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%
% Pin No.: 06293 
% Submission Number: 1404
% Title: Uniqueness results for the value function via direct
% trajectory-construction methods.
% Session Code: uqcm7 
% Session Title: New Trends on Geometric Control
% Date submitted: March 8, 2003 
%
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%  
%            This is a LateX file, containing  the paper 
%
%       UNIQUENESS RESULTS FOR THE VALUE FUNCTION VIA DIRECT
%                 TRAJECTORY-CONSTRUCTION METHODS.
%
%                       by H. J. Sussmann
%
%        This paper has been submitted for presentation at the 
%            42nd IEEE Conference on Decision and Control
%      to be held at the Hyatt Regency Maui, Hawaii, U.S.A., 
%                      December 9-12, 2003
%         and  for publication in the conference  proceedings. 
%  
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H\'ector J.\ Sussmann\\
Department of Mathematics\\
Rutgers, the State University of New Jersey\\
Hill Center---Busch Campus\\
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Piscataway, NJ 08854-8019, USA\\
{\small\tt sussmann@math.rutgers.edu}\\
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H\'ector J. Sussmann\\
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{\bf Uniqueness results for the value function via direct
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\begin{document}

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\twocolumn[\titlestuff]
\footnotetext[1]{The  author was supported in part by
NSF Grant DMS01-03901. He wrote the final version of the paper while 
he was a guest of the Institut 
Mittag-Leffler, in Djursholm, Sweden, and is very grateful to the Institute 
for its kind hospitality.  He particularly thanks Anders Lindquist 
for encouraging him to resume work on this topic and give the 
lectures that eventually resulted is this written presentation.
}

\vskip4.5cm

\ssection{Introduction}
\label{sec-intro}

\noindent The purpose of this note is to present some new \mbox{results},
together with a number of particularly \mbox{simple} and user-friendly 
versions of results obtained in \mbox{recent} years by the \mbox{author} 
and M. Malisoff, on the \mbox{uniqueness} of \mbox{solutions} of the 
Hamilton-Jacobi-Bellman \mbox{equation} (HJBE) 
for \mbox{deterministic} finite-dimensional optimal control
\mbox{problems} under \mbox{non-standard hypotheses}. Our
\mbox{approach} is completely control-theoretic and totally
self-contained, using the systematic \mbox{construction} of 
\mbox{special} trajectories of \mbox{various} kinds, and not
\mbox{involving} any PDE methods.  We will %definitely 
{\em not} assume that the \mbox{Lagrangian} is positive, or 
that the \mbox{dynamics} is Lipschitz-continuous.

We will consider autonomous Lagrangian \mbox{optimization} problems involving
a state variable $x$ which takes values in an open subset $\Omega$
of $\rr^n$,  a control
variable $u$ taking values in a control space $U$, and a target set
${\cal T}$, which is a closed subset of the closure of $\Omega$
disjoint from $\Omega$ itself.  The dynamics is given by
an ordinary differential equation
\begin{equation}\label{dyna}
\dot x=f(x,u)\,,
\end{equation}
the cost functional to be minimized is
\begin{equation}\label{lagra}
J=\int_{\tau_-(\xi)}^{\tau_+(\xi)}L(\xi(t),\eta(t))\,dt\,,
\end{equation}
(where $\tau_-(\xi)$, $\tau_+(\xi)$ are, respectively, 
the initial and \mbox{terminal} times of the trajectory $\xi$),
and the \mbox{minimization} is supposed to be, for each \mbox{initial}
state \mbox{$x\in\Omega$}, over the set $\axts$ of all pairs 
\mbox{$\Xi=(\xi,\xi_0)$} such that
\begin{itemize}
\item[(i)]  \mbox{$\Xi$} \mbox{consists}  of a trajectory
$\xi$ of (\ref{dyna}) (i.e., a \mbox{locally} \mbox{absolutely} 
continuous function  $\xi$ that satisfies $\dot\xi(t)=f(\xi(t),\eta(t))$
for almost all $t$) \mbox{corresponding} to some $U$-valued control 
$\eta$, and a ``\mbox{running} cost'' function $\xi_0$ corresponding to 
$\xi$ and $\eta$ (i.e., a \mbox{locally} absolutely continuous 
function  $\xi_0$ such that \mbox{$\dot\xi_0(t)=L(\xi(t),\eta(t))$}
for almost all $t$);
\item[(ii)]  \mbox{$\xi$} starts at $x$,  and ``ends at the target'' in a 
sense  to be defined precisely later.
\end{itemize}
We will refer to a pair $\Xi=(\xi,\xi_0)$ for which (i) above 
holds as an ``augmented trajectory'' of our \mbox{system}, 
\mbox{because} it really is a trajectory of the ``\mbox{augmented} 
\mbox{control} system''  

\begin{equation}\label{dynaaug}
\dot x=f(x,u)\,,\qquad\dot x_0=L(x,u)
\end{equation}

\medbreak\noindent obtained from (\ref{dyna}) by ``adding the cost as
an extra \mbox{variable}'' in a well known way.  We will write
$J(\Xi)$, rather than just $J$, for the left-hand side of
(\ref{lagra}), \mbox{because} it is easy to see that the natural
arguments for our cost functional $J$ are really augmented
\mbox{trajectories}, \mbox{since} (i)~one cannot just regard $J$ as a
functional of $\xi$ \mbox{only}, because the integral of (\ref{lagra})
involves the control $\eta$ as well as $\xi$, but on the other hand
(ii)~the cost is \mbox{completely} \mbox{determined} once we know
$\xi$ and a \mbox{running} cost $\xi_0$, \mbox{because} in that case
$L(\xi(t),\eta(t))=\dot\xi_0(t)$. 

The infimum $V(x)$ of the costs $J(\Xi)$ of all 
\mbox{augmented} trajectories
$\Xi\in\axts$ is the {\em value}  of our problem at $x$. (If the
set $\axts$ is itself empty, then  of course 
\mbox{$V(x)=+\infty$}.)  The function 
$V:\Omega\mapsto\rr\cup\{-\infty,+\infty\}$ is the {\em value
function} of our problem.

Our goal is to prove uniqueness theorems, \mbox{showing} that a
viscosity solution of the HJBE that satisfies an \mbox{appropriate}
boundary condition is necessarily the value function. 
``\mbox{Uniqueness}'' is to be understood as ``\mbox{uniqueness}
within a class defined by some additional properties,'' such as the
class of all functions that are continuous and bounded below.

We will work with a class of systems which is \mbox{sufficiently} general to
capture some \mbox{interesting} \mbox{phenomena} not commonly addressed in the
\mbox{literature}, and at the same time restricted \mbox{enough}
 to make it \mbox{possible} to prove strong \mbox{theorems}.  In 
particular,  we will asume that the sets 
\mbox{$\fsh(x,U)=\{(f(x,u),L(x,u)):u\in U\}$} are closed and convex,
but will not require them to be \mbox{compact}, and will instead
impose a ``local coerciveness'' \mbox{condition}, \mbox{according} to which, 
locally, an inequality of the form $\|f(x,u)\|^r\le L(x,u)+C$,
with $C>0$ and $r>1$,  holds uniformly with respect to $u$.  We will
also require $f(x,u)$ and $L(x,u)$ to be continuous 
with respect to $x$, with the continuity being uniform 
with respect to $u$ locally in $x$. 

On the other hand, we will most definitely {\em not} require 
that the dynamics $f(x,u)$ be Lipschitz-continuous with respect to $x$,
since one of the main purposes of this work is to clarify the exact
role of the Lipschitz-continuity assumptions often made in the viscosity
literature.  The answer we will propose is as follows: 
\begin{itemize}
\item[(a)] Without any Lipschitz-continuity hypotheses, one can 
prove, for continuous viscosity solutions $V$ of the HJBE, an 
{\em existence theorem for \mbox{trajectories}}, \mbox{asserting}
that, starting 
at every point of $x$ of $\Omega$, there is a maximally defined
``augmented \mbox{trajectory} of steepest descent,'' that is, 
a maximally defined pair $\Xi=(\xi,\xi_0)$ defined on a interval 
$I$ such that \mbox{$0=\min\,I$}, and having the property 
that the \mbox{inequality} 

\begin{equation}\label{steepdesc}
V(\xi(t))+\xi_0(t)\ge V(\xi(s))+\xi_0(s)
\end{equation}

\medbreak\noindent holds whenever $s,t\in I$ and $s\le t$. (This is Theorem 
\ref{maintech} below.)
\item[(b)] As a trivial corollary of the existence of \mbox{steepest} 
\mbox{descent} trajectories (applied to $-V$ and $-L$), we get the 
existence of ``DPI \mbox{trajectories}'' (where ``DPI'' is an acronym 
for ``\mbox{Dynamic} \mbox{Programming} \mbox{Inequality}''), i.e., 
\mbox{augmented}
\mbox{trajectories} \mbox{$\Xi=(\xi,\xi_0)$} along which 
the inequality 

\begin{equation}\label{steepasc}
V(\xi(t))+\xi_0(t)\le V(\xi(s))+\xi_0(s)
\end{equation}

\medbreak\noindent (that is, the exact opposite of (\ref{steepdesc}))
is satisfied.
\item[(c)] The existence result of (b) says that for every 
``\mbox{sufficiently} nice'' (e.g., piecewise constant) \mbox{control}
$\eta$ {\em there exists a trajectory of $\eta$ 
with the given \mbox{initial} condition along which the DPI holds}.
This is \mbox{almost}, but not quite, 
what is needed to prove that $V$ is bounded above by the 
value function. 
\item[(d)] The gap between the existence result for DPI 
\mbox{trajectories} and what would actually 
be needed to prove that $V$ is bounded above by the 
value \mbox{function} is that to achieve the latter goal one needs 
the DPI to hold for {\em all} trajectories, and it is not enough
to have just one DPI trajectory for every intial condition and every 
control. 
\item[(e)] The gap described in (d) clearly does not exist when 
there is uniqueness of trajectories for every given control and initial 
condition. 
\item[(f)] In particular, the gap does not exist when,
for \mbox{every} admissible control $\eta$,  the corresponding 
time-varying vector field $(x,t)\mapsto f(x,\eta(t))$ 
satisfies a Lipschitz-Carath\'eodory condition that guarantees
uniqueness of trajectories. 
\item[(g)] Naturally, the Lipschitz-Carath\'eodory condition 
can be replaced by weaker conditions that \mbox{guarantee}
uniqueness, such as the requirement that a bound
\begin{equation}\label{weakerbo}
\langle f(x,\eta(t))-f(y,\eta(t)),x-y\rangle 
\le k(t)\|x-y\|^2\,,
\end{equation}
\item[] with $k$ integrable, hold locally. 
\item[(h)] Even more generally, the only property that really matters
is that, if we pick a sequence $\{\eta_j\}_{j=1}$ of 
piecewise constant controls such that our \mbox{augmented}
trajectory $\Xi$ can be approximated by  \mbox{augmented}
trajectories $\Xi_j$ corresponding to the $\eta_j$---with, say, 
the same initial condition---then the $\Xi_j$ converge to $\Xi$ uniformly 
{\em no matter how the $\Xi_j$ are chosen}. We call such trajectories 
``uniquely limiting,'' and use this concept in the statement of our main 
theorem. 
\end{itemize}

\medbreak The following important issues will not be discussed here:

\begin{itemize}
\item[(1)] Whether  the value function itself satisfies the 
\mbox{conditions} of our main theorem, i.e., whether it is a \mbox{continuous}
viscosity solution of the HJBE and whether it is bounded below.

\item[(2)] What happens when the sets $F(x,U)$ are not closed and convex.
(This would require \mbox{considering} \mbox{relaxed} controls, and using 
trajectories for which the steepest descent property holds
\mbox{approximately} rather than exactly. It turns out to be
\mbox{possible} to extend our results  under fairly \mbox{general}
\mbox{conditions}, provided our system has 
appropriate \mbox{local} \mbox{controllability} properties.) 

\end{itemize}

\ssection{The main theorem}\label{sec-standard}
%\subssection{Systems and augmented systems.} 
\noindent If $n$ is a positive  integer, an {\em $n$-dimensional
control \mbox{system}} is a triple $\Sigma=(\Omega,U,f)$ such that
$\Omega$ (the {\em \mbox{state} space} of $\Sigma$) is an open subset 
of $\R^n$,  $U$ (the {\em control \mbox{space}} of $\Sigma$) 
is a set, and $f$ (the {\em dynamics} of $\Sigma$) 
is a map \mbox{$\Omega\times U\ni(x,u)\mapsto f(x,u)\in\R^n$}. 

An {\em $n$-dimensional augmented 
control system} is a \mbox{4-tuple} $\hat\Sigma=(\Omega,U,f,L)$ such that
$\Sigma=(\Omega,U,f,L)$ is an $n$-dimensional control system 
and $L$ (the {\em Lagrangian} of $\hat\Sigma$ is a map 
$\Omega\times U\ni(x,u)\mapsto L(x,u)\in\R$. 
(In that case, the state space, control space, and dynamics 
of $\Sigma$ are also called the state space, control space, and dynamics 
of $\hat\Sigma$.)

An augmented control system $\hat\Sigma=(\Omega,U,f,L)$ is 
{\em continuous} if the maps 
$\Omega\ni x\mapsto f(x,u)\in\R^n$ and 
$\Omega\ni x\mapsto L(x,u)\in\R$ 
are continuous for each fixed $u$. We call $\hat\Sigma$
{\em \mbox{uniformly} continuous} on a subset $S$ of $\Omega$
if there exists a function $\omega:\,]0,\infty[\mapsto[0,\infty]$ 
such that $\lim_{s\downarrow 0}\omega(s)=0$, having the property that

$$
\|f(x,u)-f(y,u)\|+|L(x,u)-L(y,u)|\le\omega(\|x-y\|)
$$

\medbreak\noindent whenever $x,y\in S$ and $u\in U$. We call 
$\hat\Sigma$ {\em \mbox{locally} \mbox{uniformly} continuous} if it is 
uniformly \mbox{continuous} on every compact subset of $\Omega$, and
{\em globally  \mbox{uniformly} \mbox{continuous}} if it is 
uniformly continuous on $\Omega$. 
We say that $\hat\Sigma$ is 
{\em Lipschitz continuous} if the maps 
\mbox{$\Omega\ni x\mapsto f(x,u)\in\R^n$} and 
$\Omega\ni x\mapsto L(x,u)\in\R$ 
are Lipschitz continuous for each fixed $u$. We call $\hat\Sigma$
{\em \mbox{uniformly} Lipschitz continuous} on a subset $S$ of $\Omega$
if there exists a positive constant $C$ such that

$$
\|f(x,u)-f(y,u)\|+|L(x,u)-L(y,u)|\le C\|x-y\|
$$

\medbreak\noindent whenever $x,y\in S$ and $u\in U$. We call $\hat\Sigma$
{\em \mbox{locally} \mbox{uniformly} Lipschitz continuous} if it is 
uniformly \mbox{Lipschitz} continuous on every compact subset of $\Omega$, and
{\em globally  uniformly Lipschitz continuous} if it is 
uniformly Lipschitz continuous on $\Omega$. 

\begin{remark}
Naturally, the concepts of continuity,   \mbox{uniform} continuity,
Lipschitz continuity,  and uniform Lipschitz continuity,  
also make sense for a control \mbox{system} $\Sigma=(\Omega,U,f)$,
by taking the same definitions given above and omitting 
the parts that refer to $L$. In order to avoid having to make 
a similar remark for other concepts to be introduced
in the future, we adopt the convention that any concept $X$
that we \mbox{define} for an augmented control system 
$\hat\Sigma=(\Omega,U,f,L)$ is \mbox{automatically}
\mbox{understood} to apply to a control system $\Sigma=(\Omega,U,f)$, 
in the sense that ``$X$ of $\Sigma$'' means 
``$X$ of the augmented system $\hat\Sigma=(\Omega,U,f,0)$.''\proofend
\end{remark}

The augmented control system $\hat\Sigma=(\Omega,U,f,L)$ is {\em
\mbox{coercive}} on a subset $S$ of $\Omega$ if there exist real constants
$r$, $A$, $C$, such that $A>0$, $C>0$, $r>1$, and 
\begin{equation}\label{iscoer}
\|f(x,u)\|^{r}\le AL(x,u)+C\,\,{\rm for\,\,all}\,\,
x\in S\,,\,u\in U\,.
\end{equation}
We call $\hat\Sigma$ {\em locally coercive} if it is coercive on every 
\mbox{compact} subset of $\Omega$, and {\em globally coercive} if 
it is \mbox{coercive} on $\Omega$.

\begin{remark}
If $\hat\Sigma$ is coercive on a set $S$, then it is 
\mbox{always} possible
to choose $C$, $r$, such that $C>0$, $r>1$, and 
(\ref{iscoer}) holds with $A=1$. Indeed, let $A$, $C$, $r$ be such that
$A>0$, $C>0$, $r>1$, and (\ref{iscoer}) holds. Pick $\rho$ such that
$1<\rho<r$,  and let$s=\frac{\rho}{r-\rho}$. 
Let $x\in\Omega$, $u\in U$. Then, 
if $\|f(x,u)\|^{r-\rho}>A$, we have 
\begin{eqnarray*}
\|f(x,u)\|^{\rho}&=&\frac{\|f(x,u)\|^{r}}{\|f(x,u)\|^{r-\rho}}\\
&\le&A^{-1}(AL(x,u)+C)\\
&=&L(x,u)+\frac{C}{A}\\
&\le&L(x,u)+\frac{C}{A}+A^s\,,
\end{eqnarray*}
while on the other hand, if $\|f(x,u)\|^{r-\rho}\le A$, we find that
\begin{eqnarray*}
\|f(x,u)\|^{\rho}&=&(\|f(x,u)\|^{r-\rho})^{s}\\
&\le& A^s\\
&\le&A^{-1}\|f(x,u)\|^{r}+A^s\\
&\le&A^{-1}(AL(x,u)+C)+A^s\\
&=&L(x,u)+\frac{C}{A}+A^s.
\end{eqnarray*}
Hence (\ref{iscoer}) holds if $r$, $C$, $A$ are replaced 
by $\rho$, $\frac{C}{A}+A^s$, and $1$.\proofend
\end{remark}

For an augmented control system $\hat\Sigma=(\Omega,U,f,L)$, 
we define a map $\fsh:\Om\times U\mapsto\R^{n+1}$
(called the {\em augmented dynamics} of $\hat\Sigma$) by 
letting $\fsh(x,u)=(f(x,u),L(x,u))$ for $x\in\Omega$, $u\in U$. 
We say that $\hat\Sigma$ {\em satisfies the convexity and 
upper semicontinuity condition} if, for each $x\in\Omega$, 
\begin{equation}\label{convus}
\fsh(x,U)=\bigcap_{\eps>0}\overline{{\rm co}}\Big(\bigcup\{\fsh(x',U):
\|x'-x\|\le\eps\}\Big)\,,
\end{equation}
where ``$\overline{{\rm co}}$'' stands for ``closed convex hull,''
and $\fsh(x,U)\equaldef\{(f(x,u),L(x,u)):u\in U\}$.

\begin{remark}
If $\hat\Sigma=(\Omega,U,f,L)$ is locally uniformly 
continuous, then $\hat\Sigma$ satisfies the convexity and 
upper semicontinuity condition if and only if 
the set  $\fsh(x,U)$ is closed and convex for every $x\in\Omega$.
Indeed, the ``only if'' assertion is trivial, since the right-hand side
of (\ref{convus}) is obviously closed and convex. To prove the 
``if'' part, we fix $x\in\Omega$ and assume that 
$\fsh(x,U)$ is closed and convex. We choose a $\delta$
such that $\delta>0$ and $B_\delta(x)\equaldef
\{x'\in\R^n:\|x'-x\|\le\delta\}\subseteq\Omega$ and a function
$\omega:]0,\infty[\mapsto[0,\infty]$ such that 
$\lim_{s\downarrow 0}\omega(s)=0$ and 
$\|\fsh(y,u)-\fsh(z,u)\|\le\omega(\|y-z\|)$ whenever 
$u\in U$ and $y,z\in B_\delta(x)$. We then pick 
$$
v\in\bigcap_{\eps>0}\overline{{\rm co}}\Big(\bigcup\{\fsh(x',U):
\|x'-x\|\le\eps\}\Big)
$$
and prove that $v\in\fsh(x,U)$. We let $\eps_k=2^{-k}$, 
and use the fact that 
\mbox{$v\in\overline{{\rm co}}\Big(\bigcup\{
\fsh(',U):\|x'-x\|\le\eps_k\}\Big)$}
to find, for each sufficiently large $k$, a member 
$v_k$ of the convex hull of 
\mbox{$\bigcup\{\fsh(x',U):\|x'-x\|\le\eps_k\}$}
such that \mbox{$\|v-v_k\|<\eps_k$}. We then write
\mbox{$v_k=\sum_{j=0}^{n+1}\alpha_{k,j}\fsh(x_{k,j},u_{k,j})$}, 
where the $x_{k,j}$ \mbox{belong} to $\Omega$ and satisfy 
$\|x_{k,j}-x\|\le\eps_k$, the $u_{k,j}$ belong to $U$,
and the $\alpha_{k,j}$ are nonnegative and satisfy 
$\sum_{j=0}^{n+1}\alpha_{k,j}=1$.  Then 
$\|\fsh(x,u_{k,j})-\fsh(x,u_{k,j})\|\le\omega(\eps_k)$. 
\mbox{Therefore},
if we let $w_k=\sum_{j=0}^{n+1}\alpha_{k,j}\fsh(x,u_{k,j})$, 
we have $\|w_k-v_k\|\le\omega(\eps_k)$. Hence 
$w_k\to v$ as $k\to\infty$. Since $\fsh(x,U)$ is convex, the 
$w_k$ belong to $\fsh(x,U)$. Since $\fsh(x,U)$ is closed, 
$v\in\fsh(x,U)$, and the proof is complete.\proofend
\end{remark}

A {\em target} for an augmented control system 
\mbox{$\hat\Sigma=(\Omega,U,f,L)$}
is a closed subset ${\cal T}$ of $\R^n$ 
such that \mbox{${\cal T}\subseteq{\rm Closure}\,\Omega$} and 
${\cal T}\cap\Omega=\emptyset$.

%\subsection{Trajectories}

A {\em trajectory} of $\hat\Sigma=(\Omega,U,f,L)$ 
is a locally absolutely continuous curve 
\begin{equation}\label{trajo}
I\ni t\mapsto\xi(t)\in\Omega\,,
\end{equation}
defined on a nonempty subinterval $I$ of $\R$, 
having the property that 
$\dot\xi(t)\in f(\xi(t),U)$ for almost every \mbox{$t\in I$}. 
An {\em augmented trajectory} of $\hat\Sigma$ is a locally 
absolutely \mbox{continuous} curve 
\begin{equation}\label{augmtrajo}
I\ni t\mapsto\Xi(t)=(\xi(t),\xi_0(t))\in\Omega\times\R\,,
\end{equation}
defined on a subinterval $I$ of $\R$,
having the property that $\dot\Xi(t)\in\fsh(\xi(t),U)$ for
almost every $t\in I$. 

The {\em initial time}, or {\em starting time} of a trajectory $\xi$ 
(\mbox{resp.} an augmented trajectory $\Xi=(\xi,\xi_0)$)  with domain $I$ 
is the time $\tau_-(\xi)\equaldef\min\,I$ (resp.
$\tau_-(\Xi)\equaldef\min\,I$)), if the \mbox{minimum} exists,
i.e., if $I$ is bounded below and \mbox{its} infimum belongs to $I$.  If
the initial time of $\xi$ (\mbox{resp.} $\Xi$) exists, 
then (i)~the point $x_-(\xi)\equaldef\xi(\tau_-(\xi))$
(\mbox{resp.} $x_-(\Xi)\equaldef\xi(\tau_-(\Xi))$)
is the {\em starting point}, or {\em \mbox{initial}  point}, of
$\xi$ (resp. $\Xi$), and (ii)~the ordered pair 
$\partial_-(\xi)\equaldef(\tau_-(\xi),x_-(\xi))$ 
(resp. $\partial_-(\Xi)\equaldef(\tau_-(\Xi),x_-(\Xi))$) is the
{\em initial condition} of $\xi$ (resp.  $\Xi$). 
If $\partial_-(\xi)=(t,x)$ (resp. 
$\partial_-(\Xi)=(t,x)$), we say that $\xi$ (resp. $\Xi$) 
{\em starts at $x$ at time $t$}. 

If ${\cal T}$ is a target for $\hat\Sigma=(\Omega,U,f,L)$, then 
a trajectory $\xi$ or augmented trajectory $\Xi=(\xi,\xi_0)$
with domain $I$ {\em ends at  ${\cal T}$} if the limit
$$
\xi(\uparrow)\equaldef\lim_{t\uparrow\sup\,I}\xi(t)
$$
exists and belongs to ${\cal T}$. 

For each $x\in\Omega$, we let $\axts$ be the set of \mbox{all}
\mbox{augmented} trajectories $\,\Xi=(\xi,\xi_0)$ of $\hat\Sigma$ such that 

\medbreak
(i)~$\partial_-(\Xi)=(0,x)$, 

\medbreak
(ii)~$\xi_0(0)=0$,

\medbreak
(iii)~$\Xi$ ends at the target,

\medbreak
(iv)~the limit 
$\xi_0(\uparrow)\equaldef\lim_{t\uparrow\sup\,{\rm domain}\,\Xi}\xi_0(t)$
exists. 

\medbreak\noindent If $\Xi=(\xi,\xi_0)\in\axts$ then the {\em cost}
of $\Xi$ is the number
\begin{equation}\label{thecost}
J(\Xi)\equaldef \xi_0(\uparrow)\,.
\end{equation}

\medbreak The {\em value function} of the optimal control \mbox{problem}
\mbox{defined}
by $\hat\Sigma$ and the target ${\cal T}$ 
is the function 
\mbox{$\vts:\Omega\cup{\cal T}\mapsto\R\cup\{-\infty,+\infty\}$}
given by
$$
\vts(x)=\left\{\begin{array}{lll}
\inf\{J(\Xi):\Xi\in\axts\}&{\rm if}&x\in\Omega\\
0&{\rm if}&x\in{\cal T}\,.\end{array}\right.
$$

\def\sdsv{{\cal SD}_{\hat\Sigma,V}}
\def\sdsvx{{\cal SD}_{\hat\Sigma,V,x}}
\def\msdsv{{\cal MSD}_{\hat\Sigma,V}}
\def\msdsvx{{\cal MSD}_{\hat\Sigma,V,x}}

If $V:\Omega\mapsto\R$ is a function, then an {\em augmented 
\mbox{trajectory} of $\hat\Sigma$ of steepest descent with respect to $V$} 
is an \mbox{augmented} trajectory $\Xi=(\xi,\xi_0)$ of $\hat\Sigma$ 
%that has a \mbox{starting} time equal to zero, and is 
such that 

\begin{eqnarray}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\xi_0(s)+V(\xi(s))&\ge&\xi_0(t)+V(\xi(t))\nonumber\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!
{\rm whenever}\,\,s,t\in{\rm domain}\,\Xi\,\,{\rm and}\,\,s\le t\,.
\label{steep}
\end{eqnarray}

\medbreak\noindent 
We use $\sdsv$ to denote the set of all  augmented 
\mbox{trajectories} of $\hat\Sigma$ of steepest descent with respect to $V$,
and $\sdsvx$ to denote the set of all  $\Xi\in\sdsv$ 
such that $\partial_-(\Xi)=(0,x)$. 

\medbreak If $x\in\Omega$, a {\em maximal augmented trajectory of
$\hat\Sigma$ from $x$ of steepest \mbox{descent} with respect to $V$}
is a $\Xi\in\sdsvx$ that \mbox{cannot} be \mbox{extended} to a
$\tilde\Xi\in\sdsvx$ defined on an interval which is strictly larger
that the domain of $\Xi$.  We use $\msdsvx$ to denote the set of all
maximal augmented \mbox{trajectories} of $\hat\Sigma$ from $x$ of steepest
descent with respect to $V$.

\begin{remark}
We have set up our definitions in such a way that the notion of 
``maximal augmented \mbox{trajectory} of steepest descent''
always means ``maximal within the class $\sdsvx$ 
of augmented \mbox{trajectories} $\Xi$ of steepest descent with
initial condition $\partial_-(\Xi)=(0,x)$, for a fixed $x\in\Omega$.''
(In other words, there is no such thing as a 
``maximal augmented \mbox{trajectory};'' there are \mbox{only}
``\mbox{maximal} augmented \mbox{trajectories} {\em from a given point $x$}.'')
Since all the members of $\sdsvx$ start at time $0$ at the point $x$, 
the only way that a $\Xi\in\sdsvx$, \mbox{defined}
on an interval $I$ (that necessarily starts at time $0$), could fail 
to \mbox{belong} to $\msdsvx$,
would be for $\Xi$ to be ``extendable to the right,'' that is, 
such that there exists an extension
$\tilde\Xi\in\sdsvx$ which is defined on an 
interval $\tilde I$ that also 
starts at $0$ and is strictly larger than $I$. Naturally, it may also happen
that $\Xi$ can be extended on the left, to an augmented 
trajectory $\hat\Xi$ of steepest descent defined on an interval
$\hat I=\,]$-$\eps,0]\cup I$ for some positive $\eps$. 
But such an extension will of course no longer start at time $0$, 
so its existence does not affect the possibility that $\Xi$ might 
belong to $\msdsvx$. \proofend
\end{remark}

\begin{definition}\label{rightunb}.
If $\Omega$ is an open subset of $\R^n$, and 
\mbox{$\xi\!:\!I\!\mapsto\!\Omega$} is a curve, we say that $\xi$ is 
\mbox{{\em right-unbounded} if}

\begin{itemize}
\vspace{-2mm}
\item[(i)] the interval $I$ is open on the right (that is, 
if \mbox{$\tau=\sup\,I$}, then either (a)~$\tau=+\infty$ or
(b)~$\tau$ is finite and does not belong to $I$),
\end{itemize}

\vspace{-2.8mm}
\noindent and 


\begin{itemize}

\vspace{-2.3mm}
\item[(ii)] if $\tau$ is finite, then for every compact subset $K$
of $\Omega$ there exists a $\tau_K$ such that $0\le\tau_K<\tau$ and
\mbox{$\xi(t)\notin K$} whenever $\tau_K<t<\tau$. 
\end{itemize}

\vspace{-2mm}\noindent (Equivalently, condition (ii) asserts 
that \mbox{$\lim_{t\uparrow\tau}\xi(t)=\infty_\Omega$}, where $\infty_\Omega$ is
the point at \mbox{infinity} of the one-point 
\mbox{compactification} of $\Omega$.)\proofend
\end{definition}

\medbreak The following observation is completely trivial
\mbox{given} our definitions, but we state it explicitly as a
\mbox{separate} \mbox{result} for future reference. We emphasize that
this \mbox{trivial} result is valid {\em under no technical hypotheses
\mbox{whatsoever} on $\hat\Sigma$ or $V$}.  The reader is warned that the 
\mbox{result} is {\em not} a true ``existence theorem'' for 
\mbox{trajectories} of \mbox{steepest} descent, even though at first sight 
it may \mbox{appear} to be, 
because the member $\Xi$ of $\msdsvx$ whose \mbox{existence}
it asserts could very well turn out to be the trivial trajectory
$\Xi_x^{triv}$, where $\Xi_x^{triv}$ is the map 
\mbox{$\Xi:\{0\}\mapsto\Omega\times\R$} such that $\Xi(0)=(x,0)$.  The true
\mbox{``existence} \mbox{theorem},'' yielding the existence of a 
\mbox{{\em nontrivial}} maximal trajectory of steepest descent and, 
in \mbox{fact}, \mbox{asserting} the stronger conclusion that 
{\em \mbox{every}} \mbox{maximal} \mbox{trajectory} of steepest descent 
is right-unbounded in the sense of Definition \ref{rightunb}.
This will be proved later (cf. Theorem
(\ref{maintech}) and, naturally, will \mbox{depend} on our technical hypotheses
on $\hat\Sigma$ and $V$. 

\begin{proposition}\label{trivialprop}
If $\hat\Sigma=(\Omega,U,f,L)$ is an \mbox{augmented} control system,
$V:\Omega\mapsto\R$ is a function, 
and $x\in\Omega$, then the set $\msdsvx$ of maximal \mbox{augmented}
\mbox{trajectories} of $\hat\Sigma$ from $x$ of steepest \mbox{descent}
with respect to $V$ is nonempty.
\end{proposition}

\begin{proof} 
Fix $x$. Let ${\cal Z}$ be the set of all pairs $(I,\Xi)$ such that
$I$ is a subinterval of $[0,\infty[$, $0\in I$, and 
\mbox{$\Xi=(\xi,\xi_0):I\mapsto\Omega\times\R$} is an augmented trajectory 
of $\hat\Sigma$ which is of steepest descent with respect
to $V$ and such that $\xi(0)=x$. We partially order  
${\cal Z}$ by stipulating that, if $(I_i,\Xi_i)\in{\cal Z}$ 
for $i=1,2$, then $(I_1,\Xi_1)\preceq(I_2,\Xi_2)$ iff
$I_1\subseteq I_2$ and $\Xi_1$ is the restriction of
$\Xi_2$ to $I_1$. 

It is clear that ${\cal Z}\neq\emptyset$, because
the pair $(\{0\},\Xi_x^{triv})$\mbox{---where} $\Xi_x^{triv}$ is the map 
defined above, in the paragraph preceding 
the statement of our proposition---belongs to 
${\cal Z}$.  If $Z$ is a totally ordered \mbox{subset}
of ${\cal Z}$, we show that $Z$ has an upper bound 
$(I_*,\Xi_*)$ in ${\cal Z}$. 
This conclusion is trivial if $Z=\emptyset$, for in that 
case we can take $(I_*,\Xi_*)=(\{0\},\Xi_x^{triv})$. Assume that 
$Z\neq\emptyset$. Let $I_*$ be the union of the intervals $I$ for all
the \mbox{members} $(I,\Xi)$ of $Z$. Then $I_*$ is obviously a subinterval
of $[0,\infty[$, and $0\in I$. If $t\in I_*$, we define $\Xi_*(t)=\Xi(t)$,
where $(I,\Xi)$ is any member of $Z$ such that $t\in I$.
Write $\Xi_*=(\xi_*,\xi_{0,*})$.  Then $\Xi_*$ is
obviously well defined, and is an augmented trajectory of $\hat\Sigma$
such that $\xi_*(0)=x$. If $t\in I_*$, then we can pick 
$(I,\Xi)\in Z$ such that $t\in I$, and then 
$\Xi_*(s)=\Xi(s)$ for all $s\in I$, and in \mbox{particular} for all
$s\in[0,t]$, since $[0,t]\subseteq I$. Furthemore, 
if we write $\Xi=(\xi,\xi_{0})$, then the  fact that 
$\Xi\in\sdsv$ \mbox{implies} that 
$V(x)\ge\xi_0(t)+V(\xi(t))=\xi_{0,*}(t)+V(\xi_*(t))$. \mbox{Since}
$t$ is an arbitrary member of $I_*$, we have shown that
$\Xi_*\in\sdsv$. Therefore $(I_*,\xi_*)\in {\cal Z}$. 
Furthermore, it is clear that
$(I_*,\Xi_*)$ is an upper bound for $Z$. So
we have shown that every totally ordered subset of ${\cal Z}$ has
an \mbox{upper} bound, and that ${\cal Z}\neq\emptyset$. Therefore Zorn's
Lemma implies that ${\cal Z}$ has a maximal element 
$(I_*,\Xi_*)$.  Clearly, such a maximal element 
is a member of $\msdsvx$, and our proof is complete.
\end{proof}


\medbreak An {\em augmented arc} is an augmented trajectory whose domain 
is a compact interval. If $\Xi=(\xi,\xi_0)$ is an \mbox{augmented} arc 
with domain $[a,b]$, then an \mbox{{\em improvement}} of $\Xi$ 
is an augmented arc  $\Xi'=(\xi',\xi'_0)$, with domain
$[a',b']$, such that $\xi'(a')=\xi(a)$, $\xi'(b')=\xi(b)$, 
and $\xi_0'(b')-\xi_0'(a')\le\xi_0(b)-\xi_0(a)$.



\medbreak If $\hat\Sigma=(\Omega,U,f,L)$ is an \mbox{augmented} 
control system, then an augmented arc $\Xi=(\xi,\xi_0)$ of
$\hat\Sigma$ with \mbox{domain} $[a,b]$ is 
{\em uniquely limiting} if there exists a sequence 
$\{\eta_j\}_{j=1}^\infty$ of piecewise constant functions
$\eta_j:[a,b]\mapsto U$ such that 
\begin{itemize}
\item[(*)] {\em if $\{\Xi_j\}_{j=1}^\infty$ is an arbitrary sequence
of \mbox{maximally} defined augmented trajectories of $\hat\Sigma$ 
such that \mbox{$a\in{\rm domain}(\Xi_j)$} and $\Xi_j(a)=\Xi(a)$
for every $j$, then $[a,b]\subseteq{\rm domain}(\Xi_j)$ if $j$ is large enough,
and $\Xi_j\to\Xi$ uniformly on $[a,b]$ as $j\to\infty$.}
\end{itemize}

\begin{example}\label{exalips}
Suppose $\Xi=(\xi,\xi_0)$  is an augmented arc  of
$\hat\Sigma$ with \mbox{domain} $[a,b]$ such that 
\begin{itemize}
\item[(\#)] {\em there exist a positive number $\delta$, 
a function \mbox{$\eta:[a,b]\mapsto U$}, 
and a function $\varphi:[a,b]\mapsto[0,\infty]$, such that
\begin{itemize}
\item[(i)] $\dot\Xi(t)=\Big(f(x,\eta(t)),L(x,\eta(t))\Big)\,\,$ for almost
\mbox{every} $t\in[a,b]$, 
\item[(ii)] the map $t\mapsto\Big(f(x,\eta(t)),L(x,\eta(t))\Big)\,\,$, on the 
\mbox{compact}
set 
\item[] $\qquad I_x\equaldef\{t:a\le t\le b\wedge \|x-\xi(t)\|\le\delta\}$, 
\item[] is measurable for each $x\in\Omega$, 
\item[(iii)] the map $x\mapsto\Big(f(x,\eta(t)),L(x,\eta(t))\Big)\,\,$, on the 
\mbox{compact}
set 
\item[] $\qquad I^t\equaldef\{x\in\Omega:\|x-\xi(t)\|\le\delta\}$, 
\item[] is measurable for each $t\in[a,b]$, 
\item[(iv)] $\varphi$ is integrable,
\item[(v)] the inequality
\item[] $\Big\langle f(x,\eta(t))-f(x',\eta(t)),x-x'\Big\rangle
\le\varphi(t)\|x-x'\|^2$ 
\item[] holds whenever $t\in[a,b]$, $\|x-\xi(t)\|\le\delta$, and
$\|x'-\xi(t)\|\le\delta$.
\item[(vi)] the inequality $|L(x,\eta(t))!\le\varphi(t)$ 
holds \mbox{whenever} $t\in[a,b]$ and $\|x-\xi(t)\|\le\delta$
\end{itemize}}\end{itemize}

\medbreak\noindent Then $\Xi$ is uniquely limiting. The proof is 
essentially as follows.  By dividing $[a,b]$ into small intervals,
we can assume that there is a fixed compact ball $B$ 
such that $B\subseteq\Omega$, $\xi$ is entirely contained in the interior
of $B$, and the bound of {\em (v)} holds whenever $t\in[a,b]$ and 
$x,x'\in B$. We then write $F^u(x)=(f(x,u,L(x,u))$
for each $x\in B$, $u\in U$, 
and observe that the set ${\cal F}=\{F^{\eta(t)}:t\in[a,b]\}$ 
is a subset of the separable Banach space $C^0(B,\rr^{n+1})$. Then 
$[a,b]\ni t\mapsto F^{\eta(t)}$ is an $L^1$ $C^0(B,\rr^{n+1})$-valued map.
Therefore one can approximate this map 
in $L^1$ by piecewise constant ${\cal F}$-valued 
maps. In other words, one can find a sequence
$\{\eta_j\}_{j=1}^\infty$ of piecewise constant 
${\cal G}$-valued functions (where 
${\cal G}=\{\eta(t):t\in[a,b]\}$), and integrable functions
$k_j:[a,b]\mapsto[0,+\infty]$,  such that 

\medbreak\noindent $\|f(x,\eta(t))\!-\!f(x,\eta_j(t))\|\!+\!
|L(x,\eta(t))\!-\!L(x,\eta_j(t))|\!\le\!k_j(t)$

\medbreak\noindent whenever $a\le t\le b$, $x\in B$, and $j\in\nn$, and 
$\lim_{j\to\infty}\int_a^bk_j(t)\,dt=0$. Then, if $a\le c\le d\le b$, 
and $\zeta:[c,d]\mapsto B$, $\theta:[c,d]\mapsto B$, 
are trajectories of $\eta$, $\eta_j$, respectively, Gronwall's inequality 
yields the bound 
$$
\|\zeta(t)-\theta(t)\|\le e^{\int_a^b\varphi(s)ds}\big(\|k_j\|_{L^1}+
\|\zeta(c)-\theta(c)\|\Big)
$$
if $t\in[c,d]$. If we apply this with $c=a$ and $\zeta=\xi$, letting
$\theta$ be any trajectory $\xi_j$ of $\eta_j$ 
\mbox{starting} at $\xi(a)$ 
at time $a$, and defined on some \mbox{subinterval} $[a,d]$ of 
$[a,b]$, we see that, as long as $j$ is large \mbox{enough}, the 
\mbox{maximum} of the $\|\xi(t)-\xi_j(t)\|$ is \mbox{bounded} by a small
\mbox{constant}. This guarantees that $\xi_j$ 
\mbox{actually} exists 
on the \mbox{whole} \mbox{interval} $[a,b]$, and then the Gronwall bound
implies that $\xi_j\to \xi$ uniformly as $j\to\infty$.
Then the \mbox{integrals} $\int_a^tL(\xi_j(s),\eta_j(s))\,ds$
differ from $\int_a^tL(\xi_j(s),\eta(s))\,ds$
by less than $\|k_j\|_{L^1}$, in view of the 
bound 
$$
|L(x,\eta(t))-L(x,\eta_j(t))|\le k_j(t)\,,
$$
and 
$\int_a^tL(\xi_j(s),\eta(s))\,ds\to\int_a^tL(\xi(s),\eta(s))\,ds$
as $j\to\infty$,
because $L(\xi_j(s),\eta(s))\to L(\xi(s),\eta(s))$ for each $s$, and 
$|L(\xi_j(s),\eta(s))|\le\varphi(s)$. \proofend
\end{example}


\medbreak An augmented trajectory $\Xi=(\xi,\xi_0)$ with domain $I$ is 
{\em locally uniquely limiting} if for every compact \mbox{subinterval} 
$I'$  of $I$ the restriction of $\Xi$ to 
$I'$ is uniquely limiting.

\medbreak\noindent
An augmented trajectory $\Xi=(\xi,\xi_0)$ with domain $I$ is {\em almost
uniquely limiting} if there exists 
a finite subset $B$ of $I$ such that the restriction of
$\Xi$ to every subinterval of $I\backslash B$ is 
locally uniquely limiting.

\medbreak\noindent If $\Omega$ is an open subset of $\R^n$, 
we say that a function $V:\Omega\mapsto\R$ {\em satisfies the inequality}

\begin{equation}\label{loweq}
\sup\{-\langle \nabla V(x),f(x,u)\rangle-L(x,u):u\in U\}\ge 0
\end{equation}

\medbreak\noindent on $\Omega$ {\em in the viscosity sense} if 

\begin{center}\begin{minipage}{7.4cm}
\begin{itemize}
\item[(V$_+$)] {\em whenever $x\in\Omega$ and $p\in\R^n$ is a 
\mbox{subdifferential} of $V$ at $x$, it follows that}

$$
\sup\{-\langle p, f(x,u)\rangle-L(x,u):u\in U\}\ge 0\,.
$$
\end{itemize}
\end{minipage}\end{center}

\medbreak\noindent(We recall that, if $\Omega$ is open in $\R^n$, then
a {\em subdifferential} of a function $V:\Omega\mapsto\R$ at a point
$\bar x\in\Omega$ is a vector $p\in\R^n$ such that

$$
\lim\inf_{x\to\bar x}\frac{V(x)-V(\bar x)-p\cdot(x-\bar x)}{
\|x-\bar x\|}\ge 0\,.\,)
$$

\medbreak Similarly, we say that $V$ {\em satisfies the inequality}

\begin{equation}\label{upeq}
\sup\{-\langle \nabla V(x),f(x,u)\rangle-L(x,u):u\in U\}\le 0
\end{equation}

\medbreak\noindent on $\Omega$ {\em in the viscosity sense} if 

\begin{center}\begin{minipage}{7.4cm}
\begin{itemize}
\item[(V$_-$)] {\em whenever $x\in\Omega$ and $p\in\R^n$ is a 
\mbox{superdifferential} of $V$ at $x$, it follows that}

$$
\sup\{-\langle p, f(x,u)\rangle-L(x,u):u\in U\}\le 0\,.
$$
\end{itemize}
\end{minipage}\end{center}

\medbreak\noindent 
(A {\em superdifferential} of $V$ at $x_*$ is a vector $p$ 
such that $-p$ is a subdifferential of $-V$ at $x_*$.) 

\medbreak We say that $V$ {\em satisfies the equation}

\begin{equation}\label{eqeq}
\sup\{-\langle \nabla V(x),f(x,u)\rangle-L(x,u):u\in U\}=0
\end{equation}

\medbreak\noindent on $\Omega$ {\em in the viscosity sense} if 
it satisfies (\ref{loweq}) and (\ref{upeq}) in the viscosity 
sense. 

\begin{remark}
The definition of ``viscosity solution'' given here 
is known to be equivalent to the more \mbox{common} one involving 
test functions, cf. \cite{BARD}.\proofend
\end{remark}

Our main result is the following theorem:

\begin{theorem}\label{mainres}
Let $\hat\Sigma=(\Omega,U,f,L)$ be an \mbox{augmented} control 
system, let ${\cal T}$ be a target for $\hat\Sigma$, and
let \mbox{$V:\Omega\cup{\cal T}\mapsto\R$} be a function. Assume that
\begin{itemize}
\item[(1)] $\hat\Sigma$ is locally uniformly continuous, locally coercive,
and such that $F_{\hat\Sigma}(x,U)$ is closed and convex for \mbox{every} 
$x\in\Omega$. 
\item[(2)] $V$ is \mbox{continuous}.
\item[(3)] $V$ satisfies (\ref{eqeq}) on $\Omega$ in the
\mbox{viscosity} sense.
\item[(4)] $V$ vanishes on ${\cal T}$.
%\item[(5)] $V$ is is bounded below.
\item[(5)] Every augmented arc has an almost locally uniquely 
limiting improvement. 
\item[(6)] Whenever $x\in\Omega$, $\Xi=(\xi,\xi_0)\in\msdsvx$, 
and $\xi$ is right-unbounded, it follows that $\Xi\in\axts$.
\end{itemize}
Then $V\equiv\vts$.
\end{theorem}

\ssection{Examples}\label{exasec}

\begin{example} {\em (Linear-quadratic optimal control)} 
\mbox{Consider} the standard linear-quadratic optimal \mbox{control} problem, in which 
$x$, $u$ take values in $\R^n$, $\R^m$, \mbox{respectively},
the dynamical law is
$$
\dot x=Ax+Bu\,,
$$
the Lagrangian is given by 
$$
L(x,u)=x^\dagger Rx+u^\dagger Su\,,
$$
the square matrices $R$, $S$ are strictly positive definite, and the pair 
$(A,B)$ is stabilizable. We take the \mbox{target} set ${\cal T}$ to consists 
of the origin of $\R^n$.  
The crucial \mbox{technical} issue here is the
fact that the Lagrangian is not bounded away from zero. 
The \mbox{hypotheses} of our main theorem (including the coerciveness,
which \mbox{follows} from the positive definiteness of $S$) are easily 
verified as long as $V$ is bounded below.  The only 
nontrivial point is the \mbox{verification} of condition (6). To prove that 
this  holds, let $\Xi:[0,\tau\,[\,\mapsto\R^n\times\R$ be a 
right-unbounded \mbox{maximal} \mbox{trajectory}
of steepest descent with respect to $V$ that does not end
at the target, and write $\Xi=(\xi,\xi_0)$ in the usual way. Then 
$\tau$ has to be infinite, because if $\tau$ was \mbox{finite} then
the boundedness of the cost (\mbox{arising} from the fact that 
$\Xi=(\xi,\xi_0)$ is of steepest descent and $V$ is bounded below)
would \mbox{trivially} imply an $L^2$ bound on the control, from which
it would follow that 
$\Xi$ can be extended to the closed \mbox{interval} $[0,\tau]$, and 
then the \mbox{assumption} that $\Xi$ does not end at the 
\mbox{target} would
\mbox{enable} us to use Proposition \ref{trivialprop} and
Theorem \ref{maintech} (\mbox{applied} with $\Omega=\R^n\backslash\{0\}$)
to \mbox{extend} $\xi$ even \mbox{further}, contradicting maximality. 
So $\tau$ is \mbox{infinite}. On the other hand, the fact that
$V$ is bounded \mbox{below} and $\Xi$ is of \mbox{steepest} descent implies
that the integral 
$$
\int_0^\infty\Big(\xi(t)^\dagger R\xi(t)+
\eta(t)^\dagger S\eta(t)\Big)dt
$$
is finite, if $\eta$ is an open-loop control that generates $\Xi$.
But then $\xi(\cdot)$ and $\eta$ are square-integrable, so the 
\mbox{condition}
that $\dot\xi=A\xi+B\eta$ implies that $\xi$ is square-integrable and has
a square-integrable derivative, and then Barbalat's lemma
implies that  $\xi$  ends at the \mbox{target}, as desired. \proofend
\end{example}

\begin{example} {\em (Fuller's problem, cf., e.g., 
\cite{ZB})}  This is the optimal control problem for the dynamical law
\begin{eqnarray*}
\dot x&=&y\,,\\
\dot y&=&u\,,
\end{eqnarray*}
with control constraint $-1\le u\le 1$. The target set ${\cal T}$ consists 
of the origin of $\R^2$.  The Lagragian is $L(x,y,u)=x^2$.
The crucial \mbox{technical} issue here is the
fact that the Lagrangian is not bounded away from zero, and in fact has a 
whole line of zeros.  The \mbox{hypotheses} of our main theorem are easily 
verified as long as $V$ is bounded below. The only 
nontrivial point is the \mbox{verification} of condition (6). To prove that 
this  holds, let $\Xi:[0,\tau\,[\,\mapsto\R^3$ be a right-unbounded
\mbox{maximal} \mbox{trajectory}
of steepest descent with respect to $V$ that does not end
at the target, and write $\Xi=(\xi,\xi_0)$ in the usual way. Then 
$\tau$ has to be infinite, because if $\tau$ was \mbox{finite} then
the boundedness of the control would \mbox{trivially} imply that 
$\Xi$ can be extended to the closed \mbox{interval} $[0,\tau]$, and 
then the assumption that $\Xi$ does not end at the target
enables us to use Proposition \ref{trivialprop} and
Theorem \ref{maintech} (applied with $\Omega=\R^2\backslash\{(0,0)\}$)
to \mbox{extend} $\xi$ even \mbox{further}, contradicting maximality. 
So $\tau$ is \mbox{infinite}. On the other hand, the fact that 
$V$ is bounded \mbox{below} and $\Xi$ is of \mbox{steepest} descent implies
that the integral $\int_0^\infty x(t)^2dt$ is finite, if we write
\mbox{$\xi(t)=(x(t),y(t))$}. But then $x(\cdot)$ is a square-integrable function
on $[0,\infty\,[\,$  whose second derivative is bounded. 
By a \mbox{straightforward}  \mbox{generalization} of Barbalat's lemma, 
this implies that 
both $x(\cdot)$ and $y(\cdot)$ go to zero, i.e., that 
$\xi$  ends at the target, as desired. \proofend
\end{example} 

\begin{example} {\em (The reflected brachistochrone problem.)} This is
the minimum time problem for the dynamical law
\begin{eqnarray*}
\dot x&=&u\sqrt{|y|}\,,\\
\dot y&=&v\sqrt{|y|}\,,
\end{eqnarray*}
with control constraint $u^2+v^2\le 1$. The target set ${\cal T}$ consists 
of a single point $B\in\R^2$.  The crucial \mbox{technical} issue here is the
fact that the dynamical law is not Lipschitz-continuous with respect to the 
state.  The \mbox{hypotheses} of our main theorem are easily verified. The only 
nontrivial point is the verification of condition (5). To prove that this 
holds, we pick an arbitrary \mbox{integral} arc $\xi:[a,b]\mapsto\R^2$,
and observe that either (i)~$\xi(t)$ never belongs to the $x$ axis
$X=\{(x,y)\in\R^2:y=0\}$ or (ii)~there exist $t_-,t_+\in[a,b]$
<such that $t_-\le t_+$, $\xi(t)\notin X$ whenever $a\le t<t_-$
or $t_+>t\le b$, $\xi(t_-)\in X$, and $\xi(t_+)\in X$. If (i) holds, 
then $\xi$ satisfies the conditions  of  Example \ref{exalips}, so 
$\xi$ is uniquely limiting.  If (ii) holds and $t_+=t_-$, then the 
restriction of $\xi$ to each of the \mbox{intervals} $[a,t_-\,[+,$, 
$\,]\,t_+,b]$, is locally uniquely limiting, so 
$\xi$ is \mbox{almost} locally uniquely limiting.  Finally, if (ii) 
holds and $t_+<t_-$, then the restriction $\tilde\xi$ 
of $\xi$ to the interval $[t_-,t_+]$ is such that
the set $S=\{t\in[a,b]:\xi(t)\notin X\}$ is the union of a finite
or countable infinite collection ${\cal I}$ of pairwise disjoint 
relatively open subintervals of $[a,b]$. If $I\in{\cal I}$, then 
the restriction $\tilde\xi_I$ of $\tilde\xi$ 
to $I$ is entirely contained in the
open upper half-plane or in the open lower half-plane. By reflecting 
$\tilde\xi_I$ with respect to $X$, if necessary, we may 
assume that $\tilde\xi_I$ is entirely contained in the
open upper half-plane for every $I\in{\cal I}$. Then 
$\tilde\xi$ is a trajectory of our system entirely 
\mbox{contained}
in the closed upper half-plane. It is well known that the problem 
in the closed upper half-plane $H_+$ 
is the famous ``brachistochrone problem,''
whose time-optimal \mbox{trajectories} $\zeta:[\alpha,\beta]\mapsto H_+$
are cycloids such that $\zeta(t)$ is an interior point of $H_+$ 
whenever $\alpha<t<\beta$.  It follows that we can always replace 
$\tilde\xi$ by a cycloid $\zeta$, thereby \mbox{obtaining} an
almost locally uniquely limiting \mbox{improvement} of $\xi$. \proofend
\end{example} 

%\newpage
\ssection{The main technical lemma}\label{maintechsec}

\def\Phids{\Phi_{\delta,\hat\Sigma}}

%\subssection{Infinitesimal steepest descent.}
\noindent Let \mbox{$\hat\Sigma=(\Omega,U,f,L)$} be an augmented control 
system. For every $x\in\Omega$ and every positive number $\delta$ such that 
${\rm dist}(x,\R^n\backslash\Omega)>\delta$, we let
$\Phids(x)$ be the closed \mbox{convex} hull of all the vectors
$\fsh(x',u)$, for all pairs $(x',u)$ such that $x'\in\Omega$,
$\|x'-x\|\le\delta$, and $u\in U$. Then $\Phids(x)$ is a closed
convex subset of $\R^{n+1}$.  Clearly, 
\mbox{$\Phids(x)\subseteq\Phi_{\delta',\hat\Sigma}(x)$} whenever
$0<\delta\le\delta'$. 

Let $V:\Omega\mapsto\bar\R$ be a real-valued function,
and let $x_*\in\Omega$. We say that $V$ {\em satisfies the infinitesimal 
steepest descent condition at $x_*$} if 

\medbreak
\begin{minipage}{8cm}
\begin{itemize}%\item[]\begin{itemize}
\item[(ISD)] {\em there exist 
sequences $\{x_j\}_{j=1}^\infty$, $\{v_j\}_{j=1}^\infty$,
$\{\lambda_j\}_{j=1}^\infty$, $\{h_j\}_{j=1}^\infty$,
$\{\gamma_j\}_{j=1}^\infty$
in $\Omega$, $\R^n$, $\R$, $\R$, and $\R$, 
\mbox{respectively}, such that 
\begin{itemize}
\item[(1)] $h_j>0$, $\gamma_j>0$, $(v_j,\lambda_j)\in\fsh(x_*,U)$, and
$\|x_j-x_*-h_jv_j\|\le h_j\gamma_j$ for all $j$, 
\item[(2)]
$h_j\downarrow 0$, $x_j\to x_*$, 
and $\gamma_j\downarrow 0$ as $j\to\infty$,
\end{itemize}
and}
\begin{itemize}
\item[(3)] $V(x_j)\le V(x_*)-h_j\lambda_j+h_j\gamma_j\,\,\,
{\rm for\,\,\, all}\,\,\,j$.
\end{itemize}\end{itemize}%\end{itemize}
\end{minipage}

\begin{theorem}\label{satinfi}
Let $\hat\Sigma=(\Omega,U,f,L)$ be an $n$-dimensional
augmented control system. 
Suppose that $\hat\Sigma$ is locally \mbox{uniformly} continuous,
locally coercive, and such that $\fsh(x)$ is closed and convex
for every $x\in\Omega$. 
Let \mbox{$V:\Omega\mapsto\R$} be a \mbox{continuous} function. Then
$V$ \mbox{satisfies} (\ref{loweq}) on $\Omega$ in the
\mbox{viscosity} sense if and only if $V$ \mbox{satisfies}
the infinitesimal steepest descent condition at every 
\mbox{$x\in\Omega$}.
\end{theorem}
\begin{proof}
First, we assume that $V$ \mbox{satisfies}
the \mbox{infinitesimal} steepest descent condition at every $x\in\Omega$,
and prove that $V$ satisfies (\ref{loweq}) on $\Omega$ in the
\mbox{viscosity} sense. For this purpose, we pick $x_*\in\Omega$
and a subdifferential $p$ of $V$ at $x_*$, and prove that
\begin{equation}\label{topro}
\sup\{-p\cdot f(x_*,u)-L(x_*,u):u\in U\}\ge 0\,.
\end{equation}
Using (ISD), 
we pick \mbox{sequences} $\{x_j\}_{j=1}^\infty$, $\{v_j\}_{j=1}^\infty$,
$\{\lambda_j\}_{j=1}^\infty$, $\{h_j\}_{j=1}^\infty$,
$\{\gamma_j\}_{j=1}^\infty$
in $\Omega$, $\R^n$, $\R$, $\R$, and $\R$, 
\mbox{respectively}, such that 
$h_j>0$, $\gamma_j>0$, \mbox{$(v_j,\lambda_j)\in\fsh(x_*)$}, 
$\|x_j-x_*-h_jv_j\|\le h_j\gamma_j$,
and \mbox{$V(x_j)\le V(x_*)-h_j\lambda_j+h_j\gamma_j$}
for all $j$, and \mbox{$h_j\downarrow 0$}, \mbox{$x_j\to x_*$}, 
and $\gamma_j\downarrow 0$ as $j\to\infty$. Let 
\mbox{$J=\{j:x_j=x_*\}$}.  Then for $j\in J$ the \mbox{inequality}
$\|x_j-x_*-h_jv_j\|\le h_j\gamma_j$ implies 
$\|v_j\|\le\gamma_j$, so $\lim_{j\to\infty,j\in J}v_j=0$. 
On the \mbox{other} hand, the \mbox{inequality}
\mbox{$V(x_j)\le V(x_*)-h_j\lambda_j+h_j\gamma_j$} 
\mbox{implies} $h_j\lambda_j\le h_j\gamma_j$, i.e., 
$\lambda_j\le \gamma_j$. Since the sequence
$\{\lambda_j\}_{j=1}^\infty$  is bounded \mbox{below} 
(for example, because the \mbox{local} coercivity \mbox{implies} a
bound $\lambda_j\ge\|v_j\|^r-C$), we may assume, after 
\mbox{replacing} $J$ by a smaller infinite set, if necessary, 
that $\lambda=\lim_{j\to\infty,j\in J}\lambda_j$ exists.
Since $\lambda_j\le \gamma_j$, $\lambda$ must be $\le 0$. 
Furthermore, the vector $(0,\lambda)$ is a limit 
of vectors $(v_j,\lambda_j)\in\fsh(x_*)$, so 
$(0,\lambda)\in\fsh(x_*)$, since $\fsh$ is closed. 
Hence there exists $\bar u\in U$ such that 
$f(x_*,\bar u)=0$ and $L(x_*,\bar u)\le 0$. But then 
$-p\cdot f(x_*,\bar u)-L(x_*,\bar u)\ge 0$, so 
(\ref{topro}) holds. 

We now consider the case when the set $J$ is finite. In this case, after 
passing to a subsequence, if necessary, we may assume that $J$ is empty, 
i.e., that $x_j\neq x_*$ for all $j$. Since $p$ is a 
subdifferential of $V$ at $x_*$, we have
\begin{equation}\label{eqa-a}
\lim\inf_{x\to x_*,x\neq x_*}\frac{V(x)-V(x_*)-p\cdot(x-x_*)}
{\|x-x_*\|}\ge 0\,.
\end{equation}
Since the $x_j$ converge to $x_*$ and are different from $x_*$, 
(\ref{eqa-a}) implies
\begin{equation}\label{eqa-b}
\lim\inf_{j\to\infty}\frac{V(x_j)-V(x_*)-p\cdot(x_j-x_*)}
{\|x_j-x_*\|}\ge 0\,.
\end{equation}
Since \mbox{$V(x_j)\le V(x_*)-h_j\lambda_j+h_j\gamma_j$}, 
(\ref{eqa-b}) implies
\begin{equation}\label{eqa-c}
\lim\inf_{j\to\infty}\frac{-h_j\lambda_j+h_j\gamma_j-p\cdot(x_j-x_*)}
{\|x_j-x_*\|}\ge 0\,.
\end{equation}
Now, $x_j-x_*=x_j-x_*-h_jv_j+h_jv_j=h_j(w_j+v_j)$, where
$w_j=h_j^{-1}(x_j-x_*-h_jv_j)$. Hence 
\begin{equation}\label{eqa-d}
\lim\inf_{j\to\infty}\frac{-\lambda_j+\gamma_j-p\cdot(w_j+v_j)}
{\|w_j+v_j\|}\ge 0\,.
\end{equation}
Hence, given a positive $\eps$ there exists a $j(\eps)$ such that 
$$
-\lambda_j+\gamma_j-p\cdot(w_j+v_j)\ge-\eps\|w_j+v_j\|\,\,{\rm whenever}
\,\,j\ge j(\eps)\,.
$$
Therefore
$$
-\lambda_j-p\cdot v_j\ge-\eps\|v_j\|-\eps\|w_j\|
+p\cdot w_j-\gamma_j\,\,{\rm if}\,\,j\ge j(\eps)\,.
$$
Since $w_j\to 0$, there is---for each $\eps$---a $j'(\eps)$ such that
\begin{equation}\label{eqa-i}
-\lambda_j-p\cdot v_j\ge-\eps\|v_j\|-\eps
\,\,{\rm whenever}\,\,j\ge j'(\eps)\,.
\end{equation}
The coercivity bound yields $\|v_j\|^r\le\lambda_j+C$, so 
\mbox{$-\lambda_j\le-\|v_j\|^r+C$}. Hence 
$$
-\|v_j\|^r+C-p\cdot v_j\ge-\eps\|v_j\|-\eps
\,\,{\rm whenever}\,\,j\ge j'(\eps)\,,
$$
so 
\begin{equation}\label{eqa-e}
-\|v_j\|^r+\eps\|v_j\|+\|p\|.\|v_j\|\ge-\eps-C
\,{\rm if}\,j\ge j'(\eps)\,.
\end{equation}
Now, if the sequence $\{\|v_j\|\}_{j=1}^\infty$ was unbounded, then 
we could pick an infinite subset $J$ of $\nn$ such that 
$\|v_j\|\to+\infty$ as $j\to\infty$ via values in $J$. But then, taking 
for example $\eps=1$, we would contradict (\ref{eqa-e}), because 
the number $-\|v_j\|^r+\|v_j\|+\|p\|.\|v_j\|$ is equal to 
\mbox{$-\|v_j\|^r\Big(1-(1+\|p\|)\|v_j\|^{1-r}\big)$}, which 
goes to $-\infty$ as $j\to\infty$ via values in $J$.  Therefore 
the sequence $\{\|v_j\|\}_{j=1}^\infty$ is bounded. Pick a
constant $K$ such that $\|v_j\|\le K$ for all 
$j$. Then, for each $\eps$, if $u_\eps$ is such that
$v_{j'(\eps)}=f(x_*,u_\eps)$ and $\lambda_{j'(\eps)}=L(x_*,u_\eps)$,
(\ref{eqa-i}) implies 
$$
-p\cdot f(x_*,u_\eps)-L(x_*,u_\eps)\ge -\eps(K+1)\,.
$$
Hence 
$$
\sup\{-p\cdot f(x_*,u)-L(x_*,u):u\in U\}\ge-\eps(K+1)\,.
$$
Since $\eps$ is arbitrary, we see that (\ref{topro}) holds. 

This concludes the proof of the ``if'' part of our statement.
We now proceed to proving the ``only if'' part.  We assume that
$V$ satisfies (\ref{loweq}) on $\Omega$ in the viscosity sense,
pick an $x_*\in\Omega$, and prove that $V$ satisfies \mbox{(ISD)}.

We will actually prove that the following condition, which is 
in principle weaker, holds. 
\medbreak
\begin{minipage}{8cm}
\begin{itemize}%\item[]\begin{itemize}
\item[(ISD')] {\em there exist 
sequences $\{x_j\}_{j=1}^\infty$, $\{v_j\}_{j=1}^\infty$,
$\{\lambda_j\}_{j=1}^\infty$, $\{h_j\}_{j=1}^\infty$,
$\{\delta_j\}_{j=1}^\infty$, $\{\gamma_j\}_{j=1}^\infty$
in $\Omega$, $\R^n$, $\R$, $\R$, $\R$, and $\R$, respectively, such that 
\begin{itemize}
\item[(1)] $h_j>0$, $\delta_j>0$, $\gamma_j>0$, 
$(v_j,\lambda_j)\in\Phi_{\delta_j}(x_*)$, $\|x_j-x_*\|\le\delta_j$, and
$\|x_j-x_*-h_jv_j\|\le h_j\gamma_j$ for all $j$, 
\item[(2)]
$h_j\downarrow 0$, $\delta_j\downarrow 0$, and 
$\gamma_j\downarrow 0$ as $j\to\infty$,
\end{itemize}
and}
\begin{itemize}
\item[(3)] $V(x_j)\le V(x_*)-h_j\lambda_j+h_j\gamma_j\,\,\,
{\rm for\,\,\, all}\,\,\,j$.
\end{itemize}\end{itemize}%\end{itemize}
\end{minipage}

\medbreak\noindent We first show that if (ISD') holds then (ISD) 
\mbox{holds} as well.  To do
this, we pick sequences $\{x_j\}_{j=1}^\infty$,
$\{v_j\}_{j=1}^\infty$, $\{\lambda_j\}_{j=1}^\infty$,
$\{h_j\}_{j=1}^\infty$, $\{\delta_j\}_{j=1}^\infty$,
$\{\gamma_j\}_{j=1}^\infty$ with the properties specified in
(ISD'). We pick $\delta$ such that

$$
B_\delta(x_*)=\{x\in\R^n:\|x-x_*\|\le\delta\}\subseteq\Omega\,,
$$

\medbreak\noindent and a \mbox{function} 
$\omega:\,\,]\,0,+\infty[\,\,\mapsto[0,+\infty]$ 
such that $\lim_{s\downarrow 0}\omega(s)=0$ and

$$
\|f(x,u)-f(x',u)\|+|L(x,u)-L(x',u)|\le\omega(\|x-x'\|)
$$

\medbreak\noindent whenever $x,x'\in B_\delta(x_*)$. We then 
pass to a \mbox{subsequence}, if necessary, and assume that 
$\delta_j\le\delta$ for all $j$. We use the fact that
$(v_j,\lambda_j)\in\Phi_{\delta_j\hat\Sigma}(x_*)$ to find a 
$(w_j,\ell_j)$ of the form $\sum_{k=0}^{n+1}\alpha_{j,k}\fsh(x_{j,k},u_{j,k})$ 
such that $\alpha_{j,k}\ge 0$, 
$\sum_{k=0}^{n+1}\alpha_{j,k}=1$, $\|x_{j,k}-x_*\|\le\delta_j$,
and $\|w_j-v_j\|+|\ell_j-\lambda_j|\le 2^{-j}$. 
We then define 
\begin{eqnarray*}
\tilde v_j&=&\sum_{k=0}^{n+1}\alpha_{j,k}f(x_*,u_{j,k})\,,\\
\tilde \lambda_j&=&\sum_{k=0}^{n+1}\alpha_{j,k}L(x_*,u_{j,k})\,,
\end{eqnarray*}
and conclude that $\|\tilde v_j-w_j\|\le\omega(\delta_j)$
and 
\mbox{$\|\tilde\lambda_j-\ell_j\|\le\omega(\delta_j)$}. We then let
$$
\tilde\gamma_j=\gamma_j+2^{-j}+\omega(\delta_j)\,,
$$
so $\tilde\gamma_j\to 0$ as $j\to\infty$. It is then clear that 
\mbox{$\|x_j=x_*-h_j\tilde v_j\|\le h_j\tilde\gamma_j$}
and \mbox{$V(x_j)\le V(x_*)-h\tilde\lambda_j+h_j\tilde\gamma_j$}
for all $j$. \mbox{Finally,} we have
$(\tilde v_j,\tilde\lambda_j)=
\sum_{k=0}^{n+1}\alpha_{j,k}\fsh(x_*,u_{j,k})$, so 
$(\tilde v_j,\tilde\lambda_j)\in\fsh(x_*,U)$, since 
$\fsh(x_*,U)$ is convex. This completes the proof that (ISD') implies
(ISD). 

\medbreak We now prove (ISD'), assuming of course that 
$V$ \mbox{satisfies} (\ref{loweq}) in the viscosity sense.
We do this by \mbox{assuming} that the sequences whose existence
is asserted by (ISD') do not exist
and deriving a contradiction.  We will \mbox{assume}, as we clearly
may \mbox{without} loss of generality, that $x^*=0$ and
$V(x^*)=0$, i.e., $V(0)=0$.

\medbreak
Since $V$ does not satisfy (ISD'), there must exist a $\bar\gamma$ such that 
$0<\bar\gamma<{\rm dist}(0,\R^n\backslash\Omega)$ and

\begin{equation}\label{traf}
V(h(v+v'))+h\lambda>h\bar\gamma
\end{equation} 

\noindent whenever $0<h\le\bar\gamma$, %$0<\delta\le\bar\gamma$, 
%$(v,\lambda)\in\Phi_{\delta,\hat\Sigma}(0)$, $h\|v\|\le\delta$, 
$(v,\lambda)\in\Phi_{\bar\gamma,\hat\Sigma}(0)$, 
\mbox{$h\|v+v'\|\le\bar\gamma$},  and $\|v'\|\le\bar\gamma$.
(Indeed, if $\bar\gamma$ did not exist, then for each sufficiently
natural number
$j$ we could \mbox{define} $\gamma_j=2^{-j}$, and find
$h_j$, %$\delta_j$,  
$v_j$, $\lambda_j$, $v'_j$, $x_j$,
such that $0<h_j\le\gamma_j$, %$0<\delta_j<2^{-j}$, 
$(v_j,\lambda_j)\in\Phi_{\gamma_j,\hat\Sigma}(0)$, 
\mbox{$h_j\|v_j+v'_j\|\le\gamma_j$},  
%$(v_j,\lambda_j)\in\Phi_{\delta_j,\hat\Sigma}(0)$, $h_j\|v_j\|<\delta_j$,  
$\|v'_j\|\le\gamma_j$, 
and \mbox{$V(x_j)+h_j\lambda_j\le\gamma_jh_j$}.  But then, if we take
\mbox{$x_j=h_j(v_j+v'_j)$, $\delta_j=\gamma_j$}, the sequence
$\{(x_j,v_j,\lambda_j,h_j,\delta_j,\gamma_j)\}_{j=1}^\infty$ 
satisfies conditions (1), (2) and (3) of (ISD'), contradicting the fact 
that $V$ does not satisfy (ISD').)

By making $\bar\gamma$ smaller, if necessary, we can assume
that there exist real numbers $C$, $r$, such that $C>0$, $r>1$ and
$\|f(x,u)\|^r\le L(x,u)+C$ whenever $\|x\|\le\bar\gamma$
and $u\in U$.  It then follows, if we let
$$
\psi(v,\lambda)=\|v\|^r-\lambda-C\,\,{\rm for}\,\,(v,\lambda)\in
\R^n\times\R\,,
$$
that 
\begin{equation}\label{itthefol}
\!\!\!\!\psi(w)\le 0\,\,{\rm whenever}\,\,w=F(x,u)\,,\|x\|\le\bar\gamma,
u\in U\,.
\end{equation} 
Since $\psi$ is convex, the inequality
$\psi(w)\le 0$ holds whenever $w\in\Phi_{\bar\gamma,\hat\Sigma}(0)$, and then
\begin{equation}\label{coer-a}
\|v\|^r\le\lambda+C\quad{\rm whenever}\,\,\,
(v,\lambda)\in\Phi_{\bar\gamma,\hat\Sigma}(0)\,.
\end{equation}
If $(v,\lambda)\in\Phi_{\bar\gamma,\hat\Sigma}(0)$, $v'\in\R^n$, and
$\|v'\|\le\bar\gamma$, then the inequality 
$(\alpha+\beta)^r\le 2^r(\alpha^r+\beta^r)$, valid 
for nonnegative $\alpha$, $\beta$, implies
\begin{eqnarray*}
\|v+v'\|^r&\le&(\|v\|+\|v'\|)^r\\
&\le&2^r(\|v\|^r+\|v'\|^r)\\
&\le&2^{r}(\lambda+C+\bar\gamma^r)\,.
\end{eqnarray*}
Pick $\tilde r$ such that $1<\tilde r<r$. Let $A=2^{\frac{r}{r-\tilde r}}$
Then
\begin{eqnarray*}
\|v+v'\|^{\tilde r}&=&\|v+v'\|^{\tilde r-r}\cdot\|v+v'\|^{r}\\
&\le& A^{\tilde r-r}\|v+v'\|^{r}\\
&\le& A^{\tilde r-r}2^{r}(\lambda+C+\bar\gamma^r)\\
&=&2^{-r}2^{r}(\lambda+C+\bar\gamma^r)\\
&=&\lambda+C+\bar\gamma^r
\end{eqnarray*}
if $\|v+v'\|^{\tilde r}\ge A$. On the other hand, 
if $\|v+v'\|^{\tilde r}\le A$, then 
$\|v+v'\|^{\tilde r}\le A\le \|v\|^r+A\le\lambda+C+A$. 
Therefore, if we let $\tilde C=C+\max(A,\bar\gamma^r)$, and then 
relabel $\tilde r$, $\tilde C$ as our new $r$ and $C$, we have shown that 
\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\Big((v,\lambda)\in\Phi_{\bar\gamma,\hat\Sigma}(0)\wedge
v'\in\R^n\wedge\|v'\|\le\bar\gamma\Big)\nonumber\\
&\,&\,\,\Rightarrow \|v+v'\|^r\le\lambda+C\,.\label{coer-ai}
\end{eqnarray}



For $0<\delta\le\bar\gamma$, we define
\begin{eqnarray*}
&\,&\Xi_\delta(0)\equaldef\Big\{(v+v',\lambda+\lambda'):\\
&\,&(v,\lambda)\in\Phi_{\delta,\hat\Sigma}(0), v'\in\R^n\,,\,
\|v'\|\le\bar\gamma\,,\,\lambda'\in\R\,,\,\lambda'\ge 0\Big\}\,.
\end{eqnarray*}
Then 
\begin{itemize}
\item[(a)]  {\em If $0<\delta\le\bar\gamma$, then $\Xi_{\delta}(0)$ is 
a closed, convex, subset of $\R^{n+1}$ with nonempty interior.}
\end{itemize}
(The fact that $\Xi_\delta(0)$ has a nonempty interior
follows because, if we pick any $(\hat v,\hat\lambda)\in
\Phi_{\delta,\hat\Sigma}(0)$, then 
$$
\Big\{(\hat v+v',\lambda):v'\in\R^n\,,\,
\|v'\|<\bar\gamma\,,\,\lambda>\hat\lambda\Big\}
\subseteq\Xi_{\delta}(0)\,.
$$
The fact that $\Xi_\delta(0)$ is closed follows because, if 
a \mbox{sequence} $\{(v_j+v_j',\lambda_j+\lambda_j')\}_{j=1}^\infty$
with the property that $(v_j,\lambda_j)\in\Phi_{\delta,\hat\Sigma}(0)$,
$v_j'\in\R^n$, $\|v_j'\|\le\bar\gamma\in\R^n$, 
$\lambda_j'\ge 0$ converges to a limit 
$(\hat v,\hat\lambda)$, then the sequence 
$\{v_j'\}_{j=1}^\infty$ is bounded, so we may assume
after passing to a \mbox{subsequence} that $\lim_{j\to\infty}v_j'=v'$
exists, and then of course $\|v'\|\le\bar\gamma\in\R^n$, 
and $\lim_{j\to\infty}v_j=v$ \mbox{exists} as well, since 
$v_j+v'_j\to \hat v$, and then $\hat v=v+v'$. Furthermore, the bound
\mbox{$\|v_j\|^r\le\lambda_j+C$}  implies 
that \mbox{$\lambda_j\ge\|v_j\|^r-C\ge -C$}, so 
$$
\lambda_j'=(\lambda_j+\lambda_j')-\lambda_j\le (\lambda_j+\lambda_j')+C\,.
$$
Hence the sequence
$\{\lambda_j'\}_{j=1}^\infty$ is bounded above,
because $\{\lambda_j+\lambda_j'\}_{j=1}^\infty$ is 
convergent. Since $\lambda_j'\ge 0$, the sequence
$\{\lambda_j'\}_{j=1}^\infty$ is bounded, so we may assume it is
convergent to a limit $\lambda'$, after passing to a subsequence.
Clearly, then, $\lambda'\ge 0$, and the limit
$\lim_{j\to\infty}\lambda_j=\lambda$ exists as well,
and satisfies $\hat\lambda=\lambda+\lambda'$. Since
$(v,\lambda)=\lim_{j\to\infty}(v_j,\lambda_j)$, 
$(v_j,\lambda_j)\in\Phi_{\delta,\hat\Sigma}(0)$, and 
$\Phi_{\delta,\hat\Sigma}(0)$ is closed, we see that 
$(v,\lambda)\in\Phi_{\delta,\hat\Sigma}(0)$. Since 
$\|v'\|\le\bar\gamma\in\R^n$, and $\lambda'\ge 0$, we see that
$(\hat v,\hat\lambda)\in\Xi_\delta(0)$. The 
convexity of $\Xi_\delta(0)$ is trivial.)

Let $\Lambda=\{0\}\times[\bar\gamma,+\infty\,[\,$, so 
$\Lambda\subseteq\R^{n+1}$.  We 
let $\Psi_{\delta}(0)$ be the convex hull of
$\Lambda\cup\Xi_\delta(0)$. We show that
\begin{itemize}
\item[(b)] {\em If $0<\delta\le\bar\gamma$ then
$\Psi_{\delta}(0)$ is a closed convex subset 
of $\R^{n+1}$ with nonempty interior.}
\item[(c)] {\em There exist real numbers $r,C$ such that 
$r>1$ and}
\begin{equation}\label{coer-ao}
\|v\|^r\le\lambda+C\quad{\rm whenever}\,\,\,
(v,\lambda)\in\Psi_{\bar\gamma}(0)\,.
\end{equation}
\item[(d)] {\em The inequality
\begin{equation}\label{traf-a}
V(hv)+h\lambda\ge h\bar\gamma
\end{equation} 
holds whenever $0<h\le\bar\gamma$, $(v,\lambda)\in\Psi_{\bar\gamma}(0)$,
and $h\|v\|\le\bar\gamma$.} 
\item[(e)] {\em $(0,\ell)\notin\Psi_{\bar\gamma}(0)$ whenever
$\ell<\bar\gamma$}
\end{itemize}
We will prove the above assertions in order, except for the statement
that $\Psi_{\delta}(0)$ is  closed if $0<\delta\le\bar\gamma$, which will 
be proved last.

The fact that $\Psi_{\delta}(0)$ is convex is trivial,
and the fact that $\Psi_{\delta}(0)$ has nonempty interior
follows from (a), \mbox{because} $\Xi_{\delta}(0)\subseteq\Psi_{\delta}(0)$.

To prove (c), we choose $r,C$ such that $r>1$, $C>0$, and 
(\ref{coer-ai}) holds,
and observe that (\ref{coer-ai}) trivially \mbox{implies} 
that the inequality $\|v\|^r\le\lambda+C$ is true 
\mbox{whenever} $(v,\lambda)\in\Xi_{\bar\gamma}(0)$.
Since $C>0$, the inequality is also 
true whenever $(v,\lambda)\in\Lambda$.
Hence  $\|v\|^r\le\lambda+C$ \mbox{whenever} $(v,\lambda)\in
\Xi_{\bar\gamma}(0)\cup\Lambda$, from which it follows that
$\|v\|^r\le\lambda+C$ whenever $(v,\lambda)\in\Psi_{\delta}(0)$,
since the function 
$(v,\lambda)\mapsto\|v\|^r-\lambda-C$ is convex.

To prove (d), we observe that (\ref{traf}) trivially 
\mbox{implies} that $V(hv)+h\lambda> h\bar\gamma$ whenever
$(v,\lambda)\in\Xi_{\bar\gamma}(0)$, \mbox{$0<h\le\bar h$},  
and $h\|v\|\le\bar\gamma$.  (Indeed, if 
$(v,\lambda)\in\Xi_{\bar\gamma}(0)$, \mbox{$0<h\le\bar h$},  
and $h\|v\|\le\bar\gamma$, then 
$(v,\lambda)=(\bar v,\bar\lambda)+(v',\lambda')$, with 
$(\bar v,\bar\lambda)\in\Phi_{\bar\gamma,\hat\Sigma}(0)$, 
$v'\in\R^n$, $\|v'\|\le\bar\gamma$, $\lambda'\in\R$, 
and $\lambda'\ge 0$. Then we can apply (\ref{traf}), with 
$\bar v$, $\bar\lambda$ in the roles of $v$, $\lambda$, and conclude that
$V(h(\bar v+v'))+h\bar\lambda>h\bar\gamma$, since 
$h\|\bar v+v'\|\le\bar\gamma$. Therefore $V(hv)+h\bar\lambda>h\bar\gamma$,
and then {\it a fortiori}  $V(hv)+h\lambda>h\bar\gamma$, since 
$\lambda=\bar\lambda+\lambda'$ and $\lambda'\ge 0$.)
Assume that $(v,\lambda)\in\Psi_{\bar\gamma}(0)$. Then we can write 
$(v,\lambda)=\alpha(v',\lambda')+(1-\alpha)(0,\ell)$, with
$(v',\lambda')\in\Xi_{\bar\gamma}(0)$, $\ell\ge\bar\gamma$,
and $0\le\alpha\le 1$. Let $h$ be such that
$0<h\le\bar\gamma$ and $h\|v\|\le\bar\gamma$. 
If $\alpha=0$, then $hv=0$ and
$\lambda=\ell\ge\bar\gamma$, so 
$V(hv)+h\lambda=V(0)+h\ell=h\ell\ge h\bar\gamma$.
If $\alpha>0$, define $\tilde h=\alpha h$. Then 
$0<\tilde h\le\bar\gamma$ and $\tilde h\|v'\|=\alpha h\|v'\|
=h\|v\|\le\bar\gamma$, since $v=\alpha v'$. Therefore 
$V(\tilde h v')+\tilde h\lambda'\ge\tilde h\bar\gamma$, since
$(v',\lambda')\in\Psi_{\bar\gamma}(0)$. On  the other hand,
$(1-\alpha)h\ell\ge(1-\alpha)h\bar\gamma$, since $\ell\ge\bar\gamma$.
Therefore 
\begin{eqnarray*}
V(hv)+h\lambda&=&V(h\alpha v')+h\alpha\lambda'+h(1-\alpha)\ell\\
&=&V(\tilde hv')+\tilde h\lambda'+h(1-\alpha)\ell\\
&\ge&\tilde h\bar\gamma+h(1-\alpha)\bar\gamma\\
&=&h\alpha\bar\gamma+h(1-\alpha)\bar\gamma\\
&=&h\bar\gamma\,,
\end{eqnarray*}
completing the proof of (d).

Statement (e) now follows easily: if $(0,\ell)\in\Psi_{\bar\gamma}(0)$, then
we can apply (d) taking $h=\bar\gamma$, $v=0$, $\lambda=\ell$, and conclude that
$\bar\gamma\ell\ge\bar\gamma^2$, so that $\ell\ge\bar\gamma$.

We now prove that $\Psi_{\delta}(0)$ is closed if $0<\delta\le\bar\gamma$.
Let $\{w_j\}_{j=1}^\infty$ be a sequence of points of
$\Psi_{\delta}(0)$ that converges to a limit
$w\in\R^{n+1}$. We will show that $w\in\Psi_{\delta}(0)$.
Let $w_j=\alpha_j(v_j,\lambda_j)+(1-\alpha_j)(0,\ell_j)$,
where $(v_j,\lambda_j)\in\Xi_\delta(0)$, $\ell_j\ge\bar\gamma$,
and $0\le\alpha_j\le 1$. Let $w=(v,\lambda)$. By 
\mbox{passing} to a
subsequence, if necessary, we may assume that the $\alpha_j$ converge to
a limit $\hat\alpha$. If the sequence $\{(v_j,\lambda_j)\}$ is bounded,
then we may pass to a subsequence and assume that 
$(\hat v,\hat\lambda)=\lim_{j\to\infty}(v_j,\lambda_j)$ exists.
Then $(\hat v,\hat\lambda)\in\Xi_\delta(0)$. Therefore
the limit $\lim_{j\to\infty}\alpha_j(v_j,\lambda_j)$ exists, so
$\mu=\lim_{j\to\infty}(1-\alpha_j)\ell_j$ exists as well, 
and $\mu\ge 0$. Clearly,
$w=\hat\alpha(\hat v,\hat\lambda)+(0,\mu)$. If $\hat\alpha=1$, then
$w=(\hat v,\hat\lambda)+(0,\mu)$, so $w\in\Xi_\delta(0)$---and {\it 
a \mbox{fortiori}} $w\in\Psi_\delta(0)$---because 
$(\hat v,\hat\lambda)\in\Xi_\delta(0)$ and $\mu\ge 0$.   If 
$\hat\alpha<1$, then 
$\ell=\lim_{j\to\infty}\ell_j$ exists and satisfies
$\ell=\frac{\mu}{1-\hat\alpha}$ and $\ell\ge\bar\gamma$. 
Then $w=\hat\alpha(\hat v,\hat\lambda)+(1-\hat\alpha)(0,\ell)$, 
and $(\hat v,\hat\lambda)$, $(0,\ell)$, belong to 
$\Xi_\delta(0)$ and $\Lambda$, respectively, so 
$w\in\Psi_\delta(0)$.  Now suppose that the sequence 
$\{(v_j,\lambda_j)\}$ is unbounded. Then (\ref{coer-ao}) implies that
$\{\lambda_j\}$ is unbounded. Since the $\lambda_j$ are bounded below,
we may assume, after passing to a subsequence, that 
$\lambda_j\to+\infty$ as  $j\to\infty$. On the other hand,
$\alpha_jv_j\to v$ and $\alpha_j\lambda_j+(1-\alpha_j)\ell_j
\to\lambda$. Since both sequences $\{\alpha_j\lambda_j\}$, 
$\{(1-\alpha_j)\ell_j\}$ are bounded \mbox{below},
we pass to a subsequence and assume that the limits
$\mu=\lim_{j\to\infty}(1-\alpha_j)\ell_j$ and 
$\nu=\lim_{j\to\infty}\alpha_j\lambda_j$ exist.
But then 
\begin{eqnarray*}
\alpha_j\|v_j\|\le\alpha_j(\lambda_j+C)^{1/r}&=&
\alpha_j\lambda_j^{1/r}\Big(1+\frac{C}{\lambda_j}\Big)^{1/r}\\
&=&\alpha_j\lambda_j\lambda_j^{1/r-1}\Big(1+\frac{C}{\lambda_j}\Big)^{1/r}\\
&\asjgoes& 0\,,
\end{eqnarray*}
since $\lambda_j\asjgoes+\infty$, $\alpha_j\lambda_j\asjgoes\nu$,
and $r>1$.  Then $v=\lim_{j\to\infty}\alpha_jv_j=0$.
This implies, in particular, that $\bar\gamma\alpha_j\|v_j\|\le\bar\gamma$ if 
$j$ is large enough. So we can apply (d) with $h=\bar\gamma\alpha_j$
and $(v_j,\lambda_j)$ in the role of $(v,\lambda)$, and conclude that 
$V(\bar\gamma\alpha_jv_j)+\bar\gamma\alpha_j\lambda_j\ge\bar\gamma\alpha_j\bar\gamma$.
On the other hand, $(1-\alpha_j)\ell_j\ge(1-\alpha_j)\bar\gamma$, because
$\ell_j\ge\bar\gamma$. Therefore
$$
V(\bar\gamma\alpha_jv_j)+\bar\gamma\alpha_j\lambda_j+\bar\gamma(1-\alpha_j)\ell_j
\ge\bar\gamma\alpha_j\bar\gamma+\bar\gamma(1-\alpha_j)\bar\gamma
=\bar\gamma^2
$$ 
for large enough $j$. If we let $j\to\infty$, and use the facts that 
$\alpha_jv_j\asjgoes 0$, $V$ is continuous, and $V(0)=0$,
we find that $\bar\gamma\lambda=\lim_{j\to\infty}0
(\bar\gamma\alpha_j\lambda_j+\bar\gamma(1-\alpha_j)\ell_j)
\ge\bar\gamma^2$, so $\lambda\ge\bar\gamma$. Therefore 
$w=(0,\lambda)$ and $\lambda\ge\bar\gamma$, so $w\in\Lambda$ 
and then $w\in\Psi_\delta(0)$.

We have now completed the proofs of (b), (c), (d) and (e). 
Since $\Psi_{\bar\gamma}(0)$ is closed and convex,
the fact that $(0,0)\notin\Psi_{\bar\gamma}(0)$ implies that there
exist a nonzero vector $p^*\in\R^{n+1}$ and a real number $\bar p$ such that 
$\bar p>0$ and $\langle p^*,w\rangle\ge\bar p$ whenever 
$w\in\Psi_{\bar\gamma}(0)$.  We fix $p^*$ and $\bar p$ from now on.

We now let $Q$ be the set of all vectors $q\in\R^{n+1}$ such that
$q=hw$ for some $h,w$ such that $h\in[0,\infty[$ and
$w\in\Psi_{\bar\gamma}(0)$, and write $\qint$ to denote the interior
of $Q$.  We will show that
\begin{itemize}
\item[(f)] {\em $Q$ is a closed convex cone with nonempty interior.}
\item[(g)] {\em $\langle p^*,q\rangle\ge 0$ for all $q\in Q$ and 
$\langle p^*,q\rangle>0$ for all $q\in\qint$.}
\item[(h)] $(0,0)\notin\qint$.
\end{itemize} 
Indeed, $Q$ is obviously a convex cone. The fact that $\qint\neq\emptyset$
follows from (b). To show that $Q$ is closed, we pick a sequence 
$\{q_j\}_{j=1}^\infty$of points of $Q$ that converges to a limit
$q\in\R^{n+1}$, and show that $q\in Q$. Write $q_j=h_jw_j$, $h_j\ge 0$,
$w_j\in\Psi_{\bar\gamma}(0)$.  If $q=(0,0)$ then $q\in Q$, so we may assume 
that $q\neq (0,0)$ and that $q_j\neq (0,0)$ for all $j$. Then $h_j\neq 0$ as well.
If the sequence $\{w_j\}_{j=1}^\infty$ is bounded, then we may pass to a subsequence
and assume that the $w_j$ converge to a limit $w$, which must belong to
$\Psi_{\bar\gamma}(0)$ because $\Psi_{\bar\gamma}(0)$ is closed. In particular,
$w$ and the $w_j$ are $\neq(0,0)$. But then
$h_j=\frac{\|q_j\|}{\|w_j\|}\asjgoes\frac{\|q\|}{\|w\|}\equaldef h$. Therefore
$q=hw$, so $q\in Q$. Now suppose that  
the sequence $\{w_j\}_{j=1}^\infty$ is unbounded.  Write 
$w_j=(v_j,\lambda_j)$, and use (c) to conclude that
$\|v_j\|^r\le\lambda_j+C$ for all $j$. Then 
the sequence $\{\lambda_j\}_{j=1}^\infty$ is unbounded, and we may assume,
after passing to a subsequence, that $\lambda_j\asjgoes+\infty$.
Since $q=\lim_{j\to\infty}(h_jv_j,h_j\lambda_j)$, the sequence
$\{h_j\lambda_j\}_{j=1}^\infty$ converges to a finite limit $\mu$,
so $h_j\asjgoes 0$. Then 
$h_j\|v_j\|\le h_j(\lambda_j+C)^{1/r}=
h_j\lambda_j\lambda_j^{1/r-1}\Big(1+\frac{C}{\lambda_j}\Big)^{1/r}\asjgoes 0$.
So $h_jv_j\asjgoes 0$. Therefore $q=(0,\mu)$, so 
$q=h(0,\bar\gamma)$, where $h=\frac{\mu}{\bar\gamma}$. 
Since $\mu\ge 0$ and $(0,\bar\gamma)\in\Psi_{\bar\gamma}(0)$, 
it is now clear that $q\in Q$. This completes the proof of (f).

The fact that $\langle p^*,q\rangle\ge 0$ for all $q\in Q$ follows trivially
from the definition of $Q$, since $\langle p^*,w\rangle\ge\bar p$ whenever 
$w\in\Psi_{\bar\gamma}(0)$. The fact that $\langle p^*,q\rangle>0$ for all 
$q\in\qint$ is then a consequence of the nontriviality of $p^*$. This proves
(g), and then (h) follows.

Next, we define a function $\sigma:Q\mapsto\R$ by letting $\sigma(q)$ be,
if $q\in Q$, the largest $h\in\R$ such that $q=hw$ for some 
$w\in\Psi_{\bar\gamma}(0)$. (The existence of such a largest
$h$ is trivial if $q=0$, for in that case the only possible value of
$h$ is $0$, since $(0,0)\notin\Psi_{\bar\gamma}(0)$. If $q\in Q$ and 
$q\neq 0$, let $H=\{h\in\R:h>0,\,h^{-1}q\in\Psi_{\bar\gamma}(0)\}$. Then 
$H$ must be bounded, for otherwise $(0,0)$ would be a limit of
points of $\Psi_{\bar\gamma}(0)\}$, and then 
$(0,0)$ would have to belong to $\Psi_{\bar\gamma}(0)$.
If $h=\sup\,H$, then the fact that $\Psi_{\bar\gamma}(0)$ is closed
implies that $h\in H$, so $\sigma(q)$ exists and is equal to $h$.)
We prove the following properties of $\sigma$.

\begin{itemize}
\item[(i)] {\em $\sigma$ is strictly positive on $Q\backslash\{0\}$.}

\vspace{-3mm}
\item[(j)] {\em $\sigma$ is positively homogeneous of degree $1$ (that is, $\sigma(q)\ge 0$ and 
$\sigma(rq)=r\sigma(q)$ whenever $q\in Q$ and $r\ge 0$).} 

\vspace{-3mm}
\item[(k)] {\em $\sigma$ is upper semicontinuous.}

\vspace{-3mm}
\item[(l)] {\em There exists a constant $\kappa\in\R$ such that
\mbox{$\sigma(q)\le\kappa\|q\|$} whenever $q\in Q$.}

\vspace{-3mm}
\item[(m)] {\em $\sigma$ is continuous at $(0,0)$.}

\vspace{-3mm}
\item[(n)] {\em $\sigma$ is concave on $Q$ (that is, the \mbox{inequality}
\begin{equation}\label{conca}
\sigma(\alpha q+(1-\alpha)q')\ge \alpha\sigma(q)+(1-\alpha)\sigma(q')
\end{equation}

\item[] holds whenever $q,q'\in Q$ and $0\le\alpha\le 1$).}

\vspace{-3mm}
\item[(o)] {\em $\sigma(\alpha q+\alpha'q')\ge \alpha\sigma(q)+
\alpha'\sigma(q')$ whenever $q,q'\in Q$ and $\alpha,\alpha'\in[0,\infty[$.}

\vspace{-3mm}
\item[(p)] {\em $\sigma$ is locally Lipschitz on $\qint$.}
\end{itemize} 
Statements (i) and (j) are immediate consequences of the definition
of $\sigma$. To show that  $\sigma$ is upper \mbox{semicontinuous},  
we assume that $\{q_j\}_{j=1}^\infty$ is a sequence of members
of $Q$ that converges to a limit $q\in Q$ and $\alpha\in\R$ is such that
$\sigma(q_j)\ge\alpha$ for all $j$, and prove that 
$\sigma(q)\ge\alpha$.   By passing to a subsequence, we may assume
that the limit $L=\lim_{j\to\infty}\sigma(q_j)$ exists
in $[0,\infty]$, and it is then clear that $L\ge\alpha$.
The conclusion that $\sigma(q)\ge\alpha$ is trivial
if $\alpha\le 0$, so we may assume that $\alpha>0$. Write
$q_j=\sigma(q_j)w_j$, $w_j\in\Psi_{\bar\gamma}(0)$. Then
$w_j=\frac{q_j}{\sigma(q_j)}$. If $L=+\infty$ then 
$\lim_{j\to\infty}w_j=0$, so $0$ (that is, the zero
point $(0,0)$ of $\R^n\times\R$) would belong to $\Psi_{\bar\gamma}(0)$,
since $\Psi_{\bar\gamma}(0)$ is closed.  Since we know that
$(0,0)\notin\Psi_{\bar\gamma}(0)$, $L$ has to be finite. 
Since $L\ge\alpha$ and we are assuming that $\alpha>0$, $L$ must be
strictly positive. Then, if we let $w=\frac{q}{L}$, we have 
$w=\lim_{j\to\infty}\frac{q_j}{\sigma(q_j)}=\lim_{j\to\infty}w_j$.
Therefore $w\in\Psi_{\bar\gamma}(0)$.  Since 
$q=Lw$, the definition of $\sigma$ implies that $\sigma(q)\ge L\ge\alpha$, concluding
the proof of (k).

Let $K=\{q\in Q:\|q\|=1\}$. Then $K$ is compact. 
Since $\sigma$ is upper semicontinuous,  $\sigma$ is bounded above 
on $K$. Let $\kappa$ be such that $\sigma(q)\le\kappa$ whenever
$q\in K$. Then the homogeneity of $\sigma$ implies that
$\sigma(q)\le\kappa\|q\|$ whenever $q\in Q$. This proves (l), and then
(m) follows immediately.

Let us prove the concavity of $\sigma$. We must show that 
$\sigma(\alpha q+(1-\alpha)q')\ge \alpha\sigma(q)+(1-\alpha)\sigma(q')$ 
whenever $q,q'$ belong to $Q$ and 
$0\le\alpha\le 1$.  The conclusion is trivial if
$q=0$, $q'=0$, $r=0$, or $r=1$. So let us suppose that
$q,q'\in Q$, $q\neq 0\neq q'$, and $0<r<1$. Then
$\sigma(q)>0$, $\sigma(q')>0$,
$q=\sigma(q)w$ and $q'=\sigma(q')w'$ for some
$w,w'\in\Psi_{\bar\gamma}(0)$.  It follows that
\begin{eqnarray*}
\alpha q+(1-\alpha)q'&=&\alpha\sigma(q)w+(1-\alpha)\sigma(q')w'\\
&=&h\tilde w\,,
\end{eqnarray*}
where 
$$
h\!=\!\alpha\sigma(q)+(1-\alpha)\sigma(q')\,,\,\,
\tilde w\!=\!\frac{\alpha\sigma(q)w\!+\!(1\!-\!\alpha)\sigma(q')w'}{h}\,.
$$
The convexity of $\Psi_{\bar\gamma}(0)$ then implies that 
\mbox{$\tilde w\in\Psi_{\bar\gamma}(0)$}. It then follows from the definition of $\sigma$ that 
\mbox{$\sigma(\alpha q+(1-\alpha)q')\ge h$}, as desired.

Finally, statement (o) follows trivially from the 
\mbox{concavity} and the
homogeneity of $\sigma$, and statement (p) is a consequence of the
well-known fact that a concave function on an open convex set is
locally Lipschitz.  This completes the proof of (f), (g), (h), (i),
(j), (k), (l), (m), (n), (o), and (p).

We now define a function $\theta:Q\mapsto\R$ by letting 
$$
\theta(q)=\bar\gamma\sigma(q)\,.
$$
Then $\theta$ is nonnegative, positively homogeneous of \mbox{degree}
$1$, concave, upper semicontinuous, and continuous at $(0,0)$, and satisfies
the bound $\theta(q)\le\kappa\bar\gamma\|q\|$ for all $q\in Q$.  In 
particular,  (o) implies that 
\begin{eqnarray}
\theta(\alpha q+\alpha'q')&\ge&\alpha\theta(q)+\alpha'\theta(q')\nonumber\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!{\rm whenever}\quad
q,q'\in Q\,\,\,{\rm and}\,\,\,\alpha,\alpha'\in[0,\infty[\,.
\label{conchomth}
\end{eqnarray}
Next, we define 
\begin{eqnarray*}
\gamma^*&\equaldef&\frac{\bar\gamma}{\max(1,\kappa)}\\
Q^*&\equaldef&\Big\{q\in Q:\|q\|\le\gamma^*\Big\}\,.
\end{eqnarray*}
It then follows immediately that the interior $\qints$ of $Q^*$ is given by
$$
\qints=\Big\{q\in\qint:\|q\|<\gamma^*\Big\}\,.
$$
Let us define a function  $W:\Omega\times\R\mapsto\R$ by letting
$W(x,x_0)=V(x)+x_0$ for $(x,x_0)\in\Omega\times\R$. We then claim that
\begin{equation}\label{traf-i}
W(q)\ge\theta(q)\quad
{\rm whenever}\,\,\,q\in Q^*\,.
\end{equation} 
To prove (\ref{traf-i}), we first observe that the inequality is clearly true
if $q=0$. Let us pick $q\in Q^*\backslash\{0\}$ and write 
$q=hw$, where $h=\sigma(q)$ and
\mbox{$w=(v,\lambda)\in \Psi_{\bar\gamma}(0)$.}
Then $0<h$, since $q\neq 0$, and $h\le\bar\gamma$, 
\mbox{because} \mbox{$\sigma(q)\le\kappa\|q\|\le\kappa\gamma^*\le\bar\gamma$.}
Furthermore, $\|q\|\le\bar\gamma$ (because $\gamma^*\le\bar\gamma$),
and $q=(hv,h\lambda)$, so 
$h\|v\|\le\bar\gamma$. It then follows from (\ref{traf}) that
$$
W(q)=V(hv)+h\lambda\ge h\bar\gamma=\bar\gamma\sigma(q)=\theta(q)\,,
$$
completing the proof of (\ref{traf-i}).

\medbreak
We now fix a point $q^{\#}\in\qint$ such that
$\sigma(q^{\#})=1$. We then define, for each positive real 
number $s$, a subset $Q_s$ of $Q$, by letting
$Q_s=\{q+sq^{\#}:q\in Q\}$. Then $Q_s$ is a closed
convex set, because it is a translate of $Q$. Moreover, it is clear that
$Q_s\subseteq\qint$. 

If $0<a<b<c<\infty$, define a function $\eta_{\,a,b,c}:\R\mapsto\R$
by letting 
$$
\eta_{\,a,b,c}(t)=\left\{\begin{array}{lcl}
t+b-a&{\rm if}&t\le a\,,\\
\frac{t(c-b)+c(b-a)}{c-a})&{\rm if}&a\le t\le c\,,\\
t&{\rm if}&c\le t\,.\end{array}\right.
$$
Then $\eta_{\,a,b,c}$ is strictly increasing, continuous,
and piecewise linear. Furthermore,, 
$$
\frac{d\eta_{\,a,b,c}}{dt\phantom{_{\,a,b,c}}}(t)=\left\{\begin{array}{lcl}
1&{\rm if}&t\le a\,\,\,\,{\rm or}\,\,\,\,t\ge c,\\
\frac{c-b}{c-a}&{\rm if}&a\le t\le c\,.\end{array}\right.
$$
If $s$, $\eps$, $\rho$, $\zeta$ are positive numbers such that 
$0<\eps<\rho$, we define 
a function $\theta_{s,\eps,\rho,\zeta}:Q\mapsto\R$ by letting 
$$
\theta_{s,\eps,\rho,\zeta}(q)=\qquad\qquad\qquad\qquad\qquad\qquad\,\,
\qquad\qquad\qquad
$$
$$\left\{\begin{array}{lcl}
\theta(q)&{\rm if}&q\in Q\backslash Q_{s}\,,\\
\max\Big(\theta(q),
\eta_{\,0,\eps,\rho}\big(\zeta .\langle p^*,q\rangle\big)\Big)
&{\rm if}&q\in Q\cap Q_{s}\,,\end{array}\right.
$$
where $p^*$ is the vector introduced earlier.
Then $\theta_{s,\eps,\rho,\zeta}$ need not be continuous. However,
\begin{itemize}
\item[(q)] {\em $\theta_{s,\eps,\rho,\zeta}$ is upper semicontinuous on $Q$.}
\end{itemize}
(Indeed, if $\alpha\!\in\!\R$ and $q\!\in\!Q$, then 
$\theta_{s,\eps,\rho,\zeta}(q)\ge\alpha$ if and only if 
either (i)~$\theta(q)\ge\alpha$ or (ii)~~$q\in Q_s$ and 
\mbox{$\eta_{\,0,\eps,\rho}\big(\zeta.\langle p^*,q\rangle\big)\ge\alpha$}. 
Suppose $q_j\in Q$, $q_j\to q$, and 
$\theta_{s,\eps,\rho,\zeta}(q_j)\ge\alpha$. Let 
$J=\{j\in\nn:\theta_{s,\eps,\rho,\zeta}(q_j)=\theta(q_j)\}$. 
If $J$ is infinite, then $\theta(q_j)\ge\alpha$ for $j\in J$, so 
$\theta(q)\ge\alpha$, and then 
$\theta_{s,\eps,\rho,\zeta}(q)\ge\alpha$. If $J$ is finite and 
$I=\nn\backslash J$, then $q_j\in Q_s$ and
$\eta_{\,0,\eps,\rho}\big(\zeta .\langle p^*,q_j\rangle\big)=
\theta_{s,\eps,\rho,\zeta}(q_j)\ge\alpha$ for $j\in I$, so $q\in Q_s$ and 
$\eta_{\,0,\eps,\rho}\big(\zeta .\langle p^*,q\rangle\big)\ge\alpha$,
from which it follows that 
$\theta_{s,\eps,\rho,\zeta}(q)\ge\alpha$.)

Next, we define $\tilde\gamma\equaldef\min\Big(\,\bar\gamma,\bar p\zeta
\Big(1-\frac{\eps}{\rho}\Big)\,\Big)$, and prove that
\begin{itemize}
\item[(r)] {\em $\theta_{s,\eps,\rho}(q+uw)\ge\theta_{s,\eps,\rho}(q)+u\tilde\gamma$
whenever $q\in Q$, $w\in\Psi_{\bar\gamma}(0)$, and $u\ge 0$.}
\end{itemize}
Fix $q\in Q$,  $w\in\Psi_{\bar\gamma}(0)$ and $u\in[0,\infty[$. If both 
$q$ and $q+uw$ belong to $Q\backslash Q_s$, then 
\begin{eqnarray*}
\theta_{s,\eps,\rho}(q+uw)&=&\theta(q+uw)\\
&\ge&\theta(q)+u\theta(w)\\
&=&\theta(q)+u\sigma(w)\bar\gamma\\
&\ge&\theta(q)+u\bar\gamma\\
&=&\theta_{s,\eps,\rho}(q)+u\bar\gamma\\
&\ge&\theta_{s,\eps,\rho}(q)+u\tilde\gamma\,,
\end{eqnarray*}
where we have used (\ref{conchomth}) and the trivial fact that 
\mbox{$\sigma(w)\ge 1$} whenever $w\in\Psi_{\bar\gamma}(0)$. 

Now suppose that both $q$ and $q+uw$ belong to $Q_s$. If $\theta(q)\ge 
\eta_{\,0,\eps,\rho}\big(\zeta.\langle p^*,q\rangle\big)$, then 
$\theta_{s,\eps,\rho,\zeta}(q)=\theta(q)$, and then
\begin{eqnarray*}
\theta_{s,\eps,\rho,\zeta}(q+uw)&\ge&\theta(q+uw)\\
&\ge&\theta(q)+u\bar\gamma\\
&=&\theta_{s,\eps,\rho,\zeta}(q)+u\bar\gamma\\
&\ge&\theta_{s,\eps,\rho,\zeta}(q)+u\tilde\gamma\,.
\end{eqnarray*}
If, on the other hand, $\eta_{0,\eps,\rho}\big(\zeta.\langle p^*,q\rangle\big)
\ge\theta(q)$, then 
\begin{eqnarray*}
\theta_{s,\eps,\rho,\zeta}(q+uw)&\ge&\eta_{\,0,\eps,\rho}\big(\zeta.\langle 
p^*,q+uw\rangle\big)\\
&\ge&\eta_{\,0,\eps,\rho}\big(\zeta.\langle p^*,q\rangle\big)
+u\zeta.\langle p^*,w\rangle\frac{\rho-\eps}{\rho}\\
&\ge&\theta_{s,\eps,\rho,\zeta}(q)+u\bar p\zeta\Big(1-\frac{\eps}{\rho}\Big)\\
&\ge&\theta_{s,\eps,\rho,\zeta}(q)+u\tilde\gamma\,,
\end{eqnarray*}
so $\theta_{s,\eps,\rho,\zeta}(q+uw)\ge\theta_{s,\eps,\rho,\zeta}(q)+u\tilde\gamma$.

\medbreak
Finally, we consider the case when one of the points $q$, $q+uw$ 
belongs to $Q_s$, while the other one is in $Q\backslash Q_s$. Then 
$q\in Q\backslash Q_s$ and $q+uw\in Q_s$, because the definition
of $Q_s$ clearly implies that $Q_s+Q\subseteq Q_s$,
so that if $q\in Q_s$ it would follow that
$q+uw\in Q_r$ as well.  Then 
$$
\theta_{s,\eps,\rho,\zeta}(q+uw)\ge\theta(q+uw)\ge\theta(q)+u\bar\gamma\ge
\theta(q)+u\tilde\gamma\,,
$$
completing the proof of (r).

We now observe that, if we let
$$
\mu\equaldef\min\{W(q):q\in Q\,,\,\|q\|=\gamma^*\}\,,
$$
then $\mu$ exists (because $W$ is continuous and the set 
\mbox{$\{q\in Q:\|q\|=\gamma^*\}$} is compact) and $\mu>0$ (because
$W(q)\ge\theta(q)>0$ whenever $q\in Q$ and $\|q\|=\gamma^*$). Also, we
let 
$$
\nu=\max\{\langle p,q\rangle:q\in Q\,,\,\|q\|=\gamma^*\}\,.
$$
We then choose $\zeta$ and $\rho$ such that 
$$
\zeta\nu<\mu\quad{\rm and}\quad\rho=\zeta\nu\,.
$$
With these choices, we will prove that
\begin{itemize}
\item[(s)] {\em $W(q)\ge\theta_{s,\eps,\rho,\zeta}(q)$ whenever $q$ 
belongs to the \mbox{boundary} of $Q^*$.}
\end{itemize}
Indeed, the conclusion is true if $q\in Q\backslash Q_s$, because in that case
$\theta_{s,\eps,\rho,\zeta}(q)=\theta(q)$. Assume that $q\in Q_s$. Then
$\|q\|=\gamma^*$, so $\langle p^*,q\rangle\le\nu$, 
$\zeta.\langle p^*,q\rangle\le\zeta\nu=\rho$, and then 
$$
\eta_{\,0,\eps,\rho}\big(\zeta.\langle p^*,q\rangle\big)
\le\eta_{\,0,\eps,\rho}(\rho)=\rho<\mu\le W(q)\,,
$$
while on the other hand (\ref{traf-i}) tells us that 
$\theta(q)\le W(q)$ as well, and then  
$$
\theta_{s,\eps,\rho,\zeta}(q)=\max\Big(\theta(q),
\eta_{\,0,\eps,\rho}\big(\zeta.\langle p^*,q\rangle\big)\Big)\le W(q)\,.
$$

We now choose $s$ such that 
$$
0<s<\gamma^*\quad{\rm and}\quad W(sq^{\#})+s<\rho
$$
(this is possible because $W(sq^{\#})+s\to 0$ as $j\downarrow 0$), and then
let 
$$
\eps= W(sq^{\#})+s\,.
$$
Then $0<\eps<\rho$. We then let $\Theta$ be the 
function $\theta_{s,\eps,\rho,\zeta}$ corresponding to our choices of
$s$, $\eps$, $\rho$ and $\zeta$. It then follows from (r) that 
\begin{itemize}
\item[(t)] {\em $\Theta(q+uw)\ge\Theta(q)+u\tilde\gamma$
whenever $q\in Q$, $w\in\Psi_{\bar\gamma}(0)$, and $u\ge 0$.}
\end{itemize}

We claim that
\begin{itemize}
\item[(u)] {\em The function $Q^*\ni q\mapsto{\cal W}(q)\equaldef W(q)-
\Theta(q)$ attains a global minimum at some point $q\in\qints$.}
\end{itemize}
(Indeed, ${\cal W}$ is lower semicontinuous on $Q$, since 
$W$ is continuous and $\Theta$ is upper semicontinuous. So ${\cal W}$ 
\mbox{attains} a minimum value on the compact set $Q^*$. 
\mbox{Furthermore}, 
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!{\cal W}(sq^{\#})\\
&=&W(sq^{\#})-\max\Big(\theta(sq^{\#}),
\eta_{\,0,\eps,\rho}\big(\zeta.\langle p^*,sq^{\#}\rangle\big)\Big)\\
&=&\min\Big(W(sq^{\#})-\theta(sq^{\#}),W(sq^{\#})\\
&\,&\,\,-\eta_{\,0,\eps,\rho}\big(\zeta.\langle p^*,sq^{\#}
\rangle\big)\Big)\\
&\le&W(sq^{\#})-\eta_{\,0,\eps,\rho}\big(\zeta.\langle p^*,sq^{\#}
\rangle\big)\\
&\le&W(sq^{\#})-\eta_{\,0,\eps,\rho}(0)\\
&=&W(sq^{\#})-\eps\\
&=&-s\,.
\end{eqnarray*}
Therefore ${\cal W}(sq^{\#})<0$. In view of (s), the minimum value of 
${\cal W}$ must be attained at an interior point of $Q^*$, and the 
proof of (u) is complete.)

Our next step is to regularize $\Theta$, and show that (t)
and (u) remain true for the regularizations $\Theta^\alpha$,
provided that the regularization parameter $\alpha$
is small enough. This has to be done with
care because $\Theta$ is not known to be continuous, so the
regularizations $\Theta^\alpha$ will not \mbox{necessarily} converge to
$\Theta$ uniformly on compact sets.

We fix a nonnegative compactly supported function
$\psi:\R^{n+1}\mapsto\R$ of class $C^\infty$ such that
$\int_{\R^{n+1}}\psi=1$ and $\,{\rm support}(\psi)\subseteq\{q\in
Q:\|q\|\le 1\}$.  We then define, for every positive real number
$\alpha$, a function \mbox{$\Theta^\alpha:Q\mapsto\R$} by letting
$$
\Theta^\alpha(q)=\int_{\R^{n+1}}\Theta(q+\alpha y)\psi(y)\,dy
$$
for $q\in Q$.  Then $\Theta^\alpha$ is continuous on $Q$ and of 
class $C^\infty$ on $\qint$ whenever $\alpha>0$.  We also define the
\mbox{regularizations} $W^\alpha$ of $W$ by letting
$$
W^\alpha(q)=\int_{\R^{n+1}}W(q+\alpha y)\psi(y)\,dy\,.
$$

\medbreak\noindent Furthermore, for each $\alpha$, we let

\medbreak
$\omega_1(\alpha)$$=$$\max\{\Theta(q)\!-\!W(q)\!:\!q\!\in\!Q,\,
d(q,Q^*\backslash\qints)\!\le\!\alpha\}\,$

$\omega_2(\alpha)$$=$$\max\{W^\alpha(q)-W(q):q\in Q^*\backslash\qints\}$

\medbreak\noindent (The maxima exist because $\Theta-W$ is upper 
\mbox{semicontinuous} and $\Theta^\alpha-W^\alpha$ is continuous.)  
It is clear that
\begin{itemize}
\item[(v)] {\em $W(q)-\Theta^\alpha(q)\ge-\omega_1(\alpha)-\omega_2(\alpha)$ whenever 
$\alpha>0$ and $q\in Q^*\backslash\qints$.}
\end{itemize}
(Indeed, 
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
W^\alpha(q)-\Theta^\alpha(q)\\
&=&\int_{\R^{n+1}}(W(q+\alpha y)-\Theta(q+\alpha y))\psi(y)\,dy\\
&\ge&-\omega_1(\alpha)\int_{\R^{n+1}}\psi(y)\,dy\\
&=&-\omega_1(\alpha)\,.
\end{eqnarray*}
Therefore
\begin{eqnarray*}
W(q)-\Theta^\alpha(q)&\ge&W(q)-W^\alpha(q)+W^\alpha(q)-\Theta^\alpha(q)\\
&\ge&-\omega_2(\alpha)-\omega_1(\alpha)\,,
\end{eqnarray*}
as desired.)  The numbers $\omega_2(\alpha)$ go to
$0$ as $\alpha\downarrow 0$,
\mbox{because} $W^\alpha\to W$ uniformly on compact sets.
On the other hand $\lim\sup_{\alpha\downarrow 0}\omega_1(\alpha)\le 0$. 
(Otherwise there would exist $\alpha_j$, $q_j$, $\beta$, such that 
$\beta>0$, $\alpha_j\downarrow 0$, 
$q_j\in Q$, $d(q_j,Q^*\backslash\qints)\le\alpha_j\}$, and 
$\Theta(q_j)-W(q_j)\ge\beta$. By \mbox{passing}
to a subsequence, we may assume that the
$q_j$ converge to a limit $q\in Q^*\backslash\qints$, and then the upper
semicontinuity  of $\Theta-W$ implies that $\Theta(q)-W(q)\ge\beta$, 
contradicting (s).) Therefore 
\begin{equation}\label{limofreg}
\liminf_{\alpha\downarrow 0}\Big(\min\Big\{W(q)\!-\!\Theta^\alpha(q)\!:\!
q\!\in\! Q^*\backslash\qints\Big\}\Big)\!\ge\!0.
\end{equation}

The value of $\Theta^\alpha$ at the point 
$sq^{\#}$ can be estimated from below
using the facts that $\psi$ is supported in $Q$ and 
$\Theta(sq^{\#}+\alpha y)\ge\eta_{\,0,\eps,\rho}(
\zeta.\langle p^*,sq^{\#}+\alpha y\rangle$ whenever $y\in Q$, because 
in that case $sq^{\#}+\alpha y$ belongs to $Q_s$. Since 
$\zeta.\langle p^*,sq^{\#}+\alpha y\rangle\ge 0$, we have
$$
\eta_{\,0,\eps,\rho}(\zeta.\langle p^*,sq^{\#}+\alpha y\rangle
\ge\eta_{\,0,\eps,\rho}(0)=\eps\,.
$$
Therefore 
\begin{equation}\label{atpoint}
\Theta^\alpha(sq^{\#})\ge\eps=
W(sq^{\#})+s\quad{\rm whenever}\quad\alpha>0\,,
\end{equation}
so that 
\begin{equation}\label{atpoint-a}
W(sq^{\#})-\Theta^\alpha(sq^{\#})\le-s\quad{\rm whenever}\quad\alpha>0\,.
\end{equation}
It follows from (\ref{limofreg}) that we can choose $\alpha$ such that 
\begin{equation}\label{atpoint-b}
W(q)-\Theta^\alpha(q)\ge-\frac{s}{2}\quad{\rm whenever}\quad \,
q\in Q^*\backslash\qints\,.
\end{equation}

Then (\ref{atpoint-a}) and (\ref{atpoint-b}) imply that
\begin{itemize}
\item[(w)] {\em The function $Q^*\ni q\mapsto{\cal W}^\alpha(q)\equaldef W(q)-
\Theta^\alpha(q)$ attains a global minimum at some point $q\in\qints$.}
\end{itemize}

Furthermore, we claim that 
\begin{itemize}
\item[(x)] {\em $\Theta^\alpha(q+uw)\ge\Theta^\alpha(q)+u\tilde\gamma$
whenever $q\in Q$, \mbox{$w\in\Psi_{\bar\gamma}(0)$}, and $u\ge 0$.}
\end{itemize}
To prove this, we write
\begin{eqnarray*}
\Theta^\alpha(q+uw)&=&\int_{\R^{n+1}}\Theta(q+uw+\alpha y)\psi(y)\,dy\\
&\ge&\int_{\R^{n+1}}\Big(\Theta(q+\alpha y)+u\tilde\gamma\Big)
\psi(y)\,dy\\
&=&\Theta^\alpha(q)+u\tilde\gamma\int_{\R^{n+1}}\psi(y)\,dy\\
&=&\Theta^\alpha(q)+u\tilde\gamma\,,
\end{eqnarray*}
where we have used (t) and the fact that 
$q+\alpha y\in Q$ whenever $\psi(y)\neq 0$ to conclude that
the inequality
\mbox{$\Theta(q+uw+\alpha y)\ge\Theta(q+\alpha y)+u\tilde\gamma$}
holds for all such $y$.

Now let $\hat q=(\hat x,\hat x_0)$ be a point of $Q^*$ where 
${\cal W}^\alpha$ has a global
minimum.  Since $\Theta^\alpha$ is everywhere differentiable,
we can let $(\pi,\pi_0)$ be the differential of
$\Theta^\alpha$ at $\hat q$. Then, if we write $q=(x,x_0)$, we have

\begin{equation}\label{ifdif}
\lim_{q\to\hat q}
\frac{\Theta^\alpha(q)-\Theta^\alpha(\hat q)-\pi\cdot(x-\hat x)
-\pi_0(x_0-\hat x_0)}{\|x-\hat x\|+|x_0-\hat x_0|}=0\,.
\end{equation}

\medbreak\noindent Taking $q=(x,\hat x_0)$, this implies 

$$
\lim_{x\to\hat x}
\frac{\Theta^\alpha(x,\hat x_0)-\Theta^\alpha
(\hat x,\hat x_0)-\pi\cdot(x-\hat x)}{\|x-\hat x\|}=0\,.
$$
Since 


\begin{eqnarray*}
\Theta^\alpha(x,\hat x_0)-\Theta^\alpha(\hat x,\hat x_0)
&\ge& W(x,\hat x_0)-W(\hat x,\hat x_0)\\
&=&V(x)-V(\hat x)\,,
\end{eqnarray*}
we conclude that 

$$
\lim\inf_{x\to\hat x}
\frac{V(x)-V(\hat x)-\pi\cdot(x-\hat x)}{\|x-\hat x\|}\ge 0\,,
$$

\medbreak\noindent so $\pi$ is a subdifferential of $V$ at $\hat x$. 

Taking  $q=(\hat x,x_0)$, (\ref{ifdif}) implies 

$$
\lim_{x_0\to\hat x_0}
\frac{\Theta^\alpha(\hat x,x_0)-\Theta^\alpha
(\hat x,\hat x_0)-\pi_0(x_0-\hat x_0)}{|x_0-\hat x_0|}=0\,.
$$
Since 

\begin{eqnarray*}
\Theta^\alpha(\hat x,x_0)-\Theta^\alpha(\hat x,\hat x_0)
&\ge& W(\hat x,x_0)-W(\hat x,\hat x_0)\\
&=&x_0-\hat x_0\,,
\end{eqnarray*}
we conclude that 

$$
\lim\inf_{x\to\hat x}\frac{x_0-\hat x_0-\pi_0(x_0-\hat x_0)}
{|x_0-\hat x_0|}\ge 0\,,
$$

\medbreak\noindent so $\pi_0=1$.

Now, if $u\in U$, then 
$w=(f(\hat x,u),L(\hat x,u))\in\Phi_{\bar\gamma,\hat\Sigma}(0)$, so 
$w\in\Psi_{\bar\gamma}(0)$, and then 
$\Theta^\alpha(\hat q+\alpha w)\ge\Theta(\hat q)+\alpha\tilde\gamma$
for all positive $\alpha$. It follows that
\begin{eqnarray*}
\pi\cdot f(\hat x,u)+L(\hat x,u)&=&
\lim_{\alpha\downarrow 0}\frac{
\Theta^\alpha(\hat q+\alpha w)-\Theta^\alpha(\hat q)}{\alpha}\\
&\ge&\lim\sup_{\alpha\downarrow 0}\frac{\alpha\tilde\gamma}{\alpha}\\
&=&\tilde\gamma\,.
\end{eqnarray*}
Since this is true for every $u\in U$, we see that

$$
\sup\{-\pi\cdot f(\hat x,u)-L(\hat x,u):u\in U\}\le-\tilde\gamma\,.
$$

\medbreak\noindent Since $\pi$ is a subdifferential of $V$ at $\hat x$, 
this contradicts our hypothesis that $V$ satisfies (\ref{loweq}) in the 
viscosity sense, and completes our proof.
\end{proof}	

\ssection{The compactness theorem.}

\noindent If \mbox{$\hat\Sigma=(\Omega,U,f,L)$} is an  augmented control
system, 
and $\eps$ is a positive \mbox{number}, an {\em $\eps$-approximate
\mbox{augmented} \mbox{trajectory} of $\hat\Sigma$} is a locally absolutely 
\mbox{continuous} curve 
\mbox{$I\ni t\mapsto\Xi(t)=(\xi(t),\xi_0(t))\in\R^{n+1}$}
having the property that there exists a measurable function 
\mbox{$I\ni t\mapsto v(t)\in\R^{n}$} such that 

\begin{itemize}
\item[(1)] $\|v(t)\|\le\eps$ for almost all $t\in I$, 

\item[(2)] $\dot\Xi(t)-(v(t),0)\in\fsh(\xi(t),U)$ 
for almost all $t\in I$.
\end{itemize}

\begin{remark}
Roughly speaking, an $\eps$-approximate
\mbox{augmented} \mbox{trajectory} of $\hat\Sigma$ is an augmented
\mbox{trajectory} of the ``$\eps$-extended system'' $\hat\Sigma^{\eps}
=(\Omega,U^{\eps},f^{\eps},L^{\eps})$ whose control space 
$U^{\eps}$ is the Cartesian product 
\mbox{$U\times\{v\in\R^{n}:\|v\|\le\eps\}$},
and whose dynamics $f^{\eps}$ and Lagrangian $L^{\eps}$ 
are given by 

$$
f^{\eps}(x,u,v)=f(x,u)+v\,,\quad L^{\eps}(x,u,v)=L(x,u)\,. 
$$

\medbreak\noindent More precisely, a curve 
\mbox{$\Xi=(\xi,\xi_0):I\mapsto\R^{n+1}$ is an} \mbox{$\eps$-approximate}
\mbox{augmented} \mbox{trajectory} of $\hat\Sigma$ if and only if 
$\,\Xi$ is locally absolutely \mbox{continuous} and there exist functions 
\mbox{$I\ni t\mapsto\eta(t)\in U$}, 
\mbox{$I\ni t\mapsto v(t)\in\{v\in\R^{n}:\|v\|\le\eps\}$}
such that $v$ is measurable and $\dot\Xi(t)\in
F_{\hat\Sigma^{\eps}}(\xi(t),U^{\eps})$ for almost every $t\in I$.
The definition of an augmented trajectory of $\hat\Sigma^{\eps}$ is
exactly the same, except that in that case the requirement
that $v$ be measurable is omitted.\proofend
\end{remark}

\begin{theorem}\label{compacttheo}
Let $\hat\Sigma=(\Omega,U,f,L)$ be an $n$-dimensional locally
uniformly continuous, locally coercive \mbox{augmented} control system such
that $\fsh(x,U)$ is closed and convex for every $x\in\Omega$.  Let $K$
be a compact subset of $\Omega$, let $T$ be a positive time, and let
$k\in\R$. Let $\{\eps_j\}_{j=1}^\infty$ be a 
\mbox{sequence} of positive
numbers such that $\eps_j\to 0$ as $j\to\infty$, and let
$\Xi^j=(\xi^j,\xi_{0}^j)$ be, for each $j$, an \mbox{$\eps_j$-approximate}
\mbox{augmented} trajectory of $\hat\Sigma$, defined on the interval $[0,T]$,
such that $\xi^{j}(t)\in K$ for all $t\in[0,T]$, $\xi_{0}^j(0)=0$,  
and $\xi_{0}^j(T)\le k$. Then there exist a \mbox{subsequence}
$\{\Xi^{j(k)}\}_{k=1}^\infty$ of the  \mbox{sequence}
$\{\Xi^j\}_{j=1}^\infty$ and an augmented
trajectory $\Xi^\infty=(\xi^\infty,\xi_{0}^\infty)$ of 
$\hat\Sigma=(\Omega,U,f,L)$ such that
\begin{itemize}
\item[(i)] $\xi_0^\infty(0)=0$, 
\item[(ii)] the sequence $\{\xi^{j(k)}\}_{k=1}^\infty$
converges uniformly to $\xi^\infty$, and 
\item[(iii)] 
$\liminf_{k\to\infty}\inf\big\{\Delta^{j(k),\infty}(t,s)\!:\!
0\!\le\!s\!\le\!t\!\le\!T\big\}\!\ge\!0$,
where, if $j,j'\in\nn\cup\{+\infty\}$, we define
\end{itemize}

\vspace{-2.5mm}
\begin{equation}\label{defodel}
\Delta^{j,j'}(t,s)\equaldef(\xi_0^j(t)-\xi_0^j(s))-
(\xi_0^{j'}(t)-\xi_0^{j'}(s))\,.
\end{equation}
\end{theorem}

\bigbreak \vspace{1mm}
\begin{proof}
Pick constants $r$, $C$, such that $r>1$, 
\mbox{$C>0$},
and $\|f(x,u)\|^r\le L(x,u)+C$ whenever $x\in K$, $u\in U$,
and a function $\omega:\,]\,0,+\infty\,[\,\mapsto[0,+\infty]$ 
such that $\lim_{s\downarrow 0}\omega(s)$ and 
$\|\fsh(x,u)-\fsh(x',u)\|\le\omega(s)$ whenever
$x,x'\in K$ and $\|x-x'\|\le s$. Choose,
for each $j$, a measurable function \mbox{$v^j:[0,T]\mapsto\R^n$} and a
function $\eta^j:[0,T]\mapsto U$ such that $\|v(t)\|\le\eps_j$,
$\dot\xi^j(t)=f(\xi^j(t),\eta^j(t))+v(t)$, and
$\dot\xi_0^j(t)=L(\xi^j(s),\eta^j(s))$ for almost all $t\in[0,T]$.

Let $C'=2^r(C+\eps^r)$. 
Then for each $j$ the coerciveness condition implies the inequality

$$
\|\dot\xi^j(t)\|^{r}\le 2^rL(\xi^j(t),\eta^j(t)) +C'\,\,
{\rm for\,\,almost\,\,all}\,\,t\,,
$$

\medbreak\noindent because 
\begin{eqnarray*}
\|\dot\xi^j(t)\|^{r}&=&\|f(\xi^j(t),\eta^j(t))+v(t)\|^r\\
&\le&\big(\|f(\xi^j(t),\eta^j(t))\|+\|v(t)\|\big)^r\\
&\le&2^r\|f(\xi^j(t),\eta^j(t))\|^r+2^r\|v(t)\|^r\\
&\le&2^r(L(\xi^j(t),\eta^j(t))+C)+2^r\eps^r\\
&=&2^rL(\xi^j(t),\eta^j(t))+C'\,.
\end{eqnarray*}

\medbreak\noindent from which it follows that
$\int_0^T\|\dot\xi^j(t)\|^{r}dt\le 2^rk+C'T$. Then the sequence
$\{\dot\xi^j(t)\}_{j=1}^\infty$ is uniformly bounded in $L^{r}$, so
we may assume, after passing to a subsequence if necessary,
that the weak $L^{r}$-limit $\zeta=\,$w-$\lim_{j\to\infty}\dot\xi^j$
\mbox{exists}.  Then the $\xi^j$ converge uniformly as $j\to\infty$
to a limit $\xi^\infty$ such that 
$\xi^\infty(t)-\xi^\infty(s)=\int_s^t\zeta(r)\,dr$ for all
$s,t\in[0,T]$.  After passing to a subsequence once more, if necessary,
we assume that

\begin{equation}\label{twoj}
\|\xi^j(t)-\xi^\infty(t)\|\le 2^{-j}\,\,{\rm for\,\,all}\,\,j\in\nn\,,
\,\,t\in[0,T]\,.
\end{equation}

\medbreak\noindent
Let $\theta^j(t)=\dot\xi_0^j(t)+C$.  Then the $\theta^j$ 
are nonnegative, because 
\mbox{$\dot\xi_0^j(t)=L(\xi^j(t),\eta^j(t))\ge-C$}.
Furthermore, the sequence $\{\theta^j\}_{j=1}^\infty$ 
is bounded in $L^1([0,T],\rr)$, \mbox{because}
\begin{eqnarray*}
\|\theta^j\|_{L^1}&=&\int_0^T\theta^j(t)\,dt\\
&=&\int_0^T(\dot\xi_0^j(t)+C)\,dt\\
&\le&k+CT\,.
\end{eqnarray*}
The space $L^1([0,T],\rr)$ can be embedded in the usual way in
$C^0([0,T],\rr)^\dagger$ (the dual of $C^0([0,T],\rr)$), which is the
space of finite Borel measures on $[0,T]$, by \mbox{means} of 
the map $\psi\mapsto\mu_\psi$ that assigns to
each function \mbox{$\psi\in L^1([0,T],\rr)$} the Borel measure $\mu_\psi$
such that 
$$
\mu_\psi(\varphi)= \int_0^T\psi(t)\varphi(t)dt\,\,{\rm for\,\,every}
\,\,\varphi\in C^0([0,T],\rr)\,.
$$

Then we may assume, after passing to a subsequence for a third time,
if \mbox{necessary}, that the weak* limit $\mu^\infty$ 
of the measures $\mu^j\equaldef\mu_{\theta^j}$ 
defined by the $\theta^j$ exists as $j\to\infty$.
The measure $\mu^\infty$ then has a decomposition
$$
\mu^\infty=\mu^{\infty,ac}+\mu^{\infty,at}+\mu^{\infty,sing}
$$
into the sum of an absolutely continuous part, an \mbox{atomic} part, and 
a singular part. Since the measure $\mu^\infty$ is 
\mbox{positive}, because 
it is a limit of positive measures, the three components
$\mu^{\infty,ac}$, $\mu^{\infty,at}$, $\mu^{\infty,sing}$ are 
positive as well.  Let $\theta^\infty$ be the Radon-Nikodym 
derivative of $\mu^{\infty}$, so $\theta^\infty$ is an integrable 
function on $[0,T]$ such that 
$$
\mu^{\infty,ac}(\varphi)=\int_0^T\theta^\infty(t)\varphi(t)\,dt\,\,
{\rm for}\,\,\varphi\in C^0([0,T],\rr)\,.
$$
Define 
\begin{equation}\label{defdef}
\xi_0^\infty(t)=-Ct+\int_0^t\theta^\infty(s)ds
\end{equation}
and then set $\Xi^\infty=(\xi^\infty,\xi_0^\infty)$. 

\medbreak We will show that 
$\Xi^\infty$ is an augmented \mbox{trajectory} of the system
$\hat\Sigma$. To see this, 
we observe first of \mbox{all} that by \mbox{construction}
the functions \mbox{$\xi^\infty:[0,T]\mapsto\rr^n$} and 
\mbox{$\xi_0^\infty:[0,T]\mapsto\rr$} are absolutely continuous,
and their
\mbox{derivatives} at $t$  are \mbox{equal} 
to $\zeta(t)$ and $\theta^\infty(t)$,
\mbox{respectively}, for all $t$ in a subset $G$ of $[0,T]$ 
such that $[0,T]\backslash G$ has \mbox{measure} zero. Let 
$A$ be the set of atoms of $\mu^{\infty, at}$, and let 
$B$ be a subset of $[0,T]$ of Lebesgue measure zero such that
\mbox{$\mu^{\infty,sing}([0,T]\backslash B)=0$}. Then the set
\mbox{$G'=G\backslash(A\cup B\cup\{T\})$} has measure $T$ and 
\mbox{$\mu^{\infty,at}(G')=\mu^{\infty,sing}(G')=0$}. Let
$G''$ be the set of points of density of $G'$ that are Lebesgue points
of $\zeta$ and $\theta^\infty$,  so 
$G''$ has measure $T$ as well. (\mbox{Recall} that a {\em Lebesgue
point} of a scalar- or vector-valued \mbox{integrable}
function $\sigma$ defined on
an interval $[a,b]$ is a point \mbox{$t\in\,]\,a,b\,[\,$} such that 
\mbox{$\lim_{h\downarrow 0}\frac{1}{h}
\int_{t-h}^{t+h}\|\sigma(s)-\sigma(t)\|\,ds=0$}.)
Let $t\in G''$,
and fix an $h$ such that \mbox{$0<h<T-t$}. Let 
\mbox{$E_{t,h}=[t,t+h]\cap G'$}, so the Lebesgue
measure $|E_{t,h}|$ of $E_{t,h}$ satisfies
$\lim_{h\downarrow 0}h^{-1}|E_{t,h}|=1$.
\mbox{Using} the \mbox{facts} that the Borel measure
$\mu^{\infty,at}+\mu^{\infty,sing}$ is \mbox{regular}
and $(\mu^{\infty,at}+\mu^{\infty,sing})(E_{t,h})=0$,
we can find a \mbox{relatively}  open subset $\tilde U_{t,h}$ 
of $[0,T]$ such that \mbox{$E_{t,h}\subseteq\tilde U_{t,h}$}
and 
\mbox{$(\mu^{\infty,at}+\mu^{\infty,sing})(\tilde U_{t,h})\le h^2$}.
We then let \mbox{$U_{t,h}=\tilde U_{t,h}\cap\,]\,t,t+h\,[\,$}, so $U_{t,h}$ 
is an open subset of $\R$, $U_{t,h}\subseteq\,]\,t,t+h\,[\,$, and 
\mbox{$(\mu^{\infty,at}+\mu^{\infty,sing})(U_{t,h})\le h^2$}.
Using the regularity of Lebesgue measure we can find a compact subset
$K_{t,h}$ of $E_{t,h}\backslash\{t,t+h\}$ such that 
\mbox{$|K_{t,h}|\ge|E_{t,h}|-h^2$}. Then of course 
\mbox{$\lim_{h\downarrow 0}h^{-1}|K_{t,h}|=1$}, and 
$K_{t,h}\subseteq U_{t,h}$ for each $h$.
Let \mbox{$\tilde\varphi_{t,h}(s)={\rm dist}(s,\R\backslash U_{t,h})$}, so
$\tilde\varphi_{t,h}:\R\mapsto\R$ is \mbox{continuous}, 
$\tilde\varphi_{t,h}(s)=0$ whenever $s\notin U_{t,h}$, 
and $\tilde\varphi_{t,h}(s)>0$ \mbox{whenever} $s\in K_{t,h}$. If we let
\begin{eqnarray*}
\beta_{t,h}&=&\min\{\tilde\varphi_{t,h}(s):s\in K_{t,h}\}\,,\\
\hat\varphi_{t,h}(s)&=&\min(\tilde\varphi_{t,h}(s),\beta_{t,h})\,,\\
\varphi_{t,h}(s)&=&\beta_{t,h}^{-1}\hat\varphi_{t,h}(s)\,,
\end{eqnarray*}
then $\varphi_{t,h}$  is  a continuous real-valued function
on $\R$ such that $0\le\varphi_{t,h}(s)\le 1$ for all $s$,
$\varphi_{t,h}(s)=1$ for all \mbox{$s\in K_{t,h}$}, and
$\varphi_{t,h}(s)=0$ for all $s\in\R\backslash U_{t,h}$.
In \mbox{particular}, $\varphi_{t,h}(s)=0$ whenever 
$s\notin\,]\,t,t+h\,[\,$.


Let $a_{t,h}=\int_t^{t+h}\varphi_{t,h}(s)\,ds=
\int_{-\infty}^{+\infty}\varphi_{t,h}(s)\,ds$. Then
$$
|E_{t,h}|-h^2\le a_{t,h}\le h\,,
$$
from which it follows that
$$
\lim_{h\downarrow 0}\frac{a_{t,h}}{h}=1\,.
$$
Let $\psi_{t,h}=a_{t,h}^{-1}\varphi_{t,h}$. Then
$$
\int_t^{t+h}\psi_{t,h}(s)\,ds=1\,.
$$
If $h>0$, write 
$$
\delta_j(h)=2^{-j}+h^{1/\rho}(2^r|k|+C'T)^{1/r}\,.
$$ 
Then, if $s\in[t,t+h]$, we have
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\|\xi^j(s)-\xi^\infty(t)\|\\
&\le&\|\xi^j(t)-\xi^\infty(t)\|+\|\xi^j(s)-\xi^j(t)\|\\
&\le&2^{-j}+\int_t^s\|\dot\xi^j(\tau)\|\,dt\\
&\le&2^{-j}+
(s-t)^{1/\rho}\Big(\int_t^s\|\dot\xi^j(\tau)\|^r\,dt\Big)^{1/r}\\
&\le&2^{-j}+
h^{1/\rho}\Big(\int_0^T\|\dot\xi^j(\tau)\|^r\,dt\Big)^{1/r}\\
&\le&2^{-j}+h^{1/\rho}\Big(\int_0^T(2^rL(\xi^j(\tau),\eta^j(\tau))
+C')\,d\tau\Big)^{1/r}\\
&\le&2^{-j}+h^{1/\rho}(2^r|k|+C'T)^{1/r}\\
&=&\delta_j(h)\,.
\end{eqnarray*}
For almost all $s\in[t,t+h]$, the derivative 
$\dot\Xi^j(s)$ \mbox{exists} and is equal to 
$\fsh(\xi^j(s),\eta^j(s))+(v^j(s),0)$. Hence
$$
\|\,\dot\Xi^j(s)-\fsh(\xi^\infty(t),\eta^j(s))\|
\le\omega(\delta_j(h))+\eps_j\,,
$$ 
from which it follows that
$$
{\rm dist}(\dot\Xi^j(s),\fsh(\xi^\infty(t),U)
\le\omega(\delta_j(h))+\eps_j\,.
$$ 
Therefore the average 
$$
A_{t,h}^j=\int_t^{t+h}\psi_{t,h}(s)\dot\Xi^j(s)\,ds
=\int_{0}^{T}\psi_{t,h}(s)\dot\Xi^j(s)\,ds
$$ 
also satisfies 
\begin{equation}\label{avesat}
{\rm dist}(A_{t,h}^j,\fsh(\xi^\infty(t),U)
\le\omega(\delta_j(h))+\eps_j\,,
\end{equation}
because $\fsh(\xi^\infty(t),U)$ is closed and convex.
As $j\to\infty$, the vector functions $\dot\xi^j$ converge weakly in $L^r$ 
to $\zeta$, so 
\begin{eqnarray*}
\int_t^{t+h}\psi_{t,h}(s)\dot\xi^j(s)\,ds
&\to&\int_t^{t+h}\psi_{t,h}(s)\zeta(s)\,ds\\
&=&\int_t^{t+h}\psi_{t,h}(s)\dot\xi^\infty(s)\,ds
\end{eqnarray*}
as $j\to\infty$, $j\in J$. The integral 
$\int_t^{t+h}\psi_{t,h}(s)\dot\xi^\infty(s)\,ds$ satisfies
\begin{eqnarray*}
\int_{t}^{t+h}\psi_{t,h}(s)\dot\xi^\infty(s)\,ds&=&
\int_{t}^{t+h}\psi_{t,h}(s)\dot\xi^\infty(t)\,ds+{\cal E}_{t,h}\\
&=&\dot\xi^\infty(t)+{\cal E}_{t,h}\\
\end{eqnarray*}
where 
$$
{\cal E}_{t,h}=\int_{t}^{t+h}\psi_{t,h}(s)(\dot\xi^\infty(s)-
\dot\xi^\infty(t))\,ds\,,
$$
and we have used the fact that $\int_{t}^{t+h}\psi_{t,h}(s)\,ds=1$.

The error term ${\cal E}_{t,h}$ satisfies
\begin{eqnarray*}
|{\cal E}_{t,h}|&\le&\max\big\{\psi_{t,h}(s):s\in[0,T]\big\}\\
&\,&\,\,\,\times 
\int_{t}^{t+h}\|\dot\xi^\infty(s)-\dot\xi^\infty(t)\|\,ds\\
&=&ha_{t,h}^{-1}\hat{\cal E}_{t,h}\\
&=&\alpha_{t,h}\hat{\cal E}_{t,h}\,,
\end{eqnarray*}
where $\alpha_{t,h}=ha_{t,h}^{-1}$, so $\alpha_{t,h}\to 1$ as
$h\downarrow 0$, and
$$
\hat{\cal E}_{t,h}=\frac{1}{h}\int_{t}^{t+h}\|\dot\xi^\infty(s)-
\dot\xi^\infty(t)\|\,ds\,.
$$
Then 
\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\lim\sup_{j\to\infty,j\in J}
\Big\|\int_t^{t+h}\psi_{t,h}(s)\dot\xi^j(s)\,ds
-\dot\xi^\infty(t)\Big\|\nonumber\\
&\le&\alpha_{t,h}\hat{\cal E}_{t,h}\,.\label{gosto-x}
\end{eqnarray}
%\lim\sup_{h\downarrow 0}\alpha_{t,h}(h+\hat E_{t,h})\,.

To analyze the behavior of the integrals 
$$
I_{t,h}^j\equaldef\int_t^{t+h}\psi_{t,h}(s)\dot\xi_0^j(s)\,ds\,,
$$
we write $\dot\xi_0^j(s)=\theta^j(s)-C$, so 
\begin{eqnarray*}
I_{t,h}^j&=&-Ch+\int_t^{t+h}\psi_{t,h}(s)\theta^j(s)\,ds\\
&=&-Ch+\int_0^{T}\psi_{t,h}(s)\theta^j(s)\,ds\\
&=&-Ch+\int_{[0,T]}\psi_{t,h}d\mu^j\,.
\end{eqnarray*} 
Hence 
\begin{equation}\label{gosto}
I_{t,h}^j\to-Ch+\int_{[0,T]}\psi_{t,h}(s)\,d\mu^\infty(s)
\,\,{\rm as}\,\,j\to\infty\,.
\end{equation}
Write 
\begin{eqnarray*}
\int_{[0,T]}\psi_{t,h}(s)\,d\mu^\infty(s)&=&
\int_0^T\psi_{t,h}(s)\theta^\infty(s)\,ds\\
&\,&\,\,\,+\int_{[0,T]}\psi_{t,h}(s)\,d\hat\mu^\infty(s)\,,
\end{eqnarray*}
where $\hat\mu^\infty=\mu^{\infty,at}+\mu^{\infty,sing}$. Since
$\dot\xi_0^\infty(s)=\theta^j(s)-C$, (\ref{gosto}) implies
\begin{eqnarray}
I_{t,h}^j&\to&\int_{[0,T]}\psi_{t,h}(s)\dot\xi_0^\infty(s)\,ds\nonumber\\
&\,&\,\,\,\,+\int_{[0,T]}\psi_{t,h}(s)\,d\hat\mu^\infty(s)
\,\,{\rm as}\,\,j\to\infty\,.\label{gosto-a}
\end{eqnarray}
The integral $\int_{[0,T]}\psi_{t,h}(s)\,d\hat\mu^\infty(s)$
that occcurs in (\ref{gosto-a}) 
is a \mbox{nonnegative} \mbox{number}, and is bounded above
by \mbox{$\max\{\psi_{t,h}(s):s\in[0,T]\}$} times
$\hat\mu^\infty(U_{t,h})$, since $\psi_{t,h}$ \mbox{vanishes} outside
$U_{t,h}$. Therefore
\begin{equation}\label{gosto-b}
0\le\int_{[0,T]}\psi_{t,h}(s)\,d\hat\mu^\infty(s)
\le h^{2}a_{t,h}^{-1}=h\alpha_{t,h}\,,
\end{equation}
On the other hand, the integral 
$\int_{[0,T]}\psi_{t,h}(s)\dot\xi_0^\infty(s)\,ds$ satisfies
\begin{eqnarray}
\int_{[0,T]}\psi_{t,h}(s)\dot\xi_0^\infty(s)\,ds&=&
\int_{t}^{t+h}\psi_{t,h}(s)\dot\xi_0^\infty(s)\,ds\nonumber\\
&=&\dot\xi_0^\infty(t))\,ds+E_{t,h}\nonumber\\
&=&\dot\xi_0^\infty(t)+E_{t,h}\,,\label{ontheot}
\end{eqnarray}
where 
$$
E_{t,h}=\int_{t}^{t+h}\psi_{t,h}(s)(\dot\xi_0^\infty(s)-
\dot\xi_0^\infty(t))\,ds\,,
$$
and we have used the fact that $\int_{t}^{t+h}\psi_{t,h}(s)\,ds=1$.

The error term $E_{t,h}$ satisfies
\begin{eqnarray*}
|E_{t,h}|&\le&\max\big\{\psi_{t,h}(s):s\in[0,T]\big\}\\
&\,&\,\,\,\times 
\int_{t}^{t+h}|\dot\xi_0^\infty(s)-\dot\xi_0^\infty(t)|\,ds\\
&=&ha_{t,h}^{-1}\hat E_{t,h}\\
&=&\alpha_{t,h}\hat E_{t,h}\,,
\end{eqnarray*}
where 
$$
\hat E_{t,h}=\frac{1}{h}\int_{t}^{t+h}|\dot\xi_0^\infty(s)-
\dot\xi_0^\infty(t)|\,ds\,.
$$


It follows from (\ref{gosto-a}), (\ref{gosto-b}), (\ref{ontheot}), 
and the bound
\mbox{$|E_{t,h}|\le\alpha_{t,h}\hat E_{t,h}$}, that
\begin{equation}\label{gosto-y}
\lim\sup_{j\to\infty,j\in J}\Big|I_{t,h}^j-\dot\xi_0^\infty(t)\Big|
\le\alpha_{t,h}(h+\hat E_{t,h})\,.
\end{equation}
If we now combine (\ref{gosto-x}) and (\ref{gosto-y}), we find that
\begin{equation}
\limsup_{j\to\infty,j\in J}\Big\|A_{t,h}^j\!-\!\dot\Xi^\infty(t)\Big\|
\!\le\!\alpha_{t,h}(h\!+\!\hat E_{t,h}\!+\!{\cal E}_{t,h}).\label{gosto-z}
\end{equation}
Then (\ref{avesat}) implies that
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
{\rm dist}(\dot\Xi^\infty(t),\fsh(\xi^\infty(t)))\\
&\,&\!\!\!\le\!\alpha_{t,h}(h\!+\!\hat E_{t,h}\!+\!{\cal E}_{t,h})\!+\!
\limsup_{j\to\infty}(\omega(\delta_j(h))\!+\!\eps_j)\,.
\end{eqnarray*}
Hence, given any $j_*$, we have 

\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!
{\rm dist}(\dot\Xi^\infty(t),\fsh(\xi^\infty(t)))\\
&\,&\!\!\!\le\!\alpha_{t,h}(h\!+\!\hat E_{t,h}\!+\!{\cal E}_{t,h})\!+\!
\sup\{\omega(\delta_j(h))\!+\!\eps_j:j\ge j_*\}\,.
\end{eqnarray*}

\medbreak\noindent Given any positive number $\beta$, we can find a positive 
$\gamma$ such that $\omega(s)<\beta$ whenever $0<s\le\gamma$, and
then find $j_*$, $h_*$ such that $\delta_j(h)\le\gamma$ whenever 
$j\ge j_*$ and $0<h\le h_*$, and $\eps_j<\beta$ whenever 
$j\ge j_*$. Then we can pick $h$ such that  $0<h\le h_*$, 
and $\alpha_{t,h}(h\!+\!\hat E_{t,h}\!+\!{\cal E}_{t,h})<\beta$.
Then 
$$
{\rm dist}(\dot\Xi^\infty(t),\fsh(\xi^\infty(t)))<3\beta\,.
$$
Since $\beta$ was arbitrary, we conclude that 
${\rm dist}(\dot\Xi^\infty(t),\fsh(\xi^\infty(t)))=0$, 
so $\dot\Xi^\infty(t)\in\fsh(\xi^\infty(t))$, because 
$\fsh(\xi^\infty(t))$ is closed. Since this is true 
for almost all $t\in[0,T]$, and $\Xi^\infty$ is absolutely
continuous, we have shown that $\Xi^\infty$ is an augmented
trajectory of $\hat\Sigma$.

By construction, the $\xi^j$ converge uniformly to $\xi^\infty$.
\mbox{Also}, it is clear from (\ref{defdef})  that $\xi_0^\infty(0)=0$.
To \mbox{conclude} our proof, we have to show that

\begin{equation}\label{conclu}
\liminf\,_{j\to\infty}\inf\big\{\Delta^{j,\infty}(t,s)\!:\!
0\!\le\!s\!\le\!t\!\le\!T\big\}\!\ge\!0\,,
\end{equation}

\medbreak\noindent where $\Delta^{j,\infty}(t,s)$ is the quantity defined
in (\ref{defodel}). 
Suppose that
\mbox{$\liminf_{j\to\infty}\inf\big\{\Delta^{j,\infty}(t,s)\!:\!
0\!\le\!s\!\le\!t\!\le\!T\big\}\!<\!0$}. 
Pick  a number  $\beta$ such that $\beta>0$ and

$$
\liminf_{j\to\infty}
\inf\big\{\Delta^{j,\infty}(t,s)\!:\!
0\!\le\!s\!\le\!t\!\le\!T\big\}\le-3\beta\,.
$$ 

\medbreak\noindent Then there exists a subsequence
$\{\Xi^{j(k)}\}_{k=1}^\infty$ of $\{\Xi^{j}\}_{j=1}^\infty$ such that
$\inf\big\{\Delta^{j,\infty}(t,s)\!:\!
0\!\le\!s\!\le\!t\!\le\!T\big\}\le-2\beta$
for all $k$.  We can then choose, for each $k$, members $s_k$ and $t_k$
of $[0,T]$ such that $s_k\le t_k$ and 
\begin{equation}\label{conclu-x}
\big(\xi_0^{j(k)}(t_k)-\xi_{0}^{j(k)}(s_k)\big)-
\big(\xi_0^{\infty}(t_k)-\xi_{0}^{\infty}(s_k)\big)\le-\beta\,.
\end{equation}
By passing to a subsequence, if necessary, we may \mbox{assume}
that the $s_k$ and the $t_k$ converge to limits $s$, $t$. Clearly, 
then, $s\le t$.

If $s=t$, then 
\begin{eqnarray*}
\xi_0^{j(k)}(t_k)-\xi_0^{j(k)}(s_k)&=&\int_{s_k}^{t_k}
\dot\xi_0^{j(k)}(v)\,dv\\
&=&\int_{s_k}^{t_k}\big(\theta^{j(k)}(v)-C)\,dv\\
&\ge&-C(t_k-s_k)\,,
\end{eqnarray*}
so the fact that 
$\lim_{k\to\infty}(\xi_0^{\infty}(t_k)-\xi_{0}^{\infty}(s_k))=0$ implies 
the inequalities
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\liminf_{k\to\infty}\Big(\big(\xi_0^{j(k)}(t_k)-\xi_{0}^{j(k)}(s_k)\big)-
\big(\xi_0^{\infty}(t_k)-\xi_{0}^{\infty}(s_k)\big)\Big)\\
\!\!\!&\!\!\!=&\!\!\!\!\liminf_{k\to\infty}\big(\xi_0^{j(k)}(t_k)\!
-\!\xi_{0}^{j(k)}(s_k)\big)
\!-\!\lim_{k\to\infty}\big(\xi_0^{\infty}(t_k)\!-\!\xi_{0}^{\infty}(s_k)\big)\\
\!\!\!&\!\!\!=&\!\!\!\!\liminf_{k\to\infty}\big(\xi_0^{j(k)}(t_k)-
\xi_{0}^{j(k)}(s_k)\big)\\
\!\!\!&\!\!\!\ge&\!\!\!\!\liminf_{k\to\infty}(-C(t_k-s_k))\\
\!\!\!&\!\!\!=&\!\!\!\!\lim_{k\to\infty}(-C(t_k-s_k))\\
\!\!\!&\!\!\!=&\!\!\!\!0\,,
\end{eqnarray*}
which clearly contradict (\ref{conclu-x}).

Now assume that $s<t$. Fix a positive number $\gamma$ such that
$2\gamma<t-s$,  and let $\Phi_{s,t,\gamma}$ be the set of all continuous 
nonnegative 
functions $\varphi:\R\mapsto\R$  that vanish outside the interval
$[s+\gamma,t-\gamma]$ and 
are such that $\varphi(v)\le 1$ for all $v\in\R$. Then

\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\liminf_{k\to\infty}\Big(\xi_0^{j(k)}(t_k)-\xi_0^{j(k)}(s_k)\Big)\\
&=&\liminf_{k\to\infty}\int_{s_k}^{t_k}\dot\xi_0^{j(k)}(v)\,dv\\
&=&\liminf_{k\to\infty}\int_{s_k}^{t_k}(\theta^{j(k)}(v)-C)\,dv\\
&=&\liminf_{k\to\infty}\Big(-C(t_k-s_k)+\int_{s_k}^{t_k}
\theta^{j(k)}(v)\,dv\Big)\\
&=&-C(t-s)+\liminf_{k\to\infty}\int_{s_k}^{t_k}\theta^{j(k)}(v)\,dv\\
&\ge&-C(t-s)+\liminf_{k\to\infty}\int_{s_k}^{t_k}
\varphi(v)\theta^{j(k)}(v)\,dv\\
&=&-C(t-s)+\liminf_{k\to\infty}\int_{[0,T]}\varphi(v)d\mu^{j(k)}(v)\\
&=&-C(t-s)+\lim_{k\to\infty}\int_{[0,T]}\varphi(v)d\mu^{j(k)}(v)\\
&=&-C(t-s)+\int_{[0,T]}\varphi(v)d\mu^\infty(v)\\
&\ge&-C(t-s)+\int_{[0,T]}\varphi(v)d\mu^{\infty,ac}(v)\\
&=&-C(t-s)+\int_{s}^t\varphi(v)\theta^\infty(v)\,dv\\
&=&\int_{s}^t\varphi(v)(\theta^\infty(v)-C)\,dv\\
&=&\int_{s}^t\varphi(v)\dot\xi_0^\infty(v)\,dv\,,
\end{eqnarray*}
where, for the first inequality, we have used the 
\mbox{fact} that $[s+\gamma,t-\gamma]\subseteq[s_k,t_k]$ 
when $k$ is large enough. 
\mbox{Therefore} 

$$
\liminf_{k\to\infty}\Big(\xi_0^{j(k)}(t_k)-\xi_0^{j(k)}(s_k)\Big)
\ge\int_{s}^t\varphi(v)\dot\xi_0^\infty(v)\,dv
$$

\medbreak\noindent for \mbox{every} \mbox{$\varphi\in\Phi_{s,t,\gamma}$}. 
Hence

\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\liminf_{k\to\infty}\Big(\xi_0^{j(k)}(t_k)-\xi_0^{j(k)}(s_k)\Big)\\
&\ge&\sup\Big\{\int_{s}^t\varphi(v)\dot\xi_0^\infty(v)\,dv:
\varphi\in\Phi_{s,t,\gamma}\Big\}\\
&=&\int_{s+\gamma}^{t-\gamma}\dot\xi_0^\infty(v)\,dv\\
&=&\xi_0^\infty(t-\gamma)-\xi_0^\infty(s+\gamma)\,.
\end{eqnarray*}

\medbreak\noindent Since $\gamma$ is arbitrary, we can let 
$\gamma\downarrow 0$, and conclude that 

\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!
\liminf_{k\to\infty}\Big(\xi_0^{j(k)}(t_k)-\xi_0^{j(k)}(s_k)\Big)\\
&\ge&\xi_0^\infty(t)-\xi_0^\infty(s)\\
&=&\lim_{k\to\infty}\Big(\xi_0^\infty(t_k)-\xi_0^\infty(s_k)\Big)\,,
\end{eqnarray*}
so 

$$
\liminf_{k\to\infty}\Big(\big(\xi_0^{j(k)}(t_k)-\xi_0^{j(k)}(s_k)\big)-
\big(\xi_0^\infty(t_k)-\xi_0^\infty(s_k)\big)\Big)\ge 0\,,
$$
contradicting (\ref{conclu-x}). This completes the proof of (\ref{conclu}). 
\end{proof}

\ssection{Trajectories of steepest descent.}

\noindent We recall from \S\ref{sec-standard} that $\msdsvx$ denotes 
the set of all maximal augmented trajectories of $\hat\Sigma$ from $x$
of \mbox{steepest} descent with respect to $V$ and that, as \mbox{explained}
in Proposition \ref{trivialprop} and the remarks preceding its 
\mbox{statement}, $\msdsvx$ is always nonempty for 
\mbox{trivial} reasons, because 
the \mbox{trivial} trajectory $\Xi_x^{triv}$ always 
\mbox{belongs} to $\sdsvx$,
and once we know that $\sdsvx\neq\emptyset$ it \mbox{follows} immediately
from Zorn's Lemma that $\sdsvx$ must have a maximal \mbox{element}.

The truly nontrivial and useful result is the \mbox{statement} that
maximal steepest descent trajectories not only \mbox{exist} but are
``large,'' in the sense that they are ``right-unbounded.'' 
Precisely, if $\xi:I\mapsto\Omega$ is a curve, we say that $\xi$ is 
{\em right-unbounded} if (i)~the interval $I$ is open on the right
(that is, if $\tau=\sup\,I$, then either 
(a)~$\tau=+\infty$ or (b)~$\tau$ is finite and does not belong to 
$I$), and (ii)~if $\tau$ is finite, then for every compact subset $K$
of $\Omega$ there exists a $\tau_K$ such that $0\le\tau_K<\tau$ and
$\xi(t)\notin K$ whenever $\tau_K<t<\tau$. (Equivalently, condition
(ii) asserts that 
$\lim_{t\uparrow\tau}\xi(t)=\infty_\Omega$, where $\infty_\Omega$ is
the point at infinity of the one-point compactification of $\Omega$.)

\begin{theorem}\label{maintech}
Let $\hat\Sigma=(\Omega,U,f,L)$ be a
locally \mbox{coercive}, locally uniformly continuous augmented control system
such that $\fsh(x,U)$ is closed and convex for 
\mbox{every} $x\in\Omega$. 
Let $V:\Omega\mapsto\R$ be a \mbox{continuous} function that  
\mbox{satisfies} (\ref{loweq}) on $\Omega$ in the
\mbox{viscosity} sense. Let $x_*\in\Omega$, and let 
$\Xi=(\xi,\xi_0)$ 
be a maximal augmented trajectory of $\hat\Sigma$  from $x_*$ 
of \mbox{steepest} 
\mbox{descent} with respect to $V$, defined on an
interval $I$. Then $\xi$ is right-unbounded. 
\end{theorem}
\begin{proof}
We assume that the conclusion is not true and derive a contradiction.
Pick a $\Xi=(\xi,\xi_0)\in{\cal MSD}_{\hat\Sigma,V,x_*}$ that violates 
the conclusion.
This means, to begin with, that $\Xi$ is defined on a bounded interval 
$I$, and, in \mbox{addition}, this interval is either of the form
$[0,\tau]$, with $0\le\tau<+\infty$ (the ``right-closed case''), 
or of the \mbox{form} $[0,\tau\,[\,$, with $0<\tau<+\infty$
(the ``right-open case''). \mbox{Furthermore}, in the right-open case
there exist a \mbox{compact} subset $K$ of $\Omega$ and a sequence
$\{t_j\}_{j=1}^\infty$ such that $t_j\in I$ and 
$\xi(t_j)\in K$ for all $j$, and $\lim_{j\to\infty}t_j=\tau$. 
In order to treat the right-open and right-closed cases together, we also 
choose a compact subset $K$ of $\Omega$ and a sequence $\{t_j\}_{j=1}^\infty$ 
in $I$  in the right-closed case, subject 
to the only  requirements that $\xi(t)\in K$ for all $t\in[0,\tau]$
and $t_j\to\tau$ as $j\to\infty$. (For example, we could just 
choose $K=\{\xi(t):0\le t\le\tau\}$, $t_j=\tau$ for all $j$.)
Then in both cases
the $t_j$ belong to $I$ and converge to $\tau$, and 
the $\xi(t_j)$ belong to $K$.
In addition, we pick a function $\eta:I\mapsto U$ such that
$\dot\Xi(t)=F(\xi(t),\eta(t))$ for almost all $t\in I$.

Fix a positive number $\delta$ such that the compact set
$$
K_\delta=\{x\in\R^n:{\rm dist}(x,K)\|\le\delta\}
$$
is contained in $\Omega$. Then use the fact that 
$\hat\Sigma$ is locally coercive to choose $r,C$ such that 
$r>1$, $C>0$, and $\|f(x,u)\|^{r}\le L(x,u)+\tilde C$ 
for all $x\in K_\delta$, $u\in U$. Let 

$$
\bar V=\max\{|V(x)|:x\in  K_\delta\}\,.
$$

\medbreak\noindent
If $\rho=\frac{r}{r-1}$, so that $\frac{1}{r}+\frac{1}{\rho}=1$,	
and we define 
$$
\bar\delta(\sigma)\equaldef
2\Big(\sigma^{1/\rho}(2|\bar V|+(C+1)\sigma)^{1/r}+\sigma\Big)\,,
$$ 
then $\bar\delta(\sigma)$ goes to zero
as $\sigma\downarrow 0$. Therefore we can pick $\sigma$ 
such that $\bar\delta(\sigma)<\delta$.

Now suppose that we are in the right-open case. \mbox{Given}  any $j$,
the point $x_j=\xi(t_j)$ belongs to $K$. Let 

$$
S_j=\Big\{t\in I:t\ge t_j\wedge(\forall s)\big((t_j\le s\le t)\Rightarrow
\xi(s)\in K_\delta\big)\Big\}\,.
$$

\medbreak\noindent Then $S_j$ is a subinterval of $I$, whose left 
endpoint is $t_j$.

\medbreak
If $t,t'\in S_j$, and $t'\le t$, then the steepest descent property 
of $\Xi$ implies that 

$$
\xi_0(t)+V(\xi(t))\le\xi_0(t')+V(\xi(t'))\,,
$$
so
$$
\xi_0(t)-\xi_0(t')\le V(\xi(t))-V(\xi(t'))\le 2|\bar V|\,,
$$

\medbreak\noindent since both $\xi(t)$ and $\xi(t')$ belong to $K_\delta$.
Furthermore, the inequality 

$$
\|f(\xi(s),\eta(s))\|^r\le L(\xi(s),\eta(s))+C
$$

\medbreak\noindent is true for all $s\in S_j$, since 
$\xi(s)\in K_\delta$ for all such $s$. Therefore

\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\|\xi(t)-\xi(t')\|\\
&=&\Big\|\int_{t'}^t\dot\xi(s)\,ds\Big\|\\
&\le&\int_{t'}^t\|\dot\xi(s)\|\,ds\\
&=&\int_{t'}^t\|f(\xi(s),\eta(s))\|\,ds\\
&\le&(t-t')^{1/\rho}
\Big(\int_{t'}^t\|f(\xi(s),\eta(s))\|^r\,ds\Big)^{1/r}\\
&\le& (t-t')^{1/\rho}
\Big(\int_{t'}^{t}(L(\xi(s),\eta(s))+C)\,ds\Big)^{1/r}\\
&=& (t-t')^{1/\rho}(C(t-t')+\xi_0(t)-\xi_0(t'))^{1/r}\\
&=& (t-t')^{1/\rho}(C(t-t')+2|\bar V|))^{1/r}\\
&\le&\frac{1}{2}\bar\delta(t-t')\,.
\end{eqnarray*}
Now pick $j$ so large that $\tau-t_j<\sigma$, and apply 
the above inequality with $t'=t_j$. Then 
\begin{equation}\label{impli-x}
\|\xi(t)-\xi(t_j)\|\le\frac{1}{2}\bar\delta(\sigma)
\le\frac{1}{2}\delta\,\,\,\,{\rm for\,\,\,all}\,\,\,\,t\in S_j\,. 
\end{equation}
Hence, if we let $\bar t=\sup\,S_j$, it is impossible
that \mbox{$\bar t<\tau$}, because if $\bar t<\tau$ then 
$\bar t\in I$,  and  (\ref{impli-x}) implies that
\mbox{$\|\xi(\bar t)-\xi(t_j)\|\le\frac{1}{2}\delta$}, from which it follows,
by \mbox{continuity}, that there exists a positive $\alpha$ such that
$\|\xi(t)-\xi(t_j)\|<\delta$ for $\bar t\le t\le\bar t+\alpha$;
but then $\xi(t)\in K_\delta$ if $\bar t\le t\le\bar t+\alpha$,
since $\xi(t)\in K$; therefore $\bar t+\alpha\in S_j$, contradicting 
the definition of $\bar t$. 

Therefore $\bar t=\tau$, and this implies
that $S_j=[t_j,\tau\,[\,$, so $\xi(t)\in K_\delta$ 
for all $t\in[t_j,\tau\,[\,$. Then the bound 
$L(\xi(s),\eta(s))+C\ge\|f(\xi(s),\eta(s))\|^r$ 
holds for almost all $s\in[t_j,\tau\,[\,$, showing in particular that
the function 
$$
[t_j,\tau\,[\,\ni s\mapsto\psi(s)\equaldef L(\xi(s),\eta(s))+C
$$
which is measurable because $L(\xi(s),\eta(s))=\dot\xi_0(s)$,
is nonnegative. Hence, to prove that 
$\psi$ is Lebesgue-integrable on $[t_j,\tau\,[\,$, 
it suffices to show that
the integrals $\int_{t_j}^t (L(\xi(s),\eta(s))+C)\,ds$,
for $t_j\le t<\tau$, are bounded above by a fixed constant. 
But, if $t_j\le t<\tau$, then

\begin{eqnarray*}
\int_{t_j}^t (L(\xi(s),\eta(s))+C)\,ds&=&\xi_0(t)-\xi_0(t_j)+C(t-t_j)\\
&\le&2|V|+C\sigma\,.
\end{eqnarray*}

\medbreak\noindent Therefore $\psi$ is Lebesgue-integrable on 
$[t_j,\tau\,[\,$,  and then $\dot\xi_0$  is also Lebesgue-integrable 
on $[t_j,\tau\,[\,$, because 
\mbox{$\dot\xi_0=\psi-C$}.  Since 
$\dot\xi_0$  is  Lebesgue-integrable on $[0,t_j]$, we conclude that
$\dot\xi_0$  is  Lebesgue-integrable on $[0,\tau\,[\,$. Hence
the limit $x_0^{\#}=\lim_{t\uparrow\tau}\xi_0(t)$ exists,
and the \mbox{extended} function $\xi_0^{\#}:[0,\tau]\mapsto\R$
defined by $\xi_0^{\#}(t)=\xi_0(t)$ if $0\le t<\tau$, 
$\xi_0^{\#}(\tau)=x_0^{\#}$, is absolutely continuous.

In addition, the bound

$$
\|\dot\xi(s)\|^r=\|f(\xi(s),\eta(s)\|^r\le L(\xi(s),\eta(s))+C=\psi(s)\,,
$$

\medbreak\noindent valid on $[t_j,\tau\,[\,$, shows that
$\dot\xi$ in $L^r$ on $[s_j,\tau\,[\,$, so {\em a \mbox{fortiori}}
$\dot\xi$ is Lebesgue-integrable on $[s_j,\tau\,[\,$, and
then the limit $x^{\#}=\lim_{t\uparrow\tau}\xi(t)$ exists
and belongs to $K_\delta$ (\mbox{because} $x^{\#}=
\lim_{\ell\to\infty}\xi(t_\ell)$ and the $\xi(t_\ell)$ belong to 
$K_\delta$), and 
the extended function $\xi^{\#}:[0,\tau]\mapsto\Omega$
defined by $\xi^{\#}(t)=\xi(t)$ if $0\le t<\tau$, 
$\xi^{\#}(\tau)=x_0^{\#}$, is absolutely continuous.  

Hence we have constructed an absolutely continuous
extension $\Xi^{\#}=(\xi^{\#},\xi_0^{\#})$ of
$\Xi$ to the closed interval $[0,\tau]$. Clearly, 
$\Xi^{\#}$ is also an augmented trajectory of $\hat\Sigma$
\mbox{starting} at $x_*$ at time $0$, and 
the fact that $\Xi$ is of steepest descent with respect to $V$
and $V$ is continuous implies that 
$\Xi^{\#}$ is of steepest descent with respect to $V$ as well. 
\mbox{Therefore} $\Xi$ is not maximal, because we have constructed 
an \mbox{extension} to a strictly larger interval. It follows that the
``right-open case'' cannot arise at all. 
 
\medbreak
We now analyze the right-closed case, and show that it cannot arise either.
We do this by constructing an extension $\Xi^{\#}$ of $\Xi$ to 
a $\Xi^{\#}\in{\cal MSD}_{\hat\Sigma,V,x_*}$ defined on the
interval $[0,\tau+\sigma]$. This will, of course, contradict 
the assumed maximality of $\Xi$, and conclude our proof.

To construct $\Xi^{\#}$, we construct $\eps$-approximate 
\mbox{augmented}
trajectories $\Xi^\eps:[\tau,\tau+\sigma]\mapsto\Omega\times\R$
of $\hat\Sigma$ 
that are ``$\eps$-approximately of steepest descent,''
and such that $\Xi^\eps(\tau)=\Xi(\tau)$. We then pass to the limit 
as $\eps\downarrow 0$, using the compactness theorem (\ref{compacttheo}),
and get an exact  augmented
trajectory $\tilde\Xi:[\tau,\tau+\sigma]\mapsto\Omega\times\R$
of $\hat\Sigma$ such that $\tilde\Xi(\tau)=\Xi(\tau)$, which is 
exactly of steepest descent. 

Write $x^{\#}=\xi(\tau)$, $x_0^{\#}=\xi_0(\tau)$. 
Fix an $\eps$ such that $0<\eps<1$.
Let ${\cal Z}_\eps$ be the set of all
triples $(I',\Xi',S')$ such that 

\begin{itemize}
\item[(1)] $I'$ is a \mbox{subinterval} of $[\tau,\tau+\sigma]$
such that $\tau\in I'$, 
\item[(2)] \mbox{$\Xi'=(\xi',\xi'_0):I'\mapsto K_\delta\times\R$} 
is an $\eps$-approximate augmented trajectory of $\hat\Sigma$, 
\item[(3)] $\Xi'(\tau)=(x^{\#},x_0^{\#})$, 
\item[(4)] $S'$ is a strongly $\eps$-dense subset of $I'$ 
such that \mbox{$\xi'_0(s)+V(\xi'(s))+\eps(t-s)\ge\xi'_0(t)+V(\xi'(t))$} 
for all $s,t\in S'$ such that $s\le t$.
\end{itemize}

\noindent (We say that a subset $S$ of an interval $J$ is {\em
strongly \mbox{$\eps$-dense}} if for every $t\in J$ there exist $s,s'\in S$
such that $s\le t\le s'$ and $s'-s\le\eps$.)

\medbreak
We partially order ${\cal Z}_\eps$ by stipulating that, if 
\mbox{$(I_i,\Xi_i,S_i)\in{\cal Z}_\eps$} for $i=1,2$, then 
\mbox{$(I_1,\Xi_1,S_1)\preceq(I_2,\Xi_2,S_2)$} iff $I_1\subseteq I_2$,
$\Xi_1$ is the \mbox{restriction} of $\Xi_2$ to $I_1$, and 
$S_1=S_2\cap I_1$. 

It is clear that ${\cal Z}_\eps\neq\emptyset$, because the triple 
$(\{\tau\},\Xi_{x^{\#},\tau},\{\tau\})$\mbox{---where} $\Xi_{x^{\#},\tau}$
is the map \mbox{$\Xi:\{\tau\}\mapsto\Omega\times\R$} such that 
$\Xi(\tau)=(x^{\#},\xi_0(\tau))$---belongs to 
${\cal Z}_\eps$.  If $Z$ is a totally ordered subset 
of ${\cal Z}_\eps$, we show that $Z$ has an upper bound 
$(I'_*,\Xi'_*,S'_*)$ in ${\cal Z}_\eps$. 
This conclusion is \mbox{trivial} if $Z=\emptyset$, for in that 
case we can take 
\mbox{$(I'_*,\Xi'_*,S'_*)=(\{\tau\},\Xi_{x^{\#},\tau},\{\tau\})$}. 
Assume that 
$Z\neq\emptyset$. Let $I'_*$ be the union of the intervals $I'$ for all
the \mbox{members} $(I',\Xi',S')$ of $Z$. Then $I'_*$ is clearly a subinterval
of $[\tau,\tau+\sigma]$, and $\tau\in I'$. 
If $t\in I'_*$, we define $\Xi'_*(t)=\Xi'(t)$,
where $(I',\Xi',S')$ is any member of $Z$ such that $t\in I$.
Write $\Xi'_*=(\xi'_*,\xi'_{0,*})$.  Then $\Xi'_*$ is
obviously well \mbox{defined}, and is an 
$\eps$-approximate augmented \mbox{trajectory} of $\hat\Sigma$
such that $\xi'_*(\tau)=x^{\#}$ and
$\xi'_{0,*}(\tau)=x_0^{\#}$. We then let 

\begin{equation}\label{showing}
S'_*=\bigcup_{(I',\Xi',S')\in Z}S'\,.
\end{equation}

We want to prove that the triple $(I'_*,\Xi'_*,S'_*)$ is 
an upper bound for $Z$ in ${\cal Z}_\eps$. If we show that 

\begin{equation}\label{showing-x}
(I'_*,\Xi'_*,S'_*)\in {\cal Z}_\eps\,,
\end{equation}

\medbreak\noindent then the fact that
$(I'_*,\Xi'_*,S'_*)$ is an upper bound for $Z$  is immediate,
so all we really need is to prove (\ref{showing-x}).

It is evident that the triple 
$(I'_*,\Xi'_*,S'_*)$ satisfies the first three of the four
conditions in the definition of ${\cal Z}_\eps$. 
Let us show that it satisfies the fourth one as well. 
If $t\in I'_*$,  we can find  $(I'',\Xi'',S'')\in Z$ such that
$t\in I''$. Since $S''$ is a strongly $\eps$-dense subset
of $I''$, there exist $s_1,s_2\in S''$ such that
$s_1\le t\le s_2$ and $s_2-s_1\le\eps$. Then 
$s_1$ and $s_2$ belong to $ S'_*$, and this establishes that
$S'_*$ is a strongly $\eps$-dense subset of $I'_*$. 
Now, if $s_1$, $s_2$ are members of $S'_*$ such that
$s_1\le s_2$, then we can find (using the fact that 
$Z$ is totally ordered) a member $(I',\Xi',S')$ 
of $Z$ such that $s_1$ and $s_2$ belong to $I'$. It then follows
easily that $s_1$ and $s_2$ must  belong to $S'$. Then,
if we write $\Xi'=(\xi',\xi'_0)$, the fact that 
$s_1\in S'$, $s_2\in S'$, $s_1\le s_2$, and 
\mbox{$(I',\Xi',S')\in{\cal Z}_\eps$},
imply that 

$$
\xi'_0(s_1)+V(\xi'(s_1))+\eps(s_2-s_1)\ge\xi'_0(s_2)+V(\xi'(s_2))\,.
$$
Hence 

$$
\xi'_{0,*}(s_1)+V(\xi'_*(s_1))+\eps(s_2-s_1)\ge
\xi'_{0,*}(s_2)+V(\xi'_*(s_2))\,.
$$
Since this is true for any two members 
$s_1$, $s_2$ of $S'_*$ such that $s_1\le s_2$, we conclude that
the fourth condition holds as well, and the proof of 
(\ref{showing-x}) is complete.

We have shown that every totally ordered subset of ${\cal Z}_\eps$ has
an upper bound in ${\cal Z}_\eps$. Therefore Zorn's
Lemma implies that ${\cal Z}_\eps$ has a maximal element 
$(I',\Xi',S')$.  We claim that $I'=[\tau,\tau+\sigma]$. 
Suppose this was not true.
Then either 
\begin{itemize}
\item[(A)] $I'=[\tau,\zeta\,[\,$ 
for some $\zeta$ such that $\tau<\zeta\le\tau+\sigma$, 
\end{itemize}
or
\begin{itemize}
\item[(B)] $I'=[\tau,\zeta]$ for some $\zeta$ such that 
$\tau\le\zeta<\tau+\sigma$.
\end{itemize}

\medbreak\noindent We shall exclude both possibilities. 

\medbreak Write $\Xi'=(\xi',\xi'_0)$. 
Let \mbox{${\cal B}_\eps=\{v\in\R^n:\|v\|\le\eps\}$.} 
Let 
$I'\ni t\mapsto(\eta'(t),v(t))\in U\times{\cal B}_\eps$ 
be a function such that the function $v(\cdot)$ is measurable,
and the equalities 
$\dot\xi'(t)=f(\xi'(t),\eta'(t))+v(t)$
and $\dot\xi'_0(t)=L(\xi'(t),\eta'(t))$ hold for almost all $t\in I'$. 

Let $t\in I'$. Then there must exist a $\bar t\in S'$ such that
$\bar t\ge t$. Also, $\tau$ must belong to $S'$, because
$\tau$ is the leftmost point of $I'$. Then 

$$
\xi'_0(\tau)+V(\xi'(\tau))+\eps(t-\tau)\ge \xi'_0(\bar t)+V(\xi'(\bar t))\,,
$$
so (since $\eps<1$) 

\begin{eqnarray*}
\xi'_0(\bar t)-\xi'_0(\tau)&\le&\eps(t-\tau)+
V(\xi'(\tau))-V(\xi'(\bar t))\\
&\le&2|\bar V|+\sigma\,.
\end{eqnarray*}

\medbreak\noindent We then have (using the fact that
$\|v(s)\|<1$, which is true because $\eps<1$), 

\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\int_\tau^{t}\|\dot\xi'(s)\|ds\\
&=&\int_\tau^{t}\|f(\xi'(s),\eta'(s))+v(s)\|ds\\
&\le&\int_\tau^{\bar t}\|f(\xi'(s),\eta'(s))+v(s)\|ds\\
&\le&\int_\tau^{\bar t}\|f(\xi'(s),\eta'(s))\|ds+\int_\tau^{\bar t}
\|v(s))\|ds\\
&\le&(\bar t-\tau)^{1/\rho}
\Big(\int_\tau^{\bar t}\|f(\xi'(s),\eta'(s))\|^rds\Big)^{1/r}
+\bar t-\tau\\
&\le& \sigma^{1/\rho}
\Big(\int_\tau^{\bar t}\big(L(\xi'(s),\eta'(s))+C\big)ds\Big)^{1/r}
+\sigma\\
&\le& \sigma^{1/\rho}
\Big(\xi'_0(\bar t)-\xi'_0(\tau)+C(\bar t-\tau)\Big)^{1/r}
+\sigma\\
&\le& \sigma^{1/\rho}\big(2|\bar V|+\sigma+C\sigma\big)^{1/r}
+\sigma\\
&=& \frac{1}{2}\bar\delta(\sigma)\\
&\le&\frac{1}{2}\delta\,.
\end{eqnarray*}

Also, if we let $\theta(s)=\dot\xi'_0(s)+C=
L(\xi'(s),\eta'(s))+C$, then $\theta$ is nonnegative 
and 
\begin{eqnarray*}
\int_\tau^{t}\theta(s)ds&\le&\int_\tau^{\bar t}\theta(s)ds\\
&=&\xi'_0(\bar t)-\xi'_0(\tau)+C(\bar t-\tau)\\
&\le& 2|\bar V|+(C+1)\sigma\,.
\end{eqnarray*}

Since $t$ is an arbitrary member of $I'$, the above
\mbox{inequalities} imply that the functions $\dot\xi'$ and $\theta$
are \mbox{integrable} on $I'$. Since $I'$ is bounded, $\dot\xi'$ is
\mbox{integrable} as well.  This implies, in particular, that the
\mbox{limits} $\lim_{t\uparrow\tau+\zeta}\xi'(t)$ and
$\lim_{t\uparrow\tau+\zeta} \xi'_0(t)$ exist. Hence, if (A) holds, we
can extend $\Xi'$ to the closed interval 
\mbox{$\tilde I'\equaldef[\tau,\tau+\zeta]$}, 
and the result is a curve $\tilde\Xi'$
in $\R^{n+1}$ which is obviously an $\eps$-approximate augmented
\mbox{trajectory} of $\hat\Sigma$. If we then define $\tilde
S'=S'\cup\{\tau+\zeta\}$, then $\tilde S'$ is a strongly $\eps$-dense
subset of $\tilde I'$. Now, if $s_1,s_2$ are \mbox{members} of $\tilde
S'$ such that $s_1\le s_2$ we have to prove that
$\tilde\xi'_0(s_1)+V(\tilde\xi'(s_1))+\eps(s_2-s_1)\ge
\tilde\xi'_0(s_2)+V(\tilde\xi'(s_2))$. This is clearly true if
$s_2<\tau+\zeta$ or if $s_1=s_2$. So the only 
\mbox{remaining} case for us to consider
is when $s_1<s_2=\tau+\zeta$. But in that case we can take a \mbox{sequence}
$\{v_\ell\}_{\ell=1}^\infty$ of points of $I'$ such that
$v_\ell\uparrow\tau+\zeta$ as $\ell\to\infty$, and $v_\ell>s_1$ for
all $\ell$. Then we can pick, for each $\ell$, a $w_\ell\in S'$ such
that $v_\ell\le w_\ell$. On the other hand, $s_1\in S'$, since
$s_1<\tau+\zeta$. Therefore $\tilde\xi'_0(s_1)+V(\tilde\xi'(s_1))
+\eps(w_\ell-s_1)\ge
\tilde\xi'_0(w_\ell)+V(\tilde\xi'(w_\ell))$ for all
$\ell$. If we let $\ell\to\infty$ and use the continuity 
of $\tilde\xi'_0$ and $V$, we find that 

$$
\tilde\xi'_0(s_1)+V(\tilde\xi'(s_1))+\eps(s_2-s_1)
\ge\tilde\xi'_0(s_2)+V(\tilde\xi'(s_2))\,,
$$

\medbreak\noindent as desired. This completes the proof that the 
\mbox{extension} $(\tilde I',\tilde\Xi',\tilde S')$ of 
$(I',\Xi',S')$ is also in ${\cal Z}_\eps$, a fact that 
of course \mbox{contradicts} the maximality of 
$(I',\Xi',S')$ if (A) holds. We have thus derived a contradiction 
from the \mbox{assumption} that (A) is true. Hence (A) is excluded. 

\medbreak We are thus left with Case (B), that is, the possibility that
$I'=[\tau,\tau+\zeta]$ and $\zeta<\sigma$.
We now \mbox{proceed} to exclude this case.
The integral calculation done above shows that 

$$
\int_\tau^{\tau+\zeta}\|\dot\xi'(t)\|dt\le\frac{\delta}{2}\,,
$$
so 
\begin{equation}\label{bbc}
\|\xi'(\tau+\zeta)-\xi'(\tau)\|\le\frac{\delta}{2}\,.
\end{equation}

\medbreak\noindent Since $\xi'(\tau)\in K$, (\ref{bbc}) implies that,
if $q=\xi'(\tau+\zeta)$, then ${\rm dist}(q,K)\le\frac{\delta}{2}$.
In particular, $q$  is an interior point of $K_\delta$. Using Theorem 
\ref{satinfi},  we construct  sequences 
$\{x_j\}_{j=1}^\infty$, $\{v_j\}_{j=1}^\infty$,
$\{\lambda_j\}_{j=1}^\infty$, $\{h_j\}_{j=1}^\infty$,
$\{\gamma_j\}_{j=1}^\infty$
in $\Omega$, $\R^n$, $\R$, $\R$, and $\R$, respectively, such that 
\begin{itemize}
\item[(1)] the inequalities
\begin{eqnarray*}
h_j&>&0\,,\\
\gamma_j&>&0\,,\\
\|x_j-q-h_jv_j\|&\le&h_j\gamma_j\,,\\
V(x_j)&\le&V(q)-h_j\lambda_j+h_j\gamma_j\,,
\end{eqnarray*}
hold for all $j$, 
\item[(2)] $(v_j,\lambda_j)\in\fsh(q,U)$ for all $j$, 
\item[(3)] $h_j\downarrow 0$, $\gamma_j\downarrow 0$, and
$x_j\to q$  as $j\to\infty$.
\end{itemize}
We let $(v_j,\lambda_j)=\fsh(q,u_j)$, $u_j\in U$, and write
\mbox{$\tilde v_j=h_j^{-1}(x_j-q)$}, so that 
\begin{equation}\label{cypr}
\|\tilde v_j-v_j\|\le\gamma_j\,.
\end{equation}

Choose $j$ so large that the following conditions are fulfilled:
\begin{eqnarray}
h_j&<&\eps\,,\\
\gamma_j&<&\frac{\eps}{2}\,,\\
\|x_j-q\|&<&\beta\,,
\end{eqnarray}
where $\beta$ is a positive number such that $\beta<\frac{\delta}{2}$ and
\mbox{$\omega(\beta)<\frac{\eps}{2}$}.


Then 
$$
h_j\|\tilde v_j\|\le\|x_j-q\|\le\beta\le\frac{\delta}{2}\,.
$$

\medbreak\noindent Define a new trajectory 
$\xi^{\#}:I^{\#}\mapsto\R^n$, where
$$
I^{\#}\equaldef[\tau,\tau+\zeta+h_j]\,,
$$ 
by letting 
$$
\xi^{\#}(s)=\left\{\begin{array}{lll}
\xi'(s)&{\rm if}&\tau\le s\le\tau+\zeta\,,\\
q\!+\!(s\!-\!\tau\!-\!\zeta)\tilde v_j
&{\rm if}&\tau+\zeta\le s\le\tau+\zeta+h_j\,,
\end{array}\right.
$$
and augment it by defining
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!
\xi_0^{\#}(s)=\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\left\{\begin{array}{lll}
\xi'_0(s)&{\rm if}&\tau\!\le\!s\!\le\!\tau\!+\!\zeta\,,\\
q_0\!+\!\int_{\tau+\zeta}^sL(\xi^{\#}(v),u_j)\,dv
&{\rm if}&\tau\!+\!\zeta\!\le\!s\!\le\!\tau\!+\!\zeta\!+\!h_j
,\end{array}\right.
\end{eqnarray*}

\medbreak\noindent where $q_0=\xi'_0(\tau+\zeta)$. 
Let $\Xi^{\#}=(\xi^{\#},\xi_0^{\#})$. Then $\Xi^{\#}$ is 
clearly \mbox{absolutely} continuous, and its restriction to 
$[\tau,\tau+\zeta]$
is an \mbox{$\eps$-approximate} augmented trajectory of $\hat\Sigma$. If 
$s$ belongs to \mbox{$[\tau+\zeta,\tau+\zeta+h_j]$}, then

$$
\|\xi^{\#}(s)-q\|\le h_j\|\tilde v_j\|\le\frac{\delta}{2}\,.
$$

\medbreak\noindent Therefore $\xi^{\#}(s)\in K_\delta$. Then

\begin{eqnarray*}
\dot\Xi^{\#}(s)&=&(v_j,L(\xi^{\#}(s),u_j))\\
&=&(f(q,u_j),L(\xi^{\#}(s),u_j))\\
&=&\fsh(\xi^{\#}(s),u_j)+(w_j(s),0)\,,
\end{eqnarray*}

\medbreak\noindent where 

$$
w_j(s)=f(q,u_j)-f(\xi^{\#}(s),u_j)\,.
$$

\medbreak\noindent Since

$$
\|\xi^{\#}(s)-q\|\le h_j\|v_j\|\le\beta\,,
$$

\medbreak\noindent we have 

$$\
\|w_j(s)\|\le\omega(\beta)\le\eps\,.
$$

\medbreak\noindent Therefore $\Xi^{\#}$ is an 
\mbox{$\eps$-approximate} augmented trajectory of $\hat\Sigma$. 

Clearly, $\xi^{\#}(\tau+\zeta+h_j)=x_j$. Therefore,

\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
V(\xi^{\#}(\tau+\zeta+h_j))\\
&=&V(x_j)\\
&\le&V(q)-h_j\lambda_j+h_j\gamma_j\\
&=&V(\xi^{\#}(\tau+\zeta))-h_jL(q,u_j)+h_j\gamma_j\\
&=&V(\xi^{\#}(\tau+\zeta))
-\int_{\tau+\zeta}^{\tau+\zeta+h_j}L(\xi^{\#}(v),u_j)\,dv\\
&\,&\,\,
-\int_{\tau+\zeta}^{\tau+\zeta+h_j}\Big(L(q,u_j)-L(\xi^{\#}(v),u_j)\Big)\,dv
+h_j\gamma_j\\
&=&V(\xi^{\#}(\tau+\zeta))+\xi_0^{\#}(\tau+\zeta)
-\xi_0^{\#}(\tau+\zeta+h_j)+E\,,
\end{eqnarray*}
where 
$$
E=
-\int_{\tau+\zeta}^{\tau+\zeta+h_j}\Big(L(q,u_j)-L(\xi^{\#}(v),u_j)\Big)\,dv
+h_j\gamma_j\,.
$$

\medbreak\noindent Then
$$
E\le h_j\omega(\beta)+h_j\gamma_j\le\eps h_j\,.
$$

\medbreak\noindent Therefore

\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!V(\xi^{\#}(\tau+\zeta+h_j))\le\\
&\,&\!\!\!\!\!V(\xi^{\#}(\tau+\zeta))+\xi_0^{\#}(\tau+\zeta)
-\xi_0^{\#}(\tau+\zeta+h_j)+\eps h_jE\,,
\end{eqnarray*}
that is, 
\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
V(\xi^{\#}(\tau+\zeta))+\xi_0^{\#}(\tau+\zeta)
+\eps h_j\nonumber\\
&\ge&V(\xi^{\#}(\tau+\zeta+h_j))+\xi_0^{\#}(\tau+\zeta+h_j)\,.\label{labla-y}
\end{eqnarray}

\medbreak\noindent This last inequality implies that, if we define

$$
S^{\#}=S'\cup\{\tau+\zeta+h_j\}\,,
$$

\medbreak\noindent then we can easily show that
$S^{\#}$ is a strongly $\eps$-dense subset
of $I^{\#}$ such that 

\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
V(\xi^{\#}(s_1))+\xi_0^{\#}(s_1)+\eps(s_2-s_1)\nonumber\\
&\ge& V(\xi^{\#}(s_2))+\xi_0^{\#}(s_2)\label{labla}
\end{eqnarray}

\medbreak\noindent whenever $s_1,s_2\in S^{\#}$ and
$s_2\ge s_1$.  The strong \mbox{$\eps$-density} follows because 
$S'$ is a strongly $\eps$-dense subset of $I'$ and $h_j<\eps$,
since $\tau+\zeta$ necessarily belongs to $S'$. Inequality (\ref{labla}) 
is clearly true if $s_1=s_2$ or both $s_1$ and $s_2$ belong to 
$S'$. To verify that it holds in the remaining case, that is, when
$s_1\in S'$ and $s_2=\tau+\zeta+h_j$, if suffices to use once again
the fact that $\tau+\zeta\in S'$, so 


\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
V(\xi^{\#}(s_1))+\xi_0^{\#}(s_1)+\eps(\tau+\zeta-s_1)\nonumber\\
&\ge&  V(\xi^{\#}(\tau+\zeta)+\xi_0^{\#}(\tau+\zeta)\label{labla-x}\,.
\end{eqnarray}

\medbreak\noindent If we add (\ref{labla-y}) and (\ref{labla-x}), and
cancel the sum 
\mbox{$V(\xi^{\#}(\tau+\zeta))+\xi_0^{\#}(\tau+\zeta)$} that
appears on both sides of the result, we get 

\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!
V(\xi^{\#}(s_1))+\xi_0^{\#}(s_1)+\eps(\tau+\zeta+h_j-s_1)\nonumber\\
&\ge&  V(\xi^{\#}(\tau+\zeta+h_j))
+\xi_0^{\#}(\tau+\zeta+h_j)\label{labla-uu}\,,
\end{eqnarray}

\medbreak\noindent that is, 

\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
V(\xi^{\#}(s_1))+\xi_0^{\#}(s_1)+\eps(s_2-s_1)\nonumber\\
&\ge&  V(\xi^{\#}(s_2))+\xi_0^{\#}(s_2)\label{labla-vv}\,.
\end{eqnarray}


\medbreak\noindent It then follows that $(I^{\#},\Xi^{\#},S^{\#})$
belongs to ${\cal Z}_\eps$. \mbox{Since}
$(I',\Xi',S')\preceq(I^{\#},\Xi^{\#},S^{\#})$ but
\mbox{$(I',\Xi',S')\neq(I^{\#},\Xi^{\#},S^{\#})$}, we have arrived at a
\mbox{contradiction}, which this time has arisen from the \mbox{assumption}
that (B) holds. Hence (B) is excluded as well.

It now follows that $I'=[\tau,\tau+\sigma]$. In other \mbox{words}, we
have shown that there exists an $\eps$-approximate \mbox{augmented}
trajectory $\Xi_\eps=(\xi_\eps,\xi_{0,\eps})$ of $\hat\Sigma$ which
is \mbox{defined} on $[\tau,\tau+\sigma]$, and is such that there is a
strongly $\eps$-dense subset $S_\eps$ of $[\tau,\tau+\sigma]$ having
the property that 

\begin{equation}
V(\xi_\eps(s_1))\!+\!\xi_{0,\eps}(s_1)\!+\!\eps(s_2\!-\!s_1)
\!\ge\! V(\xi_\eps(s_2))\!+\!\xi_{0,\eps}(s_2)\label{labla-aa}
\end{equation}


\medbreak\noindent whenever $s_1,s_2\in S_\eps$ and $s_1\le s_2$.

Clearly, the points $\tau$, $\tau+\sigma$ must belong to $S_\eps$.
If we aplly (\ref{labla-aa}) with $s_1=\tau$, $s_2=\tau+\sigma$, we find
the bound 

$$
\xi_{0,\eps}(\tau+\sigma)-\xi_{0,\eps}(\tau)\le 2|\bar V|+\sigma\,.
$$

\medbreak\noindent Therefore, if $\{\eps_j\}_{j=1}^\infty$ is a sequence
of positive numbers that converges to $0$, we can apply 
Theorem (\ref{compacttheo}) with $T=\sigma$ and
$\Xi^j=(\xi^j,\xi_{0}^{j})=(\check\xi_{\eps_j},\check\xi_{0,\eps_j})$,
where \mbox{$\check\xi_{\eps_j}(s)=\xi_{\eps_j}(s+\tau)$} and 
\mbox{$\check\xi_{0,\eps_j}(s)=\xi_{0,\eps_j}(s+\tau)-
\xi_{0,\eps_j}(\tau)$} for $s\in[0,\sigma$. 
Then, after passing to a \mbox{subsequence},
we may assume that there exists an augmented
trajectory $\Xi^\infty=(\xi^\infty,\xi_{0}^\infty)$ of 
$\hat\Sigma$ such that
\begin{itemize}
\item[(i)] $\xi_0^\infty(\tau)=0$, 
\item[(ii)] the sequence $\{\xi^{j}\}_{j=1}^\infty$
converges uniformly to $\xi^\infty$, and 
\item[(iii)] 
$\liminf_{j\to\infty}\inf\big\{\Delta^{j,\infty}(t,s)\!:\!
0\!\le\!s\!\le\!t\!\le\!T\big\}\!\ge\!0$,
where, if $j,j'\in\nn\cup\{+\infty\}$, we define
$\Delta^{j,j'}$ as in (\ref{defodel}). 
\end{itemize}

\medbreak\noindent We now show that $\Xi^\infty$ 
is an augmented  trajectory of steepest descent of
$\hat\Sigma$ from $x^{\#}$ with respect to $V$.  For this purpose,
we pick $s_1,s_2\in[0,\sigma]$ such that $s_1\le s_2$, and prove that

\begin{equation}
V(\xi^\infty(s_1))+\xi_{0}^\infty(s_1)
\ge V(\xi^\infty(s_2))+\xi_{0}^\infty(s_2)\,.\label{labla-bb}
\end{equation}

\medbreak\noindent For this purpose we pick, for each sufficiently large
$j$, points $s_1^j,s_2^j$ in $S_{\eps_j}$ such that 


$$
s_1^j\le s_1+\tau\le s_1+\eps_j<s_2^j-\eps_j\le
s_2+\tau\le s_2^j\,.
$$

\medbreak\noindent (We assume that $s_1<s_2$, because 
(\ref{labla-bb}) is trivially true if $s_1=s_2$.) 

Then 

\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
V(\xi_{\eps_j}(s_1^j))-V(\xi_{\eps_j}(s_2^j))+\eps_j(s_2^j-s_1^j)
\nonumber\\
&\ge&\xi_{0,\eps_j}(s_2^j)-\xi_{0,\eps_j}(s_1^j)
\,,\label{labla-cc}
\end{eqnarray}

\medbreak\noindent that is, 

\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
V(\xi^j(s_1^j-\tau))-V(\xi^j(s_2^j-\tau))+\eps_j(s_2^j-s_1^j)
\nonumber\\
&\ge&\xi_{0}^j(s_2^j-\tau)-\xi_{0}^j(s_1^j-\tau)\nonumber\\
&=&\xi_{0}^\infty(s_2^j-\tau)-\xi_{0}^\infty(s_1^j-\tau)
+\Delta^{j,\infty}(s_2^j-\tau,s_1^j-\tau)\nonumber\\
&\ge&\xi_{0}^\infty(s_2^j-\tau)-\xi_{0}^\infty(s_1^j-\tau)\nonumber\\
&\,&\,\,+\inf\{\Delta^{j,\infty}(t,s):0\le s\le t\le\sigma\}\nonumber
\end{eqnarray}

\medbreak\noindent Then 

\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!V(\xi^\infty(s_1))-V(\xi^\infty(s_2))
\nonumber\\
&=&\lim_{j\to\infty}\Big(
V(\xi^j(s_1^j\!-\!\tau))\!-\!V(\xi^j(s_2^j\!-\!\tau))\!+\!
\eps_j(s_2^j\!-\!s_1^j)\Big)\nonumber\\
&\ge&\liminf_{j\to\infty}\Big(
\xi_{0}^j(s_2^j-\tau)-\xi_{0}^j(s_1^j-\tau)\Big)\nonumber\\
&=&\liminf_{j\to\infty}\Big(
\xi_{0}^\infty(s_2^j-\tau)-\xi_{0}^\infty(s_1^j-\tau)\nonumber\\
&\,&\,\,\,+\Delta^{j,\infty}(s_2^j-\tau,s_1^j-\tau)\Big)\nonumber\\
&\ge&\liminf_{j\to\infty}\Big(
\xi_{0}^\infty(s_2^j-\tau)-\xi_{0}^\infty(s_1^j-\tau)\nonumber\\
&\,&\,\,+\inf\{\Delta^{j,\infty}(t,s):0\le s\le t\le\sigma\}\Big)\nonumber\\
&=&\lim_{j\to\infty}\Big(
\xi_{0}^\infty(s_2^j-\tau)-\xi_{0}^\infty(s_1^j-\tau)\Big)\nonumber\\
&\,&\,\,+\liminf_{j\to\infty}\Big(
\inf\{\Delta^{j,\infty}(t,s):0\le s\le t\le\sigma\}\Big)\nonumber\\
&=&\xi_{0}^\infty(s_2)-\xi_{0}^\infty(s_1)\nonumber\\
&\,&\,\,+\liminf_{j\to\infty}\Big(
\inf\{\Delta^{j,\infty}(t,s):0\le s\le t\le\sigma\}\Big)\nonumber\\
&\ge&\xi_{0}^\infty(s_2)-\xi_{0}^\infty(s_1)\,.\nonumber
\end{eqnarray}

\medbreak\noindent We have thus proved (\ref{labla-bb}), thereby 
establishing that $\Xi^\infty$ is an augmented trajectory of
$\hat\Sigma$ from $x^{\#}$ of steepest descent 
with respect to $V$. If we now concatenate
$\Xi$ and $\Xi^\infty$ in the obvious way, by defining

$$
\xi^{\#}(s)=\left\{\begin{array}{lll}
\xi(s)&{\rm if}&0\le s\le\tau\,,\\
\xi^\infty(s-\tau)&{\rm if}&\tau\le s\le\tau+\sigma\,,
\end{array}\right.
$$

$$
\xi_0^{\#}(s)=\left\{\begin{array}{lll}
\xi_0(s)&{\rm if}&0\le s\le\tau\,,\\
\xi_0^\infty(s-\tau)+\xi_0(\tau)&{\rm if}&\tau\le s\le\tau+\sigma\,,
\end{array}\right.
$$

\medbreak\noindent then $\Xi^{\#}$ is an augmented
trajectory of $\hat\Sigma$ from $x_*$ of steepest descent
with respect to $V$, defined on $[0,\tau+\sigma]$. This contradicts
the maximality of $\Xi$, and concludes our proof. 
\end{proof}

\vbox{

\ssection{The dynamic programming \mbox{inequality}.}

\begin{theorem}\label{dynineqtheo}
Let $\Omega$ be an open subset of $\R^n$,  and let 
$f:\Omega\mapsto\R^n$, \mbox{$L:\Omega\mapsto\R$}
be continuous maps. 
Let \mbox{$V:\Omega\mapsto\R$} be a \mbox{continuous} function that  
satisfies 

\begin{equation}\label{sativis}
-\nabla V(x)\cdot f(x)-L(x)\le 0
\end{equation}
	
\medbreak\noindent on $\Omega$ in the \mbox{viscosity} sense. Then 
for every $x_*\in\Omega$ there exists a curve $\xi$ in
$\Omega$, defined on an interval $I$ that contains $0$, such that 
\begin{itemize}
\item[(1)] $\xi$ is an integral curve of $f$ (that is, 
$\xi$ is \mbox{locally} \mbox{absolutely} continuous and $\dot\xi(t)=f(\xi(t))$
for \mbox{almost} every $t\in I$, from which it follows that 
$\xi$ is \mbox{continuously} differentiable and
$\dot\xi(t)=f(\xi(t))$ for every $t\in I$), 
\item[(2)] $\xi(0)=x_*$,
\item[(3)] $V(\xi(s))\le V(\xi(t))+\int_s^tL(\xi(v))\,dv$ 
whenever $s,t\in I$ and $s\le t$, 
\item[(4)] $\xi$  is right-unbounded.
\end{itemize}
\end{theorem}
}

\begin{proof}
Let $U$ be a set consisting of a single point $\bar u$. Let $\tilde\Sigma$ 
be the augmented control system $(\Omega,U,\tilde f,\tilde L)$,
where $\tilde f(x,\bar u)=f(x)$, $\tilde L(x,\bar u)=-L(x)$. Then 
$\tilde\Sigma$ satisfies all the hypotheses of Theorem \ref{maintech}. 

Let ${\cal V}=-V$. We claim that ${\cal V}$ satisfies 
$$
\sup\{-\nabla{\cal V}(x)\cdot\tilde f(x,u)-\tilde L(x,u):u\in U\}\ge 0
$$
in the viscosity sense. To prove this, we have to pick a
point $x\in\Omega$ and a subdifferential $p$ of ${\cal V}$ at
$x$, and show that 
$$
\sup\{-p\cdot\tilde f(x,u)-\tilde L(x,u):u\in U\}\ge 0\,,
$$
i.e., that 
\begin{equation}\label{sativis-x}
-p\cdot f(x)+L(x)\ge 0\,.
\end{equation}
But, if $p$ is a subdifferential of $-V$ at $x$, and we let 
\mbox{$\pi=-p$}, then it follows that
$\pi$ is a superdifferential of $V$ at $x$. Since $V$ satisfies 
(\ref{sativis}) in the viscosity sense, this implies that
$-\pi\cdot f(x)-L(x)\le 0$. But then 
\mbox{$p\cdot f(x)-L(x)\le 0$}, so $-p\cdot f(x)+L(x)\ge 0$, 
and (\ref{sativis-x}) has been proved.
		
We can therefore apply the trivial Proposition \ref{trivialprop}
to the augmented system $\tilde\Sigma$ and the function ${\cal V}$, 
and \mbox{conclude} 
that there exists a maximal augmented \mbox{trajectory}
$\Xi=(\xi,\xi_0)$ of $\tilde\Sigma$ from $x_*$ of steepest descent 
with respect to ${\cal V}$, 
and then use the nontrivial \mbox{Theorem} \ref{maintech} to conclude that
$\xi$ is right-unbounded. 
The fact that $\xi$ is a trajectory of $\tilde\Sigma$ 
means, of course, that $\xi$ is an \mbox{integral} curve of $f$. 
The steepest descent condition says that
$$
{\cal V}(\xi(s))\ge{\cal V}(\xi(t))+\int_s^t(-L(\xi(v)))\,dv
$$
whenever $0\le s\le t<\tau$. But this says precisely that 
$$
V(\xi(s))\le V(\xi(t))+\int_s^tL(\xi(v))\,dv
$$
whenever $0\le s\le t<\tau$. Hence $\xi$ satisfies all the desired
properties, and our proof is complete. 
\end{proof}

The following result is then a trivial corollary of 
\mbox{Theorem} \ref{dynineqtheo}.

\begin{theorem}\label{dynineqtheo-a}
Let $\hat\Sigma=(\Omega,U,f,L)$ be an 
\mbox{$n$-dimensional} augmented system
such that the map \mbox{$\Omega\ni x\mapsto (f(x,u),L(x,u)$} is continuous
for each $u\in U$. 
Let \mbox{$V:\Omega\mapsto\R$} be a \mbox{continuous} function that  
satisfies (\ref{upeq}) on $\Omega$ in the \mbox{viscosity} sense. Then 
for every $x_*\in\Omega$ and every piecewise constant
function $\eta:[0,\infty\,[\,\mapsto U$ there exists a curve 
$\xi:I\mapsto\Omega$, defined on a subinterval $I$ of $[0,\infty\,[\,$,
such that 
\begin{itemize}
\item[(1)] $0\in I$ and $\xi(0)=x_*$,
\item[(2)] $\xi$ is a trajectory for the control $\eta$ (that is, 
$\xi$ is \mbox{locally} \mbox{absolutely} continuous and 
$\dot\xi(t)=f(\xi(t),\eta(t))$ for \mbox{almost} every $t\in I$)
\item[(3)] $V(\xi(s))\le V(\xi(t))+\int_s^tL(\xi(v),\eta(v))\,dv$ 
whenever $s,t\in I$ and $s\le t$, 
\item[(4)] $\xi$ is right-unbounded. \proofend
\end{itemize}
\end{theorem}

Theorem  \ref{dynineqtheo-a} has the following immediate 
\mbox{consequence}.

\begin{theorem}\label{dynineqtheo-b}
Let $\hat\Sigma=(\Omega,U,f,L)$ be an 
\mbox{$n$-dimensional} augmented system
such that the map \mbox{$\Omega\ni x\mapsto (f(x,u),L(x,u)$} is continuous
for each $u\in U$. 
Let \mbox{$V:\Omega\mapsto\R$} be a \mbox{continuous} function that  
satisfies (\ref{upeq}) on $\Omega$ in the \mbox{viscosity} sense.
Let $\Xi(\xi,\xi_0):I\mapsto\Omega\times\R$ be a locally uniquely 
limiting augmented trajectory of $\hat\Sigma$. Then
the dynamic programming inequality 
$$
V(\xi(s))\le V(\xi(t))+\xi_0(t)-\xi_0(s)
$$
holds for all $s,t\in I$ such that $s\le t$. \proofend
\end{theorem}

\ssection{Proof of Theorem \ref{mainres}}

\noindent According to Theorem \ref{dynineqtheo-b}, the \mbox{dynamic} 
\mbox{programming} inequality holds along every almost locally
uniquely limiting augmented trajectory. The hypothesis 
that every augmented arc has an 
almost locally uniquely limiting improvement then implies that 
the dynamic \mbox{programming} inequality
holds along every augmented trajectory. If we apply 
the inequality to an augmented arc $\Xi$ that starts at a point 
$x$ and ends at the target,
and use the fact that $V=0$ on the target, we find that 
$V(x)\le J(\Xi)$. Hence $V$ is bounded above by the value function
$\vts$.

To prove that $V\ge\vts$, we pick $x\in\Omega$ and
use \mbox{Proposition} \ref{trivialprop} to conclude that
there exists a maximal augmented \mbox{trajectory}
$\Xi=(\xi,\xi_0)$ of $\tilde\Sigma$ from $x$ of \mbox{steepest} descent 
with respect to ${\cal V}$, and then use  
\mbox{Theorem} \ref{maintech} to conclude that $\xi$ is right-unbounded.  
We then invoke our hypotheses to conclude
that $\Xi$ ends at the target. Then $V(x)\ge J(\Xi)$, so
$V$ is bounded below by the value function, and our proof is complete.\proofend



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