}$. Thereby $s<P_{-,\infty}$ and so 
$E\hookrightarrow L^{s}(\Omega)$ continuously, whence we obtain the existence 
of a positive constant $C$ such that
\begin{equation}\label{cont}
|u|_{s}\leq C\|u\|_{\vec{p}(\cdot)}\quad \forall u\in E.
\end{equation}
Without loss of generality we consider that $\|u\|_{\vec{p}(\cdot)}<1$ 
for any $u\in E$. By applying the Jensen's inequality to the convex function 
$a:\mathbb{R}^{+}\to\mathbb{R}^{+}$, $a(t)=t^{P_{-}^{-}}$, $P_{-}^{-}\geq2$, 
using $\alpha_{i}$ and $\beta_{i}$ defined in Lema \ref{lim}, and 
by \eqref{g--s} and \eqref{cont} we infer that
\begin{align*}
\frac{J_0(u)}{I_0(u)}
&=\frac{\int_{\Omega}\sum_{i=1}^{N}\big(|\partial_{x_{i}}u|^{p_{i}(x)}
+|u|^{p_{i}(x)}\big)dx+\int_{\Omega}|u|^{q(x)}dx}{\int_{\Omega}g(x)|u|^{r(x)}dx}\\
&\geq\frac{\frac{\|u\|_{\vec{p}(\cdot)}^{P_{-}^{-}}}{(2N)^{P_{-}^{-}-1}}-2N}
{|g|_{\frac{s}{s-r^{-}}}|u|_{s}^{r^{-}}+|g|_{\frac{s}{s-r^{+}}}|u|_{s}^{r^{+}}}\\
&\geq\frac{\|u\|_{\vec{p}(\cdot)}^{P_{-}^{-}}-(2N)^{P_{-}^{-}}}{(2N)^{P_{-}^{-}-1}
\big(|g|_{\frac{s}{s-r^{-}}}C^{r^{-}}\|u\|_{\vec{p}(\cdot)}^{r^{-}}
+|g|_{\frac{s}{s-r^{+}}}C^{r^{+}}\|u\|_{\vec{p}(\cdot)}^{r^{+}}\big)}.
\end{align*}
Given that $r^{+}\geq r^{-}>P_{-}^{-}$ and passing to the limit in the above 
inequality it is obvious that 
$\lim_{\|u\|_{\vec{p}(\cdot)}\to\infty}\frac{J_0(u)}{I_0(u)}=+\infty$ occurs, 
and so the Lemma \ref{lemlem} is proved.
\end{proof}

\begin{lemma}\label{lem'}
Suppose that $(u_{n})$ converges weakly to $u$ in $E$. Then we have
\begin{equation}\label{I1}
\lim_{n\to\infty}\langle I_{1}'(u_{n}),u_{n}-u\rangle=0.
\end{equation}
\end{lemma}

\begin{proof}
We define $\Phi:E\to\mathbb{R}$ by
\[
\Phi(v)=\int_{\Omega}g(x)|u_{n}|^{r(x)-2}u_{n}v\, dx.
\]
Is easily seen that $\Phi$ is linear and we want to show that is also continuous. 
Indeed, by H\"{o}lder type inequality \eqref{Holder} we have
\begin{equation} \label{Phi1}
\begin{aligned}
|\Phi(v)|&=\big|\int_{\Omega}g(x)|u_{n}|^{r(x)-2}u_{n}v\, dx\big|
\leq\int_{\Omega}\left|g(x)|u_{n}|^{r(x)-2}u_{n}v\right|dx\\
&=\int_{\Omega}g(x)|u_{n}|^{r(x)-1}|v|dx
\leq|g|_{\infty}\int_{\Omega}|u_{n}|^{r(x)-1}|v|dx\\
&\leq2|g|_{\infty}\big||u_{n}|^{r(x)-1}\big|_{\frac{r(\cdot)}{r(\cdot)-1}}
|v|_{r(\cdot)}.
\end{aligned}
\end{equation}
We have $E\hookrightarrow L^{r(\cdot)}(\Omega)$ continuously, thus there
exists a constant $C>0$ such that
\[
|v|_{r(\cdot)}\leq C\|v\|_{\vec{p}(\cdot)}\quad \forall v\in E.
\]
By the above inequality and \eqref{Phi1} we obtain the continuity of $\Phi$.
Then $\Phi(u_{n})\to\Phi(u)$, or
\[
\lim_{n\to\infty}\int_{\Omega}g(x)|u_{n}|^{r(x)-2}u_{n}(u_{n}-u)dx=0
\]
which is exactly \eqref{I1}.
\end{proof}

Now, we return to the proof of Step 4. Let $\lambda_{n}\searrow\lambda_{1}$.  
Considering the Step 3 we infer that for any $n$ there exists 
$u_{n}\in E\setminus\{0\}$ so that
\begin{equation}\label{J1I1}
\langle J_{1}'(u_{n}),v\rangle=\lambda_{n}\cdot\langle I_{1}'(u_{n}),v\rangle\quad
 \forall v\in E.
\end{equation}
Making the substitution $v=u_{n}$ in \eqref{J1I1} we obtain
\begin{equation}\label{JI}
J_0(u_{n})=\lambda_{n}\cdot I_0(u_{n}),
\end{equation}
and passing to the limit as $n\to\infty$ we find that
\begin{equation}\label{0}
\lim_{n\to\infty}(J_0(u_{n})-\lambda_{n}\cdot I_0(u_{n}))=0.
\end{equation}
Now, if we suppose that $\|u_{n}\|_{\vec{p}(\cdot)}\to\infty$, 
then reasoning as in the proof of Lemma \ref{lim} we reach a contradiction 
with \eqref{0}. Hence, the sequence $(u_{n})$ is bounded in $E$. 
On the other hand, we know that $E$ is a reflexive Banach space, and due 
to this reason we deduce that there is an element $u\in E$ so that, 
up to a subsequence, labeled again $(u_{n})$, we have that 
$u_{n}\rightharpoonup u$ in $E$. Therefore, \eqref{I1} occurs.

To proceed we use the inequality
\begin{equation}\label{replace}
\left(|\xi_{i}|^{r_{i}-2}\xi_{i}-|\psi_{i}|^{r_{i}-2}\psi_{i}\right)
\left(\xi_{i}-\psi_{i}\right)
\geq2^{-r_{i}}|\xi_{i}-\psi_{i}|^{r_{i}}\quad
 \forall \xi_{i}, \psi_{i}\in\mathbb{R},\; \forall r_{i}\geq2
\end{equation}
(see \cite[inequality (2.2)]{s2}).
Replacing in the above inequality $\xi_{i}$ by $\partial_{x_{i}}u_{n}$, 
$\psi_{i}$ by $\partial_{x_{i}}u$ and $r_{i}$ by $p_{i}(x)$ , and  then 
$\xi_{i}$ by $u_{n}$,
$\psi_{i}$ by $u$ and $r_{i}$ by $p_{i}(x)$ respectively, for each
$i\in\{1, \dots, N\}$ and $x\in\Omega$, then adding the two inequalities 
obtained, and taking into account that $2^{p_{i}(x)}$ is bounded, it results 
that there exists $L_{1}>0$ such that
\begin{equation} \label{L1}
\begin{aligned}
&L_{1}\int_{\Omega}\left(|\partial_{x_{i}}u_{n}
-\partial_{x_{i}}u|^{p_{i}(x)}+|u_{n}-u|^{p_{i}(x)}\right)dx\\
&\leq\int_{\Omega}\left(|\partial_{x_{i}}u_{n}|^{p_{i}(x)-2}\partial_{x_{i}}u_{n}
-|\partial_{x_{i}}u|^{p_{i}(x)-2}\partial_{x_{i}}u\right)
(\partial_{x_{i}}u_{n}-\partial_{x_{i}}u)dx\\
&\quad +\int_{\Omega}\left(|u_{n}|^{p_{i}(x)-2}u_{n}-|u|^{p_{i}(x)-2}u\right)
(u_{n}-u)dx\quad \forall i\in\{1, \dots, N\}.
\end{aligned}
\end{equation}
Also, using again inequality \eqref{replace}, we find that there is
$L_{2}>0$ such that
\begin{equation}\label{L2}
L_{2}\int_{\Omega}|u_{n}-u|^{q(x)}dx\leq\int_{\Omega}
\left(|u_{n}|^{q(x)-2}u_{n}-|u|^{q(x)-2}u\right)(u_{n}-u)dx.
\end{equation}
Summing from $1$ to $N$ in \eqref{L1} and adding the inequality which we
obtain with \eqref{L2} we can see that
\begin{align*}
&L_{1}\int_{\Omega} \sum_{i=1}^{N}\left(|\partial_{x_{i}}u_{n}-\partial_{x_{i}}u|^{p_{i}(x)}
+|u_{n}-u|^{p_{i}(x)}\right)dx\\
&\leq\int_{\Omega}\sum_{i=1}^{N}\left(|\partial_{x_{i}}u_{n}|^{p_{i}(x)-2}\partial_{x_{i}}u_{n}
-|\partial_{x_{i}}u|^{p_{i}(x)-2}\partial_{x_{i}}u\right)(\partial_{x_{i}}u_{n}-\partial_{x_{i}}u)dx\\
&\quad +\int_{\Omega}\sum_{i=1}^{N}\left(|u_{n}|^{p_{i}(x)-2}u_{n}-|u|^{p_{i}(x)-2}u\right)(u_{n}-u)dx\\
&\quad +\int_{\Omega}\left(|u_{n}|^{q(x)-2}u_{n}-|u|^{q(x)-2}u\right)(u_{n}-u)dx \\
&=\langle J_{1}'(u_{n})-J_{1}'(u),u_{n}-u\rangle.
\end{align*}
Taking into account \eqref{I1} and \eqref{J1I1} and that $(u_{n})$ converges
weakly to $u$ in $E$, we arrive at
\begin{align*}
&L_{1}\int_{\Omega}\sum_{i=1}^{N}\left(|\partial_{x_{i}}u_{n}
-\partial_{x_{i}}u|^{p_{i}(x)}+|u_{n}-u|^{p_{i}(x)}\right)dx\\
&\leq\langle J_{1}'(u_{n})-J_{1}'(u),u_{n}-u\rangle\\
&=\langle J_{1}'(u_{n}),u_{n}-u\rangle-\langle J_{1}'(u),u_{n}-u\rangle\\
&\leq|\langle J_{1}'(u_{n}),u_{n}-u\rangle|+|\langle J_{1}'(u),u_{n}-u\rangle|\\
&=\lambda_{n}|\langle I_{1}'(u_{n}),u_{n}-u\rangle|+|\langle J_{1}'(u),u_{n}-u\rangle|\to0,
\end{align*}
as $n\to\infty$.
By \eqref{conv 0} we deduce that
\[
\sum_{i=1}^{N}\left(|\partial_{x_{i}}u_{n}
-\partial_{x_{i}}u|_{p_{i}(\cdot)}+|u_{n}-u|_{p_{i}(\cdot)}\right)
\to0
\]
or equivalently
\[
\|u_{n}-u\|_{p_{i}(\cdot)}\to0,
\]
that is,
$u_{n}\to u$  in $E$.
Passing to the limit, as $n\to\infty$ in \eqref{J1I1}, yields
\[
\langle {T_{\lambda_{1}}^{1}}'(u),v\rangle=0\quad \forall v\in E,
\]
which means that $u$ is a critical point for $T_{\lambda_{1}}^{1}$.
We intend to show that $u\neq0$ and this fact would lead us to $\lambda_{1}$
is an eigenvalue for \eqref{pb}. To this end we suppose that $u=0$.
Then $u_{n}\to0$ in $E$, that is to say, $\|u_{n}\|_{\vec{p}(\cdot)}\to0$.
Applying Lemma \ref{lemlem} we obtain
\begin{equation}\label{contradiction}
\lim_{\|u_{n}\|_{\vec{p}(\cdot)}\to0}\frac{J_0(u_{n})}{I_0(u_{n})}=+\infty.
\end{equation}
But, if we pass to the limit as $n\to\infty$ in \eqref{JI} we obtain
\[
\lim_{n\to\infty}\frac{J_0(u_{n})}{I_0(u_{n})}=\lambda_{1},
\]
which is a contradiction to \eqref{contradiction}.
So the assumption made is false, accordingly, $u\neq0$ and thus $\lambda_{1}$
is an eigenvalue for problem \eqref{pb} and Step 4 is verified.

From Steps 2--4 we obtain $\lambda_0\leq\lambda_{1}$ and thereby 
the proof of  Theorem \ref{thm1} is complete.


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\end{document}