% Dear Eduardo,
% enclosed is a part of chapter 4 in the book written with Andrea,
% which contains  a (short and readable)proof of the extension of
% Kurzweil's theorem to (time varying) differential inclusions. Here,
% the Lyapunov function is SMOOTH.  This proof is a quite important part in
% the book, which was not given with the rest of the book when submitting it,
% since it was not entirely typed. This result was mentionad in the few lines
% I sent to you at Wednesday. It may be of interest to you. In order to not
% saturate your mailbox, I put the LaTeX file, instead of a ps file.
%
% Best regards,
%
% -lionel

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\begin{document}
\section{Global asymptotic stability}
In this chapter, we are concerned with stability properties for a
differential inclusion $\dot x\in F(t,x)$. We first investigate the (uniform)
global asymptotic stability. After a brief review of the main contributions 
in this topic, we give a complete (and short) proof of a result which 
generalizes and unifies all the converses of the second Liapunov theorem
in the litterature. The second part of the chapter is devoted to the
first Liapunov theorem and its converse (given together with its 
proof) for differential inclusions. 

The reader is referred to Chapter 1 for the notations (namely,
$\Vert \cdot \Vert$, $\overline{B}_r(x)$, $h$, etc.) and the current
assumptions (($H_1$), Hausdorff continuity, etc.), and to
Chapter 3 for some definitions of stability.
\subsection{Sufficient conditions}
Let $F:\R ^+\times \R ^n\to 2^{\R ^n}$
be a multivalued map, fulfilling ($H_1$), ($H_2$), ($H_3$) and ($H_4$).
It follows that for each pair $(t_0,x_0)\in \R _+\times \R ^n$, 
the solutions $x(\cdot )$ of the differential inclusion
\begin{equation}
\label{eq:R1}
\dot x\in F(t,x),
\end{equation}
issuing from $x_0$ at $t=t_0$, exist, at least {\em locally}.
(The set of the maximal solutions of above Cauchy problem 
is still denoted by ${\cal S}_{t_0,x_0}$.)
Assume that the origin is an equilibrium
position for (\ref{eq:R1}), i.e.
$0\in F(t,0)\ \hbox{ for a.e. } \ t\ge 0$.
We aim to characterize the (possible) asymptotic stability of the origin by 
means of some (strict) Liapunov function.
According to definition 26,
(\ref{eq:R1}) is said to be (uniformly) globally asymptotically stable
(UGAS, in short),  if the origin is uniformly stable (and Lagrange stable)
and uniformly attrative. In other words, it means that
there exist two functions
$m\!\! :(0,+\infty )\to   (0,+\infty )$ and
$T\!\! :(0,+\infty )^2\to (0,+\infty )$, such that\va
(i) for each $R>0$, each pair $(t_0,x_0)\in \R _+\times \R ^n$
and each $x(\cdot )\in {\cal S}_{t_0,x_0}$ 
\begin{equation*}
\Vert x_0\Vert \le R \quad  \im \quad \Vert x(t)\Vert  < m(R) \quad \forall
t\ge t_0;
\end{equation*}
(ii) $\ \lim_{R\to 0^+}m(R) =0$;\va
(iii) for each $R>0$, each $\varepsilon >0$, each pair 
$(t_0,x_0)\in \R _+\times \R ^n$
and each $x(\cdot )\in {\cal S}_{t_0,x_0}$
\begin{equation*}
\Vert x_0\Vert \le R \quad  \im \quad \Vert x(t)\Vert  < \varepsilon 
\quad \forall t\ge t_0+T(R,\varepsilon ).
\end{equation*}
The following result is an obvious generalization of Proposition 12. Its
proof may be found in \cite[Lem. 1]{K2}. 
\begin{prop}
\label{prop10} 
The following statements are equivalent:\va
(i) The origin is UGAS for (\ref{eq:R1}); \va
(ii) there exists a function 
$\beta:[0,+\infty )\times [0,+\infty ) \to \R _+$ 
which is of class ${\cal LK}$ and such that
\begin{equation*}
\Vert x(t_0+h)\Vert \le \beta (h, \Vert x_0\Vert ) 
\end{equation*}
for each $t_0\in \R ^+$, each $x_0\in \R ^n$, each 
$x(\cdot )\in S_{t_0,x_0}$ and each $h\ge 0$.
\end{prop}
The second Liapunov theorem, as stated in Theorem 25, is
still valid in the context of differential inclusions. Notice, however, 
that for condition (iii) in Definition 30 to be fulfilled, it is 
{\em sufficient} (not necessary) that the following (strong) 
infinitesimal decrease condition holds   
\begin{equation*}
\frac{\partial V}{\partial t}(t,x)+
%\frac{\partial V}{\partial x}(t,x)\cdot v \le -\lambda V(t,x)
\langle \nabla _x V(t,x),v\rangle  \le -c(\Vert x\Vert )
\quad \hbox{ for a.e. }\  t\ge 0,\hbox{ for all }  x\in \Rn
\hbox{ and all } v\in F(t,x),
\end{equation*}
where $c$ denotes some function in ${\cal K}_0^{\infty}$.
Let us end this section by some useful result.
\begin{prop}
\label{prop11}
Let $F:\R _+\times \R ^n\to 2^{\R ^n}$ be a locally Lipschitz continuous
(for the Hausdorff distance) multivalued map, which assumes nonempty
closed values for each $(t,x)$. If (\ref{eq:R1}) is UGAS at the origin, then
the same is true for the system $\dot x\in \overline{co}\{ F(t,x)\}$.
\end{prop}
\noindent {\sc Proof.} It is sufficient to go back to the definition of
the (uniform) global asymptotic stability for (\ref{eq:R1}), expressed
in terms of $m(R)$ and $T(R,\varepsilon )$ (as above),  and to invoke
\cite[Thm. 3]{F1}.\qeed
\subsection{Time invariant systems}
In this subsection, we limit ourselves to the time independent case, i.e.
we consider differential inclusions of the form
\begin{equation}
\label{eq:R200}
\dot x \in F(x),
\end{equation}
where $F:\R ^n\to 2^{\R ^n}$ take nonempty compact values. The first 
converse of second Liapunov theorem in this context has
been given by Lin, Sontag and
Wang in \cite[Thm 2.9]{LSW}. 
\begin{thm}
\label{LSW}
Assume that the origin is UGAS for (\ref{eq:R200}), where $F$ is
a locally Lipschitz continuous multivalued map, which takes
nonempty compact values. Then there exists a positive definite smooth
strict Liapunov function $V$, which satisfies
\begin{equation*}
\langle \nabla _x V(x),v\rangle  \le -c(\Vert x\Vert ) 
\qquad \forall x\in\R ^n,\ \forall v\in F(x),
\end{equation*}
for some function $c\in {\cal K}_0^{\infty}$.
\end{thm}
\noindent 
Actually, Theorem \ref{LSW} is stated in \cite{LSW} in a somewhat different
manner: (i) $F$ takes in \cite{LSW} the special form
$F(x):=\{ f(x,d), d\in D\}$, where $f:\R ^n\times \R ^m\to \R ^n$
is a smooth function  and $D$ is a compact set in $\R ^m$; (ii) 
Theorem 2.9 in \cite{LSW} deals with the asymptotic stability 
w.r.t. to any {\em compact invariant set}, instead of the
origin. Theorem \ref{LSW} has been used in \cite{SW1} to prove 
that the converse of  Theorem 14 holds true, as well.

Another converse Liapunov theorem has been obtained 
a few years later by Clarke, Ledyaev and Stern in \cite{CLS} 
for another class of multivalued maps.
\begin{thm}
\label{CLS}
Assume that the origin is UGAS for (\ref{eq:R200}), where $F$ is
a upper semi-continuous (u.s.c.) multivalued map, which takes
nonempty compact values. Then there exists a positive definite smooth
strict Liapunov function, which satisfies
$\langle \nabla _x V(x),v\rangle  \le -W(x)$ for each 
$x\in\R ^n \setminus \{ 0 \}$ and 
each $v\in F(x)$,
for some definite positive continuous function $W$.
\end{thm}
\noindent
Here, $F$ is only supposed to be u.s.c., but $F(x)$ is 
convex for each $x$. Thanks to Proposition \ref{prop11}, Theorem
\ref{LSW} turns out to be a direct consequence of Theorem \ref{CLS}.
\subsection{Time varying systems}
In \cite{R1}, the author proves a converse theorem for global asymptotic
stability, by extending  Kurzweil's proof to Filippov's solutions of
equations with time-dependent, discontinuous  right hand side. In fact,
using the same method as in \cite{R1} and \cite[Lem. 4.1]{BR1}, it is
possible to  achieve a stronger result.
\begin{thm}
\label{Ro92}
Assume that the origin is an equilibrium position for (\ref{eq:R1}),
and that $F$ fulfills $(H_1), (H_2), (H_3)$ and $(H_4)$. If (\ref{eq:R1})
is UGAS, then there exists a strict Liapunov function, which is locally
Lipschitz continuous with respect to both variables.
\end{thm}
\noindent
Surprisingly enough, the smoothness of the Liapunov function in  
Theorem \ref{Ro92} cannot be improved in general, as it is shown by 
next result (see \cite[Prop. 1]{R2})
\begin{prop}
\label{prop12}
There exists a bounded Borel map $\lambda \! :\R \to \R$ such that the origin
is UGAS for the system
\begin{equation}
\dot{x}=\lambda (t) x
\label{eq:10}
\end{equation}
and such that there does not exist any (strict) Liapunov function
in the large {\rm (}or in the small{\rm )} \sou{of class $C^1$}.
\end{prop}
\noindent
Therefore, if we intend to
construct a {\em smooth} Liapunov function, some
additional assumption has to be made as far as the regularity of $f$
w.r.t. time is concerned. Let us introduce the following
\begin{defi}
\label{def1}
Let $f\in {\cal L}_{\rm loc}^{\infty}(\R^+\times \R ^n,\R ^m)$,
$m,n\ge 1$. The function $f$ is said to
be {\em essentially continuous w.r.t. time} ({\bf e.c.t.} in short)
if there exists
a set ${\cal N}\subset \R^+\times \R ^n$ of measure zero such that
for each
pair $(t_0,x_0)\in\R^+\times \R ^n$ and for each $\varepsilon >0$, there
exists
$\delta >0$  such that for each pair $(t,x)\in\R^+\times \R ^n$
\begin{equation}
\label{eq:13}
\left.
\begin{array}{l}
\vert t-t_0\vert +\Vert x-x_0\Vert <\delta \\
(t,x)\not\in {\cal N} \\
(t_0,x)\not \in {\cal N}
\end{array}
\right\} \im \Vert f(t,x)-f(t_0,x)\Vert < \varepsilon .
\end{equation}
\end{defi}
\noindent
Roughly speaking, the function $f$ is {\bf e.c.t.}
if the restriction of $f$ to the
complement of some set of measure zero is continuous w.r.t. time, in a
(locally) uniform way w.r.t. $x$.
Such a property is fulfilled when, for instance, $f$ takes the form
\begin{equation*}
f(t,x):=h(u(t),v(x)),
\end{equation*}
where $u\! :\R^+ \to\R ^p$ is a (locally) Riemann integrable map,
$v\in {\cal L}_{\rm loc}^{\infty}(\R^n,\R^q)$ and
$h\in C^0(\R^{p+q},\R^m )$, $m,n,p,q\ge 1$.
Indeed, a bounded function defined on a segment of
$\R$ is Riemann integrable iff the set of points at which the function fails
to be continuous is of measure zero \cite{Rd1}. Notice that
every field $f$ of the form
\begin{equation*}
f(t,x):= f_0(x) + \sum_{i=1}^p u_i(t)\, f_i(x),
\end{equation*}
where  for each
$i\ge 0$ $f_i\in {\cal L}_{\rm loc}^{\infty}(\R^n,\R^n)$ and
for each $i\ge 1$
$u_i$ is a piecewise continuous (or piecewise monotonous) function, is
concerned.

To state a converse Liapunov theorem which provides a {\em smooth}
Liapunov function
in a setvalued context, we need introduce a strenghtened
version of (H), namely the assumption
(H'): There exists a zero measure set $N_0\subset \R ^+$ such that\va
($H_1'$) For each $(t,x)\in (\R ^+\setminus N_0)\times \R ^n$,
$F(t,x)$ is a nonempty convex compact set in $\R ^n\, $; \va
($H_2'$) For each $R>0$, there exists a number $M>0$ such that 
\begin{equation*}
\left( t\in [0,R]\setminus N_0,\ \vert x\vert \le R \right) 
\im F(t,x)\subset  \overline{B}_M\, ;
\end{equation*}
($H_3'$) The restriction
$F_{\vert _{(\Rp \setminus N_0 )\times \R ^n}}$ is upper semicontinuous
(u.s.c.),
i.e. for each pair $(t_0,x_0)\in (\Rp \setminus N_0 )\times \Rn$
and each $\epsilon >0$ there exists $\delta >0$ such that for any
$(t,x)\in (\Rp \setminus N_0 )\times \Rn$ 
\begin{equation*}
\Vert (t-t_0,x-x_0)\Vert <\delta \ \im \ F(t,x)\subset
F(t_0,x_0)+\overline{B}_{\epsilon}.
\end{equation*}
The link between the essential continuity w.r.t. time and {\rm (}
${\rm H}'${\rm )} appears
clearly in next 
\begin{prop}
Let $f\in {\cal L}^{\infty}_{\rm loc}(\R ^+\times \R ^n,\R ^n)$. If $f$ is
{\bf e.c.t.}, then (${\rm H}'$) is fulfilled by
$F(t,x):=K_x\, f(t,x)$.
\end{prop}
We are now in position to state the main result of this section, which contains
Theorems 27 (in Chapter 3), and Theorems \ref{LSW} and  \ref{CLS}.
\begin{thm} {\sc (Converse of second Liapunov theorem)} 
\label{thm1}
Assume that (\ref{eq:R1}) is uniformly globally asymptotically stable (UGAS)
at the origin  and that (H')  is fulfilled. Then for any $\lambda >0$
there exist a function $V:\Rp\times \Rn\to\R _+$ of class $C^{\infty}$
and two functions $a,b$ in ${\cal K}_0^{\infty}$
such that 
\begin{equation}
\label{eq:R2}
a(\Vert x\Vert )\le V(t,x)\le b(\Vert x\Vert )\quad
\hbox{for all }\  t\ge 0 \hbox{ and } x\in \Rn\, ;
\end{equation}
\begin{equation}
\label{eq:R3}
\frac{\partial V}{\partial t}(t,x)+
%\frac{\partial V}{\partial x}(t,x)\cdot v \le -\lambda V(t,x)
\langle \nabla _x V(t,x),v\rangle  \le -\lambda\,  V(t,x)
\quad \hbox{ for all }\  t\in \Rp\setminus N_0,\ x\in \Rn 
\hbox{ and } v\in F(t,x).
\end{equation}
Furthermore, we may also require that $V$ be time-periodic if $F$ is
time-periodic (with the same period), and that $V$ be time-independent if
$F$ is time-independent.
\end{thm}
\noindent
{\sc Remark:} We infer from (\ref{eq:R2}) and (\ref{eq:R3}) the
standard estimate
\begin{equation*}
\frac{\partial V}{\partial t}(t,x)+
%\frac{\partial V}{\partial x}(t,x)\cdot v \le -c\, (\Vert x\Vert ),
\langle \nabla _x V(t,x),v\rangle  \le -c\, (\Vert x\Vert ),
\end{equation*}
where $c(\cdot )$ denotes some function in 
${\cal K}_0^{\infty}$. (Take
$c:=\lambda a$.) We may also infer from (\ref{eq:R3}) that $V$ is
exponentially decreasing along the trajectories of (\ref{eq:R1}), meaning
that for any trajectory $x(\cdot )\! :\Rp\to \Rn$ of (\ref{eq:R1})
\begin{equation*}
V\big( x(t)\big) \le e^{-\lambda t}V\big( x(0)\big)\quad\forall t\ge 0  .
\end{equation*}
The interested reader is referred to next subsection for a detailed proof.
Let us briefly outline this proof.
In the first step, using an
idea from \cite{CLS}, we show that the (uniform) global asymptotic
stability also holds true for some perturbed  system 
\begin{equation}
\label{eq:Ra3}
\dot x\in F_2(t,x),
\end{equation}
where $F(t,x)\subset F_2(t,x)$ for every $x$ and a.e. $t$,
and $F_2$ is locally
Lipschitz continuous (for the Hausdorff distance) in
$\Rp\times (\Rn\setminus \{ 0\} )$. The nice property we gain is that, for
each pair $(t_0,t_1)$ of times ,  the flow map
$x_0\mapsto \{ x(t_1): x(\cdot) 
\hbox{ solves (\ref{eq:Ra3}) and } x(t_0)=x_0\}$
is locally Lipschitz continuous for the Hausdorff distance.
In next step, we
adapt some explicit construction by Yoshizawa in \cite{Y1} 
of a locally Lipschitz continuous 
Liapunov function $V_L$ to our differential inclusion setting. In the final
step, we smooth $V_L$ by a standard method using  convolution 
with mollifiers and partition of unity.
\subsection{Proof of the converse of second Liapunov theorem}
The following set is introduced for notational convenience  
\begin{equation*}
E:=([-1, +\infty ) \setminus N_0 )\times \Rn .
\end{equation*}
Next lemma, which is a special instance of Theorem 5 (Chapter 1), 
is reported here for the comfort of the reader.
\begin{lem}
\label{lem1}
Let $F=F(t,x)$ be a multivalued map fulfilling $(H_1)$, let
$y:[T_1,T_2]\to\Rn$ denote some solution of (\ref{eq:R1}) and
let $b>0$.  Assume that $F$ is Lipschitz continuous in the region
$t\in [T_1,T_2],\ \Vert x-y(t)\Vert \le b$. It means that
there exists some constant $K>0$ such that for any $t,\bar{t}\in [T_1,T_2]$
and any $x,\bar{x}\in\Rn$ with $\Vert x-y(t)\Vert\le b$,
$\Vert \bar{x}-y(\bar{t})\Vert\le b$ we have
\begin{equation*}
h\big( F(t,x),F(\bar{t},\bar{x})\big) \, \le \, K\, 
\Vert (t-\bar{t},x-\bar{x})\Vert .
\end{equation*}
Let $(t_0,x_0)\in [T_1,T_2]\times \Rn$ be such that
$\Vert x_0-y(t_0)\Vert \le b$. Then there exists a solution $x(\cdot )$ of
(\ref{eq:R1}) such that $x(t_0)=x_0$ and
\begin{equation}
\label{eq:R4}
\Vert x(t)-y(t)\Vert \le \Vert x_0-y(t_0)\Vert e^{K\vert t-t_0\vert}
\end{equation}
as long as $\Vert x_0-y(t_0)\Vert e^{K\vert t-t_0\vert}\le b$.
\end{lem}
\noindent
The construction of $F_2$, $V_L$ and $V$  will be first
achieved in the general setting, i.e. when $F(t,x)$ is any time-dependent
setvalued map. The main modifications to be brought  in the time-periodic
or time-independent cases will be indicated after.\va\va
\begin{tabular}{|l|}
\hline
{\sc First step.} \\
\hline
\end{tabular}
$\quad$ (Regularization of $F$)\va\va
In order to facilitate the construction of a Liapunov function which is
{\em smooth up to $t=0$}, we extend $F$ on $[-1,0)\times \Rn$ by setting
$F(t,x)=\{ -x \}$. It is easily seen that $F$ fulfills (H')  on
$[-1,+\infty )\times \Rn$ (provided that $0$ is put into $N_0$)
and that (\ref{eq:R1}) remains UGAS
$[-1,+\infty )\times \Rn$. To smooth $F$ ``from the inside''
we need the following compactness lemma.
\begin{lem}
\label{lem2}
Assume that (H') holds true for $F$ on $[-1,+\infty )\times \Rn$.
Let $R>0$ be given and let $(t_1^j)_{j\ge 0}$, $(t_2^j)_{j\ge 0}$
and $(\delta ^j)_{j\ge 0}$ be sequences of numbers such that 
\begin{equation*}
\begin{array}{l}
-1\le t_1^j\le t_2^j\le  R\quad \hbox{for all }j,\\
\delta ^j\to 0\quad \hbox{as } j\to \infty .
\end{array}
\end{equation*}
Let $(x^j)_{j\ge 0}$ be a sequence of absolutely continuous functions
$x^j:[t_1^j,t_2^j]\to \overline{B}_R$ 
such that, for a.e. $t\in [t_1^j,t_2^j]$
\begin{equation*}
\dot{x}^j(t)\in \overline{co}\big\{
F\big( \overline{B}_{\delta ^j}(t,x^j(t))\cap E\big)\big\}
\end{equation*}
{\rm (}$\overline{B}_{\delta ^j}(t,x^j(t))$ denotes 
the closed ball  for the sup-norm in
$\R ^{n+1}$, which is centered at $(t,x^j(t))$ and has $\delta ^j$ as 
a radius.{\rm )}
Then there exist two numbers $t_1,t_2\in [-1,R]$, a function
$x:[t_1,t_2]\to \overline{B}_R$ and a sequence $j^k\to \infty$ such that
$x(\cdot )$ is a solution of (\ref{eq:R1}) on $[t_1,t_2]$,
$t_1^{j_k}\to t_1$, $t_2^{j_k}\to t_2$ as $k\to \infty$ and
\begin{equation*}
\lim_{k\to \infty}x^{j_k}(t_1^{j_k}) =x(t_1),\quad
\lim_{k\to \infty}x^{j_k}(t_2^{j_k}) =x(t_2).
\end{equation*}
\end{lem}
\noindent
{\sc Proof:} In what follows, we sometimes need to take convergent
subsequences of certain sequences of numbers or functions. In these
cases, for the sake of simplicity and without loss of generality, we
avoid using multiple indices: rather, we assume that the given sequence
is actually convergent. Thus, since $-1\le t_1^j\le t_2^j\le  R$,
we may assume that $t_1^j\to t_1$ and $t_2^j\to t_2$.
We may also assume that $\delta ^j\le 1$ for any $j\ge 0$.
According to ($H_2'$), there exists a constant $M>0$ such that
\begin{equation*}
t\in [-1,R+1]\setminus N_0,\ \Vert x\Vert \le R+1\ \im\
F(t,x)\subset \overline{B}_M.
\end{equation*}
For $j\ge 0$ and $t\in [-1,R]$  we set 
\begin{equation*}
\tilde{x}^j(t):=
\left\{
\begin{array}{lll}
x^j(t_1^j)\ &\hbox{ if }\  &-1\le t\le t_1^j,\\
x^j(t)\ &\hbox{ if }\  &t_1^j\le t\le t_2^j,\\
x^j(t_2^j)\ &\hbox{ if }\  &t_2^j\le t\le R.
\end{array}
\right.
\end{equation*}

Since for any $j\ge 0$, $\Vert \tilde{x}^j(t)\Vert \le R$ for
every $t\in [-1,R]$ and $\Vert \dot{\tilde x}^j(t)\Vert \le M$ for
a.e. $t\in [-1,R]$, we infer that the sequence $(\tilde{x}^j)$ is bounded
in the Sobolev space $H^1\big( (-1,R),\Rn\big)$, hence a subsequence
(also denoted by $(\tilde{x}^j)$) weakly converges in
$H^1\big((-1,R),\Rn\big)$ towards a function $x$. It follows that
$x^j\to x$ in $C^0( [-1,R],\R ^n)$ and that $\dot{\tilde x}^j\cf
\dot x$ in $L^2\big( (-1,R),\Rn\big)$ (see e.g. \cite[Thm. VIII.7]{Brezis}).
As an easy consequence, $\tilde{x}^j(t_1^j)\to x(t_1)$  and
$\tilde{x}^j(t_2^j)\to x(t_2)$. To prove that $x(\cdot )$ is
indeed a solution of (\ref{eq:R1}), i.e. that $\dot{x}(t)\in F(t,x(t))$
for a.e. $t\in [t_1,t_2]$, we consider the functional $J$ defined
on $L^2\big( (-1,R),\Rn\big)$ by
\begin{equation*}
J(w):=\int_{t_1}^{t_2}\hbox{dist}\big( w(t),F(t,x(t))\big)\, dt.
\end{equation*}
Notice that the nonnegative map
$t\mapsto \hbox{dist}\big( w(t),F(t,x(t))\big)$ is {\em measurable},
according to \cite[Prop. 3.4]{D}
\footnote{To apply this result we need to check that
the map $t\in (t_1,t_2)\mapsto F(t,x(t))$ be measurable. But the
map $t\in (t_1,t_2)\setminus N_0\mapsto (t,x(t))\in E$ is continuous and
$F_{\vert _E}$ is u.s.c., hence the composite map
$t\in (t_1,t_2)\setminus N_0 \mapsto F(t,x(t))$ is u.s.c., 
and therefore  measurable.
The same is true for the map $t\in (t_1,t_2)\mapsto F(t,x(t))$.}.
It is easily seen that the functional $J$ is well-defined, convex and
continuous for the strong topology of $L^2\big( (-1,R),\Rn\big)$,
hence it is l.s.c. for the weak topology of $L^2\big( (-1,R),\Rn\big)$
(see \cite[Cor. III.8]{Brezis}). Since
$\dot{\tilde x}^j\cf\dot x$ in $L^2\big( (-1,R),\Rn\big)$, we get
\begin{equation*}
(0\le )\ J(\dot{x})\le \liminf_{j\to \infty} J(\dot{\tilde x}^j).
\end{equation*}
To prove $J(\dot x )=0$, which is equivalent to (\ref{eq:R1}) on
$(t_1,t_2)$, we only have to show that
$\lim_{j\to \infty}J(\dot{\tilde x}^j)$ \va 
$=0$. Since
$\hbox{dist}\big( \dot{\tilde x}^j(t),F(t,x(t))\big) \le 2M$ for
a.e.  $t\in (t_1,t_2)$ and each $j$, applying Lebesgue's theorem, we are
done if we prove that for a.e. $t\in (t_1,t_2)$
\begin{equation*}
\lim_{j\to \infty}\hbox{dist}\big( \dot{\tilde x}^j(t),F(t,x(t))\big) =0.
\end{equation*}
For a.e. $t_0\in (t_1,t_2)\setminus N_0$, there exists an integer $j_0\ge 0$
such that for any $j\ge j_0$, $t_0\in (t_1^j,t_2^j)$ (hence the tilde
may be dropped) and $\dot{x} ^j(t_0)$ exists and it belongs to
$\overline{co}\big\{ F\big( \overline{B}_{\delta ^j}(t_0, x^{j}(t_0))
\cap E\big) \big\}$.
Let $\epsilon >0$. According to ($H_3'$), there exists $\delta >0$ such that
for any $(t,x)\in E$
\begin{equation*}
\Vert (t-t_0,x-x(t_0))\Vert \le \delta \im
F(t,x)\subset F(t_0,x(t_0))+\overline{B}_{\epsilon}.
\end{equation*}
By increasing $j_0$ if needed, we may also assume that for all $j\ge j_0$
\begin{equation*}
\delta ^j<\frac{\delta}{2}\ \hbox{ and } \ 
\Vert x^j(t_0)-x(t_0)\Vert<\frac{\delta}{2} \cdot
\end{equation*}
We infer that for all $j\ge j_0$,
$\ \overline{B}_{\delta ^j}(t_0,x^j(t_0)) 
\subset \overline{B}_{\delta }(t_0,x(t_0))$ and
\begin{equation*}
\overline{co}\, \big\{ 
F\big( \overline{B}_{\delta ^j}(t_0, x^{j}(t_0))\cap E\big) \big\}
\subset F(t_0,x(t_0))+\overline{B}_{\epsilon}.
\end{equation*}
It follows that
$\hbox{dist}\big( \dot{\tilde x}^j(t_0),F(t_0,x(t_0))\big) \le
\epsilon $ 
for $j\ge j_0$, as required.\qed

In what follows, $\delta =\delta (t,x)$ denotes any continuous  function
$[-1,+\infty )\times \Rn\to \R _+$  such that 
\begin{equation*}
\forall (t,x)\in [-1,+\infty )\times \Rn\quad
\delta (t,x)=0\, \iff\, x=0.
\end{equation*}
If such a function $\delta$ is given, we set for
any $(t,x)\in [-1,+\infty )\times \Rn$
\begin{equation}
\label{eq:R400}
F_1(t,x):= \overline{co}\, 
\big\{ F \big( \overline{B}_{\delta (t,x)}(t,x) \cap E\big) 
\big\} .
\end{equation}
We aim to prove that, for a suitable choice of $\delta$, the system
\begin{equation}
\label{eq:R5}
\dot x\in F_1(t,x),\quad t\ge -1,\ x\in \R ^n
\end{equation}
inherits the (uniform) global asymptotic stability of (\ref{eq:R1}).
Before proving such a property, let us observe that, for any choice of
$\delta$, the multivalued map $F_1$ fulfills (H') on
$[-1,+\infty )\times \Rn$, as well. Indeed,
\begin{equation*}
F_1(t,0)=\left\{ 
\begin{array}{ll}
F(t,0)\ &\hbox{ if }\ t\not\in N_0,\\
\emptyset\ &\hbox{ if }\  t\in N_0,
\end{array}
\right.
\end{equation*}
hence ($H_1'$) and ($H_2'$) are obvious. Next ($H_3'$) follows easily from
the upper-semicontinuity of the multivalued maps $F_{\vert _E}$ and
$(t,x)\mapsto \overline{B}_{\delta (t,x)}(t,x)\cap E \ (\in 2^E)$.

Since (\ref{eq:R1}) is UGAS, 
there exists a sequence
$(\varphi _i)_{i\in \Z}$ of continuous decreasing functions
$[0,+\infty )\to (0,+\infty )$ fulfilling the following properties:\va
(i) For any pair $(t_0,x_0)\in [-1,+\infty )\times \Rn$ and any
solution $x(\cdot )$ of (\ref{eq:R1}) such that $x(t_0)=x_0$,
\begin{equation*}
\Vert x_0\Vert \le 2^i\ \im \ 
\Vert x(t_0+h)\Vert < \varphi _i(h)\hbox{ for all } h\ge 0;
\end{equation*}
(ii) For each $i\in \Z$, $\varphi _i(t)\to 0$ as $t\to \infty$;\va
(iii) $\big( \varphi _i(0) \big) _{i\in \Z}$ is a nondecreasing sequence such
that $\varphi _i(0)\fd 0$ as $i\to -\infty$ and
$\varphi _i(0)\fd +\infty$ as $i\to +\infty$.\va
Indeed, it is sufficient to set $\varphi _i(h):=\beta (h,2^i)$, where
$\beta$ is given by Proposition \ref{prop10}. Notice that $\varphi _i(0)>2^i$.
For each $i\in \Z$, let $p_i\in \Z$ be the greatest integer
such that $\varphi _{p_i}(0)\le 2^{i-1}$.
Clearly, $p_i<i-1 $, the sequence $(p_i)$ is nondecreasing and 
$p_i\to \pm \infty$ as $i\to \pm \infty$. 
\begin{rmq}
\label{rmq1}
As a direct consequence, 
any compact set in 
$\R ^n \setminus \{ 0\} $ intersects the sets 
$\{ x\! :  2^{p_i}\le \Vert x\Vert \le \varphi _i(0)\}$
for at most a {\em finite} number of $i$.
\end{rmq}
For each $i\in \Z$, pick a number $T_i\ge 1$ such that
$\varphi _i(T_i)\le 2^{i-1}$ and set
$\hat{T}_i=\max(T_i,\max(T_j\! :\ p_j=i))$ when there exists $j\in\Z$ 
s.t. $p_j=i$,
$\hat{T}_i=T_i$ otherwise.
\begin{lem}
\label{lem3}
Let $i\in\Z$ and $k\in \{ -1\} \cup \N$. Then there exists a constant
$\delta >0$ such that, for any
solution $x(\cdot )$ of
\begin{equation}
\label{eq:R401}
\dot{x}\in \overline{co}\, \big\{ F\big( 
\overline{B}_\delta (t,x)\cap E\big)\big\}  
\end{equation}
fulfilling $\Vert x(t_0)\Vert \le 2^i$ for some time $t_0\in [k,k+1]$,
we have
\begin{equation*}
\Vert x(t_0 +h)\Vert < \varphi _i(h) \quad \forall h\in [0,\hat{T}_i].
\end{equation*}
\end{lem}
\noindent
{\sc Proof:} If the statement of Lemma \ref{lem3} is false, then there exists
a sequence of positive numbers $\delta ^j\searrow 0$ and a sequence of
absolutely continuous functions $(x^j)_{j\ge 0}$, with
$x^j:[t_0^j,t_1^j]\to\Rn$, 
$t_0^j\in [k,k+1]$, $t_0^j\le t_1^j\le t_0^j + \hat{T}_i$,
such that 
\begin{equation*}
\begin{array}{l}
\dot{x}^j(t)\in \overline{co}\{ F \big( 
\overline{B}_{\delta ^j}(t,x^j(t))\cap E\big)\}
\quad \hbox{for a.e. }\ t\in [t_0^j,t_1^j],\\
\Vert x^j(t_0^j)\Vert\le 2^i,\ \ 
\Vert x^j(t)\Vert <\varphi _i(t-t_0^j)
\hbox{ for all } t\in [t_0^j,t_1^j),\ 
\hbox{ and }\ \Vert x^j(t_1^j)\Vert  
=\varphi _i(t_1^j-t_0^j).
\end{array}
\end{equation*}
Using Lemma \ref{lem2} and extracting subsequences if needed, we may
also assume that for some numbers $t_0,t_1\in [k,k+1+\hat{T}_i]$ and
some solution $x(\cdot ):[t_0,t_1]\to \Rn$ of (\ref{eq:R1}), we have
$(t_0^j,x^j(t_0^j))\to (t_0,x(t_0))$ and
$(t_1^j,x^j(t_1^j))\to (t_1,x(t_1))$ as $j\to \infty$.
Hence $\Vert x(t_0)\Vert \le 2^i$ and
$\Vert x(t_1)\Vert =\varphi  _i(t_1-t_0)$,
which contradicts the definition of $\varphi _i$.\qed

For each pair $(i,k)\in \Z \times (\{ -1\} \cup \N )$, we denote by
$\delta _i^k$  the number $\delta >0$ given by Lemma \ref{lem3}.
Pick any continuous function
$\delta :[-1,+\infty )\times \R ^n\to \R _+$ fulfilling 
\begin{equation}
\label{eq:R6}
\delta (t,x)=0\, \iff \, x=0 \qquad \hbox{ and}
\end{equation}
\begin{equation}
\label{eq:R7}
\delta (t,x)<\min (\delta _i^k,\delta _{p_i}^k) \quad \hbox{whenever }
k\le t\ \hbox { and }\ 2^{p_i}\le \Vert x\Vert \le \varphi _i(0).
\end{equation}
Notice that such a function $\delta$ exists, due to Remark \ref{rmq1}.
Replacing (if needed) $\delta$ by the inf-convolution function
\begin{equation*}
\tilde{\delta}(t,x)=\inf_{(s,y)\in [-1,+\infty )\times \Rn}
\big( \delta (s,y)+\Vert (t-s,x-y)\Vert \big) ,
\end{equation*}
we may also require that $\delta$ be
Lipschitz continuous, with 1 as a Lipschitz
constant\footnote{
Indeed, $\tilde\delta (t,0)=0<\tilde\delta (t,x)\le \delta (t,x)$ for
$t\ge -1$, $x\ne 0$, and it is easy to see that
$\Vert \tilde\delta (t,x) -\tilde\delta (s,y)\Vert \le
\Vert (t-s,x-y)\Vert$
for all pairs
$(t,x),(s,y)\in [-1,+\infty )\times \Rn$.}.
To prove the (uniform) global  
asymptotic stability of (\ref{eq:R5}) we need the
\begin{lem}
\label{lem4} 
Let $\delta (\cdot )$ fulfill (\ref{eq:R6})-(\ref{eq:R7}), and 
let $x(\cdot )$ be any solution of (\ref{eq:R5}).
Assume that $\Vert x(t_0)\Vert \le 2^i$ for some
time $t_0\ge -1$ and some $i\in \Z$. Then 
\begin{equation*}
\begin{array}{ll}
\hbox{\rm (a)}\ &\Vert x(t_0+h)\Vert  < \varphi _i(0)\quad\forall 
h\in [0,T_i],\\
\hbox{\rm (b)}\ &\Vert x(t_0+T_i)\Vert\le 2^{i-1}
\end{array}
\end{equation*}
\end{lem}
\noindent
{\sc Proof:}
If (a) is false, then there exist two times $t_1,t_2$ such that
$t_0\le t_1<t_2\le t_0+T_i$ and 
$2^i = \Vert x(t_1)\Vert < \Vert x(t)\Vert 
< \Vert x(t_2)\Vert  =\varphi _i(0)$ 
for $t_1<t<t_2$.
Set $k:=[t_1]$. Since
$2^{p_i}\le \Vert x(t)\Vert \le \varphi _i(0)$ for
$t\in [t_1,t_2]$, we infer from (\ref{eq:R7}) that $x(\cdot )$ is
also a solution of
\begin{equation}
\label{eq:M1}
\dot x\in \overline{co}\, 
\big\{ F\big( \overline{B}_{\delta _i^k}(t,x)\cap E\big) \big\}
\end{equation}
on $[t_1,t_2]$, hence, by Lemma \ref{lem3},
$\Vert x(t_2 )\Vert <\varphi _i(t_2-t_1)<\varphi _i(0)$,
which contradicts some property
of $t_2$. We now turn to the proof of (b). Assume that
$\Vert x(t_0+T_i)\Vert  >2^{i-1}$.
If $2^{p_i}\le \Vert  x(t_0+h)\Vert \ (\le \varphi _i(0))$
for all $h\in [0,T_i]$, then
$x(\cdot )$ again solves (\ref{eq:M1}) on $[t_0,t_0+T_i]$
(with here $k:=[t_0]$), hence, by Lemma 
\ref{lem3},
$\Vert x(t_0+T_i)\Vert < \varphi _i(T_i)\le 2^{i-1}$, leading to
a contradiction. We infer that there exist two times $t_1,t_2$ such that   
$t_0\le t_1<t_2\le t_0+T_i$ and
$2^{p_i}=\Vert x(t_1)\Vert < \Vert x(t)\Vert <
\Vert x(t_2)\Vert =2^{i-1}$
for $t_1<t<t_2$. Set again $k:=[t_1]$. Since 
$\delta (t,x(t)) <\delta ^k_{p_i}$ for
$t_1\le t\le t_2$ and $t_2-t_1\le T_i\le \hat{T}_{p_i}$,
we infer from Lemma \ref{lem3} (applied with $p_i$ instead of $i$) that
$\Vert x(t_2)\Vert <\varphi _{p_i}(0)\le 2^{i-1}$, 
contradicting some  property
of $t_2$.\qeed
\begin{cor}
\label{cor1}
Let $\delta$ and $F_1$ be as above. Then (\ref{eq:R5})
is UGAS on $[-1,+\infty )$ $\times \Rn$.    
\end{cor}
\noindent
Indeed, we infer from (a) and (b) that $\Vert x(t_0+h)\Vert <
\varphi_{i-l}(0)$
for all $l\in\N$ and all $h\ge\sum_{j=i-l+1}^{i} T_j$.
\va

We are now in position to define the enlarged (and regularized) system 
$\dot x\in F_2(t,x)$. We need some suitable  partition of unity. We set
\begin{equation*} 
U:=(-1,+\infty )\times (\Rn \setminus \{ 0\} )
\end{equation*}
and, for any $(t,x)\in U$,
\begin{equation}
\label{eq:R10}
W(t,x):=
\{  (s,y)\in U:\ \Vert (s-t,y-x)\Vert <
\frac{1}{3}\delta (t,x)\}  \cdot
\end{equation}
We have the following \va\va
{\sc Claim 1.} The family 
$\big( W(t,x)\big) _{(t,x)\in U\cap E}$ is an open covering of $U$.\va\va
To prove the claim, we have to check that for any $(t_0,x_0)\in U$ there
exists some $(t,x)\in U\cap E$  such that $(t_0,x_0)\in W(t,x)$. If
$(t_0,x_0)\in U\cap  E=\big( (-1,+\infty )\setminus N_0\big) \times 
(\Rn\setminus \{ 0\} )$, then $(t_0,x_0)\in W(t_0,x_0)$ (since 
$\delta (t_0,x_0)>0$). If $(t_0,x_0)\in N_0\times (\Rn \setminus \{ 0\} )$,
since $N_0$, as a set of measure zero, has a dense complement in 
$(-1,+\infty )$, we may find a time $t\in (-1,+\infty )\setminus N_0$ such 
that $\vert t-t_0\vert <\frac{1}{6}\delta (t_0,x_0)$ and 
$\delta (t,x_0)>\frac{1}{2}\delta (t_0,x_0)$. Then $(t_0,x_0)\in W(t,x_0)$.
\qed

Let $(\psi _i)_{i\ge 1}$ be some $C^{\infty}$ partition of unity on $U$
subordinate to the open covering \va 
$\big( W(t,x)\big) _{(t,x)\in U\cap E}$ of
$U$. It means that 1) each $\psi _i$ is a nonnegative function of class
$C^{\infty}$ on $\R ^{n+1}$, with its support contained in $W(t_i,x_i)$
for some $(t_i,x_i)\in U\cap E$, 2) 
$\sum_{i=1}^{\infty}\psi _i(t,x)=1$ for any $(t,x)\in U$ and 3)
for each $(t,x)\in U$,  there exists a number $\rho >0$ such that 
$\psi _i\equiv 0$  on $\overline{B}_{\rho }(t,x)$  for all $i\ge 1$ except a 
{\em finite} number.
\va
Set for any $(t,x)\in (-1,+\infty )\times \Rn$
\begin{equation}
\label{eq:R11}
F_2(t,x):=
\left\{
\begin{array}{ll}  
\sum_{i=1}^{\infty}\psi _i(t,x)\ \overline{co}\, 
\big\{ F\big( 
\overline{B}_{\frac{1}{3}\delta (t_i,x_i)}(t_i,x_i)\cap E\big)\big\}
&\hbox{ if }x\ne 0,\\
F(t,0)&\hbox{ if }x=0.
\end{array}
\right.
\end{equation}
Since on compact subsets of $U$ the sum in (\ref{eq:R11}) is {\em finite},
we infer that $F_2$ is locally Lipschitz continuous (for the Hausdorff
distance) on $U$. The links between $F$, $F_1$ and $F_2$  appear
clearly in next claims.\va\va
{\sc Claim 2.} $\forall t\in (-1,+\infty )\setminus N_0,\
\forall x\in \Rn , \quad F(t,x)\subset F_2(t,x).$\va\va
The result is obvious for $x=0$, so let $x\ne 0$  and 
$t\in (-1,+\infty )\setminus N_0$. Let $i\ge 1$ be such that 
$\psi _i(t,x) >0$, hence $(t,x)\in W(t_i,x_i)$. By
(\ref{eq:R10}), $\Vert (t-t_i,x-x_i)\Vert <\frac{1}{3}\delta (t_i,x_i)$,
hence 
\begin{equation*}
F(t,x)\subset F\big( 
\overline{B}_{\frac{1}{3}\delta (t_i,x_i)} (t_i,x_i) \cap E\big) ,
\end{equation*} 
from which we infer that $F(t,x)\subset F_2(t,x)$.\qed
{\sc Claim 3.} $\forall (t,x)\in U\quad F_2(t,x)\subset F_1(t,x).$\va\va
To prove this claim, consider any pair $(t,x)\in  U$  and let $i\ge 1$
be such that $\psi _i(t,x)>0$. Since $\delta $ is Lipschitz continuous
with 1 as a Lipschitz constant, we have
\begin{equation}
\label{eq:R1100}
\delta (t_i,x_i)-\delta (t,x)\le \Vert (t-t_i,x-x_i)\Vert 
<\frac{1}{3} \delta (t_i,x_i)
\end{equation}
hence 
\begin{equation*}
\overline{B}_{\frac{1}{3}\delta (t_i,x_i)}(t_i,x_i) \subset 
\overline{B}_{\frac{2}{3}\delta (t_i,x_i)}(t,x) \subset 
\overline{B}_{\delta (t,x)}(t,x),
\end{equation*}
and we infer that 
$F_2(t,x)\subset\overline{co}\, 
\big\{ F\big( \overline{B}_{\delta (t,x)}(t,x)\cap E\big)\big\} =F_1(t,x)$.
\qed
\begin{rmq}
(H') is also fulfilled by $F_2$ on
$(-1,+\infty )\times \Rn$, with the same $N_0$ as for $F$.
Indeed $(H_1')$ is clear, $(H_2')$ follows from Claim 3, and
$(H_3')$ is obvious on $U$, since $F_2$ is locally Lipschitz
continuous on $U$. Finally, using the continuity of $\delta$, (\ref{eq:R6}), 
(\ref{eq:R1100}), and the fact that $F_{\vert _E}$ is 
u.s.c., we easily
check that ${F_2}_{\vert _E}$ is u.s.c. at each
$(t_0,0)$ for $t_0\in (-1,+\infty )\setminus N_0$.
\end{rmq}
\begin{cor}
\label{cor2}
The origin is UGAS on $(-1,+\infty )\times \R ^n$ 
for the system
\begin{equation}
\label{eq:R13}
\dot x \in F_2(t,x)
\end{equation}
\end{cor}
\noindent
Corollary 2 is a direct consequence of Corollary 1 and Claim 3. 
This completes the first step in the proof of Theorem 1. We now proceed to the
\va\va
\begin{tabular}{|l|}
\hline
{\sc Second step.} \\
\hline
\end{tabular}
$\quad$ (Construction of a locally Lipschitz continuous Liapunov function)
\va\va   
For any $(t_0,x_0)\in (-1, +\infty )\times \R ^n$, let  
${\cal S}_{t_0,x_0}$ denote the set of solutions 
$x(\cdot )\! :[t_0,+\infty )$ of the Cauchy problem 
\begin{equation}
\label{eq:R14}
\left\{ 
\begin{array}{l}
\dot x\in F_2(t,x)\\
x(t_0)=x_0.
\end{array}
\right.
\end{equation}
Set for any $q\in \N ^*$ and any $r\in [0,+\infty )$,
\begin{equation}
\label{eq:R15}
G_q(r):=\max \big( 0,r-\frac{1}{q} \big) .
\end{equation}
Finally, set for any $q\in \N ^*$ and any $(t,x)\in (-1,+\infty ) 
\times \R ^n$,
\begin{equation}
\label{eq:R16}
V_q(t,x):=\sup_{\varphi \in {\cal S}_{t,x} }\sup_{\tau \ge 0}
e^{2\lambda \tau }G_q(\Vert \varphi (t+\tau )\Vert ) ,
\end{equation}
where $\lambda >0$ is the number appearing in the statement of Theorem 1.
The system (\ref{eq:R13}) being UGAS, for
any $R>0$ and any $q\in \N ^*$  there exist positive numbers $m(R)$ and
$T(R,q)$ such that for each pair $(t_0,x_0)\in (-1,+\infty )\times
\R ^n$ and for each solution $\varphi \in {\cal S}_{t_0,x_0}$, if
$\Vert x_0\Vert \le R$ then $\Vert \varphi (t_0+\tau  )\Vert <m(R)$
for all $\tau \ge 0$ and $\Vert \varphi (t_0+\tau  )\Vert <
\frac{1}{q}$ for all $\tau > T(R,q)$. Furthermore, $T(R,q)$ and
$m(R)$ may be chosen in such a way that $T(\cdot ,q)$ is nondecreasing
for each $q\ge 1$ and $m(\cdot )$ is in 
${\cal K}_0^{\infty }$. As a direct consequence, we obtain
the 
\begin{lem}
\label{lem5}
Let $R>0$ and $(t,x)\in (-1,+\infty )\times \overline{B}_R$. 
Then, for each  $q\in \N ^*$, 
\begin{equation}
\label{eq:R17}
G_q(\Vert x\Vert )\le V_q(t,x) =\sup_{\varphi\in {\cal S}_{t,x}} 
\sup_{\tau \in [0,T(R,q)]} e^{2\lambda \tau} 
G_q(\Vert \varphi  (t+\tau )\Vert ) \le e^{2\lambda T(R,q)}m(R) <+\infty . 
\end{equation}
\end{lem}
\noindent
Indeed, for any $\varphi\in {\cal S}_{(t,x)}$ and any 
$\tau >T(R,q)$, $\Vert \varphi (t+\tau )\Vert <\frac{1}{q}$,
hence $G_q(\Vert \varphi (t+\tau )\Vert )=0$, which implies the equality
in (\ref{eq:R17}).
Next, we remark that $G_q(\Vert \varphi (t+\tau )\Vert )\le 
\Vert \varphi (t+\tau )\Vert \le m(R)$. 
Finally, the first inequality is obvious 
(consider $\tau =0$ in (\ref{eq:R16})). 
\qed 
Notice that $V_q(t,0)=0$ for any $t>-1$, since ${\cal S}_{t,0}$ contains
only the trivial solution $x\equiv 0$.  Another important property
of $V_q$, namely the local Lipschitz continuity, is stated in next
\begin{prop}
\label{prop1}
Let $q\in \N ^*$ and $R>0$ be given. Then there exists a positive constant
$C_q(R)$ such that for any $t_1,t_2\in [-\frac{R}{R+1},R]$ and any   
$x_1,x_2\in \overline{B}_R$,
\begin{equation}
\label{eq:R18}
\vert V_q(t_1,x_1)-V_q(t_2,x_2)\vert \le C_q(R)\, \Vert 
(t_1-t_2,x_1-x_2)\Vert .
\end{equation}
\end{prop}
\noindent 
The proof of Proposition \ref{prop1} 
is somewhat long and technical. It will be reported 
at the end of Step 2. Without loss of generality, we may assume that 
for each $q\in \N ^*$
the function $R>0\mapsto C_q(R)>0$ is nondecreasing. We are now in position 
to define a locally Lipschitz continuous (strict) Liapunov function 
for (\ref{eq:R13}). For any  
$(t,x)\in (-1,+\infty )\times \R ^n$ we set
\begin{equation}
\label{eq:R19}
V_L(t,x):=\sum_{q=1}^{+\infty }\frac{2^{-q}}{1+C_q(q)} \,
e^{-2\lambda  T(q,q)} \,  V_q(t,x) .
\end{equation}
{\sc Claim 4.} There exists a function 
$a_L(\cdot )$ in ${\cal K}_0^{\infty }$
such that 
\begin{equation}
\label{eq:R20}
a_L(\Vert x\Vert ) \le V_L(t,x)\quad \forall (t,x)\in (-1,+\infty )
\times \R ^n.
\end{equation}
To prove the claim, set for any $r\ge 0$
\begin{equation}
\label{eq:R21}
a_L(r):=\sum_{q=1}^{+\infty}
\frac{2^{-q}e^{-2\lambda T(q,q)}}{1+C_q(q)}\, G_q(r) \cdot
\end{equation}
Clearly, $a_L(\cdot )$ is a well-defined, increasing,
 Lipschitz continuous 
function (with $\sum_{q=1}^{+\infty}
\frac{2^{-q}e^{-2\lambda T(q,q)}}{1+C_q(q)}$ 
as a Lipschitz constant)
such that $\lim_{r\to +\infty}\, a_L(r)=+\infty$.
(\ref{eq:R20}) follows from (\ref{eq:R17}) and (\ref{eq:R19}).\qed
{\sc Claim 5.} Let $R>0$ be given. Then there exists a constant
$L(R)>0$ such that for any $t_1,t_2\in \big[ -\frac{R}{R+1}, R\big]$ 
and any $x_1,x_2\in \overline{B}_R$
\begin{equation}
\label{eq:R22}
\vert V_L(t_1,x_1)-V_L(t_2,x_2)\vert \le L(R)\, 
\Vert (t_1-t_2,x_1-x_2)\Vert .
\end{equation}
Indeed, 
\begin{equation}
\label{eq:R23}
\begin{array}{rl}
V_L(t_1,x_1)&\le V_L(t_2,x_2)+
\sum_{q=1}^{+\infty} \frac{2^{-q}e^{-2\lambda T(q,q)}}{1+C_q(q)}
\vert V_q(t_1,x_1)-V_q(t_2,x_2)\vert \\
&\le V_L(t_2,x_2)+ 
\big( \sum_{q=1}^{+\infty}2^{-q} \frac{C_q(R)}{1+C_q(q)}
e^{-2\lambda T(q,q)} \big) \Vert (t_1-t_2,x_1-x_2)\Vert ,
\end{array}
\end{equation}
due to Proposition \ref{prop1}. For any $R>0$, set
\begin{equation}
\label{eq:R24}
L(R):=\sum_{q=1}^{+\infty}2^{-q} \frac{C_q(R)}{1+C_q(q)}
e^{-2\lambda T(q,q)} .
\end{equation}
Let us prove that $L(R)<+\infty$ for any $R>0$. Using the fact that 
$C_q(q)\ge C_q(R)$ if $q>R$, we get 
\begin{equation*}
L(R)\le \sum_{q=1}^{[R]}2^{-q} \frac{C_q(R)}{1+C_q(q)}
e^{-2\lambda T(q,q)} +
\sum_{q=[R]+1}^{+\infty}2^{-q}\, <+\infty .
\end{equation*}
Notice also that $L(\cdot )$ is a nondecreasing function. If
$x_2=0$, then $V_L(t_2,x_2)=0$ and it follows from (\ref{eq:R23}) 
(applied with $t_2=t_1$) that $V_L(t_1,x_1)\le L(R)\, \Vert x_1\Vert
<+\infty$, hence $V_L$ is well-defined. 
Exchanging the r\^oles of $(t_1,x_1)$ and $(t_2,x_2)$ in (\ref{eq:R23}),
we get
(\ref{eq:R22}).\qed
{\sc Claim 6.} There exists a function $b_L(\cdot )$ in 
${\cal K}_0^{\infty}$ such that
\begin{equation}
\label{eq:R25}
V_L(t,x)\le b_L(\Vert x\Vert ) \quad \forall (t,x)\in 
(-1,+\infty )\times \R ^n.
\end{equation}
Indeed, it follows from (\ref{eq:R17}) that for any
$R>0$ and any $(t,x)\in (-1,+\infty )
\times \overline{B}_R$
\begin{equation*}
\begin{array}{rl}
V_L(t,x)&\le  \sum_{q=1}^{+\infty}2^{-q}
\frac{e^{2\lambda \big( T(R,q)-T(q,q)\big)}m(R)}{1+C_q(q)} \\
&\le \big[ \sum_{q=1}^{[R]}
2^{-q} \frac{e^{2\lambda \big( T(R,q)-T(q,q)\big)}}{1+C_q(q)} 
+1\big] m(R) =:{\tilde m}(R).
\end{array}
\end{equation*}
(We used the fact that $T(R,q)\le T(q,q)$ for $q\ge [R]+1$.) 
Since $\tilde m (\cdot )$ is nondecreasing and\va 
$\lim_{R\to 0^+}{\tilde m}(R)=0$, there exists a function $b_L(\cdot )$
in ${\cal K}_0^{\infty }$ such that $\tilde{m}(R)\le b(R)$ for all
$R>0$. Then $V_L(t,x)\le b_L(\Vert x\Vert  )$, as wanted. \qed
The last useful property of $V_L$ we report here is the way $V_L$ 
decreases along the trajectories of (\ref{eq:R13}). \va\va
{\sc Claim 7.} Let $(t_0,x_0)\in (-1,+\infty )\times \R ^n$ and
let $\psi \in {\cal S}_{t_0,x_0}$. Then 
\begin{equation}
\label{eq:R26}
V_L(t_0+h,\psi (t_0+h)) \le e^{-2\lambda h}\, V_L(t_0,x_0)\quad 
\forall h\ge 0.
\end{equation}
This claim is a direct consequence of next result, which will be needed
when proving Proposition \ref{prop1}. 
\begin{lem}
\label{lem6}
Let $(t_0,x_0)\in (-1,+\infty )\times \R ^n$ and let
$\psi \in {\cal S}_{t_0,x_0}$. Then, for any $q\in \N ^*$, 
\begin{equation}
\label{eq:R27}
V_q(t_0+h,\psi (t_0+h)) \le e^{-2\lambda h}V_q(t_0,x_0)\quad
\forall h\ge 0.
\end{equation}
\end{lem}
\noindent
To prove the lemma, consider for any
$\varphi \in {\cal S}_{t_0+h,\psi (t_0+h)}$
the function $\overline{\varphi}:[t_0,+\infty )\to \R ^n$ defined by 
\begin{equation*}
\overline{\varphi}(t):=\left\{
\begin{array}{ll}
\psi (t)\ &\hbox{ for } t_0\le t\le t_0+h\\
\varphi (t) \ &\hbox{ for } t_0+h \le t.  
\end{array}
\right.
\end{equation*}
Clearly $\overline{\varphi}\in {\cal S}_{t_0,x_0}$, hence 
\begin{equation*}
\begin{array}{rl}
V_q(t_0,x_0)&\ge \sup_{\tau \ge 0} e^{2\lambda \tau}
G_q\big( \Vert \overline{\varphi} (t_0+\tau )\Vert \big) \\
&\ge e^{2\lambda h}\, \sup_{\tau \ge 0} e^{2\lambda \tau} 
G_q\big( \Vert \varphi (t_0+h+\tau )\Vert \big) .
\end{array}
\end{equation*}
$\varphi \in {\cal S}_{t_0+h,\psi (t_0+h)}$ being arbitrary, 
we infer that 
$V_q(t_0,x_0) \ge e^{2\lambda h}V_q(t_0+h,\psi (t_0+h))$. \qeed
\begin{cor}
\label{cor3}
For a.e. $(t_0,x_0)\in U$,
\begin{equation*}
\forall v\in F_2(t_0,x_0),
\quad \dvt (t_0,x_0) +\langle \dvx (t_0,x_0),v\rangle
\ \le \ -2\, \lambda \, V_L(t_0,x_0).
\end{equation*}
\end{cor}
\noindent{\sc Proof:}
Since $V_L$ is locally Lipschitz continuous on $U$
(by Proposition \ref{prop1}), we infer from Rademacher's theorem
that $V_L$ is differentiable at $(t_0,x_0)$ for almost every
$(t_0,x_0)\in U$.
Therefore, it is sufficient to prove that for {\em every }
pair $(t_0,x_0) \in U$,
\begin{equation*}
\forall v\in F_2(t_0,x_0),
\quad
\limsup _{h\to 0^+}
\frac{V_L(t_0+h,x_0+hv)-V_L(t_0,x_0)}{h} \le -2\, \lambda \, V_L(t_0,x_0).
\end{equation*}
Notice first that for each $v\in F_2(t_0,x_0)$,
there exists a solution $x(t)$ of (\ref{eq:R13}) such that
$x(t_0)=x_0$ {\em and} $\dot{x}(t_0)=v$. Indeed, we may infer from
the local Lipschitz continuity of $F_2$ that the map
\begin{equation}
\label{eq:R100}
g(s,y):=\pi _{F_2(s,y)}(v),\quad (s,y)\in U,
\end{equation}
(where $\pi _{F_2 (s,y)}$ denotes the projection on the convex compact
set
$F_2(s,y)$) is {\em continuous}. 
Hence, by Peano's theorem, there exists
a solution $x(t)$ on some interval
$[{t_0},{t_0}+\varepsilon ]$ of the (classical) Cauchy problem
\begin{equation*}
\left\{
\begin{array}{rl}
\dot{x}&=g(t,x)\  (\in F_2(t,x))\\
x(t_0)&=x_0.
\end{array}
\right.
\end{equation*}
Obviously, ${\dot x}(t_0)=g(t_0,x_0)=v$.
Now there exists a constant $K>0$ such that for any $(t_1,x_1)$, $(t_2,x_2)$
in some neighborhood of
$(t_0,x_0)$,
\begin{equation*}
\vert V_L (t_1,x_1)-V_L (t_2,x_2)\vert \le K\, \Vert (t_1-t_2,x_1-x_2
)\Vert.
\end{equation*}
It follows that for
$h>0$ small enough
\begin{equation*}
\begin{array}{l}
\displaystyle\frac{V_L(t_0+h,x_0+hv)-V_L (t_0,x_0)}{h}\\
\qquad =\displaystyle\frac{V_L (t_0+h,x_0+hv)-V_L (t_0+h,x(t_0+h))}{h} +
\displaystyle\frac{V_L (t_0+h,x(t_0+h))-V_L (t_0,x_0)}{h}  \\
\qquad \le K\left\Vert\frac{x(t_0+h)-x_0}{h}-v\right\Vert 
+\frac{e^{-2\lambda h}-1}{h}\, V_L(t_0,x_0),
\end{array}
\end{equation*}
hence
\begin{equation*}
\limsup _{h\to 0^+}\frac{V_L (t_0+h,x_0+hv)-V_L (t_0,x_0)}{h} 
\le -2\lambda V_L(t_0,x_0).
\end{equation*}
\qed
It remains to prove Proposition \ref{prop1}. We first need
the following elementary lemma.
\begin{lem}
\label{lem7}
Let $V$ be a function defined on a set 
$K:=\prod_{i=1}^n[ a_i,b_i] \subset \R ^n$. Assume the existence of
a constant $L>0$ such that, for any $x_0\in K$ we may find a number 
$\eta _0 > 0$ for which  
\begin{equation}
\label{eq:R28}
\vert V(x)-V(x_0)\vert \le L\, \Vert x-x_0\Vert \quad 
\forall x\in \overline{B}_{\eta _0}(x_0)\cap K.
\end{equation}
Then $V$ is Lipschitz continuous on $K$, with $nL$ as a Lipschitz constant. 
\end{lem}
\begin{rmq} (\ref{eq:R28}) is weaker than the assumption 
$\vert V(x)-V(y)\vert \le L\, \Vert x-y\Vert \ \ \forall x,y\in 
\overline{B}_{\eta _0}(x_0)\cap K$, 
for which Lemma \ref{lem7} is classical.
\end{rmq}
\noindent{\sc Proof.} Let $x^1=(x^1_1,\ldots ,x^1_n)$ and  
$x^2=(x^2_1,\ldots ,x^2_n)$ be given in $K$. Then
\begin{equation*}
\vert V(x^1)-V(x^2)\vert \le 
\sum_{j=1}^n 
\vert V(x^1_1,\ldots ,x^1_j,x^2_{j+1},\ldots ,x^2_n) -
V(x^1_1,\ldots ,x^1_{j-1},x^2_j,\ldots ,x^2_n)\vert 
\end{equation*}
so we are done if we prove that for any $j\in [1,n]$ 
\begin{equation*}
\vert V(x^1_1,\ldots ,x^1_j,x^2_{j+1},\ldots ,x^2_n) -
V(x^1_1,\ldots ,x^1_{j-1},x^2_j,\ldots ,x^2_n)\vert
\le L \vert x^1_j-x^2_j\vert .
\end{equation*}
Fix $j\in [1,n]$ and assume for instance that $x^1_j \le x^2_j$. 
Define a function $f:[x^1_j ,x^2_j]\to \R$ by 
$f(s)=V(x^1_1,\ldots ,x^1_{j-1},s,x^2_{j+1},\ldots ,x^2_n)$. Then it
follows from (\ref{eq:R28}) that $\vert \overline{D^+}\, f(s)\vert \le L$
for all $s\in [x^1_j ,x^2_j)$, hence 
$\vert f(x^1_j )-f(x^2_j)\vert \le L\, \vert  x^1_j - x^2_j\vert$, as wanted.
\qed
Proposition \ref{prop1} is a direct consequence of Lemma \ref{lem7}
and of the following result.
\begin{prop}
\label{prop2}
Let $q\in \N ^*$ and $R>0$ be given. Then there exists a positive
constant $L>0$ such that for any $t_0\in [-\frac{R}{1+R},R]$ and
any $x_0\in \overline{B}_R$, we may find a number $\eta _0>0$ for which
\begin{equation}
\label{eq:R29}
\vert V_q(t,x)-V_q(t_0,x_0)\vert \le L\, \Vert (t-t_0,x-x_0)\Vert
\quad \forall (t,x)\in \overline{B}_{\eta _0}(t_0,x_0).
\end{equation}
\end{prop}
\noindent
{\sc Proof of Proposition 2.} We need introduce some numbers depending
only on $R$ and $q$. Let $T=T(R+1,q)$. By local Lipschitz continuity
of $F_2$ on $U$, there exists
a number $K>0$ such that for any pairs $(t_1,x_1)$, $(t_2,x_2)$ with
$-\frac{R+1}{R+2}\le t_i\le R+T+1$ and
$\frac{1}{2}m^{-1}(\frac{1}{q})\le \Vert x_i\Vert \le  m(R+2)$
for $i=1,2$, we have 
\begin{equation}
\label{eq:R30}
h\big( F_2(t_1,x_1),F_2(t_2,x_2)\big) \le K
\Vert (t_1-t_2,x_1-x_2)\Vert .
\end{equation}
Let $M\ge 1$ be such that for any 
$(t,x)\in \big( [ -\frac{R+1}{R+2},R+T+1] \setminus N_0 \big) \times 
\overline{B}_{m(R+2)}$, we have
\begin{equation}
\label{eq:R31}
F_2(t,x)\subset  \overline{B}_M.
\end{equation}
We set
\begin{equation}
\label{eq:R32}
L:=e^{2\lambda T}\big( (M+1)e^{K(T+1)}+2\lambda m(R+1)\big) .
\end{equation}
Pick a number $\overline{\eta}_0$ such that 
\begin{equation}
\label{eq:R33}
0<\overline{\eta}_0
<\min 
\left\{ \frac{R+1}{R+2} - \frac{R}{R+1},
e^{-K(T+1)}
\min\big\{ m(R+2)-m(R+1),\frac{1}{2}m^{-1}\big(\frac{1}{q}\big) \big\} 
\right\}
\end{equation}
and set
\begin{equation}
\label{eq:R34}
b:=\overline{\eta}_0 e^{K(T+1)}.
\end{equation}
We begin with the
\begin{lem}
\label{lem8}
Let $(t_0,x_0)\in [-\frac{R}{R+1},R] \times \overline{B}_R$
and $(t_1,x_1)\in \overline{B}_{\overline{\eta}_0}(t_0,x_0)$. If 
$V_q(t_1,x_1)>0$ then for any $\varphi _1\in {\cal S}_{t_1,x_1}$ 
fulfilling
$\Vert \varphi _1 (t_1+\tau )\Vert >\frac{1}{q}$ for some 
$\tau \in [0,T]$ (such functions exist by (\ref{eq:R15}) and 
(\ref{eq:R17})) we have
\begin{equation}
\label{eq:R35}
m^{-1}(\frac{1}{q}) < \Vert \varphi _1(t_1+h) \Vert
<m(R+1)\quad \hbox{ for }\  0\le h\le \tau .
\end{equation}
Furthermore, the region $t_1\le t\le t_1+\tau $, 
$\Vert x-\varphi _1(t)\Vert\le b$ is contained in the region 
$[-\frac{R+1}{R+2},R+T+1] \times [\frac{1}{2}m^{-1}(\frac{1}{q}),
m(R+2)]$ 
where $F_2$ is Lipschitz continuous  with constant $K$. 
\end{lem}
\noindent{\sc Proof.} 
Since $\overline{\eta} _0<\frac{R+1}{R+2}-\frac{R}{R+1}<1$,
we get $-\frac{R+1}{R+2}<t_1\le t_1+\tau <R+T+1$. Pick
any function $\varphi _1$ as in the statement of the lemma. 
Since $\Vert x_1\Vert \le \Vert x_0\Vert +{\eta _0}<R+1$ we have 
$\Vert \varphi _1(t_1+h)\Vert < m(R+1)\quad \forall h\ge 0$.
Since 
$\Vert \varphi _1(t_1+\tau )\Vert >\frac{1}{q}$, we 
must have $\Vert \varphi _1(t_1+h )\Vert > m^{-1}(\frac{1}{q})$ for
$0\le h\le \tau$. The remaining part of the lemma is an easy consequence
of (\ref{eq:R33})-(\ref{eq:R35}). \qed
Let $(t_0,x_0)\in [-\frac{R}{R+1},R]\times \overline{B}_R$ be fixed. In
what follows, we always assume that
\begin{equation}
\label{eq:eta}
0 < \eta _0 < \frac{\overline{\eta} _0}{2M+1}
\end{equation} 
We have to distinguish two cases: $V_q(t_0,x_0)\ne 0$ or $V_q(t_0,x_0)=0$. 
\va\va
\begin{tabular}{|l|}
\hline
{\bf First case: $V_q(t_0,x_0)\ne 0$.}\\
\hline
\end{tabular}
\va\va
We first prove that $V_q(t,x)\ne 0$ in a neighborhood of $(t_0,x_0)$. \va\va
{\sc Claim 8.} If $\eta _0$ is small enough, then $V_q(t,x)\ne 0$
for each pair $(t,x)\in \overline{B}_{\eta _0}(t_0,x_0)$.\va\va
{\sc Proof.} Let $\varphi _0\in {\cal S}_{t_0,x_0}$ be such that 
$V_q(t_0,x_0)-e^{2\lambda \tau}G_q(\Vert \varphi _0(t_0+\tau )\Vert )
< \frac{V_q(t_0,x_0)}{2}$ for some 
$\tau\in (0,T]$. Hence
$\Vert \varphi _0(t_0+\tau )\Vert >\frac{1}{q}$ and it follows from 
Lemma \ref{lem8} that 
\begin{equation}
\label{eq:R36}
m^{-1}(\frac{1}{q})\, <\, 
\Vert \varphi _0(t_0+h)\Vert \, <\, m(R+1)\quad \hbox{ for }
\ 0\le h\le \tau .
\end{equation}
By decreasing $\eta _0$ if needed, we may assume that
$\eta _0 <\tau$, that $\varphi _0$ is 
defined on $[t_0-\eta _0,t_0]$, that (\ref{eq:R36}) holds true for
$-\eta _0 \le h\le \tau$ and that 
$[t_0-\eta _0, t_0+\tau ] \subset [-\frac{R+1}{R+2},R+T+1]$. 
Let $(t,x)\in 
\overline{B}_{\eta _0}(t_0,x_0)$. 
Then $\vert t-t_0\vert \le \eta _0$
(hence $\varphi _0(t)$ is meaningful) and, by (\ref{eq:R31}), 
\begin{equation*}
\begin{array}{rl}
\Vert \varphi _0 (t)-x\Vert &\le 
\Vert \varphi _0(t)-\varphi _0(t_0)\Vert +\Vert x_0 - x\Vert \\
&\le M\vert t-t_0\vert +\eta _0\\
&\le (M+1)\eta _0<\overline{\eta} _0<b,
\end{array}
\end{equation*}
by (\ref{eq:eta}).
By Lemmas \ref{lem1} and \ref{lem8}, there
exists $\psi\in {\cal S}_{t,x}$ such that 
\begin{equation}
\label{eq:R37}
\Vert \varphi _0(t+s) -\psi (t+s)\Vert \le
\Vert \varphi _0(t)-x\Vert\,  e^{K\vert s\vert} 
\end{equation}
as long as 
$m^{-1}(\frac{1}{q}) \le \Vert \varphi _0(t+s)\Vert \le m(R+1)$ 
and the r.h.s. of (\ref{eq:R37}) is less than $b$. This occurs 
for any $s$ such that  
$t_0 \le t+s\le t_0+\tau$, since 
\begin{equation*}
\Vert \varphi _0(t)-x\Vert\,  e^{K\vert s\vert}\le 
\overline{\eta} _0e^{K(T+1)}=b.
\end{equation*}
We infer from (\ref{eq:R37}) that 
\begin{equation*}
\begin{array}{rl}
\Vert \psi (t_0+\tau )\Vert &\ge \Vert \varphi _0(t_0+\tau )\Vert  
-\Vert \varphi _0(t)-x\Vert e^{K(\eta _0+\tau )} \\
&\ge \Vert \varphi _0(t_0+\tau )\Vert -(M+1)e^{K(T+1)}\eta _0 \\
&> \frac{1}{q} ,
\end{array}
\end{equation*}
if $\eta _0$ is small enough. This implies $V_q(t,x)>0$, as required.\qed
Let $\eta _0$ be as in Claim 8, and let 
$(t_1,x_1),(t_2,x_2)\in \overline{B}_{\eta _0}(t_0,x_0)$. (\ref{eq:R29}) 
is a particular instance of next\va\va
{\sc Claim 9.} $\qquad  \vert V_q(t_1,x_1)-V_q(t_2,x_2)\vert\le
L\, \Vert (t_1-t_2,x_1-x_2)\Vert$. \va\va
Without loss of generality we may assume that $t_1\le t_2$. We need
some preliminary steps.\va\va
\sou{Step 1.} $\qquad \vert V_q(t_1,x_1)-V_q(t_1,x_2)\vert \le
e^{(2\lambda  +K)T}\, \Vert x_1-x_2\Vert $. \va\va
{\sc Proof.} By definition of $V_q(t_1,x_1)$, for each
$\sigma \in \big( 0,V_q(t_1,x_1)\big)$,
there exists a solution $\varphi _1\in {\cal S}_{t_1,x_1}$ and
$\tau\in [0,T]$ s.t.
\begin{equation*}
V_q(t_1,x_1)-\sigma <e^{2\lambda \tau}
G_q\big( \Vert \varphi _1(t_1+\tau )\Vert\big) \le V_q(t_1,x_1).  
\end{equation*}
Hence $V_q(t_1,x_1)-V_q(t_1,x_2)<
e^{2\lambda \tau}  G_q\big( \Vert \varphi _1(t_1+\tau )\Vert\big)
-V_q(t_1,x_2) +\sigma$. Since $\Vert x_1-x_2\Vert \le 2\eta _0<
\overline{\eta} _0<b$, we infer from Lemmas \ref{lem1}, \ref{lem8} and
Claim 8 that there exists a solution $\varphi _2\in {\cal S}_{t_1,x_2}$
s.t.
$\Vert \varphi _1(t)-\varphi _2(t)\Vert \le \Vert x_1-x_2\Vert
e^{K\vert t-t_1\vert}$,
as long as $m^{-1}(\frac{1}{q}) \le \Vert \varphi _1(t)\Vert \le m(R+1)$
and $\Vert x_1-x_2\Vert e^{K\vert t-t_1\vert} \le b$, hence for 
$t_1\le t\le t_1+\tau $.
Since $V_q(t_1,x_2)\ge e^{2\lambda \tau}
G_q(\Vert \varphi _2(t_1+\tau )\Vert )$ and since $G_q$ 
is Lipschitz continuous with constant $1$, we get 
\begin{equation*}
\begin{array}{rl}
V_q(t_1,x_1)-V_q(t_1,x_2)&\le e^{2\lambda \tau}
\bigg( G_q\big( \Vert \varphi _1(t_1+\tau )\Vert\big)
- G_q\big( \Vert \varphi _2(t_1+\tau )\Vert\big) \bigg) +\sigma \\
&\le e^{(2\lambda +K)T}\Vert x_1-x_2\Vert +\sigma .
\end{array}
\end{equation*}
The reasoning can be repeated by exchanging the r\^oles of $x_1$ and $x_2$. 
$\sigma$ being arbitrary, the proof of Step 1 is completed.\va\va
\sou{Step 2.} $\qquad V_q(t_2,x_2)-V_q(t_1,x_2)\le 
Me^{(2\lambda +K)T}\vert t_2-t_1\vert .$\va\va
{\sc Proof.} Pick any $\varphi \in {\cal S}_{t_1,x_2}$. Set  
$x_3=\varphi (t_2)$. Due to Lemma \ref{lem6},
\begin{equation*}
V_q(t_2,x_3)\le e^{-2\lambda (t_2-t_1)}V_q(t_1,x_2)\le V_q(t_1,x_2),
\end{equation*}
hence
\begin{equation}
\label{eq:R38}
V_q(t_2,x_2)-V_q(t_1,x_2)\le V_q(t_2,x_2)-V_q(t_2,x_3).
\end{equation}
Notice that 
\begin{equation*}
\Vert x_2-x_3\Vert \le \int_{t_1}^{t_2}\Vert \dot \varphi (t)\Vert\, dt
\le M\vert t_2-t_1\vert \le 2{\eta _0}M.
\end{equation*}
Since, by (\ref{eq:eta}),  $2{\eta _0}M<\overline{\eta} _0<b$ we may 
use the same arguments as in Step 1 to conclude that
\begin{equation*}
V_q(t_2,x_2)-V_q(t_2,x_3)\le 
e^{(2\lambda +K)T}\, \Vert x_2-x_3\Vert 
\le Me^{(2\lambda +K)T}\vert t_2-t_1\vert
\end{equation*}
which, combined with (\ref{eq:R38}), gives the desired result. \va\va
\sou{Step 3.}
$\qquad V_q(t_1,x_2)-V_q(t_2,x_2)\le 
\big( Me^{KT}+2\lambda m(R+1)\big) e^{2\lambda T}
\vert t_1-t_2\vert$.
\va\va
\noindent{\sc Proof.} 
By definition, for each $\sigma \in \big( 0, V_q(t_1,x_2) \big)$, there exists 
a solution $\psi\in{\cal S}_{t_1,x_2}$ and $\tau \in [0,T]$ such that
$V_q(t_1,x_2)\le e^{2\lambda \tau}G_q(\Vert \psi (t_1+\tau )\Vert )+\sigma$.
Let us distinguish two cases.\va\va
{\bf First case:} $t_1+\tau > t_2 \, (\ge t_1)$. \va
Set $x_4:=\psi (t_2)$. Then $\Vert x_4-x_2\Vert \le \int_{t_1}^{t_2}
\Vert \dot \psi (t)\Vert \, dt\le M\vert t_2-t_1\vert <
2M\eta _0$, hence $\Vert x_4-x_0 \Vert <(2M+1)\eta _0<\overline{\eta }_0$
(by (\ref{eq:eta})), and we get
$(t_2,x_4)\in \overline{B}_{\overline{\eta} _0}(t_0,x_0)$. 
It follows from 
Lemma \ref{lem8} (with $(t_2,x_4)$ substituted for $(t_1,x_1)$) combined with
Lemma \ref{lem1} that there exists a solution 
$\varphi\in {\cal S}_{t_2,x_2}$ such that 
\begin{equation*}
\Vert \varphi (t)-\psi (t)\Vert \le \Vert x_4-x_2\Vert \,
e^{K\vert t-t_2\vert }\qquad \hbox{ for } t_2\le t\le t_1+\tau .
\end{equation*}
Since $V_q(t_2,x_2)\ge e^{2\lambda (t_1+\tau -t_2)}
G_q(\Vert \varphi (t_1+\tau )\Vert )$, 
we may write
\begin{equation*}
\begin{array}{rl}
V_q(t_1,x_2)-V_q(t_2,x_2)
&\le e^{2\lambda \tau} G_q(\Vert \psi (t_1+\tau )\Vert )
-e^{2\lambda (\tau +t_1-t_2)} G_q(\Vert \varphi (t_1+\tau )\Vert )+\sigma\\
&\le e^{2\lambda \tau}\big\{  
\big\vert G_q\big( \Vert\psi (t_1+\tau )\Vert \big)
-G_q\big( \Vert \varphi (t_1+\tau )\Vert \big)\big\vert \\
&\qquad\ \  +(1-e^{-2\lambda \vert t_1-t_2\vert })
G_q\big( \Vert \varphi (t_1+\tau )\Vert \big) \big\} +\sigma .
\end{array}
\end{equation*}
But  
\begin{equation*}
\begin{array}{rl}
\bigg\vert G_q\big( \Vert\psi (t_1+\tau )\Vert \big) 
-G_q\big( \Vert \varphi (t_1+\tau )\Vert \big) \bigg\vert 
&\le \Vert \psi (t_1+\tau ) - \varphi (t_1+\tau )\Vert \\
&\le \Vert x_4-x_2\Vert\, e^{K\vert t_1+\tau -t_2\vert }\\
&\le Me^{KT}\vert t_2-t_1 \vert 
\end{array}
\end{equation*}
and $(1-e^{-2\lambda \vert t_1-t_2\vert }) G_q\big( \Vert 
\varphi (t_1+\tau )\Vert \big) \le 2\lambda \vert t_1-t_2\vert m(R+1)$.
Summing up we get
\begin{equation}
\label{eq:R39}
V_q(t_1,x_2)-V_q(t_2,x_2)\le e^{2\lambda T}\big( Me^{KT}+
2\lambda m(R+1)\big) \vert t_1-t_2\vert +\sigma .
\end{equation}
{\bf Second case:} $t_1+\tau \le t_2$.\va
Using $V_q(t_2,x_2)\ge G_q(\Vert x_2\Vert )$ we get 
\begin{equation*}
\begin{array}{rl}
V_q(t_1,x_2)-V_q(t_2,x_2) 
&\le e^{2\lambda \tau} G_q\big( \Vert \psi (t_1+\tau )\Vert \big) 
-G_q(\Vert x_2\Vert ) +\sigma \\
&\le e^{2\lambda \tau} \big\vert 
G_q\big( \Vert \psi (t_1+\tau )\Vert \big) -G_q(\Vert x_2\Vert )\big\vert \\ 
&\ \ +(e^{2\lambda \tau}-1)G_q(\Vert x_2\Vert )+\sigma .
\end{array}
\end{equation*}
Since $\big\vert \Vert \psi (t_1+\tau )\Vert -\Vert x_2\Vert \big\vert
\le \Vert \int_{t_1}^{t_1+\tau}\dot\psi (t)\, dt\Vert \le M\tau \le
M\vert t_2-t_1\vert$ and 
$\vert e^{2\lambda \tau}-1\vert \le e^{2\lambda \tau}\cdot 2\lambda \tau
\le 2\lambda e^{2\lambda T}\vert t_2-t_1\vert$, we finally obtain 
\begin{equation*}
\begin{array}{rl}
V_q(t_1,x_2)-V_q(t_2,x_2)
&\le e^{2\lambda T} \big( M+2\lambda m(R+1)\big)\vert t_1-t_2\vert  +\sigma \\
&\le e^{2\lambda T} \big( Me^{KT}+2\lambda m(R+1)\big)\vert t_1-t_2\vert  
+\sigma . 
\end{array}
\end{equation*}
Hence (\ref{eq:R39}) holds in both cases and the proof of Step 3 is
completed in letting $\sigma \to 0$. Finally, by Steps 1-3,
\begin{equation*}
\begin{array}{rl}
\vert V_q(t_1,x_1)-V_q(t_2,x_2)\vert 
&\le \vert V_q(t_1,x_1)-V_q(t_1,x_2)\vert +
\vert V_q(t_1,x_2)-V_q(t_2,x_2)\vert   \\
&\le e^{(2\lambda +K)T}\Vert x_1-x_2\Vert +
\big( Me^{KT}+2\lambda m(R+1)\big) e^{2\lambda T}\vert t_1-t_2\vert \\
&\le L\, \Vert (t_1-t_2,x_1-x_2)\Vert .
\end{array}
\end{equation*}
This finishes the proof of Claim 9 and the one of Proposition 2 
when $V_q(t_0,x_0)\ne 0$. We now
proceed to the \va\va
\begin{tabular}{|l|}
\hline
{\bf Second case: $V_q(t_0,x_0)=0$} \\
\hline
\end{tabular}
\va\va
It follows from (\ref{eq:eta}) that $M\eta _0<1$. We claim that
for any $(t,x)\in \overline{B}_{\eta _0}(t_0,x_0)$ and any 
$\varphi\in {\cal S}_{t,x}$, $\varphi$ is defined on 
$[t-\eta _0,+\infty )$ 
(hence at time $t_0$). Indeed, by (\ref{eq:R31}), for any
$s\in \hbox{dom}(\varphi )\cap [t-\eta _0,t]$
\begin{equation}
\label{eq:R41}
\Vert \varphi (s)\Vert \le \Vert \varphi (s) -\varphi (t)\Vert 
+\Vert x\Vert \le M\vert s-t\vert +R+\eta _0\le (M+1)\eta _0+R\le R+1,
\end{equation}
{\em as long as}
$\Vert \varphi (s)\Vert \le m(R+2)$ (a condition which must be
satisfied in order to apply (\ref{eq:R31})).
(Notice that $[t-\eta _0,t]\subset [-\frac{R+1}{R+2},R+1]$, since
$t-\eta _0\ge t_0-2\eta _0>t_0-\overline{\eta}_0>-\frac{R+1}{R+2} \cdot $)
Since $R+1<m(R+2)$, a
standard argument shows that 
$[t-\eta _0,t]\subset \hbox{dom}(\varphi )$ 
and that (\ref{eq:R41}) holds true on $[t-\eta _0,t]$. 
Pick any $(t,x)\in \overline{B}_{\eta _0}(t_0,x_0)$. If $V_q(t,x)=0$, then
(\ref{eq:R29}) is trivial. If $V_q(t,x)>0$, then for any 
$\sigma \in \big( 0,V_q(t,x) \big)$
we may find a solution $\varphi \in {\cal S}_{t,x}$ and a time 
$\tau \in [0,T]$ s.t. 
\begin{equation}
\label{eq:R4100}
V_q(t,x)\le e^{2\lambda \tau}G_q\big( \Vert \varphi (t+\tau )\Vert \big)
+\sigma .
\end{equation}
We again distinguish two cases. \va\va
{\bf First case: $t_0<t+\tau$} \va
Recall that $\varphi$ is defined on $[t-\eta _0,+\infty )$, hence
at time $t_0$.
Since $\Vert \varphi (t+\tau )\Vert >\frac{1}{q}$, we still have
$m^{-1}(\frac{1}{q}) < \Vert \varphi (s)\Vert < m(R+1)$ for any
$s\in [t-\eta _0,t+\tau ]$.
Furthermore,
\begin{equation*}
\Vert \varphi (t_0) -x_0\Vert \le \Vert \varphi (t_0)-\varphi (t)\Vert 
+\Vert x-x_0\Vert \le (M+1)\eta _0<\overline{\eta} _0,
\end{equation*}
according to (\ref{eq:eta}). By Lemmas \ref{lem1} and \ref{lem8} there
exists a solution $\psi\in {\cal S}_{t_0,x_0}$ s.t. 
\begin{equation*}
\begin{array}{rl}
\Vert \psi (t+\tau )-\varphi (t+\tau )\Vert 
&\le \Vert \psi (t_0)-\varphi (t_0)\Vert e^{K \vert t+\tau -t_0\vert} \\
&\le \big(  \Vert x_0-x\Vert +
\Vert \varphi (t)-\varphi (t_0)\Vert \big) e^{K(T+1)} \\ 
&\le (M+1)e^{K(T+1)}\, \Vert (t-t_0,x-x_0)\Vert . 
\end{array}
\end{equation*}
Notice that $G_q\big( \Vert\psi (t+\tau )\Vert \big) =0$, since 
$V_q(t_0,x_0)=0$. We obtain 
\begin{equation}
\label{eq:R42}
\begin{array}{rl}
V_q(t,x)
&\le  e^{2\lambda \tau}\big[ G_q\big( \Vert \varphi (t+\tau )\Vert \big)
-G_q\big( \Vert \psi (t+\tau )\Vert \big)\big] +\sigma \\
&\le (M+1)e^{2\lambda T+K(T+1)}\Vert (t-t_0,x-x_0)\Vert +\sigma .
\end{array}
\end{equation}
{\bf Second case: $t_0\ge t+\tau $}\va
Since
\begin{equation}
\label{eq:R43}
\begin{array}{rl}
G_q\big( \Vert \varphi (t+\tau )\Vert \big)
&= G_q\big( \Vert \varphi (t+\tau )\Vert \big)
- G_q(\Vert x_0\Vert ) \\
&\le \big\vert G_q\big( \Vert \varphi (t+\tau )\Vert \big)
-G_q(\Vert x\Vert )\big\vert  
+\big\vert G_q(\Vert x\Vert )-G_q(\Vert x_0\Vert )\big\vert \\
&\le M\tau  +\Vert x-x_0\Vert \le M\vert t-t_0\vert +\Vert x-x_0\Vert
\end{array}
\end{equation}
we have, by (\ref{eq:R4100}) and (\ref{eq:R43}) 
\begin{equation*}
V_q(t,x)\le e^{2\lambda T}(M+1)\Vert (t-t_0,x-x_0)\Vert +\sigma .
\end{equation*}
Thus, in each case, $0\le V_q(t,x)\le L\, \Vert (t-t_0,x-x_0)\Vert +\sigma$.
Letting $\sigma \to 0$ we get (\ref{eq:R29}). The proof of Proposition 2 
is achieved. This completes Step 2 in the proof of Theorem 1.\va\va
\begin{tabular}{|l|}
\hline
{\sc Third Step.}\\
\hline
\end{tabular}
$\quad$ (Regularization of $V_L$)\va\va
The regularization of $V_L$ is carried out by means of convolution with
mollifiers and of some partition of unity. Next lemma is similar to
\cite[Lem. 5.1]{CLS} and \cite[Lem. 1]{R2}.
\begin{lem}
\label{lem9}
Let $S$ be any compact set in
$U=(-1,+\infty )\times \big( \R ^n\setminus \{ 0\} \big)$ and let
$\varepsilon >0$. Then there exists a function $\overline{V}$ of
class $C^{\infty }$ and with compact support in 
$(-1,+\infty )\times \R ^n$, such that
\begin{equation}
\label{eq:R450}
\Vert \overline{V}-V_L\Vert _{L^{\infty}(S)} <\varepsilon
\quad \hbox{ and }
\end{equation}
\begin{equation}
\label{eq:R46}
\forall (t_0,x_0)\in S,\ \forall v\in F_2(t_0,x_0),\quad
\frac{\partial \overline{V}}{\partial t}(t_0,x_0) +
\langle \nabla _x  \overline{V}(t_0,x_0),v\rangle
\le -\frac{3}{2}\, \lambda \, V_L(t_0,x_0).
\end{equation}
\end{lem}
\noindent
{\sc Proof.}
Let $\rho\in C^{\infty}(\R ^{n+1}, \R )$ be a nonnegative function such
that $\hbox{supp} (\rho )\subset \overline{B}_1$
and $\int_{\R ^{n+1}} \rho (t,x)\, dt\, dx=1$.
For any $\delta >0$, set  
$\rho _{\delta }(t,x):=\frac{1}{\delta ^{n+1}}\rho
\left( \frac{t}{\delta},\frac{x}{\delta}\right)$
and
\begin{equation*}
\begin{array}{rl}
V_{\delta}(t,x):&=V_L \ast \rho _{\delta}(t,x)  \\
&=\int_{\R^{n+1}}V_L (t-s,x-y)\rho _{\delta}(s,y)\, ds\, dy \\
&=\int_{\Vert (\bar{s},\bar{y})\Vert \le 1}
V_L (t-\delta \bar{s},x-\delta\bar{y})
\rho (\bar{s},\bar{y})d\bar{s}\, d\bar{y}.
\end{array}
\end{equation*}
\def\f{V _{\delta}}
Then $\f$ is defined and of
class $C^{\infty}$ on $(-1+\delta ,+\infty) \times \R ^n$,
and $\f (t,x)\to V_L (t,x)$
{\em uniformly} on
$S$ as $\delta \to 0$.
If $\theta$ denotes a function of class $C^{\infty}$ with compact support
in $(-1,+\infty)\times \R^n$ and which assumes the value 1 on a
neighborhood of $S$, then the (smooth) function 
$\overline{V}:=\theta \cdot V_{\delta}$ 
has a compact support in $(-1,+\infty )\times \R ^n$ and it
satisfies (\ref{eq:R450}), provided that $\delta $ be small enough.
To complete the proof of the lemma, it remains to show that there exists
$\delta _0>0$  such that for any $\delta \in (0,\delta _0 )$,
\begin{equation*}
\forall (t_0,x_0)\in S,\ \forall v\in F_2(t_0,x_0)\quad
\frac{\partial \f}{\partial t}(t_0,x_0) +
\langle\nabla _x  \f (t_0,x_0),v\rangle
\le -\frac{3}{2}\lambda V_L(t_0,x_0).
\end{equation*}
Let $\delta _1>0$ be such that
$S+\overline{B}_{\delta _1}\subset U$
and let $L>0$ be such that for any pairs $(t_1,x_1),(t_2,x_2)$ in
$S+\overline{B}_{\delta _1}$ ($h$ denoting the Hausdorff distance between 
closed sets)
\begin{equation}
\label{eq:48}
h\big( F_2(t_1,x_1),F_2(t_2,x_2)\big) +
\vert V_L (t_1,x_1)-V_L (t_2,x_2)\vert  \le L\, \Vert
(t_1-t_2,x_1-x_2)\Vert .
\end{equation}
It follows that for a.e. $(t,x)\in S+B_{\delta _1}$,
$V_L$ is differentiable at $(t,x)$ and
\begin{equation}
\label{eq:480}
\Vert \big(\frac{\partial V_L}{\partial t},\nabla _x V_L\big)
(t,x)\Vert \le L.
\end{equation}
Let $\delta \in (0,\delta _1)$, $(t_0,x_0)\in S$ and 
$v\in F_2(t_0,x_0)$.
Applying Lebesgue's theorem we infer
from (\ref{eq:48}) that
\begin{equation}
\label{eq:481}
\begin{array}{l}
\frac{\partial \f}{\partial t}(t_0,x_0) +
 \langle \nabla _x  \f (t_0,x_0),v\rangle \\
\qquad =\lim_{\eta \to 0}\int_{\Vert (\bar{s},\bar{y})\Vert \le 1}
\displaystyle
\frac{V_L (t_0-\delta\bar{s}+\eta ,x_0-\delta \bar{y}+\eta v)-
V_L (t_0-\delta \bar{s},x_0-\delta \bar{y})}{\eta}\rho
(\bar{s},\bar{y})
d\bar{s}\, d\bar{y}\\
\qquad =\int_{\Vert (\bar{s},\bar{y})\Vert \le 1}
\left(
\frac{\partial V_L}{\partial t}(t_0-\delta\bar{s},x_0-\delta \bar{y}) +
\langle \nabla _x V_L (t_0-\delta\bar{s},x_0-\delta \bar{y}),v\rangle \right)
\rho (\bar{s},\bar{y})d\bar{s}\, d\bar{y}.
\end{array}
\end{equation}
$g$ denoting also the map defined by (\ref{eq:R100}), we may write,
thanks to (\ref{eq:R100}), (\ref{eq:480}), (\ref{eq:481}) and Corollary 3,
\begin{equation}
\label{eq:490}
\begin{array}{l}
\frac{\partial \f}{\partial t}(t_0,x_0) +
 \langle \nabla _x  \f (t_0,x_0),v\rangle \\
\qquad =\int_{\Vert (\bar{s},\bar{y})\Vert \le 1}
\left(
\frac{\partial V_L}{\partial t} +
\langle \nabla _x V_L ,g\rangle \right)
(t_0-\delta\bar{s},x_0-\delta \bar{y})
\rho (\bar{s},\bar{y})d\bar{s}\, d\bar{y}\\
\qquad \ \ +
\int_{\Vert (\bar{s},\bar{y})\Vert \le 1}
\langle \nabla _x V_L
(t_0-\delta\bar{s},x_0-\delta \bar{y})
, v-g(t_0-\delta\bar{s},x_0-\delta \bar{y})\rangle
\rho (\bar{s},\bar{y})d\bar{s}\, d\bar{y},\\
\qquad\le -2\, \lambda\, V_{\delta}(t_0,x_0) +\sqrt{n}\,
L\, \int_{\Vert (\bar{s},\bar{y})\Vert \le 1}
\Vert v-g(t_0-\delta\bar{s},x_0-\delta \bar{y})\Vert
\rho (\bar{s},\bar{y})d\bar{s}\, d\bar{y}.
\end{array}
\end{equation}
Now, observe that for any $\Vert (\bar{s},\bar{y})\Vert \le 1$,
\begin{equation*}
\begin{array}{rl}
\Vert v-g(t_0-\delta \bar{s},x_0-\delta \bar{y}) \Vert
&= \hbox{dist}\big( v,F_2 (t_0-\delta \bar{s},x_0-\delta \bar{y})\big)
\\
&\le h\big( F_2(t_0,x_0),F_2(t_0-\delta \bar{s},x_0-\delta \bar{y})\big)
\\
&\le L\delta ,
\end{array}
\end{equation*}
due to (\ref{eq:48}). We obtain
\begin{equation*}
\frac{\partial \f}{\partial t}(t_0,x_0) + 
\langle \nabla _x  \f (t_0,x_0),v\rangle
\le -2\, \lambda \, V_{\delta}(t_0,x_0)+\sqrt{n}\, L^2\,
\delta \le -\frac{3}{2}\, 
\lambda \, V_L(t_0,x_0),
\end{equation*}
if $\delta $ is small enough. \qed
Let $(\psi _i)_{i\ge 1}$ be a $C^{\infty}$ partition of unity on $U$,
such that for each $i\ge 1$ the support
$S_i$ of $\psi _i$ is a {\em compact} set in $U$.
For each $i\ge 1$, set
\begin{equation*}
\begin{array}{rl}
q_i&:=\sup_{(t,x)\in S_i,\, v\in F_2(t,x)} 
\vert \frac{\partial \psi _i}{\partial t}(t,x) +
\langle \nabla _x\psi _i(t,x),v\rangle \vert  <+\infty\\
\varepsilon _i&:=\frac{1}{2^{i+2}(1+q_i)}\cdot \frac{\lambda}{\lambda +1}\cdot
\min_{(t,x)\in S_i} V_L(t,x) >0.
\end{array}
\end{equation*}
Lemma \ref{lem9} provides, for each $i\ge 1$, a function $V_i$ of class
$C^{\infty}$ on $(-1,+\infty )\times \R ^n$ such that
\begin{equation*}
\forall (t,x)\in S_i,\quad \vert V_i(t,x)-V_L(t,x)\vert <\varepsilon _i
\end{equation*}
and
\begin{equation*}
\forall (t,x)\in S_i,\ \forall v\in F_2(t,x),\quad
\frac{\partial V _i}{\partial t}(t,x) + \langle \nabla _x V_i(t,x),v\rangle
\le -\frac{3}{2}\, \lambda \, V_L(t,x).
\end{equation*}
We define on $(-1,+\infty )\times \R ^n$ 
a function $\tilde V$ by
\begin{equation*}
\tilde{V}(t,x):=\left\{
\begin{array}{ll}
\sum_{i=1}^{+\infty}\, \psi _i(t,x)\, V_i(t,x)\ &\hbox{ if } x\ne 0,\\
0 \ &\hbox{ if }x=0.
\end{array}
\right.
\end{equation*}
It is clear that $\tilde V$ is of class $C^{\infty}$ on
$U$. Moreover, for each $(t,x)\in U$,
\begin{equation*}
\begin{array}{rl}
\vert \tilde{V}(t,x)-V_L(t,x)\vert 
&\le \sum _{i=1}^{+\infty} \psi _i(t,x)\vert V_i(t,x)-V_L(t,x)\vert \\
&\le \sum _{i=1}^{+\infty} \frac{1}{2^{i+2}} V_L(t,x) =
\frac{1}{4}V_L(t,x),
\end{array}
\end{equation*}
hence 
\begin{equation*}
\frac{3}{4}a_L(\Vert x\Vert )\le \frac{3}{4}V_L(t,x)\le
\tilde{V}(t,x)\le \frac{5}{4}V_L(t,x)\le \frac{5}{4} b_L(\Vert x\Vert ),
\end{equation*}
where $a_L,b_L\in {\cal K}_0^{\infty}$ are defined in Claims 4 and 6,
respectively.  Clearly, $\tilde V$ is continuous on
$(-1,+\infty )\times \R ^n$. Pick any $(t,x)\in U$ and any
$v\in F_2(t,x)$. Then
\begin{equation*}
\begin{array}{rl}
\frac{\partial \tilde V}{\partial t}(t,x) + 
\langle \nabla _x \tilde V(t,x),v\rangle  
&= \sum_{i=1}^{+\infty} \big( \frac{\partial \psi _i}{\partial t}
+ \langle \nabla _x \psi _i,v\rangle \big) (V_i-V_L) \\
&\quad + \sum_{i=1}^{+\infty} \psi _i\big( \frac{\partial V_i}{\partial t}
+ \langle \nabla _x V_i,v\rangle \big) \\
&\le \sum_{i=1}^{+\infty}
\big[\frac{q_i}{1+q_i}\cdot \frac{\lambda }{2^{i+2}}
\cdot V_L(t,x)\big] 
+ \sum_{i=1}^{+\infty} \psi _i \big( -\frac{3}{2}
\, \lambda \, V_L(t,x)\big) \\
&\le  -\frac{5}{4} \lambda V_L(t,x) \le -\lambda {\tilde V} (t,x).
\end{array}
\end{equation*}
In order to get a function which is smooth {\em up to } $x=0$, we
apply \cite[Thm. 6]{K2}, which asserts the existence of a function
$\nu\in C^{\infty}(\R )$ such that $\nu (r)=0$ for $r\le 0$,
$\nu '_{\vert _{\R _+}}$  is increasing (hence $\nu '(r)>0$, $\nu (r)>0$
for $r>0$), $\nu (r)\to +\infty $ as $r\to +\infty$  and
$\nu\circ {\tilde V}\in 
C^{\infty}\big( (-1,+\infty )\times \R ^n\big)$.
Moreover, $\partial ^{\alpha} (\nu\circ {\tilde V}) (t_0,0)=0$ for any
$t_0 >-1$ and any $\alpha \in \N ^{n+1}$.
We set $V:=\nu \circ \tilde V$ and $a(r):=\nu \big(\frac{3}{4}\, a_L(r)\big)$,
$b(r):=\nu \big(\frac{5}{4}\, b_L(r)\big)$ for all $r>0$. Clearly,
$a,b\in {\cal K}_0^{\infty}$ and (\ref{eq:R2}) holds true.
Let $(t,x)\in ([0,+\infty )\setminus N_0)\times (\R ^n\setminus \{ 0\} )$ 
and let $v\in F(t,x)$.
According to Claim 2, $v\in F_2(t,x)$, hence 
\begin{equation*}
\frac{\partial \tilde V}{\partial t}(t,x) +
\langle \nabla _x \tilde V(t,x),v\rangle \le -\lambda {\tilde V} (t,x).
\end{equation*}
On the other hand, since $\nu '$ is increasing on $\R _+$, we get for
each $r\ge 0$
\begin{equation*}
\nu (r) =\int_0^r \nu '(s)\, ds\le \int_0^r \nu '(r)\, ds =r\, \nu '(r).
\end{equation*}
It follows that 
\begin{equation*}
\begin{array}{rl}
\frac{\partial V}{\partial t}(t,x) + \langle \nabla _x V(t,x),v\rangle
& = \nu '(\tilde V(t,x))\cdot \big(
\frac{\partial \tilde V}{\partial t}(t,x) +
\langle \nabla _x \tilde V(t,x),v\rangle  \big) \\
&\le -\lambda \, \nu '(\tilde V(t,x))\, {\tilde V}(t,x)\\
&\le  -\lambda \, \nu (\tilde V(t,x))=-\lambda V(t,x).
\end{array}
\end{equation*}
The proof of Theorem \ref{thm1}  is complete in the general
setting. \qed

Let us briefly indicate the modifications to be
brought in the time periodic or time independent cases. We begin with the
\va\va
$\bullet$ {\sc Time independent case.}\va
Lemma \ref{lem3} holds true when we replace (\ref{eq:R401}) by
$\dot x\in \overline{co}\{ F\big( \overline{B}_{\delta }(x)\big) \}$
and the condition $t_0\in [k,k+1]$ by $t_0=0$. Here, $\delta$ depends
only on $i$, so we write $\delta _i$ instead of $\delta _i^k$.
Next, we pick a Lipschitz continuous function (with 1 as a Lipschitz constant)
$\delta :\R ^n\to \R _+$, such that
$\delta (x)=0\iff x=0$ and $\delta (x)<\min (\delta _i , \delta _{p_i})$
whenever $2^{p_i}\le \Vert x\Vert \le \varphi _i(0)$.
$F_1$ is defined by
$F_1(x):=\overline{co}\{ F\big( \overline{B}_{\delta (x)}(x)\big) \}$.
We set $U:=\R ^n\setminus \{ 0\}$, and, for any $x\in U$,
$W(x):= B_{\frac{1}{3}\delta (x)}(x)$. We pick a $C^{\infty}$ partition of
unity $(\rho _i )_{i\ge 1}$ subordinate to the open covering
$\big( W(x)\big) _{x\in U}$ of $U$. For each $i\ge 1$, we choose
some $x_i\ne 0$ such that $\hbox{supp}(\rho _i)\subset W(x_i)$. $F_2$ is
defined by
\begin{equation*}
F_2(x):=
\left\{
\begin{array}{ll}
\sum_{i=1}^{\infty}\rho _i(x)\ \overline{co}\,
\big\{ F\big(
\overline{B}_{\frac{1}{3}\delta (x_i)}(x_i)\big)\big\}
&\hbox{ if }x\ne 0,\\
F(0)&\hbox{ if }x=0.
\end{array}
\right.
\end{equation*}
$F_2$ is time independent. The same is true for the functions
$V_q$ and $V_L$, defined by (\ref{eq:R16}), (\ref{eq:R19}), respectively.
Lemma \ref{lem9} remains valid when $S$ is a compact set in
$\R ^n\setminus \{ 0\}$ and (\ref{eq:R46}) is replaced by
$\langle \nabla _x \overline{V}(x_0),v \rangle 
\le -\frac{3}{2}\lambda \, V_L(x_0)$
for all $x_0\in S$ and all $v\in F_2(x_0)$. Here, the
function $V$ provided by Lemma \ref{lem9} is of class $C^{\infty}$
on $\R ^n$. We set $S_i:=\hbox{supp}\, (\rho _i)$, and
$q_i,\varepsilon _i$ are defined as in the general case, with
$\rho _i (x)$ instead of $\psi _i(t,x)$. Finally, we set
\begin{equation*}
{\tilde V}(x):=\left\{
\begin{array}{ll}
\sum_{i=1}^{+\infty}\, \rho _i(x)\, V_i(x)\ &\hbox{ if } x\ne 0,\\
0 \ &\hbox{ if }x=0,
\end{array}
\right.
\end{equation*}
and we define again $V$ by $V:=\nu \circ {\tilde V}$. We now turn to the
\va\va
$\bullet$ {\sc Time periodic case.}\va
Without loss of generality, we may assume that the period is $T=1$,
that $F$ is defined on the whole space $\R _t \times\R _x^n$ and that $(H')$
holds true (on $\R ^{n+1}$) for some $N_0\subset \R$ 
of measure $0$ and such that
$N_0+1=N_0$. Here, we take $E:=(\R \setminus N_0)\times \R ^n$ and
$U:=\R \times (\R ^n\setminus \{ 0\} )$. If $x(\cdot )$ is any solution of
(\ref{eq:R1})  issuing from $x_0$  at some time $t_0\in \R$, then
${\tilde x}(t):=x(t+[t_0])$ is also a solution of (\ref{eq:R1}), issuing from
$x_0$ at time ${\tilde t}_0:=t_0-[t_0]\in [0,1]$. Therefore, we
may always assume that $t_0\in [0,1]$. Thanks to this trivial remark, we see
that Lemma \ref{lem3} is valid with a constant $\delta >0$ which
depends {\em only} on $i$  (and is denoted by $\delta _i$), whatever be
$t_0\in \R$. Next, the function $\delta (x)$ is chosen as in the
time-independent case. $F_1$ is defined by
\begin{equation*}
F_1(t,x):= \overline{co}\,
\big\{ F \big( \overline{B}_{\delta (x)}(t,x) \cap E\big)
\big\} .
\end{equation*}
Clearly, $F_1$ is still 1-periodic in time. In order to construct a
multivalued map $F_2$ which is also 1-periodic in time, we need some
partition of unity which is left ``invariant'' by the transformation 
$t\mapsto t+1$.
Let $(x_i)_{i\ge 1}$ and $(\rho _i )_{i\ge 1}$ be as in the time-independent
case.  Pick a function $\omega \in C^{\infty}([0,1],[0,1])$ such that
\begin{equation*}
{\omega }(t):=\left\{
\begin{array}{ll}
1\ &\hbox{ for } 0\le t\le \frac{1}{3},\\
0\ &\hbox{ for } \frac{2}{3}\le t\le 1.
\end{array}
\right.
\end{equation*}
Extend $\omega$ to $\R$ by setting $\omega (-t):=1-\omega (1-t)$
for $0\le t\le 1$, and $\omega (t)=0$ for $\vert t\vert >1$.
Obviously, $\omega \in {\cal D}(\R )$ and
$\big( \omega (\cdot -k)\big) _{k\in\Z}$ constitutes a
$C^{\infty}$ partition of unity
subordinate to the covering $\big( (k-1,k+1)\big) _{k\in \Z}$ of $\R$.
For each $i\ge 1$, pick an integer $M_i>\frac{6}{\delta (x_i)}$. For each
$k=0,\ldots ,M_i-1$, pick a time $t_{i,k}\in \big(
\frac{k-1}{M_i},\frac{k+1}{M_i}\big) \setminus N_0$,
and set
$t_{i,k+p\, M_i}:=t_{i,k}+p$ for any $0\le k\le M_i-1$ and
any $p\in \Z$.
Notice that
$(t_{i,k},x_i)\in U\cap E$ for each pair $(i,k)$. Set 
$\psi _{i,k}(t,x):=\omega (M_i\, t-k)\, \rho _i(x)$
for all $(t,x)\in U$. Then
$\big( \psi _{i,k}\big) _{i\ge 1, k\in \Z}$ is a
$C^{\infty}$ partition of unity subordinate to the covering
$\left( \big( \frac{k-1}{M_i},\frac{k+1}{M_i}\big) \times W(x_i)\right)
_{i\ge 1, k\in \Z}$ of $U$. 
We are now ready to define $F_2$. 
\begin{equation*}
F_2(t,x):=
\left\{
\begin{array}{ll}
\sum_{i=1}^{\infty}\sum_{k\in \Z}
\psi _{i,k}(t,x)\ \overline{co}\,
\big\{ F\big(
\overline{B}_{\frac{1}{3}\delta (x_i)}(t_{i,k},x_i)\cap E\big)\big\}
&\hbox{ if }x\ne 0,\\
F(t,0)&\hbox{ if }x=0.
\end{array}
\right.
\end{equation*}
Let us check that Claim 2 remains valid. Let $x\ne 0$ and let
$t\in \R \setminus N_0$. Pick any pair $(i,k)$ such that
$\psi _{i,k}(t,x)>0$, hence
$(t,x)\in \big( \frac{k-1}{M_i},\frac{k+1}{M_i}\big)
\times B_{\frac{1}{3}\delta (x_i)}(x_i)$. Since
$t_{i,k}\in \big(\frac{k-1}{M_i},\frac{k+1}{M_i}\big)$ and
$\frac{1}{M_i}<\frac{1}{6}\delta (x_i)$, we conclude that
$\Vert (t-t_{i,k},x-x_i)\Vert <\frac{1}{3}\delta (x_i)$, hence
$F(t,x)\subset
F\big( \overline{B}_{\frac{1}{3}\delta (x_i)}(t_{i,k},x_i)\cap E\big)$,
which implies $F(t,x)\subset F_2(t,x)$. The proof of Claim 3 is as in the
general case. Finally, let us check that $F_2$ is 1-periodic in time.
Let $x\ne 0$ and $t\in \R$. Since $\psi _{i,k}(t+1,x)=
\omega (M_i\, t+M_i-k)\, \rho _i(x)=\psi _{i,k-M_i}(t,x)$, setting $k'=k-M_i$,
we get
\begin{equation*}
F_2(t+1,x)=\sum_{i=1}^{+\infty}\sum_{k'\in \Z} 
\psi_{i,k'}(t,x)\ \overline{co}\, \big\{ F\big(
\overline{B}_{\frac{1}{3}\delta (x_i)}(t_{i,k'+M_i},x_i)\cap E\big)\big\} .
\end{equation*}
But $t_{i,k'+M_i}=t_{i,k'}+1$, hence
$F\big(
\overline{B}_{\frac{1}{3}\delta (x_i)}(t_{i,k'+M_i},x_i)\cap E\big)
= F\big(
\overline{B}_{\frac{1}{3}\delta (x_i)}(t_{i,k'},x_i)\cap E\big)$.
It follows that $F_2(t+1,x)=F_2(t,x)$. We readily infer that the functions
$V_q$ (for each $q\ge 1$) and $V_L$, defined on
$\R\times \R ^n$ by (\ref{eq:R16}) and (\ref{eq:R19}), respectively,
are 1-periodic in time, as well.
In the proof of Lemma \ref{lem9}, we observe that
$V_{\delta}(t+1,x)=V_{\delta }(t,x)$ for each $(t,x)\in \R ^{n+1}$.
This times,  we set $\overline{V}:=V_{\delta }$ (for $\delta >0$
small enough) instead of $\overline{V}:=\theta \cdot V_{\delta}$, so that
$\overline{V}$ is 1-periodic in time. For each pair $(i,k)$,
let $S_{i,k}$ denote the support
of $\psi _{i,k}$ and let $V_{i,k}$ be the (periodic) function
(associated with $S_{i,k}$  and $\varepsilon _{i,k}$), which is provided by
Lemma \ref{lem9}. Thanks to the periodicity of $V_L$ and $V_{i,k}$, we may in
fact require that $V_{i,k+pM_i}\equiv V_{i,k}$ for all $p\in \Z$.
Setting $\tilde{V}:=\sum_{i\ge 1}\sum_{k\in\Z}\psi _{i,k}V_{i,k}$
and $V:=\nu \circ  \tilde{V}$,
we observe that $\tilde V$ and $V$ are 1-periodic in time. \qed
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\end{document}

