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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 15, pp. 1--7.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/15\hfil Asymptotically periodic Schr\"odinger equations]
{Existence of solutions to asymptotically periodic Schr\"odinger equations}

\author[M. F. Furtado, R. Marchi \hfil EJDE-2017/15\hfilneg]
{Marcelo F. Furtado, Reinaldo de Marchi}

\address{Marcelo F. Furtado \newline
Universidade de Bras\'ilia,
 Departamento de Matem\'atica,
70910-900 Bras\'ilia-DF, Brazil}
\email{mfurtado@unb.br}

\address{Reinaldo de Marchi \newline
Universidade Federal do Mato Grosso,
Departamento de Matem\'atica, 78060-900 Cuiab\'a-MT, Brazil}
\email{reinaldodemarchi@ufmt.br}

\dedicatory{Communicated by Claudianor O. Alves}

\thanks{Submitted July 8, 2016. Published January 13, 2017.}
\subjclass[2010]{35J50, 35J45}
\keywords{Strongly indefinite functionals; Schr\"odinger equation;
\hfill\break\indent asymptotically periodic problem}

\begin{abstract}
 We show the existence of a nonzero solution for the semilinear 
 Schr\"odinger  equation $-\Delta u+V(x)u=f(x,u)$. 
 The potential $V$ is periodic  and $0$ belongs to a gap of $\sigma(-\Delta +V)$.
 The function $f$ is superlinear and asymptotically periodic with respect to
 $x$ variable. In the proof we apply a new critical  point theorem for
 strongly indefinite functionals proved in \cite{furtado2016asymptotically}.
\end{abstract}

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

\section{Introduction}


We consider the existence of nonzero solutions for
the semilinear Schr\"odinger equation
\begin{equation}\label{eq1}
-\Delta u+V(x)u=f(x,u),\quad x\in\mathbb{R}^N,
\end{equation}
where $V \in C(\mathbb{R}^N,\mathbb{R})$ the  nonlinearity
$f \in C(\mathbb{R}^N \times \mathbb{R},\mathbb{R})$ satisfy the following
assumptions:
\begin{itemize}
\item[(A1)]  $V(x)=V(x_1,\ldots,x_N)$ is 1-periodic in $x_1,\ldots,x_N$;

\item[(A2)] if $\sigma(-\Delta +V)$ denotes the spectrum of the operator
$-\Delta +V$, then $0\not\in\sigma(-\Delta +V)$ and
$\sigma(-\Delta+V)\cap(-\infty,0)\neq\emptyset$,

\item[(A3)] there exist $c_1,c_2>0$ and $p \in (2,2^*)$ such that
$$
|f(x,t)| \leq c_1|t| + c_2|t|^{p-1},\quad\forall
 (x,t) \in \mathbb{R}^N \times \mathbb{R};
$$

\item[(A4)] $f(x,t)t \geq 0$, for all $(x,t) \in \mathbb{R}^N \times \mathbb{R}$;

\item[(A5)] $f(x,t)=o(|t|)$, as $t \to 0$, uniformly in $x \in \mathbb{R}^N$;

\item[(A6)]  it holds
$$
\lim_{|t|\to\infty}\dfrac{F(x,t)}{t^2} = \infty,\quad\text{uniformly in }
x \in \mathbb{R}^N,
$$
where $F(x,t) := \int_0^t f(x,\tau)d\tau$.
\end{itemize}

We denote by $\mathfrak{F}$ the class of all functions
$h\in C(\mathbb{R}^N,\mathbb{R})\cap L^\infty(\mathbb{R}^N,\mathbb{R})$
such that, for every $\varepsilon>0$, the set
$\{x\in\mathbb{R}^N : |h(x)|\geq\varepsilon\}$ has
finite Lebesgue measure, and we assume that
\begin{enumerate}

\item[(A7)] there exist $p_\infty\in(2,2^*)$, $\varphi\in\mathfrak{F}$ and
$f_\infty\in C(\mathbb{R}^N\times\mathbb{R},\mathbb{R})$,
1-periodic in $x_1,\ldots,x_N$, such that, for all for all
 $(x,t)\in\mathbb{R}^N\times\mathbb{R}$,

\begin{itemize}
\item[(i)] $f_{\infty}(x,t)t \geq 0$ and $f_{\infty}(x,t)/|t|$ is not decreasing in  $\mathbb{R} \setminus \{0\}$;
\item[(ii)] $F(x,t)\geq F_\infty(x,t):=\int_0^t f_\infty(x,\tau)d\tau$;
\item[(iii)] $|f(x,t)-f_\infty(x,t)|\leq \varphi(x)|t|^{p_\infty-1}$.
\end{itemize}

\item[(A8)] there exists $\theta_0 \in (0,1)$ such that
$$
\frac{1-\theta^2}{2} tf(x,t) \geq F(x,t) - F(x,\theta t),\quad \forall
 \theta \in [0,\theta_0],\;(x,t) \in \mathbb{R}^N \times \mathbb{R}.
$$
\end{enumerate}


Our first result can be stated as follows.

\begin{theorem}\label{thm1}
Suppose that {\rm (A1)--(A8)} are satisfied.
Then  problem \eqref{eq1} has a nonzero solution.
\end{theorem}

In the proof we apply a version of the Linking Theorem due to
Li and Szulkin \cite{li2002asymptotically} to obtain a Cerami sequence
for the associated functional. Thanks to  (A8), the same argument employed
by Tang in \cite{tang2014new} provides the boundedness of this sequence.
If $f$ is periodic is sufficient to guarantee that, up to translations,
the weak limit of the sequence is a nonzero solution.
In our case we do not have periodicity and therefore the strategy
of \cite{tang2014new} fails. To overcome this difficult we use a a local
version of the Linking Theorem  proved  in \cite{furtado2016asymptotically}.

The same idea can be used to replace condition (A8)  by another one
introduced by Ding and Lee in \cite{ding2006multiple}
(see also \cite{10} for a weaker condition). More specifically, we assume that
\begin{enumerate}
\item[(A4')] $F(x,t)\geq0$ for all $(x,t)\in\mathbb{R}^N\times\mathbb{R}$;
\item[(A8')] there exist $\tau>\max\{1,N/2\}$ and positive constants
$r,a_1,R_1$  such that
\begin{gather*}
q(r):=\inf\{\widehat{F}(x,t): x\in\mathbb{R}^N\ \text{and}\ |t|\geq r\}>0, \\
|f(x,t)|^\tau\leq a_1|t|^\tau\widehat{F}(x,t),\quad \text{for all }
 x\in\mathbb{R}^N,\; |t|\geq R_1,
\end{gather*}
where $\widehat{F}(x, t):=\frac{1}{2}f(x, t)t-F(x, t)$.
\end{enumerate}

\begin{theorem}\label{thm2}
Suppose that {\rm (A1), (A2), (A4'), (A5)--(A7),  (A8')} are satified.
Then  problem \eqref{eq1} has a nonzero solution.
\end{theorem}

In this article we  denote
 $B_R(y):=\{ x \in \mathbb{R}^N : |x-y|<R\}$ and
$|A|$ for the Lebesgue measure of a set $A\subset\mathbb{R}^N$. We
write  $\int_A u$ instead of $\int_A u(x) dx $. We also omit the set
$A$ whenever $A=\mathbb{R}^N$. Also we write $|\cdot|_p$ for the norm in
$L^p(\mathbb{R}^N)$.

\section{Variational setting}

 We  denote by $S$ the  selfadjoint operator
$-\Delta+V$ acting on $L^2(\mathbb{R}^N)$ with domain
$\mathcal{D}(S):=H^2(\mathbb{R}^N)$. Under the conditions (A1) and (A2), we
have the orthogonal decomposition $L^2(\mathbb{R}^N)=L^-\oplus L^+$, with
the subspaces $L^+$ and $L^-$ being such that $S$ is negative in
$L^-$ and positive in $L^+$. If we consider the Hilbert space
$H:=\mathcal{D}(|S|^{1/2})$ with the inner product
$(u,v):=(|S|^{1/2}u,|S|^{1/2}v)_{L^2}$,
and the corresponding norm
$\|u\|:=||S|^{1/2}u|_2$, it follows from (A1) and (A2) that
$H=H^1(\mathbb{R}^N)$ and the above norm is equivalent to the usual
norm of this space. Hence, we obtain the decomposition
$$
H=H^+\oplus H^-,\ H^\pm=H\cap L^\pm,
$$
which is orthogonal with respect to $(\cdot,\cdot)_{L^2}$ and $(\cdot,\cdot)$.


Let $(e_k) \subset H$ be a total orthonormal sequence  in $H^-$.
We introduce a new topology on $H$ by setting
\begin{equation} \label{norma-tau}
\|u\|_{\tau} := \max\Big\{ \|u^+\|, \sum_{k=1}^{\infty}
\frac{1}{2^k} |\langle u^-,e_k\rangle|\Big\}.
\end{equation}
 The above norm  induces a topology in $H$ which we call $\tau$-topology.
Given a set $M \subset H$, an homotopy $h:[0,1] \times M \to H$
is said to be admissible if
\begin{itemize}

\item[(i)] $h$ is $\tau$-continuous, that is, if $t_n \to t$ and
$u_n \stackrel{\tau}{\to} u$ then $h(t_n,u_n) \stackrel{\tau}{\to} h(t,u)$;

\item[(ii)] for each $(t,u) \in [0,1]\times M$  there is a neighborhood $U$
of $(t,u)$ in the
 product topology of $[0,1]$ and $(H,\tau)$ such  that
  the set  $\{w-h(t,w): (t,w)\in U\cap([0,1]\times M)\}$ is contained in
  a finite dimensional subspace of  $H$.
\end{itemize}
When $I \in C^1(E,\mathbb{R})$ the symbol $\Gamma$  denotes the  class of  maps
\begin{align*}
\Gamma :=  \big\{& h\in  C([0,1]\times M,H): h \text{ is admissible, }
 h(0,\cdot)=\operatorname{Id}_M, \\
&  I(h(t,u)) \leq \max\{I(u),-1\} \text{ for all }(t,u) \in [0,1]\times M\big\}.
\end{align*}
The first part of the following abstract result can be found in
\cite[Theorem 2.1]{li2002asymptotically} while the last one was proved
in \cite[Theorem 2.3]{furtado2016asymptotically}.

\begin{theorem} \label{local}
 Suppose that $I \in C^1(H,\mathbb{R})$ satisfies
 \begin{itemize}
 \item[(A9)] The functional $I$ can be written as
  $$
I(u)=\frac12(\|u^+\|^2-\|u^-\|^2)-J(u),
$$
 with $J\in C^1(H,\mathbb{R})$ bounded from below, weakly sequentially
lower semicontinuous and  $J'$ is weakly sequentially continuous;

\item[(A10)] there exist $u_0\in H^+\backslash\{0\}$, $\alpha>0$ and
 $R>r>0$ such that
$$
\inf_{N_r} I \geq \alpha,\quad \sup_{\partial M} I \leq 0,
$$
where $N_r:=\{u\in H^+: \|u\|=r\}$,
$$
M_{R,u_0}=M:= \{u=u^-+\rho u_0: u^-\in H^-,\; \|u\|\leq R,\; \rho\geq0\},
$$
and $\partial M$ denotes the boundary of $M$ relative to $\mathbb{R}u_0 \oplus H^-$.
\end{itemize}
If
$$
c:=\inf_{h\in \Gamma}\sup_{u\in M}I(h(1,u)),
$$
then there exists  $(u_n) \subset H$ such that
\begin{equation} \label{cs}
I(u_n) \to c \geq \alpha,\quad (1+\|u_n\|)\|I'(u_n)\|_{H^*} \to 0.
\end{equation}
If there exists $h_0\in\Gamma$ such that
$$
 c=\inf_{h\in \Gamma}\sup_{u\in M}I(h(1,u)) =  \sup_{u \in M}  I(h_0(1,u)),
$$
then  $I$ possesses a nonzero critical point  $u_0 \in h_0(1,M)$ such that $I(u_0)=c$.
\end{theorem}

We  intend to apply the above result to obtain solutions for our equation.
To define the functional we notice that for a  given $\varepsilon>0$, we can
use (A3) and (A5) to obtain $C_{\varepsilon}>0$ such that
\begin{equation}\label{c1}
|f(x,t)| \leq \varepsilon|t| + C_{\varepsilon}|t|^{p-1},\quad
 |F(x,t)| \leq \varepsilon|t|^2 + C_{\varepsilon}|t|^p,
\end{equation}
for any $(x,t) \in \mathbb{R}^N \times \mathbb{R}$. The same inequality
holds under conditions (A5), $\widehat{(f_5)}$ and (A7)(ii)
 (see \cite[Lemma 4.1]{furtado2016asymptotically}).
Therefore, in the setting of our main theorems, we can easily conclude
that the functional  $I: H\to\mathbb{R}$ given by
$$
I(u) := \frac12\|u^+\|^2-\frac12\|u^-\|^2-\int F(x,u) ,
$$
for any $u=u^++u^-$, with $u^\pm\in H^\pm$, is well defined.
Moreover, it belongs to $C^1(H,\mathbb{R})$ and its critical points
are the weak solutions of \eqref{eq1}.

To define the linking subsets we consider the  periodic limit problem
\begin{equation*}
-\Delta u+V(x)u=f_\infty(x,u), \quad  x\in\mathbb{R}^N.
\end{equation*}
Under our conditions we can use \cite[Theorem 1.2]{tang2014new}
to conclude that it has a ground state solution
$u_\infty\in H^1(\mathbb{R}^N)$. More precisely, if
$$
I_\infty(u):=\frac12\|u^+\|^2-\frac12\|u^-\|^2-\int F_\infty(x,u) ,
$$
we have
\begin{equation}\label{c0}
I_\infty(u_\infty)=\inf\{I_\infty(u): u\in H \setminus\{0\}, \,I'_\infty(u)=0\}>0.
\end{equation}

 We set $u_0:=u_\infty^+$ and consider
$$
M:=\{u=u^-+\rho u_0: u^-\in H^-,\ \|u\|\leq R,\ \rho\geq0 \},\quad
N_r:=\{u\in H^+: \|u\|=r\}.
$$
As proved in  \cite[Proposition 39 and Theorem 40]{szulkin2010method}
 and \cite[Corollary 2.4]{9}, we have
\begin{equation}\label{u0}
\sup_{M} I_\infty(u)\leq I_\infty(u_\infty).
\end{equation}

We finish this section by stating  two technical
convergence results whose proofs can be found in
\cite[Lemmas 5.1 and 5.2]{lins2009quasilinear}, respectively.

\begin{lemma}\label{convergence1}
Suppose that {\rm (A7)} holds. Let $(u_n)\subset H^1(\mathbb{R}^N)$
be a bounded sequence and $v_n(x):=v(x-y_n)$, where $v\in H^1(\mathbb{R}^N)$
and $(y_n)\subset\mathbb{R}^N$. If $|y_n|\to\infty$, then
$[f_\infty(x,u_n)-f(x,u_n)]v_n\to0$,
strongly in $L^1(\mathbb{R}^N)$, as $n\to\infty$.
\end{lemma}

\begin{lemma}\label{convergence2}
Suppose that $h\in\mathfrak{F}$ and $s\in[2,2^*)$. If $v_n\rightharpoonup v$
weakly in $H^1(\mathbb{R}^N)$, then
$\int h(x)|v_n|^s  \to \int  h(x)|v|^s$, as $n\to\infty$.
\end{lemma}

\section{Proofs of main results}

In this section we prove Theorems \ref{thm1} and \ref{thm2}.

\begin{lemma}\label{link structure}
Under the hypothesis of our main theorems the functional $I$ satisfies
 the geometric conditions {\rm (A9)} and {\rm (A10)}.
\end{lemma}

\begin{proof}
Conditions (A5), (A8') and (A7)(ii) imply (A3). Thus, the inequalities
in \eqref{c1} holds under the assumptions of our main theorems and we
can easily conclude that $I$ satisfies (A9).
Since $N_r \subset H^+$,  for any $u \in N_r$, it holds
$I(u)=(1/2)\|u^+\|^2 - \int F(x,u)$.
Hence, it follows from \eqref{c1} that $\inf_{N_r} I \geq \alpha>0$
for some $r,\alpha>0$. For $R>r$ large we need to verify that
 $\sup_{\partial M} I \leq 0$.
We fix $u=u^-+\rho u_0\in\partial M_R$. If $\|u\|\leq R$ and $\rho=0$, we
have $u=u^-\in H^-$ and therefore $I(u) \leq 0$, since (A4) implies that
$F \geq 0$. Thus, it remains to consider $\|u\|=R$ and $\rho>0$. Arguing by
contradiction, we suppose that for some sequence $(u_n)$ such that
$u_n=u_n^-+\rho_n u_0$, $\rho_n>0$, $\|u_n\|=R_n\to\infty$ we have
that $I(u_n)>0$. Then
\[
\frac{I(u_n)}{\|u_n\|^2}=\frac12
\Big(\frac{\rho_n^2\|u_0\|^2}{\|u_n\|^2}-\frac{\|u_n^-\|^2}{\|u_n\|^2}\Big)
-\int\frac{F(x,u_n)}{\|u_n\|^2} >0.
\]
Since $F\geq0$, we must have $\rho_n\|u_0\|\geq\|u_n^-\|$. From
$$
\frac{\rho_n^2\|u_0\|^2}{\|u_n\|^2}+\frac{\|u_n^-\|^2}{\|u_n\|^2}=1,
$$
it follows that $\frac{1}{\sqrt2\|u_0\|}
\leq \frac{\rho_n}{\|u_n\|}\leq\frac{1}{\|u_0\|}$ and
$u_n^-/\|u_n\|$ is bounded. Thus, up to a subsequence, we have
$$
\frac{\rho_n}{\|u_n\|}\to\rho>0, \quad
\frac{u_n^-}{\|u_n\|}\rightharpoonup v\in H^-, \quad
\frac{u_n^-}{\|u_n\|}\to v\quad \text{a.e. for } x\in\mathbb{R}^N.
$$
This and $\|u_n\|\to\infty$ imply that $\rho_n\to\infty$. Thus, we
have
$$
\lim|u_n(x)|=\infty\ \text{ a.e. in } \Omega=\{x\in\mathbb{R}^N: \rho
u_0(x)+v(x)\neq0\}.
$$
Taking   the $\limsup$ in the inequality
$$
0<\frac{I(u_n)}{\|u_n\|^2}
\leq \frac12\Big(\frac{\rho_n^2\|u_0\|^2}{\|u_n\|^2}-
\frac{\|u_n^-\|^2}{\|u_n\|^2}\Big)
-\int_\Omega\frac{F(x,u_n)}{u_n^2}\frac{u_n^2}{\|u_n\|^2} \,\mathrm{d}x,
$$
using Fatou's Lemma and (A6), we conclude that
$$
0\leq\frac12\left(\rho^2\|u_0\|^2-\|v\|^2\right)-\int_\Omega
\liminf_{n\to\infty} \frac{F(x,u_n)}{u_n^2}(\rho
u_0+v)^2\,\mathrm{d}x =-\infty,
$$
which is a contradiction.
\end{proof}

We are ready to obtain a solution for equation \eqref{eq1}.

\subsection*{Proof of the main results}
 By  Lemma \ref{link structure} and the first part of  Theorem \ref{local}
we can obtain $(u_n)\subset H$ such that
$$
I(u_n)\to c\geq\alpha>0,\quad  (1+\|u_n\|)I'(u_n)\to0,\quad \text{as } n\to\infty.
$$
Under  condition (A8), arguing along the same lines as in
 \cite[Lemma 3.4]{tang2014new} we can prove that this sequence is bounded.
As proved in \cite[Lemma 4.3]{furtado2016asymptotically}, the same holds if
 $f$ satisfies (A4') and (A8'). We omit the (rather long) details in both cases.
Since $(u_n)$ is bounded in $H$, up to a subsequence, we have that
$u_n \rightharpoonup u$ weakly in $H$. By using (A3), (A5) and standard
calculations we can show that $I'(u)=0$.
If $u\neq0$ we are done. So, we need only to consider only the case
$u=0$.

We claim that  there exist a sequence $(y_n)\subset\mathbb{R}^N$, $R>0$,
and $\beta>0$ such that $|y_n|\to\infty$ as $n\to\infty$, and
\begin{equation}\label{d2}
\limsup_{n\to\infty}\int_{B_R(y_n)}|u_n|^2 \,\mathrm{d}x \geq\beta >0.
\end{equation}
Indeed, if this is not the case, from a result due to Lions
 \cite{lions1984concentration}  it follows  that   $|u_n|_s \to 0$ for any
$s\in(2,2^*)$. Hence, the first  inequality in \eqref{c1} implies that
 $\int F(x,u_n) \to 0$ as $n\to+\infty$. The same holds with
$\int f(x,u_n)u_n$. On the other hand
$$
c=\lim_{n\to\infty}\left(I(u_n)-\frac12 I'(u_n)u_n\right)
=\lim_{n\to\infty}\int\left(\frac12f(x,u_n)u_n-F(x,u_n)\right) =0
$$
which contradicts $c>0$.

Without loss of generality we may assume that $(y_n)\subset\mathbb{Z}^N$
(see \cite{chabrowski1999weak}).
Writing  $\widetilde{u}_n(x):=u_n(x+y_n)$ and observing that
$\|\widetilde{u}_n\|_{H^1}=\|u_n\|_{H^1}$, up to subsequence we have
 $\widetilde{u}_n\rightharpoonup\widetilde{u}$ in $H$,
$\widetilde{u}_n\to\widetilde{u}$ in $L^2_{loc}(\mathbb{R}^N)$ and for
almost every $x\in\mathbb{R}^N$.
It follows from  \eqref{d2} that  $\widetilde{u}\neq0$.

We fix $\eta\in C^\infty_0(\mathbb{R}^N)$ and define, for each  
$n \in \mathbb{N}$, the translation  $\eta_n(x):=\eta(x-y_n)$. Using \eqref{c1}, 
the Lebesgue Theorem and the periodicity  of $f_\infty$ we get
$$
I'_\infty(\widetilde{u}_n)\eta=I'_\infty(u_n)\eta_n  
=I'_\infty(\widetilde{u})\eta+o_n(1),
$$
where $o_n(1)$ stands for a quantity approaching zero as $n\to +\infty$. 
Hence, we need only to show that $I'_\infty(u_n)\eta_n=o_n(1)$. 
However, Lemma \ref{convergence1} provides
$$
I'_\infty(u_n)\eta_n=I'(u_n)\eta_n-\int[f(x,u_n)-f_{\infty}(x,u_n)]\eta_n 
= I'(u_n)\eta_n + o_n(1).
$$
Since $I'(u_n)\eta_n \to 0$ it follows that $I_{\infty}'(\widetilde{u})=0$.

We  claim that
$\liminf_{n\to\infty}\int \widehat{F}(x,\widetilde{u}_n)
\geq\int \widehat{F}_\infty(x,\widetilde{u})$.
Indeed,  from (A7) we obtain
$$
|\widehat{F}(x,u_n)-\widehat{F}_\infty(x,u_n)|
\leq\Big(\frac12+\frac{1}{p_\infty}\Big)h(x)|u_n|^{p_\infty}.
$$
Thus, by Lemma \ref{convergence2}, Fatou's lemma and periodicity of 
$\widehat{F}_\infty$,
\begin{align*}
\liminf_{n\to\infty}\int\widehat{F}(x,u_n) 
=\liminf_{n\to\infty}\int\widehat{F}_\infty(x,\widetilde{u}_n)
 \geq \int\widehat{F}_\infty(x,\widetilde{u}) .
\end{align*}
In view of the above considerations we obtain
\begin{align*}
c&=\lim_{n\to\infty}\Big(I(u_n)-\frac12 I'(u_n)u_n\Big)\\
&=\liminf_{n\to\infty}\int\widehat{F}(x,u_n) \\
&\geq \int\widehat{F}_\infty(x,\widetilde{u})
=I_\infty(\widetilde{u})-\frac12
I_\infty'(\widetilde{u})\widetilde{u}=I_\infty(\widetilde{u}),
\end{align*}
and therefore $I_\infty(\widetilde{u})\leq c$. Hence, using the 
definition of $c$, (A7) and (\eqref{u0} we obtain
$$
c\leq\sup_{u \in M}I(u)
\leq\sup_{u \in M}I_\infty(u)
\leq I_\infty(u_\infty)
\leq I_\infty(\widetilde{u})\leq c.
$$
Thus, if we define  $h_0:[0,1] \times M \to H$ by $h_0(t,u):=u$ 
for any $(t,u) \in [0,1] \times M$, the above inequality implies 
$\sup_{u \in M} I(h_0(u,1))=c$.
It follows from the last statement of Theorem \ref{local} that $I$ has 
a nonzero critical point.  
\hfill\qed

 
\subsection*{Acknowledgments}
M. F. Furtado was  partially supported by CNPq/Brazil.
R. Marchi was partially supported by CAPES/Brazil.
The authors would like to thank the anonymous referees for their
 useful suggestions.

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\end{document}