%From piccoli@triton.iacm.forth.gr Fri Jun 26 13:02:32 1998
%Date: Fri, 26 Jun 1998 20:00:53 +0300
%From: piccoli@triton.iacm.forth.gr (Piccoli Benedetto)%
%
\magnification=\magstephalf
\def\nitembf{{\global\advance\myitemno by 1}\item{{\bf\the\myitemno .}}}
\newcount\mycommentno%the comment number
\def\gnu#1{\medbreak\global\advance\mycommentno by 1
\centerline{\hbox to 6cm{\hrulefill}}
\medbreak\noindent{\bf COMMENT NO. \the\mycommentno}:\enspace #1
\medbreak
\centerline{\hbox to 6cm{\hrulefill}}
\medbreak\noindent}
\def\gnus#1{\medbreak\global\advance\mycommentno by 1
\centerline{\hbox to 6cm{\hrulefill}}
\medbreak\noindent {\bf COMMENT NO. \the\mycommentno}:\enspace #1}
\def\gnum#1{\medbreak\noindent#1}
\def\gnue#1{\medbreak\noindent#1
\medbreak
\centerline{\hbox to 6cm{\hrulefill}}
\medbreak\noindent}
\def\comment#1{}
\def\midbreak{\vskip 0.1cm}
\def\tb{{\rm TB}}    
\def\tvvf{{\rm TVVF}}
\def\admco{{\rm Adm_{C^1}}}
\def\IP{{\cal IP}}
\def\EP{{\cal EP}}
\def\traj{{\rm Traj}}
\def\mtraj{{\rm MTraj}}
\input psfig.sty
\def\frac#1#2{#1\over#2}
\def \today {\ifcase \month \or January\or February\or  
  March\or April\or May\or June\or July\or August\or 
  September\or October\or November\or December\fi\   
  \number \day, \number \year} 
\xdef\parnormal{\parindent=18pt}
\xdef\pari{\parindent=30pt}
\xdef\parii{\parindent=62pt}
\xdef\pariim{\parindent=47pt}
\def\Dom{{\rm Dom}}
\def\Adm{{\rm Adm}}
\def\Admo{{\rm Adm}^0}
%\hsize=18true cm
\def\pmb#1{\setbox0=\hbox{$#1$}%
  \kern-.025em\copy0\kern-\wd0
  \kern.05em\copy0\kern-\wd0
  \kern-.025em\raise.0433em\box0 }
\def\equaldef{{\buildrel \rm def \over =}}
\def\steero{$\sim$\kern -0.1cm$\sim$\kern -0.1cm$\sim$\kern-0.2cm$>$\kern-0.28cm$>$\kern-0.28cm$>$\kern-0.28cm$>$\kern-0.28cm$>$\kern-0.28cm$>$\kern-0.28cm$>$\kern0.1cm}
\def\steer#1#2{$#1$\steero$#2$}
\def\steert#1#2#3{\hbox{$#1$\hbox{\lower0.4ex\hbox{\steero}}\kern-0.59cm\raise0.53ex\hbox{$#2$}}\kern0.35cm$#3$}
\newcount\myitemno%the item number
\newcount\sectno%the section number
\newcount\subsectno%the subsection number
\newcount\theono%the theorem number
\newcount\theotno%the T theorem number
\newcount\theononosec%the theorem number without changing at sections
\newcount\lemmno%the lemma number
\newcount\corono%the corollary number
\newcount\propno%the corollary number
\newcount\defino%the definition number
\newcount\defidno%the D definition number
\newcount\equano%the equation number
\newcount\equanonosec%the equation number without the section
\newcount\remano%the remark number
\newcount\examno%the example number
\newcount\figuno%the figure number
 \newcount\footno%the footnote number
\def\restr{\lceil}
\sectno=0
\subsectno=0
\equanonosec=1
\footno=0
%BLACKBOARD BOLD
\def\vbar{\mathchoice{\vrule height6.3ptdepth-.5ptwidth.8pt\kern-.8pt}
   {\vrule height6.3ptdepth-.5ptwidth.8pt\kern-.8pt}
   {\vrule height4.1ptdepth-.35ptwidth.6pt\kern-.6pt}
   {\vrule height3.1ptdepth-.25ptwidth.5pt\kern-.5pt}}
\def\fudge{\mathchoice{}{}{\mkern.5mu}{\mkern.8mu}}
\def\bbc#1#2{{\rm \mkern#2mu\vbar\mkern-#2mu#1}}
\def\bbb#1{{\rm I\mkern-3.5mu #1}}
\def\bba#1#2{{\rm #1\mkern-#2mu\fudge #1}}
\def\bb#1{{\count4=`#1 \advance\count4by-64 \ifcase\count4\or\bba A{11.5}\or
   \bbb B\or\bbc C{5}\or\bbb D\or\bbb E\or\bbb F \or\bbc G{5}\or\bbb H\or
   \bbb I\or\bbc J{3}\or\bbb K\or\bbb L \or\bbb M\or\bbb N\or\bbc O{5} \or
   \bbb P\or\bbc Q{5}\or\bbb R\or\bbc S{4.2}\or\bba T{10.5}\or\bbc U{5}\or
   \bba V{12}\or\bba W{16.5}\or\bba X{11}\or\bba Y{11.7}\or\bba Z{7.5}\fi}}
%MACROS FOR SYCON PAPER
\def\thegoodheadline{{\it H.J. Sussmann\hfill\myrunninghead}\hfill{\bf\folio}}
\def\thegoodheadlinewe{
{\it Sontag--Sussmann\hfill\myrunninghead}\hfill{\bf\folio}}
\def\thegoodheadlinewk{
{\it Kokotovic--Sussmann\hfill\myrunninghead}\hfill{\bf\folio}}
\def\thegoodheadlinewkinv{
{\it Sussmann--Kokotovic\hfill\myrunninghead}\hfill{\bf\folio}}
\def\myownsyconheadline
{\ifnum\pageno>0\thegoodheadline\else \hfill\fi}
\def\myownsyconheadlinewitheduardo
{\ifnum\pageno>0\thegoodheadlinewe\else \hfill\fi}
\def\myownsyconheadlinewithkokotovic
{\ifnum\pageno>0\thegoodheadlinewk\else \hfill\fi}
\def\myownsyconheadlinewithkokotovicinv
{\ifnum\pageno>0\thegoodheadlinewkinv\else \hfill\fi}
\def\startsycona{\nopagenumbers\pageno=0
\line{\hfill Report {\sl SYCON-\theshortyear-\thesyconreportnumber}}
\vfill
\centerline{\bf \thetitleone}
\centerline{\bf \thetitletwo}
\vskip 1true cm}
\def\centermathdept{
\centerline{Department of Mathematics}
\centerline{Rutgers University}
\centerline{New Brunswick, NJ 08903}
}
\def\mynameandaddress{
\centerline{H. J. Sussmann$^{*}$}
\centermathdept
\centerline{(908)\ 445-5407}
\centerline{sussmann@hamilton.rutgers.edu\ \ {\it or}\ \ sussmann@pisces.bitnet}}
\def\mynameandaddresswithkokotovic{
\centerline{P. V. Kokotovic$^{+}$}
\centerline{Coordinated Science Laboratory}
\centerline{University of Illinois}
\centerline{1101 W. Springfield Ave.}
\centerline{Urbana, Illinois 61801}
\vskip 0.3true in
\centerline{H. J. Sussmann$^{*}$}
\centermathdept
\centerline{(908)\ 445-5407}
\centerline{sussmann@hamilton.rutgers.edu\ \ {\it or}\ \ sussmann@pisces.bitnet}}
\def\mynameandaddresswithkokotovicinv{
\centerline{H. J. Sussmann$^{*}$}
\centermathdept
\centerline{(908)\ 445-5407}
\centerline{sussmann@hamilton.rutgers.edu\ \ {\it or}\ \ sussmann@pisces.bitnet}
\vskip 0.3true in
\centerline{P. V. Kokotovic$^{+}$}
\centerline{Coordinated Science Laboratory}
\centerline{University of Illinois}
\centerline{1101 W. Springfield Ave.}
\centerline{Urbana, Illinois 61801}}
\def\mynameandaddresswitheduardo{
\centerline{E.D. Sontag$^{+}$ and H.J. Sussmann$^{*}$}
\centermathdept
\centerline{(908)\ 445-3072\ \ ---\ \ (908)\ 445-5407}
\centerline{sontag@fermat.rutgers.edu\ \ {\it or}\ \ sontag@pisces.rutgers.edu}
\centerline{sussmann@hamilton.rutgers.edu\ \ {\it or}\ \ sussmann@pisces.bitnet}}
\def\hjsacknowledgement{\noindent{\mysmall
$^{*}$\ \ This author's work was supported in part by the National
Science Foundation under NSF Grant DMS-8902994.} 
\vskip 0.05cm}
\def\hjsacknowledgementnoauthor{\noindent{\mysmall
$^{*}$\ \ This work was supported in part by the National
Science Foundation under NSF Grant DMS-8902994.} 
\vskip 0.05cm}
\def\hjsacknowledgementcaip{
\noindent{\mysmall $^{*}$\ \ This author's work was supported in part
by the National Science Foundation under Grant DMS-8902994, and by
the CAIP Center, 
Rutgers University, with funds provided by the New Jersey Commission on
Science and Technology and by CAIP's industrial members.}
\vskip 0.05cm}
\def\edsacknowledgement{
\noindent{\mysmall $^{+}$\ \ This author's work was supported in part 
by the National Science Foundation under Grant DMS-8803396, and by 
the U.S. Air Force under Grants AFOSR-85-0247 and AFOSR-88-0235}
\vskip 0.05cm}
\def\kokoacknowledgement{
\noindent{\mysmall $^{+}$\ \ This author's work was supported in part
by the National Science Foundation under  Grant ECS 87-15811 and
by the U.S. Department of Energy under Grant DE-FGD2-88 ER 13939}
\vskip 0.05cm}
\def\edsacknowledgementcaip{
\noindent{\mysmall $^{+}$\ \ This author's work was supported in part 
by the National Science Foundation under Grant DMS-8803396, by the
U.S. Air Force under Grants AFOSR-85-0247 and AFOSR-88-0235, and 
by the CAIP Center,
Rutgers University, with funds provided by the New Jersey Commission on
Science and Technology and by CAIP's industrial members.}
\vskip 0.05cm}
\def\startsyconb{
\vskip 1true cm
\centerline{\bf ABSTRACT}
\vskip 0.15true in
\noindent\theabstract
\vfill
\centerline{\sl Rutgers Center for Systems and Control,
\themonth\ \thelongyear}
\vfill}
\def\startsyconpaper{\startsycona
\mynameandaddress
\startsyconb
{\noindent {\mysmall\theextrafootnote}}
\vskip 0.05true cm
\hjsacknowledgement
\eject}
\def\startsyconpaperwitheduardo{\startsycona
\mynameandaddresswitheduardo
\startsyconb
\edsacknowledgement
\hjsacknowledgement
{\noindent {\mysmall\theextrafootnote}}
\eject}
\def\startsyconpaperwithkokotovic{\startsycona
\mynameandaddresswithkokotovic
\startsyconb
\kokoacknowledgement
\hjsacknowledgement
{\noindent {\mysmall\theextrafootnote}}
\eject}
\def\startsyconpaperwithkokotovicinv{\startsycona
\mynameandaddresswithkokotovicinv
\startsyconb
\hjsacknowledgement
\kokoacknowledgement
{\noindent {\mysmall\theextrafootnote}}
\eject}
\def\startsyconpapercaipwitheduardo{\startsycona
\mynameandaddresswitheduardo
\startsyconb
\edsacknowledgementcaip
\hjsacknowledgementcaip
{\noindent {\mysmall\theextrafootnote}}
\eject}
%MACROS FOR BOLTZMANN MACHINES PAPER
\def\rnbdw{NB\bigl(\rho,\check\cw(\cdot)\bigr)}
\def\ctrw{\ct_{\rho}\bigl(\bar\cw(\cdot)\bigr)}
\def\eps{\varepsilon}
\def\ffi{\varphi}
\def\pmath{{\rm I}\kern-.2em{\rm P}}
\def\emath{{\rm I}\kern-.2em{\rm E}}
\def\jk{{\bf J}_{k}}
\def\jkl{{\bf J}_{k}^{L}}
\def\jklr{{\bf J}_{k}^{L,r}}
\def\jkh{{\bf J}_{k}^{H}}
\def\tk{t_{k}}
\def\tkr{t_{k}^{r}}
\def\tkh{t_{k}^{H}}
\def\neqngwt{{\bf N}=({\cal N},{\cal G},\hat{\cal W},\theta)}
\def\pn{{\bf P}(\cn)}
%\def\pm{{\bf P}(\cm)}
\def\ppn{{\bf P}_{+}(\cn)}
\def\ppm{{\bf P}_{+}(\cm)}
\def\confn{I^{{\cal N}}}
\def\confh{I^{{\cal H}}}
\def\confm{I^{{\cal M}}}
\def\rrn{\rr^{\cal N}}
\def\rrg{\rr^{\cal G}}
\def\rrsg{\rr_{s}^{\cal G}}
\def\rrgn{\rr^{\cal G}\times\rr^{\cal N}}
\def\rrsgn{\rr_{s}^{\cal G}\times\rr^{\cal N}}   
\def\rrgpr{\rr^{{\cal G}'}}
\def\cg{{\cal G}}
\def\cl{{\cal L}}
\def\cn{{\cal N}}
\def\cc{{\cal C}}
\def\cm{{\cal M}}
\def\ch{{\cal H}}
\def\ck{{\cal K}}
\def\cw{{\cal W}}
\def\ct{{\cal T}}
\def\ca{{\cal A}}
\def\calat{{\cal A}_{t}}
\def\calatk{{\cal A}_{t_{k}}}
\def\calatkmin{{\cal A}_{t_{k}}^{-}}
\def\calatkr{{\cal A}_{t_{k}^{r}}}
\def\calatkh{{\cal A}_{t_{k}^{H}}}
\def\cz{{\cal Z}}
\def\cb{{\cal B}}
\def\cwbar{\bar{\cal W}}
\def\cwtil{\tilde{\cal W}}
\def\cwt{({\cal W},\theta)}
\def\cwthat{(\hat{\cal W},\theta)}
\def\gng{{\bf G}(\cn,\cg)}
\def\gngtwo{{\bf G}_{2}(\cn,\cg)}
\def\gmhg{{\bf G}(\cm,\ch,\cg)}
\def\gmhgtwo{{\bf G}_{2}(\cm,\ch,\cg)}
\def\ldata{(\bar P,\delta,\kappa,\nu,\cwtil,\tilde\tau)}
\def\rarr{\rightarrow}
\def\dcwk{\Delta\cw(k)}
\def\eijkone{E_{ij,k}^{1}}
\def\eijkthree{E_{ij,k}^{3}}
\def\eijkfour{E_{ij,k}^{4}}
\def\eijkfive{E_{ij,k}^{5}}
\def\eijkthreece{E_{ij,k}^{3,ce}}
\def\eijkfource{E_{ij,k}^{4,ce}}
\def\eijkfivece{E_{ij,k}^{5,ce}}
\def\eijkthreefl{E_{ij,k}^{3,fl}}
\def\eijkfourfl{E_{ij,k}^{4,fl}}
\def\eijkfivefl{E_{ij,k}^{5,fl}}
\def\eijkrtwo{E_{ij,k,r}^{2}}
\def\dwijk{\Delta w_{ij}(k)}
\def\dwijhk{\Delta w_{ij,H}(k)}
\def\dwijlk{\Delta w_{ij,L}(k)}
\def\dwijlrk{\Delta w_{ij,L,r}(k)}
\def\dwijlrkce{\Delta w_{ij,L,r}^{ce}(k)}
\def\dwijlkce{\Delta w_{ij,L}^{ce}(k)}
\def\dwijkce{\Delta w_{ij}^{ce}(k)}
\def\dwijlrkfl{\Delta w_{ij,L,r}^{fl}(k)}
\def\dwijlkfl{\Delta w_{ij,L}^{fl}(k)}
\def\dwijhkce{\Delta w_{ij,H}^{ce}(k)}
\def\dwijhkfl{\Delta w_{ij,H}^{fl}(k)}
\def\dwijkfl{\Delta w_{ij}^{fl}(k)}
\def\dcwkone{(\Delta\cw(k))^{1}}
\def\dwijkone{(\Delta w_{ij}(k))^{1}}
%MACROS FOR LETTER
\font\mymagfivebf=cmbx10 scaled\magstep 5
\font\mymagfiverm=cmr10 scaled\magstep 5
\font\mymagfourbf=cmbx10 scaled\magstep 4
\font\mymagfourrm=cmr10 scaled\magstep 4
\font\mymagthreebf=cmbx10 scaled\magstep 3
\font\mymagthreerm=cmr10 scaled\magstep 3
\font\mymagtwobf=cmbx10 scaled\magstep 2
\font\mymagtworm=cmr10 scaled\magstep 2
\font\mymagonebf=cmbx10 scaled\magstep 1
\font\mymagonerm=cmr10 scaled\magstep 1
\font\mymagzerobf=cmbx10 scaled\magstep 0
\font\mymagzerorm=cmr10 scaled\magstep 0
\font\mymagfivesl=cmsl10 scaled\magstep 5
\font\mymagfiveit=cmti10 scaled\magstep 5
\font\mymagfoursl=cmsl10 scaled\magstep 4
\font\mymagfourit=cmti10 scaled\magstep 4
\font\mymagthreesl=cmsl10 scaled\magstep 3
\font\mymagthreeit=cmti10 scaled\magstep 3
\font\mymagtwosl=cmsl10 scaled\magstep 2
\font\mymagtwoit=cmti10 scaled\magstep 2
\font\mymagonesl=cmsl10 scaled\magstep 1
\font\mymagoneit=cmti10 scaled\magstep 1
\font\mymagzerosl=cmsl10 scaled\magstep 0
\font\mymagzeroit=cmti10 scaled\magstep 0
\font\mynine=cmr9
\font\myninebf=cmbx9
\font\mysmall=cmr7
\font\mysmallit=cmti7
\font\myverysmall=cmr5
\def\compo{\hbox{$\,${\myverysmall o}$\,$}}
\newcount\lettermarker
%it's 0 while writing a letter and 1 for endfile and labels
\def\myownletterheadline
{\ifnum\pageno>1 \hfil\thedate\hfil P.\folio\else \hfil\fi}
\def\thephone{Phone: (908)445-5407[office]; (908)463-0667[home]}
\def\thephoneson{Phone: (908)445-3072}
\def\theemail{sussmann@hamilton.rutgers.edu\enspace---
sussmann@pisces.bitnet}
\def\theemailson{sontag@hilbert.rutgers.edu\enspace}
\def\thefax{FAX: (908)445-5530}
\def\plno#1{\hbox to 0.45true cm{\hfill #1}}
\def\pleft#1{\line{\noindent #1\hfil}}
\def\pleftskip#1{\line{\noindent #1\hfil}\vskip 0.1true cm}
\def\pleftre#1{\line{\noindent\hbox to 1.2true in{Re:\hfil}{\it #1}\hfil}}
\def\pleftrerm#1{\line{\noindent\hbox to 1.2true in{Re:\hfil}{#1}\hfil}}
\def\plefto#1{\line{\noindent\hbox to 1.2true in{\hfill}{#1}\hfil}}
\def\plright#1{\line{\noindent\hbox to 4.3true in{\hfil}#1\hfil}}
\def\plrightneural#1{\line{\noindent\hbox to 3.7true in{\hfil}#1\hfil}}
\def\pline#1{\line{\ \ \ \ \ \ #1\hfill}}
\def\pltwo#1#2{\line{\noindent\hbox to 3.8true in{#1\hfil}#2\hfil}}
\def\pltwoito#1#2{\line{\noindent\hbox to 1.8true in{#1\hfil}#2\hfil}}
\def\pltwoclose#1#2{\line{\noindent\hbox to 1.4true in{#1\hfil}#2\hfil}}
\def\toplogoa{\pageno=1 
\vglue-\voffset \vskip -1cm
\line{\noindent\mymagfivebf RUTGERS\hfill}
\line{\noindent\sl THE STATE UNIVERSITY OF NEW JERSEY\hfill}
\line{\leaders \hrule width16.5cm height10pt depth-9.6pt\hfil}}
\def\toplogob{\pageno=1 \vglue-\voffset \vskip -1cm
\centerline{\hbox to 5.2cm{\leaders\hrule height-1.2pt depth1.6pt\hfill}}
\centerline{\hbox to 5.2cm{\sl T\hfil H\hfil E\hfil\hfil\hfil\hfil\hfil
S\hfil T\hfil 
A\hfil T\hfil E\hfil\hfil\hfil\hfil\hfil\hfil U\hfil N\hfil I\hfil V\hfil
E\hfil R\hfil S\hfil I\hfil T\hfil Y}}
\vskip 0.4pt
\centerline{\underbar{\hbox to 5.2cm{\mymagfivebf  R\hfil U\hfil T\hfil G\hfil 
E\hfil R\hfil S}}}}
\def\toplogoc{\pageno=1 \vglue-\voffset \vskip -1cm
\centerline{\hbox to 5.9cm{\leaders\hrule height-1.2pt depth1.6pt\hfill}}
\vskip -0.1true cm
\centerline{\hbox to 5.9cm{\mysmall \hfill THE STATE UNIVERSITY OF NEW JERSEY
\hfill}}
\vskip 0.5pt
\centerline{{\hbox to 5.9cm{\mymagfivebf \hfil R\hfill U\hfill T\hfill G\hfill 
E\hfill R\hfill S\hfil}}}
\vskip -0.15true cm
\centerline{{\hbox to 5.9cm{\mysmall \hfill Campus at New Brunswick\hfill}}}
\vskip -0.45true cm
\centerline{\hbox to 5.9cm{\leaders\hrule height-1.2pt depth1.6pt\hfill}}}
\def\toplogom{\pageno=1 
\vglue-\voffset \vskip -1cm
\line{\noindent\mymagthreebf Mathematics of Control, Signals and Systems\hfill}
\line{\leaders \hrule width16.5cm height10pt depth-9.6pt\hfil}}
\def\nextlogoa{\vskip-9pt
\pltwo{\mysmall H. J. Sussmann}{\it Department of Mathematics}
\pltwo{\mysmall\thephone}{\it Hill Center\ -\ Busch Campus}
\pltwo{\mysmall\theemail}{\it New Brunswick, NJ 08903, USA}
\pltwo{\mysmall\thefax}{\it Telephone: (908)445-2390}
\vskip 0.2true in
\plright{\thedate}}
\def\nextlogoason{\vskip-9pt
\pltwo{\mysmall E. D. Sontag}{\it Department of Mathematics}
\pltwo{\mysmall\thephoneson}{\it Hill Center\ -\ Busch Campus}
\pltwo{\mysmall\theemailson}{\it New Brunswick, NJ 08903, USA}
\pltwo{\mysmall\thefax}{\it Telephone: (908)445-2390}
\vskip 0.2true in
\plright{\thedate}}
\def\nextlogoawithout{\vskip-9pt
\pltwo{\ \ }{\it Department of Mathematics}
\pltwo{\ \ }{\it Hill Center\ -\ Busch Campus}
\pltwo{\ \ }{\it New Brunswick, NJ 08903, USA}
\pltwo{\ \ }{\it Telephone: (908)445-2390}
\vskip 0.2true in
\plright{\thedate}}
\def\nextlogob{\vskip 0.1true cm
\centerline{\it Department of Mathematics \enspace$\bullet$\enspace Hill Center for the Mathematical Sciences} 
\centerline{\it Busch Campus\enspace$\bullet$\enspace New Brunswick, New Jersey 08903, 
USA \enspace$\bullet$\enspace Telephone: (908)445-2390}
\vskip 0.1 true cm
\centerline{\mysmall H. J. Sussmann\enspace --\enspace\thephone}
\vskip-3pt
\centerline{\mysmall Electronic Mail:\theemail}
\vskip-3pt
\centerline{\thefax}
\vskip 0.3true cm
\plright{\thedate}}
\def\nextlogom{\vskip-9pt
\pltwo{\mysmall H. J. Sussmann}{\it Department of Mathematics}
\pltwo{\mysmall\thephone}{\it Rutgers University}
\pltwo{\mysmall Electronic Mail:}{\it New Brunswick, NJ 08903, USA}
\pltwo{\mysmall\theemail}{\it Telephone: (908)445-2390}
\vskip 0.2true in
\plright{\thedate}}
\def\logoa{\toplogoa\nextlogoa}
\def\logoason{\toplogoa\nextlogoason}
\def\logoawithout{\toplogoa\nextlogoawithout}
\def\logob{\toplogob\nextlogob}
\def\logoc{\toplogoc\nextlogob}
\def\logom{\toplogom\nextlogom}
\def\sighjs{\plright{H. J. Sussmann}}
\def\sighjsneru{\pltwo{Mahesh Nerurkar}{H. J. Sussmann}}
\def\sighjsneural{\plrightneural{H. J. Sussmann}\nobreak
\plrightneural{Associate Editor, {\it Neural Networks}}}
\def\sighjswithsontag{\pltwo{E. D. Sontag}{H. J. Sussmann}}
\def\sighjswithsontagindent{\pltwo{\ \ \ \ \ \ E. D. Sontag}{H. J. Sussmann}}
\def\sighjswithmcclam{
\pltwo{\hbox to 5true cm{\hfill N. Harris McClamroch \hfill}}
{\hbox to 5true cm{\hfill H\'ector J. Sussmann \hfill}}}
\def\sighejs{\plright{H\'ector J. Sussmann}}
\def\sighjsprof{
\plright{H. J. Sussmann}\nobreak
\plright{\it Professor of Mathematics}}
\def\finhjsneural#1#2{
\vskip 0.5true in\sighjsneural\vfil#1\vfil\vfil\pleft{#2}\vfil\eject}
\def\finhjs#1#2{
\vskip 0.5true in\sighjs\vfil#1\vfil\vfil\pleft{#2}\vfil\eject}
\def\finhjsneru{
\vskip 0.3true in\sighjsneru\vfil\eject}
\def\finhjswithsontag#1#2{
\vskip 0.8true in\sighjswithsontag\vfil#1\vfil\vfil\pleft{#2}\vfil\eject}
\def\finhjswithsontagshort#1#2{
\vskip 0.4true in\sighjswithsontagindent\vfil#1\vfil\vfil\pleft{#2}\vfil\eject}
\def\finhjsson#1#2{
\vskip 0.8true in\sighjswithsontag\vfil#1\vfil\vfil\pleft{#2}\vfil\eject}
\def\finhjswithmcclam#1#2{
\vskip 0.8true in
\pltwo{\hbox to 5true cm{\hrulefill}}{\hbox to 5true cm{\hrulefill}}
\sighjswithmcclam\vfil#1\vfil\vfil\pleft{#2}\vfil\eject}
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}}}



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Rend. Sem. Mat. Univ. Padova} {\bf 95}, (1996), pp. 59-79.
}}}

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%draft: 17 Februrary 1998



\vskip 2cm

\centerline{\bf REGULAR SYNTHESIS} 
\centerline{\bf AND SUFFICIENCY CONDITIONS FOR OPTIMALITY}

\vskip 4em

\centerline{Benedetto Piccoli}
\centerline{\sl SISSA-ISAS}
\centerline{\sl via Beirut 2-4, 34014 Trieste, Italy}

\vskip 1em

\centerline{H\'ector J. Sussmann$^*$}
\centerline{\sl Department of Mathematics, Rutgers University}
\centerline{\sl 110 Frelinghuysen Rd, Piscataway, New Jersey, NJ 08854}

\vskip 21em

\noindent
{\bf Abstract.} 
The aim of this paper is to propose a new definition of "regular synthesis" 
more general than those suggested by other authors (e.g. Boltianskii
\boltcount, Brunovsky \brunacount, \brunbcount) and to give a detailed
proof of the 
corresponding sufficiency theorem, a slightly weaker version of which
had been stated in an earlier article (see \sussmcount), with
only a rough outline of the proof.
We illustrate the strength of our result showing that the synthesis for 
the famous Fuller problem \fullcount{} satisfies our hypotheses.

\gnu{This abstract is too short and uninformative.  Look at the new
abstract that I wrote.}

\vskip 1em

\noindent
{\bf Keywords.} Optimal control, regular synthesis, sufficient conditions.


\vskip 3em
\centerline{Ref. SISSA 68/96/M}

\vskip 1em
\gnu{``Ospitality'' is wrong!} 
$(^*)$ Work partially supported by NSF grant DMS95-00798. Part of this work was
done while the author was a visitor of the Centre Emile Borel at the IHP
in Paris for whose ospitality he is deeply grateful.
\vfill\eject


\parnormal


\section{Introduction}

The aim of this paper is to propose a new definition of "regular synthesis"
more general than those suggested by other authors (e.g. Boltianskii
\boltcount, Brunovsky \brunacount, \brunbcount) and to give a detailed
\gnu{In the SIAM J. Control paper, the name of VGB is spelled 
``Boltyanskii,'' so I am using the same spelling here. 
Also, ``Brunovsk\'y'' is supposed to have an accent on the ``y''.}
proof of the corresponding sufficiency theorem, a slightly weaker version
of which had been stated in an earlier article (see \sussmcount), with only
a rough outline of the proof. We illustrate the strength of our result
showing that the synthesis for the famous Fuller problem
\fullcount{} satisfies our hypotheses, but fails to obey the
requirements of other sufficiency theorems such as those of
\boltcount{} and \brunbcount, because of the technical problem arising
from the occurrence of an infinite number of switchings.  (Not
surprisingly, the optimality of Fuller's synthesis 
was originally proved without resorting to ``regular synthesis''
arguments, and using instead special symmetry properties of the
problem, cf. e.g. Zelikin-Borisov \zelikincount.)  

\gnus{One of the most terrible structural flaws of this paper is that we start
by saying that in this paper we propose a new definition of "regular
synthesis" and then we never give the definition. Anybody looking at the
introduction of the paper is going to start turning the pages to look for
this ``new definition'' that we say we are proposing. And he won't find it,
because IT'S NOT THERE. {\bf Exactly where do we give this ``definition''
that we say we are going to give? I didn't see it anywhere!!!!}  In my new
version, I have put it in. Now there are definitions clearly labeled as
such, and the reader can tell what it is that we are defining. 
Read the paper carefully and see how much difference this change
makes!!!  After reading your version, NOBODY (not even ourselves) would
have been able to answer the question: ``what is a regular synthesis
according to Piccoli and Sussmann''?  After reading my new version, I think
the question has a clear answer.  Am I right?}
\gnum{Also, notice that the conceptual structure now is much clearer. We
define ``presynthesis,'' and then we define what means for a presynthesis 
to be ``regular.''  In order to make this conceptual structure even
clearer, I have separated the boundary condition from the
``regularity'' part. Properly speaking, the condition that in your
version is called 2.1.b is not a ``regularity'' condition at all, so 
I didn't incorporate it into the regularity notion.}
\gnue{Incidentally, since in the new version I am explicitly labeling the
definition of a regular presyntheis as a definition, numbered and
everything, I thought I might as well do the same thing with other
definitions. Take a look at new product and see for yourself how it's
much clearer now.}

The purpose of the theory of regular synthesis is to turn into a rigorous,
precise theorem the vague assertion that "if a collection $\Gamma$ of
extremals covers the whole space in a sufficiently smooth way, then all the
members of $\Gamma$ are optimal". Naturally, a good definition will be one
that is stringent enough to lead to a correct proof of the sufficiency
assertion, and broad enough to cover as large as possible a class of
optimal control problem. In particular, it would be desirable for the
\gnu{It should be ``problems,'' not ``problem.''}
theory to apply to well known examples where discontinuous optimal controls
occur, and even controls with infinitely many switchings, as in Fuller's
problem.

There are essentially two different ways of defining a synthesis.
Following the approach of \boltcount, \brunacount, \cesaricount, \flemcount,
\sussacount, a synthesis is a collection of trajectories generated
by a piecewise regular feedback control. The set of interest (that is 
the set of initial points of optimal trajectories) is partitioned 
into a locally finite collection of submanifolds of various dimensions. 
More precisely, the set is stratified and
for each stratum a regular feedback control is specified.
An alternative strategy was followed in \piccacount, \sussmcount, where
a "synthesis" is just a collection of trajectories not necessarily
arising from a feedback control. With this definition 
the feedback control is not sufficient for
describing a synthesis, indeed
a discontinuous feedback control may generate too many trajectories, some
of which could fail to be optimal, as
shown in Example~1 of Section~5. 
\gnu{I don't understand the grammatical structure of the above sentence,
especially the role of ``indeed.'' You seem to to think that ``indeed''
is a conjunction, but it isn't. It's an {\bf adverb}.}
This same example illustrates another way
in which our definition of synthesis is more general than that of [2] and [6],
because we allow the cost function associated to the synthesis to be
discontinuous, whereas [2] and [6] require it to be continuous
\gnu{The definitions of [2] and [6] are different, so we should 
have written ``those of [2] and [6]'' rather than 
``that of [2] and [6].''} 
Another point that should be further investigated is the relation
between the geometrical approaches of \brunacount, \brunbcount,
\sussacount, \sussbcount,
\sussccount, \sussecount, \sussfcount, \susslcount, \sussmcount, 
and the theory of
viscosity solutions \bardicount, \flemsonercount. 
We point out that the existing results of this theory for deterministic
optimal control do not appear to cover cases such as Fuller's problem,
cfr. \bardicount, where the main result is for Lagrangian that are bounded
away from zero.



\vs

\def\nin{\noindent}
\def\vs{\vskip 1em}
\def\R{\rr}
\def\C{{\cal C}}
\def\tf{\tilde f}
\def\xb{\bar x}
\def\g{\gamma}
\def\gx{\gamma_x}
\def\gxb{\gamma_{\bar x}}
\def\e{\eta}
\def\ex{\eta_x}
\def\exb{\eta_{\bar x}}
\def\m{\mu}
\def\mx{\mu_x}
\def\mxb{\mu_{\bar x}}
\def\n{\nu}
\def\nx{\nu_x}
\def\nxb{\nu_{\bar x}}
\def\vp{\varphi}
\def\ve{\varepsilon}
\def\veb{\bar\varepsilon}
\def\pe{\psi_{\ve}}
\def\la{{\lambda}}
\def\l0{\lambda^0}
\def\tx{T_x}
\def\txb{T_{\bar x}}


% \input draft.tex

\section{A sufficiency theorem}

\gnu{I changed the title section. The new one is more informative, I
think.}

\noindent We start with some definitions that will enable us to state and
prove a mild generalization of
the theorem whose proof was outlined in \sussmcount.  The
assumptions made here will be minimal, so as to achieve maximum generality.
In subsequent sections we will impose more restrictive conditions and apply
our result to other definitions of synthesis given in the literature only
for more regular control systems.

We consider a control system:
\xdef\system{\equanosectno}
$$\dot x=f(x,u)\,,\qquad x\in\Omega\,,\qquad u\in U\,,\numeq$$
and the family $\{{\cal P}(f,L,{\cal T},\tau,x)\}_{x\in\Omega}$ of
minimization problems, where  ${\cal P}(f,L,{\cal T},\tau,x)$ is the 
problem: 
\xdef\minim{\equanosectno}
$${\rm minimize\,\,\,\,\,}\int_a^b L(\g(t),\e(t))dt+\tau(\gamma(b))
\qquad{\rm subject\,\,\,to}\,\,\,(\system)\,,\,\,\,\g(a)=x\,,\,\,\,
{\rm and}\,\,\,\g(b)\in{\cal T}\,.\numeq$$
(A more precise formulation of the problems ${\cal P}(f,L,{\cal T},\tau,x)$,
carefully specifying the space of of curves to be considered, will be given
below. Since both $f$ and $L$ have no explicit dependence on $t$, 
we can always take $b=0$.)

We assume that 

\medbreak
{\sl 
{\pari
\item{(A1)} $\Omega$ is an open subset of $\R^n$,

\item{(A2)} $U$ is a set, 

\item{(A3)} $f:\Omega\times U\rightarrow \R^n$ and $L:\Omega\times
U\rightarrow \R$ are maps,

\item{(A4)} $f(\cdot,u)$ and $L(\cdot,u)$ are
maps of class ${\cal{C}}^1$  for every fixed $u\in U$,

\item{(A5)} ${\cal T}$ is a nonempty subset of $\Omega$,

\item{(A6)} $\tau:\Omega\to\R$ is a function of class ${\cal{C}}^1$.
}
}

\medbreak\noindent
We use $T_x^cS$ to denote the {\sl contingent tangent}
cone to a set $S$ at a point 
$x\in S$, defined as follows: a vector $v$ is in $T_x^cS$ iff there exists
a sequence $\{x_j\}$ of points of $S$ and a sequence $\{h_j\}$ of strictly
positive numbers such that $h_j\to 0$ and $x_j=x+h_jv+o(h_j)$ as $j\to\infty$.
We write 
$$
\tilde T_xS=T_x^cS\cap(-T_x^cS)=\{v:v\in T_x^cS\quad{\rm and\quad}-v\in 
T_x^cS\}\,.\numeq
$$
 

\medbreak\noindent
As in Sussmann \sussmcount, we use $\tf$ to denote the $\R^{n+1}$ valued map
$(x,u)\to (f(x,u),L(x,u))$.

If $\mu:A\to B$ is a map, we use $\Dom(\mu)$ to indicate the domain
of $\mu$, i.e. the set $A$. 

A {\sl  control} is a $U$-valued map $\eta$ such that $\Dom(\eta)$
is a compact subinterval $[a,b]$ of $\R$.
A {\sl trajectory} $\gamma$ for a control $\eta$ is an absolutely
continuous map $\gamma:\Dom(\eta)\to\Omega$ that satisfies
$\dot\gamma(t)=f(\gamma(t),\eta(t))$ for almost every $t\in \Dom(\eta)$.  If
$\Dom(\eta)=[a,b]$ (so that $\Dom(\gamma)=[a,b]$ as well), and $x=\gamma(a)$,
$y=\gamma(b)$, we say that $\gamma$ (or the pair $(\gamma,\eta)$)
{\sl goes from $x$ to $y$}, and that $\eta$ {\sl steers $x$ to $y$},
and we write $\gamma^-=x$, $\gamma^+=y$.

A control $\eta:[a,b]\rightarrow U$ is {\sl admissible} if the
map $\Omega\times[a,b]\ni(x,t)\to\tf_{\eta}(x,t)=\tf(x,\eta(t))
\in\R^{n+1}$ satisfies the following ${\cal{C}}^1$ Carath\'eodory conditions:

\medbreak
{\pariim

\item{(C.i)} {\sl $t\mapsto \tf_{\eta}(x,t)$ is measurable for each $x\in\Omega$,}
\vs
\item{(C.ii)} {\sl for every compact $K\subseteq\Omega$ there exists an
integrable function $\vp_K$ such that for every $x\in K$ and $t\in[a,b]$:}
\xdef\intbound{\equanosectno}
$$\|\tf_{\eta}(x,t)\|+\|D_x\tf_{\eta}(x,t)\|\leq \vp_K(t)\,.\numeq$$

}

\medbreak\noindent If $\eta$ is an admissible control and $\gamma$ a
trajectory corresponding 
to $\eta$, then we say that $(\gamma,\eta)$ is an {\sl admissible pair} for
$\tf$.  We use $\Adm(\tf)$ to denote the set of all admissible pairs for
$\tf$, and write $\Admo(\tf)$, $\Admo(\tf,{\cal T})$, $\Admo_x(\tf,{\cal
T})$, to denote, respectively, the set of all $(\g,\e)\in\Adm(\tf)$ such
that $\Dom(\e)$ is an interval of the form $[T,0]$ with $T\le 0$, the set
of those $(\g,\e)\in\Admo(\tf)$ such that $\gamma(0)\in{\cal T}$, and the
set of those $(\g,\e)\in\Admo(\tf,{\cal T})$ such that $\g^-=x$.  For an
admissible pair $(\gamma,\eta)\in\Admo(\tf,{\cal T})$, we use
$J(\gamma,\eta)$ to denote the {\sl cost} of $(\gamma,\eta)$, so that
$$
J(\g,\e)=\int_{T}^{0} L(\g(t),\e(t))\,dt+\tau(\g^+)\,.
$$
If $\Gamma\!\subseteq\!\Admo(\tf,{\cal T})$ is an arbitrary set of admissible
pairs, we 
define a function $V_{\Gamma}\!:\!\Omega\!\mapsto\!\R\cup\{+\infty\}$,
called the 
{\sl value function of $\Gamma$}, by letting 
$$
V_{\Gamma}(x)=\inf\{J(\gamma,\eta):(\gamma,\eta)\in\Gamma,\,\gamma^-=x\}\numeq
$$ 
for $x\in\Omega$.  
(In particular, $V_{\Gamma}(x)=+\infty$ iff there is no
$(\gamma,\eta)\in\Gamma$ starting at $x$.)  If $\Gamma=\Admo(\tf,{\cal T})$,
then the function $V_{\Gamma}$ is the {\sl value function} of our problem,
and in that case we will write $V_{\tf,{\cal T}}$ rather than
$V_{\Admo(\tf,{\cal T})}$.  

We can now formulate ${\cal P}(f,L,{\cal T},\tau,x)$ precisely: 
${\cal P}(f,L,{\cal T},\tau,x)$ is the problem of finding a 
$(\g,\e)\in\Admo_x(\tf,{\cal T})$ such that 
$J(\g,\e)=V_{\tf,{\cal T}}(x)$.   Any such pair is said to be a
{\sl solution} of ${\cal P}(f,L,{\cal T},\tau,x)$.    A pair 
$(\g,\e)\in\Admo(\tf,{\cal T})$ is {\sl optimal} if it is a solution
of ${\cal P}(f,L,{\cal T},\tau,\g^-)$.

Given $\la\in\R_n$ (the space of row $n$ vectors), 
$\l0\in\R$, $x\in\Omega$ and $u\in U$, we define:
\xdef\hamil{\equanosectno}
$${\cal H}(\la,\l0,x,u)=\langle \la , f(x,u)\rangle +\l0 L(x,u)\,,\numeq$$
\xdef\minhamil{\equanosectno}
$$H(\la,\l0,x)=\inf\{{\cal H}(\la,\l0,x,u):u\in U\}\,.\numeq$$
The functions ${\cal H}:\R_n\times\R\times\Omega\times U\to\R$
and $H:\R_n\times\R\times\Omega\to\R\cup\{-\infty\}$
are known as the {\sl Hamiltonian} and the {\sl minimized Hamiltonian}
of $\tf$.

We say that the pair $(\gamma,\eta)\in \Admo(\tf,{\cal T})$ 
is {\sl extremal} if there exist an
absolutely continuous map $\la:{\rm Dom}(\g)\mapsto \R_n$
and a constant $\l0\geq 0$, that satisfy:

\medbreak


{\pariim
\item{(EX1)} {\sl $(\l0,\la(t))\neq (0,0)$ for some (and hence every) $t$.}

\midbreak

\item{(EX2)} {\sl $\dot\la = -{\partial {\cal H}\over \partial x}
(\la(t),\l0,\gamma(t),\eta(t))$ for a.e. $t\in\Dom(\e)$.}

\midbreak

\item{(EX3)} $H(\gamma(t),\la(t),\l0)={\cal H}(\gamma(t),\la(t),\l0,\eta(t))=0$
{\sl for a.e.}  $t\in \Dom(\eta)$, 

\midbreak

\item{(EX4)} $\langle\la(0),v\rangle=\l0\langle \nabla\tau(\g(0)),v\rangle
\,\,\,{\sl for\,\,\,all}\,\,\,v\in \tilde T_{\g(0)}{\cal T}$.  

}

\medbreak\noindent
Conditions (EX1), (EX2), (EX3), (EX4) are, respectively,
the {\sl nontriviality condition}, the {\sl adjoint equation},
the {\sl Hamiltonian minimization condition}, and the {\sl transversality
condition}. A pair $(\la,\l0)$ such that (EX2) holds is called
an {\sl adjoint vector for} $(\g,\e)$.  An adjoint vector for which (EX3) 
holds is a {\sl minimizing adjoint vector for} $(\g,\e)$.  

The {\sl maximum principle} says that, under some special conditions
on the target set ${\cal T}$, {\sl every optimal pair $(\g,\e)$
is extremal.}
(This is true, for example, if ${\cal T}$ is a submanifold of $\Omega$ 
of class ${\cal{C}}^1$, or a closed convex subset of $\Omega$, or
a set which is locally equivalent to a convex set via a ${\cal{C}}^1$
diffeomorphism.  More generally, it is true if the cone 
$\tilde T_{\g(0)}{\cal T}$ is convex ---in which case it is of course a 
subspace, since $v\in\tilde T_{\g(0)}{\cal T}\Rightarrow
-v\in\tilde T_{\g(0)}{\cal T}$---  and is an approximating cone 
to ${\cal T}$ at $\g(0)$.   Recall that a closed convex cone $C$ is an
{\sl approximating cone} to a set $S$ at a point $s\in S$ if there exist a
neighborhood $U$ of $0$, and a continuous map $F:U\cap C\to S$,  
such that $F(x)=s+x+o(||x||)$ as $x\to 0$ via values in $C$.) 

\medbreak

A {\sl presynthesis} for our family of problems is a subset
$\Gamma$ of $\Admo(\tf,{\cal T})$ such that,

{\pariim

\item{(PS)} {\sl Whenever $(\gamma_1,\eta_1)\in\Gamma$,
$(\gamma_2,\eta_2)\in\Gamma$, 
and $\g_1^-=\g_2^-$, it follows that $(\gamma_1,\eta_1)=(\gamma_2,\eta_2)$.}

}

The set $\Dom(\Gamma)\equaldef\{\g^-:(\g,\e)\in\Gamma\}$ is called the {\sl
domain} of $\Gamma$.  We say that $\Gamma$ is a {\sl presynthesis on} $S$ 
if $\Gamma$ is a presynthesis and $S=\Dom(\Gamma)$.  Clearly, 
giving a presynthesis on $S$  amounts to choosing, 
for each $x\in S$, a $(\g,\e)\in \Admo(\tf,{\cal T})$ such that
$\g^-=x$.  

If $\Gamma$ is a presynthesis such that $\Dom(\Gamma)$ consists of all
points that can be steered to the origin in finite time by an admissible
pair, then we say that $\Gamma$ is {\sl total}.  

Given a presynthesis $\Gamma$ and an $x\in\Dom(\Gamma)$ 
we will always use $(\gx,\ex)$ to denote the unique pair ($\g,\e)\in\Gamma$ 
such that $\g^-=x$, and will write  $T_x=\min\,\Dom(\g_x)$, so 
$T_x\le 0$ and $\Dom(\g_x)=\Dom(\e_x)=[T_x,0]$.

A presynthesis is {\sl memoryless} if
whenever $x\in S$ and $t\in \Dom(\ex)$ it follows that $y=\gx(t)$ belongs to
$S$ and $\eta_y$ is the restriction of $\ex$ to the interval $[t,0]$.  A
{\sl synthesis} is a memoryless presynthesis.  If each pair of a
presynthesis $\Gamma$ is optimal (resp. extremal) then we say that $\Gamma$
is {\sl optimal} (resp. {\sl extremal}).  In particular, a presynthesis
$\Gamma$ is a total optimal presynthesis if and only if $V_{\Gamma}\equiv
V_{\tf}$.

Our goal is to prove that, under suitable regularity assumptions on
$\Gamma$ and $V_{\Gamma}$, a total extremal presynthesis is optimal.  We
first describe these regularity assumptions in detail.

We say that a function $V:\Omega\rightarrow \R\cup \{-\infty,+\infty\}$ is
{\sl weakly upper semicontinuous} (w.u.s.c) at a point $x\in\Omega$ if
$$
\lim\inf_{y\to x} \lim\sup_{z\to y}V(z)\le V(x)\,.\numeq
$$

Equivalently, $V$ is w.u.s.c. at $x$ if 
there exist sequences $x_j$, $\ve_j$ and $\delta_j$ such that
$x_j\in\Omega$, $x_j\rightarrow x$, $\ve_j>0$, $\delta_j>0$,
$\delta_j\rightarrow 0$, having the property that $V(y)\le V(x)+\delta_j$
whenever $\|y-x_j\|<\ve_j$.

Given a locally Lipschitz vector field $X$ on $\Omega$, we say that $V$ has
the {\sl NDJ (``no downward jumps'') property along $X$} if the following
holds:

{\pariim

\medbreak

\item{(NDJ)}  {\sl For every $\gamma$ integral curve of $X$, 
if $\Dom(\gamma)=[a,b]$ then for every $t\in ]a,b]$,
$$\lim\sup_{h\downarrow 0} V(\gamma(t)) - V(\gamma(t-h))\geq 0$$.}
%if $\{t_j^-\}$, $\{t_j^+\}$ are sequences such that
%$t_j^-\leq t\leq t_j^+$, $t_j^-,t_j^+\rightarrow t$, and 
%$L=\lim_j [V(\gamma(t_j^+))-V(\gamma(t_j^-))]$, then $L\geq 0$.}


\medbreak
}


Finally we say that $V$ satisfies the {\sl weak continuity conditions} for
the control problem (\system), (\minim) if $V$ is lower semicontinuous and
weakly upper semicontinuous, and has the NDJ property along the vector
field $x\to f(x,u)$ for every $u\in U$.

We now define the ``weak differentiability conditions'' for $\Gamma$.  In
this definition it is understood that $\tf(y,\ex(t))=0$ for every $t\notin
\Dom(\ex)$.  A presynthesis $\Gamma=\{(\gamma_x,\eta_x):x\in S\}$ is {\sl
$(f,L)$--differentiable} at a point $\xb\in\Omega$ if the following 
condition is satisfied:

\medbreak

{\pari

\item{(DC)} {\sl there exist a neighborhood $N$ of $\xb$ (in $\Omega$)
such that $N\subseteq \Dom(\Gamma)$, and an interval $[T,0]$ such that 
$\Dom(\ex)\subseteq [T,0]$ for all $x\in N$,  and an $\bar\ve>0$, 
with the property that}

}

\vs

{\pariim

\item{(DC1)} {\sl there exist, for $\ve\in]0,\bar\ve]$, integrable
functions $\pe :[T,0]\rightarrow \R$ such that
$$
\lim_{\ve\to 0}\int_T^0 \pe(t)\, dt=0\,,\numeq
$$
and the inequality}

}

$$\biggl\|\tf(y,\ex(t))-\tf(\gx(t),\ex(t))-{\partial \tf\over \partial
x}\Bigl(\gxb(t),\exb(t)\Bigr)\cdot(y-\gxb(t))\biggr\|\leq\pe(t).
(||y-\gxb(t)||+||x-\bar x||)\numeq
$$

{\pariim

\item{} {\sl holds for every $y\in\Omega$, $x\in N$,  $t\in[T,0]$ such that 
$\|y-\gxb(t)\|\leq\ve$ and $||x-\bar x||\le\ve$.}

\vs
\item{(DC2)} {\sl the map $v\rightarrow\rho_v$, where $\rho_v$ is the
integrable $\R^{n+1}$ valued function on $[T,0]$ given by}
$$\rho_v(t)=\tf(\gxb(t),\eta_{\xb+v}(t))-\tf(\gxb(t),\exb(t))$$

\item{} {\sl is weak$^*$-differentiable at $v=0$, regarded as a map into the
space of $\R^{n+1}$--valued Borel measures. That is: for every continuous
function $\alpha:[T,0]\rightarrow\R$, the map 
$\R^n\ni v\to \int_T^0
\rho_v(t)\alpha(t) \, \in\R^{n+1}dt$ is differentiable at $v=0$.}

\vs
}

A subset $A$ of $\Omega$ is {\sl thin} if there exist $A_1$, $A_2$,
such that $A=A_1\cup A_2$, $A_1$ is a finite or countable union of
connected ${\cal{C}}^1$ submanifolds of positive codimension and
${\cal{}H}^{n-1}(A_2)=0$, where ${\cal{}H}^k$ denotes $k$-dimensional
Hausdorff measure.

\gnu{I completely changed the statement of the main theorem. 
As you will see, now I no longer asusme that the synthesis is total.
One of many reasons why I wanted to do this is the following:
in our previous version, we lied, because we said that the theorem here is
``slightly more general'' than that of my presynthesis paper, but in fact it
wasn't. It was exactly the same. Now, with the new conditions on the
domain, it is true that it is ``slightly more general.''}

We can now state the main theorem (see \sussmcount):
\vs
\xdef\maintheo{\the\sectno.\the\theono}
\theo let $\Omega$, $U$, $f$, $L$ be such that
Assumptions (A1) -- (A6) hold.
Let $\Gamma$ be a total extremal
presynthesis. Assume that 

{\pariim
{\sl 
\item{(\maintheo.a)} the associated value function $V_{\Gamma}$
satisfies the weak continuity conditions, 

\midbreak

\item{(\maintheo.b)} $V_{\Gamma}(x)\le\tau(x)$ for every $x\in{\cal T}$,

\midbreak

\item{(\maintheo.c)}  $\Gamma$ is
$(f,L)$--differentiable at all points in the complement of a thin set $A$.
}

}

\midbreak

\noindent {\sl Then $\Gamma$ is optimal.}

\medbreak

\beginproof
We let $S={\rm Dom}(\Gamma)$, so $S$ is exactly the set of 
all $x\in\Omega$ that can be steered to the target ${\cal T}$ in
finite time by an admissible control.

We first establish the relation between the differential of the value
function $V_{\Gamma}$ and the adjoint covector fields along the
extremal trajectories of the presynthesis.   This result is important in
its own right, so we state it as a separate theorem:

\xdef\maintheodiff{\the\sectno.\the\theono}
\theo Let $\Omega$, $U$, $f$, $L$ be such that Assumptions (A1) -- (A6) hold.
Let $\Gamma=\{(\gamma_x,\eta_x):x\in S\}$ be a 
presynthesis. Let $\bar x\in S$ be such that 
$\Gamma$ is
$(f,L)$--differentiable at $\bar x$.  Then 

{\pariim
{\sl 
\item{(\maintheodiff.a)} the function $V_{\Gamma}$ is differentiable 
at $\bar x$;

\midbreak

\item{(\maintheodiff.b)} if the pair $(\gamma_{\bar x},\eta_{\bar x})$ is
extremal, then every minimizing adjoint vector $(\la,\l0)$ for 
$(\gamma_{\bar x},\eta_{\bar x})$ satisfies the identity
$\l0 D_y V_{\Gamma}(\xb)= \la(\txb)$. In particular, 
$\l0\neq 0$, and $(\la,\l0)$ is unique up to multiplication by a positive 
constant. 

}

}


\medbreak\noindent{\sl PROOF OF THEOREM \maintheodiff.\enspace}
We use Theorem A.5 in Appendix A with $P=\R^n$, 
letting $f_x(y,t)=f(y,\ex(t))$, $L_x(y,t)=L(y,\ex(t))$, for 
$y\in\Omega$, $x\in N$, $t\in[T_x,0]$, and defining 
$f_x(y,t)=0$, $L_x(y,t)=0$, for $T\le t<T_x$.  
Let $\hat V_\Gamma(x)=\int_{T_x}^0L(\gx(t),\ex(t))\,dt$, so 
$\hat V_\Gamma$ is the ``Lagrangian part'' of our cost functional.

The map $p\mapsto \bar x_p$ of Theorem A.5 is the identity map,
which is certainly differentiable at $\bar x$.   So
Theorem A.5 tells us that the ``endpoint map''$x\mapsto
{\cal E}(x)\equaldef \gx(0)$ and the function
$\hat V_\Gamma$ are differentiable at $\bar x$,  and the identity 
$$
\lambda(0).D_y{\cal E}(\bar x).v+\lambda^0 D_y\hat V_{\Gamma}(\bar x).v
-\lambda(T_{\bar x})
\cdot v\phantom{aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa}
$$
\xdef\identi{\equanosectno}

\vskip -1.2cm
$$
\phantom{aaaaaaaaaaaaaa}=\lim_{\ve\downarrow 0}\,\,\ve^{-1}\!\int_T^0\!\Big(
H(\la(t),\l0,\gxb(t),\eta_{\bar x +\ve v}(t))-
H(\la(t),\l0,\gxb(t),\exb(t))\Big)\,dt.\numeq
$$
holds for every vector $v$ if $(\la,\l0)$ is an adjoint vector
for $(\gxb,\exb)$. (Here the function $\la$ is extended 
to $[T,0]$ by letting it equal zero on $[T,T_{\bar x}[$.
The Hamiltonian $H(\la(t),\l0,y,\ex(t))$ is 
also defined for all $t\in[T,0]$, by letting 
$H((\la(t),\l0,y,\ex(t))=\la(t).f(y,\ex(t))+\l0 L(y,\ex(t))$, 
and recalling that this expression vanishes for $t\in[T,T_x[$,
because $f(y,\ex(t))$ and $L(y,\ex(t))$ have been defined to be 
zero for $t\in[T,T_x[$.)

Now, if $(\la,\l0)$ is minimizing, then the right-hand side
of the inequality (\identi) is nonnegative, because
$H(\la(t),\l0,\gxb(t),\exb(t))=0$ and 
$H(\la(t),\l0,\gxb(t),\eta_{\bar x +\ve v}(t))\ge 0$ 
for almost all $t\in[T,0]$. So
$$
\lambda(0).D_y{\cal E}(\bar x).v+\lambda^0 D_y\hat V_{\Gamma}(\bar x).v
-\lambda(T_{\bar x})
\cdot v\ge 0\qquad{\rm for\,\,\,all\,\,\,}v\in\R^n\,.\numeq
$$
Given $v$, the curve $r\mapsto \gamma_{\bar x+rv}(0)$ is defined
for $r$ in a neighborhood of $0$ in $\R$, and is differentiable 
at $0$ with derivative $D_y{\cal E}(\bar x).v$.  On the other hand, this
curve is contained in ${\cal T}$, so the vector 
$w=D_y{\cal E}(\bar x).v$ belongs to  $\tilde T_{\g_{\bar x}(0)}{\cal T}$.  
The transversality condition (EX4) then says that 
$\langle\la(0),w\rangle=\l0\langle \nabla\tau(\g(0)),w\rangle$.
Therefore 
$$
\l0\biggl(\nabla\tau(\g(0)).D_y{\cal E}(\bar x).v+D_y\hat V_{\Gamma}(\bar
x).v\biggr) 
-\lambda(T_{\bar x})
\cdot v\ge 0\qquad{\rm for\,\,\,all\,\,\,}v\in\R^n\,.\numeq
$$
Since the left-hand side of this inequality is linear in $v$,
it has to be equal to zero. So 
$$
\l0\biggl(\nabla\tau(\g(0)).D_y{\cal E}(\bar x).v+D_y\hat V_{\Gamma}(\bar
x).v\biggr)=\lambda(T_{\bar x})\cdot v\qquad
{\rm for\,\,\,all\,\,\,}v\in\R^n\,.\numeq
$$
Clearly, the left-hand side of this equation is $\l0 D_yV_\Gamma(\bar x).v$.
Since the equality holds for all $v$, we have 
$\l0 D_yV_\Gamma(\bar x)=\lambda(T_{\bar x})$. 

In particular, $\l0$ cannot vanish, because $\l0=0$ would imply
$\lambda(T_{\bar x})=0$, contradicting (EX1). 
So we may, after multiplication by a positive constant,  
assume that $\l0=1$, and then 
$$ 
D_y V_{\Gamma}(\xb)= \la(\txb).\numeq
$$
This proves, in particular, the uniqueness of $(\lambda(\cdot),\l0)$
up to multiplication by a positive constant, since the
Cauchy problem for the adjoint equation has unique solutions. \proofend

\medbreak\noindent{\sl CONTINUATION OF THE PROOF OF THEOREM \maintheo.\enspace}
The minimization condition (EX4) of the Pontryagin Maximum Principle
implies that, for a given $x\in S\setminus A$, $u\in U$, the inequality
$$\langle \la(t),f(\gx(t),u)\rangle +L(\gx(t),u)\geq 0$$
holds for almost all $t\in [T_x,0]$, where $(\la,\l0)$ is a minimizing
adjoint vector for $(\gx,\ex)$, normalized so that $\l0=1$. 
Since $t\mapsto \langle \la(t),f(\gx(t),u)\rangle +L(\gx(t),u)$
is continuous on $[T_x,0]$, the 
inequality is valid at time $\tx$. In view of Theorem
\maintheodiff, $\la(T_x)=dV_\Gamma(x)$, so 
\xdef\difvalf{\equanosectno}
$$
\langle dV_{\Gamma}(x),f(x,u)\rangle +L(x,u)\geq 0\,.
\numeq
$$

Now consider an arbitrary $(\g,\eta)\in\Admo(\tf,{\cal T})$, with domain
$[T,0]$.   We want to prove that $J(\g,\e)\ge V_\Gamma(x)$, if $x=\g^-$. 
It follows from Lemma~4.1 of \sussncount{} that
there exists a sequence of admissible
pairs $(\gamma_j,\eta_j)$, with domain $[T,0]$, 
such that $\gamma_j(0)=\g(0)$,
$\eta_j$ is piecewise constant, $\gamma_j\rightarrow\g$ uniformly 
on $[T,0]$, and
$J(\gamma_j,\eta_j)\rightarrow J(\g,\eta)$. 
Since the function $V_{\Gamma}$ is lower semicontinuous, it follows
that, if 
$V_{\Gamma}(\gamma_j(T))\leq J(\gamma_j,\eta_j)$, then 
$V_{\Gamma}(x)\leq J(\gamma,\eta)$.
Hence it suffices to assume that  the control $\eta$ is piecewise
constant, and we make this assumption from now on. 

Let $\eta:[T,0]\to U$ be a piecewise constant control, and let
$t_0, \ldots,t_m$, $u_1,\ldots,u_m$ be such that 
$T=t_0<t_1<\ldots<t_{m-1}<t_m=0$ and $\eta(t)=u_i$ if
$t_{i-1}<t<t_i$ for $i=1,\ldots,m$.  For each $y\in\Omega$, let
$t\to\zeta^y(t)$ be the trajectory of $\eta$ that satisfies the
terminal condition $\zeta^y(0)=y$. Let $W$ be an open 
subset of $\Omega$ such that $\zeta^y(T)$ is defined for all 
$y\in W$ and $\g(0)\in W$.  We will show that 

\medbreak
\itemitem{(\#)} {\sl for almost every $y\in W$ the set
$B^y=\{t:\zeta^y(t)\in A\}$ is finite or countable.}

\medbreak\noindent Since $A$ is thin, we can write $A=A_1\cup A_2$,
where $A_1=\cup_{j}M_j$ and $\{M_j\}_{j\in J}$ is a finite or
countable family of connected submanifolds of $\Omega$ of codimension
$d_j>0$, while ${\cal{}H}^{n-1}(A_2)=0$.  After replacing each $M_j$
by a finite or countable family of open submanifolds of $M_j$, we may
assume that the $M_j$ are embedded.

For $i=1,\ldots,m$, let $\Phi_i:W\times[t_{i-1},t_i]\to\Omega$ be the
map $(y,t)\to \zeta^y(t)$. Then $\Phi_i$ is a ${\cal{C}}^1$ submersion,
because for each fixed $t$ the map $y\to\Phi_i(y,t)$ is a ${\cal{C}}^1$
diffeomorphism. Therefore each set $\tilde M_{j,i}=\Phi_i^{-1}(M_j)$
is an embedded submanifold of $W\times[t_{i-1},t_i]$ of codimension
$d_j$, and ${\cal{}H}^{n}(\Phi_i^{-1}(A_2))=0$. 

Let $\pi_i:W\times[t_{i-1},t_i]\to W$ be the projection $(y,t)\to y$.
Let $S_{j,i}$ be the set of points $s\in\tilde M_{j,i}$ such that
the restriction $\pi_i\restr\tilde M_{j,i}$ of $\pi$ to 
$\tilde M_{j,i}$ is not regular at $s$. (Recall that, if
$S_1$, $S_2$ are ${\cal{C}}^1$ manifolds, then  a ${\cal{C}}^1$ map $F:S_1\to S_2$
is {\sl regular} at $s\in S_1$ if the differential $d\pi(s)$ maps the
tangent space $T_sS_1$ surjectively onto $T_{F(s)}S_2$.)

Then ${\cal{}H}^{n}(\pi_i(\Phi_i^{-1}(A_2)))=0$ (see for example
\fedcount), and  ${\cal{}H}^{n}(\pi_i(S_{j,i}))=0$ for each $i,j$, by
Sard's Theorem. So the ``bad'' set 
$$
{\cal{}B}=\Bigl(\bigcup_{i=1}^m\pi_i(\Phi_i^{-1}(A_2))\Bigr)
\cup\Bigl(\bigcup_{j,i}\pi_i(S_{j,i})\Bigr)
$$
has measure zero. 

Now let $y\in W\backslash{\cal{}B}$. Then $\zeta^y(t)\notin A_2$ if
$T\le t\le 0$. (Otherwise, if $\zeta^y(t)\in A_2$, $t_{i-1}\le t\le
t_i$, it would follow that $(y,t)\in\Phi_i^{-1}(A_2)$, so
$y\in\pi_i(\Phi_i^{-1}(A_2))$, and then $y\in{\cal{}B}$.)  Also, we
claim that the set $\{t:\zeta^y(t)\in A_1\}$ is at most countable. To
see this, it suffices to show that, for each $j,i$, the set
$\{t:t_{i-1}\le t\le t_i\,,\,\zeta^y(t)\in M_j\}$ is finite.  Fix $i,j$,
and suppose $\zeta^y(t)\in M_j$. The manifold $\tilde M_{j,i}$ has
codimension $d_j$ in $W\times[t_{i-1},t_i]$. Since $d_j>0$, the
dimension $\nu_{j,i}$ of $\tilde M_{j,i}$ is $\le n$. Moreover,
$d\pi(y,t)$ maps $T_{y,t}\tilde M_{j,i}$ surjectively onto $\R^n$ (the
tangent space to $\Omega$ at $y$), because $y\notin{\cal{}B}$. So
$\nu_{j,i}=n$, and $d\pi(y,t)$ is injective on $T_{y,t}\tilde
M_{j,i}$.  Since $\Bigl\langle d\pi(y,t),{\partial\,\,\over\partial
t}\Bigr\rangle$ is obviously equal to $0$, it follows that 
${\partial\,\,\over\partial t}\notin T_{y,t}\tilde M_{j,i}$.   Since
$\tilde M_{j,i}$ is embedded, 
$(y,\tau)\notin\tilde M_{j,i}$ if $0<|\tau-t|\le\varepsilon$, provided
that $\varepsilon$ is sufficiently small. Therefore 
$t$ is an isolated point of $\{\tau:t_{i-1}\le\tau\le t_i\,,\,
\zeta^y(\tau)\in M_j\}$. So
$\{t:t_{i-1}\le t\le t_i\,,\,\zeta^y(t)\in M_j\}$ is finite.
\gnu{In other words, you are saying that, if every point of a subset of
$\rr$ is isolated, then the set is finite!!!! How about the set $\{{1\over
n}:n=1,2,3\ldots\}$? I am amazed that the referee allowed this to get by.
I guess he must not have been a very competent referee. But obviously we
cannot say that to Henk Nijmeijer!}
We have therefore proved that, if $y\in W\backslash{\cal{}B}$, then
$\zeta^y(t)\notin A_2$ for all $t\in[T,0]$, and the set $\{t:\zeta^y(t)\in
A_1\}$ is at most countable.  Since $A=A_1\cup A_2$, and ${\cal{B}}$ has
measure zero, ($\#$) follows.

{}From the weak upper semicontinuity of $V_{\Gamma}$, there exist $x_i$,
$\ve_i>0$, $\delta_i>0$ such that $x_i\rightarrow \gamma(0)$,
$\delta_i\rightarrow 0$ and $V_{\Gamma}(y)\leq
V_{\Gamma}(\gamma(0))+\delta_i$
whenever $\|y-x_i\|\leq \ve_i$.   
\gnu{I don't understand the above construction. How can something ``exist''
{\it from} the weak semicontinuity of something?}
Choose the $\ve_i$ so that 
$\|y-x_i\|\leq\ve_i$ implies $y\in W$. Then, by ($\#$), 
we can choose for each $i$ a $y_i$ such that $\|y_i-x_i\|\leq \ve_i$ 
and the bad set $B^{y_i}$ is at most finite or countable.
Since $\Gamma$ is total and $V_{\Gamma}(\zeta^{y_i}(0))<+\infty$,
it is clear that, for every $i$, $\zeta^{y_i}(t)\in S$ for every
$t\in[T,0]$.  Hence (\difvalf) implies that the function:
$$
\varphi_i(t)=
V_{\Gamma}(\zeta^{y_i}(t))+\int_T^t L(\zeta^{y_i}(s),\eta(s)) ds
$$
is differentiable at every $t$, except for a finite or countable number 
of values (which may include the switching times of $\eta$), 
with nonnegative derivative.
Thus, $\varphi_i$ verifies assumption (B.1.a) of Lemma~B.1 (Appendix~B).
\gnu{I don't understand how a mathematical object can ``verify'' 
an assumption.  It can SATISFY it, but not VERIFY it.  Can you say
in Italian, foe example, that a function ``verifies the assumptions
of the mean value theorem''?  I bet you the full value of a 
Luiggi di Biaggio that you cannot say it. I am sure that in Italian,
as in English, things ``satisfy'' conditions, and we humans ``verify''
that certain things ``satisfy'' certain properties!!!!!}
Moreover, the lower semicontinuity of $V_{\Gamma}$
ensures assumption (B.1.b) of Lemma~B.1 for $\varphi_i$, and the (NDJ)
condition 
ensures assumption (B.1.c). Hence we can apply Lemma~B.1 to $\varphi_i$,
and conclude that $\varphi_i(T)\le\varphi_i(0)$.  So
$$
V_{\Gamma}(\zeta^{y_i}(T))\leq \int_T^0
L(\zeta^{y_i}(t),\eta(t))dt +V_{\Gamma}(y_i)\leq
\int_T^0 L(\zeta^{y_i}(t),\eta(t))dt +V_{\Gamma}(\gamma(0))+\delta_i.
$$ Since $V_{\Gamma}(\gamma(0))\leq \tau(\gamma(0))$ and
$\zeta^{y_i}\rightarrow \gamma$ uniformly (and in particular
$\zeta^{y_i}(T)\rightarrow \gamma(T)=x$), we conclude, using the lower
semicontinuity of $V_{\Gamma}$, that $V_{\Gamma}(x)\leq
J(\gamma,\eta)$. 

So the desired inequality holds when $\eta$ is piecewise constant. 
As explained earlier, this implies that $\Gamma$ is optimal.

\vs

\noindent{\bf Remark 1.}
Notice that the weak upper semicontinuity of
$V_{\Gamma}$ everywhere is equivalent to its weak upper semicontinuity 
at the points of the target.
Indeed, let $x\in\Omega$.  If $x\notin S$, then 
$V_{\Gamma}(x)=+\infty$, and (2.8) is clearly true. Next, suppose 
$x\in S$, and let  $(\gx,\ex)$ be its associated
admissible pair.  Assume that $V_{\Gamma}$ is weakly upper semicontinuous 
at $\gx(0)$. Let
$y_j$, $\ve_j$, $\delta_j$ be such that $y_j\to \gx(0)$, $\ve_j>0$,
$\delta_j>0$, $\delta_j\to 0$, and $V_{\Gamma}(y)\le
V_{\Gamma}(\gx(0))+\delta_j$  whenever $\|y-y_j\|\le\ve_j$. Let $\zeta^{y_j}$
be the trajectory corresponding to the control $\ex$ and such that
$\zeta^{y_j}(0)=y_j$.

Since $y_j\to \gx(0)$, we may assume, by picking $j$ large enough,
that $\zeta^{y_j}$ is defined on $[\tx,0]$ for all $j$, 
and let $x_j=\zeta^{y_j}(\tx)$. Let
$\tilde\ve_j$ be such that $0<\tilde\ve_j<{1\over j}$ and, if
$\|z-x_j\|\le\tilde\ve_j$, then the trajectory $\theta^z$ of $\ex$
with initial condition $\theta^z(\tx)=z$ is defined on $[\tx,0]$ and
satisfies $\|\theta^z(0)-y_j\|<\ve_j$. Then, using the same reasoning
as in the proof of Theorem 2.2, we can show that 
\gnus{If you had used the label for this theorem, rather than 2.2,
then it would have been less work for me to renumber things!!!!!}
\gnue{In addition, it seems to me that what you really meant was Theorem
2.1, not 2.2. Right?  For example, Statement ($\#$) is in the proof 
of Theorem 2.1, not the proof of Theorem 2.2. Right?} 
\xdef\whatever{\equanosectno}
$$V_\Gamma(z)\le\int_{\tx}^0L(\theta^z(t),\ex(t))\,dt+V_{\Gamma}(\gx(0))+\delta_j\numeq
$$
whenever $\|z-x_j\|\le\tilde\ve_j$. (First approximate $\ex$ by
piecewise constant controls, $\eta^k$, and let $\theta^{z,k}$ be the
corresponding trajectories with terminal condition 
$\theta^{z,k}(0)=\theta^z(0)$.  Then for each $k$ use Statement ($\#$)
of the proof of Theorem 2.2 to approximate the terminal point 
\gnu{As I said above, I think you meant 2.1. And it would have been 
better if you had used the labels. That's what they are for.}
$\theta^{z}(0)$ by points $w^{z,k,i}$ with the property that, if 
$\tilde\theta^{z,k,i}$ is the trajectory for $\eta^k$ for which 
$\tilde\theta^{z,k,i}(0)=w^{z,k,i}$, it follows that 
$V_\Gamma(\tilde\theta^{z,k,i}(\tx))
\le\int_{\tx}^0L(\theta^{z,k,i}(t),\eta^k(t))\,dt+
V_{\Gamma}(w^{z,k,i})$. Since $\|\theta^z(0)-y_j\|<\ve_j$, 
the $w^{z,k,i}$ satisfy $\|w^{z,k,i}-y_j\|<\ve_j$ if $i$ is large
enough, so $V_\Gamma(\tilde\theta^{z,k,i}(\tx))\le
V_\Gamma(\gx(0))+\delta_j$ when $i$ is large enough. If we let
$i\to\infty$ and then let $k\to\infty$, and use the lower
semicontinuity of $V_\Gamma$, we find that (\whatever) holds.)

Let
$\tilde\delta_j=\sup\Bigl\{\|\int_{\tx}^0L(\theta^z(t),\ex(t))\,dt-
\int_{\tx}^0L(\gx(t),\ex(t))\,dt\|:\|z-x_j\|\le\tilde\ve_j\Bigr\}$.
Then $\tilde\delta_j\to 0$, because $x_j\to x$ and $\tilde\ve_j\to
0$. On the other hand, $V_\Gamma(z)\le
V_{\Gamma}(x)+\delta_j+\tilde\delta_j$ whenever
$\|z-x_j\|\le\tilde\ve_j$. So $V_{\Gamma}$ is weakly upper
semicontinuous at $x$.

\vs


\section{The Fuller phenomenon}

\medbreak

\noindent The aim of this section is to prove that the famous control
system considered by Fuller in \fullcount ---and extensively studied
by other authors, e.g. Zelikin-Borisov \zelikincount---satisfies the
hypotheses of Theorem 2.1. 
\gnu{Once again, it would have been  better if you had used the labels.}

\def\r{\rho}
\def\a{\alpha}

Consider the system:

\xdef\constr{\equanosectno}
$$\cases{ \dot x_1 =  x_2  &$\,,$\cr
          \dot x_2 =  u  &$\,,$\cr
            |u|  \leq 1  &$\,,$\cr}\numeq$$
and the minimization problem:
\xdef\minimful{\equanosectno}
$$\min \int_0^{+\infty} x_1^2(t) dt.\numeq$$

We now describe the optimal synthesis for this problem, determined by
Fuller. First define the ``switching locus'' 
$\zeta=\{(x_1,x_2):|x_1|=\xi x_2^2, x_1x_2\leq 0\}$, where $\xi$ is
the unique positive root of the polynomial $x^4+x^2/12-1/18$, and
write $\zeta=\zeta_+\cup \zeta_-$, where
$\zeta_{\pm}=\{(x_1,x_2)\in\zeta:\pm x_1\ge 0\}$.  Then the piecewise
smooth curve $\zeta$ divides the plane into two regions $A^-$ and
$A^+$ consisting, respectively, of the points that lie above and
below $\zeta$. For every $x\in\R^2$ there exist a unique optimal
trajectory $\hat\gamma_x$ and corresponding control $\hat\eta_x$, related by
$\hat\eta_x(t)=+1$ when $\hat\gamma_x(t)\in A^+$, and $\hat\eta_x(t)=-1$ when
$\hat\gamma_x(t)\in A^-$. The switchings occur along $\zeta$. The curve
$\hat\gamma_x$ reaches the origin in a finite time $\tau_x$, and then
stays there. Therefore $\hat\eta_x(t)=0$ when $\hat\gamma_x(t)=0$, which
happens iff $t\ge\tau_x$ (see Fig.~1.)

\vskip 1em


\centerline{\hbox{\psfig{figure=fuller.ps,width=3in,height=3in}}}

\medbreak
\centerline{Figure 1}

\vskip 1em


Therefore the admissible pairs $(\hat\gamma_x,\hat\eta_x)$ actually solve the
problem of reaching the origin in finite time, subject to the
dynamical constraints (\constr), and minimizing the
cost (\minimful).  However, the sufficiency theorems of \boltcount{} and
\brunbcount{} do not apply, since the controls $\hat\eta_x$ 
have an infinite number of switchings.  For this  reason, 
the optimality of the synthesis obtained by Fuller is usually
established by other means. 

For example, one can prove optimality using the {\sl symmetry properties} of
the problem.  Precisely, there exists a one parameter family
$\Delta=\{\Delta_{\rho}\}_{\rho>0}$ of maps from $\R^2$ to $\R^2$
such that, if $(\gamma,\eta)$ is an admissible pair having cost $c$, and we
let $\gamma_\rho(t)=\Delta_\rho(\gamma({t\over\rho}))$,
$\eta_\rho(t)=\eta({t\over\rho})$, then $(\gamma_\rho,\eta_\rho)$ is
an admissible pair having cost $\rho^5c$.  It then follows that the
$\Delta_\rho$ map optimal trajectories to optimal trajectories.
The proof of optimality using the dilation symmetry is given, for
example, in \zelikincount{}, and will not be repeated here. It will
turn out, however, that the dilations $\Delta_\rho$ will play a role 
in our arguments, so we recall their definition and one simple
property:

\medbreak
\itemitem{(1)}  By definition, $\Delta_\rho(x_1,x_2)=(\rho^2x_1,\rho x_2)$. 
\itemitem{(2)} It follows from (1), in particular, that 
$\Delta_\rho(x_1,x_2)-x=(\r-1)\, (2x_1,x_2)+(\r-1)^2\, (x_1,0)$
for all $\rho,x_1,x_2.$ This identity will be useful later.

\gnu{You say that we are going to ``recall their definition and one simple
property,'' and then you have items (1) and (2). So the reader is going to
assume that (1) is the ``definition'' and (2) is the ``simple property''
that you are ``recalling.''  Now: is the statement that ``This identity
will be useful later'' part of the ``simple property'' you are
``recalling''? Or is it a comment you make about the property and about the
reason for mentioning it here?}

\medbreak\noindent
In order to apply our Theorem 2.1, we first make a time translation
and define $\gamma_x(t)=\hat\gamma_x(t+\tau_x)$, 
$\eta_x(t)=\hat\eta_x(t+\tau_x)$, for $-\tau_x\le t\le 0$.   Then 
$\Gamma=\{(\gamma_x,\eta_x)\}_{x\in\R^2}$ is a memoryless family of
admissible pairs 
such that $\gamma_x$ is defined on $[-\tau_x,0]$,
$\gamma_x(-\tau_x)=x$, and $\gamma_x(0)=0$. We want to prove that 
$\Gamma$ is an optimal synthesis.  For this purpose, we will 
verify the hypotheses of our Theorem 2.1, and in particular 
the $(f,L)$--differentiability of $\Gamma$
at every point $x\in\R^2\backslash\zeta$.
Since it is very easy to check Property (DC1), we will only verify 
(DC2).

Fix a point $\xb\in\R^2\backslash\{(0,0)\}$. Then there exist
$s_0,s_1$ such that $s_0<s_1<0$, $s_1\le\theta s_0$, 
$$\exb=\pm\sum_{k=0}^{+\infty} (-1)^k \chi_{_{_{I_k}}}$$
where  $I_0=[s_0,s_1]$  and, for $k=1,2,\ldots$, 
$$I_k=[\theta^{k-1}s_1,\theta^ks_{1}]\,,\qquad
\theta=\sqrt{ {{1\over 2}-\xi \over {1\over 2}+\xi}},$$
$\chi_{_{_{S}}}$ is the indicator function of a set $S$, 
and the sign $\pm$ depends on $\xb$ as follows: it is 
$+$ if $\xb\in A^+\cup \zeta^+$, and $-$ if $\xb\in A^-\cup\zeta^-$.

Let $[T,0]$ be an interval containing $\Dom(\ex)$ for $x$ near $\bar
x$. Fix a continuous function $\a:[T,0]\rightarrow\R$.
To prove (DC2) we must study the dependence on $x$ of the 
integral
$$
J^\alpha(x)=\int_T^0\a(t).\tilde f(\gxb(t),\eta_{x}(t))\,dt\,,\numeq$$
and show that $J^\alpha$ is differentiable at $x=\xb$ for every
$\xb\in\R^2\setminus\zeta$.  We will do
this by showing the existence, and continuity with respect to $x$,
of the directional derivatives $D_{v_1(x)}J^\alpha$, 
$D_{v_2(x)}J^\alpha$,  where $v_1(x)$, $v_2(x)$ are two linearly
independent directions depending smoothly on $x$. We choose 
$v_1(x)$ to be the direction of the curve $\rho\to \Delta_\rho(x)$ at
$\rho=1$, and take $v_2(x)=f(x,\eta_x(-\tau_x))$.  We will just
determine the directional derivatives at $\xb$, and the result will
make it obvious that they depend continuously on $\xb$, so
differentiability will follow.   


Let us assume, for simplicity, that $\xb\in A^+$. (The case when 
$\xb\in A^-$ is similar.)   We first
differentiate $J^\alpha$ in the direction of $v_1(\xb)$.  
If $\rho>0$, we have (letting $s_k=\theta^{k-1}s_1$ for $k=2,3,\ldots$):
$$
\tau_{_{\Delta_\rho(\xb)}}=\rho\tau_{\xb}\,,\quad
\eta_{_{\Delta_\rho(\xb)}}=\sum_{k=0}^{+\infty} (-1)^k\chi_{_{_{I_k^\rho}}}
\,,\quad
I_k^\rho=[\r s_k,\r s_{k+1}]$$
and
\xdef\tildef{\equanosectno}
$$\tilde f(\gxb(t),\eta_{_{\Delta_\rho(\xb)}}(t))-\tilde f(\gxb(t),\exb(t))=
\left( \matrix{0\cr \eta_{_{\Delta_\rho(\xb)}}(t)-\exb(t) \cr 0\cr} \right)+R_\rho(t),\numeq$$
where $R_\rho(t)$ is a term arising because of the possibility that 
$\tau_{\xb}\neq\tau_{_{\Delta_{\rho}(\xb)}}$.

Then, if $\rho\ge 1$ and $\rho-1$ is sufficiently small,
$$\eqalign{
\int_T^0 (\eta_{_{\Delta_{\r}(\xb)}}(t)-\exb(t)) \a(t) dt&=
\int_{\rho s_0}^0 \sum_{k=0}^{+\infty} (-1)^k
(\chi_{_{_{I_k^\rho}}}(t)-\chi_{_{_{I_k}}}(t))\a(t)dt\cr
&=\sum_{k=0}^{+\infty} (-1)^k\Biggl(\int_{\rho s_k}^{\rho s_{k+1}}\a(t)dt-
\int_{s_k}^{s_{k+1}}\a(t)dt\Biggr)\cr
&=\sum_{k=0}^{+\infty} (-1)^k\Biggl(\int_{\rho s_k}^{s_k}\a(t)dt
-\int_{\rho s_{k}}^{s_{k+1}}\a(t)dt\Biggr)\cr
&=\int_{\rho s_0}^{s_0}\a(t)dt+2\sum_{k=1}^{+\infty}
(-1)^k\Biggl(\int_{\rho s_k}^{s_k}\a(t)dt\Biggr)\,.\cr}
$$
The numbers $(\rho-1)^{-1}\Big\|\int_{\rho s_k}^{s_k}\a(t)dt\|$ are
bounded by $\theta^{k-1}|s_1|.\|\a\|_{L^\infty}$, which is the general
term of a convergent series, since $\theta<1$. So we can divide by
$\rho-1$ and let $\rho\to 1$ taking the limit of each term separately.
We then conclude that 
\xdef\therefore{\equanosectno}
$$\lim_{\r\downarrow 1} {1\over \r-1}\int_T^0  (\eta_{x_{\r}}(t)-\exb(t)) \a(t)
dt=-s_0\a(s_0)-2s_1\sum_{k=1}^{+\infty}(-1)^k \theta^{k-1}\a(\theta^{k-1}s_1)\,.\numeq$$

We now find the limit of $(\rho-1)^{-1}\int_T^0\a(t).R_\rho(t)\,dt$
as $\rho\downarrow 1$.   Then
$-\tau_{_{\Delta_{\rho}(\xb)}}=-\rho\tau_{\xb}<-\tau_{\xb}$, so
$$
\int_T^0\a(t).R_\rho(t)\,dt=\int_{-\rho\tau_{\xb}}^{-\tau_{\xb}}
\a(t).\left(\matrix{\gamma_{_{\Delta_{\rho}(\xb)}}^2(t)\cr 0
\cr\Bigl( \gamma_{_{\Delta_{\rho}(\xb)}}^1(t)\Bigr)^2}\right)\,dt\,,
$$
where $ \gamma_{_{\Delta_{\rho}(\xb)}}^i(t)$, for $i=1,2$, is the 
$i$-th component of  $\gamma_{_{\Delta_{\rho}(\xb)}}(t)$. Clearly,
then, if $\xb=(\xb_1,\xb_2)$, we have 
\xdef\thereforeb{\equanosectno}
$$
\lim_{\rho\downarrow 1}(\rho-1)^{-1}
\int_T^0\a(t).R_\rho(t)\,dt=\tau_{\xb}\a(\tau_{\xb}).
\left(\matrix{\xb_2\cr 0\cr \xb_1^2}\right)\,.\numeq
$$
\gnu{This formula does not look right. According to your choices, 
$\tau_{\xb}$ is positive.  But $\a$ is a function defined on $[T,0]$. So
you cannot have $\a(\tau_{\xb})$. It has to be $\a(-\tau_{\xb})$.}
Similar computations shows that (\therefore) and (\thereforeb) hold as well
\gnu{It should be ``Similar computations show'', 
or ``A similar computation shows'', but not ``Similar computations shows''.}
for the limits as $\rho\uparrow 1$.  So the directional derivative of
$J^\a$ at $x=\xb$ in the direction of $v_1(\xb)$ exists and is given by the
sums of the right-hand sides of (\therefore) and (\thereforeb), which
depend continuously on $\xb$, since $s_0$, $s_1$ and $\tau_{\xb}$ (which is
none other than $-s_0$) depend continuously on $\xb$ as long as $\xb\in
A^+$.

A much simpler argument verifies the existence of the derivative of
$J^\a$ in the direction of $v_2(\xb)$, i.e.  of $f(\xb,1)$, whose
value turns out to be the vector with components
$\a(\tau_{\xb})\xb_2$, $\a(\tau_{\xb})$, and $\a(\tau_{\xb})\xb_1^2$.  Once again,
this expression depends continuously on $\xb$, so the proof of (DC2)
is complete.

The other assumptions of  Theorem 2.1 are easily checked. So the theorem
applies, and the optimality of the synthesis described above follows.

\vs

\section{The Boltianskii and Brunovsky definitions of synthesis}

\gnu{I changed the section title to ``Comparison with other definitions of
synthesis,'' because we also discuss a definition proposed in some of my
papers.} 

\medbreak\noindent
We start by reviewing the definition of synthesis proposed by Brunovsky
\brunbcount{}, 
and then we explain how it differs from the more restrictive
definition given by Boltianskii \boltcount.

\gnus{{\rm It is not completely true that the Boltyanskii definition 
is ``more restrictive'' than that of Brunovsk\'y (or, as you said in the
August 97 version, ``then we recall some further restrictions existing in
the definition given by Boltyanskii \boltcount''). For example: Boltianskii
does not require that the partition into cells be a stratification, but
Brunovsk\'y does.}}


\gnum{{\rm Even more dramatically: {\bf Boltyanskii does not require
the set of cells to be locally finite.}  Of course, he has a local
finiteness condition in his definition of ``piecewise smooth set,'' but
this does not imply that the set of cells, as he defines the notion of
``cell'' (a connected components of $P^i\setminus(P^{i-1}\cup N)$ for some
$i$), is locally finite.   My own guess is that he probably
intended to make that requirement, and he thought that his conditions
implied that the set of cells is locally finite. But {\it they don't}.  You
can see by looking at the Boltyanskii paper that he is very imprecise,
so it's perfectly possible that he may have made a mistake here. In the new
version of the paper, I put in an example showing that for a Boltyanskii 
regular synthesis the set of cells need not be locally finite.}}

\gnum{{\rm Also, Boltyanskii allows a set $N$ where things can be quite
bad (e.g. the ``marked trajectory'' there need not be unique) but
Brunovsk\'y does not.  So in some ways the Boltianskii definition is {\bf
less restrictive} rather that more restrictive than that of Brunovsk\'y.
Of course, in some other much more important ways the Boltianskii
definition is indeed ``more restrictive,'' but 
the only true statement we can make is that the two definitions are not
comparable.}} 

\gnum{{\rm Here again we see an example of how I made a mistake by trusting
you and expecting that you would carefully do the scholarly work involved
in checking the references and making sure that when we quoted them we did
so accurately. {\bf I had to go and read the Boltyanskii and Brunovsk\'y papers
carefully by myself in order to get a precise understanding of the
comparison between the two notions.}}}

\gnus{In this section, you seem to have completely forgotten that we are
now dealing with a new version in which the target is no longer the
origin. For that reason, you talk throughout the section about ``the
problem with target the origin.''  To fix this, I had two choices: (a)
rewriting everything in the section with the general target, or (b) say
that, ``for simplicity,'' in this section we take the target to be the
origin.  The first choice was too much work for me.}

\gnue{It turns out, however, that both Boltianskii and Brunovsky work with
a one-point target anyhow. So the best thing to do is to tell the reader
that, since they both work a one-point target, we will do the same. That
gives a totally different impression: we are not limiting ourselves to a
point target because we are lazy or we don't know how to discuss the
general case. We do it because the others did it.  On the other hand, OUR
theorem is for a general target. So here is another reason why our theorem
is better.  But we don't say so. (After all, the generalization from point
target to general target is sort of trivial.) But the reader will clearly
see the point, and find on his own another reson in favor of our paper, and
in addition respect us for our modesty in not saying so openly.}

We consider the system (\system), (\minim), with the restriction 
that $f$ is $\C^1$ as a function of $(x,u)$.

\gnu{What on Earth does it mean for $f$ to be a $C^1$ function of $x$ and
$u$ if $u\in U$ and $U$ is just a set?
In addition: we also need to require that $L$ be $C^1$, right?}
Given a $k\in\{1,2,\ldots\}\cup\{+\infty,\omega\}$, and a manifold $M$
of class ${\cal{C}}^k$, a {\sl ${\cal{C}}^k$ stratification in $M$} of a
\gnu{Now, suddenly, in the middle of a discussion of systems and 
optimal control, you start talking about stratifications without giving 
any indication to the reader of why this is relevant at this point! 
{\bf You should never switch from one subject to another without 
a continuous transition, or an explanation of why this digression is
necessary!!!!}}
subset $S$ of $M$ is a  partition $\P$ of $S$ into
connected embedded ${\cal{C}}^k$ submanifolds of $M$ such that 
$\P$ is locally finite in $M$ (i.e. every compact subset of $M$
intersects finitely many members of $\P$) and the
following ``frontier axiom'' holds:

\medbreak
\itemitem{(FA)} {\sl if $P_1,P_2\in\P$, and $P_1\cap
clos_M(P_2)\not=\emptyset$, then $P_1\subset clos_M(P_2)$ and 
$dim(P_1)<dim(P_2)$}.  

\gnu{The Frontier axiom, stated this way, is {\bf wrong}, because 
$P_1$ could be equal to $P_2$.  In fact, the way you stated it the FA is
ALWAYS false, because you can take any member  $P_1$ of $\P$ and take $P_2$
to be $P_1$, and then  the conditions 
$P_1,P_2\in\P$ and $P_1\cap clos_M(P_2)\not=\emptyset$ will hold, 
but $dim(P_1)<dim(P_2)$ will not be true.}

\medbreak\noindent A {\sl $\C^k$-stratified subset of $M$} is a
pair $(S,\P)$ such that $S\subseteq M$ and ${\cal P}$ is a $C^k$
stratification of $S$ in $M$.


Let $S\subseteq\Omega$ be an open subset such that 
%$\hat x\in\bar G$. 
the origin belongs to $S$. Brunovsky's definition (somewhat modified by us to 
correct what we thought were some minor imprecisions in the
original formulation), says that a {\sl regular synthesis in $S$ 
for the control problem (\system), (\minim) with target
%$\hx$ 
the origin},
is a 6--tuple $\Gamma=(\P,\P_1,\P_2,\Pi,\Sigma,v)$ such that

{\sl 

\vs
\item{1)} $\P$ is the union of $\{\{0\}\}$ 
%$\{\hx\}$ 
and a locally finite (in $S$) partition $\P\backslash\{\{0\}\}$ of 
$S\setminus \{
%\hx
0
\}$ into connected embedded ${\C}^1$ submanifolds (called ``cells''),

\vs
\item{2)} $\P\backslash\{\{0\}\}$ is the disjoint union of $\P_1$ (the
set of ``Type I cells'') and
$\P_2$ (the set of ``Type II cells''), 

\vs
\item{3)} $v:S\rightarrow U$ (the ``closed loop
control''), $\Pi:\P_1\rightarrow\P$, and $\Sigma:\P_2\rightarrow\P_1$ 
are maps such that the following properties are satisfied:

\vs
\itemitem{3.A}  If $\P'=\{P\in\P:{\rm dim}\;P<n\,,\,P\neq\{0\}\}$, and 
$S'=\bigcup \{P\in\P'\}$, then 
$(S',P')$ is a stratified subset of $S$ of dimension $<n$.  

\gnu{{\rm This statement is O.K. but it is extremely weird. Look at the comment
I added, with a detailed proof that I put in the long version. Of course,
this is Brunovsk\'y's fault, not yours or mine, but I think that whenever 
one can make things clearer one should.}}

\vs\itemitem{3.B} The function $v$ is $\C^1$ on each cell.

\gnu{{\rm Again, what could it possibly mean to say that a function with %
values in an abstract set $U$ is $C^1$?}}

\vs\itemitem{3.C}  If $P_1\in\P_1$ and
$x\in P_1$, then $f(x,v(x))\in T_x P_1$ (the tangent space to $P_1$ at $x$)
and there exists a time $t_x$
such that the maximally defined solution $\xi_x$ of the initial value problem
$$\dot\xi=f(\xi,v(\xi))\,,\,\,\,\xi(0)=x\,,\,\,\,\xi\in P_1$$
is such that $[0,t_x\,[\,\subseteq Dom(\xi_x)$ and 
$\lim_{t\uparrow t_x}\xi_x(t)$ exists (in $\Omega$) 
and belongs to $\Pi(P_1)$. 

\gnus{{\rm Compare my rewording of (3.C) with your original wording. Make sure
you understand why mine is better. The way you had it, there is ONE 
universal quantifier ``for all $x\in P_1$'' covering all occurrences 
of $x$. Within this quantifier, you don't know that $f(.,v(.))$ is 
a vector field tangent to $P_1$ everywhere. So you cannot talk about 
$\xi_x$ the way you do. See?}}

\gnum{{\rm Another problem with your version of 3.C is that it would allow for
the possibility that $\Pi(P_1)$ is just equal to $P_1$ and the time 
$t_x$ is always chosen to be smaller than the true exit time from
$P_1$. You don't want that. Look at the way I rewrote this. (You NEVER
said that $\Pi(P_1)$ cannot be equal to $P_1$. And you NEVER said that
$t_x$ is the exit time.  All you say is that the interval 
$[0,t_x[$ is a subset of $Dom(\xi_x)$. This does not imply that 
$t_x$ is the sup of $Dom(\xi_x)$, i.e. that $t_x$ is the 
exit time.)}}

\gnue{{\rm Finally, notice that Brunovsk\'y made a mistake in talking 
about $\xi_x$ being the solution of the IVP $\dot x=f(x,v(x))$ with 
initial condition $\xi_x(0)=x$. Once you fix $x$ to be the initial
condition, you cannot use $x$ as a variable. I have corrected that.}}

\vs\itemitem{3.D}   If $P_2\in\P_2$, then the control
$v$ is continuous on $P_2\cup \Sigma(P_2)$, and for each $x\in P_2$
there exists a unique curve $\xi_x:[0,t_x[\to\Omega$ such that 
$\xi_x\restr]0,t_x[$ is a maximally defined integral curve 
of the vector field $f(\cdot,v(\cdot))$ on $\Sigma(P_2)$,
and $\xi_x(0)=x$.

\vs\itemitem{3.E} On every cell $P$, $x\to t_x$ is a
continuously differentiable function, and $(t,x)\to\xi_x(t)$, 
$(t,x)\to u_x(t)\equaldef v(\xi_x(t))$ are
continuously differentiable functions of $t,x$ on the set
$$
E(P)\equaldef\{(t,x):x\in P\,,\,t\in[0,t_x]\}
$$
in the sense that they can be prolonged to $\C^1$ functions on
some relatively open subset of $P\times[0,\infty[$ containing 
$E(P)$. 
\gnu{{\rm It should be $[0,\infty[\times P$, not $P\times[0,\infty[$. And, in
any case, it seems to me that what we really want is $\R\times P$.}}
\vs
\itemitem{3.F} For every $x\in S\backslash\{0\}$, if we let
$\tilde\xi_x$ denote the curve obtained in an obvious
way by piecing together the trajectories on every single cell, 
and write $\tilde\eta_x(t)=v(\tilde\xi_x(t))$, then 
the admissible pair $(\tilde\xi_x,\tilde\eta_x)$
ends at the origin after passing through a finite number of cells,
and is extremal. 
\vs
\itemitem{3.G} The value function $V_{\Gamma}$ is continuous.
}


\vs
\noindent
Boltianskii's definition, as given in \boltcount{},
imposes the following additional requirements:
\gnu{The way you phrased it, you are saying that Boltyanskii 
imposes all the requirements that Brunovsky does, plus a few ``additional''
ones.  That is not true. For example, Boltyanskii never talks about
stratifications, so in particular Boltyanskii does not require that
anything be a stratification.  Moreover, as I explained earlier,
Boltyanskii never says that the collection of cells has to be locally
finite. Also, Boltyanskii has the set $N$, and you completely forgot about
it. Finally, Boltyanskii's definition is not phrased in terms of the same
object as Brunovsk\'y's---a 6--tuple
$\Gamma=(\P,\P_1,\P_2,\Pi,\Sigma,v)$---but in terms of a sequence
$P^0,P^1,\ldots, P^n$ of ``piecewise smooth sets,'' from which he gets the
``$k$-dimensional cells'' as the connecetd components of
$P^k\setminus(P^{k-1}\cup N)$.  I don't see any mention of all this in
what you wrote.}
\vs

{\sl 
\itemitem{4.A} $dim(\Pi(P))=\dim(P)-1$ whenever $P\in\P_1$.

\vs
\itemitem{4.B} The only zero dimensional cell is $\{0\}$.

\gnu{{\rm Expressions such as ``zero-dimensional'' or ``one-dimensional'' 
should be hyphenated.}}

\vs
\itemitem{4.C} Every trajectory $\xi_x$ must enter a new cell  
``at a nonzero angle''. (That is, more precisely:
whenever $P\in\P_1$ and $x\in P$, the tangent vector
to $\xi_x$ at time $t_x$ is not tangent to $\Pi(P)$ at $\xi_x(t_x)$.)

}

\vs

\noindent {\bf Remark 2.}
Another definition ---closely related to Brunovsky's---
was proposed in \sussacount{}, and used to prove an existence
theorem for stabilizing synthesis.  The definition of Sussmann
\sussmcount{} ---which is the one used in this paper--- is
more general in a number of ways, e.g.  since it is not based on the 
existence of any feedback control generating the trajectories. 

\vs

The examples of the following section show optimal syntheses
that do not match the definitions above. Let $\Gamma$ be the
\gnu{To ``match a definition''?  What's that? I have never seen it 
anywhere. Where did you get it from?}
synthesis, in Brunovsky's sense, of Example~1. 
We have that $V_{\Gamma}$ is discontinuous, hence
$\Gamma$  does not satisfy the point $3.G$ of the Brunovsky definition. 

\gnu{The previous paragraph is very bad English. See how I rewrote it.
In particular, ``we have that'' is truly awful.   And I don't see how 
something can ``satisfy a point.''  Can you talk about ``satisfying a 
point'' in Italian?  I doubt it.  It seems to me that, in all the Western
languages that I know, the kinds of things that can be ``satisfied'' 
in mathematics are properties, conditions, requirements, assumptions, but
not ``points.''}

The same example show another important feature of Brunovsky's definition. 
\gnu{``The same example show.'' It should be ``shows.''?}
It shows the necessity
\gnu{The verb ``show'' is repeated just one line later. That's
very poor writing.}
of specifying, for each cell $P_2\in{{\cal P}}_2$, a cell $\Sigma(P_2)\in
{{\cal P}}_1$ entered by every trajectory $\xi_x$ (of 3.D) for $x\in P_2$.
Indeed, if
we do not define the map $\Sigma$ for Example~1 then some not optimal
trajectories would be admissible solutions of the corresponding discontinuous
feedback $v$. 
The hypothesis of the uniqueness of the $\xi_x$ entering
$\Sigma(P_2)$ can be relaxed after adding a specification of $\xi_x$
on an initial interval of time. This was done in \sussacount{}, \susslcount{},
where for every
$x\in P_2\in\P_2$ we have an assigned trajectory $\xi_x$, defined for
an interval of time of length $\ve_x$, that enters $\Sigma(P_2)$.
\gnu{I found this whole paragraph completely incomprehensible as
written.  After reading it twice, I still wasn't able to figure out what
you were trying to say,  even though presumably I am in a better position
than most readers to understand what you are talking about. 
I am sure no reader would be able to figure it out either. It took me a
lot of effort to decipher it, and I think I have now succeeded in a way
that can actually be understood. Please compare what I have written with
what you had written, and make sure that you understand the difference.
Actually, I found the whole text, from here up to the end of the section,
disorganized, incoherent, bordering on the incomprehensible. You first
announce that we are going to give ``examples of syntheses that do not
match the definitions'', but then what you actually do is meander
back and forth between this and several other quite different things,
without much of a pattern. You start by talking about the Brunosvky
definition and referring to Example 1 of the  following section, but the
only way in which this example does not ``match the definition'' is that
the cost function is discontinuous. You then venture into the quite
different issue of explaining why in Brunovsky's definition one needs
to specify $\Sigma$. This is indeed important, but it doesn't have a damn
thing  to do with ``not matching'' the brunosvky definition: quite the
contrary, it shows why the Brunovsky definition is good. You then make a
cryptic reference to ``relaxing'' the uniqueness as I did in my papers,
and then you go back to the discussion of Example 1 and how it shows the
importance of specifying $\Sigma$.  You explanantion of how it does so is
incomprehensible. You make a mysterious mention of the fact that at the 
point $(-1,0)$ both controls $1$ and $-1$ determine the same velocity
vector, but you don't give any clue to the reader about why this
observation matters. If at this point you had told the reader
that Italy did not get to the finals of the World Cup, this observation
would not have appeared any more irrelevant to the rest of the discussion
than your remark about the two controls giving rise to the same velocity.
(There is even inconsistent notation: in this this discussion you refer to
the feedback of Example 1 as ``$v$'', but when you actually present
Example 1 in the following section you call it ``$u$.'' Is it the same
object?)  You then move, in a seemeingly randon way, to repeating the
observation made earlier that the synthesis for Fuller's problem does not
``match'' the Brunovsky definition. You then go in the other direction,
by loooking at the flows of the Boltyianskii definition. In all this,
there is nodiscernible oreder, no progression, no structure to help the
reader make sense of what he is reading. Please take
a careful look at what you had written, compare it with my rewriting,
and try to understand the difference. I know you are not going to take
seriously what I am saying, and you are going to think that I am just
complaining about your English, and I am being intolerant because you are
not a native English speaker. This is NOT my problem. I think all the
points I am raising are language-independent, and would apply equally
well if you had written this in Italian. They are matters of logic, of
organization and intellectual discipline, the things that I would call 
``pre-mathematics'', meaning things that are not yet truly mathematical
but are a precondition for mathematical thinking and communication.}
In any case it is important to specify, for every $P_2\in{{\cal P}}_2$,  
the cell the trajectory $\xi_x$, $x\in P_2$, is
supposed to enter. If we allow all the trajectories of the discontinuous
feedback $v$ then we obtain some not optimal trajectories. For example,
starting from the Type II cell $R_3=\{(-1,0)\}$ we can use the trajectory
corresponding to the control $+1$ for
time $2\pi$, reaching again the same point, and then go to the origin
using the control $-1$. The curve $\gamma_1$ defined in this way 
is an admissible trajectory of $v$, indeed at the
\gnu{Here you have two complete sentences, not connected by a conjunction,
separated by a comma. That is contrary not only to grammr, but to logic
and common sense, and I don't think this is just a problem of English. 
I bet you cannot write like that in Italian either. Can you? 
In addition, I really cannot figure out what this sentence means. 
Are you sure you understand the word ``indeed''?} 
point $(-1,0)$ the two controls $\pm 1$ determine the same
velocity vector. However, $\gamma_1$ is clearly not time optimal.
We can also define the trajectory $\gamma_2$ of $v$,
with domain $[0,+\infty[$,
corresponding to the constant control $+1$ and starting from $(-1,0)$. 
This trajectory neither reaches the origin in finite time nor
approaches it as time goes to infinity.
Notice that the trajectories $\gamma_1$ and $ \gamma_2$ satisfy the equation
$\dot\gamma(t)=f(\gamma(t),v(\gamma(t)))$ for every $t$ in their
domain. Hence they satisfy the equation corresponding to the
discontinuous feedback $v$ in any reasonable {\sl weak} sense. 
For a classical definition of
solutions to a discontinuous ODE we refer to \filicount.
\gnus{I don't understand why you mention Filippov in this context. It is 
NOT true that, if a curve $\xi$ satisfies $\dot\xi(t)=f(\xi(t))$ for almost
all $t$ (or even for all $t$), then $\xi$ is a solution of the equation in
Filippov's sense. (For example, in 1 dimension, let $f(x)=1$ if $x\neq 0$,
$f(0)=0$. Let $\xi(t)=0$. Then $\xi$ is a classical solution but not a
Filippov solution.) Yet, you say ``Notice that the trajectories $\gamma_1$
and $ \gamma_2$ satisfy the equation
$\dot\gamma(t)=f(\gamma(t),v(\gamma(t)))$ for every $t$ in their
domain. Hence they satisfy the equation corresponding to the discontinuous
feedback $v$ in any reasonable {\sl weak} sense.''  And you even add ``For
a classical definition of solutions to a discontinuous ODE we refer to
\filicount.'' This clearly tells the reader that you think that the
Filippov sense is one of these ``reasonable senses'' that you had been
talking about.}
\gnue{In my new version, I removed the Filippov reference at this point.
But I gave a much more detailed discussion of the precise relationship
between classical solutions, Filippov solutions, and the ``feedback
solutions'' of Eduardo Sontag and his collaborators. Please read this
carefully (it's {\bf your} paper, after all!) and see how things now make
sense.} 
The synthesis of Fuller example does not match the definitions above
\gnu{To ``match'' definitions? What's that? I have never seen it 
anywhere. Where did you get it from? }
despite its evident {\sl regularity}. Indeed since every trajectory has an
infinite number of switchings the synthesis does not satisfy
assumption 3.F.

More dramatic is the situation for the definition given by
Boltianskii. The synthesis of Example~1 does not match 
the requirements 4.A and 4.B: the cells $R_7$ and $R_8$ are one
\gnu{To ``match a requirement''? }
dimensional Type I cells and $\Pi(R_8)=R_7$, the cell $R_3=\{(-1,0)\}$
is a zero dimensional cell different from the origin.
\gnu{I think it is silly to use this example as one where Boltianskii's 
condition 4.A fails. Of course it fails with this particular choice of
the partition into cells, but one can eliminate this problem by 
refining the partition, adding new $0$-dimensional cells. This is a
natural thing to do, because when one does it one gets a true
stratification, so all the Brunovsky conditions are satisfied, except for
the continuity of the cost function. In my new version I use this
partition instead of yours.}
The requirements 4.A and 4.B are
not satisfied even by very simple synthesis $\Gamma$ for which $V_{\Gamma}$ is
continuous. In \piccccount, \piccdcount, \piccacount, \piccbcount, 
\sussdcount, \sussecount, \susshcount, \sussicount, \susslcount, it was
studied the time optimal synthesis and the time optimal trajectories properties
for control affine single--input two dimensional systems. There are
some singular trajectories called turnpike. In general these turnpikes
\gnu{``turnpikes'', not ``turnpike''!}
are one dimensional cells of the time optimal synthesis with
target the origin and they are not necessarily ending at the
origin but enter other one dimensional or zero dimensional cells. 
For some explicit examples see \piccbcount.

The synthesis of Example~2 does not match the assumption 4.A as it is
\gnu{Now it's ``match an assumption''!!!! You really seem to have liked
this ``match'' stuff, because you use it over and over and over. 
Where did you get it from? }
easily shown, since some trajectories leave a three dimensional cell
entering a one dimensional cell. 

Finally, in \brunbcount{} it was shown an example not satisfying
hypothesis 4.C.

\gnu{``It was shown an example'' is not English!!!!}


%
%\section{Introduction}
%
%The purpose of this note is to present an example of an optimal synthesis
%that can be computed explicitly and happens to exhibit a peculiar feature,
%not often encountered in such problems.  The fact that this feature can
%occur has deep implications regarding the proper way to define the concept
%of an ``optimal feedback,'' as we now explain.
%
%Consider an optimal control problem arising from a controlled dynamics
%of the form
%\xdef\system{\equanosectno}
%$$
%\dot x=f(x,u)\,,\,\,\,x\in\R^n\,,\,\,\,u\in U\,,\numeq
%$$
%where $U$ is a compact convex subset of $\R^m$,
%and $f:\R^n\times U\to\R^n$ is smooth.  Assume a nonvoid {\it target set}
%$S\subseteq\R^n$ is specified, and we want to study the ``time optimal
%synthesis for (\system) with target $S$.''  (More generally, the state 
%space could be an open subset $\Omega$ of $\R^n$ rather
%than $\R^n$ itself, and one could even take $\Omega$ to be 
%be a smooth manifold. The control space $U$ could be a completely abstract
%set, and very minimal regularity requirements on $f$ would suffice.
%In addition, the cost functional to be minimized could be much more general
%than time, for example a functional of the Lagrangian type.  However,
%since the point of the present note is simply to show that a phenomenon can
%happen by giving an example, and the example happens to be very 
%well behaved, there is no need here to discuss the possibility 
%of more general technical conditions.)
%

\section{Examples}

\gnu{As you will notice, I have expanded and improved the ``Examples''
section, so that we now have examples for lots of other things that we
didn't even mention before. In addition, I want you to pay attention to the
{\bf structure}: if someone reads this section, he will be able to 
know what an example is about before reading the details of the example.
The previous version was written in such a way that nobody would have been
able to figure out what was the point of Example 1.}


\noindent{\bf Example 1.}

\noindent We study the time-optimal synthesis for the planar system:
\xdef\eqoo{\equanosectno} 
$$
\dot q= F(q)+ u\,G(q)\,,\qquad |u|\leq 1\,, \numeq
$$
where
\xdef\eqot{\equanosectno} 
$$
q\dot= \left(\matrix{x\cr \cr y\cr}\right)\,\,\in\,\,\R^2\,,\qquad
F(q)\,\dot= \left(\matrix{1-{y\over 2}\cr \cr {x+1\over 2}\cr}\right)\,,\qquad
G(q)\,\dot= \left(\matrix{-{y\over 2}\cr \cr {x+1\over 2}\cr}\right)\,,
\numeq
$$
and the target is the origin of $\R^2$.

The trajectories of (\eqoo) corresponding to the constant control $u\equiv
-1$ are straight horizontal lines going from left to right, while those
corresponding to $u\equiv +1$ are circles centered at the point (-1,1),
running counterclockwise.  The optimal synthesis, to be determined below,
is shown in Fig.2.

\vskip 1em


\centerline{\hbox{\psfig{figure=benny.ps,width=3in,height=3in}}}

\medbreak
\centerline{Figure 2}

\vskip 1em

%A {\it control} is a measurable function $u:[a,b]\rightarrow [-1,1]$ for
%some $a,b\in \R$ such that $a\le b$.  (As usual, we identify measurable
%functions that are equal almost everywhere.)  A {\it trajectory} for a
%control $u$ is an absolutely continuous function
%$\gamma:[a,b]\rightarrow\R^2$ verifying
%$\dot\gamma(t)=F(\gamma(t))+u(t)G(\gamma(t))$ for almost every $t\in[a,b]$.
%A trajectory $\gamma:[a,b]\to\R^2$ is {\it optimal} if there is no
%trajectory $\gamma':[a',b']\to \R^2$ such that $\gamma'(a')=\gamma(T)$,
%$\gamma'(b')=\gamma(b)$, and $b'-a'<b-a$.



An arc $\gamma:[a,b]\to\R^2$ is {\it simple} if $\gamma$ is an injective
map. It is clear that an optimal trajectory is simple. 

Since $G(q)=0$ only for $q=(-1,0)$, it is clear that a trajectory
$\gamma:[a,b]\to\R^2$ uniquely determines the corresponding control
$u(\cdot)$, unless the set $\{t:\gamma(t)=(-1,0)\}$ is of positive
measure. In particular, for simple trajectories $\gamma$ ---the only ones
that will be of interest to us--- the control is uniquely determined by
$\gamma$.

We use $\R^k_*$ to denote the space of real row vectors with $k$ entries,
\gnu{Inconsistent notation. In earlier sections we used $\R_k$ to denote
the space of row $k$-vectors! }
while $\R^k$ denotes the space of column vectors. If $\Omega$ is an open
subset of $\R^k$ and $X:\Omega\to\R^k$ is a vector field of class $C^1$,
then $\nabla X$ will denote the Jacobian matrix of $X$, i.e.  the $k\times
k$-matrix-valued function on $\Omega$ whose rows are the gradients 
\gnu{Inconsistent notation again. We used $DX$ earlier to denote the
Jacobian of $X$!!!!}
of the components of $X$.  If $X,Y$ are vector fields of class $C^1$,
then $[X,Y]$ will denote the Lie bracket of $X$ and $Y$, so 
$[X,Y]=(\nabla Y).X-(\nabla X).Y$.

%A {\it variational covector field} (or ``adjoint vector'') along a
%trajectory $\gamma$ of (\eqoo) ---corresponding to a control $u(\cdot)$---
%is an absolutely continuous function$\lambda:[a,b]\rightarrow\R^2_*$ 
%that verifies the ``adjoint equation''
%$$
%\dot\lambda(t)=-\lambda(t)\cdot\Bigl(\nabla F(\gamma(t))+
%u(t)\nabla G(\gamma(t))\Bigr) \numeq
%$$
%for almost every $t\in[a,b]$. We call $\lambda$ {\it nontrivial}
%if $\lambda(t)\not\equiv 0$. We define the {\it Hamiltonian}
%${\cal H}:\R^2_*\times\R^2\times\R\rightarrow \R$ by:
%$$
%{\cal H}(\lambda,q,u)=\lambda\cdot\Bigl(F(q)+uG(q)\Bigr)\,.\numeq
%$$
%A covector field $\lambda$ along $\gamma$ is said to be {\it minimizing}
%if:
%\xdef\mineq{\equanosectno} 
%$$
%{\cal H}\Bigl(\lambda(t),\gamma(t),u(t)\Bigr)=\min\{{\cal
%H}(\lambda(t),\gamma(t),w):|w|\leq 1\}\,\,\,\,{\rm for\,\,\,
%a.e.}\,\,\,t\in[a,b]\,.\numeq
%$$
%If a trajectory $\gamma$ is optimal, then $\gamma$ is simple, and has to be
%{\it extremal}, meaning that, if $u(\cdot)$ is the corresponding control,
%then the pair $(\gamma,u(\cdot))$ satisfies the necessary condition given
%by the Pontryagin Maximum Principle:
%
%\medbreak
%\item{(E)} {\it there exist a nontrivial
%minimizing variational covector field
%$\lambda$ and a nonnegative constant $\lambda_0$ such that}
%$$
%\lambda_0+{\cal H}(\lambda(t),\gamma(t),u(t))=0
%\,\,\,{\rm for\,\,\,almost\,\,\,every}\,\,\,t\in[a,b]\,.\numeq
%$$



\medbreak\noindent
If $X,Y$ are $\R^2$-valued functions on a subset of $\R^2$, then 
$(X,Y)$ will denote the $2\times 2$ matrix with columns $X,Y$.
In particular, we can define the functions, see \susshcount:
$$
\Delta_{A}=det\,(F,G)\qquad\Delta_{B}=det\,(G,[F,G])\numeq
$$
where ``{\it det\/}'' stands for determinant.
\gnu{Can you explain to me the grammatical structure of the sentence
``we can define the functions, see \susshcount:
$\Delta_{A}=det\,(F,G)\qquad\Delta_{B}=det\,(G,[F,G])$''?  I don't see it. 
It seems to me that ``see \susshcount'' is an entirely separate sentence,
not connected to the rest by any connective (conjunction or preposition
or relative pronoun) that might make it an admissible part of a larger
sentence. If you don't see my problem, I suggest that you translate 
this word by word into Italian and then, once you got it in Italian, you
try to parse it using the techniques for analyzing sentences that I am
sure you learned in your Italian high school.
Use your techniques to parse the sentence, and then let me know what you
found. My guess is that you will not be able to do it, because this
sentence is grammatically wrong in Italian as well as in English. 
Please compare with the way I rewrote this, and you will see that now 
it DOES make sense, grammatically as well as logically.}
{}from the results of \susshcount, we know that every extremal trajectory 
of (\eqoo) is a finite concatenation of ``bang'' and ``singular'' arcs.
A bang arc is a trajectory for a constant control with values in
the set $\{-1,1\}$ of endpoints of the control interval. 
Singular arcs lie in the set of zeros of the function $\Delta_{B}$.
In our case,
\xdef\eqdelab{\equanosectno} 
$$
\Delta_{A}={x+1\over 2}\,\,\,\,\,{\rm \,\,\,and\,\,\,}\,\,\,\,\,
\Delta_{B}=-{y\over 4}\numeq
$$
so the only singular arcs are those entirely contained in the $x$ axis,
and these correspond to the control $-1$, so they actually are  bang-bang.
Therefore {\it all the extremals of our problem are bang-bang}.

If $\Delta_A(q)\not= 0$ then the vectors $F(q)$ and $G(q)$ are linearly
independent, and we can define the numbers $f(q)$ and $g(q)$ as the
coefficients of the linear combination:
$$
[F,G](q)=f(q)F(q)+g(q)G(q).\numeq
$$
Let $\Omega_A$,  $\Omega_B$ be, respectively, the sets 
$\{q:\Delta_A(q)\neq 0\}$, $\{q:\Delta_B(q)\neq 0\}$, and define
$\Omega_{AB}=\Omega_A\cap\Omega_B$. 
As was proved in \susshcount, if $\gamma$ is an extremal trajectory and
$\gamma(t)\in\Omega_{AB}$ for every $t$, then $\gamma$ is bang-bang with at
most one switching, and this switching can 
only be from the control value $-1$ to the value $+1$ if $f>0$,
and from $+1$ to $-1$ if $f<0$. In our case, it follows from (\eqdelab) that
$$
\Omega_A=\{(x,y):x\neq-1\}\,,\,\,\,\,\,\,\Omega_B=\{(x,y):y\neq 0\}\,,\numeq
$$
and
$$
[F,G]= \left(\matrix{0\cr {1\over 2}\cr}\right)=
{y\over 2(x+1)}F+{2-y\over 2(x+1)}G\,,\numeq
$$
so $\Omega_{AB}$  is the union of four open quadrants $\Omega_{AB,i}$,
$i=1,2,3,4$, defined by 
$$\eqalign{
\Omega_{AB,1}&=\{(x,y):x>-1\,,\,y>0\}\,,\cr
\,\,\,\Omega_{AB,2}&=\{(x,y):x<-1\,,\,y>0\}\,,\cr
\,\,\,\Omega_{AB,3}&=\{(x,y):(x,-y)\in\Omega_{AB,2}\}\,,\cr
\,\,\,\Omega_{AB,4}&=\{(x,y):(x,-y)\in\Omega_{AB,1}\}\,.\cr}\numeq
$$
\gnu{See how, in my version, I reordered the three preceding formulas. 
I think now it makes more logical sense. Agreed?}
Clearly,  $f>0$ on $\Omega_{AB,1}\cup\Omega_{AB,3}$ and 
$f<0$ on $\Omega_{AB,2}\cup\Omega_{AB,4}$.  So the control switchings
are from $-1$ to $1$ on $\Omega_{AB,1}\cup\Omega_{AB,3}$ and 
{}from $1$ to $-1$ on $\Omega_{AB,2}\cup\Omega_{AB,4}$.

%Assume now to have two trajectories $\gamma_1$, $\gamma_2$, having the
%same initial point and the same terminal point. Following [], we have
%that if $\gamma_2^{-1}*\gamma_1$, that is the concatenation of
%$\gamma_1$ with $\gamma_2$ run backwards, is a simple curve enclosing
%the region $\cal R$ then:
%$$T(\gamma_1)-T(\gamma_2)=\int \int_{\cal R}
%{\Delta_B\over(\Delta_A)^2}$$
%where $T$ is the time along a trajectory.

To construct an optimal synthesis, 
we will
actually construct a collection ${\cal F}$ of simple
trajectories such that every
optimal trajectory ending at the target is in ${\cal F}$, and
for every initial point $q$ we will find a $\gamma_q\in{\cal F}$
such that if $\delta\in{\cal F}$ steers $Q$ to $0$ and $\delta\neq\gamma_q$
\gnus{According to our definitions, which make perfect logical sense, 
a CONTROL "steers" a point to another point. A TRAJECTORY does not.}

\gnue{In addition, you must have meant $q$ rather than $Q$ in the previous
line.}
then $\delta$ is not optimal. Since the existence of optimal trajectories
{}from any initial point follows by elementary arguments, we will be able to
conclude that $\gamma_q$ is optimal.

Suppose that $\gamma:[a,b]\to\R^2$ is time-optimal, $\gamma(b)=(0,0)$,
$a<b$, and $\gamma$ arises from an open-loop control $t\to u(t)$.
Then $\gamma$is simple, and 
is a concatenation of finitely many maximal bang arcs.
Since $\gamma$ is an extremal, we can fix 
a nontrivial minimizing adjoint vector $\lambda$  along $\gamma$
and a constant  $\lambda_0\ge 0$ for which (EX4) holds. Define
the {\it switching function}
$$\varphi(t)=\lambda(t)\cdot G(\gamma(t))\,.$$
{}The minimization condition implies that
$$
\varphi(t)\not= 0\,\, \Longrightarrow \,\,u(t)=-sgn(\varphi(t))\,.\numeq
$$
Let $\delta:[c,d]\to\R^2$ be a maximal $u=-1$ piece of $\gamma$, so
\gnu{This business of ``maximal $u=-1$ piece''  sounds bad to me. 
See what I wrote instead. }
$c<d$ and $\delta$ is of the form: 
$$
\delta(t)=(x_0+t,y_0)\,.\numeq
$$
%Since $\delta$ is maximal,
%
%\medbreak
%\item{(i)}~either 
%\itemitem{(i.1)} $c=a$,
%\item{}~or 
%\itemitem{(i.2)} $c>a$ and a control switching from $1$ to $-1$ occurs at $c$,
%\item{}~and
%\item{(ii)}~either 
%\itemitem{(ii.1)} $d=b$,
%\item{}~or 
%\itemitem{(i.2)}$d<b$ and a control switching from $-1$ to $1$ occurs at $c$.

%\medbreak
\noindent For a.e. $t\in[c,d]$, we have
$\dot \lambda=-\lambda\cdot \nabla (F-G)=0$.
So $\lambda=(\lambda_1,\lambda_2)$ is constant and:
$$
\varphi(t)=-\lambda_1 {y_0\over 2}+\lambda_2 {x_0+t+1\over 2}\numeq
$$
is a linear function of $t$.  Since a linear function cannot have more than
one zero unless it vanishes identically, we conclude that
$a<c<d<b$ can only be true if $y_0=0$. 
If $d=b$, then of course $y_0=0$, since $\gamma(b)=(0,0)$. 
So one of the following two possibilities occurs:

\medbreak
\item{(I)}~$\delta$ is contained in the $x$ axis,
\item{(II)}~$a=c$. 
\medbreak

This shows that all the maximal $u=-1$ pieces of $\gamma$ are entirely 
contained in the $x$ axis, with the only possible exception that 
$\gamma$ may contain an initial horizontal segment $[a,d]$ for which $y\neq
0$. 
\gnu{$\gamma$ is a curve in the plane. How can it contain a segment
$[a,d]$? And what can it possibly mean for a subinterval $[a,d]$ of the
real line to be ``horizontal''? 
I guess what you meant was that THE DOMAIN OF $\gamma$ contains
$[a,d]$. See how I rewrote this.}

In particular there are extremal trajectories that switch from
control $+1$ to control $-1$ at a point $(-L,0)$ with $L<0$, and then
reach the origin with control $-1$.

\gnu{I deleted the previous sentence because I don't think it's needed.}

Let us now analyze the adjoint equation for a maximal $u=+1$ piece
$\delta:[c,d]\to\R^2$ of $\gamma$, such that $c<d$. Clearly, $\delta$ is of
the form: 
\xdef\eqxycircle{\equanosectno} 
$$
x(t)=-1+r \cos(t+\theta)\,,\qquad y(t)=1+r \sin(t+\theta)
\,\qquad c\le t\le d\,,\numeq
$$
where $r>0$ is a constant and $\theta\in [0,2\pi]$. (The possibility
that $r=0$ is obviously excluded, because $\gamma$ is time-optimal and
$c<d$.)  The  components $\lambda_1,\lambda_2$ of $\lambda$ satisfy
$\dot\lambda_1=-\lambda_2$ and $\dot\lambda_2=-\lambda_1$, 
\gnu{The equations cannot be 
$\dot\lambda_1=-\lambda_2$ and $\dot\lambda_2=-\lambda_1$. One of the
minus signs must be wrong!}
so there exist constants $A>0$, $\theta_1\in[0,2\pi]$, such that
$$
\lambda_1(t)=A\cos(t+\theta_1)\,,\qquad\lambda_2(t)=A\sin(t+\theta_1)
\,\qquad c\le t\le d\,.\numeq
$$
Therefore, up to multiplication by a positive constant,
\xdef\rst{\equanosectno} 
$$
\varphi(t)= -\cos(t+\theta_1)+r\sin(\theta_1-\theta)\,.\numeq
$$
Now suppose that one of the two endpoints $c,d$ of the domain 
of $\delta$ is a switching time and the corresponding point
in $\R^2$ lies in the $x$ axis.  That is, we assume that
$h\in\{c,d\}$ is such that $\varphi(h)=0$ and $y(h)=0$.  Then 
\xdef\rsintheta{\equanosectno} 
$$
r\sin(h+\theta)=-1\,,\numeq
$$
so $\varphi(t)=-\cos(t+\theta_1)
+r\Bigl(\sin(h+\theta_1)\cos(h+\theta)
-\cos(h+\theta_1)\sin(h+\theta)\Bigr)$, and then
\gnu{Are you saying that the previous equation follows from (\rsintheta)?
I don't see it. It seems to me that this equation just follows from 
(\rst) by the formula for the sine of a difference. (\rsintheta) has
nothing to with this!}
$$\varphi(t)=-\cos(t+\theta_1)+\cos(h+\theta_1)+r\sin(h+\theta_1)\cos(h+\theta)
\,.\numeq$$
In particular, 
\xdef\iszero{\equanosectno} 
$$
0=\varphi(h)=r\sin(h+\theta_1)\cos(h+\theta)\,,\numeq
$$
so $\varphi(t)= -\cos(t+\theta_1)+\cos(h+\theta)$. If $r\neq 1$
then (\rsintheta) implies $0\ge \sin(h+\theta)\neq -1$, so 
$\cos(h+\theta)\neq 0$.  But then (\iszero) implies 
$\sin(h+\theta_1)=0$, so 
$\cos(h+\theta_1)=\pm 1$.  The possibility that 
$\cos(h+\theta_1)=1$ is excluded, for in that case
$\varphi$ would be $>0$ throughout $(c,d)$, contradicting the minimization
condition. So 
$\varphi(t)= -1-\cos(t+\theta_1)$. This shows that 
$\varphi(t)<0$ for $t\in[c,d]-\{h\}$, unless $d-c=2\pi$.  But 
if $d-c=2\pi$ then $\gamma$ would not be simple.
So $d-c<2\pi$, and we have shown that the other endpoint 
$h'$ of $[c,d]$ cannot be a switching point, and therefore $h'\in\{a,b\}$.

It then follows that one of the following possibilities must occur:

\medbreak
\item{(i)} $d=b\,$;
\item{(ii)} $a=c<d<b$ and $\gamma(d)$ lies in the $x$ axis;
\item{(iii)} $a<c<d<b$ and $\gamma(d)=(-1,0)$.

\medbreak\noindent
Indeed, if $d<b$ but $\gamma(d)$ is not in the $x$ axis, then we could let
$\delta'$ be the maximal $u=-1$ piece of $\gamma$ that starts at time $d$,
and our analysis of the $u=-1$ pieces of $\gamma$, applied to $\delta'$,
would show that $\delta'$ cannot in fact occur, since $\delta'$ is not an
initial segment of $\gamma$ and is not contained in the $x$ axis.  So, if
$d<b$ then $\gamma(d)$ lies in the $x$ axis, and $d$ is a switching time of
$\gamma$. Our previous argument shows that $c=a$, unless $r=1$.  But if
$r=1$, and $\gamma(d)$ is in the $x$ axis, then $\gamma(d)=(-1,0)$.  So
$d<b$ implies that either (ii) or (iii) must hold.

We now combine the results about the two types of pieces and provide a
complete description of the optimal trajectories ending at the origin. 
Let us use $L$, $L_0$, $C$ to denote, respectively, the trajectory
types ``$u=-1$ and not contained in the $x$ axis,''
``$u=-1$ and contained in the $x$ axis,'' and ``$u=1$.'' (The symbols
$L$ and $C$ stand for ``line'' and ``circle.'')  Let us 
use $T_1T_2\ldots T_m$ to denote the trajectory type 
``$T_1$ followed by $T_2\,\,\ldots\,\,\,$followed by $T_m$''.

With this notation, we have established that a nontrivial optimal
\gnu{What do you mean, ``we have established?'' We haven't, yet!!!
In fact, we are going to establish it in the next 13 lines or so.}
trajectory $\gamma$ ending at the origin is of one of the following seven
types: $L_0$, $C$, $L_0C$, $LC$, $CL_0$, $LCL_0$, $CL_0C$.  
(Concatenations
of six or more types are ruled out because any such concatenation would
have to contain two different $C$ pieces for which (iii) above holds, so
$\gamma$ would go at least twice through the point $(-1,0)$, and would not
be simple.  excluded for the same reason.  The type $L_0CL_0$ is excluded
\gnu{I don't understand the previous line! Something must be missing! 
This seems to be an old mistake. I went back to the file you sent to me
on August 20 1997, and the same line is still there. So that means
that you haven't reread the paper carefully after that date. And it
also means that the referee didn't read it carefully either.}
because the last switching would have to happen at $(-1,0)$, so $C$ would
have to be the circle with center $(-1,1)$ and radius $1$, which intersects
the $x$ axis only at $(-1,0)$.  So the other switching would also happen at
$(-1,0)$, and $\gamma$ would not be simple.  The exclusion of $L_0CL_0$
implies in turn that the types $CL_0CL_0$, $LCL_0CL_0$ and $L_0CL_0CL_0$
are ruled out as well. The types $L_0CL_0C$ and $LCL_0C$ are excluded
because the switching from $C$ to $L_0$ would have to happen at $(-1,0)$,
so the $L_0$ piece following the switching has to be a segment going from
$(-1,0)$ to a point $(\al,0)$ and $\al$ must satisfy $-1<\al<0$, for if
$\al$ was $\ge 0$ then the $L_0$ piece would already reach the origin
before switching to the last $C$. But if $-1<\al<0$ then the $C$ arc
starting at $(\al,0)$ does not go through the origin. Finally, having
excluded $L_0CL_0C$ we are also forced to exclude $CL_0CL_0C$.)
The $LCL_0$ trajectories are further restricted by the condition that the
switching from $C$ to $L_0$ must happen at the point $(-1,0)$.  Let us call
those $LCL_0$ trajectories that satisfy this extra condition ``good.''

It turns out that the optimal synthesis involves all seven types listed
above, but not all trajectories of one of these types are optimal, so we
need a finer analysis.  We let ${\cal F}$ be the set of all simple
trajectories that end at the origin and are either of one of the 
six types $L_0$, $C$, $L_0C$, $LC$, $CL_0$, $CL_0C$, or of type
$LCL_0$ and good.

We now observe that $\dot x=1-y$ along a $u=1$ arc, so $\dot x>1$ as long
as $y<0$, $u=-1$. Therefore, if $q_1=(x_1,y_1)$ and $q_2=(x_2,y_2)$ satisfy
\gnu{I dont' understand the two lines above. Shouldn't it be $u=1$ 
before ``Therefore''?}
$y_1=y_2\le 0$, $x_1<x_2$, $x_1+x_2=-2$, so that $q_1$ and $q_2$ can be
joined both by a $C$ arc and an $L$ (or $L_0$) arc, then the $C$ arc is
contained in the lower half-plane, and is faster than the $L$ or $L_0$ arc.
In particular, if we let $L_0(\beta)$ denote, for $\beta>0$, the $L_0$ arc
going from $(-\beta,0)$ to $(0,0)$, we see that {\it $L_0(\beta)$ is not
optimal if $\beta>1$}, because the piece of $L_0$ going from $(-\beta,0)$
to $(-2+\beta,0)$ can be replaced by a faster type $C$ arc.
\gnu{It seems to me that this argument is only valid if $\beta\le 2$. Right? }

Next we show that {\it $L_0(1)$ is optimal}. It suffices
to notice that there is no trajectory from $(-1,0)$ to $(0,0)$ of any of
the types $C$, $LC$, $LCL_0$, and exactly one trajectory of each of the
types $L_0$, $L_0C$, $CL_0$, and $CL_0C$, but the last three are not
simple, so $L_0(1)$ is left as the only possible candidate.

We now study the optimal trajectories that are of one of the types $C$,
$L_0C$, $LC$, and $CL_0C$.  Let $\gamma:[a,b]\to\R^2$ be such a trajectory,
and take $b=0$, so $\gamma(0)=(0,0)$.  

If $\gamma$ is a $C$ arc, then $a>-2\pi$, for otherwise $\gamma$ would
contain a full loop. For $0<a<2\pi$, we let $C(a)$ be the $C$ arc
$\gamma:[-a,0]\to\R^2$ such that $\gamma(0)=(0,0)$.  It will be shown below
that the arcs $C(a)$ are all optimal.

Next suppose that $\gamma$ is $LC$.  Since the $C$ piece satisfies
(\eqxycircle), we must have $r=\sqrt{2}$ and $\theta=-{\pi\over 4}$. Let 
\gnu{I thought you had said earlier that $\theta\in[0,2\pi]$!
So how can you have  $\theta=-{\pi\over 4}$? }
$c$ be the time when the switching from $L$ to $C$ occurs. On
$[c,0]$, the switching function $\varphi$ is given, up to multiplication 
by a positive constant, by:
$$
\varphi(t)=\sqrt{2}\sin\Bigl(\theta_1-{\pi\over 4}\Bigr)-\cos (t+\theta_1)
=\sin{\theta_1}-\cos{\theta_1}-\cos (t+\theta_1).
$$
\gnus{The first equality of the previous formula is wrong.  In (5.14) you
had $\sin(\theta_1-\theta)$.  So you should have had 
$\sin\Bigl(\theta_1+{\pi\over 4}\Bigr)$ rather than 
$\sin\Bigl(\theta_1-{\pi\over 4}\Bigr)$.}
\gnum{This mistake in the sign of one term in the formula for $\varphi$
changes the whole analysis.  I have had to redo the whole thing, using
the new formula. Please read what I have done carefully. As you will see,
the analysis using the correct expression is much simpler and more
elegant than the wrong analysis we had in the original version.  
This is in fact how I discovered the error.  It seemed wrong to me that 
we would prove that $c$ has to be $>-\pi$ and then use a different, 
ad hoc argument (based on the fact that certain switchings are not
permitted in certain quadrants) to exclude the possibility that
$c<-{\pi\over 2}$. % It didn't make sense, because the first argument
seemed to tells us that one can construct an $LC$ arc with a switching 
function having the correct sign required by the MP, and then we pull
this other thing out of the hat to say that some of these $LC$ arcs are
not allowed, even though one can clearly construct them. The two proofs
that certain switchings are not allowed are essentially the same, so they
cannot give different results. I spent several hours thinking about
this, until I found the mistake. Luckily, once we correct the mistake 
and do things correctly we end up with the same synthesis, the proof 
becomes simpler and nicer, and the apparent contradiction that we
appeared to have reached goes away, as it should have.}
\gnum{This should teach you a lesson: WHEN I TELL YOU TO BE CAREFUL
AND PAY ATTENTION TO THE DETAILS I KNOW WHAT I AM TALKING ABOUT.
I know that you think that I am taking this whole thing too seriosly, 
and it doesn 't matter too much if there are lots of mistakes as long 
as the referees don't catch them, and that just one little sign in one
formula isn't important. I think it is, though. Thank God that this 
time the mistake in the proof did not affect the result. But it could
have.}
\gnum{Finally, in a sense this whole thing started with the ``ospitality''
misspelling. It was this that gave me the clue that you had not really
done your job carefully and thoroughly.  Armed with this basic intuition,
I went on reading what you had written expecting to find lots of
mistakes, and it seems that this was the right thing for me to do.}
\gnum{Just one more thing: If I had been a referee, I would have acted the 
same way. The ``ospitality'' spelling would have alerted me to the fact
that the authors of this paper were not serious. Then I would have looked
for mistakes all over the place, and they are not that hard to find,
as I am sure you see by now if you have read all my comments. So I would
have recommended that the paper not be accepted as it was.  I think it is
a good indication of the poor quality of current journals, and the lack 
of truly competent refereeing, that a paper like ours got accepted.}
\gnum{Unfortunately, it would not be a good strategy on our part to use 
the above considerations as arguments to prove to Henk Nijmeijer 
that the reviewer didn't really read the paper carefully, and for that 
reason when he recommends all those shortenings he should not be taken
seriously.  We'd better keep our opinion of the referee to ourselves.}
\gnue{In any case, I am sure that in  the future you will spell 
``hospitality'' correctly. And I hope you will spell other words 
correctly too. As Aesopus could have said, ``people who make small
mistakes usually make big ones as well.''}
So
$$
\varphi(t)=(1+\sin t)\sin{\theta_1}-(1+\cos t).\cos{\theta_1}
$$
In addition, $c$ cannot be $-\pi$, because  $c=-\pi$ would correspond
to $\gamma(c)=(-2,2)\in Q_{AB,2}$,
\gnu{I don't understand what you mean by the notation $Q_{AB,2}$. Could it
be the same object that you called $\Omega_{AB,2}$ before?  If so, then we
would have another case of inconsistent notation.}
and we know that a switching from $u=-1$ to $u=1$ is not permitted
at points in $Q_{AB,2}$.

If $\sin{\theta_1}=0$, then $\cos{\theta_1}=\pm 1$, so 
$\cos{\theta_1}=1$, because $\varphi(0)\le 0$.  
So $\varphi(t)=1+\cos t$, and $\varphi(c)=0$ would imply
$c=-\pi$. Since this possibility has been ruled out,
we conclude that $\sin{\theta_1}\neq 0$.  We can then define
$$
\psi(t)={1+\sin t\over 1+\cos
t}\,,\,\,\,\,\nu={\cos{\theta_1}\over\sin\theta_1}\,.
$$
Then $\psi$ is well defined and smooth except at values of 
$t$ such that $t=\pi + 2k\pi$, $k$ an integer.  In our case, $\psi$ is 
defined on the whole interval $[c,0]$, except at $t=-\pi$ (if
$-\pi\in[c,0]$).   Since $c\neq-\pi$, $\psi(c)$ is defined,
and the switching condition $\varphi(c)=0$ says that $\psi(c)=\nu$.  If
$\sin{\theta_1}>0$, then $\varphi(t)<0$ iff $\psi(t)<\nu$ and 
$\varphi(t)=0$ iff $\psi(t)=\nu$. Then $\psi(0)\le \nu$ and,
since $\lim_{t\to -\pi}\psi(t)=+\infty$, we see that $c$ must 
satisfy $-\pi<c<0$. If
$\sin{\theta_1}<0$, then $\varphi(t)<0$ iff $\psi(t)>\nu$. So
${1\over 2}=\psi(0)\ge\nu$. 
A straightforward
computation shows that ${d\psi\over dt}(t)>0$ for
$-2\pi\le t<-\pi$. Since $\psi(-2\pi)=\psi(0)\ge\nu$, we conclude 
that $\psi(t)>\nu$ for $-2\pi\le t<-\pi$. So, once again, the switching
cannot occur at a point in $[-2\pi,-\pi)$, and then 
$c\in(-\pi,0)$.

So we have shown in all possible cases that $-\pi<c<0$. If
$-\pi<c<-{\pi\over 2}$, then $\gamma(c)\in Q_{AB,2}$, so a switching
{}from $L$ to $C$ is not permitted. If $c=-{\pi\over 2}$, then 
$\gamma(c)$ lies in the $x$ axis, so the previous piece is $L_0$ rather
than $L$.   So $-{\pi\over 2}<c<0$.   If ${\pi\over 4}<c$, then 
$\gamma(c)\in Q_{AB,4}$, and the switching is not permitted either.
So the only possibility left to us is $-{\pi\over 2}<c\le-{\pi\over 4}$.

For $0<\sigma<2\pi$, $s>0$, let $LC(s,\sigma)$ denote the trajectory 
that consists of a $u=-1$ arc during time $s$, followed by a $u=1$ arc
during time $\sigma$, and ending at the origin.   We have shown that 
{\it $LC(s,\sigma)$ is not optimal unless 
${\pi\over 4}\le\sigma<{\pi\over 2}$.}

We now show that {\it when ${\pi\over 4}\le\sigma<{\pi\over 2}$ then
$LC(s,\sigma)$ is optimal.}  To see this, let $q$ be the starting point of
$LC(s,\sigma)$, for $s>0$, ${\pi\over 4}\le\sigma<{\pi\over 2}$.  It is
clear that $LC(s,\sigma)$ is the only $LC$ trajectory from $q$ to $(0,0)$.
Also, it is easy to see that there are no trajectories from $q$ to $(0,0)$
of the types $L_0$, $C$, $L_0C$, or good $LCL_0$. 
There is exactly one $CL_0$ trajectory, but it is easy to see that the time
$\hat\tau$ along this trajectory is larger than the time
$\tau$ along $LC(s,\sigma)$.  (Actually, $\tau<|x|<\hat\tau-{3\pi\over
2}$.) Finally, there is exactly one $CL_0C$ trajectory $\gamma$, 
which can also be ruled out. (If $\bar x=-1-\sqrt{2-(y+1)^2}$, so 
\gnu{I dont' understand this calculation.  Shouldn't it be $(y-1)^2$ 
rather than $(y+1)^2$? }
$\bar q=(\bar x,y)$ is the point where 
$LC(s,\sigma)$switches from $L$ to $C$, then both 
$LC(s,\sigma)$ and $\gamma$ go through $\bar q$ and coincide from that point
on. So it suffices to compare the parts of 
$LC(s,\sigma)$ and $\gamma$ up to $\bar q$. The time along 
$LC(s,\sigma)$ is $\bar x-x$, and that along $\gamma$ is 
$>\sqrt{(1+x)^2+(1-y)^2-1}-1+{3\pi\over 2}$, so it suffices to notice 
\gnu{I don't understand this. I made the calculation, based on the fact
that the time along the $L_0$ piece must be more than the difference of
the radii of the larger circle (with radius $\sqrt{(1+x)^2+(1-y)^2}$)
and the smaller one (with radius $\sqrt{2}$), and I got that the time
along $\gamma$ is $>\sqrt{(1+x)^2+(1-y)^2}-\sqrt{2}+{3\pi\over 2}$, which
isn't exactly the same thing. Can you explain to me how you got your
formulas? }
that $\sqrt{(1+x)^2+(1-y)^2-1}+{3\pi\over 2}+x-\sqrt{2-(y+1)^2}>0$.)
\gnu{Take a look at my own calculation. It came out quite different,
although the final conclusion is the same. I think that my calculation
is right, an in addition it's cleaner than yours. Am I right?
{\bf PLEASE CHECK MY NEW CALCULATION CAREFULLY!}  }
for $s>0$, we let $L_0C(s)$ denote the trajectory that consists of a $u=-1$
arc during time $s$ followed by a $u=1$ arc during time ${\pi\over 2}$, and
ending at the origin.  Then is is easily shown that {\it $L_0C(s)$ is
optimal for every $s>0$.}  (Indeed, if $q=(x,0)$ is the starting point of
$L_0C(s)$, for $s>0$, then $L_0C(s)$ is the only $L_0C$ trajectory from $q$
to $(0,0)$, and there are no simple 
trajectories from $q$ to $(0,0)$ of the types
$C$,  $LC$, $CL_0$, $LCL_0$ or $CL_0C$.  There is an $L_0$ arc, namely,
$L_0(|x|)$, but we know that this arc is not optimal, because $|x|>1$.)

Now define $R_1=\{(0,0)\}$, $R_2=(-1,0)\times\{0\}$, $R_3=\{(-1,0)\}$. 
\gnu{In the above line, the expression $(-1,0)$ occurs twice, once meaning
interval and once meaning ordered pair. This proves the wisdom of
Bourbaki, when they chose to use $]a,b[$ rather than $(a,b)$ as a
notation for open intervals. I have followed Bourbaki throughout the
paper. This has required quite a bit of work!!!!!}

For $q=(0,0)$, let $\Gamma_q$ be the trajectory $\gamma$ with domain $\{0\}$
\gnu{Weren't we using $\gamma_q$ (with LOWER CASE $\gamma$)  to denote 
the trajectories of a synthesis $\Gamma$? }
such that $\gamma(0)=(0,0)$. For $q=(x,0)\in R_2\cup R_3$, let 
$\Gamma_q=L_0(|x|)$.  Then the $\Gamma_q$, $q\in R_1\cup R_3\cup R_3$ are
optimal. 

For $r>0$, let $D(r)$ be the open disc with center $(-1,1)$ and radius $r$,
and let $\partial D(r)$ denote the boundary of $D(r)$.  Let $R_4=D(1)$,
$R_5=\partial D(1)\backslash\{(-1,0)\}$. Then a point $q\in\R^2$ can be
joined to $(0,0)$ by a good $LCL_0$ trajectory if and only if $q\in R_4$,
\gnu{This is not true.  There are lots of points that are not in $R_4$ 
and can be joined to $0$ by a good $LCL_0$ trajectory!!!}
and in that case the good trajectory is unique, and we will use $LCL_0(q)$
to denote it.  Then {\it the arcs $LCL_0(q)$, for $q\in R_4$, are
optimal}. To see this, observe that if $q\in R_4$ then the only
trajectories in ${\cal F}$ other than $LCL_0(q)$ that go from $q$ to the
origin are either $LCL_0$ but not good, or $LC$ with a switching from $L$
\gnu{Something is wrong here. The way we defined ${\cal F}$, an $LCL_0$
 trajectory can only b in ${\cal F}$ if it is good, so how can we talk 
 about  ``trajectories in ${\cal F}$ ... that ... are ... $LCL_0$ but not 
 good? }
to $C$ at a time $c$ that violates the requirement that $-{\pi\over
2}<c$. So all the alternative candidates are excluded, and $LCL_0(q)$ must
be the optimal trajectory.

A similar argument shows that if $q\in R_5$ then the unique $CL_0$
trajectory going from $q$ to $(0,0)$ ---henceforth labeled $CL_0(q)$--- is
optimal.

Next, let $R_6=D(\sqrt{2})\backslash (R_2\cup r_3\cup R_4\cup R_5)$.  
\gnus{In the above line it should be $R_3$, not $r_3$.} 
\gnue{The analysis for $R_6$ is incomplete. You failed to discuss the
possibility that $y=2$. This case actually requires some extra
work to be excluded!  I have done it in my version.}
If $q\in R_6$, then the obvious
candidate for optimality is the arc $\Gamma_q$ of type $CL_0$ obtained by
following the $C$ arc through $q$ until $R_2$ is reached, and then
continuing along $R_2$ towards the origin.  We now prove that {\it the
arcs $\Gamma_q$, $q\in R_6$, are optimal}.  To see this, observe that if
$q=(x,y)\in R_6$ and $y\neq 0$ then there are no simple 
arcs from $q$ to $(0,0)$
of the types $L_0$, $C$, $L_0C$, $CL_0C$.  There is an $LC$ arc, but (i)~if
$y<0$ then this arc switches from $L$ to $C$ in $Q_{AB,4}$, so it is not
optimal; (ii)~if $y>0$, then the arc is $LC(s,\sigma)$ for some
$\sigma\in\pi,2\pi)$, so it is not optimal either.  If $y<0$ or $y>2$ or
\gnu{Even if you use the standard $(a,b)$ notation for open intervals
rather than Bourbaki's, there is a parenthesis missing in the previous
line. }
$x>-1$, then there is no good $LCL_0$ arc. If $0<y<2$ and $x<-1$ then there
is a good $LCL_0$ arc, but it switches from $L$ to $C$ at a point in
$Q_{AB,2}$, so this arc is not optimal.  Finally, if $y<0$ then $\Gamma_q$
is the only $CL_0$ arc, and if $y>0$ then there are two such arcs, but the
other one is not optimal, because it contains $L_0(\beta)$ for some
$\beta>1$.  This proves the optimality of the arcs $\Gamma_q$ for $q\in
R_6$, $y\neq 0$. A similar analysis works for $y=0$.  (Alternatively, if
$q=(x,0)$ then $\Gamma_q$ is a cofinal piece of $\Gamma_{q'}$ for some
$q'=(x',y')\in R_6$ such that $y'\neq 0$, so $\Gamma_q$ is optimal.)

Next, we let $R_7=\partial D(\sqrt{2})\backslash\{(0,0)\}$.  For $q\in R_7$
there is a unique trajectory $\Gamma_q$ of type $C$ going from $q$ to
$(0,0)$. The optimality of $\Gamma_q$ for $q\in R_7$ is proved by
considerations similar to those used for $R_6$.

We let $R_8=(-\infty,-2)\times\{0\}$, $R_9=\{(x,y):1-\sqrt{2}<y<0\,,\,
x<-1\,,\,(x+1)^2+(1-y)^2>2\}$, $R_{10}=(-\infty,-1)\times\{ 1-\sqrt{2}\}$.
Then $R_9\cup R_{10}$ is exactly the set of starting points $q$ of the
trajectories $LC(s,\sigma)$, for $s>0$, ${\pi\over 4}\le\sigma<{\pi\over
2}$, and we know that for each $q\in R_9\cup R_{10}$
the $LC(s,\sigma)$ trajectory is unique and optimal. We define $\Gamma_q$ 
to be this trajectory. A similar fact is true for $R_8$, with 
$L_0C(s)$ rather than $LC(s,\sigma)$, and $\Gamma_q$ is then defined 
to be the optimal trajectory.   

Finally, we let $R_{11}=\R^2\backslash(R_1\cup\ldots\cup R_{10})$.  For
$q\in R_{11}$, we let $\Gamma_q$ be the unique simple 
$CL_0C$ trajectory from $q$
to $(0,0)$. We show that {\it $\Gamma_q$ is optimal for all $q\in R_{11}$}.
Let $\gamma\in {\cal F}$ be an extremal trajectory starting at $q$ and
verifying $\gamma(0)=0$. There is no such trajectory of type $L_0$,$C$
or $L_0C$. The case $LC$ is impossible, otherwise $\gamma$ would have
a switching at time $c\in [-5\pi/4,-\pi/2[$. 
\gnu{I don't see why the switching time $c$ would have to satisfy 
$c\ge -5\pi/4$.   It doesn't matter, anyhow, because the thing that
suffices to rule this case out is $c<-\pi/2$.}
If $\gamma$ is of type $CL_0$ then it contains a not optimal
\gnu{``a not optimal trajectory'' is not acceptable in English.}
trajectory $L_0(\beta)$, $\beta>1$.
\gnu{I don't understand the grammar of the above sentence. It seems
to me that ``$\beta>1$'' is a separate sentence, since you didn't connect
it to anything else with words like ``such that'' or ``with'' of ``for
which.'' But then you cannot just adjoin it to the rest with a comma. (You
cannot do this in Italian either, I bet. Can you? Please tell me.) Look at
the way I rewrote this sentence.}

Finally, if $\gamma$ is of type $LCL_0$ good then $\gamma(t)=p\in
R_6$ for some $t$, but $\Gamma_p$ is not the restriction of $\gamma$
\gnu{I rewrote this. Please check.}
to $[t,0]$. Hence, if $\gamma$ is optimal then it is of type $CL_0C$.

Since we have considered all the extremal trajectories 
\gnu{The sentence ``we have considered all the extremal trajectories''
is extremely vague. Look at how I rewrote this. }
this concludes the construction.
\gnu{The construction of WHAT?  The discussion of this example begins 
with the words ``we study the optimal synthesis for the problem ...''
So you did not ANNOUNCE at the beginning that you were going to
``construct'' the optimal synthesis. I have fixed that, by changing both
the beginning and the ending of the discussion. }

This ends the analysis and proves that the synthesis of Fig.1 is the
optimal one.

\gnu{Isn't it Figure 2 rather than Figure 1?}

Let define the feedback control $u(x,y)$ to be equal to $0$ on 
\gnu{``Let define'' should be ``Let us define.''}
%
\gnu{Is this feedback $u$ the same one that you called $v$ in the previous
section? If so, why change the letter?} 
$R_1$, to $-1$ on the set
$R_2\cup R_3\cup R_4\cup R_8\cup R_9\cup R_{10}$,  
and to $+1$ on $R_5\cup R_6\cup R_7\cup R_{11}$. 
Then the $+1$ trajectory starting from the point $(-1,0)$ is an admissible
trajectory for this feedback control but it is not optimal.

\bigbreak


%%%%%%%%%%%%%%SECOND EXAMPLE

\noindent{\bf Example 2.}

\noindent The following example provides a synthesis that is optimal but
does not match the restriction 4.A of Boltianskii's definition.
\gnus{``Matching a restriction is awful.''  See how I rewrote this.}

\gnus{Example 2 is obviously wrong.  I didn't even attempt to
follow your calculations, but it's clear to me that the answer that you get
cannot be right.}
\gnum{Let me explain.  As you state it, the problem that you are looking at is
one of finding the distance function to the origin and the optimal arcs for
lenght minimization for a perfectly ordinary smooth Riemannian metric $g$
on $\R^3$.  Moreover, it's a fairly simple metric, given by
$$
  ds^2 = A dx^2 + B dy^2 + C dz^2   \eqno{1)}
$$
where $A$, $B$, $C$ are smooth strictly positive functions. (I'll show you
below what $A$, $B$, $C$ are in this case.)   When you have a Riemannian
metric given by (1), then an arc $t\to \xi(t)=(x(t),y(t),z(t))$ is
parametrized by arc-length if  $ds$ is just $dt$, i.e. if
$$
A{dx\over dt}^2+B{dy\over dt}^2+C{dz\over dt}^2=1  \eqno{(2)}
$$
If you write  the evolution equations as
$$
\dot x={u\over \sqrt{A}} ,  \dot x={v\over \sqrt{B}},
\dot y={w\over \sqrt{C}} ,
$$
then (2) corresponds exactly to $u^2+v^2+w^2=1$. Naturally, if you use
instead the control constraint $u^2+v^2+w^2 \le 1$, you get the same
minimizers.}
 
\gnue{Your example 2 arises if we take
$$
A = B = {1+x^2+y^2\over x^2+y^2}\,,\,\,\,
C =  1+x^2+y^2
$$
So the optimal control problem of your Example 2 is just a nice ordinary
Riemannian problem.  For such a problem the minimizers are the GEODESICS.
And the geodesics are ALL SMOOTH. So I don't see how you could possibly get
the solutions that you say you do.}
 

Consider the system:
$$\cases{\dot x = u {\sqrt{x^2+y^2}\over \sqrt{1+x^2+y^2}} &\cr 
         \dot y = v {\sqrt{x^2+y^2}\over \sqrt{1+x^2+y^2}} & 
                                           $u^2+v^2+w^2\leq 1$ \cr
         \dot z = w {1\over \sqrt{1+x^2+y^2}} &\cr}.$$
and the problem of stabilization to the origin in minimum time.
Consider the following regions:
\gnu{Too many ``consider''. }
$$R_0=\{(0,0,0)\},\qquad  R_1=\{(x,y,z): x=y=0, \, z>0\},\qquad
R_2=\{(x,y,z):x=y=0,\, z<0\},$$
$$R_3=\{(x,y,z):x^2+y^2>0,\, z=0\}, \qquad\qquad
R_4=\{(x,y,z):x^2+y^2>0,\,z>0\},$$  
$$R_5=\{(x,y,z):x^2+y^2>0,\, z<0\}.$$
\gnu{The above formula looks ugly.  I realigned it. }
We assign a feedback control $(u_i(x,y,z),v_i(x,y,z),w_i(x,y,z))$ for
every region $R_i$ in the following way:
$$u_0=v_0=w_0=0,\qquad u_1=v_1=0,\;w_1=-1\qquad
u_2=v_2=0,\;w_2=1,$$
$$u_3=u_4=u_5=-{x\over\sqrt{x^2+y^2}},\;v_3=v_4=v_5=-{y\over\sqrt{x^2+y^2}},
\;w_3=w_4=w_5=0.$$
\gnu{Again, the above formula looks ugly.  I realigned it. }
Hence starting from a point $(x,y,z)\in R_3\cup R_4\cup R_5$ we first reach
\gnu{I don't understand this ``hence.''   ``Hence'' means 
``for this reason.''  I don't see what ``reason'' you have in mind.}
the $z$ axis using control $(u_3,v_3,w_3)$, and then, if we are not yet at
the origin,  we reach the origin running on
the $z$ axis. In this way we have constructed a synthesis $\Gamma$.
It is possible to prove that this synthesis $\Gamma$ is optimal applying
\gnu{``we have constructed a synthesis $\Gamma$'' immediately followed 
by ``It is possible to prove that this synthesis $\Gamma$ ...'' !!!!!
That's very poor writing. Repetitive!!! }
Theorem~2.1. However, in this case it is immediate to verify that the
\gnu{I wish you had used the labels rather that the number 2.1. That's 
what labels are for!!!!!}
associated value function
$$V_{\Gamma}=|z|+\sqrt{x^2+y^2}$$
satisfies the Hamilton--Jacobi equation on the set
$\{(x,y,z):x^2+y^2>0,z\not=0\}$. Indeed:
$$\inf_{u^2+v^2+z^2\leq 1}\left(
{\partial V\over \partial x}\,u\,{\sqrt{x^2+y^2}\over
\sqrt{1+x^2+y^2}}+
{\partial V\over \partial y}\,v\,{\sqrt{x^2+y^2}\over
\sqrt{1+x^2+y^2}}+
{\partial V\over \partial z}\,w\,{1 \over
\sqrt{1+x^2+y^2}} \right)=$$
$$=-\left\|\left({x\over \sqrt{1+x^2+y^2}},
{y\over \sqrt{1+x^2+y^2}},
{1\over \sqrt{1+x^2+y^2}} \right)\right\|=-1.$$

We can deduce that equation (2.16) holds at every point of $R_4\cup R_5$.
\gnu{My God! Why didn't you use labels?  How am I going to figure out,
after all the changes, how exactly which equation is (2.16)? 
Do you realize that by being lazy and not doing the job properly you end
up making this much worse and making me work double the time just to
figure out what these stupid numbers stand for?  }
The other assumptions of Theorem~2.1, necessary to conclude the proof, are
easily checked.
\gnus{I don't understand the grammatical structure of the above sentence,
and I don't understand the meaning. Are you trying to say that 
one can check ALL the other assumptions, or only those assumptions that
are needed to conclude the proof? (In English, no matter which one was
the case, you would NOT say it the way you did. And I bet you cannot say
it that way in Italian either!!!)}
\gnue{Even more seriously, I don't understand the LOGICAL structure of the
above paragraph, and in fact I cannot figure out what you are saying:
you seem to say first that we could use Theorem 2.1 but it's easier 
to do it without using it, via the HJB equation; but then you go ahead 
and talk ---very confusingly--- about ``veryfying the assumptions of
Theorem 2.1''.   So, are we applying Theorem 2.1, yes or no? I cannot
tell? How do you expect the reader to understand this?}
Notice that every trajectory starting from a point of the three
dimensional cell $R_4$, respectively $R_5$, reaches the one dimensional cell
$R_1$, respectively $R_2$.
Every trajectory starting from the two dimensional cell $R_3$ reaches the
zero dimensional cell $R_0$.  Hence, the synthesis $\Gamma$ does not match
the restriction 4.A of section~4.
\gnu{``Match''?} 

\bigbreak
\appen{A}{Differentiation of trajectories with respect to a parameter} 

\noindent The purpose of this appendix is to prove a general theorem on
differentiation with respect to a parameter $p$, taking values in a normed
space  $P$, of a family of trajectories $x_p:[a,b]\mapsto\R^n$
of vector fields $f_p$.  We will state and prove the theorem under
minimal hypotheses, much weaker than what is actually needed for the main 
results of this paper, because the theorem has other applications, such as
the theory of envelopes, where the more general statement is useful.

If $\dot x_p(t)=f_p(x_p(t),t)$, then formal differentiation with respect
to $p$ at $p=p_0$ in some direction $v$ yields the variational equation
\xdef\varformal{\equanoapno}
$$
\dot y_{p_0,v}(t)={\partial f_{p_0}\over \partial{x}\,\,\,}
\Bigl(x^{p_0}(t),t\Bigr).y_{p_0,v}(t) 
+\lim_{\ve\downarrow 0}{w_{p_0+\ve v}(t)-w_{p_0}(t)\over \ve}\,,\numeqap
$$
where $y_{p_0,v}(t)$ is the directional derivative of $x_p(t)$ at $p_0$ in
the $v$-direction, and 
$w_p(t)=f_p(x_{p_0}(t),t)$.   The technical problem that will concern
us here is to make (\varformal) rigorous even when the limit in the
right-hand side only exists in a weak sense. 

%For example, suppose 
%that $n=1$ and we are looking at the  control system $\dot x=u$, and 
%a family of ``bang-bang'' controls $u_s$ depending on $s\in[0,\infty[$,
%such that $u_s(t)=1$ for $t<s$, and $u_s(t)=-1$ for $t\ge s$.  Let 
%$x_s:[0,\infty[\,\mapsto\R$ be the solution of $\dot x=u_s$, $x(0)=0$. Then 
%$x_s(t)=t$ for $t\le s$, and $x_s(t)=2s-t$ for $t\ge s$. Given any
%$T>0$, the limit $\lim_{\ve\to 0}{u_{s+\ve}^T-u_s^T\over\ve}$
%---where $u_s^T$ is the restriction of $u_s$ to $[0,T]$---
%does not exist pointwise, but exists in the weak$^*$ sense 
%---regarding the $u_s$ as members of the dual of the space of continuous
%functions on $[0,T]$--- and equals $2\delta_s$, where $\delta_s$ is the 
%Delta function at $s$.  If we let $y_s(t)$ be the derivative of $x_s(t)$
%with respect to $s$ ---so $y_{s}(t)=y_{s,1}(t)$--- then $y_s$ 
%satisfies the differential equation $\dot y_s(t)=2\delta_s(t)$, which makes
%better sense in integrated form: $y_s(t)=2\chi_{[s,\infty[}(t)$.
%This says, in particular, that $y_s(T)=2$ for $s<T$, and $y_s(T)=0$
%for $s>T$.    
%As this example shows, 

The variational equation that arises
when the limit of (\varformal) exists in a weak sense is a linear equation
$\dot y(t)=A(t)y(t)+w(t)$ where the ``input'' $w$ is the formal derivative
of a function $W$ which is not necessarily absolutely continuous or even
continuous. So our first task will be to study the solutions of linear 
systems with ``generalized inputs'' $w$.

\medbreak
Let $-\infty<a<b<\infty$, and let $n$ be a positive integer.  
If $A\in L^1([a,b];\R^{n\times n})$, and $\bar Y\in\R^n$,
then it is well known that the initial value problem
\xdef\inhom{\equanoapno}
$$
\dot Y(t)=A(t)\cdot Y(t)+w(t)\,,\,\,\,\,Y(a)=\bar Y\numeqap
$$
has a unique solution for every initial condition and every map 
$w\in L^1([a,b],\R^n)$.  Because (\inhom) can also be written formally as
\xdef\inhomfor{\equanoapno}
$$
dY(t)=A(t)\cdot Y(t)\,dt+w(t)\,dt\,,\,\,\,\,Y(a)=\bar Y\numeqap
$$
we will use $Y_{A,w\,dt,\bar Y}$ to denote the unique solution of (\inhom). 

As explained before, it will be important for us to be able to solve
(\inhom) for ``inputs'' $w$ more general than integrable functions. The
appropriate class of inputs turns out to be that of {\sl formal derivatives
of bounded measurable functions}.  Precisely, we let $BM([a,b],\R^n)$
denote the space of all bounded Lebesgue measurable functions
$W:[a,b]\to\R^n$, and use $BM_0([a,b],\R^n)$ to denote the set of those
$W\in BM([a,b],\R^n)$ such that $W(a)=0$.  (In $BM([a,b],\R^n)$ we do {\it
not} identify two functions that are equal almost everywhere.)  We endow
$BM([a,b],\R^n)$ with the norm 
$$||W||_{BM}=\sup\{||W(t)||:t\in[a,b]\}\,.\numeqap
$$
Then $BM([a,b],\R^n)$ is a Banach space, and $BM_0([a,b],\R^n)$ is a closed
subspace of $BM([a,b],\R^n)$.  The space $C^0([a,b],\R^n)$ of continuous
maps $W:[a,b]\mapsto\R^n$ is of course a closed subspace of $BM([a,b],\R^n)$,
but $L^\infty([a,b],\R^n)$ is {\sl not} a subspace of $BM([a,b],\R^n)$,
because in $L^\infty([a,b],\R^n)$ we identify functions that are equal
a.e., but in $BM([a,b],\R^n)$ we do not.

If $W\in BM_0([a,b],\R^n)$ is absolutely continuous, and $w=\dot W$, 
so that, formally,  $dW=w\,dt$,  then we can rewrite (\inhom) formally as
\xdef\inhomint{\equanoapno}
$$
dY(t)=A(t)\cdot Y(t)\, dt+dW(t)\,\,\,\,\,\,\,\,\,\,Y(a)=\bar Y\,.\numeqap
$$
A solution of (\inhomint) is then a continuous map $t\to Y(t)$ such 
that 
\xdef\inhomintsol{\equanoapno}
$$
Y(t)=\bar Y+\int_a^tA(s)\cdot Y(s)\,ds+W(t)\,\qquad{\rm for
\quad all\quad} t\in[a,b]\,.\numeqap
$$ 
Now (\inhomintsol) makes sense for arbitrary $Y\in BM([a,b],\R^n)$,
$W\in BM_0([a,b],\R^n)$, even if $W$ is not absolutely continuous.
So we turn it into a definition:  for a $W\in BM_0([a,b],\R^n)$, a 
{\sl solution} of (\inhomint) is a map
$Y\in BM([a,b],R^n)$ such that (\inhomintsol) holds.     It then follows that 

\medbreak


{\pariim

\item{(EU)} {\sl for every $W\!\!\in\!\!BM_0([a,b],\R^n)$  and every 
$\bar Y\!\!\in\!\!\R^n$ there exists a unique solution $Y$ of (\inhomint).}

}

\medbreak\noindent
Indeed, uniqueness is trivial, since the difference $Y^1-Y^2$ of two solutions
is a solution of the homogeneous problem (i.e. of (\inhom) with $w\equiv 0$
and $\bar Y=0$),  so $Y^1-Y^2\equiv 0$.  To prove existence, we observe that 
(\inhomintsol) is equivalent to 
\xdef\inhomintsolequiv{\equanoapno}
$$
Y(t)-W(t)=\bar Y+\int_a^tA(s)\cdot (Y(s)-W(s))\,ds+\int_a^tA(s)\cdot W(s)
\,ds\,\qquad{\rm for\quad all\quad} t\in[a,b]\,,\numeqap
$$ 
i.e. to $dZ=AZ\,dt+AW\,dt$, $Z(a)=\bar Y$, with $Z=Y-W$.  Since $AW\in
L^1$, because  
$A\in L^1$ and $W$ is bounded and measurable, the equation
$\dot Z(t)=A(t)\cdot Z(t)+A(t)W(t)$ has a unique solution $Z^*$ for which 
$Z^*(a)=\bar Y$.  If we let $Y(t)=Z^*(t)+W(t)$,  and recall that 
$W(a)=0$,  we see that $Y$ is the desired solution of (\inhomint).

We will use $Y_{A, dW,\bar Y}$ to denote the solution of (\inhomint),
whose existence  and uniqueness has just been established.   
In addition to proving (EU), the preceding argument also gives an explicit 
formula for $Y_{A,dW,\bar Y}$, namely:
\xdef\inhomintsolfla{\equanoapno}
$$
Y_{A,dW,\bar Y}=W+Y_{A,AW\,dt,\bar Y}\,.\numeqap
$$
Clearly, the function $Y_{A,AW\,dt,\bar Y}$ is absolutely continuous,
so the regularity properties of $Y_{A,dW,\bar Y}$ ---modulo absolutely
continuous  functions--- are exactly the same as those of $W$. 
(For example: $Y$ is continuous,  right-continuous, or left-continuous 
at a point $t\in[a,b]$ if and only if $W$ is; $Y$ is of bounded variation 
iff $W$ is; $Y$ is absolutely continuous iff $W$ is.)

Gronwall's inequality, applied to (\inhomintsolequiv), tells us that
$$
||Y_{A,dW,\bar Y}(t)-W(t)||\le e^{||A||_{L^1}}\Bigl(||\bar Y||+
||A||_{L^1}||W||_{BM}\Bigr)\,.\numeqap
$$
Therefore
\xdef\ineqyw{\equanoapno}
$$
||Y_{A,dW,\bar Y}(t)||_{BM}\le ||W||_{BM}+e^{||A||_{L^1}}\Bigl(||\bar Y||+
||A||_{L^1}||W||_{BM}\Bigr)\,.\numeqap
$$
On the other hand, (\inhomintsol) also implies that
\xdef\ineqwy{\equanoapno}
$$
||W||_{BM}\le||\bar Y||+||Y_{A,dW,\bar Y}||_{BM}\Bigl(1+||A||_{L^1}\Bigr)\,.\numeqap
$$   
\xdef\theolin{{\rm \appenlet}.\the\theono}
\theoap
Let $\{W_j\}$,  $\{\bar Y_j\}$, $\{A_j\}$ be sequences in 
$BM_0([a,b],\R^n)$, $\R^n$, $L^1([a,b],\R^{n\times n})$, respectively.
Assume that $\{A_j\}$ is bounded in $L^1([a,b],\R^{n\times n})$. 
Let $Y_j=Y_{A_j,dW_j,\bar Y_j}$. Then

\itemitem{(\theolin.1)}  {\sl $\{Y_j\}$ is bounded in $BM([a,b],\R^{n})$ if 
and only if $\{W_j\}$ is bounded in $BM([a,b],\R^{n})$ and $\{\bar Y_j\}$ 
is bounded in $\R^{n}$.}

\itemitem{(\theolin.2)} {\sl if $\{A_j\}$ converges in $L^1([a,b],\R^n)$
to a limit  $A$, $\bar Y\in\R^n$, $\bar Y_j\to\bar Y$, 
$\{W_j\}$ is bounded in $BM([a,b],\R^n)$, 
$W\in BM_0([a,b],\R^n)$, 
and $Y=Y_{A,dW,\bar Y}$, then $Y_j\to Y$ almost everywhere iff $W_j\to W$ 
almost everywhere.  In that case,  $Y_j-W_j$ converges to $Y-W$ uniformly on 
$[a,b]$. In particular, for every $t\in[a,b]$, $Y_j(t)\to Y(t)$ iff 
$W_j(t)\to W(t)$.} 

\beginproof
Statement (\theolin.1) follows from the bounds (\ineqyw) and (\ineqwy).  To
prove (\theolin.2), we first observe that under the hypotheses of
(\theolin.2) we can apply (\theolin.1) and conclude that $\{Y_j\}$ is
bounded in $BM([a,b],\R^n)$.

Let $\tilde Y_j=Y_j-W_j=Y_{A_j,A_jW_j\,dt,\bar Y_j}$, 
$\tilde Y=Y-W=Y_{A,AW\,dt,\bar Y_j}$.  Then 
$$
\tilde Y_j(t)=\bar Y_j+\int_a^tA_j(s)Y_j(s)\,ds
=\bar Y_j+\int_a^tA_j(s)\tilde Y_j(s)\,ds+\int_a^tA_j(s)W_j(s)\,ds\,,\numeqap
$$ 
so
\xdef\forgrony{\equanoapno}
$$
\tilde Y_j(t)-\tilde Y(t)=\int_a^t A(s)\Bigl(\tilde Y_j(s)-\tilde Y(s))\,ds+
R_j(t)\,,\numeqap
$$
where 
$$
R_j(t)=\bar Y_j-\bar Y+\int_a^t\Bigl(A_j(s)-A(s)\Bigr)\tilde Y_j(s)\,ds
-\int_a^t A(s)\Bigl(W_j(s)-W(s)\Bigr)\,ds\,,\numeqap
$$
\gnu{This formula seems to be wrong! I think the correct formula is 
$$
R_j(t)=\bar Y_j-\bar Y+\int_a^t\Bigl(A_j(s)-A(s)\Bigr) Y_j(s)\,ds
+\int_a^t A(s)\Bigl(W_j(s)-W(s)\Bigr)\,ds\,.
$$
Please check.  I know that you are going to say that this is the part that
I wrote, so you are not responsible for my mistakes. But I cannot accept
that! As you may have noticed by now, I did check carefully everything you
wrote. So you should have checked carefully everything I wrote.}
and also
\xdef\forgronw{\equanoapno}
$$
\tilde Y_j(t)-\tilde Y(t)=\bar Y_j-\bar Y+\int_a^t\Bigl(A_j(s)-A(s)\Bigr)
Y_j(s)\,ds
+\int_a^t A(s)\Bigl(Y_j(s)-Y(s)\Big)\,ds\,.\numeqap
$$
If $W_j-W\to 0$ a.e., then, since the sequences $\{\tilde Y_j\}$, $\{Y_j\}$
and $\{W_j\}$ are uniformly bounded, and $A_j\to A$ in $L^1$, it is clear
that the first two terms of the sum defining $R_j(t)$ go to zero uniformly
as $j\to\infty$.  Also,
$$
\biggl\|\int_a^t A(s)\Bigl(W_j(s)-W(s))\,ds\biggr\|\le \int_a^b||A(s)||\cdot
||W_j(s)-W(s)||\,ds\,,\numeqap
$$
which goes to zero because
$s\to ||A(s)||$ is integrable and the sequence $\{W_j-W\}$ is uniformly
bounded and goes to zero almost everywhere.  So, if we let 
$K_j=\sup\{||R_j(t)||:a\le t\le b\}$, we see that $K_j\to 0$ as
$j\to\infty$. Then Gronwall's inequality, applied to (\forgrony), gives 
$||\tilde Y_j(t)-\tilde Y(t)||\le K_je^{||A||_{L^1}}$.
So $\tilde Y_j-\tilde Y\to 0$  uniformly.

If $Y_j-Y\to 0$ a.e., then the first two terms of the right-hand side of (\forgronw) 
go to zero uniformly, and the third one  is bounded by 
$\int_a^b||A(s)||\cdot||Y_j(s)-Y(s)||\,ds$, which also goes to zero. So
$\tilde Y_j-\tilde Y\to 0$  uniformly.

We have thus shown that $\tilde Y_j-\tilde Y\to 0$  uniformly if either
$W_j\to W$ a.e. or $Y_j\to Y$ a.e. This clearly implies  all the conclusions
of (\theolin.2).\proofend
 
A very important subspace of $BM([a,b],\R^n)$ is $BV([a,b],\R^n)$, the space of
all functions $W:[a,b]\mapsto\R^n$ such that (a)~$W$ is of bounded
variation, (b)~$W$ is right-continuous at every point of $]a,b]$, and
(c)~$W(a)=0$.  There is a canonical correspondence between $BV([a,b],\R^n)$ and
the dual space $C^0([a,b],\R_n)^*$ of the Banach space $C^0([a,b],\R_n)$ of
continuous functions $[a,b]\mapsto\R_n$. (We write $R_n$ rather than $R^n$
because we want to think of the members of $C^0([a,b],\R_n)$ as {\sl
row-vector-valued functions}, cf. below.)

The members of $C^0([a,b],\R_n)^*$ are the {\sl $\R^n$-valued Borel
measures on $[a,b]$}.  When $n=1$, they are the {\sl finite signed Borel
measures on $[a,b]$}.  
%Clearly, $C^0([a,b],\R_n)^*$ is canonically
%identified with $\Bigl(C^0([a,b],\R)^*\Bigr)^n$, so the members of
%$C^0([a,b],\R_n)^*$ can be thought of as $n$-tuples of finite signed Borel
%measures.  
The identification map from $BV([a,b],\R^n)$ to
$C^0([a,b],\R_n)^*$ is the one that assigns to a function $W\in
BV([a,b],\R^n)$ the unique Borel measure $\mu_W$ such that
$\mu_W([a,t])=W(t)$ for $a<t\le b$. 
%(The measure of a single-point set
%$\{t\}$ is then $\mu_W(t+)-\mu_W(t-)$, which is equal to
%$\mu_W(t)-\mu_W(t-)$ if $t>a$ and to $\mu_W(a+)$ if $t=a$.  In particular,
%$\mu_W(\{t\})=0$ iff $\mu_W$ is continuous at $t$.)

{}From now on we identify a function $W\in BV([a,b],\R^n)$ with its
corresponding measure $\mu_W$, and write $\int\alpha\cdot dW$ for
$\int\alpha\cdot d\mu_W$.  
%As explained above, it is convenient to think of
%the members of $C^0([a,b],\R_n)$ as row-vector-valued functions
%$\alpha=(\alpha_1,\ldots,\alpha_n)$, and of those of $BV([a,b],\R^n)$ as
%column-vector-valued functions $W=(W^1,\ldots,W^n)$, so the expression
%$\alpha\cdot dW$ stands for the inner product $\sum_k\alpha_k\,dW^k$.
Clearly, $\int\alpha\cdot dW$ is defined, more generally, for an arbitrary
bounded Borel measurable $\R_n$-valued map $\alpha$ on $[a,b]$.

Naturally,  the integral $\int\alpha\,dW$ of a scalar bounded Borel measurable
function $\alpha:[a,b]\mapsto\R$ is also well defined, and is a
vector in $\R^n$.   
%We emphasize that, if
%$W$ is not continuous, then for an integral such as $\int_s^tdW$ to be well
%defined we have to be careful to specify whether the integral is to be
%interpreted as $\int \chi_{[s,t]}\,dW$ or $\int \chi_{]s,t]}\,dW$ or $\int
%\chi_{[s,t[}\,dW$ or $\int \chi_{]s,t[}\,dW$, so we will never write
%$\int_s^t\,dW$ unless $W$ is continuous at $s$ and $t$.
 
On $BV([a,b],R^n)$ there are at least three important topologies , namely,
(a)~the one induced by the norm $||\ldots||_{BM}$, (b)~the strong topology
of $BV([a,b],\R^n)$ as the dual of the Banach space $C^0([a,b],\R_n)$,
(c)~the weak$^*$ topology, also corresponding to the duality with
$C^0([a,b],\R_n)$.  
%The first one will not be of direct interest to us,
%except for the fact that, in order to apply Theorem \theolin{}, we will
%often need to know that a set is bounded in the norm $||\ldots||_{BM}$. 
The second one is induced by the {\sl total variation norm}:
$$
||W||_{TV}=\sup\biggl\{\Bigl|\int \alpha\cdot dW\Bigr|:\alpha\in
C^0([a,b],\R_n)\,,\,||\alpha||_{sup} \le 1\biggr\}\,.\numeqap
$$ 
%It is well known that $||W||_{TV}$ is the supremum of the sums
%$S_{\pi}(W)$, ranging over all partitions
%$\pi\in\Pi([a,b])=\cup_{m}\Pi_m([a,b])$, where $\Pi_m([a,b])$ is the set of
%all $m+1$-tuples $(t_0,t_1,\ldots,t_m)$ of members of $[a,b]$ such that
%$a=t_0<t_1<\ldots<t_m=b$ and, for $\pi=(t_0,t_1,\ldots,t_m)\in \Pi([a,b])$,
%$S_{\pi}(W)\equaldef\sum_{i=1}^m||W(t_i)-W(t_{i-1})||$.  In particular, 
It is clear that $||W||_{BM}\le ||W||_{TV}$ for all $W$ in $BV([a,b],\R^n)$.
Therefore, {\sl a subset of $BV([a,b],\R^n)$ which is bounded in $TV$ norm
is also bounded in $BM$ norm}.

We will need the following
\xdef\lemweakconv{{\rm \appenlet}.\the\lemmno}
\lemmap
If a sequence $\{W_j\}$ in $BV([a,b],\R^n)$ weak$^*$-converges to a $W\in
BV([a,b],\R^n)$, then (a)~$\{W_j\}$ is bounded in $TV$ norm, (b)~$W_j(a)\to
W(a)$ and $W_j(b)\to W(b)$, (c)~every subsequence of $\{W_j\}$ has a
subsequence that converges to $W$ at all but at most countably many points
of $[a,b]$, (d)~$\int_a^b\beta(s)||W_j(t)-W(t)||\,dt\to 0$ for every
function $\beta\in L^1([a,b],\R)$.

\beginproof
Statement (a) follows from the uniform boundedness theorem, and Statement
(b) is trivial, since $W_j(a)=0=W(a)$, $W_j(b)=\int 1\,dW_j$, and
$W(b)=\int 1\,dW$.  Statement (d) clearly follows from (c), using the
dominated convergence theorem together with the fact that boundedness in
$TV$ norm implies boundedness in $BM$ norm.

So all we need to do is prove (c).  If $Z^i$ is the $i$-th component of $Z$
for $Z\in BM([a,b],\R^n)$, $i=1,\ldots,n$, then $\{W_j\}$
weak$^*$-converges to $W$ iff $\{W_j^i\}$ weak$^*$-converges to $W^i$ for
each $i$.  From this it clearly follows that it suffices to prove (c) for
$n=1$.

If $n=1$, then every finite signed Borel measure $\mu$ on $[a,b]$ 
admits a unique decomposition $\mu=\mu^+-\mu^-$,  where $\mu^+$ and $\mu^-$
are nonnegative Borel measures that are mutually singular (that is:
there exist two $\mu$-measurable sets $E$ and $F$, such that
$[a,b]=E\cup F$, $E\cap F=\emptyset$, $\mu^+(E)=\mu^-(F)=0$).
This gives, using the identification of $BV([a,b],R)$ with
$C^0([a,b],R)^*$, a decomposition $Z=Z^+-Z^-$ of every $Z\in BV([a,b],\R)$
into a difference of two members $Z^+$, $Z^-$ of  
${\cal M}(a,b)$, the space of all monotonically
nondecreasing functions $W:[a,b]\mapsto\R$ such that $W(a)=0$ and $W$ is
right-continuous at every $t\in]a,b]$.    It is well known that
$||Z||_{TV}=Z^+(b)+Z^-(b)$.   

Now let $\{W_{j(k)}\}_{k=1}^\infty$ be an arbitrary subsequence of $\{W_j\}$, 
and write  $Z_k=W_{j(k)}$. Letting
$Z_k=Z_k^+-Z_k^-$, we have $||Z_k^+||_{TV}=Z_k^+(b)\le||Z_k||_{TV}$, so 
the sequence  $\{Z_k^+\}$ is bounded in $BV([a,b],\R)$. Similarly,  
$\{Z_k^-\}$ is bounded in $BV([a,b],\R)$.   Since the closed balls
of $BV([a,b],R)$ are weak$^*$-compact and metrizable,  there exists
a subsequence $\{Z_{k(\ell)}\}$ of $\{Z_k\}$ such that  
$Z_{k(\ell)}^+$ and $Z_{k(\ell)}^-$ weak$^*$-converge to 
limits $U$, $V$. Then of course  $Z_{k(\ell)}=Z_{k(\ell)}^+-Z_{k(\ell)}^-$ 
weak$^*$-converges  to  $U-V$, and also to $W$,  so $U-V=W$.  (But this does
{\sl not} mean that $U=W^+$ and $V=W^-$.) 

Since the $Z_{k(\ell)}^+$ are in ${\cal M}(a,b)$, it is well known (and
very easy to prove directly) that $\{Z_{k(\ell)}^+(t)\}$ converges 
to $U(t)$ at every point
of continuity of $U$.  Since $U$ is monotonically nondecreasing,
$Z_{k(\ell)}^+(t)\to U(t)$ at all but countably many $t$'s in $[a,b]$.  A
similar conclusion holds for $Z_{k(\ell)}^-$ and $V$. So $Z_{k(\ell)}(t)\to
W(t)$ for all but countably many $t$'s, concluding our proof. \proofend


As a corollary of Lemma \lemweakconv{}, we get the integration by parts 
formula:
\xdef\lemparts{{\rm \appenlet}.\the\lemmno}
\lemmap
If $\lambda:[a,b]\mapsto\R_n$ is absolutely continuous, and $W\in
BV([a,b],\R^n)$, then 
\xdef\forparts{\equanoapno}
$$
\lambda(b)\cdot W(b)=\int \lambda\cdot dW+\int_a^b \dot\lambda(t)
\cdot W(t)\,dt\,.\numeqap
$$

\beginproof
The result is obviously true, from the standard product rule for
derivatives, if $W$ is also absolutely continuous. In the general case, we
use the fact that an arbitrary $W\in BV([a,b],\R^n)$ is weak$^*$-limit of a
sequence $\{W_j\}$ of absolutely continuous functions.  Then
(i)~$\lambda(b)\cdot W_j(b)\to \lambda(b)\cdot W(b)$, because $W_j(b)\to
W(b)$, (ii)~$\int \lambda\cdot dW_j\to\int \lambda\cdot dW$, because
$\{W_j\}$ weak$^*$-converges to $W$, and $\lambda$ is continuous, and
(iii)~$\int_a^b \dot\lambda(t)\cdot W_j(t)\,dt\to
\int_a^b \dot\lambda(t)\cdot W(t)\,dt$, because Part (d) of Lemma
\lemweakconv{} implies that 
$\int_a^b ||\dot\lambda(t)||.||W_j(t)-W(t)||\,dt\to 0$,  since
$\dot\lambda\in L^1$.   \proofend

An important consequence of Lemma \lemparts{} is the following
\xdef\lemadjo{{\rm \appenlet}.\the\lemmno}
\lemmap
Let $W\in BV([a,b],\R^n)$, $A\in L^1([a,b],R^{n\times n})$,
$\bar Y\in\R^n$, and write $Y=Y_{A,dW,\bar Y}$. Suppose
$\lambda:[a,b]\to\R_n$ is absolutely continuous and satisfies
$\dot\lambda(t)=-\lambda(t)A(t)$ for almost all $t$.  Then
\xdef\foradjo{\equanoapno}
$$
\lambda(b)\cdot Y(b)-\lambda(a)\cdot Y(a)=\int\lambda\cdot dW\,.\numeqap
$$

\beginproof
Write $Y(t)=V(t)+W(t)$, where $V(t)=Y(a)+\int_a^tA(s).Y(s)\,ds$.  
Then 
$$
\lambda(b)\cdot Y(b)-\lambda(a)\cdot Y(a)=
\lambda(b)\cdot V(b)-\lambda(a)\cdot V(a)+\lambda(b)\cdot W(b)\,.\numeqap
$$
The function $t\to\lambda(t)\cdot V(t)$ is absolutely continuous
and its derivative is $\dot\lambda(t)\cdot V(t)+\lambda(t)\cdot\dot V(t)$,
which equals $-\lambda(t)A(t)Y(t)+\lambda(t) A(t)Y(t)-\dot\lambda(t)\cdot W(t)$,
i.e. $-\dot\lambda(t)\cdot W(t)$. So 
$$
\lambda(b)\cdot Y(b)-\lambda(a)\cdot Y(a)=\lambda(b)\cdot W(b)
-\int_a^b\dot\lambda(t)\cdot W(t)\,dt\,.\numeqap
$$
Equation (\foradjo) follows from this and the integration by parts formula
(\forparts). \proofend

It is clear that, if $A\in L^1([a,b],\R^{n\times n})$, $\bar Y\in\R^n$, and 
$W\in BM_0([a,b],\R^n)$, then $Y_{A,dW,\bar Y}-W$ is absolutely continuous,
so $W\in BV([a,b],\R^n)$ iff $Y_{A,dW,\bar Y}$ is of bounded variation,
which happens iff $Y_{A,dW,\bar Y}-\bar Y\in BV([a,b],\R^n)$.

\xdef\theolinbv{{\rm \appenlet}.\the\theono}
\theoap
Let $\{W_j\}$,  $\{\bar Y_j\}$, $\{A_j\}$ be sequences in 
$BV([a,b],\R^n)$, $\R^n$, $L^1([a,b],\R^{n\times n})$, respectively.
Assume that $\{A_j\}$ is bounded in $L^1([a,b],\R^{n\times n})$  and 
$\{\bar Y_j\}$ is bounded in $\R^{n}$. Let $Y_j=Y_{A_j,dW_j,\bar Y_j}$ . Then

\itemitem{(\theolinbv.1)}  {\sl $\{Y_j-\bar Y_j\}$ is bounded in 
$BV([a,b],\R^{n})$ if and only if $\{W_j\}$ is bounded in
$BV([a,b],\R^{n})$.}

\itemitem{(\theolinbv.2)} {\sl If $\{A_j\}$ converges in $L^1([a,b],\R^n)$
to a limit $A$, $\bar Y_j\to\bar Y$, $W\in BV([a,b],\R^n)$, and we write
$Y=Y_{A,dW,\bar Y}$, then $\{W_j\}$ weak$^*$-converges to $W$ iff
$\{Y_j-\bar Y_j\}$ weak$^*$-converges to $Y-\bar Y$.  In that case,
$Y_j-W_j-\bar Y_j$ converges to $Y-W-\bar Y$ strongly in
$BV([a,b],\R^n)$. In particular, (a)~$W_j(b)\to W(b)$, (b)~$Y_j(b)\to
Y(b)$, (c)~for every $t\in[a,b]$, $Y_j(t)\to Y(t)$ iff $W_j(t)\to W(t)$.}



\beginproof
We know from theorem \theolin{} that $\{Y_j-\bar Y_j\}$ is bounded in $BM$
norm iff $\{W_j\}$ is.  Since boundedness in $BV$ norm implies boundedness
in $BM$ norm, we see that boundedness of one of the two sequences in $BV$
norm implies that there is a constant $C$ such that $||Y_j(t)||\le C$ for
all $j,t$, and then the integral $I_j=\int_a^b||A_j(t)Y_j(t)||\,dt$ is
bounded by a fixed constant $K$.  But $I_j=||Y_j-\bar Y_j-W_j||_{BV}$.  So
boundedness in $BV$ of one of the sequences $\{W_j\}$, $\{Y_j\}$ implies
boundedness in $BV$ of the other one.

If either $W_j$ weak$^*$-converges to $W$, or $Y_j-\bar Y_j$
weak$^*$-converges to $Y-\bar Y$, then the uniform boundedness theorem
implies that either $\{W_j\}$ or $\{Y_j-\bar Y_j\}$ is bounded in $BV$
norm, so both sequences are bounded in $BV$ norm ---and hence also in
$BM$--- norm by a constant $C$.  Writing $Z_j=Y_j-\bar Y_j-W_j$, $Z=Y-\bar
Y-W$, we have then $Z_j-Z=R_j^1+R_j^2$, where
$$
R_j^1(t)=\int_a^t\Bigl(A_j(s)-A(s)\Bigr)Y_j(s)\,ds\,,\qquad
R_j^2(t)=\int_a^tA(s)\Bigl(Y_j(s)-Y(s)\Bigr)\,ds\,.\numeqap
$$
Then $||R_j^1||_{BV}\le C||A_j-A||_{L^1}$, so $||R_j^1||_{BV}\to 0$.  Also,
$$||R_j^2||_{BV}\le||A||_{L^1}.||\bar Y_j-\bar Y||+
\int_a^b||A(t)||.||Y_j(t)-\bar Y_j-(Y(t)-\bar Y)||\,dt\,.\numeqap
$$  

If $\{Y_j-\bar Y_j\}$ weak$^*$-converges to $Y-\bar Y$, then Lemma
\lemweakconv{} implies that $||R_j^2||_{BV}\to 0$, so $Z_j\to Z$ in
$BV([a,b],\R^n)$.

If $\{W_j\}$ weak$^*$-converges to $W$, then we rewrite $R_j^2$ as
$$
R_j^2(t)=\int_a^tA(s)\Bigl(Z_j(s)-Z(s)\Bigr)\,ds+R_j^3(t)+R_j^4(t)\,,\numeqap
$$
where  $R_j^3(t)=\int_a^tA(s)\Bigl(\bar Y_j-\bar Y\Bigr)\,ds$ and
$R_j^4(t)=\int_a^tA(s)\Bigl(W_j(s)-W(s)\Bigr)\,ds$.  We then 
have $||R_j^3||_{BV}\le||A||_{L^1}.||\bar Y_j-\bar Y||$ and
$||R_j^4||_{BV}\le\int_a^b||A(s)||.||W_j(s)-W(s)||\,ds$.  Lemma
\lemweakconv{} then implies that $||R_j^4||_{BV}\to 0$.  So
$Z_j(t)-Z(t)=\int_a^tA(s).(Z_j(s)-Z(s))\,ds+R_j(t)$, where
$R_j=R_j^1+R_j^3+R_j^4$, so $||R_j||_{BV}\to 0$.  Then Gronwall's
inequality implies that $||Z_j-Z||_{sup}\le e^{||A||_{L^1}}||R_j||_{BV}$.
If $Q_j(t)=\int_a^tA(s).(Z_j(s)-Z(s))\,ds$, then $||Q_j||_{BV}\le
||A||_{L^1}||Z_j-Z||_{sup}$, so $||Q_j||_{BV}\to 0$.  Since
$Z_j-Z=Q_j+R_j$, we see that $||Z_j-Z||_{BV}\to 0$.

So we have shown that if one of the two sequences $\{W_j-W\}$,
$\{Y_j-\bar{Y}_j-(Y-\bar{Y})\}$ weak$^*$-converges to $0$, then
$Y_j-\bar{Y}_j-W_j\to Y-\bar{Y}-W$ strongly in $BV([a,b],\R^n)$.  This
implies all the conclusions of (\theolinbv.2).\proofend
  
\medbreak
Recall that, if $P,Q$ are normed spaces, and $F:\Omega\to Q$ is a map
defined on a neighborhood $\Omega$ of a point $p$ of $P$, then $F$ is said
to be {\sl Fr\'echet differentiable} at $p$ if there exists a bounded
linear map $L$ ---called the {\sl Fr\'echet differential} of $F$ at $p$---
such that $F(p')=F(p)+L(p'-p)+o(||p'-p||)$ as $p'\to p$.  If $X$ is a Banach
space and $Q=X^*$, then we call $F$ {\sl weak$^*$ Fr\'echet differentiable}
at $p$ if for every $x\in X$ the map $\Omega\ni p'\mapsto\langle
F(p'),x\rangle\in\R$ is Fr\'echet differentiable at $p$.  (We use $\langle
x^*,x\rangle$, for $x\in X$, $x^*\in X^*$, as an alternative notation for
$x^*(x)$.)

%It is easy to see that in general weak$^*$ Fr\'echet differentiability does
%not imply Fr\'echet differentiability.  (For example, if $X=C^0([0,1],\R)$,
%$P=\R$, $\Omega=[0,1]$, and $F(t)=\chi_{[0,t]}$ for $t\in\Omega$, then for
%every $\alpha\in X$ the map $t\mapsto \langle F(t),\alpha\rangle$ is just
%the map $t\mapsto\int_a^t\alpha(s)\,ds$, which is differentiable at every
%$t\in]0,1[$.  The weak$^*$-limit $\lim_{h\to 0}h^{-1}(F(t+h)-F(t))$ is
%$\delta_t$, the delta function at $t$ (i.e. the functional
%$\alpha\to\alpha(t)$), but $||h^{-1}(F(t+h)-F(t))-\delta_t||_{BV}=2$ if
%$h\neq 0$ and $t,t+h\in[0,1]$.)

Suppose $P$ is a normed space, $X$ is a Banach space, $p\in P$,
$F:\Omega\to X^*$ is a map defined on a neighborhood $\Omega$ of $p$, and
$F$ is weak$^*$ Fr\'echet differentiable at $p$.  Then for each $x\in X$
there is a bounded linear functional $\theta_x$ on $P$ such that $\langle
F(p+v),x\rangle=\langle F(p),x\rangle+\theta_x(v)+o(||v||)$ as $v\to
0$ in $P$. Clearly, $\theta_x$ is unique, and is given by the formula
$$
\theta_x(v)=\lim_{h\to 0}h^{-1}\langle F(p+hv)-F(p),x\rangle\,.\numeqap
$$
For any given $v\in P$, use $\theta^v$ to denote the map
$x\to\theta_x(v)$. Then $\theta^v$ is a limit of linear maps, so it is
linear.  If $\{h_j\}$ is any sequence going to $0$, then the sequence of
linear functionals $\nu_j$ on $X$ given by $\nu_j(x)=h_j^{-1}\langle
F(p+h_jv)-F(p),x\rangle$ converges pointwise to $\theta^v$. Since each
$\nu_j$ is a bounded linear functional on $X$, it follows for the uniform
boundedness theorem that $\theta^v$ is bounded.  Since
$\theta^v(x)=\theta_x(v)$, which is linear with respect to $v$, we can
conclude that $v\mapsto \theta_v$ is a linear map from $P$ to $X^*$. We use
$D^w\!F(p)$ to denote this map, and $D^w\!F(p).v$ to denote its value for a
particular $v$, so $D^w\!F(p).v=\theta^v$, and 
\xdef\defofdwf{\equanoapno}
$$
\langle D^w\!F(p).v,x\rangle=\lim_{h\to 0}h^{-1}\langle
F(p+hv)-F(p),x\rangle\,.\numeqap
$$
\xdef\isbounded{{\rm \appenlet}.\the\lemmno}
\lemmap
If $P$ is a normed space, $X$ is a Banach space, $p\in P$,
$F:\Omega\to X^*$ is a map defined on a neighborhood $\Omega$ of $p$,
and $F$ is weak$^*$ Fr\'echet differentiable at $p$.  Then: 

{\pari
\item{(\isbounded.i)} {\sl there is a $C>0$ such that} 
\xdef\bdforf{\equanoapno}
$$
||F(p+v)-F(p)||\le C||v|| \qquad{\sl for \,\,\,all\,\,\,sufficiently
\,\,\,small \qquad} v\in P\,,\numeqap
$$

\item{(\isbounded.ii)} {\sl the linear map $D^w\!F(p):P\mapsto X^*$ 
defined by (\defofdwf) is bounded.} 

}


\beginproof
Suppose (\isbounded.i) is not true. Then there exists a sequence
$\{v_j\}$ in $P$, converging to $0$ and such that
$||F(p+v_j)-F(p)||>j||v_j||$. Define, for each $j$, a  
member $\sigma_j$ of $X^*$,  by 
$$
\sigma_j={F(p+v_j)-F(p)-\theta^{v_j}\over ||v_j||}\,.\numeqap
$$
Then $\sigma_j(x)\to 0$ for each $x$, so the sequence $\{\sigma_j\}$ is
pointwise bounded and  hence uniformly bounded.   
    
On the other hand, for each $x\in X$, the numbers ${\theta^{v_j}(x)\over
||v_j||}$ are bounded, because $\theta^{v_j}(x)=\theta_x(v_j)$, and
$\theta_x$ is a bounded linear functional on $P$,  so
$|\theta^{v_j}(x)|\le||\theta_x||.||v_j||$. So, using the uniform
boundedness theorem once again, we conclude that the sequence
$\Bigl\{{\theta^{v_j}\over ||v_j||}\Bigr\}$ is bounded in $X^*$.  Since
$\{\sigma_j\}$ is 
bounded, we conclude that the sequence
$\Bigl\{{F(p+v_j)-F(p)\over ||v_j||}\Bigr\}$ is
bounded, contradicting the fact that $||F(p+v_j)-F(p)||>j||v_j||$.
So (\bdforf) holds for some $C>0$, and we have proved (\isbounded.i). 

It follows from  (\isbounded.i) that $|\langle
F(p+v)-F(p),x\rangle|\le C||x||.||v||$ for all $x\in X$, if $v\in P$ is
small enough.  Applying this with $hv$ in the role of $v$, and letting
$h\downarrow 0$, we find that
$$
|\langle D^w\!F(p).v,x\rangle|\le C||x||.||v||\qquad{\rm for\,\,\,all\qquad}
x\in X\,,\,\,\,v\in P\,.\numeqap
$$
Therefore $||D^w\!F(p).v||\le C||v||$, so $D^w\!F(p)$ is a bounded linear 
operator from $P$ to $X^*$.  \proofend


We are now ready to study the differentiability properties with respect to
a parameter $p$, belonging to a normed space $P$, of a family of
trajectories $x_p:[a,b]\to\R^n$ of time-varying vector fields $f_p$.  We
will prove, under minimal technical hypotheses, that if the initial
condition map $p\mapsto x_p(a)$ is differentiable at $p=0$, and the map
$p\mapsto f_p$ satisfies a ``weak differentiability condition'' at the
trajectory $x_0$, then the map $p\mapsto x_p$, is weak$^*$- differentiable
\gnu{There should not be a comma after $x_p$.} 
at $p=0$, as a map into $BV([a,b],\R^n)$, and in particular the endpoint
map $p\mapsto x_p(b)$ is differentiable at $p=0$.

A {\sl curve in $\R^n$} is a continuous map $\xi:I\mapsto\R^n$, whose
domain $I={\rm Dom}(\xi)$ is a nonempty interval. A curve $\xi$ is an {\sl
arc} if ${\rm Dom}(\xi)$ is compact.  The {\sl graph} $G(\xi)$ of a curve
$\xi$ is the set $G(\xi)=\{(\xi(t),t):t\in{\rm Dom}(\xi)\}$.  If
$\xi:[a,b]\mapsto\R^n$ is an arc in $\R^n$, and $\ve>0$, then the
$\ve$-{\sl tube about $\xi$} is the set ${\cal
T}(\xi,\ve)=\{(x,t):x\in\R^n,a\le t\le b,||x-\xi(t)||\le\ve\}$.

A {\sl time-varying vector field} on $\R^n$ is an $\R^n$-valued map $f$
whose domain ${\rm Dom}(f)$ is a ---possibly empty--- subset $S$ of the
product $\R^n\times\R$. We will use $TVVF(\R^n)$ to denote the set of all
time-varying vector fields on $\R^n$.  A {\sl trajectory}, or {\sl integral
curve}, of an $f\in TVVF(\R^n)$ is a locally absolutely continuous curve
$\xi$ in $\R^n$ such that $G(\xi)\subseteq{\rm Dom}(f)$ and
$\dot\xi(t)=f(\xi(t),t)$ for almost all $t\in{\rm Dom}(\xi)$.  We use
$\traj(f)$ to denote the set of all trajectories of $f$, and $\traj_c(f)$
to denote the set of all $\xi\in\traj(f)$ such that ${\rm Dom}(\xi)$ is
compact.

If $f\in TVVF(\R^n)$ and $\xi\in\traj(f)$, we say that $\xi$ is a {\sl
maximal} trajectory of $f$ if it cannot be extended to a trajectory
$\tilde\xi:\tilde I\mapsto\R^n$ of $f$ defined on an interval $\tilde I$
such that $I\subseteq\tilde I$ but $I\neq\tilde I$. We use $\mtraj(f)$ to
denote the set of all maximal trajectories of $f$.  Given $\bar x\in\R^n$,
$\bar t\in\R$, we use $\traj(f,\bar x,\bar t)$, $\mtraj(f,\bar x,\bar t)$
to denote, respectively, the set of all $\xi\in\traj(f)$ such that
$\xi(\bar t)=\bar x$, and the set $\traj(f,\bar x,\bar t)\cap\mtraj(f)$.
Zorn's lemma implies that every $\xi\in\traj(f,\bar x,\bar t)$ can be
extended to a $\tilde\xi\in\mtraj(f,\bar x,\bar t)$, so
$\traj(f,\bar{x},\bar t)\neq\emptyset$ iff
$\mtraj(f,\bar{x},\bar{t})\neq\emptyset$. Clearly,
$\traj(f,\bar{x},\bar{t})\neq\emptyset$ iff
$(\bar{x},\bar{t})\in{\rm{}Dom}(f)$.  (Indeed, if
$(\bar{x},\bar{t})\in{\rm{}Dom}(f)$, then the map $\xi$ with domain
$\{\bar{t}\}$ such that $\xi(\bar{t})=\bar x$ is in $\traj(f)$.)
 


We say that an $f\in TVVF(\R^n)$ has the {\sl forward existence property}
at a point $(\bar x,\bar t)$ if for some $\ve>0$ there exists a
$\xi\in\traj(f,\bar x,\bar t)$ with domain $[\bar t,\bar t+\ve]$.  We say
that $f$ has the {\sl forward limit property} on a subset $S$ of
${\rm{}Dom}(f)$ if, whenever $\xi:[a,b[\,\mapsto\R^n$ is a trajectory of $f$
which is contained in a compact subset of $S$, it follows that
$\lim_{t\uparrow b}\xi(t)$ exists.

Suppose we are given the following data:

\medbreak

{\pari

\item{(D.1)} a  normed space $P$,

\item{(D.2)} a family ${\bf f}=\{f_p\}_{p\in P}$ of
time-varying vector fields on $\R^n$, depending on a parameter $p\in P$,

\item{(D.3)} a $p_0\in P$, 

\item{(D.4)} a trajectory $\xi:[a,b]\mapsto\R^n$ of $f_{p_0}$.

}

\medbreak\noindent We then define functions $w_p:[a,b]\mapsto\R^n$  by 
\xdef\defofwp{\equanoapno}
$$
w_p(t)=f_p(\xi(t),t)-f_{p_0}(\xi(t),t)\quad{\rm for\quad}t\in[a,b]\,,\numeqap
$$
so $w_p$ is defined on $[a,b]$ whenever $G(\xi)\subseteq{\rm Dom}(f_p)$.
If $w_p$ is integrable, we let 
\xdef\defofWp{\equanoapno}
$$
W_p(t)=\int_a^tw_p(s)\,ds\qquad{\rm for\qquad}t\in[a,b]\,.\numeqap
$$
We then let ${\cal W}$ be the map  that sends $p$ to ${\cal W}(p)=W_p$,
with domain  
\xdef\defofdomw{\equanoapno}
$$
{\rm Dom}({\cal W})=\{p\in P: G(\xi)\subseteq{\rm Dom}(f_p),
\,\,w_p\in L^1([a,b],\R^n)\}\,.\numeqap
$$
We regard ${\cal W}$ as a map from ${\rm Dom}({\cal W})$ to 
$BV([a,b],\R^n)$. 

\medbreak\noindent  We then say that ${\bf f}$ is {\sl weakly
differentiable at $p_0$ along $\xi$} if there exists $\bar\ve>0$ such that

\medbreak
{\pariim 

\item{(WD1)} {\sl ${\cal T}(\xi,\bar\ve)\subseteq{\rm Dom}(f_p)$ for every 
$p$ such that $||p-p_0||\le\bar\ve$.}

\item{(WD2)} {\sl For every $p$ such that $||p-p_0||\le\bar\ve$,  $f_p$ has
the forward limit property on ${\cal T}(\xi,\bar\ve)$ and the forward
existence property at every $(x,t)$ such that $a\le t<b$ and
$||x-\xi(t)||<\bar\ve$.}

}

{\pariim

\item{(WD3)}  {\sl There exist $A\!\in\!L^1([a,b],\R^{n\times n})$, 
and functions $\psi_\ve\!\in\!L^1([a,b],\R)$,  for 
$\ve\!\in ]0,\bar\ve]$, such that}
\xdef\fpxt{\equanoapno}
$$
||f_p(x,t)-f_p(\xi(t),t)-A(t).(x-\xi(t))||\le\psi_\ve(t)(||x-\xi(t)||
+||p-p_0||)\numeqap
$$
{\sl whenever $a\le t\le b$, $||x-\xi(t)||\le\ve$,   $||p-p_0||\le\ve$,  and} 
\xdef\limpsive{\equanoapno}
$$
\lim_{\ve\downarrow 0}\int_a^b\psi_\ve(t)\,dt=0\,.\numeqap
$$


\item{(WD4)}  {\sl The function $w_p$ defined by (\defofwp)
is integrable for every $p$ such that $||p-p_0||\le\bar\ve$ and, if
${\cal{}W}$ is defined as above, then ${\cal W}$ is weak$^*$ Fr\'echet
differentiable at $p_0$.}

}


\parnormal

\medbreak\noindent 
The second part of condition (WD4) says that for each $\alpha\in
C^0([a,b],\R_n)$ the map $p\to \int\alpha\cdot dW_p$ is Fr\'echet
differentiable at $p_0$.  Lemma \isbounded{} then implies 
---since $W_{p_0}=0$--- that there exists a bounded linear map $D^w{\cal
W}(p_0):P\mapsto BV([a,b],\R^n)$ having the property that
\xdef\flafordifwp{\equanoapno}
$$
\int\alpha\cdot dW_p=\langle{\cal W}(p)-{\cal W}(p_0),\alpha\rangle=
\int\alpha\cdot\Bigl(D^w{\cal W}(p_0).(p-p_0)\Bigr)+o(||p-p_0||)
\qquad{\rm as}\qquad p\to p_0\numeqap
$$
for each $\alpha\in C^0([a,b],\R_n)$, and also that there is a constant
$C$ such that
\xdef\constantc{\equanoapno}
$$
||W_p||_{TV}\le C||p-p_0||\qquad{\rm for\,\,\,all\,\,\,}p\in P{\rm
\,\,\,such\,\,\,that\,\,\,}\|p-p_0\|{\rm \,\,\,is\,\,\,
sufficiently \,\,\,small}.\numeqap
$$
\xdef\theodiftraj{{\rm  \appenlet}.\the\theono}
\theoap
Let $P$, ${\bf f}$, $p_0$, $\xi$ be data as in (D.1,2,3,4). 
Assume that ${\bf f}$ is weakly differentiable at $p_0$ along $\xi$.  Let
$\bar\ve>0$ be such that (WD1) to (WD4) hold.  Then

{\pariim

\item{(\theodiftraj.a)} {\sl  For every $\ve\in]0,\bar\ve]$ there exists an 
$\delta\in]0,\ve[$ such that, whenever $||p-p_0||\le\delta$, $||\bar
x-\xi(a)||\le\delta$, and $\zeta\in\mtraj(f_p,\bar x,a)$, it follows that
$[a,b]\subseteq{\rm Dom}(\zeta)$ and $||\zeta(t)-\xi(t)||\le\ve$ for all
$t\in[a,b]$.}

\item{(\theodiftraj.b)} {\sl If $\{\bar x_p\}_{||p-p_0||\le\bar\ve}$ is a
family of points of $\R^n$ such that the map $p\mapsto \bar x_p$ is
Fr\'echet differentiable at $p_0$ and $\bar x_{p_0}=\xi(a)$, then, if
$\{\xi_p\}_{||p-p_0||\le\bar\ve}$ is a family of curves such that
$\xi_p\in\mtraj(f_p,\bar x_p,a)$, and $x_p$ denotes the restriction of
$\xi_p$ to $[a,b]$ ---which is well defined for small enough $||p-p_0||$ by
(\theodiftraj.a)--- then the map ${\cal X}$ that sends each $p$ to the
function $x_p$ is weak$^*$ Fr\'echet differentiable at $p_0$ as a map from
$P$ to $BV([a,b],\R^n)$.  Moreover, $D^w{\cal X}(p_0)$ can be computed as
follows.  For $v\in P$, 
$$
D^w{\cal{}X}(p_0).v=Y_{A,\,d(D^w{\cal{}W}(0).v),
\,L.(p-p_0)}\,,\numeqap
$$
where $L:P\to\R^n$ is the differential at $p_0$
of the map $p\to \bar x_p$.} 

}

\beginproof
We will assume, as we clearly may without loss of generality, that
$p_0=0$. Also, we know that there is a $C>0$ such that the bound
(\constantc) holds.  So we can assume, by making $\bar\ve$ smaller, if
necessary, that $||W_p||_{TV}\le C||p||$ whenever $||p||\le\bar\ve$.  By
making $C$ larger and $\bar\ve$ smaller, if necessary, we may also assume
that $C\ge 1$, and
\xdef\alsoassume{\equanoapno}
$$
8e^{||A||_{L^1}}||\psi_\ve||_{L^1}<1 \qquad{\rm for\qquad}
\ve\in]0,\bar\ve]\,. \numeqap
$$ 
Pick $\ve\in]0,\bar\ve[$, and choose $\delta>0$ such that  
\xdef\choicea{\equanoapno}
$$
C\delta<\ve\,.\numeqap
$$  
Notice that $\delta<\ve$, since $C\ge 1$. 

Let $p$, $\bar x$ be such that $||p||\le\delta$ and
$||\bar{x}-\xi(a)||\le\delta$.  Let $\zeta\in\mtraj(f_p,\bar x,a)$, and let
$I={\rm Dom}(\zeta)$.  Then $a\in I$. Let $c$ be the supremum of those $t$
such that $a\le t\le b$ and $||\zeta(s)-\xi(s)||\le\ve$ for $a\le s\le t$.
Then $a<c\le b$, $c\in I$, and either $c=b$ or $||\zeta(c)-\xi(c)||=\ve$.
(The fact that $a<c$ follows from the local existence property, since
$||\bar x-\xi(a)||\le\delta<\ve<\bar\ve$, and $\zeta$ is continuous.  The
fact that $c\in I$ follows from the forward limit property, since the graph
of the restriction of $\zeta$ to $[a,c[$ is contained in ${\cal
T}(\xi,\ve)$.  If $c<b$ and $||\zeta(c)-\xi(c)||<\ve$, then the forward
existence property implies that $\zeta$ is not maximal.)

We will estimate $||\zeta(t)-\xi(t)||$ for $t\in[a,c]$, and show that ---if
$\ve$ is small enough, and $\delta$ is suitably chosen---
$||\zeta(t)-\xi(t)||<\ve$ for all such $t$, including $t=c$.  This will
imply that $c=b$,
 
\gnu{The paragraph should end with a period, not a comma.}

{}From now on, we work on the interval $[a,c]$.  We have  
$$
\zeta(t)-\xi(t)=\bar x-\xi(a)+\int_a^t\Bigl(f_p(\zeta(s),s)-f_0(\xi(s),s)\Bigr)
\,ds\,,\numeqap
$$
so 
$$
\zeta(t)-\xi(t)=\bar x-\xi(a)+W_p(t)+
\int_a^t\Bigl(f_p(\zeta(s),s)-f_p(\xi(s),s)\Bigr)\,ds\,.\numeqap
$$
Let $r_p(t)=f_p(\zeta(t),t)-f_p(\xi(t),t)-A(t).(\zeta(t)-\xi(t))$.
Then 
$$
\zeta(t)-\xi(t)=\bar x-\xi(a)+W_p(t)+\int_a^tA(s).(\zeta(s)-\xi(s))\,ds
+\int_a^tr_p(s)\,ds\,,\numeqap
$$
so
$$
||\zeta(t)-\xi(t)||\le||\bar x-\xi(a)||+
||W_p||_{sup}+\int_a^t||A(s)||.||\zeta(s)-\xi(s)||\,ds+\int_a^t
r_p(s)\,ds\,,\numeqap
$$
and then 
$$
||\zeta(t)-\xi(t)||\le||\bar x-\xi(a)||+||W_p||_{sup}+
2\ve||\psi_\ve||_{L^1}+\int_a^t||A(s)||.||\zeta(s)-\xi(s)||\,ds\,,\numeqap
$$
because $||r_p(s)||\le 2\ve\psi_{\ve}(s)$ for $s\in[a,c]$, since 
$||\zeta(s)-\xi(s)||\le\ve$ and $||p||\le\ve$.

Since $||\bar x-\xi(a)||\le\delta$, and  
$||W_p||_{sup}\le||W||_{TV}\le C||p||\le C\delta$, Gronwall's inequality
implies  
$$
||\zeta(t)-\xi(t)||\le
e^{||A||_{L^1}}\Bigl((C+1)\delta+2\ve||\psi_\ve||_{L^1}\Bigr)\,.\numeqap
$$
In view of (\alsoassume),we have  
$$
2\ve e^{||A||_{L^1}}||\psi_\ve||_{L^1}<{\ve\over 4}\,.\numeqap
$$
Now suppose that 
\xdef\choiceb{\equanoapno}
$$
\delta\le\kappa\ve\,,\qquad{\rm where}\qquad\kappa=(4C+4)^{-1}e^{-||A||_{L^1}}
\,.\numeqap
$$
Then  
$$
||\zeta(t)-\xi(t)||\le{\ve\over 2}\qquad{\rm for\qquad all}\qquad
t\in[a,c]\,.\numeqap
$$   
This is our desired estimate, from which it follows, as explained above, 
that $c=b$, and then $||\zeta(t)-\xi(t)||\le\ve$ for all $t\in[a,b]$.  This 
completes the proof of (\theodiftraj.a).  

Besides proving (\theodiftraj.a), we have also shown that, if $\kappa$ is
the constant given by (\choiceb), then the conclusion of (\theodiftraj.a)
holds for every $\delta$ such that $0<\delta\le\kappa\ve$.  (Notice that if
$\delta\le\kappa\ve$ then (\choicea) holds.)

Now suppose we are given a map $P\ni p\mapsto\bar x_p\in\R^n$ which is
defined for $||p||\le\bar\delta$ for some $\bar\delta>0$.  Assume that this
map is Fr\'echet differentiable at $0$, with differential $L$, and is such
that $\bar x_0=\xi(a)$.  Write $z_p=Y_{A,dV_p,L.p}$, where $V_p=D^w{\cal
W}(0).p$.

It then follows that there exist a constant $C'\ge 1$ and a 
$\bar\delta'\in]0,\bar\delta]$ such that  (i)~$C'\bar\delta'\le\kappa\bar\ve$,
and (ii)~$||\bar x_p-\xi(a)||\le C'||p||$ for  $||p||\le\bar\delta'$. 

Pick in an arbitrary fashion $\xi_p\in\mtraj(f_p,\bar x_p,a)$ for
$||p||\le\bar\delta'$. Then, if $||p||\le\bar\delta'$, it follows that
(i)~$\xi_p$ is defined, (ii)~$\xi_p\in\mtraj(f_p,\bar x,a)$ for an $\bar x$
---namely, $\bar x_p$--- such that $||\bar x-\xi(a)||\le C'\bar\delta'
\le\kappa\bar\ve$.    Since $C'\ge 1$, we also have 
$||p||\le\kappa\bar\ve$. Therefore we can conclude ---using $\ve=\bar\ve$ and
$\delta=\kappa\bar\ve$,  that $[a,b]\subseteq{\rm Dom}(\xi_p)$ 
and $(\xi_p(t),t)\in{\cal T}(\xi,\bar\ve)$ for $t\in[a,b]$. 

Let $x_p$ be the restriction of $\xi_p$ to $[a,b]$.  For
$||p||\le\bar\delta'$, we have
$$
x_p(t)-\xi(t)-z_p(t)=\bar
x_p-\xi(a)-L.p+\int_a^tA(s).(x_p(s)-\xi(s)-z_p(s))\,ds 
+W_p(t)-V_p(t)+R_p(t)\,,\numeqap
$$ 
where $R_p(t)=\int_a^tr_p(s)\,ds$.  If $p\neq 0$, let 
$$
Y_p=||p||^{-1}(x_p-x_0-z_p)\,,\qquad U_p=||p||^{-1}(W_p-V_p+R_p)\,,
\qquad \bar Y_p=||p||^{-1}(\bar x_p-x_0(a)-L.p)\,.\numeqap
$$
Then $Y_p=Y_{A,dU_p,\bar Y_p}$.

We have to prove that $Y_p$ weak$^*$ converges to $0$ as $p\to 0$.  Since
$\bar Y_p\to 0$, our conclusion will follow if we show that $U_p$ weak$^*$
converges to $0$.  Since $V_p=D^w{\cal W}(0).p$, it is clear that
$||p||^{-1}(W_p-V_p)$ weak$^*$-converges to $0$.  So all we need is to
prove that $||p||^{-1}R_p$ weak$^*$-converges to $0$.  If $||p||$ is
sufficiently small, then we can apply the previous estimates using
$\delta=||p||$, $\ve={\delta\over\kappa}$.  We conclude that $||r_p(s)||\le
2\ve\psi_{\ve}(s)$ for all $s\in[a,b]$. This implies that
$$
||r_p||_{L^1}\le{2||p||\over K}||\psi_{\kappa^{-1}||p||}||_{L^1}\,.\numeqap
$$
This shows that $||R_p||_{TV}=||r_p||_{L^1}=o(||p||)$ as $p\to 0$.
So $||p||^{-1}R_p$ converges to $0$ strongly ---and 
{\it a fortiori} weakly$^*$--- in $BV([a,b],\R^n)$. This completes 
our proof. \proofend

If $f\in TVVF(\R^n)$ and  $\xi:[a,b]\mapsto\R^n$ is a trajectory of $f$, 
we say that {\sl $f$ is differentiable along $\xi$} if the map 
$x\mapsto f(x,t)$ is differentiable at $\xi(t)$ for almost every 
$t$.  In that case, we write 
$D_{\xi}f(t)\equaldef{\partial f\over\partial x}(\xi(t),t)$, and refer 
to the map $[a,b]\!\ni\! t\!\mapsto\!D_\xi f(t)\!\in\!\R^{n\times n}$ as the {\sl differential
of $f$ along $\xi$.}

\xdef\lemdifisadj{{\rm \appenlet}.\the\lemmno}
\lemmap
If ${\bf f}$ is weakly differentiable at $p_0$ along a trajectory 
$\xi:[a,b]\to\R^n$ of $f_{p_0}$, then $f_{p_0}$ is differentiable along 
$\xi$ and the map $A$ of (WD3) is such that $A(t)=D_\xi f_{p_0}(t)$ 
for almost all $t\in[a,b]$. 

\beginproof
We begin by showing that the functions $\psi_\ve$ for which (WD3) holds can
be chosen so that $\psi_\ve(t)$ is a monotonically nonincreasing function
of $\ve$ for each $t$.  To see this, start with some family
$\{\psi_\ve\}_{\ve\in]0,\bar\ve]}$ of functions for which (WD3) holds, and
define $\hat\psi_{\ve}=\psi_{2^{-k}\bar\ve}$ whenever
$2^{-k-1}\bar\ve<\ve\le 2^{-k}\bar\ve$ and $k$ is a nonnegative integer.
Then (WD3) also holds if the $\hat\psi_\ve$ are substituted for the
$\psi_\ve$.  Next define
$\bar\psi_\ve(t)=\min\{\hat\psi_{\ve'}(t):\ve\le\ve'\le\bar\ve\}$.  Then
the $\bar\psi_\ve$ are well defined integrable functions of $t$, because
for each $\ve$ the set of functions
$\{\hat\psi_{\ve'}:\ve\le\ve'\le\bar\ve\}$ is finite.  It is clear that
(WD3) also holds with the $\bar\psi_\ve$ instead of the $\hat\psi_\ve$, and
in addition $\hat\psi_\ve(t)$ is nonincreasing as a function of $\ve$.

Now, if we pick the $\psi_\ve$ to be monotonically nonincreasing with
respect to $\ve$, it follows that the limit $\rho(t)=\lim_{\ve\downarrow
0}\psi_\ve(t)$ exists for every $t$.  By the dominated convergence theorem,
$\int_a^b\rho(t)dt=0$.  Since $\rho(t)\ge 0$ for all $t$, it follows that
$\rho(t)=0$ almost everywhere.

Let $t$ be such that $\rho(t)=0$.  Let $\theta(x)=\psi_{||x-\xi(t)||}(t)$.
Then $\lim_{x\to\xi(t)}\theta(x)=0$.  On the other hand, (WD3) implies that
\xdef\saysprec{\equanoapno}
$$
||f_{p_0}(x,t)-f_{p_0}(\xi(t),t)-A(t).(x-\xi(t))||\le\theta(x)
||x-\xi(t)||\,.\numeqap
$$
Since $\theta(x)\to 0$ as $x\to\xi(t)$,  (\saysprec) says precisely
that $A(t)$ is the differential at $\xi(t)$ of the map 
$x\mapsto f_{p_0}(x,t)$, completing our proof. \proofend
 
If $f\in TVVF(\R^n)$ is differentiable along a trajectory 
$\xi:[a,b]\to\R^n$ of $f$, and the matrix-valued function $D_\xi f$ 
is integrable, then the ordinary differential equation 
\xdef\adjequa{\equanoapno}
$$
\dot\lambda(t)=-\lambda(t).D_{\xi}f(t)\,,\numeqap
$$
for an $\R_n$-valued function $\lambda$ on $[a,b]$, is the {\sl adjoint
variational equation} ---or, simply, the {\sl adjoint equation}--- for $f$
along $\xi$.  A solution of the adjoint equation (i.e. an absolutely
continuous function $\lambda:[a,b]\mapsto\R_n$ such that (\adjequa) holds
for almost all $t$) is known as an {\sl adjoint vector for $f$ along
$\xi$.}

\xdef\theodifwithlambda{{\rm \appenlet}.\the\theono}
\theoap  
Let $P$, ${\bf f}=\{f_p\}_{p\in P}$, $p_0\in P$, $\xi:[a,b]\to\R^n$ be as
in Theorem \theodiftraj. Let $P\ni p\mapsto\bar x_p\i\R^n$ be a map which
is Fr\'echet differentiable at $p_0$ and such that $\bar x_{p_0}=\xi(a)$,
and let $L_a$ be its differential at $p_0$.  Let $\xi_p$ be, for $p$ close
enough to $p_0$, maximal trajectories of $f_p$ such that
$\xi_p(a)=\bar{x}_p$, and let $x_p$ be the restriction of $\xi_p$ to
$[a,b]$. Then the map $p\mapsto x_p(b)$ is differentiable at $p_0$. Let
$L_b$ be its differential at $p_0$.  Let $\lambda:[a,b]\to\R^n$ be an
adjoint vector for $f_{p_0}$ along $\xi$.  Then
$$
\lambda(b)\cdot L_b(v)-\lambda(a)\cdot L_a(v)
=\lim_{\ve\downarrow 0}\,\ve^{-1}\!\int_a^b\Bigl\langle\lambda(t),f_{p_0+\ve v}
(\xi(t),t)-f_{p_0}(\xi(t),t)\Bigr\rangle dt\,.\numeqap
$$

\beginproof
Assume, without loss of generality, that $p_0=0$. According to Theorem 
\theodiftraj{} and Lemma \lemdifisadj, the map $p\mapsto x_p$ is weak$^*$
Fr\'echet differentiable at $0$ ---as a map from $P$ to $BV([a,b],\R^n)$---
and its differential at $0$ is the map that sends $v\in P$ to
$Y_{A,d(U.v),L_a(v)}$, where $U=D^w{\cal W}(0)$, and ${\cal W}$ is the map
$p\mapsto W_p$, where
$W_p(t)=\int_a^t\Bigl(f_{p}(\xi(s),s)-f_{0}(\xi(s),s)\Bigr)dt$, and
$A=D_{\xi}f$.  Then $p\mapsto x_p(b)$ is Fr\'echet differentiable at $0$,
and the differential $L_b$ is given by $L_b(v)=Y_{A,d(U.v),L_a(v)}(b)$.
Clearly, $L_a(v)=Y_{A,d(U.v),L_a(v)}(a)$.  Since $\lambda$ satisfies
$\dot\lambda=-\lambda.A$, Lemma \lemadjo{} tells us that
$$
\lambda(b)\cdot L_b(v)-\lambda(a)\cdot L_a(v)=\int_a^b\lambda.d(U.v)\,.
$$
Since $U.v$ is the weak$^*$-limit of $\ve^{-1}W_{\ve v}$ as $\ve\downarrow
0$, our conclusion follows. \proofend

We now turn to the differentiation of a Lagrangian cost functional.  

If $f\in TVVF(\R^n)$, a {\sl Lagrangian} for $f$ in $\R^n$ is a real-valued 
function $L$  such that

\medbreak

{\pari

\item{(L.1)}  ${\rm Dom}(f)\subseteq{\rm Dom}(L)$;

\item{(L.2)} the function $t\mapsto L(\xi(t),t)$ 
is locally integrable for every $\xi\in\traj(f)$.

}

\medbreak\noindent
If $f\in TVVF(\R^n)$ has components $f^1,\ldots,f^n$, and $L$ is a
Lagrangian for $f$, then we can consider the time-varying vector field
$[L;f]$ in $\R^{n+1}$ whose components are $(f^0,f^1,\ldots,f^n)$, where
$f^0\equiv L$.  A trajectory of $[L;f]$ can then be regarded a pair
$(\xi^0,\xi)$, where $\xi\in\traj(f)$ and $\xi^0$ is a real-valued locally
absolutely continuous function on ${\rm Dom}(\xi)$ such that
$\dot\xi^0(t)=L(\xi(t),t)$ for a.e. $t$.  Any such function $\xi^0$ will be
called a {\sl running $L$-cost along $\xi$}.  Our definition of a
Lagrangian implies that a running $L$-cost along $\xi$ exists for all
$\xi\in\traj(f)$, and it is clear that any two running $L$-costs along
$\xi$ must differ by a constant. If $\xi$ is such that ${\rm Dom}(\xi)$ is
of the form $[a,b]$ or $[a,b[$ or $[a,\infty[$, then there is a canonical
choice of running $L$-cost along $\xi$, obtained by letting
$\xi^0(t)=\int_a^tL(\xi(s),s)\,ds$.  In that case, we will write $\xi^L$ to
denote the curve $t\mapsto(\xi^0(t),\xi(t))$, where $\xi^0$ is the canonical
running $L$-cost.

Now suppose we are given data
$P$, ${\bf f}$, $p_0$, $\xi$  as in (D.1,2,3,4), as well as  

\medbreak

{\pari

\item{(D.5)} a family ${\bf L}=\{L_p\}_{p\in P}$ such that each $L_p$
is a Lagrangian for $f_p$.  

}

\medbreak\noindent
Given a trajectory $\xi:[a,b]\mapsto\R^n$ of
$f_{p_0}$, we say that the pair $({\bf f},{\bf L})$ is {\sl weakly
differentiable at $p_0$ along $\xi$} if the family $\{[L_p;f_p]\}_{p\in P}$
is weakly differentiable along $\xi^{L_{p_0}}$.

It is easy to see that $({\bf f},{\bf L})$ is weakly differentiable at
$p_0$ along $\xi$ iff there exists $\bar\ve>0$ such that  (WD1,2,3,4)
hold, and in addition

\medbreak
{\pariim

\item{(WD5)}  {\sl There exists $h\in L^1([a,b],\R_n)$ such that
\xdef\lpxt{\equanoapno}
$$
||L_p(x,t)-L_p(\xi(t),t)-h(t).(x-\xi(t))||\le\psi_\ve(t)(||x-\xi(t)||
+||p||)\numeqap
$$
whenever $a\le t\le b$, $||x-\xi(t)||\le\ve$ and
$||p||\le\ve$.}  

\gnu{It should be $||p-p_0||$, not $||p||$.}

\item{(WD6)}  {\sl If we let $w_p^0(t)=L_p(\xi(t),t)-L_{p_0}(\xi(t),t)$,
$W_p^0(t)=\int_a^tw_p^0(s)\,ds$, then the map
${\cal W}^0$ that sends $p$ to ${\cal W}^0(p)=W_p^0\in BV([a,b],\R)$ is
weak$^*$ Fr\'echet differentiable at $p_0$.} 

}

\medbreak\noindent
If $f\in TVVF(\R^n)$, $\xi:[a,b]\mapsto\R^n$ is a trajectory of $f$, and
$L$ is a Lagrangian for $f$, we say that {\sl $L$ is differentiable along
$\xi$} if the map $x\mapsto L(x,t)$ is differentiable at $\xi(t)$ for almost
every $t$.  In that case, we write $\nabla_{\xi}L(t)\equaldef
\nabla_xL(\xi(t),t)$, and refer to the map $[a,b]\ni t\to \nabla_\xi
L(t)\in\R^{n\times n}$ as the {\sl differential of $L$ along $\xi$.}
 
Lemma \lemdifisadj{}, applied to the curve $\xi^{L_{p_0}}$,  yields

\xdef\lemdifisadjwithl{{\rm \appenlet}.\the\lemmno}
\lemmap
If $({\bf f},{\bf L})$ is weakly differentiable at $p_0$ along a trajectory 
$\xi:[a,b]\to\R^n$ of $f_{p_0}$, then $f_{p_0}$ and $L_{p_0}$ are
differentiable along $\xi$ and the maps $A$, $h$ of (WD3,5) are such that 
$A(t)=D_\xi f(t)$ and $h(t)=\nabla_\xi L_{p_0}(\xi(t),t)$ for almost all 
$t\in[a,b]$. \ \proofendno

Under the hypothesis of Lemma \lemdifisadjwithl{}, we can apply Theorem
\theodifwithlambda{} to the family $\{[L_p;f_p]\}_{p\in P}$ and the curve
$\xi^{L_{p_0}}$.   Define
$$
J_p(\zeta)=\int_a^b L_p(\zeta(t),t)\,dt\,,\numeqap
$$
if $\zeta:[a,b]\to\R^n$ is a curve such that the integral exists.  (We will
only be using this when $\zeta\in\traj(f_p)$, in which case $J_p(\zeta)$ is
well defined.)  An adjoint vector for $[L_{p_0};f_{p_0}]$ along
$\xi^{L_{p_0}}$ is then a pair $(\lambda^0,\lambda)$ such that
$\lambda^0\in\R$, $\lambda:[a,b]\mapsto\R_n$ is absolutely continuous, and
\xdef\adjeqwithlzero{\equanoapno}
$$
\dot\lambda(t)=-\lambda(t).A(t)-\lambda^0h(t)\qquad{\rm for\qquad a.e.\qquad}
t\in[a,b]\,.\numeqap
$$
Theorem \theodifwithlambda{} then gives:
\xdef\theodifwithlambdalag{{\rm \appenlet}.\the\theono}
\theoap
Assume that we are given data 
$P$, ${\bf f}$, $p_0$, $\xi$, ${\bf L}$  as in (D.1,2,3,4,5).
Assume that $({\bf f},{\bf L})$ is weakly differentiable at
$p_0$ along $\xi$.  Let $P\ni p\mapsto\bar x_p\i\R^n$ be a map which is
Fr\'echet differentiable at $p_0$ and such that $\bar x_{p_0}=\xi(a)$, and
let $L_a$ be its differential at $p_0$.  Let $\xi_p$ be, for $p$ close
enough to $p_0$, maximal trajectories of $f_p$ such that
$\xi_p(a)=\bar{x}_p$, and let $x_p$ be the restriction of $\xi_p$ to
$[a,b]$. Then the maps $p\mapsto x_p(b)$ and $p\mapsto J_p(x_p)$ are
differentiable at $p_0$. Let $L_b$, $L_b^0$ be their differentials at
$p_0$.  Then, if $(\lambda^0,\lambda)$ is an adjoint vector for
$[L_{p_0},f_{p_0}]$ along $\xi$, it follows that 
$$
\lambda(b)\cdot L_b(v)+\lambda^0 L_b^0(v)
-\lambda(a)\cdot L_a(v)=\lim_{\ve\downarrow 0}\ve^{-1}\!\int_a^b\!\Big(
H_{p_0+\ve v}(\xi(t),\lambda(t),\lambda_0,t)-
H_{p_0}(\xi(t),\lambda(t),\lambda_0,t)\Big)\,dt\,,\numeqap
$$
where 
$$
H_p(x,z,z^0,t)=\langle z,f_{p}(x,t)\rangle+z^0 L_p(x,t)\numeqap
$$
for $x\in\R^n$, $z\in\R_n$, $z^0\in\R$, $t\in[a,b]$. \proofendno

\appen{B}{A lemma}

\xdef\lemincreasing{{\rm \appenlet}.\the\lemmno}
\lemmap
Let $f$ be a real-valued function on a compact interval $[a,b]$. Assume that
there exists a finite or countable subset $E$ of $[a,b]$ with the following 
properties:

{\pariim

{\sl

\item{(\lemincreasing.a)} $\lim\inf_{h\downarrow 0}{f(x+h)-f(x)\over h}\ge 0$
for all $x\in[a,b[\setminus E$, 

\item{(\lemincreasing.b)} $\lim\inf_{h\downarrow 0}{f(x+h)-f(x)}\ge 0$
for all $x\in[a,b[$,

\item{(\lemincreasing.c)} $\lim\sup_{h\downarrow 0}{f(x)-f(x-h)}\ge 0$
for all $x\in]a,b]$.

}

}

\noindent{\sl Then $f(b)\ge f(a)$.}

\beginproof  Let $\ve>0$ be arbitrary. We will prove that $f(b)\ge f(a)-\ve$. 
Let $\ve'={\ve\over 1+b-a}$ and $E=\{x_k\}_{k\in N}$, $N\subset {I\!\! N}$. 

For each $x\in[a,b]$, let $I(x)=\{k:x_k<x\}$. Define $g:[a,b]\mapsto\R$ by 
$$
g(x)=\sum_{k\in I(x)}2^{-k}\ve'\,.
$$
Then $g$ is left-continuous and monotonically nondecreasing. 

Let $S$ be the set of all $x\in[a,b]$ such that 
$f(y)\ge f(a)-(y-a)\ve'-g(y)$ for all $y\in[a,x]$.   It is clear that 
$a\in S$, and also that $S$ is an interval. So $S$ is of the form 
$[a,c[$ or $[a,c]$, with $c\in[a,b]$.   

Suppose $S=[a,c[$.   Then $c>a$.  By  (\lemincreasing.c), given 
$\delta>0$ there is a sequence
$\{z_j\}$ in $[a,b]$ such that $z_j<c$, $z_j\to c$, and 
$f(c)-f(z_j)\ge -\delta$.  Then $z_j\in S$, so 
$f(z_j)\ge f(a)-(z_j-a)\ve'-g(z_j)\ge f(a)-(c-a)\ve'-g(c)$. It then follows that
$f(c)\ge f(a)-(c-a)\ve'-g(c)-\delta$. Since $\delta$ is arbitrary,  we have
$f(c)\ge f(a)-(c-a)\ve'-g(c)$, so $c\in S$, which is a contradiction.

Therefore $S=[a,c]$ for some $c\in[a,b]$.   We now show that $c=b$.
Assume that $c<b$, and consider separately the cases  $c\notin E$,  $c\in E$.

If $c\notin E$, then (\lemincreasing.a) implies that there exists
a $c'>c$ such that $f(x)-f(c)\ge -(x-c)\ve'$ for $x\in[c,c']$.  But then, since
$f(c)\ge f(a)-(c-a)\ve'-g(c)$, we have $f(x)\ge f(a)-(x-c)\ve'-g(c)\ge
f(a)-(x-a)\ve'-g(x)$ for $x\in[c,c']$.  Therefore $c'\in S$, contradicting
the fact that $S=[a,c]$.   
   
Now suppose $c\in E$. Then $c=x_k$ for some $k$, and $x_k\neq b$. By 
(\lemincreasing.b),  there exists $h>0$ such that $c+h\le b$ and 
$f(y)-f(c)\ge-2^{-k}\ve'$ for all $y\in[c,c+h]$.  If $y\in]c,c+h]$, then 
$k\notin I(c)$, and $I(c)\cup\{k\}\subseteq I(y)$. Therefore  
$g(y)\ge g(c)+2^{-k}\ve'$.   Since $f(c)\ge f(a)-(c-a)\ve'-g(c)$,
we have $f(y)\ge f(c)-2^{-k}\ve'\ge f(a)-(c-a)\ve'-2^{-k}\ve'-g(c)\ge
f(a)-(y-a)\ve'-g(y)$. So $c+h\in S$, which is a contradiction. 

We have therefore proved that $c=b$. Then 
$$
f(b)\ge f(a)-(b-a)\ve'-g(b)\ge f(a)-(b-a)\ve'-\ve'=f(a)-\ve\,.
$$
Since $\ve$ is arbitrary, we have shown that $f(b)\ge f(a)$, as 
desired.   \proofend  

%\remap
%If $f$ is lower semicontinuous then (\lemincreasing.b) automatically 
%holds. \proofend   



\bigbreak
\centerline{{\bf REFERENCES}}

\gnu{The puctuation of your bibliographical references is wrong. 
In English, you put the comma or period before and not after you close the
quotation marks. I fixed that all over the place. Also, I corrected other
punctuation errors, for example the omission of the blank space 
before ``and'' in the Bardi reference.} 

\bardi

\gnu{I guess you didn't bother to update the bibliography, did you?
The book by Martino and Italo HAS appeared, as I am sure you know.}

\bolt

\piccc

\piccd

\gnu{I presume you have been so terribly busy that you didn't even have
the time to update the items in the bibliography that were your own
publications. THESE TWO PAPERS HAVE APPEARED, HAVEN'T THEY?}

\bruna

\brunb

\cesari

\fed

\fili

\flem

\flemsoner

\full

\loja

\picca

\piccb

\sussa

\sussb

\sussc

\sussd

\susse

\sussf

\sussg

\sussh

\sussi

\sussl

\sussn

\sussm

\zelikin

\bye

