\documentclass{amsproc}

%%%%%PACKAGES%%%%%%%%%%%%%

\usepackage{enumerate}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{euscript}
\usepackage{epsf}
\usepackage{amscd}
%\usepackage{dvips}{epsfig}

%%%%%PAPER%%%%%%%%%%%%%%%%%%%%%%%%%%

%\newcommand{\ben}{\begin{enumerate}}
%\newcommand{\een}{\end{enumerate}}
\newtheorem{theorem}{Theorem}[section]
\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}
%\newtheorem{ttheorem}{Theorem*}
%\newtheorem{etheorem}{Theorem*}
\newtheorem{qtheorem}{Theorem*}
\def\ca{\mathcal{A}}
\def\ci{\mathcal{I}}
\def\cu{\mathcal{U}}
\def\ct{\mathcal{T}}
\def\cb{\mathcal{B}}
\def\calc{\mathcal{C}}
\def\cx{\mathcal{X}}
\def\cw{\mathcal{W}}
\def\cg{\mathcal{G}}
\def\cd{\mathcal{D}}
\def\cf{\mathcal{F}}
\def\cua{\mathcal{U}^{a}}
\def\cv{\mathcal{V}}
\def\ce{\mathcal{E}}
\def\cp{\mathcal{P}}
\def\cq{\mathcal{Q}}
\def\cs{\mathcal{S}}
\def\cy{\mathcal{Y}}
\def\cz{\mathcal{Z}}
\def\cale{\mathcal{E}}
%\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\ontop#1#2{{\buildrel #1\over#2}}
\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}}

\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{ANONYMOUS}
\def\guo{SECOND-ORDER CONDITIONS}

\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{\vje}{\eps}
\newcommand{\bqrm}{\begin{Quoterm}}
\newcommand{\eqrm}{\end{Quoterm}}
\newcommand{\bqsl}{\begin{Quotesl}}
\newcommand{\eqsl}{\end{Quotesl}}
\parindent 6mm
\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 Mysterious author}}
\newcommand{\hdright}{{\rm \hfil Hamilton and Weierstrass}}

{\Large 

\noindent Hi Urszula and Heinz! 

Let me tell you how I see things now. I think I have made some
progress, and we have material for a nice piece of joint work, maybe a
major paper or series of papers.  I still have a lot of stuff that
needs to be checked, but what I have so far definitely looks
promising, provided that it turns out to be right and that there isn't
a mistake somewhere that invalidates the whole thing.

I include below my notes with what I have so far. In these comments,
I will be referring always to the main theorem of AEMP (that is, $p=2$
only; I do not want to get involved with the case of a $p>2$ until I have
thoroughly gone over the $p=2$ case). And it's always for the fixed
terminal state problem, although my guess is that what I say, if it is
right, will work for more general terminal conditions.

It seems that, using the methods I know (Brouwer fixed point
theorem stuff, no Lyusternik or Avakov or Dubovitskii-Milyutin) I can prove
your theorem WITH $\rho=0$ AND WITHOUT THIRD DERIVATIVES.

This is not what I had originally set out to prove, but it is what seems to
come out naturally by applying the methods I know and like.

My original idea was that by doing things my way I would either 

\begin{itemize}
\item[a.] reach a point where I would be stuck and would not be able 
to go any further, and then I would have (reluctantly) to bring in the 
Avakov+Dubovitskii-Milyutin machinery in order to get unstuck,
\end{itemize}
or
\begin{itemize}
\item[b.] end up proving your result with these other methods, in which case I 
would finally get to understand in my own terms why $\rho$ is needed 
and why one needs those third derivatives which seem so out of place in
a second-order result; in particular, at some point in my pursuit the
new multiplier would pop out, and I would finally see why it's got to 
be there. 
\end{itemize}

To my surprise, what I seem to have 
found is that the proof can indeed be pursued all
the way to the end, and I can get your result by doing things my way,
but $\rho$ is still nowhere to be seen. If my theorem is true and my proof 
is right, then your theorem can be derived from it by just taking
$\rho=0$. 

When I realized this, I was incredulous at first, because my theorem
sort of contradicts yours. (Formally, there is no contradiction. Your
theorem would follow from my theorem by just taking $\rho=0$. So if my
theorem is true then your theorem is also true, and everything is
fine. Yet, it seems clear to me that one of the main points of your
theorem is the new multiplier $\rho$, and that if $\rho$ can always be
taken to be zero then we are back to the old stuff of a single
multiplier, and a lot of what makes the new result surprising is
lost.)  So I thought that there must be something wrong with my proof,
and went back to it trying to find a mistake.  I didn't find anything,
so I decided to ask you to show me an example proving that $\rho$
cannot always be taken to be zero.  Such an example would have
persuaded me that I was on the wrong track, and that my proof had to
be wrong even if I could not find a mistake. 

That's why I sent you the message with Question 2, thinking that you
would be able to come up with an example.  When you replied that you
did {\em not} have an example, I regained confidence in my own proof,
and now I am beginning to believe that it could be right, after all. I
have written it down, trying to be very careful, and I have gone over
it quite a few times. I think it is right, it feels right, I do not
see how it could be wrong. But I have reached the point of diminishing
returns by now. I think it would be useless for me to check again.
Now it's your turn to do some checking. 

So I am sending what I have to you, hoping that you will check it 
quite thouroughly. I am sure you will. 

The theorem that you should compare with your result is Theorem 4.1.
You may read it by thinking that $L=1$, that is, that there is only one 
pair $(h,v)$.  (I did it with several such pairs because I wanted 
something that would also include the Goh and Legendre-Clebsch 
conditions. Theorem 4.1 does the job, but it would take some extra 
work to prove it.  I do not want to do this extra work until I am sure
that Theorem 4.1 is right. That's why I would ask you to try to 
go over my proof as fast as you can.)   The ``abstract'' necessary 
condition for a minimum from which that theorem follows is
Theorem 3.1. To make things clearer, I have also included
(in Theorem 3.2) a restatement of Theorem 3.1 just for the case
of a single $\hat v$. You may find it useful to look at Theorem
3.2 first. I am sure that you will recognize it as the abstract result that 
suffices to get your theorem with $\rho=0$.
In my way of doing things, a necessary condition such as the one of Theorem
3.1 is just a special case of a necessary condition for separation.
So I am getting Theorem 3.1 as a corollary of the separation
result, namely, Theorem 2.2 and its ``multiplier'' version,
Corollary 2.3.   The crux of the matter is the proof of Theorem 
2.2. Technically, the main point of the proof is the use of the Brouwer
fixed point theorem, which appears near the end of \S 2. This theorem
plays in my framework the same role as the Lyusternik and Avakov
theorems play in yours, namely, that of allowing us to solve
{\em exactly} certain equations that we know how to solve approximately
to first or second order. 

You may also notice that in my approach one does not need to work 
with Banach spaces. Normed spaces are enough. (Of course, some completeness
is needed somewhere, but in my way of doing things the only 
completeness one uses is that of {\em finite-dimensional spaces}. 
That completeness is undoubtedly an essential ingredient of the 
Brouwer fixed point theorem, but it's not needed otherwise.)
I like to work with normed spaces that are not necessarily complete, because,
for example, sometimes it may be technically
convenient to use $L^\infty$ controls but endow the space of controls
with some much weaker norm (so that, for example, every control
will be a limit of bang-bang controls).

Also, the convex set $\cu$ over which we carry out the constrained
minimization in the ``abstract'' setting is {\em not} assumed to have
nonempty interior. Notice, in particular, that my set of ``second
order feasible directions'' (called $var_{2,1}(u_*,v,\cu)$ in my
notes) is more general than your set, and in particular can be
nonempty even if the set $\cu$ has empty interior.

If it turns out that I am right, then I suggest we write a joint paper
with this stuff plus some more things that you may want to put in
(e.g. $p>2$, the ``global'' rather than ``local'' maximum principle,
Goh and Legendre-Clebsch, whatever). It could be a very nice paper,
showing how lots of things such as your extended maximum principle and
many other results can be incorporated into one general unified
framework. (It would be a {\em single multiplier} framework,
naturally.) I am sure we can write it in a very positive, upbeat tone.
The only non-upbeat thing that I would like to put in the paper, if
you don't mind, would be a little remark to the effect that the
Brouwer fixed point methods have proved themselves to be superior to
the Dubovitskii-Milyutin approach, at least for this specific
finite-dimensional type of problem.

I look forward to receiving your comments. Please do not hesitate to
be critical, even harsh if necessary. As I said, I have checked and
rechecked my proof many times, and I don't see how it could be wrong,
but maybe you will see things that I do not see. And, if it turns out
that I was wrong, only you and I will know that I made this mistake.
And you are good friends, so I know that I have nothing to worry
about.

Cordially,

\bigbreak
\medbreak

$\qquad$ H\'ector
}
\newpage
\thispagestyle{empty}
$\qquad$

\newpage
\setcounter{page}{1}
\thispagestyle{empty}
\begin{center}
{\Large{\bf A NOTE ON SECOND-ORDER CONDITIONS}}
\end{center}

\bigbreak

\begin{center}
{\huge April 18, 2001}
\end{center}

\bigbreak

{\large\it \bigbreak\noindent NOTE: this note is a very preliminary
tentative draft.  It's probably full of minor mistakes such as typos,
and perhaps even major mistakes, such as, for example, the possibility
that the whole thing is completely wrong. The author has checked the
proofs to the best of his ability, and since he has worked so damn
hard on this, he flatly refuses to accept any responsibility for any errors
that might still be there.  All such errors should be blamed on bad
luck, the weather, the author's wife who repeatedly distracted him, 
or whatever, but the author is not responsible because he has
done his best, and if he made any mistakes, he definitely didn't mean
to, so these mistakes do not count (or, at least, should still entitle
him to a lot of partial credit), as the author's undergraduate
students have explained to him many times.

The only people in the whole universe who are authorized to read this
note are Urszula Ledzewicz, Heinz Sch\"attler, and the
author (who reserves the right to re-read this any time he wants to).
Anyone else reading this stuff will be indicted,
prosecuted, convicted, drawn
and quartered and sentenced to roast in Hell for all eternity.

Nobody, not even Heinz and Urszula, is allowed to quote from this 
text, except possibly in conversations between Heinz and Urszula and
between one or both of them and the author. 

Until the validity of the results and the proofs have been thouroughly 
checked, the author wishes to remain anonymous, which is why his name does 
not appear anywhere in the text.}

\bigbreak

\section{Preliminaries} 

\noindent If $A$ is a set and $j$ is a positive integer, we use $A^j$ to
denote the Cartesian product of $j$ copies of $A$.
If $a$ is any object and $j$ is a positive integer, 
we use $a^{<j>}$ to denote the $j$-tuple $(a,a,\ldots,a)$, consisting of 
$j$ items all of them equal to $a$.

If $\cx$, $\cy$ are normed spaces, and $k$ is a nonnegative
integer, then a {\em polynomial map of degree $\le k$ from 
$\cx$ to $\cy$} is a map 
\begin{equation}\label{inequa}
\cx\ni x\mapsto\cp(x)=P_0+\sum_{j=1}^k\frac{1}{j!}P_j(x^{<j>})\in\cy
\end{equation}
such that
\begin{itemize}
\item[1.] $P_0\in\cy$,
\item[2.] for each $j\in\{1,\ldots,k\}$, 
$$
\cx^j\ni(x_1,\ldots,x_j)
\mapsto P_j(x_1,\ldots,x_j)\in\cy
$$
is a symmetric multilinear map.
\end{itemize}

The maps $P_j$ are the {\em Taylor coefficients} of
$\cp$. If $P_0=0$ then we say that $\cp$ is a polynomial map
{\em without constant term}. If $P_k\neq 0$, then we say 
that  $\cp$ is a polynomial map {\em of degree $k$}. 

It is easy to see that
\begin{Fact}
The Taylor coefficients of a map $\cp$ given by (\ref{inequa})
are uniquely determined by the map $\cp$, \proofend
\end{Fact}
\noindent and
\begin{Fact}
The map $\cp$ given by (\ref{inequa})
is continuous if and only if all the Taylor coefficients 
$P_j$ are  bounded, that is, if and only if there exists a constant
$C\in\rr$ such that
$$
\|P_j(x_1,\ldots,x_j)\|\le C\prod_{i=1}^j\|x_i\|
\,\,{\rm whenever}\,\, j\in\{1,\ldots,k\} 
\,\,{\rm and}\,\,(x_1,\ldots,x_j)\in\cx^j\,.\quad\diamondsuit
$$ 
\end{Fact}


\begin{definition}
{\rm Assume that 
\begin{itemize}
\item[1.] $\cx$, $\cy$ are normed spaces, 
\item[2.] $\cs$ is a subset of $\cx$,
\item[3.] $x_*\in\cs$,
\item[4.] $\cf$ is a map from $\cs$ to $\cy$,
\item[5.] $k$ is a positive integer. 
\end{itemize}

\medbreak\noindent We say that {\em $\cf$ is of class $D^k$ at $x_*$
in the direction of $\cs$} if 
\begin{itemize}
\item[1.] $\cf$ is continuous on $N\cap\cs$ for some neighborhood
$N$ of $x_*$ in $\cx$,
\item[2.] there exists a continuous polynomial map $\cp$ of degree
$\le k$ without constant term such that
$$
\quad\qquad\qquad\qquad
\lim_{x\to x_*,x\in\cs,x\neq x_*}\frac{\cf(x)-\cf(x_*)-\cp(x-x_*)}
{\|x-x_*\|^k}=0\,.\qquad\qquad\qquad\quad\diamondsuit
$$
\end{itemize}}
\end{definition}
\begin{definition}\label{defvaria}
{\rm Let $\cx$ be a normed space, let $\cs$ be a subset of $\cx$,
and let $x_*\in\cs$.
A {\em variation of $x_*$ in $\cs$} is a continuous curve
\begin{equation}\label{varia}
[0,\bar\eps_\eta]\ni\eps\mapsto \eta(t)\in\cs\,,
\end{equation}
defined on the interval $[0,\bar\eps_\eta]$ for some positive real number
$\bar\eps_\eta$, having the property that $\eta(0)=x_*$.} \proofend
\end{definition}
We use $Var(x_*,\cs)$ to indicate the set of all variations
of $x_*$ in $\cs$. 
\begin{definition}
{\rm Let $\cx$, $\cs$, $x_*$ be as in Definition \ref{defvaria}.
A {\em first-order variational direction of $\cs$ at $x_*$} is a 
vector \mbox{$v\in\cx$} such that there exists a variation 
$\eta\in Var(x_*,\cs)$ with the property that
$$
\qquad\qquad\qquad\qquad\qquad\eta(\eps)=x_*+\eps v+o(\eps)\quad{\rm as}\quad
\eps\downarrow 0\,.\qquad\qquad\qquad\qquad\qquad
$$
A {\em limiting first-order variational direction of $\cs$ at $x_*$} is a 
vector \mbox{$v\in\cx$} which is the limit of a sequence $\{v_j\}_{j\in\nn}$
of first-order variational directions of $\cs$ at $x_*$.}\proofend
\end{definition}
We use $var_1(x_*,\cs)$, $\overline{var}_1(x_*,\cs)$, 
to denote, respectively, 
the set of all first-order variational 
directions of $\cs$ at $x_*$.
and the set of all limiting first-order variational directions of 
$\cs$ at $x_*$. 
Clearly, $\overline{var}_1(x_*,\cs)$ is the closure of $var_1(x_*,\cs)$
in $\cx$.

\begin{definition}
{\rm Let $\cx$ be a normed space. A {\em cone} in $\cx$ is a
nonempty subset $C$  of $\cx$ such that $r\cdot x\in C$ whenever
$x\in C$, $r\in\rr$, and $r\ge 0$}.\proofend
\end{definition}

\begin{definition}\label{boul}
{\rm Let $\cx$ be a normed space, let $\cs$ be a subset of $\cx$,
and let $x_*\in\cs$.  The {\em Bouligand tangent cone} to 
$\cs$ at $x_*$ is the set of all vectors $v\in\cx$ such that
there exist sequences $\{v_j\}_{j\in\nn}$, $\{\alpha_j\}_{j\in\nn}$
for which}
\begin{itemize}
{\rm 
\item[1.] $v_j\in\cx$ for every $j$,
\item[2.] $\alpha_j\in\rr$ and $\alpha_j>0$ for every $j$,
\item[3.] $x_*+\alpha_jv_j\in\cs$ for every $j$,
\item[4.] $\lim_{j\to\infty}\alpha_j=0$,}
\item[5.] {\rm $\lim_{j\to\infty}v_j=v$.}\proofend
\end{itemize}
\end{definition}
If $\cx$, $\cs$, $x_*$ are as in Definition \ref{boul}, we use
$Boul(x_*,\cs)$ to denote the Bouligand tangent cone to $\cs$ at $x_*$.


\begin{Fact}
Let $\cx$ be a normed space, let $\cs$ be a convex subset of $\cx$,
and let $x_*\in\cs$.  Then 
$$
\overline{var}_1(x_*,\cs)=Boul(x_*,\cs)\,.
$$
\end{Fact}

\medbreak{\sc Proof}. Let $v\in\overline{var}_1(x_*,\cs)$, $j\in\nn$.
Then we can find $w_j\in var(x_*,\cs)$ such that 
$\|v-w_j\|\le 2^{-j}$. Since $w_j\in var(x_*,\cs)$, there 
exist a positive $\bar\eps_j$ and a curve 
$\eta_j:[0,\bar\eps_j]\mapsto\cs$ such that 
$\eta_j(\eps)=x_*+\eps w_j+o(\eps)$ as $\eps\downarrow 0$. 
Choose $\alpha_j$ such that $0<\alpha_j<\min(\bar\eps_j,2^{-j})$
and $\|\eta_j(\alpha_j)-x_*-\alpha_jw_j\|\le\alpha_j2^{-j}$.
Then $\lim_{j\to\infty}\alpha_j=0$.

Let 
$$
v_j=\frac{\eta_j(\alpha_j)-x_*}{\alpha_j}\,.
$$
Then $x_*+\alpha_jv_j=\eta_j(\alpha_j)\in\cs$. In addition,
$$
\|v_j-w_j\|=\Big\|\frac{\eta_j(\alpha_j)-x_*}{\alpha_j}-w_j\Big\|
=\frac{1}{\alpha_j}\|\eta_j(\alpha_j)-x_*-\alpha_jw_j\|\le 2^-j\,.
$$
Therefore $v_j-w_j\to 0$ as $j\to\infty$. So 
$\lim_{j\to\infty}v_j=v$.  This proves that $v$
belongs to $Boul(x_*,\cs)$.

Conversely, suppose that $v\in Boul(x_*,\cs)$.  Let 
$\{v_j\}_{j\in\nn}$, $\{\alpha_j\}_{j\in\nn}$ be sequences having 
the five properties of Definition \ref{boul}. Define a map
$\eta_j:[0,\alpha_j]\mapsto\cx$ by letting 
\begin{equation}\label{strong}
\eta_j(\eps)=x_*+\eps v_j
\end{equation}
for $\eps\in[0,\alpha_j]$. 
Then $\eta_j(\eps)$ is, for each $\eps$, a convex combination 
of $x_*$ and $x_*+\alpha_jv_j$. Since $x_*\in\cs$, 
$x_*+\alpha_jv_j\in\cs$, and $\cs$ is convex, we can conclude that
$\eta_j(\eps)\in\cs$, so $\eta_j$ is a curve in $\cs$. Clearly,
$$
\eta_j(\eps)=x_*+\eps v_j+o(\eps)\qquad{\rm as}\qquad\eps\downarrow 0\,,
$$
because the much stronger condition (\ref{strong}) holds. So
$v_j\in var_1(x_*,\cs)$ for every $j\in\nn$, and then 
$v\in\overline{var}_1(x_*,\cs)$. \proofend

\bigbreak

\begin{Fact}
Let $\cx$, $\cs$, $x_*$ be as in Definition \ref{defvaria}.
Then the sets $var_1(x_*,\cs)$ and $\overline{var}_1(x_*,\cs)$ 
are cones in $\cx$, and $\overline{var}_1(x_*,\cs)$ is closed. 
If $\cs$ is convex, then $var_1(x_*,\cs)$  
and $\overline{var}_1(x_*,\cs)$ are convex cones.   
\end{Fact}

{\sc Proof.} 
It is clear that $0\in var(x_*,\cs)$, since $x_*\in\cs$.
Assume that $v$ belongs to $var_1(x_*,\cs)$ and $\alpha>0$. Let 
$\eta\in Var(x_*,\cs)$ be such that 
$\eta(\eps)=x_*+\eps v+o(\eps)$ as $\eps\downarrow 0$. Let $\tilde\eta$
be the reparametrized curve 
$$
\Big[0,\frac{\bar\eps_\eta}{\alpha}\Big]\ni\eps\mapsto\eta(\alpha\eps)
\equaldef\tilde\eta(\eps)\in\cs\,.
$$
Then
$$
\tilde\eta(\eps)=x_*+\eps\cdot(\alpha v)+o(\eps)
\quad{\rm as}\quad\eps\downarrow 0\,.
$$
Therefore $\alpha v\in var_1(x_*,\cs)$. 
So $var_1(x_*,\cs)$ is a cone. The other statements are obvious.\proofend

\begin{definition}
{\rm Let $\cx$, $\cs$, $x_*$ be as in Definition \ref{defvaria}.
A {\em second-order variational direction of $\cs$ at $x_*$} is an
ordered pair $(v,w)$ of vectors in $\cx$ such that there exists a variation 
$\eta\in Var(x_*,\cs)$ with the property that}
$$
\qquad\qquad\qquad\qquad\qquad
\eta(\eps)=x_*+\eps v+\eps^2 w+o(\eps^2)\quad{\rm as}\quad\eps\downarrow 0\,.
\,\qquad\qquad\qquad\qquad\diamondsuit
$$
%A {\em limiting second-order variational direction of $\cs$ at $x_*$} is an 
%ordered pair $(v,w)$ of vectors in $\cx$ which is the limit
%of a sequence $\{(v_j,w_j)\}_{j\in\nn}$
%of second-order variational directions of $\cs$ at $x_*$.}\proofend
\end{definition}
We use $var_2(x_*,\cs)$ 
%, $\overline{var}_2(x_*,\cs)$, 
to denote
%, respectively, 
the set of all second-order variational
directions of $\cs$ at $x_*$.
%and the set of all limiting second-order variational directions of 
%$\cs$ at $x_*$. 
%Clearly, $\overline{var}_2(x_*,\cs)$ is the closure of $var_2(x_*,\cs)$
%in $\cx\times\cx$.
In addition, if $v_*\in\cx$, we use $var_{2,1}(x_*,v_*,\cs)$ to denote
the set of all $w\in\cx$ such that $(v_*,w)\in var_2(x_*,\cs)$. 
%Also, we use $\overline{var}_{2,1}(x_*,v_*,\cs)$ to denote
%the set of all $w\in\cx$ such that $(v_*,w)\in\overline{var}_2(x_*,\cs)$. 

\begin{Fact}\label{factvari2}
Let $\cx$, $\cs$, $x_*$ be as in Definition \ref{defvaria}. Then
%, if $V$ is one of the sets $var_2(x_*,\cs)$,
%$\overline{var}_2(x_*,\cs)$, it follows that 
\begin{itemize}
\item[1.] $(0,0)\in var_2(x_*,\cs)$, 
\end{itemize}
and 
\begin{itemize}
\item[2.] $var_2(x_*,\cs)$ has the {\em dilation invariance property}:
\begin{itemize}
\item[(DIP)] $(\rho v,\rho^2w)\in var_2(x_*,\cs)$
whenever $(v,w)\in var_2(x_*,\cs)$ and $\rho>0$.
\end{itemize}\end{itemize}
If, in  addition, $\cs$ is convex, then 
\begin{itemize}
\item[3.] $var_2(x_*,\cs)$ is convex,
\item[4.] $var_2(x_*,\cs)$ satisfies:
\begin{itemize}
\item[(4.\#)] $(v,\rho w)\in var_2(x_*,\cs)$ whenever 
$(v,w)\in var_2(x_*,\cs)$ and $\rho\ge 1$,
\end{itemize}
\item[5.] if $v_*\in\cx$, 
%and $W$ is one of the sets
%$var_{2,1}(x_*,v_*,\cs)$, $\overline{var}_{2,1}(x_*,v_*,\cs)$,
then 
\begin{itemize}
\item[(5.\#)] $var_{2,1}(x_*,v_*,\cs)$ is convex,
\item[(5.*$\,$)] 
%$var_{2,1}(x_*,v_*,\cs)$ has the following invariance property:
$\rho w\in var_{2,1}(x_*,v_*,\cs)$ whenever $w\in var_{2,1}(x_*,v_*,\cs)$
and $\rho\ge 1$,
\item[(5.\&)] if $0\in var_{2,1}(x_*,v_*,\cs)$ then 
$var_{2,1}(x_*,v_*,\cs)$ is a convex cone.
\end{itemize}\end{itemize}
\end{Fact}

{\sc Proof.} 
The fact that $(0,0)\in var_2(x_*,\cx)$ is obvious.

Assume that $(v,w)\in var_2(x_*,\cs)$ and $\rho>0$. Let 
$\eta\in Var(x_*,\cs)$ be such that 
$\eta(\eps)=x_*+\eps v+\eps^2w+o(\eps^2)$ as 
$\eps\downarrow 0$. Let $\tilde\eta$
be the reparametrized curve 
$$
\Big[0,\frac{\bar\eps_\eta}{\rho}\Big]\ni\eps\mapsto\eta(\rho\eps)
\equaldef\tilde\eta(\eps)\in\cs\,.
$$
Then
$$
\tilde\eta(\eps)=x_*+\eps\cdot(\rho v)+\eps^2\rho^2w+o(\eps^2)
\quad{\rm as}\quad\eps\downarrow 0\,.
$$
Therefore $(\rho v,\rho^2 w)\in var_2(x_*,\cs)$. 
So (DIP) holds.
% if $V=var_2(x_*,\cs)$. It then follows easily that (DIP) also holds if 
%$V=\overline{var}_2(x_*,\cs)$.

Now assume that the set
$\cs$ is convex. Pick pairs $(v_1,w_1)\in var_2(x_*,\cs)$,
$(v_2,w_2)\in var_2(x_*,\cs)$, and $\alpha$ such that 
$0\le\alpha\le 1$. Let $\eta_1\in Var(x_*,\cs)$,
$\eta_2\in Var(x_*,\cs)$, be such that, for $i=1,2$, 
$$
\eta_i(\eps)=x_*+\eps v_i+\eps^2w_i+o(\eps^2)\quad{\rm as}\quad
\eps\downarrow 0\,.
$$
Let $\eta(\eps)=\alpha\eta_1(\eps)+(1-\alpha)\eta_2(\eps)$..
Then $\eta(\eps)\in\cs$ if $\eps$ is small enough, and
$$
\eta(\eps)=x_*+\eps(\alpha v_1+(1-\alpha)v_2) 
+\eps^2(\alpha w_1+(1-\alpha)w_2)+o(\eps^2)\quad{\rm as}\quad
\eps\downarrow 0\,.
$$
Therefore $\alpha(v_1,w_1)+(1-\alpha)(v_2,w_2)\in
var_2(x_*,\cs)$. This proves that $var_2(x_*,\cs)$ is convex.
%The convexity of $\overline{var}_2(x_*,\cs)$ then follows
%immediately. 

If $(v,w)\in var_2(x_*,\cs)$, and $\rho\ge 1$, then 
$(\rho v,\rho^2w)\in var_2(x_*,\cs)$ by dilation invariance.
It follows that
$$
(v,\rho w)=\frac{1}{\rho}(\rho v,\rho^2 w)
=\Big(1-\frac{1}{\rho}\Big)(0,0)+
\frac{1}{\rho}(\rho v,\rho^2 w)\in var_2(x_*,\cs)\,,
$$ 
because $(0,0)\in var_2(x_*,\cs)$, $0\le\frac{1}{\rho}\le 1$, 
and $var_2(x_*,\cs)$ is convex. So (4.\#) holds.
% if $V=var_2(x_*,\cs)$,
%and this easily implies that (4.\#) also holds if  
%$V=\overline{var}_2(x_*,\cs)$.

Now pick $v_*\in\cx$, and let $W=var_{2,1}(x_*,v_*,\cs)$.
%, $\overline{var}_{2,1}(x_*,v_*,\cs)$.
Let $w_1\in W$, $w_2\in W$. Let $\alpha$ be such that 
$0\le\alpha\le 1$. Then 
$$
(v_*,\alpha w_1+(1-\alpha)w_2)=\alpha(v_*,w_1)+(1-\alpha)(v_*,w_2)\in 
var_{2}(x_*,v_*,\cs)\,,
$$
because $var_{2}(x_*,v_*,\cs)$ is convex. So $\alpha w_1+(1-\alpha)w_2\in W$. 
This proves that $W$ is convex. 

Property (5.*) follows easily from (4.\#). Finally, we assume that
$0\in W$ and show that $W$ is a convex cone. We already know that W 
is convex and is closed under multiplication by real numbers $\rho$ such 
that $\rho\ge 1$. So it suffices to show that
$\rho w\in W$ whenever $w\in W$ and $0\le\rho\le 1$. But, if 
$w\in W$ and $0\le\rho\le 1$, then 
$\rho w=(1-\rho)\cdot 0+\rho w$, which belongs to $W$
because $0\in W$ and $W$ is convex. 
So $W$ is indeed a convex cone. \proofend

\medbreak We let $\pi_{2,1}^\cx$ be the projection map
$$
\cx\times\cx\ni(v,w)\mapsto \pi_{2,1}^\cx(v,w)\equaldef v\in\cx\,.
$$
Then the following observation is trivial:
\begin{Fact}
If $\cx$, $\cs$, $x_*$ are as in Definition \ref{defvaria},
then 
$$
\quad\quad\quad\quad\quad\quad\quad\quad\quad
\quad\pi_{2,1}^\cx\Big(var_2(x_*,\cs)\Big)\subseteq var_1(x_*,\cs)\,.
%\quad{\rm and}\quad\pi_{2,1}^\cx\Big(\overline{var}_2(x_*,\cs)\Big)
%\subseteq\overline{var}_1(x_*,\cs)\,.
\qquad\quad\quad\quad\quad\quad\quad\quad\quad\diamondsuit
$$
\end{Fact}

\section{The separation theorem}


\noindent We assume that

\begin{itemize}
\item[(A1)] {\em $\cua$ (the ``ambient source space'')
and $Y$ (the ``target space'') are normed \mbox{spaces}, 
$Y$ is finite-dimensional,
$\cu$ is a convex subset of $\cua$, $u_*$ is a point 
of $\cu$,  \mbox{$\cf:\cu\mapsto Y$} is a map,
$\omega\in Y$ and $\omega\neq 0$.}
\end{itemize}

\begin{definition}
{\rm Let
$\cx$ be a normed space, let $\cs$ be a subset of $\cx$, and let
$\cf$ be a map from $\cs$ to another normed space $\cy$. We say that
a subset $\cs_0$ of $\cs$ is {\em $(\cs,\cf)$-abundant}
if there exists a sequence $\{\sigma_j\}_{j\in\nn}$ of
maps $\sigma_j:\cs\mapsto\cs_0$ such that 
\begin{itemize} 
\item[1.] the map $\cf\circ\sigma_j$ is continuous for each $j$,
\item[2.] $\cf\circ\sigma_j$ converges to $\cf$ 
uniformly on compact subsets of $\cs$ as $j\to\infty$.\proofend
\end{itemize}}
\end{definition}

We assume that
\begin{itemize}
\item[(A2)] {\em $\ca$ is an $(\cu,\cf)$-abundant 
subset of $\cu$ such that $u_*\in\ca$.}
\end{itemize}



We write
\begin{equation}\label{ystar}
y_*=\cf(u_*)\,,
\end{equation}
and let $\Lambda$ be the closed half-line
\begin{equation}\label{laline}
\Lambda\equaldef\{y_*+t\omega:t\in\rr,t\ge 0\}\,.
\end{equation}
Clearly, $y_*\in\cf(\ca)\cap\Lambda$. We will look for second-order
necessary conditions for the following {\em separation property}:

\medbreak
\begin{itemize}%\item[]\begin{itemize}
\item[(SEP)] $\qquad\qquad\qquad\qquad\qquad\cf(\ca)\cap\Lambda=\{y_*\}$\,. 
\end{itemize}%\end{itemize}

We assume that
\begin{itemize}
\item[(A3)] {\em $\cf$ is of class $D^2$ at $u_*$.}
\end{itemize}

We let $\cv$ be the set of all ``first-order variational directions.''
That is, we let
\begin{equation}\label{firstovar}
\cv=var_1(u_*,\cu)\,.
\end{equation}
We consider, in addition, a finite set $\hat\cv$  of
``independent first-order variational directions with second-order effects.'' 
This means, to begin with, that
\begin{itemize}
\item[(A4)] $\hat\cv$ is a finite subset of $\cv$.
\end{itemize}

We will need additional assumptions on $\hat\cv$. In order to state these
assumptions, we use (A3) and choose $P_1$, $P_2$ such that
\begin{itemize}
\item[(A5)] {\em $P_1$ is a continuous linear map 
from $\cua$ to $Y$, $P_2$ is a continuous bilinear symmetric  map 
from  $\cua\times\cua$ to $Y$, and}
$$
\cf(u_*+v)=\cf(u_*)+P_1\cdot v+\frac{1}{2}P_2(v,v)+
o(\|v\|^2)\,\,\,{\rm as}\,\,\,v\to 0\,,\,\,u_*+v\in\cu\,.
$$ 
\end{itemize}
We define
\begin{equation}\label{lambdao}
\Lambda_0=\{t\omega:t\ge 0\}\,,
\end{equation}
so that 
$$
\Lambda=y_*+\Lambda_0=\{y_*+z:z\in\Lambda_0\}\,.
$$
We then assume that
\medbreak
\begin{itemize}
\item[(A6)] {\em the vector $P_1\cdot v$ belongs to $\Lambda_0$
whenever $v\in\hat\cv$,}
\end{itemize}
as well as the {\em cross term condition}:

\medbreak
\begin{itemize}
\item[(A7)] {\em whenever $v_1,v_2\in\hat\cv$ are such that
$v_1\neq v_2$, the vector $P_2(v_1,v_2)$ 
belongs to $\Lambda_0$.}
\end{itemize}

Finally, we need the ``independence condition.''  The purpose
of this condition is to be able to add variations 
$$
\eps\mapsto\eta_v(\eps)=u_*+\tilde\eta_v(\eps)=
u_*+\eps v+o(\eps)
$$
of $u_*$ in $\cu$ corresponding to different members $v$ of $\hat\cv$,
and still obtain a variation of $u_*$ in $\cu$. We will guarantee this
by requiring that the values $\tilde\eta_v(\eps)$ belong to sets 
$\cu_v^0$ such that $u_*+\sum_{\ell}\cu_v^0\subseteq\cu$, 
from which it will follow that 
$u_*+\sum_{v}\tilde\eta_v(\eps)\in\cu$. 

The sets $\cu_v^0$ will actually play a role in the statement of
Theorem \ref{maintheor}. Therefore, rather than merely assume that
they exist, we will actually regard them as additional data
for our separation problem. If we write 
$\cu_v=\cu_v^0+u_*$, the ``independence condition'' becomes:

\medbreak
\begin{itemize}
\item[(A8)] {\em $\{\cu_v\}_{v\in\hat\cv}$ is a family of 
convex subsets of $\cua$ such that
\begin{itemize}
\item[1.] $u_*\in\cu_v$ for every $v\in\hat\cv$, 
\item[2.] $u_*+\sum_{v\in\hat\cv}\Big(\cu_v-u_*)\subseteq\cu$,
\item[3.] $v\in var_1(u_*,\cu_v)$ for each $v\in\hat\cv$. 
\end{itemize}}\end{itemize}

We will prove that if a {\em second-order surjectivity
condition}---called (SUR-2)---holds, then (SEP) fails. 
So, for (SEP) to hold, it will be a necessary condition that 
\mbox{(SUR-2)} not be true. Rewriting this in terms of multipliers
will give us a second-order Maximum Principle. 

\medbreak\noindent We define the cone $C_1(\cv)$ to be the smallest 
closed convex cone in $Y$ 
that contains all the vectors $P_1\cdot v$, for $v\in\cv$.

If $v\in\hat\cv$, we define 
$$
\cw(v)\equaldef\{w\in\cua:(v,w)\in var_2(u_*,\cu_v)\}\,,
$$
that is, 
\begin{equation}\label{secondovar}
\cw(v)\equaldef var_{2,1}(u_*,v,\cu_v)\,.
\end{equation}
We then let $C_2(\cv,\hat\cv)$ be the smallest closed convex cone
in $Y$ that contains $C_1(\cv)$ as well as all the vectors 
$\frac{1}{2}P_2(v,v)+P_1\cdot w$, for  all $(v,w)$ such that
$v\in\hat\cv$ and $w\in \cw(v)$.  
The {\em second-order surjectivity condition} then says that

\medbreak
\begin{itemize}\item[]\begin{itemize}
\item[(SUR-2)] {\em $\omega$ is an interior point of the cone 
$C_2(\cv,\hat\cv)$.}
\end{itemize}\end{itemize}

We are going to prove the following result.
\begin{theorem}\label{maintheor}
Assume that the data $\cua$, $Y$, $\cu$, $u_*$, $\cf$, $\omega$,
$\ca$, $\hat\cv$, $P_1$, $P_2$, $\{\cu_v\}_{v\in\hat\cv}$ 
satisfy Assumptions (A1)-(A8), where $y_*$,
$\Lambda$, $\cv$, $\Lambda_0$ are defined by (\ref{ystar}),
(\ref{laline}), (\ref{firstovar}), (\ref{lambdao}).  Then the
second-order surjectivity condition (SUR-2) implies that 
the separation property (SEP) does not hold.
\end{theorem}

An obvious consequence of Theorem \ref{maintheor} is the following
``multiplier'' result:
\begin{corollary}\label{maincoro}
Assume that the hypotheses of Theorem \ref{maintheor} hold. Then
a necessary condition for the separation property (SEP) is that there 
exist a nonzero covector $\bar\psi\in Y^*$ such that
\begin{eqnarray}
\langle\bar\psi,P_1\cdot v\rangle&\ge& 0\,\,\,{\rm for\,\,\,all}
\,\,\,v\in\overline{var}_1(u_*,\cu)\,,\label{equatio1}\\
\langle\bar\psi,\frac{1}{2}P_2(v,v)+P_1\cdot w\rangle&\ge& 0\,\,\,
{\rm for\,\,\,all}\,\,\,v\in\hat\cv,\,w\in var_{2,1}(x_*,v,\cu_v)
\,,\label{equatio2}\\
\langle\bar\psi,\omega\rangle&\le& 0\,.\label{equatio3}
\end{eqnarray}
Condition (\ref{equatio2}) implies, in particular, that
\begin{equation}\label{equatio}
\left.\begin{array}{rccllll}
\qquad\qquad
\langle\bar\psi,P_1\cdot w\rangle&\ge& 0&\,\,\,&{\rm whenever}&
\,\,\,&v\in\hat\cv,w\in var_{2,1}(x_*,v,\cu_v)\,.\end{array}\right.
\end{equation}
\end{corollary}

\def\cint{C^{^{^{\!\!\!\!o}}}}


\medbreak{\sc Proof that Theorem \ref{maintheor} implies Corollary
\ref{maincoro}.} First, we observe that (\ref{equatio}) 
is a consequence of (\ref{equatio2}) because, if $v\in\hat\cv$,
$w\in\cw(v)$, then $\rho\cdot w\in\cw(v)$ whenever
$\rho\ge 1$, in view of condition (5.*) of Fact \ref{factvari2}, so that
$$
\left.\begin{array}{rccllll}
\langle\bar\psi,\frac{1}{2}P_2(v,v)+\rho.P_1\cdot w\rangle&\ge& 0&\,\,\,&\,
{\rm whenever}&\,\,\,&\rho\ge 1\,,\end{array}\right.
$$
and then, dividing by $\rho$ and letting $\rho\uparrow+\infty$, 
(\ref{equatio}) follows.

Write 
\begin{equation}\label{c-cint}
C=C_2(\cv,\hat\cv)\,,\qquad\cint\equaldef{\rm Interior}(C)\,.
\end{equation}
If $\cint=\emptyset$ then the linear subspace of $Y$ spanned by $C$
is of positive codimension in $Y$, so there exists a nonzero covector
$\bar\psi\in Y^*$ such that $\langle\bar\psi,v\rangle=0$ for all
$v\in C$. After replacing $\bar\psi$ by $-\bar\psi$, if necessary, we may 
assume that $\langle\bar\psi,\omega\rangle\le 0$, and then 
$\bar\psi$ satisfies all our conditions. 

If $\cint\neq\emptyset$, then $\cint$ and $\Lambda_0$ are disjoint
convex subsets of $Y$, $\cint$ is open, and $\Lambda_0$ is closed.
The Hahn-Banach theorem then implies that there exist 
$\bar\psi\in Y^*\backslash\{0\}$ and $\alpha\in\rr$ such 
that $\langle\bar\psi,t\omega\rangle\le\alpha$ for all
$t\in[0,+\infty[$, and 
$\langle\bar\psi,z\rangle\ge\alpha$ for all $z\in\cint$. 
The first condition tells us in particular that 
$\langle\bar\psi,\omega\rangle\le\alpha$ and 
$0\le\alpha$. The second condition implies that
$\langle\bar\psi,z\rangle\ge\alpha$ for all $z\in C$, since 
$C$ is the closure of $\cint$, thanks to the fact that 
$\cint\neq\emptyset$. Since $0\in C$, we have $\alpha\le 0$.
Then $\alpha=0$, and our conclusion follows. (The fact 
that the inequality of (\ref{equatio1}) holds for all
$v\in\overline{var}_1(u_*,\cu)$ follows because any such
$v$ is the limit of a sequence $\{v_j\}_{j\in\nn}$ of 
members of $\cv$, so that $P_1\cdot v_j\in C$ for every 
$j$, and then $\langle\bar\psi,P_1\cdot v_j\rangle\ge 0$,
so  $\langle\bar\psi,P_1\cdot v\rangle\ge 0$.) \proofend

\bigbreak {\sc Proof of Theorem \ref{maintheor}.} Use $C$, 
$\cint$ as in (\ref{c-cint}). Assume that (SUR-2) holds,
so that $\omega\in\cint$. Let $\calc$ be the set consisting
of all vectors
$P_1\cdot v$, $v\in\cv$, together with all the vectors
$\frac{1}{2}P_2(v,v)+P_1\cdot w$, for  $(v,w)$ such that
$v\in\hat\cv$ and $w\in \cw(v)$.  Then every member of 
$C$ is a limit of finite linear combinations of members of $\calc$ 
with nonnegative coefficients.

Enumerate the members of $\hat\cv$, by writing
$$
\hat\cv=\{\hat v_1,\ldots,\hat v_L\}\,,\qquad
{\rm with}\qquad v_\ell\neq v_{\ell'}\qquad
{\rm whenever}\qquad \ell\neq \ell'\,.
$$
Also, write 
$$
\cu_\ell\equaldef\cu_{\hat v_\ell}\,,\quad\quad\quad
\cu_\ell^0\equaldef\cu_{\ell}-u_*\,.
$$

\medbreak\noindent Let $n={\rm dim}\,Y$. Let 
$\tilde e_1,\ldots,\tilde e_n$ be a basis of $Y$, and define
$\tilde e_0=-(\tilde e_1+\ldots+\tilde e_n)$.
Since $\omega\in\cint$, we can choose $\bar s$ such that 
$\bar s>0$ and the vectors $\omega+\bar s\tilde e_i$ belong to $\cint$ for
$i=0,\ldots,n$. Let $\hat e_i=\bar s\tilde e_i$ for 
$i=0,\ldots,n$.  Then the vectors $\hat e_1,\ldots,\hat e_n$ 
constitute a basis of $Y$, $\hat e_0=-(\hat e_1+\ldots+\hat e_n)$, and
$$
\omega+\hat e_i\in\cint\,\,\,{\rm for}\,\,\,i=0,\ldots,n\,.
$$
Let $\hat\omega_i=\omega+\hat e_i$, for $i=0,\ldots,n$. Then each 
$\hat\omega_i$ is the limit of a sequence $\{\hat\omega_i^j\}_{j\in\nn}$
of finite linear combinations of members of $\calc$ 
with nonnegative coefficients.  
Write
$$
\Omega=(\omega,1)\,,\quad
\hat\omega_i^\infty=\hat\omega_i\,,\quad\Omega_i=\Omega_i^\infty=
(\hat\omega_i,1)\,,\quad
\Omega_i^j=(\hat\omega_i^j,1)\,.
$$
Then $\Omega$ and the  $\Omega_i^j$---for $i=0,\ldots,n$,
$j\in\bar\nn\equaldef\nn\cup\{+\infty\}$---are members of 
$Y\times\rr$. Moreover, $\Omega_0,\Omega_1,\ldots,\Omega_n$ 
are linearly independent. (Reason: 
if $\sum_{i=0}^nr_i\Omega_i=0$ then 
$\sum_{i=0}^nr_i\hat\omega_i=0$ and $\sum_{i=0}^nr_i=0$, so 
$$
\sum_{i=0}^nr_i\hat e_i=\sum_{i=0}^nr_i(\omega+\hat e_i)=
\sum_{i=0}^nr_i\hat\omega_i=0\,,
$$
and then $\sum_{i=1}^n(r_i-r_0)\hat e_i=0$,
since $\hat e_0=-(\hat e_1+\ldots+\hat e_n)$. But then 
the linear independence of the vectors
$\hat e_1,\ldots,\hat e_n$ implies that
$$
r_1-r_0=r_2-r_0=\ldots=r_n-r_0=0\,.
$$ 
Therefore $0=\sum_{i=0}^nr_i=(n+1)r_0$, so $r_0=0$, and then
$r_1=r_2=\ldots=r_n=0$.)  

It follows that the $n+1$-tuple
$(\Omega_0^j,\Omega_1^j,\ldots,\Omega_n^j)$ is linearly 
independent for large $j$, so that we may assume, after passing 
to a subsequence if necessary, that 
$(\Omega_0^j,\Omega_1^j,\ldots,\Omega_n^j)$ is a linearly 
independent $n+1$-tuple for all $j\in\bar\nn$. Therefore, 
if $j\in\bar\nn$ 
then every vector $\xi\in Y\times\rr$ can be expressed
in a unique way as a linear combination 
$\sum_{i=0}^n\tilde s_i^j(\xi)\Omega_i^j$
of $\Omega_0^j,\Omega_1^j,\ldots,\Omega_n^j$. Clearly,
the coefficients $\tilde s_i^j(\xi)$ depend continuously on $\xi$ 
for each $i,j$ and, moreover,  
\mbox{$\lim_{j\to\infty}\tilde s_i^j(\xi)=\tilde s_i^\infty(\xi)$}
for each $i,\xi$, uniformly with respect to $\xi$ for $\xi$ in a compact 
subset of $Y\times\rr$.   If $y\in Y$, define 
$s_i^j(y)=\tilde s_i^j(y,1)$. It follows that
$$
\sum_{i=0}^ns_i^j(y)\hat\omega_i^j=y\quad\quad
{\rm and}\quad\quad \sum_{i=0}^ns_i^j(y)=1\quad\quad{\rm whenever}
\quad y\in Y\,.
$$
Moreover,
$$
\omega=\frac{1}{n+1}(\hat\omega_0+\hat\omega_1+\ldots+\hat\omega_n)\,,
$$
so that
$$
s_0^\infty(\omega)=s_1^\infty(\omega)=\ldots+s_n^\infty(\omega)=
\frac{1}{n+1}\,.
$$
It follows that 
\begin{equation}\label{henry}
s_0^j(\omega)>0\,,\,\,
s_1^j(\omega)>0\,,\,\,\ldots\,\,\,s_n^j(\omega)>0\,,
\end{equation}
if $j$ is large enough. 

Now fix a $j\in\nn$ such that (\ref{henry}) holds. Write
$$
\hat\omega_i=\hat\omega_i^j\qquad{\rm for}\qquad i=0,\ldots,n\,.
$$
Then every $y\in Y$ has a unique expression as a linear
combination of $\hat\omega_0,\ldots,\hat\omega_n$ with coefficients
adding up to $1$, and the coefficients of this combination 
are the numbers $\hat s_i(y)\equaldef s_i^j(y)$. These coefficients
depend continuously on $y$, and 
$\hat s_i(\omega)>0$ for $i=0,\ldots,n$. Moreover, each 
$\hat\omega_i$ is a finite linear combination with nonnegative
coefficients of members of $\calc$. It follows
(using the fact that the sets $\cw(v)$ are convex for each 
$v\in\hat\cv$)  that there exist
\begin{itemize}
\item[1.] a finite subset $\{v_1,\ldots,v_J\}$ of $\cv$, 
\item[2.] members $w_{i\ell}$ of $\cw(\hat v_\ell)$ for
$i\in\{0,\ldots,n\}$, $\ell\in\{1,\ldots,L\}$, 
\item[3.] nonnegative real numbers 
$\alpha_{ij}$, for $i=0,\ldots,n$, $j=1,\ldots,J$,
\item[4.] nonnegative real numbers $\hat\beta_{i\ell}$,
for $i=0,\ldots,n$, $\ell=1,\ldots,L$,
\end{itemize}
such that 
$$
\hat\omega_i=\sum_{j=1}^J\alpha_{ij}P_1\cdot v_j
+\sum_{\ell=1}^L\hat\beta_{i\ell}\Big(P_1\cdot w_{i\ell}
+\frac{1}{2}P_2(\hat v_{\ell},\hat v_{\ell})\Big)\,\,\,
{\rm for}\,\,\,i=0,\ldots,n\,.
$$

Now, if $\delta>0$, define
$$
\omega_i^\delta=\sum_{j=1}^J\alpha_{ij}P_1\cdot v_j
+\sum_{\ell=1}^L(\hat\beta_{i\ell}+\delta)\Big((1-\delta)
P_1\cdot w_{i\ell}+\frac{(1-\delta)^2}{2}P_2(\hat v_{\ell},
\hat v_{\ell})\Big)\,,
$$
for $i=0,\ldots,n$.  Then $\omega_{i}^\delta\to\hat\omega_i$ 
as $\delta\downarrow 0$. It follows that, if $\delta$ is sufficiently
small, then 
\begin{itemize}
\item[(\&)] every $y\in Y$ has a unique expression as a linear
combination of $\omega_0^\delta,\ldots,\omega_n^\delta$ with coefficients
adding up to $1$, and the coefficients $s_{i,\delta}(y)$
of this combination depend continuously on $y$, and 
are such that $s_{i,\delta}(\omega)>0$ for $i=0,\ldots,n$. 
\end{itemize}


Fix a $\delta$ for which (\&) happens, such that $0<\delta<1$. Call it 
$\bar\delta$, and write 
$$
\omega_i=\omega_i^{\bar\delta}\,,\qquad
s_i(y)=s_{i,\bar\delta}(y)\,,\qquad
\beta_{i\ell}=\hat\beta_{i\ell}+\bar\delta\,.
$$
Then
\begin{equation}\label{eqonomi}
\omega_i=\sum_{j=1}^J\alpha_{ij}P_1\cdot v_j
+\sum_{\ell=1}^L\beta_{i\ell}\Big((1-\bar\delta)
P_1\cdot w_{i\ell}+\frac{(1-\bar\delta)^2}{2}P_2(\hat v_{\ell},
\hat v_{\ell})\Big)\,,
\end{equation}
for $i=0,\ldots,n$, and there exist continuous functions
$Y\ni y\mapsto s_i(y)\in\rr$, for $i=0,\ldots,n$, such that
\begin{equation}\label{eqony}
y=\sum_{i=0}^ns_i(y)\omega_i\,\,\,{\rm and}\,\,\,
\sum_{i=0}^ns_i(y)=1\,\,\,{\rm for\,\,\,all}\,\,\,y\in Y\,.
\end{equation}
Moreover, the numbers $\alpha_{ij}$ are nonnegative, the
$\beta_{i\ell}$ are strictly positive, and 
\begin{equation}\label{henryo}
s_0(\omega)>0\,,\,\,
s_1(\omega)>0\,,\,\,\ldots\,\,\,s_n(\omega)>0\,.
\end{equation}

Let $\Sigma$ be the convex hull of the vectors
$\omega_i$, $i=0,\ldots,n$, so $\Sigma$ is the simplex 
\begin{eqnarray*}
&\,&\Sigma=\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\Big\{y\in Y:
(\exists s_0,\ldots,s_n\in\rr)\Big(
y=\sum_{i=0}^ns_i\omega_i\,,\,\,s_i\ge 0\,\,{\rm for}
\,\,i=0,\ldots,n\,,\,\,\sum_{i=0}^ns_i=1\Big)\Big\},
\end{eqnarray*}
that is,
$$
\Sigma=\Big\{\,y\in Y:s_i(y)\ge 0\,\,{\rm for}
\,\,i=0,\ldots,n\,\Big\}\,.
$$


Then the following are true:
\begin{itemize}
\item[1.] $\omega$ is an interior point of $\Sigma$,
\item[2.] every $y\in\Sigma$ has an expression
$$
y=\sum_{i=0}^ns_i(y)\omega_i\,,
$$
\end{itemize}
where the functions $\Sigma\ni y\mapsto s_i(y)\in\rr$ are 
continuous, nonnegative, and such that 
$\sum_{i=0}^ns_i(y)=1$ for every $y\in\Sigma$.

\medbreak We now fix a $\rho$ such that 
$$
\rho>0\qquad{\rm and}\qquad\cb \subseteq\Sigma\,,\qquad{\rm where}
\qquad\cb=\{y\in Y:\|y-\omega\|\le\rho\}\,.
$$
We then use (A6) and (A7) 
to pick nonnegative numbers $\tau_\ell$, $\theta_{\ell,\ell'}$
such that
\begin{eqnarray}
P_1\cdot\hat v_{\ell}&=&
\tau_\ell\omega\,\,\,\,{\rm for}\,\,\,\ell=1,\ldots,L\,,\label{ineq-uu}\\
P_2(\hat v_{\ell},\hat v_{\ell'})&=&
\theta_{\ell,\ell'}\omega\,\,\,\,{\rm for}\,\,\,
\ell,\ell'=1,\ldots,L\,,\,\ell\neq\ell'\,.\label{ineq-vv}
\end{eqnarray}

\begin{lemma}\label{illem} Let $\delta$ be such that $0<\delta<1$.
Then there exist an $\bar\eps$ such 
that $\bar\eps>0$, and continuous curves 
$$
\eta^y:[0,\bar\eps]\mapsto\cu\,,
$$
such that 
\begin{itemize}
\item[1.] the map
$$
[0,\bar\eps]\times\Sigma\ni(\eps,y)\mapsto\eta^y(\eps)\in\cu
$$
is continuous,
\item[2.] as $\eps\downarrow 0$,
\begin{eqnarray}
\eta^y(\eps)&=&u_*+\eps\Biggl((1-\delta)\sum_{\ell=1}^L\Biggl(
\sqrt{\sum_{i=0}^n s_{i}(y)\beta_{i\ell}}\Biggr)\hat v_\ell
\Biggr)
+\eps^2\Biggl(\sum_{j=1}^J\sum_{i=0}^ns_i(y)\alpha_{ij}v_{j}
\Biggr)\nonumber\\
&\,&\,\,+\eps^2(1-\delta)\Biggl(\sum_{\ell=1}^L\sum_{i=0}^ns_i(y)\beta_{i\ell}
w_{i\ell}\Biggr)
+o(\eps^2)\,,\label{equ-b}
\end{eqnarray}
uniformly with respect to $y\in\Sigma$.
\end{itemize}
\end{lemma}


\medbreak{\sc Proof of Lemma \ref{illem}.} 
The fact that the $v_j$ belong
to $\cv$ implies that there exist continuous curves
$\tilde\zeta_j:[0,\tilde\eps_j]\mapsto\cu$ such 
that $\tilde\zeta_j(\eps)=u_*+\eps v_j+o(\eps)$.
Let 
$$
\bar\alpha=\max\{\alpha_{ij}:i=0,\ldots,n\}\,.
$$ 
Define 
$$
\zeta_j^y(\eps)=\tilde\zeta_j\Big(\frac{J\eps^2}{\delta}
\sum_{i=0}^ns_i(y)\alpha_{ij}\Big)\,.
$$
Then the curves $\zeta_j^y$ are certainly defined for 
$\eps\in[0,\bar\eps_1]$, where
$$
\bar\eps_1=\sqrt{\frac{\delta
\min\{\tilde\eps_j:j=1,\ldots,J\}}{1+J\bar\alpha}}\,.
$$
Moreover, the $\zeta_j^y$ are $\cu$-valued, and satisfy
\begin{equation}\label{equzeta}
\zeta_j^y(\eps)=u_*+\frac{J\eps^2}{\delta}
\Big(\sum_{i=0}^ns_i(y)\alpha_{ij}\Big)
v_j+o(\eps^2)\quad{\rm as}\quad\eps\downarrow 0\,,
\end{equation}
the various $o(\eps^2)$s being uniform with respect to $y$.

Then define 
$$
\zeta^y(\eps)=\frac{1}{J}\Big(\sum_{j=1}^J\zeta_j^y(\eps)\Big)\,.
$$
The map $[0,\bar\eps_1]\times\Sigma\ni(\eps,y)\mapsto\zeta^y(\eps)$
is clearly continuous and $\cu$-valued.  Moreover, 
$$
\zeta^y(\eps)=u_*+\frac{\eps^2}{\delta}
\Big(\sum_{i=0}^n\sum_{j=1}^Js_i(y)\alpha_{ij}
v_j\Big)+o(\eps^2)\,.
$$

Next we use the fact that the pairs $(\hat v_\ell,w_{i\ell})$ belong
to $var_2(u_*,\cu_\ell)$ to conclude that there exist
continuous curves $\hat\theta_{i\ell}:[0,\hat\eps_{i\ell}]\mapsto\cu_\ell$,
defined on intervals $[0,\hat\eps_{i\ell}]$, with $\hat\eps_{i\ell}>0$, 
such that 
$$
\hat\theta_{i\ell}(\eps)=u_*+\eps\hat v_\ell+\eps^2w_{i\ell}+o(\eps^2)
\quad{\rm as}\quad\eps\downarrow 0\,.
$$
Define
\begin{equation}\label{problo}
\tilde\theta_{\ell}^y(\eps)=\sum_{i=0}^n
\frac{s_i(y)\beta_{i\ell}}{\sum_{i'=0}^n s_{i'}(y)\beta_{i'\ell}}
\hat\theta_{i\ell}^y(\eps)\,.
\end{equation}
Then
$$
\tilde\theta_{\ell}^y(\eps)=u_*+\eps\hat v_\ell
+\eps^2\sum_{i=0}^n
\frac{s_i(y)\beta_{i\ell}w_{i\ell}}{\sum_{i'=0}^n s_{i'}(y)\beta_{i'\ell}}
+o(\eps^2)\quad{\rm as}\quad\eps\downarrow 0\,.
$$
Define 
$$
\theta_{\ell}^y(\eps)=\tilde\theta_{\ell}^y\Biggl(
\eps\sqrt{\sum_{i=0}^n s_{i}(y)\beta_{i\ell}}\Biggr)\,.
$$
Then all the curves $\theta_{\ell}^y$ are defined on the interval
$[0,\bar\eps_2]$, where 
$$
\bar\eps_2=\frac{\min\{\hat\eps_{i\ell}:i=0,\ldots,n,\ell=1\,\ldots,L\}}
{\sqrt{\max\{\beta_{i\ell}:i=0,\ldots,n,\ell=1\,\ldots,L\}}}\,\,.
$$
Moreover,
$$
\theta_{\ell}^y(\eps)=u_*+
\Biggl(\eps\sqrt{\sum_{i=0}^n s_{i}(y)\beta_{i\ell}}\Biggr)v_\ell
+\eps^2\Big(\sum_{i=0}^n s_{i}(y)\beta_{i\ell}w_{i\ell}\Big)
+o(\eps^2)\quad{\rm as}\quad\eps\downarrow 0\,.
$$

Now, the sets $\cu_\ell$ are convex, the curves
$\hat\theta_{i\ell}$ are $\cu_\ell$-valued, the 
vectors $\bar\theta_{\ell}^y(\eps)$ are, for each $y,\eps$,
convex combinations of the $\hat\theta_{i\ell}(\eps)$,
and the curves $\theta_{\ell}^y$ are just reparametrized versions of the 
$\bar\theta_{\ell}^y$. This implies that the curves 
$\theta_{\ell}^y$ are $\cu_\ell$-valued. It follows that 
$$
\theta_{\ell}^y(\eps)-u_*\in\cu_\ell^0\quad
{\rm whenever}\quad\eps\in[0,\bar\eps_2]\,,\,\,y\in\Sigma\,.
$$

Define
$$
\theta^y(\eps)=u_*+\sum_{\ell=1}^L\Big(\theta_{\ell}^y(\eps)-u_*\Big)\,.
$$
Then the curves $\theta^y$ take values in the set
$u_*+\cu_1^0+\cu_2^0+\cdots+\cu_L^0$, which by our assumptions
is a subset of $\cu$. So the $\theta^y$ are $\cu$-valued curves.
Moreover, 
$$
\theta^y(\eps)=u_*
+\Biggl(\eps\sum_{\ell=1}^L
\sqrt{\sum_{i=0}^n s_{i}(y)\beta_{i\ell}}\Biggr)v_\ell
+\eps^2\Big(\sum_{\ell=1}^L\sum_{i=0}^n s_{i}(y)\beta_{i\ell}w_{i\ell}\Big)
+o(\eps^2)\quad{\rm as}\quad\eps\downarrow 0\,.
$$
It is clear from our construction that the map
$[0,\bar\eps_2]\times\Sigma\ni(\eps,y)\mapsto\theta^y(\eps)$ 
is continuous.  (The only step which in principle could have caused
some problems is the definition of $\tilde\theta_\ell^y(\eps)$ 
in (\ref{problo}), where the denominator 
$\sum_{i'=0}^n s_{i'}(y)\beta_{i'\ell}$ occurs. Notice, 
however, that we have taken the precaution to make sure that all
the numbers $\beta_{i\ell}$ are strictly positive, and this ensures
that the denominator never vanishes, since 
$\sum_{i'=0}^n s_{i'}(y)=1$.)

Finally, we let
$$
\eta^y(\eps)=\delta\zeta^y(\eps)+(1-\delta)\theta^y(\eps)
\quad{\rm for}\quad y\in\Sigma\,,\,\eps\in[0,\bar\eps]\,,
$$
where $\bar\eps=\min(\bar\eps_1,\bar\eps_2)$.


Then the $\eta^y$   satisfy all the desired conditions, and the proof
of Lemma \ref{illem} is complete.\proofend

\bigbreak{\sc Continuation of the proof of Theorem 
\ref{maintheor}.} We apply Lemma \ref{illem}, taking $\delta$ to be the 
number $\bar\delta$ introduced earlier. The lemma allows us to choose
a family $\{\eta^y\}_{y\in\Sigma}$ of continuous $\cu$-valued
curves on an interval $[0,\bar\eps]$, $\bar\eps>0$, such that  
$\eta^y(\eps)$ is jointly continuous in $\eps,y$, and 
(\ref{equ-b}) holds. 

We then define, for $\eps\in[0,\bar\eps]$, $y\in\Sigma$, a vector
$\mu_\eps(y)\in Y$ by letting
$$
\mu_\eps(y)=\frac{1}{\eps^2}\Biggl(\cf\Big(\eta^y(\eps)\Big)
-\cf(u_*)-\Big(\eps^2+\xi_\eps(y)\Big)\omega\Biggr)\,,
$$
where 
\begin{eqnarray*}
\xi_\eps(y)&\equaldef&
\Biggl(\eps(1-\delta)\sum_{\ell=1}^L\tau_\ell
\sqrt{\sum_{i=0}^n s_{i}(y)\beta_{i\ell}}\Biggr)\\
&\,&\,\,+\Biggl(\frac{\eps^2(1-\delta)^2}{2}\sum_{1\le\ell,\ell'\le L,
\ell\neq\ell'}\theta_{\ell,\ell'}\sqrt{\sum_{i=0}^n s_{i}(y)\beta_{i\ell}}
\sqrt{\sum_{i=0}^n s_{i}(y)\beta_{i\ell'}}
\Biggr)\,.
\end{eqnarray*}
Then  $\mu_\eps$ is a continuous map from $\Sigma$ to $Y$,
and $\xi_\eps$ is a continuous map from $\Sigma$ to $\rr$
with nonnegative values.

The next step is to use (\ref{equ-b}) to expand 
$\mu_\eps(y)$ to second powers of $\eps$. For this purpose,
we first expand $\cf\Big(\eta^y(\eps)\Big)$. 
The result is as follows.

\begin{eqnarray*}
\cf\Big(\eta^y(\eps)\Big)
&=&\cf(u_*)+\eps\Biggl((1-\delta)
\sum_{\ell=1}^L\Biggl(\sqrt{\sum_{i=0}^n s_{i}(y)\beta_{i\ell}}\Biggr)P_1\cdot
\hat v_\ell\Biggr)\\
&\,&\!\!+\eps^2\Biggl(\sum_{j=1}^J\sum_{i=0}^ns_i(y)\alpha_{ij}P_1\cdot v_{j}
\Biggr)+\eps^2(1-\delta)\Biggl(\sum_{\ell=1}^L\sum_{i=0}^ns_i(y)\beta_{i\ell}
P_1\cdot w_{i\ell}\Biggr)\\
&\,&\!\!+\frac{\eps^2}{2}\Biggl((1-\delta)^2\sum_{\ell=1}^L\sum_{\ell'=1}^L
\sqrt{\sum_{i=0}^n s_{i}(y)\beta_{i\ell}}
\sqrt{\sum_{i=0}^n s_{i}(y)\beta_{i\ell'}}
\Biggr)P_2(\hat v_\ell,\hat v_{\ell'})\\
&\,&\!\!+o(\eps^2)\,.
\end{eqnarray*}

We now use (\ref{ineq-uu}) and (\ref{ineq-vv}) to replace 
$P_1\cdot\hat v_{\ell}$ by $\tau_\ell\omega$ and 
$P_2(\hat v_{\ell},\hat v_{\ell'})$ by 
$\theta_{\ell,\ell'}\omega$ when $\ell\neq\ell'$. This yields:

\begin{eqnarray*}
\cf\Big(\eta^y(\eps)\Big)
&=&\cf(u_*)+\Biggl(\eps(1-\delta)\sum_{\ell=1}^L\tau_\ell
\sqrt{\sum_{i=0}^n s_{i}(y)\beta_{i\ell}}\Biggr)\omega\\
&\,&\!\!+\eps^2\Biggl(\sum_{j=1}^J\sum_{i=0}^ns_i(y)\alpha_{ij}P_1\cdot v_{j}
\Biggr)+\eps^2(1-\delta)\Biggl(\sum_{\ell=1}^L\sum_{i=0}^ns_i(y)\beta_{i\ell}
P_1\cdot w_{i\ell}\Biggr)\\
&\,&\!\!+\Biggl(\frac{\eps^2(1-\delta)^2}{2}\sum_{\ell=1}^L
\Big(\sum_{i=0}^n s_{i}(y)\beta_{i\ell}\Big)
\Biggr)P_2(\hat v_\ell,\hat v_{\ell})\\
&\,&\!\!+\Biggl(\frac{\eps^2(1-\delta)^2}{2}\sum_{1\le\ell,\ell'\le L,
\ell\neq\ell'}\theta_{\ell,\ell'}\sqrt{\sum_{i=0}^n s_{i}(y)\beta_{i\ell}}
\sqrt{\sum_{i=0}^n s_{i}(y)\beta_{i\ell'}}
\Biggr)\omega\\
&\,&\!\!+o(\eps^2)\,.
\end{eqnarray*}

It follows that
\begin{eqnarray*}
\cf\Big(\eta^y(\eps)\Big)&=&\cf(u_*)+\xi_\eps(y)\omega+o(\eps^2)\\
&\,&\,\,+\eps^2\sum_{i=0}^ns_i(y)
\Biggl(\sum_{j=1}^J\alpha_{ij}P_1\cdot v_{j}
+(1-\delta)\Big(\sum_{\ell=1}^L\beta_{i\ell}P_1\cdot w_{i\ell}\Big)\\
&\,\,&\,\,\,\,\,+\frac{(1-\delta)^2}{2}\Big(\sum_{\ell=1}^L
\beta_{i\ell}P_2(\hat v_\ell,\hat v_{\ell}\Big)\Biggr)\\
&=&\cf(u_*)+\xi_\eps(y)\omega+\eps^2\sum_{i=0}^ns_i(y)\omega_i
+o(\eps^2)\\
&=&\cf(u_*)+\xi_\eps(y)\omega+\eps^2y+o(\eps^2)\,.
\end{eqnarray*}

Therefore,
\begin{eqnarray*}
\mu_\eps(y)&=&
\frac{1}{\eps^2}\Biggl(\cf\Big(\eta^y(\eps)\Big)-
\cf(u_*)-(\eps^2+\xi_\eps(y))\omega\Biggr)\\
&=&\frac{1}{\eps^2}\Big(\eps^2(y-\omega)+o(\eps^2)\Big)\\
&=&y-\omega+o(1)\,,
\end{eqnarray*}
as $\eps\downarrow 0$, the $o(1)$ being uniform with respect 
to $y\in\Sigma$. 

Fix an $\eps$ such that $\eps>0$ and 
\begin{equation}\label{wing}
\|\mu_\eps(y)-(y-\omega)\|\le\frac{\rho}{2}\qquad
{\rm for\,\,\,all}\qquad y\in\Sigma\,.
\end{equation}
Let $K=\{\eta^y(\eps):y\in\Sigma\}$ so 
$K$ is a compact subset of $\cu$. Let $\{\sigma_j\}_{j\in\nn}$ be a 
sequence of maps from $\cu$ to $\ca$ such that 
$\cf\circ\sigma_j$ is continuous for each $j$ and 
$\cf\circ\sigma_j\to\cf$ uniformly on compact subsets of $\cu$. Pick 
$j$ such that 
$$
\|\cf(\sigma_j(u))-\cf(u)\|\le\frac{\eps^2\rho}{2}\qquad
{\rm whenever}\qquad u\in K\,.
$$
Define
$$
\mu_\eps^j(y)=
\frac{1}{\eps^2}\Biggl(\cf\Big(\sigma_j(\eta^y(\eps))\Big)-
\cf(u_*)-(\eps^2+\xi_\eps(y))\omega\Biggr)\,.
$$
Then $\|\mu_\eps^j(y)-\mu_\eps(y)\|\le\frac{\rho}{2}$ for all
$y\in\Sigma$. Therefore 
\begin{equation}\label{wingo}
\|\mu_\eps^j(y)-(y-\omega)\|\le\rho\qquad
{\rm for\,\,\,all}\qquad y\in\Sigma\,.
\end{equation}

Now, if $y\in\cb$, we have 
$$
\|y-\mu_\eps^j(y)-\omega\|=\|\mu_\eps^j(y)-(y-\omega)\|\le\rho\,,
$$
so that $y-\mu_\eps(y)^j\in\cb$. Therefore the continuous map
$$
\cb\ni y\mapsto y-\mu_\eps^j(y)\in Y
$$
is in fact a map from $\cb$ to $\cb$. By the Brouwer fixed point theorem,
this map has a fixed point. Therefore, there exists a
$y\in\cb$ such that $y-\mu_\eps^j(y)=y$, and then
this $y$ satisfies $\mu_\eps^j(y)=0$. 
We have thus shown that {\em for every sufficiently small positive 
$\eps$ there exists a $j(\eps)\in\nn$ such that 
the equation $\mu_\eps^j(y)=0$ has a solution
$y^{j,\eps}\in\Sigma$ for every $j\in\nn$ such that $j\ge j(\eps)$}.
 
Now, if $y^{j,\eps}$ is a solution for an $\eps$ such that $\eps>0$
and a $j\in\nn$, then the fact that $\mu_\eps^j(y^{j,\eps})=0$ says that 
$$
\cf(\sigma_j(\eta^{y^{j,\eps}}(\eps)))=
y_*+(\eps^2+\xi_\eps(y^{j,\eps}))\omega\,.
$$
So 
$$
\cf(\sigma_j(\eta^{y^{j,\eps}}(\eps)))\in\cf(\ca)\cap\Lambda\,.
$$
Moreover,
$$
\cf(\sigma_j(\eta^{y^{j,\eps}}(\eps)))\neq y_*\,,
$$
because $\eps>0$, $\xi_\eps(y^{j,\eps})\ge 0$, and $\omega\neq 0$. 
Therefore (SEP) fails to hold.

Our proof of Theorem \ref{maintheor} is thus complete.\proofend


\section{Second-order necessary condition for a minimum}


\begin{theorem}\label{theoopti}
Assume that 
\begin{itemize}
\item[1.] $\cua$ and $Z$ are normed \mbox{spaces}, 
\item[2.] $Z$ is finite-dimensional,
\item[3.] $\cu$ is a convex subset of $\cua$, 
\item[4.] $u_*\in\cu$,  
\item[5.] \mbox{$\ce:\cu\mapsto Z$} is a map,
\item[6.] \mbox{$\ci:\cu\mapsto \rr$} is a function,
\item[7.] $\ca$ is an $(\cu,\cf)$-abundant 
subset of $\cu$ such that $u_*\in\ca$, where $\cf$ is the
map 
$$
\cu\ni u\mapsto\cf(u)\equaldef(\ce(u),\ci(u))\in Y\equaldef Z\times\rr\,.
$$
\item[8.] $\ce$ and $\ci$ are of class $D^2$ at $u_*$.
\item[9.] $P_1$, $p_1$ are continuous linear maps
from $\cua$ to $Z$, $\rr$, respectively, 
$P_2$, $p_2$ are continuous bilinear symmetric  maps 
from  $\cua\times\cua$ to $Z$, $\rr$, 
$$
\ce(u_*+v)=\ce(u_*)+P_1\cdot v+\frac{1}{2}P_2(v,v)+
o(\|v\|^2)\,\,\,{\rm as}\,\,\,v\to 0\,,\,\,u_*+v\in\cu\,,
$$ 
and 
$$
\ci(u_*+v)=\ci(u_*)+p_1\cdot v+\frac{1}{2}p_2(v,v)+
o(\|v\|^2)\,\,\,{\rm as}\,\,\,v\to 0\,,\,\,u_*+v\in\cu\,,
$$
\item[10.] $\hat\cv$ is a finite subset of $var_1(u_*,\cu)$.
\item[11.] $P_1\cdot v=0$ and $p_1\cdot v\le 0$ 
whenever $v\in\hat\cv$,
\item[12.] $P_2(v_1,v_2)=0$  and $p_2(v_1,v_2)\le 0$  
whenever $v_1,v_2\in\hat\cv$ are such that
$v_1\neq v_2$,
\item[13.] $\{\cu_v\}_{v\in\hat\cv}$ is a family of 
convex subsets of $\cua$ such that
\begin{itemize}
\item[i.] $u_*\in\cu_v$ for every $v\in\hat\cv$, 
\item[ii.] $u_*+\sum_{v\in\hat\cv}\Big(\cu_v-u_*)\subseteq\cu$,
\item[iii.] $v\in var_1(u_*,\cu_v)$ for each $v\in\hat\cv$. 
\end{itemize}\end{itemize}
Let
\begin{equation}\label{zstar}
z_*=\ce(u_*)\,.
\end{equation}
Assume that 
\begin{equation}\label{optiassum}
\ci(u_*)\le\ci(u)\qquad{\rm for\,\,\,all}\qquad u\in\ca\qquad
{\rm such\,\,\,that}\qquad\ce(u)=z_*\,.
\end{equation}
Then there exist a covector $\psi\in Z^*$ and a real number $\psi_0$
such that
\begin{itemize}
\item[a.] $\psi_0\ge 0$,
\item[b.] $\|\psi\|+\psi_0>0$,
\item[c.] $\langle\psi,P_1\cdot v\rangle+\psi_0p_1\cdot v\ge 0$
for all $v\in\overline{var}_1(u_*,\cu)$, 
\item[d.] $\langle\psi,P_1\cdot w\rangle+\psi_0p_1\cdot w
+\frac{1}{2}\Big(\langle\psi,P_2(v,v)\rangle+\psi_0p_2(v,v)\Big)\ge 0$ 
for all $(v,w)$ such that $v\in\hat\cv$ and
$w\in var_{2,1}(u_*,v,\cu_v)$.
\item[e.] $\langle\psi,P_1\cdot w\rangle+\psi_0p_1\cdot w$
for all $(v,w)$ such that $v\in\hat\cv$ and
$w\in var_{2,1}(u_*,v,\cu_v)$.
\end{itemize}
\end{theorem}

\medbreak{\sc Proof.} Define
$$
c_*=\ci(u_*)\,,\qquad y_*=(z_*,c_*)=\cf(u_*)\,.
$$
Let 
$$
\omega=(0,-1)\in Z\times\rr=Y\,.
$$
Let $\Lambda$ be the closed half-line
\begin{equation}\label{laline-bb}
\Lambda\equaldef\{y_*+t\omega:t\in\rr,t\ge 0\}\,.
\end{equation}
Then $y_*\in\cf(\ca)\cap\Lambda$. Assumption (\ref{optiassum}) says precisely
that $\cf(\ca)\cap\Lambda=\{y_*\}$, i.e., that the separation condition
of the previosu section is satisfied. 

If we then apply Corollary \ref{maincoro}, we get a nonzero covector
$\bar\psi$ that satisfies the inequalities of that corollary.
Write $\bar\psi=(\psi,\psi_0)$. Then the inequality 
$\langle\bar\psi,\omega\rangle\le 0$ implies that $\psi_0\ge 0$.
All the other desired conclusions follow immediately from
the properties of $\bar\psi$ stated in Corollary \ref{maincoro}.\proofend

\bigbreak We now restate Theorem \ref{theooptione} for the case when 
$\hat\cv$ consists of a single vector.

\begin{theorem}\label{theooptione}
Assume that 
\begin{itemize}
\item[1.] $\cua$ and $Z$ are normed \mbox{spaces}, 
\item[2.] $Z$ is finite-dimensional,
\item[3.] $\cu$ is a convex subset of $\cua$, 
\item[4.] $u_*\in\cu$,  
\item[5.] \mbox{$\ce:\cu\mapsto Z$} is a map,
\item[6.] \mbox{$\ci:\cu\mapsto \rr$} is a function,
\item[7.] $\ca$ is an $(\cu,\cf)$-abundant 
subset of $\cu$ such that $u_*\in\ca$, where $\cf$ is the
map 
$$
\cu\ni u\mapsto\cf(u)\equaldef(\ce(u),\ci(u))\in Y\equaldef Z\times\rr\,.
$$
\item[8.] $\ce$ and $\ci$ are of class $D^2$ at $u_*$.
\item[9.] $P_1$, $p_1$ are continuous linear maps 
from $\cua$ to $Z$, $\rr$, respectively, 
$P_2$, $p_2$ are continuous bilinear symmetric  maps 
from  $\cua\times\cua$ to $Z$, $\rr$, 
$$
\ce(u_*+v)=\ce(u_*)+P_1\cdot v+\frac{1}{2}P_2(v,v)+
o(\|v\|^2)\,\,\,{\rm as}\,\,\,v\to 0\,,\,\,u_*+v\in\cu\,,
$$ 
and 
$$
\ci(u_*+v)=\ci(u_*)+p_1\cdot v+\frac{1}{2}p_2(v,v)+
o(\|v\|^2)\,\,\,{\rm as}\,\,\,v\to 0\,,\,\,u_*+v\in\cu\,,
$$
\item[10.] $\hat v$ is a member of $var_1(u_*,\cu)$,
\item[11.] $P_1\cdot\hat v=0$ and $p_1\cdot\hat v\le 0$.
\end{itemize}
Let
\begin{equation}\label{zstarb}
z_*=\ce(u_*)\,.
\end{equation}
Assume that 
\begin{equation}\label{optiassumb}
\ci(u_*)\le\ci(u)\qquad{\rm for\,\,\,all}\qquad u\in\ca\qquad
{\rm such\,\,\,that}\qquad\ce(u)=z_*\,.
\end{equation}
Then there exist a covector $\psi\in Z^*$ and a real number $\psi_0$
such that
\begin{itemize}
\item[a.] $\psi_0\ge 0$,
\item[b.] $\|\psi\|+\psi_0>0$,
\item[c.] $\langle\psi,P_1\cdot v\rangle+\psi_0p_1\cdot v\ge 0$
for all $v\in\overline{var}_1(u_*,\cu)$, 
\item[d.] $\langle\psi,P_1\cdot w\rangle+\psi_0p_1\cdot w
+\frac{1}{2}\Big(\langle\psi,P_2(\hat v,\hat v)\rangle+
\psi_0p_2(\hat v,\hat v)\Big)\ge 0$ 
for all vectors \mbox{$w\in var_{2,1}(u_*,\hat v,\cu)$},
\item[e.] $\langle\psi,P_1\cdot w\rangle+\psi_0p_1\cdot w$
for all \mbox{$w\in var_{2,1}(u_*,v,\cu_v)$}.
\end{itemize}
\end{theorem}


\medbreak{\sc Proof.} It suffices to apply Theorem \ref{theooptione}
with $\hat\cv=\{\hat v\}$, and to observe that the cross-term
condition 12 of the statement of Theorem \ref{theooptione}
is vacuously true, and the independence condition 13 can be satisfied
by just taking $\cu_{\hat v}=\cu$.\proofend

\section{Optimal control}

\noindent We consider a control system
\begin{equation}\label{cosys}
\dot x=f(x,u,t)\,,\,\,\,x\in\rr^n\,,\,\,\,u\in U\,,a\le t\le b\,,
\end{equation}
and a Lagrangian 
\begin{equation}\label{cosyslag}
\rr^n\times U\times[a,b]\ni(x,u,t)\mapsto L(x,u,t)\in\rr\,.
\end{equation}
We assume that 
\begin{itemize}
\item[B1.] {\em $m\in\nn$ and $U$ is a closed convex subset of $\rr^m$,}
\item[B2.] {\em the partial maps 
$$
\rr^n\times U\ni(x,u)\mapsto(f(x,u,t),L(x,u,t))\in\rr^{n+1}
$$
are of class $C^2$ for every $t\in[a,b]$,}
\item[B3.] {\em the partial  maps 
$$
[a,b]\ni t\mapsto(f(x,u,t),L(x,u,t))\in\rr^{n+1}
$$
are measurable for every $(x,u)\in\rr^n\times U$,} 
\item[B4.] {\em the maps of condition B2 are bounded, together with 
their first 
and second derivatives with respect to $(x,u)$, on every compact subset
of $\rr^n\times U\times[a,b]$.}
\end{itemize}

\bigbreak\noindent{\sc Warning!}. As stated, the conditions do no guarantee
that trajectories do not explode. So, strictly speaking, the trajectory 
map defined below need  not be everywhere defined, and the same is true,
naturally, of $\ce$ and $\ci$. This 
does not cause any problems, but in the final writing of this note this 
issue has to be taken care of. At this point, we may, if we wish to be 
absolutely sure that everything is O.K., assume that 
$f_x$ is {\em globally bounded}. \proofend

\bigbreak



We let $\cua=L^\infty([a,b],\rr^m)$, and let $\cu$ be the set
of those controls $u(\cdot)\in\cua$ that are $U$-valued.

A {\em trajectory-control pair} is a pair $(x(\cdot),u(\cdot))$ 
consisting of an absolutely continuous curve 
$x(\cdot):[a,b]\mapsto\rr^n$ and a control 
$u(\cdot):[a,b]\mapsto U$ such that 
$$
\dot x(t)=f(x(t),u(t),t)\qquad{\rm for\,\,\,almost\,\,\,
all}\quad t\in[a,b]\,. 
$$

We fix an initial state $\bar x$ and a terminal state $\bar x$,
and let $TCP(\bar x,\hat x)$ denote the set of all trajectory-control
pairs $(x(\cdot),u(\cdot))$ such that $x(a)=\bar x$ and $x(b)=\hat x$.

We assume that we are given a pair $(x_*(\cdot),u_*(\cdot))$
such that 
\begin{itemize}
\item[B5.] {\em $(x_*(\cdot),u_*(\cdot))\in TCP(\bar x,\hat x)$,
and $(x_*(\cdot),u_*(\cdot))$ minimizes the functional}
$$
TCP(\bar x,\hat x)\ni(x(\cdot),u(\cdot))\mapsto
J(x(\cdot),u(\cdot))\equaldef\int_a^bL(x(t),u(t),t)dt\,.
$$
\end{itemize}

We assume that we are also given pairs $(h_1,v_1),\ldots,(h_L,v_L)$ 
that are trajectory-control pairs of the linearized equation 
about $(x_*(\cdot),u_*(\cdot))$, supported by 
pairwise disjoint subintervals $I_1,\ldots,I_\ell$.
More precisely, we assume that 
\begin{itemize}
\item[B6.] $v_1,\ldots,v_L\in\cu^a$, 
$h_1,\ldots,h_L$ {\em are absolutely continuous maps from $[a,b]$ to $\rr^n$
such that 
$$
h_1(a)=\ldots=h_L(a)=0
$$
and 
$$
\dot h_\ell(t)=\frac{\partial f}{\partial x}
(x_*(t),u_*(t),t)\cdot h_\ell(t)+\frac{\partial f}{\partial u}
(x_*(t),u_*(t),t)\cdot v_\ell(t)
$$
for a.e. $t\in[a,b]$, for $\ell=1,\ldots,L$.}
\end{itemize}

Moreover, we assume that 
\begin{itemize}
\item[B7.] $h_1(b)=\ldots=h_L(b)=0$, 
\item[B8.] $\int_a^bL_x(x_*(t),v_*(t),t)\cdot h_\ell(t)
+\int_a^bL_u(x_*(t),v_*(t),t)\cdot v_\ell(t)\le 0$
{\em for} $\ell=1,\ldots,L$, 
\item[B9.] $I_1,\ldots,I_\ell$ {\em are pairwise disjoint subintervals 
of $[a,b]$ such that} 
\begin{equation}\label{disjov}
\{t\in[a,b]:v_\ell(t)\neq 0\}\subseteq I_\ell\qquad
{for}\qquad \ell=1,\ldots,L\,.
\end{equation}
\end{itemize}

\medbreak We remark that conditions B6, B7 and B9 imply that
\begin{equation}\label{disjoh}
\{t\in[a,b]:h_\ell(t)\neq 0\}\subseteq I_\ell\qquad
{\rm for}\qquad \ell=1,\ldots,L\,.
\end{equation}

We let $\ct$ be the {\em trajectory map}, i.e., the map 
that assigns to each $u\in\cu$ the curve $\ct(u)$ which is
the unique solution $x(\cdot)$ of the initial value problem
$$
\left\{\begin{array}{rcl}
\dot x(t)&=&f(x(t),u(t),t)\quad{\rm for\,\,\,a.e.}\quad t\in[a,b]\,,\\ 
x(a)&=&\bar x\,.\end{array}\right.
$$
Then $\ct$, regarded as a map from $\cu$ to the space $C^0([a,b],\rr^n)$,
is twice Fr\'echet differentiable. The derivatives of $\ct$ at $u_*$ 
are easily computed by observing that 
$$
\ct(u_*+\eps v)=x_*+\eps\ct'(u_*)\cdot v+
\frac{\eps^2}{2}\ct''(u_*)(v,v)+o(\eps^2)\,.
$$
If we let 
$$
\ct(u_*+\eps v)=x^\eps\,,\quad 
y^\eps(t)=\frac{d\,\,}{d\eps}\Big(x^\eps(t)\Big)\,,\quad
z^\eps(t)=\frac{d\,\,}{d\eps}\Big(y^\eps(t)\Big)\,,
$$
we have
$$
x^\eps(t)=\bar x+\int_a^tf(x^\eps(s),u_*(s)+\eps v(s),s)\,ds\,,
$$
so that
$$
y^\eps(t)=\int_a^tf_x(x^\eps(s),u_*(s)+\eps v(s),s)\cdot y^\eps(s)\,ds
+\int_a^tf_u(x^\eps(s),u_*(s)+\eps v(s),s)\cdot v(s)\,ds\,,
$$
and
\begin{eqnarray*}
z^\eps(t)&=&\int_a^tf_x(x^\eps(s),u_*(s)+\eps v(s),s)\cdot z^\eps(s)\,ds\\
&\,&+\int_a^tf_{xx}(x^\eps(s),u_*(s)+\eps v(s),s)(y^\eps(s),
y^\eps(s))\,ds\\
&\,&+2\int_a^tf_{xu}(x^\eps(s),u_*(s)+\eps v(s),s)(y^\eps(s),
v(s))\,ds\\
&\,&+\int_a^tf_{uu}(x^\eps(s),u_*(s)+\eps v(s),s)(v(s),
v(s))\,ds\,.
\end{eqnarray*}
It is clear that $\ct'(u_*)\cdot v=y^0$, and 
$\ct''(u_*)(v,v)=z^0$. Therefore
\begin{eqnarray*}
\Big(\ct'(u_*)\cdot v\Big)(t)&=&
\int_a^tf_x(x_*(s),u_*(s),s)\cdot\Big(\ct'(u_*)\cdot v\Big)(s)\,ds\\
&\,&+\int_a^tf_u(x_*(s),u_*(s),s)\cdot v(s)\,ds\,,
\end{eqnarray*}
and
\begin{eqnarray*}
&\,&\Big(\ct''(u_*)(v,v)\Big)(t)=\int_a^tf_x(x_*,u_*(s),s)\cdot
\Big(\ct''(u_*)(v,v)\Big)(s)\,ds\\
&\,&+\int_a^tf_{xx}(x_*(s),u_*(s),s)\Big(\Big(\ct'(u_*)\cdot v\Big)(s),
\Big(\ct'(u_*)\cdot v\Big)(s)\Big)\,ds\\
&\,&+2\int_a^tf_{xu}(x_*(s),u_*(s),s)
\Big(\Big(\ct'(u_*)\cdot v\Big)(s),v(s)\Big)\,ds\\
&\,&+\int_a^tf_{uu}(x_*(s),u_*(s),s)(v(s),v(s))\,ds\,.
\end{eqnarray*}
It then follows easily that the symmetric bilinear form 
$(v,\tilde v)\mapsto\ct''(u_*)(v,\tilde v)$ satisfies
\begin{eqnarray*}
&\,&\Big(\ct''(u_*)(v,\tilde v)\Big)(t)=\int_a^tf_x(x_*,u_*(s),s)\cdot
\Big(\ct''(u_*)(v,\tilde v)\Big)(s)\,ds\\
&\,&+\int_a^tf_{xx}(x_*(s),u_*(s),s)\Big(\Big(\ct'(u_*)\cdot v\Big)(s),
\Big(\ct'(u_*)\cdot\tilde v\Big)(s)\Big)\,ds\\
&\,&+\int_a^tf_{xu}(x_*(s),u_*(s),s)
\Big(\Big(\ct'(u_*)\cdot v\Big)(s),\tilde v(s)\Big)\,ds\\
&\,&+\int_a^tf_{xu}(x_*(s),u_*(s),s)
\Big(\Big(\ct'(u_*)\cdot\tilde v\Big)(s),v(s)\Big)\,ds\\
&\,&+\int_a^tf_{uu}(x_*(s),u_*(s),s)(v(s),\tilde v(s))\,ds\,.
\end{eqnarray*}
In particular, 
\begin{eqnarray*}
&\,&\Big(\ct''(u_*)(v_\ell,v_{\ell'})\Big)(t)=
\int_a^tf_x(x_*,u_*(s),s)\cdot
\Big(\ct''(u_*)(v_\ell,v_{\ell'})\Big)(s)\,ds\\
&\,&+\int_a^tf_{xx}(x_*(s),u_*(s),s)\Big(h_\ell(s),h_{\ell'}(s)\Big)\,ds\\
&\,&+\int_a^tf_{xu}(x_*(s),u_*(s),s)
\Big(h_\ell(s),v_{\ell'}(s)\Big)\,ds\\
&\,&+\int_a^tf_{xu}(x_*(s),u_*(s),s)
\Big(h_{\ell'}(s),v_\ell(s)\Big)\,ds\\
&\,&+\int_a^tf_{uu}(x_*(s),u_*(s),s)(v_\ell(s),v_{\ell'}(s))\,ds\,.
\end{eqnarray*}
If $\ell\neq\ell'$, then the last four integrals vanish
because of (\ref{disjov}), (\ref{disjoh}), and the fact that
$I_\ell\cap I_{\ell'}=\emptyset$. It follows that
\begin{equation}\label{crossissat}
\ct''(u_*)(v_\ell,v_{\ell'})=0\qquad{\rm if}\qquad\ell\neq\ell'\,.
\end{equation}
We let $\calc$ be the {\em running cost map}, that is, the map
$u\mapsto\zeta_u$, where 
$$
\zeta_u(t)=\int_a^tL(\ct(u)(s),u(s),s)\,ds\,.
$$
Then a similar calculation shows that
\begin{equation}\label{crossissatin}
\calc''(u_*)(v_\ell,v_{\ell'})=0\qquad{\rm if}\qquad\ell\neq\ell'\,.
\end{equation}

We now let $\ce$, $\ci$ be, respectively,
the {\em terminal point map} and the {\em terminal cost map}, i.e.,
the maps defined by
$$
\ce(u)=\ct(u)(b)\,,\qquad\ci(u)=\calc(u)(b)\,.
$$
Then all the conditions of Theorem \ref{theoopti} are satisfied,
and we obtain the following result.
\begin{theorem}\label{theocontrol}
If B1-B9 hold, then there exist an absolutely comtinuous
field of covectors $[a,b]\ni t\mapsto\psi(t)\in\rr_n$ 
(where we are writing $\rr_n$ to denote the space of real
$n$-dimensional  row vectors) and a real number $\psi_0$ such that
\begin{itemize}
\item[1.] $\psi_0\ge 0$,
\item[2.] $\psi_0+\|\psi(b)\|>0$,
\item[3.] $\dot\psi(t)=
-\psi(t)\cdot f_x(x_*(t),u_*(t),t)-\psi_0 L_x(x_*(t),u_*(t),t)$ 
for almost all $t\in[a,b]$,
\item[4.] for almost all $t\in[a,b]$, the inequality 
$$
\Big(\psi_0L_u(x_*(t),u_*(t),t)+
\psi(t)\cdot f_u(x_*(t),u_*(t),t)\Big)\cdot(u-u_*(t))\ge 0
$$ 
holds for all $u\in U$,
\item[5.] for every $\ell\in\{1,\ldots,L\}$, the inequalities
$$
\int_a^b\Big(\psi_0L_u(x_*(t),u_*(t),t)+
\psi(t)\cdot f_u(x_*(t),u_*(t),t)\Big)\cdot w(t)\,dt\ge 0
$$ 
and
\begin{eqnarray*}
&\,&\int_a^b\Big(\psi_0L_u(x_*(t),u_*(t),t)+
\psi(t)\cdot f_u(x_*(t),u_*(t),t)\Big)\cdot w(t)\,dt\\
&\,&+\frac{1}{2}\Biggl(
\int_a^b\Big[\psi(t)\cdot f_{xx}(x_*(t),u_*(t),t)\Big(h_\ell(t),
h_\ell(t)\Big)\\
&\,&\,\,\,+\psi_0\cdot L_{xx}(x_*(t),u_*(t),t)\Big(h_\ell(t),
h_\ell(t)\Big)\Big]\,dt\\
&\,&+2\int_a^b\Big[\psi(t)\cdot f_{xu}(x_*(t),u_*(t),t)\Big(h_\ell(t),
v_\ell(t)\Big)\\
&\,&\,\,\,+\psi_0\cdot L_{xu}(x_*(t),u_*(t),t)\Big(h_\ell(t),
v_\ell(t)\Big)\Big]\,dt\\
&\,&+\int_a^b\Big[\psi(t)\cdot f_{uu}(x_*(t),u_*(t),t)\Big(v_\ell(t),
v_\ell(t)\Big)\\
&\,&\,\,\,+\psi_0\cdot L_{uu}(x_*(t),u_*(t),t)\Big(v_\ell(t),
v_\ell(t)\Big)\Big]\,dt\Biggr)\ge 0
\end{eqnarray*}
\end{itemize}
hold for every $w\in var_{2,1}(u_*,v_\ell,\cu)$ such that
$\{t:w(t)\neq 0\}\subseteq I_\ell$.
\end{theorem}

\medbreak{\sc Proof.} We apply Theorem \ref{theoopti}
to the maps $\ce$ and $\ci$, observing that 
\begin{itemize}
\item[1.] the cross-term condition
(Hypothesis 12 of the statement of Theorem \ref{theoopti}) is 
automatically satisfied, because of (\ref{crossissat})
and (\ref{crossissatin}),
\item[2.] the independence condition
(Hypothesis 13 of the statement of Theorem \ref{theoopti}) is 
satisfied if we take 
\begin{equation}\label{chocc}
\cu_\ell=\{u\in\cu:u(t)=u_*(t)\,\,{\rm whenever}\,\,t\notin I_\ell\}\,.
\end{equation}
More precisely: if we define $\cu_\ell$ by (\ref{chocc}), then either
$v_\ell\in var_1(u_*,\cu)$, in which case 
Hypothesis 13 of the statement of Theorem \ref{theoopti} is valid,
or \mbox{$v_\ell\notin var_1(u_*,\cu)$}, in which case the set 
$var_{2,1}(u_*,v_\ell,\cu)$ is empty, and we can simply omit 
$v_\ell$. 
\end{itemize}
The conclusions follow by applying the formulas for $\ct'$,
$\ct''$, $\ci'$ and $\ci''$ obtained earlier, except for the fact 
that condition 4 is obtained in an integrated form, as the statement that
$$
\int_a^b\Big(\psi_0L_u(x_*(t),u_*(t),t)+
\psi(t)\cdot f_u(x_*(t),u_*(t),t)\Big)\cdot\Big(u(t)-u_*(t)\Big)\,dt\ge 0
$$ 
for all $u(\cdot)\in\cu$. To go from this to the form given in our 
conclusion no. 4, it suffices to take $t$ to be a Lebesgue point
of the functions 
$$
s\mapsto \psi_0L_u(x_*(s),u_*(s),s)+\psi(s)\cdot f_u(x_*(s),u_*(s),s)
$$ 
and 
$$
s\mapsto\Big(
\psi_0L_u(x_*(s),u_*(s),s)+\psi(s)\cdot f_u(x_*(s),u_*(s),s)\Big)
\cdot u_*(s)\,,
$$ 
and let $u(\cdot)$ be an arbitrary constant control.\proofend

\section{Comparison with the result of Ledzewicz and Sch\"attler}

\noindent If we take $\ell=1$, then Theorem \ref{theocontrol}
contains as a special case Theorem 4.1 of \cite{ls}, with the following 
additional features:

\begin{itemize}
\item[1.] the additional multiplier $\rho$ of 
Theorem 4.1 of \cite{ls} can be taken to be zero,
\item[2.] in Hypothesis (A1)  of Page 166 of \cite{ls}, the 
requirement that $f$ be of class $C^3$ in $x$ and $u$ can be weakened
by changing ``$\,C^3\,$'' to ``$\,C^2\,$,''
\item[3.] Hypothesis (c)  in Page 168 of \cite{ls} is not needed,
\item[4.] the ``second-order feasible set'' $FC^{(2)}(\cu,u_*,v)$ 
of \cite{ls} is replaced by the set $var_{2,1}(u_*,v,\cu)$, which  
in general is much larger,
\item[5.] Hypothesis (e)  in Page 168 of \cite{ls}, on the convexity
of $FC^{(2)}(\cu,u_*,v)$, is not needed.
\end{itemize}

\section{Things to be done}

\begin{itemize}
\item[1.] Show in detail how the Goh and Legendre-Clebsch conditions
follow from our theorem,
\item[2.] bring in the ``abundant set'' stuff to show, via relaxed
controls, that Theorem \ref{theocontrol} yields a {\em global}
as well as a local maximum principle,
\item[3.] take care of the possibility
of explosions mentioned in the ``Warning'' in \S 5; a related
issue is that of taking the state space to be an open subset 
of $\rr^n$ rather than all of $\rr^n$, or a manifold; these things are 
of course trivial, but writing them down will take some work,
\item[4.] discuss in detail the case of a nonconvex control set,
\item[5.] bring in more general terminal conditions; the appropriate
setting should be a terminal condition $x(b)\in S$, where $S$ is a
fairly general set having a tangent cone (in some adequate 
sense, e.g., Boltyanskii approximating cone) $C$ at $x_*(b)$; the conclusion
would then contain the transversality condition 
$\langle\psi(b),v\rangle\le 0$ for all $v\in C$, 
\item[6.] do the higher-order case, $p>2$.
\end{itemize} 

\section{A big mystery}

\noindent {\Large Will the high-order multipliers also go away
for $p>2$? This author's guess is ``yes.'' But it's only a guess.

Stay tuned.}

\begin{thebibliography}{A}

\bibitem{ls} Ledzewicz, U., and H. Sch\"attler, ``An extended maximum 
principle.'' {\it Nonlineanr Analysis, Theory, Methods and Applications}
{\bf 29}, 1997, pp. 159-183.
\end{thebibliography}

\end{document}










