\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
2005-Oujda International Conference on Nonlinear Analysis.
\newline {\em Electronic Journal of Differential Equations},
Conference 14, 2006, pp. 173--180.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or
http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}
\setcounter{page}{173}

\begin{document}

\title[\hfilneg EJDE/Conf/14 \hfil A generalization of Ekeland's variational principle]
{A generalization of Ekeland's variational principle with
applications}

\author[A. R. El Amrouss,  N. Tsouli \hfil EJDE/Conf/14 \hfilneg]
{Abdel R. El Amrouss,  Najib Tsouli}  % in alphabetical order

\address{Abdel R. El Amrouss \newline
University Mohamed 1er, Faculty of sciences,
 Department of Mathematics, Oujda, Morocco}
\email{amrouss@sciences.univ-oujda.ac.ma}

\address{Najib Tsouli  \newline
University Mohamed 1er, Faculty of sciences,
 Department of Mathematics, Oujda, Morocco}
\email{tsouli@sciences.univ-oujda.ac.ma}

\date{}
\thanks{Published September 20, 2006.}
\subjclass[2000]{58E05, 35J65, 49B27}
\keywords{Ekeland's principle variational;
 Palais-Smale condition; optimization}

\begin{abstract}
 In this paper, we establish a variant of Ekeland's variational
 principle. This result suggest  to introduce a generalization of
 the famous Palais-Smale condition. An example is provided showing
 how it is used to give the existence of minimizer for functions
 for which the Palais-Smale condition and the one introduced by
 Cerami are not satisfied.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}

\section{Introduction}

Let $E$ be a complete metric space with metric $d$ and
$\Phi : E\to \mathbb{R} \cup \{\infty\} $  a lower  semicontinuous
function which is bounded
from below and not identically to  $+\infty$.
 The  Ekeland's variational principle, see \cite{E1}, allows for each
$\varepsilon > 0$,   each $\delta  > 0$   and each $x \in E$
  such as
$$
\Phi(x) \leq \inf_E \Phi + \varepsilon,
$$
to build an element $v \in E$ minimizing the functional $\Phi_v$  given by
$$
\Phi_v(x) = \Phi(x) + \frac{\varepsilon}{\delta}d(x,v).
$$
 This principle has wide applications in optimization and nonlinear
analysis  \cite{E1, E2, C}.

If $E$ is a Banach space and $\Phi: E \to \mathbb{R}$ is G\^ateaux
differentiable, lower semi-continuous and bounded from below, then
the Ekeland's variational principle  provides the existence of a
minimizing sequence $(u_n)$ such as $\Phi'(u_n) \to 0$, when $n
\to \infty$. It is well known that if $\Phi$ satisfies the
Palais-Smale condition then $\Phi$ reaches its minimum. But, it is
possible to find  a minimizing sequence $(u_n)$ such as
$\Phi'(u_n) \to 0$, when $n \to \infty$,  not having any
convergent subsequence. Let us take the example of the function
$\Phi(s) = \arctan(s)$.


Ekeland \cite{E2} prove that if $\Phi$ is bounded below and
satisfies the  Cerami condition for every $c \in \mathbb{R}$,
introduced by \cite{Ce}, then $\Phi$ has a minimal point.

In this note, we prove a variant of Ekeland's variational
principle. This result suggest  to introduce a generalization of
the classical Palais-Smale condition. An example is provided
showing how it is used to give the existence of minimizer for
functions for which the Palais-Smale condition or the  Cerami
condition are not satisfied. We also generalize some results cited
in \cite{E1},  \cite{CS}, which the Palais-Smale condition or
Cerami condition has failed.

\section{Variants of Ekeland's variational principle}

In this section we will prove the following variant of Ekeland's
variational principle. We start with a definition.

\begin{definition} \rm
We say that $\alpha: [0,\infty[ \to ]0,\infty[$ is  a comparison
function of order $k$ if for every  $q \geq k$ there exist $c,d
\geq 0$ such that
$$
\frac{\alpha((t+1)s)}{\alpha(t)}  \leq
cs^q + d, \forall t,s \in \mathbb{R}^+.
$$
{\bf Examples:}
\begin{enumerate}
\item $\alpha(s) = (1+s)^k$
\item $\alpha(s) = (1+s)^k Log(2+s)$
\end{enumerate}
\end{definition}

Let $(E,d)$ be a complet space metric and $u \in E$. Denote by
 $\bar{B}(u,r) = \{x \in E \mid d(u,x) \leq r\}$
the closed  boule and $B(u,r) = \{x \in E \mid d(u,x) < r\}$ the
open boule.

\begin{theorem} \label{thm2.1}
Let $(E,d)$ be a complete space metric, $x_0 \in E$ fixed, $ \Phi:
E \to \mathbb{R}$ a lower  semi-continuous and bounded below. Let
$\alpha: [0,\infty[ \to ]0,\infty[$ be a comparison function of
order $k$
 continuous nondecreasing. Thus for each $\varepsilon > 0$, each
$\delta >0$ and each $u \in E$ such that
$$\Phi(u) \leq \inf_E \Phi + \varepsilon$$
 there exists a convergent sequence $(z_n)_{n\geq 1}$ of $E$ satisfies:
\begin{itemize}
\item[(i)] $z_1 = u, z_n \in \bar{B}(u, \gamma(u))$
with $\gamma(u)$ be a positive constant such that
$u \mapsto \frac{\gamma(u)}{1+d(x_0,u)}$ is bounded in $E$

\item[(ii)] The sequence $(d(x_0,z_n))_{n\geq 1}$ is nondecreasing

 \item[(iii)] $\sum_{n=1}^{j} \frac{d(z_n,z_{n+1})}{\alpha(d(x_0,z_{n+1}))} <
2\delta$, for all $j \geq 1$

\item[(iv)] for $v =\lim_{n\to\infty} z_n, \Phi(v) \leq \Phi(u)$

\item[(v)] $d(u,v) \leq min\{\delta\alpha(d(x_0,v)), \gamma(u)\}$

\item[(vi)] for every $w \in   \bar{B}(u, \gamma(u)) \setminus B(u, d(x_0,u))$,
$$
\Phi(w) \geq \Phi(v) - \frac{\varepsilon}{\delta\alpha(d(x_0,w))}d(v,w).
$$
\end{itemize}
\end{theorem}

\begin{proof}  Let us define a partial order in $E$
by letting
\begin{equation}\label{e0}
\Phi(r) \leq \Phi(s) - \frac{\varepsilon}{\delta\alpha(d(x_0,r))}d(r,s)
\end{equation}
and
\begin{equation}\label{e1}
d(x_0,r)  \geq d(x_0,s).
\end{equation}
This relation is easily seen to be reflexive, antisymmetry and
transitive. Indeed, it is clear that
 $r \prec r$, for every $ r \in E$.
The partial order $\prec$ is antisymmetry. Indeed, if  $r \prec s$ and
$s \prec r$ then $d(x_0,r) = d(x_0,s)$,
$$
\Phi(r)
\leq \Phi(s) - \frac{\varepsilon}{\delta\alpha(d(x_0,r))}d(r,s)
\leq \Phi(r) - \frac{2\varepsilon}{\delta\alpha(d(x_0,r))}d(r,s).
$$
However $d(r,s) = 0$ and so $r = s$.
$\prec$ is transitive, because if $r \prec s$ and  $s \prec t$
then $d(x_0,r) \geq d(x_0,s) \geq d(x_0,t)$ and
\begin{equation}\label{e2}
\Phi(r) \leq \Phi(s) - \frac{\varepsilon}{\delta\alpha(d(x_0,r))}d(r,s),
\quad
\Phi(s) \leq \Phi(t) - \frac{\varepsilon}{\delta\alpha(d(x_0,s))}d(t,s).
\end{equation}
 From $d(t,s) \leq d(t,r) + d(r,s)$, (\ref{e2}) becomes
$$
\Phi(r) \leq \Phi(s) - \frac{\varepsilon}{\delta\alpha(d(x_0,r))}[d(t,r)
- d(t,s)], \quad
\Phi(s) \leq \Phi(t) - \frac{\varepsilon}{\delta\alpha(d(x_0,s))}d(t,s).
$$
This implies
$$
\Phi(r) \leq \Phi(t) + \Big[\frac{\varepsilon}{\delta\alpha(d(x_0,r))}
- \frac{\varepsilon}{\delta\alpha(d(x_0,s))}\Big]d(t,s)
 - \frac{\varepsilon}{\delta\alpha(d(x_0,r))}d(r,t).
$$
Since  $\alpha(.)$ is nondecreasing and $d(x_0,r) \geq d(x_0,s)$,
we obtain
\begin{align*}
r \prec s \mbox{ and } s \prec t
&\Rightarrow \left\{
\begin{array}{l}
\Phi(r) \leq \Phi(t) - \frac{\varepsilon}{\delta\alpha(d(x_0,r))}d(r,t) \\
d(x_0,r) \geq  d(x_0,t)
\end{array}\right.    \\
&\Rightarrow r \prec t.
\end{align*}
Moreover, if we denote  $S = \{r \in E \mid r \prec s\}$, by
lower semi-continuity of $E$, $S$ is closed.

Let $\varepsilon$, $\delta$, $u$ and $\gamma(u)$ given by theorem.
 Now we define  a sequence  $S_n$ of subsets as follows. Start
with $z_1 = u$ and define
$$
S_1 = \{w \in E \mid w \prec z_1\} \cap \bar B(u, \gamma(u)),
$$
and inductively
$$
S_{n} = \{w \in E \mid w \prec z_{n} \} \cap \bar B(u, \gamma(u)),
z_{n+1} \in S_n
$$
such that
\begin{equation}\label{e3}
\Phi(z_{n+1}) \leq \inf_{S_n} \Phi + \frac{1}{(n+1)\alpha(d(x_0,z_n))}.
\end{equation}
Clearly by transitivity of $\prec$ the sequence  $(S_n)_n$ is a
decreasing sequence of non empty  closed sets. Hence also
$(d(x_0,z_n))_n$ is a bounded  nondecreasing sequence and
converges in  $[d(x_0,u), d(x_0,u)+ \gamma(u)]$.

Now we prove that the diameters of these sets go to zero: $diamS_n
\to 0$. Indeed, on one hand  $w \in S_{n+1}$ implies
$$
\Phi(w) \leq \Phi(z_{n+1})  - \frac{\varepsilon}{\delta \alpha(d(x_0,w))}
d(w,z_{n+1}) \quad\text{and}\quad
 d(x_0,w) \geq d(x_0,z_{n+1}).
$$
From (\ref{e3}), it results
$$
\Phi(w) \leq  \inf_{S_n} \Phi + \frac{1}{(n+1)\alpha(d(x_0,z_n))}
- \frac{\varepsilon}{\delta \alpha(d(x_0,w))} d(w,z_{n+1}).
$$
This implies
$$
d(w,z_{n+1}) \leq
\frac{\delta}{\varepsilon(n+1)} \frac{\alpha(d(x_0,w))}{\alpha(d(x_0,z_n))}. \\
$$
On the other hand, we have that $w$ belongs to
$\bar B(u, \gamma(u))$, we obtain
\begin{equation}\label{e4}
d(w,z_{n+1})
\leq \frac{\delta}{\varepsilon (n+1)}
\frac{\alpha(\gamma(u)+d(x_0,u))}{\alpha(d(x_0,z_n))}.
\end{equation}
Since the function  $u \mapsto \frac{\gamma(u)}{1+d(x_0,u)}$ is
bounded, then there exists $M > 0$ such that
\begin{equation}\label{e5}
\gamma(u) \leq M(1 + d(x_0,u)).
\end{equation}
 From (\ref{e4}), (\ref{e5}) and $\alpha(.)$ is a nondecreasing
function, it results
\begin{equation}\label{e6}
d(w, z_{n+1}) \leq
 \frac{\delta}{ \varepsilon (n+1)}
 \frac{\alpha((M+1)(1+d(x_0,z_n)))}{\alpha(d(x_0,z_n))}.
\end{equation}
By (\ref{e6}) and  $\alpha(.)$ is a comparison function of order
$k$,  there exist $c, d > 0$ such that
$$
d(w, z_{n+1}) \leq
 \frac{\delta}{ \varepsilon (n+1)}(c(M+1)^k+d), n\in \mathbb{N}
$$
which gives $\mathop{\rm diam} S_{n+1}$ go to $0$, when $n \to \infty$.

Now we claim that the unique point $v \in E$ in the intersection
of the $S_n$'s satisfies conditions (iii)--(vi) of Theorem \ref{thm2.1}. Let
then $\cap_n S_n = \{v\}$ and $z_n$ converges to $v$. Since $z_j
\prec z_{j-1}\prec \dots \prec z_1$; and by  (\ref{e0}), we have
 \begin{align*}
 \Phi(z_{j+1}) &\leq \Phi(z_j)
 - \frac{\varepsilon}{\delta \alpha(d(x_0,z_{j+1}))} d(z_j,z_{j+1})  \\
&\leq \Phi(z_1) - \sum_{n=1}^{j} \frac{\varepsilon
d(z_j,z_{j+1})}{\delta\alpha(d(x_0,z_{j+1}))}
\end{align*}
or
 \begin{align*}
\sum_{n=1}^{j} \frac{\varepsilon
d(z_j,z_{j+1})}{\delta\alpha(d(x_0,z_{j+1}))}
&\leq \Phi(u) - \Phi(z_{j+1}) \\
&\leq \inf_{E} \Phi + \varepsilon - \Phi(z_{j+1})  \leq \varepsilon.
\end{align*}
 Thus  assertion (iii) is shown. Since  $v \in S_1$, (iv) is clear. It also
results from it that
$$
\frac{\varepsilon}{\delta\alpha(d(x_0,v))}d(v,u) \leq \Phi(u) -
\Phi(v) \leq \inf_E \Phi + \varepsilon -\Phi(v) \leq \varepsilon.
$$
The assertion (v) is shown.

Finally, we prove   (vi), let $w \in E$ such that $w \prec v$ and
$w \in \bar B(u, \gamma(u))$, then we have $ w \prec z_n$ for
every $n $. This gives  $w \in \cap_n S_n$ and  $w = v$,
which means that $v$ be an minimal element in  $\bar B(u,
\gamma(u))$, i.e.
$$
w \in \bar B(u, \gamma(u)) \quad\text{and}\quad w\prec v \Rightarrow w = v.
$$
Consequently,
$$
\Phi(w) > \Phi(v) - \frac{\varepsilon}{\delta\alpha(d(x_0,w))}d(v,w)
$$
for every $ w \in \bar B(u, \gamma(u)) \setminus B(x_0, d(x_0,v))$.
 The proof is complete.
\end{proof}

\section{Applications}
In this section $H$ denotes a Hilbert space, recall that a
function
 $ \Phi: H \to \mathbb{R}$ is called
G\^ateaux differerentiable if at every point $x_0$, there exists a
continuous linear functional $f'(x_0)$ such that, for every
$e \in X$,
$$
\lim_{t \to 0}\frac{f(x_0+te) - f(x_0)}{t} =
\langle f'(x_0), e\rangle .
$$
We always assume that $\alpha: [0,\infty[ \to ]0,\infty[$ is a continuous
nondecreasing comparison function of
order  $k$. For the rest of the  text we will write
$$
\Phi^c =\{u \in H : \Phi(u) \leq c\},
$$
for the sublevel sets as usual.

\begin{definition} \rm
We  say that  $\Phi$ satisfies  $(C^{\alpha}_c)$ if:
Every sequence  $(u_n)_n \subset H$ such that
$ \Phi(u_n) \to c$ and $\Phi'(u_n) \alpha(\|u_n\|) \to 0$ possesses a
convergent subsequence.
\end{definition}

\begin{remark} \label{rmk3.1}\rm
Note that if $\alpha(s) = cte$, the  $(C^{\alpha}_c)$ condition is
just the famous Palais-Smale condition and if $\alpha(s) = s + 1$,
$(C^{\alpha}_c)$ is $(C)$ condition introduced by Cerami in
\cite{Ce}.
\end{remark}

We can now state the following result.

\begin{theorem} \label{thm3.1}
Let  $H$ be a  Hilbert space, $ \Phi: H \to \mathbb{R}$ lower
semi-continuous, bounded below and  G\^ateaux differentiable. Let
$\alpha: [0,\infty[ \to ]0,\infty[$ a continuous nondecreasing
comparison function of order $k$. Assume that  for every
$\varepsilon > 0$,
$$
\Phi^{a + \varepsilon} \cap K \neq \emptyset,
$$
with $K$ is bounded in $H$ and $\Phi$ satisfies
$(C^{\alpha}_a)$, with $a = \inf_{H}\Phi$, then $\Phi$ has a
minimal point.
\end{theorem}

For the proof of this theorem we will use the following lemmas.

\begin{lemma} \label{lem3.1}
Under the conditions of Theorem \ref{thm3.1},
for every $\varepsilon > 0$, every $u \in H$ such that $\Phi(u)
\leq \inf_H \Phi + \varepsilon$ and every $\delta > 0$ such that
$$
\delta \leq \frac{\|u\|+1}{2\alpha(3(1+\|u\|))}
$$
 there exists  $v \in H$ that satisfies
\begin{enumerate}
\item $\Phi(v) \leq \Phi(u)$ \item $\frac{\|v-u\|}{\alpha(\|v\|)}
\leq \delta$ \item for every $h \in H, t \in \mathbb{R}$ such
that$ \|h\| = 1, |t| \leq 1$  and $t<v,h> \geq 0$ we have
$$
\Phi(v + th) \geq \Phi(v) - \frac{\varepsilon}{\delta\alpha(\|v + th\|)}|t|.
$$
\end{enumerate}
\end{lemma}

\begin{proof} Let in Theorem \ref{thm2.1}, $x_0 = 0, \gamma(u) = 2(\|u\| +
1)$ and $d(x,y) =\|x - y\|$ for every $x, y \in H$. Then, by $iv)$
and  $v)$ of Theorem \ref{thm2.1}, there exists $v \in H$ ( $v = \lim_{n
\to \infty} z_n, (z_n)$ the sequence built in theorem \ref{thm2.1}) such
that
\begin{equation}\label{e7}
\Phi(v) \leq \Phi(u) ~~{\rm and}~ \|v-u\| \leq
\delta\alpha(\|v\|). \end{equation}
Thus  assertions 1. and  2. follow.

Now we  prove the assertion 3. Let $h \in H$ such that $\|h\|
= 1$ and $|t| \leq 1$ we have $v \in  \bar B(u, \|u\|+1)$. Indeed,
if not $\|u\|+1 < \|v - u \|$. Since $\alpha(.)$ is nondecreasing,
$ \delta \leq \frac{\|u\|+1}{2\alpha(3(1+\|u\|))} $ and by
(\ref{e7}), it results
$$
\|u\|+1 < \|v- u\| \leq \delta \alpha(\|v\|)
\leq \delta \alpha(3(1+\|u\|))) \leq \frac{\|u\|+1}{2}.
$$
This is a contradiction.
Furthermore,   we have
\begin{align*}
\|v + th\| & \leq \|v\| + |t|\|h\| = \|v\| + |t| \\
&\leq 2\|u\| + 1 + 1 = \gamma(u).
\end{align*}
On the other hand, it is clear , since $t\langle v,h\rangle \geq 0$, that
$$
\|v +th\| = [\|v\|^2+\|th\|^2+2t<v,h>]^{1/2} \geq \|v\|.
$$
Thus, by  (iv), (v), (vi) of Theorem \ref{thm2.1}, assertions 1, 2, 3 of the lemma
follow.
\end{proof}

\begin{lemma} \label{lem3.2}
Under the conditions of Theorem \ref{thm3.1}, we have
\begin{equation}\label{e8}
|\langle \Phi'(v),h\rangle | \leq \frac{\varepsilon}{\delta\alpha(\|v\|)},
\quad \forall h \in H, \|h\| = 1.
\end{equation}
\end{lemma}

\begin{proof}  Let $h \in H$ such that $\|h\| = 1$ and consider
 two cases:\\
Case 1. If $\langle v,h\rangle  \geq 0$  and $t > 0$, from 3. of
Lemma \ref{lem3.1}
and $\Phi$ being  G\^ateaux differentiable, letting   $t$ approach $0$, we
obtain
$$
\langle \Phi'(v),h\rangle  \geq - \frac{\varepsilon}{\delta\alpha(\|v\|)}.
$$
 Case 2. In the similar way, if $ \langle v,h \rangle \leq 0$  and
$t < 0$ goes to $0$, we have
$$
\langle \Phi'(v),h\rangle  \leq  \frac{\varepsilon}{\delta\alpha(\|v\|)},
\quad \forall h, \|h\| = 1.
$$
Thus the Lemma \ref{lem3.2} follows.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm3.1}]
 For  $\varepsilon = \frac{1}{n}$, with
$n \geq 1$, there exists a sequence  $(u_n) \subset K$ such that
$$
\Phi(u_n) \leq a + \frac{1}{n}
$$
and, since $(u_n)$ is bounded, there exists $\delta > 0$ such that
$$
\delta \leq \frac{\|u_n\|+1}{\alpha(3(1+\|u_n\|))}, \forall n \geq 1.
$$
Consequently, by Lemma \ref{lem3.1} and Lemma \ref{lem3.2}, there exists a
sequence $(v_n)$ satisfying
\begin{itemize}
\item[(i)] $\Phi(v_n) \leq \Phi(u_n)$
\item[(ii)]
$\|\Phi'(v_n)\|\alpha(\|v_n\|) \to 0$, as $n \to \infty$.
\end{itemize}
The $(C^\alpha_a)$ condition implies that  $(v_n)$ has a
subsequence $(v_{n_k})$ convergent to some point $u$. Since $\Phi$
is lower semi-continuous, we get
$$
\inf_{H}\Phi \leq \Phi(u) \leq \liminf_{k\to
\infty}\Phi(v_{n_k}) \leq \inf_{H}\Phi.
$$
Therefore, $\Phi(u) = \inf_{H}\Phi$.
\end{proof}

Now, we illustrate Theorem \ref{thm3.1} by an example where the function
$\Phi$ checks the conditions of Theorem \ref{thm3.1}, but the
Palais-Smale condition and  Cerami condition do not hold.

\noindent{\bf Example.}  Consider
 $$
f(s) = \begin{cases} \arctan(s)& \text{if } s \leq 0 \\
\sin(s)  & \text{if }  0 \leq s \leq 2\pi \\
\arctan(s-2\pi) & \text{if } s \geq 2\pi.
\end{cases}
$$
and $\Phi(u) = f(2\pi + Log(\|u\|^2+1) - (\|u\|^2 +
1)^\frac{1}{2})$ for  $ u \in H$. It is clear  that $\Phi$
is  $C^1$ functional and  $a = \inf_H\Phi = -1$. Take
$$
K = \{ u\in H : \log(\|u\|^2+1) - (\|u\|^2 + 1)^\frac{1}{2} \in [-2\pi,
0]\},
$$
it is easy to see that $\Phi^{-1+\varepsilon} \cap K \neq \emptyset$
for every $\varepsilon > 0$. On the other hand, $\Phi$ satisfies
$(C^{\alpha}_c)$, with $\alpha(s) = s^2+1$, and by Theorem \ref{thm3.1},
$\Phi$ has a minimal point $u_0$ which $\Phi'(u_0) = 0$.

\begin{theorem} \label{thm3.2}
Let $\Phi : H \to \mathbb{R}$ be G\^ateaux differentiable and bounded below,
 says $a \inf_{H}\Phi$. Assume that
$\alpha: [0,\infty[ \to ]0,\infty[$ be a continuous nondecreasing
 function  such that  $\int_{1}^{\infty} \frac{1}{\alpha(s)} \,ds
= +\infty$. If $ \Phi$ satisfies $(C^\alpha_a)$ then the set
$\Phi^{a+\beta}$ is bounded, for some $\beta > 0$.
\end{theorem}

The main point to prove Theorem \ref{thm3.2} is the following.

\begin{lemma} \label{lem3.3}
Under the conditions of Theorem \ref{thm3.2},
for every $\varepsilon > 0$, every $u \in H$ such that $\Phi(u)
\leq \inf_H \Phi + \varepsilon$ and every $\delta > 0$ there
exists  $v \in H$ satisfies
\begin{enumerate}
\item $\Phi(v) \leq \Phi(u)$
\item $\frac{\|v-u\|}{\alpha(\|v\|)} \leq \delta$
\item for every $h \in H$ such that $ \|h\| = 1$, we
have
$$
|\langle \Phi'(v),h\rangle | \leq \frac{\varepsilon}{\delta\alpha(\|v\|)}.
$$
\end{enumerate}
\end{lemma}

\begin{proof}  Let in Theorem \ref{thm2.1}, $x_0 = 0$ and
$d(x,y) = \|x - y\|$ for every $x, y \in H$. From theorem \ref{thm2.1}
there exists a sequence  $(z_n)_{n\geq 1}$ satisfying  $(\|z_n\|)$ is
nondecreasing and
\begin{equation}\label{e9}
 \sum_{n=1}^{j} \frac{\|z_n-z_{n+1}\|}{\alpha(\|z_{n+1}\|)} < 2\delta, \quad
 \forall j \geq 1.
 \end{equation}
  However, since   $\int_{1}^{\infty} \frac{1}{\alpha(s)} \,ds =
+\infty$ there exists  $\gamma > 0$ such that
\begin{equation}\label{e10}
\delta \leq
\frac{1}{2}\int_{\|u\|}^{\|u\|+\gamma} \frac{1}{\alpha(s)}\,ds.
\end{equation}
 Put  $v = \lim_{n\to\infty} z_n$ and $\gamma(u) = 2\|u\| + \gamma
 +1$ in Theorem \ref{thm2.1}.
Thus, by  (iv)-(v) of Theorem \ref{thm2.1}, we obtain
 $$
\Phi(v) \leq \Phi(u) \quad\text{and}\quad
\|v-u\| \leq \delta\alpha(\|v\|).
$$
 For the proof of assertion  3, it is enough  to verify that
 $h \in H$ such that  $\|h\| = 1$ we have $v + th \in \bar B(u,\gamma(u))$
 for every  $t$  sufficiently small. Now we prove that
\begin{equation}\label{e11}
\|z_n\| \leq \|u\|+\gamma, \quad \forall n \geq 1.
\end{equation}
 If not, there exists  $j \geq 1$ such that $\|z_{j+1}\| > \|u\|+\gamma$.
However, by (\ref{e10}) and $\alpha$ is nondecreasing, we
obtain
\begin{align*}
2\delta &\leq  \int_{\|z_1\|}^{\|z_{j+1}\|} \frac{1}{\alpha(s)} \,ds  \\
 &\leq  \sum_{n=1}^{j} \int_{\|z_n\|}^{\|z_{n+1}\|} \frac{1}{\alpha(s)} \,ds \\
&\leq  \sum_{n=1}^{j} \frac{\|z_{n+1}\| - \|z_n\|}{\alpha(\|z_{n+1}\|)}   \\
 &\leq \sum_{n=1}^{j} \frac{\|z_n - z_{n+1}\|}{\alpha(\|z_{n+1}\|)}.
\end{align*}
This contradicts (\ref{e9}).
Using  (\ref{e11}), we have
\begin{equation}\label{e12}
\|v - u\| \leq 2\|u\| + \gamma.
\end{equation}
Thus, for  $|t| \leq 1$ and  $h \in H$ such that $ \|h\| = 1$ and
by (\ref{e12}), it results
$$\|v + th - u\| \leq 2\|u\| + \gamma +1 = \gamma(u).$$
Finally, the Lemma \ref{lem3.2} allows to conclude. The proof is complete.
\end{proof}

\begin{proof}[Proof of theorem \ref{thm3.2}]
Suppose, by contradiction, that $\Phi^{a + \beta}$ is unbounded for
 all $\beta > 0$. Then, there exists
$(u_n)$ such that $\|u_n\| \geq n$ and
$$
\Phi(u_n) \leq a + \frac{1}{n}.
$$
 and Lemma \ref{lem3.3} with  $\varepsilon = (\frac{1}{n})^2, \delta = \frac{1}{n}$
implies the existence of  $(v_n)$ satisfying
\begin{itemize}
\item[(i)] $\Phi(v_n) \leq \Phi(u_n)$
\item[(ii)]$ \|v_n - u_n\|
\leq \frac{1}{n} \alpha(\|v_n\|)$
\item [(iii)] $\|\Phi'(v_n)\|\alpha(\|v_n\|) \to 0$, as $n \to \infty$.
\end{itemize}
We reach a contradiction with $(C^\alpha_a)$, since (i)-(iii) give
respectively
\begin{enumerate}
\item $\Phi(v_n) \to a$, as $n \to \infty$,
\item $ \|v_n\| \to \infty$, as $n \to \infty$,
\item $\|\Phi'(v_n)\|\alpha(\|v_n\|) \to 0$, as $n \to \infty$.
\end{enumerate}
\end{proof}

As an immediate  consequence of the above results we have the following
result.

\begin{corollary}
Let  $H$ be a  Hilbert space, $ \Phi: H \to \mathbb{R}$ lower
semi-continuous, bounded below and  G\^ateaux differentiable.
Assume that $\alpha: [0,\infty[ \to ]0,\infty[$ be a continuous
nondecreasing  function  such that
 $\int_{1}^{\infty} \frac{1}{\alpha(s)} \,ds = +\infty$.
If $ \Phi$ satisfies $(C^\alpha_a)$, with
$a = \inf_{H}\Phi$, then $\Phi$ has a minimal point.
\end{corollary}

\begin{thebibliography}{00}

\bibitem{E1} Ekeland I., \emph{On the variational principle},
  J. Math. Anal. Applic., 47 (1974) 324-357.

\bibitem{E2} Ekeland I., \emph{Convexity methods in Hamiltonian mechanics},
  Springer, Berlin, (1990).

\bibitem{Ce}  Cerami G., \emph{Un criterio de esistenza per i punti critici
su variet\'a ilimitate},  Rc. Ist. Lomb. Sci. Lett. 121,(1978) 332-336.

\bibitem{C} Clark, F. H. \emph{Optimisation and Nonsmooth Analysis},
  Wiley, NewYork1983.

\bibitem{CS} Costa, D. G., Elves De B. eSilva;
\emph{The Palais-Smale condition versus coercivity},
 Nonlinear Analysis, T. M. A. 16 (1991) 371-381.

\end{thebibliography}


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