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% Pin No.: 06293
% Paper ID: CDC02-INV3201
% Title:  Path-integral generalized differentials
% Paper Accepted Category: REGULAR INVITED
% Session Title: Nonsmooth Analysis Methods in Control Theory, II
% Session Code: TuP11-1
% Presentation Format: ORAL
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% The full Author's Kit is available for you at the CDC02 web site:
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%            This is a LateX file, containing  the paper 
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%              PATH-INTEGRAL GENERALIZED DIFFERENTIALS 
%
%                       by H. J. Sussmann
%
%        This paper has been submitted for presentation at the 41th IEEE 
%        Conference on Decision and Control, to be held in Las Vegas,
%        December, 2002,  and  for publication in the conference  proceedings. 
%  
%               Date of this version: February 21, 2002
%
%            THIS VERSION SUPERSEDES ALL PREVIOUS VERSIONS
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%                  The paper is 13 printed pages long.%
%       You should run Latex twice on this file. After the second run,
%      there will still be several ``underfull  hbox'' messages.  Just
%             ignore them and print the .dvi file.
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$\quad\quad\,\,$\mbox{\begin{minipage}{12.2cm}
#1u
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\def\pigd{\mbox{path}-\mbox{integral} \mbox{generalized} \mbox{differential}}
\def\pigds{path-integral generalized differentials}
\def\Pigd{\mbox{Path}-\mbox{integral} \mbox{generalized}
\mbox{differential}}
\def\Pigds{\mbox{Path}-\mbox{integral} \mbox{generalized}
\mbox{differentials}}
\def\pigdab{PIGD}
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\vspace{3cm}

{\Large  

\noindent This is the back of {\Huge Page \arabic{pagono}}\\
of the paper

\vspace{1cm}
\begin{center}
{\huge Path-integral\\
generalized differentials}

$\quad$\\

by\\

$\quad$\\ 
{\huge H\'ector J. Sussmann}\\

\end{center}
}

\newpage 

\null

\vspace{3.3cm}

\noindent Paper ID: $\,\,\,${\huge  CDC02-INV3201} \\

\noindent Session Title:\\

\medbreak
{\Large \noindent Nonsmooth Analysis Methods\\

\vspace{-3mm}
in Control Theory, II}


\vspace{4mm}
Session Code: {\Huge TuP11-1} \\

Paper Category: REGULAR INVITED

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{\Large\em Dedicated to Jack Warga on his 80th birthday}\\
$\,$\\
H\'ector J.\ Sussmann\\
Department of Mathematics\\
Rutgers, the State University of New Jersey\\
Hill Center---Busch Campus\\
%110 Frelinghuysen Road\\
Piscataway, NJ 08854-8019, USA\\
{\small\tt sussmann@math.rutgers.edu}\\
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H\'ector J. Sussmann\\
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Department of Mathematics; Rutgers, the State University of New Jersey\\
Hill Center---Busch Campus, 110 Frelinghuysen Road, Piscataway, NJ
08854-8019, USA\\ 
\vskip5pt
E-mail: sussmann@hamilton.rutgers.edu\quad\quad
Web site: http://www.math.rutgers.edu/$\widetilde{\,\,}$sussmann\\
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{\bf
Path-integral generalized differentials$\,^{*}$}
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\begin{document}

\abovedisplayskip 4pt plus2pt minus5pt\belowdisplayskip \abovedisplayskip
\twocolumn[\titlestuff]
\footnotetext[1]{Supported in part by
NSF Grant DMS98-03411-00798.}

\vskip4.5cm


\ssection{Introduction}\label{secintrod}
\noindent This is the second of a series of two papers
on generalized differentiation theories (abbr. GDTs). The first paper
discussed the definition of the GDT concept, 
\mbox{presented} several GDTs (the Warga derivate \mbox{containers},
weak multidifferentials, and generalized differential \mbox{quotients})
and compared them showing, in particular, that none of these theories
contains all the others. 

In this paper, we introduce another \mbox{concept}
of generalized differential---the ``\pigd,'' abbreviated \pigdab---that
achieves the desired unification.

\ssection{Preliminaries}\label{prelimpigds}

\noindent If $n,m\in\zz_+$, $\alpha:[0,1]\to\rr^n$ is a Lipschitz function,
and $h:[0,1]\to\rr^{m\times n}$ is integrable,
we use $h*\alpha$ to denote the ``chronological product'' of
$h$ and $\alpha$, that is, the absolutely continous 
function $\beta:[0,1]\to\rr^m$ given
by $\beta(t)=\int_0^th(s)\cdot\dot\alpha(s)\,ds$.

The following lemma says that the chronological product operation
$\czer{[0,1]}{\rr^{n}}\times L^1([0,1],\rr^{m\times n})\ni$
$(\alpha,h)\mapsto h*\alpha\in\czer{[0,1]}{\rr^{m}}$
is jointly continuous, as long as the function $\alpha$ 
varies in a uniformly Lipschitz subset of $\czer{[0,1]}{\rr^n}$.
The proof is very simple and will be omitted.

\begin{lemma}\label{jointcontlemma}
Let $n,m\in\zz_+$. Let $\{(\alpha_j,h_j)\}_{j=1}^\infty$ be a sequence
of members of the product space 
\mbox{$S=\czer{[0,1]}{\rr^n}\times L^1([0,1],\rr^{m\times n})$} 
that converges in $S$ to a limit
$(\alpha_\infty,h_\infty)$. Assume that the sequence $\{\alpha_j\}$ 
is uniformly Lipschitz (that is, there exists a constant $r\in\rr$ such that
$\|\alpha_j(t)-\alpha_j(s)\|\le r|t-s|$ for all $j\in\nn$ and all
$t,s\in[0,1]$).  Then 
$$
h_j*\alpha_j\to h_\infty*\alpha_\infty\quad{\rm in}
\quad\czer{[0,1]}{\rr^{m}}\quad{\rm as}\quad j\to\infty\,.\diamondsuit
$$
\end{lemma}

Let $n\in\zz_+$, and let $S$ be a subset of $\rr^n$.  We write $\aof{S}$
to denote the subset of $\czer{[0,1]}{\rr^n}$ consisting of all
absolutely continuous curves $\alpha:[0,1]\to\rr^n$ such that
$\alpha(0)=0$ and $\dot\alpha(t)\in S$ for almost all $t\in[0,1]\,$.


\medbreak\noindent If $\mathcal{S}\subseteq\czer{[0,1]}{\rr^n}$,
we write $\tau(\mathcal{S})$ to denote the  set
$\tau(\mathcal{S})\equaldef\{\,\alpha(1):\alpha\in\mathcal{S}\,\}$,
so $\tau(\mathcal{S})$ is the set of all terminal  points
of curves in $\mathcal{S}$.

The following is then an immediate consequence of our
definitions.

\begin{proposition}\label{immeobs}
If $K$ is a compact convex subset of $\rr^n$, then $\aof{K}$ is a compact 
convex subset of $\czer{[0,1]}{\rr^n}$, and 
$\tau\Big(\mathcal{A}(K)\Big)=K$. \proofend
\end{proposition}

If $m\in\zz_+$ and  $v\in\rr^m$, we use $\xi_v$ to denote the curve
\begin{equation}\label{xivee}
[0,1]\ni t\to tv\,\,\equaldef\,\,\xi_v(t)\in\rr^m\,.
\end{equation}
If 
\begin{equation}\label{seattle}
S\subseteq\rr^n\,\,{\rm and}\,\,G:\aof{S}\tto 
\czer{[0,1]}{\rr^{m\times n}}\times\rr^m\,, 
\end{equation}
then we can define  set-valued maps
$$
\mathcal{I}_G:\aof{S}\tto\czer{[0,1]}{\rr^m}\,,\,\,\,\,\,
\Phi_G:S\tto \rr^m\,, 
$$
by letting 
\begin{eqnarray*}
\mathcal{I}_G(\alpha)&=&\Big\{\,h*\alpha+\xi_v: (h,v)\in G(\alpha)\,\Big\}\,,\\
\Phi_G(x)&=&\Big\{\,y\in \rr^m:(\,\exists(\alpha,h,v)\in\gr(G)\,)\\
&\,&\,\,\,(\,\alpha(1)=x\,\vee\,(h*\alpha)(1)+v=y\,)\,\Big\}\,.
\end{eqnarray*}

The following fact is then trivial.
\begin{proposition}\label{einsteinfact}
Let $n,m\in\zz_+$, and let $S$, $G$ be such that 
(\ref{seattle}) holds. Then:
\ben
\item[1.] $ \dom(\mathcal{I}_G)=\dom(G)$, so in particular
$\mathcal{I}_G$ is everywhere defined if and only if $G$ is.
\item[2.] If $G$ is everywhere defined then 
$\Phi_G$ is everywhere defined.
\item[3.] If $G$ is single-valued at a particular
$\alpha\in\aof{S}$, then $\mathcal{I}_G$ is single-valued
at $\alpha$; in particular, if $G$ is single-valued and everywhere defined,
then $\mathcal{I}_G$ is single-valued and everywhere defined. \proofend
\een
\end{proposition}
\begin{lemma}\label{einsteinlem}
Let $n,m\in\zz_+$, and let $S$, $G$ be such that (\ref{seattle}) holds. 
Assume that $S$ is compact and convex. Then:
\ben
% \item[4.] If $\rho>0$, $H$ takes values in the ball
% $\{h\in\czer{[0,1]}{\rr^{m\times n}}:\|h\|_{sup}\le\rho\}$, and $D$ 
% is a closed convex cone in $\rr^m$ such that $h(t)\cdot C\subseteq D$ 
% whenever $\alpha\in\mathcal{A}(C,r)$ and $h\in H(\alpha)$, 
% then $\mathcal{I}_H$ takes values in $\mathcal{A}(D,r\rho)$.
\item[1.] If $\gr(G)$ is compact, then $\gr(\mathcal{I}_G)$ and
$\gr(\Phi_G)$ are compact.
\item[2.] If $G$ is single-valued, everywhere defined, and \mbox{continuous},
then $\mathcal{I}_G$ is single-valued, everywhere defined, and continuous.
\item[3.] If $G$ is regular, then $\mathcal{I}_G$ and 
$\Phi_G$ are regular.
\een
\end{lemma}
\begin{proof}
To prove the first statement, assume that $G$
has a compact graph.  We want to show that
$\gr(\mathcal{I}_G)$ and $\gr(\Phi_G)$ are compact.

Let $\{(\alpha_j,\beta_j)\}_{j=1}^\infty$ be a sequence in
$\gr(\mathcal{I}_G)$.  We \mbox{want} to extract a subsequence 
that converges to a limit 
$(\alpha_\infty,\beta_\infty)\in\gr(\mathcal{I}_G)$.  
Since $\beta_j\in\mathcal{I}_G(\alpha_j)$, there exist $(h_j,v_j)\in
G(\alpha_j)$ such that $\beta_j=h_j*\alpha_j+\xi_{v_j}$ for $j\in\nn$.  Then
$(\alpha_j,h_j,v_j)\in\gr(G)$.  Since $\gr(G)$ is
compact, we may assume, after passing to a subsequence, that
(i)~the sequences $\{\alpha_j\}_{j=1}^\infty$, $\{h_j\}_{j=1}^\infty$,
converge uniformly to \mbox{limits} $\alpha_\infty$, $h_\infty$, 
(ii)~$\{v_j\}_{j=1}^\infty$ converges in $\rr^m$ to a limit $v_\infty$,
(iii)~$\alpha_\infty\in\aof{S}$, and
(iv)~$(h_\infty,v_\infty)\in G(\alpha_\infty)$.

\skipopag

\noindent Let $\beta_\infty=h_\infty*\alpha_\infty+\xi_{v_\infty}$.  
Then $\beta_\infty\in\mathcal{I}_G(\alpha_\infty)$.  Lemma
\ref{einsteinlem} implies that $\beta_j\to\beta_\infty$ uniformly as
$j\to\infty$.  So $\gr(\mathcal{I}_G)$ is compact.

Now $(x,y)\in\gr(\Phi_G)$ if and only if there exists a pair 
$(\alpha,\beta)\in\gr(\mathcal{I}_G)$ such that 
$\alpha(1)=x$ and $\beta(1)=y$. So 
$\gr(\Phi_G)$ is the image of $\gr(\mathcal{I}_G)$ under the projection
\begin{eqnarray*}
&\,&\czer{[0,1]}{\rr^n}\times\czer{[0,1]}{\rr^m}\ni\\
&\,&\qquad\,\,(\alpha,\beta)\to(\alpha(1),\beta(1))\in\rr^n\times\rr^m\,.
\end{eqnarray*}
Since this projection is continuous, 
$\gr(\Phi_G)$ is compact, and the proof of the first statement is complete.

If  $G$ is single-valued,  everywhere defined, and 
\mbox{continuous},
then $\mathcal{I}_G$ is single-valued and everywhere 
\mbox{defined} and,
moreover, the graph $\gr(\mathcal{I}_G)$ is compact,  because
$\gr(G)$ is compact. This implies that
$\mathcal{I}_G$ is 
\mbox{continuous}, and the second statement is proved. 

Finally, let us prove the third statement. Assume that $G$ is regular. We
want to show that $\mathcal{I}_G$ and $\Phi_G$ are regular.
This requires that we prove that
\ben
\item[(a)]  the graphs $\gr(\mathcal{I}_G)$ and $\gr(\Phi_G)$ are compact, 
\item[(b)] $\mathcal{I}_G$ and $\Phi_G$ can be approximated 
in the sense  of inward graph convergence by sequences of single-valued 
continuous maps. 
\een
Part (a) follows from the fact that $\gr(G)$ is \mbox{compact}. We now prove 
part (b). 
Using the regularity of $G$, let $\{G_j\}_{j=1}^\infty$ be a sequence
of single-valued, \mbox{everywhere} \mbox{defined} continuous maps from
$\aof{S}$ to $\czer{[0,1]}{\rr^{m\times n}}\times\rr^m$ such that
$G_j\togr{}G$ as $j\to\infty$. Then the $\mathcal{I}_{G_j}$ are
single-valued, everywhere defined, and continuous.  

We show that $\mathcal{I}_{G_j}\togr\mathcal{I}_{G}$ as $j\to\infty$. 
Let 
$$
\delta_j=\sup\Big\{\,{\rm dist}\Big((\alpha,\beta),\gr(\mathcal{I}_{G})\Big):
(\alpha,\beta)\in\gr(\mathcal{I}_{G_j})\,\Big\}\,.
$$
We want to show that $\delta_j\to 0$ as $j\to\infty$. Assume this is
not true.  Then there exists an infinite subset $J$ of $\nn$ and a
strictly positive number $\theta$ such that $\delta_j\ge 2\theta$ for
all $j\in J$.  We can therefore pick members $(\alpha_j,\beta_j)$ of
$\gr(\mathcal{I}_{G_j})$ for $j\in J$ such that
\begin{equation}\label{kame}
{\rm dist}\Big((\alpha_j,\beta_j),\gr(\mathcal{I}_{G})\Big)
\ge\theta\,\,\,{\rm whenever}\,\,\,j\in J\,.
\end{equation}
If $j\in J$, then $\beta_j\in\mathcal{I}_{G_j}(\alpha_j)$, so 
we can pick pairs $(h_j,v_j)\in G_j(\alpha_j)$ such that 
$\beta_j=h_j*\alpha_j+\xi_{v_j}$. 
Since $G_j\togr G$, we may assume, after making $J$ smaller, if necessary,
that the limit 
$$
(\alpha_\infty,h_\infty,v_\infty)=\lim_{j\to\infty,j\in J}(\alpha_j,h_j,v_j)
$$
exists and belongs to $\gr(G)$.  Let 
$\beta_\infty=h_\infty*\alpha_\infty+\xi_{v_\infty}$. 
Then $\beta_\infty\in\mathcal{I}_{G}(\alpha_\infty)$.
Lemma \ref{einsteinlem} implies that $\beta_j\to\beta_\infty$ uniformly 
as $j\to\infty$ via values in $J$. But then
$$
(\alpha_\infty,\beta_\infty)=\lim_{j\to\infty,j\in J}(\alpha_j,\beta_j)\,.
$$
Since $(\alpha_\infty,\beta_\infty)\in\gr(\mathcal{I}_{G})$, 
we have shown that 
$$
\lim_{j\to\infty,j\in J}
\,{\rm dist}\Big((\alpha_j,\beta_j),\gr(\mathcal{I}_{G})\Big)=0\,,
$$
contradicting (\ref{kame}).  Therefore  $\delta_j\to 0$ as $j\to\infty$, 
and we have completed the proof that $\mathcal{I}_G$ is regular.  

We must now show that $\Phi_G$ is regular. For each 
\mbox{$x\in S$}, let 
$\alpha_x$ be the curve given by 
$$
\alpha_x(t)=tx\quad{\rm for}\quad t\in[0,1]\,\,.
$$
Then $\alpha_x\in\aof{S}$, and the map $S\ni x\to\alpha_x\in\aof{S}$
is continuous. Define 
$$
\Phi^j(x)=\mathcal{I}_{G_j}(\alpha_x)(1)\quad{\rm for}\quad x\in S\,.
$$
Then $\Phi^j$ is a continuous map from $S$ to $\rr^m$. (Continuity
follows because the map $\mathcal{I}_{G_j}:\aof{S}\to
\czer{[0,1]}{\rr^m}$ is continuous, and the maps $x\to\alpha_x$ and
$\beta\to\beta(1)$ are continuous as well.  The continuity of
$\mathcal{I}_{G_j}$ follows from Lemma \ref{jointcontlemma}.)

We now show that $\Phi^j\togr\Phi_G$.  Let $(x_j,y_j)\in\gr(\Phi^j)$. 
We want to extract a subsequence of $\{(x_j,y_j)\}_{j=1}^\infty$ that converge
to a limit $(x,y)\in\gr(\Phi_G)$.   Pick 
$\beta_j\in\mathcal{I}_{G_j}(\alpha_{x_j})$. Then 
$$
{\rm dist}\Big(\,(\alpha_{x_j},\beta_j),\gr(\mathcal{I}_{G})\,\Big)
\to 0\quad{\rm as}\quad j\to\infty\,,
$$
because $\mathcal{I}_{G_j}\togr\mathcal{I}_{G}$. Since 
$\gr(\mathcal{I}_{G})$ is compact we may assume, after passing to a susequence,
that there exists a pair $(\alpha,\beta)\in\gr(\mathcal{I}_{G})$
such that $\alpha_{x_j}\to\alpha$ and $\beta_j\to\beta$. If we let 
$x=\alpha(1)$, then $x_j\to x$.  Therefore 
\mbox{$\alpha_{x_j}\to\alpha_x$}, so
$\alpha=\alpha_x$. Let $y=\beta(1)$. Then $y\in\Phi_G(x)$, and
\mbox{$(x_j,y_j)\to (x,y)$}. So our proof is complete.
\end{proof}

\ssection{The main definition}

If $n\in\zz_+$, $C$ is a cone in $\rr^n$, and $r\in\,]\,0,\infty\,[\,$, 
we write $C(r)$ to denote the set  $C\cap r\nball$, that is
$$
C(r)\equaldef\{\,x\in C:\|x\|\le r\,\}\,.
$$
Then $C(r)$ is compact convex if $C$ is a closed convex cone.

\begin{definition}\label{defipigd}
Let $n,m$ be nonnegative integers, let $F$ be a set-valued map
from $\rr^n$ to $\rr^m$, and let $C$ be a closed convex cone in $\rr^n$.
We say that $\Lambda$ is a {\bem \pigd{} of $F$ at $(0,0)$ in the direction 
of $C$,} and write $\Lambda\in\pigdab(F,C)$, if
(1)~$\Lambda$ is a nonempty compact subset of $\rr^{m\times n}$, 
and (2)~for every positive real number $\delta$ there exists a 
\mbox{number}
$R\in\,]\,0,\infty\,[\,$ with the property that 
for every $r\in\,]\,0,R]$ there exists a regular 
set-valued map
$G:\aof{C(r)}\tto\czer{[0,1]}{\rr^{m\times n}}\times\rr^m$
such that
\ben
\medbreak\item[(2a)] $h(t)\in\Lambda^\delta$ and $\|v\|\le\delta r$ 
whenever 
\mbox{$\alpha\in\aof{C(r)}$}, $(h,v)\in G(\alpha)$,  $t\in[0,1]$,
\item[(2b)] $\gr(\Phi_{G})\subseteq\gr(F)$. \proofend
\een
\end{definition}


\ssection{The chain rule}

\begin{theorem}
Let $n_1$, $n_2$, $n_3$ be nonnegative integers, and let $F_i$ be, 
for $i=1,2$, set-valued maps from $\rr^{n_i}$ to $\rr^{n_{i+1}}$.
Assume that
\ben
\item[1.] $C_i$ is a closed convex cone in $\rr^{n_i}$ for $i=1,2$,
\item[2.]  $C_2$ is polyhedral,
\item[3.]  $\Lambda_i\in\pigdab(F_i,C_i)$ for $i=1,2$,
\item[4.] $F_1(C_1)\subseteq C_2$,
\omitshort{\item[and]}
\item[5.] $\Lambda_1\cdot C_1\subseteq C_2$  (that is,
$L\cdot C_1\subseteq C_2$  for every $L\in \Lambda_1$). 
\een
Then 
$$
\Lambda_2\compo\Lambda_1\in\pigdab(F_2\compo F_1,C_1)\,.
$$
\end{theorem}

\skipopag

\omitlong{\noindent{\it Outline of the proof}.}
\omitshort{\noindent{\it Proof}.}
The crucial point is that, since the cone $C_2$ is polyhedral, 
it is possible to pick a vector
$\bar w\in{\rm Int}_S(C_2)$ and a positive constant $\bar k$  such that 
the following ``error correction property'' holds

\vspace{-1mm}
\begin{center}
\begin{minipage}{8cm}
(ECP){\bem \ If $m\in\nn$, $u_1,\ldots,u_m\in C_2$, $w\in C_2$, and
\mbox{$u_1+\ldots+u_m=w+\bar w$}, then there exist \mbox{vectors}
$c_1,\ldots,c_m$ such that $\|c_1\|\!+\!\ldots\!+\!\|c_m\|\!\le\!\bar k$,
\mbox{$c_1\!+\!\ldots\!+\!c_m\!=\!\bar w$}, and 
the \mbox{conditions} $\|c_i\|\le\bar k\|u_i\|$ and
$u_i-c_i\in C_2$ hold for $i=1,\ldots,m\,$.
} 
\end{minipage}\end{center}

\omitlong{\vspace{-1mm}%\medbreak
\noindent(This is not easy to prove, and we will omit the proof
for lack of space.)}

\omitshort{\vspace{-1mm}%\medbreak
\noindent(This follows from Theorem \ref{hasecp}.)}

We then may---and will---assume, without loss of generality,
that $\|\bar w\|\le 1$. We then fix a number
$\rho\in\,]\,0,\infty\,[\,$ such that
\begin{equation}\label{cond-on-rho}
\bar w+\rho\bar\bmath^{n_2}\subseteq C_2\,.
\end{equation} 
Then the vector $\bar w$ and the number $\rho$ satisfy 
\begin{equation}\label{cond-on-rho-useful}
\Big(s>0\wedge u\in S\wedge\|u\|\le s\Big)
\Rightarrow u+\rho^{-1}s\bar w\in C_2\,.
\end{equation}

\omitshort{\noindent(Indeed, let $y=\frac{\rho}{s}u$. 
Then $\|y\|\le\rho$. Therefore
(\ref{cond-on-rho}) implies $y+\bar w\in C_2$. Since 
$u+\frac{s}{\rho}\bar w=\frac{s}{\rho}(y+\bar w)$, and $C_2$ is a cone, 
$u+\frac{s}{\rho}\bar w\in C_2$, as stated.)}

Write $F=F_2\compo F_1$, $\Lambda=\Lambda_2\compo\Lambda_1$, 
$n=n_1$, $m=n_3$, $C=C_1$, and let
\begin{eqnarray*}
S&=&{\rm linear\,\,\,span\,\,\,of}\,\,\,C_2\,\,\,{\rm in}\,\,\,\rr^{n_2}\,,\\
\Pi&=&{\rm the\,\,\,orthogonal\,\,\,projection\,\,\,from}\,\,\,
\rr^{n_2}\,\,\,{\rm  to}\,\,\,S\,,\\
\kappa_i&=&\sup\{\,\|L\|:L\in\Lambda_i\,\}\quad{\rm for}\quad i=1,2\,.
\end{eqnarray*}
Then $F:\rr^n\tto\rr^m$, and $\Lambda$ is a nonempty compact 
\mbox{subset} of $\rr^{m\times n}$.

\medbreak\noindent Fix a positive real number $\delta$.  We want to
find a number $R\in\,]\,0,\infty\,[\,$ satisfying Property 2\#{} of
Definition \ref{defipigd}.

Define a function $\Psi:[0,\infty\,[\,\times[0,\infty\,[\,
\to[0,\infty\,[\,$ by 
\begin{eqnarray}
\Psi(\delta_1,\delta_2)&\equaldef&
(1+\bar k)\Big(\kappa_1+2\delta_1(1+\rho^{-1})\Big)\delta_2
\nonumber\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\!\!\!\!\!\!\!\!\!\!\!\!+(\kappa_2+\delta_2)\delta_1
+2\delta_1\Big(1+\rho^{-1}(1+\bar k)\Big)(\kappa_2+\delta_2)\,.
\label{defofPhi} 
\end{eqnarray}
Then
\begin{equation}\label{limofPhi} 
\lim_{(\delta_1,\delta_2)\to(0,0)}\Psi(\delta_1,\delta_2)=0\,.
\end{equation}


We let $\delta_1$, $\delta_2$, be positive numbers such that
\begin{equation}\label{used-b}
\Psi(\delta_1,\delta_2)\le\frac{\delta}{2}\,.
\end{equation}
For $i=1,2$, using the fact that $\Lambda_i\in\pigdab(F_i,C_i)$,
choose $R_i\in\,]\,0,\infty\,[\,$ with the property that for every
$r_i\in\,]\,0,R_i]$ there exists $G_i$ such that
\ben
\item[(1)] \mbox{$G_i\!\in\!\regu{\aof{C_i(r_i)}}
{\czer{[0,1]}{\rr^{n_{i+1}\times n_i}}\!\times\!\rr^{n_{i+1}}}$},
\item[(2)] $\gr(\Phi_{G_i})\subseteq\gr(F_i)\,$,
\item[(3)] $(h_i(t),v_i)\in\Lambda_i^{\delta_i}\times
(\delta_ir_i\bar\bmath^{n_{i+1}})$ 
whenever $t\in[0,1]$, $\zeta\in\aof{C_i(r_i)}$, 
and $(h_i,v_i)\in G_i(\zeta)$.
\een
Inequality (\ref{used-b}) implies in particular that the inclusion
\begin{equation}\label{incluimp}
\Lambda_2^{\delta_2}\compo\Lambda_1^{\delta_1}\subseteq\Lambda^{\delta}
\end{equation}
holds. \omitshort{({\bem Proof.}\ If $L_i\in\Lambda_i^{\delta_i}$ for 
$i=1,2$, and \mbox{$L=L_2\compo L_1$}, then we can write 
$L_i=L_i^0+E_i$, where $L_i^0\in\Lambda_i$ and $\|E_i\|\le\delta_i$. 
Let $L^0=L_2^0\compo L_1^0$, so $L^0\in\Lambda$. Then 
\mbox{$L=L^0+E$}, where 
$E=L_2^0\compo E_1+E_2\compo L_1^0+E_2\compo E_1$, so 
$$
\|E\|\le\kappa_2\delta_1+\delta_2\kappa_1+\delta_2\delta_1\,.
$$
It follows easily from (\ref{used-b}) that
\begin{equation}\label{used-a}
\kappa_2\delta_1+\delta_2\kappa_1+\delta_2\delta_1\le\delta\,.
\end{equation}
Therefore $\|E\|\le\delta$, showing that $L\in\Lambda^{\delta}$, as stated.)}

\omitlong{
We let 
$\theta_0=\kappa_1+2\delta_1\Big(1+\frac{1}{\rho}\Big)$, 
$\theta=(1+\bar k)\theta_0$,
and choose 
$R=\min\Big(\,R_1,\frac{R_2}{\theta}\,\Big)$.
We then have to show that, with this choice of $R$, the property of 
Definition \ref{defipigd} is satisfied. For this purpose, we  
pick $r\in\rr$ such that $0<r\le R$, and prove the existence of
a $G$ satisfying the conditions of Definition \ref{defipigd}.}

\omitshort{We let 
$$
\theta_0=\kappa_1+2\delta_1\Big(1+\frac{1}{\rho}\Big)\,,\qquad
\theta=(1+\bar k)\theta_0\,,
$$
and choose 
$$
R=\min\Big(\,R_1,\frac{R_2}{\theta}\,\Big)\,.
$$
We then have to show that, with this choice of $R$, the property of 
Definition \ref{defipigd} is satisfied. For this purpose, we  
pick $r\in\rr$ such that $0<r\le R$, and prove the existence of
a $G$ satisfying the conditions of Definition \ref{defipigd}.}

\omitshort{
Let 
\begin{eqnarray*}
r_1&=&r\,,\\
r_2&=&\theta r\,,
\end{eqnarray*}
and observe that $0<r_1\le R_1$ and $0<r_2\le R_2$.  Pick $G_1$, 
$G_2$, such that (1)-(2)-(3) hold.

Let 
\begin{eqnarray*}
\omega&=&2\delta_1r_1\,,\\
\tilde r&=&\kappa_1r\,,\\
\hat r&=&\theta_0r_1\,,
\end{eqnarray*}
so that
\begin{equation}\label{incluig-roos}
r_2=(1+\bar k)\hat r\,.
\end{equation}}

\omitlong{
Let  $r_1=r$, $r_2=\theta r$,
and observe that $0<r_1\le R_1$ and $0<r_2\le R_2$.  Pick $G_1$, 
$G_2$, such that (1)-(2)-(3) hold. Let 
$\omega=2\delta_1r_1$, $\tilde r=\kappa_1r$, $\hat r=\theta_0r_1$,
so that
\begin{equation}\label{incluig-roos}
r_2=(1+\bar k)\hat r\,.
\end{equation}}
Let $K$ be the $\omega$-neighborhood of $C_2(\tilde r)$, that is,
$$
K=\{\,x\in\rr^{n_2}:{\rm dist}(x,C_2(\tilde r))\le\omega\,\}\,.
$$


\omitshort{\noindent We claim that}
\omitlong{\noindent It is easy to see tat}
\begin{equation}\label{incluig}
\mathcal{I}_{G_1}\Big(\,\aof{C_1(r_1)}\,\Big)\,\subseteq\,\aof{K}\,.
\end{equation}

\omitshort{\noindent To see this, let
$\beta\in\mathcal{I}_{G_1}\Big(\,\aof{C_1(r_1)}\,\Big)$.  Pick
\mbox{$\alpha\in\aof{C_1(r_1)}$} such that
$\beta\in\mathcal{I}_{G_1}(\alpha)$. Then choose $(h_1,v_1)\in
G_1(\alpha)$ such that $\beta=h_1*\alpha+\xi_{v_1}$.  It is clear 
that $\beta(0)=0$, and $\beta$ is absolutely continuous. So 
(\ref{incluig}) will be proved if we show that $\dot\beta(t)\in K$
for almost all $t\in[0,1]$.


Let $E$ be a subset of $[0,1]$ such that ${\rm meas}(E)=1$, having the 
property that for every $t\in E$
\ben
\item[(I)] the derivative $\dot\alpha(t)$ exists and belongs to $C_1(r_1)$, 
\item[(II)] $\dot\beta(t)$ exists and is equal to $h_1(t)\cdot\dot\alpha(t)+v_1$.  
\een
Let $t\in E$. Since $h_1(t)\in\Lambda_1^{\delta_1}$, we can write 
$h_1(t)=L+\tilde L$, with $L\in\Lambda_1$ and $\|\tilde L\|\le\delta_1$.  Then
$$
\dot\beta(t)=L\cdot\dot\alpha(t)+\tilde L\cdot\dot\alpha(t)+v_1\,,
$$
and 
$$
\|\tilde L\cdot\dot\alpha(t)+v_1\|\le 2\delta_1r_1=\omega\,.
$$
We now use the hypothesis that $\Lambda_1C_1\subseteq C_2$ to conclude that
$L\cdot\dot\alpha(t)\in C_2$. Moreover,  
$\|L\cdot\dot\alpha(t)\|\le\kappa_1r_1=\tilde r$. Therefore
$L\cdot\dot\alpha(t)\in C_2(\tilde r)$. It then follows that 
\begin{equation}\label{itnofollo}
{\rm dist}(\,\dot\beta(t),C_2(\tilde r)\,)\le\omega\,.
\end{equation}
So $\dot\beta(t)\in K$. Since this is true for all $t\in E$, we have shown 
that  $\dot\beta(t)\in K$ for almost all $t$, completing the proof of
(\ref{incluig}).
}


We are now ready to begin the long process of defining the set-valued map
$$
G:\aof{C(r)}\tto\czer{[0,1]}{\rr^{m\times n}}\times\rr^m\,.
$$
The first step will be to assign, to each triple
$(\alpha,h_1,v_1)$ such that 
$\alpha\in\aof{C(r)}$ and $(h_1,v_1)\in G_1(\alpha)$, 
a curve $\beta_{\alpha,h_1,v_1}:[0,1]\to\rr^{n_2}$.

Pick $\alpha\in\aof{C(r)}=\aof{C_1(r_1)}$, and 
\mbox{$(h_1,v_1)\in G_1(\alpha)$}.
\omitshort{Let  
$$
\beta_{\alpha,h_1,v_1}\,\equaldef\,h_1*\alpha+\xi_{v_1}\,.
$$}
\omitshort{Let  $\beta_{\alpha,h_1,v_1}\,\equaldef\,h_1*\alpha+\xi_{v_1}$.}
Then $\beta_{\alpha,h_1,v_1}\in\mathcal{I}_{G_1}(\alpha)$, so
(\ref{incluig}) implies that 
\mbox{$\beta_{\alpha,h_1,v_1}\in\aof{K}$}. Moreover,
\begin{equation}\label{incluiog-beta}
\beta_{\alpha,h_1,v_1}(1)\in C_2\,,
\end{equation}
\omitshort{
because 
$$
\beta_{\alpha,h_1,v_1}(1)\in\Phi_{G_1}(\alpha(1))\subseteq F_1(\alpha(1))
\subseteq C_2\,.
$$}
\omitlong{because 
$\beta_{\alpha,h_1,v_1}(1)\in\Phi_{G_1}(\alpha(1))\subseteq F_1(\alpha(1))
\subseteq C_2$.}

The second step is to correct the error arising from the fact
that $\beta\in\aof{K}$ rather than in $\aof{C_2(\tilde r)}$. For this
purpose we define, whenever $\alpha$ belongs to 
$\aof{C(r)}$ and $(h_1,v_1)\in G_1(\alpha)$,
a new curve $\gamma_{\alpha,h_1,v_1}\in\czer{[0,1]}{\rr^{n_2}}$ by letting 
$\gamma_{\alpha,h_1,v_1}(t)=\Pi(\beta_{\alpha,h_1,v_1}(t))
+\frac{\omega t}{\rho}\bar w$ for $t\in[0,1]$.

We claim that 
\begin{equation}\label{incluig-gamma}
\gamma_{\alpha,h_1,v_1}\in\aof{C_2(\hat r)}\,.
\end{equation}
To see this, write $\beta=\beta_{\alpha,h_1,v_1}$,  
$\gamma=\gamma_{\alpha,h_1,v_1}$, and observe that 
$\gamma$ is an absolutely continuous curve, and $\gamma(0)=0$.
In addition, for almost all $t\in[0,1]$, $\dot\gamma(t)$
exists and is equal to $\Pi(\dot\beta(t))+\frac{\omega}{\rho}\bar w$, and
$\dot\beta(t)\in K$.  Let $E$ be the set of all $t\in[0,1]$ for which
this is true.  Then ${\rm meas}(E)=1$.  Moreover,
\begin{equation}\label{incluig-gamma-a}
\dot\gamma(t)\in C_2(\hat r)\quad{\rm whenever}\quad t\in E\,.
\end{equation}
({\bem Proof}. Fix $t\in E$. Since $\dot\beta(t)\in K$, we can write 
\mbox{$\dot\beta(t)=b_1+b_2$}, with $b_1\in C_2(\tilde r)$ and 
$\|b_2\|\le\omega$. Then  $\Pi(\dot\beta(t))=\Pi(b_1)+\Pi(b_2)=b_1+\Pi(b_2)$,
since $b_1\in S$.  Moreover,  $\|\Pi(b_2)\|\le\omega$, because 
$\|b_2\|\le\omega$ and $\Pi$ is an 
orthogonal projection.   Then (\ref{cond-on-rho-useful}) implies that 
\mbox{$\Pi(b_2)+\rho^{-1}\omega\bar w\in C_2$}.  Since $b_1\in C_2$, and
\mbox{$\dot\gamma(t)=b_1+\Pi(b_2)+\rho^{-1}\omega\bar w$},
we conclude that $\dot\gamma(t)\in C_2$. Furthermore,
\begin{eqnarray*}
\|\dot\gamma(t)\|
&\le&\|\Pi(\dot\beta(t))\|+\rho^{-1}\omega\|\bar w\|
\le\|\dot\beta(t)\|+\rho^{-1}\omega\\
&\le&\tilde r+\omega+\rho^{-1}\omega
=(\kappa_1+2\delta_1+2\rho^{-1}\delta_1)r_1\\
&=&\theta_0r_1=\hat r\,.
\end{eqnarray*}
So $\dot\gamma(t)\in C_2(\hat r)$, and the proof of 
(\ref{incluig-gamma-a}) is complete.) Therefore 
(\ref{incluig-gamma}) holds.

\medbreak The third step is to make a piecewise linear 
\mbox{approximation} of the curves 
$\gamma_{\alpha,h_1,v_1}$, by first choosing a large positive integer
$N$ as follows.  The fact that $G_2$ is regular implies that the set
$G_2(\aof{C_2(r_2)})$ is compact in $\czer{[0,1]}{\rr^{n_3\times n_2}}\times\rr^{n_3}$. Let $H$ be the set of those
$h_2\in\czer{[0,1]}{\rr^{n_3\times n_2}}$ such that
$(h_2,v_2)\in G_2\Big(\aof{C_2(r_2)}\Big)$ for some $v_2\in\rr^{n_3}$. Then
$H$ is compact as well.  Hence $H$ is uniformly 
\mbox{equicontinuous},  so 
we can choose $N\in\nn$ such that
$\|h_2(t)-h_2(s)\|\le \frac{\delta}{4\theta_0}$ 
\omitlong{\skipopag}

\noindent whenever
$h_2\in H$, $t,s\in[0,1]$, and $|t-s|\le\frac{1}{N}$.
With this choice of $N$, we define
\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\eta_{\alpha,h_1,v_1}(t)=
(j-Nt)\gamma_{\alpha,h_1,v_1}(N^{-1}(j-1))
\nonumber\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!+
(Nt+1-j)\gamma_{\alpha,h_1,v_1}(N^{-1}j)\label{incluig-gamma-uu}
\end{eqnarray}
whenever $\alpha\in\aof{C(r)}$, $(h_1,v_1)\in G_1(\alpha)$,
$\frac{j-1}{N}\le t\le \frac{j}{N}\,$, $j\in\nn$, $1\le j\le N$.

Then $\eta_{\alpha,h_1,v_1}\in\czer{[0,1]}{\rr^{n_2}}$, and 
\begin{eqnarray}
\eta_{\alpha,h_1,v_1}(0)&=&0\,,\label{undes-aa}\\
\eta_{\alpha,h_1,v_1}(1)&=&\gamma_{\alpha,h_1,v_1}(1)\nonumber\\
&=&\beta_{\alpha,h_1,v_1}(1)
+\rho^{-1}\omega\bar w\nonumber\\
&\in& C_2+\rho^{-1}\omega\,.\label{undes-bb}
\end{eqnarray}
Moreover, the map $\eta_{\alpha,h_1,v_1}$ is linear on each interval
$I_j=[N^{-1}(j-1),N^{-1}j\,]$.
The derivative $\dot\eta_{\alpha,h_1,v_1}(t)$ of 
$\eta_{\alpha,h_1,v_1}$ is equal, for $t\in I_j$, to  
$u_{\alpha,h_1,v_1,j}^N$, where 
\begin{eqnarray}
u_{\alpha,h_1,v_1,j}^N\!\!\!\!&=&\!\!\!\!N\gamma_{\alpha,h_1,v_1}(N^{-1}j)
\!-\!N\gamma_{\alpha,h_1,v_1}(N^{-1}(j-1))\nonumber\\
\quad\!\!\!\!&=&\!\!\!\!
N\int_{\frac{j-1}{N}}^{\frac{j}{N}}\dot\gamma_{\alpha,h_1,v_1}(t)\,dt\,.
\label{incluig-gamma-a-oink-u}
\end{eqnarray}
Since $\dot\gamma_{\alpha,h_1,v_1}(t)\in C_2(\hat r)$ for almost all $t$, 
the vectors $u_{\alpha,h_1,v_1,j}^N$ belong to $C_2(\hat r)$ as well.
So $\dot\eta_{\alpha,h_1,v_1}(t)\in C_2(\hat r)$ for almost all $t\in[0,1]$, 
and then (\ref{undes-aa}) implies that
\begin{equation}\label{incluig-gamma-a-oink-uiu}
\eta_{\alpha,h_1,v_1}\in\aof{C_2(\hat r)}\,.
\end{equation}

The fourth step is to take care of the 
\mbox{undesirable} fact that 
$\eta_{\alpha,h_1,v_1}$ satisfies (\ref{undes-bb}), and produce a curve
whose terminal point is $\beta_{\alpha,h_1,v_1}(1)$ rather than 
\mbox{$\beta_{\alpha,h_1,v_1}(1)+\rho^{-1}\omega\bar w$}. For this purpose,
we define $\tilde u_{\alpha,h_1,v_1,j}^N=
\frac{\rho}{N\omega}u_{\alpha,h_1,v_1,j}^N$.  Then
$$
\sum_{j=1}^N\tilde u_{\alpha,h_1,v_1,j}^N=
\frac{\rho}{\omega}\beta_{\alpha,h_1,v_1}(1)+\bar w\,.
$$
It then follows from (ECP) that there exists an $N$-tuple
$\tilde{\bf c}=(\tilde c_1,\ldots,\tilde c_N)$
of vectors that satisfies
\begin{eqnarray}
\tilde u_{\alpha,h_1,v_1,j}^N-\tilde c_j&\in& C_2
\phantom{\bar k\|\tilde u_{\alpha,h_1,v_1,j}^N\|}
\label{incluig-gamma-a-oink-u-aaa}\\
\tilde c_1+\cdots+\tilde c_N&=&\bar w\,,\label{incluig-gamma-a-oink-u-bbb}\\
\|\tilde c_1\|+\cdots+\|\tilde c_N\|&\le&\bar k\,,
\label{incluig-gamma-a-oink-u-ccc}\\
\|\tilde c_j\|&\le&\bar k\|\tilde u_{\alpha,h_1,v_1,j}^N\|\phantom{C_2}
\,,\label{incluig-gamma-a-oink-u-ddd}
\end{eqnarray}
for $j=1,\ldots,N$. Define
$c_j=\rho^{-1}N\omega\tilde c_j$ and \mbox{${\bf c}=(c_1,\ldots,c_N)$}.
Then, for $j=1,\ldots,N$, 
\begin{eqnarray}
u_{\alpha,h_1,v_1,j}^N-c_j&\in& C_2\phantom{\bar k
\|u_{\alpha,h_1,v_1,j}^N\|}\,,\label{incluig-gamma-a-oink-u-aaa-rep}\\
c_1+\cdots+c_N&=&\rho^{-1}N\omega\bar w\,,
\label{incluig-gamma-a-oink-u-bbb-rep}\\
\|c_1\|+\cdots+\|c_N\|&\le&\rho^{-1}N\omega\bar k\,,
\label{incluig-gamma-a-oink-u-ccc-rep}\\
\|c_j\|&\le&\bar k\|u_{\alpha,h_1,v_1,j}^N\|\phantom{C_2}
\,.\label{incluig-gamma-a-oink-u-ddd-rep}
\end{eqnarray}


Let $\mu_{{\bf c}}$ be the function such that
\begin{eqnarray}
\mu_{{\bf c}}&\in&\czer{[0,1]}{\rr^{n_2}}\,,\label{propxi-a}\\ 
\mu_{{\bf c}}(0)&=&0\,,\label{propxi-b}\\
\dot\mu_{{\bf c}}&\equiv&c_j\quad
{\rm on}\quad I_j\quad{\rm for}\quad j=1,\ldots,N\,.\label{propxi-c}
\end{eqnarray}
Then 
\begin{equation}\label{propxi-d}
\int_0^1\|\dot\mu_{{\bf c}}(t)\|\,dt=\frac{1}{N}(\,\|c_1\|+\cdots+\|c_N\|\,)
\le\frac{\omega\bar k}{\rho}
\end{equation}
and
\begin{equation}\label{propxi-e}
\mu_{{\bf c}}(1)=\frac{1}{N}(\,c_1+\cdots+c_N\,)=\frac{\omega}{\rho}\bar w\,.
\end{equation}
Define a curve $\zeta_{\alpha,h_1,v_1,{\bf c}}$ by letting
\begin{equation}\label{propxi-d-uau}
\zeta_{\alpha,h_1,v_1,{\bf c}}(t)=
\eta_{\alpha,h_1,v_1}(t)-\mu_{{\bf c}}(t)\quad
{\rm for}\quad t\in[0,1]\,.
\end{equation}
Then $\zeta_{\alpha,h_1,v_1,{\bf c}}$ satisfies
\begin{eqnarray}
\zeta_{\alpha,h_1,v_1,{\bf c}}&\in&\czer{[0,1]}{\rr^{n_2}}\,,\label{undes-aa-rip-rip}\\
\zeta_{\alpha,h_1,v_1,{\bf c}}(0)&=&0\,,\label{undes-aa-rip}\\
\zeta_{\alpha,h_1,v_1,{\bf c}}(1)&=&\eta_{\alpha,h_1,v_1}(1)-
\mu_{{\bf c}}(1)\nonumber\\
&=&\beta_{\alpha,h_1,v_1}(1)\,.\label{undes-bb-rip}
\end{eqnarray}
Moreover, $\zeta_{\alpha,h_1,v_1,{\bf c}}$ is linear on each interval $I_j$,
and the derivative $\dot\zeta_{\alpha,h_1,v_1,{\bf c}}(t)$ of 
$\zeta_{\alpha,h_1,v_1,{\bf c}}$ is equal, for $t\in I_j$, to  
$u_{\alpha,h_1,v_1,j}^N-c_j$.
It follows from (\ref{incluig-gamma-a-oink-u-aaa-rep}) that
the vectors $v_j=u_{\alpha,h_1,v_1,j}^N-c_j$ belong to $C_2$. Moreover, the bound
(\ref{incluig-gamma-a-oink-u-ddd-rep}) implies that 
$$
\|v_j\|\le(1+\bar k)\|u_{\alpha,h_1,v_1,j}^N\|\le(1+\bar k)\hat r=r_2\,.
$$
Therefore
\begin{equation}\label{propxi-d-uau-vv}
\dot\zeta_{\alpha,h_1,v_1,{\bf c}}\in C_2(r_2)\quad
{\rm for\,\,\,a.e.}\,\,\,t\in[0,1]\,, 
\end{equation}
and then (\ref{undes-aa-rip}) implies that
\begin{equation}\label{incluig-gamma-a-oink-uiu-gug}
\zeta_{\alpha,h_1,v_1,{\bf c}}\in\aof{C_2(r_2)}\,.
\end{equation}

We have now finally succeeded in producing, for each curve $\alpha\in\aof{C(r)}$
and each pair $(h_1,v_1)\in G_1(\alpha)$, a curve
$\zeta_{\alpha,h_1,v_1,{\bf c}}\in\aof{C_2(r_2)}$ whose terminal 
point is exactly $\beta_{\alpha,h_1,v_1}(1)$.  Moreover, this curve 
is ``close'' to $\beta_{\alpha,h_1,v_1}$, in the sense
that it is close to $\eta_{\alpha,h_1,v_1}$, which is 
close to $\gamma_{\alpha,h_1,v_1}$, which is 
close to $\beta_{\alpha,h_1,v_1}$.  

This curve need not be unique, because ${\bf c}$ may fail to be unique, so 
this nonuniqueness will have to be taken care of. As a first step 
in that direction, we introduce the notation
${\bf C}_{\alpha,h_1,v_1}$ to refer to the set of all $N$-tuples
${\bf c}=(c_1,\ldots,c_N)$ that belong to $(\rr^{n_2})^N$ and satisfy 
(\ref{incluig-gamma-a-oink-u-aaa-rep}), 
(\ref{incluig-gamma-a-oink-u-bbb-rep}), 
(\ref{incluig-gamma-a-oink-u-ccc-rep}), 
and (\ref{incluig-gamma-a-oink-u-ddd-rep}).


Given a curve $\alpha\in\aof{C(r)}$, a pair $(h_1,v_1)\in G_1(\alpha)$, and
a ${\bf c}\in{\bf C}_{\alpha,h_1,v_1}$, 
pick  $(h_2,v_2)\in G_2(\zeta_{\alpha,h_1,v_1,{\bf c}})$.  Define 
\begin{eqnarray*}
h_{\alpha,h_1,v_1,{\bf c},h_2,v_2}(t)&=&h_2(t)\cdot h_1(t)
\,\,\,{\rm for}\,\,\,
t\in[0,1]\,,\\
v_{\alpha,h_1,v_1,{\bf c},h_2,v_2}\phantom{(t)}&=&
v_2+\Big(\int_0^1h_2(t)\,dt\Big)\cdot v_1\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
-\int_0^1h_2(t)\cdot(\dot\beta(t)-\dot\zeta(t))\,dt\,.
\end{eqnarray*}
We let $G(\alpha)$ be the set of all pairs 
$(h_{\alpha,h_1,v_1,{\bf c},h_2,v_2},v_{\alpha,h_1,v_1,{\bf c},h_2,v_2})$, as 
$\alpha$ varies over $\aof{C(r)}$,
$(h_1,v_1)$ varies over all members of $G_1(\alpha)$, 
${\bf c}$ varies over all members of ${\bf C}_{\alpha,h_1,v_1}$, 
and the pair $(h_2,v_2)$ 
varies over all members of $G_2(\zeta_{\alpha,h_1,v_1,{\bf c}})$.

With this definition, it is clear that $G$ is a set-valued map from
$\aof{C(r)}$ to $\czer{[0,1]}{\rr^{m\times n}}\times\rr^m$.  
\mbox{Moreover},
if $(h,v)\in G(\alpha)$ for an $\alpha\in\aof{C(r)}$, then the following 
\omitshort{three facts are true.}
\omitlong{can be verified}
\ben
\item[(F1)] {\sl The matrix-valued function $h$ 
takes values in $\Lambda^\delta$}. 
\item[(F2)] {\sl If $x=\alpha(1)$, $\sigma=h*\alpha+\xi_v$, 
and $z=\sigma(1)$, then $z\in F(x)$}. 
\item[(F3)] $\|v\|\le\delta r$.
\een

\omitshort{
\medbreak\noindent {\bem Proof of (F1).\/}  Write
\begin{equation}\label{ifwewrite-a}
(h,v)=(h_{\alpha,h_1,v_1,{\bf c},h_2,v_2}\,,\,
v_{\alpha,h_1,v_1,{\bf c},h_2,v_2})\,. 
\end{equation}
Then, if $t\in[0,1]$, (\ref{incluimp}) implies that $h_2(t)\cdot
h_1(t)\in\Lambda^{\delta}$, since $h_1(t)\in\Lambda_1^{\delta_1}$ and
\mbox{$h_2(t)\in\Lambda_2^{\delta_2}$}.}

\omitshort{
\medbreak\noindent{\bem Proof of (F2).\/}  Write
$\beta=\beta_{\alpha,h_1,v_1}$, $\gamma=\gamma_{\alpha,h_1,v_1}$, 
\mbox{$\eta=\eta_{\alpha,h_1,v_1}$}, $\zeta=\zeta_{\alpha,h_1,v_1,{\bf c}}$,
$\lambda=\beta-\zeta$, $y=\beta(1)=\zeta(1)$,
$\nu=h_2*\zeta+\xi_{v_2}$.}

\omitshort{
Then $y\in\Phi_{G_1}(x)$, because
$\beta\in\mathcal{I}_{G_1}(\alpha)$, so $\beta(1)\in\Phi_{G_1}(x)$.
Therefore $y$ belongs to $F_1(x)$, because 
$\gr(\Phi_{G_1})\subseteq\gr(F_1)$. Now,
$z=\int_0^1h(t)\cdot\dot\alpha(t)\,dt+v$, so
\begin{eqnarray*}
z&=&\int_0^1h_2(t)\cdot h_1(t)\cdot\dot\alpha(t)\,dt+
\int_0^1(h_2(t)\cdot v_1)\,dt+v_2\\
&\,&-\int_0^1h_2(t)\cdot\dot\lambda(t)\,dt\\
&=&\int_0^1h_2(t)\cdot\Big(h_1(t)\cdot\dot\alpha(t)+v_1\Big)\,dt+v_2\\
&\,&-\int_0^1h_2(t)\cdot\dot\lambda(t)\,dt\\
&=&\int_0^1h_2(t)\cdot\dot\beta(t)\,dt+v_2\\
&\,&-\int_0^1h_2(t)\cdot(\dot\beta(t)-\dot\zeta(t))\,dt\\
&=&\int_0^1h_2(t)\cdot\dot\zeta(t)\,dt+v_2\\
&=&\nu(1)\,.
\end{eqnarray*}
But $\nu\in\mathcal{I}_{G_2}(\zeta)$, because 
$\nu=h_2*\zeta+\xi_{v_2}$ and \mbox{$(h_2,v_2)\in G_2(\zeta)$}. 
Therefore 
$$
z=\nu(1)\in\Phi_{G_2}(\zeta(1))=\Phi_{G_2}(y)\,.
$$ 
Hence $z$ belongs to $F_2(y)$, because $\gr(\Phi_{G_2})\subseteq\gr(F_2)$. 
Since $y\in F_1(x)$ and $z\in F_2(y)$, the conclusion that 
$z\in F(x)$ follows.}

\omitshort{
\medbreak\noindent {\bem Proof of (F3).\/}  Using the notations introduced 
above, we have
\begin{equation}\label{ele-a}
v=v_2+\Big(\int_0^1h_2(t)\,dt\Big)\cdot v_1+
\int_0^1h_2(t)\cdot\dot\lambda(t)\,dt\,,
\end{equation}
so
\begin{eqnarray}
\|v\|&\le&\delta_2r_2+(\kappa_2+\delta_2)\delta_1r_1+E\nonumber\\
&\le&\Big(\theta\delta_2+(\kappa_2+\delta_2)\delta_1\Big)r+E\,,\label{estima-za}
\end{eqnarray}
where
\begin{equation}\label{estima-zb}
E=\Biggl\|\int_0^1h_2(t)\cdot\dot\lambda(t)\,dt\Biggr\|\,.
\end{equation}
To estimate $E$, we write
\begin{equation}\label{estima-zc}
\lambda=(\beta-\beta^*)+(\beta^*-\gamma)+(\gamma-\eta)+(\eta-\zeta)\,
\end{equation}
where $\beta^*=\Pi\compo\beta$. Then $\dot\beta^*=\Pi\compo\dot\beta$,
because $\Pi$ is linear.  It follows from (\ref{itnofollo})
that 
$$
\|\dot\beta(t)-\dot\beta^*(t)\|\le\omega\quad{\rm for\,\,\,a.e.}\quad
t\in[0,1]\,,
$$
and then
\begin{eqnarray}
\!\!\!\!\!\!\!\!\!\!\!\!
\Big\|\int_0^1h_2(t)\cdot(\dot\beta(t)-\dot\beta^*(t))\,dt\Big\|
\!\!\!\!&\le&\!\!\!\!(\kappa_2+\delta_2)\omega\nonumber\\
\!\!\!\!\!\!\!\!\!\!\!\!\quad
\!\!\!\!&=&\!\!\!\!
2\delta_1(\kappa_2+\delta_2)r\,.\label{estima-a}
\end{eqnarray}

The function $\dot\beta^*-\dot\gamma$ has a constant value, equal to
$\frac{\omega}{\rho}\bar w$. Therefore
\begin{eqnarray}
\!\!\!\!\!\!\!\!\!\!\!\!
\Big\|\int_0^1h_2(t)\cdot(\dot\beta^*(t)-\dot\gamma(t))\,dt\Big\|
\!\!\!\!\!&\le&\!\!\!\!\!
\rho^{-1}(\kappa_2+\delta_2)\omega\|\bar w\|\nonumber\\
\!\!\!\!\!&\le&\!\!\!\!\!
2\rho^{-1}\delta_1(\kappa_2\!+\!\delta_2)r\,.\label{estima-b}
\end{eqnarray}
The function $\eta-\zeta$ is $\mu_{{\bf c}}$. Then (\ref{propxi-d}) implies
the bound
\begin{eqnarray}
\!\!\!\!\!\!\!\!\!\!\!\!
\Big\|\int_0^1h_2(t)\cdot(\dot\eta(t)-\dot\zeta(t))\,dt\Big\|
\!\!\!\!\!&\le&\!\!\!\!\!
\rho^{-1}\bar k(\kappa_2+\delta_2)\omega\nonumber\\
\!\!\!\!\!&\le&\!\!\!\!\!
2\rho^{-1}\delta_1\bar k(\kappa_2\!+\!\delta_2)r\,.\label{estima-c}
\end{eqnarray}

Finally, we have to estimate the integral 
\mbox{$\sigma=\int_0^1h_2(t)\cdot(\dot\gamma(t)-\dot\eta(t))\,dt$}.  Let 
$\hat h_2$ be the \mbox{piecewise} constant function on $[0,1]$ whose 
value on each interval $I_j$ is equal to $h_2(\frac{j-1}{N})$. Then
\begin{eqnarray*}
\sigma&=&
\int_0^1(h_2(t)-\hat h_2(t))\cdot(\dot\gamma(t)-\dot\eta(t))\,dt\\
&\,&\,\,+\int_0^1\hat h_2(t)\cdot(\dot\gamma(t)-\dot\eta(t))\,dt\,.
\end{eqnarray*}
Since $\gamma$ and $\eta$ belong to $\aof{C_2(\hat r)}$, 
the estimate
\begin{equation}\label{estima-d-dada}
\|\dot\gamma(t)-\dot\eta(t)\|\le 2\hat r
\end{equation}
is true for almost all $t$. Our choice of $N$ implies the estimate
\begin{equation}\label{estima-d-dodo}
\|h_2(t)-\hat h_2(t)\|\le\frac{\delta}{4\theta_0}=\frac{\delta r}{4\hat r}\,.
\end{equation}
Then (\ref{estima-d-dada}) and (\ref{estima-d-dodo}) imply
\begin{equation}\label{estima-d}
\Biggl\|\int_0^1(h_2(t)-\hat h_2(t))\cdot(\dot\gamma(t)-\dot\eta(t))\,dt\Biggr\|
\le \frac{\delta r}{2}\,.
\end{equation}
Finally, 
\begin{equation}\label{estima-e}
\int_0^1\hat h_2(t)\cdot(\dot\gamma(t)-\dot\eta(t))\,dt=0\,,
\end{equation}
because $\hat h_2$ is constant on each interval $I_j$, and 
$\dot\eta$ is also constant on $I_j$, and equal to the average
of $\dot\gamma$ over $I_j$.

If we combine (\ref{estima-za}), (\ref{estima-zb}), (\ref{estima-zc}), 
(\ref{estima-a}), (\ref{estima-b}), (\ref{estima-c}), (\ref{estima-d}), 
and (\ref{estima-e}), we end up with the bound
\begin{equation}\label{estima-f}
\|v\|\le\Big(\bar\psi+\frac{\delta}{2}\Big)r\,,
\end{equation}
where
\begin{eqnarray*}
\bar\psi&=&\theta\delta_2+(\kappa_2+\delta_2)\delta_1
+2\delta_1\Biggl(1+\frac{1+\bar k}{\rho}\Biggr)(\kappa_2+\delta_2)\\
&=&(1+\bar k)\Biggl(\kappa_1+2\delta_1\Big(1+\frac{1}{\rho}\Big)\Biggr)
\delta_2+(\kappa_2+\delta_2)\delta_1\\
&\,&+2\delta_1\Biggl(1+\frac{1+\bar k}{\rho}\Biggr)
(\kappa_2+\delta_2)\\
&=&\Psi(\delta_1,\delta_2)\\
&\le&\frac{\delta}{2}\,. 
\end{eqnarray*}
The proof of (F3) is thus complete.}

\omitlong{\skipopag}

In view of (F1), (F2) and (F3), our conclusion will follow if we prove that 
$G$ is regular.
To prove the regularity of $G$, we express $G$ as a composite of
regular maps.

We define
\begin{eqnarray*}
\mathcal{U}&=&\czer{[0,1]}{\rr^{n_2\times n}}\,,\\
\hat\mathcal{U}&=&\Big\{\,h_1\in\mathcal{U}:\|h_1(t)\|\le\kappa_1+
\delta_1\,\,\,
{\rm for\,\,\,all}\,\,\,t\in[0,1]\,\Big\}\,,\\
\mathcal{Y}&=&\czer{[0,1]}{\rr^{n_2}}\,,\\
\mathcal{B}&=&\Big\{\,\beta\in\mathcal{Y}:(\kappa_1+2\delta_1)
\|\beta(t)-\beta(s)\|\le|t-s|\\
&\,&\,\,\,\,\,\,\,\,\,\,{\rm whenever}\,\,\,t,s\in[0,1]\,\Big\}\,,\\
\tilde\mathcal{Y}&=&(\rr^{n_2})^N\,,\\
\mathcal{Z}&=&\Big\{\,\eta\in\mathcal{Y}:\eta\,\,\,{\rm is
\,\,\,linear\,\,\,on}\,\,\,I_j\,\,\,{\rm for\,\,\,}j=1,\ldots,N\,\Big\}\,,\\
\tilde\mathcal{Z}&=&\mathcal{Z}\cap\aof{C_2(r_2)}\,,\\
\mathcal{W}&=&\czer{[0,1]}{\rr^{m\times n_2}}\,,\\
\hat\mathcal{W}&=&\Big\{\,h_2\in\mathcal{W}:\|h_2(t)\|\le\kappa_2+\delta_2\,\,\,
{\rm for\,\,\,all}\,\,\,t\in[0,1]\,\Big\}\,,\\
\mathcal{Q}&=&\czer{[0,1]}{\rr^{m\times n}}\,,\\
\mathcal{X}_1&=&\aof{C(r)}\times\hat\mathcal{U}\times\rr^{n_2}\,,\\
\mathcal{X}_2&=&\aof{C(r)}\times\hat\mathcal{U}\times\rr^{n_2}
\times\mathcal{B}\times\mathcal{Y}\times\mathcal{Z}\,,\\
\mathcal{X}_3&=&\aof{C(r)}\times\hat\mathcal{U}\times\rr^{n_2}
\times\mathcal{B}\times\mathcal{Y}\times\mathcal{Z}\times\tilde\mathcal{Y}\,,
\end{eqnarray*}

\noindent We then let
$\Gamma_1:\aof{C(r)}\tto\mathcal{X}_1$
be the set-valued map that sends $\alpha\in\aof{C(r)}$ to the set
$\{\alpha\}\times G_1(\alpha)$, so that
$$
\Gamma_1(\alpha)\!=\!\Big\{(\alpha,h_1,v_1)\!:\!(h_1,v_1)\!\in\!
G_1(\alpha)\Big\}\,\,{\rm if}\,\,\alpha\!\in\!\aof{C(r)}.
$$
Then
\begin{equation}\label{reguof1}
\Gamma_{1}\in\regu{\aof{C(r)}}{\mathcal{X}_1}\,,
\end{equation}
because of the identity
$\Gamma_1=(\id_{\aof{C(r)}}\times G_1)\circ\Delta_1$,
where $\Delta_1:\aof{C(r)}\to\aof{C(r)}\times\aof{C(r)}$ is the diagonal map
(i.e., the map that sends $\alpha\in\aof{C(r)}$ to the pair 
$(\alpha,\alpha)$). 

\omitshort{
We then let $\Gamma_2:\mathcal{X}_1\to\mathcal{X}_2$
be the ordinary map that sends each triple $(\alpha,h_1,v_1)\in\mathcal{X}_1$ 
to the  6-tuple 
$$
(\alpha,h_1,v_1,\beta,\gamma,\eta)\in\mathcal{X}_2\,,
$$
where 
\begin{eqnarray*}
\beta&=&h_1*\alpha+\xi_{v_1}\,,\\
\gamma(t)&=&\Pi(\beta(t))
+\frac{\omega t}{\rho}\bar w\quad{\rm for}\quad t\in[0,1]\,,
\end{eqnarray*}
and
\begin{equation}\label{incluig-gamma-uu-new}
\eta(t)=
(j-Nt)\gamma\Big(\frac{j-1}{N}\Big)+(Nt+1-j)\gamma\Big(\frac{j}{N}\Big) 
\end{equation}
whenever $\frac{j-1}{N}\le t\le \frac{j}{N}$, $j\in\nn$, and $1\le j\le N\,$.
}

\omitlong{
We then let $\Gamma_2:\mathcal{X}_1\to\mathcal{X}_2$
be the ordinary map that sends each triple $(\alpha,h_1,v_1)\in\mathcal{X}_1$ 
to the  6-tuple 
$(\alpha,h_1,v_1,\beta,\gamma,\eta)\in\mathcal{X}_2$,
where $\beta=h_1*\alpha+\xi_{v_1}$,
\mbox{$\gamma(t)=\Pi(\beta(t))
+\frac{\omega t}{\rho}\bar w\quad{\rm for}\quad t\in[0,1]$,}
and
\begin{equation}\label{incluig-gamma-uu-new}
\eta(t)=
(j-Nt)\gamma\Big(\frac{j-1}{N}\Big)+(Nt+1-j)\gamma\Big(\frac{j}{N}\Big) 
\end{equation}
whenever $\frac{j-1}{N}\le t\le \frac{j}{N}$, $j\in\nn$, and $1\le j\le N\,$.
}


We then let $\Gamma_3^0:\mathcal{Z}\to\tilde\mathcal{Y}$ 
be the map that sends each $\eta\in\mathcal{Z}$ to the $N$-tuple
\begin{equation}\label{incluig-gamma-a-oink-u-eta}
\Gamma_3^0(\eta)=(u_1,\ldots,u_N)\in \tilde\mathcal{Y}\,,
\end{equation}
where 
\begin{equation}\label{incluig-gamma-a-oink-u-eta-u}
\!\!\!\!\!\!\!\!\!
u_{j}=N\Biggl(\,\eta\Big(\frac{j}{N}\Big)-
\eta\Big(\frac{j-1}{N}\Big)\,\Biggr)\,\,\,{\rm for}\,\,\,j=1,\ldots,N\,.
\end{equation}
Next, we let $\Gamma_3:\mathcal{X}_2\to\mathcal{X}_3$ 
be the map that sends each 6-tuple $(\alpha,h_1,v_1,\beta,\gamma,\eta)$
to the 7-tuple 
\begin{equation}\label{incluig-gamma-a-oink-u-eta-ripp}
\Gamma_3(\alpha,h_1,v_1,\beta,\gamma,\eta)=
(\alpha,h_1,v_1,\beta,\gamma,\eta,\Gamma_3^0(\eta))\,.
\end{equation}

Next, we define $\tilde\mathcal{K}_0=C_2(\hat r)^N$, 
and let $\tilde\mathcal{K}$ be the set of all 
$(u_1,\ldots,u_N)\in\tilde\mathcal{K}_0$ such that 
$N(u_1+\ldots+u_N)=w_0+\frac{\omega}{\rho}\bar w$
for some $w_0\in C_2$. Then $\tilde\mathcal{K}_0$ and $\tilde\mathcal{K}$ are 
compact convex subsets of $\tilde\mathcal{Y}$. 

We let $\Gamma_4^0$ be a continuous retraction from $\tilde\mathcal{Y}$ onto
$\tilde\mathcal{K}$, and define
\omitshort{
$$
\mathcal{X}_4=\aof{C(r)}\times\hat\mathcal{U}\times\rr^{n_2}
\times\mathcal{B}\times\mathcal{Y}\times\mathcal{Z}\times\tilde\mathcal{K}\,.
$$}
\omitlong{
$\mathcal{X}_4=\aof{C(r)}\times\hat\mathcal{U}\times\rr^{n_2}
\times\mathcal{B}\times\mathcal{Y}\times\mathcal{Z}\times\tilde\mathcal{K}$.}
We then let 
$\Gamma_4:\mathcal{X}_3\to\mathcal{X}_4$ 
be the map that sends each 7-tuple 
$(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u})$
to the 7-tuple 
\begin{equation}\label{incluig-gamma-a-oink-u-eta-ripp-a}
\!\!\!\!\!\!\!
\Gamma_4(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u})\!=\!
(\alpha,h_1,v_1,\beta,\gamma,\eta,\Gamma_4^0({\bf u}))\,.
\end{equation}

We now define a set-valued map 
$\Gamma_5^0:\tilde\mathcal{K}\tto\tilde\mathcal{Y}$, 
by letting $\Gamma_5^0(u_1,\ldots,u_N)$ be, 
if $(u_1,\ldots,u_N)\in\tilde\mathcal{K}$, 
he set of all $N$-tuples $(c_1,\ldots,c_N)$ that satisfy,
for $j=1,\ldots,N$, the conditions
\begin{eqnarray}
u_j-c_j&\in& C_2\phantom{\bar k\|u_{\alpha,h_1,v_1,j}^Na\|}
\,,\label{incluig-gamma-a-oink-u-aaa-rep-ripo}\\
c_1+\cdots+c_N&=&\frac{N\omega}{\rho}\bar w\,,
\label{incluig-gamma-a-oink-u-bbb-rep-ripo}\\
\|c_1\|+\cdots+\|c_N\|&\le&\frac{N\omega\bar k}{\rho}\,,
\label{incluig-gamma-a-oink-u-ccc-rep-ripo}\\
\|c_j\|&\le&\bar k\|u_j\|\phantom{C_{2,\alpha,h_1,v_1,j}}
\,.\label{incluig-gamma-a-oink-u-ddd-rep-ripo}
\end{eqnarray}
Then $\Gamma_5^0$ has convex values and a compact graph. 
\mbox{Moreover}, (ECP) implies  that the values of $\Gamma_5^0$ 
are nonempty.  It then follows from 
Theorem 5.2 of \cite{sussother} that 
\begin{equation}\label{reguof123-c}
\Gamma_5^0\in\regu{\tilde\mathcal{K}}{\tilde\mathcal{Y}}\,. 
\end{equation}
We then  define  
$$
\mathcal{X}_5=\aof{C(r)}\times\hat\mathcal{U}\times\rr^{n_2}
\times\mathcal{B}\times\mathcal{Y}\times\mathcal{Z}\times\tilde\mathcal{K}
\times\tilde\mathcal{Y}\,,
$$
and let $\Gamma_5:\mathcal{X}_4\tto\mathcal{X}_5$ be the set-valued map that 
sends each 7-tuple 
$(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u})\in\mathcal{X}_4$
to the set 
\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\Gamma_5(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u})\nonumber\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\,\,=\Big\{\,(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u},{\bf c}):{\bf c}
\in\Gamma_5^0
({\bf u})\,\Big\}\subseteq\mathcal{X}_5\,.\label{incluig-gamma-a-oink-u-eta-ripp-aa}
\end{eqnarray}
Then 
\begin{equation}\label{reguof123-cc-o-u}
\Gamma_5=\Big(\id_{\mathcal{X}_4}\times\Gamma_5\Big)\circ\Delta_2\,,
\end{equation}
where 
$\Delta_2:\mathcal{X}_4\to\mathcal{X}_4\times\tilde\mathcal{K}$ 
is the map that sends a 7-tuple $(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u})$
to the 8-tuple $(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u},{\bf u})$.
It follows from (\ref{reguof123-c}) and (\ref{reguof123-cc-o-u}) that
\omitshort{
\begin{equation}\label{reguof123-cc-o}
\Gamma_5\in\regu{\mathcal{X}_4}{\mathcal{X}_5}\,.
\end{equation}}
\omitlong{
$\Gamma_5\in\regu{\mathcal{X}_4}{\mathcal{X}_5}$.}


Next, define       
$$
\mathcal{X}_6=\aof{C(r)}\times\hat\mathcal{U}\times\rr^{n_2}
\times\mathcal{B}\times\mathcal{Y}\times\mathcal{Z}\times\tilde\mathcal{K}
\times\tilde\mathcal{Y}\times\mathcal{Z}\times\mathcal{Z}\,,
$$
and let $\Gamma_6:\mathcal{X}_5\to\mathcal{X}_6$ be the ordinary map that 
sends each 8-tuple
$(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u},{\bf c})\in\mathcal{X}_5$
to the 10-tuple

\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\Gamma_6(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u},{\bf c})\\
&\,&=(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u},{\bf c},\mu_{{\bf c}},
\eta-\mu_{{\bf c}})\in\mathcal{X}_6\,.
\end{eqnarray*}

\medbreak\noindent We now observe that the real linear space 
$\mathcal{Z}$ is finite-dimensional,
and $\tilde\mathcal{Z}$ is a nonempty compact convex 
\mbox{subset} of $\mathcal{Z}$.
Let $\Gamma_7^0$ be a continuous retraction from $\mathcal{Z}$ onto
$\tilde\mathcal{Z}$. Define
$$
\mathcal{X}_7=\aof{C(r)}\times\hat\mathcal{U}\times\rr^{n_2}
\times\mathcal{B}\times\mathcal{Y}\times\mathcal{Z}\times\tilde\mathcal{K}
\times\tilde\mathcal{Y}\times\mathcal{Z}\times\tilde\mathcal{Z}\,,
$$
and let $\Gamma_7:\mathcal{X}_6\to\mathcal{X}_7$ be 
the \omitshort{ordinary} map that sends each 10-tuple
$(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u},{\bf c},\mu,\zeta)\in\mathcal{X}_6$
to the 10-tuple
\begin{eqnarray}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\Gamma_7(\alpha,h,v,\beta,\gamma,\eta,{\bf u},{\bf c},\mu,\zeta)\nonumber\\
&\,&\,\,\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
=(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u},{\bf c},\mu,\Gamma_7^0(\zeta))
\in\mathcal{X}_7\,.\label{incluig-gamma-a-oink-u-eta-ripp-aaaa}
\end{eqnarray}

Next, we let $\Gamma_8^0$ be the set-valued map 
$\tilde\mathcal{Z}\tto\hat\mathcal{W}\times\rr^m$ that sends 
$\zeta\in\tilde\mathcal{Z}$ to the set $G_2(\zeta)\subseteq\hat\mathcal{W}\times\rr^m$. 
\omitshort{Then
$$
\Gamma_8^0=G_2\circ\iota_{_{\aof{C_2(r_2)},\tilde\mathcal{Z}}}\,,
$$
where $\iota_{_{\aof{C_2(r_2)},\tilde\mathcal{Z}}}$ is the inclusion map
from $\tilde\mathcal{Z}$ to $\aof{C_2(r_2)}$. Therefore
$\Gamma_8^0\in\regu{\tilde\mathcal{Z}}{\hat\mathcal{W}\times\rr^m}$.}
\omitlong{Then
$\Gamma_8^0=G_2\circ\iota_{_{\aof{C_2(r_2)},\tilde\mathcal{Z}}}$,
where $\iota_{_{\aof{C_2(r_2)},\tilde\mathcal{Z}}}$ is the inclusion map
from $\tilde\mathcal{Z}$ to $\aof{C_2(r_2)}$. So
$\Gamma_8^0\in\regu{\tilde\mathcal{Z}}{\hat\mathcal{W}\times\rr^m}$.}

\omitlong{\skipopag}
Define $\mathcal{X}_8$ to be the product

\noindent$\aof{C(r)}\times\hat\mathcal{U}\times\rr^{n_2}
\times\mathcal{B}\times\mathcal{Y}\times\mathcal{Z}\times\tilde\mathcal{K}
\times\tilde\mathcal{Y}\times\mathcal{Z}\times\tilde\mathcal{Z}\times
\hat\mathcal{W}\times\rr^m$,

\noindent and let $\Gamma_8:\mathcal{X}_7\tto\mathcal{X}_8$ be the 
set-valued map that sends each 10-tuple
$(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u},{\bf c},\mu,\zeta)\in\mathcal{X}_7$
to the set 
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!
\Gamma_8(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u},{\bf c},\mu,\zeta)\\
\,\!\!\!\!\!
&\!=\!&\!\!\!\!\Big\{\,(\alpha,h_1,v_1,\beta,\gamma,\eta,
{\bf u},{\bf c},\mu,\zeta)\,\Big\}\times G_2(\zeta)\subseteq\mathcal{X}_8\\
\,\!\!\!\!\!
&\!=\!&\!\!\!\!\Big\{(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u},
{\bf c},\mu,\zeta,h_2,v_2)\!:\!
(h_2,v_2)\!\in\!G_2(\zeta)\Big\}\,.
\end{eqnarray*}
It is then clear that
$\Gamma_8=\Big(\id_{\mathcal{X}_7}\times G_2\Big)\circ\Delta_3$,
where \mbox{$\Delta_3:\mathcal{X}_7\to\mathcal{X}_7\times\tilde\mathcal{Z}$}
is the map that sends a 10-tuple
$(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u},{\bf c},\mu,\zeta)\in\mathcal{X}_7$
to the 11-tuple
$$
(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u},{\bf c},\mu,\zeta,\zeta)\in\mathcal{X}_7
\times\tilde\mathcal{Z}\,.
$$
\omitlong{Therefore $\Gamma_8\in\regu{\mathcal{X}_7}{\mathcal{X}_8}$.}

\omitshort{Therefore
\begin{equation}\label{regugam8}
\Gamma_8\in\regu{\mathcal{X}_7}{\mathcal{X}_8}\,.
\end{equation}}

Finally, we define an ordinary map 
$\Gamma_9:\mathcal{X}_8\to\mathcal{Q}\times\rr^m$ 
by letting
\begin{eqnarray*}
&\,&\Gamma_9(\alpha,h_1,v_1,\beta,\gamma,\eta,{\bf u},{\bf c},\mu,\zeta,h_2,v_2)\\
&\,&=v_2+\Big(\int_0^1h_2(t)\,dt\Big)\cdot v_1
-\Big(h_2*(\beta-\zeta)\Big)(1)\,.
\end{eqnarray*}

\omitshort{
We have thus defined nine set-valued maps
\begin{eqnarray*}
\Gamma_{1}&\in&\regu{\aof{C(r)}}{\mathcal{X}_1}\,,\\
\Gamma_2&\in&\czer{\mathcal{X}_1}{\mathcal{X}_2}\,,\\
\Gamma_3&\in&\czer{\mathcal{X}_2}{\mathcal{X}_3}\,,\\
\Gamma_4&\in&\czer{\mathcal{X}_3}{\mathcal{X}_4}\,,\\
\Gamma_5&\in&\regu{\mathcal{X}_4}{\mathcal{X}_5}\,,\\
\Gamma_6&\in&\czer{\mathcal{X}_5}{\mathcal{X}_6}\,,\\
\Gamma_7&\in&\czer{\mathcal{X}_6}{\mathcal{X}_7}\,,\\
\Gamma_8&\in&\regu{\mathcal{X}_7}{\mathcal{X}_8}\,,\\
\Gamma_9&\in&\czer{\mathcal{X}_8}{\mathcal{Q}\times\rr^m}\,.
\end{eqnarray*}}
\omitlong{
We \ have \ thus \ defined \ nine $\,$set-valued $\,$maps
\mbox{$\Gamma_{1}\!\in\!\regu{\aof{C(r)}}{\mathcal{X}_1}$,
$\Gamma_2\!\in\!\czer{\mathcal{X}_1}{\mathcal{X}_2}$},
$\Gamma_3\!\in\!\czer{\mathcal{X}_2}{\mathcal{X}_3}$,
$\Gamma_4\in\czer{\mathcal{X}_3}{\mathcal{X}_4}$,
$\Gamma_5\in\regu{\mathcal{X}_4}{\mathcal{X}_5}$,
$\Gamma_6\in\czer{\mathcal{X}_5}{\mathcal{X}_6}$,
$\Gamma_7\in\czer{\mathcal{X}_6}{\mathcal{X}_7}$,
$\Gamma_8\in\regu{\mathcal{X}_7}{\mathcal{X}_8}$,
$\Gamma_9\in\czer{\mathcal{X}_8}{\mathcal{Q}\times\rr^m}$.}
\omitshort{
(The regularity of $\Gamma_1$, $\Gamma_5$ and $\Gamma_8$ has already been proved, cf.
(\ref{reguof1}), (\ref{reguof123-cc-o}), (\ref{regugam8}). The continuity of
$\Gamma_2$, $\Gamma_3$, $\Gamma_4$, $\Gamma_6$ and $\Gamma_7$ follows 
trivially from 
their \mbox{definitions}. The continuity of $\Gamma_9$ follows from Lemma
\ref{jointcontlemma} because, if $\{\xi^j\}_{j\in\nn}$ is a sequence
of points in $\czer{\mathcal{X}_8}{\mathcal{Q}\times\rr^m}$ that converges in 
$\czer{\mathcal{X}_8}{\mathcal{Q}\times\rr^m}$ to a limit $\xi^\infty$, and we write
$$
\xi^j=(\alpha^j,h_1^j,v_1^j,\beta^j,\gamma^j,\eta^j,{\bf u}^j,{\bf c}^j,
\mu^j,\zeta^j,h_2^j,v_2^j)\,,
$$
then the $\beta_j$ belong to $\mathcal{B}$ and the 
$\zeta_j$ belong to $\tilde\mathcal{Z}$, so the sequence 
$\{\beta^j-\zeta^j\}_{j\in\nn}$ is uniformly Lipschitz, and then 
the facts that $\beta^j\to\beta^\infty$, $\zeta^j\to\zeta^\infty$, 
$h_2^j\to h_2^\infty$, imply that 
$h_2^j*(\beta^j-\zeta^j)\to h_2^\infty*(\beta^\infty-\zeta^\infty)$.)}

\omitshort{
On the other hand,
$$
G=\Gamma_9\circ\Gamma_8\circ\Gamma_7\circ\Gamma_6\circ\Gamma_5\circ\Gamma_4
\circ\Gamma_3\circ\Gamma_2\circ\Gamma_1\,.
$$
Therefore $G$ is regular, and our proof is complete. \proofend}

\omitlong{
Since 
$G=\Gamma_9\circ\Gamma_8\circ\Gamma_7\circ\Gamma_6\circ\Gamma_5\circ\Gamma_4
\circ\Gamma_3\circ\Gamma_2\circ\Gamma_1$, it follows that
$G$ is regular, and our proof is complete. \proofend}



\medbreak
\ssection{The open mapping theorem}\label{sec-open-pigds}

\noindent We now show that \pigds{} have the directional open mapping 
property.

\begin{theorem}
Assume that $n,m\in\zz_+$, $F\!\in\!\mathcal{SVM}(\rr^n,\rr^m)$, $\bar
w\!\in\!\rr^m$,
\mbox{$C$ is a polyhedral} \mbox{convex} cone in $\rr^n$, 
$\Lambda$ \mbox{belongs} to $PIGD(F;0,0;C)$, and 
\mbox{$\bar w\!\in\!\bigcap_{L\in\Lambda}{\rm Int}(LC)$}. Then 
there \mbox{exists} a closed convex cone $D$ in $\rr^m$ such that
$\bar w\in{\rm Int}(D)$, having the property that
for every $\delta\!\in]0,\infty[$ there exists
an $\eps(\delta)\!\in]0,\infty[$ such that
\fbox{$D\!\cap\!\{y\!\in\!\rr^m\!:\!\|y\|\!\le\!\eps(\delta)\}\!\subseteq\!
F(C\!\cap\!\{x\!\in\!\rr^n\!:\!\|x\|\!\le\!\delta\})$}.
\end{theorem}

\medbreak\noindent {\bem Proof.\/}  Let us assume that $\bar w\neq 0$. 
Pick a closed convex cone $\hat D$ in $\rr^n$ such that
$\bar w\in{\rm Int}(\hat D)$, a compact neighborhood
$\Lambda'$ of $\Lambda$ such that
$\hat D\subseteq LC$ for every $L\in\Lambda'$,
and a continuous map $\Lambda'\times\hat D\ni(L,y)\mapsto\eta(L,y)\in C$
which is positively homogeneous of degree $1$ with respect
to $y$ and such that $L\cdot\eta(L,y)=y$ for all 
$(L,y)\in\Lambda'\times\hat D$. Pick $\theta$ such that
$0<\theta<\|\bar w\|$ and the ball $\{y\in\rr^m:\|y-\bar w\|\le 2\theta\}$ 
is entirely
contained in $\hat D$. Let $D$ be the smallest closed convex cone 
that contains the ball  $\{y\in\rr^m:\|y-\bar w\|\le \theta\}$. Then 
$D$ satisfies:
\begin{equation}\label{itskater}
\Big(y\in D\wedge\|z\|\le\frac{\theta\|y\|}{\|\bar w\|+\theta}\Big)
\Rightarrow y+z\in\hat D\,.
\end{equation}
(Indeed, if $y=0$ then $z=0$, so $y+z=0\in\hat D$.
Assume that $y\neq 0$. Then we can write 
$y=su$ with $s>0$ and $\|u-\bar w\|\le\theta$. 
Then $\|u\|\le\|\bar w\|+\|u-\bar w\|\le\|\bar w\|+\theta$.
Therefore $\|y\|\le s(\|\bar w\|+\theta)$, so
$s\le\frac{\|y\|}{\|\bar w\|+\theta}$. Furthermore,
$y+z=s\tilde u$, where $\tilde u=u+\frac{z}{s}$. Then
$$
\|\tilde u-\bar w\|=\|u-\bar w+\frac{z}{s}\|
\le\|u-\bar w\|+\frac{\|z\|}{s}
\le\theta+\frac{\|z\|}{\|y\|}(\|\bar w\|+\theta)\,,
$$
so $\|\tilde u-\bar w\|\le 2\theta$, and then $y+z\in\hat D$.)

Let $M=\sup\{\|\eta(L,y)\|:L\in\Lambda',y\in\hat D,\|y\|\le 1\}$.
Then $\|\eta(L,y)\|\le M\|y\|$ whenever 
$L\in\Lambda'$ and $y\in\hat D$.

\vspace{1mm}
Fix a $\delta\!\in]0,\infty[$. Let $\delta'$ be such that
$0<\delta'\le\delta$, $\Lambda^{\delta'}\subseteq\Lambda'$, 
$2M\delta'\le\frac{\theta}{\|\bar w\|+\theta}$. 
Then choose $R\in\rr$ such that $R>0$ and a family 
$\{G_r:0<r\le R\}$ of regular set-valued maps
$G_r:\aof{C(r)}\tto\czer{[0,1]}{\rr^{m\times n}}\times\rr^m$
such that
\ben
\medbreak\item[(a)] $h(t)\in\Lambda^{\delta'}$ and $\|v\|\le\delta' r$ 
whenever 
\mbox{$\alpha\in\aof{C(r)}$}, $(h,v)\in G_r(\alpha)$,  $t\in[0,1]$,
\item[(b)] $\gr(\Phi_{G_r})\subseteq\gr(F)$.
\een
By making $R$ smaller, if necessary, we may assume that
$R\le\delta$.


We choose $\eps$ such that $2M\eps<R$, and show that this choice
of $\eps$ satisfies our requirements. 

Fix $y$ such that $0<\|y\|\le\eps$. Let $\rho=\|y\|$, and choose
$r=2M\rho$.  Then $r<R$. We will show that there is an $x\in C$ 
such that $\|x\|\le\delta$ and  $y\in F(x)$. For this purpose,
it suffices to find a triple $(\alpha,h,v)\in\gr(G_r)$ such that
$v+\int_0^1h(t)\cdot\dot\alpha(t)\,dt=y$ and $\|\alpha(1)\|\le\delta$.
The first equality, in turn, will follow if $\alpha$ satisfies
\begin{equation}\label{skate}
h(t)\cdot\dot\alpha(t)=y-v\quad{\rm for\,\,a.e.}\quad t\,, 
\end{equation}
as well as $\|\alpha(1)\|\le\delta$. If $y-v\in D$, then (\ref{skate})
will follow if $\dot\alpha(t)=\eta(h(t),y-v)$ for a.e. $t$, i.e.,
if 
\begin{equation}\label{skater}
\alpha(t)=\int_0^t\eta(h(s),y-v)\,ds\quad{\rm for\,\,all}\quad t\,.
\end{equation}
Let $\Sigma$ be the set-valued map on $\mathcal{A}(C(r))$ that asigns to
each $\alpha\in\mathcal{A}(C(r))$ the set $\Sigma(\alpha)$ 
of all paths $\beta=\beta_{\alpha,h,v}$ for all
$(h,v)\in G_r(\alpha)$, where 
$\beta_{\alpha,h,v}(t)=\int_0^t\eta(h(s),y-v)\,ds$. Then $\Sigma$ 
is well defined and 
takes values in $\mathcal{A}(C(r))$. ({\em Proof.}
Let $\alpha\in\mathcal{A}(C(r))$ and $(h,v)\in G_r(\alpha)$.
Then $\|v\|\le\delta'r=2M\delta'\rho=2M\delta'\|y\|
\le\frac{\theta\|y\|}{\|\bar w\|+\theta}$. Therefore 
(\ref{itskater}) implies that $y-v\in\hat D$. Since 
$h(t)\in\Lambda^{\delta'}\subseteq\Lambda'$  for each $t$,
$\eta(h(t),y-v)$ is defined for each $t$. Since 
the map $t\mapsto\eta(h(t),y-v)$ is continuous, 
$\beta_{\alpha,h,v}$ is well \mbox{defined}. Moreover, 
$\dot\beta_{\alpha,h,v}(t)=\eta(h(t),y-v)\in C$.
On the other hand, 
\mbox{$\|\dot\beta_{\alpha,h,v}(t)\|=\|\eta(h(t),y-v)\|\le
M\|y-v\|$}. But $\|y-v\|\le\|y\|+\|v\|\le\rho+\delta'r$,
and this implies that
$\|\dot\beta_{\alpha,h,v}(t)\|\le M\rho+M\delta'r$.
But $M\rho=\frac{r}{2}$, and we know that $M\delta'r\le 
\frac{\theta r}{2(\|\bar w\|+\theta)}<\frac{r}{2}$. So 
$\|\dot\beta_{\alpha,h,v}(t)\|\le r$. \mbox{Therefore}
$\dot\beta_{\alpha,h,v}(t)\in C(r)$. Hence
$\beta_{\alpha,h,v}\in\mathcal{A}(C(r))$.) 

It is easy to see that $\Sigma$ is regular. Hence $\Sigma$ is a regular 
map from the compact convex set $\mathcal{A}(C(r))$ to itself. 
By the obvious extension of Schauder's fixed point theorem
to regular maps, $\Sigma$ has a fixed point $\alpha_*$. 
Then, for some $(h,v)\in G_r(\alpha_*)$, we have 
$h(t)\cdot\dot\alpha_*(t)=y-v$ for almost all $t$. Hence
if we let $x=\alpha(1)$, we have $y=v+\int_0^1h(t)\cdot\dot\alpha_*(t)\,dt$,
so that $y\in\Phi_{G_r}(x)$, and then $y\in F(x)$. Finally
the fact that $\alpha\in\mathcal{A}(C(r))$ implies that
$\|x\|\le r=2M\rho\le 2M\eps<R\le\delta$. This completes the proof.\proofend

%\bibliographystyle{amsalpha}
%\begin{thebibliography}{A}

\vspace{-3mm}
\begin{thebibliography}{99}

\bibitem{sussother}{Sussmann,~H.J., 
``Warga derivate containers and other generalized differentials.''  
Submitted for publication in \textit{Proc. 41st IEEE Conf. Decision
and Control, Las Vegas, \mbox{December} 2002.}}
 
\end{thebibliography}

\omitlong{\skipopag}

\end{document}
\ssection{A property of polyhedral cones}

Let $C$ be a closed convex cone in a finite-dimensional normed linear
space $X$. Let $S_C$ be the linear subspace of $X$ spanned by $C$,
so the interior ${\rm Int}_{S_C}(C)$ of $C$ relative to $S_C$ is
nonempty.

Suppose we are trying to add several vectors $u_1 ,\ldots,u_m$
belonging to $C$ so as to obtain a vector $w\in C$, but instead of
achieving this desired result we produce a sum $\tilde w$, which is
``larger'' than the target 
\mbox{value} $w$ in the sense of the partial ordering
induced by the cone, i.e., such that $\tilde w=w+e$ for some $e\in
C$. We would like to correct this error $e$ by subtracting 
\mbox{correction}
terms $c_i$---not necessarily belonging to $C$---to the vectors $u_i$,
in such a way that the new vectors $v_i=u_i-c_i$ belong to $C$ and add up to
$w$. Moreover, we want to be able to do this while keeping the total
error norm---i.e., the sum $E=\|c_1\|+\ldots+\|c_m\|$---bounded by a
fixed constant $\bar k$, independently of $m$, the $u_i$'s, and
$w$. And, in addition, we want the $c_i$'s to be bounded by the
$u_i$'s.  Equivalently, we want to subtract off the ``error'' $e$ from
the sum $u_1+\ldots+u_m$ by expressing $e$ as a sum $e=c_1+\ldots+c_m$
in such a way that after subtracting each $c_i$ from its corresponding
$u_i$ the resulting vectors still belong to $C$, and we want to do this
with a bound on the sum $E$, and with the the $c_i$'s bounded by
the $u_i$'s.

Whether this ``error correction'' is possible for a 
\mbox{particular} choice
of $e\in C$ is a property---the ``error-correcting property,''
abbreviated ``ECP''---of the pair $(C,e)$.  Naturally, the ECP holds
if $e=0$ but, as we shall see, it is important for the ECP to hold for
some $e\in{\rm Int}_{S_C}(C)$, so the only case when $e=0$ is an
acceptable choice is when $C$ is a linear subspace, i.e., $C=S_C$.

The purpose of this subsection is to show that $e$ can be chosen to
belong to ${\rm Int}_{S_C}(C)$ if $C$ is a polyhedral cone.  We
will also exhibit an example showing that for more general cones it
may happen that there do not exist any $e\in C\backslash\{0\}$
such that $(C,e)$ has the ECP.

It will be clear, both from the proof of the positive result for
polyhedral cones, and from the counterexample for a cone which is not
polyhedral, that the crucial property of polyhedral cones, for the
purpose of establishing the ECP, is the fact that {\bem the set of all possible
tangent cones $T_xC$, as $x$ varies over $C$, is finite.}

First, we give a formal definition of the ECP.

\begin{definition}\label{defofecp}
Let $X$ be a finite-dimensional normed linear space, 
and let $C$ be a closed convex cone in $X$. Let $e\in C$. Then
\ben
\item[1.] If $\bar k\in\rr$, $\bar k\ge 0$, we say that the pair 
$(C,e)$ has the {\bem error-correction property}---abbr. 
``ECP''---{\bem with constant $\bar k$} if
\medbreak
\ben
\item[(ECP)] for every $m\in\nn$, every $w\in C$, and every 
$m$-tuple $(u_1,\ldots,u_m)$ of members of $C$ such that 
$$
u_1+\ldots+u_m=w+e
$$
there exist vectors $c_1,\ldots,c_m$ in $X$ such that 
\een\een
\begin{eqnarray}
u_i\!-\!c_i&\!\in\!&C\phantom{\bar k\|u_i\|}\,\,{\rm for}\,\,
i\!=\!1,\ldots,m\,,\label{ecp-eq-a}\\
c_1\!+\!\ldots\!+\!c_m&\!=\!&e\,,\label{ecp-eq-b}\\
\|c_1\|\!+\!\ldots\!+\!\|c_m\|&\!\le\!&\bar k\,,\label{ecp-eq-c}\\
\|c_i\|&\!\le\!&\bar k\|u_i\|\phantom{C}\,\,{\rm for}\,\,i\!=\!1,\ldots,m\,.
\label{ecp-eq-d}
\end{eqnarray}
\ben
\item[2.] We say that $(C,e)$ has the {\bem error-correction property}---abbr. 
``ECP''---if there exists a constant $\bar k$ such that 
$(C,e)$ has the error-correction property with constant $\bar k$.
\item[3.] We say that $C$ is {\bem ECP-good} if there exists a vector $e$
belonging to the interior ${\rm Int}_{S_C}(C)$ of $C$ relative to the 
linear span $S_C={\rm span}(C)$ such that the pair $(C,e)$ has the 
ECP. \proofend
\een
\end{definition}

\begin{remark}\label{oncorrec-uu}
The ECP property does not depend on the choice of the norm on $X$.
Moreover, {\bem the ECP \mbox{property} is an intrinsic property of
the cone $C$, and is independent of the space $X$ in which $C$ is
\mbox{embedded}.} (\mbox{Precisely}, this means that {\bem $C$ is
ECP-good as a cone in $X$ if and only if it is ECP-good as a cone in
$S_C$}. The proof is straightforward. First, it is trivial that if $C$
is ECP-good as a cone in $S_C$ then it is ECP-good as a cone in
$X$. To establish the reverse 
\mbox{implication} we \mbox{observe} that,
\mbox{whenever} $m$, $w$, $u_1,\ldots,u_m$ are given as in
\mbox{Definition} \ref{defofecp}, if it is possible to choose
$c_1,\ldots,c_m$ in $X$ such that (\ref{ecp-eq-a}), (\ref{ecp-eq-b}),
(\ref{ecp-eq-c}), (\ref{ecp-eq-d}) hold, then the $c_i$ can also be
chosen to belong to $S_C$, since one can \mbox{replace} the $c_i$ by
their projections $\Pi(c_i)$, where $\Pi$ is the \mbox{orthogonal}
\mbox{projection} from $X$ onto $S_C$ relative to some inner
\mbox{product} on $X$.)\proofend
\end{remark}

\begin{remark}\label{oncorrec}
If $X$, $C$, $e$, $w$, $m$, $u_1,\ldots,u_m$ are as in Definition
\ref{defofecp}, then it is always possible to choose 
\mbox{correction} terms
$c_i$ such that $c_1+\ldots+c_m=e$ and $u_i-c_i\in C$. For example,
one can choose 
\mbox{$c_i=u_i-\lambda_iw$}, where the $\lambda_i$ are
arbitrary nonnegative numbers such that
$\lambda_1+\ldots+\lambda_m=1$.  Then, if we let 
\mbox{$v_i=u_i-c_i$}, we
have $v_i=\lambda_iw$, so the $v_i$ belong to $C$ and
\mbox{$v_1+\ldots+v_m=w$}. It follows that $c_1+\ldots+c_m=e$, so
(\ref{ecp-eq-a}) and (\ref{ecp-eq-b}) are true.  This, however, does
not \mbox{suffice} to guarantee that the ECP holds, because the error bounds
(\ref{ecp-eq-c}) and (\ref{ecp-eq-d}) need not be satisfied. \proofend
\end{remark}


It is easy to see that
\begin{proposition}
If $X$ is a finite-dimensional normed linear space,
$C$ is a closed convex cone in $X$, and
$e=0$, then $(C,e)$ has the error-correction property. In 
\mbox{particular},
if $C$ is a linear subspace of $X$, then $C$ is ECP-good.
\end{proposition}
\begin{proof}
If $e=0$, and $w$, $m$, $u_1,\ldots,u_m$ are as in 
\mbox{Definition} \ref{defofecp},
then we can choose $c_i=0$, and all the conditions of the definition
hold, with $\bar k=0$. 

If $C$ is a linear subspace, then $0\in{\rm Int}_{S_C}(C)$, and 
$(C,0)$ has the ECP. Therefore $C$ is ECP-good. 
\end{proof}

\begin{remark}\label{nocorrec}
The following example shows that {\em not all closed convex
cones are ECP-good}.

Let $X=\rr^3$, and let $C$ be the cone 
$$
C=\{\,(x,y,z)\in\rr^3:z\ge 0\wedge z^2\ge x^2+y^2\,\}\,.
$$
Then {\bem $C$ is not ECP-good}.  To see this, we will actually
show that {\bem the pair $(C,e)$ does not have the ECP for any
$e\in C$ such that $e\neq 0$.} 

Fix  $e\in C$ such that $e\neq 0$. Let $e=(\alpha,\beta,\gamma)$, so 
$\gamma\ge 0$, $\gamma^2\ge \alpha^2+\beta^2$, and $(\alpha,\beta,\gamma)\neq(0,0,0)$.
After a rotation of the $x$ and $y$ axes,
we may assume that $\alpha=0$ and $\beta\ge 0$. Then $e=(0,\beta,\gamma)$, 
$\gamma\ge\beta\ge 0$, and $\gamma>0$. 

For $N\in\rr$, $N>0$, let 
$w^N=(2N,0,2N)$, so $w^N\in C$.  Then all the $w^N$  belong to the ray
$$
R=\{\,(r,0,r):r\ge 0\,\}\,.
$$
Clearly, $R\subseteq C$, and $R$ is an  extreme ray
(that is, if $z_1\in C$, $z_2\in C$, and $z_1+z_2\in R$, then 
$z_1\in R$ and $z_2\in R$). 

Let $m=2$, and let 
$$
u_1^N=\Big(N,g_N,N+\frac{\gamma}{2}\Big)\,,\quad
u_2^N=\Big(N,h_N,N+\frac{\gamma}{2}\Big)\,,
$$
where 
$$
g_N=\frac{1}{2}\sqrt{\gamma^2+4\gamma N}\,,\qquad
h_N=\beta-g_N\,.
$$
Then 
\begin{eqnarray*}
N^2+g_N^2&=&N^2+\frac{1}{4}(\gamma^2+4\gamma N)\\
&=&N^2+\gamma N+\frac{\gamma^2}{4}\\
&=&\Big(N+\frac{\gamma}{2}\Big)^2\,,
\end{eqnarray*}
and
\begin{eqnarray*}
N^2+h_N^2&=&N^2+(\beta-g_N)^2\\
&=&N^2+\beta^2+g_N^2-2\beta g_N\\
&=&\Big(N+\frac{\gamma}{2}\Big)^2+\beta^2-2\beta g_N\\
&=&\Big(N+\frac{\gamma}{2}\Big)^2+\beta(\beta-2g_N)\\
&\le&\Big(N+\frac{\gamma}{2}\Big)^2\,,
\end{eqnarray*}
where the last inequality follows because 
$$
0\le\beta\le\gamma\le\sqrt{\gamma^2+4\gamma N}= 2g_N\,.
$$
Therefore $u_1^N\in C$ and $u_2^N\in C$. Moreover, it is clear that
$$
u_1^N+u_2^N=w^N+e\,.
$$
Fix $N$, and let $c_1$, $c_2$ be such that 
$v_1=u_1^N-c_1\in C$, $v_2=u_2^N-c_2\in C$, and $c_1+c_2=e$. 
Then $v_1+v_2=w^N$. Since the ray $R$ is extreme,
the vectors $v_1$ and $v_2$ must belong to $R$.
Therefore, if we write $v_i=(v_{i,1},v_{i,2},v_{i,3})$, 
$c_i=(c_{i,1},c_{i,2},c_{i,3})$, we must have $v_{i,2}=0$, and then
$c_{1,2}=g_N$, $c_{2,2}=h_N$. Therefore 
$$
\|c_1\|+\|c_2\|\ge g_N+|h_N|\,.
$$
Since $N$ is arbitrary, and the quantity 
$g_N+|h_N|$ goes to $+\infty$ as $N\uparrow+\infty$ (because
$\gamma>0$),  the constant $\bar k$ of
Definition \ref{defofecp} cannot exist. Therefore the pair $(C,e)$ 
does not have
the ECP. \proofend
\end{remark} 

Recall that 
\begin{itemize}
\item{} A {\bem convex polyhedron} in a  linear space $X$ is a subset of $X$ 
which is the convex hull of a finite set.
\item{} The {\bem convex cone generated by a subset $S$} of a linear space $X$
is the smallest convex cone in $X$ that contains $S$. 
\item{} The {\bem closed convex cone generated by a subset $S$} of a linear space $X$
is the smallest closed convex cone in $X$ that contains $S$. 
\item{} A {\bem polyhedral cone} in $X$ is a cone in $X$ which is the convex cone
generated by a finite subset of $X$.
\item{} If $K$ is a compact convex subset of a normed linear space $X$, and
$x\in X$, the {\bem tangent cone to $K$ at $x$} is the closed 
convex cone in $X$ generated by the set of all vectors $y-x$, $y\in K$.
\end{itemize}

If $X$, $K$, $x$ are as above, we use $T_xK$ to denote the tangent cone to
$K$ at $x$. The following facts are then well known and easy to prove.

\begin{lemma}\label{lemmonpoly}
Let $X$ be a normed linear space. Then
\ben
\item[(1)] Every convex polyhedron in $X$ is compact.
\item[(2)] Every polyhedral cone  in $X$ is closed.
\item[(3)] If $C$ is a polyhedral cone $F$, and $F$ is a finite 
\mbox{subset}
of $X$ such that $C$ is the convex cone 
\mbox{generated} by $F$, then there exists
a positive constant $\kappa$ such that 
\mbox{every} $v\in C$ can be expressed
as a linear \mbox{combination}
\begin{equation}\label{orlando-aa}
v=\sum_{f\in F}\alpha_ff\,,\quad 0\le\alpha_f\le\kappa\|v\|\quad{\rm for\,\,\,all}
\quad f\in F\,.
\end{equation}
\item[(4)] If $K$ is a convex polyhedron, and $x\in K$, then 
\ben
\item[(4.a)] $T_xK$ is a polyhedral cone,
\item[(4.b)] there exists a $\rho\in\rr$ such that $\rho>0$ and
\mbox{$x+v\in K$} whenever 
$v\in T_xK$ and $\|v\|\le\rho$.
\een
\item[(5)] If $K$ is a convex polyhedron, then 
$\{T_xK:x\in K\}$ is a finite set.\proofend
\een
\end{lemma}

\begin{theorem}\label{hasecp}
Let $X$ be a finite-dimensional real linear space, and let 
$C$ be a polyhedral cone in $X$. Then $C$ is ECP-good.
\end{theorem} 
\begin{proof}
Recall that a cone $D$ is {\bem pointed} if there is no nonzero 
vector $v$ such that $v\in D$ and $-v\in D$. 

We first show that 
\ben
\item[(*)] {\bem we may assume, without loss of generality, that $C$ is pointed
and has nonempty interior.}
\een

To prove (*), we assume that every pointed polyhedral cone with
nonempty interior in any finite-dimensional real linear space $Y$
is ECP-good, and prove that if $C$ is an arbitrary
polyhedral cone in $X$ then $C$ is ECP-good.
For this purpose, since the ECP property does not depend on the choice
of norm, we may assume that $X$ is Euclidean.

Let $C$ be a polyhedral cone in $\rr^n$. Let $E$ be the ``edge''
of $C$, that is, the set of all $v\in C$ such that $-v\in C$. Then 
$E$ is a linear subspace of $X$. Let $E'$ be the orthogonal 
complement of $E$ in $X$, and let $C'=C\cap E'$. Then $C'$ is 
a closed convex cone. Moroever, $C'$ is clearly polyhedral and pointed.

Let $E''$ be the linear span of $C'$, so $E''\subseteq E'$. Then, as a
subset of $E''$, $C'$ is a pointed polyhedral cone with nonempty
interior. So $C'$ is ECP-good. Let $e\in{\rm Int}_{E''}(C')$ be such
that $(C',e)$ has the ECP. We show that $e\in{\rm Int}_{S_C}(C)$ and
$(C,e)$ has the ECP.  

First, we observe that $C\subseteq E+E''$. ({\bem Proof.} If $v\in C$, then
$v=w+w'$, $w\in E$, $w'\in E'$. Moreover, $-w\in E\subseteq C$, 
so $w'=v-w\in C+C$, and then $w'\in C$. Then $w'\in E'\cap C'=E''$, so
$v\in E+E''$.) 

Since $E\subseteq C$, the cone $C$ is in fact equal to the sum 
$E+(C\cap E'')$. Also, 
$$
C\cap E''=C\cap E''\cap E'=C'\cap E''=C'\,,
$$
since $C'\subseteq E''\subseteq E'$. So 
$$
C=E+C'\,.
$$
This clearly implies that 
$$
S_C=E+E''\,.
$$
Moreover, the sum $E+E''$ is orthogonal, so $E+E''$ is isomorphic to the product 
$E\times E''$, under the map 
$\mu:E\times E''\to E+E''$ given  by $\mu(v,v'')=v+v''$. Clearly,
$\mu(E\times C')=E+C'=C$. Since $e\in{\rm Int}_{E''}(C')$ and 
$0\in{\rm Int}_{E}(E)$, we have $(0,e)\in{\rm Int}_{E\times E''}(E\times C')$, 
so $\mu(0,e)\in{\rm Int}_{\mu(E\times E'')}(\mu(E\times C'))$, 
that is, 
$$
e\in{\rm Int}_{E+E''}(C)={\rm Int}_{S_C}(C)\,.
$$

Now let $\bar k$ be such that $(C',e)$ has the ECP with constant $\bar k$. 
Let $m\in\nn$, $w\in C$, $(u_1,\ldots,u_m)\in C^m$ be such that 
$$
u_1+\ldots+u_m=w+e\,.
$$
Write
$$
u_j=\tilde u_j+u_j''\,,\quad \tilde u_j\in E\,,\quad u_j''\in C'\,,
$$
and also 
$$
w=\tilde w+w''\,,\quad \tilde w\in E\,,\quad w''\in C'\,.
$$
Then
\begin{eqnarray*}
\tilde u_1+\ldots+\tilde u_m&=&\tilde w\,,\\
u_1''+\ldots+u_m''&=&w''+e\,.
\end{eqnarray*}
Since $(C',e)$ has the ECP with constant $\bar k$, 
there exist vectors $c_1,\ldots,c_m$ in $E''$ such that 
\begin{eqnarray}
u_i''-c_i&\in&C'\phantom{\bar k\|u_i\|}\quad{\rm for}\quad  
i=1,\ldots,m\,,\label{ecp-eq-a-lu}\\
c_1+\ldots+c_m&=&e\,,\label{ecp-eq-b-lubbb}\\
\|c_1\|+\ldots+\|c_m\|&\le&\bar k\,,\label{ecp-eq-c-bbbbbb}\\
\|c_i\|&\le&\bar k\|u_i''\|\phantom{C}\quad{\rm for}\quad i=1,\ldots,m\,.
\label{ecp-eq-d-dddddddd}
\end{eqnarray}
But then $u_i-c_i=\tilde u_i+u_i''-c_i\in C$, and $\|u_i''\|\le\|u_i\|$, 
because the sum $E+E''$ is orthogonal. Therefore
\begin{eqnarray}
u_i-c_i&\in&C\phantom{\bar k\|u_i\|}\quad{\rm for}\quad  
i=1,\ldots,m\,,\label{ecp-eq-a-lu-buuu}\\
\|c_i\|&\le&\bar k\|u_i\|\phantom{C}\quad{\rm for}\quad i=1,\ldots,m\,,
\label{ecp-eq-d-dddddddd-galaxy}
\end{eqnarray}
and (\ref{ecp-eq-b-lubbb}), (\ref{ecp-eq-c-bbbbbb}) hold. Hence 
$(C,e)$ has the ECP, and the proof of (*) is complete.

From now on, we assume that $C$ is pointed and has a nonempty interior
in $X$.  We choose an arbitrary interior point $e$ of $C$, and prove
that $(C,e)$ has the ECP. 

Let $\bar\zeta\in X^\dagger$ be a nontrivial linear functional such that
$$
\bar\zeta(v)>0\quad{\rm whenever}\quad v\in C\,,\,\,v\neq 0\,.
$$
({\bem Proof of the existence of $\bar\zeta$.} This follows from 
the fact that $C$ is pointed.
Indeed, let $C^\dagger$ be the polar cone of $C$.  Then $C^\dagger$ 
has nonempty 
interior in $X^\dagger$, because otherwise there would exist 
a nontrivial linear functional $\xi$ on $X^\dagger$ that vanishes
identically on $C^\dagger$, and this would yield---using the
canonical identification of  $X$ with $X^{\dagger\dagger}$ 
and the identity $C^{\dagger\dagger}=C$---a nonzero vector $v\in X$
such that $\zeta(v)=0$ for all $\zeta\in C^\dagger$. But  then
$v\in C$ and $-v\in C$, contradicting the fact that $(C,e)$ 
is pointed. If $\zeta$ is an interior point 
of $C^\dagger$, and $v\in C$, $v\neq 0$, then
$\zeta(v)<0$, because we know that $\zeta(v)\le 0$, and if
$\zeta(v)=0$ then the nontrivial linear map $X^\dagger\ni z\to z(v)\in\rr$
would have a local maximum at $\zeta$. So 
$\zeta(v)<0$ for all $v\in C\backslash\{0\}$, and then
$\bar\zeta=-\zeta$ has the desired property.)

Since $e\neq 0$ and $e\in C$, we may normalize $\bar\zeta$ so that
$\bar\zeta(e)=1$. Since the ECP property does not
\mbox{depend} on the norm,
we are entitled to assume that $X$ is 
\mbox{Euclidean}, and we may choose the
inner product so that 
\mbox{$\|e\|=1$} and $e$ is orthogonal to the kernel of
$\bar\zeta$. Under the canonical identification of $X$ with
$X^{\dagger}$ arising from the inner product, $\bar\zeta$ will then
correspond to a vector $\bar z\in X$ such that $e\perp\bar z^\perp$.
But then $e=r\bar z$ for some $r\in\rr$, and the facts that
$\bar\zeta(e)=1$ and $\|e\|=1$ imply $1=\langle\bar z,e\rangle=r\|e\|^2=r$, so
$r=1$ and then $\bar z=e$.  


Let $n={\rm dim}(X)$. We may assume that $n>0$, because if $n=0$ then
our conclusion is trivial. Write $n=\nu+1$, $\nu\in\zz_+$.  By
choosing an orthonormal basis of $X$ whose first member is $e$, we may
identify $X$ with $\rr\times\rr^\nu$ in such a way that $e=(1,0)$.
Then $C$ is a polyhedral cone in $\rr\times\rr^\nu$, having the
property that
$$
(r,x)\in C\backslash\{0\}\Longrightarrow r>0\,.
$$
(This follows because, if $v=(r,x)\in C\backslash\{0\}$, then 
\mbox{$r=\langle e,v\rangle=\langle\bar z,v\rangle=\bar\zeta(v)>0$}.)

Since $C$ is polyhedral, we may fix a finite set 
\mbox{$P=\{p_1,\ldots,p_N\}$}
of points of $C$ such that $C$ is the 
\mbox{smallest} convex cone 
containing $P$, i.e., the set of all linear combinations
$v=\alpha_1p_1+\ldots+\alpha_Np_N$ such that 
$\alpha_i\ge 0$ for $i=1,\ldots,N$.
Clearly, we may assume that all the $p_i$ are nonzero,
and then we can write 
$$
p_i=(\rho_i,q_i)\,,\quad \rho_i>0\,,\quad q_i\in\rr^\nu\,. 
$$
Then, after multiplication by $\rho_i^{-1}$, we may assume that $\rho_i=1$ for 
$i=1,\ldots,N$. 
Let 
$$
K=\{\,x\in\rr^\nu:(1,x)\in C\}\,,\quad Q=\{q_1,\ldots,q_N\}\,.
$$
It then follows that 
\begin{eqnarray}
K&=&{\rm convex\,\,\,hull\,\,\,of}\,\,\,Q\,,\label{combios-a}\\
0&\in&{\rm Int}_{\rr^\nu}K\,,\label{combios-b}\\
C&=&\Big\{\,(r,rx):x\in K\,,\,r\ge 0\,\Big\}\,.\label{combios-c}
\end{eqnarray}

Let $\tmath$ be the set of all the tangent cones $T_xK$, as $x$
varies over all points of $K$. Then $\tmath$ is a finite set by
Lemma \ref{lemmonpoly}. Let $M$ be the number of members of
$\tmath$.  For each $k\in\nn$, let $\mathcal{T}_k$ be the set of
all $k$-tuples ${\bf C}=(C_1,\ldots,C_k)$ of different members of
$\tmath$ such that 
\begin{equation}\label{noonef-sam}
C_1+\ldots+C_k=\rr^\nu\,.
\end{equation}
Then $\mathcal{T}_k=\emptyset$ if $k>M$.  Let
$\mathcal{T}=\cup_{k\in\nn}\mathcal{T}_k$. Then the set $\mathcal{T}$
is finite.

Choose, for each ${\bf C}=(C_1,\ldots,C_k)\in\mathcal{T}$, a continuous map
$$
\rr^\nu\!\ni\!x\!\mapsto\!\Theta^{{\bf C}}(x)\!=\!
(\theta_1^{{\bf C}}(x),\ldots,\theta_k^{{\bf C}}(x))\!\in
\! C_1\!\times\!\cdots\!\times\!C_k
$$
which is positively homogeneous of degree $1$ and satisfies
$$
\theta_1^{{\bf C}}(x)+\cdots+\theta_k^{{\bf C}}(x)=
x\quad{\rm for\,\,\,all}\quad x\in\rr^\nu\,.
$$
({\bem Proof that $\Theta^{{\bf C}}$ exists.} Fix ${\bf C}$.
Let $(b_1,\ldots,b_\nu)$ be the 
canonical basis of $\rr^n$, so $b_i$ is the vector 
$(b_i^1,\ldots,b_i^\nu)$, where $b_i^j=0$ if $i\neq j$, $b_i^i=1$.
Let $b_0=-(b_1+\cdots+b_\nu)$. Then every vector $x\in\rr^\nu$ 
can be expressed 
in a unique way as an affine combination of $b_0,b_1,\ldots,b_\nu$, i.e., as
a linear combination
\begin{equation}\label{noonef}
x\!=\!\alpha_0b_0\!+\!\alpha_1b_1\!+\!\cdots\!+\!\alpha_\nu b_\nu,\,\,
\alpha_0\!+\!\alpha_1\!+\!\cdots\!+\!\alpha_\nu\!=\!1\,,
\end{equation}
as can be seen by observing that, if 
\mbox{$\tilde b_i=(b_i,1)\in\rr^{\nu+1}$}, 
$\tilde x=(x,1)\in\rr^{\nu+1}$, then $(\tilde b_0,\tilde b_1,\cdots,
\tilde b_\nu)$ is a basis of $\rr^{\nu+1}$, and 
(\ref{noonef}) holds if and only if 
$$
\tilde x=\alpha_0\tilde b_0+\alpha_1\tilde b_1+\cdots+\alpha_\nu\tilde b_\nu\,.
$$
For $x\in\rr^\nu$, let $\alpha_i(x)$, $i=0,1,\ldots,\nu$, be the unique 
coefficients $\alpha_i$ such that (\ref{noonef}) holds. Then the functions
$\alpha_i:\rr^\nu\to\rr$ are obviously continuous. Clearly,
$$
\alpha_0(0)=\alpha_1(0)=\cdots=\alpha_\nu(0)=\frac{1}{\nu+1}\,.
$$
So the continuity of the $\alpha_i$ implies that there \mbox{exists}
\mbox{$r\in\rr$}
such that $r>0$ and $\alpha_i(x)\ge 0$ whenever $i\in\{0,1,\ldots,\nu\}$,
$x\in\rr^\nu$ and $\|x\|\le r$.
Define, for $i=1,\ldots,\nu$,
$$
\beta_i(x)=\left\{\begin{array}{lcl}
\frac{\|x\|}{r}\,\cdot\,\alpha_i\Big(\frac{rx}{\|x\|}\Big)
&{\rm if}&x\in\rr^\nu\backslash\{0\}\,,\\
0&{\rm if}&x=0\in\rr^\nu\,,\end{array}\right.
$$  
so the $\beta_i$ are continuous real-valued functions on $\rr^\nu$ that
are everywhere nonnegative and  positively homogeneous of degree $1$.
Using (\ref{noonef-sam}), write 
$$
b_i=c_{i,1}+c_{i,2}+\cdots+c_{i,k},\,\,c_{i,j}\in C_j\,\,
{\rm for}\\,\,j=1,2,\ldots,k\,.
$$
Then, let $\theta_j(x)=\beta_0(x)c_{0,j}+\beta_1(x)c_{1,j}+\cdots+
\beta_\nu(x)c_{\nu,j}$
for $x\in\rr^\nu$, $\,j=1,\ldots,k$.
Then each $\theta_j$ is a continuous map from $\rr^\nu$ to $C_j$, which is 
positively homogeneous of degree $1$. Moreover, 
if $x\in\rr^\nu$ and $x\neq 0$, then
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!
\sum_{j=1}^k\theta_j(x)=\sum_{j=1}^k\sum_{i=1}^\nu\beta_i(x)c_{i,j}
=\sum_{i=1}^\nu\beta_i(x)\Big(\sum_{j=1}^kc_{i,j}\Big)\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!
=\sum_{i=1}^\nu\beta_i(x)b_i
=\frac{\|x\|}{r}\sum_{i=1}^\nu\alpha_i\Bigg(\frac{rx}{\|x\|}\Biggr)b_i
=\frac{\|x\|}{r}\,\cdot\,\frac{rx}{\|x\|}=x\,.
\end{eqnarray*}
Obviously, the identity $\sum_{j=1}^k\theta_j(x)=x$ is also valid when $x=0$.
Therefore, if we let $\Theta^{{\bf C}}$ be the map 
$x\to(\theta_1(x),\ldots,\theta_k(x))$, then all the desired properties are 
satisfied.)

Let $\kappa_1=\max\Big\{\,\|x\|:x\in K\,\Big\}$, 
$\kappa_2=\sqrt{1+\kappa_1^2}$.
For each ${\bf C}=(C_1,\ldots,C_k)\in\mathcal{T}$, define
$$
\kappa^{{\bf C}}=\sup\Big\{\,\|\theta_j^{{\bf C}}(x)\|:x\in\rr^\nu\,,\,
\|x\|=1\,,\,j=1,\ldots,k\,\Big\}\,.
$$
Then define
$\kappa_3=\max\Big\{\,\kappa^{{\bf C}}:{\bf C}\in\mathcal{T}\,\Big\}$.
We now choose the constant $\bar k$ by letting 
$\bar k=1+\kappa_2+2M\kappa_1\kappa_3$, 
and prove that the pair $(C,e)$ has the ECP with constant $\bar k$.

Pick $m\in\nn$, $w\in C$, and an $m$-tuple $(u_1,\ldots,u_m)$ of 
members of $C$ such that 
\begin{equation}\label{combios-aa}
u_1+\ldots+u_m=w+e\,,
\end{equation}
and write 
\begin{equation}\label{combios-aa-uaoo}
w=(R,R\omega)\,,\,\,\,R\ge 0\,,\,\,\,\omega\in K\,.
\end{equation}
We have to find vectors $c_1,\ldots,c_m$ in $X$ such that 
\begin{eqnarray}
u_i\!-\!c_i&\!\in\!&C\phantom{\bar k\|u_i\|}\,\,{\rm for}\,\,  
i\!=\!1,\ldots,m,\label{ecp-eq-a-gug}\\
c_1\!+\!\ldots\!+\!c_m&\!=\!&e,\label{ecp-eq-b-gug}\\
\|c_1\|\!+\!\ldots\!+\!\|c_m\|&\!\le\!&\bar k,\label{ecp-eq-c-gug}\\
\|c_i\|&\!\le\!&\bar k\|u_i\|\phantom{C}\,\,{\rm for}\,\,i\!=\!1,\ldots,m.
\label{ecp-eq-d-gug}
\end{eqnarray}
For this purpose, we find a Lipschitz curve 
\begin{equation}\label{combios-aaa}
J\ni s\to\gamma(s)=(c_1(s),\ldots,c_m(s))\in(\rr^n)^m\,,
\end{equation}
defined on $J=[0,1]$ and such that  
\begin{equation}\label{ecp-eq-a-gugs-aa}
c_i(0)=0\phantom{e}\quad{\rm for}\quad  
i=1,\ldots,m\,,
\end{equation}
\begin{equation}\label{ecp-eq-a-gugs}
u_i-c_i(s)\in C\phantom{\,}\quad{\rm for}\quad  
i=1,\ldots,m\,,\quad s\in J\,,
\end{equation}
\begin{equation}\label{ecp-eq-b-gugs}
c_1(s)+\ldots+c_m(s)=se\quad{\rm for}\quad s\in J\,,
\end{equation}
\begin{equation}\label{ecp-eq-d-gugs}
\langle e,\dot c_i(s)\rangle\ge 0\quad{\rm for}\quad 
i=1,\ldots,m\,,\quad {\rm a.e.}\quad s\in J\,,
\end{equation}
\begin{equation}\label{ecp-eq-c-gugs}
\|\dot c_1(s)\|+\ldots+\|\dot c_m(s)\|\le\bar k\phantom{e}
\quad{\rm for}\quad {\rm a.e.}\quad s\in J\,.
\end{equation}

\medbreak Let $\Gamma$ be the set of all pairs $(J,\gamma)$ such that
(1)~$J$ is a subinterval of $[0,1]$, (2)~$0\in J$, and (3)~$\gamma$ is
a Lipschitz curve of the form (\ref{combios-aaa}) such that
(\ref{ecp-eq-a-gugs-aa}), (\ref{ecp-eq-a-gugs}),
(\ref{ecp-eq-b-gugs}), (\ref{ecp-eq-d-gugs}), (\ref{ecp-eq-c-gugs})
hold.  Partially order $\Gamma$ by letting
$(J_1,\gamma_1)\preceq(J_2,\gamma_2)$ iff $J_1\subseteq J_2$ and
$\gamma_1=\gamma_2\restr J$.  It then follows from Zorn's Lemma that
$\Gamma$ has a member $(J,\gamma)$ which is maximal with respect to
$\preceq$. Let $\bar s=\sup{J}$, so $0\le\bar s\le 1$, and either
$J=[0,\bar s\,[\,$ or $J=[0,\bar s]$. If $J=[0,\bar s\,[\,$ then,
since $\gamma$ is Lipschitz, it can be extended to a curve
$\hat\gamma$ defined on $[0,\bar s]$, and then the pair $([0,\bar
s],\hat\gamma)$ satisfies $([0,\bar s],\hat\gamma)\in\Gamma$,
$(J,\gamma)\preceq([0,\bar s],\hat\gamma)$, and
$(J,\gamma)\neq([0,\bar s],\hat\gamma)$, thus 
\mbox{contradicting} the
maximality of $(J,\gamma)$. So the possibility that $J=[0,\bar s\,[\,$
is excluded. Hence $J=[0,\bar s]$.

We now prove that $\bar s=1$. We do this by showing that if $\bar s<1$ then 
$\gamma$ can be extended to a curve $\hat\gamma:[0,\bar s+\eps]$ 
for some positive $\eps$ such that $\eps\le 1-\bar s$, in such a way that 
the pair $([0,\bar s+\eps],\hat\gamma)$ still belongs to $\Gamma$.

Let $\bar y=w+(1-\bar s)e$.
Then $\bar y\in{\rm Int}_X(C)$, because $w\in C$, $e\in{\rm Int}_X(C)$ 
and $\bar s<1$. Write 
$\bar y=(\bar r,\bar r\bar x)$, with $\bar r\ge 0$ and $\bar x\in K$.
Then $\bar y=(\bar r,R\omega)$,
$\bar r=R+1-\bar s>0$, $\bar x=\frac{R}{\bar r}\omega$,
and $\bar x\in{\rm Int}_{\rr^\nu} K$.
Next, write
$$
u_i-c_i(\bar s)=\bar v_i=(\bar r_i,\bar r_i\bar x_i),
$$
with $\bar r_i\ge 0$ and $\bar x_i\in K$ for $i=1,\ldots m$.
Then 
\begin{eqnarray*}
\bar v_1+\cdots+\bar v_m&=&\bar y\,,\\ 
\bar r_1+\cdots+\bar r_m&=&\bar r\,,\\ 
\bar r_1\bar x_1+\cdots+\bar r_m\bar x_m&=&\bar r\bar x=R\omega\,,
\end{eqnarray*}
so
\begin{equation}\label{combios-aaa-u}
\bar x=\bar\rho_1\bar x_1+\cdots+\bar\rho_m\bar x_m\,,
\end{equation}
where 
$$
\bar\rho_i=\frac{\bar r_i}{\bar r}=\frac{\bar r_i}{R+1-\bar s}\,,
$$
so that
$$
\bar\rho_i\ge 0\quad{\rm for}\quad i=1,\ldots m\,,
$$
and
$$
\bar\rho_1+\cdots+\bar\rho_m=1\,.
$$
It then follows that 
\begin{equation}\label{combios-aaa-uau}
\bar\rho_1T_{\bar x_1}K+\cdots+\bar\rho_mT_{\bar x_m}K=\rr^\nu\,.
\end{equation}
({\bem Proof.} Using the fact that $\bar x\in{\rm Int}_{\rr^\nu} K$,
pick a real number $\delta$ such that $\delta>0$ and 
$\|y\|\le\delta\Longrightarrow \bar x+y\in K$.
If $\|y\|\le\delta$, and we let $z=\bar x+y$, then $z\in K$, and
\begin{eqnarray*}
y&=&z-\bar x=z-\sum_{i=1}^m\bar\rho_ix_i=\sum_{i=1}^m\bar\rho_i(z-\bar x_i)\\
&\in&\bar\rho_1T_{\bar x_1}K+\cdots+\bar\rho_mT_{\bar x_m}K\,.
\end{eqnarray*}
Therefore the cone $\bar\rho_1T_{x_1}K+\cdots+\bar\rho_1T_{x_m}K$ contains a 
neighborhood of $0$ in $\rr^\nu$, and (\ref{combios-aaa-u}) follows.)

Clearly, (\ref{combios-aaa-uau}) remains true if we eliminate from the
sum those terms $\bar\rho_iT_{\bar x_i}K$ for which $\bar\rho_i=0$.
Once this is done, we can replace each remaining term
$\bar\rho_iT_{\bar x_i}K$ by $T_{\bar x_i}K$, because the $T_{\bar x_i}K$ are
cones. Finally, we can eliminate repeated terms, after which we will
be left with at most $M$ terms.

Hence there exist $\mu\in\nn$ and  ${\bf i}=(i_1,\ldots,i_\mu)\in\nn^\mu$
such that (i)~$1\le i_1<i_2<\cdots<i_\mu\le m$, 
(ii)~$T_{\bar x_{i_j}}K\neq T_{\bar x_{i_\ell}}K$
whenever $j\neq\ell$ and $1\le j,\ell\le\mu$,
(iii)~$\bar\rho_{i_j}>0$ whenever $1\le j\le\mu$, and
(iv)~$T_{\bar x_{i_1}}K+\cdots+T_{\bar x_{i_\mu}}K=\rr^\nu$.

Now let 
$$
y(h)=\bar y-he=(\bar r-h,R\omega\Big)
\quad{\rm for}\quad h\ge 0\,.
$$
Then 
\begin{eqnarray*}
y(h)&=&\Big(\bar r-h,(\bar r-h)\frac{R}{\bar r-h}\omega\Big)\\
&=&\Biggl(\bar r-h,(\bar r-h)\Big(\bar x+\frac{R}{\bar r-h}\omega-
\bar x\Big)\Biggr)\\
&=&\Biggl(\bar r-h,(\bar r-h)\Big(\bar x+\frac{R}{\bar r-h}
\omega-\frac{R}{\bar r}\omega\Big)\Biggr)\\
&=&\Biggl(\bar r-h,(\bar r-h)\Big(\bar x+\Big(\frac{R}{\bar r-h}
-\frac{R}{\bar r}\Big)\omega\Big)\Biggr)\\
&=&\Biggl(\bar r-h,(\bar r-h)\Big(\bar x+\frac{hR}{\bar r(\bar r-h)}
\omega\Big)\Biggr)\\
&=&\Big(\bar r-h,(\bar r-h)x(h)\Big)\,, 
\end{eqnarray*}
where 
$$
x(h)=\bar x+\frac{hR}{\bar r(\bar r-h)}\omega\,.
$$
Let ${\bf T}$ be the $\mu$-tuple 
$$
{\bf T}=\Big(\,T_{\bar x_{i_1}}K\,,\,T_{\bar x_{i_2}}K\,,\,
\ldots\,,\, T_{\bar x_{i_\mu}}K\,\Big)\,.
$$
Then ${\bf T}\in\mathcal{T}_\mu$, so $\mu\le M$.  Let
$$
\rr^\nu\ni x\to\Theta(x)=(\theta_1(x),\ldots,\theta_\mu(x))
$$ 
be the map $\Theta^{{\bf T}}$, so 
$\Theta:\rr^\nu\mapsto
T_{\bar x_{i_1}}K\times T_{\bar x_{i_2}}K\times\cdots T_{\bar x_{i_\mu}}K$. 
Then $\Theta$ 
is positively homogeneous of degree $1$ and such that 
$\theta_1(x)+\cdots+\theta_\mu(x)=x$
for all $x\in\rr^\nu$.  Moreover, 
\begin{eqnarray}
\|\theta_j(x)\|&\le&\kappa^{{\bf T}}\|x\|\le\kappa_3\|x\|\le\kappa_1\kappa_3
\nonumber\\
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
{\rm whenever}\quad j=1,\ldots,\mu\,\,\,{\rm and}\,\,\,x\in K\,.
\label{combios-aaa-uauo-uu-vvv-auau}
\end{eqnarray}
Define
$$
\psi_j(h)=\frac{hR}{\bar\rho_{i_j}\bar r(\bar r-h)}\theta_j(\omega)\,\,\,
{\rm for}\,\,\,j=1,\ldots,\mu\,.
$$
Then 
\begin{equation}\label{combios-aaa-uauo-uu-vvv}
\bar\rho_{i_1}\psi_1(h)+\cdots+\bar\rho_{i_\mu}\psi_\mu(h)
=\frac{hR}{\bar r(\bar r-h)}\omega=x(h)-\bar x\,.
\end{equation}
Next, let
\begin{eqnarray}
\!\!\!r_i(h)\!\!\!&\!=\!&\!\!\!\bar r_i-\bar\rho_i h=
\bar\rho_i(\bar r-h)
\,\,\,{\rm if}\,\,\,i=1,\ldots,m\,,\label{combios-aaa-uauo-uu-vvv-aio}\\
\!\!\!x_i(h)\!\!\!&\!=\!&\!\!\!\left\{\begin{array}{lcl}
\!\!\!\bar x_{i_j}\!+\!\psi_{j}(h)&{\rm if}&i\!=\!i_j\,,\,\,
j\!\in\!\{1,\ldots,\mu\},\\
\!\!\!\bar x_{i}&{\rm if}&i\notin\{i_1,\ldots,i_\mu\}\,.\end{array}\right.
\label{combios-aaa-uauo-uu-vvv-ai}\\
\!\!\!v_i(h)\!\!\!&\!=\!&\!\!\!(r_i(h),r_i(h)x_i(h))
\,\,\,{\rm if}\,\,\,i=1,\ldots,m\,.\label{combios-aaa-uauo-uu-vvv-aiu}
\end{eqnarray}
Then
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
 r_{1}(h)x_1(h)+\cdots+r_{m}(h)x_m(h)\\
&=&(\bar r-h)\Big(\bar\rho_{1}x_1(h)+\cdots+\bar\rho_{m}x_m(h)\Big)\\
&=&(\bar r-h)\Big(\bar\rho_{1}\bar x_1+\cdots+\bar\rho_{m}\bar x_m\Big)\\
&\,\,&\,\,+(\bar r-h)\Big(\bar\rho_{i_1}\psi_{1}(h)+
\cdots+\bar\rho_{i_\mu}\psi_{\mu}(h)\Big)\\
&=&(\bar r-h)\bar x+(\bar r-h)(x(h)-\bar x)\\
&=&(\bar r-h)x(h)\,,
\end{eqnarray*}
and 
\begin{equation}\label{combios-aaa-uauo-uu-aaa}
r_{1}(h)+\cdots+r_{m}(h)=\bar r-h\,.
\end{equation}
Therefore
\begin{equation}\label{combios-aaa-uauo-uu-aaa-vv}
v_{1}(h)+\cdots+v_{m}(h)=(\bar r-h,(\bar r-h)x(h))=y(h)\,.
\end{equation}

If $j\in\{1,\ldots,\mu\}$, then Lemma \ref{lemmonpoly} implies that
there is a positive constant $\delta_j$
such that $\bar x_{i_j}+z\in K$ whenever $z\in T_{\bar x_{i_j}}K$ and 
$\|z\|\le\delta_j$. Since $\psi_j(h)$ is a multiple of $\theta_j(\omega)$,
with a factor that goes to zero as $h\downarrow 0$, and 
$\theta_j(\omega)\in T_{\bar x_{i_j}}K$, there exists 
a positive constant $\eps_j$ such that $\bar x_{i_j}+\psi_j(h)\in K$ whenever 
$0\le h\le\eps_j$. Let
\begin{equation}\label{combios-aaa-uauo-uu-ggg-eps}
\eps=\min\Big(\,1-\bar s\,,\,\frac{\bar r}{2}\,,\,\eps_1\,,\,\ldots\,,
\,\eps_\mu\,\Big)\,.
\end{equation}
Then $\bar s+\eps\le 1$, and 
\begin{equation}\label{combios-aaa-uauo-uu-ggg-xih}
x_{i}(h)\in K\,\,{\rm if}\,\, i\in\{1,\ldots,m\}\,,\,\,0\le h\le\eps\,.
\end{equation}
We now extend the functions $c_i:[0,\bar s]\to\rr^n$ to the interval 
$[0,\bar s+\eps]$  by letting 
$c_{i}(\bar s+h)=c_i(\bar s)+\bar v_i-v_i(h)$ 
if $i=1,\ldots,m$ and $0\le h\le\eps$.  Then 
\begin{eqnarray*}
&\,&\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!
c_{1}(\bar s+h)+\cdots+c_{m}(\bar s+h)\\
&=&c_{1}(\bar s)+\cdots+c_{m}(\bar s)+\bar v_{1}+\cdots+\bar v_{m}\\
&\,&\,\,\,-(v_{1}(h)+\cdots+v_{m}(h))\\
&=&\bar se+\bar y-y(h)\\
&=&\bar se+he\\
&=&(\bar s+h)e\,,
\end{eqnarray*}
so that
\begin{equation}\label{combios-aaa-uauo-uu-gggauua}
c_{1}(\bar s+h)+\cdots+c_{m}(\bar s+h)=(\bar s+h)e
\end{equation}
if $0\le h\le\eps$.
Moreover, if $h\in[0,\eps]$, and $i\in\{1,\ldots,m\}$, then
\begin{eqnarray*}
u_i-c_i(\bar s+h)&=&u_i-\Big(c_i(\bar s)+\bar v_i-v_i(h)\Big)\\
&=&u_i-c_i(\bar s)-\bar v_i+v_i(h)\\
&=&\bar v_i-\bar v_i+v_i(h)\\
&=&v_i(h)\\
&=&(r_i(h),r_i(h)x_i(h))\\
&=&r_i(h)(1,x_i(h))\,.
\end{eqnarray*}
Since $x_i(h)\in K$ by (\ref{combios-aaa-uauo-uu-ggg-xih}), the point 
$(1,x_i(h))$ belongs to $C$,  and then 
$r_i(h)(1,x_i(h))$ also belongs to $C$. So we have shown that 
\begin{equation}\label{combios-aaa-uauo-uu-gggauua-isin}
u_i-c_i(\bar s+h)\in C \quad{\rm if}\quad i=1,\ldots,m\,,\quad
0\le h\le\eps\,.
\end{equation}

We now study the derivatives $\dot c_i$ of the functions $c_i$ on the 
interval $[\bar s,\bar s+\eps]$. We have (using the fact that 
$v_i(h)=r_i(h)(1,x_i(h))\,\,$)
\begin{eqnarray}
\dot c_i(\bar s+h)&=&-\dot v_i(h)\nonumber\\
&=&-\dot r_i(h)(1,x_i(h))-r_i(h)(0,\dot x_i(h))\nonumber\\
&=&\bar\rho_i(1,x_i(h))-r_i(h)(0,\dot x_i(h))\label{righto}\,.
\end{eqnarray}
This proves, first of all, that the first component of 
\mbox{$\dot c_i(\bar s+h)$} is equal to the nonnegative number 
$\bar\rho_i$, so 
\begin{equation}\label{combios-aaa-uauo-uu-aaa-angle}
\langle e,\dot c_i(\bar s+h)\rangle\ge 0\quad{\rm for}\quad
i=1,\ldots,m\,,\quad h\ge 0\,.
\end{equation}
Moreover, the norm of the first term of the right-hand side of 
(\ref{righto}) is bounded by $\bar\rho_i\kappa_2$, since $x_i(h)\in K$.

If $i\notin\{i_1,\ldots,i_\mu\}$, then the second term of the right-hand 
side of 
(\ref{righto}) vanishes, and we get the bound 
\begin{equation}\label{combios-aaa-uauo-uu-ggg-boa}
\|\dot c_i(\bar s+h)\|\le\bar\rho_i\kappa_2\,\,\,{\rm if}\,\,\,
i\notin\{i_1,\ldots,i_\mu\}\,.
\end{equation}

If $i\in\{i_1,\ldots,i_\mu\}$, then the second term of the right-hand side of 
(\ref{righto}) need not vanish, so we have to estimate it. Assume 
$i=i_j$, $j\in\{1,\ldots,\mu\}$. Then 
$$
\dot x_i(h)=\dot\psi_j(h)
=\frac{R}{\bar\rho_{i}\bar r(\bar r-h)}\theta_j(\omega)
+\frac{hR}{\bar\rho_{i}\bar r(\bar r-h)^2}\theta_j(\omega),
$$
so (\ref{combios-aaa-uauo-uu-vvv-aio}) implies that
\begin{eqnarray*}
r_i(h)\dot x_i(h)&=&
\frac{\bar\rho_i(\bar r-h)R}{\bar\rho_{i}\bar r(\bar r-h)}
\theta_j(\omega)\,\,\,
+\frac{h\bar\rho_i(\bar r-h)R}{\bar\rho_{i}\bar r(\bar r-h)^2}
\theta_j(\omega)\\
&=&\Big(\frac{R}{\bar r}+\frac{hR}{\bar r(\bar r-h)}\Big)\theta_j(\omega)\,,
\end{eqnarray*}
and then 
$$
\|r_i(h)\dot x_i(h)\|\le\Big(1+\frac{h}{\bar r-h}\Big)
\|\theta_j(\omega)\|\le 2\kappa_1\kappa_3\,,
$$
because (1)~$\omega\in K$, so 
(\ref{combios-aaa-uauo-uu-vvv-auau}) can be used to 
conclude that $\|\theta_j(\omega)\|\le\kappa_1\kappa_3$, 
(2)~$R\le\bar r$, and (3)~the bound $h\le\eps\le\frac{\bar r}{2}$ 
implies $h\le \bar r-h$, and then $\frac{h}{\bar r-h}\le 1$.

This gives the estimate	
\begin{equation}\label{combios-aaa-uauo-uu-ggg-boa-uf}
\|\dot c_i(\bar s+h)\|\le\bar\rho_i\kappa_2+2\kappa_1\kappa_3\,\,\,
{\rm if}\,\,\,i\in\{i_1,\ldots,i_\mu\}\,.
\end{equation}
Then  (\ref{combios-aaa-uauo-uu-ggg-boa}) and 
(\ref{combios-aaa-uauo-uu-ggg-boa-uf}) imply (since $\bar\rho_1+\cdots+
\bar\rho_{m}=1$) that 
\begin{equation}\label{combios-aaa-uauo-uu-ggg-boa-uf-est}
\sum_{j=1}^\mu\|\dot c_j(\bar s+h)\|
\!\le\!\kappa_2\!+\!2\mu\kappa_1\kappa_3\!\le\!
\kappa_2\!+\!2M\kappa_1\kappa_3\!\le\!\bar k\,.
\end{equation}
It follows from (\ref{combios-aaa-uauo-uu-gggauua}), 
(\ref{combios-aaa-uauo-uu-gggauua-isin}), 
(\ref{combios-aaa-uauo-uu-aaa-angle}), 
and (\ref{combios-aaa-uauo-uu-ggg-boa-uf-est}), that the new curve
$[0,\bar s+\eps]\ni s\to\hat\gamma(s)=(\hat c_1(s),\ldots,\hat c_m(s))$,
where the $\hat c_i$ are 
the components $c_i$ of $\gamma$, extended as above to the interval 
$[0,\bar s+\eps]$, satisfies the conditions 
(\ref{ecp-eq-a-gugs-aa}), (\ref{ecp-eq-a-gugs}), (\ref{ecp-eq-b-gugs}), 
(\ref{ecp-eq-d-gugs}), (\ref{ecp-eq-c-gugs}) that characterize the set 
$\Gamma$. Hence the pair $([0,\bar s+\eps],\hat\gamma)$ belongs to $\Gamma$.
Therefore the pair $([0,\bar s],\gamma)$ is not a maximal
member of $\Gamma$, and we have reached a contradiction.  This happened
because we assumed that $\bar s<1$. Hence
$\bar s=1$.

We have therefore established that there exists a curve (\ref{combios-aaa})
which is defined on the full interval $J=[0,1]$ and is such that  
(\ref{ecp-eq-a-gugs-aa}), (\ref{ecp-eq-a-gugs}), (\ref{ecp-eq-b-gugs}), 
(\ref{ecp-eq-d-gugs}), (\ref{ecp-eq-c-gugs}) hold.  Fix such a curve,
and define
$$
c_i=c_i(1)\quad{\rm for}\quad i=1,\ldots,m\,.
$$
Then (\ref{ecp-eq-a-gugs}) implies that (\ref{ecp-eq-a-gug}) is satisfied,
(\ref{ecp-eq-b-gugs}) implies (\ref{ecp-eq-b-gug}), and 
(\ref{ecp-eq-c-gugs}) together with (\ref{ecp-eq-a-gugs-aa})
imply (\ref{ecp-eq-c-gug}).

To complete our proof, we have to show that (\ref{ecp-eq-d-gug}) is
true as well. At this point, we will use (\ref{ecp-eq-d-gugs}).
The vector $v_i=u_i-c_i$ belongs to $C$, and if we write 
$u_i=(\hat r_i,\hat r_i\hat x_i)$, $v_i=(r_i,r_ix_i)$, with
$\hat x_i, x_i\in K$, $\hat r_i\ge 0$, $r_i\ge 0$,
then (\ref{ecp-eq-d-gugs}) implies that $r_i\le\hat r_i$. Therefore
$$
\|v_i\|^2=r_i^2(1+\|x_i\|^2)\le r_i^2\kappa_2^2
\le\hat r_i^2\kappa_2^2\le\|u_i\|^2\kappa_2^2\,,
$$
since $\hat r_i\le\|u_i\|$.
Hence $\|v_i\|\le\kappa_2\|u_i\|$, and then, since $c_i=u_i-v_i$, we have
$\|c_i\|\le(1+\kappa_2)\|u_i\|\le\bar k\|u_i\|$, 
showing that (\ref{ecp-eq-d-gug}) is indeed true, and completing our 
proof.
\end{proof} 

