\documentstyle[twoside]{article} \input amssym.def % used for R in Real numbers \pagestyle{myheadings} \markboth{\hfil Radial Symmetric Solutions \hfil EJDE--1996/07}% {EJDE--1996/07\hfil C.O. Alves, D.C.de Morais \& M.A.S. Souto \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol.\ {\bf 1996}(1996), No.\ 07, pp. 1--12. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp (login: ftp) 147.26.103.110 or 129.120.3.113} \vspace{\bigskipamount} \\ Radially Symmetric Solutions for a Class of Critical Exponent Elliptic Problems in $\Bbb R^N$ \thanks{ {\em 1991 Mathematics Subject Classifications:} 35A05, 35A15 and 35J20. \newline\indent {\em Key words and phrases:} Radial solutions, critical Sobolev exponents, \newline\indent Palais-Smale condition, Mountain Pass Theorem.\newline\indent \copyright 1996 Southwest Texas State University and University of North Texas.\newline\indent Submitted July 04, 1996. Published: August 30, 1996.\newline\indent Second and third author are partially supported by CNPq/Brazil } } \date{} \author{C. O. Alves, D. C. de Morais Filho \& M. A. S. Souto} \date{} \maketitle \newtheorem{theo}{Theorem} \newtheorem{remark}{Remark} \newtheorem{lemma}{Lemma} \begin{abstract} We give a method for obtaining radially symmetric solutions for the critical exponent problem $\left\{ \begin{array}{c} -\Delta u+a(x)u=\lambda u^q+u^{2^{*}-1}\mbox{ in }\Bbb R^N \\ u>0\mbox{ and }\int_{\Bbb R^N}|\nabla u|^2<\infty \end{array} \right.$ where, outside a ball centered at the origin, the non-negative function $a$ is bounded from below by a positive constant $a_o>0$. We remark that, differently from the literature, we do not require any conditions on $a$ at infinity. \end{abstract} \section{Introduction} Our purpose in this paper is to solutions for the semi-linear elliptic problem: $$\left\{ \begin{array}{c} -\Delta u+a(x)u=\lambda u^q+u^{2^{*}-1}\mbox{ in }\Bbb R^N \\ u>0\mbox{ and }\int_{\Bbb R^N}|\nabla u|^2<\infty . \end{array} \right. \label{global}$$ where $a:\Bbb R^N\to \Bbb R$ is a non-negative radially symmetric $C^1$ function, \linebreak $2*=2N/(N-2)$; $10$, and $N\geq 3$. Several researchers have studied variants of problem (\ref{global}). Among others, we can cite the article by Br\ezis \& Nirenberg \cite{BN1} which treats the case $a\equiv 0$ in bounded domains. Azorero \& Alonzo in \cite{AA1} and \cite{AA3} generalize some similar results for the $p-$Laplacian operator in bounded domains. Egnell \cite{E1} also generalizes some results in \cite{BN}. In the case of unbounded domains, Rabinowitz \cite{R} considers a more general non-linearity, but he does not treat the Sobolev critical exponent case. Benci \& Cerami \cite{BC} consider the problem (\ref{global}) when $\lambda =0,$ and \cite{AGM} deals with the case where $\lambda$ is replaced by an integrable function. In \cite{CMP}, a variation of this problem, with $a$ constant, was solved for the biharmonic operator. To finish citations we list the following works: \cite{GA} Alves \& Gon\c calves, \cite{GM} Gon\c calves \& Miyagaki, \cite{J} Jianfu, \cite{JX} Jianfu \& Xiping. All the last results in unbounded domains are obtained under the crucial hypothesis that $a$ is a coercive function or that $\lim\limits_{|x|\rightarrow +\infty }a(x)$ exists. We improve their results, relaxing the coerciveness of $a$ and the existence of the above limit. As in \cite{R}, we shall use variational method to solve problem (\ref {global}). To describe precisely our results, we present below the hypotheses on the function $a$: $(A_o)$ $a\in C^1(\Bbb R^N)$ is a radially symmetric function and there are $a_o,$ $R>0$ such that $a(x)\geq a_o,$ for all $|x|\geq R$ . Let us consider the following $W^{1,2}(\Bbb R^N)$ Hilbert subspace: $H_{\rm rad}^1(\Bbb R^N)=\{u\in W^{1,2}(\Bbb R^N):u\mbox{ radially symmetric }\}.$ Our main result is the following. \begin{theo} If ($A_o$) is satisfied, then problem (\ref{global}) possesses a nontrivial classical solution $u\in H_{\rm rad}^1(\Bbb R^N),$ for all $\lambda >0$ and $10,$ there is a constant $a_1>0,$ such that $|f(x,s)|\leq \varepsilon |s|+a_1|s|^p,\mbox{ for all }x\in \Bbb R^N, \quad s\in \Bbb R,$ where $1\leq p<2^{*}-1.$ \item{$(f_2)$} There is $\mu >2,$ such that $0<\mu F(x,s)\dot \leq sf(x,s),\mbox{ for all }x\in \Bbb R^N,\mbox{ }s\in \Bbb R\backslash \{0\},$ where $F(x,s)=\int_o^sf(x,t)dt.$ \end{description} Then (\ref{global2}) possesses a nontrivial classical solution $u\in W^{1,2}(\Bbb R^N).$ \end{theo} \section{Proof of Theorem 1} First, let us formulate a proper framework to solve problem (\ref{global}). Define the Hilbert space $E=\{u\in H_{\rm rad}^1(\Bbb R^N):\int_{\Bbb R^N}a(x)u^2<\infty \},$ endowed with the inner product $\langle u,v\rangle=:\int_{\Bbb R^N}(\nabla u\nabla v+a(x)uv)$ and the norm $||u||^2=:\int_{\Bbb R^N}(|\nabla u|^2+a(x)u^2$ Now we present two lemmas that will be used in the proof of the Theorem~1. \begin{lemma} Let $w$ be a $W_{\rm loc}^{1,s}(\Bbb R^N)$ function satisfying $$-\Delta w=h,\mbox{ } \label{Lwh}$$ in $\Bbb R^N\backslash \{0\}$ in the weak sense, where $h$ is a $% L_{\rm loc}^1(\Bbb R^N)$ function and $s\geq \frac N{N-1}$. Then (\ref{Lwh}) is weakly satisfied in the whole $\Bbb R^N.$ \end{lemma} \paragraph{Proof:} In order to prove this result, consider $\varphi \in C^\infty (\Bbb R^N)$ such that $\varphi (x)=0$ in $|x|\leq 1$ and $% \varphi (x)=1$ in $|x|\geq 2.$ For each $\varepsilon >0,$ define $\psi _\varepsilon (x)=\varphi (\frac x\varepsilon ).$ Fix a function $\phi \in C_c^\infty (\Bbb R^N).$ As $\psi _\varepsilon \phi \in C_c^\infty (\Bbb R^N\backslash \{0\})$ we have that $\int_{\Bbb R^N}\nabla w\nabla (\psi _\varepsilon \phi )=\int_{\Bbb R^N}h(x)(\psi _\varepsilon \phi ),$ and then $$\int_{\Bbb R^N}\psi _\varepsilon \nabla w\nabla \phi +\int_{\Bbb R^N}\phi \nabla w\nabla \psi _\varepsilon =\int_{\Bbb R^N}h(x)(\psi _\varepsilon \phi ). \label{epsil}$$ Using the dominated convergence theorem, we obtain the limits \begin{eqnarray} \lim_{\varepsilon \rightarrow 0}\int_{\Bbb R^N}\psi _\varepsilon \nabla w\nabla \phi &=&\int_{\Bbb R^N}\nabla w\nabla \phi \mbox{ } \label{limites} \\ \lim_{\varepsilon \rightarrow 0}\int_{\Bbb R^N}h(x)(\psi _\varepsilon \phi ) &=&\int_{\Bbb R^N}h(x)\phi . \nonumber \end{eqnarray} We claim that the limit of the second term on the left side of (\ref{epsil}) is zero. In fact, $\left| \int_{\Bbb R^N}\phi \nabla w\nabla \psi _\varepsilon \right| \leq ||\phi ||_{L^\infty (\Bbb R^N)}\int_{|x|\leq 2\varepsilon }|\nabla w||\nabla \psi _\varepsilon |.$ Using H\"older's inequality in the above inequality with $\frac 1s+\frac 1q =1,$ we obtain that $\left| \int_{\Bbb R^N}\phi \nabla w\nabla \psi _\varepsilon \right| \leq ||\phi ||_{L^\infty (\Bbb R^N)}\left( \int_{|x|\leq 2\varepsilon }|\nabla w|^s\right) ^{1/s}\left( \int_{|x|\leq 2\varepsilon }|\nabla \psi _\varepsilon |^q\right) ^{1/q}$ and then $\left| \int_{\Bbb R^N}\phi \nabla w\nabla \psi _\varepsilon \right| \leq ||\phi ||_{L^\infty (\Bbb R^N)}||\nabla \varphi ||_{L^q(\Bbb R^N)}\left( \int_{|x|\leq 2\varepsilon }|\nabla w|^s\right) ^{1/s}\varepsilon ^{ \frac{N-q}q}.$ Observe that $N\geq q$ and passing to the limit in this last inequality we prove the claim. Finally using the claim and the limits (\ref{limites}) in (\ref{epsil}) we have that $\int_{\Bbb R^N}\nabla w\nabla \phi =\int_{\Bbb R^N}h(x)\phi .$ \begin{remark} The above result is not valid for $W_{\rm loc}^{1,1}(\Bbb R^N)$ functions. The function $w=|x|^{2-N}$(if $N\geq 3,$ or $w=\log |x|,$ if $N=2$) belongs to $% W_{\rm loc}^{1,1}(\Bbb R^N),$ satisfies $-\Delta w=0$ in $\Bbb R^N\backslash \{0\},$ but if $v$ is a radially symmetric function in $C_c^\infty \left( \Bbb R^N\right)$ such that $v(0)\neq 0,$ we have that $\int_{\Bbb R^N}\nabla w\nabla v=\frac{\omega _N}{2-N}\int_o^\infty r^{N-1}r^{1-N}v^{\prime }(r)dr=\frac{\omega _N}{2-N}v(0)\neq 0,\mbox{ if }% N\geq 3$ or $\int_{\Bbb R^N}\nabla w\nabla v=2\pi v(0)\neq 0,\mbox{ if }N=2.$ \end{remark} \begin{lemma} Let $f:\Bbb R^N\times \Bbb R\rightarrow \Bbb R$ be a $C^1$ function satisfying $(f_o)$ such that $|f(x,s)|\leq c|s|+|s|^{2^{*}-1}\mbox{ for all }x\in \Bbb R^N,\mbox{ }s\in \Bbb R;$ and let $a$ be a radially symmetric function. Suppose that $u\in E$ satisfies $\int_{\Bbb R^N}(\nabla u\nabla v+a(x)uv)=\int_{\Bbb R^N}f(x,u)v,\mbox{ for all }v\in E\,.$ Then $u\in C^2(\Bbb R^N)$ and $-\Delta u(x)+a(x)u(x)=f(x,u(x))$ for all $x\in \Bbb R^N$. \end{lemma} \paragraph{ Proof:} Since $a$ and $f$ are radially symmetric we rewrite the above expression as $$\int_o^\infty r^{N-1}(u^{\prime }v^{\prime }+a(r)uv)dr=\int_o^\infty r^{N-1}f(r,u)vdr, \label{EDOinte}$$ for all $v \in E$. We have that $h(r):=-a(r)u(r)+f(r,u(r))$ is in $C^{o ,\alpha }(\Bbb R^N\backslash \{0\}),$ since $H_{\rm rad}^1(\Bbb R^N)$ is contained in $C^{o,\alpha }(\Bbb R^N\backslash \{0\})$. Hence $\int_o^\infty r^{N-1}u^{\prime }\psi ^{\prime }dr=\int_o^\infty r^{N-1}h(r)\psi dr,\mbox{ for all }\psi \in C_c^\infty (0,+\infty ),$ and $$\int_o^\infty u^{\prime }(r^{N-1}\psi )^{\prime }dr=\int_o^\infty [\frac{N-1}% ru^{\prime }+h(r)](r^{N-1}\psi )dr,\mbox{ } \label{phipsi}$$ for all $\psi \in C_c^\infty (0,+\infty ).$ For each $\varphi \in C_c^\infty (0,+\infty ),$ considering $\psi =r^{1-N}\varphi$ in (\ref{phipsi}) we conclude that $u$ is a weak solution of $-u^{\prime \prime }=\frac{N-1}ru^{\prime }+h(r),\mbox{ for }r>0.$ Since $u^{\prime }\in L_{\rm loc}^2(0,+\infty ),$ it follows that $u\in H_{\rm loc}^2(0,+\infty )$, $u^{\prime }\in H_{\rm loc}^1(0,+\infty )$, and $$u\in H_{\rm rad}^1(\Bbb R^N)\cap C^2(\Bbb R^N\backslash \{0\})\,.$$ Moreover for $|x|>0$ , the function $u$ satisfies (\ref{global}) in the classical sense. \hfill $\diamondsuit$ \paragraph{Proof of Theorem 1}. This proof consists of using variational methods to get critical points of the Euler-Lagrange functional associated to (1) and defined on $E$: $I(u)=\frac 12\int_{\Bbb R^N}(|\nabla u|^2+a(x)u^2)-\frac \lambda {q+1}\int_{\Bbb R^N}\left( u^{+}\right) ^{q+1}-\frac 1{2^{*}}\int_{\Bbb R^N}\left( u^{+}\right) ^{2^{*}}$ where $u^+(x)=\max \{u(x),0 \}$ and $u^+(x)=\min \{-u(x),0 \}$. The critical points of $I$ are precisely the weak solutions of (1). These solutions may be regularized. The Hilbert space $E$ is immersed continuously in $W^{1,2}(\Bbb R^N).$ This assertion comes from $(A_o)$ and the following inequalities \begin{eqnarray*} \left( \int_{|x|\leq R}u^2\right) ^{1/2}& \leq& c_1\left( \int_{|x|\leq R}\left| u\right| ^{2^{*}}\right) ^{1/2^*}\\ & \leq& c_1\left(\int_{\Bbb R^N}\left| u\right| ^{2^{*}}\right) ^{1/2^*}\\ &\leq& c_2\left( \int_{\Bbb R^N}|\nabla u|^2\right) ^{1/2}\,. \end{eqnarray*} We also have that $H_{\rm rad}^1(\Bbb R^N)\subset L^p(\Bbb R^N)$ continuously if $2 \leq p\leq 2^{*}$ and compactly if $2 \leq p<2^{*}$(see \cite{K}). Using these results one has the following lemma: \begin{lemma} The Banach space $E$ is continuously immersed in $L^p(\Bbb R^N)$ if % \linebreak $2 \leq p\leq 2^{*}$ and compactly if $2 \leq p<2^{*}$. \end{lemma} Using lemma 3 we verify that $I$ is a well-defined $C^1(E)$ functional - see \cite{RZ}. It is easy to verify that $$\frac \lambda {q+1}\int_{\Bbb R^N}(u^+)^{q+1}+\frac 1{2^{*}}\int_{\Bbb R^N}(u^+)^{2^{*}}=o(||u||^2)\mbox{ as }u\rightarrow 0, \label{opeqeno}$$ and hence that $I$ has a local minimum at the origin. This is not a global minimum. If $u\in E\backslash \{0\}$ , $u\geq 0,$ we have that $I(tu)=\frac{t^2}2\int_{\Bbb R^N}(|\nabla u|^2+a(x)u^2)-\frac{\lambda t^{q+1}% }{q+1}\int_{\Bbb R^N}(u^+)^{q+1}-\frac{t^{2^{*}}}{2^{*}}\int_{\Bbb R^N} (u^+)^{2^{*}}.$ Since $\int_{\Bbb R^N}(u^+)^{2^{*}}\neq 0,$ we conclude that $I(tu)\rightarrow -\infty$ as $t\rightarrow \infty$. So, we have just seen that $I$ has the\textit{\ Mountain Pass Theorem Geometry.} Let $e\in E$ such that $I(e)<0,$ and $\Gamma =\{g:[0,1]\rightarrow E:g(0)=0\mbox{ and }g(1)=e\}$ and $c=\inf_{g\in \Gamma }\max_{0\leq t\leq 1}I(g(t)).$ Thus $c\$ is the mountain pass minimax value associated to $I$. At this moment, it is important to notice that $c$ is not the minimax value associated to the Euler Lagrange functional of problem (\ref{global}) defined in the whole $W^{1,2}(\Bbb R^N).$ Assertion (\ref{opeqeno}) implies $c>0.$ Using an application of the Ekeland Variational Principle (Theorem 4.3 of \cite{MW}), there exists a sequence $\{u_m\}\subset E$ such that $$I(u_m)\rightarrow c,\mbox{ }I^{\prime }(u_m)\rightarrow 0. \label{PSseq}$$ \begin{lemma} The above sequence $\{u_m\}$ is bounded. \end{lemma} \paragraph{Proof:} Notice that $I(u_m)-\frac 1{q+1}I^{\prime }(u_m)u_m=\left( \frac 12-\frac 1{q+1}\right) ||u_m||^2+\left( \frac 1{q+1}-\frac 1{2^{*}}\right) \int_{\Bbb R^N}\left( u_m^{+}\right) ^{2^{*}},$ then $I(u_m)-\frac 1{q+1}I^{\prime }(u_m) u_m \geq \left( \frac 12-\frac 1{q+1} \right) ||u_m||^2.$ Combining this last inequality with $I(u_m)-\frac 1{q+1}I^{\prime }(u_m)u_m \leq 1+c+||u_m||$ for large $m$, we conclude the proof. \hfill $\diamondsuit$ The following lemma shows that we can choose a vector $e\in E\backslash \{0\}$ in the definition of $\Gamma ,$ such that $I(e)<0$ and 00$and one of the following conditions is satisfied: \begin{description} \item{(i)}$N\geq 4;$\item{(ii)}$N=3$and$30,$consider the function $\phi _\varepsilon (x)=\frac{\left[ N(N-2)\varepsilon \right] ^{(N-2)/4}}{% \left( \varepsilon +|x|^2\right) ^{(N-2)/2}}.$ The functions$\phi _\varepsilon $satisfy the problem $\left\{ \begin{array}{c} -\Delta u=u^{2^{*}-1},\mbox{ in }\Bbb R^N \\ u>0,\mbox{ }\int_{\Bbb R^N}|\nabla u|^2<\infty \end{array} \right.$ and $\int_{\Bbb R^N}|\nabla \phi _\varepsilon |^2=\int_{\Bbb R^N}|\phi _\varepsilon |^{2^{*}}=S^{N/2}$ (see \cite{Ta}, lemma 2 - pp. 364). Now, consider$v_\varepsilon =\varphi \phi _\varepsilon $where$\varphi \in C_o^\infty (\Bbb R^N),0\leq \varphi (x)\leq 1$and $\varphi (x)=\left\{ \begin{array}{ll} 1&\mbox{ if }x\in B_1 \\ 0&\mbox{ if }x\notin B_1 \end{array} .\right.$ Using arguments due to \cite{Mi} there is$\varepsilon >0$such that $\sup_{t\geq 0}I(tv_\varepsilon )<\frac 1NS^{N/2}.$ If$t_\varepsilon >0$is such that$I(t_\varepsilon v_\varepsilon )<0,$we choose$e=t_\varepsilon v_\varepsilon $and the proof is complete.~$\diamondsuit $In order to complete the proof of Theorem 1, let us consider$e\in E\backslash \{0\}$given by lemma 5. Let$\{u_m\}$be the sequence in$E$satisfying (\ref{PSseq}). From Lemmas 3 and 4, we may assume that \begin{eqnarray*} &u_m\rightharpoonup u\mbox{ in }E&\\ &u_m\rightarrow u\mbox{ in }L^s(\Bbb R^N),\, 2 \leq s<2^* &\\ &u_m(x)\rightarrow u(x)\mbox{ a.e. in }\Bbb R^N\,.& \end{eqnarray*} The above limits with an observation in Br\ezis \& Lieb \cite{BLieb} yield that$u$must be a critical point of$I$in$E,$that is, $I^{\prime }(u)=0.$ We claim that$u\neq 0$. In fact, if$u\equiv 0$and taking$l\geq 0$such that $\int_{\Bbb R^N}|\nabla u_m|^2\rightarrow l,$ then $\int_{\Bbb R^N}\left( u_m^{+}\right) ^{2^{*}}\rightarrow l$ for the reason that$I^{\prime }(u_m)\rightarrow 0$and$E\subset L^{q+1}(\Bbb R^N)$compactly. Since$I(u_m)\rightarrow c,$we get $$Nc=l. \label{ncl}$$ From the definition of$S$, $\int_{\Bbb R^N}|\nabla u_m|^2\geq S\left( \int_{\Bbb R^N}\left| u_m\right| ^{2^{*}}\right) ^{\frac 2{2^{*}}}\geq S\left( \int_{\Bbb R^N}\left( u_m^{+}\right) ^{2^{*}}\right) ^{\frac 2{2^{*}}}.$ Taking the limit in the last inequalities, we achieve that $l\geq Sl^{ 2/ 2^*}$ and by (\ref{ncl}) that $c\geq \frac 1NS^{N/2}>c$ which contradicts the above choice of$e$, and thus the claim is proved. Observe that$I^{\prime }(u)u^{-}=0$implies$\int_{\Bbb R^N}|\nabla u^{-}|^2+a(x)\left( u^{-}\right) ^2=0$and then$u^{-}\equiv 0$which implies$u\geq 0.$Notice that at this moment we do not know if$u$satisfies (\ref{global}) in the$W^{1,2}(\Bbb R^N)$sense but, thanks to lemma 2,$u$is a nontrivial classical solution of (\ref{global}) with$u\geq 0$. The Hopf maximum principle assures that$u>0$. Theorem 1 is proved. \hfill$\diamondsuit $We conclude this section by justifying Remark 1 in the beginning of Section~1. The argument we are going to use is due to Azorero \& Alonzo \cite{AA3}. \paragraph{Justification of Remark 1.} Fix$\varphi \in C_o^\infty (\Bbb R^N\backslash \{0\})$,$\varphi (x)\geq 0$. Notice that the real function$I(t\varphi )$possesses a positive maximum value . Suppose that this maximum value is assumed for$t=t_\lambda .$Thus $\frac d{dt}\left. I(t\varphi )\right| _{t=t_\lambda }=0$ then $||\varphi ||^2=t_\lambda ^{q-1}\lambda \int_{\Bbb R^N}\varphi ^{q+1}+t_\lambda ^{2^{*}-2}\int_{\Bbb R^N}\varphi ^{2^{*}}\geq t_\lambda ^{q-1}\lambda \int_{\Bbb R^N}\varphi ^{q+1}.$ From the last inequality we have that$t_\lambda \rightarrow 0$as$\lambda \rightarrow \infty .$On the other hand $\sup_{t\geq 0}I(t\varphi )\leq \frac{t_\lambda }2\int_{\Bbb R^N}|\nabla \varphi |^2$ and for large enough$\lambda >0$we get $\sup_{t\geq 0}I(t\varphi )<\frac 1NS^{\frac N2}.$ Using the same arguments employed in the proof of Theorem 1 we conclude the justification. We have just finished the proof of Theorem 1. Our next step is the proof of Theorem 2 \section{Proof of Theorem 2} Let$\left( E,||\cdot ||\right) $be the same defined in the proof of Theorem 1 and consider $$I(u)=\frac 12\int_{\Bbb R^N}(|\nabla u|^2+a(x)u^2)-\int_{\Bbb R^N}F(x,u) \label{Euler2}$$ defined in$E$, as the associated Euler-Lagrange functional to problem (\ref {global2}), which is$C^1-$see \cite{RZ} . Under hypothesis$(f_1)$, it is easy to verify that $$\int_{\Bbb R^N}F(x,u)=o(||u||^2)\mbox{ as }u\rightarrow 0, \label{opeqeno2}$$ and hence that$I$has a local minimum at the origin. Hypothesis$(f_2)$implies that $$F(x,s)\geq a_2|s|^\mu \label{Fmu}$$ for large$|s|$. Then, by (\ref{opeqeno2}) and (\ref{Fmu}),$I$has the% \textit{\ Mountain Pass Theorem Geometry. }Let $\Gamma =\{g:[0,1]\rightarrow E:g(0)=0\mbox{ and }I(g(1))\leq 0\}$ and $c=\inf_{g\in \Gamma }\max_{0\leq t\leq 1}I(g(t)).$ As in the proof of Theorem 1,$c>0$and there is a sequence$\{u_m\}\subset E $satisfying (\ref{PSseq}). Using standard arguments,$\left( f_2\right) $implies that$||u_m||$is a bounded sequence. Therefore, along a subsequence,$u_m$converges weakly in$E$and strongly in$L^p(\Bbb R^N)$,$2\leq p<\frac{2N}{N-2},$to a function$u\in E$which is a weak solution of (% \ref{global2}). We claim that$u\neq 0.$In fact, for large$m$, $\frac c2\leq I(u_m)-\frac 12I^{\prime }(u_m)u_m=\int_{\Bbb R^N}[\frac 12f(x,u_m)u_m-F(x,u_m)].$ Taking$m\rightarrow \infty ,$in the above expression we obtain that $\int_{\Bbb R^N}[\frac 12f(x,u)u-F(x,u)]\geq \frac c2$ contradicting a possible vanishing of$u.$Then the claim is proved. 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Souto} (marco@dme.ufpb.br)\newline \noindent Departamento de Matem\'atica e Estat\'\i stica \newline Centro de Ci\^encias e Tecnologia \newline Universidade Federal da Para\'\i ba \newline Caixa Postal 10044 \newline CEP 58.109-970 - Campina Grande - Para\'\i ba - Brazil\newline \end{document}