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 \newcommand{\ibf}[1]{{\bf #1}}
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 \title{\bf Global Minimizers of \\ Autonomous Lagrangians}
 
 
 \author{{\bf Gonzalo Contreras} \\{\bf  Renato Iturriaga} \\ 
         \\{\sc cimat} \\{\sc mexico, gto.}}

  
 
 
 
 \begin{document}

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 \date{\copyright \; 2000}
 
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 \quad
 
 
 \newpage
 
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%
% 
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%         \it\qquad
%           Queremos agradecer a Edilsa, Lourdes, 
%           \linebreak
%           Andreia, Santiago y Sebasti\'an
%           por el apoyo y paciencia que tuvieron
%           con nosotros durante la redacci\'on
%           de este libro.
%%          Queremos agradecer \`as nossas familias
%%          pelo apoio e a paciencia que tiveram conosco
%%          durante a reda\c c\~ao deste livro.
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%         

 \tableofcontents
 
 
 
 

 




 \chapter{Introduction.}
 

 \renewcommand{\thepage}{\arabic{page}}
 
 \section{Lagrangian Dynamics.}
 
 \medskip
 
 Let $M$ be a boundaryless $n$-dimensional complete
 riemannian manifold.
 An (autonomous) Lagrangian on $M$ is a smooth function $L:TM\to\re$
 satisfying the following conditions:
 
 \begin{itemize}
 \item[(a)]{\it Convexity:} \index{convexity}
           \index{condition!convexity}
	   \index{hypothesis on $L$}
           The Hessian 
           $\displaystyle\frac{\partial^2 L}{\partial v_i\partial
           v_j}(x,v)$,
           calculated in linear coordinates on the fiber $T_xM$, is
           uniformly positive definite for all $(x,v)\in TM$, i.e.
           there  is $A>0$ such that
           $$
           w\cdot L_{vv}(x,v)\cdot w \ge  A\; |w|^2
           \quad \text{ for all $(x,v)\in TM$\text{ and }$w\in
           T_xM$.}
           $$
 \item[(b)]{\it Superlinearity:} 
           \index{superlinearity|ibf}
           \index{growth!superlinear}
           \index{condition!superlinearity}
	   \index{hypothesis on $L$}
           $$
           \lim_{|v|\to+\infty}\frac{L(x,v)}{|v|}=+\infty,
           \quad\text{ uniformly on }x\in M,
           \label{sup0}
           $$
           equivalently, for all $A\in\re$ there is $B\in\re$ such
           that
           \begin{equation*}\label{superlinearity}
           L(x,v)\ge A\,|v|-B\quad\text{ for all }(x,v)\in TM.
           \end{equation*}           
 \item[(c)]{\it Boundedness\footnote{The Boundedness condition~(c)
           is equivalent to the condition that the associated
	   hamiltonian $H$ is convex and superlinear, see
	   remark~\ref{Rboundedness}. This condition is immediate
	   when the manifold $M$ is compact.}:} \index{boundedness}
           \index{condition!boundedness}  
	   \index{hypothesis on $L$}
           For all $r\ge 0$,
           \begin{align}
           \ell(r) &= \sup_{\stackrel{(x,v)\in TM,}{|v|\le r}}
           L(x,v) < +\infty.   
           \label{ell}      \\
           g(r) &= \sup_{\stackrel{|w|=1}{|(x,v)|\le r}}
           w\cdot L_{vv}(x,v)\cdot w < +\infty.
           \label{g(r)}
           \end{align}
            
 \end{itemize}
 
 
 \medskip

 The {\it Euler-Lagrange equation\/}
 \label{ELeq}\index{Euler-Lagrange!equation}
 associated to a lagrangian $L$ is (in local coordinates)
 \begin{equation}\label{EL}
 \frac{d\,}{dt}\frac{\partial L}{\partial v}\bigl(x,\dx)
 = \frac{\partial L}{\partial x}(x,\dx).
 \tag{E-L}
 \end{equation}
 The condition~(c) implies that the Euler-Lagrange equation~\eqref{EL}
 defines a complete flow $\vr_t$ on $TM$ (proposition~\ref{complete}),
 called the 
  \index{Euler-Lagrange!flow}
 \index{flow!Euler-Lagrange}
 \index{lagrangian!flow}
 {\it Euler-Lagrange flow},
 by setting
 $\vr_t(x_0,v_0)=\bigl(x_v(t),\dx_v(t)\bigr)$, where $x_v:\re\to M$ is
 the solution of~\eqref{EL} with $x_v(0)=x_0$ and $\dx_v(0)=v_0$.
 \index{$x_v$} 
 
%  When the manifold $M$ is compact, it is enough to ask only for the
%  convexity and superlinearity of the lagrangian, because the
%  boundedness follows from the continuity of $L$ and the completeness
%  follows from the compactness of the energy level and the fact that
%  the lagrangian vector field obtained from the Euler-Lagrange
%  equation~\eqref{EL} is Lipschitz on the energy level.
 
 We shall be interested on coverings $p:N\to M$ of a compact manifold
 $M$ and the lifted Lagrangian $\L=L\circ dp:TN\to\re$ of a convex
 superlinear lagrangian $L$ on $M$. The lagrangian $\L$ then satisfies
 (a)--(c) and its flow $\psi_t$ is the lift of $\vr_t$.

 

 Observe that when we add a {\it closed} 1-form $\om$ 
 \index{closed form} 
 \index{form!closed}
  to the
  lagrangian $L$, the new lagrangian $L+\om$ also satisfies 
 the hypothesis (a)-(c) and has the same Euler-Lagrange
 equation as $L$. This can also be seen using the variational
 interpretation of the Euler-Lagrange equation (see~\ref{R1form}). 


 \newpage
 
 \section{The Euler-Lagrange equation.}
 
  
 The action of a differential curve $\gamma :[0,T]\to M$ is defined by 
 \begin{equation*}
   A_L(\gamma)=\int_0^TL(\gamma (t),\dot\gamma (t)) \; dt
   \index{$A_L(\gamma)$}\index{action!of a curve}
 \end{equation*}
 
 One of the main problems of the calculus of variations
 is to find and to study the curves that minimize the action. 
 Denote by $C^k(q_1,q_2;T)$\index{$C^k(q_1,q_2;T)$} 
 the set of $C^k$-differentiable  curves 
 $\gamma:[0,T]\to M$ such that $\gamma (0)=q_1$ and $\gamma (T)=q_2$.
 
 
  \begin{Prop}
 \label{prop.I.lag.1}
  If a curve $x(t)$ in the space $C^k(q_1,q_2;T)$ is a critical
  point of the action functional on $C^k(q_1,q_2;T)$, 
  then  $x$ satisfies the Euler-Lagrange equation
 \begin{equation}
   \label{ec.I.lag.2}
   \frac {d}{dt}L_v(x(t),\dot x(t))=L_x(x(t),\dot x(t))
   \tag{E-L}
 \end{equation}
 in local coordinates.
 Consequently, this equation does not depend on the coordinate system.
 \end{Prop}

 \begin{proof}
 Choose a coordinate system $(x_1,\ldots,x_n)$ about $x(t)$.
 Let $h(t)$ a differentiable curve such that  $h(0)=h(T)=0$.
 Then for every $\e  $, sufficiently small the curve
 $y_{\e }=x+\e  h$ is on 
 $C^k(q_1,q_2;T)$ and contained in the
 coordinate system. Define 
 \begin{equation*}
   g(\e )= A_L(y_{\e })
 \end{equation*}
 
 Then $g$ has a minimum in zero and
 \begin{align*}
   \lim_{\e \to 0}\frac{g(\e )-g(0)}{\e }  
   &=\lim_{\e \to 0}\int_0^T\frac{L(x+ \e  h, \dot
        x+\e  \dot h)-L(x,\dot x)}{\e }\;dt \\
   &= \int_0^T\lim_{\e \to 0}\frac{\e  L_x h+ \e 
        L_v \dot h +o(\e  )}{\e } \;dt\\
   &= \int_0^T L_x h+L_v \dot h\; dt \\
   &= \int_0^T( L_x -\tfrac{d}{dt}L_v) \, h\; dt+ L_v\,h\big|_0^T \\
   &= \int_0^T( L_x -\tfrac{d}{dt}L_v) \, h\; dt.
 \end{align*}
 Hence
 \begin{equation*}
   0=\int_0^T\left[ L_x(x(t),\dot x(t)) -\frac{d}{dt}L_v(x(t),\dot
   x(t))\right]\,  h\;dt,
 \end{equation*}
 for any function $h\in C^k(0,0;T)$. This implies that $x(t)$ satisfies
 the  Euler Lagrange equation \eqref{ec.I.lag.2}.
 \end{proof}
 
 
 
 
 
 The Euler Lagrange equation is a second order differential equation on
 $M$, but the convexity hypothesis ($L_{vv}$
 invertible) implies that this equation can also be seen as a first
 order
 differential equation on $TM$:
 \begin{align*}
   \dot x &=  v, \\
  \dot v &=  (L_{vv})^{-1}(L_x-L_{vx}v).
 \end{align*}
 
 The associated vector field $X$ on $TM$ is called the lagrangian
 vector field and its flow $\vr_t$ the lagrangian flow. Observe that
 $X$ is of the form
 $$
 X(x,v)=(v,\cdot).
 $$
 
 
 
  \begin{remark}\label{IsSol}
  
  It is possible to do the same thing in the space
  $\C_T(p,q)$,\index{$C_T(x,y)$} the set of absolutely continuous curves 
  $\gamma:[0,T]\to M$ such that $\gamma (0)=p$ and $\gamma (T)=q$. 
  A priori minimizers do not have to be differentiable and there 
  are examples where they are  not, see  Ball \& Mizel~\cite{BM}. 
  However when the lagrangian flow is complete 
  (cf. proposition~\ref{complete}), every  absolutely continuous  
  minimizers is $C^2$ and satisfies the  Euler-Lagrange equation. 
  See Mather~\cite{Mat5}.
   \end{remark}
   
 
 \begin{remark}\label{R1form}
 If we add a closed 1-form $\om$ to the lagrangian $L$,
 the lagrangian $L+\om$ also satisfies the hypothesis 
 (a)-(c). Moreover, the action functional $A_{L+\om}$
 on a neighbourbood of a curve $\ga\in C^k(x_1,x_2,T)$
 satisfies
 $$
 A_{L+\om}(\eta) = A_{L}(\eta) + \oint_\ga \om,
 $$
 because the curve $\eta$ is homologous to $\ga$.
 Therefore, the critical points for $A_{L+\om}$ and for 
 $A_L$ are the same. This implies that the Euler-Lagrange
 equations for $L$ and $L+\om$ are the same.
 But since the values of $A_{L+\om}$ and $A_L$ are
 different, minimizers of these two actions may be
 different.

 \end{remark}
 
 
 
 \newpage
 
 \section{The Energy function.}
 
  The {\it energy function\/} of the lagrangian $L$ is $E:TM\to\re$,
 defined by
 \begin{equation}\label{energy}
 \index{energy!function|(}
  E(x,v) = \frac{\partial L}{\partial v}(x,v)\cdot v - L(x,v).
 \end{equation}
 Observe that if $x(t)$ is a solution of the Euler-Lagrange
 equation~\eqref{EL}, then
 $$
 \tfrac{d\,}{dt} \, E(x,\dx) 
 = \left(\tfrac{d\,}{dt} L_v - L_x\right)\cdot\dx =0.
 $$
 Hence $E:TM\to\re$ is an {\it integral\/}\label{integral}  
 (i.e. invariant function\footnote{The energy is invariant only for
 {\it autonomous\/} (i.e. time-independent) lagrangians.}) 
 for the lagrangian flow $\vr_t$ and its level
 sets, called {\it energy levels\/}\label{EnLev}
 are invariant under $\vr_t$. Moreover, the convexity implies that
 $$
 \tfrac{d\,}{ds}\,E(x,sv)\vert_{s=1}
 =  v\cdot L_{vv}(x,v)\cdot v>0.
 $$
 Thus 
 $$
 \min_{v\in T_xM}E(x,v)=E(x,0)=-L(x,0).
 $$
 Write
 \begin{equation}\label{e0}
 e_0:=\max_{x\in M} E(x,0)=-\min_{x\in M}L(x,0)>-\infty,
 \end{equation}
 \index{$e_0$}
 by the superlinearity $e_0>-\infty$, then
 \index{surjectivity!of the projection of an energy level}
 \begin{equation}\label{e02}
 e_0=\min\,\{\,k\in\re\;|\;\pi:E^{-1}\{k\}\to M\text{ is surjective
 }\}.
 \end{equation}
 
 By the  uniform convexity, and the  boundedness condition, 
 $$
 A:=\inf_{\stackrel{(x,v)\in TM}{|w|=1}}
 w\cdot L_{vv}(x,v)\cdot w > 0,
 $$
 and then using~\eqref{ell} and~\eqref{g(r)},
 \begin{align}
 E(x,v) &=E(x,0)+\int_0^{|v|}
           \tfrac{d\,}{ds}\,E\bigl(x,s\tfrac{v}{|v|}\bigr)\; ds
           \notag \\
       &\ge -\ell(0)+ \tfrac 12\; A\;|v|^2.
           \label{E>}  \\
 \intertext{Similarly, using~\eqref{g(r)},}
 E(x,v) &\le e_0 + g(|v|)\;|v|.
         \label{E<}
 \end{align}
 \index{bound!for the energy}
 
 
 
 Hence
 \begin{remark}\label{Ecomp}\quad\newline\indent
 If $k\in\re$ and $K\subseteq M$ is compact, then
 $E^{-1}\{k\}\cap T_KM$ is compact. 
 \label{EComp}
 \index{energy!level!compactness}
 %\index{compact!energy level}
 \end{remark}

 \begin{proposition}\label{complete}
 The Euler-Lagrange flow is complete.
 \end{proposition}

  \begin{proof}
  Suppose that $]\a,\be[$ is the maximal interval of definition
  of $t\mapsto \vr_t(v)$, and $-\infty<\a$ or $\be<+\infty$.
  Let $k=E(v)$. Since $E(\vr_t(v))\equiv k$, by~\eqref{E>},
  there is $a>0$ such that $0\le|\vr_t(v)|\le a$ for 
  $\a\le t\le \be$. Since $\vr_t(v)$ is of the form 
  $(\ga(t),\dga(t))$, then $\vr_t(v)$ remains in the interior 
  of the compact set
  $$
  Q:=\bigl\{\,(y,w)\in TM\,\big\vert\,
  d(y,x)\le a\,\big[|\be-\a|+1\big],\;\,
  |w|\le a+1\,\bigr\}\,,
  $$ 
  where $x=\pi(v)$. The Euler-Lagrange vector field
  is uniformly Lipschitz on $Q$. Then by the theory
  of ordinary differential equations, we can extend
  the interval of definition $]\a,\be[$ of 
  $t\mapsto\vr_t(v)$.
  \end{proof}

 \index{energy!function|)}
 \newpage


\section{Hamiltonian Systems.}

Let $T^*M$ be the cotangent bundle of $M$. Define the {\it Liouville's
1-form\/}\label{1-form}
\index{Liouville's 1-form}
\index{form!Liouville's 1-form}
 $\Theta$ on $T^*M$ as
\begin{equation*}\label{E-1-form}
\Theta_p (\xi) = p\,(d\pi\, \xi) 
\qquad \text{ for } \xi\in T_p (T^*M),
\end{equation*}
where $\pi:T^*M\to M$ is the projection. The {\it canonical symplectic
form\/}\label{SympForm}
\index{form!canonical symplectic --}
\index{symplectic form}
on $T^*M$ is defined as $\om = d\Theta$.

A local chart $\bx=(x_1,\ldots,x_n)$ of $M$ induces a local chart
$(\bx,\bp)=(x_1,\ldots,x_n;p_1,\ldots,p_n)$ of $T^*M$ writing
$\bp\in T^*M$ as $\bp=\Sigma_i\; p_i\;dx_i$. In these coordinates
the forms $\Theta$ and $\om$ are written
\begin{align*}
\Theta &= \bp\cdot d\bx = \tsum_i\; p_i\; dx_i\,, \\
\om &= d\bp\wedge d\bx = \tsum_i\; dp_i\wedge dx_i\,.
\end{align*}


 A {\it hamiltonian}\index{hamiltonian} is a smooth function $H:T^*M\to\re$.
 The {\it hamiltonian vector field\/}\index{hamiltonian!vector field}
 \index{vector field!hamiltonian} $X_H$ associated to $H$
 is
 defined by
 \begin{equation}
 \om(X_H,\cdot\,)=dH.
 \label{e.ham}
 \end{equation}
 In local charts, the hamiltonian vector field defines the differential
 equation
 \begin{equation}
 \begin{aligned}
 \dx &= &&H_p \, ,\\
 \dot{p} &= - &&H_x \,,
 \end{aligned}
 \end{equation}
 where $H_x$ and $H_p$ are the partial derivatives of $H$
 with respect to $x$ and $p$.
 Let $\psi_t$ be the {\it hamiltonian flow}\index{hamiltonian!flow}.
 Observe that it preserves $H$, because
 $$
 \tfrac{d}{dt} H = H_x\,\dx+H_p\,\dot p = 0.
 $$
 Moreover, it preserves the symplectic form $\om$, 
 because\footnote{$\cL_X$ is the {\it Lie derivative},\index{Lie
 derivative} defined on forms $\eta$ by 
 $\cL_X\eta=d i_X\eta+i_X d\eta$, where $i_X\eta = \eta(X,\cdot)$
 is the {\it contraction} by $X$.\index{contraction!by a vector field}
 The Lie derivative satisfies $\cL_X\eta=\tfrac{d}{dt}\psi_t^*
 \eta\vert_{t=0}$, where $\psi_t$ is the flow of $X$.}
 \begin{align*}
 \tfrac{d}{dt}(\psi_t^* \om) =\cL_{X_H}\om
 =d\,i_{X_H}\om+i_{X_H}\om 
 =d(dH)+i_{X_H}(0)=0.
 \end{align*}
 
 We shall be specially interested in hamiltonians obtained by
 the Fenchel transform 
 \index{Fenchel transform}\index{transform!Fenchel}
 of a lagrangian:
 $$
 H(x,p)=\max_{v\in T_xM} \; p\, v - L(x,v).
 $$
 
 Observe that $H=E\circ \cL^{-1}$, where $E$ is the energy
 function~\eqref{energy} and $\cL(x,v)=(x,L_v(x,v))$ 
 is the Legendre transform
 \index{Legendre transform}\index{transform!Legendre}
 of $L$. Moreover
 
 \begin{proposition}
 The Legendre transform $\cL:TM\to T^*M$,
 $\cL(x,v)=(x,L_v(x,v))$ is a conjugacy between the lagrangian
 flow and the hamiltonian flow.
 \end{proposition}
 
 \begin{proof}
 By corollary~\ref{T.2}, the convexity and superlinearity hypothesis
 imply that $L=L^{**}=H^*$. So if $p=L_v(x,v)$ then $v=H_p(x,p)$.
 With this notation:
 \begin{align*}
 H(x,p)&=v\cdot L_v(x,v)-L(x,v)=E\circ\cL^{-1}
       \\
       &=p\cdot H_p(x,p) - L(x,H_p(x,p)).
 \end{align*}
 Thus $H_x = -L_x$, and the Euler-Lagrange equation
 \begin{alignat*}{4}
 \dx &= \tfrac{d\,}{dt}\;x &&=v &&=&&H_p\,,  \\
 \dot{p} &=\tfrac{d\,}{dt}\;L_v &&= L_x &&=-&&H_x\,,
 \end{alignat*}
 is the same  as the hamiltonian equations.
 \end{proof}

 
 \begin{remark}\label{Rboundedness}
 Using that $L^*=H$ and $H^*=L$, from proposition~\ref{T.2}
 in the appendix we obtain that the boundedness condition
 is equivalent to
 \begin{itemize}
 \item[(c)] {\it Boundedness:} $H=L^*$ is convex and superlinear.
 \end{itemize}
 \index{boundedness}\index{hypothesis on $L$}
 \index{condition!boundedness}
 \end{remark} 
 
 
 \bigskip
 
 We say that an energy level $H^{-1}(k)$ is {\it regular},
 \index{energy!level!regular|ibf}\index{regular!energy level|ibf}
 if $k$ is a regular value of $H$, i.e. $dH(x,p)\ne 0$
 whenever $H(x,p)=k$.
 
 
 \begin{proposition}\label{repara}
 Two hamiltonian flows restricted to a same regular energy level
 are reparametrizations of each other. 
 \end{proposition}
 
 \begin{proof}
 Suppose that $H,G:T^*M\to\re$ are two hamiltonians with
 $H^{-1}(k)=G^{-1}(\ell)$ and $k$, $\ell$ are regular values
 for $H$ and $G$ respectively.
 Then, if $H(x,p)=k$,
 $$
 \ker d_{(x,p)}H=T_{(x,p)}H^{-1}(k
 )=T_{(x,p)}H^{-1}(k)=\ker d_{(x,p)}G.
 $$
 Thus there exists $\la(x,p)>0$ such that
 $d_{(x,p)}H=\la(x,p)\, d_{(x,p)}G$.
 Equation~\eqref{e.ham} implies that $X_H=\la(x,p)\,X_G$
 when $H(x,p)=k$.
 \end{proof}
 
 \bigskip
 
 We shall need the following estimate on the norm of the
 partial derivative $L_v(x,v)$.
 
 \begin{lemma}\label{CotaLv}
 \index{bound!for $L_v$}
 \index{bound!for the Legendre transform}
 There is a function $f:[0,\infty[\to\re^+$ such that
 $\lV L_v(x,v)\rV\le f(|v|)$ for all $(x,v)\in TM$.
 \end{lemma}
 
 \begin{proof}
 The convexity condition implies that the maximum in
 $$
 H(x,p)=\max_{w\in T_xM}p\cdot w-L(x,w)
 $$
 is attained at $w=v_0$ with $p=L_v(x,v_0)$. 
 Since $H(x,L_v(x,v))=E(x,v)$, 
 %then
 $$
 L_v(x,v)\cdot w \le E(x,v)+L(x,w), \qquad\forall v,w\in T_xM, 
 \;\forall x\in M.
 $$
 Applying this inequality to $-w$, we get that
 $$
 L_v(x,v)\cdot w \ge -E(x,v)-L(x,-w).
 $$
 Thus using~\eqref{E>}, \eqref{E<} and~\eqref{ell}, for $|v|\le r$, we
 have
 \begin{align*}
 \lV L_v(x,v)\rV
  &\le |E(x,v)|+\max\{\,|L(x,v)|,|L(x,-v)|\,\} \\
  &\le \max\{|\ell(0)|+\tfrac 12 Ar^2,\;e_0+g(r)r\}
      +|\ell(r)| =:f(r).
 \end{align*}
 \end{proof}

 \newpage




 
 
 
 \section{Examples.}
 
 We give here some basic examples of lagrangians.
 
 \bigskip
 
 \noindent{\bf  Riemannian Lagrangians:}
 
 Given a riemannian metric $g=\langle\cdot,\cdot\rangle_x$ on $TM$,
 the riemannian lagrangian\label{RiemannianLagrangian}
 \index{riemannian!lagrangian}
 \index{lagrangian!riemannian}
 on $M$ is given by the 
 {\it kinetic energy\/}
 \label{KineticEnergy}\index{energy!kinetic}
 \index{kinetic energy}
 \begin{equation}\label{E-RiemannianLagrangian}
 L(x,v) = \tfrac 12 \; \lV v\rV^2_x.
 \end{equation}
 Its Euler-Lagrange equation~\eqref{EL} is the equation
 of the geodesics of $g$: 
 \begin{equation}\label{EL-geo}
 \index{Euler-Lagrange!equation!riemannian lagrangian}
 \index{riemannian lagrangian!E-L equation}
 \tfrac{D\,}{dt} \,\dx \equiv 0,
 \end{equation}
 and its Euler-Lagrange flow is the geodesic flow.
 \index{flow!riemannian geodesic}
 \index{geodesic flow!riemannian}
 Its corresponding hamiltonian is
 \begin{equation*}\label{Riemannianhamiltonian}
 \index{hamiltonian!riemannian}
 \index{riemannian!hamiltonian}
 H(x,p)=\tfrac 12 \lV p\rV^2_x.
 \end{equation*}
 
 Analogous to the riemannian lagrangian is the {\it Finsler
 lagrangian\/}\label{FinslerLagrangian},
 \index{Finsler!lagrangian}
 \index{lagrangian!Finsler}
  given also by
 formula~\eqref{E-RiemannianLagrangian}, but where $\lV \cdot\rV_x$ is
 a {\it Finsler metric\/},\label{FinslerMetric} 
 \index{Finsler!metric}
 \index{metric!Finsler}
 i.e. $\lV \cdot\rV_x$
 is a (non necessarily symmetric\footnote{i.e. 
 $\lV \la\,v\rV_x=\la\,\lV v\rV_x$ only for $\la\ge 0$})
 norm on $T_xM$ which varies smoothly on $x\in M$.
 The Euler-Lagrange flow of a Finsler lagrangian is called
 the {\it geodesic flow\/}
 \index{geodesic flow!of a Finsler metric}
 \index{flow!Finsler geodesic --}
  of the Finsler metric $\lV\cdot\rV_x$.
 
 
 \bigskip
 
 \noindent{\bf  Mechanic Lagrangians:}
 
 The {\it mechanic lagrangian\/},
 \label{MechanicLagrangian}
 \index{mechanic lagrangian}
 \index{lagrangian!mechanic}
  also called {\it natural lagrangian\/},\label{NaturalLagrangian}
  \index{natural lagrangian}
  \index{lagrangian!natural}
  is given by the kinetic energy
  \label{KineticEnergy2}\index{energy!kinetic}
  \index{kinetic energy}
   minus the {\it potential energy\/}\label{PotentialEnergy}
   \index{energy!potential --}\index{potential!energy}
   $U:M\to\re$,
  \begin{equation}\label{EL-mec}
  \index{Euler-Lagrange!equation!mechanic lagrangian|}
  \index{mechanic lagrangian!E-L equation}
  L(x,v) = \tfrac 12 \lV v\rV^2_x - U(x).
  \end{equation}
  Its Euler-Lagrange equation is
  \begin{equation*}
  \tfrac{D\,}{dt}\, \dx = - \nabla U(x),
  \end{equation*}
  where $\tfrac{D}{dt}$ is the covariant derivative
  and $\nabla U$ is the gradient\label{gradient}
  \index{gradient}
  of $U$ with respect to the riemannian metric $g$, i.e.
  $$
  d_xU(v) = \langle \nabla U(x),v\rangle_x 
  \quad
  \text{ for all } (x,v)\in TM.
  $$
  Its energy function and its hamiltonian are given
  by the kinetic energy plus potential energy:
  \begin{align*}
  E(x,v) &= \tfrac 12 \lV v\rV^2_x + U(x),  \\
  H(x,p) &= \tfrac 12 \lV p\rV^2_x + U(x).
  \end{align*}
  
  \bigskip
  
  \noindent{\bf Symmetric Lagrangians.}
  
  \index{symmetric lagrangian}
  \index{lagrangian!symmetric}
  The symmetric lagrangians is a  class of lagrangian systems 
  which includes the riemannian and mechanic lagrangians. These are
  the lagrangians which satisfy
  \begin{equation}\label{SymmetricLagrangian}
  L(x,v)=L(x,-v) \qquad \text{ for all } (x,v)\in TM.
  \end{equation}
  Their Euler-Lagrange flow is {\it reversible} 
  in the sense that $\vr_{-t}(v)=-\vr_t(-v)$.
  \index{reversible flow}\index{flow!reversible}
  \index{Euler-Lagrange!flow!reversible}
  \index{lagrangian!flow!reversible}

   
  \bigskip 
  
  \noindent{\bf Magnetic Lagrangians.}
  
  If one adds a closed 1-form $\om$ to a lagrangian,
  $\L(x,v)=L(x,v)+\om_x(v)$, the Euler-Lagrange flow
  does not change. This can be seen by first observing that
  the solutions of the Euler-Lagrange equation are the
  critical points of the action functional on curves
  on $\C(x,y,T)$ (with fixed time interval and fixed endpoints). 
  Since $\om$ is closed, the action functional of $\L$ and $L$
  on $\C(x,y,T)$ differ by a constant and hence they have
  the same critical points.
  
  But adding a non-closed 1-form to a lagrangian does change
  the Euler-Lagrange flow. We call a {\it magnetic  lagrangian\/}
  \label{MagneticLagrangian} 
  \index{magnetic lagrangian}
  \index{lagrangian!magnetic}
  a lagrangian of the form
  \begin{equation}\label{EML}
  L(x,v) = \tfrac 12 \;\lV v\rV_x + \eta_x(v) - U(x),
  \end{equation} 
  where $\lV\cdot\rV_x$ is a riemannian metric,
  $\eta$ is a 1-form on $M$ with $d\eta\ne 0$, and
  $U:M\to\re$ a smooth function. If $Y:TM\to TM$ is the
  bundle map such that 
  $$
  d\eta(u,v)=\langle Y(u),v\rangle
  $$
  then the Euler-Lagrange equation of~\eqref{EML}
  is
  \begin{equation}\label{EL-ML}
  \index{Euler-Lagrange!equation!magnetic lagrangian}
  \index{magnetic lagrangian!E-L equation!}
  \tfrac{D\;}{dt} \;\dx = Y_x(\dx) -\nabla U(x).
  \end{equation}
  This models the motion of a particle with unit mass
  and unit charge under the effect of a magnetic field
  with Lorentz force $Y$ and potential energy $U(x)$.
  The energy functional is the same as that of the
  mechanical lagrangian but its hamiltonian changes
  because of the change in the Legendre transform:
  \begin{align*}
  E(x,v)&=\tfrac 12 \;\lV v\rV_x^2 + U(x), \\
  H(x,p)&=\tfrac 12 \;\lV p-A(x)\rV^2_x+U(x),
  \end{align*}
  where $A:M\to TM$ is the vector field given by
  $\eta_x(v)=\langle A(x),v\rangle_x$.
  
  
  \bigskip
  
  \noindent{\bf Twisted geodesic flows.}
  
  \index{flow!twisted geodesic}
  \index{twisted geodesic flow}
  \index{geodesic flow!twisted --}
  The twisted geodesic flows correspond to the motion of a particle
  under the effect of a magnetic field with no potential energy.
  This can be modeled as the Euler-Lagrange flow of a lagrangian
  of the form $L(x,v)=\tfrac 12 \lV v\rV^2_x + \eta_x(v)$, where
  $d\eta \not\equiv 0$. But the Euler-Lagrange equations depend only
  on the riemannian metric and $d\eta$. A generalization of these
  flows can be made using a non-zero 2-form $\Om$ instead of $d\eta$
  and not requiring $\Om$ to be exact. This is better presented in the
  hamiltonian setting.
  
  Fix a riemannian metric $\langle\; ,\,\rangle$  and a 2-form
  $\Om$ on $M$. Let $K:TTM\to TM$ be the connection map
 $K\xi=\nabla_{\dot{x}} v$, where
 $\xi=\frac{d\,}{dt}\bigl(x(t),v(t)\bigr)$.
 Let $\pi:TM\to M$ be the canonical projection. Let $\om_0$ be the
 symplectic  form in $TM$ obtained by pulling back the canonical symplectic form
 via the  Legendre transform associated to the riemannian metric, i.e.
 $$
 \om_0(\xi,\zeta)=
        \langle d\pi\, \xi\,,\, K\zeta\rangle
      - \langle d\pi\,\zeta\,,\, K\xi\rangle\,.
 $$
 The coordinates $T_\theta TM \ni\xi\longleftrightarrow
 (d\pi\,\xi,K\xi)\in T_{\pi(\theta)}M\oplus
 T_\pi(\theta)M=H(\theta)\oplus V(\theta)$ are the standard way of
 writing the horizontal and vertical components of a vector $\xi\in
 T_\theta TM$ for a riemannian manifold $M$ (see
 Klingenberg~\cite{Kli}).
 
 Define a new symplectic form $\om_\Om$ on $TM$ by
 $$
 \om_\Om = \om_0 + \pi^*\Om\, .     
 $$
 This is called a {\it twisted symplectic structure} on $TM$.
 Let $H:TM\to\re$ be the hamiltonian
 $$
 H(x,v)=\tfrac 12 \lV v\rV_x^2\, .
 $$
 Consider the hamiltonian vector field $X_F$ corresponding
 to $(H,\om_\Om)$, i.e.
 \begin{equation}\label{E:X.4}
 \om_\Om\bigl(X_\Om(\theta),\cdot\bigr)  = dH\,.
 \end{equation}
 Define $Y:TM\to TM$ as the bundle map such that
 \begin{equation}\label{E:X.5}
 \Om_x(u,v) = \langle Y(u), v\rangle_x\,.
 \end{equation}
 The hamiltonian vector field $X_\Om(\theta)\in T_\theta TM$ is given
 by $X_\Om(\theta)=(\theta,Y(\theta))\in H(\theta)\oplus V(\theta)$.
 Hence the hamiltonian equation is
 $$
 \tfrac{D\,}{dt}\;\dx = Y_x(\dx),
 $$
 recovering equation~\eqref{EL-ML} with $U\equiv 0$, but where $\Om$
 doesn't need to be exact. 
 
 If $H^1(M,\re)=0$, both approaches
 coincide, and any twisted geodesic flow is the lagrangian flow of a
 magnetic lagrangian of the form $L(x,v)=\tfrac 12 \, \lV
 v\rV^2_x+\eta_x(v)$, with $d\eta=\Om$. For example if $N$ is a
 compact manifold $\Om$ is a 2-form in $N$ and $M$ is the abelian
 cover or the universal cover of $N$; if $\Om$ is not exact, then
 the corresponding twisted geodesic flow is a lagrangian flow on 
 $M$ but not on $N$ (where it is locally a lagrangian flow). This
 lagrangian flow on $M$ is actually the lift of the twisted geodesic
 flow on $N$.
  
  
  \bigskip
  
  \newpage
  
  \noindent{\bf Embedding flows:}
  
  \index{flow!embedded}
  There is a way to embed the flow of any bounded
  vector field on a lagrangian system.
  Given a smooth bounded vector field $F:M\to TM$, let
  \begin{equation}\label{EmbeddedFlow}
  L(x,v) = \tfrac 12\, \lV v - F(x)\rV^2_x.
  \end{equation}
   Since $F(x)$ is bounded, then the lagrangian $L$ is convex,
   superlinear and satisfies the boundedness condition. 
   The lagrangian $L$ on a fiber $T_xM$ is minimized
   at $(x,F(x))$, hence the integral curves of the vector field,
   $\dx=F(x)$, are solutions to the Euler-Lagrange equation.
   
 
 

 
 

 
 \chapter{Ma\~n\'e's critical value.}
 
 \section{The action potential and the critical value.}
 
 \bigskip
 
 We shall be interested on action minimizing curves with {\it free}
 time interval.  Unless otherwise stated, {\it all the curves will be
 assumed to be absolutely continuous.}
 For $x,y\in M$, let 
 $$
 \C(x,y)=\{\,\ga:[0,T]\to M\,|\,T>0,\,\ga(0)=x,\,\ga(T)=y\,\}.
 \label{C(x,y)}\index{$C(x,y)$}
 $$  
 For $k\in\re$ define the {\it action potential\/} $\Phi_k:M\times
 M\to\re\cup\{-\infty\}$, by
 $$
 \Phi_k(x,y)=\inf_{\ga\in\C(x,y)}A_{L+k}(\ga).
 \label{Phi}\index{action!potential}
 \index{$\Phi_k(x,y)$}
 \index{potential!action --}
 $$
 Observe that if there exists a closed curve $\ga$ on $N$ with
 negative $L+k$ action, then $\Phi_k(x,y)=-\infty$ for all $x,y\in N$,
 by going round $\ga$ many times.
 
 
 %  \bigskip
 %  \bigskip
 %  \centerline{\bf FIGURA}
 %  \bigskip
 %  \bigskip
 
  Define the {\it critical level\/} $c=c(L)$ as
  $$
  c(L)=\sup\{\,k\in\re\,|\,\exists\text{ closed curve $\ga$ with }
  A_{L+k}(\ga)<0\,\}.
  \label{c(L)}\index{critical value|ibf}\index{$c(L)$}
  $$
  Observe that the function $k\mapsto \Phi_k(x,y)$ is
  increasing. The superlinearity
  implies that $L$ is bounded below.
  \index{superlinearity}
  \index{bound!lower -- for a lagrangian}
  \index{lower bound!for a lagrangian}
  \index{lagrangian!lower bound}
  \index{growth!superlinear}
  Hence there is $k\in\re$ such that $L+k\ge 0$. Thus $c(L)<+\infty$.
  Since $k\mapsto A_{L+k}(\ga)$ is increasing for any $\ga$,
  we have that
  $$
  c(L) = \inf\{\,k\in\re\,|\,A_{L+k}(\ga)\ge
         0\;\forall\;\text{closed curve }\ga\,\}.
         \label{c(L)2}\index{critical value}
  $$ 
 
  \begin{proposition}\quad\label{p.c.1}
  %\renewcommand{\theenumi}{(\arabic{enumi})}
  \index{potential!action --!properties}
  \begin{enumerate}
  \item     \label{p.c.1.1}
            (a) For $k<c(L)$,\; $\Phi_k(x,y)=-\infty$\; for all $x,y\in M$.\\
            (b) For $k\ge c(L)$,\; $\Phi_k(x,y)\in\re$\; \quad for all $x,y\in M$.
  \item \label{p.c.1.2}
             For $k\ge c(L)$,\;  
             $\Phi_k(x,z)\le\Phi_k(x,y)+\Phi_k(y,z)$,\;
             \index{triangle inequality!for the action potential}
	     \index{action potential!triangle inequality}
             $\forall\; x,y,z\in M$.             
  \item \label{p.c.1.3}
          {\tiny\quad}\!\!\!\!\phantom{For $k\ge c(L)$,\;}
                 $\Phi_k(x,x)=0$,\;\;$\forall\; x\in M$.           
  \item \label{p.c.1.4}
                 {\tiny\quad}\!\!\!\!\phantom{For $k\ge c(L)$,\;}
                 $\Phi_k(x,y)+\Phi_k(y,x)\ge 0$\;\;
                 $\forall\;x,y\in M$.\\
             For $k>c(L)$,\,\, 
                 $\Phi_k(x,y)+\Phi_k(y,x)>0$\;\;
                 if $x\ne y$.   
  \item \label{p.c.1.5}
            \index{continuity!of the action potential}
             For $k\ge c(L)$ the action potential $\Phi_k$ is
             Lipschitz.
             \index{Lipschitz!property!action potential's}
             \index{action potential!Lipschitz property}           
  \end{enumerate}
  \renewcommand{\theenumi}{{\it \arabic{enumi}}}
  \end{proposition}
  
  \medskip
  
  \begin{remark}\label{r.c.2}\quad
             The action potential $\Phi_k$ is not symmetric in
             general, but items~\ref{p.c.1.2}, \ref{p.c.1.3},
	     \ref{p.c.1.4} imply that
             $$
             d_k(x,y)=\Phi_k(x,y)+\Phi_k(y,x)
             \label{dk(x,y)}\index{$d_k(x,y)$}
             $$
             is a metric for $k>c(L)$ and a pseudo-metric for $k=c(L)$
             [i.e. perhaps $d_c(x,y)=0$ for some $x\ne y$ and
             $c=c(L)$].
  \end{remark}
  
  \medskip
  
  %\begin{proof}\quad
  \noindent{\bf Proof:}
  \begin{itemize}
  \item[\ref{p.c.1.2}.] We first prove~\ref{p.c.1.2} for all $k\in\re$. Since
             $\Phi_k(x,y)\in\re\cup\{-\infty\}$, the inequality in
             item~\ref{p.c.1.2}
	     makes sense for all $k\in\re$. If $\ga\in\C(x,y)$,
             $\eta\in\C(y,z)$, then $\ga*\eta\in\C(x,z)$ and hence
             $$
             \Phi_k(x,z)\le A_{L+k}(\ga*\eta)\le
             A_{L+k}(\ga)+A_{L+k}(\eta).
             $$
             Taking the infima on $\ga\in\C(x,y)$ and
             $\eta\in\C(y,z)$, we obtain~\ref{p.c.1.2}.
   \item[\ref{p.c.1.1}.] 
            (a) If $\ga$ is a closed curve with $A_{L+k}(\ga)<0$ and
                  $\ga(0)=z$, then 
                  $$
                  \Phi_k(z,z)\le\lim_{N\to\infty}
                  A_{L+k}(\ga*\overset{N}\cdots*\ga)
                  =\lim_N N\; A_{L+k}(\ga)
                  =-\infty.
                  $$       
                  For $x,y\in M$, item~\ref{p.c.1.2} implies that
                  $$
                  \Phi_k(x,y)
                  \le\Phi_k(x,z)+\Phi_k(z,z)+\Phi_k(z,y)
                  =-\infty.
                  $$  
                  Since the function $k\mapsto\Phi_k(x,y)$ is
                  increasing, then item~\ref{p.c.1.1}(a) follows.
                
                \noindent
                (b) Conversely, if $\Phi_k(x,y)=-\infty$ for some
                $k\in\re$ and $x,y\in M$, then
                $$
                \Phi_k(x,x)\le\Phi_k(x,y)+\Phi_k(y,x)=-\infty.
                $$  
                Thus there is $\ga\in\C(x,x)$ with $A_{L+k}(\ga)<0$.
                Then $k\le c(L)$. Observe that the set
                $\{\,k\in\re\,|\,A_{L+k}(\ga)<0\text{ for some closed
                curve }\ga\,\}$ is open. Hence $\Phi_k(x,y)=-\infty$
                actually implies that $k<c(L)$. This proves 
		item~\ref{p.c.1.1}(b).
  \item[\ref{p.c.1.3}.]
             Let $k\in\re$ by the boundedness condition there exists
             $Q>0$ be such that 
             \begin{equation}\label{bounded}
             |L(x,v)+k|\le Q \quad\text{ for } |v|\le 2.
             \end{equation}
             Now let $\ga:[0,\e]\to M$ be a differentiable curve with
             $|\dga|\equiv 1$ and $\ga(0)=x$. Then
             \begin{align*}
             \Phi_k(x,x) &\le \Phi_k(x,\ga(\e))+\Phi_k(\ga(\e),x)\\
                         &\le A_{L+k}\left(\ga\vert_{[0,\e]}\right)
                            + A_{L+k}\left(\ga(t-\e)\vert_{[0,\e]}\right)
                            \\
                         &\le 2\,Q\,\e.     
             \end{align*}
             Letting $\e\to 0$ we get that $\Phi_k(x,x)\le 0$. But the
             definition of $c(L)$ and the monotonicity of
             $k\mapsto\Phi_k(x,x)$ imply that $\Phi_k(x,x)\ge 0$ 
             for all $k\ge c(L)$.
  
  \openup -0.5pt
  
  \item[\ref{p.c.1.5}.] Let $k\ge c(L)$.
             Given $x_1,x_2\in M$ we have that
             $$
             \Phi_k(x_1,x_2)\le A_{L+k}(\ga)\le Q\; d_M(x_1,x_2),
             $$
             where $\ga:[0,d(x_1,x_2)]\to N$ is a unit speed
             minimizing geodesic joining $x_1$ to $x_2$ and $Q>0$ is 
             from~\eqref{bounded}. If $y_1,y_2\in M$, then
             the triangle inequality implies that
             \begin{align*}
             \Phi_k(x_1,y_1)-\Phi_k(x_2,y_2)
               &\le \Phi_k(x_1,x_2)+\Phi_k(y_2,y_1) \\
               &\le Q\; [d_M(x_1,x_2)+d_M(y_1,y_2)].
             \end{align*}          
             Changing the roles of $(x_1,y_1)$ and $(x_2,y_2)$ we get
             item~\ref{p.c.1.5}.
             
  \item[\ref{p.c.1.4}.] The first part of item~\ref{p.c.1.4}
            follows from items~\ref{p.c.1.2}
             and~\ref{p.c.1.3}. 
	     Now suppose that $k>c(L)$, $x\ne y$ and 
             $d_k(x,y)=0$. Let $\ga_n:[0,T_n]\to M$, $\ga_n\in\C(x,y)$ 
             be such that $\Phi_k(x,y)=\lim_n A_{L+k}(\ga_n)$. We
             claim that $T_n$ is bounded below.
             
             Indeed, suppose that $\lim_n T_n=0$.
             Let $A>0$, from the superlinearity there is $B>0$ such that
             $L(x,v)\ge A\,|v|-B,\; \forall\;(x,v)\in TM$. Then 
             \begin{align*}
             \Phi_k(x,y)&=\lim_n\int_0^{T_N} L(\ga_n,\dga_n)+k \\
                        &\ge\lim_n \;A\,\textstyle\int|\dga| +(k-B)\, T_n \\
                        &= A\; d_M(x,y)
             \end{align*} 
             Letting $A\to+\infty$ we get that $\Phi_k(x,y)=+\infty$
             which is false.
             
             Now let $\eta_n:[0,S_n]\to M$, $\eta_n\in\C(y,x)$ with
             $\lim_nA_{L+k}(\eta_n)=\Phi_k(y,x)$. Choose
             $0<T<\lim\inf_n T_n$ and $0<S<\lim\inf_nS_n$.
             Then for $c=c(L)<k$,
             \begin{align*}
             \Phi_c(x,x) &\le \lim_n A_{L+c}(\ga_n*\eta_n) \\
                         &\le \lim_n A_{L+k}(\ga_n) + (c-k)T
                                     + A_{L+k}(\eta_n) + (c-k) S
                           \\
                          &\le\lim_n \Phi_k(x,y)+\Phi_k(y,x) + (c-k)(T+S)
                           \\
                          &\le (c-k)(T+S) <0,           
             \end{align*}
             which contradicts item~\ref{p.c.1.3}.               
             \qed
   \end{itemize}
   %\end{proof}
  
   \openup 0.5pt
 
   \newpage
   
   \openup -0.5pt
  
 \section{Continuity of the critical value.}
  
  
  \begin{lemma}\label{C.Cont}
  \index{continuity!of the critical value}
  The function 
 $C^\infty(M,\re)\ni\psi\mapsto c(L+\psi)$ is continuous in the
 topology induced by the supremum norm.
 \label{uno}
 \end{lemma}
 
 \begin{proof}
  Suppose that $\psi_{n}\to\psi$ and let $c_{n}:=c(L+\psi_{n})$
 and $c:=c(L+\psi)$. We will prove that $c_n\to c$.
 
 Fix $\e>0$. Since $c-\e<c$, by the definition of critical value
 there exists a closed curve $\ga:[0,T]\to M$ such that
 $A_{L+\psi+c-\e}(\ga)<0$,
 hence for all $n$ sufficiently large
 $$A_{L+\psi_{n}+c-\e}(\ga)<0.$$
 %\newline
 Therefore for  $n$ sufficiently large
 $c-\e<c_n$,
 and thus
 $c-\e\leq \liminf_n c_n$.
 \newline
 Since $\e$ was arbitrary we have that
 $c\leq \liminf_n c_n$.

 We show now that $\limsup_n c_n\leq c$. Suppose that
 $c<\limsup_n c_n$. Take $\e$ such that
 \begin{equation}
 c<c+\e<\limsup\nolimits_n c_n.
 \label{ch}
 \end{equation}
 Since $\psi_{n}\to\psi$, there exists $n_0$ such that
 for all $n\geq n_0$,
 \begin{equation}
 -\e\leq\psi-\psi_n\leq\e.
 \label{ch1}
 \end{equation}
 By (\ref{ch}), there exists $m\geq n_0$ such that
 \[c<c+\e<c_m.\]
 By the definition of critical value
 there exists a closed curve $\ga:[0,T]\to M$ such that
 \[A_{L+\psi_{m}+c+\e}(\ga)<0,\]
 and hence using (\ref{ch1}) we have
 \[A_{L+\psi+c}(\ga)\leq A_{L+\psi_{m}+c+\e}(\ga)<0,\]
 which yields a contradiction to the definition of the
 critical value $c$.
 \newline  \end{proof}

 This proof also shows that $L\mapsto c(L)$ is continuous
 if we endow the set of lagrangians $L$ with the
 topology induced by the supremum norm on compact subsets
 of $TM$.

 \openup 0.5pt

 \newpage
 
 \section{Holonomic measures.}\label{HolMea}

 
 Let $C^0_\ell$\label{C0l} 
 \index{$C^0_\ell$}
 be the set of continuous functions 
 $f:TM\to\re$ having linear growth,
 \index{growth!linear}\label{lingrowth} i.e.
 $$
 \lV f\rV_\ell:=\sup_{(x,v)\in TM}\frac{|f(x,v)|}{1+\lV v\rV}<+\infty.
 $$
 \index{$\lV\cdot\rV_\ell$}
 
 
  Let $\cM_\ell$\label{Ml}\index{$\cM_\ell$}
  be the set of Borel probabilities $\mu$ on
 $TM$ such
 that
 $$
 \int_{TM}\lV v\rV\;d\mu<+\infty,
 $$
 endowed with the topology such that $\lim_n\mu_n=\mu$ if and only if
 \begin{equation}\label{defmulim}
 \lim_n\int f\;d\mu_n=\int f\;d\mu
 \end{equation}
 for all $f\in C^0_\ell$.
 
 Let $(C^0_\ell)'$ the dual of $C^0_\ell$. Then  $\cM_\ell$ is
 naturally embedded in $(C^0_\ell)'$ and its topology coincides 
 with that  induced by the weak* topology on $(C^0_\ell)'$. 
 
 We shall see that this topology is metrizable.
 Let $\{f_n\}$ be a sequence of functions with compact support 
 on  $C^0_\ell$ which is dense on $C^0_\ell$ in the topology 
 of uniform convergence on compact sets of $TM$. 
 Define a metric $d(\cdot,\cdot)$ on $\cM_\ell$ by
 \begin{equation}
   \label{distmedidas}
   d(\mu_1,\mu_2)=
   \lv\int |v| \;d\mu_1-\int|v|\;d\mu_2\rv
   +\sum_n\frac{1}{2^n}\frac{1}{c_n}
   \lv\int f_n\; d\mu_1-\int f_n\; d\mu_2\rv
 \end{equation}
 where $c_n= \sup_{(x,v)}|f_n(x,v)|$. 

 \begin{exercises}\label{ex231}\quad
 \begin{enumerate}
 \item Construct $\mu\in\cM_\ell$ such that $\int |v|^2\;d\mu=+\infty$.
 \item Show that the first term in~\eqref{distmedidas} is necessary.
 \end{enumerate}
 \end{exercises}



 \begin{proposition}\quad\newline\indent
 The metric $d(\cdot,\cdot)$ induces the weak* topology on
 $\cM_\ell\subset (C^0_\ell)'$.
 \end{proposition}


 \begin{proof}

  We prove that $d(\cdot,\cdot)$ generates the weak* topology on
  $\cM_\ell$.
  Suppose that
  $$
  \int f\;d\mu_n\to \int f\;d\mu,\qquad \forall f \in C^0_\ell.
  $$
  Given $\e>0$, choose $M>0$ such that $\sum_{m\ge
  M}\frac1{2^m}\cdot2<\e$, and choose $N>0$ such that
  \begin{align*}
  \lv\int f_m\;d\mu_n-\int f_m\;d\mu\rv&<\e,
    \qquad \text{for } 0\le m\le M,\;\;n\ge N;
  \\
  \lv\int|v|\; d\mu_n-\int|v|\;d\mu\rv&<\e,
  \qquad\text{for }n\ge N.  
  \end{align*}
  Since $\frac{\lV f_n\rV_\infty} {c_n}=1$, then for $n>N$ we have that
  $$
  d(\mu_n,\mu)\le\e+\textstyle\sum\limits_{m=1}^M\frac{1}{2^m}\cdot\e
  +\sum\limits_{m\ge M+1}\tfrac{1}{2^m}
  \cdot 2\cdot\tfrac{\lV f_m\rV}{c_m}
  =3\,\e.
  $$
  Thus $d(\mu_n,\mu)\to 0$.
 
 Now suppose that $d(\mu_n,\mu)\to 0$.
 Let $K_m$ be compact sets such that
 $K_m\subset K_{m+1}$ and that $TM=\cup K_m$. 
 Then
  \begin{align*}
  \int_{K_m}f\;d\mu_n&\longrightarrow \int_{K_m}f\;d\mu,\qquad
  \forall f\in C^0_\ell,\quad \forall m;
  \\
  \int|v|\;d\mu_n&\longrightarrow  \int|v|\;d\mu.
  \end{align*}
 This implies that
 \begin{equation}
  \label{convnormafuera}
   \lim_{n\to\infty}\int_{TM-K_m}|v|\;d\mu_n= \int_{TM-K_m}|v|\;d\mu,
   \qquad \forall\,m.
 \end{equation}
 
 Given $\e>0$, choose $m(\e)>0$ such that
 \begin{align*}
 \int_{TM- K_{m(\e)}} (1+|v|)\;d\mu&<\frac\e{4},
 \intertext{and $N$ such that}
 \int_{TM- K_{m(\e)}}(1+|v|)\;d\mu_n&<\frac{\e}2,
 \qquad\forall n>N.
 \end{align*}

 Fix $f\in C^0_\ell$. Choose $N>0$ such that
 $$
 \lv\int_{K_{m(\e)}}f\;d\mu_n-\int_{K_{m(\e)}} f\;d\mu\rv<\e,
 \qquad\forall\; n>N.
 $$
 Then
 $$
 \int_{TM-K_{m(\e)}}\lv f\rv\;d\mu_n
 \le\lV f\rV_\ell\;
 \int_{TM-K_{m(\e)}}\negthickspace\negthickspace
 (1+|v|)\;d\mu_n
 \le\lV f\rV_\ell\;\frac\e 2,\quad\forall\; n>N.
 $$
 Using a similar estimate for $\mu$ we obtain that
 $$
 \lv\int f\;d\mu_n-\int f\;d\mu\rv
 \le \e +\lV f\rV_\ell\;(\tfrac\e2+\tfrac\e4).
 $$
 \end{proof}
 
 
 \bigskip
 
 
  If $\ga:[0,T]\to M$ is a closed 
  absolutely continuous curve, 
 let $\mu_\ga\in\cM_\ell$
 \label{muga}\index{$\mu_\ga$} be
 defined by
 $$
 \int f\; d\mu_\ga =
 \frac 1T\int_0^Tf\bigl(\ga(t),\dot{\ga}(t)\bigr)\;dt
 $$
 for all $f\in C^0_\ell$. 
 Observe that $\mu_\ga\in\cM_\ell$ because if $\ga$
 is absolutely continuous then $\int|\dga(t)|\;dt<+\infty$.
 Let $\C(M)$\label{C(M)}\index{$\C(M)$}
 be the set of such $\mu_\ga$'s and let $\ov{\C(M)}$
 be its closure in $\cM_\ell$.
  Observe that the set $\ov{\C(M)}$ is convex.
 We call $\ov{\C(M)}$ the set of {\it holonomic measures} on $M$.
 \index{holonomic measure}

 \newpage
 
 \section{Invariance of minimizing measures.}
 \index{minimizing measure!invariance of --}
 \index{measure!minimizing!invariance of --}

  
 
  Given a Borel probability measure $\mu$ in $TM$
 define its {\it action\/} by
 $$
 A_L(\mu)=\int_{TM}L\; d\mu.
 \index{action!of a measure}\index{$A_L(\mu)$}
 $$
 Since by the superlinearity the lagrangian $L$ is bounded below, 
 this action is well defined.
 Observe that $L\notin C^0_\ell$ and that $A_L(\mu)=+\infty$ 
 for some $\mu\in\cM_\ell$
 (see exercise~\ref{ex231}).
 
 
  Let $\cM(L)$\index{$\cM(L)$}
 \index{invariant measures $\cM(L)$}
 \index{probability!invariant}
  be the set of $\vr_t$-invariant probabilities
 on $TM$.

 

 
 \bigskip
 
 %\newpage
 
 \begin{theorem} [Ma\~n\'e~\cite{Ma6}, prop. 1.1, 1.3, 1.2]
 \label{E.2}\quad
 \begin{enumerate}
 \item\label{E.2.1} $\cM(L)\subseteq\ov{\C(M)}\subseteq\cM_\ell$.
 \item\label{E.2.2} If $\mu\in\ov{\C(M)}$ satisfies
            $$
            A_L(\mu)=\min\{\,A_L(\nu)\,|\,\nu\in\ov{\C(M)}\,\},
            $$
            then $\mu\in\cM(L)$.
 \item\label{E.2.3} If $M$ is compact and $a\in\re$, then the set
            $\{\,\mu\in\ov{\C(M)}\,|\, A_L(\mu)\le a\,\}$ is
            compact.
 \end{enumerate}
 \end{theorem}                      
            
  Observe that item~\ref{E.2.3} implies the existence of a minimizer
  as in item~\ref{E.2.2}.
 
  The inclusion $\cM(L)\subseteq\ov{\C(M)}$ follows from 
  Birkhoff's ergodic theorem and the fact that $\ov{\C(M)}$ is
  convex. Taking $f=\lV v\rV$ in equation~\eqref{defmulim}
  we see that $\cM_\ell$ is closed, so that 
  $\ov{\C(M)}\subseteq \cM_\ell$.
  
  
  \medskip
  
  \noindent{\bf Proof of item~\ref{E.2}.\ref{E.2.3}:}
 
  Since $\ov{\C(M)}$ is closed, it is enough to prove that
  the set 
  $$
  \cA(a):=\big\{\,\mu\in\cM_\ell\,\vert\,
  A_L(\mu)\le a\,\big\}
  $$
  is compact in $\cM_\ell$.
  First we prove that $\cA(a)$ is closed.
  Let $k>0$ and define $L_k:=\min\{L,\,k\}$.
  Let
  $$
  \cB_k:=\big\{\,\mu\in\cM_\ell\,\vert\,
  \textstyle\int L_k\,d\mu\le a\,\big\}.
  $$
  Since $L_k\in\C^0_\ell$, then $\cB_k$ 
  is closed in $\cM_\ell$. Since 
  $\cA(a)=\cap_{k>0}\cB_k$, then $\cA(a)$
  is closed.
  
  In order to prove the compactness, consider
  a sequence $\{\mu_n\}\subset\cA(a)$.
  Applying the Riesz' theorem~\ref{Riesz}, 
  \index{Riesz theorem}
  taking a subsequence
  we can assume that there exists a measure
  $\mu$ on the Borel $\sigma$-algebra of $TM$ such
  that
  \begin{equation}\label{MesConvfi}
  \int f_i\;d\mu_n\longrightarrow\int f_i\;d\mu,
  \end{equation}
  for every $f_i$ in the sequence used for the definition
  of $d(\cdot,\cdot)$. Approximating the function 1 by the
  functions $f_i$ we see that $\mu$ is a probability.
  
  Approximating $L_k$ by functions $f_i$ we
  have that
  \begin{align*}
  \int L_k\;d\mu=\lim_n \int L_k\;d\mu_n
                \le\liminf_n \int L\;d\mu_n\le a.
  \end{align*}
  Letting $k\uparrow+\infty$, by the monotone convergence theorem,
  we get that
  \begin{equation}\label{MuIncAa}
  A_L(\mu)\le a.
  \end{equation}
  Let $B>0$ be such that $|v|<L(x,v)+B$
  for all $(x,v)\in TM$. Then
  \begin{equation}\label{MuInMell}
  \int |v|\;d\mu\le A_L(\mu) + B \le a+B<+\infty.
  \end{equation}
  So that $\mu\in\cM_\ell$. 
  
  We now prove that 
  $\lim_n\int |v|\;d\mu_n\longrightarrow
  \int|v|\;d\mu$.
  Let $\e>0$.
  By adding a constant we may assume that $L>0$.
  Choose $r>0$ such that $L(x,v)>a\,\e^{-1}\,|v|$
  for all $|v|>r$. Then
  $$
  \int_{|v|>r}|v|\;d\mu_n
  \le \frac{\e}{a}\int_{|v|>r}L\;d\mu_n
  \le\frac{\e}{a}\int L\;d\mu_n\le \e.
  $$
  Similarly, by~\eqref{MuIncAa},
  $$
  \int_{|v|>r}|v|\;d\mu\le\e.
  $$
  From~\eqref{MesConvfi} we obtain that there is
  $N>0$ such that
  $$
  \lv\int_{|v|\le r}|v|\;d\mu 
  -\int_{|v|\le r}|v|\;d\mu_n\rv
  <\e, \quad \text{for }n>N.
  $$
  Adding these inequalities we get that
  $$
  \lv \int|v|\;d\mu_n-\int|v|\;d\mu\rv
  \le 3\,\e.
  $$
  \qed
  
  \bigskip
  
  
  The proof of item~\ref{E.2}.\ref{E.2.2} requires
  some preliminary results which we present now.
  Item~\ref{E.2}.\ref{E.2.2} is proved at the 
  end of the section.
  
  The following proposition is needed to show that
  the minimum of the action in $\ov{\C(M)}$
  is the same as the minimum on $\C(M)$.   
  

 
  \begin{proposition}\quad\label{MuLim}
 \newline\indent
 Given $\mu\in\ov{\C(M)}$, 
 there are $\mu_{\eta_n}\in\C(M)$ such that
 $\mu_{\eta_n}\to\mu$ and 
 $$
 \lim_n\int L\;d\mu_{\eta_n}=\int L\;d\mu.
 $$
 \end{proposition}
 
 
 
 \begin{remark}
  The statement of proposition~\ref{MuLim} is not  trivial.
  It is easy to see that the function $A_L:\ov{\C(M)}\to\re$  is
  always lower semicontinuous (see the last argument of the
  proof of~\ref{MuLim}),  but in general it is not continuous. 
  It is possible to give  
  a sequence $\mu_{\ga_n}\in\C(M)$ such that $\mu_{\ga_n}\to\mu$
  in $\ov{\C(M)}$ but $\liminf_n A_L(\mu_{\ga_n})>A_L(\mu)$ for a
  quadratic lagrangian $L$.
  
  This can be made by calibrating the high speeds in $\ga_n$
  so that $\int_{[|v|>R]}|v|\,d\mu_{\ga_n}\to 0$ but
  $a:=\liminf_n\int_{[|v|>R]} L\,d\mu_{\ga_n}>0$. Then
  the limit measure $\mu$ will have support on $[|v|\le R]$
  and ``will not see'' the remnant $a$ of the action.
 \end{remark} 
 
  \medskip 
   
 
  \begin{proof}
 
 Let $A>1$ and let $\ga:[0,T]\to M$ be a closed absolutely continuous
 curve.
 We reparametrize $\ga$ to a curve $\eta:[0,S]\to M$ such
 that $\deta=\dga$ when $|\dga|<A$ and $\deta=\frac{\dga}{|\dga|}\,A$
 when $|\dga|>A$. So that $|\deta|\le A$. 
 Write $\eta(s(t))=\ga(t)$, $w(s)=|\deta(s)|$ and
 $v(t)=|\dga(t)|$. We want
 $$
 \int_0^{s(t)} w(s)\; ds = \int_0^t v(t)\;dt,
 $$
 so that
  $$
  s'(t)=\frac{v(t)}{w(s(t))}=\begin{cases}
                            \;1  &\text{when }v(t)\le A.  \\
                            \;\tfrac{v(t)}{A}&\text{when }v(t)\ge A.
                            \end{cases}
  $$
  Then
  \begin{align}
  S(T)&=\int_{[v(t)\le A]}dt+\int_{[v(t)\ge A]}\frac{v(t)}{A}\;dt,
      \notag
      \\
  \frac{S(T)}{T}
      &= \mu_\ga([|v|\le A])
            +\int_{[|v|\ge A]}\frac{|v|}{A}\;d\mu_\ga,    
      \notag
      \\      
   \left|\frac{S(T)}{T}-1\right|
      &\le\int_{[|v|>A]}d\mu_\ga
          +\int_{[|v|\ge A]}\frac{|v|}{A}\;d\mu_\ga  
        \notag
        \\  
       &\le 2 \int_{[|v|>A]}|v|\;d\mu_\ga. 
        \label{MuDist.0}  
  \end{align}
  
  Suppose that $f:TM\to\re$ is $\mu_\eta$-integrable.
  Since $\tfrac{ds}{dt}=\tfrac{v(t)}{A}$ when $v(t)>A$ then
  $$
  \int_{[|\dga(t(s))|>A]}
   f(\eta(s),\tfrac{\dga(t(s))}{|\dga(t(s))|}A)\;ds
  =\int_{[|\dga(t)>A]}
  f\big(\ga(t),\tfrac{\dga(t)}{|\dga(t)|}A\big)
  \,\frac{|\dga(t)|}{A}\;dt.
  $$ 
  Then
  \begin{align*}
  \int &f\;d\mu_{\eta}
             =\frac 1{S(T)}\int f(\eta(s),\deta(s))\;ds
             \\
             &=\frac1{S(T)}\left[\,
             \int_{[|\dga|\le A]} f(\ga(t),\dga(t))\;dt
             +\int_{[|\dga(t(s))|>A]}\negthickspace\negthickspace
             f\big(\eta(s),\tfrac{\dga(t(s))}{|\dga(t(s))|} A\big)\;ds
             \;\right]
             \\
             &=
             \frac{T}{S(T)}\;\left[\,
             \int_{[|v|\le A]}f(v)\;d\mu_\ga(v)
             +\int_{[|v|>A]}f\big(\tfrac{v}{|v|}A\big)\,\frac{|v|}{A}
             \;d\mu_\ga\,\right]
 \end{align*}
 For $A>1$ big enough,
 \begin{equation}\label{MuDist.E}
 \int_{[|v|>A]}|v|\;d\mu_\ga < \e<\tfrac{1}{4}.
 \end{equation}
 Define
 $$
 f_A(v):=\begin{cases}
         f(v) &\text{ if $|v|\le A$,}\\
         f\big(\tfrac{v}{|v|} A\big)\,\frac{|v|}{A} 
                 &\text{ if $|v|>A$.}
         \end{cases}        
 $$
 Then
 \begin{equation}\label{MuDist.IfA}
 \int f\;d\mu_\eta=\tfrac{T}{S(T)}\,\int f_A\;d\mu_\ga.
 \end{equation}
 Observe that from~\eqref{MuDist.0} and~\eqref{MuDist.E}, we have that
 
 \begin{equation}\label{MuDist.TS}
 \left|\tfrac{T}{S(T)}-1\right|\le 4\,\e.
 \end{equation}
 Then
 \begin{equation}\label{2.14}
 \left|\int f\;d\mu_\eta-\int f_A\;d\mu_\ga\right|
 \le \left|\tfrac{T}{S(T)}-1\right| \int |f_A|\;d\mu_\ga
 \le 4\,\e \,\int|f_A|\;d\mu_\ga.
 \end{equation}
 If $\lV f\rV_\infty\le 1$, then
 \begin{align*}
  \int |f_A|\,d\mu_\ga
       &\le
       \int_{[|v|\le A]} |f|\;d\mu_\ga+\int_{[|v|>A]}|f-f_A|\,d\mu_\ga 
        \\
       &\le 1 +\int_{[|v|>A]} \frac{|v|}{A}\,d\mu_\ga \le 1+\e.
 \end{align*}

 

\begin{align*}
 \left|\int f\;d\mu_\eta-\int f\;d\mu_\ga\right|
 &=\left|\tfrac{T}{S(T)}\int f_A\;d\mu_\ga 
     -\int f\;d\mu_\ga\right|
 \\
 &\le\left|\tfrac{T}{S(T)}-1\right|\,\int |f_A|\;d\mu_\ga
      +\int|f-f_A|\;d\mu_\ga
\\
 &\le 4\,\e\,(1+\e)+\e\le 6\, \e.
\end{align*}
 Also, using \eqref{2.14},
 \begin{align*}
 \left|\int|v|\;d\mu_\eta-\int|v|\;d\mu_\ga\right|
 =\left|\,\tfrac{T}{S(T)}-1\,\right|\,\int|v|\;d\mu_\ga
 \le 4\,\e\;\int|v|\;d\mu_\ga.
 \end{align*}
 Hence
 \begin{equation}\label{MuDist.1}
 d_{\cM_\ell}(\mu_\eta,\mu_\ga)\le 6\,\e\,\int (|v|+1)\;d\mu_\ga.
 \end{equation}
 
 Now let $\mu\in\ov{\C(M)}$. Let
 \begin{equation}\label{MuDist.K}
 K:=1+\int(|v|+1)\;d\mu.
 \end{equation}
 For $R>0$, define
 $$
 L_R(v):=\begin{cases}
         L(v) &\text{if $|v|\le R$.} \\
         L\big(\tfrac{v}{|v|} R\big)\,\tfrac{|v|}{R} &\text{if
         $|v|>R$.}
         \end{cases}
 $$
 
 \noindent{\it Claim:} 
 If $E(v)>0$ for all $|v|\ge R$, then
 $$
 L_{R}(v)\le L(v) \quad \text{ for all }v\in TM.
 $$
 
 \noindent{\it Proof:} \quad

 If $|v|\le R$ then $L_R(v)=L(v)$.
  Suppose that $|v|=R$. For $s\ge 1$ let
 \begin{equation*}
 f(s):= L(sv)-L_R(sv) = L(sv)-sL(v).
 \end{equation*}
 It is enough to prove that $f(s)\ge 0 $ for all $s\ge 1$. We have
 \begin{align*}
  f'(s)&=v\cdot L_v(sv)-L(v)\\
  f''(s)&=v\cdot L_{vv}(sv)\cdot v >0.
 \end{align*}
  We have that $f(1)=0$, $f'(1)=E(v)>0$, $f''(s)>0$ for all $s\ge 1$.
  This implies  that $f(s)\ge 0$ for all $s\ge 1$.

  \hfill{$\diamondsuit$}


  Given $N>0$, choose $R=R(N)>1$ such that
 \begin{equation}\label{MuDist.R}
  E(v)>0 \quad\text{ if $|v|\ge R$}
 \qquad\text{ and }\qquad
 \int_{|v|>R}|v|\;d\mu<\tfrac 1{N}.
 \end{equation}

 
 Observe  $L_{R(N)}$ has linear growth. 
 Choose $\mu_{\ga_N}\in\C(M)$ such that
 \begin{align}
 d_{\cM_\ell}(\mu_{\ga_N},\mu)&<\tfrac 1{N},
 \label{MuDist.2}
 \\
 \int L_{R(N)}\;d\mu_{\ga_N} &\le \int L_{R(N)}\;d\mu+\tfrac 1{N},
 \notag
 \intertext{and}
 \int_{[|v|\le R(N)]}|v|\;d\mu_{\ga_N}
      &\ge \int_{[|v|\le R(N)]} |v|\;d\mu-\tfrac 1N.
 \label{MuDist.b}     
 \end{align}
 Then
 \begin{alignat}{2}
 \int (|v|+1)\;d\mu_{\ga_N}&\le K 
 \qquad&&\text{ from~\eqref{MuDist.K} and~\eqref{MuDist.2},}
 \label{MuDist.a}\\
 \int_{[|v|>R(N)]}|v|\;d\mu_{\ga_N}&<\tfrac 3{N}
 \label{MuDist.e}
 &&\text{ from~\eqref{MuDist.R},~\eqref{MuDist.2}
 and~\eqref{MuDist.b}.}
 \end{alignat}
 Construct $\eta_N$ as above for $\ga_N$ and $A=R(N)$. Then
 from~\eqref{MuDist.E}, \eqref{MuDist.1}, \eqref{MuDist.a} 
 and~\eqref{MuDist.e},
 $d_{\cM_\ell}(\mu_{\eta_N},\mu_{\ga_N})<\frac{18}{N}\,K$.
 From~\eqref{MuDist.2},
 $$
 d_{\cM_\ell}(\mu_{\eta_N},\mu)<\tfrac {18}{N}\,K+\tfrac 1{N}.
 $$
 Thus $\mu_{\eta_N}\overset{N}\longrightarrow\mu$ in $\ov{\C(M)}$.
 Moreover, from~\eqref{MuDist.IfA},~\eqref{MuDist.TS},
 \eqref{MuDist.E} and the claim,
 \begin{align*}
 \int L\;d\mu_{\eta_N}
   &=\tfrac{T_N}{S(T_N)}\int L_{R(N)}\;d\mu_{\ga_N}
   \le \tfrac{T_N}{S(T_N)}\,
      \left[\,\int L_{R(N)}\;d\mu +\tfrac 1{N}\,\right]
   \\   
   &\le \big(1+\tfrac{12}{N}\big)\,
      \left[\,\int L\;d\mu+\tfrac 1{N}\,\right].
 \end{align*}
 Hence
 $$
 \limsup_N\int L\;d\mu_{\eta_N}\le \int L\;d\mu.
 $$

  
 Fix $R>0$ such that $E>0$ on $|v|>R$.
 Then $L_R$ has linear growth and by the claim $L_R\le L$.
 Therefore
 $$
 \liminf_N\int L\;d\mu_{\eta_N}
 \ge\lim_N\int L_R\;d\mu_{\eta_N}=\int L_R\;d\mu.
 $$
 Letting $R\uparrow+\infty$, by the dominated convergence theorem we
 get that
 $$
 \liminf_N\int L\;d\mu_{\eta_N}\ge\int L\;d\mu.
 $$
 \end{proof}


 \bigskip
  
  Given $x,\,y\in M$, define
  \index{$S(x,y;T)$}
  $$
  S(x,y;T):=\inf_{\ga\in \C_T(x,y)} A_L(\ga).
  $$
  Observe that $S(x,y;T)>-\infty$ because
  $L$ is bounded below.
  If $\ga\in\Cac([0,T],M)$, define
  \index{$S^+(\ga)$}
  $$
  S^+(\ga):=A_L(\ga)-S(\ga(0),\ga(T);T).
  $$ 
  The absolutely continuous curves $\ga$
  with $S^+(\ga)=0$ are called {\it Tonelli minimizers}.
  \index{Tonelli!minimizer}
  Observe that a Tonelli minimizer is a solution 
  of~\eqref{EL}.
   Given $\ga_1,\,\ga_2\in\Cac([0,T],M)$, the
  {\it absolutely continuous distance}
   \index{absolutely continuous!metric}
   \index{metric!absolutely continuous}
   \index{distance!absolutely continuous}
   \index{$d_1(\ga_1,\ga_2)$}
   $d_1(\ga_1,\ga_2)$ is
  defined by
  $$
  d_1(\ga_1,\ga_2):=\!\sup_{t\in[0,T]}
  d\big(\ga_1(t),\ga_2(t)\big)
  +\int_0^T\!\!\!d_{TM}\big(
  [\ga_1(t),\dga_1(t)],[\ga_2(t),\dga_2(t)]\big)\,dt.
  $$
 

 
 \begin{proposition}
 \label{anc}
 Given a compact subset $K\subseteq M$ 
 and given $C$, $\e>0$ 
 there exist $\de >0$ such that if $\ga :[0,T]\to M $
 is absolutely continuous and satisfies 
 \renewcommand\theenumi{\roman{enumi}}
 \begin{enumerate}
 \item\label{anc.1} \quad $1\le T\le C$.
 \item\label{anc.2} \quad $A_L(\ga ) \le C$.
 \item\label{anc.3} \quad $S^+(\ga)\le \de$.
 \end{enumerate}
 \renewcommand\theenumi{{\it\arabic{enumi}}}
 Then either $\ga([0,T])\cap K=\0$ or there exists a Tonelli
 minimizer $\ga_0 :[0,T]\to M $ such that
 $d_1(\ga_0,\ga)\le \e$.
 \end{proposition}
 
 \begin{proof}

  If such $\de$ does not exists
  then there is a sequence $\ga_n\in\Cac([0,T_n],M)$
  such that $\ga_n([0,T_n])\cap K\ne \0$,
  $1\le T_n\le C$, $S^+(\ga_n)\to 0$,
  $A_L(\ga_n)\le C$ and $d_1(\ga_n,\eta)\ge\e$
  for any Tonelli minimizer $\eta$.
  

  
  Adding a constant we can assume that $L>0$.
  Let $B>0$ be such that $L(x,v)>|v|-B$ for all $(x,v)\in TM$.
  Choose $s_0\in[0,T_n]$ such that $\ga_n(s_0)\in K$.
  Then
  \begin{align*}
  d(K,\ga_n(t))&\le d(\ga_n(s_0),\ga_n(t))
  \le\int_{[s_0,t]} |\dga_n|
  \\
  &\le \int_{[s_0,t]} \big[\,L(\ga_n,\dga_n)+B\,\big]
  \le C+B\,C.
  \end{align*}
  Let $Q:=\{\, y\in M\,|\,d(y,K)\le C+B\,C\;\}$.
  Then we have that 
  $\ga_n([0,T_n])\subseteq Q$.
  
  
  We can assume that  $T_n\to T$, 
  $\ga_n(0)\to x\in Q$ and $\ga_n(T_n)\to y\in Q$.
  Moreover, we can assume that $T_n\equiv T$,
  $\ga_n(0)\equiv x$ and $\ga_n(T)\equiv y$.
  By theorem~\ref{MaTon}, the set
  $\cA(b)=\big\{\ga\in\C_T(x,y)\,|\,A_L(\ga)\le b\,\big\}$
  is compact in the $C^0$-topology.
  Then we can assume that there is $\ga_0\in\C_T(x,y)$
  such that $\ga_n\to\ga_0$ in the $C^0$-topology.
  Since the action functional is lower semicontinuous,
  then
  $A_L(\ga_0)\le\liminf_n A_L(\ga_n)=S(x,y;T)$,
  because $S^+(\ga_n)\to 0$. Thus $\ga_0$ is a 
  Tonelli minimizer.
  Moreover, we have that $A_L(\ga_n)\to A_L(\ga_0)$.
  By proposition~\ref{AdTon}, $\ga_n\to \ga_0$
  in the $d_1$-topology. 
  \newline
  \end{proof}
  

  
  Let\index{$\cH$}
  $$
  \cH:=\big\{\,h:TM\to\re\,\big\vert\, \lV f\rV_\infty\le 1,\;
  [h]_\Lip\le 1,\; h \text{ with compact support}\;\big\},
  $$
  where 
  $$
  [h]_\Lip = \sup_{(x,v)\ne (y,w)}
  \frac{|h(x,v)-h(y,w)|}{d_{TM}\big((x,v),(y,w)\big)}
  $$
  \index{$[h]_\Lip$}
  is the smallest Lipschitz constant
  \index{Lipschitz!smallest -- constant} for $h$. 

 \begin{corollary}
 \label{Corquasiinvar}
 \quad\newline\indent
 Given $h\in\cH$ and $C>0$ there exist 
 $\de=\de(C,h)>0$ such that if 
 $\ga :[0,T]\to M $ satisfies 
 conditions~\ref{anc}.\ref{anc.1},~\ref{anc}.\ref{anc.2},~\ref{anc}.\ref{anc.3} 
 then
 \begin{equation}
   \label{quasiinvarian}
   \lv\oint_\ga h -\oint_{\ga} h\circ\vr_1\rv\le 5.
 \end{equation}
 \end{corollary}
 
 \begin{proof}
 Let $K=\pi\big(\supp(h)\cup\vr_{-1}(\supp(h))\big)$.
 Given $C>0$ and
 $\e>0$ let $\de=\de(C,\e)>0$ and $A>0$ be given by 
 proposition~\ref{anc} then if 
 $\ga :[0,T]\to M $ satisfies
 conditions~\ref{anc}.\ref{anc.1},
 ~\ref{anc}.\ref{anc.2},~\ref{anc}.\ref{anc.3} 
 we have that either $\ga ([0,T])\cap K=\0$,
 or we can take $\ga_0$ minimizing such that
 $d_1(\ga_0,\ga)\le \e$.
 
 Observe that if $\ga ([0,T])\cap K=\0$,
 then $h(\ga,\dga)\equiv 0$ and 
 $h\circ\vr_1(\ga,\dga)\equiv 0$.
 This implies~\eqref{quasiinvarian}.
 Suppose then that $d_1(\ga_0,\ga)\le \e$.
 
 We have that
 \begin{align*}
 \lv\oint_{\ga} h -\oint_{\ga_0} h\rv
  &\le[h]_\Lip\;d_1(\ga,\ga_0)
   \le 1\cdot 1\cdot \e,
 \intertext{where $[h]_\Lip$ is the
 smallest Lipschitz constant of $h$. 
 Let $Q(h):=\vr_{-1}(\supp(h))$, then}
 \lv\oint_{\ga} h\circ\vr_1
   \!-\!\oint_{\ga_0} h\circ\vr_1\rv  
   &\le [h]_\Lip\;[\vr_1\vert_{Q(h)}]_\Lip\;d_1(\ga,\ga_0)
   \le 1\cdot [\vr_1\vert_{Q(h)}]_\Lip\cdot \e.
 \end{align*}
 
 Since $\ga_0$ is a solution of~\eqref{EL},
 we have that
 \begin{align*}
 \lv\oint_{\ga_0}h-\oint_{\ga_0} h\circ\vr_1\rv
  &=\lv\int_0^T h\bigl(\ga_0(t),\dga_0(t)\bigr)
     -h\bigl(\ga_0(t+1),\dga_0(t+1)\bigr)\;dt\rv
  \\
  &\le \int_0^1 |h(\ga_0,\dga_0)|\;dt
    +  \int_T^{T+1}|h(\ga_0,\dga_0)|\;dt
    \le 2.
 \end{align*}
 Hence
 $$
 \lv\oint_\ga h-\oint_\ga h\circ \vr_1\rv
 \le \e \big( 1 + [\vr_1\vert_{Q(h)}]_\Lip \big) + 2.
 $$
 \end{proof}
 



 \bigskip

 \noindent{\bf Proof of item~\ref{E.2}.\ref{E.2.2}:}

 Observe that to prove that $\mu$ is invariant
 it is enough to prove that
 \begin{equation}\label{EquivToInvar}
 \int h\;d\mu=\int h\;d(\vr_1^*\,\mu)
 \qquad\text{for all }h\in\cH.
 \end{equation}

 By proposition~\ref{MuLim}, there exists a sequence 
 $\mu_{\ga_n}\in\C(M)$ such that
 $\mu_{\ga_n}\to\mu$ and
 \begin{equation}\label{ConvAcK}
 \lim_n A_L(\mu_{\ga_n})=A_L(\mu)
 =\min\big\{\,A_L(\nu)\,\vert\,\nu\in\ov{\C(M)}\,\big\}
 =:k.
 \end{equation}
 Let $T_n$ be a period of the curve $\ga_n:\re\to M$.
 Take an integer $N>0$. By joining a constant curve
 if necessary, we can assume that every $T_n$ is 
 a multiple of $N$ and that 
 $\lim_{n\to\infty}T_n=+\infty$.
 Given $C>0$ let
 $$
 \cB_n(C):=\big\{\,j\in\na\,\vert\,
 1\le j\le \tfrac{T_n}{N},\;
 A_L(\ga_{n,j})\ge C\,\big\},
 $$
 where
 $$
 \ga_{n,j}:=\ga_n\vert_{[jN,(j+1)N]}.
 $$
 By the superlinearity $L$ is bounded below,
 adding a constant we can assume that $L>0$.
 Then we can assume that
 $$
 A_L(\mu_{\ga_n})
 =\frac 1{T_n}\int_0^{T_n}
 L(\ga_n,\dga_n)\;dt \le 2\,k\qquad \forall n.
 $$
 Hence
 $$
 2\,k\,T_n
 \ge\sum_{j\in\cB_n(C)}A_L(\ga_{n,j})
 \ge C\,\#\cB_n(C).
 $$
 Thus
 \begin{equation}\label{NroC}
 \frac {\# \cB_n(C)}{T_n}\le \frac {2k}C.
 \end{equation}
 Given $\de>0$, let 
 $$
 \cB_n'(\de):=\big\{\,j\in\na\,\vert\,
 1\le j\le\tfrac{T_n}N-1,\;\;
 S^+(\ga_{n,j})>\de\,\big\}.
 $$
 Then
 $$
 S^+(\ga_n)\ge\sum_{j=1}^{({T_n}/N)-1}S^+(\ga_{n,j})
 \ge\de\;\#\cB_n'(\de).
 $$
 Moreover,
 $$
 k\le\frac1{T_n}\;S(\ga_n(0),\ga(T_n);T_n)
 =A_L(\mu_{\ga_n})-\frac 1{T_n}\;S^+(\ga_n).
 $$
 Hence
 $$
 S^+(\ga_n)\le T_n \big( A_L(\mu_{\ga_n})-k\big).
 $$
 Therefore
 \begin{equation}\label{NroDe}
 \frac{\#\cB_n'(\de)}{T_n}\le\frac1{\de}\;
 \big(A_L(\mu_{\ga_n})-k\big).
 \end{equation}
 
 Now fix $h\in\cH$. Then
 $$
 \lv\int h\;d\mu_{\ga_n}-\int h\;d(\vr_1^*\,\mu_{\ga_n})\rv
 \le\frac 1{T_n}\,\sum_{j=0}^{(T_n/N)-1}
 \lv\oint_{\ga_{n,j}} h-\oint_{\ga_{n,j}}h\circ\vr_1\rv.
 $$
 Denote $\cB_n'':=\cB_n(C)\cup\cB_n'(\de)$.
 Since $\sup|h|\le 1$, then
 $$
 \lv\int h\;d\mu_{\ga_n}-\int h\;d(\vr_1^*\,\mu_{\ga_n})\rv
 \le\frac 1{T_n}\,\sum_{j\notin\cB_n''}
 \lv\oint_{\ga_{n,j}}\!\!\!\!\!h
 -\oint_{\ga_{n,j}}\!\!\!\!\!h\circ\vr_1\rv
 +\frac1{T_n}\,2 N\,\#\cB_n''.
 $$
 Now choose $C\ge N^2$ and $\de=\de(C,h)>0$ 
 from corollary~\ref{Corquasiinvar}.
 Using equations~\eqref{NroC},~\eqref{NroDe} and
 corollary~\ref{Corquasiinvar} we obtain that
 \begin{align*}
 \lv\int h\;d\mu_{\ga_n}-\int h\;d(\vr_1^*\,\mu_{\ga_n})\rv
   &\le \frac5{T_n}\;\left(\frac{T_n}N-\#\cB_n''\right)
     +\frac1{T_n}\,2N\,\#\cB_n''
   \\  
   &\le \frac 5N+2N\left(\frac{2k}C+\frac 1\de
    \Big(A_L(\mu_{\ga_n})-k\Big)\right).
 \end{align*}
 Now let $n\to\infty$. Using equation~\eqref{ConvAcK}
 and that $C\ge N^2$, $\mu_{\ga_n}\to\mu$ and 
 $h,\,h\circ\vr_1\in C^0_\ell$ (because they
 have compact support), we obtain that
 $$
 \lv\int h\;d\mu-\int h\;d(\vr_1^*\,\mu)\rv
 \le\frac5{N}+\frac{4k}N.
 $$
 Since $N$ is arbitrary, this difference is zero and
 we get~\eqref{EquivToInvar}.
 \qed



 


 \newpage
 
 \section{Ergodic characterization of the critical value.}
 
 \medskip
 

 Given a Borel probability measure $\mu$ in $TM$
 define its {\it action\/} by
 $$
 A_L(\mu)=\int_{TM}L\; d\mu.
 \index{action!of a measure}\index{$A_L(\mu)$}
 $$
 Since by the superlinearity\index{superlinearity}
  \index{bound!lower -- for a lagrangian}
  \index{lower bound!for a lagrangian}
  \index{lagrangian!lower bound}
  \index{growth!superlinear}
 the lagrangian $L$ is bounded below, this action is well defined.
 
  Let $\cM(L)$\index{$\cM(L)$}
 \index{invariant measures $\cM(L)$}
 \index{probability!invariant}
  be the set of $\vr_t$-invariant probabilities
 on $TM$.

 
 \begin{theorem} [Ma\~n\'e~\cite{Ma3}]\label{E.1}
 If $M$ is compact, then
 $$
 c(L)=-\min\{\,A_L(\mu)\,\vert\,\mu\in\cM(L)\,\}.
 \label{c(L)3}\index{critical value}
 \index{$c(L)$}
 $$
 \end{theorem}
 
 We will obtain theorem~\ref{E.1} from theorem~\ref{E.3} below, which
 also applies to the non-compact case. If $M$ is non-compact, 
 theorem~\ref{E.1} may not hold, as seen in
 example~\ref{EmpresarioPobre}. 
 
 
 
  Recall that if $\ga:[0,T]\to M$ 
  is a closed absolutely continuous curve, 
  the measure $\mu_\ga\in\cM_\ell$\index{$\mu_\ga$} 
  is defined by
 $$
 \int f\; d\mu_\ga =
 \frac 1T\int_0^Tf\bigl(\ga(t),\dot{\ga}(t)\bigr)\;dt
 $$
 for all $f\in C^0_\ell$, and that $\ov{\CM}$
 \index{$\C(M)$}
 is the closure of the set of such $\mu_\ga$'s in $\cM_\ell$.
 
 \bigskip
 
 
 \begin{theorem}\label{E.3}
 \begin{equation}\label{e.e.1}\index{critical value}
 \begin{aligned}
 c(L) &= -\inf\{\,A_L(\mu)\,|\,\mu\in\ov{\C(M)}\,\}\\
      &= -\inf\{\,A_L(\mu)\,|\,\mu\in\C(M)\,\}.
 \end{aligned}
 \end{equation}
 \end{theorem}
 
 \bigskip
 
 \begin{Definition} We say that a holonomic measure $\mu\in\C(M)$ is
 (globally) {\it minimizing} if $A_L(\mu)=-c(L)$.
 \index{minimizing measure}\index{measure!minimizing}
 \end{Definition}
  
 \newpage
 
 \begin{remarks}\quad
 \begin{enumerate}
 \item Recall that by theorem~\ref{E.2}, if a minimizing measure
       exists, then it is invariant under the lagrangian flow.
 \item The equality between the two infima in theorem~\ref{E.3}
       is non-trivial and follows from proposition~\ref{MuLim}.
 \item Theorems~\ref{E.3} and~\ref{E.2} imply theorem~\ref{E.1}.
 \item If $p:N\to M$ is a covering,
            \index{covering}
            $M$ is compact and $\L=L\circ dp$
	    is the lifted lagrangian, then theorems~\ref{E.3}
            and~\ref{E.2} imply that
            \begin{equation}\label{e.e.2}
            c(\L)=-\min\{\,A_L(\mu)\,|\,\mu\in\cM(L)\cap\ov{dp_*\C(N)}\,\},
            \end{equation}
            by noticing that $A_L(dp_*\,\nu)=A_\L(\nu)$ for
            $\nu\in\ov{\C(N)}$. Here $dp_*\C(N)$ is the set of
            probabilities $\mu_\ga$ on $TM$ where $\ga$ is a curve on
            $M$ whose lifts to $N$ are closed. The compactness
            property on theorem~\ref{E.2}(3) allows to obtain a
            minimum on~\eqref{e.e.2} instead of the infimum
            on~\eqref{e.e.1} which may not be attained in the
            non-compact case.
  \item The statement for coverings in equation~\eqref{e.e.2}
            allows to obtain minimizing measures which don't appear
            in the Mather's theory. For example if $c_u$ 
            \index{$c_u(L)$}
            \index{critical value!of the universal cover}
            is the critical value of the universal cover $\tM$ of $M$
            and $c_0$ is the critical value of the abelian cover $\hM$
            of $M$; the minimizing measures on a fixed homology class
            (corresponding to Mather's theory) all have action
            $A(\mu)\ge -c_0$, (see equation~\eqref{E:c0} 
            and proposition~\ref{P.ca}),
            while the minimizing measures
            for $\tM$ have action $c_u<c_0$.

            The measures
            for $\tM$ correspond to ``minimizing in the zero homotopy
            class'' while the measures for $\hM$ are minimizing in the
            zero homology class.
            
             

            
            The drawback of this approach is that we obtain 
            honest minimizing invariant measures on $TM$ which may not
            lift to finite measures on the covering $TN$.
 \end{enumerate}
 \end{remarks}
 
 \bigskip
 
 \newpage
 
 \noindent{\bf Proof of theorem~\ref{E.3}:}
 
 If $\mu_\ga\in\C(M)$, then $A_{L+c(L)}(\mu_\ga)\ge 0$. Hence
 $A_{L}(\mu_\ga)\ge -c(L)$. Thus 
 $$
 -c(L)\le \inf\{\, A_L(\mu)\,|\,\mu\in\C(M)\,\}
 =\inf\{\, A_L(\mu)\,|\,\mu\in\ov{\C(M)}\,\},
 $$
 where the last equality follows from
 proposition~\ref{MuLim}.
 
 If $k<c(L)$ then there is a closed absolutely 
 continuous curve $\ga$ on $M$
 such that $A_{L+k}(\ga)<0$.  
 Thus $\mu_\ga\in\C(M)$ and 
 $$
 -k>A_L(\mu_\ga)\ge\inf\{\,A_L(\mu)\,|\,\mu\in\ov{\C(M)}\}.
 $$
 Now let $k\uparrow c(L)$.
 \qed


 \newpage
 
 \section{The Aubry-Mather Theory.}
 \index{Mather!theory|(}
 \index{Aubry-Mather theory|(}


 
 
 Through this section we shall assume that $M$ is compact.
 
 

%  
%   \subsection{Holonomic measures.}
%  
%  
%  A probability $\mu$ on $TM$ is said {\it holonomic\/}\label{holon} 
%  \index{probability!holonomic}
%  \index{measure!holonomic}
%  if $\int_{TM} |v|\; d\mu <+\infty$ and
%  \begin{equation*}\label{E-holon}
%  \int_{TM}df\; d\mu = 0 \quad
%  \text{ for all } f\in C^\infty(M,\re).
%  \end{equation*}
%  
%  \begin{proposition}
%  
%  The set of holonomic measures
%  \end{proposition}
%  
%  The set of holonomic measures is the natural class of measures
%  where the homology of a measure can be defined. Given a holonomic
%  probability, we define its  {\it homology class\/} as


  
   \subsection{Homology of measures.}
  
 Observe that since $M$ is compact, 
 any $1$-form is in $C^0_\ell$.
 By definition, an holonomic  probability $\mu\in\ov{\CM}$  
 \index{probability!holonomic}
 \index{measure!holonomic}
 satisfies
 $\int_{TM} |v|\; d\mu <+\infty$ and
 \begin{equation*}\label{E-holon}
 \int_{TM}df\; d\mu = 0 \quad
 \text{ for all } f\in C^\infty(M,\re).
 \end{equation*}
  Then we can define its {\it homology class\/} as
 $\rho(\mu)\in H_1(M,\re)\approx H^1(M,\re)^*$ by
 \index{$\rho(\mu)$}
 \begin{equation}\label{rho}
 \index{rotation of a measure}
 \index{homology!of a measure}
 \bigl\langle\, \rho(\mu)\,,[\om]\,\bigr\rangle =
 \int_{TM}\om\;d\mu\,,
  \end{equation}
 for any closed 1-form $\om$\label{CF}\index{closed form}
 \index{form!closed}
 on $M$, where $[\om]\in
 H^1(M,\re)$ is the cohomology class of $\om$. Here we have used
 the identification\footnote{In fact,
        $H^1(M,\re)\approx\hom\bigl(H_1(M,\re),\re\bigr)=H_1(M,\re)^*$
        by the universal coefficient theorem. Since $M$ is compact,
        then $H_1(M,\re)$ is a finite dimensional vector space and
        hence
        it is naturally isomorphic to its double dual $H^1(M,\re)^*$.}
 $H_1(M,\re)\approx H^1(M,\re)^*$ and
 equation~\eqref{rho} shows how the homology class $\rho(\mu)$
 acts on $H^1(M,\re)$. Since $\mu$ is holonomic, the integral
 in~\eqref{rho} depends only on the cohomology class of $\om$.
 The class $\rho(\mu)$
 is called the {\it homology\/} of $\mu$ or the {\it rotation\/}
 of $\mu$ by analogy to the twist map theory.
  
 Using a finite basis $\{\,[\om_1],\ldots,[\om_k]\,\}$ for 
 $\Co$ and the topology of $\ov{\C(M)}$, we have that
 
 
 
 \begin{lemma}\label{RhoCont}
 The map $\rho:\ov{\C(M)}\to\Ho$ is continuous.
 \end{lemma}
 \index{continuity!of the homology function $\rho$}

 

 
 \subsection{The asymptotic cycle.}
 

 
 Given a differentiable flow $\vr_t$ on a compact manifold
 $N$ and a $\vr_t$-invariant probability $\mu$, the
 Schwartzman's~\cite{Schw}
 \index{Schwartzman}
 {\it asymptotic cycle\/}\label{AsCi}\index{asymptotic cycle}
  of an invariant probability $\mu$
  is defined to be the homology class 
  $A(\mu)\in H_1(N,\re)\approx H^1(N,\re)^*$ 
  such that 
  \begin{equation*}\label{AS}
  \bigl\langle\,A(\mu)\,, [\om]\,\bigr\rangle
  = \int_N\om(X)\;\;d\mu\,,
  \end{equation*}
  for any closed 1-form $\om$, where $[\om]\in H^1(N,\re)$ is the
  cohomology class of $\om$ and $X$ is the vector field of $\vr_t$.
  This integral depends only on the cohomology class of $\om$ because
  the integral of a coboundary by an invariant measure is zero:
  in fact, if $df$ is an exact 1-form, then define
  $$
   F(y):=\lim_{T\to+\infty}
    \frac 1T \int_0^T df\bigl(X(\vr_t\,y)\bigr)\; dt
   =\lim_{T\to+\infty}
   \tfrac 1T\;\bigl[f(\vr_t\, y) - f(y) \,\bigr]
   =0,
  $$
  by Birkhoff's theorem,
  $$
  \int_N df(X)\; d\mu = \int_N F\; d\mu =0.
  $$
  
  If $\mu$ is ergodic and $x\in N$ is a {\it generic
  point\/}\label{GenP}\index{generic!point}
  \index{point!generic}\footnote{i.e. 
      $\lim\limits_{T\to+\infty}\frac 1T\int_0^T f(\vr_t\, x)\; dt
      =\int f\; d\mu$ for all $f\in C^0(N,\re)$.}
 for $\mu$, then    
  \begin{equation*}
  \bigl\langle\,A(\mu)\,, [\om]\,\bigr\rangle
  = \lim\limits_{T\to+\infty}\frac 1T
  \int_0^T\om\bigl(X(\vr_t\,x)\bigr)\;dt\,.
  \end{equation*}
  Applying this to a basis $\{\om_1,\ldots,\om_k\}$ for $H^1(N,\re)$,
  we get that
 $$
 A(\mu)=\lim_{T\to+\infty}\tfrac 1T\;[\,\ga_T * \de_T\,]\in
 H_1(N,\re),
 $$
 where $\ga_T(t)=\vr_t(x)$,~$t\in[0,T]$, the curve $\de_T$ is a unit
 speed
 geodesic from $\vr_T(x)$ to $x$, and the limit is on the finite
 dimensional vector space $H_1(N,\re)$. 

  In the case of a lagrangian flow, the phase 
  space\label{PhSp.2}~$N=TM$\index{phase space}
  is not compact, 
  but it has the same homotopy type as 
  the configuration space\label{ConfSp.2}~$M$
  \index{configuration space}
   because
  $M$ is a deformation retract\label{DefRet}
  \index{deformation retract} of $TM$
  (contracting $TM$ along the fibers to the zero section $M\times 0$).
  Moreover, the ergodic components of an invariant measure
  of a lagrangian flow are contained in a unique energy level,
  which is a compact submanifold of $TM$ by remark~\ref{Ecomp}.

  We see that the homology of an invariant probability and its
  asymptotic cycle coincide under the identification
  $H_1(TM,\re)\overset{\pi_*}\approx H_1(M,\re)$.
  

  
  \begin{proposition}
  $$
  \pi_*\bigl(A(\mu)\bigr) = \rho(\mu)
  \quad\text{ for all } \mu\in\cM(L),
  $$
  where $\pi_*:H_1(TM,\re)\to H_1(M,\re)$ is the map induced 
  by the projection $TM\overset{\pi}\to M$.
  \end{proposition}
  
  
  \begin{proof}
  If $\om$ is a closed 1-form on $M$, then
  $$
  (\pi^*\,\om)(X(x,v)) = \om\bigl[d\pi\bigl(X(x,v)\bigr)\bigr]
  =\om_x(v)\,,
  $$
  because the lagrangian vector field $X$ has the form $X(x,v)=(v,*)$.
  Then
  \begin{align*}
  \bigl\langle\, \pi_*\, A(\mu)\,,[\om]\,\bigr\rangle
  &=\bigl\langle\, A(\mu)\,,\pi^*\,[\om]\,\bigr\rangle  
  =\int_{TM} (\pi^*\,\om)(X)\;\;d\mu   \\
  &=\int_{TM} \om\; d\mu
  =\bigl\langle\, \rho(\mu)\,,[\om]\,\bigr\rangle\,.
  \end{align*}
  \end{proof}

 
  \begin{lemma}\label{RhoSurj}
  \index{surjectivity!of the homology function $\rho$}
 The map $\rho:\cM(L)\to H_1(M,\re)$ is surjective.
 \end{lemma}
 
 
 \begin{proof}
 
 Let $h\in H_1(M,\Z)$ be an integer homology class. 
 Let $\eta:[0,1]\to M$ be a closed curve with homology class $h$.
 Let $\ga$ be a minimizer of the action of $L$ among the set
 of absolutely continuous curves $[0,1]\to M$ with 
 the same homotopy class as $\eta$.
 Then by remark~\ref{IsSol},
 $\ga$ is a periodic orbit for the lagrangian flow with period 1.
 The invariant measure $\mu_\ga$ satisfies $\rho(\mu_\ga)=h$.
 
 The map $\rho$ is affine and $\cM(L)$ is convex; 
 hence $\rho\bigl(\cM(L)\bigr)$ is convex and, in particular, 
 it contains the convex hull of $H_1(M,\Z)$.
 Thus, $H_1(M,\re)\subseteq \rho\bigl(\cM(L)\bigr)$.
 \end{proof}



 
 \subsection{The alpha and beta functions.}
 
 
  The action functional $A_L:\cM(L)\to\re$  is lower 
  semicontinuous\footnote{$A_L$ is lower semicontinuous 
  iff $\liminf_nA_L(\nu_n)\ge A_L(\mu)$
  when $\nu_n\to\mu$.}\index{semicontinuous}
  and the sets
 \index{$\cM(h)$} 
 \begin{equation*}\label{M(h)}
 \cM(h):=\{\,\mu\in\cM(L)\,\vert\,\rho(\mu)=h\,\}
 \end{equation*} 
 are closed.
 Hence we can define the {\it Mather's beta function\/} 
 \index{Mather!beta function}
 $\beta:H_1(M,\re)\to\re$, as
 \begin{equation*}\index{beta function}\index{$\beta(h)$}
 \beta(h):=\min_{\mu\in\cM(h)}A_L(\mu).
 \end{equation*}

 We shall prove below that the $\beta$-function is convex.
 The {\it Mather's alpha function\/}\index{alpha function}
 $\a=\be^*:H^1(M,\re)\to\re$ 
 is the convex dual\label{Conv*2}
 \index{convex!dual}
 \index{dual!convex --}
 of the $\beta$-function:
 \begin{align}
 \a([\om]) 
    &= \max_{h\in H_1(M,\re)}
       \bigl\{\,\langle[\om]\,,h\rangle - \be(h)\,\bigr\}  
       \notag                                              \\
    &=-\min_{\mu\in\cM(L)}\bigl\{\,A_L(\mu)
            - \langle[\om]\,,\rho(\mu)\rangle\,\bigr\}   
       & \text{using \ref{RhoSurj},}                          
       \notag                                         \\
    &=-\min_{\mu\in\cM(L)} A_{L-\om}(\mu)                  
            \notag                                         \\
    &= c(L-\om)\,,    &\text{ by \ref{E.1}}.         
            \label{alfa}\index{critical value}
            \index{$c(L)$}
            \index{$\a(\om)=\be^*(\om)$}
 \end{align}
 Observe that since $L-\om$ is also a convex superlinear lagrangian,
 then $\a([\om])$ is finite.
 

 
 \begin{theorem}\label{ABConvex}
 \index{convexity!of the $\beta$ function}
 \index{convexity!of the $\alpha$ function}
 \index{superlinearity!of the $\beta$ function}
 \index{superlinearity!of the $\alpha$ function}
 \index{growth!superlinear} 
  \index{convex!function}
  \index{function!convex}
 The $\a$ and $\beta$ functions are convex and superlinear.
 
 \end{theorem}
 
 
 \begin{proof}
 We first prove that $\beta$ is convex.
 Let $h_1,\,h_2\in H_1(M,\re)$ and $0\le\la\le 1$.
 Let $\mu_1,\,\mu_2\in\cM(L)$ be such that
 $\rho(\mu_i)=h_i$ and $A_L(\mu_i)=\beta(h_i)$ for $i=1,2$.
 The probability $\nu=\la\,\mu_1+(1-\la)\,\mu_2$ 
 satisfies $\rho(\nu)=\la\,h_1+(1-\la)\,h_2$.
 Hence
 \begin{align*}
 \beta\bigl(\la\,h_1\!+\!(1-\la)\, h_2\bigr)
    \le A_L\bigl(\la\,\mu_1\!+\!(1-\la)\, \mu_2\bigr)
    =\la\,\beta(h_1)+(1-\la)\,\beta(h_2).
 \end{align*}
 
 By proposition~\ref{RhoSurj}, $\rho$ is surjective, and hence $\be$
 is finite.
 By proposition~\ref{T.1} on the appendix, $\a$ is superlinear.
 By~\ref{T.1}, $\a$ and $\be$ are convex. Formula~\eqref{alfa},
 implies that $\a$ is finite and then by~\ref{T.1}, $\be$ is
 superlinear.
 \end{proof}
 
 \bigskip
 
    For $h\in H_1(M,\re)$ and $\om\in H^1(M,\re)$, write
  \begin{align*}
  \cM_h(L)&:=\{\,\mu\in\cM(L)\;|\;\rho(\mu)=h,\;A_L(\mu)=\be(h)\,\}\,,
  \\
  \cM^{\om}(L)&:=\{\,\mu\in\cM(L)\;|\;A_{L-\om}(\mu)=-c(L-\om)\,\}.
  \end{align*}
 
  Since the $\be$-function has a supporting hyperplane at each
  homology class $h$, if $\om\in\partial\be(h)$, then
  $\cM_h(L)\subseteq\cM^\om(L)$. Conversely, since by 
  corollary~\ref{T.2} $\a^*=\be$, then $\cM^\om(L)\subseteq\cM_h(L)$ 
  if $h\in\partial\a(\om)$. Thus
  $$
  \tcup_{h\in\Ho}\back\cM_h(L)\;\;
  =\!\!\!\tcup_{\om\in\Co}\back\cM^\om(L).
  $$
  We call these measures {\it Mather minimizing} measures
  \index{Mather!minimizing measures}
  \index{measure!Mather minimizing}
  and the set
%   $$
%   \cM:=
%   \bigcup_{\om\in H^1(M,\re)}\bigcup_{\mu\in\cM^\om(L)}\supp(\mu)
%   =\bigcup_{h\in H_1(M,\re)}\bigcup_{\mu\in\cM_h(L)}\supp(\mu)
%   $$
   $$
   \cM:=\cM^0(L)=\{\mu\in\cM(L)\,\vert\,A_{L}(\mu)=-c(L)\,\},
   $$
  the {\it Mather set}.
  \index{$\cM$}
  \index{Mather!set}
  \index{set!Mather}
  
  Define the {\it strict critical value\/}\label{strict} as
  \index{critical value!strict ($c_0$)}
  \index{strict critical value ($c_0$)}
  \index{$c_0(L)$}
  \begin{equation}\label{E:c0}
  \begin{aligned}
  c_0(L):&=\min_{\om\in\Co}c(L-\om) 
         =\min_{\om\in\Co}\a(\om)  \\
         &=-\be(0).
  \end{aligned}
  \end{equation}
  By corollary~\ref{E(mu)} the strict critical value is 
  the lowest energy level which supports Mather minimizing measures 
  and since 
  $c_0(L)=-\be(0)$, these minimal energy Mather minimizing measures
  have trivial homology.
  

 \index{Mather!theory|)}
 \index{Aubry-Mather theory|)}



 \newpage
  
 \section{Coverings.}
 \index{covering|(}
 
 
 We shall deal mainly with compact manifolds $M$, but there are some
 important non-compact cases, for example the coverings of $M$. 
 Particularly interesting are the abelian cover $\hM$, the universal
 cover $\tM$ and the finite coverings.
 
  The {\it abelian cover}
 \index{abelian cover}
 \index{covering!abelian}
 $\hM$ of $M$ is the covering whose fundamental
 group is the kernel of the Hurewicz homomorphism 
 \index{Hurewicz homomorphism}
 $\pi_1(M)\to H_1(M,\Z)$. Its deck transformation group is
 $H_1(M,\Z)$ and $H^1(\tM,\Z)=\{0\}$. 
 When $\pi_1(M)$ is abelian, $\hM=\tM$.
 A closed curve in $\hM$ projects to a closed curve
 in $M$ with trivial homology.
 
 If $M_1\overset{p}\longrightarrow M$ is a covering,
 denote by $L_1:=L\circ dp:TM_1\to\re$ the lifted lagrangian to
 $TM_1$.
 
 \begin{lemma}\label{CoverIneq}
 If  $M_1\overset{p}\longrightarrow M$ is a covering, 
 then $c(L_1)\le c(L)$.
 \index{critical value!of a covering}
 \end{lemma}
 
 \begin{proof}
 The lemma follows form the fact that closed curves on $N$ project to
 closed curves on $M$.
 \end{proof}
 
 \begin{proposition}
 If $M_{1}$ is a {\it finite} covering of $M_{2}$ then
 $c(L_{1})= c(L_{2}). $  \label{igual}\index{covering!finite}
 \end{proposition}
 
 \begin{proof}
  We know that $c(L_{1})\leq c(L_{2})$. Suppose that
 the strict inequality holds and let $k$ be such that
 $c(L_{1})<k<c(L_{2})$. Hence there exists a closed curve
 $\ga$ in $M_{2}$ with negative $(L_{2}+k)$-action. Since $M_{1}$
 is a finite covering of $M_{2}$ some iterate of $\ga$ lifts to
 a closed curve in $M_{1}$ with negative $(L_{1}+k)$-action
 which contradicts $c(L_{1})<k$.
 \end{proof}
 


 \begin{proposition}{\bf \cite{PP4}}\label{P.ca}
  \index{critical value!of the abelian cover $(c_a)$}  
  \index{$c_a(L)$}
  \index{$c_0(L)$}\index{strict critical value ($c_0$)}
  \index{critical value!strict ($c_0$)}
  $$
  c_0(L)=c_a(L)=\text{critical value of the abelian cover.}
  $$
  \end{proposition}

   Then we have
  $$
  c_u(L)\le c_a(L)=c_0(L)\le c(L-\om)\qquad\forall [\om]\in\Co,
  $$
  \openup -0.3pt
  where $c_u$
  \index{$c_u(L)$}
  \index{critical value!of the universal cover}
  \index{covering!universal}
   is the critical value of the lift of 
  the lagrangian to the universal cover.  
  When $c_u(L)<c_0(L)$, the method in equation~\eqref{e.e.2}
  gives some minimizing measures which are not Mather minimizing.
  For symmetric lagrangians $c(L)=e_0=c_0(L)=c_u(L)$.
  \index{symmetric lagrangian}
  \index{lagrangian!symmetric}
  Ma\~n\'e~\cite{Ma7} gives an example 
  \index{example!$c(L)>c_0(L)>e_0$}in which $e_0<c_a(L)=c_0(L)<c(L)$.
  G. Paternain and M. Paternain~\cite{PP4} give an example in which
  \index{example!$c_0(L)>c_u(L)$}
  $c_u(L)<c_a(L)$.
 




  
  \begin{proof}
  Let $\om$ be a closed form in $M$. 
  Since $H_1(M,\re)=\{0\}$, the lift $\widehat{\om}$ 
  of $\om$ to $\hM$ is exact, then
  $$
  c_a(L):=c(\hL)=c(\hL-\widehat{\om})\le c(L-\om).
  $$
  Hence
  $$
  c_a(L)\le\min_{\om\in\Co}c(L-\om)=c_0(L).
  $$
  Moreover,
  \begin{align*}
  -c_a(L)&=\inf\Big\{\,A_\hL(\mu)\,|\,\mu\in\ov{\C(\hM)}\,\Big\}
         \\
         &=\inf\{\,A_\hL(\mu_{\ga})\,|\,
          \mu_\ga\in\C(\hM)\,\}
          &\text{by proposition~\ref{MuLim},}
         \\
         &=\inf\{\,A_L(\mu_\ga)\,|\,\mu_\ga\in\CM,\,\rho(\mu)=0\,\},
  \end{align*}       
   because a closed curve $\ga$ on $M$ has homology $[\ga]=0$ if and only if
   it has a closed lift to $\hM$. 
   Then
%   because if $\mu\in\ov{\CM}$, by proposition~\ref{MuLim}
%   there are $\mu_{\ga_n}\in\CM$ with $A_\hL(\mu_{\ga_n})\to
%   A_\hL(\mu)$. Since $\ga_n$ is closed, 
%   its projection $p\circ\ga_n$
%   has homology $[p\circ\ga_n]=0$. Then
%   $p^*(\mu_{\ga_n})=\mu_{p\circ\ga_n}\to p^*\mu$,
%   $A_L(p^*\mu_{\ga_n})\to A_L(p^*\mu)=A_\hL(\mu)$ 
%   and by proposition~\ref{RhoCont}
%   $\rho(p^*\mu)=\lim_n\rho(p^*\mu_{\ga_n})=0$. 
%   
%   Since
%   $\cM(L)\subseteq\ov{\cM}$, we get that    
  \begin{align*}
    -c_a(L)&=\inf\{\,A_L(\mu)\,|\,\mu\in\ov{\CM},\,\rho(\mu)=0\,\},
    &\text{by~\ref{MuLim} and~\ref{RhoCont}}
     \\
     &\le\min\{\,A_L(\mu)\,|\,\mu\in\cM(L),\,\rho(\mu)=0\,\}
      &\text{because $\cM(L)\subset\ov{\CM}$}
           \\
           &=-\be(0)=c_0(L). 
  \end{align*}
  \end{proof}
 
 
  The {\it real abelian cover} 
  \index{abelian cover!real --}
  \index{covering!real abelian}
  \index{real abelian cover}
  is the covering $\kM$ of $M$
  with $h:\pi(M)\to H_1(M,\re)$ 
  is the Hurewicz homomorphism.
  \index{Hurewicz homomorphism}
  It is an intermediate covering
  $\tM\to\kM\to M$ and the deck transformations
  of $\hM\to\kM$ are given by the torsion\footnote{i.e.
  the elements of finite order
  $\Z_{n_1}\oplus\cdots\oplus\Z_{n_k}\subseteq H_1(M,\Z)$.}
  \index{torsion}
  of $H_1(M,\Z)$. Hence $\hM\to\kM$ is a finite cover
  so that they have the same critical value $c_a(L)=c_0(L)$.

  \openup 0.3pt

   \index{covering|)}
   
%   \newpage
   
%    \section{Symplectomorphisms.}
%    
%    The critical value >of the universal cover? $c_u$
%    is invariant under symplectomorphisms.

  \chapter{Globally minimizing orbits.}
 
  
 
 \section{Tonelli's theorem.}
 
  Given $x,\,y\in M$ and $T>0$, let
  $$
  \C_T(x,y):=\bigl\{\,\ga\in\Cac([0,T],M)\;|\;
  \ga(0)=x,\;\ga(T)=y\,\bigr\}.
  \label{CT}\index{$C_T(x,y)$}
  $$
  We say that $\ga\in\C_T(x,y)$ is a {\it Tonelli
  minimizer\/}\label{TonMin}
  \index{Tonelli!minimizer}
  if 
  $$
  A_L(\ga)=\min_{\eta\in\C_T(x,y)}A_L(\eta)\,.
  $$

  In this section we shall prove

  \begin{Tonellitheorem}\label{TonTh}
  \index{theorem!Tonelli's}
  \index{Tonelli!theorem}\quad\newline
  For all $x,\,y\in M$ and $T>0$ there exists
  a Tonelli minimizer on $\C_T(x,y)$.
  \end{Tonellitheorem}

  The only difference in the proof of this theorem when $M$ is
  noncompact is corollary~\ref{C.Fc}. An independent proof of
  this corollary is given in remark~\ref{R.Fc}.

  The idea of Tonelli's theorem is to prove that the sets
  \begin{equation}\label{Ac}
  \cA(c):=\bigl\{\, \ga\in\C_T(x,y)\;|\; A_L(\ga)\le c\;\}
  \end{equation}
  are compact in the $C^0$-topology. 
  Then a Tonelli minimizer will be a curve in
  $$
  {\textstyle\bigcap\limits_{c\ge\a}}\cA(c)\ne \nulo\,,
  $$
  where $\a=\inf_{\eta\in\C_T(x,y)}A_L(\eta)\ge\inf L  > -\infty$.
  
  An addendum to Tonelli's theorem due to Mather~\cite{Mat5}
  states that these sets are compact in the topology of 
  absolutely continuous curves. 
  Given $\ga_1,\,\ga_2\in\Cac([0,T],M)$ define their
  {\it absolutely continuous distance}
  \index{$d_1(\ga_1,\ga_2)$}
  \index{absolutely continuous!metric}
  \index{metric!absolutely continuous}
  \index{distance!absolutely continuous}
  by
  $$
  d_1(\ga_1,\ga_2):=\!\!\!\sup_{t\in[0,T]}\!\!
  d_M\big(\ga_1(t),\ga_2(t)\big)
  +\int_0^T\!\!\!\!\!d_{TM}\big(
  [\ga_1(t),\dga_1(t)],[\ga_2(t),\dga_2(t)]\big)\,dt.
  $$

  \begin{theorem}[Mather~\cite{Mat5}]
  \label{MaTon}
  For any $x,\,y\in M$, $T>0$, $b\in\re$, the set
  $$
  \cA(b):=\big\{\,\ga\in\C_T(x,y)\,|\,A_L(\ga)\le b\,\big\}
  $$
  is compact in the $C^0$-topology.
  \end{theorem}  
  
  This theorem is proved in~\ref{TonProof}. We quote here
  the following proposition (addendum on page 175 of Mather~\cite{Mat5}).

  


 \begin{proposition}{(\bf Mather~\cite{Mat5})}\label{AdTon}
 \quad\newline\indent
 If $N\subseteq M$ is a compact subset and 
 $\ga_1,\,\ga_2,\,\ldots$ is a sequence in
 $\Cac([a,b],N)$ which converges $C^0$ to $\ga$ 
 and $A_L(\ga_i)$ converges to $A_L(\ga)$,
 then $\ga_1,\,\ga_2,\ldots$ converges in 
 the $d_1$-topology to $\ga$.
 \end{proposition}

 In fact, the set $\cA(b)$ in proposition~\ref{MaTon}
 is not compact in the $d_1$-topology as the following
 example shows. Since the action functional is only lower
 semicontinuous on $\C_T(x,y)$ a priori it is not possible to
 ensure the convergence of the action in a $C^0$-convergent
 sequence obtained from theorem~\ref{MaTon}.
 Unless, for example, if the action is converging to the
 minimal action (e.g. proposition~\ref{anc}).
 
 \newpage
 
 \begin{example}{\it $\cA(b)$ is not compact in the $d_1$-topology.}
 
 Let $L=\tfrac 12 |v|^2$ be the riemannian lagrangian on $\re^2$
 with the flat metric. Let $\eta(t)=(t,0)$ and
 $\ga_n(t)=(t, \tfrac 1n \sin( 2\pi n t))$, $t\in[0,1]$. 
 The action $A_L(\ga_n)=1+2\pi^2$ is bounded. $\ga_n\to\eta$
 in the $C^0$-topology. 
 
 The length of $\ga_n$ is bounded below by a polygonal
 curve joining the maxima and minimums of its second component.
 Hence
 \break
  $\ell(\ga_n)\ge 2n\,\sqrt{ 4/{n^2}+1/({4n^2})}>4$.
 Therefore $\ell(\ga_n)\not\to\ell(\eta)$, and thus
 $\ga_n\not\to\eta$ in the $d_1$-topology.
 Moreover, since a reparametrization preserves length,
 there is no reparametrization of the $\ga_n$'s
 which converges in the $d_1$ topology to $\eta$.
 \end{example} 
 
 
  \bigskip
  
  We shall split the proof of Tonelli's theorem in several parts:
  
  \begin{Definition}
  \index{absolutely equicontinuous}
  \index{equicontinuous!absolutely}
  \label{AbsEqCont}
  \quad\newline\indent
  A family $\cF\subseteq C^0([a,b],M)$ is {\it absolutely
  equicontinuous\/} if  $\forall\;\e>0$ $\;\exists\;\de>0$ 
  such that
  $$
  \tsum_{i=1}^N |t_i-s_i|<\de \quad\then\quad 
  \tsum_{i=1}^N d(x_{s_i},x_{t_i})<\e,
  $$
  whenever $]s_1,t_1[,\ldots,]s_N,t_N[$ are disjoint intervals in
  $[a,b]$.  
  \end{Definition}
  
  \begin{remark}\quad
  \begin{itemize}
  \item[(i)] An absolutely equicontinuous family is equicontinuous.
  \item[(ii)] A uniform limit of absolutely equicontinuous
              functions is absolutely continuous.
  \end{itemize}
  \end{remark}
  
  \begin{lemma}\label{L.Fc}
  For all $c\in\re$ and $T>0$, the family
  $$
  \cF(c):=\bigl\{\,\ga\in\Cac([0,T],M)\;|\;A_L(\ga)\le c\,\bigr\}
  $$
  is absolutely equicontinuous.
  \end{lemma}
  
  \begin{proof}
  Since by the superlinearity, the lagrangian $L$ is bounded below;
  by adding a constant we may assume that $L\ge 0$. For $a>0$ let
  \begin{equation}\label{Ka}
  K(a):=\inf\Big\{\,\frac{L(x,v)}{|v|}\;\Big|\;
  (x,v)\in TM,\,|v|\ge a\,\Big\}.
  \end{equation}
  The superlinearity implies that $\lim_{a\to+\infty}K(a)=+\infty$.
  Given $\e>0$ let $a>0$ be such that 
  $$
  \frac{c}{K(a)}<\frac{\e}2.
  $$
  Let $0\le s_1<t_1\le\cdots\le s_N<t_N\le T$, 
  $J:=\cup_{i=1}^N[s_i,t_i]$ and $E:=J\cap\bigl[|\dx|>a\bigr]$, 
  then $L(x_s,\dx_s)\ge K(a)\,|\dx_s|$ for $s\in E$. We have that 
  \begin{align*}
  K(a)\; \tsum_{i=1}^N d(x_s,x_t)
    &\le K(a) \int_E|\dx|+K(a)\int_{J\setminus E}|\dx|
    \\
    &\le\int_EL(x,\dx)+a\cdot K(a)\;m(J)
    \\
    &\le c +a\cdot K(a)\;m(J),
     \qquad\text{(because $L\ge 0$),}
  \end{align*}
  where $m$ is the Lebesgue measure on $[0,T]$
  \begin{equation}\label{blong}
  \tsum_{i=1}^N d(x_{s_i},x_{t_i})\le \tint_J |\dx_s|
            \le\tfrac{c}{K(a)}+a\;m(J)
            \le\tfrac{\e}2+a\;m(J).   
  \end{equation}
  This implies the absolute equicontinuity of $\cF(c)$.
  
  \end{proof}

  In order to apply the Arzela-Ascoli theorem we need a compact 
  range, for this we have:
  
  \begin{corollary}\label{C.Fc}
  For all $c\in\re$ and $T>0$ there is $R>0$ such that for all
  $x,\,y\in M$,
  $$
  \cA(c)\subseteq \Cac\bigl([0,T],\ov{B}(x,R)\bigr),
  $$
  where $\ov{B}(x,R):=\{\,z\in M\;|\;d_M(x,z)\le R\,\}$.
  \end{corollary}
  
  \begin{proof}
  Inequality~\eqref{blong} for $N=1$ and $J=[s,t]$ is
  $d(x_s,x_t)\le\tfrac\e2+a\,|t-s|$. 
  It is enough to take $R=\tfrac{\e}2+a\, T$.
  \end{proof}
  
 
  
  \begin{remark}\label{R.Fc}\quad
  \newline\indent
  {\small
  Corollary~\ref{C.Fc} is the only difference for the proof
  of Tonelli's theorem when $M$ is non-compact. Another proof
  for corollary~\ref{C.Fc} is the following:
  
  Adding a constant we may assume that $L\ge 0$. There is $B>0$
  such that $L(x,v)\ge |v| - B$ for all $(x,v)\in TM$. Then
  for $0\le s\le t\le T$, we have that
  $$
  d(x_s,x_t)\le \int_s^t|\dx|\le BT+\int_s^tL(x,\dx)\le BT+c.
  $$
  \qed}
  \end{remark}
    
  \bigskip
  
  Recall that
  $$
  \cF(c):=\bigl\{\,\ga\in\Cac([0,T],M)\;|\;A_L(\ga)\le c\,\bigr\}
  $$
  \begin{theorem}\label{Cle}\quad
  
  If $\ga_n\in\cF(c)$ and 
  $\ga_n\to\ga$ in the uniform topology,
   then $\ga\in\cF(c)$.
  \end{theorem}
  
  We shall need the following lemma.  We may assume that
  $M=\re^n$,
  
  
  \begin{lemma}\label{LemCle}
  Given $K$ compact, $a>0$ and $\e>0$, there exists $\de>0$
  such that if $x\in K$, $\lv x-y\rv\le\de$, $|v|\le a$ and $w\in\re^n$,
  then
  \begin{equation}\label{E.LemCle}
  L(y,w)\ge L(x,v)+L_v(x,v)\;(w-v) -\e.
  \end{equation}
  \end{lemma}
  
  \bigskip
  
  \noindent{\bf Proof of theorem~\ref{Cle}:}
  
  It is immediate from the definition~\ref{AbsEqCont} that a uniform
  limit of absolutely equicontinuous curves is absolutely continuous.
  We may assume that $\ga_n([0,T])$ is contained in a compact
  neighbourhood $K$ of $\ga([0,T])$.
  By the superlinearity we may assume that $L\ge 0$.
  Let $\e>0$ and $E=\bigl[|\dga|\le a\bigr]$, then
  by lemma~\ref{LemCle}, for $n$ large,
  \begin{equation}\label{ecle}
  \int_{E} 
  \bigl[ L(\dga)+L_v(\dga)(\dga_n-\dga)-\e\bigr]
  \le\int_{E}L(\dga_n)\le c
  \qquad\text{(since $L\ge0$).}
  \end{equation}
      
  \noindent{\bf Claim:}
  %\qquad \qquad $\displaystyle  
   $$\lim_n\int_{E}L_v(\dga)\;(\dga_n-\dga)=0.
   $$
   

  Letting $n\to+\infty$ on~\eqref{ecle}, we have that
  $$
  \int_{E}L(\dga)-\e\; T\le c.
  $$
  Since $E\uparrow [0,T]$ when $a\to+\infty$ and $L\ge 0$, then
  $$
  \int_0^TL(\dga)
  =\lim_{a\to+\infty}\int_{E}L(\dga)
  \le c+\e\; T.
  $$
  Now let $\e\to 0$.

 %\qed



   We now prove the claim.   
   We have to prove that 
   \begin{equation}\label{psint}
    \lim_n\int \psi\,\dz_n = 0,
   \end{equation}
   where $\psi=L_v(\dga)\cdot 1_E$ is bounded, $z_n=\ga_n-\ga$ and $1_E$ is the characteristic
   function of $E$. Since $\lV z_n\rV_\infty\to 0$,  if $A=[a,b]$ is an interval then
   \begin{equation}\label{zInt0}
   \lim_n\int_A\dz_n=\lim_n z_n\Big\vert_a^b = 0.
   \end{equation}
   Since $\ga$ is absolutely continuous, $\dga\in\cL^1$. 
   From $\dga\in\cL^1$ and  inequality \eqref{blong}, given $\e>0$ there is $\de>0$ such that 
   \begin{equation}\label{mDe}
   m(D)<\de \quad\Longrightarrow\quad \int_D\lv\dz_n\rv<\e,\qquad\forall n\in\na.
   \end{equation}
   Equations \eqref{zInt0} and \eqref{mDe} imply that \eqref{zInt0} holds for any 
   Borel set $A$, approximating $A$ by a finite union of intervals.
   Hence  \eqref{psint} holds if $\psi$ is  a simple function $\sum_k a_k\,1_{A_k}$.
  
   Let $f$ be a simple function such that $\lV f\rV_\infty\le 2\lV\psi\rV_\infty$ 
   and $\int\lv\psi-f\rv<\e^2$.   Let $B= [\lv\psi-f\rv>\e]$, then $m(B)<\e$.
   $$
   \lv\int(\psi-f)\,\dz_n\rv\le\e\,\int_{B^c}\lv\dz_n\rv+3\lV\psi\rV_\infty\int_B\lv\dz_n\rv
   \le h(\e).
   $$
   By \eqref{mDe} one can take $h(\e)$ with $\lim_\e h(\e)=0$. Then
   $$
   \lv\lim_n\int\psi\,\dz_n\rv
   \le \lim_n\lv\int f\,\dz_n\rv
   +\limsup_n\lv\int(\psi-f)\,\dz_n\rv
   \le 0 + h(\e)
   \overset{\e}\longrightarrow 0.
   $$
    \qed




  
%   \begin{lemma}\label{cL1}
% 
%   If $z_n\in C^0([0,T],\re^d)$ is an absolutely equicontinuous family such that
%   $\lV z_n\rV_\infty\to 0$ then $\dz_n\to 0$ in $\cL^1$.
%   \end{lemma}
%   
%   \begin{proof}
%   {\bf Using $\psi=\tfrac{\dz_n}{|\dz_n|}$, it is enough to prove that for any bounded measurable function $\psi$}
%   $$
%   \lim_n\int\psi \,\dot z_n =0.
%   $$
%   
%   Given $\e>0$ we will construct a Borel set $Q\subset[0,T]$ and 
%   $\psi_\e\in C^1([0,T],\re^d)$ such that $m(Q^c)<\e$, 
%   $\lV\psi_\e\rV_\infty\le\lV\psi\rV_\infty$ and
%   $$
%   \lv\psi_\e(t)-\psi(t)\rv\le\e \qquad \forall\, t\in Q.
%   $$
%   With this,
%   \begin{align*}
% \lv  \int\psi_\e\,\dz_n\rv
% &\le\lv\psi_\e\,z_n\big\vert_0^T -\int\dot\psi_\e\,z_n\;\rv\\
% &\le \big(\lV\psi_\e\rV_\infty+\Vert\dot\psi_\e\Vert_\infty\big)\,T\,\lV z_n\rV_\infty
% \overset{n}{\longrightarrow} 0
%   \end{align*}
%   By inequality \eqref{blong}, given $\de>0$ there is $\rho>0$ such that 
%   \begin{equation*}\label{zUnifInt}
%   m(D)<\rho \quad\Longrightarrow\quad \int_D\lv\dz_n\rv <\de.
%   \end{equation*}
%   Then 
%   $$
%   \lv\int(\psi-\psi_\e)\,\dz_n\rv\le \e\,\int_Q\lv\dz_n\rv
%   +2\,\lV\psi\rV_\infty\int_{Q^c}\lv\dz_n\rv\le f(\e)\overset{\e}\longrightarrow 0
%   $$
%   (uniformly in $n$). Therefore
%   $$
%   \limsup_n\lv\int\psi\,\dz_n\rv\le\lim_n\int\psi_\e\,\lv\dz_n\rv+f(\e)= f(\e)\overset{\e}\longrightarrow 0.
%   $$
%   
%   Now we construct  $Q$ and $\psi_\e$. By Lusin theorem there is a compact $Q\subset[0,T]$
%   such that $\psi\vert_Q$ is continuous and $m(Q^c)<\e$.
%   
%   There is a continuous function $\phi\in C^0([0,t],\re^d)$ such that $\phi|_Q=\psi|_Q$. 
%   Indeed, the complement $Q^c=[0,T]\setminus Q$ is a union of intervals and it is 
%   enough to extend $\psi|_Q$ linearly on each maximal interval $]a,b[$ in $Q^c$, i.e.
%   $$
%   \phi(s)= \tfrac{b-s}{b-a}\,\psi(a)+\tfrac{s-a}{b-a}\, \psi(b), \qquad s\in ]a,b[.
%   $$
%   Also $\lV\phi\rV_\infty\le\lV\psi|_Q\rV_\infty$.
%   Let $\de>0$ be such that 
%   \begin{equation}
%   \lv s-t\rv\le \de \quad \Longrightarrow \quad \lv\phi(s)-\phi(t)\rv<\e.
%   \label{PhiUnifCont}
%   \end{equation}
%   Let $k\in C^\infty(\re,[0,1])$ be such that $\int_\re k= 1$, $k\big\vert_{[-1,1]^c}\equiv 0$.
%   Let 
%   $$
%   \psi_\e(t):=\int\frac 1\de\;  k\!\left(\tfrac{s-t}\de\right)\,\phi(s)\;ds.
%   $$
%   Observe that $f_t(s):=\frac 1\de\;k\left(\frac{s-t}\de\right)$ is a probability density supported 
%   on $[t-\de,t+\de]$ and that  $\psi_\e$ is $C^\infty$. If $t\in Q$ then $\psi(t)=\phi(t)$ 
%   and by \eqref{PhiUnifCont},
%   $$
%   \lv\psi_\e(t)-\psi(t)\rv = \lv\int_\re f_t(s)\,\big[\phi(s)-\phi(t)\big]\;ds\rv\le \e,
%   \qquad t\in Q.
%   $$
%   Also $\lV\psi_\e\rV_\infty\le\lV\phi\rV_\infty\le\lV\psi\rV_\infty$.
%   
%   \end{proof}
%   
%   
  
  
%   We now prove the claim.
%   By inequality~\eqref{blong}, for $\e>0$ there is $\de>0$ such that
%   \begin{equation}\label{GaUnifInt}
%   m(D)<\de \;\;\Longrightarrow\; \int_D|\dga_n| <\e.
%   \end{equation}
%   where $m$ is the Lebesgue measure.
%   If $A=[s,t]$, then
%   \begin{equation}\label{GaInt0}
%   \lim_{n\to\infty}\int_A(\dga_n-\dga)
%       =\lim_{n\to\infty}(\ga_n-\ga)\Big\vert_s^t=0.
%   \end{equation}
%   Equations~\eqref{GaUnifInt} and~\eqref{GaInt0} imply
%   that~\eqref{GaInt0} holds for any Borel set $A$,
%   approximating $A$ by finite unions of intervals.
%   Then
%   $$
%   \lim_n \int_{[0,T]}(\dga_n-\dga)\;\psi = 0,
%   $$
%   for any $\psi\in\cL^\infty$.
%   This implies  that $\dga_n\to\dga$ in $\cL^1$.
%   \qed
  

   \openup -1pt

  \bigskip
   



  %\newpage 
   
  \refstepcounter{Thm}\label{TonProof}   
  \noindent{\bf \ref{TonProof}.\quad Proof of Tonelli's theorem:}
  
  By lemma~\ref{L.Fc}, the family $\cA(c)$ in~\eqref{Ac} is
  equicontinuous, and by corollary~\ref{C.Fc}, the curves in
  $\cA(c)$ have a uniform compact range. By Arzel\'a-Ascoli's
  theorem and theorem~\ref{Cle}, $\cA(c)$ is compact. Then
  $$
  \ga\in\bigcap_{c\ge\inf_{\C_T(x,y)}A_L}\cA(c)
  $$
  is a Tonelli's minimizer on $\C_T(x,y)$.
  \qed  
  
  \bigskip
  
  \noindent{\bf Proof of lemma~\ref{LemCle}:}
  
   Let
   \begin{align*}
   C_1&:=\sup\bigl\{\, |L_v(x,v)|\;|\;x\in K,\;|v|\le a\,\bigr\}
      \\
   C_2&:=\sup\bigl\{\,L(x,v)-L_v(x,v)\cdot v\;|\;x\in K,\;|v|\le
   a\,\bigr\}.
   \end{align*}
   
   Let $b>0$ be such that 
   $$
   K(b)\cdot r \ge C_2+C_1\; r \qquad\text{ for all }r\ge b,
   $$
   where $K(b)$ is from~\eqref{Ka}. Then if $y\in M$ and $|w|>b$,
   \begin{align*}
   L(y,w) 
       &\ge K(b)\; |w|      \\
       &\ge C_2+C_1\;|w|    \\
       &\ge C_2+L_v(x,v)\cdot w  \\
       &\ge L(x,v)+L_v(x,v)\cdot(w-v) 
              \qquad\text{ for }|w|\ge b.
   \end{align*} 
   This gives \eqref{E.LemCle} when $|w|\ge b$.
   Since $L$ is convex,
   $$
   L(x,w)\ge L(x,v) +L_v(x,v)\cdot(w-v)  \qquad\forall\, w\in\re^n.
   $$
   Then there is $\de>0$ such that for $|x-y|\le\de$, $|v|\le a$
   and $|w|\le b$ inequality~\eqref{E.LemCle} holds.
   
   \qed
   
   
   
   
   \openup 1pt
   
 \newpage
 
 
 \section{A priori compactness.}
 
 The following lemma, due to Mather~\cite{Mat5} for Tonelli
 minimizers in the non-autonomous case, 
 will be very useful. 
 In the autonomous case its proof is very
 simple.
 
 \begin{lemma}\label{MatLem}\quad
 \newline\indent
 For $C>0$ there exists $A=A(C)>0$ such that if 
 $x,\,y\in M$ and $\ga\in\C_T(x,y)$ is a
 solution of the Euler-Lagrange equation
 with $A_L(\ga)\le C\,T$, then
 $|\dga(t)|<A$ for all $t\in[0,T]$.
 \end{lemma}
 \index{bound!a priori -- for speed}
 \index{priori@a priori bound for speed}
 

 \begin{proof}
 By the superlinearity there is $D>0$ such that $L(x,v)\ge |v|-D$
 for all $(x,v)\in TM$. Since $A_L(\ga)\le C\, T$, the mean value
 theorem implies that there is $t_0\in ]0,T[$ such that
 $$
 |\dga(t_0)|\le D+C.
 $$
 The conservation of the energy and the uniform bounds~\eqref{E<} 
 and~\eqref{E>} imply that there is $A=A(C)>0$ such that
 $|\dga|\le A$.

 \end{proof}
 
  \medskip
 
  For $k\ge c(L)$ and $x,\,y\in M$, define
  $$
  \Phi_k(x,y;T):=\inf_{\ga\in\C_T(x,y)}A_{L+k}(\ga).
  \label{PhiT}\index{$\Phi_k(x,y;T)$}
  \index{action!potential!of finite time}
  \index{finite! -- time potential}
  $$    

 
 \begin{corollary}\label{TonPri}
  Given $\e>0$, there are constants $A(\e),\,B(\e),\,C(\e)>0$
 such that if $T\ge \e$, $d(x,y)<R$ and $k\in\re$, then
 \begin{itemize}
 \item[\quad (i)] $\;\;\Phi_k(x,y;T)\le [C(\e,R)+k]\;T$.
 \item[\quad (ii)] If $\ga\in\C_T(x,y)$ is a solution of the
                   Euler-Lagrange equation such that 
                   $A_{L+k}(\ga)\le [C(\e,R)+k]\,T+1$, then 
                   $|\dga|\le A(\e,R)$ and $E(\ga,\dga)\le B(\e,R)$.
 \end{itemize}
 \end{corollary}
 
 \begin{proof}
 Comparing with the action of a geodesic on $\C_T(x,y)$, we 
 get (i) with 
 $$
 C(\e,R)=\sup\bigl\{\,L(v)\;\big|\;|v|\le \tfrac{d(x,y)}{\e}
 \le\tfrac R\e\,\bigr\}.
 $$
 Using (i) and lemma~\ref{MatLem},we obtain $A(\e,R)$. Using $A(\e,R)$
 and inequality~\eqref{E<} we obtain $B(\e,R)$.
 \end{proof}

 
 \medskip
 
 \begin{lemma}\label{priori}\quad
 \index{bound!a priori -- for speed}
 \index{priori@a priori bound for speed}
 \newline\indent
 There exists $A>0$ such that if $x,\,y\in M$ and $\ga\in\C_T(x,y)$ is
 a solution of the Euler-Lagrange equation with 
 $$
 A_{L+c}(\ga) \le \Phi_c(x,y) + d_M(x,y),
 $$
 then
 \parbox[t][1.25cm][b]{6cm}{
 \openup -5pt
 \begin{itemize}
 \item[(a)]\;\; $T\,>\,\tfrac 1A\; d_M(x,y)$.
 \item[(b)]\;\; $|\dga(t)|\,<\,A$ for all $t\in[0,T]$.
 \end{itemize}
 \openup 5pt
 }
  \end{lemma}

 \begin{proof}
 Let $\eta:[0,d(x,y)]\to M$ be a minimal geodesic with 
 $|\deta|\equiv 1$. Let $\ell(r)$ be from~\eqref{ell} and
 $D=\ell(1)+c +2$. From the superlinearity condition there is 
 $B>0$ such that
 $$
 L(x,v) + c > D\, |v| - B, \qquad \forall (x,v)\in TM.
 $$
 Then
 \begin{align}
 [\ell(1)+c]\;d(x,y) 
    &\ge A_{L+c}(\eta) \ge \Phi_c(x,y)         
    \label{ap-1}\\
    &\ge A_{L+c}(\ga)-d(x,y)
    \label{ap-2}\\
    &\ge\int_0^T\bigl( D\;|\dga|-B\,\bigr)\,dt - d(x,y)
    \notag\\
    &\ge D\;d(x,y) - B\,T - d(x,y).
    \notag
 \end{align}
 Hence
 $$
 T \ge \tfrac{D-\ell-c-1}{B}\; d(x,y) \ge \tfrac 1B \; d(x,y).
 $$
 From~\eqref{ap-1} and~\eqref{ap-2}, we get that
 \begin{align*}
 A_L(\ga)&\le \bigl[\,\ell(1)+c+1\,\bigr]\;d(x,y)-c\,T ,
        \\
         &\le \bigl\{\,B\,[\,\ell(1)+c+1\,]-c\,\bigr\}\;T.
 \end{align*}
 Then lemma~\ref{MatLem} completes the proof.
 \end{proof}

 \newpage
 
 \section{Energy of time-free minimizers.}
 
 A curve $\ga\in\C(x,y)$ is a {\it global minimizer\/}
 or {\it time free } minimizer for $L+k$ if
 $k\ge c(L)$ and $A_{L+k}(\ga)=\Phi_k(x,y)$.
 
 \begin{proposition}\label{EGM}
 \index{energy!of a time-free minimizer}
 \index{time-free minimizer!energy of}
 A time-free minimizer for $L+k$ has energy $E\equiv k$.
 \end{proposition} 
 
 
  We need the following
 
 \begin{lemma}\label{Ltime1}
 Let $:x:[0,T]\to M$ be an absolutely continuous curve
 and $k\in\re$.
 For $\la>0$, let $x_\la(t):=x(\la t)$ and 
 $\cA(\la):=A_{L+k}(x_\la)$. Then
 $$
 \cA'(1)=\int_0^T\bigl[\,E(x,\dx)-k\,\bigr]\;dt.
 $$
 \end{lemma}
 
  \begin{proof}
  Since $\dx_\la(t)=\la\;\dx(\la t)$, then
  $$
  \cA(\la)=\int_0^{\frac T\la}
  \bigl[\,L(x(\la t),\,\la\,\dx(\la t))+k\,\bigr]
  \;dt.
  $$
  Differentiating $\cA(\la)$ and evaluating at $\la=1$, we have that
  $$
  \cA'(1)=-T\,\bigl[\,L(x(T),\dx(T))+k\,\bigr]
          +\int_0^T\bigl[\,L_x\,t\,\dx+L_v\,(\dx+t\,\ddx)\,\bigr]\;\,dt.
  $$
  Integrating by parts the term
  $(L_x\,\dx+L_v\ddx)\,t=\bigl(\tfrac{d\,}{dt}L\bigr)\,t$,
  we have that
    \begin{align*}
  \cA'(1)&=-T\,\bigl[\,L(x(T),\dx(T))+k\,\bigr]
          +L\,t\,\vert_0^T+\int_0^T\bigl(\,L_v\,\dx-L\bigr)\;dt
          \\
         &=-T\,k+\int_0^TE(x,\dx)\;dt
         =\int_0^T\bigl[\,E(x,\dx)-k\,\bigr]\;dt.
  \end{align*}
  \end{proof}
 
  %\medskip
  
  \noindent{\bf Proof of proposition~\ref{EGM}.}
  
    Since $\ga$ is a solution of the Euler-Lagrange equation
  its energy $E(\ga,\dga)$ is constant. Since it minimizes
  with free time, the derivative in lemma~\ref{Ltime1} must
  be zero. So that $E(\ga,\dga)\equiv k$.
  \qed
 
 
 %\medskip
 
 \begin{corollary}\label{Ltime2}\quad
 \newline\indent
 Let $x\in\C^{ac}([0,1],M)$ and $k>0$. For $T>0$, write
 $y_T(t)=x(\frac{t}T):[0,T]\to M$ and
 $\cB(T)=A_{L+k}(y_T)$. Then
 $$
 \cB'(T)=-\frac 1T\int_0^T\bigl[\,E(y_T,\dy_T)-k\,\bigr]\;dt.
 $$ 
 \end{corollary}
 
 \begin{proof}
 Using $\la=\frac{T}S$ on lemma~\ref{Ltime1}, we have that
 $\frac{d\,}{dS}=-\frac{T}{S^2}\;\frac{d\,}{d\la}$.
 Thus
 $$
 \tfrac{d\,}{dS}\big\vert_{S=T}\cB
 =-\frac 1T\;\tfrac{d\,}{d\la}\big\vert_{\la=1}\cA
 =-\frac 1T \int_0^T\bigl[\,E(y_T,\dy_T)-k\,\bigr]\;dt.
 $$ 
  \end{proof}




 \newpage
 
 \section{The finite-time potential.}
 
 Recall that if $k\ge c(L)$, $x,\,y\in M$  and $T>0$,
 $$
  \Phi_k(x,y;T):=\inf_{\ga\in\C_T(x,y)}A_{L+k}(\ga).
 $$    
 Here we shall prove the following proposition.
 See also corollary~\ref{PhiToH}.

 \begin{proposition}\label{P.PhiT}\quad
 \begin{enumerate}
 \item\label{P.PhiT.I1}
  For $k\in\re$, a compact subset $R\subseteq M$
  and $\e>0$, the function
 $(x,y,t)\mapsto\Phi_k(x,y;t)$ is Lipschitz on 
 $R\times R\times[\e,+\infty[$ for any $\e>0$. 
 \index{Lipschitz!property!finite-time potential's}

 \item\label{P.PhiT.I2}
  $\Phi_k(x,y;T)=\Phi_c(x,y;T)+(k-c)\,T$,
     \quad for $k\ge c(L)$,\;\;$x,\, y\in M$.
 \item\label{P.PhiT.I3}
 $\lim_{\e\to 0^+}\Phi_k(x,y;\e)
     =+\infty, \qquad\text{ for }\;\;k\ge c(L),\;\; x\ne y$.
     %\notag\\
 \item    
 $\lim_{T\to +\infty}\Phi_k(x,y;T) 
     =+\infty, \qquad\text{ for }\;\;k> c(L),\;\; x,y\in M.$  
     \label{P.PhiT.e1} 
 \item
  $\lim_{T\to +\infty}\Phi_k(x,y;T) 
     =-\infty, \qquad\text{ for }\;\;k< c(L),\;\; x,y\in M.  $
     \label{P.PhiT.e2}
 
 \item When $M$ is compact, the limits in items~\ref{P.PhiT.e1}
  and~\ref{P.PhiT.e2} 
   are uniform in $(x,y)$.
   \end{enumerate}
 \end{proposition}

 

 \medskip

 
 
 %\noindent{\bf Proof of proposition~\ref{P.PhiT}}
 
 \begin{proof}
 Item~\ref{P.PhiT.I2} follows from the fact that both action
 potentials satify an equivalent variational principle.
 We now compute the limits.
 
 \ref{P.PhiT.I3}. Given $A>0$, let $B>0$ be such that $L(x,v)>A\,|v|-B$.
 Then
 \begin{align*}
 \Phi_k(x,y;\e)&=\inf_{\ga\in\C_\e(x,y)}A_{L+k}(\ga)
              \ge\inf_\ga \int_0^\e A\,|\dga|-B+k
              \\
              &\ge A\,d(x,y)+(k-B)\,\e.
 \end{align*}
 Thus, $\liminf_{\e\to 0^+} \Phi_k(x,y;\e)\ge A\, d(x,y)$.            
 Now let $A\to+\infty$.
 
 \ref{P.PhiT.e1}. Observe that if $k>c(L)$,
 $$
 \lim_{T\to+\infty}\Phi_k(x,y;T)
 \ge\lim_{T\to+\infty}\bigl[\,\Phi_c(x,y)+(k-c)\,T\,\bigr]
 =+\infty.
 $$
 When $M$ is compact, this limit is uniform because
 $$
 \Phi_k(x,y;T)\ge\inf_{x,y\in M}\Phi_c(x,y)+(k-c)\,T.
 $$
 
 

  \ref{P.PhiT.e2}. For $k<c(L)$ there exists a
 closed curve $\ga\in\C_S(z,z)$, $S>0$, $z\in M$ such that
 $A_{L+c}(\ga)<0$. Then for $T=nS+\tau$, with $S\le\tau<2S$, we have
 that
 $$
 \Phi_k(x,y;T)
     \le\Phi_k(x,z;\tfrac\tau 2)
        +n\,A_{L+c}(\ga)
	+\Phi_k(z,y;\tfrac{\tau}2).
 $$
 This implies that $\lim_{T\to +\infty}\Phi_k(x,y;T)=-\infty$.
 The uniformity of this limit for $(x,y)$ on compact subsets  
 follows from the Lipschitz condition of item~\ref{P.PhiT.I1}
 that we now prove.
 
 %\noindent\ref{P.PhiT.I1}.
 Fix $\e>0$.
 We prove now that the function $]\e,+\infty[\ni t\mapsto
 \Phi_k(x,y;t)$ is uniformly Lipschitz.
 If $T>\e$ and $\ga\in\C_T(x,y)$ is a Tonelli minimizer,
 from corollary~\ref{TonPri} and inequality~\eqref{E>}
 there exists $D=D(\e,R)>0$
 such that $|E(\ga,\dga)-k|\le D(\e,R)+|k|$.
 Denote $h(s):=\Phi_k(x,y;s)$. 
 If $\ga_s(t):=\ga(\tfrac Ts t)$, $t\in[0,s]$, then
 $h(s)\le A_{L+k}(\ga_s)=:\cB(s)$.
 Using corollary~\ref{Ltime2} we have that
 \begin{align*}
 f(T):&=\limsup_{\de\to 0}\frac{h(T+\de)-h(T)}{\de}
      \\
      &\le\cB'(T)
      =\frac 1T\int_0^T\big[\,k - E(\ga,\dga)
                           \,\big]\;dt
      \\
      &\le \,D(\e,R)+|k|\,.                     
 \end{align*}
 If $S,\,T>\e$ we have that
 \begin{align*}
 \Phi_k(x,y;S)&\le\Phi_k(x,y;T)+\int_T^Sf(t)\;dt
              \\
              &\le\Phi_k(x,y;T)+\big[\,D(\e,R)+|k|\,\big]\;|T-S|.
 \end{align*}
 Since we can reverse the roles of $S$ and $T$, 
 this implies the uniform Lipschitz condition for 
 $T\mapsto\Phi_k(x,y;T)$.
 
 Now we prove the Lipschitz condition on a neighbourhood of $y_0$
 for $y\mapsto \Phi_k(x_0,y;T_0)$. The proof for a neighbourhood
 of $x_0$ is similar. Since $R$ is compact, this is enough
 to show that the function is Lipschitz on $R\times
 R\times]\e,+\infty[$.

 Fix $0<\de<\tfrac\e2$ small. 
 Let $\la\in\C_T(x_0,y_0)$ be a Tonelli minimizer.
 Observe that $\la$ realizes $\Phi_k(x_0,y_0,T)$.
 Let $\ga:[0,1]\to M$ be a length
 minimizing geodesic joining $y_0$ to $y$, with $|\dga|\equiv 1$.
 Let $\eta(s,t)$, $s\in[0,1]$, $t\in[T-\de,T]$ be a variation
 by solutions of~\eqref{EL} such that $\eta(s,T-\e)=\la(T-\de)$,
 $\eta(s,T)=\ga(s)$ and $\eta(0,t)=\la(t)$.
 Let
 $$
 \cE(s):=
 A_{L+k}\bigl(\la|_{[0,T-\de]}*\eta(s,\cdot)|_{[T-\de,T]}\bigr).
 $$
 Then
 \begin{align*}
 \cE'(s)
   &=\frac{d\;}{ds}\int_{T-\de}^TL
      \big(\tfrac{\partial\eta}{\partial t}(s,t)\big)\;dt 
   =\int_{T-\de}^T L_x\;\tfrac{\partial\eta}{\partial s}   
      +L_v\;\tfrac{\partial^2\eta}{\partial s \,\partial t}
    \\
    &=L_v\;\tfrac{\partial\eta}{\partial s}\Big|_{T-\de}^T
      +\int_{T-\de}^T \left(L_x-\tfrac{d}{dt}L_v\right)
       \tfrac{\partial\eta}{\partial s}
    \\
    &=L_v\big(\ga(s),\tfrac{\partial\eta}{\partial t}(s,T)\big)
      \cdot \dga(s).   
 \end{align*}
 By Weierstrass theorem~\ref{Weierstrass}, if $\de$ and $d(y,y_0)$
 are small enough, the curves $t\mapsto\eta(s,t)$ are Tonelli
 minimizers --- and thus they realize $\Phi_k(y_0,y,\de)$. 
 Then by corollary~\ref{TonPri}, there exists $A(\de,R)>0$
 such that $|\tfrac{\partial\eta}{\partial t}(s,T)|<A(\de,R)$.
 By lemma~\ref{CotaLv},
 %\index{bound!for the Legendre transform}
 $\Vert L_v(x,\tfrac{\partial\eta}{\partial t}(s,T))\Vert
 \le f(A(\de,R))$.
 Thus
 \begin{align*}
 \Phi_k(x_0,y;T)&\le\Phi_k(x_0,y_0;T)+\int_0^1 \cE'(s)\;ds \\
 &\le \Phi_k(x_0,y_0;T)+\int_0^1 f(A(\de,R)) \;\lv\dga(s)\rv\;ds \\
 &= \Phi_k(x_0,y_0;T)+ f(A(\de,R)) \; d(y,y_0).
 \end{align*}
 The value od $\de$ can be taken locally constant on a neighbourhood
 of $y_0$. Changing the roles of $y$ and $y_0$ we obtain that
 $f(A(\de,R))$ is a local Lipschitz constant for
 $y\mapsto\Phi_k(x,y;T)$, $x,y\in R$.
 \end{proof}
 
 
 

  \newpage
  

  
   \section{Global Minimizers.}
 
 
 Here we construct curves that realize the action potential.
 \newline
 For $k<c(L)$, $\Phi_k\equiv-\infty$, so there are no minimizers.
 
 
 \begin{proposition}\label{PhikMin}\quad
 \newline\indent
  If $k>c(L)$ and $x,\;y\in M$, $x\ne y$, then there is
  $\ga\in\C(x,y)$
 such that 
 $$
 A_{L+k}(\ga)=\Phi_k(x,y).
 $$
 Moreover, the energy of $\ga$ is $E(\ga,\dga)\equiv k$.
 \end{proposition}
 
 \begin{proof}
 Let $f(t):=\Phi_k(x,y;t)$. By proposition~\ref{P.PhiT},
 $f(t)$ is continuous and $f(t)\to+\infty$ when $t\to0^+$ or
 $t\to+\infty$. Hence it attains its minimum at some $T>0$.
 Moreover, $\Phi_k(x,y)=\inf_{t>0}\Phi_k(x,y;t)=\Phi_k(x,y;T)$.
 Now take a Tonelli minimizer $\ga$ on $\C_T(x,y)$.
 From lemma~\ref{Ltime1}, the energy of $\ga$ is $k$.
 \end{proof}

 \bigskip

 We now study minimizers at $k=c(L)$.
 Observe that for $c=c(L)$ and any absolutely continuous curve
 $\ga\in\C(x,y)$, we have that
 \begin{equation}\label{GM.1}
 A_{L+c}(\ga)\ge\Phi_c(x,y)
           \ge -\Phi_c(y,x).
 \end{equation}
 
 \begin{Definition}
 Set $c=c(L)$. 
 
 An absolutely continuous curve $\ga\in\C(x,y)$
 is said {\it semistatic\/}\label{semistatic}
 \index{semistatic curve}
 \index{curve!semistatic}
 if
 $$
 A_{L+c}(\ga)=\Phi_c(x,y).
 $$

 An absolutely continuous curve $\ga\in\C(x,y)$
 is said {\it static\/}\label{static}
 \index{static!orbit}
 \index{curve!static}
 if
 $$
 A_{L+c}(\ga)=-\Phi_c(y,x).
 $$
 \end{Definition}

 %\bigskip
 
 These names are justified by the following remark:
 For mechanic lagrangians $L=\tfrac 12|v|_x^2-U(x)$,
 static orbits are the fixed points  $(x_0,0)$ 
 of the lagrangian flow where $U(x)$ is maximal;
 and semistatic orbits lie in the stable or unstable 
 manifolds of those fixed points.
 
 
 
 By the triangle inequality for $\Phi_c$ the definition of semistatic
 curve $x:[a,b]\to M$ is equivalent to
 \begin{equation}\label{SemiAB}
 A_{L+c}\bigl(x\vert_{[s,t]}\bigr)=\Phi_c\bigl(x(s),x(t)\bigr),
 \qquad \forall\, a\le s\le t\le b.
 \end{equation}
 Inequality~\eqref{GM.1} implies that
 static curves are semistatic. 
 
 %\newpage
 
 Moreover, a curve 
 $\ga\in\C(x,y)$ is {\it static\/}
  if
 \begin{itemize}
 \item[(a)] \quad $\ga$ is semistatic, and
 \item[(b)] \quad $d_c(x,y)=\Phi_c(x,y)+\Phi_c(y,x)=0$.
 \end{itemize}

 

 \noindent
 From proposition~\ref{EGM} we get
 
 \begin{corollary}\label{statE}
 \index{semistatic curve!energy of}
 \index{energy!of a semistatic curve}
 \index{curve!semistatic!energy of}
 \index{static!orbit!energy of}
  Semistatic curves have energy $E\equiv c(L)$.
 \end{corollary}
 
 \begin{Definition}
 \begin{align*}
 \cM&=\cup\bigl\{\,\supp(\mu)\;\big|\;\mu\in\cM(L),\;\,A_L(\mu)=-c(L)\,\bigr\}\\
 \cNH=\Si(L)&:=
  \bigl\{\,w\in TM\,\big|\, x_w:\re\to M\text{ is semistatic }\bigr\}
       \\
 \cA=\hS(L)&:=
  \bigl\{\,w\in TM\,\big|\, x_w:\re\to M\text{ is static }\bigr\}
       \\
    \Si^-(L)&:=     
  \bigl\{\,w\in TM\,\big|\, x_w:]-\infty,0]\to M\text{ is semistatic }\bigr\}
       \\    
  \Si^+(L)&:=     
  \bigl\{\,w\in TM\,\big|\, x_w:[0,+\infty[\to M\text{ is semistatic }\bigr\}
 \end{align*}
 \index{$\cNH=\Si(L)$}\index{$\hS(L)$}\index{$\Si^+(L)$}\index{$\Si^-(L)$}
  We call $\cM$ the {\it Mather set},\index{Mather set}\index{set!Mather}  
  $\cNH$ the {\it Ma\~n\'e set},
  \index{set!Ma\~n\'e}\index{Ma\~n\'e!set}
  $\cP=\pi(\hS(L))$ the {\it Peierls 
  set}\footnote{The name is justified by
 proposition~\ref{P.PSet}.\ref{IPSet}.} 
 \index{Peierls set $\cP$}
 \index{set!Peierls|ibf}
 \index{$\cP=\pi(\hS(L))$}
 and $\cA=\hS(L)$ the {\it Aubry set}. 
 \index{Aubry set}
 \index{set!Aubry}
 \index{$\cA$}
 \end{Definition}

 Using the characterization of minimizing measures~\ref{T.CMM} 
 and corollary~\ref{statE} we have that\footnote{The typographical
 relationship was observed by Albert Fathi.}
 \begin{equation}\label{mane}
 \cM\subseteq\cA\subseteq\cNH\subseteq\acute{\cE}\;,
 \end{equation}
 where $\cM$ is the Mather set,
 \index{Mather!set}\index{set!Mather}
 $\cA$ is the Aubry set,
 $\cNH$ is the Ma\~n\'e set and $\acute{\cE}$ is
 the energy level $\acute{\cE}=[E\equiv c(L)]$.
 All these inclusions can be made proper constructing
 examples of embedded flows as in equation~\eqref{EmbeddedFlow}
 and adding a properly chosen potential $\phi(x)$.
 
 \medskip
  \noindent
  {\bf Denote by $\a(v)$\index{$\a(v)$}\index{alpha limit}
  and $\om(v)$\index{$\om(v)$}\index{omega limit}
  the $\a$~and $\om$-limits of $v$
  under the Euler-Lagrange flow.}
  
 \begin{proposition}\label{LocStat}\quad
 \newline\indent
 \index{local!static curve}
 \index{static!curve!local}
 A local static is a global static, i.e.  
 if $x_v\vert_{[a,b]}$ is static then
 $v\in\hS(L)$ 
 \newline
 (i.e. the {\sl whole\/} orbit is static).
 \end{proposition}
 
 \begin{proof}
 Let $\eta(t)=\pi\,\vr_t(v)$ and let
 $\ga_n\in\C_{T_n}(\eta(b),\eta(a))$
 be solutions of~\eqref{EL} 
 with
 $$
 A_{L+c}(\ga_n)<\Phi_c(\eta(b),\eta(a))+\tfrac 1n.
 $$
 By the a priori bounds~\ref{priori} $|\dga_n|<A$. We can assume that 
 $\dga_n(0)\to w$. Let $\xi(s)=\pi\,\vr_s(w)$. If $w\ne \deta(b)$
 then the curve $\eta|_{[b-\e,b]}*\xi|_{[0,\e]}$ is not $C^1$, and
 hence by remark~\ref{IsSol}, it
 can not be a Tonelli minimizer. Thus
 $$
 \Phi_c(\eta(b-\e),\xi(\e))
        < A_{L+c}(\eta|_{[b-\e,b]})
        + A_{L+c}(\xi|_{[0,\e]}).
 $$
 \begin{align*}
 &\Phi_c(\eta(a),\eta(a))
    \le \Phi_c(\eta(a),\eta(b-\e))
         +\Phi_c(\eta(b-\e),\xi(\e))
         +\Phi_c(\xi(\e),\eta(a))
    \\
    &\;< A_{L+c}(\eta_{[a,b-\e]})
       +A_{L+c}(\eta|_{[b-\e,b]})
       +A_{L+c}(\xi|_{[0,\e]})
       +\liminf_n A_{L+c}(\ga_n|_{[\e,T_n]})
    \\
    &\;\le A_{L+c}(\eta|_{[a,b]})
      +\lim_n\bigl(\,\ga_n|_{[0,\e]}*\ga_n|_{[\e,T_n]}\bigr)       
    \\
    &\;\le -\Phi_c(\eta(b),\eta(a))+\Phi_c(\eta(b),\eta(a))
    =0,  
 \end{align*}
 which contradicts proposition~\ref{p.c.1}(3).
 Thus $w=\deta(b)$ and similarly $\lim_n\dga_n(T_n)=\deta(a)$.
 
 If $\limsup T_n<+\infty$, we can assume that $\tau=\lim_n T_n>0$
 exists. In this case $\eta$ is a (semistatic)
 periodic orbit of period $\tau+b-a$ and then it is static.
 
 Now suppose that $\lim_n T_n=+\infty$.
 If $s>0$, we have that
 \begin{align*}
 A_{L+c}&(\eta|_{[a-s,b+s]})
   +\Phi_c(\eta(b+s),\eta(a-s))\le
   \\
   &\le\lim_n\big\{\,A_{L+c}(\ga_n|_{[T_n-s,T_n]})
       +A_{L+c}(\eta)
       \begin{aligned}[t]
       &+A_{L+c}(\ga_n|_{[0,s]})\,\big\}\\
       &+\Phi_c(\eta(b+s),\eta(a-s))
       \end{aligned}
    \\
    &\le\begin{aligned}[t]
    &\Phi_c(\eta(a),\eta(b)) \\
    &+\lim_n \big\{\,A_{L+c}(\ga_n|_{[0,s]})
                    +A_{L+c}(\ga_n|_{[s,T_n-s]})
                    +A_{L+c}(\ga_n|_{[T_n-s,T_n]})
             \,\big\}
    \end{aligned}         
    \\
    &\le \Phi_c(\eta(a),\eta(b))+\Phi_c(\eta(b),\eta(a))              
    =0.
 \end{align*}
 Thus $\eta_{[a-s,b+s]}$ is static.
 \end{proof}
 
 \medskip
  
  \begin{Definition} Set $c=c(L)$. 
  \index{ray}
  \newline
  An absolutely continuous curve 
  $\ga:[0,+\infty[\to M$ (resp. $\eta:]-\infty,0]\to M$)
  is a {\it ray} if  $\ga|_{[0,t]}$ 
  (resp. $\eta|_{[-t,0]}$) is a Tonelli minimizer for all $t>0$;
  i.e. 
  $$
  A_{L+c}(\ga|_{[0,t]})=\Phi_c(\ga(0),\ga(t);t)
  \quad\text{ for all } t>0.
  $$
  \end{Definition}
  
  Clearly a semi-infinite semistatic curve is a ray.
  We shall see in corollary~\ref{Ray=Semi} that rays are semistatic.
 
 
  \bigskip
  \hrule
  \bigskip
  
   
  
  \bigskip
  \hrule
  \bigskip

 
 
  %    \bigskip
 %  \hrule
 %  \bigskip
 %  \centerline{\bf Fuera de lugar.}
 %  \bigskip
 %  \hrule
 %  \bigskip

  \begin{proposition}\label{OmSemi}\quad
  \index{semistatic curve!$\a$ and $\om$-limits}
  \index{static!class}
  \index{omega limit!of a semistatic}
  \index{alfa limit!of a semistatic}
  \newline\indent
  If $v\in\Si$ is semistatic, then $\a(v)\subset\hS(L)$ and
  $\om(v)\subset\hS(L)$. Moreover $\a(v)$~and $\om(v)$ are each
  contained in a static class.
  \end{proposition}
  



  \begin{proof}
  We prove only that $\om(v)\subset\hS$. Let $\ga(t)=\pi\,\vr_t(v)$.
  Suppose that $t_n\to+\infty$ and $\dga(t_n)\to w\in TM$.
  Let $\eta(t)=\pi\,\vr_t(w)$. Since $\ga$~and $\eta$ are solutions 
  the Euler-Lagrange equation, then
  $\ga\vert_{[t_n-s,t_n+s]}
  \underset{C^1}\longrightarrow
  \eta\vert_{[-s,s]}$.
  Then
  \begin{align*}
  A_{L+c}\bigl(\eta\vert_{[-s,s]}\bigr)
    &+\Phi_c(\eta(s),\eta(-s))=
    \\
    &= \lim_n\bigl\{\,A_{L+c}(\ga\vert_{[t_n-s,t_n+s]})
     +\lim_m A_{L+c}(\ga\vert_{[t_n+s,t_m-s]})\,\bigr\}
    \\
    &=\lim_n\;\lim_m\;A_{L+c}(\ga\vert_{[t_n-s,t_m-s]})
    \\
    &=\lim_n\;\lim_m\;\Phi_c(\ga(t_n-s),\ga(t_m-s))
    \\
    &=\Phi_c(\eta(-s),\eta(-s))=0.
  \end{align*}
  Thus $w\in\hS(L)$. 
  Let $u\in\om(v)$. We may assume that
  $\dga(s_n)\to u$ with $t_n<s_n<t_{n+1}$.
  Then
  \begin{align*}
  d_c(\pi w,\pi u)
   &= \Phi_c(\pi w,\pi u)+\Phi_c(\pi u,\pi w)
   \\
   &=\lim_n \;A_{L+c}(\ga\vert_{[t_n,s_n]})
          +A_{L+c}(\ga\vert_{[s_n,t_{n+1}]})
   \\       
   &=\lim_n\;A_{L+c}(\ga_n\vert_{[t_n,t_{n+1}]})
          =\Phi_c(\pi w,\pi w)=0.
  \end{align*}
  Thus $w$ and $u$ are in the same static class.
  \end{proof}
  
  

 
 
     \newpage
  


 
 \section{Characterization of minimizing measures.}
 
 
 Recall that minimizing measures $\mu$ satisfy $A_L(\mu)=-c(L)$
 and that by theorem~\ref{E.2} they are invariant.
 \index{minimizing measure}\index{measure!minimizing}
 
\begin{theorem}[Ma\~n\'e~\cite{Ma7}]\label{T.CMM}\quad
\index{minimizing measure!support of}
\newline\indent
 $\mu\in\cM(L)$ is a minimizing measure if and only if
 $\supp(\mu)\subseteq\hS(L)$.
 \end{theorem} 

 \begin{proof}
 
 Since $\hS$ is closed,
 it is enough to prove the theorem for ergodic measures.
 Suppose that $\mu\in\cM(L)$ is ergodic and
 $\supp(\mu)\subseteq\Si(L)$.
 Since $\mu$ is finite, by Birkhoff's theorem there is a set
 of total $\mu$-measure $A$ such that if $\theta\in A$ then
 $\liminf_{T\to+\infty} d_{TM}(\theta,\vr_T\theta)=0$ and 
 \begin{align*}
 \int_M L+c\;d\mu&=\lim_{T\to+\infty}\frac 1T\int_0^T
 \bigl[\,L(\vr_t\,\theta)+c\,\bigr]\;dt
 \\
 &\le \liminf_{T\to+\infty}\,\tfrac
 1T\,\Phi_c\bigl(\pi(\vr_T\,\theta),\pi(\theta)\bigr)=0.
 \end{align*}
 Now suppose that $\mu\in\cM(L)$ is minimizing. 
  Then $\int L+c\;d\mu=0$.
 Applying corollary~\ref{C.FancyBirk} to $F=L+c$ and $X=TM$,
 we get that  there is a set $A\subset TM$ of total
 $\mu$-measure 
 such that if $\theta\in A$ then there is a sequence $T_n\to+\infty$ 
 such that $d(\theta,\vr_{T_n}\,\theta)\to 0$ and
 $$
 \lim_n\int_0^{T_n}\bigl[\,L(\vr_t\,\theta)+c\,\bigr]\;dt=0.
 $$
 Then
 \begin{align*}
 0\le d_c(\pi\,\theta,\,\pi\vr_1\,\theta)
    &=\lim_n \bigl[\,\Phi_c(\pi\,\theta,\pi\,\vr_1\,\theta)
        +\Phi_c(\pi\,\vr_1\,\theta,\pi\,\vr_{T_n}\,\theta)\,\bigr]
   \\
   &\le\lim_n\int_0^{T_n}\bigl[\,L(\vr_t\,\theta)+c\,\bigr]\;dt
   =0.             
 \end{align*}
 This implies that $\theta$ is static. Since $\hS(L)$ is closed and
 $A$ is dense in $\supp(\mu)$ then $\supp(\mu)\subseteq\hS(L)$.
 % \hrule
 % {\bf OJO! si se puede probar que el sop. de minimizante
 % esta en semiestaticas ya no se usa el Fancy Birkhoff}
 % \hrule
 \end{proof}
 
 \newpage
 
 It follows from theorem~\ref{E.1} that
 
 \begin{corollary}\label{C.CMM}\quad
 \newline\indent
 If $M$ is compact then $\hS(L)\ne\nulo$.
 \end{corollary}
 \index{Peierls set $\cP$!non-emptyness}
 \index{static!orbit!existence}
 
 
 
 Combining theorem~\ref{T.CMM} with corollary~\ref{statE},
 we get

 \begin{corollary}[Dias Carneiro~\cite{Carneiro}]\label{E(mu)}
 \index{minimizing measure!energy of --}
 \index{energy!of a minimizing measure}
 %\quad\newline
 If $\mu$ is a minimizing measure then
 it is supported in the energy level
 $E(\supp(\mu))=c(L)$.
 \end{corollary}
 
 \begin{lemma} \label{FancyBirk}
 Let $(X, \fB, \nu)$ be a probability space, $f$ an ergodic measure
 preserving map and $F:X\to \re$ an integrable function. Given $A\in
 \fB$
 with $\nu(A)>0$ denote by $\hat{A}$ the set of point $p\in A$ such
 that for all
 $\e>0$ there exists an integer $N>0$ such that $f^N(p)\in A$ and
 $$
 \Big| \tsum_{j=0}^{N-1} F\bigl( f^j(p) \bigr) - N\, 
  \dint F \, d\nu\, \Big|
 < \e \, .
 $$
 Then $\nu(\hat{A})=\nu(A)$.
 \end{lemma}
 
 
 
 \begin{proof}
 Without loss of generality we can assume that $\int F\, d\nu =0$. For
 $p\in X$ denote
 $$
 S_NF(p) := \tsum_{n=0}^{N-1} F\bigl(f^n(p)\bigr)\, .
 $$
 Let
 $$
 A(\e) := \left\{ \, p\in A\, \vert \, \exists N>0 \text{ such that }
 f^N(p)\in A \text{ and } \lv S_N F(p)\rv < \e\, \right\}\, .
 $$
 It is enough to prove that $\nu( A(\e)) = \nu(A)$, because
 $\hat{A} =\bigcap_n A(\frac 1n)$. 
 Let $\hat{X}$
 be the set of
 points for which the Birkhoff's theorem holds for $F$ and the
 characteristic
 functions of $A$ and of $A(\e)$. Take $x\in A\cap\hat{X}$ and let
 $N_1<N_2<\cdots$ be the integers for which $f^{N_i}(x)\in A$. Define
 $\de(k)$ by
 $$
 N_k \, \de(k) = \lv S_{N_k} F(x)\rv\, .
 $$
 Since $x\in\hat{X}$ we have that $\lim_{k\to +\infty} \de(k) =
 0$.
 Set
 \begin{align*}
 c_k  &:= S_{N_k} F(x) \, , \\
 \calS &:=\bigl\{\,k\in\na\;\big\vert\;
          \forall \ell > k ,\;\lv c_\ell - c_k \rv \ge \e \,
	  \bigr\},\\ 
 S(k) &:= \bigl\{ \, 1\le j\le k-1\;\big\vert \; 
           j\in\calS \, \bigr\} \, .
 \end{align*}
  Observe that if $j\notin S(k)$, then there is an $\ell >j$ such that
 $\lv c_\ell - c_j\rv \le\e$, so that
 $$
 \lv S_{N_\ell - N_j} F\left( f^{N_j}(x) \right) \rv = \lv c_\ell -
 c_j\rv \le
 \e \, .
 $$
 Hence
 $$
 j\notin S(k) \Longrightarrow f^{N_j}(x)\in A(\e)\, .
 $$
 This implies that
 \begin{align}
 \nu\bigl(A - A(\e)\bigr) 
   &=\lim_{k\to +\infty} \frac 1{N_k}\; 
   \#\bigl\{ \, 0\le j < N_k \, \vert\,  f^j(x) \in A-A(\e)\,\bigr\}
   \notag\\
   &\le \frac 1{N_k} \; \# S(k)  \, .
   \label{FancyBirk.1}
 \end{align}
 
 If $\calS$ is finite, from inequality~\eqref{FancyBirk.1}
 we get that $\nu\big(A-A(\e)\big)=0$,
 concluding the proof of the lemma. 
 
 Assume that $\calS$ is infinite. This implies that
 the set $\{\,c_k\,\vert\,k\in\calS\,\}$ is unbounded.
 Choose an infinite sequence $\cK$ in $\calS$ such that
 for all $k\in\cK$ we have,
 $$
 |c_k|=\max_{j\in S(k)}|c_j|.
 $$
 Then, for $k\in\cK$,
 $$
 \tfrac 12\;(2\e)\;\#S(k)\le|c_k|=\de(k)\,N_k.
 $$
 From~\eqref{FancyBirk.1}, we get that
 $$
 \nu\big(A-A(\e)\big)\le \lim_{k\in\cK}\frac{\de(k)}\e=0.
 $$
 \end{proof}
 

 
 \begin{corollary}\label{C.FancyBirk}
 If besides the hypothesis of lemma~\ref{FancyBirk}, $X$ is
 a complete separable metric space, 
 and $\fB$ is its Borel $\sigma$-algebra, then for a.e.
 $x\in X$
 the following property holds: for all $\e>0$ there exists $N>0$ such
 that
 $d( f^N(x), x)<\e$ and
 $$
 \lv {\tsum_{j=0}^{N-1}} F\bigl( f^j(x)\bigr) - N \int F\, d\nu\rv<\e
 $$
 \end{corollary}
 
 \begin{proof}
  Given $\e>0$ let $\{ V_n(\e)\}$ be a countable basis of
 neighbourhoods with
 diameter $<\e$ and let $\hat{V}_n$ be associated to $V_n$ as in 
 lemma~\ref{FancyBirk}.
 Then the full measure subset $\underset m\cap \underset n\cup
 \hat{V}_n(\frac1m)$
 satisfies the required property.
 \end{proof}
 

 \newpage


  \section{The Peierls barrier.}
  
  \index{Peierls barrier|(}\index{barrier!Peierls'|(}
  For $T>0$ and $x,\;y\in M$ define 
  $$
  h_T(x,y)=\Phi_c(x,y;T):=\inf_{\ga\in\C_T(x,y)}A_{L+c}(\ga).
  \label{hT}
  $$\index{$h_T(x,y)=\Phi_c(x,y;T)$}
  So that the curves which realize $h_T(x,y)$ are the
  Tonelli minimizers\label{TonMin2} on $\C_T(x,y)$. 
  Define the {\it Peierls barrier\/}\label{Peierls}
  as
  $$
  h(x,y) := \liminf_{T\to+\infty} h_T(x,y).
  \label{hxy}
  $$\index{$h(x,y)$}
  The difference between the action potential and the Peierls
  barrier is that in the Peierls barrier the curves must be
  defined on large time intervals. Clearly \newline
  \centerline{$h(x,y)\ge \Phi_c(x,y)$.}
  

  
  
  
  \renewcommand{\theenumi}{{\it \arabic{enumi}}}

  \begin{proposition}\label{P.PSet}\quad
  \newline\indent
  If $h:M\times M\to\re$ is finite, then 
  \begin{enumerate}
  \item \label{hLip} $h$ is Lipschitz.
  \index{Lipschitz!property!Peierls barrier's} 
  \index{Peierls barrier!Lipschitz property}
  \item \label{hPos}
        $\forall\;x,\,y\in M$, $h(x,x)\ge \Phi_c(x,y)$, 
        in particular 
        $h(x,x)\ge 0$, $\forall\; x\in M$.
  \item \label{Trh}
        \index{triangle inequality!for the Peierls barrier}
        $h(x,z)\le h(x,y)+h(y,z)$, \qquad $\forall\,x,\,y,\,z\in M$.
  \item\label{TRH}
        $h(x,y)\le\Phi_c(x,p)+h(p,q)+\Phi_c(q,y)$,\qquad
        $\forall\;x,\,y,\,p,\,q\in M$.      
  \item \label{IPSet}  
  \index{Peierls set $\cP$}
  \index{set!Peierls}\index{$\cP=\pi(\hS(L))$}
          $h(x,x)=0\Longleftrightarrow x\in\pi(\hS)=\cP$.
  \item \label{HBounded}\index{bound!for the Peierls barrier}
          If $\hS\ne\nulo$,
          $h(x,y)\le\inf_{p\in\pi(\hS)}\Phi_c(x,p)+\Phi_c(p,y)$.        
  \end{enumerate}
  \end{proposition}
  
  \begin{proof}
  Item~\ref{hPos} is trivial.
  Observe that for all $S,\,T>0$ and $y\in M$,
  $$
  h_{T+S}(x,z)\le h_T(x,y)+h_S(y,z).
  $$
  Taking $\liminf_{T\to+\infty}$ we get that
  $$
  h(x,z)\le h(x,y)+h_S(y,z), \qquad\text{ for all }S>0.
  $$
  Taking $\liminf_{S\to+\infty}$, we obtain item~{\it\ref{Trh}}.
  
  \ref{hLip}. Taking the infimum on 
  $S>0$, we get that
  \begin{align*}
  h(x,z)&\le h(x,y)+\Phi_c(y,z) \qquad \forall\,x,\,y,\,z\in M.
  \\
  &\le h(x,y)+ A\; d_M(y,z),
  \end{align*}
  where $A$ is a Lipschitz constant for $\Phi_c$. 
  Changing the roles of $x,\,y,\,z$, we obtain that $h$ is Lipschitz.

  \ref{TRH}.
  Observe that 
  $$
  \inf_{S>T} h_S(x,y)\le \Phi_c(x,p)+h_T(p,q)+\Phi_c(q,x).
  $$
   Taking $\liminf_{T\to+\infty}$ we get item~\ref{TRH}.
   
  \ref{IPSet}. 
  We first prove that if $p\in\cP=\pi(\hS)$, then $h(p,p)=0$.
  Take $v\in\hS$ such that $\pi(v)=p$
  and $y\in\pi(\om\text{-limit}(v))$.
  Let $\ga(t):=\pi\,\vr_t(v)$ and choose
  $t_n\uparrow+\infty$
  such that  $\ga(t_n)\to y$.
  Then
  \begin{align*}
  0\le h(p,p)&\le h(p,y)+\Phi_c(y,p)   \\
        &\le \lim_n A_{L+c}(\ga\vert_{[0,t_n]})+\Phi_c(y,p)  \\
        &\le \lim_n -\Phi_c( \ga(t_n),p) +\Phi_c(y,p) =0.
  \end{align*}

  
  Conversely, if $h(x,x)=0$, then there exists a 
  sequence of Tonelli minimizers
  $\ga_n\in\C(x,x;T_n)$ with $T_n\to+\infty$ and 
  $A_{L+c}(\ga_n)\overset{n}\to 0$. By lemma~\ref{priori},
  $|\dga|$ is uniformly bounded.
  Let $v$ be an accumulation point of $\dga_n(0)$ and
  $\eta(t):=\pi\,\vr_t(v)$. Then if $\dga_{n_k}(0)\overset{k}\to v$,
  for any $s>0$ we have that
  \begin{align*}
  0&\le \Phi_c(x,\pi\,\vr_s\,v)+\Phi_c(\pi\,\vr_s\,v,x)
    \\
    &\le A_{L+c}\bigl(\eta\vert_{[0,s]}\bigr)+\Phi_c(\pi\,\vr_s\,v,x)
    \\
    &\le\lim_k A_{L+c}\bigl(\ga_{n_k}\vert_{[0,s]}\bigr)
     + A_{L+c}\bigl(\ga_{n_k}\vert_{[s,T_n]}\bigr)
     \\
    &=0. 
  \end{align*}
  Thus $v\in\hS$.
  
  \ref{HBounded}. 
  Using items~\ref{TRH} and~\ref{IPSet}, we get that
  $$
  h(x,y)\le\inf_{p\in\pi(\hS)}
  \bigl[\,\Phi_c(x,p)+0+\Phi_c(p,y)\,\bigr].
  $$
  
  \end{proof}

  
  \begin{proposition}\label{hP}
  If $M$ is compact, then
  $$
  h(x,y) = \inf_{p\in\pi(\hS)}\bigl[\Phi_c(x,p)+\Phi_c(p,y)\bigr].
  $$
  \end{proposition}
  \index{Peierls barrier!characterization!with action potential}
  
  \begin{proof}\quad
  \newline\indent
  From proposition~\ref{P.PSet}.\ref{HBounded} we have that
  $$
   h(x,y)\le\inf_{p\in\pi(\hS)}
  \bigl[\,\Phi_c(x,p)+\Phi_c(p,y)\,\bigr].
  $$
  In particular $h(x,y)<+\infty$ for all $x,\,y\in M$. Now let
  $\ga_n\in\C_{T_n}(x,y)$ with $T_n\to+\infty$ and 
  $A_{L+c}(\ga_n)\to h(x,y)<+\infty$. Then
  $\tfrac 1TA_{L+c}(\ga_n)\to 0$.
  Let $\mu$ be a weak limit of a subsequence of the measures
  $\mu_{\ga_n}$. Then $\mu$ is minimizing. Let
  $q\in\pi\bigl(\supp(\mu)\bigr)$ and $q_n\in\ga_n([0,T_n])$
  be such that $\lim_n q_n=q$. Then,
  \begin{align*}
  \Phi_c(x,q)+\Phi_c(q,y)
     &\le \Phi_c(x,q_n)+\Phi_c(q_n,y)+2 A\, d(q_n,q) 
     \\
     &\le A_{L+c}(\ga_n)+ 2 A\, d(q_n,q).
  \end{align*}
  Letting $n\to\infty$, we get that
  $$
  \Phi_c(x,q)+\Phi_c(q,y)\le h(x,y).
  $$
  \end{proof}
  
   


  
  \index{Peierls barrier|)}\index{barrier!Peierls'|)}

  



 \newpage
 
 \section{Graph Properties.}
 \index{graph!properties|(}
 \index{properties!graph --|(}
 
 In this section we shall see that the projection $\pi:\hS\to M$ is
 injective. We shall call $\cP:=\pi(\hS)$ the {\it Peierls
 set}.\index{Peierls set $\cP$|ibf}\index{set!Peierls|ibf}
 \label{cP}\footnote{This name is justified by
 proposition~\ref{P.PSet}(\ref{IPSet}).}
 Thus the projection $\pi\vert_\hS$ gives an
 identification $\cP\approx \hS$.
 
 For $v\in TM$, write $x_v(t)=\pi\,\vr_t(v)$. Given $\e>0$, let
 $$
 \Sigma^\e :=\{w\in TM \,\vert\, x_w:[0,\e)\to M \text{ or }
             x_w:(-\e,0]\to M \text{ is semistatic }\}.
 $$
 
 \begin{theorem} {\bf (Ma\~n\'e) \cite{Ma7}}\quad
 \newline\indent
  For all $p\in \pi(\hS)$ there exists a unique $\xi(p)\in T_pM$
  such that $(p,\xi(p))\in\Sigma^\e$, in particular
  $(p,\xi(p))\in\widehat{\Sigma}$ and $\widehat{\Sigma}=
  \text{graph}(\xi)$. 
  \newline\indent
  Moreover, the map $\xi:\pi(\hS)\to \Sigma$  
  is Lipschitz. 
  \index{Lipschitz!graph}
  \label{GraphProperty}
 \end{theorem}
 
 The proofs of the injectivity of $\pi$ in this book only need that
 the solutions of the Euler-Lagrange equation are
 differentiable\footnote{and hence a non-differentiable curve can not
 be a Tonelli minimizer.}. The reader may provide those proofs as
 exercises. The proof of the Lipschitz condition need the following
 lemma, due to Mather. For the proof see~\cite{Mat5} or
 Ma\~n\'e \cite{Ma4}.
 
 
 \begin{MatherCrossinglemma}
 \cite{Mat5}\label{L:MATHER2}
 \quad\index{Mather!crossing lemma}
  
  Given $A>0$  there exists
  $K>0$  $\e_1>0$ and  $\de>0$ with the following property: if
 $\lv v_i\rv <A$,
 $(p_i,v_i)\in TM$, $i=1,2$ satisfy $d(p_1,p_2)<\de$ and
 $d((p_1,v_1),(p_2,v_2))\ge K^{-1}d(p_1,p_2)$ then, if  $a\in\re$ and
 $x_i:\re\to M$, $i=1,2$, are the solutions of $L$ with $x_i(a)=p_i$,
 $\dot{x}_i(a)=v_i$, there exist solutions $\ga_i:[a-\e,a+\e]\to M$
 of $L$
 with $0<\e<\e_1$, satisfying
 \begin{align*}
 \ga_1(a-\e)=&x_1(a-\e) \quad , \quad \ga_1(a+\e)=x_2(a+\e) \, ,\\
 \ga_2(a-\e)=&x_2(a-\e) \quad , \quad \ga_2(a+\e)=x_1(a+\e) \, , \\
 A_L(x_1\vert_{[a-\e,a+\e]}&)
 +A_L(x_2\vert_{[a-\e,a+\e]})>A_L(\ga_1)+A_L(\ga_2)
 \end{align*}
 \end{MatherCrossinglemma}

 \newpage

 \noindent{\bf Proof of theorem~\ref{GraphProperty}:}
 
 We prove that if  $(p,v)\in\hS$, $(q,w) \in \Sigma^\e$, 
 and $d(p,q)<\de$, then
 $$
 d_{TM}\bigl((p,v),(q,w)\bigr)<K\, d_M(p,q)\, .
 $$
 Observe that this implies the theorem. For simplicity, we only
 prove the case in which $x_v\vert_{[-\e,0]}$ is semistatic.
 Suppose it is false. Then by
 lemma~\ref{L:MATHER2} there exist $\a$, $\be:[-\e,\e]\to M$ such that
 \begin{alignat*}{4}
 \a(-\e) &= x_w(-\e) &&=:q_{-\e }\;,\qquad \qquad &\a(0)  &= p\,, \\
 \be(-\e)&= x_v(-\e) &&=:p_\e\;, \qquad \qquad &\be(0) &= q\,,
 \end{alignat*}
 and
 $$
 A_L(\a) +A_L(\be) <A_L(x_w\vert_{[-\e,0]})
 +A_L(x_v\vert_{[-\e,0]}).
 $$
 So
 \begin{align*}
 \Phi_c(q_{-\e},p)+\Phi_c(p_{-\e},q) &<\Phi_c(q_{-\e},q)+
 \Phi_c(p_{-\e},p) \\
   & =\Phi_c(q_{-\e},q)-\Phi_c(p,p_{-\e})
 \end{align*}
 \begin{figure}[h]
 \centerline{\epsfxsize=6cm\epsfbox{6b.eps}}
 \caption{\small \sc graph property.}
 \end{figure}
 Thus
 $$
 \Phi_c(q_{-\e},q)\le\Phi_c(q_{-\e},p)+\Phi_c(p,p_{-\e})+
 \Phi_c(p_{-\e},q) < \Phi_c(q_{-\e},q)
 $$
 which is a contradiction.
 \qed
 
 
  \newpage
  
  \noindent
  Using the graph property~\ref{GraphProperty} 
  we can define an equivalence relation on
  $\hS$ by  
  \begin{equation*}\label{E:SatClass}
  u,v\in\hS, \quad u\equiv v \qquad\iff\qquad d_c(\pi(u),\pi(v)) =0.
  \end{equation*}
  The equivalence classes are called {\it static
  classes}.\label{SatClass}
  \index{class!static}
  \index{static!class|ibf}
  The continuity of the pseudo-metric $d_c$ implies that
  a static class is closed, and it is invariant by
  proposition~\ref{LocStat}.
  
  For $v\in TM$ denote by $\om(v)$ its $\om$-limit.
  Let $\Ga$ be a static class, the set\index{$\Ga^+$}
  $$
  \Ga^+=\bigl\{\,v\in\Si^+(L)\;\vert\;\om(v)\subseteq\Ga\,\bigr\}
  $$ 
  is called the (forward) {\it basin}\index{basin!of a static class} of $\Ga$.
  Clearly $\Ga^+$ is forward invariant. Let\index{$\Ga^+_0$}
  \begin{align*}
  \Ga^+_0&={\textstyle\bigcup_{t>0}}\;\vr_t(\Ga^+)
         \\
        &={\textstyle\bigcup_{\e>0}}\;
       \bigl\{\,v\in TM\;\vert\; x_v\vert_{]-\e,+\infty[} 
       \text{ is semistatic and $\om(v)\subseteq\Ga$ }\bigr\}.
  \end{align*}
  The set $\pi(\Ga^+\setminus\Ga^+_0)$
  is called the {\it cut locus}\label{CutLocus} 
  of $\Ga^+$. \index{cut locus!of a static class}
 
  \begin{theorem}{\bf (Ma\~n\'e)\cite{Ma7}}\label{GPga}\quad
  \newline\indent
  For every static class $\Ga$,
  the projection $\pi:\Ga^+_0\to M$ is injective 
  with Lipschitz inverse.
  \index{Lipschitz!inverse}
  \end{theorem}
  
  The projection $\pi:\Ga^+\to M$ may not be surjective.
%
%
%  as seen in example\ref{ex.NotSutj}.  
% 
%
  But when $M$ is
  compact for generic lagrangians 
  $\pi(\Ga^+)=\pi(\Ga^-)=M$
  because there is only one static 
  class~(cf. theorem~\ref{Thm.4.1}.(B)). 
  But $\pi$ may not be injective on 
  $\Ga^+\setminus\Ga^+_0$ even for generic lagrangians.
 
  \begin{proof}
  We prove that for $K$ as in lemma~\ref{L:MATHER2}, if 
  $v\,,w\in\Ga^+_0$ then
  \begin{equation}\label{1.GPga}
  d_{TM}(v,w)\le K\, d_M(\pi(v),\pi(w)).
  \end{equation}
  Suppose it is false. Then there are $v,\,w\in\Ga^+_0$
  such that inequality~\eqref{1.GPga} does not hold.
  Let $\e>0$ be such that $x_v\vert_{[-\e,+\infty[}$
  and $x_w\vert_{[-\e,+\infty[}$ are semistatic.
  By lemma~\ref{L:MATHER2}, there exist
  $\a\in\C_{2\e}(x_v(-\e),x_w(\e))$ and
  $\be\in\C_{2\e}(x_w(-\e),x_v(\e))$ such that
  \[
  A_{L+c}(\a)+A_{L+c}(\be)+\de
  <A_{L+c}(x_v\vert_{[-\e,\e]})
  +A_{L+c}(x_w\vert_{[-\e,\e]}),
  \]
  for some $\de>0$.
  Let $p,\,q\in\pi(\Ga)$ and $s_n\,t_n\to+\infty$ be such that 
  $x_v(s_n)\to+\infty$ and $x_v(s_n)\to p$, $x_w(t_n)\to q$.
  Then
  \begin{align*}
  \Phi_c(x_v(-\e),x_w(t_n))
     &+\Phi_c(x_w(-\e),x_v(s_n))+\de
     \\
     &\le A_{L+c}(\a*x_w\vert_{[\e,t_n]})
          + A_{L+c}(\be*x_v\vert_{[\e,s_n]})
          +\de
     \\
     &< A_{L+c}(x_v\vert_{[-\e,s_n]})
        + A_{L+c}(x_w\vert_{[-\e,t_n]})
     \\   
   &= \Phi_c(x_v(-\e),x_v(s_n))+\Phi_c(x_w(-\e),x_w(t_n)).          
  \end{align*}
  Letting $n\to\infty$ and adding $d_c(p,q)=0$, we have that
  \begin{align*}
  \Phi_c(x_v(-\e),p)
       &+\Phi_c(x_w(-\e),q)
       \\
       &\le
       \Phi_c(x_v(-\e),q)+\Phi_c(x_w(-\e),p)
          +\Phi_c(q,p)+\Phi_c(p,q)
       \\
       &<\Phi_c(x_v(-\e),p)+\Phi_c(x_w(-\e),q).      
  \end{align*}
  
  \end{proof}


 \index{graph!properties|)}
 \index{properties!graph --|)} 

 \newpage
 
 \section{Coboundary Property.}
 
 \index{coboundary!properties|(}
 \index{properties!coboundary --|(}
 
 The coboundary property was first presented by R. Ma\~n\'e
 in~\cite{Ma6} and further developed in~\cite{Ma7} and by
 A. Fathi. 
 %
 % 
 % The more natural approach to the coboundary property
 % nowadays is using weak KAM solutions and is presented in 
 % section~\S\ref{S.}. We present here the version in
 % Ma\~n\'e~\cite{Ma7} to show the simplicity of his approach.
 %
 %

 \begin{theorem}{\bf (Ma\~n\'e) \cite{Ma7}}
 \label{coboundary}\quad
 \newline\indent
 If $c=c(L)$, then $(L+c)\big\vert_\hS$ is a Lipschitz coboundary. More
 precisely, taking any $p\in M$ and defining $G:\hS\to\re$ by
 \index{coboundary!Lipschitz}
 \index{Lipschitz!coboundary}
 $$
 G(w)=\Phi_c\bigl(p,\pi(w)\bigr),
 $$
 then
 $$
 (L+c)\big\vert_\hS = \frac{dG}{df} \, ,
 $$
 where
 $$
 \frac{dG}{d\vr}(w):= \lim_{h\to 0}\;\frac 1h
 \bigl[\, G(\vr_h(w)-G(w)\, \bigr]\, .
 $$
 \end{theorem}
 
 
 
 
 
 
 \begin{proof}
 
 Let $w\in\hS$ and define $F_w(v):=\Phi_c(\pi(w),\pi(v))$. 
 We have that
 \begin{align*}
 \left.\frac{dF_w}{d\vr}\right\vert_w
     &= \lim_{h\to 0}\;\tfrac 1h\, \bigl[\, F_w(\vr_h\, w) - F_w(w)\,\bigr]
     \\
     &= \lim_{h\to 0}\;\tfrac 1h\, \bigl[\, \Phi_c(\pi w, \pi \vr_h\, w)
           - \Phi_c(\pi w, \pi w)\, \bigr] \\
     &= \lim_{h\to 0}\;\tfrac 1h\,
            S_{L+c}\left(x_w\vert_{[0,h]}\right)  \\
     &= \lim_{h\to 0} \;\frac 1h\,
           \int_0^h \bigl[L\bigl( x_w(s),
           \dot{x}_w(s)\bigr)\,+c\bigr]\, ds \\
     &= L(w)+c.
 \end{align*}
 We claim that for any $p\in M$ and any $w\in\hS$, $h\in\re$,
 \begin{align}
 G(\vr_h\, w) &= \Phi_c\bigl(p,\pi(\vr_h\, w) \bigr)
                = \Phi_c\bigl( p,\pi(w)\bigr)
                + \Phi_c\bigl(\pi(w),\pi(\vr_h \,w)\bigr) \notag  \\
 G(\vr_h\, w) &= \Phi_c\bigl( p,\pi(w)\bigr) 
                + F_w\bigl( \vr_h(w)\bigr).
 \label{E:IX.1}
 \end{align}
 This is enough to prove the theorem because then
 $$
 \left.\frac{dG}{d\vr}\right\vert_w
      = \left.\frac{d\,}{dh} F_h(\vr_h\, w)\right\vert_{h=0}
      = \left.\frac{F_w}{d\vr}\right\vert_w
      = L(w)\, +c,
 $$
 and $G$ is Lipschitz by proposition~\ref{p.c.1}.
 
 We now prove \eqref{E:IX.1}. Let $q:=\pi(w)$, $x:=\pi(\vr_h\, w)$. We
 have to
 prove that
 \begin{equation}\label{E:IX.2}
 \Phi_c(p,x)=\Phi_c(p,q)+\Phi_c(q,x).
 \end{equation}
 Since the points $q$ and $x$ can be joined by the static curve
 $x_w\vert_{[0,h]}$, then
 $$
 \Phi_c(x,q) = - \Phi_c(q,x).
 $$
 Using twice the triangle inequality for $\Phi_c$ we get that
 $$
 \Phi_c(p,q)\le \Phi_c(p,x)+\Phi_c(x,q)=\Phi_c(p,x)-\Phi_c(q,x)\le
 \Phi_c(p,q).
 $$
 This implies \eqref{E:IX.2}.
 \end{proof}
 




 \index{coboundary!properties|)}
 \index{properties!coboundary --|)}
 
 \newpage
 
 \section{Covering Properties.}
 \index{covering!properties|(}
 \index{properties!covering --|(}
 
 
 \begin{theorem}\label{CoveringProperty}
 $\pi(\Si^+(L))=M$.
 \end{theorem}
 
 But in general $\pi:\Si^+(L)\to M$ is not injective.
 
 \begin{proof}
 First suppose that $\hS(L)\ne\nulo$. Take $p\in\pi(\hS)$. Given $x\in
 M\setminus\pi(\hS)$, take a Tonelli minimizer $\ga_n\in\C_{T_n}(x,p)$
 such that 
 $$
 A_{L+c}(\ga_n)<\Phi_c(x,p)+\tfrac 1n.
 $$
 By the a priori bounds~\ref{priori}, $|\dga_n|<A$ and 
 $T_n>\tfrac 1A\,d(x,p)$. Let $v=\lim_k\dga_{n_k}(0)$ 
 be an accumulation point of $\langle\dga_n(0)\rangle$.
 Let $\eta(t):=\pi\,\vr_t(v)$. Then, if $0<s<\liminf_kT_{n_k}$,
 we have that
 \begin{align*}
 A_{L+c}\bigl(\eta\vert_{[0,s]}\bigr)
     &=\lim_k A_{L+c}\bigl(\ga_{n_k}\vert_{[0,s]}\bigr)
     \\
     &\le\lim_k\bigl[\,\Phi_c\bigl(\ga_{n_k}(0),\ga_{n_k}(s)\bigr)
            +\tfrac 1{n_k}\,\bigr]
     \\
     &=\Phi_c\bigl(\eta(0),\eta(s)\bigr).
 \end{align*}
 Then $\eta$ is semistatic on $[0,S]$, where $S=\liminf_kT_{n_k}$.
 If $S<+\infty$ then $\eta(S)=\lim_k\ga_{n_k}(T_k)=p$.
 Since $x\notin\pi(\hS)$, this contradicts the graph
 property~\ref{GraphProperty}; hence $S=+\infty$.
 Thus $\eta\vert_{[0,+\infty[}$ is semistatic and $v\in\Si^+$.
 
 If $\hS=\nulo$, then by corollary~\ref{C.CMM}, $M$ is non-compact.
 Let $x\in M$ and $\langle y_n\rangle\subseteq M$ such that 
 $d_M(x,y_n)\to+\infty$. Let $\ga_n\in \C_{T_n}(x,y_n)$ be a Tonelli
 minimizer such that 
 $$
 A_{L+c}(\ga_n)<\Phi_c(x,y_n)+\tfrac 1n.
 $$
 Then by lemma~\ref{priori}, $|\dga_n|<A$, and hence $T_n\to+\infty$.
 The rest of the proof is similar to the case above.
 \end{proof}   
 
 By corollary~\ref{statE} $E(\Si^+)=c(L)$, using~\eqref{e02}
 we get that $c(L)\ge e_0$ and then 
 \begin{equation}\label{e0<c0}
  e_0\le c_u\le c_a\le c_0 \le c(L).
 \end{equation}
 \index{$e_0$}
 \index{inequality!$c(L)\ge e_0$}
 \index{critical value!strict}
 \index{critical value!$e_0\le c_u\le c_a\le c_0\le c(L)$}

 
 \index{covering!properties|)}
 \index{properties!covering --|)}
 \newpage

 \section{Recurrence Properties.}
 \index{recurrence!properties|(}
 \index{properties!recurrence --|(}

  Let $\boldsymbol{\La}$ be the set of static classes.
 Define a reflexive partial order $\preceq$
 \index{$\preceq$}\index{partial order}
  in $\boldsymbol{\La}$ by
%  \newline\indent (a)\quad $\preceq$ is reflexive.
%  \newline\indent (b)\quad $\preceq$ is transitive.
%  \newline\indent (c)\quad If there is $v\in\Sigma$ with the
%                    $\alpha$-limit set $\a(v)\subseteq\La_i$ 
%                     \newline\indent\phantom{(a)}\quad 
%                     and $\om$-limit set
%                    $\om(v)\subseteq\La_j$, then $\La_i\preceq\La_j$.        
  \begin{itemize}
  \item[(a)] $\preceq$ is reflexive.
  \item[(b)] $\preceq$ is transitive.
  \item[(c)] If there is $v\in\Sigma$ with the $\alpha$-limit set
             $\a(v)\subseteq\La_i$ 
             \newline and $\om$-limit set
             $\om(v)\subseteq\La_j$, then $\La_i\preceq\La_j$.
  \end{itemize}
 
 
 
 
 \begin{theorem}\label{multibump}\quad
 \newline\indent
  Suppose that $M$ is compact and  the number
 of static classes is finite.
 Then given $\La_i$ and $\La_j$ in $\boldsymbol{\La}$,
 we have that $\La_i\preceq\La_j$.
 \end{theorem}
 
 
 % \vspace{.5cm}
 \begin{figure}[h]
      \special{psfile=homfig2.eps hoffset=70 voffset=-115
               hscale=49 vscale=35 angle=0}
 \vspace{4cm} 
 \caption{\small {\sc connecting orbits between static classes.}}
 \medskip
 \centerline{\parbox{9.05cm}{ \small
  The three closed curves represent the static classes and 
  the other curves represent semistatic orbits connecting them.
  }}
 \label{fig1}
 \end{figure}
  
  Theorem~\ref{multibump} could be restated by saying that if
  the cardinality of $\boldsymbol{\La}$ is finite, then given two
  static classes $\La_i$ and $\La_j$ there exist classes
 $\La_i=\La_1,\dots,\La_n=\La_j$
 and semistatic vectors $v_1,\dots,v_{n-1}\in \Sigma$ such that
 for all $1\leq k\leq n-1$ we have that $\alpha(v_{k})\subseteq \La_k$
 and $\omega(v_k)\subseteq\La_{k+1}$.
 In other words, between two static classes there exists a chain of
 static classes connected by heteroclinic semistatic orbits
 (cf. figure~\ref{fig1}).

 \newpage 

    
   A proof of the following theorem can be found in~\cite{CDI}
   
   \begin{theorem} 
   If $M$ is compact, then
   \begin{enumerate}
   \item $\Si(L)$ is chain transitive.
         \index{chain!transitive}
   \item $\hS(L)$ is chain recurrent.
         \index{chain!recurrent}
         \index{Peierls set $\cP$!chain recurrency}
         \index{set!Peierls!chain recurrency}
   \end{enumerate}
   \end{theorem}
   
   \bigskip
   
   Now we proceed to prove theorem~\ref{multibump}. Assume for the
   rest of this section that $M$ is compact.

  
 %    \bigskip
 %  \hrule
 %  \bigskip
 %  \centerline{\bf Fuera de lugar.}
 %  \bigskip
 %  \hrule
 %  \bigskip

  \begin{proposition}\label{OmSemi}\quad
  \index{semistatic curve!$\a$ and $\om$-limits}
  \index{static!class}
  \index{omega limit!of a semistatic}
  \index{alfa limit!of a semistatic}
  \newline\indent
  If $v\in\Si$ is semistatic, then $\a(v)\subset\hS(L)$ and
  $\om(v)\subset\hS(L)$. Moreover $\a(v)$~and $\om(v)$ are each
  contained in a static class.
  \end{proposition}
  
  \begin{proof}
  We prove only that $\om(v)\subset\hS$. Let $\ga(t)=\pi\,\vr_t(v)$.
  Suppose that $t_n\to+\infty$ and $\dga(t_n)\to w\in TM$.
  Let $\eta(t)=\pi\,\vr_t(w)$. Since $\ga$~and $\eta$ are solutions 
  the Euler-Lagrange equation, then
  $\ga\vert_{[t_n-s,t_n+s]}
  \underset{C^1}\longrightarrow
  \eta\vert_{[-s,s]}$.
  Then
  \begin{align*}
  A_{L+c}\bigl(\eta\vert_{[-s,s]}\bigr)
    &+\Phi_c(\eta(s),\eta(-s))=
    \\
    &= \lim_n\bigl\{\,A_{L+c}(\ga\vert_{[t_n-s,t_n+s]})
     +\lim_m A_{L+c}(\ga\vert_{[t_n+s,t_m-s]})\,\bigr\}
    \\
    &=\lim_n\;\lim_m\;A_{L+c}(\ga\vert_{[t_n-s,t_m-s]})
    \\
    &=\lim_n\;\lim_m\;\Phi_c(\ga(t_n-s),\ga(t_m-s))
    \\
    &=\Phi_c(\eta(-s),\eta(-s))=0.
  \end{align*}
  Thus $w\in\hS(L)$. 
  Let $u\in\om(v)$. We may assume that
  $\dga(s_n)\to u$ with $t_n<s_n<t_{n+1}$.
  Then
  \begin{align*}
  d_c(\pi w,\pi u)
   &= \Phi_c(\pi w,\pi u)+\Phi_c(\pi u,\pi w)
   \\
   &=\lim_n \;A_{L+c}(\ga\vert_{[t_n,s_n]})
          +A_{L+c}(\ga\vert_{[s_n,t_{n+1}]})
   \\       
   &=\lim_n\;A_{L+c}(\ga_n\vert_{[t_n,t_{n+1}]})
          =\Phi_c(\pi w,\pi w)=0.
  \end{align*}
  Thus $w$ and $u$ are in the same static class.
  \end{proof}
  
  
  \newpage

  
  \begin{proposition}\label{ConnectedClass}
  Every static class is connected.
  \end{proposition}
   
   \begin{proof}   
  Let $\La$ be a static class and suppose that it is not connected. 
    Let $U_1,U_2$ be disjoint open sets such that $\La\subseteq
    U_1\cup U_2$
    and $\La\cap U_i\ne\nulo$, $i=1,2$. Let $p_i\in\pi(U_i\cap\La)$,
    $i=1,2$.
  Since $U_{1}$ and $U_{2}$ are disjoint sets we can take
  a solution
  $x_{v_n}:[a_n,b_n]\to M$, $a_n<0<b_n$ of (E-L)
  such that $x_{v_n}(0)\notin \pi(U_1\cup U_2)$,
    $x_{v_n}(a_n)=p_1$, $x_{v_n}(b_n)=p_2$ and
 \begin{equation}
    A_{L+c}(x_{v_n})\le\Phi_c(p_1,p_2)+\tfrac 1n.  \label{zwei}
 \end{equation}
    Let $u$ be a limit point of $v_n$,
    then $x_u:\re\to M$ is semistatic (see the proof of claim~2 item
    (a)). 
 Then, for $a_n\le s\le t\le
    b_n$,
 $$d_c(p_1,p_2) 
      \leq\Phi_c(p_1,x_{v_n}(s))+\Phi_c(x_{v_n}(s),x_{v_n}(t))
           +\Phi_c(x_{v_n}(t),p_2)+\Phi_c(p_2,p_1),$$
 therefore
    \begin{align*}
 d_c(p_1,p_2) 
      \leq
      &\;\Phi_c(p_2,p_1) \\
      &+\liminf_n\left[\Phi_c(p_1,x_{v_n}(s))
             +\Phi_c(x_{v_n}(s),x_{v_n}(t))+\Phi_c(x_{v_n}(t),p_2)\right]
                \\
      \leq &\;\Phi_{c}(p_{2},p_{1})+\liminf_{n} A_{L+c}(x_{v_{n}})\\
      \leq &\;d_{c}(p_{1},p_{2})=0,
 \end{align*}
 where in the last inequality we used (\ref{zwei}).
 Hence
 $$\Phi_c(p_1,x_{u}(s))+\Phi_c(x_{u}(s),x_{u}(t))
           +\Phi_c(x_{u}(t),p_2)+\Phi_c(p_2,p_1)=0.$$
 Combining the last equation with the triangle inequality
   we obtain
    \begin{align*}
    d_c&(x_u(s),x_u(t))\le   \\
    &\le\Phi_c(x_u(s),x_u(t))
          +\left[\Phi_c(x_u(t),p_2)+\Phi_c(p_2,p_1)
          +\Phi_c(p_1,x_u(s))\right]
          =0.
    \end{align*}
  So that $u\in\widehat{\Sigma}$. Moreover, for $s=0,t=1$:
  \begin{align*}
  d_c&(x_u(0),p_1)\le  \\
  &\le\Phi_c(p_1,x_u(0))
        +\left[\Phi_c(x_u(0),x_u(1))+\Phi_c(x_u(1),p_2)
        +\Phi_c(p_2,p_1)\right]=0.
  \end{align*}
  Hence $x_{u}(0)\in\pi(\La)$. On the other hand $x_{u}(0)\notin
  \pi(U_{1}\cup U_{2})$.
  This contradicts the fact that $\La\subseteq U_1\cup U_2$.
   \end{proof}      
    
   
  {\noindent \bf Proof of theorem~\ref{multibump}.}
 
  Given $v\in TM$ denote by $\a(v)$ and $\om(v)$ its $\a$ and
  $\om$-limits respectively.
 By proposition~\ref{ConnectedClass}
 the static classes are connected.
 Hence if we assume
 that there are only finitely many of them, 
  the connected components of $\hS$ are
  finite and must coincide with the static classes.
  For $\e>0$, let $\hS(\e)$ be the $\e$-neighborhood of
  $\hS$, i.e.
  $$
  \hS(\e) :=\{ \, v\in TM \, \vert \,
  d_{TM}(v,\hS)<\e\,\}.
  $$
  Fix $\e>0$ small enough such that the connected components of
  $\hS(\e)$ are the $\e$-neighborhoods of the static classes.
  So that for $0<\de<\e$, $\hS(\de)=\sum_{i=1}^{N(\e)}\La_i(\de)$,
  where $\La_i(\de)$ are disjoint open sets containing exactly one
  static class and the number of components $N(\e)$ is fixed for all
  $0<\de<\e$.
  
  Now suppose that the theorem is false. Let $\La_i$,
  $\La_k\in\boldsymbol{\La}$ be such that $\La_i\not\preceq\La_k$.
  Let
  $$
  {\Bbb A}:=\bigcup_{\{\,\La_j\in\boldsymbol{\La}\,\vert\,
  \La_i\preceq\La_j\,\}}\La_j
  \quad , \quad
  {\Bbb B} :=\bigcup_{ \{\,\La_j\in\boldsymbol{\La}\,\vert\,
  \La_i\not\preceq\La_j\,\}}\La_j.
  $$
 
   Given $v\in\Sigma$ with $\a(v)\subseteq{\Bbb A}$ and $0<\de<\e$,
  define inductively $s_k(v)$, $t_k(v)$, $T_k(v)$ as follows. Let
  \begin{align*}
  s_1(v) &:= \inf \{\, s\in\re\,\vert\, f_s(v)\notin{\Bbb A}(\e)\,\}
  \in\re\cup\{+\infty\}.\\
  \intertext{If $s_k(v)<+\infty$, $k\geq 1$, define }
  t_k(v) &:= \sup\{\, t<s_k(v)\,\vert\, f_t(v)\in{\Bbb A}(\de)\,\},  \\ 
  T_k(v) &:= \inf\{\,t>s_k(v)\,\vert\,f_t(v)\in{\Bbb A}(\de)\,\}.   \\          
  \intertext{Observe that $s_k(v)<+\infty$ implies that
  $T_k(v)<+\infty$ because by the definition of ${\Bbb B}$ and the 
  transitivity of $\preceq$ we have that $\om(v)\subseteq {\Bbb A}$.
  Define}
  A_k=A_k(\de) &:= \sup \{\,\vert\,T_k(v)-t_k(v)\,\vert\;:\;
                 v\in\Sigma,\;\a(v)\subseteq{\Bbb A},\,
		 s_k(v)<+\infty\,\},
  \end{align*}
  if $s_k(v)=+\infty$ for all  $v\in\Sigma$ with $\a(v)\subseteq{\Bbb
  A}$, write $A_\ell(\de)\equiv 0$ for all $\ell\ge k$. Now set:
  \[s_{k+1}(v):= \inf\{\,s>T_k(v)\,\vert\,f_t(v)\notin{\Bbb A}(\e)\,\}.\]
  Observe that $s_k(v)$, $t_k(v)$ and $T_k(v)$ are invariant under $f_{t}$.

 
  We split the rest of the proof of theorem~\ref{multibump}
  into the following claims:
 
  
  \noindent{\bf Claim 1.} 
  {\it $A_k(\de)<+\infty$ for all $k=1,2,\ldots$ and all $0<\de<\e$.}
   
  Define
  $$
  {\Bbb M} := \{\,v\,\vert\,\;v\in\Sigma,\;\a(v)\subseteq{\Bbb A}\,\}.
  $$
  
  \noindent{\bf Claim 2.}
  {\it\begin{itemize}
  \item[(a)] $\ov{\Bbb M}\cap{\Bbb B}\ne \nulo$.
  \item[(b)] $\limsup_k A_k(\de)=\sup_k A_k(\de)=+\infty$.
  \end{itemize}}
   
  \noindent{\bf Claim 3.}
  {\it There exist sequences $v_n\in\Sigma$, $0<s_n<t_n$ such that
  $v_n\to u_1\in{\Bbb A}$, $f_{s_n}(v_n)\to u_2\notin{\Bbb A}(\e)$,
  $f_{t_n}(v_n)\to u_3\in{\Bbb A}$ and $d_c(\pi u_1, \pi u_3) = 0$.} 
  
  \medskip
  
  We now use claim~3 to complete the proof of theorem~\ref{multibump}.
  If
  $u_1\in\La_j\subseteq{\Bbb A}$, we shall prove that
  $u_2\in\La_j\setminus{\Bbb A}(\e)$, obtaining a contradiction
  and thus proving theorem~\ref{multibump}. 
  It is enough to show
  that $d_c(\pi u_1,\pi u_2)=0$. Indeed
  \begin{align*}
  d_c(\pi u_1,\pi u_2) 
     &= \Phi_c(\pi u_1,\pi u_2) + \Phi_c(\pi u_2,\pi u_1) \\
     &\le \Phi_c(\pi u_1,\pi u_2) + \Phi_c(\pi u_2, \pi u_3)
          + \Phi_c(\pi u_3, \pi u_1) \\
     &\le \lim_n\big[\Phi_c(\pi v_n,\pi f_{s_n}(v_n))
          \begin{aligned}[t]
          + \Phi_c(\pi f_{s_n}(v_n)&,\pi f_{t_n}(v_n))\big]
          \\
          &+ \Phi_c(\pi u_3,\pi u_1)
          \end{aligned} \\
     &=\lim_n \Phi_c(\pi v_n, \pi f_{t_n}(v_n))
          + \Phi_c(\pi u_3,\pi u_1) \\
     &= d_c(\pi u_1, \pi u_3) = 0,
   \end{align*}
   where the fourth equation holds because $v_n$ is a semistatic
   vector.
  
   \qed           
   
   
    

   
   \noindent{\bf Proof of claim 1:}
   
    Suppose that $A_i<+\infty$ for $i=1,\ldots,k-1$ and $A_k=+\infty$.
   The case $k=1$ is similar. Then there exists $v_n\in\Sigma$, with
   $\a(v_n)\subset{\Bbb A}$ and $T_k(v_n)-t_k(v_n)\to +\infty$. We can
   assume that $t_k(v_n)=0$ and that $v_n$ converges ($\Sigma$ is
   compact). Let
   $u= \lim_n v_n\in\partial{\Bbb A}(\de)$. 
   Then for all $n$, we have that
   $$
   m\{\, t<0\,\vert\,f_t(v_n)\notin {\Bbb
   A}(\e)\,\}\le\tsum_{i=1}^{k-1}A_i\,,
   $$
   where $m$ is the Lebesgue measure on $\re$. This implies that
   $$
   m\{\, t<0\,\vert\, f_t(u)\notin{\Bbb
   A}(\e)\,\}\le\tsum_{i=1}^{k-1}A_i
   $$ 
   and hence $\a(u)\cap\ov{{\Bbb A}(\e)}\ne\nulo$.
   By proposition~\ref{OmSemi}, 
   $\a(u)\subset{\Bbb A}$. Since $f_t(v_n)\notin{\Bbb A}(\e)$
   for $0<t<T_k(v_n)$ and $T_k(v_n)\to+\infty$, then
   $f_t(u)\notin{\Bbb A}(\e)$ for all $t>0$ and hence
   $\om(u)\subseteq{\Bbb B}$. But then the orbit of $u$ contradicts
   the
   definition of ${\Bbb B}$.
   
   \qed
   

   %\medskip    
   
   {\noindent \bf Proof of claim 2:}
   
   
   
   (a) Let $p\in\pi{\Bbb A}$, $q\in\pi{\Bbb B}$. For $n>0$, let
   $x_{v_n}:[a_n,b_n]\to M$ be a solution of (E-L) such that
   $x_{v_n}(a_n)=p$, $x_{v_n}(b_n)=q$ and
   $$
   A_{L+c}(x_{v_n})\le \Phi_c(p,q) + \tfrac 1n.
   $$
   This implies that
   \begin{equation}\label{E:m.1}
   A_{L+c}\left(x_{v_n}\vert_{[s,t]}\right)\le
   \Phi_c\bigl(x_{v_n}(s),x_{v_n}(t)\bigr)+\tfrac 1n
   \end{equation}
   for all $a_n\le s\le t\le b_n$. We can assume that
   $$
   \inf\{\, s>a_n\,\vert\,x_{v_n}(s)\in{\Bbb B}(\de)\,\}=0,
   $$
   and that the sequence $v_n$ converges (cf. lemma~\ref{priori}).
   Let $u=\lim_n v_n\in\pi^{-1}(\partial\,\pi\,{\Bbb B}(\de))$.  
    Taking limits in
   \eqref{E:m.1} we obtain that $x_u\vert_{[s,t]}$ is
   semistatic for all $\lim\inf_n a_n \le s\le t \le \lim\sup_n b_n$.
   
   Any limit point $w$ of $\dot{x}_{v_n}(a_n)=f_{a_n}(v_n)$ satisfies
   $\pi(w)=p\in\pi{\Bbb A}$, and by the graph
   property (theorem~\ref{GraphProperty}), $w\in{\Bbb A}$. Similarly,
   any limit
   point of $f_{b_n}(v_n)$ is in ${\Bbb B}$. Since ${\Bbb A}\cup {\Bbb
   B}$ is invariant and $u\notin{\Bbb A}\cup{\Bbb B}$, then $\lim_n
   a_n=-\infty$, $\lim_n b_n=+\infty$. Hence $u\in\Sigma$. Since
   $f_t(v_n)\notin{\Bbb B}(\de)$ for all $a_n\le t<0$ and $a_n\to
   -\infty$, then $f_t(u)\notin{\Bbb B}(\de)$ for all $t<0$. 
   Hence
   $\a(u)\subseteq{\Bbb A}$ and thus $u\in {\Bbb M}$. Since 
   $u\in\pi^{-1}(\partial\,\pi\,{\Bbb B}(\de))$ there exists $z\in{\Bbb B}$
   such that $d_{M}(\pi(u),\pi(z))\leq \de$. Since $z\in \hS$ and
   $u\in\Sigma$,  by theorem \ref{GraphProperty} we have that
   $$
   d_{TM}\big((\pi(u),u),(\pi(z),z)\big)\leq K\,\de.
   $$
   Thus $u\in{\Bbb M}\cap {\Bbb B}(K\de)$. 
   Letting $\de\to 0$, we obtain that
   $\ov{\Bbb M}\cap{\Bbb B}\ne\nulo$.
   
                  
   \medskip               
                  
   (b) By claim~1 it is enough to show that $\sup_k A_k(\de)=+\infty$.
   If $\sup_k A_k(\de)<T$, then ${\Bbb M}\subseteq{\Bbb M}(\de,T)$,
   where 
   $$
   {\Bbb M}(\de,T)=\{\, v\in\Sigma\,\vert\, f_{[-T,T]}(v)\cap{\Bbb
   A}(\de)\ne \nulo\,\}.
   $$
   Then ${\Bbb M}\cap{\Bbb B}\subseteq{\Bbb M}(\de,T)\cap{\Bbb
   B}=\nulo$, because ${\Bbb B}$ is invariant and ${\Bbb
   B}\cap{\Bbb A}(\de)=\nulo$. This contradicts item (a).
   
   \qed
   
    
   
   \noindent{\bf Proof of claim~3:}
   
   Given $0<\de<\e$, by claim~2(b) there exists $k>N(\e)$ such that
   $A_k(\de)>0$. Hence there is $v=v_\de\in\Sigma$ with
   $\a(v)\subset{\Bbb A}$, such that the orbit of $v$ leaves ${\Bbb
   A}(\e)$ and returns to ${\Bbb A}(\de)$ at least $k$ times. Since
   $k>N(\e)$ there is one component $\La_j(\de)\subseteq{\Bbb A}(\de)$
   with two of these returns, i.e. there exist
   $\tau_1(\de)<s(\de)<\tau_2(\de)$ with $f_{\tau_1}(v)\in\La_j(\de)$,
   $f_s(v)\notin{\Bbb A}(\e)$ and $f_{\tau_2}(v)\in\La_j(\de)$. 
   We can choose $v_\de$ so that $\tau_1(\de)=0$. Now,
   there exists a sequence such that the repeated component
   $\La_j\subset\La_j(\de_n)$ is always the same. Let
   $s_n:=s(\de_n)$, $t_n:=\tau_2(\de_n)$
   and choose a subsequence 
   such that $v_{n}$, $f_{s_n}(v_n)$ and
   $f_{t_n}(v_n)$ converge. Let $u_1=\lim_n v_n\in \cap_n
   \La_j(\de_n)=\La_j$, $u_3=\lim_n f_{t_n}(v_n)\in\La_j$ and
   $u_2=\lim_n f_{s_n}(v_n)\notin{\Bbb A}(\e)$. Since
   $u_1,u_3\in\La_j$, then $d_c(\pi u_1,\pi u_3)=0$.
   
   \qed
   
   
   \bigskip
   
  
   \index{recurrence!properties|)}
   \index{properties!recurrence --|)}
 



 \chapter{The Hamiltonian viewpoint.}
 
 \section{The Hamilton-Jacobi equation.}

  
  Let $\om$ \index{symplectic form!canonical} be the  
 canonical symplectic form on $T^*M$. A subspace 
 $\la $ of $T_pT^*M$ is called {\it isotropic} 
 \index{isotropic subspace}
 \index{subspace!isotropic} if $\om (X,Y)=0$ for all $X,Y$ on
 $\la $. Since $\om$ is nondegenerate, the isotropic
 subspaces have dimension $\le n$, half of the dimension of
 $T^*M$. Isotropic  spaces of dimension $n$ are called {\it
 lagrangian}
 subspaces \index{lagrangian!subspace} \index{subspace!lagrangian}
 We say that a submanifold $W\subset T^*M$
  is {\it lagrangian\/}\label{LagrangianSubmanifold}
  \index{lagrangian!submanifold}
  \index{submanifold!lagrangian}
   if at each point
  $\theta\in W$, its tangent space $T_\theta W$ is a lagrangian
  subspace of $T_\theta T^*M$. In particular, $\dim W = \dim M=n$.
   
  
  \begin{theorem}[Hamilton-Jacobi]\quad
  \index{Hamilton-Jacobi!theorem}
  \index{theorem!Hamilton-Jacobi}
  \newline\indent
  If the hamiltonian $H$ is constant on a lagrangian submanifold $N$,
  then $N$ is invariant under the hamiltonian flow.
  
  \end{theorem}
  
  
  
  \begin{proof}
  We only have to show that the hamiltonian vector field $X$ is tangent
  to $N$. Since $H$ is constant on $N$, then $dH\vert_{TN}\equiv 0$.
  Since $\om(X,\cdot)=dH$, then $\om(X,\xi)=0$ for all $\xi\in TN$.
  Since the tangent spaces to $N$ are lagrangian, they are maximal
  isotropic subspaces, therefore $X\in TN$.
  \end{proof}

  
  
    Some distinguished $n$-dimensional manifolds on $T^*M$ are the
  graph submanifolds,\label{GraphSubmanifold}
  \index{graph!submanifold}
  \index{submanifold!graph}
   which are of the form
  \begin{equation}\label{graph}
  G_\eta = \{ \;(x,\eta_x)\,\vert\, x\in M\;\}\subset T^*M,
  \end{equation}
  where $\eta_x$ is a 1-form on $M$. A {\it lagrangian
  graph\/}\label{LagrangianGraph}
  \index{graph!lagrangian}
  \index{lagrangian!graph}
  is a lagrangian graph submanifold.
   In fact, 
  \begin{lemma}\label{L-LagGraph}
  \index{closed form}
  \index{form!closed|ibf}
  %\index{graph!lagrangian}
  %\index{lagrangian!graph}
  $G_\eta$ is a lagrangian graph if and only if
  the form $\eta$ is closed:
  $$
  G_\eta \text{ is lagrangian }\Longleftrightarrow\;
  d\eta \equiv 0
  $$
  \end{lemma}
  
  \begin{proof}
  Choose local coordinates $q_1,\ldots,q_n$ of $M$.
  Then $\eta(q) = \sum_k \, p_k(q)\, dq_k$. A basis of the 
  tangent space to the graph $G_\eta$ is given by
  $E_i=\left(\tfrac{\partial\;}{\partial q_i}, \sum_k
  \frac{\partial p_k}{\partial q_i}\;
  \frac{\partial\;}{\partial p_k}\right)$.
  Applying $\om=dp\wedge dq$, 
  \begin{equation*}
  \om(E_i,E_j)=\tfrac{\partial p_i}{\partial q_j} 
              -\tfrac{\partial p_j}{\partial q_i}.
  \end{equation*}
  Since
  $$
  d\eta = \sum_{i<j} \left(\tfrac{\partial p_i}{\partial q_j}
                     -\tfrac{\partial p_j}{\partial q_i}\right)
                     \; dq_j\wedge dq_i\,,
  $$
  then $\om\vert_{TG_\eta}\equiv 0 
  \Longleftrightarrow d\eta \equiv 0$.
  
  \end{proof}
 
  \bigskip
   
  Thus, we can associate a cohomology class $[\eta]\in H^1(M,\re)$
  \index{lagrangian!graph!cohomology class of}
  to each lagrangian graph $G_\eta$. Lagrangian graphs with zero
  cohomology class are the graphs of the exact 1-forms: $G_{df}$,
  with $\eta=df$ and $f:M\to\re$ a smooth function. These are
  called {\it exact lagrangian graphs}.\index{lagrangian!graph!exact}

  The Hamilton-Jacobi equation for autonomous hamiltonians is
  \begin{equation}\label{HJ}
  H(x,d_xu) = k, \qquad u:M\to\re.
  \tag{H-J}
  \end{equation}
  Thus a smooth solution of the Hamilton-Jacobi equation
  corresponds to an exact invariant lagrangian graph.


  \newpage
  
  \section{Dominated functions.}
  \index{dominated function|(}
  
  We say that a function $u$ is {\it dominated\/}\label{domi}
  \index{dominated $(\prec)$}
  by $L+k$, and write $u\prec L+k$~\index{$\prec$} ~if
  \begin{equation*}\label{e.domi}
  u(y)-u(x)\le \Phi_k(x,y) \quad\text{ for all } x,y\in M.
  \end{equation*}
  
  The triangle inequality implies that the
  functions $u(x)=\Phi_k(y,x)$ and $v(x)=-\Phi_k(x,y)$
  are dominated, for any $y\in M$.
  
 
  \medskip
  
  \begin{lemma}\label{Domd}\quad
  \begin{enumerate}
  \item\label{ULip} If $u\prec L+k$, then $u$ is Lipschitz with the
                    same Lipschitz constant as $\Phi_c$.
                    In particular, a family of dominated functions
                    is equicontinuous.
  \item\label{UH<k}
  If $u\prec L+k$ then $H(x,d_xu)\le k$ at any differentiability point
  $x$ of $u$.
  \end{enumerate}
   \end{lemma}
  
  
  
  \begin{proof}\quad
  
  \ref{ULip}. We have that $u(y)-u(x)\le \Phi_c(x,y)\le A\;d_M(x,y)$,
              where $A$ is a Lipschitz constant for $\Phi_c$.
              Changing the roles of~$x$ and~$y$, we get that
              $u$ is Lipschitz.
              
  \ref{UH<k}. We have that 
  $$
  u(y)-u(x)\le\int_\ga L(\ga,\dga)+k
  $$
  for all curves $\ga\in\C(x,y)$. This implies that
  $$
  d_xu\cdot v \le L(x,v)+k
  $$
  for all $v\in T_xM$ when $u$ is differentiable at $x\in M$.
  Since 
  $$
  H(x,d_xu)=\sup\{\, d_xu\cdot v-L(x,v)\,|\,v\in T_xM\,\},
  $$
  then $H(x,d_xu)\le k$.
  \end{proof}
  
  \begin{exercises}\label{ExerSubsol}\quad
  \begin{enumerate}  
  \item If $u:M\to\re$ is differentiable, then
        $$
	H(x,d_xu)\le k \iff u\prec L+k.
	$$
   \item \label{ExerSubsol.2} Fix $x_0\in M$.
         For $k>c(L)$ choose $f_k\in C^\infty(M,\re)$
	 such that $f(x_0)=0$ and $H(df_k)<k$ 
	 (cf.~\ref{gafa<}). Let
	 $u(x):=\limsup_{k\to c(L)}f_k(x)$.
	 \newline
         Then $u<+\infty$, $u$ is Lipschitz and
	 $H(x,d_xu)\le c(L)$ for a.e. $x\in M$.	
  \end{enumerate}
  \end{exercises}
  
  \bigskip
  
  
  \begin{Definition}
  Given a dominated function $u\prec L+k$, we say that
  an absolutely continuous curve $\ga:[a,b]\to M$
  {\it realizes} $u$, if
  \begin{equation}\label{RealizeU}
  u(\ga(t))-u(\ga(s))=A_{L+k}(\ga|_{[s,t]}),
  \quad\text{ for all } a\le s\le t\le b.
  \end{equation}
  \end{Definition}
  
  Observe that such realizing curves must be 
  global minimizers. In particular for $k=c(L)$,
  they are semistatic.
  
  The following proposition shows that we actually get a solution
  of~\eqref{HJ} if there are (semistatic) curves which realize
  a dominated function $u$.
  

  \begin{proposition}\label{Udiff}\quad
  Suppose that $u\prec L+k$.
  \begin{enumerate}
  \item\label{Udiff.1} If $\ga:]-\e,\e[\to M$ realizes $u$, then
        $u$ is differentiable at $\ga(0)$.
  \item\label{Udiff.2} If $\ga:]-\e,0]\to M$ or $\ga:[0,\e[\to M$
        realizes $u$ and $u$ is differentiable
	at $x=\ga(0)$, then
	$d_xu=L_v(x,\dga(0))$ and $H(x,d_xu)=k$.	
  \end{enumerate}
  \end{proposition}
  
  \begin{remarks}\label{RUdiff}\quad
  \begin{enumerate}
%   \item\label{RUdiff.0}
%   The domination condition implies that the curve $\ga$ is semistatic.
  \item\label{RUdiff.1}
  Equation $d_x u = L_v(x,\dga(0))$
  means that the tangent vector $(x,\dga(0))$ of 
  any a.c. curve $\ga$ realizing $u$ is sent by the Legendre transform
  to $d_xu$.
  \index{Legendre transform}
  \index{transform!Legendre}
  
  \item\label{RUdiff.2}
        In particular, since the functions $u(x)=\Phi_k(p,x)$
        (resp. $v(x)=-\Phi_c(x,p)$) are dominated, 
        then they are
        differentiable at any point which is not at the
	(backward) (resp. forward) $(L+k)$-cut locus of $p$.
	\index{cut locus!of a point}
	
  \item\label{RUdiff.3}
        Observe that the energy $E(x,\dga(0))=H(x,d_xu)$.
	In proposition~\ref{Busemann}, we show that
	any semistatic orbit realizes some dominated function.
	Thus we obtain another proof 
	 for $\Sigma\subset E^{-1}\{c\}$, i.e. that the
	    semistatic orbits have energy $c(L)$.
  \end{enumerate}	
  \end{remarks}

  \bigskip
  
  \begin{proof}
  \ref{Udiff.1}. Let $w\in T_xM$ and let $\eta(s,t)$ be a variation of $\ga$ fixing
  the endpoints $\ga(-\e)$, $\ga(\e)$ such that $\eta(0,t)=\ga(t)$ and 
  $\tfrac{\partial\,}{\partial s}{\eta}(0,0)=w$. Define
  \begin{equation*}
  \cA(s) := \int_{-\e}^0 L(\tfrac{\partial}{\partial t}\eta(s,t))+k \;\; dt.
  \end{equation*}
  Then, integrating by parts and using the Euler-Lagrange
  equation~\eqref{EL}, 
  \begin{equation*}
  \cA'(0) = L_v\;\xi\vert_{-\e}^0 
            +\int_{-\e}^0\left[L_x-\tfrac{d\,}{dt}L_v\right]\,\xi\; dt
	 = L_v(x,\dga(0))\cdot w\,,
  \end{equation*}
  where $\xi(t):=\tfrac{\partial\,}{\partial s}\eta(0,t)$. Also
  \begin{align*}
  \tfrac 1{s}\bigl[\, u(\eta(s,0))-u(x)\,\bigr]
     &=  \tfrac 1s \bigl[
       u(\eta(s,0))-u(\ga(-\e))+u(\ga(-\e))-u(\ga(0))\,\bigr]
       \\
     &\le\tfrac 1s \left[\, \cA(s) - \cA(0)\,\right] \,,        
  \end{align*}
  where we used that $u\prec L+k$
  and~\eqref{RealizeU}. Hence
  \begin{equation}\label{deriv<}
  \lim_{s\to 0}\sup\tfrac 1s \,\bigl[\, u(\eta(s,0))-u(x)\,\bigr]
  \le \cA'(0).
  \end{equation}
  Similarly, if
  $\cB(s):=A_{L+k}\bigl(\eta(s,\cdot)\vert_{[0,\e]}\bigr)$, then
  $$
  u(\ga(\e))-u(\eta(s,0))-u(\ga(\e))+u(x)
    \le \cB(s)-\cB(0),
  $$
  $$
  \limsup_{s\to 0}\tfrac 1s\,\bigl[ \, u(x)-u(\eta(s,0))\,\bigr]
  \le\cB'(0) = - L_v(x,\dga(0))\cdot w.
  $$
  Hence 
  \begin{equation}\label{deriv>}
  \liminf_{s\to 0}\tfrac 1s\,\bigl[ \, u(\eta(s,0))-u(x)\,\bigr]
  \ge  L_v(x,\dga(0))\cdot w.
  \end{equation}
  From~\eqref{deriv<} and~\eqref{deriv>} we get that
  $u$ is differentiable at $x=\ga(0)$.
   and $d_xu=L_v(x,\dga(0))$.
  
  \ref{Udiff.2}. Now assume that $\ga:]-\e,0]\to M$ realizes $u$
  and that $u$ is differentiable at $x=\ga(0)$. The same argument as 
  in~\ref{deriv<} shows that 
  $$
  d_xu\cdot w\le L_v(x,\dga(0))\cdot w
  \qquad\text{ for all }w\in T_xM.
  $$
  Applying this inequality to $-w$ and combining both inequalities
  we get that $d_xu=L_v(x,\dga(0))$.
  
  

  Now, since $u\prec L+k$, by lemma~\ref{Domd}, $H(x,d_xu)\le k$.
  Since
  $$
  u(\ga(0))-u(\ga(t))=A_{L+k}(\ga\vert_{[t,0]})
  =\int_t^0\bigl[\, L(\ga(s),\dga(s))+k\,\bigr]\;ds,
  $$
  then 
  $$
  d_xu\cdot\dga(0)=L(\ga(0),\dga(0))+k.
  $$
  Hence
  $$
  H(x,d_xu)=\sup_{v\in T_xM}\{\,d_xu\cdot v -L(x,v)\,\}\ge k.
  $$
  \end{proof}

  \index{dominated function|)}


  \newpage
  
   \section[Weak solutions of the Hamilton-Jacobi equation.]
   {Weak solutions of the Hamilton-Jacobi \\ equation.}
   
  \index{weak!solution of H-J|(}  
  \index{Hamilton-Jacobi!equation!weak solution|(}
  \index{solution!weak -- of H-J equation|(}

 
 We shall see in corollary~\ref{NoWeak} that
 there are no weakly differentiable subsolutions
 of~\eqref{HJ} for $k<c(L)$. 
 
 In the next proposition we show that when $k\ge c(L)$ there are
 always Lipschitz solutions of~\eqref{HJ}.
 On the other hand, in theorem~\ref{NWk>c} we show that
 when $M$ is compact the only energy level that supports
 a differentiable solution is $k=c(L)$. 
 {\bf When $M$ is non-compact there may be differentiable 
 solutions on $k>c(L)$, as example~\ref{} shows}.
 
 \begin{proposition}\label{WSol}
 If $k\ge c(L)$, then for any $y\in M$, the 
 function $u(x)=\Phi_k(y,x)$ satisfies
 $H(x,d_xu)=k$ for a.e.~$x\in M$.
 \end{proposition}
 

 
 \begin{proof}
 Since $u$ is Lipschitz, by by Rademacher's theorem~\cite{EG},
 it is differentiable at Lebesgue-almost every point.
 Since $u$ is dominated by proposition~\ref{Udiff}.\ref{Udiff.2},
 it is enough to see that $u$ is one-sided realized at every point.
 
 If $k>c(L)$, by proposition~\ref{PhikMin},
 for all $x\in M$, $x\ne y$, there exists a
 finite-time global minimizer $\ga\in\C_T(y,x)$ with
 $A_{L+k}(\ga)=\Phi_k(y,x)$. By the triangle inequality,
 the function $\de(t)=A_{L+k}(\ga|_{[0,t]})-\Phi_k(y,\ga(t))$
 is increasing. Also $\de(t)\ge 0$ and $\de(T)=0$. So that
 $\de(t)\equiv 0$ and  hence $\ga$ backward-realizes $u$ at $x$.
 
 If $k=c(L)$ then $u$ may be realized by an infinite
 semistatic orbit as follows. Let $\ga_n\in\C(y,x)$,
 $\ga_n:[-T_n,0]\to M$ be a Tonelli minimizer such that
 $$
 \Phi_c(y,x) \le A_{L+c}(\ga_n)\le \Phi_c(y,x)+\tfrac 1n.
 $$
 This implies that
 $$
 \Phi_c(y,\ga(s)) \le A_{L+c}(\ga_n|_{[-T_n,s]})
 \le \Phi_c(y,\ga(s))+\tfrac 1n,
 $$
 for all $-T_n\le s\le 0$. Thus
 \begin{equation}\label{WSol.1}
 \lv u(x)-u(\ga_n(s)) - A_{L+c}(\ga_n|_{[s,0]}) \rv \le
 \tfrac 1n.
 \end{equation}
 By proposition~\ref{priori}, $\lv\ga_n\rv<A$ for all $n$.
 Hence, if $x\ne y$, $\liminf_n T_n>0$. We can assume
 that $\dga_n(0)\to v\in T_xM$ and $T_n>\e>0$.
 Let $\la(t)=\pi\,\vr_t(v)$. Then $\ga_n|_{[-\e,0]}\to
 \la|_{[-\e,0]}$ in the $C^1$-topology. 
 Letting $n\to\infty$ on~\eqref{WSol.1},
 we get that $\la$ realizes $u$.
 
 \end{proof}
 
 
  \index{weak!solution of H-J|)}  
  \index{Hamilton-Jacobi!equation!weak solution|)}
  \index{solution!weak -- of H-J equation|)}
  
  
  \newpage
  
 
  \section{Lagrangian graphs.}
  
  
  
  We say that a function $u:M\to\re$ is a {\it
  subsolution}\index{subsolution of the Hamilton-Jacobi equation}
  of the Hamilton-Jacobi equation if
  $$
  H(x,d_xu)\le k,
  $$
   We shall prove that for $k>c(L)$ there is always a
  $C^\infty$ subsolution of the Hamilton-Jacobi equation
  and for $k<c(L)$ there are no (weakly) differentiable subsolutions.
  Hence
  
  \begin{theorem}\label{gafaA}
  If $M$ is any covering of a closed
  \index{covering}
  manifold, then 
 \begin{align*}
 c(L) &=\inf_{f\in C^{\infty}(M,\re)}\sup_{x\in M}H(x,d_{x}f)\\
      &=\inf\{k\in\re:\;\mbox{\rm there exists $f\in C^{\infty}(M,\re)$
 such that $H(df)<k$}\}. 
% \\
%       &=\inf\{k\in\re:\;\exists f\in\Lip(M,\re),\;
%         H(x,d_xf)\le k \text{ for a.e. }x\in M\}.
 \end{align*}
 where $H$ is the hamiltonian associated with $L$.
 \end{theorem}
  
 Theorem~\ref{gafaA} could be restated by saying that $c(L)$
 is the infimum of the values of $k\in\re$ for which
 $H^{-1}(-\infty,k)$ contains an exact lagrangian graph. This is
 a very geometric way of describing the critical value.
 
 
   
 In exercise~\ref{ExerSubsol}.\label{ExerSubsol.2}
 there is an elementary construction of a weak 
 subsolution for $k=c(L)$.  
 Theorem~\ref{gafaA} is an immediate consequence 
 of lemma~\ref{gafa>} and proposition~\ref{gafa<} below.
 
 \begin{lemma}\label{gafa>} %\quad\newline
 If there exists a $C^{1}$ function $f:M\to\re$
 such that $H(df)\le k$, then $k\geq c(L)$.  
 \end{lemma}
 
 \begin{proof}
  Recall that
 $$
 H(x,p)=\max_{v\in T_{x}M}\{p(v)-L(x,v)\}.
 $$
 Since $H(df)\le k$ it follows that for all $(x,v)\in TM$,
 $$
 d_{x}f(v)-L(x,v) \le k.
 $$
 Therefore, if $\ga:[0,T]\to M$ is any absolutely continuous closed
 curve with $T>0$, we have
 $$
 \int_{0}^{T}(L(\ga,\dot{\ga})+k)\,dt
 =\int_{0}^{T}\big[\,L(\ga,\dot{\ga})+k-d_{\ga}f(\dot{\ga})\,\big]\;dt\ge 0,
 $$
 and thus $k\geq c(L)$.
 \end{proof}
 
 \begin{remark}
 The utility of a differentiable subsolution of the Hamilton-Jacobi
 equation can be seen in lemma~\ref{gafa>}. If $H(du)\le c(L)$, then
 the lagrangian $L$ can be replaced by the lagrangian
 $$
 \L(x,v):=L(x,v)-d_xu(v) +c(L)\ge 0
 $$
 The new lagrangian $\L$ is positive, 
 has the same minimizing measures as $L$,
 its $\a$ and $\be$ functions are translates of those for $L$.
 The static set $\hS(\L)$ is contained in the level set $\L=0$. 
 \end{remark}
 
 
 
 
 
 
 
 
 \bigskip
 
 \begin{proposition}\label{gafa<}\quad\newline\indent
 For any $k>c(L)$ there 
 exists $f\in C^\infty(M,\re)$ such that
 $H(df)<k$.                    
 \end{proposition}
 
 \begin{proof}
 Set $c=c(L)$.
 Fix $q\in M$ and let $u(x):=\Phi _{c}(q,x)$. By the triangle
 inequality, $u\prec L+c$.
  By lemma~\ref{Domd}.\ref{UH<k}, $H(d_{x}u)\le c$ at any point $x\in M$ where
 $u(x)$ is differentiable. 
 
 We proceed to regularize $u$.
 Since $u$ is Lipschitz, by Rademacher's theorem (cf.~\cite{EG}) 
 \index{Rademacher's theorem}
 \index{Lipschitz!Rademacher's theorem}
 it is differentiable at Lebesgue almost every point.
 Moreover it is weakly differentiable (cf. \cite[Section
 4.2.3]{EG}), that is,
 for any $C^\infty$ function $\vr:U\to\re$ with compact support,
 equation~\eqref{Wdiff} holds.
 The next lemma completes the proof.
 \end{proof}
 
 

\begin{lemma}\label{mollification}
\index{mollification}\index{convolution}
 Let $M$ be a riemannian covering of a compact manifold
 and suppose that 
 $\sup_{|v|\le k}\lV\tfrac{\partial L}{\partial x}(x,v)\rV<+\infty$.
  If $u:M\to\re$ is weakly differentiable and 
  $$
  H(x,d_xu)\le k \text{ for a.e. }x\in M,
  $$
  then  for all $\de>0$
  there exists $f\in C^\infty(M,\re)$ such that $H(x,d_xf)<k+\de$
  for all $x\in M$.
 \end{lemma}
 
 \begin{remark}
 If $M$ is a covering $M\overset{\rho}\to N$
 of a compact manifold $N$ and the lagrangian $L$ on $M$
 is  lifted  from a lagrangian $\ell$ on $N$:
 $L=\ell\circ d\rho$, then the condition
 on $\tfrac{\partial L}{\partial x}$ follows.
 In general it may not be true as the
 lagrangian $L(x,v)=\tfrac 12 |v|^2 +\sin(x^2)$
 on $\re$ shows.
 
 The condition on $M$ can be replaced by some
 bounds in the riemannian metric of $M$,
 see appendix~\ref{AConv}.
 \end{remark}

 \begin{proof}
  We shall explain first how to prove the proposition
 in the case in which $M$ is compact and then we 
 will lift the construction
 to an arbitrary covering $M$.

 
%   {\bf By Nash theorem~\cite{Nash}, we can assume that
%   $M$ is isometrically embedded in some $\re^N$. }

  We can assume that $M\subset\re^N$.
  Let $U$ be a tubular neighbourhood of $M$ in $\re^N$, and 
  $\rho:U\to M$ a $C^\infty$ projection along the normal bundle. 
  Extend $u(x)$ to $U$
 by $\ov{u}(z)=u\bigl(\rho(z)\big)$. Then $\ov{u}(z)$ is also
 weakly differentiable.
 
 Extend the lagrangian to $U$ by 
 $$
 \ov{L}(z,v) = L\bigl(\rho(z),d_{z}\rho(v)) 
 +\tfrac 12\,\left\vert v - d_{z}\rho(v)\right\vert^2.
 $$
 Then the corresponding hamiltonian satisfies
 $\ov{ H}(z,p\circ d_{z}\rho) = H\bigl(\rho(z),p\bigr) $
 for $p\in T^{*}_{\rho(z)}M$. At any point
 of differentiability of $\ov{u}$, we have that
 $d_{z}\ov{u} = d_{\rho(z)}u\circ d_{z}\rho$, and
 $\ov{ H}(d_{z}\ov{u}) = H( d_{\rho(z)}u) \le k$.
 
 
 
 Let $\e > 0$ be such that 
 \begin{itemize}
 \item[(a)]  The  $3\e$-neighbourhood of $M$ in $\re^N$ is
            contained in $U$.
 \item[(b)] If $x\in M$, $(y,p)\in  T^*\re^N= \re^{2N}$,
            $\ov{ H}(y,p)\le k$ and $d_{\re^N}(x,y)<\e$,
            then $\ov{ H}(x,p)<k+\de$.
 \end{itemize}
 Let $\psi:\re\to\re$ be a $C^\infty$ function such that $\psi(x)\ge
 0$,
 $\text{support}(\psi)\subset (-\e,\e)$ and
 $\int_{\re^N}\psi(\vert x\vert)\, dx =1$. Let
 $K:\re^N\times\re^N\to\re$
 be $K(x,y)=\psi(\vert x- y \vert)$. Let $N_\e$ be the
 $\e$-neighbourhood of $M$ in $\re^N$. Define
 $f:N_\e\to\re$ by
 $$
 f(x) = \int_{\re^N} \ov{u}(y)\; K(x,y)\; dy\,.
 $$
 Then $f$ is $C^\infty$ on $N_\e$.
 
 Observe that $\partial_x K(x,y) = - \partial_y K(x,y)$.
 Since $\ov{u}(y)$ is weakly differentiable,
 for any $C^\infty$ function $\vr:U\to\re$ with compact support
 \begin{equation}\label{Wdiff}
 \int_{\re^N} (\vr\;d\ov{u} + \ov{u}\; d\vr)\; dx = 0\,.
 \end{equation}
 Hence
 \[- \int_{\re^N} \ov{u}(y)\, \partial_y K(x,y)\; dy=
 \int_{\re^N}K(x,y)\; d_{y}\ov{u}\; dy.\]
 Now, since
 \[d_{x}f =\int_{\re^N}\ov{u}(y)\, \partial_x K(x,y)\; dy,\]
 we obtain
 \[d_{x}f = \int_{\re^N} K(x,y)\;d_{y}\ov{u}\; dy.\]
 
   From the choice of $\e>0$ we have that
 $\ov{H}(x,d_{y}\ov{u})< k+\de$ for almost every
 $y\in\text{supp}K(x,\cdot)$ and $x\in M$. Since $K(x,y)\,dy       $ is
 a
 probability measure, by Jensen's inequality
 $$
 H(d_{x}f)
     \le \ov{ H}( d_{x}f)
     \le \int_{\re^N} \ov{H}(x, d_{y}\ov{u})\, K(x,y)\; dy
     < k+\de\,.
 $$
 for all $x\in M$.
 
 
 Now, suppose that $M$ is a covering of a compact manifold $N$ with
 covering projection $p$. 
 Assume that $N\subseteq \re^{N}$. 
 %Fix $q\in M$ and set $u(x):=\Phi_{c(L)}(q,x)$.
 We can regularize our function $u$ similarly as we shall
 now explain.
 For $\tx\in M$ let $x$ be the projection of $\tx$ 
 to $N$ and let $\mu_{x}$ be
  the Borel probability measure on $N$ defined by
 \[\int_{N}\vr\,d\mu_{x}=\int_{\re^{N}}(\vr\circ\rho)(y)\;K(x,y)\;dy,\]
 for any continuous function $\vr:N\to \re$.
 Then the support of $\mu_{x}$ satisfies
  \[\text{supp}(\mu_{x})\subset \{y\in
  N\,:\,d_{N}(x,y)<\varepsilon\}.\]
 Let $\widehat{\mu}_{\tx}$ be the Borel probability measure on $M$
  uniquely defined
 by the conditions:
 $\text{supp}(\widehat{\mu}_{\tx})\subset \{\widehat{y}\in
 M\,:\,d_{M}(\tx,\widehat{y})<\varepsilon\}$ and
 $p_{*}\widehat{\mu}_{\tx}=\mu_{x}$.
 Then we have
 \[\frac{d}{d\tx}\int_{M}\vr\,d\widehat{\mu}_{\tx}=
 \int_{M}d_{\widehat{y}}\vr\, d\widehat{\mu}_{\tx}(\widehat{y}),\]
 for any weakly differentiable function $\vr:M\to\re$.
 The condition~(b) above is now granted by the
 bound on $\tfrac{\partial L}{\partial x}$.
 Then the same arguments as above show that 
 \[f(\tx)=\int_{M}u(\widehat{y})\,d\widehat{\mu}_{\tx}(\widehat{y}),\]
 satisfies $H(d_{\tx}f)<k$.
 \end{proof} 
 
  
 Combining lemma~\ref{mollification} with lemma~\ref{gafa>},
 we obtain:
   
 \begin{corollary}\label{NoWeak}\quad\newline
 There are no weakly differentiable
 subsolutions of~\eqref{HJ} for $k<c(L)$.
 \index{Lipschitz!subsolutions of H-J}
 \index{weak!subsolution of H-J}
 \end{corollary}
 
 
 
 
 
 If we consider lagrangian graphs with other cohomology
 classes, we obtain Mather's alpha function:
 
 \begin{corollary} If $M$ is compact,
 \[\alpha(\kappa)= \inf_{[\omega]=\kappa}\; \sup_{x\in M} H(x,\omega(x)).\]
 \end{corollary}
 
 In particular the critical value of the abelian cover 
 $c_0=\min_\kappa \a(\kappa)$ is the infimum of the
 energy levels which  contain a lagrangian graph of any cohomology
 class in is interior.

 \begin{proof}
 Let us fix a closed one form $\omega_{0}$
 such that $[\omega_{0}]=\kappa$. By equality \eqref{alfa} we have
 that $\alpha(\kappa)=c(L-\omega_{0})$. Hence, it suffices to show that
 \begin{equation}
 c(L-\omega_{0})=\inf_{[\omega]=\kappa} \sup_{x\in M}
 H(x,\omega(x)).\label{fin}
 \end{equation}
 It is straightforward to check that the hamiltonian
 associated with $L-\omega_{0}$ is $H(x,p+\omega_{0}(x))$.
 Since all the closed one forms in the class
 $\kappa$ are given by $\omega_{0}+df$ where $f$
 ranges among all smooth functions, equality (\ref{fin}) is now
 an immediate consequence of theorem~\ref{gafaA}.

 \end{proof}
  
 \newpage
 
   \section{\bf Contact flows.}
   
   {\bf Let $N$ be a $2n+1$ smooth manifold and let $\a$ be
   a non-degenerate 1-form, i.e. $\a\wedge (d\a)^n$
   is a volume form for $N$.
   
   $\dim\ker d\a\equiv 1$,
   where
   $$
   \ker d_x\a:=\{\,v\in T_xN\,|\,
   d_x\a(v,w)=0,\;\forall w\in T_x N\,\},
   \quad x\in N.
   $$
   Define a vector field $Y$ on $N$ by $Y(x)\in\ker d_x\a$,
   
   \bigskip
   \hrule
   \bigskip
   
   \centerline{etc}
   
   \bigskip
   \hrule
   \bigskip
   

   \begin{proposition}\label{Contact}
   The hamiltonian flow on the energy level $[H\equiv k]$
   is a reparametrization of a $\Theta$-preserving flow
   if and only if $k>c(L)$.
   \end{proposition}
   
   \bigskip
   \hrule
   \bigskip
   
   >Tambi\'en se tiene desigualdad estricta cuando $M$ no es compacta?
   
   \bigskip
   \hrule
   \bigskip
   
   }
   \begin{proof}
   Let $\Sigma=[H\equiv k]$.
   Observe that 
   \begin{align*}
   \tfrac{d}{dt}\psi^*_t\Theta
   =\cL_X \Theta \vert_\Sigma
   =i_X\,d\Theta + d\, i_X \Theta 
   %\\ 
   =dH|_\Sigma+ d\Theta(X)
   =d\Theta(X)|_\Sigma.
   \end{align*}
   Thus a reparametrization of $\psi_t$ preserves $\Theta$
   if and only if its vector field 
   is a constant multiple of $Y=\tfrac 1{\Theta(X)}\, X$.
   Such reparametrization exists if and only if
   $\Theta(X)\ne 0$ on $\Sigma$.
   
   Since $H$ is convex, the sets $\Sigma_x:=T^*_xM\cap\Sigma$
   have compact convex interiors. The outward normal vector to
   $\Sigma_x$ at $(x,p)$ is $H_p(x,p)$.
   Observe that $\Theta(X)=p\cdot H_p$.
   Then if the point $(x,0)$ lies on the exterior of $\Sigma_x$
   and $(x,p)$ is the tangency point of a tangent line to
   $\Sigma_x$ passing through $(x,0)$, we have that
   $\Theta(X(x,p))=0$. Thus $\Theta(X)\ne 0$ on $\Sigma$ implies that
   the zero section of $T^*M$ lies in $H <k$. Since the zero section
   is the derivative of a constant function, by theorem~\ref{gafaA},
   $k\ge c(L)$.
   
   Moreover, if the zero section lies inside $H<k$, then
   $\Theta(X)>0$.
   
   Now suppose that $k>c(L)$. By theorem~\ref{gafaA},
   there exists $f\in C^\infty(M,\re)$ with $H(df)<k$.
   Define the new convex hamiltonian $\H(x,p):= H(x,p+d_xf)$.
   Then the energy level $[\H\equiv k]$ contains
   the zero section. Thus, if $\X$ is its hamiltonian
   vector field, then $\Theta(\X)>0$.
   Let $\L:=L-df$. Let $\H$ be the hamiltonian of $\L$
   and $\X$ its hamiltonian flow. Then 
   
   
   
   \end{proof}


  \newpage
  \section{Finsler metrics.}
   \index{Finsler!metric|(} 
   
     In this section we prove that if $k>c(L)$
  then the Euler-Lagrange flow on the energy
  level $E\equiv k$ is a reparametrization of
  the geodesic flow on the unit tangent bundle
  of a Finsler metric. This allows to borrow
  theorems from Finsler geometry.
  
  We give first a lagrangian proof and
  afterwards a hamiltonian proof, with a
  flavor of symplectic geometry.

 When $k>c(L)$, the subsolution $H(df)<k$
 obtained in proposition~\ref{gafa<} can
 be used to replace the lagrangian $L$
 by the lagrangian $\L=L-df$, then $\L+k>0$. 
 These lagrangians have
 the same energy function and equivalent variational
 principles. Hence they have the same lagrangian flow, 
 minimizing orbits, and the same action functional on
 closed curves and invariant measures. Their action
 potentials are related by
 $$
 {\mathbb P}_k(x,y) = \Phi_k(x,y) +f(y) - f(x). 
 $$

  Given a Finsler metric $\sqrt{F}$ and an absolutely
 continuous curve $\ga$, define its Finsler length as 
 $$
 l_F(\ga)=\int \sqrt{F(\dot\ga)}.
 $$
 Observe that since the Finsler metric is homogeneous of degree one, 
 the definition does not depend on the parametrization of the curve. 
 Define the Finsler distance as
 $$
 D_F(x,y)=\inf_{\ga\in\C(x,y)} l_F(\ga ).
 $$
  

   
   
 \begin{theorem}\cite{CIPP,IS1}\label{Finsler}
 If $k>c(L)$ then the lagrangian flow on the energy
 level $E\equiv k$ is a reparametrization of
 the geodesic flow of a Finsler metric on its
 unit tangent bundle.
 
 Moreover, if $f\in C^\infty(M,\re)$ is such that
 $H(df)<k$, the Finsler lagrangian $F$ can be taken
 to be 
 $$
 \sqrt{F(x,v)}= L(x,v) +k -d_xf(v)
 $$
 on $E(x,v)=k$, and then
 \begin{equation}\label{Finsler.2}
 \Phi_k(x,y)=D_F(x,y)+f(y)-f(x),\quad \text{ for all }x,y\in M.
 \end{equation}
 If $k>-\inf L$ then $f$ can  chosen $f\equiv 0$.
 \end{theorem}
 \index{geodesic flow!Finsler}
 \index{flow!Finsler geodesic --}

  \begin{proof}
 By theorem~\ref{gafaA}, if $k>c(L)$ then there exists
 $f\in C^\infty(M,\re)$ such that $H(df)<k$.
 Observe that
 $$
 H(x,0)=\max_{v\in T_xM}\big[\,0\cdot v-L(x,v)\,\big]
 =-\inf_{v\in T_xM}L(x,v).
 $$
 So that if $k>-\inf L$, we can choose $f\equiv 0$.
 
  Observe that if $H(df)<k$ then
 $k>c(L)>e_0$. So that the zero section 
 $M\times 0\subset E^{-1}\{k\}\subset TM$. Also, $L+k-df>0$.
 Then we can define a Finsler metric on $TM$ by
 $\sqrt{F}=L+k-df$ on $E^{-1}\{k\}$ and extend
 it by homogeneity. Since $k>c(L)=c(L-df)$, by
 proposition~\ref{PhikMin}, for any $x\ne y$ there
 exists a global minimizer on $\C(x,y)$ for $L+k-df$
 which has energy $k$. By the homogeneity of $\sqrt{F}$,
 we can restrict the curves in the definition of $D_F$ to those with
 energy $k$. Thus $\Phi_k-\Delta f=D_F$.
 
 To show that the lagrangian flow on $E\equiv k$
 is a reparametrization of the geodesic flow on the unit tangent
 bundle of $\sqrt{F}$, we only need to prove that sufficiently
 small Euler-Lagrange solutions with energy $k$ are geodesics of 
 $\sqrt{F}$.  
   Let $\L=L-df$. 
  The equality~\eqref{Finsler.2} implies that any $(\L+k)$-global
 minimizer is a geodesic for $\sqrt{F}$. So it is enough to prove that
 sufficiently small orbits with energy $k$ are global minimizers.
 
 Fix $x\in M$ and a small neighbourhood $\cN(x)$ of $x$ 
 such that for all $y\in\cN(x)$ there exists a unique Euler-Lagrange
 solution contained in $\cN(x)$, with energy $k$  and joining $x$ to $y$. 
 Let $\bP_k$ be the action potential for $\L$ and let
 \begin{gather*}
 \e=\inf\{\,\bP_k(x,z)\,|\, z\notin \cN(x)\,\}>0,\\
 \cM(x)=\{\,y\in M\,|\,\bP_k(x,y)<\tfrac\e2\,\}. 
 \end{gather*}
 Then $\cM(x)$ is a neighborhood of $\cN(x)$ with
 $\cM(x)\subset\cN(x)$. By the triangle inequality,
 any $(\L+k)$-global minimizer joining $x$ to a point $y\in\cM(x)$
 must be contained in $\cN(x)$. By proposition~\ref{PhikMin}
 such minimizer exists. Hence all the small solutions
 contained in $\cN(x)$ joining $x$ to points $y\in\cM(x)$
 are global minimizers.
  \end{proof}

 
 Now we shall give a hamiltonian proof of
 theorem~\ref{Finsler}.
  
  First we need a hamiltonian characterization of 
  Finsler lagrangians. 
  We say that a function $f:\re^n\to\re$ is a
  {\it Finsler 
  energy}\index{Finsler!energy}\index{energy!Finsler}
  if $f(x)>0$ when $x\ne 0$ and $f$ is positively homogeneous of
  order 2. Observe that $L$ is a Finsler lagrangian
  if and only if it is a Finsler energy on
  each tangent space $T_xM$


 \begin{lemma}\label{Homogeneous}
 Let $f\in C^2(M,\re)$ be strictly convex and superlinear. 
 Then $f$ is a Finsler energy if and only if 
 its convex dual $f^*$ is a Finsler energy.
  \end{lemma}
 
 \begin{proof}
 If $f$ is homogeneous, then, writing $v=\la\,w$,
 $\la>0$, we have
 $$
 f^*(\la\,p)=\max_v\big[\,\la p\; v-f(v)\,\big]
 =\max_w\big[\, \la^2 p\; w-\la^2 f(w)\,\big]
 =\la^2f^*(p).
 $$
 Let $\cL_f$ be the Legendre transform of $f$.
 Observe that 
 $$
 f^*\circ \cL_f(p) = f'(p)\cdot p-f(p)
 =\tfrac{d}{dt}f(tp)\big\vert_{t=1}-f(p)
 =2\,f(p)-f(p)=f(p).
 $$
 So that $f^*>0$.
 
 Since $f$ is strictly convex and superlinear,
 then $f^{**}=f$. Thus the argument above shows that
 if $f^*$ is homogeneous then so is $f$.
 \end{proof}

 \begin{lemma}\label{repara}
 If two convex hamiltonians have a common 
 regular level set $\Sigma$,
 then their hamiltonian vector fields 
 are parallel on $\Sigma$.
 \end{lemma}
 
 \begin{proof}
 Suppose that $H^{-1}\{k\}=G^{-1}\{\ell\}=\Sigma$.
 If $p\in\Sigma$, then $d_pH=\ker d_p G=T_p\Sigma$.
 Since $\Sigma$ is a regular energy level, these
 derivatives are nonzero. Moreover, since
 $H$ and $G$ are convex, they are positive
 on vectors pointing outwards $\Sigma$.
 Thus $d_pH=\la(p)\,d_pG$ for some $\la(p)>0$.
 Also,
 $$
 \om(X_H,\cdot)=d_pH=\la(p)\,d_pG=\la(p)\,\om(X_G,\cdot)
 $$
 So that $X_H=\la(p)\,X_G$.
 \end{proof}
 
 \begin{lemma}\label{Fmultiples}
 If $F$ is a Finsler lagrangian, then the orbits of its lagrangian
 flow are reparametrizations of the unit speed geodesics of $F$.
 \end{lemma}
 
 \begin{proof}
 We prove it for a Finsler hamiltonian $G$. The result
 follows because the Euler-Lagrange flow of $F$ is the
 hamiltonian flow of its energy function on $TM$
 with respect to the symplectic form $\cL_F^*(\om)$, where
 $\om$ is the canonical symplectic form on $T^*M$.
 
 Let $G$ be a Finsler hamiltonian. Since $G(x,\la p)=\la G(x,p)$,
 then
 $$
 d_{\la p}G(w)=\tfrac{d}{ds}G(x,\la p+sw)\big\vert_{s=1}
 =\tfrac{d}{ds}\la^2\,G(x,p+s\tfrac w\la)\big\vert_{s=1}
 =\la\,d_pG(w).
 $$
 If $X$ is the hamiltonian flow of $G$, then $\om(X(p),\cdot)=d_pG$.
 Therefore $X(\la\, p)=\la X(p)$.
 \end{proof}
 

 
  
 \noindent
 {\bf Hamiltonian proof of theorem~\ref{Finsler}:}
 
  If $k>c(L)$, by theorem~\ref{gafaA},
  there exists $f\in C^\infty(M,\re)$ with
 $H(df)<k$.
 Let $\H(x,p)\df H(x,p+d_xf)$.
 Then $\H^{-1}(]-\infty,k[)$ contains 
 the zero section of $T^*M$.
  Define a new hamiltonian
 $G:T^*M-M\times 0\to\re$ by $G\equiv\tfrac14$ on $\H^{-1}\{k\}$
 and $G(x,\la p)=\la^2\, G(x,p)$ for all $\la\ge 0$.
 By lemma~\ref{Homogeneous}, the convex dual $G^*$ 
 of $G$ is a Finsler metric on $TM$.
 
 Since by definition $G^{-1}\{\tfrac14\}=\H^{-1}\{k\}$, 
 it follows from lemma~\ref{repara}
 that the hamiltonian flows of $G$ and $\H^{-1}\{k\}$ 
 coincide up to reparametrization on the energy level
  $G^{-1}\{\tfrac14\}=\H^{-1}\{k\}$.
  
  The Legendre transforms $\cL_G(x,p)=(x,G_p)$ and
  $\cL_\H(x,p)=(x,\H_p)$ on $\H^{-1}\{k\}$ satisfy
  \begin{equation}\label{FinsLeg}
  \begin{aligned}
  G_p\cdot w &= 0 = \H_p\cdot p,\qquad &&\text{for all }w\in
  T_p(T^*_XM\cap\H^{-1}\{k\}); \\
  G_p\cdot p &= 2\,G(p)=\tfrac12>0,\quad &&\H_p\cdot p =\Theta(\X)>0,
  \;\;\text{ (by~\ref{Contact})}.
  \end{aligned}
  \end{equation}
  So that $\cL_G(x,p)=\la\,\cL_\H(x,p)$ for some $\la(x,p)>0$.
  Also $\cL_\H$ conjugates the hamiltonian flow on
  $\H^{-1}\{k\}$ to the lagrangian flow of $L$ on 
  $E^{-1}\{k\}$.
  By lemma~\ref{Fmultiples} the orbits of the Euler-Lagrange flow of $L$
  on $E^{-1}\{k\}$ are reparametrization of unit speed geodesic of
  $G^*$.
  
  We now compute $G^*$.  
  From~\eqref{FinsLeg},
  $\H_p=2\,\Theta(\X)\, G_p$, thus
  \begin{align*}
   G^*(x,H_p) &=G^*(x,2\,\Theta(\X)\,G_p)
          =4\,\Theta(\X)^2\, G^*(x,G_p) \\
          &=4\,\Theta(\X)^2\, G(p) 
          =\Theta(\X)^2=(p\cdot\H_p)^2\\
          &=(\L_v\cdot v)^2=(\L+k)^2.
  \end{align*}
  \qed
  
  Let $h:T^*M\to T^*M$ be the map
 $h(x,p)=(x,p+d_xf)$. Then the
 hamiltonian flows $\phi_t$ of $H$ and
 $\psi_t$ of $H_{df}(x,p)=H(x,p+d_xf)$
 are related by $h\circ\phi_t=\psi_t\circ h$.
 Thus, the hamiltonian flow of $H$ is 
 {\it conjugate} to a Finsler hamiltonian flow. 
 
 A hamiltonian level set can be made a
 Finsler level set if and only if it
 contains the zero section. On the other 
 hand, a lagrangian energy level $[E=k]$
 with $k>e_0$ always contains the zero section.
 
 
  \index{Finsler!metric|)}
  
  

 
 
  \newpage

 

 \section{Anosov energy levels.}

 An {\it Anosov energy level} 
 \index{Anosov energy level}
 \index{energy!level!Anosov}
 \index{energy!level!regular}
 is a regular energy level on
 which the flow $\phi_{t}$ is an Anosov flow.


 \begin{theorem}\label{gafaB}
 If the energy level $\E^{-1}(k)$ is Anosov, then
 \[k> c_{u}(\la).\] 
 \end{theorem} 
 \index{covering!universal}
 %\index{universal!cover}
  
 \begin{proof}
 
 
 Suppose that the energy level $k$ is Anosov
 and set $\Sigma \df \bH^{-1}(k)$. Let $\pi :T^{*}N\to N$
 denote the canonical projection.
 G.P. Paternain and M. Paternain proved in \cite{PP1} that $\Sigma$
 must project onto the whole manifold $N$ and that
  the weak stable foliation ${\cal W}^{s}$ of $\phi^{*}_{t}$ is
 {\it transverse} to the fibers of the fibration by $(n-1)$-spheres
 given by
 \[\pi|_{\Sigma}:\Sigma\to N.\]
 Let $\widetilde{N}$ be the universal covering of $N$.
 Let $\widetilde{\Sigma}$ denote the energy level $k$
 of the lifted hamiltonian $H$.
 We also have a fibration by $(n-1)$-spheres
 \[\widetilde{\pi}|_{\widetilde{\Sigma}}:\widetilde{\Sigma}\to
 \widetilde{N}.\]
 Let $\widetilde{{\cal W}}^{s}$ be the lifted foliation which is in
 turn
 a weak stable foliation for the hamiltonian flow
  of $H$
 restricted to $\widetilde{\Sigma}$.
 The foliation $\widetilde{{\cal W}}^{s}$ is also transverse to the
 fibration 
 $\widetilde{\pi}|_{\widetilde{\Sigma}}:\widetilde{\Sigma}\to
 \widetilde{N}$.
 Since the fibers are compact a result of Ehresmann (cf. \cite{CLN})
 implies that
 for every $(x,p)\in \widetilde{\Sigma}$ the map
 \[\widetilde{\pi}|_{\widetilde{{\cal W}}^{s}(x,p)}:\widetilde{{\cal
 W}}^{s}(x,p)\to \widetilde{N},\]
 is a covering map. Since $\widetilde{N}$ is simply connected, 
 $\widetilde{\pi}|_{\widetilde{{\cal W}}^{s}(x,p)}$ is in fact a
 diffeomorphism
 and $\widetilde{{\cal W}}^{s}(x,p)$ is simply connected.
 Consequently, $\widetilde{{\cal W}}^{s}(x,p)$ intersects
 each fiber of the fibration 
 $\widetilde{\pi}|_{\widetilde{\Sigma}}:\widetilde{\Sigma}\to
 \widetilde{N}$ 
 at just one point. In other words, each leaf $\widetilde{{\cal
 W}}^{s}(x,p)$
 is the graph of a one form. On the other hand it is well
 known that the weak stable leaves of an Anosov energy level are
 lagrangian
 submanifolds. Since any closed one form in the universal covering
 must be exact, it follows that each leaf $\widetilde{{\cal
 W}}^{s}(x,p)$ is
 an exact lagrangian graph. The theorem now
 follows from lemma~\ref{gafa>} and the fact that by 
 structural stability
  there exists $\varepsilon>0$
 such that for all $k'\in (k-\varepsilon,k+\varepsilon)$ the
 energy level $k'$ is Anosov.
 \end{proof}
 
 

 
 
 For $e\in\re$, let $\cA_e$ be the set of $\phi\in C^\infty(M)$ such
 that the 
 flow of $H+\phi$ is Anosov in $(H+\phi)^{-1}(e)$ and let $\cB_e$ be
 the set of $\phi\in C^\infty(M)$ such that $(H+\phi)^{-1}(e)$ contains
 no 
 conjugate points. As is well known  $\cA_e$ is open in $C^k$ topology
 and 
 $\cB_e$ is closed. On the other hand G. and M. Paternain~\cite{P1} have
 shown 
 that  $\cA_e$ is contained in $\cB_e$. It is proved in~\cite{CIS1} the
 following 
 
 \begin{theorem}\label{TipoOswaldo}
 The interior of $\cB_e$ in the $C^2$ topology is $\cA_e$.
 \end{theorem}
 This theorem is an extension to the Hamiltonian setting 
 of a result of  R. O. Ruggiero  for the geodesic flow~\cite{Rug}. 
 Theorems~\ref{TipoOswaldo} and~\ref{gafaB} have as corollary:
 
 



 
 \begin{corollary} Given a convex superlinear lagrangian $L$,
 $k<c_{u}(L)$ and $\varepsilon>0$ there exists a smooth function
 $\psi:N\to \re$ with $|\psi|_{C^{2}}<\varepsilon$ and such that the
 energy level
 $k$ of $L+\psi$ possesses conjugate points. 
 \end{corollary}

 
 \begin{proof}
 Suppose now that there exists $\epsilon>0$ such that for every $\psi$
 with
 $|\psi|_{C^2} <\epsilon $, the energy level $k$ of $\la+\psi$ has no
 conjugate 
 points. The main result in \cite{CIS1} says that in this case
 the energy level $k$ of $\la$ must be Anosov thus contradicting
 theorem~\ref{gafaB}. 
 \end{proof}
 
 
 \begin{proposition} \label{conj}
 If $k$ is a regular value of the energy such
 that $k<e$, then the energy level $k$ has conjugate points.

 \end{proposition}
 
 \begin{proof}
  If an orbit does not have conjugate points
 then there exist along it two subbundles called the {\it Green
 subbundles}.
 They have the following properties: they are invariant, lagrangian and
 they have dimension $n=\dim N$. Moreover, they
 are contained in the same energy level as the orbit and they
 do not intersect the vertical subbundle (cf. \cite{CI}).
 If $k$ is a regular value of the energy with $k<e$,
  then $\pi(\E^{-1}(k))$ is a manifold with boundary and
 at the boundary the vertical subspace is completely contained
 in the energy level. Therefore the
 orbits that begin at the boundary must have conjugate points, because
 at the boundary two $n$-dimensional subspaces contained in the energy
 level (which is $(2n-1)$-dimensional) must intersect.
 \end{proof}
 
 
 


 \newpage
 
 
  \bigskip
 \section{The weak KAM Theory.}
 \bigskip

  In the rest of this chapter we develop the theory
  of global weak KAM solutions, discovered by Albert Fathi.
  Recall that for an autonomous hamiltonian 
  $H:T^*M\to\re$, the Hamilton-Jacobi
  equation is 
  \begin{equation}
  H(x,d_xu)=k,
  \tag{H-J}
  \end{equation}
  where $u:U\subseteq M\to\re$.
   Here we are interested on {\it
  global\/}\label{gsHJ} solutions of~\eqref{HJ}, i.e.
  $u:M\to\re$ satisfying~\eqref{HJ}. 
  
  
  It may not be possible to obtain
  a smooth global solution of~\eqref{HJ}. Instead, for certain values
  of $k$, we shall find {\it weak\/}\label{wsHJ} solutions 
  of~\eqref{HJ}, which are Lipschitz. 
  By Rademacher's theorem~\cite{EG},
  a Lipschitz function is Lebesgue
  almost everywhere differentiable 
  so that~\eqref{HJ} makes sense in a.e. point.
  
  In fact, we have seen in corollary~\ref{NoWeak}
  that there are no weakly differentiable\footnote{In particular,
  by Rademacher's theorem, there are no Lipschitz global solutions.}
  global solutions for $k\le c(L)$.
  In theorem~\ref{NWk>c} we shall see that when
  $M$ is compact, there are no $C^{1+\Lip}$ solutions
  unless $k=c(L)$. In proposition~\ref{Busemann} we show
  that there are always Lipschitz solutions for $k=c(L)$,
  and in proposition~\ref{wKAMk} we show that when $M$ is non-compact
  there are Lipschitz solutions for $k>c(L)$.
  
 

  Given a dominated function $u\prec L+c$ define
  the sets
  $$
 \Ga^+_0(u) :=\{\,v\in\Sigma^+\,\vert\,
        u(x_v(t))-u(x_v(0))=\Phi_c(x_v(0),x_v(t)),\;\forall t>0\,\},
  $$
  $$      
  \Ga^-_0(u) :=\{\,v\in\Sigma^-\,\vert\,
        u(x_v(0))-u(x_v(t))=\Phi_c(x_v(t),x_v(0)),\;\forall t<0\,\},
  $$
  $$        
 \Ga^+(u) := {\textstyle\bigcup\limits_{t>0}}         
               \vr_t\bigl(\Ga^+(u)\bigr)
 \quad ,
  \qquad      
  \Ga^-(u) := {\textstyle\bigcup\limits_{t<0}}         
               \vr_t\bigl(\Ga^+(u)\bigr),
  $$ 
 where $x_v(t)=\pi\,\vr_t(v)$.
 We call $\Ga^+(u)$ (resp. $\Ga^+(u)$) the {\it basin\/} 
 \index{basin!of a weak KAM}
  of $u$ and 
 $\pi\big(\Ga^+_0(u)\setminus\Ga^+(u)\big)$
 (resp.  $\pi\big(\Ga^+_0(u)\setminus\Ga^+(u)\big)$) 
 \index{cut locus!of a weak KAM}
 the {\it cut locus\/} of $u$.
 
 

  
  
  \begin{Definition}\label{wKAMdef}
  \quad
  \newline\noindent
    A function $u_-:M\to\re$ is a {\it backward weak KAM\/} solution
    \index{KAM!solution|ibf}
    \index{weak!KAM solution|ibf}
  of~\eqref{HJ}  if
  \begin{enumerate}
  \item  $u_-\prec L+c$. 
  \item\label{wK-d.2} $\pi\big(\Ga^-_0(u_-)\big)=M$.
  \end{enumerate}
  
    \noindent
    A function $u_+:M\to\re$ is a {\it forward weak KAM\/} solution
  of~\eqref{HJ} if
  \begin{enumerate}
  \item  $u_+\prec L+c$. 
  \item\label{wK+d.2} $\pi\big(\Ga^+_0(u_+)\big)=M$.
  \end{enumerate}
  \end{Definition}

  \begin{remark}\label{diffWKAM}\quad\newline
        From the domination condition it follows that $u$ 
        is Lipschitz and that the curve $\ga$ is semistatic. 
        From proposition~\ref{Udiff},  at an interior point 
        $x$ of such a curve $\ga$, the function
	$u$ is differentiable and
        $H(x,d_xu)=c$. Moreover, item~\ref{Udiff.2} in
        proposition~\ref{Udiff} shows that if $u$ is differentiable
        at an endpoint of a curve $\ga$, then $H(x,d_xu)=c$.
        By Rademacher's theorem~\cite{EG}, $u$ is differentiable
        at (Lebesgue) almost every point in $M$. So that
        $u$ is indeed a weak solution of the Hamilton-Jacobi
        equation for $k=c(L)$.
  \end{remark}  


  \begin{theorem}\label{TWK}\quad\newline\indent
  If $u\in\cS^+$ (resp. $u\in\cS^-$)
   is a weak KAM solution, then
  \begin{enumerate}
  \item\label{TWK.1}
          \index{properties!Lipschitz --}
	  \index{Lipschitz!properties!weak KAM solution's}
          $u$ is Lipschitz and hence 
          differentiable (Lebesgue)-almost everywhere.
  \item\label{TWK.2}
         \index{properties!domination --}
	 \index{domination!properties}
        $u\prec L+c$.
  \item\label{TWK.2.5}
        $H(x,d_xu)=c(L)$ at any differentiability point $x$ of $u$.
  \item \label{TWK.3}
         \index{properties!covering --}
	 \index{covering!properties}
        {\bf Covering Property:} $\pi(\Ga^+_0(u))=M$.
  \item \label{TWK.4}
         \index{properties!graph --}
	 \index{graph!properties}
        {\bf Graph Property:}
        $\pi:\Ga^+(u)\to M$ is injective
        and its inverse is Lipschitz, with Lipschitz constant
        depending only on $L$.
  \item \label{TWK.5}
        \index{properties!smoothness --}
	\index{smoothness!properties}
        {\bf Smoothness Property:}
         $u$ is differentiable on $\Ga^+(u)$ and
        its derivative $d_xu$ is the image of
        $(\pi\vert_{\Ga^+(u)})^{-1}(x)$ under the
        Legendre transform $\cL$ of $L$.
        In particular, the energy of $\Ga_0^+(u)$ is $c(L)$.      
  \end{enumerate}
  \end{theorem}  
  
  \begin{proof}
  Items~\ref{TWK.2} and~\ref{TWK.3} are the definition of $u\in\cS^+$.
  Item~\ref{TWK.1} follows from proposition~\ref{Domd}.\ref{ULip} and
  Rademacher's theorem~\cite{EG}.
  Item~\ref{TWK.2.5} follows from proposition~\ref{Udiff}.\ref{Udiff.2} and
  the fact that by item~\ref{TWK.3}, $u$ is one-sided realized
  at every point.
  Item~\ref{TWK.5} follows from proposition~\ref{Udiff} and
  remark~\ref{RUdiff}. 

    
  We prove item~\ref{TWK.4}. Let $(z_1,v_1)$,~$(z_2,v_2)\in\Ga^+(u)$
  and suppose that $d_{TM}(v_1,v_2)>K\,d_M(z_1,z_2)$, where $K$ is
  from lemma~\ref{L:MATHER2} and the $A$ that we input 
  on lemma~\ref{L:MATHER2} is from
  lemma~\ref{priori}. 
  Let $0<\e<\e_1$, (with $\e_1$ from lemma~\ref{L:MATHER2})
  be such that $\vr_{-\e}(z_i,v_i)\in\Ga^+(u)$.
  Let $y_i=x_{v_i}(\e)$, $i=1,2$,
  then $u(y_i)=u(x_i)+\Phi_c(x_i,y_i)$, $i=1,2$.
  Then lemma~\ref{L:MATHER2} implies that
  $$
  \Phi_c(x_1,y_2)+\Phi_c(x_2,y_1)<\Phi_c(x_1,y_1)+\Phi_c(x_2,y_2).
  $$
  Adding $u(y_1)+u(y_2)$ and using that $u\prec L+c$, we get that
  \begin{align*}
  u(x_1)+u(x_2)
  &\le\Phi_c(x_1,y_2)+u(y_2)+\Phi_c(x_2,y_1)+u(y_1)
  \\
  &<\Phi_c(x_1,y_1)+u(y_1)+\Phi_c(x_2,y_2)+u(y_2)
  \\
  &=u(x_1)+u(x_2),
  \end{align*}
  which is a contradiction. This proves item~\ref{TWK.4}.
    
 \begin{figure}[h]
 \centerline{\epsfysize=4cm\epsfbox{crucel.eps}}
 \caption{\small \sc graph property.}
 \end{figure}
  \end{proof}
 
 
 \bigskip
 
 \begin{theorem}[Fathi~\cite{Fa4}]\label{NWk>c}\quad\newline\indent
 If $u\in C^{1+\text{Lip}}(M,\re)$, $M$ is compact and
 \begin{equation}\label{HJk}
 H(x,d_xu)=k,\qquad\forall x\in M.
 \end{equation}
 Then $k=c(L)$ and $u$ is a weak KAM solution $u\in\cS^-\cap\cS^+$.
 
 Conversely, if $u\in\cS^+\cap\cS^-$, then $u\in C^{1+\text{Lip}}$.
 \end{theorem}
 
 \begin{proof}
 From~\eqref{HJk} we get that 
 \begin{equation}\label{maxdu}
 \max_{v\in T_xM}d_xu\cdot v - L(x,v)=k.
 \end{equation}
 The strict convexity of $L$ on $T_xM$ implies that the maximum is
 attained at a unique vector $\xi(x)\in T_xM$. The implicit function
 theorem implies that the vector field $x\mapsto\xi(x)$ is Lipschitz.
 Thus it can be integrated to obtain a flow $\psi_t$ on $M$.
 
 From~\eqref{maxdu}, $u\prec L+k$ and the flow lines of $\psi_t$
 realize $L+k$, i.e.
 \begin{equation}\label{upsi}
 u\big(\psi_t(x)\big)=u\big(\psi_s(x))
 +\int_s^t\big[L(\psi_\tau(x),\tfrac{d\,}{d\tau}\psi_\tau(x))
 +k\big]\;d\tau,
 \;\forall s<t,\;\forall x\in M.
 \end{equation}
 
 From~\eqref{maxdu}, $L+k-du\ge 0$. This implies that
 $$
 c(L-du)=c(L)\ge k.
 $$
 Let $\mu$ be and invariant measure for the flow $\psi_t$.
 Observe that the measure $\nu:=\xi_*(\mu)$ is holonomic
 and $L+k=du$ on $\supp(\nu)$. Then
 $$
 \int (L+k)\; d\nu = \int(du) \; d\nu = 0.
 $$
 Hence $k\le c(L)$ and thus $k=c(L)$.
 Then by~\eqref{upsi}, $u\in\cS^-\cap\cS^+$.
 
 Conversely, if $u\in\cS^-\cap\cS^+$, by~\ref{TWK}.\ref{TWK.5}, 
 $u$ is differentiable and $\cL^{-1}(d_xu)=\xi(x)$. Moreover,
 $\Ga^\mp(u)=M$ and by~\ref{TWK}.\ref{TWK.4}, $\xi$ is Lipschitz.
 
 \end{proof}
 
  
  \newpage
  
  
  
  \section{Construction of weak KAM solutions}
  %\bigskip
  
  
  
  In this section we present three ways to construct
  weak KAM solutions: when the Peierls set is
  non-empty (in remark~\ref{RWK.1}.\ref{RWK.1.4}), 
  when the Peierls barrier is
  finite (in proposition~\ref{WK.1}),
  and the general case (in proposition~\ref{Busemann}).
  In the horocycle flow (example~\ref{Horocycle}), 
  the Peierls barrier is finite, but the Peierls set is empty. 
  In example~\ref{EmpresarioPobre}, $h_c=+\infty$ and also
  $\cP=\0$.
   
  When $M$ is compact we characterize all weak KAM solutions
  in terms of their values on each static class.

  
  We begin by observing that 
  
  \begin{lemma}\label{LInf}\quad
  \begin{enumerate}
  \item\label{LInf.1}
  If $\cU\subseteq\cS^-$ is such that
  $v(x):=\inf_{u\in\cU} u(x) >-\infty$, for all $x\in M$;
  then $v\in\cS^-$.
  \item\label{LInf.2}
  If $\cU\subseteq\cS^+$ is such that
  $v(x):=\sup_{u\in\cU} u(x) <+\infty$, for all $x\in M$;
  then $v\in\cS^+$.
  \end{enumerate}
  \end{lemma}  

 \begin{proof}
 We only prove item~\ref{LInf.1}.
 Since $u\prec L+c$ for all $u\in\cU$, then
 \begin{equation}\label{InfDom}
 v(y)=\inf_{u\in\cU}u(y)\le
 \inf_{u\in\cU}u(x)+\Phi_c(x,y)
 =v(x)+\Phi_c(x,y).
 \end{equation}
 Thus $v\prec L+c$.
 
 Let $x\in M$ and choose $u_n\in\cU$ such that $u_n(x)\to v(x)$.
 Choose $w_n\in\Ga^-(u_n)\cap T_xM$. Since by lemma~\ref{priori} $|w_n|<A$,
 we can assume that $w_n\to w\in T_xM$. By
 lemma~\ref{Domd}.\ref{ULip}, all the functions $u\in\cU$ have the
 same Lipschitz constant $K$ as $\Phi_c$.
 For $t<0$, we have that
 \begin{align*}
 v(x_w(t))&\le \liminf_n u_n(x_{w_n}(t))+K\,d_M(x_w(t),x_{w_n}(t))\\
          &=\liminf_n u_n(x)-\Phi_c(x_{w}(t),x)
                       +K\,d_M(x_w(t),x_{w_n}(t))  \\
          &=v(x)-\Phi_c(x_w(t),x) 
          \le v(x_w(t)), \qquad\text{because $v\prec L+c$}.             
 \end{align*}
 Hence $w\in\Ga^-(v)$.
 \end{proof}
 
 \subsection{Finite Peierls barrier.}
  
  \begin{proposition}\label{WK.1}
  If $h_c<+\infty$ and  $f:M\to\re$ is a continuous function.
  Suppose that
  \begin{align*}
  v_-(x)&:=\inf_{z\in M}f(z)+h_c(z,x)>-\infty,
        \\
  v_+(x)&:=\sup_{z\in M} f(z)-h_c(x,z)>-\infty.        
  \end{align*}
  Then  $v_-\in\cS^-$ and $v_+\in\cS^+$.
  \end{proposition}

 \begin{proof}
 We only prove that $v_-\in\cS^-$. By lemma~\ref{LInf} it is enough
 to prove that the functions $u(x)\mapsto h_c(z,x)$ are in $\cS^-$
 for all $z\in M$.
 
  By proposition~\ref{P.PSet}.\ref{TRH}, $u\prec L+c$.
  Now fix $x\in M$.  
  Choose  Tonelli minimizers $\ga_{n}:[T_{n},0]\to M$ such that
  $\ga_{n}\in\C(z,x)$, $T_{n}<-n$ and
  $$
  A_{L+c}\big(\ga_{n}\vert_{[T_{n},0]}\big)
  \le h_c(z,x)+\tfrac 1n.
  $$
  By lemma~\ref{priori}, $|\dga_{n}(0)|<A$ for all $n$.
  We can assume that $\dga_{n}(0)\overset{n}\to w\in T_xM$.
  If $-n\le s\le 0$, then $s>T_{n}$ and
  \begin{align*}
  A_{L+c}\big(\ga_{n}&\vert_{[T_{n},s]}\big)
    +\Phi_c(\ga_{n}(s),x)\le
    \\
    &\le A_{L+c}\big(\ga_{n}\vert_{[T_{n},s]}\big)
     +A_{L+c}\big(\ga_{n}\vert_{[s,0]}\big)
    \\
    &\le h_c(z,x)+\tfrac 1n
    \\
    &\le h_c(z,\ga_{n}(s))+\Phi_c(\ga_{n}(s),x)+\tfrac 1n,
    \quad\text{ for }-n\le s<0. 
  \end{align*}
  Taking $\liminf_{n\to\infty}$, we get that
  $$
  h_c(z,x_w(s))+A_{L+c}\big(x_{w}\vert_{[s,0]}\big)
  = h_c(z,x).
  $$
  Hence $w\in\Ga^-(u)$.
  \end{proof}
  

  


 \begin{remarks}\label{RWK.1}\quad\newline
 \begin{enumerate}
 \item \label{u<L+c}
        Observe that, since $\Phi_c(x,x)=0$, 
       \begin{equation*}
       u\prec L+c \iff u(x)=\inf_{z\in M}u(z)+\Phi_c(z,x).
       \end{equation*}
 \item Item~\ref{RWK.1}.\ref{u<L+c} implies that
       the function $h_c$ in proposition~\ref{WK.1}
       can not be replaced by $\Phi_c$. 
       In fact, the function $u_z(x)=\Phi_c(z,x)$ satisfies
       $u_z\prec L+c$, but in general $u\notin\cS^-$, if
       $z$ is not properly chosen.
 \item \label{HisU}
       For any $z\in M$ the function $u_z(x)=h_c(z,x)\in\cS^-$
       and $v_z(x):=-h_c(x,z)\in\cS^+$.
       \index{Peierls barrier!is weak KAM}
 \item \label{RWK.1.4}
       If $p\in\cP$ then $u_p(x):=\Phi_c(p,x)\in\cS^-$, because
       $$
       \Phi_c(p,x)\le h_c(p,x)
       \le h_c(p,p)+\Phi_c(p,x)\le\Phi_c(p,x).
       $$
       Similarly, $v_p(x):=-\Phi_c(x,p)\in\cS^+$.
 \end{enumerate}
 \end{remarks}


 \subsection{The compact case.}


  In the next theorem we characterize all weak KAM
  solutions when $M$ is compact. They are determined
  by their values on one point of each static class.



 Let $\bGa=\cP/d_c$ be the set of static classes of $L$.
 For each $\ga\in\bGa$ choose $p_\Ga\in\bGa$ and let
 $\bP=\{\,p_\Ga\,\vert\,\Ga\in\bGa\,\}$. We say that 
 a function $f:\bP\to\re$ is {\it dominated\/}\index{dominated}
 ($f\prec L+c$) \index{$\prec$}
 if \quad $f(p)\le f(q)+\Phi_c(q,p)$, \quad for all $p,\,q\in\bP$.
 
 
 \begin{theorem}\label{CorBij}\quad\newline\indent
 If $M$ is compact, the maps
 $\{\,f:\bP\to\re\,\vert\, f\prec L+c\,\}\to\cS^-$,
 $$
 f\longmapsto u_f(x):=\inf_{p\in\bP}f(p)+\Phi_c(p,x),
 $$
 and  $\{\,f:\bP\to\re\,\vert\, f\prec L+c\,\}\to\cS^+$,
 $$
 f\longmapsto v_f(x):=\sup_{p\in\bP}f(p)-\Phi_c(x,p),
 $$
 are bijective isometries in the {\rm sup} norm.
 \end{theorem}
 
 

 \begin{proof}
 
 We only prove it for $f\mapsto u_f$.
 
  The domination condition $f\prec L+c$ implies that $u_f>-\infty$.
 Then remark~\ref{RWK.1}.\ref{RWK.1.4} and lemma~\ref{LInf}.\ref{LInf.1}
 imply that $u_f\in\cS^-$.
 
 The injectivity follows from
 $$
 u_f(p)=\min_{q\in\bP}f(q)+\Phi_c(q,p)=f(p)\qquad\forall p\in\bP,
 $$
 because $f$ is dominated.
 
 To prove the surjectivity, let $u\in\cS^-$ and let $f=u\vert_{\bP}$.
 Given $x\in M$ choose $w\in\Ga_0^-(u)\cap T_xM$ and let
 $\ga(t)=\pi\,\vr_t(w)$. So that
  \begin{equation}\label{WK.2.1}
  u(x)-u(\ga(t))=A_{L+c}(\ga\vert_{[t,0]})=\Phi_c(\ga(t),x)
  \quad\text{ for }t<0.
  \end{equation}
  Choose $q\in\pi[\a\text{-lim}(v)]\subset\cP$ (by
  proposition~\ref{OmSemi}), and $t_n\to-\infty$ such that
  $\ga(t_n)\overset{n}\to q$. Using $t=t_n$ on
  equation~\eqref{WK.2.1}, in the limit
  we have that
  \begin{equation}\label{CorBij.1}
  u(x)=u(q)+\Phi_c(q,x).
  \end{equation}
  Now take $p\in\bP$ such that $d_c(p,q)=0$.
  Since $u\prec L+c$, 
  then 
  $$
  u(q)\le u(p)+\Phi_c(p,q)
  \le u(q)+\Phi_c(q,p)+\Phi_c(p,q)=u(q).
  $$
  So that 
  \begin{equation}\label{CorBij.3}
  u(q)=u(p)+\Phi_c(p,q).
  \end{equation}
  By the triangle inequality
  \begin{align*}
  \Phi_c(p,x)&\le\Phi_c(p,q)+\Phi_c(q,x) \\
  &\le \Phi_c(p,q)+\Phi_c(q,p)+\Phi_c(p,x)=\Phi_c(p,x).
  \end{align*}
  So that
  \begin{equation}\label{CorBij.2}
  \Phi_c(p,x)=\Phi_c(p,q)+\Phi_c(q,x).
  \end{equation}
  Combining equalities~\eqref{CorBij.1}, \eqref{CorBij.2}
  and~\eqref{CorBij.3},
  we have that 
  $$
  u(x)=u(p)+\Phi_c(p,x),
  $$
  with $p\in\bP$. So that $u_f\le u$.
  But since $u\prec L+c$ and $f=u\vert_\bP$, using
  remark~\ref{RWK.1}.\ref{u<L+c}, we have that $u\le u_f$.

  Now we see that $f\mapsto u_f$ is an isometry in the
  supremum norm $\lV\;\;\rV_0$. \index{$\lV\;\;\rV_0$}
   Given $x\in M$, choose $p_n\in\bP$
  such that $u_g(x)=\lim_n f(p_n)+\Phi_c(p_n,x)$. Since
  $u_f\prec L+c$, we have that
  \begin{align*}
  u_f(x)-u_g(x)
  &\le \liminf_n f(p_n)+\Phi_c(p_n,x)-g(p_n)-\Phi_c(p_n,x) \\
  &\le \lV f-g\rV_0.
  \end{align*}
  Changing the roles of $f$ and $g$ we get that
  $\lV u_f-u_g\rV_0\le \lV f-g\rV_0$.
  Since $u_f|_\bP=f$ and $u_g|_\bP=g$, then
  $\lV u_f-u_g\rV_0\ge \lV f-g\rV_0$.
   

 \end{proof}


 \begin{corollary}\label{Unique}
 There is only one static class
 if and only if  $\cS^-$ (resp. $\cS^+$)
 is unitary modulo an additive constant. 
 \end{corollary} 


   This characterization of weak KAM solutions allows us to recover
 the following theorem: We say that two weak KAM solutions
 $u_-\in\cS^-$ and $u_+\in\cS^+$ are {\it conjugate} if
 $u_-=u_+$ on $\cP$ and denote it by $u_-\sim u_+$.
 
 
 
 \begin{corollary}{\bf (Fathi~\cite{Fa2})}\label{HFathi}
  If $M$ is compact, then
 $$
 h(x,y) =\sup_{\stackrel{u_\mp\in\cS^\mp}{u_-\sim u_+}}
 \{\,u_-(y)-u_+(x)\,\}.
 $$
 \end{corollary} 
 \index{Peierls barrier!characterization!with KAM}
  
  
 \begin{proof}
 If $u_+\sim u_-$ and $p\in\cP$, from the domination we get
 \begin{align*}
 u_+(p) &\le u_=(x)+\Phi_c(x,p), \\
 u_-(y) &\le u_-(p)+\Phi_c(p,y).
 \end{align*}
 Adding these equations and using that $u_=(p)=u_-(p)$, we get
 $$
 u_-(y)-u_+(x)\le \Phi_c(x,p)+\Phi_c(p,y).
 $$
 Taking $\inf_{p\in\cP}$ and then $\sup_{u_=\sim u_-}$ we obtain
 $$
 \sup_{u_+\sim u_-}\big\{\,u_-(y)-u_+(x)\,\big\}\le h(x,y).
 $$
 On the other hand, let $u_=(z):=-h(z,y)$ and
 \begin{align}
 u_-(z):&=\min_{q\in\cP}\big\{\,u_=(q)+\Phi_c(q,z)\,\}
        \label{HFathi.1}
	\\
	&=\min_{q\in\cP}\big\{\,-h(q,y)+\Phi_c(q,z)\,\big\}
	\notag 
	\\
	&=\min_{q\in\cP}\big\{\,-\phi_c(q,y)+\Phi_c(q,z)\,\}
	\label{HFathi.2}
 \end{align}
 From remark~\ref{RWK.1}.\ref{HisU} and corollary~\ref{CorBij},
 $u_\pm\in\cS^\pm$. Since $u_+$ is dominated, from~\eqref{HFathi.1} we
 get that $u_+\sim u_-$. From~\eqref{HFathi.2}, $u_-(y)=0$ and hence
 $u_-(y)-u_+(x)=h(x,y)$.
 
 \end{proof} 




  
  \subsection{Busemann weak KAM solutions.}


 \index{Peierls barrier!infinite}
 When $h_c=+\infty$, we use another method to obtain weak KAM
 solutions, resembling the constructions of Busemann functions
 in riemannian geometry. 
 By proposition~\ref{CoveringProperty},
 $\Si^+\ne\0$ and $\Si^-\ne\0$ even when $M$ is non-compact.
 We call the functions of proposition~\ref{Busemann}
 {\it weak KAM Busemann} functions.
 \index{Busemann weak KAM}
 \index{weak!KAM solution!Busemann --}
 
 
 

 \begin{proposition}\label{Busemann}
 \begin{enumerate}
 \item[]
 \item\label{iBus.1}
       If $w\in\Si^-(L)$ and $\ga(t)=x_w(t)$, then
       \begin{align*}
       u_w(x) &= \inf_{t<0} \left[\Phi_c(\ga(t),x)
       -\Phi_c(\ga(t),\ga(0))\right]
       \\
       &=\lim_{t\to-\infty}\left[\Phi_c(\ga(t),x)
       -\Phi_c(\ga(t),\ga(0))\right]
       \end{align*}
       is in $\cS^-$.
 \item\label{iBus.2}
       If $w\in\Si^+(L)$ and $\ga(t)=x_w(t)$, then 
       \begin{align*}
       u_w(x) &= \sup_{t>0} \left[\Phi_c(\ga(0),\ga(t))
       -\Phi_c(x,\ga(t))\right]
       \\
       &=\lim_{t\to+\infty}\left[\Phi_c(\ga(0),\ga(t))
       -\Phi_c(x,\ga(t))\right]
       \end{align*}
       is in $\cS^+$.  
 \end{enumerate}
 \end{proposition}


 \begin{proof}
  We only prove item~\ref{iBus.1}.
 We start by showing that the function
 $$
 \de(t)=\Phi_c(\ga(t),x)-\Phi_c(\ga(t),0)
 $$ is increasing.
 If $s<t$, then
 \begin{align*}
 \de(t)-\de(s)&=\Phi_c(\ga(t),x)-\Phi_c(\ga(s),x)
                +\big[\,\Phi_c(\ga(s),\ga(0))
                -\Phi_c(\ga(t),\ga(0))\,\big]
              \\
              &=\Phi_c(\ga(t),x)-\Phi_c(\ga(s),x)
               +\Phi_c(\ga(s),\ga(t))
              \\
              &\ge 0,   
 \end{align*}
 where the  last inequality follows from the triangle
 inequality applied to the triple $(\ga(s),\ga(t),x)$.
 By the triangle inequality, $\de(t)\le \Phi_c(\ga(0),x)$,
 hence  $\lim_{t\downarrow-\infty}\de(t)=\inf_{t<0}\de(t)$ 
 and this limit is finite.
 
 \noindent Since
 \begin{align*}
 u(y)&=\inf_{t<0}\Phi_c(\ga(t),y)-\Phi_c(\ga(t),\ga(0))
     \\
     &\le \inf_{t<0}\Phi_c(\ga(t),x)
         +\Phi_c(x,y)-\Phi_c(\ga(t),\ga(0))
     \\    
     &=u(x)+\Phi_c(x,y),
 \end{align*}
 then $u\prec L+c$.

   Suppose that $x\in\cP\ne\0$.  Let $(x,v)\in\hS$ and $t<0$. 
   Let $p=x_v(t)$ and $y\in M$. Since
 $d_c(x,p)=0$, then
  \begin{align*}
 \Phi_c(y,x) &= \Phi(y,x)+\Phi_c(x,p) + \Phi_c(p,x)  \\
             &\ge \Phi_c(y,p) + \Phi_c(p,x) \ge \Phi_c(y,x).
 \end{align*}
 Hence $\Phi_c(y,x)=\Phi_c(y,p)+\Phi_c(p,x)$. For $y=\ga(s)$ (and 
 $p=x_v(t)$), we have that
 \begin{align*}
 u(x)-u(x_v(t)) 
  &= \lim_{s\to+\infty}\left[ \Phi_c(\ga(s),x)-\Phi_c(\ga(s),x_v(t))\right]
   = \Phi_c(x_v(t),x) \\
   &= A_{L+c}\left(x_v\vert_{[t,0]}\right).
 \end{align*}

 
 
 Now let $x\in M\setminus\cP$ and choose $y_n:[T_n,0]\to M$ a
 Tonelli
 minimizer such that $y_n(T_n)=\ga(\text{-}n)$, $y_n(0)=x$ and 
 $$
 A_{L+c}(y_n\vert_{[T_n,0]})\le\Phi_c(\ga(\text{-}n),x)+\tfrac 1n.
 $$
 This implies that 
 \begin{equation}\label{Busemann.1}
 A_{L+c}(y_n\vert_{[s,t]})\le\Phi_c(y_n(s),y_n(t))+\tfrac 1n,
 \quad\text{ for }T_n\le s<t\le 0.
 \end{equation}
 By lemma~\ref{priori}, $|\dy_n|<A$. We can assume that
 $\dy_n(0)\to v\in T_xM$. Then 
 \begin{equation}\label{inSiE}
 A_{L+c}(x_v\vert_{[t,0]})=\Phi_c(\ga(t),x)
 \quad\text{ for }\liminf_n T_n\le t\le 0.
 \end{equation}
 We prove below that $\lim_nT_n=-\infty$. Then
 $v\in\Si^-(L)$. Observe that for $T_n\le s\le 0$ we have that
 \begin{align*}
 \Phi_c(\ga(\text{-}n),x)
 &\le\Phi_c(\ga(\text{-}n),y_n(s))+\Phi_c(y_n(s),x) \\
 &\le A_{L+c}(y_n\vert_{[T_n,0]})
 \le\Phi_c(\ga(\text{-}n),x)+\tfrac 1n.
 \end{align*} 
 Since $y\mapsto \Phi_c(z,y)$ is uniformly Lipschitz, we obtain that
 \begin{align*}
 u(x)&=\lim_n\Phi_c(\ga(\text{-}n),x)-\Phi_c(\ga(\text{-}n),\ga(0))
     \\ 
     &=\lim_n\Phi_c(\ga(\text{-}n),x_v(s))
        +\Phi_c(x_v(s),x)-\Phi_c(\ga(\text{-}n),\ga(0))
     \\   
     &= u(x_v(s)) +\Phi_c(x_v(s),x)
       \qquad\text{ for all $s<0$.}
     \\  
     &= u(x_v(s))+A_{L+c}(x_v\vert_{[s,0]})
       \qquad\text{ because $v\in\Si^-$.}
 \end{align*}
 
  Now we prove that $\lim_nT_n=-\infty$. Suppose, for simplicity,
  that $\lim_n T_n = T_0>-\infty$. Since $\dy_n(0)\to v$, then
  $y_n\vert_{[T_n,0]}\overset{C^1}\longrightarrow x_v\vert_{[T_0,0]}$ and
  hence $\ga(\text{-}n)=y_n(T_n)\to x_v(T_0)=:p$. Since by lemma~\ref{priori}
  $|\dga|$ is bounded, we can assume that $\lim_n\dga(\text{-}n)=(p,w_1)$.
  By lemma~\ref{OmSemi}, $w_1\in\a\-lim(\dga)\subseteq\hS$. 
  From~\eqref{inSiE}, $\dx_v(T_0)\in\Si^\e$. 
  Since $\pi(w_1)=x_v(T_0)=p$, then
  lemma~\ref{GraphProperty} implies that $\dx_v(T_0)\in\hS$.
  Since $\hS$ is invariant, then $v\in\hS$ and hence
  $x=\pi(v)\in\pi(\hS)=\cP$. This contradicts the hypothesis
  $x\in M\setminus\cP$.
  \end{proof} 


  \begin{corollary}\label{cSne0}
  $\cS^-\ne\0$ and $\cS^+\ne\0$.  
  \end{corollary}

 \newpage
 \section{Higher energy levels.}
 
 The method in proposition~\ref{Busemann} allows us to obtain
 analogous weak KAM solutions on energy levels $k>c(L)$ when
 $M$ is non-compact.
 
 Let
 $$
 \Si^-(k):=\big\{\,v\in TM\,|\, A_{L+k}\big(x_v\vert_{[t,0]}\big)
 =\Phi_k\big(x_v(t),x_v(0)\big),\quad\forall t\le 0\,\big\}.
 $$
 \index{$\Si^\pm(k)$}
 For $u\prec L+k$ define
 \begin{equation}\label{GaUK}
 \Ga^-(u,k)=\big\{\,v\in TM\,|\, A_{L+k}\big(x_v\vert_{[t,0]}\big)
 = u\big(x_v(0)\big)-u\big(x_v(t)\big),\quad\forall t\le 0\,\big\}.
 \end{equation}
 \index{$\Ga^\pm(u,k)$}
 Let
 $$
 \cS^-(k)=\big\{\,u:M\to\re\,|\,u\prec L+k\text{ and }
 \pi\big(\Ga^-(u,k)\big)=M\,\big\}.
 $$
 \index{$\cS^\pm(k)$}
 Define similarly $\Si^+(k)$, $\Ga^+(u,k)$ and $\cS^+(k)$.
 Then the functions $u\in\cS^\pm$ satisfy Lipschitz,
 covering, graph and smoothness properties analogous 
 to theorem~\ref{TWK}.
  \index{properties!Lipschitz --}
  \index{Lipschitz!property!weak KAM solution's}
  \index{properties!domination --}
  \index{domination!properties}
  \index{properties!covering --}
  \index{covering!properties}
  \index{properties!graph --}
  \index{graph!properties}
  \index{properties!smoothness --}
  \index{smoothness!properties}
  
  Observe that we are requiring that the global minimizers 
  in~\eqref{GaUK} are defined in the whole ray $]-\infty,0]$.
  In the following lemma we show that there are not 
  such weak KAM solutions when $M$ is compact and $k>c(L)$.
  If we look for weak solutions of~\eqref{HJ} with realizing
  curves defined only on {\it finite} intervals then the
  action potential $u(x)=\Phi_k(p,x)$, $v(x)=-\Phi_k(p,x)$
  give such examples.
  
  \begin{proposition}\label{noWKAM}\quad\newline
  \noindent
  If $M$ is compact, then $\Si^\pm(k)=\0$ and hence
  $\cS^\pm(k)=\0$ for all $k>c(L)$.
  
  \noindent
  If $M$ is non-compact then $\Si^\pm(k)\ne\0$ for all $k>c(L)$.
  \newline
  Moreover, if $v\in\Si^\pm(k)$, then $\om\-lim(v)=\0$
  (resp. $\a\-lim(v)=\0$).
  \end{proposition}
  
  \begin{proof}
  Suppose that $v\in\Si^+(k)\ne\0$ and that $w\in\om\-lim(v)\ne\0$. 
  Observe that the orbit of $w$ can not be a fixed point
  (i.e. $w\ne 0$) because by~\eqref{e0<c0}, $k>c(L)\ge e_0$.
  Let $p=\pi(w)$ and choose $s>0$ such that $q=\pi(\vr_s(w))\ne p$.
  Then the same argument as in proposition~\ref{OmSemi} shows that
  $$
  d_k(p,q)=\Phi_k(p,q)+\Phi_l(q,p)=0,
  $$ 
  which contradicts proposition~\ref{p.c.1}.\ref{p.c.1.4}.

  This proves that $\Si^\pm(k)=\0$ when $M$ is compact.
  Observe that if $u_+\in\cS^+(k)$ then $\0\ne\Ga^+(u,k)\subseteq\Si^+(k)$.
  Hence $\cS^\pm(k)=\0$ when $M$ is compact.
  
 Assume now that $M$ is non-compact.
 We show that $\Si^+(k)\ne 0$.
 Let $x\in M$ and $\langle y_n\rangle\subseteq M$ such that 
 $d_M(x,y_n)\to+\infty$. Let $\ga_n\in \C_{T_n}(x,y_n)$ be a Tonelli
 minimizer such that 
 \begin{equation}\label{wKAMk.1}
 A_{L+c}(\ga_n)<\Phi_k(x,y_n)+\tfrac 1n.
 \end{equation}
 Then by lemma~\ref{priori}, $|\dga_n|<A$, and hence $T_n\to+\infty$.
 Let $v$ be a density point of $\{\dga_n(0)\}$.
 Since $\ga_n\vert_{[0,t]}\overset{C^1}\longrightarrow x_v\vert_{[0,t]}$ for
 all $t>0$, from~\eqref{wKAMk.1} we obtain that $v\in\Si^+(k)$.


  \end{proof}

  By corollary~\ref{NoWeak}, there are no weak solutions of~\eqref{HJ}
  for $k<c(L)$. We complete the picture with the following:

  \begin{proposition}\label{wKAMk}\quad
    
  If $M$ is non-compact, then $\cS^\pm(k)\ne\0$ for all $k\ge c(L)$.
  
  If $M$ is compact, then $\cS^\pm(k)=\0$ for all $k>c(L)$.
  \end{proposition} 
  
  \begin{proof}
  Proposition~\ref{noWKAM} proves the case $M$ compact.


 Now we reproduce the proof of proposition~\ref{Busemann}.
 Let $w\in\Si^+(k)\ne\0$, write $\ga(t):=x_w(t)$ and
 $$
 u_w(x):=\lim_{t\to+\infty} \left[\Phi_k(\ga(0),\ga(t))
       -\Phi_k(x,\ga(t))\right]
 $$
 The limit exists because by the triangle inequality
 for $\Phi_k$, the function
 $$
 \de(t)=\Phi_k(\ga(0),\ga(t))-\Phi_k(x,\ga(t))
 $$ 
 is non-decreasing and it is bounded above by $\Phi_k(\ga(0),x)$.
 The triangle inequality implies that $u\prec L+k$.
 
 Now let $x\in M$ and choose $y_n:[0,T_n]\to M$ a
 Tonelli minimizer such that $y_n(0)=x$, $y_n(T_n)=\ga(n)$  and 
 $$
 A_{L+c}(y_n\vert_{[0,T_n]})\le\Phi_k(x,\ga(n))+\tfrac 1n.
 $$
 This implies that for all $0<t<T_n$, 
 \begin{align}\label{wKAMk.2}
 \Phi_k(x,\ga(n))&\le 
    A_{L+k}\big(y_n\vert_{[0,t]}\big) + \Phi_k(y_n(t),\ga(n))
    \le \Phi_k(x,\ga(n))+\tfrac 1n.
 \end{align}
 By lemma~\ref{priori}, $|\dy_n|<A$.
 Since $y_n(T_n)=\ga(n)$ and by proposition\ref{noWKAM},
 $\ga(n)\to\infty$,
 then $\liminf_n T_n=+\infty$.
 Let $v\in T_xM$ be a density point of $\dy_n(0)$.
 
 Since $y_n|_{[0,t]}\overset{C^1}{\longrightarrow} x_v|_{[0,t]}$, 
 using~\eqref{wKAMk.2} we get that
 \begin{align*}
 u(x_v(t))&=\lim_n\Phi_k(\ga(0),\ga(n))-\Phi_k(x_v(t),\ga(n))
          \\
	  &=\lim_n \Phi_k(\ga(0),\ga(n)) 
	    -\Phi_k(x,\ga(n)) + A_{L+k}(y_n|_{[0,t]})
	  \\
	  &= u(x)+A_{L+k}(x_v|_{[0,t]}),  
 \end{align*}
 for all $0<t<\liminf_n T_n=+\infty$.
 
  \end{proof}

  \newpage
  
  \section{The Lax-Oleinik semigroup.}
  \index{Lax-Oleinik semigroup|(}
  \index{operator!Lax-Oleinik}
  \index{semigroup!Lax-Oleinik}
  
  The Lax-Oleinik semigroup was used to obtain weak
  KAM solutions by {\bf Lions, Papanicolaou and
  Varadhan}~\cite{LPV} on tori $\T^n$ and later by Fathi~\cite{Fa1} 
  for compact manifolds. 
  
  {\bf  Nahman and Roquejoffre~\nocite{NaRoque} }
  
  {\bf Roquejoffre~\nocite{Roque2}}
  
  Through this section we shall assume that
  $M$ is compact.
  The Lax-Oleinik semigroup $\langle T^-\rangle_{t\ge 0}$
  is the semigroup of operators $T^-_t:C^0(M,\re)\to C^0(M,\re)$
  defined by
  \begin{align*}
  T^-_t u(x) &= \inf_{\ga}\,\{\,u(\ga(0))
             +\int_0^t L(\ga(s),\dga(s))+c \;  ds\,\} \\
             &= \min_{y\in M}\,\{\,u(y)+\Phi_c(y,x;t)\,\},
  \intertext{where the infimum is taken on all 
  piecewise differentiable curves
  $\ga:[0,T]\to M$ with $\ga(T)=x$. Similarly, define}
  T^+_t u(x)&= \max_{z\in M}\,\{\,u(z)-\Phi_c(x,z;t)\,\}.
  \end{align*}
  
  \begin{proposition}\label{lax.1}\quad
  If $M$ is compact.
  
  \begin{enumerate}
  \item\label{lax.1.1}
        The unique $c\in\re$ for which the semigroup $\langle
        T^-_t\rangle_{t\ge0}$ has a fixed point is the critical
        value $c=c(L)$.
  \item\label{lax.1.2}\index{fixed point!Lax-Oleinik semigroup}
        \index{Lax-Oleinik semigroup!fixed point}
	\index{semigroup!Lax-Oleinik!fixed point}
        $u\in\cS^-$ $\Longleftrightarrow$ $u$ is a fixed point
        of the semigroup $\langle T^-_t\rangle_{t\ge 0}$.
  \item\label{lax.1.3}
       For all $\e>0$ there exists a Lipschitz constant $K=K(\e)>0$
        such that $T^-_t(C^0(M,\re))\subseteq \Lip_K(M,\re)$
	for all $t>\e$.
	\index{$\Lip_K(M,\re)$}
	\index{Lipschitz!images of Lax-Oleinik}
	\index{Lax-Oleinik semigroup!Lipschitz images}
	\index{operator!Lax-Oleinik!Lipschitz images}	
	\index{semigroup!Lax-Oleinik!Lipschitz images}	
  \item\label{lax.1.4}
       $T^-_t$ is a weak contraction: For all $t\ge0$ and
        all $u,v\in C^0(M,\re)$,
        $\lV T^-_t u-T^-_t v\rV_0\le \lV u-v\rV_0$.
	\index{weak!contraction}
	\index{Lax-Oleinik semigroup!weak contraction prop.}
	\index{operator!Lax-Oleinik!weak contraction}
	\index{semigroup!Lax-Oleinik!weak contraction}
	\index{contraction!weak}
	\index{properties!weak contraction --}
  \item\label{lax.1.5} For $\e>0$ there exists $K(\e)>0$
        such that if $u,v\in C^0(M,\re)$ and $s,\,t> \e$,
	then $\lV T^-_s u- T^-_t v\rV_0\le \lV u-v\rV_0+K(\e)\,|s-t|$.
  \end{enumerate}
  \end{proposition}
 
 \begin{proof}\quad
 
 \ref{lax.1.1}. By proposition~\ref{P.PhiT},
 $\lim_{t\to+\infty}\Phi_k(x,y;t)=\pm\infty$ when $k\ne c(L)$,
 uniformly in $x,\,y\in M$. This implies that there are no fixed
 points for $T^\pm_t$ when $c\ne c(L)$. The existence of a fixed
 point is given by item~\ref{lax.1.2}.
 
 \ref{lax.1.2}. Let $u\in\cS^-$. For $x\in M$ take 
 $v\in\Ga^-(u)\cap T_xM$ and let $\ga(t):=\pi\,\vr_t(v)$.
 For $t>0$ we have that
 \begin{align*}
 u(x)&=u(\ga(-t))+A_{L+c}\big(\ga|_{[-t,0]}\big)\\
     &=u(\ga(-t))+\Phi_c(\ga(-t),x;t).
 \end{align*}
 But since $u\prec L+c$, then
 $$
 u(x)\le u(y)+\Phi_c(y,x)\le u(y)+\Phi_c(y,x;t).
 $$
 Hence $T^-_t u = u$ for all $t>0$.
 
 Now suppose that $T^-_tu=u$ for all $t\ge 0$. Then
 $$
 u(x)\le u(z)+\Phi_c(z,x;t) \qquad\forall t\ge 0,\;\;\;\forall z\in M.
 $$
 Taking the infimum on $t\ge 0$, we get that $u\prec L+c$.
 Moreover,
 $$
 u(x)\le\min_{z\in M}u(z)+\Phi_c(z,x)\le\min_{z\in M}u(z)+h_c(z,x).
 $$
 For $t>0$ let $z_t$ be such that
 $u(x)=u(z_t)+\Phi_c(z_t,x;t)$.
 Let $t_n\to+\infty$ be a sequence such that the limit 
 $\lim_nz_{t_n}=y$ exists. Then 
 $$
 u(x)=u(y)+\lim_n\Phi_c(z_{t_n},x;t_n)
 \ge u(y)+h_c(y,x),
 $$
 where the last inequality follows from the definition of $h_c$
 and the Lipschitz property for $\Phi_c(x,y;t)$ in
 proposition~\ref{P.PhiT}.\ref{P.PhiT.I1}.
 Thus $u(x)=\min_{z\in M}u(z)+h_c(z,x)$, by
 proposition~\ref{WK.1},
 $u\in\cS^-$.
 
 \ref{lax.1.3}. Using proposition~\ref{P.PhiT}, 
 let $K=K(\e)$ be a Lipschitz constant for
 $(x,y;t)\mapsto \Phi_c(x,y;t)$ on $M\times M\times [\e,+\infty[$.
 Given $u\in C^0(M,\re)$, 
 $x,\,y\in M$ and $t>\e$, let $z\in M$ be such that
 $T^-_tu(y)=u(z)+\Phi_c(z,y;t)$. Then
 \begin{align*}
 T^-_tu(x)&\le u(z)+\Phi_c(z,x;t) \\
     &\le u(z)+\Phi_c(z,y;t)+\lv\Phi_c(z,x;t)-\Phi_c(z,y;t)\rv\\
     &\le T^-_tu(y) + K(\e)\;d(x,y).
 \end{align*}
 Changing the roles of $x$ and $y$ we get that
 $T^-_tu\in\Lip_K(M,\re)$.
 
 \ref{lax.1.5}. Observe that item~\ref{lax.1.5} implies
 item~\ref{lax.1.4}.
 Let $u,\,v\in C^0(M,\re)$, $s,\,t>\e$ and $x\in M$.
 Let $K=K(\e)$ be as in item~\ref{lax.1.3}. Choose
 $z\in M$ such that $T^-_su(x)=u(z)+\Phi_c(z,x;s)$. Then
 \begin{align*}
 T^-_tv(x)&\le v(z)+\Phi_c(z,x;t) \\
          &\le T^-_su(x)+\lv v(z)-u(z)\rv
	  +\lv\Phi_c(z,x;t)-\Phi_c(z,x;s)\rv \\
	  &\le T^-_tu(x)+\lV u-v\rV_0 + K(\e)\;|s-t|.
 \end{align*}
 Changing the roles of $u$ and $v$ we get item~\ref{lax.1.5}.
 \end{proof}  


 \begin{remark}{\it Fixed Points for the Lax-Oleinik semigroup.}
 \index{fixed point!Lax-Oleinik semigroup}
 \index{Lax-Oleinik semigroup!fixed point}
 \index{semigroup!Lax-Oleinik!fixed point}
 \index{operator!Lax-Oleinik!fixed point}
 
 
 {\bf Another proof of the existence of a fixed point
 in \ref{LaxConv}}
 
 We sketch here another proof of existence of a fixed 
 point for the Lax-Oleinik semigroup using properties~\ref{lax.1.3}
 and~\ref{lax.1.4} of proposition~\ref{lax.1}.
 
 
 Consider the semigroup $\la T^-_t$, with $0<\la<1$
 acting on the space $C^0(M,\re)/_\re$ of continuous
 functions modulo an additive constant.
 Then $\la T^-_t$ is a contraction whose image
 is in the compact space $\Lip_K(M,\re)/_\re$,
 independent of $\la$ or $t$.
 Let $u_{\la,t}$ be a fixed point for $\la T^-_t$. 
 Choosing a sequence $\la_n\to 1$ and a subsequence 
 such that $u_{\la_n,t}$ converges in $\Lip_K(M,\re)/_\re$,
 one obtains a fixed point $u_t$ for $T^-_t$.
 Now let $t_n\to 0$ and choose a subsequence such that
 $u_{t_n}$ converges to some $v$ in $\Lip_K(M,\re)/_\re$.
 Fix $s\in\re$ and let $N_n\in\Z$ be such that
 $\lim_n N_n t_n=s$. Since by the semigroup property,
 $u_{t_n}$ is also a fixed point for $T^-_{N_n t_n}$,
 using proposition~\ref{lax.1}.\ref{lax.1.5}, we get that
 in the $C^0$-topology,
 $$
 T^-_s v=\lim_n T^-_{N_n t_n} u_{t_n} =\lim_n u_{t_n} = v.
 $$
 So that $v$ is a fixed point for the whole semigroup
 $\langle T^-_t\rangle_{t\ge 0}$.
 \end{remark}
   
   
 \begin{theorem}[Fathi~\cite{Fa4}]\label{LaxConv}
 \index{Lax-Oleinik semigroup!convergence}
 \index{semigroup!Lax-Oleinik!convergence}
 \index{operator!Lax-Oleinik!convergence}
 If $M$ is compact, \newline
 for all $u\in C^0(M,\re)$ the uniform limit 
 $\lim\limits_{t\to+\infty}T^-_t u$ exists.
 \end{theorem} 
 
 To prove this theorem we shall need some lemmas.
 
 \begin{lemma}\label{Phi/T}\quad
 If $M$ is compact then
 $\lim_{t\to+\infty}\frac 1t\Phi_c(x,y;t)=0$,
 uniformly on $(x,y)\in M\times M$.
 \end{lemma}
 
 \begin{proof}
 Write $h_t(x,y):=\Phi_c(x,y;t)$. Then
 $$
 \inf_{M\times M} h_t + 2\;\inf_{M\times M} h_1
 \le h_{t+2}
 \le \inf_{M\times M} h_t + 2\;\sup_{M\times M} h_1.
 $$
 Then, writing $C=2\;(\sup h_1 -\inf h_1)$, we have that
 $$
 \forall t>2,\quad\sup_{M\times M} h_t-\inf_{M\times M} h_t \le 2\, C.
 $$
 Now let $u$ be a weak KAM solution. Since $T^-_tu=u$, we have
 that for all $t>0$,
 $$
 u(y)=\min_{x\in M} u(x)+\Phi_c(x,y;t).
 $$
 Thus, for all $(x,y)\in M\times M$ and $t> 2$,
 $$
 \lv \Phi_c(x,y;t)\rv\le \sup_{x\in M}u(x)-\inf_{x\in M}u(x)+2\,C.
 $$
 This implies the lemma.
 
 \end{proof}
 
 
 \begin{lemma}\label{EtoC}
 For all $\e>0$, there exists $T_0>0$ such that if $T>T_0$
 and $\ga:[0,T]\to M$ is a Tonelli minimizer, then
 $\lv E(\ga,\dga)-c(L)\rv <\e$.
 \end{lemma}
 
 \begin{proof}
 Let $T_n\to+\infty$ and let $\ga_n:[0,T_n]\to M$ be a Tonelli minimizer.
 Then by lemma~\ref{TonPri} there is $A>0$ such that $\lv \dga_n\rv<A$
 for all $n$. Then there exists a subsequence such that the
 probabilities $\mu_{\ga_n}$ converge weakly* to a limit $\mu$.
 Since $L$ is bounded on $|v|\le A$, and using lemma~\ref{Phi/T},
 $$
 A_{L+c}(\mu)=\lim_nA_{L+c}(\mu_{\ga_n})
 =\lim_n\tfrac 1{T_n}\Phi_c(\ga_n(0),\ga_n(T_n);T_n)
 =0.
 $$
 Hence $\mu$ is a minimizing measure and $E(\supp\mu)\equiv c(L)$.
 Thus $E(\ga_n,\dga_n)\to c(L)$.
 \end{proof}
 
 For $f:T^*M\to\re$ measurable and $u:M\to\re$ Lipschitz,
 define
 $$
 \esssup_{p\in T^*M} f\index{$\esssup$}\index{ess sup}
 =\inf_{\stackrel{A\subseteq T^*M}{\Leb(T^*M\setminus A)=0}}
 \sup_{p\in A}f(p),
 $$ 
 and
 $$
 \bH(f)=\esssup_{x\in M} H(x,d_xf).
 $$\index{$\bH(u)$}
 
 \begin{lemma}\label{u<bH}
 \index{mollification}
 \index{convolution}
 If $u:M\to\re$ is Lipschitz, then $u\prec L+\bH(u)$.
 \index{$\prec$}
 \end{lemma}
 
 \begin{proof}
 Using a convolution argument as in lemma~\ref{mollification},
 we can approximate both $\bH(u)$ and $u$ in the $C^0$ topology
 by a $C^\infty$ function. Hence we can assume that $u$ is $C^1$.
 Then for all $(x,v)\in TM$,
 $$
 d_xu\cdot v\le L(x,v)+H(x,d_xu)\le L(x,v)+\bH(u).
 $$
 If $\ga\in\C(x,y)$, then
 $$
 u(y)-u(x)\le\oint_\ga du\le\oint_\ga L+\bH(u).
 $$
 \end{proof}
 
 \begin{lemma}\label{bHtoC}
 If $u\in C^0(M,\re)$, then $\lim\limits_{t\to+\infty}\bH(T^-_t u)=c$.
 \end{lemma}
 
 \begin{proof}
 By proposition~\ref{lax.1}.\ref{lax.1.3},
 $T^-_tu$ is Lipschitz. If $x$ is a differentiability
 point for $T^-_tu$, then
 $$
 d_x (T^-_t u)=L_v(\ga(t),\dga(t)),
 $$
 where $\ga:[0,t]\to M$ is a Tonelli minimizer satisfying
 $\ga(t)=x$, and
 $T^-_t u(x)=u(\ga(0))+A_{L+c}(\ga|_{[0,t]})$.
 Thus $H(d_x(T^-_t u))=E(\ga(t),\dga(t))$.
 By lemma~\ref{EtoC}, $E(\ga,\dga)$ converges to $c(L)$
 uniformly on $t$.
 \end{proof}
 
 \noindent{\bf Proof of theorem~\ref{LaxConv}:}
 
 Let $u\in C^0(M,\re)$. 
 By proposition~\ref{lax.1}.\ref{lax.1.2}
 the family $\langle T^-_tu\rangle_{t\ge 1}$
 is equilipschitz. By Arzela-\'Ascoli theorem
 there exists a sequence $t_n\to+\infty$ such that
 the uniform limit 
 $u_\infty=\lim_n T^-_{t_n}u$ exists.
 {\bf By lemma~\ref{u<bH},} $T^-_tu\prec L+\bH(T^-_tu)$ and
 {\bf by lemma~\ref{bHtoC},} $\bH(T^-_tu)\to c$.
 This implies that $u_\infty\prec L+c$.
 
 Now we prove that $u_\infty$ is a fixed point of the
 semigroup $T^-_t$. Since $u_\infty\prec L+c$, then
 $u_\infty\le T^-_tu_\infty$ for all $t\ge0$.
 Since $T^-_t$ preserves the order, we get that
 $u_\infty\le T^-_su_\infty\le T^-_tu_\infty$
 for all $s\le t$. It is enough to show that
 there is a sequence $s_n\to+\infty$ such that
 $T_{s_n}u_\infty\to u_\infty$. 
 Write $s_n=t_{n+1}-t_n$, we can assume that
 $s_n\to+\infty$.
 Then, using~\ref{lax.1}.\ref{lax.1.4},
 \begin{align*}
 \lV T^-_{s_n}u_\infty-u_\infty\rV_0
 &\le \lV T^-_{s_n}u_\infty -T^-_{s_n}T^-_{s_n}u\rV_0
 +\Vert T^-_{t_{n+1}}u-u_\infty\Vert_0
 \\
 &\le \lV u_\infty-T_{t_n}u\rV_0
 +\Vert T^-_{t_{n+1}}u-u_\infty\Vert_0
 \to 0.
 \end{align*}
 
 It remains to show that the fixed point $u_\infty$
 of $T^-_t$ is the limit of $T^-_tu$.
 But for all $s\ge 0$,
 $$
 \lV T^-_{t_n+s}u-u_\infty\rV_0=
 \lV T^-_sT^-_{t_n} u - T^-_su_\infty\rV_0
 \le\lV T^-_{t_n} u -u_\infty\rV_0
 \overset{n}\longrightarrow 0.
 $$
 
 \qed
 
 The following corollaries are also due to A. Fathi.
 
 \begin{corollary}\label{PhiToH} %[\cite{Fa4}]
 If $M$ is compact, then
 $\lim\limits_{t\to\infty}\Phi_c(x,y;t)=h_c(x,y)$,
 uniformly in $x\in M$ and also in $y\in M$.
 \end{corollary}
 
 \begin{proof}
 The uniform limit on $y\in M$ follows from the
 equality $\Phi_c(x,y;t)=T^-_{t-1} u(y)$, where
 $u(z)=\Phi_c(x,z;1)$. And the uniform limit on
 $x\in M$ follows from $-\Phi_c(x,y;t)=T^+_{t-1}v$,
 where $v(z)=-\Phi_c(z,y;1)$.
  \end{proof}
  
  Recall that a {\it ray} is a curve $\ga(t)$
  such that it is a Tonelli minimizer for all $t\ge 0$
  (resp. $t\le 0$).
 
 \begin{corollary}\label{Ray=Semi}\index{ray}
 If $M$ is compact then the rays are semistatic.
%  If $\ga:]0,+\infty[\to M$ satisfies
%  $$
%  A_{L+c}(\ga|_{[0,t]})=\Phi_c(\ga(s),\ga(t);t)
%  \quad\text{ for all } t>0.
%  $$
%  Then $\ga$ is semistatic.
 \end{corollary}
 
   \begin{proof}
  For an a.c. curve $\ga$,
  and $s\le t$ in its domain, define\index{$\de(s,t)$}  
  $$
  \de(s,t):= A_{L+c}(\ga_{[s,t]})-\Phi_c(\ga(s),\ga(t)).
  $$
  Observe that $\de(s,t)\ge 0$. We show that
  $$
  [s_1,t_1]\subseteq[s_2,t_2]\implies \de(s_1,t_1)\le\de(s_2,t_2).
  $$
  Indeed, by the triangle inequality,
  $$
  -\Phi_c(\ga(s_2),\ga(s_1))
  -\Phi_c(\ga(s_1),\ga(t_1))
  -\Phi_c(\ga(t_1),\ga(t_2))
  \le -\Phi_c(\ga(s_2,t_2).
  $$
  Adding the equality
  $$
  A_{L+c}(\ga|_{[s_2,s_1]})
  +A_{L+c}(\ga|_{[s_1,t_1]})
  +A_{L+c}(\ga|_{[t_1,t_2]})
  =A_{L+c}(\ga|_{[s_2,t_2]})
  $$
  and using that $\de(s,t)\ge 0$, we get that
  $$
  \de(s_1,t_1)\le 
  \de(s_2,s_1)+\de(s_1,t_1)+\de(t_1,t_2)
  \le\de(s_2,t_2).
  $$
  
  Now suppose that $\ga:[0,+\infty[\to M$ is a ray.
  Observe that 
  \begin{align}\label{dstIS}
  \de(s,t)=\Phi_c(\ga(s),\ga(t);t-s)-\Phi_c(\ga(s),\ga(t)).
  \end{align}
  By theorem~\ref{LaxConv}, the map $t\mapsto\de(0,t)$ is bounded.
  Define
  $$
  \de(s,+\infty):=\lim_{t\to+\infty}\de(s,t)
  =\sup_{t\ge s}\de(s,t)\le \de(0,t).
  $$
  Since the map $s\mapsto\de(s,+\infty)$ is increasing,
  we have that
  $$
  \lim_{s\to+\infty}\de(s,+\infty)=0.
  $$
  
  Now we prove that the $\om$-limit vectors of a ray
  are static. Write $\ga(t)=\pi\vr_t(v)$, $t\ge 0$.
    Suppose that $t_n\to+\infty$ and $\dga(t_n)\to w\in TM$.
  Let $\eta(t)=\pi\,\vr_t(w)$. Since $\ga$~and $\eta$ are solutions 
  the Euler-Lagrange equation, then
  $\ga\vert_{[t_n-s,t_n+s]}
  \underset{C^1}\longrightarrow
  \eta\vert_{[-s,s]}$.
  We have that
    \begin{align*}
  A&_{L+c}\bigl(\eta\vert_{[-s,s]}\bigr)
    +\Phi_c(\eta(s),\eta(-s))=
    \\
    &= \lim_n\bigl\{\,A_{L+c}(\ga\vert_{[t_n-s,t_n+s]})
     +\lim_m \Phi_c(\ga(t_n+s),\ga(t_m-s))\,\bigr\}
    \\
    &= \lim_n\bigl\{\,A_{L+c}(\ga\vert_{[t_n-s,t_n+s]})
     +\lim_m A_{L+c}(\ga\vert_{[t_n+s,t_m-s]})
     -\de(t_n+s,t_m-s)\,\bigr\}
    \\
    &=\lim_n\;\lim_m\;A_{L+c}(\ga\vert_{[t_n-s,t_m-s]})
    -\de(t_n+s,t_m-s)
    \\
    &=\lim_n\;\lim_m\;\Phi_c(\ga(t_n-s),\ga(t_m-s))
    +\de(t_n-s,t_m+s)-\de(t_n+s,t_m-s)
    \\
    &=\Phi_c(\eta(-s),\eta(-s))
    +\lim_n\de(t_n-s,+\infty)-\de(t_n+s,+\infty)
    \\
    &=0.
  \end{align*}
  Thus $w\in\hS(L)$. 
  
  Finally, we prove that $\ga$ is semistatic.
   Let $t_n\to+\infty$ be such that
 the limit $p=\lim_n\ga(t_n)$ exists.
 Then  $p\in\cP=\pi(\hS)$. 
 Using~\eqref{dstIS}, we have that corollary~\ref{PhiToH},
 implies that 
 $$
 \de(s,+\infty)=\lim_n\de(s,t_n)=h_c(\ga(s),p)-\Phi_c(\ga(s),p)=0
 $$
 for all $s\ge 0$. Thus $\de(s,t)\equiv 0$ for all $0\le s\le t$,
 i.e. $\ga$ is semistatic.
 \end{proof}


  


 \newpage
 
 %\bigskip
 \section{The extended static classes.}
 %\bigskip
 
 The method in proposition~\ref{Busemann} resembles 
 the construction of Busemann functions in complete 
 manifolds of non-positive curvature. In that case,
 Ballmann, Gromov and Schroeder~\cite{BGS}
 proved that the manifold can be compactified 
 adjoining the sphere at infinity that can be 
 defined in terms of Busemann functions. 
 
 
 
 Here we emulate that construction to obtain a compactification
 of the manifold that identifies the points in the Peierls set
 which are in the same static class and adjoins what we call
 the {\it extended Peierls set} $\ffP^\mp$.  
 By definition of Busemann function, the extended static classes
 in $\ffB^\mp$ correspond to the $\a$-limits (resp. $\om$-limits)
 of semistatic orbits in the compactification.
 But as we shall see in example~\ref{ExNoBus} the classes in
 $\ffP^\mp\setminus\ffB^\mp$ do not correspond to
 $\a$ or $\om$ limits of orbits in $TM$.
 
 
 
 On $C^0(M,\re)$ we use the topology of uniform convergence 
 on compact subsets. Consider the equivalence relation on
 $C^0(M,\re)$ defined by $f\sim g$ if $f-g$ is constant.
 Let $\cF:=C^0(M,\re)/_\sim$ 
 with the quotient topology. 
 
 Let $\fM^-$ be the closure in $\cF$ of $\{\,f(x)=\Phi_c(z,x)\,|\,z\in
 M\,\}/_\sim$ and $\fM^+$ the closure in $\cF$ of
 $\{\,g(x)=\Phi_c(x,z)\,|\,z\in M\,\}$. Fix a point 
 $0\in M$. We can identify 
 $$\cF\approx\{\,f\in C^0(M,\re)\,|\,f(0)=0\,\}.$$
 
\begin{lemma} $\fM^-$ and $\fM^+$ are compact.
\end{lemma}
 
\begin{proof}
Observe that the functions in $\fM^-$ and $\fM^+$ are dominated.
By lemma~\ref{Domd}.\ref{ULip} the families $\fM^-$ and $\fM^+$
are equicontinuous. Since $M$ is separable by Arzel\'a-Ascoli theorem
$\fM^-$ and $\fM^+$ are compact in the topology of uniform convergence
on compact subsets.
\end{proof} 
 
 Then $\fM^-$ is the closure of the classes of the functions
 $$
 f_z(x):=\Phi_c(z,x)-\Phi_c(0,x),\qquad \forall\;z\in M
 $$ and $\fM^+$ is the closure
 of the classes of 
 $$
 g_z(x):=\Phi_c(x,z)-\Phi_c(0,z),\qquad \forall\;z\in M.
 $$
 
  \begin{lemma}\label{EquiDc}\quad
  \begin{enumerate}
  \item\label{EquiDc.1} 
  If $f_w(x)=
  %\Phi_c(w,x)-\Phi_c(w,0)=\Phi_c(z,x)-\Phi_c(z,0)=
  f_z(x)$ 
  for all $x\in  M$,
  then $d_c(w,z)=0$.
  \item\label{EquiDc.2} 
  If $g_w(x)=
  %\Phi_c(x,w)-\Phi_c(0,w)=\Phi_c(x,z)-\Phi_c(0,z)=
  g_z(x)$ 
  for all $x\in  M$,
  then $d_c(w,z)=0$.
  \end{enumerate}
  \end{lemma}
  
  \begin{proof}
  We only prove item~\ref{EquiDc.1}.
  Suppose that $f_z=f_w$.
  From $f_z(z)=f_w(z)$ we get that
  $$
  \Phi_c(w,z) = \Phi_c(w,0) -\Phi_c(z,0),
  $$
  and from $f_z(w)=f_z(w)$ we get
  $$
  \Phi_c(z,w) =-\Phi_c(w,0)+\Phi_c(z,0).
  $$
  Adding these equations we get that $d_c(z,w)=0$.

  Conversely, if $d_c(z,w)=0$ and $x\in M$, then
  $$
  \Phi_c(w,x)\le d_c(w,z)+\Phi_c(z,x)
  =\Phi_c(z,x)-\Phi_c(z,w)
  \le \Phi_c(w,x).
  $$
  Thus $\Phi_c(w,x)=\Phi_c(w,z)+\Phi_c(z,x)$ for 
  all $x\in M$. This implies that $f_z=f_w$.
  \end{proof}
 
  Then we have embeddings $M/_{d_c}\hookrightarrow\fM^-$,
  by $z\mapsto[f_z]\in\cF$
  and $M/_{d_c}\hookrightarrow\fM^+$ by $z\mapsto [g_z]\in\cF$,
  where $M/_{d_c}$ is the quotient space under the equivalence
  relation $x\equiv y$ if $d_c(x,y)=0$. Let $\fB^-$ be the functions
  defined in proposition~\ref{Busemann}.\ref{iBus.1}
  and $\fB^+$ those of~\ref{Busemann}.\ref{iBus.2}.
  Let
  $\ffB^+=\fB^+/_\sim$ and $\ffB^-=\fB^-/_{\sim}$.
  
  \begin{remark}\label{LB}
            By proposition~\ref{Busemann}, 
          if $p\in\cP\ne\0$ then
          $u_-(x):=\Phi_c(p,x)\in\ffB^-$ and
          $u_+(x):=-\Phi_c(x,p)\in\ffB^+$
          (modulo an additive constant).
 \end{remark}
  
  
  
  Observe that $d_c(z,w)=0$ if and only if $z=w$ or $z,\;w\in\cP$ and they
  are in the same static class. Under the identifications
  $M\hookrightarrow\fM^\mp$ we have that
  $\ffB^\mp\cup(M\setminus\cP)\subseteq\fM^\mp$ respectively.
  But this inclusion may be strict as the following example shows:
  
  \begin{example}\label{ExNoBus}
  $\ffB^-\cup(M\setminus\cP)\neq\fM^-$.
  \end{example} 
   
  Let $M=\re$ and $L(x,v):= \tfrac 12 v^2 - \cos(2\pi x)$,
  corresponding to the universal cover of the simple pendulum
  lagrangian. Then $c(L)=1$, and the static orbits
  are the fixed points $(2k+1,0)\in T\re$, $k\in\Z$.
  Moreover, $H(x,p)=\tfrac 12 p^2 +\cos(2\pi x)$ and
  the Hamilton-Jacobi equation $H(x,d_xu)=c(L)$ gives
  $d_xu=\pm 2\,\sqrt{1-\cos(2\pi x)}$. The function
  $$
  u(x)=\int_0^x 2\,\sqrt{1-\cos(2\pi s)}\;ds,
  $$
  with $d_xu\equiv +2\,\sqrt{1-\cos(2\pi x)}$, is
  in $\cS^-$, is the limit of $u_n(x):=\Phi_c(-n,x)-\Phi_c(-n,0)$
  but it is not a Busemann function associated to a semistatic orbit
  $\ga$ because if $\ga(-\infty)=2k+1\in\Z$ is the $\a$-limit of
  $\ga$, then the Busemann function $b_\ga$ associated to $\ga$
  satisfies
  \begin{equation}\label{BusEx}
  d_xb_\ga=\begin{cases}
         +2\,\sqrt{1-\cos(2\pi x)} &\text{if }\; x\ge\ga(-\infty), \\      
         -2\,\sqrt{1-\cos(2\pi x)} &\text{if }\; x\le\ga(-\infty).
         \end{cases}
  \end{equation}
  Similarly a function $v:\re\to\re$ with $d_xv\equiv
  -2\sqrt{1-\cos(1\pi x)}$ is in $\cS^+$ but it is not a 
  Busemann function. 
  
  
  
  Observe that in the Busemann function in~\eqref{BusEx},
  at the point $y=\ga(-\infty)+3$ the semistatic orbit
  $\eta(t)$ with $\deta(0)=\Ga^-(u)\cap T_yM$ has $\a$-limit
  $\eta(-\infty)=\ga(-\infty)+2\ne\ga(-\infty)$. Moreover,
  the Busemann function $b_\eta$ associated to $\eta$ satisfies
   \begin{equation*}
  d_xb_\eta=\begin{cases}
         +2\,\sqrt{1-\cos(2\pi x)} &\text{if }\; x\ge\ga(-\infty)+2, \\ 
         -2\,\sqrt{1-\cos(2\pi x)} &\text{if }\; x\le\ga(-\infty)+2;
         \end{cases}
  \end{equation*}
  so that $b_\eta\ne b_\ga$. In fact, there is no semistatic orbit
  passing through $y$ with $\a$-limit $\ga(-\infty)$. This implies that
  the Busemann functions can not be parametrized just by a
  (semistatic) vector based on a unique point $0\in M$ as in the
  riemannian case. In particular, it may not be possible to choose
  a single point $q_\a\equiv 0\in M, \; \forall \a\in\ffB^-$ in
  the construction for theorem~\ref{ThB}.
  
  \bigskip
  
  
  
    The functions in $\fB^-$ and $\fB^+$ are special
  among the weak KAM solutions. They  are ``directed'' towards a
  single static class and they are the most regular
  in the following sense:
  
  \begin{lemma}\label{BusReg}\quad
  \begin{enumerate}
  \item\label{BusReg.1}
      If $w\in\Si^-$ and $u_w\in\fB^-$
      is as in proposition~\ref{Busemann}.\ref{iBus.1}, 
      then
      $$
      u_w(x)=\max\{\,u(x)\,|\,
      u\in\cS^-,\; u(\pi(w))=0,\;
      w\in\Ga^-(u)\,\}.
      $$
  \item\label{BusReg.2}
      If $w\in\Si^+$ and $u_w\in\fB^+$
      is as in proposition~\ref{Busemann}.\ref{iBus.2}, 
      then
      $$
      u_w(y)=\min\{\,u(y)\,|\,
      u\in\cS^+,\; u(\pi(w))=0,\; 
      w\in\Ga^+(u)\,\}.
      $$
  \end{enumerate}
  \end{lemma}
  
  By the remark~\ref{LB}, this also holds
  for the functions 
  \break
  $u_-(x)=\Phi_c(p,x)$ and
  $u_+(x)=-\Phi_c(x,p)$ (modulo an additive constant),
  for any $p\in\cP$.
  
  \begin{proof}
  We prove item~\ref{BusReg.1}.
  Let $x:=\pi(w)$ and  $v\in\cS^-$ with $v(x)=u_w(x)=0$
  and $w\in\Ga^-(v)$.
  Let $x_w(t)=\pi(\Phi_t(w))$.
  Since $v\prec L+c$ and $w\in\Ga^-(v)$,
  then for $t<0$, we have that
  \begin{align*}
  v(y)&\le v(x_w(t))+\Phi_c(x_w(t),y)
      \\
      &= v(x)-\Phi_c(x_w(t),x)+\Phi_c(x_w(t),y).
  \end{align*}
  Since $v(x)=u_w(x)=0$, letting $t\downarrow-\infty$,
  we get that $v(y)\le u_w(y)$ for all $y\in M$.
  On the other hand, $u_w$ is in the set of such $u$'s,
  so that the maximum is realized by $u_w$.
  \end{proof}

 %  \newpage
  
  \bigskip
  
  Define
  $$
  \ffP^-:=\fM^- \setminus (M-\cP) 
  \quad,\qquad
  \ffP^+:=\fM^+ \setminus (M-\cP) .
  $$
  
  
  \bigskip
  
  \begin{proposition}\label{PPisWK}\quad
  
  The functions in $\ffP^-$  and $\ffP^+$ are weak KAM solutions.
  \end{proposition}
  
  \begin{proof}
  Let $u\in \fM^-\setminus (M\setminus\cP)$. 
  Since $u$ is dominated, we only have to prove the
  condition~\ref{wKAMdef}.\ref{wK-d.2}.
  Adding a
  constant, we can assume that $u(0)=0$. Then there is a sequence
  $z_n\in M$ such that $u(x)=\lim_n\Phi_c(z_n,x)-\Phi_c(z_n,0)$.
  Let $x\in M$ and let $\ga_n\in\C_{T_n}(z_n,x)$ be a Tonelli
  minimizer such that $T_n<0$, $\ga_n(0)=x$, $\ga_n(T_n)=z_n$ and
  $A_{L+c}(\ga_n)\le\Phi_c(z_n,x)+\tfrac 1n$.
  In particular
  $$
  A_{L+c}(\ga_n\vert_{[t,0]})\le \Phi_c(\ga_n(t),x)+\tfrac 1n,
  \qquad \forall\;\; T_n\le t \le 0.
  $$
  Since $u\in\ffP^-$, then we can assume that either
  $d_M(z_n,x)\to \infty$ or $z_n\to p\in\cP$. Since by
  lemma~\ref{priori} $|\dga_n|<A$ and $h_c(p,p)=0$ for $p\in\cP$,
  in either case we can assume that $T_n\to -\infty$.
  
  We can assume that $\dga_n(0)\to v\in T_xM$. Then for $t\le 0$
  \begin{align*}
  u(x_v(x))-u(x_v(t))
   &=\lim_n\Phi_c(z_n,x)-\Phi_c(z_n x_v(t)) \\
   &=\lim_n\Phi_c(z_n,x)-\Phi_c(z_n,\ga_n(t))+ K\,d_c(\ga_n(t),x_v(t))
   \\
   &\le \lim_n A_{L+c}\bigl(\ga_n\vert_{[t,0]}\bigr)+\tfrac 1n \\
   &\le A_{L+c}\bigl(x_v\vert_{[t,x]}\bigr),
  \end{align*}
  where $K$ is a Lipschitz constant for $\Phi_c$.
  
  \end{proof}


   For $p\in\ffB$ and $z\in M$ let $x\mapsto b_{p,z}(x)$ be the
   function in the class $p\in\ffB$ such that $b_{p,z}(z)=0$,
   i.e.
   $$
   b_{p,z}(x)=\lim_{y\to p}\Phi_c(y,x)-\Phi_c(y,z).
   $$
   We now give a characterization of weak KAM solutions similar to
   that of corollary~\ref{CorBij}. For each $\a\in\ffB^-$ choose
   $q_\a\in M$ such that there is a unique semistatic vector
   $v\in\Si^-$ such that $\pi(v)=q$ and the $\a$-limit of $v$
   is in the static class $\a$. This can be done by the graph 
   property~\ref{TWK}.\ref{TWK.4}. Moreover, choose them such that
   the map $\ffB^-\ni\a\mapsto q_\a\in M$ is injective. Let 
   $\bP:=\{\, q_\a\,|\, \a\in\ffB^-\,\}$. We say that a function
   $f:\bP\to M$ is {\it strictly dominated} if
   $$
   f(q_a) < f(q_\be) + b_{\be, q_\be}(q_\a) 
   $$
   for all $a\ne\be$ in $\ffB^-$. And we say that $f$ is {\it
   dominated} if $f(q_a) \le f(q_\be) + b_{\be, q_\be}(q_\a)$ for all 
   $a\ne\be$ in $\ffB^-$.

   
   \begin{theorem}\label{ThB}\quad
   
   \noindent
   The map 
   $\{\,f:\bP\to\re\,|\, f\;\text{\sl strictly dominated}\,\}\to 
   \{u\in\cS^-\,|\,u|_\bP\;\text{\sl strictly dominated}\,\}$, 
   $f\mapsto u_f$, given by
   $$
   u_f(x) :=\inf_{\a\in\ffB^-}f(q_\a)+b_{\a,q_\a}(x),
   $$
   is a bijection.
   \end{theorem}
   
   \begin{proof}
   We first prove that $u_f$ is bounded below. 
   The domination condition implies that $u_f(q_\a)=f(q_\a)$
   for all $\a\in\ffB^-$. Then the same argument as in
   formula~\eqref{InfDom}, shows that $u_f\prec L+c$.
   Fix $\a\in\ffB^-$, then for all $x\in M$,
   \begin{equation}\label{ufBounded}
   u_f(x)\ge u_f(q_\a)-\Phi_c(q_\a,x)=f(q_\a)-\Phi_c(q_\a,x)>-\infty.
   \end{equation}
   
   Since $u_f>-\infty$  and it is an infimum of
   weak KAM solutions, from lemma~\ref{LInf} we get that
   $u_f\in\cS^-$. Since $u_f(q_\a)=f(q_\a)$ for all $\a\in\ffB^-$,
   the map $f\mapsto u_f$ is injective.
   
   We now prove the surjectivity. 
   Suppose that $u\in\cS^-$ and $u\vert_\bP$
   is strictly dominated. Let
   $$
   v(x):=\min_{\a\in\ffB^-} u(q_\a) + b_{\a,q_\a}(x).
   $$
   Observe that the domination condition implies that
   \begin{equation}\label{uqa}
   v(q_\a)=u(q_a) \quad\text{ for all } \a\in\ffB^-.
   \end{equation}
   Given $x\in M$, let $\theta\in\Ga^-(u)\cap T_xM$ and let $\a\in\ffB^-$
   be the $\a$-limit of $\theta$. Then,
   $$
   u(x)=u(x_\theta(s))+\Phi_c(x_\theta(s),x)
   \qquad\text{ for all } s<0.
   $$
   Since $u$ is dominated, $u(q_\a)\le u(x_\theta(s))+\Phi_c(x_\theta(s),q_\a)$.
   Hence
   $$
   u(x)\ge u(q_a)-\Phi_c(x_\theta(s),q_a)+\Phi_c(x_\theta(s),x)
   \qquad\text{ for all } s<0.
   $$
   Taking the limit when $s\to-\infty$, we get that
   \begin{equation}\label{u>v}
   u(x) \ge v(x) \qquad\text{ for all } x\in M.
   \end{equation}
   
   
   
   Now we prove that $u=v$ on the projection of the backward
   orbits of vectors in $\Ga^-(u)$ ending at the points $q_\a$,
   $\a\in\ffB^-$. Let $\xi\in\Ga^-(u)\cap T_{q_\a}M$ and
   let $\be\in\ffB^-$ be the $\a$-limit of $\xi$.
    From the definition of $v(x)$
   for all $\e>0$ and $s<0$ 
   there exists $\ga=\ga(s,\e)\in\ffB^-$ such that
   \begin{align*}
   v(x_\xi(s))\ge v(q_\ga)+b_{\ga,q_\ga}(x_\xi(s)) -\e.
    \end{align*}
    Since  $\xi\in\Ga^-(u)\cap T_{q_\a}M$, then for $s<0$,  
    \begin{align}
      u(q_\a) &=u(x_\xi(s))+\Phi_c(x_\xi(s),q_\a) 
      \label{20-}\\
         &\ge v(x_\xi(s))+\Phi_c(x_\xi(s),q_\a) 
           \qquad\qquad\text{by~\eqref{u>v}}
	   \label{20}\\
	 &\ge v(q_\ga)+b_{\ga,q_\ga}(x_\xi(s)) -\e
      	   +\Phi_c(x_\xi(s),q_\a)
	   \notag \\  
         &=\lim_{t\to-\infty}
		   v(q_\ga)+\Phi_c(x_\xi(t),x_\xi(s))
           +\Phi_c(x_\xi(s),q_\a)-\Phi_c(x_\xi(t),q_\ga)-\e
	   \notag\\
	 &\ge v(q_\ga)
	   +\lim_{t\to-\infty}\Phi_c(x_\xi(t),q_\a)
	   -\Phi_c(x_\xi(t),q_\ga)-\e
	   \notag
	   \\
	   &\ge v(q_\ga) +b_{\ga,q_\ga}(q_\a) -\e
	   \notag
	   \\
	   &\ge v(q_\a)-\e \label{21}
	   \\
	   &= u(q_\a) -\e. \qquad\qquad\text{by~\eqref{uqa}.}
	   \notag
    \end{align}
  Letting $\e\downarrow 0$, from the equality between~\eqref{20} and~\eqref{21}
  we get that 
   \begin{equation}\label{inGav}
   v(q_\a)=v(x_\xi(t))+\Phi_c(x_\xi(t),q_\a)
   \qquad\text{ for all } t<0.
   \end{equation}
   But then 
   $$
   v(q_\be)\le v(x_\xi(t))+\Phi_c(x_\xi(t),q_\be)
   = v(q_\a)-\Phi_c(x_\xi(t),q_\a)+\Phi_c(x_\xi(t),q_\be).
   $$
   Equivalently
   $$
   v(q_\a)\ge v(q_\be)+\Phi_c(x_\xi(t),q_\a)-\Phi_c(x_\xi(t),q_\be).
   $$
   Taking the limit when $t\to-\infty$, we get that
   $v(q_\a)\ge v(q_\be)+b_{\be,q_\be}(q_\a)$.
   This contradicts the strict domination, hence $\be=\a$.
   Then, from the equality 
   between~\eqref{20-} and~\eqref{20}, we have that
   \begin{equation}\label{u=von0}
   u(x_\xi(t))=v(x_\xi(t))\quad\text{for all }t<0,
   \text{ and }\xi\in\Si^-\cap T_{q_\a}M,\;\a\-lim(\xi)=\a.
   \end{equation}

%    By the domination condition
%    \begin{equation}\label{QaQb}
%    v(q_a)=v(q_\be)+b_{\be,q_\be}(q_\a).
%    \end{equation}
   
  
  Now let $x\in M$ and $\a\in\ffB^-$. Let $\xi\in\Si^-\cap
  T_{q_a}M$ with $\a\-lim(\xi)=\a$.  Then for $t<0$,
  \begin{align*}
  u(x)&\le u(x_\xi(t))+\Phi_c(x_\xi(t),x) \\ 
      &= v(x_\xi(t))+\Phi_c(x_\xi(t),x)  &\text{by \eqref{u=von0}} \\
      &= v(q_\a) - \Phi_c(x_\xi(t),q_\a) + \Phi_c(x_\xi(t),x),
        &\text{by~\eqref{inGav}.}
  \end{align*}
  Letting $t\to-\infty$, we have that
  $$
  u(x)\le v(q_\a)+b_{\a,q_\a}(x)=u(q_\a)+b_{\a,q_\a}(x).
  $$
  Since $\a\in\ffB^-$ is arbitrary, from the definition of $v$
  we get that $u\le v$.
  \end{proof}





  \begin{theorem}\label{thC}\quad
  
  Given $u\in\cS^-$, for all $\a\in\ffB^-(u)$ choose
  $q_\a\in\pi[\La^-_0(\a)\cap\Ga^-_0(u)]$, and
  let 
  $\bP(u):=\{ q_\a\,|\,\a\in\ffB^-(u)\,\}$.
  Then
  $$
  u(x) =\inf_{q_\a\in\bP(u)} u(q_\a)+b_{\a,q_\a}(x)
  \qquad\text{for all }x\in M.
  $$
  \end{theorem}


  
   \begin{proof}
   Let $u\in\cS^-$. For all $\a\in\ffB^-(u)$, choose
   $q_\a\in\pi(\La^-_0(\a)\cap \Ga_0^-(u))$.
   Let $\bP(u):=\{\,q_\a\,|\a\in\ffB^-(u)\,\}$. We show 
   that $u|_{\bP(u)}$ is dominated. 
   Let $\a,\be\in\ffB^-(u)$ and let $\theta\in
   T_{q_\be}M\cap\La^-(\be)\cap\Ga^-(u)$. Then
   for $t<0$,
   \begin{align*}
   u(q_\a)&\le u(x_\theta(t))+\Phi_c(x_\theta(t),q_\a)
          \\
	  &=u(q_\be)-\Phi_c(x_\theta(t),q_\be)
	            +\Phi_c(x_\theta(t),q_\a).
   \end{align*}
   Letting $t\to-\infty$, we get that
   $u(q_\a)\le u(q_\be)+b_{\be,q_\be}(q_\a)$,
   for all $\a,\be\in\ffB^-(u)$.
   
   Let
   \begin{equation}\label{DefV}
   v(x):=\inf_{q_\a\in\bP(u)}u(q_\a)+b_{\a,q_\a}(x).
   \end{equation}
   The same arguments as in equation~\eqref{ufBounded} show that
   $v>-\infty$ and by lemma~\eqref{LInf} $v\in\cS^-$.
   
   Given $x\in M$, let $\theta\in\Ga^-(u)\cap T_xM$ and let
   $\a\in\ffB^-(u)$ be the $\a$-limit of $\theta$. Then,
   \begin{align*}
   u(x)&=u(x_\theta(s))+\Phi_c(x_\theta(s),x)
   &\text{ for all } s<0,
   \\
   u(x)&\ge u(q_a)-\Phi_c(x_\theta(s),q_a)+\Phi_c(x_\theta(s),x)
   &\text{because $u$ is dominated.}
   \end{align*}
   Since $\a\in\ffB^=(u)$, taking the limit when $s\to-\infty$, we get that
   \begin{equation}
   u(x) \ge v(x) \qquad\text{ for all } x\in M.
   \end{equation}

   
 
  Now let $x\in M$ and $q_\a\in\bP(u)$. Let
  $\xi\in\La(\a)\cap\Ga^-(u)\cap T_{q_\a}M$.  Then for $s<0$,
  \begin{align*}
  u(x)&\le u(x_\xi(s))+\Phi_c(x_\xi(s),x) \\
      &= u(q_\a) - \Phi_c(x_\xi(s),q_\a) + \Phi_c(x_\xi(s),x),
        &\text{because $\xi\in\Ga^-(u)\cap T_{q_\a}M$.}
  \end{align*}
   Since $\xi\in\La^-(\a)$, letting $s\to-\infty$,  we have that
  $$
  u(x)\le u(q_\a)+b_{\a,q_\a}(x).
  $$
  Since $q_\a\in\bP(u)$ is arbitrary, we get that
  $u\le v$.
  \end{proof}

 %\newpage

  
 \chapter{Examples}
 

%  \begin{Example}\label{ExRiem}
%   Riemannian Lagrangians.
%  \end{Example}
%  
%  \addcontentsline{toc}{section}{{\numberline{\theexample}Riemannian
%   Lagrangians.}}

%    \newcounter{ex}
%    \stepcounter{ex}
%    \addcontentsline{toc}{section}{\numberline{\thechapter-\theex}Riemannian Lagrangians.}
  

   \section{\large Riemannian Lagrangians.}
   %\bigskip

 
 For a riemannian lagrangian $L(x,v)=\tfrac 12 \lV v\rV_x^2$,
 we have that 
 $$
 e_0=c_u=c_0=c(L)=0.
 $$
 The static orbits are the fixed points $(x,0)$, $x\in M$.
 The only weak KAM solutions are the constant functions
 and so are the only subsolutions of~\eqref{HJ} for $k=c(L)$.
 
 Nevertheless, since the $\a$ function is superlinear,
 from~\eqref{alpha} when we add some closed forms to $L$ we
 obtain $c(L-\om)>c(L)=0$. Moreover, the semistatic 
 static orbits are not fixed points because by~\ref{StatE}
 they have energy $c(L-\om)>0$. By the graph properties,
 the static set of $L-\om$, the support of it minimizing
 measures and the basins of weak KAM solutions are
 geodesic laminations. Since semistatic geodesics are
 minimizing, they don't have conjugate points.
   
%  \begin{Example}
%  Mechanic Lagrangians.
%  \end{Example}

 \section{\large Mechanic Lagrangians.} 
 
 For a mechanic lagrangian 
 $L(x,v)=\tfrac 12 \lV v_x\rV^2 -U(x)$,
 we have that 
 $$
 c(L)=c_0=c_u=e_0=\min_{x\in M}E(x,0)=\max_{x\in M}U(x),
 $$
 the maximum of the potential energy.
 The static orbits are the fixed points $(x,0)$,
 where $U(x)$ is maximal. If $U$ is non-degenerate
 at a minimum $x$ (i.e. $d^2_x U$ is non-singular),
 then $x$ is a hyperbolic saddle point.
 The constant functions are always subsolutions
 of~\eqref{HJ} for $k=c(L)=e_0$.
 
%  \begin{Example} Symmetric Lagrangians.
%  \end{Example}

 \section{\large Symmetric Lagrangians.}
 
 For a symmetric lagrangian $L(x,v)=L(x,-v)$,
 and then $H(x,p)=H(x,-p)$.
 Since $v\mapsto L(x,v)$ is convex, it attains
 its minimum at $v=0$. Thus, if $M$ is compact,
 $$
 L(x,v)+e_0=L(x,v)-\max_{x\in M}L(x,0)\ge 0,
 $$
 and it is $0$ exactly at the fixed points $(x,0)$
 which maximize $L(x,0)$. This implies that
 $c(L)=e_0$. In fact, the constant functions are
 subsolutions of the Hamilton-Jacobi equation~\eqref{HJ}
 for $k=e_0$. The static orbits are the fixed points
 $(x,0)$ where $x$ maximizes $L(x,0)$.
 
 If $M$ is non-compact, the constant functions
 are still subsolutions of~\eqref{HJ} for $k=e_0$,
 and thus
 $$
 c(L)=c_0=c_u=e_0,
 $$ 
 but there may be no maximizers of $L(x,0)$ and 
 then the static set $\hS(L)=\0$ 
 (see example~\ref{EmpresarioPobre}).
 
 
%  \begin{Example}
%  The simple pendulum.
%  \end{Example}

  \section{\large Simple Pendulum.}
 
 The simple pendulum is the mechanic lagrangian
 $L:TS^1\to\re$ given by
 $$
 L(\theta,\dth)=\tfrac 12 \;|\dth|^2 -cos\theta,
 $$
 identifying the circle $S^1=[0,2\pi] \mod 2\pi$.
 Hence $c(L)=e_0=1$. The static set is $\hS(L)=\{(\pi,0)\}$.
 Since there is a unique static class, there are only
 one weak KAM solutions on $\cS^+$ and $\cS^-$ modulo
 an additive constant. But $\cS^-\ne\cS^+$ because the
 solutions are not differentiable. Their cut locus is
 $x=0\in S^1$. For $u\in\cS^+$ its is the whole
 $\Si^(L)$ and it is shown in figure~\ref{Pendulum}.
 
    \begin{figure}[ht]\label{Pendulum}
    \centerline{
    \epsfysize=4cm\epsfbox{pendulum.eps}
    }
    \caption{\sc simple pendulum.}
    \end{figure}
 
 
 If $\om\ne 0$ is a closed 1-form, $c(L-\om)>e_0=1$
 and the static set $\hS(L-\om)$ is a whole component
 of the energy level $E=c(L)$, oriented by $\om>0$.
 Again, there is only one static class and $\#\cS^\pm=1$
 modulo constants. Since $\pi(\hS(L-\om))=S^1$,
 the functions on $\cS^\pm$ are differentiable.
 
%  \begin{Example}
%  The flat Torus $\T^n$.
%  \end{Example}
 
 \section{\large The flat Torus $\T^n$.}
 
 Also called the harmonic oscillator, it is
 the riemannian lagrangian for $\T^n=\re^n/\Z^n$
 with the riemannian metric induced by the 
 euclidean metric on $\re^n$. The solutions 
 of~\eqref{EL} are the projection of straight
 lines in $\re^n$ parametrized with constant speed.
 If $\om(x,v)=\langle A,v\rangle$ with $A\in\re^n$ fixed,
 then $L-\om$ is globally minimized at all the vectors
 $(x,A)$. Since the orbits of this vector field
 are recurrent, they must be static. These vectors 
 are sent to $0$ by the Legendre transform.
 By the smoothness property~\ref{TWK}.\ref{TWK.5},
 $du=0$ and thus any weak KAM solution must be constant.
 By corollary~\ref{Unique}, there is only one static class.
 
 
 \bigskip
 \hrule
 \bigskip
 
%  \begin{Example}[\cite{Dias-Lopes}]
%  Flat domain for the $\be$-function.
%  \end{Example}

 \section{\large Flat domain for the $\be$-function.}

  This example is a special case of those studied by
  Carneiro and Lopes~\cite{CarLo}.
 Let $L:T\T^2\to\re$ be the lagrangian
 $$
 L\big((x,y),(u,v)\big)
 =\tfrac 12 (u^2+v^2) + v\,\sin(x),
 $$
 where $\T^2=[0,2\pi]_{\mod 2\pi}$.
 Then the lagrangian is minimized at the vectors
 $x=-\tfrac\pi 2$,~$v=1$ and $x=\tfrac\pi 2$,~$v=-1$.
 These vectors are tangent to the closed
 curves $\ga_+\;:\;x\equiv-\tfrac\pi 2$,~$\dy\equiv 1$ and
 $\ga_-\;:\;x\equiv\tfrac\pi 2$,~$\dy\equiv-1$.
 Since the curves $\ga_+$ and $\ga_-$ are closed,
 they must be static curves. Since their tangents
 are exactly all the vectors which minimize $L$,
 the Peierls set is just
 $\cP=\big(\{-\tfrac\pi 2\}\times[0,2\pi]\big)
 \cup\big(\{\tfrac\pi 2\}\times[0,2\pi]\big)$.
 The invariant measures $\mu_+:=\mu_{\ga_+}$,
 $\mu_-:=\mu_{\ga_-}$ supported on the closed
 orbits $\dga_+$, $\dga_-$ are minimizing measures
 with homology $\rho_+=(0,\tfrac 1{2\pi})$,
 $\rho_-=(0,-\tfrac 1{2\pi})$.
 
 Since Mather's $\be$-function is convex and attains 
 its minimum at $\rho_+$ and $\rho_-$, then the line
 $$
 [\rho_-,\rho+]
 =\big\{\,(0,t)\in H_1(\T^,\re)\approx\re^2\,|\,
 -\tfrac1{2\pi}\le t\le \tfrac 1{2\pi}\,\big\}
 $$
 must be a flat domain for the $\be$-function,
 in fact $\be\vert_{[\rho_-,\rho_+]}\equiv c(L)$.
 Since the static set $\hS(L)$ contains only the
 support of the measures $\mu_+$, $\mu_-$, then for 
 any homology $h\in[\rho_-,\rho_+$, $h=t\rho_-+(1-t)\rho_+$,
 the unique minimizing measure in homology $h$ is
 $\cM(h)=\{t\mu_-+(1-t)\mu_+\}$.
 
 
 
 \bigskip
 \hrule
 \bigskip
 
%  \begin{Example}\label{EmpresarioPobre}
%  A Lagrangian with Peierls barrier $h=+\infty$.
%  \end{Example}

  \section{\large A Lagrangian with Peierls barrier $h=+\infty$.}
  \label{EmpresarioPobre}
  \index{Peierls barrier!infinite}
  \index{Peierls set $\cP$!empty}
  

 Let $L:T\re^2\to\re$ be $L(x,v)=\tfrac 12\,|v|^2+\psi(x)$,
 where $|\cdot|$ is the euclidean metric on $\re^2$ and $\psi(x)$
 is a smooth function with $\psi(x)=\tfrac 1{|x|}$ for $|x|\ge 2$,
 $\psi\ge 0$ and $\psi(x)=2$ for $0\le|x|\le 1$.
 
 Then 
 $$
 c(L)=-\inf\psi=0,
 $$
 because if $\ga_n$ is a smooth closed curve with length
 $\ell(\ga_n)=1$, $|\ga_n(t)|\ge n$ and energy $E(\ga_n)=
 \tfrac 12\,\dga_n^2-\psi(\ga_n)\equiv 0$, then
 \begin{align*}
 c(L) &\ge -\inf_{n>0}A_L(\ga_n)=-\int_0^{T_n}\tfrac
 12\,\dga_n^2+\psi(\ga_n) \\
      &=-\int_0^{\frac 1{|\dga_n|}}|\dga_n|^2
       =-|\dga_n|\le -\sqrt{\tfrac 2n}\longrightarrow 0.
 \end{align*}
 On the other hand,
 $$
 c(L)=-\inf\,\{A_L(\ga)\,|\,\ga\text{ closed}\,\}\le 0,
 $$
 because $L\ge 0$.
 
 Observe that since $L>0$ and on compact subsets of $\re^2$,
 $L>a>0$, then  we have that
 $$
 d_c(x,y)=\Phi_c(x,y)>0 \text{ for all } x,y\in\re^2.
 $$
 Hence $\hat\Sigma(L)=\0$.
 
 Suppose that $h(0,0)<+\infty$. Then $u(x):=h(x,0)$ is in $\cS^+$.
 Let $\xi\in\Ga^+(u)\cap T_0\re^2$ and write
 $x_\xi(t)=(r(t),\theta(t))$ in polar coordinates about the origin
 $0\in\re^2$. Then $\liminf_{t\to+\infty}r(t)=+\infty$ because
 otherwise the orbit of $\xi$ would lie on a compact subset $E\equiv
 0$ and then $\0\ne\om\-lim(\xi)\subseteq\Sigma(L)=\0$, then
 $$
 |\dx_\xi(t)|=\sqrt{\tfrac 2{r(t)}}
 $$
 and
 $$
 L(\vr_t\,\xi)=|\dx_\xi(t)|^2=\sqrt{\tfrac 2{r(t)}}\;|\dx_\xi(t)|.
 $$
 Let $T_n\to+\infty$ be such that $r(T_n)\to+\infty$. Since $L+c=L\ge
 0$, then
 \begin{align*}
 h(0,0) &\ge\int_0^{+\infty}L(\vr_t(\xi))+c(L)
        =\int_0^{+\infty}\sqrt{\tfrac	2{r(t)}}
	\;\big[\,|\dr|+r\,|\dth|\,\big]\;dt
	\\
	&\limsup_{T_n}\int_0^{T_n}\sqrt{\tfrac 2r}\,\dr\;dt
	=\limsup_n\int_0^{r(T_n)}\sqrt{\tfrac 2 r}\;dr
	=+\infty.
 \end{align*}
 \qed
 
 \bigskip
 
   \begin{figure}[ht]
    \epsfysize=4cm\epsfbox{horo.eps}
    \hskip 2cm
    \epsfysize=4cm\epsfbox{horo1.eps}
%    \caption{\sc example~\ref{exbus}.}
   \end{figure}
    

 
%   \begin{Example}\label{exbus}
%   Peierls barrier $0<h<+\infty$, 
%   $\hS=\0$ and differentiable 
%  Busemann functions $u$ with $\fB^-(u)=\fB^+(u)=\{\a\}$.
%  \end{Example}

 \section{\large Horocycle flow.}\label{Horocycle}
 \index{Horocycle flow}\index{twisted geodesic flow}
 \index{Peierls set $\cP$!empty}
 
 {\it  Peierls barrier $0<h<+\infty$, 
   $\hS=\0$ and differentiable 
   Busemann 
   \linebreak 
   functions $u$ with $\fB^-(u)=\fB^+(u)=\{\a\}$.}
   
  \bigskip 

 Let $\bH:=\re\times]0,+\infty[$ with the Poincar\'e metric
 $ds^2=\tfrac 1{y^2}\,(dx^2+dy^2)$. Let $L:T\bH\to\re$ be a Lagrangian
 of the form
 $$
 L(x,v)=\tfrac 12\;\lV v\rV_x^2+\eta_x(v),
 $$
 where $\eta_x$ is a 1-form on $\bH$ such that $d\eta(v)$ is the area
 form an $\lv\cdot\rv_x$ is the Poincar\'e metric. The Euler-Lagrange
 equation is
 \begin{equation}\label{2ex.1}
 \tfrac{Dt}{dt}\,\dx=Y_x(\dx)=\dx^\perp,
 \end{equation}
 where $Y_x:T\bH\to T\bH$ is a bundle map such that
 $$
 d\eta_x(u,v)=\langle Y_x(u),v \rangle.
 $$
 The energy function is $E(x,v)=\tfrac 12\,\lV v\rV_x^2$.
 On the energy levels $E<\tfrac 12$ the solutions of~\eqref{2ex.1}
 are closed curves, and on $E=\tfrac 12$ the solutions are the
 horospheres parametrized by arc length.
 
 Choose the form $\eta(x,y)=\frac{dx}y$, where
 $(x,y)\in\bH=\re\times]0,+\infty[$. Then
 $$
 L\big((x,y),(\dx,\dy)\big) = \frac 1{2 y^2}
 \left( \dx^2+\dy^2\right)+\frac{\dx}y \cdot
 $$
 Observe that the form $\eta$ is bounded in the Poincar\'e metric, so
 that the Lagrangian is superlinear and satisfies the boundedness
 condition.
 
 It can be seen directly from the Euler-Lagrange equation that
 the curves $\dx=-y$, $\dy=0$ are solutions with
 \begin{equation}\label{2ex.2}
 L(\dx=-y,\dy=0)+\tfrac 12 \equiv 0.
 \end{equation}
 The images of these curves are the stable horospheres associated to
 the geodesic $x=0$, $\dy=y$, parametrized by arc length.
 
%  We show that $c(L)\ge\tfrac 12$. Let $\ga$ be the curve whose image
%  is the boundary of the rectangle $R=[-X,0]\times[1,Y]$ oriented
%  clockwise and with hyperbolic speed $\lV\dga\rV\equiv a$. Since 
%  $E(\ga)=\tfrac 12 \, a^2$, then
%  \begin{align}
%  \int_\ga L+\tfrac 12 a^2
%     &= \int_\ga v\cdot L_v=\int_\ga \lV v\rV^2+\int_R dA
%     = a\cdot \text{length}(\ga)-\text{area}(R),
%     \notag
%     \\
%     \text{area}(R) &= X\;\int_1^Y \frac 1{y^2}\; dy
%     = X \; (1-\tfrac 1 Y) 
%     \notag\\
%     \text{length}(\ga)&=X+\frac XY + 2\int_1^Y\frac 1y\;dy
%     =X\;(1+\tfrac 1Y)+2\; \log Y,
%     \notag\\
%     \int_\ga L +\tfrac{a^2}2 &= \big[X\,(1+\tfrac 1Y)+2\;\log Y\big]
%     \; a - X\, (1-\tfrac 1Y) 
%     \notag\\
%     &= X\; \big[ a+\tfrac {2a}Y -1\big] + 2a \;\log Y.
%     \label{2ex.3}
%  \end{align}
%  
%  If $a<1$ choose $Y>1$ such that $a+\tfrac {2a}Y-1<0$.
%  Then for $X$ large enough the formula~\eqref{2ex.3}
%  is negative. Hence $c(L)\ge \tfrac 12$.
%  
%  Observe that if $v=(\dx,\dy)$, $\dx<0$, then
%  \begin{equation}\label{2ex.4}
%  L=\tfrac 12 \; \lV v\rV^2 - \lV(\dx,0)\rV
%  \ge \tfrac 12 \;\lV v \rV^2 - \lV v \rV \ge - \tfrac 12.
%  \end{equation}
%  Hence $L+\tfrac 12\ge 0$ and then $c(L)\le \tfrac 12$.
%  
 
   We show that $c(L)=\tfrac 12$ and $h_c<+\infty$.
 Observe that if $v=(\dx,\dy)$, $\dx<0$, then
 \begin{equation}\label{2ex.4}
 L=\tfrac 12 \; \lV v\rV^2 - \lV(\dx,0)\rV
 \ge \tfrac 12 \;\lV v \rV^2 - \lV v \rV \ge - \tfrac 12.
 \end{equation}
 Hence $L+\tfrac 12\ge 0$ and then $c(L)\le \tfrac 12$.
   
   
  Now fix $x\in\bH$. For $r>0$ let $D_r$ be a geodesic disc
  of radius $r$ such that $x\in\partial D_r$.
  Let $\ga_r$ be the curve whose image
 is the boundary of $D_r$ oriented
 clockwise and with hyperbolic speed $\lV\dga\rV\equiv a$. Since 
 $E(\ga)=\tfrac 12 \, a^2$, then
 \begin{align}
 \int_{\ga_r} L+\tfrac 12 a^2
    &= \int_{\ga_r} v\cdot L_v=\int_{\ga_r} \lV v\rV^2+\int_{D_r} dA 
    \notag\\
    &= a\cdot \text{length}(\ga_r)-\text{area}(D_r),
    \notag\\ 
    &= a \cdot 2\pi\,\sinh(r) - 2\pi\,\cosh(r),
    \notag \\
    &= 2\pi \bigl[ \tfrac 12 \,(a-1)\, e^r - e^{-r} \bigr] +2\pi.
    \label{2ex.3}
 \end{align}
 
 
 If $a<1$, for $r>0$ large, formula~\eqref{2ex.3}
 is negative. Hence $c(L)\ge \tfrac 12$ and $c(L)=\tfrac 12$. 
 Moreover,
 $$
 h(x,x)\le\liminf_{r\to+\infty}A_{L+\frac12}(\ga_r)=2\pi<+\infty.
 $$

 

 
 


  We prove that $\hS=\0$. This implies that $h>0$.
 First observe that if $T$
%  $$
%  T(z)=\frac{az+b}{cz+d}
%  $$
 is an isometry of $\bH$, then $d(T_*\eta)$ is also the area form, so
 that $T_*\eta$ is cohomologous to $\eta$. This implies that given
 any two points $x,y\in\D$, there is a constant $b=b(x,y)\in\re$ such that
 for all $\ga\in\C(x,y)$,
 $$
 A_L(\ga)=A_L(T\circ\ga)+b(x,y).
 $$
 In particular, the map $dT$ leaves $\sigma(L)$ and $\hS(L)$
 invariant. Since a horocycle $h_1$ can be sent by an isometry 
 to another horocycle $h_2$ with $h_1\cap h_2\ne\0$, then the
 horocycles can not be static because it would contradict the 
 graph property.

 
 The constant function $u:\bH\to\{0\}$ satisfies $u\prec L+\tfrac 12$
 because $L+\tfrac 12\ge 0$ and by~\eqref{2ex.2} the vectors
 $v=(-y,0)\in\Ga^+(u)=\Ga^-(u)\in\Sigma^-$ are semistatic.
 Its derivative $du=0$ is sent by the inverse of the 
 Legendre transform $v\mapsto L_v=\langle
 v,\cdot\rangle_x+\tfrac{dx}y$ 
 to
 $$
 \tfrac 1{y^2}\;\langle v,\cdot\rangle_{\text{eucl}}
 =-\tfrac{dx}y,
 $$
 that is $v=(-y,0)$. Also
 $$
 H(du)=\tfrac 12\,\big\Vert du-\tfrac{dx}y\big\Vert^2
 =\tfrac 12\; \big\Vert\tfrac{dx}y \big\Vert^2=\tfrac 12.
 $$
 


 Let $T:\bH\hookleftarrow$ be an isometry of the hyperbolic metric.
 Write $\eta=\tfrac{dx}{y}$. Then $d\eta=A$ is the hyperbolic area
 2-form. Since $T$ is an isometry, then 
 $$
 d(T^*\eta)=T^*(d\eta)=d\eta.
 $$
 Hence the form $T^*\eta-\eta$ is exact on $\bH$ and there is a smooth
 function $v:\bH\to\re$ such that 
 $$
 T^*\eta-\eta=-dv.
 $$
 We show that $v$ is a weak KAM solution. Observe that
 \begin{align}
 L\circ dT(x,v)&=\tfrac 12\;\lV v\rV^2_x+T^*\eta(x,v) 
     \notag \\
               &=\tfrac 12\;\lV v\rV_x^2+\eta(x,v)-dv(x,v)
     \notag \\
	       &=L(x,v)-dv(x,v).
     \label{2ex.6}	       
 \end{align}
 Since by~\eqref{2ex.4} $L\circ dT+\tfrac 12\ge 0$, then
 \begin{equation}\label{2ex.5}
 dv \le L+\tfrac 12.
 \end{equation}
 Hence $v\prec L+\tfrac 12$. Moreover, the equality in~\eqref{2ex.5}
 holds exactly when $L\circ dT(x,v)+\tfrac 12=0$, i.e. when
 $dT(v)=(-y,0)\in T_{(x,y)}\bH$.
 
 Since the isometries send horospheres to horospheres, they are
 self-conjugacies of the hamiltonian flow and hence the curves
 $\ga(t)=T^{-1}(x-ty,y)$ realize $v$, i.e.
 $$
 v(\ga(t))-v(\ga(s))=\oint_\ga dv = \oint_\ga L+\tfrac 12.
 $$
 Here $v$ is the Busemann weak KAM solution associated to the class
 $T(\infty)\in\partial \bH$, on the sphere at infinity of $\bH$.
 
 We now show a picture of a non-Busemann weak KAM solution.
 We use the isometry $T:\bH\hookleftarrow$, $T(z)=-\tfrac 1z$,
 $z=x+iy\in\co$. The isometry $T=T^{-1}$ sends the line
 $t\mapsto -ty+iy$ to a horosphere with endpoint $0\in\co$, oriented
 clockwise. Choose $v:\bH\to\re$ such that $dv=\eta-T^*\eta$.
 Since $T$ leaves the line $\Re z=0$ invariant and $\eta=0$ on
 vertical vectors, hence $v$ is constant on $\Re z=0$.
 
 Now we describe the weak KAM solution
 $$
 w(z):=\min\{\,u(z),v(z)\,\}\in\cS^-.
 $$  
 Let $\ga(t)=-ty+iy$. Then, using~\eqref{2ex.6},
 \begin{align*}
 v(T^{-1}\ga(t))&=v(T^{-1}\ga(0))+\oint_{T^{-1}\circ\ga}dv \\
           &=0+\int_0^t\big[\,L\circ dT\circ\dga+\tfrac 12\,\big]
	      -\int_0^t\big[\,L\circ\dga+\tfrac 12\,\big] \\
	   &=0+\int_0^t\big[\,L\circ dT\circ\dga+\tfrac 12\,\big]-0.             
 \end{align*}
 Since by~\eqref{2ex.5} $L(x,v)+\tfrac 12>0$ when $v\ne -y+i0$, then
 $v(z)>0$ on $\Re z>0$ and $v(z)<0$ on $\Re z<0$. Thus
 \begin{gather}
 w(z)=\begin{cases}
 0 = u(z) &\text{if } \Re z>0. \\
 v(z)     &\text{if } \Re z<0.
 \end{cases} 
 \end{gather}
 The cut locus of $w$ is $\Re z =0$ and the basin 
 of $w$ is $\Ga^-(w)=A\cup dT(A)$ where $A$ is the set of vectors
 $(y,0)\in T_{x+iy}\bH$.


 \bigskip
 
 
 \chapter{Generic Lagrangians.}
 
  In \cite{Ma6}, Ma\~n\'e introduced the concept of {\it generic
 property of a lagrangian } $L$. A property $P$ is said to be generic for the
 lagrangian $L$ if there exists a residual set  ${\cal O}$  
  on  $C^\infty(M,\re)$ such that if $\psi$ is in
 ${\cal O}$ then $L+\psi$ has the property $P$. 

 A set is called {\it residual}\index{residual} if it contains a countable intersection
 of open and dense subsets. We recall which topology is used in $C^\infty(M,\re)$.
 Given $u\in C^\infty(M,\re)$, denote by $\lv u\rv_k$ its $C^k$-norm. Define
 \index{$\lV\cdot\rV_\infty$}
 $$
 \lV u\rV_\infty:=\sum_{k\in\na}\tfrac  {\text{arctan}(\lV u\rV_k)}{2^k}.
 $$
 Note that $\lV\,\cdot\,\rV_\infty$ is not a norm.
 Endow $C^\infty(M,\re)$ with the translation-invariant metric $\lV u-v\rV_\infty$.
 This metric is complete, hence the Baire property holds: any residual subsets of
 $C^\infty(M,\re)$ is dense.
 
 
 {\bf One of Ma\~n\'e's objectives was to show that Mather's theory of minimizing
 measures becomes much more accurate and stronger if one restricts
 it to generic lagrangians. In this chapter we shall prove
 }
 
 \begin{theorem}\quad\label{tgen}
 
 
 For every lagrangian $L$ there exist a residual set $\cO\subset C^\infty(M,\re)$
 such that if $\psi\in\cO$ 
 \begin{enumerate}
 \item\label{tgen1} The lagrangian $L+\psi$ has a unique minimizing measure $\mu$,
  and this measure is uniquely ergodic. \index{ergodic!uniquely}\index{uniquely ergodic}
  \item\label{tgen2} $\supp(\mu)=\hat\Si(L+\psi)=\Si(L+\psi)$.
  \item\label{tgen3} When $\mu$ is supported on a periodic orbit or a fixed point, 
  this orbit (point) $\Ga$ is hyperbolic and if its stable and unstable manifolds 
  intersect then they do it transversally.
 \end{enumerate}
 \end{theorem}

 \bigskip
 
   \bigskip
  
  Ma\~n\'e conjectures in  \cite{Ma3} that there exists a
  generic set ${\cal O}$ such that this unique minimizing measure
  is supported on a periodic orbit or an equilibrium point.
  
  Item~\ref{tgen1} of theorem~\ref{tgen}  was proved by Ma\~n\'e in \cite{Ma2}.
 Item~\ref{tgen2} is proved in \cite{CP} and item~\ref{tgen3} in 
 \cite{CI}.  Th proof of item~\ref{tgen1} presented here is extracted from
 a more general result in \cite{BC} about families of lagrangian systems.
 
 We will prove only the first part and give only the ideas
 of the proofs of the last part.



  \bigskip
  \bigskip
 
 \section{Generic Families of  Lagrangians.}
 
In this chapter the manifold $M$ is compact. 
Denote by $\fM(L)$ the set of minimizing measures of the lagrangian $L$:
$$
\fM(L):=\{\,\mu\in\cM(L)\;:\;A_L(\mu)+c(L)=0\;\}.
$$ 
Observe that $\fM(L)$ is a simplex whose extremal points are the ergodic minimizing measures
(see exercises~\ref{esimplex}). In general $\fM(L)$ may be infinite dimensional.

 In this section we shall prove the following

\begin{theorem}\label{TFAM}
Let $A$ be a finite dimensional convex family of lagrangians. Then there 
exists a residual subset $\cO$ of $C^\infty(M,\re)$ such that
$$
u\in \cO, \quad L\in A \quad \Longrightarrow\quad  \dim\fM(L-u)\le \dim A.
$$
In other words, there exist at most $1+\dim A$ ergodic minimizing measures for $L-u$.
\end{theorem}


\bigskip


\begin{exercises}\label{esimplex}\quad
\begin{enumerate}
\item Let $V$ be a real vector space.  A set $Y\subset V$ is 
{\it affinely independent}\index{affinely independent}\index{independent!affinely}
if for any finite subset $F=\{ v_0,v_1,\cdots,v_n\}\subset Y$, 
every point in the convex hull of $F$ is uniquely expressible as a  convex combination
$\la_0\, v_0+\cdots +\la_n\,v_n$, $0\le \la_i\le 1$ of the elements of $F$. Show that 
a finite set $F$ is affinely independent if and only if $\{v_1-v_0,\cdots,v_n-v_0\}$ is linearly independent, 
and in this case the dimension of the convex hull $\dim(\text{conv}(F))=n$.

\item Let $K\subset V$ be a closed convex subset. An extreme or extremal point of $K$ 
\index{point!extreme}\index{point!extremal}\index{extreme point}\index{extremal point}
is a point in $K$ which is not in the (relative) interior of any segment contained in $K$. Prove that
$K$ is the convex closure of its extremal points.
A {\it simplex}\index{simplex} is a subset $\Si\subset V$ such that
 the set of its extremal points is affinely independent.
 
\item Show that if $\mu$ and  $\nu$ are ergodic invariant Borel probabilities of a dynamical system, then 
 they are either equal or mutually singular (i.e. there is a Borel set $A$ such that $\mu(A)=1$ and $\nu(A)=0$). 
 
\item Prove  any  set of ergodic invariant Borel probabilities is affinely independent.
 
\item Prove that the set of minimizing measures $\fM(L)$ of a lagrangian $L$ is a simplex in the vector space of signed Borel measures on $TM$ whose extremal points are the ergodic minimizing measures.

\end{enumerate}
\end{exercises}



\bigskip



Fix $B>0$ and let $\HB$ be the set of holonomic measures supported on $[|v|\le B]$:
$$
\HB:=\{\,\mu\in\ov{\C(M)}\;:\;\mu([\,|v|>B\,])=0\;\}.
$$
Consider a lagrangian $L$ as a functional $L:\HB\to\re$ given by $L(\mu)=A_L(\mu)$. Let 
$$
\fM_\HB(L):=\arg\min_\HB L
$$
be the set of measures $\mu\in\HB$ which minimize $L|_\HB$. Let $\cP=\cP(M)$ be 
the set of probability measures on $M$ endowed with the weak* topology.
A compatible metric (cf.  on $\cP(M)$ is defined as follows: fix a countable 
dense set $\{\,f_n\;|\;n\in\na\,\}$ in $C^\infty(M,\re)$ and let
$$
d(\mu,\nu)=\sum_n \frac 1{2^n}\,\frac1 {c_n}\;\lv\int f_n\,d\mu-\int f_n\,d\nu\rv.
$$
Since $M$ is compact, under this topology the set $\cP(M)$ is compact
Let 
$\pi_*:\HB\to\cP$ be the push-forward induced by the projection $\pi:TM\to M$. Let 
$$
\fM_\cP(L,B):=\pi_*\big(\fM_\HB(L)\big).
$$


\begin{proposition}\label{PHP}
Let $A$ be a finite dimensional convex family of lagrangians on $M$ and let $B>0$. 
There is a residual subset $\cO(A,B)\subset C^\infty(M,\re)$ such that 
\begin{equation}\label{ephp}
L\in A, \quad u\in\cO(A,B) \quad \then \quad
\dim\fM_\cP(L,B)\le \dim A.
\end{equation}

\end{proposition}


\noindent{\bf Proof of theorem~\ref{TFAM}:}

The set $\cO(A):=\cap_{B\in\na}\cO(A,B)$ is residual in $C^\infty(M,\re)$.
By corollary~\ref{E(mu)}, $E(\fM(L))=\{c(L)\}$. Then inequality~\ref{E>} implies that
there is $B_0(L)>0$ such that $\fM(L)\subset\HB$ for all $B>B_0(L)$.
From theorem~\ref{E.2}.\ref{E.2.2}
$$
\fM(L)=\arg\min_\HB L=\fM_\HB(L) \qquad \text{ for all }\quad B>B_0(L).
$$
By the graph property~\ref{GraphProperty} and \ref{T.CMM}
$$
\dim\fM(L)=\dim\pi_*\big(\fM(L)\big)=\dim\fM_\cP(L,B) \quad \text{ for all }
B>B_0(L).
$$
These remarks, together with proposition~\ref{PHP} prove theorem~\ref{TFAM}.

\qed

\noindent{\bf Proof of proposition~\ref{PHP}:}

Define the $\e$-neighbourhood $W_\e$ of a subset $W\subset \cP(M)$ as the union of all the 
open balls in $\cP(M)$ which have radius $\e$ and are centered in $W$.
Given $D\subset A$, $k\in\na$, $\e>0$, denote by $\cO(D,\e,k)$ the set of
potentials $u\in C^\infty(M,\re)$ such that for all $L\in D$, the convex set 
$\fM_\cP(L,B)$ is contained in the $\e$-neighbourhood of some $k$-dimensional 
convex subset of $\cP(M)$.

We shall prove that the proposition holds with 
$$
\cO(A,B)=\bigcap_{\e>0}\cO(A,\e,\dim A).
$$
Indeed, if $u\in\cO(A,B)$ then the inequality in \eqref{ephp}  holds.
Because otherwise for some $L\in A$ the convex set $\fM_\cP(L,B)$ would contain 
a ball of dimension $\dim A +1$, and, if $\e$ is small enough, such  a ball is not contained in the 
$\e$-neighbourhood of any convex set of dimension $\dim A$.
 
 So we have to prove that $\cO(A,B)$ is residual. It is enough to prove that $\cO(A,\e,\dim A)$ is open and dense 
 for any compact subset $D\subset A$ and any $\e>0$. In \ref{open} we prove that it is open and in \ref{dense} 
 that it is dense.
 
 \qed
 
 
 \medskip
 
 \subsection{Open.}\label{open}
 We prove that for all $k\in\na$, $\e>0$ and any compact $D\subset A$, the set $\cO(D,\e,k)$ is open.
 We need a lemma:
 
 
 %\pagebreak
 %\newpage
 
 \begin{lemma}\quad
 
 The set valued map $(L,u)\mapsto \fM_\HB(L-u)$ is upper semi-continuous on $A\times C^\infty(M,\re)$.
 This means that for any open subset $U$ of $\HB$, the set 
 $$
 \{(L,u)\in A\times C^\infty(M,\re)\;:\;\fM_\HB(L-u)\subset U\,\}\subset A\times C^\infty(M,\re)
 $$
 is open in $A\times C^\infty(M,\re)$. Consequently, the set-valued map
 $(L,u)\mapsto \fM_\cP(L-u,B)$ is also upper semi-continuous.
 \end{lemma}
  \begin{proof}
  The lemma is a consequence of the continuity of the map
  $$
  A\times C^\infty(M,\re)\times\HB\ni (L,u,\mu)\longmapsto
  (L-u)(\mu)=L(\mu)-\tint u\;d\mu.
  $$
  \end{proof}
  
  Now let $u_0\in \cO(D,\e,k)$. For each $L\in D$ there is a $k$-dimensional
  convex subset $V\subset \cP(M)$  such that $\fM_\cP(L-u_0,B)\subset V_\e$.
 Then the open sets in $D\times C^\infty(M,\re)$ of the form
 $$
 \{\,(L,u)\in D\times C^\infty(M,\re)\;|\;
 \fM_\cP(L-u,B)\subset V_\e\;\},
 $$ 
 where $V$ is a $k$-dimensional convex subset of $\cP(M)$,
 cover the compact subset $D\times\{ u_0\}$.
 This implies that there is a finite subcover of $D\times\{ u_0\}$ 
 by open sets of the form $\Om_i\times U_i$, where $\Om_i$
 is an open set in $A$  and $U_i\subset \cO(\Om_i,\e,k)$ is
 an open set in  $C^\infty(M,\re)$ containing $u_0$.
 We conclude that the open set $\cap  U_i$ is contained in 
 $\cO(D,\e,k)$ and contains $u_0$.
 
 \qed
 
 
 \bigskip
 
 \subsection{Dense.}\label{dense}
 
 We prove the density of  $\cO(D,\e,\dim A)$ in $C^\infty(M,\re)$ for any $\e>0$. 
 Let $w\in C^\infty(M,\re)$. We want to prove that $w$ is in the closure of 
 $\cO(D,\e,\dim A)$.  We consider a function   $w\in C^\infty(M,\re)$ as a 
 linear functional $w:\cP(M)\to\re$ as $w(\nu)=\int w\,d\nu$
 or $w:\cH_B\to\re$ as $w(\mu)=\int (w\circ\pi)\,d\mu$.
 
 \begin{lemma}\label{dlema}
 There is an integer $m$ and a continuous map 
 $$
 T_m=(w_1,\ldots,w_m):\cP(M)\to\re^m,
 $$
 with $w_i\in C^\infty(M,\re)$ such that 
 \begin{equation}\label{edlema}
 \forall x\in \re^m\qquad \diam T_m^{-1}\{x\}<\e. 
 \end{equation}
 \end{lemma}
 
 
 
 \begin{proof}
 To each function $w\in C^\infty(M,\re)$ we associate the open set
 $$
 U_w=\big\{\,  (\nu,\mu)\in \cP(M)\times\cP(M)\;\big |\;
 \tint w\,d\nu\ne \tint w \,d\mu  \,\big\}.
 $$
 The open sets $U_w$ cover the complement of the diagonal in $\cP(M)\times\cP(M)$.
 One can extract a countable subcover $U_{w_k}$, $k\in \na$. This amounts 
 to say that the sequence $w_k$ separates $\cP(M)$. Defininig
 $T_m=(w_1,\ldots,w_m)$, we have to prove that \eqref{edlema} holds 
 for $m$ large enough. Otherwise, we would have some $\e>0$ and two sequences 
 $\nu_m$ and  $\mu_m$ in $\cP(M)$ such that
 $$
 T_m(\nu_m)=T_m(\mu_m) \qquad \text{and} \qquad
 d(\nu_m,\mu_m)\ge \e.
 $$
 By extracting a subsequence, we can assume that the sequences
 $\nu_m$ and $\mu_m$ have different limits $\nu$ and $\mu$, which satisfy
 $d(\nu,\mu)\ge \e$. Take  $m$ large enough so that $T_m(\nu)\ne T(\mu)$.
 Such $m$ exists because the sequence $w_k$ separates $\cP(M)$. 
 We have that 
 $$
 T_m(\nu_k)=T_m(\mu_k) \quad \text{ for }\quad k\ge m.
 $$
 Hence at the limit $T_m(\nu)=T_m(\mu)$. This is a contradiction.
 
 \end{proof}
 
 \medskip
 
   \index{convex!function}
  \index{function!convex}
 Define the convex function $F_m:A\times \re^m\to\re\cup\{+\infty\}$ as
 $$
 F_m(L,x)=\min_{\stackrel{\scriptstyle\mu\in \cH_B}
                  {T_m\circ \pi(\mu)=x}
                 }
(L-w)(\mu),
 $$
 when $x\in T_m(\cH_B)$ and $F_m(L,x)=+\infty$ if 
 $x\notin T_m(\cH_B)$. For $y=(y_1,\ldots,y_m)\in\re^m$ let
 $$
 M_m(L,y):=\argmin_{x\in\re^m}
 \,[F(L,x)-y\cdot x]
 \subset \re^m
 $$
 be the set of points wich minimize the function $x\mapsto F(L,x)-y\cdot x$.
 We have that 
 $$
 \fM_\cP(L-w-y_1 w_1-\cdots-y_m w_m,B)\subset T_m^{-1}(M_m(L,y)).
  $$
 
 Let 
 $$
 \cO_m(w):=\{\,y\in\re^m\,|\,
 \forall L\in A\;:\; 
 \dim M_m(L,y)\le \dim A \,\}.
 $$
 From lemma~\ref{dlema} it follows that
 $$
 y\in\cO_m(w) \quad \then \quad 
 w +y_1\, w_1 +\cdots + y_m\,w_m\in
 \cO(A,\e,\dim A).
 $$
 Therefore, in order to prove that $w$ is in the closure
 of $\cO(A,\e,\dim A)$ it is enough to prove that 
 $0$ is in the closure of $\cO_m(w)$, which follows from the next proposition.
 
 \begin{proposition}
 The set $\cO_m(w)$ is dense in $\re^m$.
 \end{proposition}
 
 \begin{proof}
 Consider the Legendre transform of $F_m$ with respect to the second variable
 \begin{align}
 G_m(L,y) &= \max_{x\in\re^m} \,y\cdot x - F_m(L,x)\\
 &=  \max_{\mu\in \cH_B}
 \int \big(\,w+y_1 w_1+\cdots+y_m w_m -L\,\big) 
  \,d\mu
 \end{align}
 It follows from the second expession that the function $G_m$ is convex and 
 finite-valued,  hence continuous on $A\times \re^m$.
 
 Let $\partial G_m$ be the subdifferential \index{subdifferential} of $G_m$ and let 
 $$
 \widetilde\Si=\{\, (L,y)\in A\times\re^m\;|\;\dim\partial G_m(L,y)\ge \dim A+1\,\}\,.
 $$
 By proposition~\ref{cxprop} the Hausdorff dimension
 $$
 HD(\widetilde\Si) \le (m+\dim A) - (\dim A +1)=m-1.
 $$
 Consequently, the projection $\Sigma$ of the set $\tilde \Sigma$
on the second factor $\re^m$
also has Hausdorff dimension at most $m-1$.
Therefore, the complement of $\Sigma$ is dense in $\re^m$.
So it is enough to prove that
$$
y\notin\Si \quad\then\quad \forall L\in A\;:\quad
\dim M_m(L,y)\leq \dim A.
$$
 Since we know by definition of $\Sigma$
 that if $y\notin\Si$, $\dim \partial G_m(L,y)\leq \dim A$,
 it is enough to observe that
 $$
 \dim M_m(L,y)\leq \dim \partial G_m(L,y).
 $$
The last inequality follows from the fact that
the set $M_m(L,y)$
is contained in the subdifferential of the convex function
$$
\re^m\ni z\mapsto G_m(L,z)
$$
at the point $y$.

%   \index{convex!function|)}
% \index{function!convex|)}

 
   \end{proof}
 
 
 
 
 
 
 
 
 \chapter{Generic Lagrangians.}
 
 In  \cite{Ma2}, Ma\~n\'e introduced the concept of {\it generic
 property of
 a lagrangian } $L$. A property $P$ is said to be generic for the
 lagrangian $L$ if there exists a generic set  ${\cal O}$  (in the
 Baire sense) on the set $C^\infty(M,\re)$ such that if $\psi$ is in
 ${\cal O}$ then $L+\psi$ has the property $P$. One of 
 Ma\~n\'e's objectives was to show that Mather's theory of minimizing
 measures becomes much more accurate and stronger if we restrict
 ourselves to generic lagrangians. The main  purpose of this chapter
 is to proof the following 
 
 \setcounter{Thm}{0}
 
 \begin{theorem}
 \label{Thm.4.1}
  For every lagrangian $L$ there exists a generic set 
 ${\cal O}\subseteq C^\infty(M,\re)$ such that
 
 (A)  If $\psi$ is in ${\cal O}$ then $L+\psi$ has a 
  unique  minimizing measure, $\mu $ and this measure is uniquely
  ergodic.
 
  (B)Moreover $supp (\mu)=\hat{\Sigma} (L+\psi)= \Sigma (L+\psi) $.
 
  (C) When $\mu $ is supported on a periodic orbit or a fixed point,
  this orbit (point) $\Ga$ is hyperbolic and its stable and unstable
  manifolds  if intersect they do it  transversally. 
 
 \end{theorem}
 
  \bigskip
  
  On Ma\~n\'e \cite{Ma3}, it is conjectured that there exists a
  generic set ${\cal O}$ such that this unique minimizing measure
  is supported on a periodic orbit or an equilibrium point.
  
  The first statement of this was proved by Ma\~n\'e in \cite{Ma2}.
 The second statement is proved in \cite{CP} and the third one in 
 \cite{CI}. We will prove only the first part and give only the ideas
 of the proofs of the last part.
 
 
 
 
  
   
  \bigskip
  \bigskip

 
 
 
 
 \section{Generic Lagrangians.}
 
 
 
 Proof of (A)
 
   Given a potential $\psi$ on  $C^\infty(M,\re)$ define 
 
 \begin{eqnarray*}
   m(\psi) & = & \min_{\nu \in \overline{C}}\int L+\psi d\nu, \\
    M(\psi) & = & \{ \nu \in \overline{C} :\int L+\psi d\nu= m(\psi)\}
 \end{eqnarray*}
 
 Where $ \overline{C}$ is the set of holonomic measures.
 For $\epsilon >0$ let 
 $${\cal O}_\epsilon= \{\psi: \diam\;M(\psi)<\epsilon\}$$
 
 This set is open, in fact if $\nu_n$ is in 
 $ M(\psi_n)$ then
 \begin{equation}
   \label{eq.4.1}
   \int L+\psi d\nu_n \le m(\psi)+2||\psi-\psi_n||_{C^0}
 \end{equation}
 
 
 
 
 So by theorem~\ref{E.2} in chapter 2 if $\psi_n\to \psi$ then the
 sequence 
 $\nu_n$ is precompact and the limit is in $M(\psi)$. From this
 follows
 that ${\cal O}_\epsilon$ is open.
 
 It remains to prove that it is also dense. 
 
 Given a compact convex set $K_0$ on $\overline{C}$ and potential 
 $\psi$ on  $C^\infty(M,\re)$  define
 
 \begin{eqnarray*}
   m_0(\psi) & = & \min_{\nu \in K_0}\int \psi d\nu, \\
    M_0(\psi) & = & \{ \nu \in K_0 :\int \psi d\nu= m_0(\psi)\}
 \end{eqnarray*}
 
 
 \begin{lemma}
 \label{lemma.4.1}
 Let $K_0$ as before then if $\mu $ is an extremal point  of $K_0$,
 for
 all $\epsilon >0 $there exists $\psi$ on  $C^\infty(M,\re)$ such that
 \begin{eqnarray*}
  \diam\; M_0(\psi) & < & \epsilon \\
  d(\mu , M_0(\psi))& < & \epsilon
 \end{eqnarray*}
 \end{lemma}
 
 Proof
 
 Denote by $D$ the diagonal of $K_0\times K_0$ for each pair 
  $(\mu,\nu)$ in $K_0\times K_0-D$ take a potential $\psi_{(\mu,\nu)}$
  such 
 that $\int \psi_{(\mu,\nu)}d\mu \neq\int \psi_{(\mu,\nu)}d\nu$, 
 then there is a neigbourhood
 $U(\mu,\nu)$ contained on $K_0\times K_0$ such that
 $\int \psi_{(\mu,\nu)}d\mu' \neq\int \psi_{(\mu,\nu)}d\nu'$
 for
 every 
 $(\mu', \nu')$ in $U(\mu,\nu)$.
 
 Take a covering $\{U(\mu_n,\nu_n)\}$ of $K_0\times K_0-D$
  and set $\psi_n=\psi_{(\mu_n,\nu_n)}$ then
  if $(\mu,\nu)$ in $K_0\times K_0-D$ there exist $n$ such that
 
  \begin{equation}
     \label{eq.4.2}
    \int \psi_nd\mu \neq\int \psi_nd\nu 
   \end{equation}
 
 
 Define $T_n:\overline{C}\to\re^n$ as
 $$
 T_n(\mu)=(\int \psi_1d\mu,...,\int \psi_nd\mu)$$
 
 Using (\ref{eq.4.2}) and the compactness of $K_0\times K_0$ it is
 easy
 to
 see that given $\epsilon $ there exist $\delta>0 $ and $n>0$ such
 that
 
 \begin{equation}
     \label{eq.4.3}
   S\subset \re^n, \diam\;S<\delta \Rightarrow \diam\;T^{-1}S< \epsilon
   \end{equation}
 
 Let $B=T_n (k_0)$ then $B$ is a compact convex set, let $f:\re^n\to
 \re$ a linear function such that its minimum restricted to $B$ is
 attained in only one point $p$. 
 
 Define $\psi=\sum_i\la_i \psi_i$ where $f=\sum_i\la_ip_i$,
 then
 
 $$f\circ T_n=\sum_i\la_i\int \psi_i$$
 and so
 $$M_0(\psi)=T_n^{-1}(p)$$
 Then by (\ref{eq.4.3})
 we get 
 
 $$
 \diam\;M_0(\psi) <  \epsilon
 $$
 
 
 
 
 
 The following lemma proves the density of ${\cal O}_\epsilon$ and
 hence the
 first part of (A).
 
 
 \begin{lemma}
 \label{lemma.4.2}
 If $\psi$ is on  $C^\infty(M,\re)$ and $\mu $ is an extremal point of
 
 $M(\psi)$ then for every neighbourhood $U$ of $\psi$ and every
 $\epsilon >0$
 there exists $\psi_1$ on $U$ such that
 
 $$\diam\;M(\psi_1) = \epsilon $$
 \end{lemma}
 
 Proof
 
 For $K_0= M(\psi)$ applying the previous lemma, we can find given
 $\epsilon$ a $\psi_1$  such that $\int \psi_1d\nu$ attains its
 minimum, say
 $m_1$ for all measures $\nu$ on $K_0= M(\psi)$ on a set
 $S=M_0(\psi_1)$
 such that $d(\mu , S)<\epsilon $. Set
 
 \begin{eqnarray*}
   m_0 &  = & m(\psi)\\
   f_0(\nu)&  = & \int L+\psi- m_0 d\nu \\
   f_1(\nu)&  = & \int \psi_1- m_1 d\nu
 \end{eqnarray*}
 
 Then 
 \begin{equation}
   \label{eq.4.gener.4}
   f_1(\nu)=f_0(\nu) {\mbox { if }} \nu\in S
 \end{equation}
 
 \begin{equation}
   \label{eq.4.gener.5}
   f_1(\nu) \ge 0 {\mbox { if }} \nu\in M(\psi)
 \end{equation}
 
 \begin{equation}
   \label{eq.4.gener.6}
   \nu\in M(\psi), f_1(\nu) = 0 \Rightarrow \nu\in S
 \end{equation}
 
 
 \begin{equation}
   \label{eq.4.gener.7}
   \nu\in \overline{C}, f_0(\nu) = 0 \Rightarrow \nu\in M(\psi)
 \end{equation}
 
 
 For $\la>0$ define 
  $$f_\la=f_0+\la f_1 $$
  and set
  \begin{eqnarray*}
   m(\la) & = & \min_{\nu \in \overline{C}}f_\la(\nu), \\
    M(\la) & = & \{ \nu \in \overline{C}: f_\la(\nu)=
    m(\la) \}.
 \end{eqnarray*}
 
 
 We claim that 
  \begin{equation}
   \label{eq.4.gener.8}
   \lim_{\la\to 0} \diam\;( M(\la), \{\mu\})\le
   \epsilon
 \end{equation}
 
 This proves the lemma since 
 $$M(\la)=M(\psi+\la \psi_1)$$
 
 
 Proof of the claim
 
 Suppose otherwise that there exist $\la_n \to 0$ and 
 $\mu_{\la_n} ,\nu_{\la_n} $ on $M(\la_n)=M(\psi+\la_n
 \psi_1)$ 
 such that
 $d(\mu_{\la_n} ,\nu_{\la_n})>\epsilon$.  Then by 
 (\ref{eq.4.1}) 
 $\{\mu_{\la_n}\}$ and $\{\nu_{\la_n}\}$ are precompact and as
 in the proof of the open property we
 may assume that
 $\mu_{\la_n}\to\mu \in M(\psi)$ and $\nu_{\la_n}\to\nu \in
 M(\psi)$. 
 Naturally
 $d(\mu,\nu)\ge\epsilon$.
 
 Now 
 Because of (\ref{eq.4.gener.4}) we have that $ m(\la)\le 0$
 \begin{eqnarray*}
  0 & \ge & m(\la_n) =\\
    & = & f_0(\mu_{\la_n})+\la f_1(\mu_{\la_n})\\
  & \ge &\la_n f_1(\mu_{\la_n})
 \end{eqnarray*}
 
 So $f_1(\mu_{\la_n})\le 0$ and hence $f_1(\mu)\le 0$, since $\mu
 \in
 M(\psi)$ by (\ref{eq.4.gener.5}) $f_1(\mu)= 0$ and then by 
 (\ref{eq.4.gener.6}) $\mu $ is in
 $S$. Similarly $\nu $ is in $S$. This is a contradiction with the
 fact that the diameter of $S$ is less than $\epsilon$.
 
  The fact that $\mu$ is uniquely
  ergodic follows from the fact that ergodic components of a
  minimizing measure are also minimizing. And the proof of (A) is
  complete.
 
 
 
 
 
 
 
 It is worth to remark that the proof 
 presented here is a particular
 case of Ma\~n\'e's  original~\cite{Ma6} 
 more general setting:
 
 
 
  Let $E$, $F$ be real convex spaces, $K$
 contained on $F$ a metrizable convex subset and $\phi:E\to F'$,
 $L:F\to \re $ linear maps satisfying 
  \begin{itemize}
 \item (a) The map $E\times F\to \re$ defined by 
 $(w,x)\mapsto \phi(w)(x)$ is continuous.
 \item (b) For any $x\neq y$ in $K$ there exists $w $ in  $E$ such
 that
   $\phi(w)(x)\neq \phi(w)(y)$.
 \item (c) For all $w$ in $E$ and $c$ in $\re$  the set 
     $$\{ x \in K : L(x)+\phi(w)(x)\le c\}$$
      is compact.
 \end{itemize}
 
 Denote by 
 
 $$m(w)=\min_{x\in K}L(x)+\phi(w)(x)$$
 which exists by (c). And
 
 $$M(w)=\{ x \in K : L(x)+\phi(w)(x)=m(w)$$
 
 \begin{Prop}
 \label{Prop.4.1}
 If $E$ is a Frechet space then there exists a residual set 
 ${\cal O}$ contained on $E$ such that if $w$ is on $E$ then 
 $M(w)$ has only one element.
 \end{Prop}
 
 The reader can verify that with the following choices, we get the
 desired result.
 \begin{itemize}
 \item (1) Let $E$ be the Banach space $C^\infty(M,\re)$
 \item (2) As in section~\ref{HolMea} let $C_\ell^0$ be the set of
 continuous
   functions $f$ on $TM$ such that $\sup \frac{f(x,v)}{1+|v|}\le
   \infty $, and $\overline{C}$ the set of holonomic probabilities. 
   Let $K$ be $\overline{C}$ and  $F$ be the subspace of $(C_\ell^0)^*$
    spanned by  $\overline{C}$.
 \item (3) Finally let $L:F\to \re $ is the linear map such that if
   $\mu$ is in  $\overline{C}$ then $L(\mu )=\int Ld\mu$; and 
 for $\psi$ in  $C^\infty(M,\re)$ $\phi(\psi) $ is the restriction to 
 $F$ of the linear map on  $(C_\ell^0)^*$ such that $w\mapsto
 <w,\psi>$.
 \end{itemize}
 
 
  This general setting has some other applications see theorems A, C D
  in \cite{Ma3} and also \cite{CLT}
 
 
 
 
 
 
 
 
 
 Proof of (C)
 
  Let ${\cal O}$ be the residual given by (A).
  Let $\cA$ be the subset of ${\cal O}$ of
  potentials $\psi$ for which the measure on ${\cal M}(L+\psi)$ 
  is supported on a periodic orbit. Let ${\BB}:= {\cal O}\setminus
  {\cA}$ 
  and let ${\cA}_1$ be the subset of $\cA$ on which the minimizing 
  periodic orbit is hyperbolic. We prove that $\cA_1$ is relatively 
  open on $\cA$. For, let $\psi\in \cA_1$ and
  $$
  {\cal M}(L+\psi) = \{ \mu_\ga\}
  $$
  where $\mu_\ga$ is the invariant probability measure supported on
  the
  hyperbolic periodic orbit $\ga$ for the flow of $L+\psi$. We claim
  that if
  $\phi_k\in\cA$, $\phi_k \to \psi$ and ${\cal
  M}(L+\phi_k)=\{\mu_{\eta_k}\}$,
  then $\eta_k\to\ga$. Indeed, since $L$ is superlinear, the
  velocities
  in the  support of the minimizing measures $\mu_k:=\mu_{\eta_k}$ are
  bounded
  (cf. corollary~\ref{E(mu)} and inequality~\ref{E>}), and
  hence there exists a subsequence $\mu_k\to\nu$ converging weakly* to
  a
  some invariant measure $\nu$ for $L+\psi$. Then if $\nu\ne\mu_\ga$,
  \begin{equation}\label{E:D.0}
  \lim_k S_{L+\phi_k}(\mu_k)= S_{L+\psi}(\nu) > S_{L+\psi}(\mu_\ga)\,
  .
  \end{equation}
  Thus if $\de_k$ is the analytic continuation of the hyperbolic
  periodic orbit
  $\ga$ to the flow of $L+\phi_k$ in the original energy level
  $c(L+\psi)$,
  since $\lim_k S_{L+\psi_k}(\mu_{\delta_k})=S_{L+\psi}(\mu_\ga)$,
  for $k$ large we have that,
  $$
  S_{L+\phi_k}(\mu_{\de_k})< S_{L+\phi_k}(\mu_{\eta_k})\, ,
  $$
  which contradicts the choice of $\eta_k$. Therefore $\nu=\mu_\ga$.
  For energy levels $h$ near to $c(L+\psi)$ and potentials $\phi$ near
  to
  $\psi$,
  there exist hyperbolic periodic orbits $\ga_{\phi,h}$ which are the
  continuation of $\ga$. Now, on a small neighbourhood of a hyperbolic
  orbit
  there exists a unique invariant measure supported on it, 
  and it is in fact supported in the periodic orbit. 
  Thus, since $\eta_k\to \ga$, then  $\eta_k$ is hyperbolic. 
  Hence $\phi_k\in\cA_1$ and $\cA_1$ contains a neighbourhood of
  $\phi$
  in $\cA$.
  
  Let ${\cal U}$ be an open subset of $C^\infty (M,\re)$ such that
  $\cA_1={\cal
  U}\cap\cA$. We shall prove below that $\cA_1$ is dense in $\cA$.
  This
  implies 
   that $\cA_1\cup{\BB}$ is generic. For, let
  ${\cal \psi}:=\text{int }( C^\infty(M,\re)\setminus{\cal U})$, then
  ${\cal
  U}\cup{\cal \psi}$ is open and dense in $C^\infty(M,\re)$. Moreover,
  ${\cal \psi}\cap\cA=\0$ because
  $\cA\subseteq \ov{\cA_1}\subseteq \ov{\cal U}$ and
  ${\cal \psi}\cap \cA\subseteq \cA\setminus\ov{\cal U}=\0$.
  Since ${\cal O}=\cA\cup{\BB}$ is generic and
  \begin{align*}
  ({\cal U}\cup{\cal \psi})\cap(\cA\cup{\BB})
          &= ({\cal U}\cap\cA) \cup\bigl(({\cal U}\cup{\cal
                   \psi})\cap{\BB}\bigr) \\
          &\subseteq \cA_1\cup{\BB}\,,
  \end{align*}
  then $\cA_1\cup\BB$ is generic.                         
  
 
 The perturbation to achieve hyperbolicity in a fixed point is easy.
 Is very much as the mechanic case: $L=\tfrac{1}{2}<v,v>_x-U(x)$. The
 reader can verify that if $x_0=\max U$  then the Dirac measure
 supported on the point $(x_0, 0)$ is minimizing. And it is well known
 that this critical point is hyperbolic if and only if the maximum has
 
 non degenerate quadratic form. 
 
 In fact,  from the Euler-Lagrange equation~\eqref{EL}
  we get that
   $L_x(x_0,0)=0$. Differentiating the energy function
  \eqref{energy} we see that $(x_0,0)$ is a singularity of the energy
  level
  $c(L)$. Moreover, the minimizing property of $\mu$ implies that
  $x_0$
  is a
  minimum of the function $x\mapsto L_{xx}(x,0)$. In particular,
  $L_{xx}(x_0,0)$ is positive semidefinite in linear coordinates in
  $T_{x_0}M$.
  And it is hyperbolic if and only if it is positive definite. So to
  achieve hyperbolicity we must just add a small quadratic form.
 
 
 
  The perturbation needed  in the case of a periodic orbit
  the same spirit; Because of the graph property the projection of the
  orbit $\Gamma $, $\pi(\Gamma )$ is a simple closed curve. We add a
  $C^\infty$-small non negative potential $\psi $, which is zero if
  and
  only if $x$ is on $\pi(\Gamma )$ that is nondegenerate in the
  transversal direction. It follows that $\Gamma $ is also a
  minimizing solution of the perturbed lagrangian $L+\psi $.
 
  To prove that it actually  is hyperbolic is much more
  difficult. The reason is that the linearization of the flow (the
  Jacobi equation) is along
  the periodic orbit and hence non autonomous as in the case of a
  singularity.
 
 
 To explain the idea of the proof we need some definitions.
 Let $H$ be the associated hamiltonian by the Legendre transformation
 on $T^*M$ and $\psi$ its flow. Denote by 
  $\pi: T^*M\to M$ be the canonical projection and define the
  {\it vertical subspace\/} on $\theta\in T^*M$ by $\psi(\theta)=
  \ker(d\pi)$.
  Two points $\theta_1$, $\theta_2\in T^*M$ are said to be
  {\it conjugate\/} if  
  $\theta_2= \psi_{\tau}(\theta_1)$ for some $\tau\ne 0$ and
  $d\psi_\tau (\psi(\theta_1)) \cap \psi(\theta_2) \ne \{0\}$.
  
 
 A basic property of orbits without conjugate points is given by the
 following
 
 \begin{proposition}
 \label{Prop.4.gen.1}
 Suppose that the orbit of $\theta\in T^*M$ does not contain conjugate
  points and $H(\theta)=e$ is a regular value of $H$.
  Then there exist two $\vr$-invariant lagrangian subbundles
  $\E$, $\F \subset T(T^*M)$  along the orbit of $\theta$ given by
  
  \begin{align*}
  \E(\theta) &= \lim_{t\to +\infty}
  d\psi_{-t}\bigl(\psi(\psi_t(\theta))\bigr)\; ,\\
  \F(\theta) &= \lim_{t\to +\infty} d\psi_{t}
  \bigl(\psi(\psi_{-t}(\theta))\bigr)\;.
  \end{align*}
  Moreover, $\E(\theta)\cup\F(\theta)\subset T_\theta\Sigma$,
  $\E(\theta)\cap \psi(\theta) = \F(\theta) \cap \psi(\theta) =
  \{0\}$,
  $\langle X(\theta) \rangle \subseteq \E(\theta)\cap \F(\theta)$
  and $\dim \E(\theta) = \dim \F(\theta) = \dim M$, 
  where $X(\theta)=(H_p,-H_q)$ is the
  hamiltonian vector field and $\Sigma = H^{-1}\{e\}$.
  \end{proposition}
  
  
  These bundles where constructed  for disconjugate geodesics of
  riemannian metrics by Green \cite{Green} and of Finsler metrics by
  Foulon \cite{Foulon}. In the general case where constructed in
  \cite{CI}
 
 
 We will only sketch the proof;
 
 
  
  Fix a riemannian metric on $M$ and the corresponding induced metric
  on $T^*M$.
  Then $T_\theta T^*M$ splits as a direct sum of two lagrangian
  subspaces:
  the vertical subspace $\psi(\theta)= \text{\rm ker\,}(d\pi(\theta))$
  and
  the
  horizontal subspace $H(\theta)$ given by the kernel of the
  connection
  map.
  Using the isomorphism $K:T_{\theta}T^*M\to T_{\pi(\theta)}M\times
  T_{\pi(\theta)}^*M$, $\xi\mapsto (d\pi(\theta)\, \xi\,,\,
  \nabla_\theta(\pi\, \xi))$, we can identify $H(\theta) \approx
  T_{\pi(\theta)}M\times\{0\}$ and $\psi(\theta)\approx \{0\}\times
  T^*_{\pi(\theta)}M \approx T_{\pi(\theta)}M$. If we choose local
  coordinates
  along $t\mapsto \pi\,\psi_t(\theta)$ such that $t\mapsto
  \frac{\partial\,}{\partial q_i} (\pi\,\psi_t(\theta))$ are parallel
  vector fields, then this identification becomes
  $\xi\leftrightarrow (dq(\xi),\, dp(\xi))$.
  Let $E\subset T_\theta T^*M$ be an $n$-dimensional subspace
  such that $E\cap \psi(\theta)= \{ 0\}$. Then $E$ is a graph of some
  linear map
  $S:H(\theta)\to \psi(\theta)$. It can be checked that $E$ is
  lagrangian  if and  only if in symplectic coordinates $S$ is symmetric.
  
 
 
  Take $\theta\in T^*M$ and $\xi=(h,v)\in T_\theta T^*M = H(\theta)
  \oplus
  \psi(\theta) \approx T_{\bpi(\theta)}M\oplus T_{\bpi(\theta)}M$.
  Consider a
  variation
  $$
  \a_s(t) = \bigl( q_s(t),p_s(t)\bigr)
  $$
  such that for each $s\in ]-\e,\e[$, $\a_s$ is a solution of the
  hamiltonian $H$
  such that $\a_0(0)=\theta$ and $\frac{d}{ds}\a_s(0)\vert_{s=0}=\xi$.
  
  Writing $d\psi_t(\xi) = \bigl( h(t), v(t)\bigr)$, we obtain the
  hamiltonian
  Jacobi equations
  \begin{alignat}{2} \label{E:G.Jacobi}
  &\dot{h}\, &= \quad H_{pq}\, h &+ H_{pp}\, v \, ,\notag\\
  &\dot{v}\, &= - H_{qq}\, h     &- H_{qp}\, v \, ,
  \end{alignat}
  where the covariant derivatives are evaluated along $\bpi(\a_o(t))$,
  and
  $H_{qq}$, $H_{qp}$, $H_{pp}$ and $H_{qq}$ are linear operators on
  $T_{\bpi(\theta)}M$, that in local coordinates coincide with the
  matrices of
  partial derivatives
  $\left( \frac{\partial^2 H}{\partial q_i \partial q_j}\right)$,
  $\left( \frac{\partial^2 H}{\partial q_i\, \partial p_j}\right)$,
  $\left( \frac{\partial^2 H}{\partial p_i\, \partial p_j}\right)$
  and
  $\left(\frac{ \partial ^2 H}{\partial p_i\,\partial q_j}\right)$.
  Moreover, since the hamiltonian $H$ is convex, then $H_{pp}$ is
  positive
  definite.
  
  We derive now the Riccati equation. Let $E$ be a lagrangian subspace
  of
  $T_\theta T^*M$. Suppose that for $t$ in some interval $]-\e,\e[$ we
  have that
  $d\psi_t(E)\cap \psi(\psi_t(\theta))=\{0\}$. Then we can write
  $d\psi_t(E)=\text{graph }S(t)$, where $S(t): H(\psi_t \,\theta) \to
  \psi(\psi_t
  \,\theta)$ is a symmetric map. That is, if $\xi\in E$ then
  $$
  d \psi_t(\xi) = \bigl( h(t), S(t) \, h(t) \bigr)\, .
  $$
  Using equation~\eqref{E:G.Jacobi} we have that
  $$
  \dot{S} h  + S(H_{pq}h + H_{pp}Sh) = - H_{qq}h - H_{qp}Sh \, .
  $$
  Since this holds for all $h\in H(\psi_t(\theta))$ 
  we obtain the Riccati equation:
  \begin{equation}\label{E:Riccati}
  \dot{S} + S H_{pp}S + S H_{pq} + H_{qp}S + H_{qq} = 0\, .
  \end{equation}
  
 
 
 
 
 
 
  Let
  $K_c(\theta):H(\theta)\to \psi(\theta)$ be the symmetric linear map
  such that
  $\text{graph}(K_c(\theta))=d\psi_{-c}(\psi(\psi_c(\theta)))$.
  Define a partial order on the symmetric isomorphisms
  of $T_{\pi(\theta)}M$ by writing $A\succ B$ if $A-B$ is positive
  definite.
  
  The following proposition based essentially on the convexity of $H$
  proves \ref{Prop.4.gen.1}.
  
  \begin{proposition}\label{P:G.1.5}
  
  For all $\e>0$,
  \begin{itemize}
  \item[(a)] If $d>c>0$ then $K_{-\e} \succ  K_d  \succ  K_c$.
  \item[(b)] If $d<c<0$ then $K_\e \prec K_d  \prec K_c$.
  \item[(c)] $\lim\limits_{d\to +\infty} K_d = \SS$, \;
             $\lim\limits_{d\to -\infty} K_d = \U$.
  \item[(d)] $\SS \preceq \U.$
 \item[(e)] The graph of $\SS $ is the stable green bundle $\E $ and
 the
   graph of $\U $ is the unstable green bundle $\F $
 
 
 \end{itemize}
  \end{proposition}
  
  
  
 
 
  An example of the relationship between the transversality of the
  Green
  subspaces and hyperbolicity appears in the following
  
  
  \begin{proposition}
 \label{Prop.4.gen.2}
   Let $\Ga$ be a periodic orbit of $\psi_t$ without conjugate points.
   Then
  $\Ga$ is hyperbolic (on its energy level) if and only if
  $\E(\theta)\cap
  \F(\theta)=\langle X(\theta) \rangle$ for some $\theta\in \Ga$,
  where
  $\langle X(\theta)\rangle$ is the 1-dimensional subspace generated
  by the hamiltonian vector field $X(\theta)$. In this case $\E$ and
  $\F$ are
  its stable and unstable subspaces.
 \end{proposition}
 
 
 This proposition follows ideas of Eberlein \cite{Eberlein} and
 Freire,
 \cite{Freire}.
 
 
 
 It is known that minimizing orbits do not have conjugate points. So
 by proposition \ref{Prop.4.gen.1} and \ref{Prop.4.gen.2} to prove the
 density of hyperbolicity  it is enough to perturb to make the Green
 bundles transverse. This is done using two formulas for the index.
 One
 in the lagrangian setting and another one in the hamiltonian setting.
 
 
  
  Let $\Om_T$ be the set of continuous piecewise $C^2$ vector fields
  $\xi$ along $\ga_{[0,T]}$.
  Define the {\it index form\/} on $\Om_T$ by
  \begin{equation}\label{eq.4.gen.C.1}
  I(\xi,\eta)=\int_0^T \left(\dot{\xi}\, L_{vv}\, \dot{\eta}
                      + \dot{\xi}\, L_{vx}\, \eta
                      + \xi\, L_{xv}\, \dot{\eta}
                      +\xi\, L_{xx}\, \eta \right)\, dt\, ,
  \end{equation}
  which is the second variation of the action functional for
  variations $f(s,t)$ with $\frac{\partial f}{\partial s}\in\Om_T$.
  For general results on this form see Duistermaat \cite{Duist}.
  
 From this formula it is easy to compare the index of the original and
 the perturbed lagrangian along the same solution $\Gamma $.
 
 Finally we use the following 
 transformation of the index form. It 
  is  taken from Hartman \cite{Hartman} and originally due to Clebsch
  \cite{Clebsch}
 see also \cite{CI}.
  Let $\theta\in T^*M$ and suppose that the orbit of $\theta$,
  $\psi_t(\theta)$,
  $0\le t\le T$ does not have conjugate points.
  Let $E\subset T_\theta T^*M$ be a lagrangian subspace such that
  $d\psi_t(E)\cap \psi(\psi_t(\theta))=\{0\}$ for all $0\le t \le T$.
 
  
 
  Let $E(t):=d\psi_t(E)$ and
  let $H(t),\; \psi(t)$ be a
   matrix solution of the hamiltonian Jacobi
   equation~\eqref{E:G.Jacobi}
  such that $\det H(t)\ne 0$ and
  $E(t)=\text{Image}\, (H(t),\psi(t))\subset T_{\psi_t(\theta)}(T^*M)$
  is a lagrangian subspace. 
 For $\xi=H\zeta\in\Om_T$, $\eta=H\rho\in\Om_T$, we obtain (see
 \cite{CI}) 
  \begin{equation}\label{E:I.7}
  I(\xi,\eta)=\int_0^T (H\zeta')^* \,(H_{pp})^{-1}\,(H\rho')\, dt
             + (H\zeta)^*(V\rho)\vert_0^T\!.
  \end{equation}
  
 
 
 
 
  
  Define $\bN(\theta) := \{\, w\in T_{\pi(\theta)} M \,\vert \,
  \langle w,\dot{\ga}\rangle=0\,\}$. Then $\bN(\theta)$ is the
  subspace
  of
  $T_{\pi(\theta)}M$ generated by the vectors
  $\frac{\partial \,}{\partial x_2}$,$\ldots$,
  $\frac{\partial \,}{\partial x_n}$.
  Let $v_o\in \bN(\theta)$, $\lv v_o\rv=1$ and let $\xi^T(t) :=
  \tZ_T(t)\,v_o$.
  Denote by $\tI_T$ and $I_T$ be the index forms on $[0,T]$ for $\tL$
  and $L$
  respectively. Using the solution $(\tZ_T,\, \tV_T)$ on formula
  \eqref{E:I.7},
  we obtain that
  \begin{equation}\label{E:D.1}
  \tI_T(\xi^T,\, \xi^T) = - (\tZ_T(0)\, v_o)^*\, (\tV_T(0)\, v_o)
                       = - {v_o}^*\, \tK_T(0)\, v_o\, .
  \end{equation}
  Moreover, in the coordinates 
  $(x_1,\ldots,x_n;\frac{\partial\;}{\partial x_1},
  \ldots ,  \frac{\partial\;}{\partial x_n})$
  on $TU$ we have that
  \begin{align}
  \tI_T(\xi^T,\, \xi^T) &= \int_0^T \big(\dot{\xi}^T\, \tL_{vv} \,
  \dot{\xi}^T
                                  + 2\, \dot{\xi}^T\, \tL_{xv}\, \xi^T
                                  + \xi^T\, \tL_{xx}\, \xi^T\big) \,
                                  dt
  \notag\\
                       &= \int_0^T \big(\dot{\xi}^T\, L_{vv} \,
                       \dot{\xi}^T
                                  + 2\, \dot{\xi}^T\, L_{xv}\, \xi^T
                                  \begin{aligned}[t]
                                  &+ \xi^T\, L_{xx}\, \xi^T\big) \, dt
                                  \\
                           &+ \int_0^T \e\, \sum_{i=2}^n
                           \lv\xi_i^T\rv^2\, dt\, .
                           \end{aligned}
  \label{E:D.1.5}
  \end{align}
  
  We have that $\tZ_T(0)=I$ and for all $t>0$,
  $\lim_{t\to\infty} \tZ_T(t)=\widetilde{\h}(t)$
  with $\widetilde{\h}(t)$ the solution of the
  Jacobi equation for $\tH$ corresponding
  to the stable Green bundle.
  Writing $\pi_N(\xi)=(\xi_2,\xi_3,\ldots \xi_n)$ then 
  $\lv\pi_N \widetilde{\h}(0)\,v_0\rv=\lv v_0\rv=1$ because 
  $v_0 \in {\bf N}(\theta)$.
  Hence there exists $\la>0$ and $T_0>0$ such that 
  $\lv\pi_N \xi^T(t)\rv=\lv\pi_N\tZ_T(t)\,v_0\rv>\frac{1}{2}$
  for all $0\le t\le\la$ and $T>T_0$. Therefore
  \begin{equation}\label{E:D.2}
  \tI_T(\xi^T,\,\xi^T)\ge I_T(\xi^T,\, \xi^T) +\frac{\e\la}{4}\, .
  \end{equation}
  
  Let $(\boldsymbol{h}(t),\, \boldsymbol{v}(t))=
  d\psi_t\circ\left(d\pi\vert_{\E(\theta)}\right)^{-1}$ be the
  solution
  of the
  Jacobi equation for $H$ corresponding to the stable
  Green subspace $\E$ and let
  $\SS (\psi_t(\theta)) = \boldsymbol{v}(t)\, \boldsymbol{h}(t)^{-1}$
  be the
  corresponding solution of the Riccati equation, with
  $\text{graph}[\SS (\psi_t(\theta)]=\E(\psi_t(\theta))$. 
  Using formula \eqref{E:I.7}, and writing 
  $\xi^T(t)= \boldsymbol{h}(t)\, \zeta(t)$, we have that
  \begin{align}
  I_T(\xi^T,\, \xi^T)
          &= \int_0^T (\boldsymbol{h}\, \dot{\zeta})^*\,
                   H_{pp}^{-1}\, (\boldsymbol{h}\, \dot{\zeta})\, dt
          + 0
          - \bigl(\boldsymbol{h}(0)\, \zeta(0)\bigr)^{*}
                    \bigl(\boldsymbol{v}(0)\,\zeta(0)\bigr)
  \notag\\
  I_T(\xi^T,\, \xi^T)
          &\ge - v_o^* \, \SS (\theta)\, v_o\, .
  \label{E:D.3}
  \end{align}
  
  From \eqref{E:D.1}, \eqref{E:D.2} and \eqref{E:D.3}, we get that
  $$
  v_o^* \, \SS (\theta)\, v_o \ge v_o^*\, \tK_T \, v_o +
  \frac{\e\la}4\, .
  $$
  From proposition~\ref{P:G.1.5}, we have that
  $\lim_{T\to +\infty} \tK_T(0) = {\widetilde{\SS }}(\theta)$, where
  $\text{graph}(\widetilde{\SS }(\theta))=\widetilde{\E}(\theta)$,
  the stable Green bundle for $\tH$. Therefore
  \begin{equation}\label{E:D.4}
  v_o^* \, \SS (\theta)\, v_o
     \ge v_0^*\, \widetilde{\SS }(\theta)\, v_o +\frac{\e\la}4\, .
  \end{equation}
  
  Similarly, for the unstable Green bundles we obtain that
  \begin{equation}\label{E:D.5}
  v_o^*\, \U(\theta)\, v_o + \la_2
  \le v_o^* \, \widetilde{\U}(\theta) \, v_o
  \quad \text{ for } v_o \in\bN(\theta), \lv v_o\rv=1.
  \end{equation}
  for some $\la_2>0$ independent of $v_o$.
  
  From proposition~\ref{P:G.1.5} we have that
  $\U(\theta)\succeq \SS (\theta)$. From \eqref{E:D.4} and
  \eqref{E:D.5} we get that
  $\widetilde{\U}\vert_\bN \succ \U\vert_\bN \succeq \SS \vert_\bN
  \succ \widetilde{\SS }\vert_\bN$. Since
  $\widetilde{\E}(\theta) = \text{graph}(\widetilde{\SS }(\theta))$
  and $\widetilde{\F}(\theta)=\text{graph}(\widetilde{\U}(\theta))$,
  we
  get that $\widetilde{\E}(\theta)\cap \widetilde{\F}(\theta)
  \subseteq
  \langle \widetilde{X}(\theta)\rangle$. Then proposition B shows that
  $\Ga$ is a hyperbolic periodic orbit for $L+\phi$.
  
  
  
  This proves that $\cA_1$ is dense in $\cA$.
 
  Let $\cA_2$ be the subset of
  $\cA_1$ of potentials $\psi$ for which the minimizing hyperbolic
  periodic
  orbit $\Ga$ has transversal intersections. The proof that $\cA_2$ is
  open and dense in $\cA_1$ is similar to the previous proof, see
  \cite{CI}.
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \section{Homoclinic Orbits.}
 
  
 Assume in this section that $\hS$ contains only 
 {\it one} static class.
 By theorem~\ref{Thm.4.1}.(A), 
 this is true for generic lagrangians.
 By proposition~\ref{ConnectedClass},
 the static classes are always connected, thus if we assume
  that there is only one static class, $\hS$ must be connected.
 
 Given $\varepsilon>0$, let $U_{\varepsilon}$ 
 be the $\varepsilon$-neighbourhood of $\pi(\hS)$. 
 Since $\hS$ is connected, the open set $U_{\varepsilon}$ 
 is connected for $\varepsilon$ sufficiently small. 
  Let $H_{1}(M,U_{\varepsilon},\re)$ denote the first
  relative singular homology group of the pair 
 $(M,U_{\varepsilon})$ with real coefficients. 
 
 We shall say that an orbit of $L$ is {\it homoclinic} 
 \index{homoclinic orbit!to a static class}
 \index{homoclinic orbit!homology of --}
 \index{omega limit}
 \index{alfa limit}
 to a closed invariant set $K\subset TM$ if
 its $\alpha$ and $\omega$-limit sets are contained in $K$.
 
 
 
 Observe that to each homoclinic orbit $x:\re\to M$
  to the set of static orbits $\hS$ we can associate
 a homology class in $H_{1}(M,U_{\varepsilon},\re)$. 
 \index{homology!of a homoclinic orbit}
 Indeed, since there exists $t_{0}>0$ such that
 for all $t$ with $|t|\geq t_{0}$, $x(t)\in U_{\varepsilon}$, 
 the class of $x|_{[-t_{0},t_{0}]}$ defines an element 
 in $H_{1}(M,U_{\varepsilon},\re)$.
 Let us denote by ${\cal H}$ the subset of 
 $H_{1}(M,U_{\varepsilon},\re)$ given by all the classes 
 corresponding to homoclinic orbits to $\hS$.
 
 \begin{theorem}\label{T.Homo.1}
  Suppose that $\hS$ contains
 only one static class.
  Then for any $\varepsilon$ sufficiently small
 the set ${\cal H}$ generates over $\re$ the relative homology
 $H_{1}(M,U_{\varepsilon},\re)$.
 In particular, there exist at least 
 ${\rm \dim}\,H_{1}(M,U_{\varepsilon},\re)$
  homoclinic orbits to the set of static orbits $\hS$.
 \end{theorem}

 Let $U_\e$ be an $\e$-neighbourhood of $\supp(\mu)$.
 From theorems~\ref{T.Homo.1} and~\ref{Thm.4.1}
 we obtain:

 \begin{corollary}\label{Homo.c2} 
  Given a
 lagrangian $L$ there exists 
 a generic set ${\cal O}\subset C^{\infty}(M,\re)$
 such that if $\psi\in{\cal O}$ the lagrangian $L+\psi$ has a unique
 minimizing measure $\mu$ in ${\cal M}^{0}(L+\psi)$ and this measure
 is uniquely ergodic.
 For any $\varepsilon$ sufficiently small
 the set ${\cal H}$ of homoclinic orbits to ${\rm supp}(\mu)$ 
 generates over $\re$ the relative homology
 $H_{1}(M,U_{\varepsilon},\re)$.
 In particular, there exist at least 
 ${\rm \dim}\,H_{1}(M,U_{\varepsilon},\re)$
  homoclinic orbits to ${\rm supp}(\mu)$. 
 \end{corollary}

 To prove theorem~\ref{T.Homo.1}
 we consider finite coverings 
 \index{finite!covering}\index{covering!finite}
 $M_{0}$ of $M$ whose group of deck transformations
 \index{deck transformation}
 is given by the quotient of the torsion free part
 of $H_{1}(M,U_{\varepsilon},\Z)$ by a finite index 
 subgroup.
 Using that the lifted lagrangian $L_{0}$ has the same critical value
 as $L$,
 we conclude that the number of static classes of
 $L_0$ must be finite.
 Hence we can apply theorem~\ref{multibump} to
  $L_{0}$ to deduce that the group generated by the homoclinic orbits
  to the set
 of static orbits of $L$ coincides with
 $H_{1}(M,U_{\varepsilon},\re)$.

 We note that the homoclinic orbits that we obtain in 
 theorem~\ref{T.Homo.1} and corollary~\ref{Homo.c2} 
 have energy $c$ but they are not semistatic
 orbits of $L$.
 However, they are semistatic for
 lifts of $L$ to suitable finite covers.
 
 
 
 Combining corollary~\ref{Homo.c2}, theorem~\ref{Thm.4.1}
 and lemma~\ref{Homo.l4}, we obtain
 
 
 
 \begin{corollary}
  Let $M$ be a closed manifold with first Betti
 number $\geq 2$.
  Given a
 lagrangian $L$ there exists a generic set ${\cal O}\subset
 C^{\infty}(M,\re)$
 such that if $\psi\in{\cal O}$ the lagrangian $L+\psi$ has a unique
 minimizing measure in ${\cal M}^{0}(L+\psi)$ and this measure
 is uniquely ergodic. When this measure is supported on a periodic
 orbit, this
 orbit is hyperbolic and the stable and unstable manifolds have
 transverse homoclinic intersections. 
 \end{corollary}  
 
 
 \begin{lemma}\label{Homo.l4}
 Let $M$ be a closed manifold with first Betti
 number $b_{1}(M,\re)\geq 2$. Then if $A\subset M$ is a closed
 submanifold diffeomorphic to $S^{1}$ and $U_{\varepsilon}$ denotes
 the $\varepsilon$ neighborhood of $A$, we have that
 $H_{1}(M,U_{\varepsilon},\re)$
 is non zero for all $\varepsilon$ sufficiently small.  \label{2}
 \end{lemma}
 
 \begin{proof}
  Since $A$ is diffeomorphic to a circle, the singular homology 
 of the pair $(M,U_{\varepsilon})$ coincides with the singular
 homology
 of the pair $(M,A)$ and therefore
 the vector space $H_{1}(M,U_{\varepsilon},\re)$ must have dimension
 $\geq b_{1}(M,\re)-1\geq 1$.
 \end{proof}
 
 
 \bigskip
 
   For the proof of theorem~\ref{T.Homo.1} we shall need the following
   lemma:
   
 \begin{lemma}\label{lifts}
 Let $p:M_{1}\to M_{2}$ be a covering such that $c(L_{1})=c(L_{2})$.
 Then any lift of a semistatic curve of $L_{2}$ is a semistatic curve
 of $L_{1}$. Also the projection of a static curve of $L_{1}$ is
 a static curve of $L_{2}$. If in addition, $p$ is a finite covering,
 then any lift of a static curve of $L_{2}$ is a static curve
 of $L_{1}$.
 \end{lemma}
 \index{semistatic curve!lift of}
 \index{static!orbit!lift of}
 
 \begin{proof} Observe first that for any $k\in\re$ we have that
 \[\Phi_{k}^{1}(x,y)\geq \Phi_{k}^{2}(px,py),\]
 for all $x$ and $y$ in $M_{1}$. Hence if we write
 $c=c(L_{1})=c(L_{2})$
 we have
 \begin{equation}
 \Phi_{c}^{1}(x,y)\geq \Phi_{c}^{2}(px,py),\label{eins}
 \end{equation}
 for all $x$ and $y$ in $M_{1}$.
 
 Suppose now that $x_{2}:\re\to M_{2}$ is a semistatic curve of
 $L_{2}$
 and let $x_{1}:\re\to M_{1}$ be any lift of $x_{2}$ to $M_{1}$.
 Using (\ref{eins}) and the fact that $x_{2}$ is semistatic we
 have for $s\leq t$,
 
 \begin{align*}
 \Phi_{c}^{1}(x_{1}(s),x_{1}(t)) 
     & \leq A_{L_{1}+c}(x_{1}|_{[s,t]})=A_{L_{2}+c}(x_{2}|_{[s,t]})\\
     & =\Phi_{c}^{2}(x_{2}(s),x_{2}(t))\leq
     \Phi_{c}^{1}(x_{1}(s),x_{1}(t)).
 \end{align*}
 Hence $x_{1}$ is semistatic for $L_{1}$.
 
 Suppose now that $x_{1}:\re\to M_{1}$ is a static curve of $L_{1}$
 and let $x_{2}:\re\to M_{2}$ be $p\circ x_{1}$.
 Using (\ref{eins}) and the fact that $x_{1}$ is static we
 have for $s\leq t$,
 \begin{align*}
 -\Phi_{c}^{1}(x_{1}(t)&,x_{1}(s))=\Phi_{c}^{1}(x_{1}(s),x_{1}(t)) 
   = A_{L_{1}+c}(x_{1}|_{[s,t]})=A_{L_{2}+c}(x_{2}|_{[s,t]})\\
     & \geq \Phi_{c}^{2}(x_{2}(s),x_{2}(t)) \geq
     -\Phi_{c}^{2}(x_{2}(t),x_{2}(s)) \geq
     -\Phi_{c}^{1}(x_{1}(t),x_{1}(s)).
 \end{align*}
 Hence $x_{2}$ is static for $L_{2}$.
 
 Suppose now that $p$ is a {\it finite} covering 
 \index{finite!covering}\index{covering!finite}and let
 $x_{2}:\re\to M_{2}$ be a static curve of $L_{2}$. Let
 $x_{1}:\re\to M$ be any lift of $x_{2}$ to $M_{1}$. Since $x_{2}$
 is static, given $s\leq t$ and $\e>0$, there exists a curve
 $\alpha:[0,T]\to M_{2}$ with $\alpha(0)=x_{2}(t)$,
 $\alpha(T)=x_{2}(s)$
 such that 
 \[A_{L_{2}+c}(x_{2}|_{[s,t]})+A_{L_{2}+c}(\alpha)\leq \e.\]
 Since $p$ is a finite covering, there exists a positive integer $n$,
 bounded from above by the number of sheets of the covering,
 such that the $n$-th iterate
 of $x_{2}|_{[s,t]}*\alpha$ lifts to $M_{1}$ as a closed curve.
 Hence, there exists a curve $\beta$ joining $x_{1}(t)$ to $x_{1}(s)$
 such that
 \[A_{L_{1}+c}(x_{1}|_{[s,t]})+A_{L_{1}+c}(\beta)\leq n\e,\]
 and thus $x_{1}$ is static for $L_{1}$.
 \end{proof}
 
 
 
 
 
 \noindent{\bf Proof of theorem~\ref{T.Homo.1}:}
 
 
 
 Let $U\df U_{\varepsilon}$ denote the $\varepsilon$-neighborhood of
 $\pi(\hS(L))$,
 where $\hS(L)$ is the set of static vectors of $L$.
 Since we are assuming that $\hS(L)$ contains only one static class,
 the set $U$ is also connected
 for small $\varepsilon$. Let $i:U\to M$ be the inclusion map. 
 The vector space $H_{1}(M,U,\re)$ is isomorphic to the quotient
 of $H_{1}(M,\re)$ by $i_{*}(H_{1}(U,\re))$.
 
 Let $H$ denote the torsion free part of $H_{1}(M,\Z)$ and let
 $K$ denote the torsion free part of $i_{*}(H_{1}(U,\Z))$.
 Let us write $G\df H/K=\Z\oplus\overset{k}{\ldots}\oplus\Z$
 where $k={\rm \dim}\,H_{1}(M,U,\re)$.
 Let $J$ be a finite index subgroup of $G$. 
 There is a surjective homomorphism $j:G\to G/J$ given by the
 projection.
 
 If we take the Hurewicz map\index{Hurewicz homomorphism}
 \[\pi_{1}(M)\mapsto H_{1}(M,\Z),\]
 and we compose it with the projections $H_{1}(M,\Z)\mapsto H$,
 $H\mapsto G$ and
 $j:G\to G/J$, we obtain a surjective homomorphism
 \[\pi_{1}(M)\mapsto G/J,\]
 whose kernel will be the fundamental group of a finite covering
 \index{finite covering}\index{covering!finite}
 $M_{0}$ of $M$ with covering projection map $p:M_{0}\to M$
 and group of deck transformations given by the finite abelian group
 $G/J$.
 
 Since $J$ is a subgroup of $G=H/K$, $G/J$ acts transitively
 and freely on the set of connected components of $p^{-1}(U)$ which
 coincides with the set of connected components of
 $p^{-1}(\pi(\hS(L)))$.
 Therefore we have
 
 \begin{lemma}There is a one to one correspondence between elements
 in $G/J$ and connected components of $p^{-1}(\pi(\hS(L)))$. 
 \label{components}
 \end{lemma}
 
 Observe that to each homoclinic orbit $x:\re\to M$
  to $\hS(L)$ we can associate
 a homology class in $H/K$. Indeed, since there exists $t_{0}>0$ such
 that
 for all $t$ with $|t|\geq t_{0}$, $x(t)\in U$, the class of
 $x|_{[-t_{0},t_{0}]}$ defines an element in $H_{1}(M,U,\Z)$.
 Let us denote by ${\cal H}$ the subset of $H/K$ given by all the
 classes
 corresponding to homoclinic orbits to $\hS(L)$.
 
 
 \begin{figure}
 \special{psfile=bump1.eps hoffset=70 hscale=50 vscale=54 angle=-90}
 \vspace{7cm} 
 \caption{Creating homoclinic connections with finite coverings.}
 \label{fig2}
 \end{figure}
 
 \begin{lemma} For any $J$ as above, the image of $\langle{\cal
 H}\rangle$
 under $j$ is precisely $G/J$.  \label{j}
 \end{lemma}
 
 \begin{proof}
  Let $L_{0}$ denote the lift of the lagrangian $L$ to $M_{0}$.
 Observe first that by proposition~\ref{igual}, $c(L)=c(L_{0})$ and
 therefore
  by lemma~\ref{lifts} we have
 \begin{equation} 
 \pi_{0}(\hS(L_{0}))= p^{-1}(\pi(\hS(L))),  \label{Igualdad}
 \end{equation}
 where $\pi_{0}:TM_{0}\to M_{0}$ is the canonical projection
 of the tangent bundle $TM_{0}$ to $M_{0}$.
 
 Let us prove now that $L_{0}$ satisfies the hypothesis of
 theorem~\ref{multibump}, that
 is, the number of static
 classes of $L_{0}$ is finite. In fact, we shall show that the
 projection
 to $M_{0}$ of a static
 class of $L_{0}$ coincides with a connected component of
 $p^{-1}(\pi(\hS(L)))$.
 Using (\ref{Igualdad}) and proposition \ref{ConnectedClass} we 
 see that the projection of a static class of $L_0$ to $M_0$
 must be contained in a single connected component of
 $p^{-1}(\pi(\hS(L)))$.
 Hence, it suffices to show that if $x$ and $y$ belong to a
 connected component of $p^{-1}(\pi(\hS(L)))$
 then $d^{0}_{c}(x,y)=0$. Since we are assuming that $\hS(L)$ contains
 only one static class we have that $d_{c}(px,py)=0$.
 Since $p:M_{0}\to M$ is a finite covering there are lifts $x_{1}$ of
 $px$
 and $y_{1}$ of $py$ such that $d_{c}^{0}(x_{1},y_{1})=0$.
 Since static classes are connected $x_{1}$ and $y_{1}$ must belong to
 the same connected component of $p^{-1}(\pi(\hS(L)))$ and thus
 there is a covering transformation taking $x_{1}$ into $x$ and
 $y_{1}$
 into $y$ which implies that $d^{0}_{c}(x,y)=0$ as desired.
 
 Now theorem~\ref{multibump}
  and (\ref{Igualdad}) imply that every covering transformation
 in $G/J$ can be written as the composition of covering
 transformations
 that arise from elements in ${\cal H}$, that is, 
 $j(\langle{\cal H}\rangle)=G/J$.
 \end{proof}
 
 
 
 We shall need the following algebraic lemma.
 
 \begin{lemma}Let $G=\Z\oplus\overset{k}{\ldots}\oplus\Z$.
 Given a finite index subgroup $J\subset G$ let us denote by
 $j:G\to G/J$ the projection homomorphism.
 
 Let $A$ be a subgroup of $G$. If $A$ has the property
 that for all $J$ as above $j(A)=G/J$, then
 $A=G$.  \label{fato}
 \end{lemma}
 
 
 \begin{proof}
  The hypothesis readily implies that
 \begin{equation}
 A/A\cap J\;\mbox{ is isomorphic to}\;G/J          \label{isom}
 \end{equation}
 \begin{itemize}
 \item If the rank of $A$ is strictly less than the rank of $G$,
 one can easily construct a subgroup $J\subset G$ with finite index
 such that $A\subseteq J$ and $G/J\neq\{0\}$.
 But this contradicts (\ref{isom}) because $A/A\cap J=\{0\}$.
 \item If the rank of $A$ equals the rank of $G$, then $A$ has finite
 index in $G$ and by (\ref{isom}) $G/A=\{0\}$ and thus $G=A$.
 \end{itemize}
 \end{proof}
  
 
 
 
 
 Observe now that any set ${\cal H}$ of a free abelian group $G$ of
 rank $k$ such that the group generated by ${\cal H}$ is $G$ must have
 at least
 $k$ elements.
 Therefore if we combine lemma~\ref{j} and lemma~\ref{fato} with
 $\langle{\cal H}\rangle=A$ we deduce
 that the set ${\cal H}$ of classes corresponding to homoclinic orbits
 generates $G$ and
 must have at least $k$ elements thus concluding the proof
 of theorem~\ref{T.Homo.1}.
 \qed
           
  
   
   
 %   \appendix
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 %  \chapter{}
   
   \chapter{Appendix.}  
     
     
   \renewcommand{\theequation}{\thesection.\arabic{equation}}  
   \renewcommand{\thesection}{\Alph{section}}
   
   \section{Absolutely continuous functions.}
   \index{absolutely continuous!functions|(}
   
   \begin{Definition}\label{D.ac}
   A function $f:[a,b]\to\re$ is  {\it absolutely
   continuous\/}, if $\forall\, \e>0$ $\exists\, \de>0$
   such that
   $$
   \sum_{i=1}^N |t_i-s_i|<\de
   \quad\Longrightarrow\quad
   \sum_{i=1}^N |f(t_i)-f(s_i)|<\e,
   $$
   whenever $]s_1,t_1[,\ldots,]s_N,t_N[$ are 
   disjoint intervals in $[a,b]$.
   \end{Definition}
   
   \begin{proposition}\label{P.ac}\quad
   
   The function $f:[a,b]\to\re$ is absolutely continuous if and only if
   \begin{itemize}
   \item[(i)] The derivative $f'(t)$ exists for a.e. $t\in[a,b]$.
   \item[(ii)] $f'\in\cL^1([a,b])$.
   \item[(iii)] $f(t)=f(a)+\int_a^t f'(s)\;ds$.
   \end{itemize}
   \end{proposition}
   
   
   \begin{proof}
   Define
   $$
   \mu([s,t]):=f(t)-f(s).
   $$
   We claim that $\mu$ defines a finite signed Borel measure on
   $[a,b]$. Indeed, let $\cA$ be the algebra of finite unions of
   intervals. The function $\mu$ can be extended to a $\sigma$-additive
   function on $\cA$. Moreover, if $B$ is a Borel set and
   $\{A_n\}_{n\in\na}\subset\cA$ is a family with $A_n\downarrow B$,
   then $\mu(B):=\lim_n\mu(A_n)$ exists because 
   $\mu(B_n\setminus B_m)\to 0$ when $n,m\to+\infty$.
   
   Observe that the properties of external measures and the
   absolute continuity of $f$ imply that $\mu\ll m$,
   where $m$ is the Lebesgue measure. Let $g=\frac{d\mu}{dm}$ be the
   Radon-Nikodym derivative. Then $g\in\cL^1$ and
   $$
   f(t)-f(a)=\mu([a,t])=\int_a^t g(s)\;ds.
   $$
   By the Lebesgue differentiation theorem 
   $$
   \lim_{h\to 0}\frac{f(t+h)-f(t)}{h}
   =\lim_{h\to 0}\;\frac 1h\int_t^{t+h}\!\!\!g
   =g(t)
   \quad\text{ for a.e. }t\in[a,b].
   $$
   
   Conversely, suppose that (i)-(iii)~hold. Using~(ii), let
   $\mu(A)=\int_A f' \, dm$. Then by~(iii),
   $$
   \mu([s,t])=f(t)-f(s)\quad\text{ for }s,\,t\in[a,b].
   $$
   Then $\mu\ll m$ implies\footnote{If $\mu$ is finite,
   then $\mu\ll m$ is equivalent, using the Borel-Cantelli lemma, to
   $$
   \forall\, e>0\;\;\exists\,\de>0,\;\;
   m(A)<\de \Longrightarrow |\mu|(A)<\e.
   $$
   }
   that $f$ is absolutely continuous.
   \end{proof}
   
   The Lebesgue differentiation theorem gives the following
   characterization.
   
   \begin{corollary}\label{CAC.2}\quad
   
     The function $f:[a,b]\to\re$ is absolutely continuous if and only
     if 
     \newline\indent
     there exists $g\in\cL^1([a,b])$ such that
     $f(t)=f(a)+\int_a^t g(s)\;ds$.
   \end{corollary}
 
  \index{absolutely continuous!functions|)} 
  \newpage
 
 
  \section{Measure Theory}
  
  \begin{RieszTheorem}\cite{Halmos}\label{Riesz}
  
  \index{Riesz theorem}
  Let $X$ be a locally compact Hausdorff topological space
  and let $C_b(X)$ the vector space of continuous functions
  $f:X\to\re$ with compact support. Then any positive 
  linear functional $I:P(X)\to\re$ defines a unique
  Borel measure $\mu$ on $X$ such that
  $I(f)=\int f\,d\mu$ for all $f\in C_b(X)$.
  \end{RieszTheorem}
  
  \newpage
 
  \section{Convex functions.}
 
  \index{convex!function|(} 
  A function $f:\re^n\to\re$ is said {\it convex\/}\label{cf}
  \index{function!convex!(}
  if 
  $$
  f\bigl(\la\,x+(1-\la)\,y\bigr) \le \la\,f(x)+(1-\la)\,f(y)
  $$
  for all $0\le\la\le 1$ and $x,\,y\in\re^n$. Equivalently, if the
  set $\{\,(x,r)\in\re^n\times\re\,\vert\,r\ge f(x)\,\}$ is convex.
  
  For $x_0\in\re^n$ the {\it subdifferential} \index{subdifferential}
  of $f$ at $x_0$ is the set  
  $$
  \partial f(x_0):=\{\,p\in\re^{n^*}\,\vert\,f(x)\ge
  p\,(x-x_0)+f(x_0)\,\}.
  \label{E.SubD}\index{$\partial f(x)$}
  $$
  Its elementst
  are called {\it subderivatives\/}\label{SubD}
  \index{subderivative}
  or {\it subgradient}
  \index{subgradient}
  of $f$ at $x_0$, and the planes
  $$
  \{\,(x,r)\in\re^n\times\re\,\vert\,r=p\,(x-x_0)+f(x_0)\,\}
  \label{E.SupH}
  $$
  are called {\it supporting hyperplanes\/}\label{SupH}
  \index{plane!supporting}
  \index{supporting hyperplane}
  \index{hyperplane!supporting}
  for $f$ at $x_0$. The functional $p\in\re^{n^*}$ is called
  the {\it slope\/}\label{slope}
  \index{slope of a hyperplane|ibf}
  \index{hyperplane!slope of --}
   of the hyperplane.
   
   For the proof of the following proposition 
   see Rockafellar~\cite{Rocky}.
  
  \begin{proposition}\quad
  \begin{itemize}
  \item[(a)] $\partial f(x)\ne\nulo$ for every $x\in\Dom(f)$.
  \item[(b)] A finite convex function is continuous
             and Lebesgue almost everywhere differentiable.
  \item[(c)] If $\partial f(x)=\{p\}$ then $f$ is differentiable at $x$
  and
             $f'(x)=p$.\label{f'}\index{$f'(x)$}
  \end{itemize}
  
  \end{proposition}
  
 
 
  \newpage
 
 
  \section{The Fenchel and Legendre Transforms.}
 
  Given a convex function $f:\re^n\to \re$ 
  \index{convex!function}
  \index{function!convex}
  the  
  {\it Fenchel Transform\/}\label{FenchTra} 
  \index{Fenchel transform|ibf}
  \index{transform!Fenchel|ibf}
  (or the {\it convex dual\/}\label{Conv*}
  \index{convex!dual|ibf}
  \index{dual!convex --|ibf}
  of $f$ is the function  $f:(\re^{n})^*\to\re\cup\{+\infty\}$
  defined by
  \begin{equation}\label{E-LegTra}
  f^*(p)=\max_{x\in \re^n}\,[\,p\,x-f(x)\,]
  \end{equation}
  The function $f$ admits a supporting hyperplane\label{SupH.2} 
   \index{plane!supporting}
   \index{supporting hyperplane}
   \index{hyperplane!supporting}
  with slope\label{slope2}
  \index{slope of a hyperplane}
  $p\in\re^{n^*}$ if and only if $f^*(p)\ne+\infty$.
  If $f$ is superlinear, then $f^*$ is finite on all $\re^n$.
  
  
  \begin{proposition}\label{T.1}\quad
  \index{superlinear}
  \index{growth!superlinear}
  \index{superlinearity!of convex duals}
  \begin{enumerate}
  \item If $f$ is convex then $f^*$ is convex.
  \item\label{T.1.f**} If $f$ and $f^*$ are superlinear then $f^{**}=f$.          
  \item\label{Sup>Bound}
        $f$ is superlinear if and only if $f^*$ is bounded on
             balls, more explicitly,
            \begin{equation*}
            f(x)\ge A\,\lv x\rv - B(A)\,,\quad \forall x\in\re^n\;
            \Longleftrightarrow
            \; f^*(p) \le B(\lv p\rv)\,, \quad \forall p\in\re^{n^*}
            \end{equation*}
  \item\label{T.1.max}
        If $f$ is superlinear, the maximum~\ref{E-LegTra} is attained
        at some point $x\in\re^n$. 
  \end{enumerate}
  \end{proposition}
  
  
  \begin{proof}
  \quad
  \begin{enumerate}
  \item Given $0\le\la \le 1$ and $p_1,p_2\in\re^{n^*}$ we have
        that
             \begin{align*}
             f^*\bigl(\la\, p_1&+(1-\la)\, p_2\bigr) 
             =\max_{x\in \re^n}\bigl[\bigl(\la\, p_1+(1-\la)\,
              p_2\bigr)\,x-f(x)\bigr] \\
             &\le  \la \,\max_{x\in \re^n}\,[\,p_1\, x-f(x)\,]
              +(1-\la)\,\max_{x\in \re^n}\,[\,p_2\, x-f(x)\,]\\
             &= \la\, f^*(p_1)+(1-\la)\,f^*(p_2).
             \end{align*}
 
  \item[\ref{T.1.f**}.]
             From~\eqref{E-LegTra} we get that
             $$
             f(x)\ge p\, x - f^*(p) \qquad\text{ for all } x\in\re^n,\;
             p\in(\re^n)^*
             $$
             Hence,
             $$
             f(x) \ge \sup_{p\in\re^{n^*}}\,[\,p\,x
             -f^*(p)\,]=f^{**}(x).
             $$
             Let $p_x\in\partial f(x)\ne\nulo$.
             Then
             $$
             f(y)\ge f(x) + p_x\,(y-x), \qquad \forall \,y\in\re^n.
             $$
             Hence
             $$
             f^*(p_x)=\max_{y\in\re^n}\big[\,p_x\,y-f(y)\,\big]
             =p_x\,x-f(x).
             $$
             And
             $$
             f(x)=p_x\,x-f^*(p_x)
             \le\max_{p\in\re^{n^*}}\big[\,p\,x-f(x)\,\big]
             =f^{**}(x).
             $$
             
  \item[\ref{Sup>Bound}.]
             We have that
             \begin{align*}
             f_x^*(p) &=  \max_{v\in \re^n}\,[\,p\,v-f_x(v)\,] \\
             &\le  \max_{v\in \re^n}\,[\,p\,v-|p|\,v\,] +B(|p|)\\
             &= B(|p|).
             \end{align*}
   
              Conversely, suppose that $f_x^*(p)\le B(|p|)$. 
             Given $A\in\re$ and $x\in\re^n$ there exists
             $p_x\in\re^{n^*}$
             such that $|p_x|=A$ and $p_x\,v=|p_x|\,|x|=A\,|x|$. 
             Then 
             \begin{align*}
             f(x)&=  \max_{p\in \re^{n^*}}\,[\,p\,x-f^*(p)\,] \\
             &\ge  p_x\,v-B(|p_x|)
              =  A\,|x|-B(A).
             \end{align*}
             
  \item[\ref{T.1.max}.]
          By item~\ref{Sup>Bound}, $f^*$ is finite.
          Let $p\in\re^{n^*}$.
          If $b>0$ is such that $f(x)> (|p|+1)\,|x|-b$, then
          $$
          p\,x-f(x)<b-|x|<f^*(p)-1\quad\text{ for }|x|>b+1-f^*(p).
          $$
          Hence
          $$
          f^*(p)=\max_{|x|\le b+1-f^*(p)}[\,p\,x -f(x)\,],
          $$
          and the maximum is attained at some interior 
          point $x_p$ in the closed ball
          $|x|\le b+1-f^*(p)$.  
  \end{enumerate}
  \end{proof}
  
  \medskip
  
  \begin{corollary}\label{T.2}\quad
  \newline\indent
  If $f:\re^n\to\re$ is convex and superlinear
  then so is $f^*:\re^{n^*}\to\re$.
  In this case $f^{**}=f$.
  \end{corollary}
  
 % \begin{proof}
 % From proposition~\ref{T.1}, $f^*$ is finite, superlinear, bounded
 % on balls and $f^{**}=f$. It is enough to prove that $f^*$ is
 % continuous.
 % Let $p,q\in\re^{n^*}$, if
 % $f^*(p)=p\,x_o -f(x_o)$, then
 % $$
 % f(q)-f(p)\ge q\,x_o-f(x_o)-p\,x_o+f(x_o)
 %            \ge (q-p)\, x_o.
 % $$
 % Thus, 
 % \begin{equation}\label{f*isc}
 % \liminf_{q\to p} f^*(q)\ge f(p).
 % \end{equation}
 % 
 % If $|q-p|\le 1$, then
 % \begin{align*}
 % q\,x-f(x)
 %        &\le q\,x-2\bigl(|p|+1\bigr)\,|x|+B\bigl(|p|+1\bigr) \\
 %        &\le -|p|\,|x| + B\bigl(|p|+1\bigr)                   \\
 %        &\le f^*(p)-2
 %          \qquad\text{ if $|x|\ge C:=\tfrac{B(|p|+1)+2-f^*(p)}{|p|}$.}
 % \end{align*}
 % Since by~\eqref{f*isc}, $f^*$ is lower semi-continuous; 
 % then there is $0<\e<1$ such that if $|q-p|<\e$ then $f^*(q)>f^*(p)-2$.
 % Thus, for $|q-p|<\e$,
 % $$
 % f^*(q)=\max_{|x|\le C}\,\bigl[\;q\,x-f(x)\;\bigr]
 %       =\; q\,x_q -f(x_q).  
 % $$
 % \begin{align*}
 % f^*(p)-f^*(q) &\ge p\,x_q-f(x_q)-q\,x_q+f(q)        \\
 %              &\ge |p-q|\; C.
 % \end{align*}
 % Therefore,
 % $$
 % \limsup_{q\to p}f^*(q)\le f^*(p).
 % $$
 % 
 % 
 %            
 % \end{proof}
 % 
  
  
  
  \bigskip
  
  
  Observe that in this case we have
  $$
  f^*(0) = -\min_{x\in\re^n} f(x) 
  \qquad\text{ and }\qquad
  f(0)=-\min_{p\in\re^{n^*}}f^*(p).
  $$
  
  \bigskip
 
 
  If $f:\re^n\to\re$ is  convex and superlinear we define
  the {\it Legendre Transform\/}\index{Legendre transform|ibf}
  \index{transform!Legendre|ibf}
  $\cL:\re^n\to 2^{\re^{n^*}}$ of $f$, 
  by
  \begin{align}
  \cL(x) &= \{\,p\in\re^{n^*}\,\vert\, p\,x=f(x)+f^*(p)\,\}, 
  \end{align}
  
  
  
  
  \begin{proposition}
  If $f:\re^n\to\re$ is $C^2$ and there is $a>0$,
  such that 
  $$
  y\cdot f''(x)\cdot y\ge a\,|y|^2 \quad\text{ for all }
  x,\,y\in\re^n,
  $$
  then the Legendre transform $\cL:\re^n\to\re^{n^*}$ is
  a $C^1$ diffeomorphism given by $\cL(x)=d_xf$.
  \end{proposition}
  
  
  \begin{proof}
   The function $f$ is convex and it is superlinear because
  \begin{align*}
  f(x) &= f(0) + \int_0^1 f'(s x)\; ds \\
       &= f(0) + \int_0^1 \int_0^1 sx\,f''(ts\,x)\,x\;dt\;ds \\
       &\ge f(0) +\tfrac 12\,a\,|x|^2.
  \end{align*}
  
  From~\eqref{E-LegTra} we get that
  \begin{equation}\label{T.4.le}
   p\,x\le f(x) + f^*(p) \qquad\text{ for all }
   x\in\re^n,\;p\in\re^{n^*}.
  \end{equation}
  By proposition~\ref{T.2}, $f^*$ is superlinear.
  Then item~\ref{T.1.max} in proposition~\ref{T.2} implies that
  $$
  \cL(x)=\arg\max_{p\in\re^{n^*}}\{\,p\,x-f^*(p)\,\}\ne\nulo.
  $$
  Moreover, from~\eqref{T.4.le},
  if $p\in\cL(x)$ then
  $x=\arg\max_{x\in\re^n}\{\,p\,x-f(x)\,\}$.
  Thus $p=d_xf=\cL(x)$. This proves that $\cL$ differentiable 
  and singled valued.
  Moreover since $d_x\cL=f''(x)$ is non-singular, then $\cL$ is a
  local $C^1$ diffeomorphism.
  
  Since
  $$
  (y-x)\cdot [\,d_yf-d_xf\,]
  =\int_0^1 f''\bigl(s x+(1-s) y\bigr)\;ds >0,
  $$
  then $x\mapsto d_xf=\cL(x)$ is injective.
  We now prove that $\cL$ is surjective.
  By item~\ref{T.1.max} in proposition~\ref{T.1} the maximum
  \begin{equation*}
  f^*(p)=\max_{x\in \re^n}\,[\,p\,x-f(x)\,]
  \end{equation*}
  is attained at some $x_p\in\re^n$.
  Then $p\in\cL(x_p)$. 
  \end{proof}

   
  
  \newpage
  
  \section{Singular sets of convex funcions.}
  
   Let $f:\re^n\to\re$ be a convex function. Recall that its
   subdifferential \index{subdifferential|(}  at  $x\in\re^n$ is the set
$$
\partial f(x):=\{\,\ell:\re^n\to\re\;\text{ linear }|\; f(y)\ge f(x)+\ell\,(y-x)\,,\;\forall y\in\re^n\;\}.
$$

Then the sets $\partial f(x)\subset\re^n$ are convex. If $k\in\na$, let
$$
\Si_k(f):=\{\,x\in\re^n\;|\;\dim \partial f(x)\ge k\;\}.
$$
\begin{proposition}\label{cxprop}
If $f:\re^n\to\re$ is a convex function then for all $0\le k\le n$ the Hausdorff dimension $HD(\Si_k(f))\le n-k$.
\end{proposition}
We recall here an elegant proof due to Ambrosio and Alberti,
see \cite{AA}. More can be said on the structure
of $\Sigma_k$, see \cite{alberti,Zajcek} for example.


By adding $\lv x\rv^2$ if necessary
(which does not change $\Si_k$)
 we can assume that $f$ is superlinear and that
\begin{equation}\label{ellineq}
f(y)\ge f(x)+\ell(y-x)+\tfrac 12\,\lv y-x\rv^2\quad
\forall x,y\in\re^n,\quad\forall\ell\in\partial f(x).
\end{equation}


\begin{lemma}
 \qquad $\ell\in\partial f(x)$, \quad $\ell'\in\partial f(x')$\quad $\then$ \quad $\lv x-x'\rv\le \lV\ell-\ell'\rV$.
\end{lemma}

\begin{proof}

From inequality~\eqref{ellineq} we have that 
\begin{align*}
f(x')&\ge f(x)+\ell(x'-x)\;+\,\tfrac 12\,\lv x'-x\rv^2,\\
f(x)&\ge f(x')+\ell'(x-x')+\tfrac 12\lv x-x'\rv^2.
\end{align*}
Then 
\begin{gather}
0\ge (\ell'-\ell)(x-x')+\lv x-x'\rv^2 \\
\lV \ell-\ell'\rV \,\lv x-x'\rv\ge (\ell-\ell')(x-x')\ge\lv x-x'\rv^2.
\end{gather}
Therefore $\lV\ell-\ell'\rV\ge\lv x-x'\rv$.
\end{proof}


Since $f$ is superlinear, the subdifferential $\partial f$ 
is surjective and we have:
\begin{corollary}\quad

There exists a Lipschitz function $F:\re^n\to\re^n$ such that
 $$
 \ell\in\partial f(x) \quad \then \quad x=F(\ell).
 $$ 
\end{corollary}


\noindent{\bf Proof of Proposition~\ref{cxprop}:}

Let $A_k$ be a set with $HD(A_k)=n-k$ such that $A_k$ intersects any convex subset of dimension $k$. 
\index{convex!set}
\index{set!convex}
For example 
$$
A_k=\{ x\in\re^n\, |\, x \text{ has at least $k$ rational coordinates }\}.
$$
Observe that 
$$
x\in\Si_k\quad \then \quad \partial f(x) \text{ intersects } A_k 
\quad \then \quad x\in F(A_k). 
$$
Therefore $\Si_k\subset F(A_k)$.
Since $F$ is Lipschitz,  we have that $HD(\Si_k) \le HD(A_k)=n-k$.

\qed
 
  
  \index{subdifferential|)} 
  
 \index{convex!function|)}
 \index{function!convex|)} 
  
  \newpage
  
  \section{Symplectic Linear Algebra.}
 
   \newpage


  
       \nocite{AM}
       \nocite{Kli}
       \nocite{Carneiro}
       \nocite{CDI}
       \nocite{CI}
       \nocite{CIS1}
       \nocite{CIPP}
       \nocite{CP}
       \nocite{CLT}
       \nocite{Duist}
       \nocite{Fa1}
       \nocite{Fa2}
       \nocite{Fa3}
       \nocite{Fa4}
       \nocite{Hartman}
       \nocite{Fre-Ma}
       \nocite{IS1}      
       \nocite{Ma1}
       \nocite{Ma2}      	     
       \nocite{Ma3}
       \nocite{Ma4}
       \nocite{Ma5}
       \nocite{Ma6}
       \nocite{Ma7}
       \nocite{Massart1}
       \nocite{Massart2}
       \nocite{Mat1}
       \nocite{Mat2}
       \nocite{Mat3}
       \nocite{Mat4}
       \nocite{Mat5}
       \nocite{Mat6}
       \nocite{Mat7}
       \nocite{Mat8}
%       \nocite{Mat9}
       \nocite{Mo1}
       \nocite{Mo2}
       \nocite{Palais}
       \nocite{P1}
       \nocite{P2}
       \nocite{PP0}
       \nocite{PP1}
       \nocite{PP2}
       \nocite{PP3}
       \nocite{PP4}
       \nocite{P9}
       \nocite{Schw}
       
       
       \index{A@{\Large\bf A}\phantom|phantom}
       \index{B@{\Large\bf B}\phantom|phantom}
       \index{C@{\Large\bf C}\phantom|phantom}
       \index{D@{\Large\bf D}\phantom|phantom}
       \index{E@{\Large\bf E}\phantom|phantom}
       \index{F@{\Large\bf F}\phantom|phantom}
       \index{G@{\Large\bf G}\phantom|phantom}
       \index{H@{\Large\bf H}\phantom|phantom}
       \index{I@{\Large\bf I}\phantom|phantom}
  %        \index{J@{\Large\bf J}\phantom|phantom}
       \index{K@{\Large\bf K}\phantom|phantom}
       \index{L@{\Large\bf L}\phantom|phantom}
       \index{M@{\Large\bf M}\phantom|phantom}
       \index{N@{\Large\bf N}\phantom|phantom}
       \index{O@{\Large\bf O}\phantom|phantom}
       \index{P@{\Large\bf P}\phantom|phantom}
  %        \index{Q@{\Large\bf Q}\phantom|phantom}
       \index{R@{\Large\bf R}\phantom|phantom}
       \index{S@{\Large\bf S}\phantom|phantom}
       \index{T@{\Large\bf T}\phantom|phantom}
  %        \index{U@{\Large\bf U}\phantom|phantom}
  %        \index{V@{\Large\bf V}\phantom|phantom}
          \index{W@{\Large\bf W}\phantom|phantom}
  %        \index{X@{\Large\bf X}\phantom|phantom}
  %        \index{Y@{\Large\bf Y}\phantom|phantom}
  %        \index{Z@{\Large\bf Z}\phantom|phantom}

       
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 \backmatter
 
% \listoftables
% \listoffigures

  \addtocontents{toc}{\contentsline{chapter}{\numberline
  {}Index.}{\thepage}}


 \printindex
 
%  \newpage
%  
%  \section{falta}
%  
%  \begin{itemize}
%  \item Corregir topologia $d_1$ en "Tonelli's theorem" y "Holonomic
%        measures"
%        
%  \item La proposici\'on~\ref{anc} habla de minimizantes Tonelli
%        antes de definirlos.
%              
%  \item La cuenca es el conjunto de puntos de diferenciabilidad de u?
%  
%  \item Remark~\ref{diffWKAM}
%  
%  \item corregir prueba de las connecting orbits.
%  
%  \item minimizantes en nivel de energia fijo.
%        Teoria de Morse.       
%  \end{itemize}
 
 \end{document}
