\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2011 (2011), No. 160, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2011 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2011/160\hfil Ume's u-distance]
{Ume's u-distance and its relation with both (PS)-condition
and coercivity}

\author[G. Goga\hfil EJDE-2011/160\hfilneg]
{Georgiana Goga}

\address{Georgiana Goga \newline
Nicolae Rotaru College, 2, Ion Corvin St., Constanta, Romania}
\email{georgia\_goga@yahoo.com}

\thanks{Submitted August 15, 2011. Published November 30, 2011.}
\subjclass[2000]{49K27, 54E35, 54E50, 58E30}
\keywords{$u$-distance; Palais-Smale condition;
 coercivity; \hfill\break\indent Zhong's variational principle}

\begin{abstract}
 In this article, we study the connection between the
 $u$-distance and a new Palais-Smale condition of compactness.
 We compare this Palais-Smale condition with the coercivity.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks


\section{Introduction and preliminaries}


In 1997, Zhong \cite{z1,z2} generalized the Ekeland variational
principle and proved the existence of minimal points for
 G\^ateaux-differentiable functions under weak (PS) conditions. The
following theorem is well-known and we name it Zhong's variational
principle (ZVP).

\begin{theorem}[\cite{z1,z2}] \label{thm1.1} Let
$(X,d)$ be a complete metric space, $x_0\in X$
fixed and $f:X\to (-\infty ,\infty ]$
a proper lower semicontinuous function which is bounded from below.
Let $h:[0,\infty )\to [0,\infty )$ be
a nondecreasing continuous function such that
\begin{equation*}
\int_0^{\infty }\frac{1}{1+h(r)}\,dr=+\infty .
\end{equation*}
 Then, for every $\varepsilon >0$, every $y\in X$
such that
\begin{equation*}
f(y)<\int_{x\in X} f(x) +\varepsilon ,
\end{equation*}
 and $\lambda >0$, there exists some point $z\in X$
 such that
\begin{itemize}
\item[(i)] $f(z)\leq f(y)$,

\item[(ii)] $d(x_0,z)\leq r_0+r^{\ast}$,

\item[(iii)] $f(x)\geq f(z)-\frac{\varepsilon }{\lambda (1+h(d(x_0,z))
)}\cdot d(z,x)$, for all $x\in X$,
where $r_0=d(x_0,y)$, and $r^{\ast }$ is such
that
\begin{equation*}
\int_{r_0}^{r_0 +r^{\ast }}\frac{1}{1+h(t)}\,dt\geq \lambda .
\end{equation*}
\end{itemize}
\end{theorem}

In 2010, Ume \cite{u1} introduced a new concept of distance
called $u$-distance, which generalizes some distances anterior
studied (see e.g., $\omega$-distance \cite{k1,w1},
Tataru's distance \cite{t1},  $\tau $-distance \cite{s2})
and expanded the celebrated Ekeland's
variational principle.

In Section 2, we present a generalization of Zhong's
variational principle using  Ume's $u$-distance.
In Section 3, we define a new Palais-Smale condition related
to above variational principle and we study the existence of
the minimal point for G\^{a}teaux-differentiable functions.
In the last section, we deal with the relation between new
Palais-Smale condition and the coercivity, following a techniques
which is based on $u$-distance. Our results extend and improve
other known results due to Zhong \cite{z1,z2}, Ekeland \cite{e1,e2}
and Costa \& Silva \cite{c3}.

For the beginning, we present some  results needed in
our approach. First, we recall Ume's \cite{u1} concept of generalized
distance in metric spaces.



\begin{definition} \label{def1.2} \rm
Let $(X,d)$ be a metric space. A function
$p:X\times X\to\mathbb{R}_{+}$ is called $u$-distance on $X$
if there exists a map
$\Theta :X\times X\times\mathbb{R}_{+}\times\mathbb{R}_{+}
\to\mathbb{R}_{+}$ such that the following conditions hold:
\begin{itemize}
\item[(U1)] $p(x,z)\leq p(x,y) +p(y,z)$, for all  $x,y,z\in X$;

 \item[(U2)] $\Theta (x,y,0,0)=0$
and $\Theta (x,y,s,t)\geq \min \{ s,t\} $
for all $x,y\in X$, $s,t\in\mathbb{R}_{+}$, and for every
$x\in X$ and $\varepsilon >0$, there is $\delta >0$ such that
\begin{equation*}
| \Theta (x,y,s,t)-\Theta (
x,y,s_0,t_0)| <\varepsilon
\end{equation*}
if $| s-s_0| <\delta $, $|t-t_0| <\delta $,
$s,s_0,t,t_0\in\mathbb{R}_{+}$ whenever $y\in X$;

\item[(U3)] $\lim_nx_n=x$  and $\limsup_n
\{ \Theta (w_n,z_n,p(w_n,x_{m})
,p(z_n,x_{m})):m\geq n\} =0$ imply
\begin{equation*}
p(y,x)\leq  \liminf_{n\to \infty }p(y,x_n)\quad\text{ for }
y\in X;
\end{equation*}


\item[(U4)] The  four equalities
\begin{gather*}
 \limsup_n \{ p(x_n,w_{m}):m\geq n\} =0, \quad
 \limsup_n \{ p(y_n,z_{m}):m\geq n\} =0, \\
 \lim_n\Theta (x_n,w_n,s_n,t_n)=0, \quad
 \lim_n\Theta (y_n,z_n,s_n,t_n)=0
\end{gather*}
 imply $\lim_n\Theta (w_n,z_n,s_n,t_n)=0$;
or the  four equalitires
\begin{gather*}
\limsup_n \{ p(w_{m},x_n):m\geq n\} =0, \quad
\limsup_n \{ p(z_{m},y_n):m\geq n\} =0, \\
\lim_n\Theta (x_n,w_n,s_n,t_n)=0,\quad
\lim_n\Theta (y_n,z_n,s_n,t_n)=0
\end{gather*}
 imply $\lim_n\Theta (w_n,z_n,s_n,t_n)=0$;

\item[(U5)] The  two equalities
\begin{gather*}
\lim_n\Theta (w_n,z_n,p(w_n,x_n),p(
z_n,x_n))=0, \\
\lim_n\Theta (w_n,z_n,p(w_n,y_n),p(
z_n,y_n))=0
\end{gather*}
imply $\lim_nd(x_n,y_n)=0$;
or the two equalities
\begin{gather*}
\lim_n\Theta (a_n,b_n,p(x_n,a_n),p(
x_n,b_n))=0, \\
\lim_n\Theta (a_n,b_n,p(z_n,a_n),p(
y_n,b_n))=0
\end{gather*}
imply $\lim_nd(x_n,y_n)=0$.
\end{itemize}
\end{definition}


\begin{example}[\cite{u1}] \label{examp1.3} \rm
 Let $X$ be a space with norm $\|\cdot\|$. Then
the function $p:X\times X\to\mathbb{R}_{+}$ defined by
$p(x,y)=\|x\|$
is a $u$-distance on $X$, but it is not a $\tau$-distance on $X$,
in Suzuki's sense \cite{s2}.
\end{example}


\begin{example}[\cite{u1}] \label{examp1.4}\rm
Let $p$ be a $u$-distance on a metric space $(X,d)$ and
let $c$ be a real positive number. Then a function
$q:X\times X\to \mathbb{R}_{+}$ defined by
$q(x,y)=cp(x,y)$
 for every $x,y\in X$ is also a $u$-distance on
$X$.
\end{example}

By means of the generalized $u$-distance, Ume obtained in
\cite{u1} the following version of Ekeland's variational principle.
This result will play a crucial role in the proof of our variational
principle.

\begin{theorem}[\cite{u1}] \label{thm1.5}
Let $(X,d)$ be a complete metric space, let
$f:X\to (-\infty ,\infty ]$ be a proper lower semicontinuous
function which is bounded from below, and let
$p:X\times X\to\mathbb{R}_{+}$ be a $u$-distance on $X$.
Then the following two statements hold:
\begin{itemize}

\item[(1)] For each $x\in X$ with
$f( x)<\infty $, there exists $v\in X$ such that
$f(v)\leq f(x)$ and $f(w) >f(v)-p(v,w)$,  for all
$w\in X\backslash \ v\}$.

\item[(2)] For each $\varepsilon >0$, $\lambda>0 $ and
$x\in X$ with $p(x,x)=0$
and $f(x)<\inf_{a\in X} f(a)+\varepsilon $, there exists
$v\in X$ such that
\begin{gather*}
f(v)\leq f(x), \quad p(x,v)\leq \lambda, \\
 f(w)>f(v)-\frac{\varepsilon }{\lambda }
\cdot p(v,w),\quad \text{for all } w\in X\backslash \{v\} .
\end{gather*}
\end{itemize}
\end{theorem}

\section{A generalization of Zhong's variational principle}

We start this section by extending a result by Suzuki \cite{s3},
using the $u$-distance.

\begin{proposition} \label{prop2.1}
Let $(X,d)$ be a
complete metric space and let $p:X\times X\to\mathbb{R}_{+}$ be a
$u$-distance on $X$. Let $q:X\times X\to\mathbb{R}_{+}$ be a function
such that
\begin{itemize}

\item[(a)] $q(x,z)\leq q(x,y) +q(y,z)$ for all $x,y,z\in X$;

\item[(b)] $q$ is lower semicontinuous in its
second argument;

\item[(c)] $q(x,y)\geq p(x,y)$  for all $x,y\in X$.
\end{itemize}
 Then $q$ is also a $u$-distance.
\end{proposition}

\begin{proof}
Assumption (a) is equivalently with
(U1)$_q$. Let $\Theta :X\times X\times\mathbb{R}_{+}\times\mathbb{R}
_{+}\to\mathbb{R}_{+}$ be a function satisfying
(U2)--(U5). Clearly, (U3)$_q$ follows from (b). Now,
we assume that
\begin{equation}
\begin{gathered}
\limsup_n \{ q(x_n,w_{m}):m\geq n\} =0, \\
\limsup_n \{ q(y_n,z_{m}):m\geq n\} =0, \\
\lim_n\Theta (x_n,w_n,s_n,t_n)=0,   \\
\lim_n\Theta (y_n,z_n,s_n,t_n)=0.
\end{gathered} \label{e2.1}
\end{equation}
By \eqref{e2.1} and (c), we have
\begin{gather*}
\limsup_n\{ p(x_n,w_{m}):m\geq n\}=0,\\
\limsup_n\{ p(y_n,z_{m}):m\geq n\}=0.
\end{gather*}
Therefore, by (U4), we find $\lim_n\Theta(w_n,z_n,s_n,t_n)=0$,
and derive (U4)$_q$.

Next, we assume that
\begin{gather}
\lim_n\Theta (w_n,z_n,q(w_n,x_n),q(
z_n,x_n))=0,  \label{e2.2} \\
\lim_n\Theta (w_n,z_n,q(w_n,y_n),q(
z_n,y_n))=0.  \label{e2.3}
\end{gather}
Applying again (c) in \eqref{e2.2} and \eqref{e2.3},
we obtain
\begin{gather*}
\lim_n\Theta (w_n,z_n,p(w_n,x_n),p(z_n,x_n))=0,\\
\lim_n\Theta (w_n,z_n,p(w_n,y_n),p(z_n,y_n))=0.
\end{gather*}
By  (U5), we have $\lim_n d(x_n,y_n)=0$, and (U5)$_q$,
is also verified.
\end{proof}

Next, we establish a more general variational principle
\cite{b1,t2}, which is an extension of both Ekeland's and
Zhong's variational principles.

\begin{theorem} \label{thm2.2}
Let $(X,d)$ be a complete metric space, $a\in X$ be a fixed point
and let $ p:X\times X\to\mathbb{R}_{+}$ be a $u$-distance on
$X$ lower semicontinuous in its second argument.
Let $f:X\to (-\infty ,\infty ]$ be a proper lower semicontinuous
function which is bounded from below and let
$b:[0,\infty )\to (0,\infty )$ be a non-increasing continuous
function such that
\begin{equation*}
B(t)=\int_0^{t}b(r)dr,
\end{equation*}
where $B$  is a $C^{1}$ function from
$\mathbb{R}_{+}$ to itself and $B(\infty )=+\infty $.
 Let $y\in X$ be such that $p(y,y)=0$ and
\begin{equation}
f(y)>\inf_{x\in X} f(x).  \label{e2.4}
\end{equation}
Then, for $\epsilon _0>0$, there exists $z\in X$ such that
\begin{itemize}

\item[(i)] $f(z)\leq f(y)$,

\item[(ii)] $p(a,z)\leq \beta (y)+\beta ^{\ast }$,

\item[(iii)] $f(x)>f(z)-\frac{\epsilon _0}{\lambda }b(\beta (z))p(
z,x)$, for all $x\in X$
 where $\beta (.)=p(a,.)$, and $\beta ^{\ast }$
 is such that
\begin{equation}
\int_{\beta (y)}^{\beta (y) + \beta^{\ast }}b(t)\,dt
\geq \alpha (y),  \label{e2.5}
\end{equation}
with $\alpha (y)=f(y)-\inf{x\in X} f(x)\geq \lambda >0$.
\end{itemize}
\end{theorem}

\begin{proof}
First, we define the function $q:X\times X\to \mathbb{R}_{+}$ by
\begin{equation*}
q(x,y):=\int_{p(a,x)}^{p(a,x) +  p(x,y)}b(t)\,dt.
\end{equation*}
Since $b$ is non-increasing, for $(x,z)\in X\times X$, we deduce
\begin{align*}
q(x,z)&= \int_{p(a,x)}^{p(a,x) + p(x,z)}b(t)\,dt\\
&\leq \int_{p(a,x)}^{p(a,x) + p(x,y) + p(y,z)}b(t)\,dt \\
&= \int_{p(a,x)}^{p(a,x) + p(x,y)}b(t)\,dt+\int_{p(a,x)
 + p(x,y)}^{p(a,x) + p(x,y)  + p(y,z)}b(t)\,dt \\
&\leq \int_{p(a,x)}^{p(a,x) + p(x,y)}b(t)\,dt+\int_{p(a,y)}^{p(a,y)
 + p(y,z)}b(t)\,dt \\
&= q(x,y)+q(y,z).
\end{align*}
In addition, $q$ is obviously lower semicontinuous in its second
variable.
On the other hand, we have
\begin{equation} \label{e2.6}
\begin{aligned}
q(x,y)&= \int_{p(a,x)}^{p(a,x) + p(x,y)}b(t)\,dt   \\
&= B(p(a,x) + p(x,y)) -B(p(a,x)) \\
&\geq  b(p(a,x) + p(x,y))p(x,y).
\end{aligned}
\end{equation}
Taking into account the definition of function $b$, we obtain
boundedness from below,
\begin{equation}
b(p(a,x) + p(x,y)) >b(\infty )\geq M\geq 0  \label{e2.7}
\end{equation}
Combining \eqref{e2.6} and \eqref{e2.7}, we deduce
\begin{equation*}
q(x,y)\geq Mp(x,y).
\end{equation*}
Since $Mp(x,y)$ is a $u$-distance and the assumptions of
Proposition \ref{prop2.1} are verified, $q(x,y)$ is also $u$-distance.

Now, from \eqref{e2.4} and \eqref{e2.5}, we obtain
\begin{equation} \label{e2.8}
\begin{split}
0 &< \lambda \leq f(y)-\underset{x\in X}{\inf }f(x)
=\alpha (y)\\
&\leq \int_{\beta (y)}^{\beta (y) + \beta ^{\ast }}b(t)\,dt
= \int_0^{ \beta ^{\ast }}b(u+\beta (y)) \,du\\
&\leq \int_0^{ \beta ^{\ast }}b(u)du=B(\beta^{\ast }).
\end{split}
\end{equation}
So by the above inequality,
\begin{equation*}
f(y)\leq  \inf_{x\in X} f(x) +B(\beta ^{\ast }),
\end{equation*}
and the Theorem \ref{thm1.5} is applicable to $q(x,y)$ for
 $ \varepsilon =B(\beta ^{\ast })>0$ and
$\lambda =\alpha ( y)>0$. Therefore, there exists $z\in X$
 such that
\begin{gather}
f(z)\leq f(y),  \label{e2.9}\\
q(y,z)\leq \alpha (y)\label{e2.10} \\
f(x)>f(z)-\frac{B(\beta ^{\ast })}{
\alpha (y)}\cdot q(z,x),\quad \forall x\neq z,\;
x\in X.  \label{e2.11}
\end{gather}
By (U1), we know that
\begin{equation}
\ p(a,z)\leq p(a,y)+p(y,z)
= \beta (y)+p(y,z). \label{e2.12}
\end{equation}
On the other hand, from \eqref{e2.5} and \eqref{e2.10} it
follows that
\[
B(\beta (y) +p(y,z))-B(\beta (y))
\leq \alpha (y)
\leq B(\beta (y) +\beta ^{\ast } )-B(\beta (y)).
\]
Thereby, we find that
\begin{equation}
p(y,z)\leq \beta ^{\ast },  \label{e2.13}
\end{equation}
because $B$ is a nondecreasing function. Thus, (ii) follows
from \eqref{e2.12} and \eqref{e2.13}. Moreover, since
\begin{equation}
q(z,x)= \int_{p(a,z)}^{p(a,z)+ p(z,x)}b(t)\,dt
\leq b(p(a,z))p(z,x)
= b((\beta (z)))p(z,x); \label{e2.14}
\end{equation}
multiplying by $(-1)$ and, using
\eqref{e2.8} and \eqref{e2.11}, for
$0<B(\beta ^{\ast})\leq \epsilon _0$, we obtain
\begin{align*}
f(x)
&>&f(z)-\frac{B(\beta ^{\ast })}{\alpha (y)}\cdot q(z,x)
&\geq f(z)- \frac{\epsilon _0}{\lambda }q(z,x)
&\geq f(z)-\frac{\epsilon }{\lambda }b((\beta(z)))p(z,x),
\end{align*}
for all $x\in X$, and (iii) is verified. This completes the
proof.
\end{proof}


\begin{remark} \label{rmk3.3} \rm
 Let $a$, $f$, $b$, $p$, $\alpha (y)$, $\beta (y)$, $\beta ^{\ast }$,
and $X$ be as in Theorem \ref{thm2.2}.
\begin{itemize}
\item[(i)] When $a=y$, $b(t)\equiv 1$, $\beta^{\ast }=\lambda $,
$\epsilon _0>\alpha (y)\geq \lambda >0$,
and $p(x,y)=d(x,y)$, Theorem \ref{thm2.2} reduces
to Ekeland's variational principle (EVP) \cite{e1,e2}.

\item[(ii)] Take $a=x_0$,
\begin{equation*}
b(t)=\frac{1}{1+h(t)},
\end{equation*}
where $h:[0,\infty )\to [0,\infty )$ is a
continuous nondecreasing function such that
\begin{equation*}
\int_0^{\infty }\frac{1}{1+h(r)}dr=+\infty ,
\end{equation*}
$\epsilon _0>\alpha (y)\geq \lambda >0$,
$\beta (y)=d(x_0,y)=r_0$, $\beta ^{\ast }=r^{\ast }$ and
$p(x,y)=d(x,y)$. Therefore, Theorem \ref{thm2.2} implies
Theorem \ref{thm1.1}.
\end{itemize}
\end{remark}

\section{The b-(PS) condition and the existence of a minimal point}

Throughout this section $X$ denotes a Banach space. We recall that a
function $f:X\to (-\infty ,\infty ]$ is called  G\^{a}teaux
differentiable at $x\in X$ with $f(x)<\infty $ if
there exists a continuous linear functional $f'(x)$
such that
\begin{equation*}
\lim_{t\to 0} \frac{f(x+ty)-f(x)
}{t}=\langle f'(x),y\rangle
\end{equation*}
holds for every $y\in X$.

In the following, we assume that
$f:X\to (-\infty ,\infty ]$ is G\^{a}teaux differentiable.


\begin{theorem} \label{thm3.1}
Let $a\in X$ be fixed and $p:X\times X\to\mathbb{R}_{+}$ a
$u$-distance on $X$  lower semicontinuous
in its second argument. Let $f:X\to (-\infty ,\infty ]$
be a proper lower semicontinuous function which is
bounded from below and let
$b:[0,\infty )\to(0,\infty)$ be a nonincreasing continuous
function such that
\begin{equation*}
B(t)=\int_0^{t}b(r)dr,
\end{equation*}
where $B$ is a $C^{1}$ function from
$\mathbb{R}_{+}$ to itself such that $B(\infty )=+\infty$.
Let $y\in X$ be such that $p(y,y)=0$ and
\begin{equation*}
f(y)>\inf_{x\in X} f(x).
\end{equation*}
Then, for every $\epsilon >0$, there exists
$z\in X$ such that
\begin{itemize}
\item[(i')]   $f(z)\leq f(y)$,

\item[(ii')]  $\beta (z)\leq \beta (y)+\beta ^{\ast }$,

\item[(iii')] $\|f'(z)\| /b(\beta (z))\leq \epsilon $
for all $x\in X$,
where $\beta (.)=p(a,.)$, and $\beta ^{\ast }$ is a real number
such that
\begin{equation*}
\int_{\beta (y)}^{\beta (y) + \beta
^{\ast }}b(t)\,dt\geq \alpha (y),
\end{equation*}
 with $\alpha (y)=f(y)-\inf_{x\in X} f(x)>0$.
\end{itemize}
\end{theorem}

\begin{proof}
We have the hypotheses of Theorem \ref{thm2.2}.
So, applying this theorem, we
obtain (i') and (ii') from (i) and (ii). Moreover,
(iii) guaranties that there exists  $z\in X$ such that
\begin{equation}
f(x)\geq f(z)-\frac{\epsilon }{\lambda }b(
\beta (z))p(z,x),\quad \text{for all }
x\in X,  \label{e3.1}
\end{equation}
where $0<\lambda \leq \alpha (y)$.
Choose $x=z+ty$ with $ \| y\| =1$ in \eqref{e3.1} and obtain
\begin{equation}
\frac{f(z+ty)-f(z)}{t}\geq - \frac{\epsilon
}{\lambda }\frac{b(\beta (z))p(z,z+ty)}{t},  \label{e3.2}
\end{equation}
for every $t>0$. Let $\lambda $ be such that
\begin{equation}
\lim_{t\to 0} \frac{p(z,z+ty)}{t}\leq \lambda.  \label{e3.3}
\end{equation}
Then, letting $t\to 0$ in \eqref{e3.2} and using \eqref{e3.3},
we conclude that
\begin{equation}
\langle f'(z),y\rangle \geq -\epsilon \cdot b(\beta (z)),  \label{e3.4}
\end{equation}
for all $y\in X$ with $\| y\| =1$. Since \eqref{e3.4}
 is true for $\pm y$, we deduce that
\begin{equation}
| \langle f'(z),y\rangle| \leq \epsilon \cdot b(\beta (z)).
\label{e3.5}
\end{equation}
Now, from \eqref{e3.5}, we obtain
\begin{equation*}
\| f'(z)\| =\sup_{y\in X,\| y\| =1}
\frac{| \langle f'(z),y\rangle | }{\| y\| }\leq
\epsilon \cdot b(\beta (z)),
\end{equation*}
and the claim (iii') holds.
\end{proof}

\begin{corollary} \label{coro3.2}
Suppose that the hypotheses of Theorem \ref{thm3.1}
are verified. Then there exists a minimizing sequence
$\{z_n\} _n$ of $f$ such that
\begin{gather*}
f(z_n)<\inf_{x\in X} f(x)+\epsilon,\\
\| f'(z_n)\| /b(\beta (z_n))\to 0.
\end{gather*}
\end{corollary}

The proof follows form  taking $\epsilon =\frac{1}{n}$,
$n=1,2,\dots$ in Theorem \ref{thm3.1}.


Let $\mathcal{B}$ be the set of all non-increasing and strictly
positive continuous functions $b:[0,\infty )\to(0,\infty )$
such that
\begin{equation*}
\int_0^{\infty }b(t)\,dt=\infty .
\end{equation*}
Let $p:X\times X\to\mathbb{R}_{+}$ be a $u$-distance on $X$
lower semicontinuous in its second
variable with $p(x,x)=0$ $\forall x\in X$, $a\in X$ a fixed
point and $\beta :X\to \mathbb{R}_{+}$ defined by
$\beta (x)=p(a,x)$.



\begin{definition} \label{def3.3} \rm
Let $f:X\to (-\infty ,+\infty ]$ be a $C^{1}$ function,
$c\in \mathbb{R}$ and $b\in\mathcal{B}$.
\begin{itemize}
\item $f$ is said to satisfy the b-(PS)
condition if any sequence $\{ x_n\} _n$ in $X$
such that $\{ f(x_n)\} $ is bounded and
$\| f'(x_n)\| /b( \beta (x_n))\to 0$  has a convergent
subsequence.

\item $f$ is said to satisfy the b-(PS)$_c$
 condition if any sequence $\{ x_n\} _n$ in $X$
 such that $f(x_n)\to c$  and
$\| f'(x_n)\| /b(\beta (x_n))\to 0$  has a convergent subsequence.
\end{itemize}
\end{definition}


\begin{remark} \label{rmk3.4} \rm
Suppose that $\beta (x)=d(a,x)$.
\begin{itemize}
\item Then the b-(PS) condition is the Schechter-(PS) condition
\cite{s1}.

\item If $b$ is constant, then the b-(PS) condition is the
usual $(PS)$ condition.

\item If $b(t)=1/(1+t)$, then the b-(PS)
condition is the Cerami-(PS) condition \cite{c2}.

\item If $b(t)=1/(1+h(t))$, where $h:[0,\infty )\to [0,\infty )$
is a non-decreasing function, then the b-(PS) condition
is the Zhong-(WPS) condition \cite{z1,z2}.
\end{itemize}
\end{remark}

\begin{theorem} \label{thm3.5}
If $f$ is bounded below and
satisfying the {\rm b-(PS)} condition, then $f$
has a minimal point.
\end{theorem}

\begin{proof}
By Corollary \ref{coro3.2}, there is a minimizing sequence
$\{ z_n\} _n$ in $X$ such that
$f(z_n)<\inf_{x\in X} f(x)+\epsilon $ and
$\| f'(z_n) \| /b(\beta (z_n))\to 0$. The
b-(PS) condition implies that $\{z_n\} _n$ has a subsequence
$\{ z_{n_{k}}\} _{k}$ convergent to some point $z^{\ast }$.
Since $f$ is lower semicontinuous, we obtain
\begin{equation*}
\inf_{X} f\leq f(z^{\ast })\leq \liminf_{k\to \infty }
f(z_{n_{k}})\leq \inf_{X} f.
\end{equation*}
Therefore, $f(z^{\ast })=\inf_{X}f$.
\end{proof}

\section{The b-(PS) condition versus coercivity}


Using the method of gradient flows, Li \cite{l1} first observed that
the (PS) condition implies the coercivity for $C^{1}$ functionals
bounded from below. Using Ekeland's variational principle,
Caklovic, Li and Willem \cite{c1} proved the same result for
a G\^ateaux differentiable functional which is lower semicontinuous.
The same conclusion was also proved by Costa and Silva \cite{c3}
and Brezis and Nirenberg \cite{b2} for $C^{1}$functionals by also
employing Ekeland's principle. Using ZVP, Zhong \cite{z1} studied
the connection between (WPS) and coercivity. A similar result was
established by Suzuki \cite{s2}, using $\tau$-distance.

In this section, we discuss the relation between the b-(PS)
 condition  and coercivity. We recall that a function
$f:X\to (-\infty ,\infty ]$ is said to be coercive if
\begin{equation*}
\underset{r\to \infty }{\lim }\underset{\| x\| \geq
r}{\inf }f(x)=\infty.
\end{equation*}
For our aim, we first prove the following lemma.


\begin{lemma} \label{lem4.1}
Let $p:X\times X\to \mathbb{R}_{+}$ be a
$u$-distance on $X$  and $f:X\to\mathbb{R}$ is a G\^{a}teaux
differentiable function. Suppose that there are
$\xi \geq 0$, $\delta >0$ and either of the following
conditions is satisifed:
\begin{itemize}
\item $f(y)\geq f(x)-\xi p(x,y)$
 for all $y\in X$ with $0<p(x,y)<\delta $;
 or
\item $f(y)\leq f(x)+\xi p(x,y)$
 for all $y\in X$ with $0<p(x,y)<\delta $.
\end{itemize}
Then $\| f'(x)\| \leq \xi $.
\end{lemma}

\begin{proof}
 Assume that
\begin{equation}
f(y)\geq f(x)-\xi p(x,y)\label{e4.1}
\end{equation}
for all $y\in X$ with $0<p(x,y)<\delta $. Set
$y=x+\delta z$ in \eqref{e4.1}, and infer that
\begin{equation}
f(x+\delta z)-f(x)\geq -\xi p(x,x+\delta
z)>-\xi \delta  \label{e4.2}
\end{equation}
Then,
\begin{equation}
\frac{f(x+\delta z)-f(x)}{\delta }>-\xi .
\label{e4.3}
\end{equation}
Taking the limit as $\delta \to 0$, we obtain
\begin{equation}
\langle f'(x),y\rangle \geq -\xi . \label{e4.4}
\end{equation}
As \eqref{e4.4} holds for both of $\pm y$, we derive
\begin{equation}
| \langle f'(x),y\rangle | \leq \xi .  \label{e4.5}
\end{equation}
Then, for all $y\in X$ with $\| y\| =1$, the inequality
\eqref{e4.5} implies that
\[
\| f'(x)\|
= \sup_{y\in X,\| y\| =1} \frac{| \langle f'(x),y\rangle | } {\| y\| }
= \sup_{y\in X,\| y\| =1} |\langle f'(x),y\rangle | \leq \xi,
\]
and the desired claim holds.
\end{proof}

Next, we consider a more suitable version of Theorem \ref{thm1.5},
for our purpose.

\begin{theorem} \label{thm4.2}
Let $(X,d)$ be a complete metric spaces, let
$f:X\to (-\infty ,\infty ]$  be a proper lower semicontinuous
function which is bounded from below, and let
$p:X\times X\to\mathbb{R}_{+}$ be a $u$-distance on $X$
lower semicontinuous in its second argument.
Then for $\varepsilon >0$  and $x\in X$  with $f(x)<\infty $  and
$p(x,x)=0$, there exists $v\in X$ such that
\begin{itemize}
\item[(i)]  $f(v)\leq f(x)-\varepsilon p(x,v)$;

\item[(ii)] $f(w)>f(v)-\varepsilon p(v,w)$,  for all
 $w\in X\backslash \{ v\}$.
\end{itemize}
\end{theorem}

For the sake of completeness, we supply a proof of the equivalence
between Theorems \ref{thm1.5} and \ref{thm4.2}.

\begin{proof}
$\Leftarrow $ Let the assumptions of Theorem \ref{thm1.5} be satisfied.
Obviously, the conclusion of (1) follows by Theorem \ref{thm4.2}. For
(2), applying again Theorem \ref{thm4.2} with $\varepsilon =\frac{e}{\lambda }$,
we deduce that
\begin{equation*}
p(x,v)\leq \frac{\lambda }{e}(f(x)-f(
v))\leq \frac{\lambda }{e}e\leq \lambda .
\end{equation*}
Hence the conclusion of Theorem \ref{thm1.5} is valid.

$\Rightarrow $ Now, suppose that Theorem \ref{thm1.5} holds.
Let $x\in X$ with $ f(x)<\infty $ \ and $\varepsilon >0$ be given.
Fix any $ e>f(x)-\inf_{a\in X} f(a)$ and set
$ \lambda =\frac{e}{\varepsilon }$. Consider
\begin{equation*}
M(x)=\{ v\in X\mid f(v)\leq f(x)-\varepsilon p(x,v)\} .
\end{equation*}
By the lower semicontinuity of $f$ and $p(x,.)$, the set
$M(x)$ is closed. Furthermore, $M(x)$ is nonempty
as $x\in M(x)$. Applying Theorem \ref{thm1.5} (2) for the
chosen $e$, $\lambda $ and for $M(x)$ instead of $X$ one finds
$v\in X$ such that
\begin{gather*}
f(v)\leq f(x), \quad
p(x,v)\leq \lambda , \\
 f(w)>f(v)-\frac{e}{\lambda }\cdot
p(v,w),\quad \text{for all }w\in M(x)\backslash \{v\} .
\end{gather*}
Since $v\in M(x)$, then (i) holds.

To show (ii) it is sufficient to check that
\begin{equation*}
f(w)>f(v)-\frac{e}{\lambda }\cdot p(
v,w),\quad \text{for all }w\notin M(x).
\end{equation*}
By the definition of $M(x)$, the property
$w\notin M(x)$ means that
\begin{equation*}
f(w)>f(x)-\varepsilon p(x,w).
\end{equation*}
From this and (i) we easily deduce (ii) and
then obtain Theorem \ref{thm4.2}.
\end{proof}

 We are in position to state the main result of this section. The
proof follows a technique developed by Suzuki in \cite{s2}.

\begin{theorem} \label{thm4.3}
Let $X$ be a Banach space, $a\in X$  fixed, and let
$p:X\times X\to\mathbb{R}_{+}$ be a symmetric
$u$-distance on $X$, lower semicontinuous in its second
argument and such that $p(x,x)=0$  for all $x\in X$.
 Let $f:X\to (-\infty ,\infty ]$ be
a proper lower semicontinuous function which is bounded from below
and let $b:[0,\infty )\to (0,\infty )$ be a non-increasing continuous
function such that
\begin{equation*}
B(t)=\int_0^{t}b(r)\,dr,
\end{equation*}
where $B$ is a function from $\mathbb{R}_{+}$  to itself such that
$B(\infty )=+\infty$.
 Let $a\in X$ be fixed and $\beta :X\to\mathbb{R}_{+}$ defined by
$\beta (x)=p(a,x)$.
 Assume that $f$ is G\^{a}teaux differentiable at every
point $x\in X$ with $f(x)\in\mathbb{R}$. If
\begin{equation*}
\alpha =\liminf_{\beta (y)\to \infty } f(y)\in\mathbb{R},
\end{equation*}
then there exists a sequence $\{ z_n\} _n$ in $X$ such that
\begin{itemize}
\item[(a)] $\lim_{n\to \infty } \beta (z_n)=\infty $;

\item[(b)] $\lim_{n\to \infty } f(z_n) =\alpha $;

\item[(c)] $\lim_{n\to \infty } \| f'(z_n)\| /b(\beta (z_n))=0$.
\end{itemize}
\end{theorem}

\begin{proof}
We shall  show only the following:
for every $\varepsilon >0$, there exists $v\in X$ satisfying
$\beta (v)\geq \frac{1}{\varepsilon }$,
 $ | f(v)-\alpha | \leq \varepsilon $ and
$\| f'(v)\| /b(\beta(v))\leq \varepsilon $.
Fix $\varepsilon >0$ and define a function
$\chi :[0,\infty )\to [0,\infty )$ by
\begin{equation}
\chi (t)=\frac{1}{2}b(t+1)\label{e4.6}
\end{equation}
for $t\in [0,\infty )$. Then $\chi $ is non-increasing, and
\[
\int_0^{\infty }\chi (t)\,dt
= \frac{1}{2}\int_0^{\infty}b(t+1)\,dt
= \frac{1}{2}\int_{1}^{\infty }b(t)\,dt=\infty .
\]
We also determine a function $h:X\to (-\infty,+\infty ]$ by
\begin{equation}
h(x)=\max \{ f(x),\alpha -2\varepsilon\}  \label{e4.7}
\end{equation}
for $x\in X$. Then it is obvious that $h$ is proper lower
semicontinuous and bounded from below. We choose
$r,r'\in\mathbb{R}$ with $\frac{1}{\varepsilon }<r<r'$, $1<r$,
\begin{gather}
\inf_{\beta (y)\geq r} f(y)>\alpha -\varepsilon ,  \label{e4.8}\\
\int_{r}^{r'}\chi (t)\,dt=3.  \label{e4.9}
\end{gather}
We also choose $u\in X$ with
\begin{equation}
\beta (u)>r',\quad f(u)<\alpha +\varepsilon .  \label{e4.10}
\end{equation}
We note that $h(u)=f(u)$ because $\beta (u)>r'>r$.
We know from the earlier that the function
$q:X\times X\to\mathbb{R}_{+}$, defined by
\begin{equation}
q(u,v)=\int_{\beta (u)}^{\beta (u) + p(u,v)}\chi (t)\,dt  \label{e4.11}
\end{equation}
is a $u$-distance. So, by Proposition \ref{prop2.1}, the function
$s:X\times X\to\mathbb{R}_{+}$, defined by
\begin{equation}
s(u,v)=q(u,v)+q(v,u)\label{e4.12}
\end{equation}
is also a $u$-distance. Thereby, by Theorem \ref{thm4.2}, there
exists $v\in X$ such that
\begin{gather}
h(v)\leq h(u)-\varepsilon s(u,v), \label{e4.13} \\
h(w)>h(v)-\varepsilon s(v,w),\quad
\forall w\neq v.  \label{e4.14}
\end{gather}
Arguing by contradiction, we assume that $\beta (v)<r$.
Moreover, we have
\begin{equation}
\beta (v)<r<r'<\beta (u).  \label{e4.15}
\end{equation}
Also, from \eqref{e4.11}, \eqref{e4.12} and
\eqref{e4.13}, we successively obtain
\begin{equation} \label{e4.16}
\begin{split}
\alpha -2\varepsilon
&\leq h(v)\leq h(u) -\varepsilon \int_{\beta (u)}^{\beta (u)
 + p(u,v)}\chi (t)\,dt-\varepsilon \int_{\beta (v)}^{\beta (v)
 + p(u,v)}\chi (t)\,dt   \\
&\leq h(u)-\varepsilon \int_{\beta (v)}^{\beta(v)
 + p(u,v)}\chi (t)\,dt \\
&\leq h(u)-\varepsilon (H(\beta (v)
 + p(u,v))-H(\beta (v))),
\end{split}
\end{equation}
where $H$ is a primitive of $\chi$. Using  that $H$
is nondecreasing in \eqref{e4.16}, we obtain
\begin{equation} \label{e4.17}
\alpha -2\varepsilon
\leq h(u)-\varepsilon (H(\beta (u) )-H(\beta (v)))
= h(u)-\varepsilon \int_{\beta (v)}^{\beta (u) }\chi (t)\,dt.
\end{equation}
Then, by \eqref{e4.15}, \eqref{e4.17} and \eqref{e4.10},
we obtain
\[
\alpha -2\varepsilon
\leq h(u)-\varepsilon \int_{r}^{r' }\chi (t)\,dt
= f(u)-3\varepsilon <\alpha -2\varepsilon ,
\]
which is a contradiction. Therefore,
\begin{equation*}
\beta (v)\geq r>\frac{1}{\varepsilon },
\end{equation*}
and (a) holds. Thus, we have $h(v)=f(v)$ and
\begin{equation*}
\alpha -\varepsilon <\underset{\beta (y)\geq r}{\inf }f(
y)\leq f(v)\leq f(u)<\alpha +\varepsilon .
\end{equation*}
This implies
\begin{equation*}
| f(v)-\alpha | \leq \varepsilon ,
\end{equation*}
that is (b). For $(c)$, from \eqref{e4.11}
$, \eqref{e4.12}$ and \eqref{e4.14} and the non-increasing
property of $\chi $, we infer
\begin{equation} \label{e4.18}
\begin{split}
h(w)&> h(v)-\varepsilon \int_{\beta (
v)}^{\beta (v) + p(v,w)}\chi
(t)\,dt-\varepsilon \int_{\beta (w)}^{\beta (
w) + p(v,w)}\chi (t)\,dt \\
&\geq h(v)-\varepsilon (\chi (\beta (
v))+\chi (\beta (w)))\cdot p(v,w),
\end{split}
\end{equation}
for $w\in X$, $w\neq v$. Since $f$ is lower semicontinuous
 and $f(v)>\alpha -2\varepsilon $, there exists $\delta \in (0,1)$
 such that $f(w)>\alpha -2\varepsilon $ for $w\in X$ with
$p(v,w)<\delta $. Hence, for $w\in X$ with $0<p(v,w)<\delta $,
since $h(w)=f(w)$ and
\[
\beta (w)= p(a,w)\geq p(a,v)-p(w,v)
>\beta (v)-\delta >\beta (v)-1>0,
\]
we derive
\begin{equation} \label{e4.19}
\begin{split}
f(w)
&> f(v)-\varepsilon (\chi (\beta (v))+\chi (\beta (v)-1))
\cdot p(v,w) \\
&\geq f(v)-2\varepsilon \chi (\beta (v)-1)\cdot p(v,w) \\
&= f(v)-\varepsilon b(\beta (v))\cdot p(v,w).
\end{split}
\end{equation}
By means of Lemma \ref{lem4.1}, we reach
\begin{equation*}
\| f'(v)\| \leq \varepsilon b(\beta (v)),
\end{equation*}
and (c) is verified too. The proof is complete.
\end{proof}

\begin{corollary} \label{coro4.4}
Let $X$ be a Banach space. Let $f:X\to (-\infty ,\infty ]$
be a proper lower semicontinuous function which is bounded from below.
Assume that $f$  is G\^ateaux differentiable at every point
$x\in X$ with $f(x)\in\mathbb{R}$.
If $f$ satisfies the b-(PS)$_c$ condition for all $c\in\mathbb{R}$,
then $f$ is coercive; i.e., $f(x) \to \infty $ as
$\beta (x)\to \infty$.
\end{corollary}

\begin{proof}
Suppose the contrary; then $\alpha =\lim\inf_{\beta (x)\to \infty }
f(x)\in\mathbb{R}$. By Theorem \ref{thm4.3}, there exists a sequence
$\{ z_n\} _n$ in $X$ such that $\beta (z_n)\to \infty $,
$f(z_n)\to \alpha $ and $\| f'(z_n)\| /b(\beta (z_n))\to 0$.
Then, the b-(PS)$_{\alpha }$ condition implies
that $\{ z_n\} _n$ has a convergent subsequence, which
clearly leads to a contradiction.
\end{proof}

\begin{remark} \label{rmk4.5} \rm
Corollary \ref{coro4.4} generalizes the result proved by \cite{l1}
using a gradient flow, by Costa-Silva \cite{c3},
Caklovic-Li-Willem \cite{c1} and Brezis-Nirenberg \cite{b2}
 using EVP, and by Zhong \cite{z1} using ZVP.
\end{remark}

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\end{document}
