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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 33, pp. 1--8.\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}}

\begin{document}
\title[\hfilneg EJDE-2006/33\hfil Second-order nonconvex differential inclusion]
{Existence of solutions for nonconvex second-order
differential inclusions in the infinite dimensional space}
\author[T. Haddad, M. Yarou \hfil EJDE-2006/33\hfilneg]
{Tahar Haddad, Mustapha Yarou}  % in alphabetical order

\address{Tahar Haddad \newline
Department of Mathematics,
Faculty of Sciences, Jijel University, Algeria}
\email{haddadtr2000@yahoo.fr}

\address{Mustapha Yarou \newline
Department of Mathematics,
Faculty of Sciences, Jijel University, Algeria}
\email{mfyarou@yahoo.com}


\date{}
\thanks{Submitted December 12, 2005. Published March 16, 2006.}
\subjclass[2000]{34A60, 49J52}
\keywords{Nonconvex differential inclusions; uniformly regular functions}

\begin{abstract}
 We prove the existence of solutions to the differential
 inclusion
 $$
 \ddot{x}(t)\in F(x(t),\dot{x}(t))+f(t,x(t),\dot{x}(t)), \quad
  x(0)=x_{0}, \quad \dot{x}(0)=y_{0},
 $$
 where $f$ is a Carath\'{e}odory function and $F$
 with nonconvex values in a Hilbert space such that
 $F(x,y)\subset \gamma (\partial g(y))$, with $g$ a regular
 locally Lipschitz function and $\gamma $ a linear operator.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{example}[theorem]{Example}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\numberwithin{equation}{section}

\section{Introduction}

In the present paper we consider the Cauchy problem for
second-order differential inclusion
\begin{equation}
\begin{gathered}
\ddot{x}(t)\in F(x(t),\dot{x}(t))+f(t,x(t),\dot{x}(t)),\\
x(0)=x_{0},\quad \dot{x}(0)=y_{0}
\end{gathered} \label{e1.1}
\end{equation} where $F(\cdot ,\cdot )$ is a given set-valued map and $f$ is
a Carath\'{e}odory function. Second order differential inclusions
have been studied by many authors, mainly in the case when the
multifunction is convex valued. Several existence results may be
found in \cite{as,by,ch,m,r}.

Recently in \cite{l1} and \cite{l2}, the situation when the
multifunction is not convex valued is considered, the existence of
solution for the problem \eqref{e1.1} was obtained in the finite
dimensional case by assuming $F(\cdot ,\cdot )$ upper
semicontinuous, compact valued multifunction such that
$F(x,y)\subset \partial g(y)$ for some convex proper lower
semicontinuous function $g$. In this paper we extend this result
in two ways: we consider the infinite dimensional case and we
relax the convexity assumption on the function $g$, namely we
suppose that $g$ is uniformly regular and so the usual
subdifferentials will be replaced by the Clarke subdifferentials.
The class of proper convex lower semicontinuous functions and the
class of lower-$C^{2}$ functions (see examples \ref{exa1}, \ref{exa2})
are strictly
contained within the class of uniformly regular functions. The
paper is organized as follows: in Section 2 we recall some
preliminary facts that we need in the sequel and in Section 3 we
prove our main result.

\section{Preliminaries}

Let $\mathbb{H}$ be a real separable Hilbert space with the norm
$\| \cdot \| $ and scalar product $\langle \cdot ,\cdot \rangle$.
We denote by $\mathbb{B}:=\mathbb{B}(0,1)$ the unit open ball of
$\mathbb{H}$ and let $\overline{\mathbb{B}}$ be its closure. We
denote by $\delta ^{\ast }(.,A)$ the support function of $A$, by
$d(x,A)$ the distance from $x\in \mathbb{H}$ to $A$. for any two
subsets $A,B$ of $\mathbb{H}$, $d_{\mathbb{H}}(A,B)$ stands to the
Hausdorff distance between $A$ and $B$.

Let $\sigma $ the weak topology in $\mathbb{H}$. Let us
$(e_{n})_{n\geq 1} $ be a dense sequence in
$\overline{\mathbb{B}}$ and we consider the linear application
$\gamma :\mathbb{H}\to \mathbb{H}$ defined by
$$
\forall x\in \mathbb{H},\quad
\gamma (x)={\sum_{n=1}^{\infty}2^{-n}\langle x,e_{n}\rangle e_{n}}.
$$
Note that this series is absolutely convergent. According
to the specialists of the theory of linear operators the
application $\gamma $ belongs to the class of the nuclear
operators of $\mathbb{H}$. Further, $\gamma $ satisfies the two
following properties:

(a) The restriction of $\gamma $ to $\overline{\mathbb{B}\text{ }}$ is
continuous from $(\overline{\mathbb{B}\text{ }},\sigma )$ into $\mathbb{H}$.

(b) For all $x\in \mathbb{H}\backslash \{0\}$,
$\langle x,\gamma(x)\rangle >0$.
\\
 Indeed b) is obvious. This condition is equivalent to
$$
x\in \mathbb{H}\mapsto \langle x,\gamma (x)\rangle
$$
is a strictly convex function (see \cite{v}).

In the sequel we note by $\Gamma (\mathbb{H)}$ the set of linear
applications $\gamma :\mathbb{H\to H}$ verifying the conditions a)
and b). $\Gamma (\mathbb{H)}$ $\subset \mathbb{K(}\mathbb{H)}$ the
space of compact operators of $\mathbb{H}$. If $\mathbb{H=R}^{m}$
then $\Gamma ( \mathbb{H)}$ coincides with the set of the
automorphism of $\mathbb{R}^{m}$ associated to positive definite
matrices.

\begin{definition}[\cite{b2}] \label{def1} \rm
Let $f:\mathbb{H}\to R\cup \{+\infty \}$ be a lower semicontinuous
 function and let $\Omega \subset dom f$ be a nonempty open
subset. We will say that $f$ is uniformly regular over $\Omega $
if there exists a positive number $\beta \geq 0$ such that for all
$x\in \Omega $ and for all $\xi \in
\partial ^{P}f(x)$ one has
$$
\langle \xi ,x'-x\rangle \leq f(x')-f(x)+\beta \Vert x'-x\Vert
^{2}\quad \text{for all }x'\in \Omega .
$$
\end{definition}
Here $\partial ^{P}f(x)$ denotes the proximal subdifferential of $f$
at $x$ (for its definition the reader is refereed for instance to
\cite{bt}). We will say that $f$ is uniformly regular over closed
 set $S$ if there exists an open
set $O$ containing $S$ such that $f$ is uniformly regular over $O$. The
class of functions that are uniformly regular over sets is so large. We
state here some examples.

\begin{example} \label{exa1} \rm
Any lower semicontinuous  proper convex function $f$ is uniformly
regular over any nonempty subset of its domain with $\beta =0$.
\end{example}

\begin{example} \label{exa2}\rm
Any lower-$C^{2}$ function $f$ is uniformly regular over any
nonempty convex compact subset of its domain. Indeed, let $f$ be a
lower-$C^{2}$ function over a nonempty convex compact set
$S\subset dom f$. By Rockafellar's result ( see for instance
\cite[Theorem 10.33]{r}) there exists a positive real number
$\beta $ such that $g:=f+\frac{\beta }{2}\| .\| ^{2}$ is a convex
function on $S$. Using the definition of the subdifferential of
convex functions and the fact that the Clarke subdifferential of
$f$ is $\partial ^{C}f(x)=\partial g(x)-\beta x$ for any $x\in S$,
we get the inequality in definition \ref{def1} and so $f$ is
uniformly regular over $S$.
\end{example}

The following proposition summarizes some important properties for uniformly
regular locally Lipschitz functions over sets needed in the sequel. For the
proof of these results we refer the reader to \cite{b1,bh}.

\begin{proposition} \label{prop1}
Let $g:\mathbb{H}\to \mathbb{R}$ be a locally Lipschitz function
and $\Omega $ a nonempty open set. If $f$ is uniformly regular
over $\Omega $, then the following hold:

\begin{itemize}
\item[(i)] The proximal subdifferential of $g$ is closed over $\Omega $,
that is, for every $x_{n}\to x\in \Omega $ with $x_{n}\in \Omega $
and every $\xi _{n}\to \xi $ with $\xi _{n}\in \partial
^{P}g(x_{n})$ one has $\xi \in \partial ^{P}g(x)$

\item[(ii)] The proximal subdifferential of $g$ coincides with $\partial
^{C}g(x)$ the Clarke subdifferential for any point $x$ (see for instance \cite{bt}
for the definition of $\partial ^{C}g$)

\item[(iii)] The proximal subdifferential of $g$ is upper hemicontinuous
over $S$, that is, the support function $x\mapsto \langle v,\partial
^{P}g(x)\rangle $ is u.s.c. over $S$ for every $v\in \mathbb{H}$

\item[(iv)] For any absolutely continuous map $x:[0,T]\to \Omega $
for which $\dot{x}(t)$ is absolutely continuous one has
$$
\frac{d}{dt}(f\circ \dot{x})(t)=\langle \partial
^{C}f(\dot{x}(t));\ddot{x} (t)\rangle .
$$
\end{itemize}
\end{proposition}

For a multifunction $F:\Omega _{1}\times \Omega _{2}\subset
\mathbb{H\times H\to }$ $2^{\mathbb{H}}$ and for any
$(x_{0},y_{0})\in \Omega
_{1}\times \Omega _{2}$ we consider the Cauchy problem
$$
\ddot{x}(t)\in
F(x(t),\dot{x}(t))+f(t,x(t),\dot{x}(t)),x(0)=x_{0},\dot{x}
(0)=y_{0}
$$
under the following assumptions:
\begin{itemize}
\item[(H1)] $\Omega _{1},\Omega _{2}$ are open subsets in
$\mathbb{H}$ and $F:\Omega _{1}\times \Omega _{2}\to
2^{\mathbb{H}}$ is upper semicontinuous (i.e for all $\epsilon >0$
there exists $\delta >0$ such that $\| z-z'\| \leq \delta $
implies $F(z')\subset F(z)+\epsilon \mathbb{B}$) with compact
values.

\item[(H2)] There exist $\gamma \in \Gamma (\mathbb{H)}$ and a
locally Lipschitz $\beta $-uniformly regular function
$g:\mathbb{H}\to \mathbb{R}$ over $\Omega _{2}$ such that
\begin{equation}
F(x,y)\subset \gamma (\partial ^{C}g(y))\quad \text{  for all }
(x,y)\in \Omega _{1}\times \Omega _{2}.  \label{e2.1}
\end{equation}

\item[(H3)] $f:\mathbb{R}^{+}\times \mathbb{H}\times
\mathbb{ H}\to \mathbb{H}$ is a Carath\'{e}odory function,
(i.e. for every $x,y\in \mathbb{H}, t\longmapsto f(t,x,y)$ is
measurable, for $t\in \mathbb{R}^{+}, (x,y)\longmapsto f(t,x,y)$
is continuous) and for any bounded subset $B$ of
$\mathbb{H}\times \mathbb{H}$, there is a compact set $K$ such that
$f(t,x,y)\in K$ for all $(t,x,y)\in \mathbb{R}^{+}\times B$.
\end{itemize}
By a solution of problem \eqref{e1.1} we mean an absolutely continuous
function $x(.):[0,T]\to \mathbb{H}$ with absolutely continuous
derivative $\dot{x}(.)$ such that $x(0)=x_{0},\dot{x}(0)=y_{0}$
and $\ddot{x} (t)\in F(x(t),\dot{x}(t))+f(t,x(t),\dot{x}(t))$ a.e.
on $[0,T]$. For more details on differential inclusions, we refer
to \cite{ac}.

\section{Main result}

Our main result is the following.

\begin{theorem} \label{thm1}
Consider $F:\Omega _{1}\times \Omega _{2}\to
2^{\mathbb{H }}$,  $f:\mathbb{R}  \times \mathbb{H}\times
\mathbb{H}\to  \mathbb{H}$, $g:\mathbb{H}  \to \mathbb{R}$ and
$\gamma \in \Gamma (\mathbb{H)}$ satisfy Hypotheses
(H1)-(H3). Then, for every
$(x_{0},y_{0})\in \Omega_{1}\times \Omega _{2}$ there exist $T>0$
and $x(.):[0,T]\to \mathbb{H}$ solution to problem \eqref{e1.1}.
\end{theorem}

\begin{proof}
Let $r>0$ be such that $\bar{\mathbb{B}(}y_{0},r)\subset \Omega_{2}$
and $g$ is $L$-Lipschitz on $\bar{\mathbb{B}}(y_{0},r)$.
Then we have that $\partial ^{C}g(y)\subset L\bar{\mathbb{B}}$,
whenever $y\in \bar{\mathbb{B}(}y_{0},r)$.
 By our assumption (H3), there is a positive constant $m$ such that
$f(t,x,y)\in K \subset m\mathbb{B}$ for all $(t,x,y)\in
\mathbb{R}^{+}\times  \bar{\mathbb{B}}(x_{0},r)\times
\bar{\mathbb{B}}(y_{0},r)$. Moreover, since
$\gamma \in \Gamma (\mathbb{H})$, the set $K_1:=\gamma (L\bar{\mathbb{B})}$
is convex compact in $\mathbb{H}$ and so there exists $m_1>0$ such
that $K_1 \subset m_1\mathbb{B}$. Choose $T$ such that
$$
0<T < \min \big\{\frac{r}{m_1 +m}, \frac {r}{r+\Vert y_0\Vert }\big\}
$$
Set $I:=[0,T]$. For each integer $n\geq 1$ and for $1\leq i\leq n-1$
we set $t_{i}^{n}:=\frac{ iT}{n}$,
$I_{i}^{n}:=[t_{i-1}^{n},t_{i}^{n}[$
and $t_n^n = T$, $I_n^n = {T}$. Let define the following
approximate sequences
\begin{gather*}
y_{n}(t) =y_{n}(t_{i}^{n})+
\int_{t_{i}^{n}}^{t}[f(s,x_{n}(t_{i}^{n}),y_{n}(t_{i}^{n}))+u_{i}^{n}]ds
\\
x_{n}(t) =x_{n}(t_{i}^{n})+\int_{t_{i}^{n}}^{t}y_{n}(s)ds
\end{gather*}
whenever $t\in I_{i+1}^{n},0\leq i\leq n-1$, where
$x_{n}(0)=x_{0}$, $y_{n}(0)=y_{0}$, and
$u_{i}^{n}\in F(x_{n}(t_{i}^{n}),y_{n}(t_{i}^{n}))$.

 For every $0\leq i\leq n-1$, take
 $z_{i}^{n}\in \partial ^{C}g(y_{n}(t_{i}^{n}))$
such that $u_{i}^{n}=\gamma (z_{i}^{n})$. Now let us define the
step functions from $[0,T]$ to $[0,T]$ by
$$
\theta _{n}(t)=t_{i}^{n}\text{,\ }u_{n}(t)=u_{i}^{n}\text{, }
z_{n}(t)=z_{i}^{n}\text{\ \ \ }t\in I_{i+1}^{n}\text{.}
$$
Then, for all $n\in \mathbb{N}^{\ast }$ and all $t\in [0,T]$,
we have the following properties:
\begin{gather}
0\leq t-\theta _{n}(t)\leq \frac{T}{n}  \label{e3.1} \\
y_{n}(t)=y_{0}+\int_{0}^{t}[f(s,x_{n}(\theta _{n}(s)),y_{n}(\theta
_{n}(s)))+u_{n}(s)]ds  \label{e3.2} \\
x_{n}(t)=x_{0}+\int_{0}^{t}y_{n}(s)ds  \label{e3.3}\\
u_{n}(t)\in F(x_{n}(\theta _{n}(t)),y_{n}(\theta _{n}(t)))
\label{e3.4} \\
z_{n}(t)\in \partial ^{C}g(y_{n}(\theta _{n}(t)))  \label{e3.5} \\
u_{n}(t)=\gamma (z_{n}(t)).  \label{e3.6}
\end{gather}
Observe that $y_{n}(t)\in \bar{\mathbb{B}(}y_{0},r)$ and
$x_{n}(t) \in \bar{\mathbb{B}}(x_{0},r)$ for all
 $n\in \mathbb{N}^{\ast }$ and all $t\in [0,T]$. Indeed it is obvious that
$$
\Vert y_{n}(t)-y_{0} \Vert
= \Vert \int_{0}^{t}[f(s,x_{n}(\theta _{n}(s)),y_{n}(\theta
_{n}(s)))+u_{n}(s)]ds \Vert \leq (m_1 + m)T <r
$$
and
$$
\Vert x_{n}(t)-x_{0} \Vert =\Vert \int_{0}^{t}y_{n}(s)ds \Vert \leq
(\Vert y_0\Vert + r )T <r
$$
Hence
$$
\| y_{n}(t)-y_{n}(t')\| \leq ( m_1 + m) \vert t'-t\vert
$$
whenever $0\leq t\leq t'\leq T$ and $n\in \mathbb{N}^{\ast }$. On
the other hand, we have
\[
 \Vert x_{n}(t)-x_{n}(t')\Vert \leq  {\int_{t}^{t'}\| y_{n}(s)\| ds}
 \leq (r+\| y_{0}\| )\left\vert
t-t'\right\vert
\]
whenever $0\leq t\leq t'\leq T$ and $n\in \mathbb{N}^{\ast }$.
 Hence $(x_{n})_{n\in \mathbb{N}^{\ast }}$ and
 $(y_{n})_{n\in \mathbb{N}^{\ast }}$ are equi-Lipschitz subsets of
 $C([0,T],\mathbb{H)}$. The sets
$\{x_{n}(t):n\in \mathbb{N}^{\ast }\}$ and
$\{y_{n}(t):n\in \mathbb{N}^{\ast }\}$ are relatively compact in
$\mathbb{H}$ for every $t\in [0,T]$.
Indeed we have for all $n\in \mathbb{N}^{\ast }$ and all $t\in [0,T]$
\begin{equation*}
y_{n}(t)\in y_{0}+ [0,T] \{K_1 + K\}:= K_2
\end{equation*}
which is compact. We have also
\begin{equation*}
x_{n}(t)\in x_{0}+ [0,T]K_{2}
\end{equation*}
for all $n\in \mathbb{N}^{\ast }$ and all $t\in [0,T]$.
Then by Ascoli's theorem, $(x_{n})_{n\in \mathbb{N}^{\ast }}$ and
$(y_{n})_{n\in \mathbb{N}^{\ast }}$ are relatively compact in the
Banach space $C([0,T],\mathbb{ H)}$. Further, the sequences
$(u_{n})_{n\in \mathbb{N}}$ and $(z_{n})_{n\in \mathbb{N}}$ are
relatively $\sigma (L^{1}([0,T],\mathbb{H)}$;
$L^{\infty}([0,T],\mathbb{H))}$-compact and
$\sigma (L^{\infty }([0,T],\mathbb{H)} ;L^{1}([0,T],\mathbb{H))}$-compact
respectively since we have a.e.
\begin{equation*}
\forall n\in \mathbb{N}^{\ast }\quad u_{n}(t)\in K_1
\quad\text{and}\quad z_{n}(t)\in L\bar{\mathbb{B}}\,.
\end{equation*}
Therefore, by extracting subsequences if necessary, we can assume
that there exist $x$ in $C([0,T],\mathbb{H)}$, $y$ in
$C([0,T],\mathbb{H)}$, $u$ in $L^{1}([0,T],\mathbb{ H)}$ and $z$
in $L^{\infty }([0,T],\mathbb{H)}$ such that $x_{n}\to x$ in
$C([0,T],\mathbb{H)}$, $y_{n}\to y$ in $C([0,T],\mathbb{H)}$,
$u_{n}\to u$ for $\sigma (L^{1}([0,T],\mathbb{H})$; $L^{\infty
}([0,T], \mathbb{H))}$-topology and $z_{n}\to z$ for $\sigma
(L^{\infty }([0,T],\mathbb{H)}$;
$L^{1}([0,T],\mathbb{H))}$-topology. \newline Also, we have
$f(.,x_{n}(\theta _{n}(.)),y_{n}(\theta_{n}()))\to f(.,x(.),y(.))$
in the norm of the space $L^{1}([0,T],\mathbb{H)}$. Consequently,
for all $t\in [0,T]$,
\begin{equation*}
x_{0}+\int_{0}^{t}\dot{x}(s)ds=x(t)=\lim_{n\to \infty
}x_{n}(t)=x_{0}+\lim_{n\to \infty
}\int_{0}^{t}y_{n}(s)ds=x_{0}+\int_{0}^{t}y(s)ds
\end{equation*}
which gives the equality
\begin{equation}
\dot{x}(t)=y(t)\quad \text{for almost }t\in [0,T].  \label{e3.7}
\end{equation}
Now we assert that $u=\gamma (z)$ a.e. Indeed, for any $w\in
\mathbb{H}$ and any measurable set $A$ in $[0,T]$, one has
\begin{align*}
\langle w,\int_{A}u(\eta )d\eta \rangle
&=\int_{A}\langle w,u(\eta )\rangle d\eta  \\
&=\lim_{n\to\infty} \int_{A}\langle w,u_{n}(\eta )\rangle d\eta  \\
&=\lim_{n\to\infty} \langle w,\int_{A}\gamma (z_{n}(\eta ))d\eta \rangle  \\
&=\lim_{n\to\infty} \langle w,\gamma (\int_{A}z_{n}(\eta )d\eta )\rangle  \\
&=\langle w,\gamma (\int_{A}z(\eta )d\eta )\rangle  \\
&=\langle w,\int_{A}\gamma (z(\eta )d\eta )\rangle .
\end{align*}
Hence $u(t)=\gamma (z(t))$ for almost every $t\in [0,T]$.
Note that $\lim_{n\to\infty} x_{n}(\theta
_{n}(t))=x(t)$ and $\lim_{n\to\infty} y_{n}(\theta
_{n}(t))=y(t)$, for all $t\in [0,T]$ where
$y(t)={y_{0}+\int_{0}^{t}\dot{y}(s)ds}$, for all
$t\in [0,T]$. Then it follows from \eqref{e3.2} that
$\dot{y}(t)=f(t,x(t),y(t))+u(t)$  for almost $t\in [0,T]$
and by \eqref{e3.7} we obtain that
$$
\ddot{x}(t)=f(t,x(t),y(t))+u(t)
$$
for almost $t\in [0,T]$. By construction, we have
for a.e $t\in [0,T]$,
\begin{equation}
\begin{aligned}
\ddot{x}_{n}(t)-f_{n}(t)
&=u_{n}(t)\in F(x_{n}(\theta _{n}(t)),y_{n}(\theta _{n}(t)))\\
&\subset \gamma (\partial ^{C}g(y_{n}(\theta _{n}(t)))) \\
&=\gamma (\partial ^{P}g(y_{n}(\theta _{n}(t)))).
\end{aligned} \label{e3.8}
\end{equation}
Since $y_{n}(\theta _{n}(t)\in \bar{\mathbb{B}(}y_{0},r)\subset
\Omega _{2}$, the last equality follows from the uniform
regularity of $g$ over $\Omega _{2}$ and the part (ii) in
proposition \ref{prop1}. The convergence of $z_{n}$ to $z$ for
$\sigma (L^{\infty }([0,T],\mathbb{H)};L^{1}([0,T],\mathbb{H))}$-topology
and Mazur's Lemma entails
\begin{equation*}
z \in \bigcap_{n}\overline{co}^{\sigma }\{z_{m}:m\geq n\}, \quad
\text{for a.e. }t\in [0,T]
\end{equation*}
(here $\sigma = \sigma(L^{\infty }([0,T],\mathbb{H});L^{1}([0,T],\mathbb{H}))$. Fix any such $t$ and consider
any $\xi \in \mathbb{H}$. Then, the last
relation above yields
\begin{equation*}
\langle \xi ,z(t)\rangle \leq \inf_n \sup_{m\geq n}
\langle \xi ,z_{n}(t)\rangle
\end{equation*}
and by proposition \ref{prop1} part (iii) and \eqref{e3.5} yield
\begin{align*}
\langle \xi ,z(t)\rangle
&\leq \limsup_n  \delta^*
(\xi ,\partial ^{P}g(y_{n}(\theta _{n}(t)))) \\
&\leq \delta^*(\xi ,\partial ^{p}g(y(t)))\quad \text{for any }\xi \in H,
\end{align*}
So, by \cite[Theorem VI.4]{cv}, the convexity and the closeness of
the set $\partial ^{p}g(y(t))$ ensures
\begin{equation*}
z(t)\in \partial ^{p}g(y(t))
\end{equation*}
Now, since $g$ is uniformly regular over $\Omega _{2}$ and
$\dot{x} (t)=y(t)\in \bar{\mathbb{B}(}y_{0},r)\subset \Omega _{2}$
for all $t\in [0,T]$ we have by proposition \ref{prop1} part (iv)
\begin{align*}
\frac{d}{dt}(g\circ \dot{x})(t)
&=\langle \partial ^{p}g(\dot{x}(t)),
\ddot{x}(t)\rangle =\langle z(t),\ddot{x}(t)\rangle  \\
&=\langle z(t),f(t,x(t),y(t))+u(t)\rangle .
\end{align*}
Consequently,
\begin{equation}
g(\dot{x}(T))-g(y_{0})=\int_{0}^{T}\langle
z(t),f(t,x(t),y(t))\rangle dt+\int_{0}^{T}\langle z(t),u(t)\rangle
dt  \label{e3.9}
\end{equation}
On the other hand, since $y_{n}(\theta _{n}(t)\in
\bar{\mathbb{B}(} y_{0},r)\subset \Omega _{2}$ and by
\eqref{e3.8} and definition \ref{def1}, we have for all $i\in \{0,\dots ,n-1\}$
\begin{align*}
& g(y_{i+1}^{n})-g(y_{i}^{n}) \\
&\geq \langle
z_{i}^{n},y_{i+1}^{n}-y_{i}^{n}\rangle -\beta \|
y_{i+1}^{n}-y_{i}^{n}\| ^{2} \\
&=\langle z_{i}^{n},\int_{t_{i}^{n}}^{t_{i+1}^{n}}\ddot{x}
_{n}(s)ds\rangle -\beta \| y_{i+1}^{n}-y_{i}^{n}\| ^{2} \\
&=\langle
z_{n}(t),\int_{t_{i}^{n}}^{t_{i+1}^{n}}[f(s,x_{n}(\theta
_{n}(s)),y_{n}(\theta _{n}(s)))+u_{n}(s)]ds\rangle -\beta \|
y_{i+1}^{n}-y_{i}^{n}\| ^{2} \\
&\geq \int_{t_{i}^{n}}^{t_{i+1}^{n}}\langle
z_{n}(s),f(s,x_{n}(\theta _{n}(s)),y_{n}(\theta _{n}(s)))\rangle
ds+\int_{t_{i}^{n}}^{t_{i+1}^{n}}\langle z_{n}(s),u_{n}(s)\rangle
ds \\
&\quad -  \beta ( m_1 + m )^{2}(t_{i+1}^{n}-t_{i}^{n})^{2}
\end{align*}
By adding, we obtain
\begin{equation}
\begin{aligned}
&g(\dot{x}_{n}(T))-g(y_{0})\\
&\geq \int_{0}^{T}\langle
z_{n}(s),f(s,x_{n}(\theta _{n}(s)),y_{n}(\theta _{n}(s)))\rangle
ds+\int_{0}^{T}\langle z_{n}(s),u_{n}(s)\rangle ds-\varepsilon
_{n}
\end{aligned}\label{e3.10}
\end{equation}
 with
$$
\varepsilon _{n}=\frac{\beta ( m_1+ m )^{2}T^2}{n}\to 0
$$
as $n\to \infty$.  We have also have
$$
\lim_{n\to\infty} \int_{0}^{T}\langle
z_{n}(s),f(s,x_{n}(\theta _{n}(s)),y_{n}(\theta _{n}(s)))\rangle
ds=\int_{0}^{T}\langle z(s),f(s,x(s),y(s))\rangle ds.
$$
Indeed, for all $t\in [0,T]$ and all $n\in \mathbb{N}^{\ast }$,
$$
\langle z_{n}(t),f(t,x_{n}(\theta _{n}(t)),y_{n}(\theta
_{n}(t)))\rangle -\langle z(t),f(t,x(t),y(t))\rangle =\alpha
_{n}(t)+\beta _{n}(t)
$$
where
\begin{gather*}
\alpha _{n}(t)=\langle z_{n}(t),f(t,x_{n}(\theta
_{n}(t)),y_{n}(\theta _{n}(t)))-f(t,x(t),y(t))\rangle,\\
\beta _{n}(t)=\langle z_{n}(t)-z(t),f(t,x(t),y(t))\rangle
\end{gather*}
 Since $z_{n}(t)-z(t)\to 0$ for $\sigma (L^{\infty }([0,T],\mathbb{H)}
;L^{1}([0,T],\mathbb{H))}$,
$$
\int_{0}^{T}\beta _{n}(s)ds\to 0
$$
and $f_{n}\to f$ strongly in $L^{1}([0,T],\mathbb{H)}$
which implies
$$
\int_{0}^{T}\alpha _{n}(s)ds\to 0.
$$
Taking the limit superior in \eqref{e3.10} when $n\to \infty $ and
using the continuity of $g$, we obtain
$$
g(\dot{x}(T))-g(y_{0})\geq \int_{0}^{T}\langle
z(s),f(s,x(s),y(s))\rangle ds+\lim_n \sup
\int_{0}^{T}\langle z_{n}(s),u_{n}(s)\rangle ds
$$
This inequality compared with \eqref{e3.9} yields
\begin{equation}
\limsup_n \int_{0}^{T}\langle
z_{n}(s),u_{n}(s)\rangle ds\leq \int_{0}^{T}\langle
z(s),u(s)\rangle ds  \label{e3.11}
\end{equation}
The values of the function $z_{n}$ are in the convex weakly
compact $C:=L \bar{\mathbb{B}}$, further the application $\Lambda
:(\mathbb{H},\sigma )\to [0.+\infty ]$ defined by
\begin{equation*}
\Lambda (\alpha )=\begin{cases}
\langle \alpha ,\gamma (\alpha )\rangle &\text{if }\alpha \in C \\
+\infty & \text{otherwise}
\end{cases}
\end{equation*}
is lower semicontinuous and strictly convex on $C$ (According to
a) and b) ). The condition \eqref{e3.11} is equivalent to
\begin{equation*}
\limsup_n  \int_{0}^{T}\Lambda (z_{n}(s))ds\leq
\int_{0}^{T}\Lambda (z(s))ds\,.
\end{equation*}
Then \cite[Proposition 3.2]{bcs} yields
\begin{equation*}
z(t)\in \bigcap_{n}\overline{co}^{\sigma }\{z_{m}(t):m\geq
n\},\quad \text{for a.e } t\in [0,T].
\end{equation*}
Hence there is a negligible $N$ such that for $t\notin N$, we have
\begin{gather*}
u(t) =\gamma (z(t)) \\
z(t) \in \bigcap_{n}\overline{co}^{\sigma }\{z_{m}(t):m\geq n\}.
\end{gather*}
Now let $t\notin N$ be fixed. Then we can extract from
$(z_{n}(t))_{n\in \mathbb{N}}$ a subsequence
$(z_{n_{k}}(t))_{k\in \mathbb{N}}$, such that
 $z_{n_{k}}(t)\rightharpoonup z(t)$ weakly
in $\mathbb{H}$ so that $\gamma (z_{n_{k}}(t))\to \gamma (z(t))$
for the norm topology since $\gamma \in \Gamma (\mathbb{H)}$. By
\eqref{e3.4} and \eqref{e3.6}, recalling that
\begin{equation*}
u_{n}(t)=\gamma (z_{n}(t))\in F(x_{n}(\theta _{n}(t)),y_{n}(\theta _{n}(t)))
\end{equation*}
for every $t\in [0,T]$ and every $n\in \mathbb{N}^{\ast }$, that
$\lim_{n\to\infty} x_{n}(\theta _{n}(t))=x(t)$, \break
$\lim_{n\to\infty} y_{n}(\theta _{n}(t))=y(t)=\dot{x}(t)$,
for all $t\in [0,T]$ and that the
graph of $F$ is closed, we obtain
\begin{equation}
u(t)=\gamma (z(t))\in F(x(t)),\dot{x}(t))\quad \text{ a.e. }
\label{e3.12}
\end{equation}
Since $\ddot{x}(t)=f(t,x(t),y(t))+u(t)$ for almost $t\in [0,T]$ it
follows from \eqref{e3.12} that
$$
\ddot{x}(t)\in F(x(t),\dot{x}(t))+f(t,x(t),\dot{x}(t))\quad
\text{ a.e. on }[0,T].
$$
Therefore, differential inclusion \eqref{e1.1} admits a solution.
\end{proof}

\begin{remark} \label{rmk1} \rm
An inspection of the proof of Theorem \ref{thm1} shows that the uniformity
of the constant $\beta $ was needed only over the ball
$B(y_{0},\rho )$ and so it was not necessary over all the set
$\Omega _{2}$.
\end{remark}

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\end{document}