\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 141, pp. 1--??.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2010 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2010/141\hfil Concentration-compactness principle] {Concentration-compactness principle for variable exponent spaces and applications} \author[J. Fern\'andez Bonder, A. Silva \hfil EJDE-2010/141\hfilneg] {Juli\'an Fern\'andez Bonder, Anal\'ia Silva} % in alphabetical order \address{Juli\'an Fern\'andez Bonder \newline Departamento de Matem\'atica, FCEyN, Universidad de Buenos Aires, Pabell\'on I, Ciudad Universitaria (1428), Buenos Aires, Argentina} \email{jfbonder@dm.uba.ar, http://mate.dm.uba.ar/$\sim$jfbonder} \address{Anal\'{\i}a Silva \newline Departamento de Matem\'atica, FCEyN, Universidad de Buenos Aires, Pabell\'on I, Ciudad Universitaria (1428), Buenos Aires, Argentina} \email{asilva@dm.uba.ar} \thanks{Submitted August 8, 2009. Published October 5, 2010.} \subjclass[2000]{35J20, 35J60} \keywords{Concentration-compactness principle; variable exponent spaces} \begin{abstract} In this article, we extend the well-known concentration - compactness principle by Lions to the variable exponent case. We also give some applications to the existence problem for the $p(x)$-Laplacian with critical growth. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction} When dealing with nonlinear elliptic equations with critical growth (in the sense of the Sobolev embeddings) the concentration - compactness principle by Lions, see \cite{Lions}, have been proved to be a fundamental tool for proving existence of solutions. Just to cite a few references, we have \cite{Alves, Alves-Ding, Bahri-Lions, Drabek-Huang, FB, GAP} but there is an impressive list of references on this topic. Recently in the analysis of some new models, that are called electrorheological fluids, the following equation has been studied \begin{equation}\label{pdex} -\Delta_{p(x)} u = f(x,u) \quad \text{in }\Omega. \end{equation} The operator $\Delta_{p(x)}u := \operatorname{div}(|\nabla u|^{p(x)-2} \nabla u)$ is called the $p(x)$-Laplacian. When $p(x)\equiv p$ is the well-known $p$-Laplacian. In recent years a vast amount of literature that deal with the existence problem for \eqref{pdex} with different boundary conditions (Dirichlet, Neumann, nonlinear, etc) have appeared. See, for instance \cite{Cabada-Pouso, Dinu, FZ, Mihailescu, Mihailescu-Radulescu} and references therein. However, up to our knowledge, no results are available for \eqref{pdex} when the source term $f$ is allowed to have critical growth at infinity (see the remark after the introduction for more on this). That is, $$ |f(x,t)|\le C(1+|t|^{q(x)}) $$ with $q(x)\le p^*(x):= N p(x)/(N-p(x))$ (if $p(x)0\\ \mu \geq |\nabla u|^{p(x)} + \sum_{i\in I} \mu_i \delta_{x_i} \quad \mu_i>0\\ S \nu_i^{1/p^*(x_i)} \leq \mu_i^{1/p(x_i)} \quad \forall i\in I. \end{gather} where $\{x_i\}_{i\in I}\subset \mathcal{A}$ and $S$ is the best constant in the Gagliardo-Nirenberg-Sobolev inequality for variable exponents, namely $$ S = S_q(\Omega) :=\inf_{\phi\in C_0^{\infty}(\Omega)} \frac{\| |\nabla \phi| \|_{L^{p(x)}(\Omega)}}{\| \phi \|_{L^{q(x)}(\Omega)}}. $$ \end{theorem} We remark that in Theorem \ref{ccp} is not required the exponent $q(x)$ to be critical {\em everywhere} and that the point masses are located in the {\em criticality set} $\mathcal{A} = \{x\in \Omega\colon q(x)=p^*(x)\}$. Now, as an application of Theorem \ref{ccp}, following the techniques in \cite{GAP}, we prove the existence of solutions to \begin{equation}\label{gap} \begin{gathered} -\Delta_{p(x)} u = |u|^{q(x)-2}u + \lambda(x) |u|^{r(x)-2}u \quad \text{in }\Omega\\ u=0 \quad \text{on }\partial\Omega. \end{gathered} \end{equation} In the spirit of \cite{GAP}, we have two types of results, depending on $r(x)$ being smaller or bigger that $p(x)$. More precisely, we prove the following two theorems. \begin{theorem}\label{r0$ depending only on $p,q,r,N$ and $\Omega$ such that if $\lambda(x)$ verifies $0<\inf_{x\in\Omega}\lambda(x)\le \|\lambda\|_{L^{\infty}(\Omega)} <\lambda_1$, then there exists infinitely many solutions to \eqref{gap} in $W^{1,p(x)}_0(\Omega)$. \end{theorem} \begin{theorem}\label{r>p} Let $p(x)$ and $q(x)$ be as in Theorem \ref{ccp} and let $r(x)$ be continuous. Moreover, assume that $\max_{\overline{\Omega}} p < \min_{\overline{\Omega}} r$ and that there exists $\eta>0$ such that $r(x)\le p^*(x) - \eta$ in $\Omega$. Then, there exists $\lambda_0>0$ depending only on $p, q, r, N$ and $\Omega$, such that if $$ \inf_{x\in A_\delta}\lambda(x)>\lambda_0 \quad \text{ for some } \delta>0, $$ problem \eqref{gap} has at least one nontrivial solution in $W^{1,p(x)}_0(\Omega)$. Here, $\mathcal{A}_\delta$ is the $\delta$-tubular neighborhood of $\mathcal{A}$, namely $$ \mathcal{A}_\delta := \cup_{x\in\mathcal{A}} (B_\delta(x) \cap \Omega). $$ \end{theorem} \subsection*{Organization of this article} After finishing this introduction, in Section 2 we give a very short overview of some properties of variable exponent Sobolev spaces that will be used throughout the paper. In Section 3 we deal with the main result of the paper. Namely the proof of the concentration - compactness principle (Theorem \ref{ccp}). In Section 4, we begin analyzing problem \eqref{gap} and prove Theorem \ref{r>p}. Finally, in Section 5, we prove Theorem \ref{rp} our approach allows us to consider $\lambda(x)$ not necessarily a constant and the restriction that $\lambda$ is large is only needed in an $L^\infty$-norm in the criticality set. We believe that these improvements are significant and made our result more flexible that those in \cite{Fu}. \section{Results on variable exponent Sobolev spaces} The variable exponent Lebesgue space $L^{p(x)}(\Omega)$ is defined as $$ L^{p(x)}(\Omega) = \{u\in L^1_{\text{loc}}(\Omega) \colon \int_\Omega|u(x)|^{p(x)}\,dx<\infty\}. $$ This space is endowed with the norm $$ \|u\|_{L^{p(x)}(\Omega)}=\inf\{\lambda>0: \int_\Omega|\frac{u(x)}{\lambda}|^{p(x)}\,dx\leq 1\} $$ The variable exponent Sobolev space $W^{1,p(x)}(\Omega)$ is defined as $$ W^{1,p(x)}(\Omega) = \{u\in W^{1,1}_{\text{loc}}(\Omega) \colon u\in L^{p(x)}(\Omega) \text{ and } |\nabla u |\in L^{p(x)}(\Omega)\}. $$ The corresponding norm for this space is $$ \|u\|_{W^{1,p(x)}(\Omega)}=\|u\|_{L^{p(x)}(\Omega)} +\| |\nabla u| \|_{L^{p(x)}(\Omega)} $$ Define $W^{1,p(x)}_0(\Omega)$ as the closure of $C_0^\infty(\Omega)$ with respect to the $W^{1,p(x)}(\Omega)$ norm. The spaces $L^{p(x)}(\Omega)$, $W^{1,p(x)}(\Omega)$ and $W^{1,p(x)}_0(\Omega)$ are separable and reflexive Banach spaces when $1<\inf_\Omega p \le \sup_\Omega p <\infty$. As usual, we denote $p'(x) = p(x)/(p(x)-1)$ the conjugate exponent of $p(x)$. Define $$ p^*(x)=\begin{cases} \frac{Np(x)}{N-p(x)} & \text{if } p(x)0$ then, the embedding is compact. \end{proposition} \begin{proposition}[Poincar\'e inequality]\label{Poincare} There is a constant $C>0$, such that $$ \|u\|_{L^{p(x)}(\Omega)}\leq C\| |\nabla u| \|_{L^{p(x)}(\Omega)}, $$ for all $u\in W^{1,p(x)}_0(\Omega)$. \end{proposition} \begin{remark} \rm By Proposition \ref{Poincare}, we know that $\| |\nabla u| \|_{L^{p(x)}(\Omega)}$ and $\|u\|_{W^{1,p(x)}(\Omega)}$ are equivalent norms on $W_0^{1,p(x)}(\Omega)$. \end{remark} In this article, the following notation will be used: Given $q\colon \Omega\to\mathbb{R}$ bounded, we denote $$ q^+ := \sup_\Omega q(x), \quad q^- := \inf_\Omega q(x). $$ The following proposition is also proved in \cite{Fan} and it will be very useful here. \begin{proposition}\label{norma.y.rho} Set $\rho(u):=\int_\Omega|u(x)|^{p(x)}\,dx$. For $u,\in L^{p(x)}(\Omega)$ and $\{u_k\}_{k\in\mathbb{N}}\subset L^{p(x)}(\Omega)$, we have \begin{gather} u\neq 0 \Rightarrow \Big(\|u\|_{L^{p(x)}(\Omega)} = \lambda \Leftrightarrow \rho(\frac{u}{\lambda})=1\Big).\\ \|u\|_{L^{p(x)}(\Omega)}<1 (=1; >1) \Leftrightarrow \rho(u)<1(=1;>1).\\ \|u\|_{L^{p(x)}(\Omega)}>1 \Rightarrow \|u\|^{p^-}_{L^{p(x)}(\Omega)} \leq \rho(u) \leq \|u\|^{p^+}_{L^{p(x)}(\Omega)}.\\ \|u\|_{L^{p(x)}(\Omega)}<1 \Rightarrow \|u\|^{p^+}_{L^{p(x)}(\Omega)} \leq \rho(u) \leq \|u\|^{p^-}_{L^{p(x)}(\Omega)}.\\ \lim_{k\to\infty}\|u_k\|_{L^{p(x)}(\Omega)} = 0 \Leftrightarrow \lim_{k\to\infty}\rho(u_k)=0.\\ \lim_{k\to\infty}\|u_k\|_{L^{p(x)}(\Omega)} = \infty \Leftrightarrow \lim_{k\to\infty}\rho(u_k) = \infty. \end{gather} \end{proposition} \section{concentration compactness principle} Let $\{u_j\}_{j\in\mathbb{N}}$ be a bounded sequence in $W_0^{1,p(x)}(\Omega)$ and let $q\in C(\overline{\Omega})$ be such that $q\le p^*$ with $\{x\in\Omega\colon q(x)=p^*(x)\} \neq \emptyset$. Then there exists a subsequence, still denoted by $\{u_j\}_{j\in\mathbb{N}}$, such that \begin{itemize} \item $u_j\rightharpoonup u\quad \text{ weakly in } W_0^{1,p(x)}(\Omega)$, \item $u_j\to u \quad \text{ strongly in } L^{r(x)}(\Omega)\quad\forall1\leq r(x)0$ such that $$ \|\phi\|_{L_\nu^{r(x)}(\Omega)}\leq C\|\phi\|_{L_\mu^{p(x)}(\Omega)} $$ Then, there exist $\{x_j\}_{j\in J}\subset\overline{\Omega}$ and $\{\nu_j\}_{j\in J}\subset (0,\infty)$, such that $$ \nu=\Sigma\nu_i\delta_{x_i} $$ \end{lemma} For the proof of the lemma above, we need a couple of preliminary results. \begin{lemma}\label{Lema 2} Let $\nu$ be a non-negative bounded measure. Assume that there exists $\delta>0$ such that for all $A$ Borelian, $\nu(A)=0$ or $\nu(A)\geq\delta$. Then, there exist $\{x_i\}$ and $\nu_i>0$ such that $$ \nu=\sum \nu_i\delta_{x_i} $$ \end{lemma} The proof of the above lemma is elementary and is omitted. \begin{lemma}\label{Lema 3} Let $\nu$ be non-negative and bounded measures, such that $$ \|\psi\|_{L_\nu^{r(x)}(\Omega)}\leq C\|\psi\|_{L_\nu^{p(x)}(\Omega)} $$ Then there exist $\delta>0$ such that for all $A$ Borelian, $\nu(A)=0$ or $\nu(A)\geq\delta$. \end{lemma} \begin{proof} First, observe that if $\nu(A)\geq1$, $$ \int_\Omega\Big(\frac{\chi_{A}(x)}{\nu(A)^\frac{1}{p-}}\Big)^{p(x)}\, d\nu\leq \int_\Omega \Big(\frac{\chi_{A}(x)} {\nu(A)^\frac{1}{p(x)}}\Big)^{p(x)}\, d\nu = 1. $$ Then $\nu(A)^\frac{1}{p-}\geq\|\chi_{A}\|_{L_\nu^{p(x)}}$. On the other hand, $$ \int_\Omega \Big(\frac{\chi_{A}(x)} {\nu(A)^\frac{1}{r+}}\Big)^{r(x)}\, d\nu \geq \int_\Omega\frac{\chi_A(x)}{\nu(A)}\, d\nu = 1. $$ Then $\nu(A)^\frac{1}{r+}\leq\|\chi_{A}\|_{L_\nu^{r(x)}}$. So we conclude that $$ \nu(A)^\frac{1}{r+}\leq C\nu(A)^\frac{1}{p-}. $$ Now, if $\nu(A)<1$, we obtain $$ \nu(A)^\frac{1}{r-}\leq C\nu(A)^\frac{1}{p+}. $$ Combining all these facts, we arrive at $$ \min\{\nu(A)^\frac{1}{r-},\nu(A)^\frac{1}{r+}\}\leq C \max\{\nu(A)^\frac{1}{p-}, \nu(A)^\frac{1}{p+}\}. $$ Now, if $\nu(A)\leq1$, we have $$ \nu(A)^\frac{1}{r-}\leq C\nu(A)^\frac{1}{p+}. $$ Then, $\nu(A)=0$ or $$ \nu(A)\geq(\frac{1}{C})^{\frac{p^+r^-}{r^- -p^+}}. $$ Finally, $$ \nu(A)\geq\min\{(\frac{1}{C})^{\frac{p^+r^-}{r^- -p^+}},1\} $$ This completes the proof. \end{proof} In the rest of the proofs we will use the following notation: Given a Radon measure $\mu$ in $\Omega$ and a funcion $f\in L^1_{\mu}(\Omega)$ we denote the restriction of $\mu$ to $f$ by $$ \mu\lfloor f(E) := \int_E f\, d\mu. $$ \begin{proof}[Proof of Lemma \ref{Lema 1}] By reverse H\"older inequality \eqref{RH}, the measure $\nu$ is absolutely continuous with respect to $\mu$. As consequence there exists $f\in L_\mu^1(\Omega)$, $f\geq0$, such that $\nu=\mu\lfloor f$. Also by \eqref{RH}, we have $$ \min\big\{\nu(A)^\frac{1}{r-},\nu(A)^\frac{1}{r+}\big\} \leq C\max\big\{\mu(A)^\frac{1}{p-},\mu(A)^\frac{1}{p+}\big\} $$ for any Borel set $A\subset\Omega$. In particular, $f\in L^\infty_\mu(\Omega)$. On the other hand the Lebesgue decomposition of $\mu$ with respect to $\nu$ gives us $$ \mu=\nu\lfloor g + \sigma,\text{ where } g\in L^1_\nu(\Omega), g\geq0 $$ and $\sigma$ is a bounded positive measure, singular with respect to $\nu$. Now consider \eqref{RH} applying the test function $$ \phi=g^\frac{1}{r(x)-p(x)}\chi_{\{g\leq n\}}\psi. $$ We obtain \begin{align*} &\|g^\frac{1}{r(x)-p(x)}\chi_{\{g\leq n\}}\psi\|_{L^{r(x)}_\nu}\\ &\leq C\|g^\frac{1}{r(x)-p(x)}\chi_{\{g\leq n\}}\psi\|_{L^{p(x)}_\mu}\\ &= C\|g^\frac{1}{r(x)-p(x)}\chi_{\{g\leq n\}}\psi\|_{L^{p(x)}_{g d\nu+d\sigma}}\\ &\leq C\|g^\frac{r(x)}{p(x)(r(x)-p(x))} \chi_{\{g\leq n\}}\psi\|_{L^{p(x)}_\nu} + C\|g^\frac{1}{r(x)-p(x)}\chi_{\{g\leq n\}}\psi\|_{L^{p(x)}_\sigma} \end{align*} Since $\sigma\perp\nu$, we have $$ \|g^\frac{1}{r(x)-p(x)}\chi_{\{g\leq n\}}\psi\|_{L^{r(x)}_\nu} \leq C\|g^\frac{r(x)}{p(x)(r(x)-p(x))}\chi_{\{g\leq n\}} \psi\|_{L^{p(x)}_\nu} $$ Hence calling $d\nu_n=g^\frac{r(x)}{(r(x)-p(x))}\chi_{g\leq n} d\nu$ the following reverse H\"older inequality holds $$ \|\psi\|_{L^{r(x)}_{\nu_n}}\leq C \|\psi\|_{L^{p(x)}_{\nu_n}}. $$ By Lemma \ref{Lema 2} and Lemma \ref{Lema 3}, there exists $x_i^n$ and $K_i^n>0$ such that $\nu_n = \sum_{i\in I} K_i^n\delta_{x_i^n}$. On the other hand, $\nu_n\nearrow g^\frac{r(x)}{r(x)-p(x)}\nu$. Then, the points $x_i^n$ are in fact independent of $n$, and there will denoted by $x_i$, and the numbers $K_i^n$ are monotone in $n$. Then, we have $$ g^\frac{r(x)}{r(x)-p(x)}\nu = \sum_{i\in I} K_i\delta_{x_i} $$ where $K_i=g^\frac{r(x_i)}{r(x_i)-p(x_i)}(x_i)\nu(x_i)$. This finishes the proof. \end{proof} The following Lemma follows exactly as in the constant exponent case and the proof is omitted. \begin{lemma}\label{lema 4} Let $f_n\to f$ a.e and $f_n\rightharpoonup f$ in $L^{p(x)}(\Omega)$ then $$ \lim_{n\to\infty}\Big(\int_\Omega|f_n|^{p(x)}dx -\int_\Omega|f-f_n|^{p(x)}dx\Big)=\int_\Omega|f|^{p(x)}dx $$ \end{lemma} Now we are in position to prove the main results. \begin{proof}[Proof of Theorem \ref{ccp}] Given any $\phi\in C^\infty (\Omega)$, we write $v_j=u_j-u$ and by Lemma \ref{lema 4}, we have $$ \lim_{j\to\infty}\Big(\int_\Omega |\phi|^{q(x)}|u_j|^{q(x)} -\int_\Omega|\phi|^{q(x)}|v_j|^{q(x)} dx\Big) =\int_\Omega |\phi|^{q(x)}|u|^{q(x)} dx. $$ On the other hand, by reverse H\"older inequality \eqref{RH} and Lemma \ref{Lema 1}, taking limits we obtain the representation $$ \nu = |u|^{q(x)} + \sum_{j\in I} \nu_j\delta_{x_j} $$ Let us now show that the points $x_j$ actually belong to the {\em critical set} $\mathcal{A}$. In fact, assume by contradiction that $x_1\in \Omega\setminus \mathcal{A}$. Let $B=B(x_1,r) \subset\subset \Omega-\mathcal{A}$. Then $q(x)0$ in $\overline{B}$ and, by Proposition \ref{embedding}, The embedding $W^{1,p(x)}(B)\hookrightarrow L^{q(x)}(B)$ is compact. Therefore, $u_j\to u$ strongly in $L^{q(x)}(B)$ and so $|u_j|^{q(x)}\to |u|^{q(x)}$ strongly in $L^1(B)$. This is a contradiction to our assumption that $x_1\in B$. Now we proceed with the proof. Applying \eqref{poincare} to $\phi u_j$ and taking into account that $u_j\to u$ in $L^{p(x)}(\Omega)$, we have $$ S \|\phi\|_{L^{q(x)}_\nu(\Omega)} \leq \|\phi\|_{L^{p(x)}_\mu(\Omega)} + \|(\nabla \phi) u\|_{L^{p(x)} (\Omega)}. $$ Consider $\phi\in C^\infty_c(\mathbb{R}^n)$ such that $0\leq\phi\leq1$, $\phi(0)=1$ and supported in the unit ball of $\mathbb{R}^n$. Fixed $j\in I$, we consider $\varepsilon>0$ be arbitrary. We denote by $\phi_{\varepsilon,j}(x):= \varepsilon^{-n}\phi((x-x_j)/\varepsilon)$. By decomposition of $\nu$, we have: \begin{align*} \rho_\nu(\phi_{i_0,\varepsilon}) &:= \int_\Omega|\phi_{i_0,\varepsilon}|^{q(x)}\,d\nu \\ &= \int_\Omega |\phi_{i_0,\varepsilon}|^{q(x)}|u|^{q(x)}\, dx + \sum_{i\in I} \nu_i\phi_{i_0,\varepsilon}(x_i)^{q(x_i)} \geq \nu_{i_0}. \end{align*} For the rest of this article, we will denote \begin{gather*} q^+_{i,\varepsilon}:=\sup_{B_\varepsilon(x_i)}q(x),\quad q^-_{i,\varepsilon}:=\inf_{B_\varepsilon(x_i)}q(x),\\ p^+_{i,\varepsilon}:=\sup_{B_\varepsilon(x_i)}p(x),\quad p^-_{i,\varepsilon}:=\inf_{B_\varepsilon(x_i)}p(x). \end{gather*} If $\rho_\nu(\phi_{i_0,\varepsilon})<1$ then $$ \|\phi_{i_0,\varepsilon}\|_{L^{q(x)}_\nu(\Omega)} = \|\phi_{i_0,\varepsilon}\|_{L^{q(x)}_\nu (B_\varepsilon(x_{i_0}))} \ge \rho_\nu(\phi_{i_0,\varepsilon})^{1/q^-_{i,\varepsilon}} \ge \nu_{i_0}^{1/q^-_{i,\varepsilon}}. $$ Analogously, if $\rho_\nu(\phi_{i_0,\varepsilon})>1$, then $$ \|\phi_{i_0,\varepsilon}\|_{L^{q(x)}_\nu(\Omega)} \ge \nu_{i_0}^{1/q^+_{i,\varepsilon}}. $$ Then $$ \min\{\nu_i^\frac{1}{q^+_{i,\varepsilon}}, \nu_i^\frac{1}{q^-_{i,\varepsilon}}\} S \leq \|\phi_{i,\varepsilon}\|_{L^{p(x)}_\mu(\Omega)} + \|(\nabla\phi_{i,\varepsilon}) u\|_{L^{p(x)}(\Omega)}. $$ By Proposition \ref{norma.y.rho}, $$ \|(\nabla\phi_{i,\varepsilon}) u\|_{L^{p(x)}(\Omega)} \le \max\{\rho((\nabla\phi_{i,\varepsilon}) u)^{1/p^-}; \rho((\nabla\phi_{i,\varepsilon}) u)^{1/p^+}\}. $$ Then, by H\"older inequality, we have \begin{align*} \rho((\nabla\phi_{i,\varepsilon}) u) & = \int_{\Omega} |\nabla\phi_{i,\varepsilon}|^{p(x)} |u|^{p(x)}\, dx\\ & \leq \| |u|^{p(x)}\|_{L^{\alpha(x)}(B_\varepsilon(x_i))} \||\nabla \phi_{i,\varepsilon}|^{p(x)} \|_{L^{\alpha'(x)} (B_\varepsilon(x_i))}, \end{align*} where $\alpha(x) = n/(n-p(x))$ and $\alpha'(x) = n/p(x)$. Moreover, using that $\nabla\phi_{i,\varepsilon} =\nabla\phi\left(\frac{x-x_i}{\varepsilon}\right)\frac{1}{\varepsilon}$, we obtain $$ \||\nabla \phi_{i,\varepsilon}|^{p(x)} \|_{L^{\alpha'(x)} (B_\varepsilon(x_i))} \le \max\{\rho(|\nabla \phi_{i,\varepsilon}|^{p(x)})^{p^+/n}; \rho(|\nabla \phi_{i,\varepsilon}|^{p(x)})^{p^-/n}\}, $$ and \begin{align*} \rho(|\nabla \phi_{i,\varepsilon}|^{p(x)}) & = \int_{B_\varepsilon(x_i)} |\nabla\phi_{i,\varepsilon}|^n\, dx\\ & = \int_{B_\varepsilon(x_i)} |\nabla\phi(\frac{x-x_i}{\varepsilon})|^n \frac{1}{\varepsilon^n}\, dx \\ & = \int_{B_1(0)} |\nabla\phi(y)|^n\, dy. \end{align*} Then $\nabla\phi_{i,\varepsilon} u\to 0$ strongly in $L^{p(x)}(\Omega)$. On the other hand, $$ \int_\Omega |\phi_{i,\varepsilon}|^{p(x)}\,d\mu\leq \mu(B_{\varepsilon}(x_i)). $$ Therefore, \begin{align*} \|\phi_{i,\varepsilon}\|_{L^{p(x)}(\Omega)} &= \|\phi_{i,\varepsilon}\|_{L^{p(x)}(B_\varepsilon(x_i))} \\ &\leq \max \{ \rho_\mu (\phi_{i,\varepsilon} )^{1/p^+_{i,\varepsilon}}, \rho_\mu(\phi_{i,\varepsilon} )^{1/p^-_{i,\varepsilon}}\}\\ &\le \max \{ \mu(B_\varepsilon(x_i))^{1/p^+_{i,\varepsilon}}, \mu(B_\varepsilon(x_i))^{1/p^-_{i,\varepsilon}}\}, \end{align*} so we obtain, $$ S\min \{ \nu_i^\frac{1}{q^+_{i,\varepsilon}}, \nu_i^\frac{1}{q^-_{i,\varepsilon}}\} \leq \max \{ \mu(B_\varepsilon(x_i))^{1/p^+_{i,\varepsilon}}, \mu(B_\varepsilon(x_i))^\frac{1}{p^-_{i,\varepsilon}}\}. $$ As $p$ and $q$ are continuous functions and as $q(x_i) = p^*(x_i)$, letting $\varepsilon\to 0$, we get $$ S \nu_i^{1/p^*(x_i)} \le \mu_i^{1/p(x_i)}, $$ where $\mu_i := \lim_{\varepsilon\to 0}\mu(B_\varepsilon(x_i))$. Finally, we show that $\mu\geq|\nabla u|^{p(x)}+\Sigma\mu_i\delta_{x_i}$. In fact, we have that $\mu\geq\sum\mu_i\delta_{x_i}$. On the other hand $u_j\rightharpoonup u$ weakly in $W_0^{1,p(x)}(\Omega)$ then $\nabla u_j\rightharpoonup\nabla u$ weakly in $L^{p(x)}(U)$ for all $U\subset\Omega$. By weakly lower semi continuity of norm we obtain that $d\mu\geq|\nabla u|^{p(x)}\,dx$ and, as $|\nabla u|^{p(x)}$ is orthogonal to $\mu_1$, we conclude the desired result. This completes the proof. \end{proof} \section{Applications} In this section, we apply Theorem \ref{ccp} to study the existence of nontrivial solutions of the problem \begin{equation}\label{aplicacion} \begin{gathered} -\Delta_{p(x)} u=|u|^{q(x)-2}u+\lambda(x)|u|^{r(x)-2}u \quad \text{in }\Omega,\\ u=0 \quad \text{on }\partial\Omega, \end{gathered} \end{equation} where $r(x)p}} We begin by proving the Palais-Smale condition for the functional $\mathcal{F}$, below certain level of energy. \begin{lemma}\label{acotada} Assume that $r\le q$. Let $\{u_j\}_{j\in\mathbb{N}}\subset W_0^{1,p(x)}(\Omega)$ a Palais-Smale sequence then $\{u_j\}_{j\in\mathbb{N}}$ is bounded in $W_0^{1,p(x)}(\Omega)$. \end{lemma} \begin{proof} By definition $\mathcal{F}(u_j)\to c$ and $\mathcal{F}'(u_j)\to 0$. Now, we have $$ c+1 \geq \mathcal{F}(u_j) = \mathcal{F}(u_j) - \frac{1}{r-} \langle \mathcal{F}'(u_j), u_j \rangle + \frac{1}{r-} \langle \mathcal{F}'(u_j), u_j \rangle, $$ where $$ \langle \mathcal{F}'(u_j), u_j \rangle = \int_\Omega \big(|\nabla u_j|^{p(x)} - |u_j|^{q(x)} - \lambda(x) |u_j|^{r(x)}\big)\, dx. $$ Then, if $r(x)\leq q(x)$, we conclude that $$ c+1 \geq \big(\frac{1}{p+} - \frac{1}{r-}\big) \int_\Omega |\nabla u_j|^{p(x)}\, dx - \frac{1}{r-} |\langle \mathcal{F}'(u_j), u_j \rangle|. $$ We can assume that $\|u_j\|_{W_0^{1,p(x)}(\Omega)}\geq 1$. As $\|\mathcal{F}'(u_j)\|$ is bounded we have that $$ c+1 \geq \big(\frac{1}{p+} - \frac{1}{r-}\big) \|u_j\|^{p^-}_{W_0^{1,p(x)}(\Omega)} - \frac{C}{r-}\|u_j\|_{W_0^{1,p(x)}(\Omega)}. $$ We deduce that $u_j$ is bounded. This completes the proof. \end{proof} From the fact that $\{u_j\}_{j\in\mathbb{N}}$ is a Palais-Smale sequence it follows, by Lemma \ref{acotada}, that $\{u_j\}_{j\in\mathbb{N}}$ is bounded in $W_0^{1,p(x)}(\Omega)$. Hence, by Theorem \ref{ccp}, we have \begin{gather} |u_j|^{q(x)}\rightharpoonup \nu =|u|^{q(x)} + \sum_{i\in I} \nu_i\delta_{x_i} \quad \nu_i>0,\\ |\nabla u_j|^{p(x)}\rightharpoonup \mu \geq |\nabla u|^{p(x)} + \sum_{i\in I} \mu_i \delta_{x_i}\quad \mu_i>0,\\ S \nu_i^{1/p^*(x_i)} \leq \mu_i^{1/p(x_i)}. \end{gather} Note that if $I=\emptyset$ then $u_j\to u$ strongly in $L^{q(x)}(\Omega)$. We know that $\{x_i\}_{i\in I}\subset \mathcal{A}$. Let us show that if $c < \big(\frac{1}{p^+} - \frac{1}{q^-_\mathcal{A}}\big)S^n$ and $\{u_j\}_{j\in\mathbb{N}}$ is a Palais-Smale sequence, with energy level $c$, then $I=\emptyset$. In fact, suppose that $I \neq \emptyset$. Then let $\phi\in C_0^\infty(\mathbb{R}^n)$ with support in the unit ball of $\mathbb{R}^n$. Consider, as in the previous section, the rescaled functions $\phi_{i,\varepsilon}(x) = \phi(\frac{x-x_i}{\varepsilon})$. As $\mathcal{F}'(u_j)\to 0$ in $(W_0^{1,p(x)}(\Omega))'$, we obtain that $$ \lim_{j\to\infty} \langle \mathcal{F}'(u_j), \phi_{i,\varepsilon}u_j \rangle = 0. $$ On the other hand, $$ \langle \mathcal{F}'(u_j), \phi_{i,\varepsilon} u_j \rangle = \int_\Omega |\nabla u_j|^{p(x)-2}\nabla u_j\nabla (\phi_{i,\varepsilon}u_j) - \lambda(x) |u_j|^{r(x)} \phi_{i,\varepsilon} - |u_j|^{q(x)}\phi_{i,\varepsilon}\, dx $$ Then, passing to the limit as $j\to\infty$, we obtain \begin{align*} 0 = &\lim_{j\to\infty} \Big(\int_\Omega |\nabla u_j|^{p(x)-2} \nabla u_j \nabla(\phi_{i,\varepsilon}) u_j\, dx\Big)\\ & + \int_\Omega \phi_{i,\varepsilon}\, d\mu - \int_\Omega \phi_{i,\varepsilon}\, d\nu - \int_\Omega\lambda(x) |u|^{r(x)}\phi_{i,\varepsilon}\, dx. \end{align*} By H\"older inequality, it is easy to check that $$ \lim_{j\to\infty} \int_\Omega |\nabla u_j|^{p(x)-2} \nabla u_j \nabla(\phi_{i,\varepsilon}) u_j\, dx = 0. $$ On the other hand, \begin{gather*} \lim_{\varepsilon\to 0} \int_\Omega \phi_{i,\varepsilon}\, d\mu = \mu_i\phi(0),\quad \lim_{\varepsilon\to 0} \int_\Omega \phi_{i,\varepsilon}\, d\nu = \nu_i\phi(0), \lim_{\varepsilon\to 0} \int_\Omega \lambda(x)|u|^{r(x)} \phi_{i,\varepsilon}\, dx = 0. \end{gather*} So, we conclude that $(\mu_i-\nu_i)\phi(0)=0$; i.e., $\mu_i=\nu_i$. Then $$ S \nu_i^{1/p^*(x_i)} \leq \nu_i^{1/p(x_i)}; $$ so it is clear that $\nu_i = 0$ or $S^n\leq\nu_i$. On the other hand, as $r^->p^+$, \begin{align*} c &= \lim_{j\to\infty} \mathcal{F}(u_j) = \lim_{j\to\infty} \mathcal{F}(u_j) - \frac{1}{p+} \langle \mathcal{F}'(u_j), u_j \rangle\\ &= \lim_{j\to\infty} \int_\Omega \big(\frac{1}{p(x)} - \frac{1}{p+}\big) |\nabla u_j|^{p(x)}\, dx + \int_\Omega \big(\frac{1}{p+} - \frac{1}{q(x)}\big) |u_j|^{q(x)}\, dx\\ &\quad + \lambda \int_\Omega \big(\frac{1}{p+}-\frac{1}{r(x)}\big) | u_j|^{r(x)}\, dx\\ &\geq \lim_{j\to\infty} \int_\Omega \big(\frac{1}{p+}-\frac{1}{q(x)}\big) |u_j|^{q(x)}\, dx\\ &\geq \lim_{j\to\infty} \int_{\mathcal{A}_\delta} \big(\frac{1}{p+}-\frac{1}{q(x)}\big) |u_j|^{q(x)}\, dx\\ &\geq \lim_{j\to\infty} \int_{\mathcal{A}_\delta} \big(\frac{1}{p+}-\frac{1}{q^-_{\mathcal{A}_\delta}}\big) |u_j|^{q(x)}\, dx\,. \end{align*} However, \begin{align*} \lim_{j\to\infty} \int_{\mathcal{A}_\delta} \big(\frac{1}{p+} - \frac{1}{q^-_{\mathcal{A}_\delta}}\big) |u_j|^{q(x)}\, dx &= \big(\frac{1}{p+}-\frac{1}{q^-_{\mathcal{A}_\delta}}\big) \Big(\int_{\mathcal{A}_\delta}|u|^{q(x)}\, dx + \sum_{j\in I} \nu_j\Big)\\ &\geq \big(\frac{1}{p+} - \frac{1}{q^-_{\mathcal{A}_\delta}}\big) \nu_i\\ &\geq \big(\frac{1}{p+}-\frac{1}{q^-_{\mathcal{A}_\delta}}\big) S^n. \end{align*} As $\delta$ is positive and arbitrary, and $q$ is continuous, we have $$ c\ge \big(\frac{1}{p+}-\frac{1}{q^-_{\mathcal{A}}}\big) S^n. $$ Therefore, if $$ c < \big(\frac{1}{p+} - \frac{1}{q^-_{\mathcal{A}}}\big)S^n, $$ the index set $I$ is empty. Now we are ready to prove the Palais-Smale condition below level $c$. \begin{theorem}\label{Lemma.PS} Let $\{u_j\}_{j\in\mathbb{N}}\subset W_0^{1,p(x)}(\Omega)$ be a Palais-Smale sequence, with energy level $c$. If $c < \big(\frac{1}{p+} - \frac{1}{q^-_{\mathcal{A}}}\big) S^n$, then there exist $u\in W_0^{1,p(x)}(\Omega)$ and $\{u_{j_k}\}_{k\in\mathbb{N}}\subset \{u_j\}_{j\in\mathbb{N}}$ a subsequence such that $u_{j_{k}}\to u$ strongly in $W_0^{1,p(x)}(\Omega)$. \end{theorem} \begin{proof} We have that $\{u_j\}_{j\in\mathbb{N}}$ is bounded. Then, for a subsequence that we still denote $\{u_j\}_{j\in\mathbb{N}}$, $u_j\to u$ strongly in $L^{q(x)}(\Omega)$. We define $\mathcal{F}'(u_j):=\phi_j$. By the Palais-Smale condition, with energy level c, we have $\phi_j\to 0$ in $(W_0^{1,p(x)}(\Omega))'$. By definition $\langle \mathcal{F}'(u_j), z \rangle = \langle \phi_j, z \rangle$ for all $z\in W_0^{1,p(x)}(\Omega)$; i.e., $$ \int_\Omega |\nabla u_j|^{p(x)-2}\nabla u_j\nabla z\, dx - \int_\Omega |u_j|^{q(x)-2} u_j z\, dx - \int_\Omega\lambda(x) |u_j|^{r(x)-2} u_j z\, dx = \langle \phi_j, z \rangle. $$ Then, $u_j$ is a weak solution of the following equation. \begin{equation} \begin{gathered} -\Delta_{p(x)}u_j=|u_j|^{q(x)-2}u_j+\lambda(x)|u_j|^{r(x)-2}u_j +\phi_j=: f_j \quad \text{in }\Omega,\\ u_j = 0 \quad \text{on }\partial\Omega. \end{gathered} \end{equation} We define $T\colon (W_0^{1,p(x)}(\Omega))' \to W_0^{1,p(x)}(\Omega)$, $T(f):=u$ where $u$ is the weak solution of the equation \begin{equation} \begin{gathered} -\Delta_{p(x)}u=f \quad \text{in } \Omega,\\ u = 0 \quad \text{on } \partial\Omega. \end{gathered} \end{equation} Then $T$ is a continuous invertible operator. It is sufficient to show that $f_j$ converges in $(W^{1,p(x)}_0(\Omega))'$. We only need to prove that $|u_j|^{q(x)-2}u_j \to |u|^{q(x)-2}u$ strongly in $(W_0^{1,p(x)}(\Omega))'$. In fact, \begin{align*} \langle|u_j|^{q(x)-2}u_j-|u|^{q(x)-2}u,\psi\rangle &=\int_\Omega(|u_j|^{q(x)-2}u_j-|u|^{q(x)-2}u)\psi\,dx\\ &\leq\|\psi\|_{L^{q(x)}(\Omega)}\|(|u_j|^{q(x)-2}u_j -|u|^{q(x)-2}u)\|_{L^{q'(x)}(\Omega)}. \end{align*} Therefore, \begin{align*} &\|(|u_j|^{q(x)-2}u_j - |u|^{q(x)-2}u)\|_{(W_0^{1,p(x)}(\Omega))'}\\ &= \sup_{\genfrac{}{}{0cm}{}{\psi\in W^{1,p(x)}_0(\Omega)} {\|\psi\|_{W^{1,p(x)}_0(\Omega)}=1}} \int_\Omega (|u_j|^{q(x)-2}u_j-|u|^{q(x)-2}u)\psi\, dx\\ &\leq \|(|u_j|^{q(x)-2}u_j - |u|^{q(x)-2}u)\|_{L^{q'(x)}(\Omega)} \end{align*} and now, by the Dominated Convergence Theorem this last term approaches zero as $j\to\infty$. The proof is complete. \end{proof} We are now in position to prove Theorem \ref{r>p}. \begin{proof}[Proof of Theorem \ref{r>p}] In view of the previous result, we seek for critical values below level $c$. For that purpose, we want to use the Mountain Pass Theorem. Hence we have to check the following condition: \begin{enumerate} \item There exist constants $R,r>0$ such that when $\|u\|_{W^{1,p(x)}(\Omega)}=R$, then $\mathcal{F}(u)>r$. \item There exist $v_0\in W^{1,p(x)}(\Omega)$ such that $\mathcal{F}(v_0)r$ for some $R,r>0$. This proves (1). Now (2) is immediate as for a fixed $w\in W_0^{1,p(x)}(\Omega)$ we have $$ \lim_{t\to \infty}\mathcal{F}(tw) = -\infty. $$ Now the candidate for critical value according to the Mountain Pass Theorem is $$ c = \inf_{g\in \mathcal{C}} \sup_{t\in[0,1]} \mathcal{F}(g(t)), $$ where $\mathcal{C}=\{g:[0,1]\to W_0^{1,p(x)}(\Omega)\colon g \text{ continuous and } g(0)=0, g(1)=v_0\}$. We will show that, if $\inf_{x\in\mathcal{A}_\delta}\lambda(x)$ is big enough for some $\delta>0$ then $c < \big(\frac{1}{p+} - \frac{1}{q^-_{\mathcal{A}}}\big) S^n$ and so the local Palais-Smale condition (Theorem \ref{Lemma.PS}) can be applied. We fix $w\in W_0^{1,p(x)}(\Omega)$. Then, if $t<1$, we have \begin{align*} \mathcal{F}(tw) &\leq \int_\Omega t^{p(x)}\frac{|\nabla w|^{p(x)}}{p-} - t^{q(x)} \frac{|w|^{q(x)}}{q+} - \lambda(x) t^{r(x)} \frac{| w|^{r(x)}}{r+}\, dx\\ &\leq \frac{t^{p-}}{p-} \int_\Omega |\nabla w|^{p(x)}\, dx - \frac{t^{r+}}{r+} \int_\Omega \lambda(x)|w|^{r(x)}\, dx\\ &\leq \frac{t^{p-}}{p-} \int_\Omega |\nabla w|^{p(x)}\, dx - \frac{t^{r+}}{r+} \int_{\mathcal{A}_\delta} \lambda(x)|w|^{r(x)}\, dx\\ &\leq \frac{t^{p-}}{p-} \int_\Omega |\nabla w|^{p(x)}\, dx - \frac{t^{r+}}{r+} \int_{\mathcal{A}_\delta}(\inf_{x\in \mathcal{A}_\delta} \lambda(x))|w|^{r(x)}\, dx \end{align*} We define $g(t) := \frac{t^{p-}}{p-} a_1 -(\inf_{x\in \mathcal{A}_\delta} \lambda(x))\frac{t^{r+}}{r+} a_3$, where $a_1$ and $a_2$ are given by $a_1 = \| |\nabla w|^{p(x)}\|_{L^1(\Omega)}$ and $a_3 = \| |w|^{r(x)}\|_{L^1(\mathcal{A}_\delta)}$. The maximum of $g$ is attained at $t_\lambda = \big(\frac{a_1 }{(\inf_{x\in \mathcal{A}_\delta} \lambda(x)) a_3}\big)^\frac{1}{r+-p-}$. So, we conclude that there exists $\lambda_0>0$ such that if $(\inf_{x\in \mathcal{A}_\delta} \lambda(x))\geq\lambda_0$ then $$ \mathcal{F}(tw) < \big(\frac{1}{p+} - \frac{1}{q^-_{\mathcal{A}}}\big) S^n $$ This completes the proof. \end{proof} \begin{remark} \label{rmk4.3} \rm Observe that if $\lambda(x)$ is continuous it suffices to assume that $\lambda(x)$ is large in the {\em criticality set} $\mathcal{A}$. \end{remark} \subsection{Proof of Theorem \ref{r 1$. Then we have, by Proposition \ref{norma.y.rho} and by Poincar\'e inequality, \begin{align*} c+1 &\geq \big(\frac{1}{p^+} - \frac{1}{q^-}\big) \| |\nabla u_j| \|^{p^-}_{L^{p(x)}(\Omega)} +\|\lambda\|_\infty \big(\frac{1}{q^-} - \frac{1}{r^-}\big) \|u_j\|^{r^+}_{L^{r(x)}(\Omega)} \\ &\quad - \frac{1}{q^-}\|u_j\|_{W_0^{1,p(x)} (\Omega)}\varepsilon_j\\ &\geq \big(\frac{1}{p^+} - \frac{1}{q^-}\big) \| |\nabla u_j| \|^{p^-}_{L^{p(x)}(\Omega)} + \|\lambda\|_\infty\big(\frac{1}{q^-} - \frac{1}{r^-}\big) C \| |\nabla u_j| \|^{r^+}_{L^{p(x)}(\Omega)}\\ &\quad - \frac{1}{q^-}\|u_j\|_{W_0^{1,p(x)}(\Omega)} \end{align*} from where it follows that $\|u_j\|_{W_0^{1,p(x)}(\Omega)}$ is bounded (recall that $p^+\leq q^-$ and $r^+J_1(m)$. For these values $x_0$ and $x_1$ we can choose a smooth function $\tau_1(x)$ such that $\tau_1(x)=1$ if $x\leq x_0$, $\tau_1(x)=0$ if $x\geq x_1$ and $0\leq\tau_1(x)\leq 1$. If $\| |\nabla u| \|_{L^{p(x)}(\Omega)}>1$, we argue similarly and obtain \begin{align*} \mathcal{F}(u) &\geq\frac{1}{p^+}\| |\nabla u| \|^{p^-}_{L^{p(x)}(\Omega)} - \frac{C}{q^-} \| |\nabla u| \|^{q^+}_{L^{p(x)}(\Omega)} - \frac{\|\lambda\|_{L^\infty(\Omega)} C}{r^-} \| |\nabla u| \|_{L^{p(x)}(\Omega)}^{r^+} \\ &=: J_2(\| |\nabla u| \|_{L^{p(x)}(\Omega)}) \end{align*} where $$ J_2(x)=\frac{1}{p^+}x^{p^-}-\frac{C}{q^-}x^{q^+} -\frac{\|\lambda\|_{L^\infty(\Omega)} C}{r^-}x^{r^+}. $$ As in the previous case, $J_2$ attains a local but not a global minimum. So let $x_0,x_1$ be such that $mJ_2(m)$. For these values $x_0$ and $x_1$ we can choose a smooth function $\tau_2(x)$ with the same properties as $\tau_1$. Finally, we define $$ \tau(x)=\begin{cases} \tau_1(x) &\text{if } x\leq 1\\ \tau_2 (x) &\text{if } x>1. \end{cases} $$ Next, let $\varphi(u) = \tau(\| |\nabla u| \|_{L^{p(x)}(\Omega)})$ and define the truncated functional as follows, $$ \tilde{\mathcal{F}}(u) = \int_\Omega \frac{|\nabla u|^{p(x)}}{p(x)}\, dx - \int_\Omega \frac{|u|^{q(x)}}{q(x)}\varphi(u)\, dx - \int_\Omega\frac{\lambda(x)}{r(x)}|u|^{r(x)}\, dx $$ Next we state a Lemma that contains the main properties of $\tilde{\mathcal{F}}$. \begin{lemma} \label{lem4.6} $\tilde{\mathcal{F}}$ is $C^1$, if $\tilde{\mathcal{F}}(u)\leq 0$ then $\|u\|_{W_0^{1,p(x)}(\Omega)}0$ such that if $0<\|\lambda\|_{L^{\infty}(\Omega)} < \lambda_1$ then $\tilde{\mathcal{F}}$ satisfies a local Palais-Smale condition for $c\leq 0$. \end{lemma} \begin{proof} We have to check only the local Palais-Smale condition. Observe that every Palais-Smale sequence for $\tilde{\mathcal{F}}$ with energy level $c\leq 0$ must be bounded, therefore by Lemma \ref{elijo c} if $\lambda$ verifies $$ 0<\big(\frac{1}{p+}-\frac{1}{q^-}\big)S^n + K \min \{\|\lambda\|_{L^\infty(\Omega)}^{\frac{(q/r)^-}{(q/r)^- -1}}, \|\lambda\|_{L^\infty(\Omega)}^{\frac{(q/r)^+}{(q/r)^+ -1}}\}, $$ then there exists a convergent subsequence. \end{proof} The following Lemma gives the final ingredients needed in the proof. \begin{lemma}\label{genero} For every $n\in \mathbb{N}$ there exists $\varepsilon>0$ such that $$ \gamma(\mathcal{\tilde{F}}^{-\varepsilon})\geq n $$ where $\mathcal{\tilde{F}}^{-\varepsilon}=\{u\in W^{1,p(x)}_0(\Omega) \colon \mathcal{\tilde{F}}(u)\leq -\varepsilon\}$ and $\gamma$ is the Krasnoselskii genus. \end{lemma} \begin{proof} Let $E_n\subset W_0^{1,p(x)}(\Omega)$ be a $n$-dimensional subspace. Hence we have, for $u\in E_n$ such that $\|u\|_{W_0^{1,p(x)}(\Omega)}=1$, \begin{align*} \mathcal{\tilde{F}}(tu)&= \int_\Omega \frac{|\nabla (tu)|^{p(x)}}{p(x)}\, dx - \int_\Omega \frac{|tu|^{q(x)}}{q(x)}\varphi(tu)\, dx - \int_\Omega \frac{\lambda(x)}{r(x)}|tu|^{r(x)}\, dx\\ &\le \int_\Omega \frac{|\nabla (tu)|^{p(x)}}{p^-}\, dx - \int_\Omega \frac{|tu|^{q(x)}}{q^+}\varphi(tu)\, dx - \int_{\Omega}\frac{\lambda(x)}{r^+}|tu|^{r(x)}\, dx. \end{align*} If $t<1$, then \begin{align*} \tilde{\mathcal{F}}(tu) &\leq \int_\Omega\frac{t^{p^-}|\nabla u|^{p(x)}}{p^-}\, dx - \int_\Omega\frac{t^{q^+}|u|^{q(x)}}{q^+}\, dx - \int_{\Omega} \frac{\inf_{x \in\Omega}\lambda(x)}{r^+}t^{r^+}|u|^{r(x)}\, dx\\ &\leq \frac{t^{p^-}}{p^-} - \frac{t^{q^+}}{q^+}a_n -\inf_{x \in\Omega}\lambda(x)\frac{t^{r^+}}{r^+}b_n, \end{align*} where \begin{gather*} a_n=\inf\Big\{\int_\Omega|u|^{q(x)}\,dx\colon u\in E_n, \|u\|_{W_0^{1,p(x)}(\Omega)}=1\Big\},\\ b_n=\inf\Big\{\int_{\Omega}|u|^{r(x)}\,dx\colon u\in E_n, \|u\|_{W_0^{1,p(x)}(\Omega)}=1\Big\}. \end{gather*} Then $$ \tilde{\mathcal{F}}(tu)\leq\frac{t^{p^-}}{p^-} -\frac{t^{q^+}}{q^+}a_n-\inf_{x\in\Omega}\lambda(x) \frac{t^{r^+}}{r^+}b_n\leq\frac{t^{p^-}}{p^-} -\inf_{x \in\Omega}\lambda(x)\frac{t^{r^+}}{r^+}b_n\,. $$ Observe that $a_n>0$ and $b_n>0$ because $E_n$ is finite dimensional. As $r^+