%10418.22
% Slemrod: 3287 (San Martin, Room 107), 253, ext. 4305
% AMS-LaTeX 1.2 sample file for book proceedings, based on amsproc.cls.
%
\documentclass[twoside]{amsproc}
%\font\mynine=cmr9
\def\quocar{We follow \cite{carath}, pages 210-212.}
\newcounter{gnucount}
\newcommand{\Ft}[1]{\footnote{\mynine\baselineskip 8pt #1}}
\setcounter{gnucount}{0}
\newcommand{\gnu}[1]{\refstepcounter{gnucount}
\medbreak
\begin{center}
\fbox{\begin{minipage}{11cm}
\noindent GNU NO. \thegnucount.\ \ #1\ GNU-END
\end{minipage}}\end{center}
\medbreak}
%Uncomment the following  line to run file without printing the gnus
%\renewcommand{\gnu}[1]{}
\usepackage{epsf}
\input epsf
\usepackage{amssymb}
%\usepackage{makeidx}
\parskip=0.8mm
\def\comment#1{}
\def\snell{Snell}
\def\traj{{\rm Traj\,}}
\def\llist#1{
\vspace{-4.5mm}
\begin{center}
$\quad\quad\,\,$\mbox{\begin{minipage}{12.2cm}
#1
\end{minipage}}\end{center}}

\def\lllist#1{
\vspace{-4.5mm}
\begin{center}
$\quad\quad\quad$\mbox{\begin{minipage}{11.9cm}
#1
\end{minipage}}\end{center}}

\newcommand{\compo}{\circ}
\def\transv{{\cap\kern-5pt\lower -2.2pt\hbox{$|$}\kern-4.9pt\lower
-7pt\hbox{$-$}}} 
\def\baba{|\kern-1.7pt|}
\def\stransv{{\cap\kern-5.3pt\lower -2.2pt\hbox{$\baba$}\kern-5.3pt\lower
-7pt\hbox{$-$}}} 
\newcommand{\eps}{\varepsilon}
\def\verylongrightarrow{\raise2.2pt\hbox to 25pt{\hrulefill}\kern-4pt\longrightarrow}
\def\verylongline{\raise2.2pt\hbox to 37pt{\hrulefill}}
\def\ovarrow#1{{\stackrel{#1}{\verylongrightarrow}}}
\def\ovline#1{{\stackrel{#1}{\verylongline}}}
\def\ovlarrow#1{{\stackrel{#1}{\verylongleftarrow}}}
\def\verylongleftarrow{\longleftarrow\kern-4pt\raise2.2pt\hbox to 25pt{\hrulefill}}
\def\longsearrow{\setlength{\unitlength}{0.5cm}
\begin{picture}(1.2,1.2)
\put(-0.2,1.2){\vector(1,-1){2}}
\end{picture}}
\def\longnearrow{\setlength{\unitlength}{0.5cm}
\begin{picture}(1.2,1.2)
\put(-0.2,-1.2){\vector(1,1){2}}
\end{picture}}
\def\longdownarrow{\setlength{\unitlength}{0.25cm}
\begin{picture}(0.6,1)
\put(0.2,1){\vector(0,-1){2}}
\end{picture}}
\def\longuparrow{\setlength{\unitlength}{0.25cm}
\begin{picture}(1.2,1.2)
\put(0.2,-0.5){\vector(0,1){2}}
\end{picture}}
\def\veceps{\vec{\eps}} 
\def\vic{\pmb}
\def\Is{I^{\#}}
\def\Xis{\Xi^{\#}}
\def\xis{\xi^{\#}}
\def\etas{\eta^{\#}}
\def\zetas{\zeta^{\#}}
\def\qs{q^{\#}}
\def\ts{t^{\#}}
\def\taus{\tau^{\#}}
\def\nus{\nu^{\#}}
\def\restr{\,\lceil\,}
\def\corestr{\,\rceil\,}
\def\equaldef{{\buildrel \rm def \over =}}
\def\mybox{$\lower3.5pt\vbox{\hrule height2.6pt depth-2pt\hbox{\vrule width0.6pt\kern1pt\vbox{\kern1pt\phantom{nj}\kern1pt}\kern1.5pt\vrule width0.6pt}\hrule height2.6pt depth-2pt}$\medbreak}
\def\proofendbox{\enspace\hfill \mybox\bigskip}
\def\proofendsq{\hfill $\square$\medskip}
\def\proofend{\hfill$\diamondsuit\,\,$}
\def\pmb#1{\setbox0=\hbox{$#1$}%
  \kern-.025em\copy0\kern-\wd0
  \kern.05em\copy0\kern-\wd0
  \kern-.025em\raise.0433em\box0 }
\def\pmbo#1{\setbox0=\hbox{$#1$}%
  \kern-.055em\copy0\kern-\wd0
  \kern.05em\copy0\kern-\wd0
  \kern-.055em\raise.0433em\box0 }
%\def\bem{\bf\em }
\def\bem{\em }
\def\bemu{{\bem U}}
\def\tto{\longrightarrow\kern-9pt\longrightarrow}
\def\zz{\mathbb{Z}}
\def\qq{\mathbb{Q}}
\def\cc{\mathbb{C}}
\def\rr{\mathbb{R}}
\def\nn{\mathbb{N}}
\def\kk{\mathbb{K}}
\def\bmath{\mathbb{B}}
\def\smath{\mathbb{S}}
\def\zmath{\mathbb{Z}}
\newcommand{\nball}{{\bmath}^n}
\font\mynino=cmr9 scaled\magstephalf
\def\tub{\mathcal{T}}
\def\tubx{\mathcal{T}^X}
\def\btub{\bar\mathcal{T}}
\def\btubx{\bar\mathcal{T}^X}
\def\linx{{\rm Lin}(X,X)}
\def\linxxp{{\rm Lin}(X,X')}
\def\id{\kern.2em{\rm I}\kern-.56em{\rm 1}}
\def\idnpball{\id_{\partial\bmath_n}}
\def\so{{\rm So}}
\def\ta{{\rm Ta}}
\def\gr{{\rm Gr}}
\def\dom{{\rm Do}}
\def\im{{\rm Im}}
\def\togr{\,\,{\buildrel \rm igr \over \longrightarrow}\,\,}
\def\caF{\mathcal{F}}
%\def\spaco{\ \ \ \ \ \ }
\newcommand{\ssection}[1]{
%\ifodd \thepage%\clearpage
%\thispagestyle{empty}\hbox{$\,$}%\clearpage
%\thispagestyle{empty}\else%\clearpage
%\thispagestyle{empty}\fi
\hbox{$\,$}\vskip 0.3cm\section{#1}\vskip0.5cm}
\newcommand{\ssubsection}[1]{\subsection{#1}
\addcontentsline{toc}{section}{$\,\,\,\,$\arabic{section}.\arabic{subsection}
\ #1}}
\newcommand{\ben}{\begin{enumerate}}
\newcommand{\een}{\end{enumerate}}
\newtheorem{theorem}{Theorem}[subsection]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{Fact}[theorem]{Fact}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{example}[theorem]{Example}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{xca}[theorem]{Exercise}

\numberwithin{equation}{section}

%    Absolute value notation
\newcommand{\abs}[1]{\lvert#1\rvert}

%    Blank box placeholder for figures (to avoid requiring any
%    particular graphics capabilities for printing this document).
\newcommand{\blankbox}[2]{%
  \parbox{\columnwidth}{\centering
%    Set fboxsep to 0 so that the actual size of the box will match the
%    given measurements more closely.
    \setlength{\fboxsep}{0pt}%
    \fbox{\raisebox{0pt}[#2]{\hspace{#1}}}%
  }%
}

\setcounter{tocdepth}{1}

\begin{document}


\newcommand{\hoo}{\raise 1.2pt\hbox{$\stackrel{o}{H}$}}
\newcounter{scholco}
\setcounter{scholco}{0}
\newcounter{apco}
\setcounter{apco}{0}
\newcounter{figurco}
\setcounter{figurco}{0}
\renewcommand{\thescholco}{\Roman{scholco}}
\newcommand{\scholium}[1]{\refstepcounter{scholco}
\subsection*{$\!\!\!\!\!\!\!\!$SCHOLIUM \thescholco.\ \ #1}
\addcontentsline{toc}{section}{$\qquad${Scholium \thescholco.\
    \ #1}}}
\newcommand{\figurecap}[1]{\refstepcounter{figurco}\bigskip
\begin{center}{\sc Figure \arabic{figurco}}.\ #1
\end{center}\bigskip}
\newcommand{\ssubsectionap}[1]{\refstepcounter{apco}
\subsection*{$\!\!\!\!\!\!\!\!$A.\arabic{apco}.\ \ #1}
\addcontentsline{toc}{section}{$\qquad$A.\arabic{apco}.\
    \ #1}}

%\section*{This is an unnumbered first-level section head}
%This is an example of an unnumbered first-level heading.

%\specialsection*{This is a Special Section Head}
%This is an example of a special section head%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\footnote{Here is an example of a footnote. Notice that this footnote
%text is running on so that it can stand as an example of how a footnote
%with separate paragraphs should be written.
%\par
%And here is the beginning of the second paragraph.}%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\def\gua{H.~J.~SUSSMANN}
\def\guo{THE MAXIMUM PRINCIPLE}

\pagestyle{myheadings}
\markboth{\gua}{\guo}


\def\restr{\lceil}
\def\pmb#1{\setbox0=\hbox{$#1$}%
  \kern-.025em\copy0\kern-\wd0
  \kern.05em\copy0\kern-\wd0
  \kern-.025em\raise.0144em\box0 
  \kern-.025em\raise.0288em\box0 
  \kern-.025em\raise.0433em\box0 }
%\renewcommand{\thesubsubsection}{\S$\,$\arabic{subsubsection}$\,$}
\def\equaldef{\,{\buildrel \rm def \over =}\,}
\def\hham{\mathcal{ H}}
\font\mytenrm=cmr10 scaled\magstephalf
\font\mytensl=cmsl10 scaled\magstephalf
\font\mytenit=cmti10 scaled\magstephalf
\font\forsection=cmbx10 scaled\magstep 0
\font\mynine=cmr9
\font\mynino=cmr9 scaled\magstephalf
\font\mit=cmti9
\font\myninebf=cmbx9
\font\mysmall=cmr7
\font\mysmallit=cmti7
\font\myverysmall=cmr5
\def\eps{\varepsilon}
\def\smalltyperm{\baselineskip=12.2pt \rm}
\def\smalltypesl{\baselineskip=12.2pt \sl}
\newcommand{\bi}{\begin{itemize}}
\newcommand{\ei}{\end{itemize}}
\newenvironment{Quoterm}{
\bigskip\begin{center}\begin{minipage}{11cm}\smalltyperm\noindent}
{\end{minipage}\end{center}\bigskip}
\newenvironment{Quotesl}{
\bigskip\begin{center}
\begin{minipage}{11cm}\smalltypesl\noindent}{\end{minipage}
\end{center}\bigskip}

\newcommand{\bqrm}{\begin{Quoterm}}
\newcommand{\eqrm}{\end{Quoterm}}
\newcommand{\bqsl}{\begin{Quotesl}}
\newcommand{\eqsl}{\end{Quotesl}}
\parindent 6mm
\newtheorem{ttheorem}{Theorem*}
\newtheorem{etheorem}{Theorem*}
\newtheorem{qtheorem}{Theorem*}
\newtheorem{myquestion}[qtheorem]{Open Problem}
\newenvironment{question}{\begin{myquestion}$\!\!\!$.\ \rm}{\end{myquestion}}
\newcommand{\bques}{\begin{question}}
\newcommand{\eques}{\end{question}}
\thispagestyle{empty}
\newcommand{\hdleft}{{\rm H. J. Sussmann}}
\newcommand{\hdright}{{\rm \hfil Hamilton and Weierstrass}}

\setcounter{page}{17}
\setcounter{section}{3}

\begin{center}
{\Huge{\bf TRANSVERSALITY,\\
$\,$\\
\vspace{-5mm}
SET SEPARATION,\\
$\,$\\
\vspace{-5mm}
AND THE PROOF OF\\
$\,$\\
\vspace{-4mm}
THE MAXIMUM PRINCIPLE,\\
$\,$\\
\vspace{-4mm}
PART II}}\\
$\,\,$\\
$\,\,$\\
{\Large{\it H\'ector J. Sussmann} --- January 10, 2001}
\end{center}

\vspace{1cm}
{\large\noindent{\em This is the continuationd of the notes of 
January 3. In particular, the page and section numbering 
continue those of the previous set of notes.}}

\vspace{1cm}

\begin{center}
{\huge\bf CONTENTS}
\end{center}

\medbreak

\thispagestyle{empty}

\medbreak

{\large 

\noindent 4.\ END OF THE PROOF OF THE MAIN TECHNICAL LEMMA\hfill 18

\medbreak
\noindent 5.\ TRANSVERSALITY OF CONES AND SET SEPARATION\hfill 21

\noindent \ 5.1\ Transversality and strong transversality of cones\hfill 23

\noindent \ 5.2\ Transversality vs. strong transversality\hfill 24

\noindent \ 5.3\ The main topological lemma\hfill 25

\noindent \ 5.4\ The transversal intersection theorem\hfill 25

\medbreak
\noindent 6.\ COMPLETION OF THE  PROOF\hfill 30

\noindent \ 6.1\ Application of the transversal intersection theorem\hfill 30

\noindent \ 6.2\ The compactness argument \hfill 31

\noindent \ 6.3\ The momentum\hfill 32

\noindent \ 6.4\ Elimination of the continuity assumption on the reference
control \hfill 33



\newpage

\section{END OF THE PROOF OF THE MAIN TECHNICAL LEMMA}
We have to show that
\begin{equation}\label{ourfor}
\omega(\vec{\eps})=o(\|\vec{\eps}\|)\quad{\rm as}\quad\vec{\eps}\to 0\,,
\end{equation}
where
$$
\omega(\vec{\eps})\equaldef\sup\{\|R^{\vec{\eps}}(s)\|:a\le s\le b\}\,,
$$
\begin{eqnarray*}
R^{\vec{\eps}}(t)&\equaldef&\int_{a}^t\Big(\Delta_1^{\vec{\eps}}(s)-
A(s)\cdot(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(s)-\xi_*(s))\Big)\,ds\\
&\,&\,\,+\int_{[a,t]\cap E(\vec{\eps})}\Big(\Delta_2^{\vec{\eps}}(s)-
\theta^{\vec{\eps}}(s)\Big)\,ds\,,
\end{eqnarray*}
%GUGU
$$
\Delta_1^{\vec{\eps}}(t)\equaldef f(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t),
\eta_*(t),t)-f(\xi_*(t),\eta_*(t),t)\,,
$$
$$
\Delta_2^{\vec{\eps}}(t)\equaldef f(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t),
\eta_*^{\vec{\tau},\vec{u},\vec{\eps}}(t),t)-
f(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t),\eta_*(t),t)\,,
$$
$$
A(t)=\frac{\partial f}{\partial x}(\xi_*(t),\eta_*(t),t)\,,
$$
and
$$
\theta^{\vec{\eps}}(t)=\left\{\begin{array}{lcl}
0&{\rm if}&t\notin E(\vec{\eps})\,,\\
f(x_j,u_j,\tau_j)-f(x_j,u_{*,j},\tau_j)&{\rm if}&t\in[\tau_j,\tau_j+\eps_j]\,.
\end{array}\right.
$$

First, we have
$$
\Delta_1^{\vec{\eps}}(t)=
\int_0^1\frac{\partial f}{\partial x}\Big(\xi_*(t)+\nu
(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t)-\xi_*(t)),\eta_*(t),t\Big)
\cdot\Big(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t)-\xi_*(t)\Big)\,d\nu\,,
$$
so that
$$
\Delta_1^{\vec{\eps}}(t)-A(t)\cdot\Big(\xi_*^{\vec{\tau},\vec{u},
\vec{\eps}}(t)-\xi_*(t)\Big)=\int_0^1 B^{\vec{\eps}}(\nu,t)
\cdot\Big(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t)-\xi_*(t)\Big)\,d\nu\,,
$$
where
$$
B^{\vec{\eps}}(\nu,t)=
\frac{\partial f}{\partial x}\Big(\xi_*(t)+\nu
(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t)-\xi_*(t)),\eta_*(t),t\Big)
-\frac{\partial f}{\partial x}\Big(\xi_*(t),\eta_*(t),t\Big)\,.
$$
Since the function 
$$
Q\times U\times[a,b]\ni(x,u,t)\mapsto
\frac{\partial f}{\partial x}(x,u,t)
$$
is continuous, the quantity
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\lambda(r)\equaldef\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\sup\Big\{\Big\|\frac{\partial f}{\partial x}(x+v,u,t)
\!-\!\frac{\partial f}{\partial x}(x,u,t)\Big\|:(x,u,t)\!\in\!K,
(x+v,u,t)\!\in\!K,\|v\|\!\le\! r\Big\}
\end{eqnarray*}
satisfies 
$$
\lim_{r\downarrow 0}\lambda(r)=0\,.
$$
Using the bound 
\begin{equation}\label{condiafafa}
\|\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t)-\xi_*(t)\|\le
2\kappa\|\vec{\eps}\|e^{\kappa(b-a)}
\end{equation}
(cf. equation (7.4)) we get
$$
\|B^{\vec{\eps}}(\nu,t)\|\le
\lambda\Big(2\kappa\|\vec{\eps}\|e^{\kappa(b-a)}\Big)\quad{\rm whenever}
\quad t\in[a,b],0\le\nu\le 1\,.
$$
Then
$$
\|\Delta_1^{\vec{\eps}}(t)-A(t)\cdot\Big(\xi_*^{\vec{\tau},\vec{u},
\vec{\eps}}(t)-\xi_*(t)\Big)\|\le
\lambda\Big(2\kappa\|\vec{\eps}\|e^{\kappa(b-a)}\Big)
\cdot 2\kappa\|\vec{\eps}\|e^{\kappa(b-a)}
$$
whenever $t\in[a,b]$. Therefore
\begin{eqnarray}
&\,&\!\!\!\!\!\Big\|\int_{a}^t\Big(\Delta_1^{\vec{\eps}}(s)-
A(s)\cdot(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(s)-\xi_*(s))\Big)\,ds\Big\|
\label{esti1}\\
&\le&\lambda\Big(2\kappa\|\vec{\eps}\|e^{\kappa(b-a)}\Big)
\cdot 2(b-a)\kappa\|\vec{\eps}\|e^{\kappa(b-a)}\nonumber
\end{eqnarray}
whenever $t\in[a,b]$. This provides the desired estimate for the first
of the two terms in the definition of $R^{\vec{\eps}}(t)$. 

To estimate the second term, we fix $j$, and then 
$t\in[\tau_j,\tau_j+\eps_j]$, and write
$$
\Delta_2^{\vec{\eps}}(t)=f(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t),
u_j,t)-f(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t),\eta_*(t),t)\,.
$$
Then
$$
\Delta_2^{\vec{\eps}}(t)-\theta^{\vec{\eps}}(t)=
\sigma_1^{\vec{\eps}}(t)-\sigma_2^{\vec{\eps}}(t)\,,
$$
where
\begin{eqnarray*}
\sigma_1^{\vec{\eps}}(t)&\equaldef&
f(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t),u_j,t)-
f(\xi_*(\tau_j),u_j,\tau_j)\,,\\
\sigma_2^{\vec{\eps}}(t)&\equaldef&
f(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t),\eta_*(t),t)-
f(\xi_*(\tau_j),\eta_*(\tau_j),\tau_j)\,.
\end{eqnarray*}
Let 
$$
\hat\lambda(r)\equaldef\sup\{\|F(s_1)-F(s_2)\|:
s_1\in[a,b],s_2\in[a,b],|s_1-s_2|\le r\}\,,
$$
where
$$
F(s)\equaldef f(\xi_*(s),\eta_*(s),s)\,.
$$
Then 
\begin{equation}\label{bi3}
\lim_{r\downarrow 0}\hat\lambda(r)=0\,,
\end{equation}
because $F$ is continuous, since $f$, $\xi_*$  and $\eta_*$ are continuous.

Then 
\begin{eqnarray}
\|f(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t),\eta_*(t),t)-
f(\xi_*(t),\eta_*(t),t)\|&\le&\kappa
\|\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t)-\xi_*(t)\|\label{bi1}\\
&\le&2\kappa^2\|\vec{\eps}\|e^{\kappa(b-a)}\,,\nonumber
\end{eqnarray}
and 
\begin{equation}\label{bi2}
\|f(\xi_*(t),\eta_*(t),t)-f(\xi_*(\tau_j),\eta_*(\tau_j),\tau_j)\|
\le\hat\lambda(\|\vec{\eps}\|)\,.
\end{equation}
Therefore
$$
\sigma_2^{\vec{\eps}}(t)\le
2\kappa^2\|\vec{\eps}\|e^{\kappa(b-a)}+
\hat\lambda(\|\vec{\eps}\|)\,.
$$

To estimate $\sigma_1^{\vec{\eps}}(t)$, we define
\begin{eqnarray*}
&\,&\tilde\lambda(r)\equaldef\\
&\,&\!\!\!\!\!\!\!\!\!\sup\{\|F_j(s_1)-F_j(s_2)\|:
s_1\in[a,b],s_2\in[a,b],|s_1-s_2|\le r,j=1,\ldots,k\}\,,
\end{eqnarray*}
where
$$
F_j(s)\equaldef f(\xi_*(s),u_j,s)\,.
$$
Then 
$$
\lim_{r\downarrow 0}\tilde\lambda(r)=0\,,
$$
because the $F_j$ are continuous, since $f$ and $\xi_*$ are continuous.

It follows that
\begin{eqnarray*}
\|f(\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t),u_j,t)-
f(\xi_*(t),u_j,t)\|&\le&\kappa
\|\xi_*^{\vec{\tau},\vec{u},\vec{\eps}}(t)-\xi_*(t)\|\\
&\le&2\kappa^2\|\vec{\eps}\|e^{\kappa(b-a)}\,,
\end{eqnarray*}
and 
$$
\|f(\xi_*(t),u_j,t)-f(\xi_*(\tau_j),u_j,\tau_j)\|
\le\tilde\lambda(\|\vec{\eps}\|)\,.
$$
Therefore
$$
\sigma_1^{\vec{\eps}}(t)\le
2\kappa^2\|\vec{\eps}\|e^{\kappa(b-a)}+
\tilde\lambda(\|\vec{\eps}\|)\,.
$$
Hence
$$
\|\Delta_2^{\vec{\eps}}(t)-\theta^{\vec{\eps}}(t)\|\le
4\kappa^2\|\vec{\eps}\|e^{\kappa(b-a)}+
\hat\lambda(\|\vec{\eps}\|)+\tilde\lambda(\|\vec{\eps}\|)\,.
$$
Then
\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\|\int_{[a,t]\cap E(\vec{\eps})}\Big(\Delta_2^{\vec{\eps}}(s)-
\theta^{\vec{\eps}}(s)\Big)\,ds\|\label{esti2}\\
&\le&
\Big(4\kappa^2\|\vec{\eps}\|e^{\kappa(b-a)}+
\hat\lambda(\|\vec{\eps}\|)+\tilde\lambda(\|\vec{\eps}\|)\Big)\cdot
\|\vec{\eps}\|\,.\nonumber
\end{eqnarray}
If we combine (\ref{esti1}) and (\ref{esti2}), we get the bound
$$
\|R^{\vec{\eps}}(t)\|\le\mu(\vec{\eps})\cdot\|\vec{\eps}\|\,,
$$
where
\begin{eqnarray*}
&\,&\mu(\vec{\eps})\equaldef\\
&\,&\!\!\!\lambda\Big(2\kappa\|\vec{\eps}\|e^{\kappa(b-a)}\Big)
\cdot 2(b-a)\kappa e^{\kappa(b-a)}
+4\kappa^2\|\vec{\eps}\|e^{\kappa(b-a)}+
\hat\lambda(\|\vec{\eps}\|)+\tilde\lambda(\|\vec{\eps}\|)\,.
\end{eqnarray*}
It follows that
$$
\|\omega(\vec{\eps})\|\le\mu(\vec{\eps})\cdot\|\vec{\eps}\|\,.
$$
Since $\mu(\vec{\eps})\to 0$ as $\vec{\eps}\to 0$, (\ref{ourfor})
is proved. \proofend






\section{TRANSVERSALITY OF CONES AND SET SEPARATION}

\noindent This section presents in detail the ``topological argument'' that 
plays a crucial role in the proof of the maximum principle.   The purpose
of the argument is to extract information from the facts that 

\medbreak\begin{itemize}\item[1.] 
the reachable set $\mathcal{R}_{[a,b]}(Q,U,f;x)$ and the set $S$
are locally separated at $\hat x$,
\item[2.] $S$ has a Boltyanskii tangent cone $C$,
\item[3.] $\mathcal{R}_{[a,b]}(Q,U,f;x)$ contains the image 
of a neighborhood of $0$ relative to the positive orthant 
$\rr_+^k$ of $\rr^k$ under the endpoint map 
$\mathcal{E}^{\vec{\tau},\vec{u}}$,
\item[4.] the map $\mathcal{E}^{\vec{\tau},\vec{u}}$ 
is continuous near $0$ and differentiable at $0$.
\end{itemize}

\medbreak\noindent Facts 2, 3 and 4 tell us that we have 

\medbreak\begin{itemize}\item[a.] 
two subsets $S_1$ and $S_2$ (given by $S_1=\mathcal{R}_{[a,b]}(Q,U,f;x)$ 
and $S_2=S$) of a finite-dimensional real linear space $Y=\rr^n$, 
\item[b.] two convex cones $C_1$ and $C_2$ (given by 
$C_1=\rr_+^k$ and $C_2=C$), in linear spaces $X_1$, $X_2$
(where $X_1=\rr^k$ and $X_2=\rr^n$),
\item[c.] neighborhoods $U_1$, $U_2$ of the origin in $X_1$, $X_2$, 
respectively, 
\item[d.] continuous maps $F_1:U_1\cap C_1\mapsto S_1$ and
$F_2:U_2\cap C_2\mapsto S_2$ (given by 
$F_1=\mathcal{E}^{\vec{\tau},\vec{u}}$ and $F_2=\varphi$, where 
$\varphi$ is the approximating map that occurs in the definition of a 
Boltyanskii cone),
\item[e.] linear maps $L_1:X_1\mapsto Y$, $L_2:X_2\mapsto Y$
such that $F_1(x)=L_1x+o(\|x\|)$ as $x\to 0$, $x\in C_1$, and 
$F_2(x)=L_2x+o(\|x\|)$ as $x\to 0$, $x\in C_2$.
\end{itemize}

\medbreak\noindent Our goal is then to conclude that ``the separation
of the sets $S_1$ and $S_2$ implies that the linear approximations
$L_1C_1$, $L_2C_2$ to these sets are separated in the linear sense.'' 
Here ``linear separation'' of two cones $K_1,K_2$ in $Y$ means that
``there exists a nontrivial linear functional $\lambda:Y\mapsto\rr$
such that $\lambda(v)\ge 0$ for $v\in K_1$ and 
$\lambda(v)\le 0$ for $v\in K_2$.''  

Unfortunately, {\em separation of two sets does not imply linear
separation of their approximating cones}. This can be see most easily
by considering the following trivial example. Let $Y=\rr^2$, and take
$S_1$, $S_2$ to be the $x$ axis and the $y$ axis, respectively. Then
$S_1$ and $S_2$ are separated at the origin. On the other hand, it is
clear that $S_1$ and $S_2$ are their own linear approximations at $0$.
Yet $S_1$ and $S_2$ are {\em not} linearly separated.

The true corespondence between separation of two sets and linear separation
of their approximating cones is given by a property which is slightly 
{\em weaker} than linear separation of the approximating cones.
Precisely:

\medbreak\begin{itemize}\item[i.] linear separation of two cones
is equivalent to the property that the cones are not ``transversal,''
\item[ii.] there is another property, called ``strong transversality,''
which is slightly stronger than transversality,
\item[iii.] therefore the property of {\em not being strongly transversal}
is slightly {\em weaker} than non-transversality, i.e., slightly 
weaker than linear separation, 
\item[iv.] strong transversality of the approximating cones
implies that the sets are not separated, that is 
{\em separation of the sets implies that the approximating 
cones are not strongly 
transversal.}
\end{itemize}

\medbreak\noindent The notion of ``transversality'' of cones
is a natural extension of the well known notion of transversality 
of linear subspaces.  One says that two linear subspaces $A_1$, $A_2$ 
of a finite-dimensional real linear space $Y$ are {\em transversal} 
if the sum $A_1+A_2$ (that is, the set of all sums $a_1+a_2$,
$a_1\in A_1$, $a_2\in A_2$) is the whole space $Y$. Naturally, 
when the $A_i$ are subspaces we could equally well have used
the {\em difference} $A_1-A_2$ (that is, the set of all differences
$a_1-a_2$, $a_1\in A_1$, $a_2\in A_2$). It turns out that, {\em once we use
set difference rather than set sum, the resulting notion 
of ``transversality'' is the one that works for cones as well.}

The general philosophy of transversality theory is that, if two objects
$B_1$ and $B_2$ have linear approximations $A_1$, $A_2$ at a point $\bar x$, 
then $B_1\cap B_2$ looks, near $\bar x$, like $A_1\cap A_2$, {\em
if $A_1$ and $A_2$ are transversal}. For example, suppose 
$B_1$, $B_2$ are smooth submanifolds of $\rr^n$ of dimensions 
$n_1$, $n_2$, $\bar x\in B_1\cap B_2$, 
and $A_1$, $A_2$ are the tangent spaces to $B_1$, $B_2$ at $\bar x$. 
Then, if $A_1$ and $A_2$ are transversal, it follows that 
$n_1+n_2\ge n$, and the intersection
$A_1\cap A_2$ is a subspace of dimension $\nu=n_1+n_2-n$. 
The implicit function theorem then implies that, near $\bar x$, 
$B_1\cap B_2$ is a $\nu$-dimensional submanifold of $\rr^n$.
(Transversality is essential here! For example, if we take 
$n=2$, and let $B_1$ be the $x$ axis, and $B_2$ be the parabola
$\{(x,y):y=x^2\}$, we see that $B_1$ and $B_2$ intersect at the origin,
and their tangent spaces $A_1,A_2$ at $(0,0)$ both coincide with $x$-axis. 
So $A_1\cap A_2$ is one-dimensional, but $B_1\cap B_2$ consists 
of a single point, so $B_1\cap B_2$ does not look at all like
$A_1\cap A_2$. This shows that the principle that 
``$B_1\cap B_2$ looks near $\bar x$ like $A_1\cap A_2$'' can fail
if $A_1$ and $A_2$ fail to be transversal.)

In our case, the ``general philosophy'' suggest the following possibility:
if the approximating cones $C_1$ and $C_2$ are transversal, then 
$S_1\cap S_2$ will contain a nontrivial set if $C_1\cap C_2$ does.
Therefore, to guarantee that $S_1\cap S_2$ contains a nontrivial set,
we have to require 
\medbreak\begin{itemize}\item[(a)] that $C_1$ and $C_2$ be transversal, 
\end{itemize}and 
\begin{itemize}\item[(b)] that $C_1\cap C_2$ be nontrivial, that is, that 
$C_1\cap C_2\neq\{0\}$.  
\end{itemize}

\medbreak\noindent The conjunction of these two conditions
is precisely what we are going to call ``strong transversality.''
It will then be a rigorous theorem that 
\medbreak\begin{itemize}\item[($\#$a)] {\em strong tranversality of the 
approximating cones implies  non-separation of the sets}, 
\end{itemize}
that is, that 
\begin{itemize}\item[($\#'$)] {\em separation of the sets implies that the 
approximating cones are not strongly transversal.}
\end{itemize}

Using ($\#'$), we will conclude, in the proof of the maximum principle,
that the cones $D\mathcal{E}^{\vec{\tau},\vec{u}}(0)(\rr_+^k)$ and 
$C$ are not strongly transversal. We will then want to draw the stronger
conclusion that the cones are not transversal. This will follow
from the observation (proved in Lemma \ref{trastrong})
that, as long as the cones under consideration  are not 
both linear subspaces, transversality of two cones is in fact equivalent
to strong transversality.  That is, ``transversality and
strong transversality are essentially the same thing,'' except only 
in the case when both cones are linear subspaces.

\ssubsection{Transversality and strong transversality of cones}

If $S_1$, $S_2$ are subsets of a real linear space $X$,
and  $S_3\subseteq\rr$,  we write
\begin{eqnarray*}
S_1+S_2&\equaldef&\{s_1+s_2:s_1\in S_1,s_2\in S_2\}\,,\\
S_1-S_2&\equaldef&\{s_1-s_2:s_1\in S_1,s_2\in S_2\}\,,\\
S_3\cdot S_1&\equaldef&\{s_3\cdot s_1:s_1\in S_1,s_3\in S_3\}\,.
\end{eqnarray*}
When one of the sets consists of a single point $x$, we will write
``$x$'' rather than ``$\{x\}$'' in the above formulae. Thus, for example,
if $S\subseteq X$ and $x\in X$, then $x+S$ is the translate 
of $S$ by $x$, i.e., the set $\{x+s:s\in S\}$. Similarly, 
if $S\subseteq X$ and $r\in\rr$, then $r\cdot S$ is 
the set $\{r\cdot s:s\in S\}$. In particular, if $B$ is the closed unit
ball of $X$ centered at $0$ (with respect to some norm on $X$) 
and $x\in X$, $r\in\rr$, $r>0$, then $x+r\cdot B$ is the closed
ball of radius $r$ centered at $x$. 
\begin{definition}
{\rm Let $C_1$, $C_2$ be cones in a finite-dimensional real linear space
$X$.  We say that $C_1$ and $C_2$ are {\em transversal}---and write
$C_1\transv C_2$---if} 
$$
\qquad\quad\qquad\qquad\qquad C_1-C_2=X\,.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\diamondsuit
$$
\end{definition}
\begin{remark}
{\rm {\em If $C_1$ and $C_2$ are convex, then $C_1\transv C_2$ if and only if 
there does not exist a nonzero linear functional $\lambda:X\mapsto\rr$
such that} 
\begin{eqnarray*}
\lambda(c)&\ge&0\quad{\rm whenever}\quad c\in C_1\,,\\
\lambda(c)&\le&0\quad{\rm whenever}\quad c\in C_2\,.
\end{eqnarray*}
Equivalently, 
\begin{equation}\label{tosee}
C_1\transv C_2\Longleftrightarrow
(-C_1)^\perp\cap(C_2)^\perp=\{0\}\,.
\end{equation}
To see that (\ref{tosee}) holds, assume first that 
$C_1\transv C_2$. Let $\lambda\in(-C_1)^\perp\cap(C_2)^\perp$. 
Let $x\in X$. Write $x=c_1-c_2$, $c_1\in C_1$, $c_2\in C_2$. 
Then $\lambda(c_1)\ge 0$, because $\lambda\in(-C_1)^\perp$, and 
then $\lambda(c_2)\le 0$, because $\lambda\in(C_2)^\perp$. 
Therefore $\lambda(c_1-c_2)\ge 0$. So 
$\lambda(x)\ge 0$. Since this inequality is true for all $x\in X$, we 
can take an $x\in X$ and apply the inequality to $-x$, thereby 
concluding that $\lambda(-x)\ge 0$, i.e., that 
$\lambda(x)\le 0$. So $\lambda(x)=0$ for all $x\in X$, i.e., $\lambda=0$.

To prove the converse, we assume that $(-C_1)^\perp\cap(C_2)^\perp=\{0\}$ and 
try to prove that $C_1\transv C_2$. Suppose it is not true that
$C_1\transv C_2$. Let $D=C_1-C_2$. Then $D$ is a convex 
cone in $X$, and $D\neq X$. So $D$ is a proper convex cone 
in $X$. Let $E$ be the closure of $D$ in $X$. Then $E$ is a closed convex cone
in $X$, and $E\neq X$. (Here we are using the fact that $X$ is 
finite-dimensional, to conclude  that $E\neq X$ from the fact that
$D\neq X$. In infinite dimensions this can fail, because there 
are proper convex cones that are dense.) So by the Hahn-Banach Theorem
there exists a nonzero linear functional $\lambda:X\mapsto\rr$
such that $\lambda(x)\ge 0$ for all $x\in E$. In particular, 
if $c\in C_1$ then $c\in E$, so $\lambda(c)\ge 0$. Therefore
$\lambda\in(-C_1)^\perp$. Also, if $c\in C_2$ then $-c\in E$, 
so $\lambda(-c)\ge 0$, and then $\lambda(c)\le 0$. Therefore
$\lambda\in(C_2)^\perp$. So 
$\lambda\in(-C_1)^\perp\cap(C_2)^\perp=\{0\}$. Since
$\lambda\neq 0$, this contradicts the assumption that 
$(-C_1)^\perp\cap(C_2)^\perp=\{0\}$. Therefore 
$C_1\transv C_2$.}\proofend
\end{remark}

\begin{definition}
{\rm Let $C_1$, $C_2$ be cones in a finite-dimensional real linear space
$X$.  We say that $C_1$ and $C_2$ are {\em strongly transversal}---and write
$C_1\stransv C_2$---if $C_1\transv C_2$ and $C_1\cap C_2\neq\{0\}$.}\proofend
\end{definition}

\ssubsection{Transversality vs. strong transversality}
\begin{lemma}\label{trastrong}
Let $C_1$, $C_2$ be two convex cones in a finite-dimensional real linear
space $X$. Then the following two conditions are equivalent:
\begin{itemize}
\item[1.] $C_1\transv C_2$,
\item[2.] either 
\begin{itemize}
\item[a.] $C_1\stransv C_2$
\end{itemize}
or
\begin{itemize}
\item[b.] $C_1$ and $C_2$ are both linear subspaces and $X=C_1\oplus C_2$.
\end{itemize}\end{itemize}
\end{lemma}
\begin{proof}
It is clear that 2$\,\Longrightarrow\,$1. Let us show that 
1$\,\Longrightarrow\,$2. Assume that $C_1\transv C_2$ but
$C_1$ is not strongly transversal to $C_2$. We have to show that
Condition b holds. Clearly, our assumptions imply that
$C_1\cap C_2=\{0\}$.

First, we show that $C_2$ is a linear subspace. Let $c\in C_2$. 
Since $C_1\transv C_2$, we can write $c=c_1-c_2$, $c_1\in C_1$, $c_2\in C_2$.
Then $c+c_2=c_1$, so $c_1\in C_1\cap C_2$. Therefore $c_1=0$, so 
$c+c_2=0$. It follows that $-c=c_2$, so $-c\in C_2$. We have thus shown that
\begin{equation}\label{implo}
(\forall c)(c\in C_2\Longrightarrow -c\in C_2)\,.
\end{equation}
Since $C_2$ is a convex cone, (\ref{implo}) implies that it is a linear 
subspace. 

Second, a similar argument shows that $C_1$ is a linear subspace as well.

It then follows that $C_1+C_2=X$, because $C_1-C_2=X$, since
$C_1\transv C_2$. Moreover, we know that $C_1\cap C_2=\{0\}$.
So the sum is direct, that is, $X=C_1\oplus C_2$. \proofend
\end{proof}


\ssubsection{The main topological lemma}
\begin{lemma}\label{topolem}
Let $r\in\rr$, $r>0$, and let $B$ be the ball of radius $r$ in a 
finite-dimensional normed space $Y$. Let $\rho$ be such that
$0<\rho<r$, and let $F:B\to Y$ be a continuous 
map such that 
$$
\|F(y)-y\|\le\rho\quad{\rm whenever}\quad y\in B\,.
$$
Then 
$$
(r-\rho)B\subseteq F(B)\,.
$$
\end{lemma}
\begin{proof}
Fix $z\in(r-\rho)B$. We want to find $y\in B$ such that
$F(y)=z$. Now, the equation $F(y)=z$ is equivalent to
$y=y-F(y)+z$.  Let 
$$
G(y)=y-F(y)+z\,,
$$
for $y\in B$. 
If $y\in B$, then $\|y-F(y)\|\le\rho$. Since $\|z\|\le r-\rho$,
we can conclude that $\|G(y)\|\le r$. So $G$ is a continuous map
from $B$ to $B$. By the Brouwer fixed point theorem, 
$G$ has a fixed point. That is, there exists $y\in B$ such that
$G(y)=y$. But then this $y$ satisfies $F(y)=z$, and our proof is 
complete.
\end{proof}

\ssubsection{The transversal intersection theorem}

\begin{theorem}
Assume that the following conditions hold. 
\begin{itemize}
\item[1.] $X_1$, $X_2$, $Y$ are finite-dimensional real linear spaces. 
\item[2.] $C_1$, $C_2$ are convex cones in $X_1$, $X_2$.
\item[3.] $U_1$, $U_2$ are neighborhoods of $0$ in $X_1$, $X_2$.
\item[4.] $F_1:U_1\cap C_1\mapsto Y$, $F_2:U_2\cap C_2\mapsto Y$, 
are continuous maps such that $F_1(0)=F_2(0)=0$.
\item[5.] $F_1$ and $F_2$ are differentiable at $0$ along $C_1$, $C_2$ with 
differentials $L_1$, $L_2$. (That is, for $i=1,2$, 
$L_i$ is a linear map from $X_i$ to $Y$ such that 
$F_i(x)=L_ix+o(\|x\|)$ as $x\to 0$ via values in $C_i$.) 
\item[6.] $L_1(C_1)\stransv L_2(C_2)$. 
\end{itemize}
Then the sets 
$F_1(U_1\cap C_1)$ and $F_2(U_2\cap C_2)$ are not locally separated at $0$. 
That is, 
$F_1(U_1\cap C_1)\cap F_2(U_2\cap C_2)\cap V\neq\{0\}$ for every 
neighborhood $V$ of $0$ 
in $Y$.
\end{theorem}
\begin{remark}
{\rm The statement of the theorem involves (in Item 5) norms on the spaces
$X_1$, $X_2$, $Y$, but it is easy to see that the validity of the condition
on Item 5 does not depend on the choice of the norms}.\proofend
\end{remark}
\begin{proof}
We fix once and for all norms on $X_1$, $X_2$, $Y$. 

Let $(e_1,\ldots,e_n)$ be a basis of $Y$. Write $e_0=-(e_1+\ldots+e_n)$. 
Then every $y\in Y$ can be written uniquely as an affine combination
of $e_0,e_1,\ldots,e_n$. (Recall that an {\em affine combination} of
vectors $v_1,\ldots,v_m$ is a linear combination $a_1v_1+\ldots+a_mv_m$
with scalar coefficients $a_j$ such that $a_1+\ldots+a_m=1$. If $y\in Y$, then 
we can write $y=a_1e_1+\ldots+a_ne_n$ and, using 
$0=e_0+e_1+\ldots+e_n$, we can pick an arbitrary scalar $r$ and write 
$y=b_0e_0+b_1e_1+\ldots+b_ne_n$, with $b_0=r$ and 
$b_j=a_j+r$ for $j=1,\ldots,n$. Then $b_0+b_1+\ldots+b_n=
a+(n+1)r$, where $a=a_1+\ldots+a_m$. By choosing 
$r=\frac{1-a}{n+1}$, we get $b_0+b_1+\ldots+b_n=1$. 
This yields an expression of the desired form for $y$. To prove uniqueness,
assume $y=b_0e_0+b_1e_1+\ldots+b_ne_n=c_0e_0+c_1e_1+\ldots+c_ne_n$, with 
$b_0+b_1+\ldots+b_n=c_0+c_1+\ldots+c_n=1$. Then 
$(b_0-c_0)e_0+(b_1-c_1)e_1+\ldots+(b_n-c_n)e_n=0$, so 
$(b_1-c_1-(b_0-c_0))e_1+\ldots+(b_n-c_n-(b_0-c_0))e_n=0$. Since the 
vectors $e_1,\ldots,e_n$ are linearly independent, 
we find 
$b_1-c_1-(b_0-c_0)=\ldots=b_n-c_n-(b_0-c_0)=0$. Therefore
$b_0-c_0=b_1-c_1=\ldots=b_n-c_n$. But $\sum_{j=0}^nb_j=\sum_{j=0}^nc_j=1$,
and then $\sum_{j=0}^n(b_j-c_j)=0$. So the $n+1$ numbers $b_j-c_j$ are 
equal and add up to zero. Therefore $b_j-c_j=0$ for $j=0,1,\ldots,n$.)
Use $a_j(y)$ to denote the coefficients of this affine combination, so 
$$
y\in Y\Longrightarrow \Big(\,y=\sum_{j=0}^n a_j(y)e_j
\,\,\,{\rm and}\,\,\,\sum_{j=0}^n a_j(y)=1\,\Big)\,.
$$
It is easy to see that the functions $Y\ni y\mapsto a_j(y)\in\rr$ are 
continuous. Since 
$$
0=e_0+e_1+\ldots+e_n=\frac{1}{n+1}\cdot e_0+\frac{1}{n+1}\cdot e_1+\ldots
+\frac{1}{n+1}\cdot e_n\,,
$$
we have 
$$
a_j(0)=\frac{1}{n+1}\,\,\,{\rm for}\,\,\,j=0,1,\ldots,n\,.
$$
Since the functions $a_j$ are continuous, we can fix a positive number $\delta$
such that 
$$
\Big(\,y\in Y\,\,{\rm and}\,\,\|y\|\le\delta\,\Big)\Longrightarrow 
\Big(\,a_j(y)\ge 0\,\,{\rm for}\,\,j=0,1,\ldots,n\,\Big)\,.
$$
Let $B$ be the closed ball $\{y\in Y:\|y\|\le\delta\}$. Then every member 
$y$ of $B$ is a convex combination $\sum_{j=0}^n a_j(y)e_j$ of 
$e_0,e_1,\ldots,e_n$. 

The assumption that the cones $L_1(C_1)$ and $L_2(C_2)$ are 
transversal tells us that we can find vectors
$\tilde e_{1,0},\tilde e_{1,1},\ldots,\tilde e_{1,n}$ in $C_1$, 
$\tilde e_{2,0},\tilde e_{2,1},\ldots,\tilde e_{2,n}$ in $C_2$, such that
$$
e_j=L_1\tilde e_{1,j}-L_2\tilde e_{2,j}\quad{\rm for}\quad j=0,1,\ldots.n\,.
$$
The assumption that $L_1(C_1)$ and $L_2(C_2)$ are strongly transversal
implies that there exists a nonzero vector $\bar y\in Y$ such that
$\bar y\in L_1(C_1)\cap L_2(C_2)$. Write 
$$
\bar y=L_1v_1=L_2v_2\,,\,\,v_1\in C_1\,,\,\,v_2\in C_2\,.
$$
Then, if $r\in\rr$ is arbitrary, we have
$$
e_j=L_1(\tilde e_{1,j}+rv_1)-L_2(\tilde e_{2,j}+rv_2)\quad{\rm for}
\quad j=0,1,\ldots.n\,,
$$
because $L_1v_1-L_2v_2=0$.

We will choose $r$ in a special way, and then define 
$$
e_{1,j}=\tilde e_{1,j}+rv_1\,,\,\,\,e_{2,j}=\tilde e_{2,j}+rv_2\quad{\rm for}
\quad j=0,1,\ldots.n\,.
$$
The choice of $r$ is made by first fixing a linear functional 
$\mu:Y\mapsto\rr$ such that $\mu(\bar y)=1$, and observing that
the resulting vectors $e_{i,j}$ will then satisfy
$$
\mu(L_1e_{1,j})=\mu(L_1\tilde e_{1,j})+r\,,\quad
\mu(L_2e_{2,j})=\mu(L_1\tilde e_{2,j})+r\,.
$$
We choose $r$ such that $r>0$ and all the numbers 
$\mu(L_1\tilde e_{1,j})+r$, $\mu(L_1\tilde e_{2,j})+r$ are $\ge 1$. 

With this choice of $r$, the vectors $e_{1,j}$ belong to $C_1$, and
the $e_{2,j}$ belong to $C_2$. So we now have
$$
\left.\begin{array}{lrcl}
\,&e_j&=&L_1 e_{1,j}-L_2 e_{2,j}\,,\\
\,&e_{1,j}&\in&C_1\,,\\
\,&e_{2,j}&\in&C_2\,,\\
\,&\mu(L_1e_{1,j})&\ge&1\,,\\
{\rm and}\quad&\mu(L_2e_{2,j})&\ge&1\end{array}\right\}
{\rm for}\qquad j=0,1,\ldots.n\,.
$$

We now define a positive number $\bar r$ and a map 
$$
]0,\bar r]\times B\ni(r,y)\mapsto H(r,y)\in Y
$$
by letting
$$
H(r,y)=\frac{1}{r}\Biggl(
F_1\Big(r\theta_1(y)\Big)-F_2\Big(r\theta_2(y)\Big)\Biggr)
$$
whenever $0<r\le\bar r$ and $y\in B$. Here
\begin{eqnarray*}
\theta_1(y)&\equaldef&a_0(y)e_{1,0}+a_1(y)e_{1,1}+\ldots+a_n(y)e_{1,n}\,,\\
\theta_2(y)&\equaldef&a_0(y)e_{2,0}+a_1(y)e_{2,1}+\ldots+a_n(y)e_{2,n}\,.
\end{eqnarray*}
The number $\bar r$ is chosen so that 
$$
\bar r(\|e_{i,0}\|+\|e_{i,1}\|+\ldots+\|e_{i,n}\|)\le\delta_i\quad
{\rm for}\quad i=1,2\,,
$$
where the numbers $\delta_i$ are such that
$$
\delta_i>0\quad{\rm and}\quad(\forall x)\Biggl(\,\Big(\,x\in X_i\,\,
{\rm and}\,\,\|x\|\le\delta_i\,\Big)\Longrightarrow x\in U_i\,\Biggr)\,.
$$
It follows from our choice of $\bar r$ that 
$\|r\theta_i(y)\|\le\delta_i$ whenever $i=1,2$, $y\in B$,
and $0<r\le\bar r$.  (Recall that the coefficients $a_j(y)$ satisfy 
$0\le a_j(y)\le 1$ whenever $y\in B$.) Therefore $H$ is well defined on 
the set $]0,\bar r]\times B$.  It is then clear that $H$ is continuous. 

Now use the assumption that 
$$
F_i(x)=L_ix+o(\|x\|) \quad {\rm as}\quad x\to 0\,,\,\,x\in C_i\,,
$$
to write 
$$
F_i\Big(r\theta_i(y)\Big)=L_i\Big(r\theta_i(y)\Big)+o(r)
\,\,\,{\rm as}\,\,\,r\downarrow 0\,,
$$
since $\|\theta_i(y)\|$ is bounded by a fixed constant. Then
$$
F_i\Big(r\theta_i(y)\Big)=rL_i\Big(\theta_i(y)\Big)+o(r)
\,\,\,{\rm as}\,\,\,r\downarrow 0\,,
$$
since $L_i$ is linear. So 
$$
H(r,y)=L_1\Big(\theta_1(y)\Big)-L_2\Big(\theta_2(y)\Big)+o(1)
\,\,\,{\rm as}\,\,\,r\downarrow 0\,.
$$
But
\begin{eqnarray*}
L_1\Big(\theta_1(y)\Big)-L_2\Big(\theta_2(y)\Big)&=&
L_1\Big(\sum_{j=0}^na_j(y)e_{1,j}\Big)-L_2\Big(\sum_{j=0}^na_j(y)e_{2,j}\Big)\\
&=&\sum_{j=0}^na_j(y)(L_1e_{1,j}-L_2e_{2,j})\\
&=&\sum_{j=0}^na_j(y)e_j\\
&=&y\,.
\end{eqnarray*}
Therefore 
$$
H(r,y)=y+o(1)\,\,\,{\rm as}\,\,\,r\downarrow 0\,.
$$
In other words, if we define $H_r(y)=H(r,y)$, we find
$$
\lim_{r\downarrow 0}H_r(y)=y\,,\,\,\,
{\rm uniformly\, \,with\,\,respect\,\,to}\,\, y\in B\,.
$$
Pick any $\alpha$ such that $0<\alpha<\delta$. Then choose 
$\hat r$ such that $0<\hat r\le\bar r$ having the property that
$\|H_r(y)-y\|\le\alpha$ whenever $0<r\le\hat r$. Then 
Lemma \ref{topolem} implies that 
$$
(\delta-\alpha)B\subseteq H_r(B)\quad{\rm whenever}\quad
0<r\le\hat r\,.
$$
In particular, 
$$
0\in H_r(B)\quad{\rm whenever}\quad 0<r\le\hat r\,.
$$
This means that, for each $r\in]0,\hat r]$, we can find a point
$y_r\in B$ such that 
$$
H_r(y_r)=0\,.
$$
It then follows from the definition of $H_r$ that 
$$
F_1(r\theta_1(y_r))=F_2(r\theta_2(y_r))\,.
$$
So, if we let $w_r\equaldef F_1(r\theta_1(y_r))=F_2(r\theta_2(y_r))$, we 
have shown that 
$$
w_r\in F_1(U_1\cap C_1)\cap F_2(U_2\cap C_2)
\quad{\rm whenever}\quad 0<r\le\hat r\,.
$$
Moreover, it is clear that
$$
\lim_{r\downarrow 0}w_r=0\,,
$$
since $r\theta_1(y_r)\to 0$ and $F_1$ is continuous. It follows that,
to conclude our proof, it suffices to show that 
$$
w_r\neq 0\,\,\,{\rm if}\,\,\,r\,\,\,{\rm is\,\,small\,\, enough}\,.
$$
To see this, we estimate $\mu(w_r)$. We have 
\begin{eqnarray*}
w_r&=&F_1(r\theta_1(y_r))\\
&=&L_1(r\theta_1(y_r))+o(r)\\
&=&L_1\Big(r\sum_{j=0}^na_j(y_r)e_{1,j}\Big)+o(r)\\
&=&\Big(r\sum_{j=0}^na_j(y_r)L_1e_{1,j}\Big)+o(r)\,,
\end{eqnarray*}
and then 
\begin{eqnarray*}
\mu(w_r)&=&\mu\Big(r\sum_{j=0}^na_j(y_r)L_1e_{1,j}\Big)+o(r)\\
&=&r\sum_{j=0}^na_j(y_r)\mu(L_1e_{1,j})+o(r)\\
&\ge& r\sum_{j=0}^na_j(y_r)+o(r)\\
&=& r+o(r)\\
&>&0\,\,\,{\rm if}\,\,\,r\,\,\,{\rm is\,\,small\,\, enough}\,,
\end{eqnarray*}
since $\mu(L_1e_{1,j})\ge 1$.   This concludes our proof.
\end{proof}
\begin{remark}
{\rm The preceding proof is somewhat unusual, in that 
we actually end up proving more than is really needed. 
All we need to conclude that the sets $F_i(U_i\cap C_i)$ are not separated
at $0$ is to find a nonzero point in $F_1(U_1\cap C_1)\cap F_2(U_2\cap C_2)$.
To establish the stronger conclusion that the $F_i(U_i\cap C_i)$ are not 
locally separated at $0$ it suffices to find nonzero points of 
$F_1(U_1\cap C_1)\cap F_2(U_2\cap C_2)$ arbitrarily close to zero.
Yet, our proof actually yields a whole ``continuous'' one-parameter
family $\{w_r\}_{r>0,r\,\,{\rm small}}$ of such points! This is clearly an 
anomalous situation, suggesting that in fact the ``true conclusion''
of the theorem ought to be stronger, saying not only that 
these points exist, but that that there is a whole continuum of them, perhaps 
a continuous curve. 

It turns out that the correct answer is {\em not} that the set
$F_1(U_1\cap C_1)\cap F_2(U_2\cap C_2)$ contains a whole continuous
curve $r\mapsto w_r$ of nonzero points.  Such a strong conclusion can
fail to be true, but the weaker conclusion that 
$F_1(U_1\cap C_1)\cap F_2(U_2\cap C_2)$ contains a {\em nontrivial 
connected set containing $0$} is true, even though {\em this set may fail to 
be path-connected}. The proof of this stronger conclusion is based on a
theorem of Leray-Schauder on connected sets of zeros of a homotopy.
The issue is discussed in detail in the paper

\medbreak
\begin{center}
\begin{minipage}{11.5cm}
Sussmann, H.~J.,~``Transversality conditions and a strong maximum principle 
for systems of differential inclusions.'' In {\it Proceedings of the 37th
IEEE Conference on Decision and Control, Tampa, FL, Dec. 1998}. IEEE 
publications, New York, 1998, pp. 1-6. 
\end{minipage}\end{center}

\medbreak\noindent A Postscript version of the paper (compressed with gzip) 
can be downloaded from the author's Web page: 
\medbreak

$\qquad\qquad\qquad$
http://www.math.rutgers.edu/$\,\widetilde{\,}$sussmann}\proofend
\end{remark}

%\newpage

\section{COMPLETION OF THE  PROOF}

\ssubsection{Application of the transversal intersection theorem}
Fix $k$, $\vec{\tau}$, $\vec{u}$ as above.

Using the fact that $C$ is a Boltyanskii tangent cone to $S$ at
$\hat x$, we pick a neighborhood $V$ of the origin in $\rr^n$ 
and a continuous map $\varphi:V\cap C\mapsto S$ such that 
$\varphi(v)=\hat x+v+o(\|v\|)$ as $v\to 0$, $v\in C$. 

Since the image of $\rr_+^k$ under the endpoint map
$\mathcal{E}^{\vec{\tau},\vec{u}}$ is contained in 
the reachable set $\mathcal{R}_{[a,b]}(Q,U,f;\bar x)$, 
the image of $C$ under
$\varphi$ is contained in$S$, and the sets 
$\mathcal{R}_{[a,b]}(Q,U,f;\bar x)$ and $S$ are locally separated
at $\hat x$, it follows that the sets 
$\mathcal{E}^{\vec{\tau},\vec{u}}(\rr_+^k)$ and 
$\varphi(C)$ are locally separated at $\hat x$. The transversal 
intersection theorem then implies that
the cones $D\mathcal{E}^{\vec{\tau},\vec{u}}(0)(\rr_+^k)$ and 
$C$ are not strongly transversal. Since $C$ is not a linear subspace,
we can conclude from Lemma \ref{trastrong} that
the cones $D\mathcal{E}^{\vec{\tau},\vec{u}}(0)(\rr_+^k)$ and 
$C$ are not transversal. Therefore there exists a nonzero 
$\bar\pi\in\rr_n$ such that 
$$
\bar\pi\cdot v\le 0\quad{\rm whenever}\quad v\in
D\mathcal{E}^{\vec{\tau},\vec{u}}(0)(\rr_+^k)
$$
and 
\begin{equation}\label{condifa}
\bar\pi\cdot v\ge 0\quad{\rm whenever}\quad v\in C\,.
\end{equation}

The formula for $D\mathcal{E}^{\vec{\tau},\vec{u}}(0)(\rr_+^k)$ given
by the main technical lemma implies that
$$
M(b,\tau_j)\cdot\Big(f(x_j,u_j,\tau_j)-f(x_j,u_{*,j},\tau_j)\Big)\in 
D\mathcal{E}^{\vec{\tau},\vec{u}}(0)(\rr_+^k)\quad
{\rm for}\quad j=1,\ldots,k\,.
$$
Therefore 
$$
\bar\pi\cdot M(b,\tau_j)\cdot\Big(f(x_j,u_j,\tau_j)-f(x_j,u_{*,j},\tau_j)\Big)
\le 0\quad{\rm for}\quad j=1,\ldots,k\,,
$$
and then 
\begin{eqnarray}
\bar\pi\cdot M(b,\tau_j)\cdot f(x_j,u_j,\tau_j)&\le&
\bar\pi\cdot M(b,\tau_j)\cdot f(x_j,u_{*,j},\tau_j)\label{condifb}\\
&\,&{\rm for}\quad j=1,\ldots,k\,.\nonumber
\end{eqnarray}
Clearly, we can  also assume  that 
\begin{equation}\label{condifc}
\|\bar\pi\|=1\,.
\end{equation}

\ssubsection{The compactness argument}
In the previous subsection, we proved that 
for every choice of $k$, $\vec{\tau}$, $\vec{u}$, there exists
a $\bar\pi$ for which (\ref{condifa}),  (\ref{condifb}),  (\ref{condifc}) 
hold. 

Given any subset $W$ of $[a,b]\times U$, let
$\Pi(W)$ be the set of all $\bar\pi\in\rr_n$ that satisfy
(\ref{condifa}),  (\ref{condifc}), and
\begin{eqnarray}
\bar\pi\cdot M(b,\tau)\cdot f(\xi_*(\tau),u,\tau)&\le&
\bar\pi\cdot M(b,\tau)\cdot f(\xi_*(\tau),\eta_*(\tau),\tau)\label{condifd}\\
&\,&{\rm for\,\,all}\quad (\tau,u)\in W\,.\nonumber
\end{eqnarray}
We have established that $\Pi(W)$ is nonempty whenever 
$W$ is a finite set $\{(\tau_1,u_1),\ldots,(\tau_k,u_k)\}$ such that
$a\le \tau_1<\tau_2<\ldots<\tau_k<b$. Now suppose that 
$W$ is any  finite subset of $[a,b[\times U$.
Let 
$$
W=\{(\tau_1,u_1),\ldots,(\tau_k,u_k)\}\,,\,\,\,
a\le \tau_1\le\tau_2\le\ldots\le\tau_k<b\,.
$$
Let 
$W_\ell$ be the set
$$
\{(\tau_1,u_1),(\tau_2+\frac{1}{\ell},u_2),(\tau_3+\frac{2}{\ell},u_3),
\ldots,(\tau_k+\frac{k-1}{\ell},u_k)\,.
$$
Then $\Pi(W_\ell)\neq\emptyset$. Pick $\bar\pi_\ell\in
\Pi(W_\ell)$. Then the $\bar\pi_\ell$ belong to the unit sphere of
$\rr^n$, which is compact. So there is a subsequence
$\{\bar\pi_{\ell(\nu)}\}_{\nu\in\nn}$ that converges to a limit
$\bar\pi$.  Then $\bar\pi\in\Pi(W)$, so $\Pi(W)\neq\emptyset$.

A similar limiting argument then shows that $\Pi(W)\neq\emptyset$
if $W$ is any finite subset of $[a,b]\times U$.

Now, it is clear that the set $\Pi(W)$ is compact for every $W$.
Moreover, 
$$
\Pi(W_1)\cap\ldots\cap \Pi(W_s)=\Pi(W_1\cup\ldots\cup W_s)\neq\emptyset
$$
if $W_1,\ldots,W_s$ are finite subsets of $[a,b]\times U$.

Let $\mathcal{W}$ be the set of all finite subsets of 
$[a,b]\times U$. Then
$$
\Big\{\Pi(W)\Big\}_{W\in\mathcal{W}}
$$
is a family of compact subsets of the unit sphere of $\rr^n$ having
the property that every finite intersection of members of the family 
is nonempty. It follows that
$$
\bigcap\Big\{\,\Pi(W):W\in\mathcal{W}\,\Big\}\neq\emptyset\,.
$$
Therefore
$$
\Pi([a,b]\times U)\neq\emptyset\,.
$$
This means that there exists a $\bar\pi\in\rr_n$ that satisfies
(\ref{condifa}),  (\ref{condifc}), and is such that
\begin{eqnarray}
\bar\pi\cdot M(b,\tau)\cdot f(\xi_*(\tau),u,\tau)&\le&
\bar\pi\cdot M(b,\tau)\cdot f(\xi_*(\tau),\eta_*(\tau),\tau)\label{condifu}\\
&\,&{\rm for\,\,all}\quad (\tau,u)\in[a,b]\times U\,.\nonumber
\end{eqnarray}

\ssubsection{The momentum}
Let $\bar\pi$ be such that (\ref{condifa}),  (\ref{condifc}), and
(\ref{condifu}) hold.  Define
$$
\pi(t)=\bar\pi\cdot M(b,t)\,.
$$
Then $\pi$ satisfies the adjoint equation
$$
\dot\pi(t)=-\pi(t)\cdot A(t)\,.
$$
({\it Proof.} We know that 
$$
\frac{\partial M}{\partial t}(t,s)=A(t)\cdot M(t,s)\,.
$$
Moreover, 
$$
M(t,s)\cdot M(s,t)=1_n\,.
$$
If we differentiate this with respect to $s$, we get
$$
\frac{\partial M}{\partial s}(t,s)\cdot M(s,t)
+M(t,s)\cdot A(s)\cdot M(s,t)=0\,.
$$
Therefore 
$$
\frac{\partial M}{\partial s}(t,s)+M(t,s)\cdot A(s)=0\,.
$$
So 
$$
\frac{\partial M}{\partial s}(t,s)=-M(t,s)\cdot A(s)\,.
$$
Then 
$$
\dot\pi(s)=
\bar\pi\cdot\frac{\partial M}{\partial s}(b,s)
=-\bar\pi\cdot M(b,s)\cdot A(s)=-\pi(s)\cdot A(s)\,,
$$
as desired.)

Condition (\ref{condifa}) says that
$$
-\pi(b)\in C^\perp\,.
$$

Condition (\ref{condifc}) implies that $\pi$ is nonzero. 

Finally, Condition (\ref{condifu}) says that
\begin{eqnarray}
\pi(\tau)\cdot f(\xi_*(\tau),u,\tau)&\le&
\pi(\tau)\cdot f(\xi_*(\tau),\eta_*(\tau),\tau)\label{condifuu}\\
&\,&{\rm for\,\,all}\quad (\tau,u)\in[a,b]\times U\,.\nonumber
\end{eqnarray}
Therefore
\begin{eqnarray*}
H(\xi_*(\tau),u,\pi(\tau),\tau)&\le&
H(\xi_*(\tau),\eta_*(\tau),\pi(\tau),\tau)\\
&\,&{\rm for\,\,all}\quad (\tau,u)\in[a,b]\times U\,.
\end{eqnarray*}
It then follows that 
$$
H(\xi_*(\tau),\eta_*(\tau),\pi(\tau),\tau)
=\max\{H(\xi_*(\tau),u,\pi(\tau),\tau):u\in U\}
$$
for all $\tau\in[a,b]$.
 
We have thus shown that $\pi$ satisfies all the desired conditions.
Our proof is therefore complete, {\em under the extra assumption
that $\eta_*$ is continuous}.

\ssubsection{Elimination of the continuity assumption on the reference
control} The assumption that $\eta_*$ is continuous was made to
simplify some steps of the proof, but is not really necessary. We now
explain how to avoid this hypothesis, and assume only that $\eta_*$ is
a bounded, measurable $U$-valued function on $[a,b]$.

The continuity of $\eta_*$ was used in the proof exactly four times.
We will now show how, in each case, the hypothesis can be avoided.

\medbreak
\begin{itemize}
\item[1.] In page 11, we defined the matrix-valued function
$[a,b]\ni t\mapsto A(t)$, and asserted that this function was continuous. 
If $\eta_*$ is only bounded and measurable, then the set 
$\{(\xi_*(t),\eta_*(t),t):a\le t\le b\}$ is contained in a compact subset
of $Q\times U\times[a,b]$, so $A$ turns out to be bounded and measurable,
though not necessarily continuous. This, however, is good enough, and the 
properties of the fundamental solution $M$ are the same.

\medbreak
\item[2.] In page 13, we defined the set $U_0$ and asserted that $U_0$ is 
compact.  If $\eta_*$ is not continuous, the conclusion that $U_0$ is 
compact no longer follows. On the other hand, we can define $U_0$ in this case
to be the {\em closure} of the set of page 11, and this new set is compact.
Using the new $U_0$ instead of the original one, all the arguments 
where $U_0$ occurs are still valid.

\medbreak
\item[3.] In page 19, we used the fact that the function $F$ is continuous,
which depended very strongly on the continuity of $\eta_*$. 
This problem is much more serious than the previous ones, and to solve it
we need some basic facts from the theory of functions of a real variable.

\item[] Recall that, if $[a,b]\ni t\mapsto\psi(t)\in\rr^n$ is an integrable 
function, a {\em Lebesgue point} of $\psi$ is a $\tau\in]a,b[$ such that
$$
\lim_{h\downarrow 0}\frac{1}{h}
\int_{\tau -h}^{\tau+h}\|\psi(t)-\psi(\tau)\|\,dt=0
$$
It is then a theorem that {\em almost every point} of $[a,b]$ is a Lebesgue 
point of $\psi$. 

\item[] To avoid invoking the fact that $F$ is continuous, we make the further
restriction that {\em the times $\tau_j$ where our needle 
variations are made are Lebesgue points of the function $F$}.

\item[] The argument is then modified as follows.  Estimate
(\ref{bi1}) is still valid and useful, but (\ref{bi2})---which is still 
valid---is no longer useful, because there is no reason now for (\ref{bi3})
to hold. 

\item[] On the other hand, if we let 
\begin{eqnarray*}
&\,&\zeta_0(h)\equaldef\\
&\,&\!\!\!\!\!\!\frac{1}{h}\max\Big\{\int_{\tau_j-h}^{\tau_j+h}\|f(\xi_*(t),
\eta_*(t),t)-f(\xi_*(\tau_j),\eta_*(\tau_j),\tau_j)\|\,dt:j=1\ldots,k\Big\}
,\\
&\,&\zeta(h)\equaldef\sup\{\zeta_0(h'):0<h'\le h\}
\end{eqnarray*}
then the condition that the $\tau_j$ are Lebesgue points says precisely
that 
\begin{equation}\label{bi4}
\lim_{h\downarrow 0}\zeta(h)=0\,.
\end{equation}
We then get the integral estimate
\begin{eqnarray*}
&\,&\Big\|\int_{[a,t]\cap E(\vec{\eps})}\Big(\Delta_2^{\vec{\eps}}(s)-
\theta^{\vec{\eps}}(s)\Big)\,ds\Big\|\le\\
&\,&\Big(4\kappa^2\|\vec{\eps}\|e^{\kappa(b-a)}+
k\zeta(\|\vec{\eps}\|)+\tilde\lambda(\|\vec{\eps}\|)\Big)\cdot
\|\vec{\eps}\|\,,
\end{eqnarray*}
which replaces (\ref{esti2}). The rest of proof of the main technical 
lemma remains unchanged. 
\medbreak
\item[4.] In the compactness argument in page 31, we approximated 
the $\tau_j$, that could in principle be equal, by points 
$\tau_{j,\ell}$ that are all different.  In our new situation
this argument has to be refined, because the approximating
$\tau_{j,\ell}$ have to be Lebesgue points of $F$, and the passage to the limit
in (\ref{condifd}) requires that 
$$
\lim_{\ell\to\infty}f(\xi_*(\tau_{j,\ell}),\eta_*(\tau_{j,\ell}),\tau_{j,\ell})
=f(\xi_*(\tau_{j}),\eta_*(\tau_{j}),\tau_{j})\,.
$$
To take care of this we use {\em Lusin's theorem}, which guarantees
that for every positive $\beta$ there exists a compact subset
$J_\beta$ of $[a,b]$ such that ${\rm meas}([a,b]\backslash J_\beta)\le\beta$
having the property that the restriction of $F$ to $J_\beta$ is continuous. 

\item[] We pick sets $J_{\beta_\nu}$ as above, corresponding to a
sequence $\{\beta_\nu\}$ that converges to zero and then, for each
$\nu$, we let $\tilde G_\nu$ be the set of all points of
$J_{\beta_\nu}$ that are also Lebesgue points of $F$, and let $G_\nu$
be the set of all points of density of $G_\nu$. (Recall that a {\em
point of density} of a measurable set $E$ is a point of $E$ which is a
Lebesgue point of the indicator function of $E$.) We let 
$G=\bigcup_{\nu\in\nn}G_\nu$. Then $G$ is a subset of full measure 
of $[a,b]$, and it is easy to see that the limiting argument 
works when the $\tau_j$ belong to $G$. 
\end{itemize}

\medbreak\noindent The proof of the maximum principle, thus modified,
works exactly as before, provided that the needle variations are made
using points $\tau_j\in G$.  The result now is the same as before,
except that the Hamiltonian maximization condition is only proved for
$t\in G$. This explains why Condition E4 of Theorem 2.1.1 says that
the maximization holds ``almost everywhere'' rather than ``everywhere.''


}\end{document}

