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\begin{document}


\title[Lagrangian Flows]
      {\large \bf Lagrangian Flows:   The Dynamics of  \\
        Globally Minimizing Orbits - II}




\author[G. Contreras]{Gonzalo Contreras}
\address{Depto. de Matem\'atica. PUC-Rio\\
         R. Marqu\^es de S\~ao Vicente, 225 \\
         22453-900 Rio de Janeiro \\
         Brasil.}
\email{gonzalo@@mat.puc-rio.br}



\author[J. Delgado]{Jorge Delgado}
\address{Depto. de Matem\'atica. PUC-Rio\\
         R. Marqu\^es de S\~ao Vicente, 225 \\
         22453-900 Rio de Janeiro \\
         Brasil.}
\email{jdelgado@@mat.puc-rio.br}



\author[R. Iturriaga]{Renato Iturriaga}
\address{CIMAT  \\
         A.P. 402, 3600 \\
         Guanajuato. Gto. \\
         M\'exico.}
\email{renato@@fractal.cimat.mx}

\thanks{G. Contreras was partially supported by CNPq-Brazil.}
\thanks{R. Iturriaga was partially supported by Conacyt-Mexico,
 grant 3398E9307.}


\maketitle

\noindent{\it To the memory of Ricardo Ma\~n\'e.}
\bigskip

\noindent
{\large  \bf Introduction.}
\bigskip


In this work we prove most of the theorems of Ma\~n\'e's 
unfinished work ``Lagrangian  Flows the dynamics of 
Globally Minimizing Orbits'',~\cite{Ma3}.
Exceptions are theorem III,
whose proof is divided in~\cite{Ma2} and~\cite{CI}
and theorem IV which was proved in \cite{Ma3}. Also, we provide proofs
for slightly different statements of theorems VII, XI and XIV.
We would like to emphasize that all the theorems in this paper are due
to  Ma\~n\'e and all the
responsibility of the proofs is ours.


We encourage the reader to use Ma\~n\'e's original paper \cite{Ma3}
as the introduction of this work.
In  section 1 we prove theorems I and II, in section 2 we prove theorem V,
in section 3 we prove theorems VI, VII, VIII and IX, in section 4 we prove
theorems X and XI, and  in  section 5 we prove theorems XII, XIII and XIV.

The first and second authors want to thank the hospitality of CIMAT.

We want to use
this space to say how much we admired Ricardo's clearness and
 brightness and how grateful we are to his enormous generosity. This paper is
 to his memory.








\stepcounter{section}




%{\large \bf  Section \thesection. }
\bigskip
\bigskip

\centerline{\large \bf \thesection. Basic properties of the critical value.}

\bigskip

Let $M$ be a smooth closed manifold. We say that a smooth function
$L:TM\to \re$ is a {\it Lagrangian} if it satisfies the following conditions:
\begin{itemize}
\item[(a)]{\it Convexity:\/} For all $x\in M$, $v\in T_xM$, the Hessian matrix
          $\frac{\partial^2 L}{\partial v_i \partial v_j}(x,v)$
          (calculated with respect to linear coordinates on $T_xM$)
          is positive definite.
\item[(b)] {\it Superlinearity:\/}
           $\lim\limits_{\lV v\rV\to +\infty} 
           \frac{L(x,v)}{\lV v\rV}=+\infty$.
\end{itemize}
 Given an
absolutely continuous curve $x:[0,T]\to M$ define its $L$-action by
$$
S_L(x) := \int_a^b L\left(x(t), \dot{x}(t)\right) \, dt \, .
$$
Fixing $p,q\in M$ and $T>0$, the critical points of the action functional
on the set
$$
AC(p,q,T):=\{ \, x:[0,T]\to M\,\vert\, x(0)=p,\, x(T)=q,\, x \text{ absolutely
continuous }\}.
$$
are solutions of the Euler-Lagrange equation, which in local coordinates
is given by
\begin{equation*}
\frac{d}{dt}L_v=L_x \,. \tag{E-L}
\end{equation*}

Because of the convexity of the Lagrangian this equation can be thought
as a first order differential equation on $TM$. The {\it Lagrangian flow}
$f_t$ on $TM$ is defined by $f_t(x,v)=(\ga(t),\dot{\ga}(t))$, where $\ga$ is
the solution of (E-L) with $\ga(0)=x$ and $\dot{\ga}(0)=v$.
Define the {\it energy \/} function $E:TM\to\re$ as
$$
E(x,v) := \frac{\partial L}{\partial v}(x,v) \cdot v - L \,.
$$
%where $v = \frac{d\,}{dt}(tv)\vert_{t=1}$.
It can be seen that the value of $E(x,v)$ is constant along the orbits
of $f_t$. The superlinearity condition implies that the level sets of the
energy function have bounded velocities and hence that  are compact.
This, in turn, implies that the solutions of (E-L) are defined for all values
$t\in\re$, i.e., that the flow $f_t$ is {\it complete}.

Finally, for $p,q\in M$, let
$$
AC(p,q) := \bigcup_{T>0} AC(p,q,T)\, ,
$$
and define the {\it action potential \/} as
$$
\Phi_k(p_1,p_2) := \inf \,\left\{ \, S_{L+k}(x) \, \vert \,
x\in AC(p_1,p_2)\, \right\}\, , \, k\in \re\, .
$$

\bigskip

\noindent{\bf Theorem I.}
{\it
There exists $c(L)\in \re$ such that
\begin{itemize}

\item[(a)] $k< c(L) \then \Phi_k(p_1,p_2) = -\infty\, ,
\quad \forall p_1\, ,\, p_2\in M$.

\item[(b)] $k\ge c(L) \then \Phi_k(p_1,p_2) > -\infty \, ,
\quad \forall p_1\, ,\, p_2\in M\,$ and $\Phi_k$ is Lipschitz.

\item[(c)] $k \ge c(L) \then$
\begin{align*}
&\Phi_k(p_1,p_3) \le \Phi_k(p_1,p_2) + \Phi_k(p_2,p_3)
\quad \forall p_1\, ,\, p_2\, ,\, p_3 \in M \\
&\Phi_k(p_1,p_2) + \Phi_k(p_2,p_1) \ge 0
\quad \forall p_1\, ,\, p_2 \in M
\end{align*}


\item[(d)] $k>c(L) \then \Phi_k(p_1,p_2)+\Phi_k(p_2,p_1) > 0
\quad \forall p_1\, \neq \, p_2\,$.
\end{itemize}
}

\medskip

\begin{remark}      This theorem, with the same proof, holds for coverings
$\pi: \widehat{M} \to M$ of a compact manifold $M$, with the lifted
Lagrangian $\widehat{L}=L\circ \pi$.
\end{remark}

\noindent{\bf Proof:}

We first prove that if for some $p_1$, $p_2\in M$, 
$\Phi_k(p_1,p_2) = - \infty$, then 
$\Phi_k(q_1,q_2) = -\infty$ for all $q_1, q_2\in M$. Let $\ga$, $\eta$:
$[0,1]\to M$, $\ga \in AC( q_1,p_1)$, $\eta\in AC(p_2,q_2)$. Let
$x_n\in AC(p_1,p_2)$ be such that $\lim\limits_{n\to\infty} 
S_{L+k}(x_n) = -\infty$. Then
$$
\lim_{n\to\infty}S_{L+k}(\ga * x_n * \eta) = S_{L+k}(\ga) + S_{L+k}(\eta)
+\lim_{n\to\infty} S_{L+k}(x_n) = -\infty\, .
$$

Thus the number
$$
c(L) :=\inf\,\left\{ \, k\in\re \, \vert\, \Phi_k(p,q) > -\infty\, \right\}
$$
does not depend on $(p,q)$. We have to see that $-\infty < c(L) < +\infty$.
Since the function $k\mapsto \Phi_k(p,q)$ is nondecreasing, it is enough
to see that there exist $k_1$, $k_2\in\re$ such that $\Phi_{k_1}(p,q)=-\infty$
and $\Phi_{k_2}(p,q)>-\infty$. We first prove the existence of $k_1$.
Since $L$ is bounded on $\left\{ (x,v)\in TM \, \vert \, v\in T_x M\, ,\,
\lv v\rv \le 2\, \right\}\,$, there exists $B>0$ such that
\begin{equation}\label{E:I.1}
\lv L(x,v)\rv < B \quad \text{ if } \quad \lv v \rv <2\, .
\end{equation}
Let
$x_n:[0,n]\to M$, $x_n\in AC(p,q)$ be such that $|\dot x_n|\le 2$.
Then $S_L(x_n)\le B\, n$ and hence, for $k_1 = - B -1$, we have that
\begin{align*}
\Phi_{k_1}(p,q) &\le \lim\inf_n S_{L-B-1}(x_n) \le \lim\inf_n
\left( \int_0^n L(x_n,\dot{x}_n)\, dt - (B+1)n\right)\\
&\le\lim\inf_n \bigl( Bn - (B+1)n\bigr) = -\infty.
\end{align*}

Now we prove the existence of $k_2$. The superlinearity hypothesis
implies that $L$ is bounded below. Let $A$ be a lower bound for $L$ on $TM$.
We claim that  it is enough to take $k_2> -A+1$. Indeed
$$
S_{L+k_2}(x) \ge \int_0^T (A + k_2) \, dt \ge 0 
\quad\text{ for all } x\in AC(p,q).
$$
hence $\Phi_{k_2}(p,q) \ge 0$.

It remains to prove that $\Phi_c(p,q)>-\infty$ for all $p$, $q\in
M$, where $c=c(L)$.
Suppose not. Take $p\in M$, then $\Phi_c(p,p)=-\infty$. Let 
$\ga\in AC(p,p)$ be such that $S_{L+c}(\ga)<-a<0$. 
Then there exists $\e>0$ such that $S_{L+c+\e}(\ga)<-\frac 12 a <0$. 
Let $\de_N:=\ga *\overset{N}\ldots *\ga$, then
$$
\Phi_{c+\e}(p,p)\le \lim_N S_{L+c+\e}(\de_N)\le\lim_N
-{\textstyle{\frac 12}} \, a \, N = -\infty\, .
$$

This contradicts the definition of $c(L)$. In particular, we have also proven
that $\Phi_c(p,p)\ge 0$ for all $p\in M$. By taking $\ga\in AC(p,p)$ with
bounded velocities and arbitrarily small parameter intervals, we have that
\begin{equation}  \label{E:I.phi_c(p,p)=0}
\Phi_c(p,p)=0\quad\text{ for all } p\in M\, .
\end{equation}
Similarly
\begin{equation}  \label{E:I.phi_k(p,p)=0}
\Phi_k(p,p)=0\quad\text{ for all } p\in M\, \text{ for all } k\ge c(L).
\end{equation}


We now prove (c). Let $k\ge c(L)$ and $p_1$, $p_2$, $p_3\in M$. Let
$x^{ij}_n :[0,T_n]\to M$, $x^{ij}_n\in AC(p_i,p_j)$, be such that
$$
\lim_n S_{L+k}(x^{ij}_n) = \Phi_k (p_i, p_j) \quad i,j\in\{1,2,3\}.
$$
Then $x^{12}_n*x^{23}_n\in AC(p_1,p_3)$ and
$$
\Phi_k(p_1,p_3)\le S_{L+k}(x^{12}_n * x^{23}_n) =
S_{L+k}(x^{12}_n) + S_{L+k}(x^{23}_n).
$$
Taking the limit when $n\to\infty$ we get that
\begin{equation}\label{E:I.3}
\Phi_k(p_1,p_3)\le \Phi_k(p_1,p_2) + \Phi_k(p_2,p_3)\, .
\end{equation}
Finally using \eqref{E:I.phi_k(p,p)=0} and \eqref{E:I.3}  we obtain that
\begin{equation}
\label{4}
0=\Phi_k(p_1,p_1)\le \Phi_k(p_1,p_2)+\Phi_k(p_2,p_1)
\end{equation}
when $k\ge c(L)$.

We prove that $\Phi_k$ is Lipschitz when $k\ge c(L)$.
Let $\ga:[0,d(p,q)]\to M$ be a geodesic joining $p$ and $q$, using
\eqref{E:I.1} we obtain
\begin{align}
\Phi_k(p,q) &\le \int_0^{d(p,q)}
\left[ L\left( \ga(t),\dot{\ga}(t)\right)+k\right]\, dt  \notag\\
\Phi_k(p,q) &\le \left( B+k\right) \,\, d(p,q)
\quad \text{ for }\quad k\ge c(L)\, .\label{E:I.5}
\end{align}
Therefore if $k\ge c(L)$,
\begin{align*}
\Phi_k(p_1,p_2) - \Phi_k(q_1,q_2)
&\le \Phi_k(p_1,q_1) + \Phi_k(q_1,p_2) - \Phi_k(q_1,q_2) \\
&\le \Phi_k(p_1,q_1) + \Phi_k(q_1,q_2) + \Phi_k(q_2,p_2) - \Phi_k(q_1,q_2)  \\
&\le \Phi_k(p_1,q_1) +\Phi_k(q_2,p_2) \\
&\le (B +k)\bigl( d(p_1,q_1) + d(p_2,q_2) \bigr).
\end{align*}
Changing the roles of $p_i$ and $q_i$, $i=1,2$ we get that
$$
\lv \Phi_k(p_1,p_2) - \Phi_k(q_1,q_2)\rv \le (B+k) \, \bigl( d(p_1,q_1) +
d(p_2,q_2) \bigr)\, .
$$

We now prove that if $k\ge c(L)$ and $p\ne q$, then the function
$k\mapsto \Phi_k(p,q)$ is strictly increasing. By \eqref{4} this
implies (d). Let $p\ne q$ and $\ell>k\ge c(L)$. Let $x_n:[0,T_n]\to
M$, $x_n\in AC(p,q)$ be such that $\lim_n S_{L+\ell}(x_n)
= \Phi_\ell (p,q)$. We have that
\begin{align*}
S_{L+\ell}(x_n) &= S_{L+k}(x_n) + (\ell - k) T_n \\
\Phi_\ell (p,q) &\ge \Phi_k(p,q) + (\ell - k) \, \lim\inf_n T_n
\end{align*}
It is enough to prove that $\lim\inf_n T_n > 0$, because then
$\Phi_\ell(p,q) > \Phi_k (p,q)$. Suppose that $\lim\inf_n T_n = 0$.
By the superlinearity of $L$, for all $B>0$ there exists $A>0$ such that
$$
\lv L(x,v)\rv \ge B\, \lv v\rv - A\, .
$$
Then
\begin{align*}
\Phi_\ell(p,q) &=\lim_n S_{L+\ell}(x_n) \ge \lim\inf_n\left[
\int_0^{T_n}B\,\lv \dot{x}_n\rv \, dt + (k-A)\, T_n\,\right] \\
& \ge B\, d(p,q) + 0\, ,
\end{align*}
for all $B>0$. Therefore $\Phi_{L+\ell}(p,q) = +\infty$, which contradicts
\eqref{E:I.5}.

\qed

Through the rest of the paper we shall neeed the following results:

\begin{Theorem}{\bf (Mather \cite{Mather})}\label{L:MATHER1}

For all $C>0$ there exists $A_1=A_1(C)$ such that if $T>0$, $p,\, q\in M$ and
$x\in AC(p,q,T)$ satisfy
\begin{itemize}\begin{itemize}\begin{itemize}
\item[(a)]  $S_L(x) = \min\{\, S_L(y)\,\vert\, y\in AC(p,q,T)\,\}$.
\item[(b)]  $S_L(x)\le C\, T$.
\end{itemize}\end{itemize}\end{itemize}
Then
\begin{itemize}\begin{itemize}\begin{itemize}
\item[(c)]  $\lV \dot{x}(t)\rV < A_1$ for all $t\in [0,T]$.
\item[(d)]  $x\vert_{[0,T]}$ is a solution of (E-L).
\end{itemize}\end{itemize}\end{itemize}
\end{Theorem}

\medskip

\begin{Corollary}\label{lemmavel}

There exists $A>0$ such that if $T>1$, $p,q\in M$ and $x\in AC(p,q,T)$
satisfy
$$
S_L(x) = \min\{\, S_L(y)\, \vert\, y\in AC(p,q,T)\,\}\,,
$$
then $\lV \dot{x}(t)\rV<A$ for all $t\in [0,T]$.
\end{Corollary}

\noindent{\bf Proof:}
Let
$$
C:= \sup\{\,\lv L(x,v)\rv\,\vert\, \lV v\rV<\text{diam}\, (M)\,\}\,.
$$
There exists a geodesic $\ga\in AC(p,q,T)$ with
$$
\lv\dot{\ga}\rv \equiv  \frac{d(p,q)}T<\text{diam}(M)
\quad \text{ and } \quad S_L(\ga)<C\, T \,.
$$
Then the corollary follows from theorem \ref{L:MATHER1}.

\qed


\begin{Corollary}\label{corvel2}
There exists $A>0$ such that if $p,q\in M$ and $x\in AC(p,q,T)$
satisfy
\begin{itemize}\begin{itemize}\begin{itemize}
\item[(a)]  $S_L(x) = \min\{\, S_L(y)\,\vert\, y\in AC(p,q,T)\,\}$.
\item[(b)]  $S_L(x)<\Phi_c(p,q) + d_M(p,q)$.
\end{itemize}\end{itemize}\end{itemize}
Then
\begin{itemize}\begin{itemize}\begin{itemize}
\item[(c)]  $T>\frac 1A\, d_M(p,q)\,.$
\item[(d)]  $\lV\dot{x}(t)\rV <A$ for all $t\in [0,T]$.
\end{itemize}\end{itemize}\end{itemize}
\end{Corollary}

\noindent{\bf Proof:}

Let $B$ be from \eqref{E:I.1} and let $\ga:[0,d_M(p,q)]\to M$ be a minimal
geodesic with $\lv\dot{\ga}\rv\equiv 1$ joining $p$ and $q$. By the
superlinearity of $L$, for $D=2\, B + 1>0$ there exists $E>0$ such that
$$
\lv L(x,v)\rv > D\, \lv v\rv -E \quad \text{for all } (x,v)\in TM.
$$
We have that
\begin{align*}
B\, d_M(p,q) &> S_L(\ga) \ge \Phi_c(p,q) \ge S_L(x)-d_M(p,q) \\
             &> \int_0^T (D\,\lv\dot{x}(t)\rv - E)\, dt - d_M(p,q) \\
             &> D\, d_M(p,q) - E\, T - d_M(p,q)\,.
\end{align*}
Hence
\begin{equation}\label{E:C2.1}
T> \frac{(D-B-1)}{E} \, d_M(p,q) = \left(\frac BE\right)\, d_M(p,q)\,.
\end{equation}
Now let
$$
C:= \max\{\,\lv L(x,v)\rv\,\vert\, \lv v\rv\le \tfrac EB\, \}\,.
$$
Let $\eta\in AC(p,q,T)$ be a minimal geodesic. Then, by \eqref{E:C2.1},
$$
\lv \dot{\eta}\rv \equiv \frac{d_M(p,q)}{T} < \frac EB
$$
and hence
$S_L(\eta)\le C\, T$. Let $A_1=A_1(C)$ be from theorem \ref{L:MATHER1}, then
it is enough to use
$$
A= \max\left\{\, A_1(C),\, \tfrac EB\, \right\}.
$$

\qed


\smallskip


Let ${\cal M}(L)$ be the set of invariant borel probability
measures for the Lagrangian flow.



\medskip



\noindent{\bf Theorem II.}
$$
c(L) = - \min \left\{\left. \int L \, d\mu \, \right\vert\,
\mu\in{\cal M}(L)\, \right\}.
$$

\noindent{\bf Proof:}

Let $\mu\in{\cal M}(L)$ be ergodic. Let $(p,v)\in TM$ be such that
$$
\lim_{T\to\infty} \frac 1T \int_0^T L\bigl( f_t(p,v)\bigr) \, dt = \int L \, d\mu\, .
$$
Let $B>0$ be such that
$$
\lv L(x,v) \rv < B \quad \text{ if } \lv v \rv \le 2 \, .
$$
For $N>0$ let $q_N := \pi\bigl(f_N(p,v)\bigr)$ and let
$\ga_N:[0,d(p,q_N)]\to M$ be a geodesic joining $q_N$ to $p$. Let
$x_N:[0,N]\to M$ be defined by $x_N(t) = \pi \bigl( f_t (p,v) \bigr)$.
Then $f_t(p,v) = \bigl( x_N(t), \dot{x}_N(t)\bigr)$. For $k\in \re$,
we have that
\begin{align*}
S_{L+k}(\ga_N) &= \int_0^{d(p,q_N)} L
\bigl( \ga_N(t), \dot{\ga}_N(t) \bigr)\, dt
\le (B+k)\, \text{diam}(M)\, .\\
\lim_N \frac 1N S_{L+k}(x_N * \ga_N) &= \lim_N \frac 1N S_{L+k}(x_N) +0 \\
&= S_{L+k}(\mu) = S_L(\mu) + k \, .
\end{align*}
If $k< - S_L(\mu)$, then
$$
\Phi_k(p,p) \le \lim_N S_{L+k}(x_N * \ga_N) = -\infty\, .
$$
Hence $k\le c(L)$. Therefore
\begin{align*}
c(L) &\ge \sup\, \left\{ - S_L(\mu) \, \vert \,
\mu\in {\cal M}_{\text{erg}}(L)\, \right\} \\
&\ge - \min \left\{ S_L(\mu) \, \vert \, \mu \in{\cal M}(L)\, \right\}\, .
\end{align*}

Now let $k<c(L)$ and  $p,q\in M$. Then $\Phi_k(p,q)=-\infty$ and
there exists a sequence of paths $x_n:[0,T_n]\to M$, $x_n \in AC(p,q)$
such that
\begin{equation}\label{E:II.1}
\lim_n S_{L+k}(x_n) = -\infty\, .
\end{equation}
Since $L$ is bounded below, we have that
\begin{equation}\label{E:II.2}
\lim_n T_n = + \infty\, .
\end{equation}
Let $y_n:[0,T_n]\to M$, $y_n\in AC(p,q)$ be a minimizer of the  action
among the curves in $AC(p,q,T_n)$. Then $\bigl( y_n(t),\dot{y}_n(t)\bigr)$
is an orbit segment of $f_t$  and by \ref{lemmavel},  $|\dot y|$ is
bounded.
Let $\nu_n$ be the probability measure defined by
\begin{align*}
\int h\, d\nu_n
&= \frac 1{T_n} \int_0^{T_n} h ( y_n(t), \dot{y}_n(t) )\, dt\\
&= \frac 1{T_n} \int_0^{T_n}
h\bigl( f_t( y_n(0), \dot{y}_n(0))\bigr)\, dt\,\, ,
\end{align*}
for all $h:TM\to \re$ continuous. There exists a subsequence
$\nu_{n_i}$ which converges weakly* to a probability measure $\mu$.
By \eqref{E:II.2}, $\mu$ is invariant under the flow $f_t$.
Since the velocities are bounded we have that
$\lim\limits_{n_i} \frac 1{T_{n_i}} S_{L+k}(y_{n_i}) = S_{L+k}(\mu) = S_L(\mu)
+k\,$.
Since $\lim_n S_{L+k}(y_n) = \Phi_k(p,q) = -\infty$ and $T_n>0$ for all $n$,
then $S_L(\mu)+k\le 0$.
For any $k<c(L)$ we found an invariant measure $\mu$ such that  $k\le -
S_L(\mu)$.  Therefore
\begin{align*}
c(L) &\le \sup \left \{ - S_L(\mu) \, \vert \, \mu\in {\cal M}(L) \,
\right\}\\
&\le - \min \left\{ \, S_L(\mu)\, \vert \, \mu\in{\cal M}(L)\, \right\}\, .
\end{align*}
\qed

We state now theorem III. The proof of theorem III is split in
Ma\~n\'e \cite{Ma2} and \cite{CI}.
We say that a property holds for a {\it generic Lagrangian\/} if given any
Lagrangian $L$, there exists a residual set ${\cal O}\subseteq C^\infty(M,\re)$
such that the property holds for all the Lagrangians $L+\psi$ with
$\psi\in{\cal O}$. We say that an invariant measure is {\it uniquely ergodic}
if it is the only invariant measure on its support. A periodic orbit for
the Lagrangian flow is said {\it hyperbolic \/} if it is a hyperbolic
periodic orbit of the flow restricted to its energy level.

\medskip
\noindent{\bf Definition.} We say that $\mu\in{\cal M}(L)$ is a
{\it minimizing measure\/} if
$$
\int L \, d\mu = -c(L).
$$

Denote by $\widehat{\cal M}(L)$ the set of minimizing measures in ${\cal M}(L)$.

\medskip
\noindent{\bf Theorem III.}

{\it
\begin{itemize}
\item[(a)] For generic $L$, $\widehat{\cal M}(L)$ contains a single measure
          and this measure is uniquely ergodic.
\item[(b)] When this measure is supported on a periodic orbit, this
          orbit is hyperbolic.
\end{itemize}
}

\bigskip

Item (a) is proved in Ma\~n\'e \cite{Ma2} and item (b) is proved in \cite{CI}.

\medskip

\noindent{\bf Conjecture. (Ma\~n\'e)}

 For a  {\it generic}  $L$, $\widehat{\cal M}(L)$  consists on a single measure
{\it supported in a periodic orbit.}






\stepcounter{section}

\bigskip
\bigskip

\centerline{\large \bf \thesection. Recurrence properties.}


\bigskip


The prerequisite of the following definition is this remark:
since $d_k \ge 0$ for $k\ge c(L)$, then for any absolutely
continuous curve $x:[a,b]\to M$ and $k\ge c(L)$ we have that
\begin{equation}\label{E.S2.1}
S_{L+k}(x) \ge \Phi_k(x(a),x(b))\ge - \Phi_k(x(b),x(a))\, .
\end{equation}

\noindent{\bf Definition.} Set $c=c(L)$. We say that $x:[a,b]\to M$ is
a {\it semistatic curve\/} if it is absolutely continuous and:
\begin{equation}\label{E.S2.2}
S_{L+c}\left( x\vert_{[t_0,t_1]}\right) = \Phi_c\left(x(t_0),x(t_1)\right)\,,
\end{equation}
for all $a<t_0\le t_1<b$; and that it is a {\it static\/} curve if
\begin{equation}\label{E.S2.3}
S_{L+c}\left(x\vert_{[t_0,t_1]}\right) = - \Phi_c(x(t_1),x(t_0))
\end{equation}
for all $a<t_0\le t_1<b$.

By~\eqref{E.S2.1}, equality \eqref{E.S2.3} implies~\eqref{E.S2.2}. 
Hence static curves are semistatic. Semistatic curves 
are solutions of (E-L) because
they minimize the action of $L+c$ in $AC(x(t_0),x(t_1),t_1-t_0)$. If
$p,\,q\in M$ are on a {\it static\/} curve then $d_c(p,q)=0$.

If $w\in TM$, denote by $x_w:\re\to M$ the solution of (E-L) with
$\dot{x}_w(0) = w$.

\medskip
\noindent{\bf Definition.}
\begin{align*}
\Sigma(L) &:= \{\, w\in TM\, \vert \,
              x_w:\re\to M \text{ is semistatic }\} \, , \\
\widehat{\Sigma}(L) &:= \{\, w\in TM\, \vert \,
              x_w:\re\to M \text{ is static } \}\, , \\
\Sigma^+(L) &:= \{\, w\in M\, \vert\,
              x_w\vert_{[0,\infty[} \text{ is semistatic }\}\, .
\end{align*}

\noindent{\bf Remark.}
Replacing $c$ by any other real number in the definition of semistatic
solution, the set $\Sigma^+(L)$ (and then $\widehat{\Sigma}(L)\subset
\Sigma(L)\subset \Sigma^+(L)$) becomes empty.

For $k>c(L)$ this remark follows from the following estimates:
\begin{align*}
+\infty > \max_{p,q\in M} \Phi_k(p,q)
       &\ge S_{L+k}(x\vert_{[0,T]}) = \Phi_k(x(0),x(T)) \\
       &\ge \Phi_c(x(0),x(T)) + (k-c)\, T \\
       &\ge \max_{p,q\in M}\Phi_c(p,q) + (k-c)\, T.
\end{align*}


The following theorem is proven in Ma\~n\'e \cite{Ma3}. 

\medskip

\noindent{\bf Theorem IV. (Characterization of minimizing measures)}

{\it A measure $\mu\in{\cal M}(L)$ is minimizing if and only if
$\text{supp}(\mu)\subseteq \widehat{\Sigma}(L)$.
}

\medskip
We include a proof below, using theorem V(c).

\bigskip


Given an $f_t$ invariant subset $\La\subseteq TM$, and
 $\e>0$, $T>0$, an {\it $(\e,T)$-chain joining
$\xi_a\in\La$ and $\xi_b\in\La$} is a finite sequence
$\{(\zeta_i,t_i)\}_{i=1}^N\subset \La\times\re$ such that
$\zeta_1=\xi_a$, $\zeta_{N+1}=\xi_b$, $t_i>T$ and
$d(f_{t_i}(\zeta_i),\zeta_{i+1})<\e$ for $i=1,\ldots,N$.
We say that a set $\La\subset TM$ is {\it chain transitive}
if for all $\xi_a,\,\xi_b\in\La$, and all $\e>0$, $T>0$ there exists
and $(\e,T)$-chain in $\La$ joining $\xi_a$ and $\xi_b$.
When this condition holds only for all $\xi_a=\xi_b\in\La$ we say 
that $\La$ is {\it chain recurrent}.
\bigskip

\noindent{\bf Theorem V. (Recurrence Properties)}
{\it
\begin{enumerate}
\item[a)] $\Sigma(L)$ is chain transitive.

\item[b)] $\widehat{\Sigma}(L)$ is chain recurrent.

\item[c)] The $\alpha$ and $\om$-limit set of a semistatic orbit 
          is contained in $\widehat{\Sigma}(L)$.
\end{enumerate}
}

\noindent {\bf Proof:}

Lets first prove (c).
Let $w\in\Sigma$ and let $u\in \om(w)$.
We prove that $\om(w)\subseteq \widehat{\Sigma}(L)$, the proof
that $\a(w)\subseteq \widehat{\Sigma}(L)$ is similar.
It is enough to prove that for all
$T>0$, $x_u\vert_{[0,T]}$ is static. Let $p:= x_u(0)$, $q:=x_u(T)$. Let
$p_n=x_w(s_n)$, $q_n=x_w(t_n)$ be sequences of points in $M$ with
$s_n<s_n+T=t_n<s_{n+1}$, $s_n,\, t_n\underset{n}{\to}+\infty$ and
$\dot{x}_w(s_n)\underset{n}{\to}\dot{x}(0)$,$\,\,\dot{x}_w(t_n)
\underset{n}{\to}\dot{x}_w(T)$. We have that


\begin{align*}
&S_{L+c} \left( x_w\vert_{[s_n,t_n]}\right) = \Phi_c(p_n,q_n)\, , \\
&x_w\vert_{[s_n,t_n]} \overset{C^1}{\longrightarrow} x_u\vert_{[0,T]}\, ,\\
&\begin{aligned}
S_{L+c}\left( x_u\vert_{[0,T]} \right)
&= \lim_n S_{L+c}\left( x_w\vert_{[s_n,t_n]}\right) \\
&= \lim_n \Phi_c(p_n,q_n) \\
&= \Phi_c(p,q) \, .
\end{aligned}
\end{align*}

Moreover, by the continuity of $\Phi_c$, we have that
\begin{align*}
\Phi_c(p,q) + \Phi_c(q,p)
     &= \lim_n  \Phi_c(p_{n+1},q_{n+1}) + \Phi_c(q_n,p_{n+1})  \\
     &=\lim_n \left\{ S_{L+c}\left( x_w\vert_{[t_n,s_{n+1}]}\right)
        + S_{L+c}\left( x_w\vert_{[s_{n+1},t_{n+1}]}\right)\right\} \\
     &= \lim_n \left\{ S_{L+c}\left( x_w\vert_{[t_n, t_{n+1}]}\right)\right\}\\
     &= \lim_n \Phi_c(q_n,q_{n+1}) \\
     &= \Phi_c(q,q)=0 \, .
\end{align*}




\smallskip

\noindent {\bf Proof of theorem IV:}

Suppose that $\mu\in{\cal M}(L)$ and $\text{supp}(\mu)\in\hat{\Sigma}(L)$.
Let $\theta$ be a generic poinmt for $\mu$ and $L+c$. Then $\theta$ is
static and if $x_\theta(t) = \pi\circ\vr_t(\theta)$, we have
\begin{align*}
\int (L+c)\, d\mu 
     &= \lim_{T\to +\infty} \frac 1T \int_0^T \bigl[
             L(x_\theta,\dot{x}_\theta) + c \bigr] \, dt \\
     &= \lim_{T\to +\infty} \frac 1T \,
             \Phi_c\bigl(x_\theta(0),x_\theta(T)\bigr)   \\
     &\le \lim_{T\to +\infty} \frac 1T \, 
             \sup_{p,q\in M} \Phi_c(p,q) = 0.
\end{align*}
where the second equality is because $\theta$ is (semi)static.
Hence $\mu$ is minimizing.

Now suppose that $\mu\in{\cal M}(L)$ is minimizing.
Applying lemma II.2 in Ma\~n\'e \cite{Ma2}, we get that for
$\mu$-almost every $\theta$ there is a sequence $T_j=T_j(\theta)\to
+\infty$ such that
\begin{gather}
\lim_{j\to +\infty} d_{TM}\bigl(\theta,\vr_{T_j}(\theta)\bigr) = 0\,,
\label{E:IV.a}\\
\lim_{j\to +\infty} \int_0^{T_j}
\bigl[L( x_\theta,\dot{x}_\theta) + c \bigr]\, dt = 0\,.
\label{E:IV.b}
\end{gather}
By theorem V(c), it is enough to prove that 
$\text{supp}(\mu)\subseteq\Sigma(L)$.
Since $\Phi_c$ is continuous, it is enough to prove that
$\mu$-almost every $\theta$ is semistatic. Now let $\theta$ 
satisfy~\eqref{E:IV.a},~\eqref{E:IV.b} and define
$$
\de_\theta(T):= S_{L+c}\left(x_\theta\vert_{[0,T]}\right)
                - \Phi_c\bigl(x_\theta(0),x_\theta(T)\bigr)\,.
$$
Then $\de_\theta(t)$ is non-derecasing and $\de_\theta(t)\ge 0$.

By the continuity of $\Phi_c$ and~\eqref{E:IV.a} we have that
\begin{equation}\label{E:IV.c}
\lim_{j\to +\infty} \Phi_c\bigl( x_\theta(0),x_\theta(T_j)\bigr) =
0\,.
\end{equation}
 \, From~\eqref{E:IV.c} and~\eqref{E:IV.b} we have that $\lim_{j\to +\infty}
\de_\theta(T_j)=0$. Since $\de_\theta(t)$ is non-decerasing, then
$\de_\theta(t)\equiv 0$ for all $t\ge 0$. Hence $\theta$ is
semistatic.

\qed

                

            








\noindent {\bf Proof of theorem V(a):}


Given $\e>0$, $S>0$, $u,\, v\in\Sigma(L)$, we have to find an $(\e,S)$-chain in
$\Sigma(L)$ joining $u$ to $v$. It is easy to see that such 
$(\e,S)$-chain exists if $\om(u)\cap\a(v)\ne\o$. Let
\begin{equation}\label{E:A&O}
{\A} := \a(v) \quad , \quad \Om := \om(u)
\end{equation}
and suppose that $\A\cap\Om =\o$. Let
$p\in\pi(\Om)$, $q\in\pi(\A)$.
 Let $\eta_n:=[0,T_n]\to M$ be such that
$$
p := \eta_n(0)\in\pi(\Om)
\quad ,\quad
q := \eta_n(T_n)\in\pi(\A)
$$
\begin{equation}\label{estrella}
\SLC (\eta_n) \le \Phi_c(p,q) + \frac 1n
\end{equation}
By corollary \ref{corvel2}, we can assume that
$\eta_n$ is a solution of (E-L) and  satisfies
\begin{equation}\label{E:EtaIsBounded-1}
\lv \dot{\eta}_n(t) \rv < \A
\quad \text{ for all }
0\le t \le T_n
\end{equation}

Given $\frac \e 2 > \de_0>0$  there exists $0<\de<\de_0$ such that
if $\lv v\rv, \; \lv w\rv <\A$ and $d_{TM}(v,w)<\de$ then
\begin{equation}\label{sol}
d_{TM}(f_t(v),f_t(w))<\de_0 \text{ for all } \lv t \rv\le S.
\end{equation}


Let $\M$ be the union of $\A$, $\Om$ and the set  of accumulation points
of the tangent vectors of the
$\eta_n$'s:
\begin{equation*}
\M := \A \cup \Om \cup  \left\{\text{$v$} \in TM \, \left\vert \,
\begin{array}{lc}
\exists\, \langle n_k    \rangle\subseteq\na &0\le t_{n_k}\le T_{n_k}   \\
\exists\, \langle t_{n_k}\rangle\subseteq\re &v=\lim_k\dot{\eta}_{n_k}(t_{n_k})
\end{array}
\right.\right\}\,.
\end{equation*}
Then
\begin{equation}\label{E:N5.6}
{\Bbb M} \subseteq \{\, v\in TM\,\vert\, \lV v\rV \le A\, \}\,.
\end{equation}

We shall need the following lemmas

\begin{Lemma}\label{L:MinSigma-1}
\end{Lemma}

{\it
\begin{itemize}
\item[(a)] $\M\subseteq \Sigma(L)$.
\item[(b)] $\M$ is invariant.
\end{itemize}
}

Let ${\Bbb K}$ be the set of vectors which are on the $\om$-limit of vectors
of ${\Bbb M}$.
$$
{\Bbb K} := \bigcup_{v\in{\Bbb M}}\bigl[\om(v)\bigr]\, .
$$
Since $\Bbb M$ is closed and foward invariant, then the closure $\ov{\Bbb
K}\subseteq {\Bbb M} \subseteq \Sigma(L)$. Moreover, the vectors in ${\Bbb K}$
are chain recurrent. By \eqref{E:N5.6}  ${\ov{\Bbb K}}$ is
compact. Given $\de>0$ let
$$
{\Bbb K}_\de := 
\bigl\{\, v\in TM\, \big\vert\, d_{TM}(v,\ov{\Bbb K})<\de\,\bigr\}\,.
$$
Since $\ov{\Bbb K}$ is compact, the number of connected 
components of ${\Bbb K}_\de$ is finite. 
Let $\La_i=\La_i(\de)$, $i=0,1,\ldots,N$, $\Om\subseteq \La_0$, be the
connected components of ${\Bbb K}_\de$.


\begin{Lemma}\label{L:N5.1}
Each component $\La_i$ is $(2\de,T)$-chain transitive for any $T>0$.
\end{Lemma}



If $\A\subset \La_0$ the proposition is proved. Suppose that $\A \cap \La_0
=\o$. Consider the oriented graph $\boldsymbol{\Gamma}$ with vertices
$\La_i$, $i=0,1,\ldots,N$
and an edge $\La_i\to\La_j$ if there exists $v\in \M$ and $t_i<t_j$ such that
$f_{t_i}(v)\in\La_i$ and $f_{t_j}(v)\in\La_j$.



\begin{Lemma}\label{L:N5.2}
There exists a path in the graph joining $\La_0\supseteq\Om$ to $\La_j\supseteq \A$.
\end{Lemma}



We need that the connecting orbits between the $\La_j$'s have  time
intervals greater than $S$. Let $\La_i\to\La_j$ be an edge of the graph.
Take a connecting orbit
$f_t(v)$, $t_i<t_j$, $v\in\M$, from $\La_i(\de)$ to $\La_j(\de)$. Since
$f_{t_j}(v)$ is  in a $\de$-neighbourhood of $\ov{\Bbb K}\cap
\La_j(\de)$ and $\ov{\Bbb K}$ is invariant,  \eqref{sol}
implies  that $f_{t_j+S}(v)$ is  in a $\de_0$-neighbourhood of $\ov{\Bbb K}\cap
\La_i(\de)$.


Now we can construct an $(\e,S)$-chain joining $u\in\A$ to $v\in\Om$ by using
the connecting orbits between the $\La_i$'s and joining them by $(\e,S)$-chains
inside each $\La_i$. This completes the proof of theorem V(a).

\qed


Now we prove the lemmas

\bigskip

\noindent{\bf Proof of lemma \ref{L:MinSigma-1}(a):}


Equation \eqref{estrella} implies that
$$
S_{L+c}(\eta_n\vert_{[\a,\be]})\le \Phi_c(\eta_n(\a),\eta_m(\be)) + \frac 1n
$$
for all $0<\a<\be<T_n$. If $0<a_{n_k}< T_{n_k}$,
$\dot{\eta}_{n_k}(a_{n_k})\overset{k}{\to} v$, $b_{n_k}-a_{n_k}\overset{k}\to
\tau$, then
$$
\bigl( \eta_{n_k}(t),\dot{\eta}_{n_k}\bigr)
\to \bigl( x_v(t),\dot{x}_v(t)\bigr) = f_t(v)
\text{ for } 0\le t < b_n - a_n
$$
and by the continuity of $\Phi_c$ we have that
$$
S_{L+c}(x_v\vert_{[0,T]}) \le \Phi_c(x_v(0),x_v(T))\, .
$$
Hence $x_v\vert_{[0,T]}$ is semistatic.
\qed





\bigskip

\noindent{\bf Proof of \ref{L:MinSigma-1}(b):}

Since the set $A\cup \Om \subseteq \M$ is invariant, we only have to prove that
$\M\setminus(A\cup\Om)$ is invariant.
Let
$\om\in\M\setminus(\A\cup\Om)$ and  $\langle k\rangle\subseteq\na$,
$\langle t_{n_k}\rangle\subseteq\re^+$ be such that
$\dot{\eta}_{n_k}(t_{n_k})\underset k{\to} w$.
It is enough to show that
\begin{align}
    \lim\inf_k t_{n_k} = +\infty \, ,  \label{E:-infty}\\
    \lim\inf_k \lv T_{n_k} - t_{n_k}\rv = + \infty \, .\label{E:+infty}
\end{align}
 We need the folowing claim


\noindent{\bf Claim.}
{\it
The limit points of $\bigl\{\, \dot{\eta}_n(0)\,\vert\, n\in N\,\bigr\}$
and $\bigl\{\, \dot{\eta}(T_n)\, \vert\, n\in\na\, \bigr\}$ are in
$\Om$ and $\A$ respectively.
}


\medskip
\noindent {\bf Proof:}

We prove it only for limit points $\dot{\eta}_n (0) \to u\in TM$. Since
$\eta_n(0)\in\pi(\Om)$, then $\pi(u)\in\pi(\Om)$.
Since $\Om\subseteq \widehat{\Sigma}(L)$ and $\M\subseteq \Sigma^+(L)$, then
the claim  follows from part (b) of theorem VI.

\qed


We only prove \eqref{E:-infty}.  Suppose that
$$
\lim\inf_k t_{n_k} = a <+\infty \, .
$$
 Then there exists a
convergent subsequence
$\langle\eta_{n_\ell}\vert_{[0,a]}\rangle$ in the $C^1$
topology, to a semistatic solution $\ga$ such that
$$
\dot{\ga}(a) = \lim_\ell \dot{\eta}_{n_\ell}(a) = w \, .
$$
Then by the claim we have that
$$
\dot{\ga}(0)= f_{-a}(w) = \lim_\ell \dot{\eta}_{n_\ell}(0)\in\Om\, .
$$
and hence $w\in\Om$. This contradicts our assumption that
$w\in\M\setminus(\A\cup\Om)$.

\qed

\bigskip

\noindent{\bf Proof of lemma \ref{L:N5.1}:}


Let $T>0$. Since $\La_i = \La_i(\de)$ is open and connected, it is pathwise
connected. Let $\xi$, $\zeta\in\La_i$ and let $\Ga:[0,S]$ be a continuous path
such that $\Ga(0)=\xi$, $\Ga(s)=\zeta$. Let $0=s_0<s_1<\cdots<s_n=S$ be such
that $d_{TM}(\Ga(s_i),\Ga(s_{i+1}))<\frac{\de}2$. For each $s_i$ let
$v_i\in\La_i(\de)\cap{\ov{\Bbb K}}$ be such that $d(\Ga(s_i),v_i)<\de$. Since
the orbits of $\ov{\Bbb K}$ are recurrent, there exist $\tau_i>T$ such that
$d(f_{\tau_i}(v_i),v_i)<\frac{\de}2$. We have that
$$
d(f_{\tau_i}(v_i),v_{i+1})
      <d(f_{\tau_i}(v_i),\Ga(s_i))  +  d(\Ga(s_i),\Ga(s_{i+1}))
         +  d(\Ga(s_{i+1}),v_{i+1})<2\,\de.
$$
The orbit segments $\{\,f_t(v_i)\,\vert\, 0\le t\le \tau_i,\,
i=0,\ldots,n-1\,\}$ give the required $(2\,\de, T)$-chain.

\qed


\bigskip


In order to prove lemma \ref{L:N5.2} we need first

\begin{Lemma}\label{L:N5.4}

For all $\de>0$ there exists $N=N(\de)>0$ and $S=S(\de)>0$ 
such that if $n>N$, then for all $c\in[0,T-n]$ we have that
$$
{\Bbb K}_\de \cap \bigl\{(\eta_n(t),\dot{\eta}_n(t))\, \vert \,
t\in[c-S,c+S]\cap [0,T_n]\,\bigr\} \ne \o \,.
$$
\end{Lemma}

\medskip

\noindent{\bf Proof:}
Suppose it is not true. Then there exists $\delta >0$ and  sequences
$n_k\to\infty$ and $c_k\in[0,T_{n_k}]$, such that $0<c_k-k<c_k+k<T_{n_k}$
and
$$
{\Bbb K}_\de \cap \bigl\{\, (\eta_{n_k}(t),\dot{\eta}_{n_k}(t))\,
\vert\, t\in [c_k-k,c_k+k]\,\bigr\} = \o\, .
$$
Let
$\{\, v\in TM\,\vert\, \lV v\rV\le A,\, v\notin{\Bbb K}_\de\,\}=B$,
then $(\eta_{n_k}(t),\dot{\eta}_{n_k}(t))\in B$ for all $k\in{\Bbb N}$
and $t\in[c_k-k,c_k+k]$. Consider the measures $\mu_k$ defined by
$$
\int_B\phi\, d\mu_k :=
\frac 1{2k}\int_{c_k-k}^{c_k+k} \phi(\eta_{n_k}(t),
\dot{\eta}_{n_k}(t))\, dt\, .
$$
Since the $\eta_{n_k}$'s are solutions of (E-L), $B$ is compact and $k\to
+\infty$, then there exists a convergent subsequence
$\nu_k\to\nu$ in the weak*
topology to an invariant measure $\nu$ of the Lagrangian flow, with
$\text{supp}(\nu)\subset B$. Moreover, $\text{supp}(\nu)\subset \M$.
Let $v\in \text{supp}(\nu)$. Since $B$ is compact, we have that
the $\om$-limit $\o\ne \om(v)\subset {\Bbb K}$ and
$$
\o \ne \om(v) \subseteq \text{supp}(\nu)\subset B\cap {\Bbb K}
\subseteq B \cap {\Bbb K}_\de =\o\, .
$$
This is a contradiction.

\qed

\bigskip
%\newpage

\noindent{\bf Proof of lemma \ref{L:N5.2}:}


Suppose that $\A\cap\La_0=\o$, otherwise there is nothing to prove.
 For each $n\in{\Bbb N}$ let
\begin{align*}
a_n &:= \sup\{\,t\in[0,T_n]\,\vert\,
         (\eta_n(t),\dot{\eta}_n(t))\in\La_0\,
        \}\,,           \\
a_n + s_n &:= \inf\{\, t\in]a_n,T_n]\,\vert\,
         (\eta_n(t),\dot{\eta}_n(t))\in{\Bbb K}_\de\,\}\,.
\end{align*}
By lemma \ref{L:N5.4}, for $n>N(\de)$ we have that  $0<s_n<S(\de)$.
Choose a sequence $\eta^1_n$ such that $(\eta^1_n(a_n),\dot{\eta}^1_n(a_n))$
converges. Renumbering the $\La_i$'s if necessary, we can assume that
$s_n\to s^1\in [0,S(\de)]$ and $(\eta^1_n(a_n+s^1),\dot{\eta}_1^n(a_n+s^1))\in
\ov{\La_1}$. By lemma \ref{L:MinSigma-1}(a), the sequence
$(\eta_n^1,\dot{\eta}_n^1)\vert_{[a_n,a_n+s^1]}(s-a_n)$ converges to a
semistatic solution of (E-L) $(\ga^1,\dot{\ga}^1)\vert_{[0,s^1]}$, whose points
are in ${\Bbb M}$ and such that $\ga^1(0)\in\La_0$, $\ga^1(s^1)\in\La_1\ne
\La_0$. Hence
there is and edge $\La_0\longrightarrow \La_1$ in the graph ${\boldsymbol
\Ga}$.

Suppose that $\La_1\cap\A=\o$, otherwise the lemma is proved.
Let
$$
b_n:=\sup\{\,t\in[0,T_n]\,\vert\,
     (\eta^1_n(t),\dot{\eta}^1_n(t))\in \La_0\cap\La_1\,\}\, .
$$
The same arguments show that there exists $0<s^2\le S(\de)$ and a subsequence
$\eta^2_n\vert_{[b_n,b_n+s^2]}$ such that $\eta^2_n(b_n+s^2)\in\La_2$,
$\La_2\ne\La_i$, $i=0,1$, and an edge $\La_1\longrightarrow \La_2$ or
$\La_0\longrightarrow \La_2$. We can repeat this argument each time that
the final $\La_j$ does not contain $\A$. Since the number of the $\La_j$'s is
finite, we obtain a path in the graph ${\boldsymbol \Ga}$ from $\La_0$ to $\La_j$
with $\A\subseteq\La_j$.
\qed



\bigskip


\noindent
{\bf Proof of theorem V(b):}

Let $w\in \HSL$ and let $\A=\alpha (\text{$w$})$, $\Om=\om(w)$. 
It is easy to see that if $\A\cap \Om \ne \o$ then $w$ 
is chain recurrent in $\HSL$. Suppose that
$\A\cap\Om = \o$. For all $s<t$ and all $\e>0$ there exist $T=T(\ga)>0$ and
$\ga = \ga_{\e,s,t}:[0,T]\to M$ such that $\ga(0)=x_w(t)$, $\ga(T)=x_w(s)$ and
$$
\SLC (\ga) \le -\Phi_c\bigl( x_w(t),x_w(s) \bigr) + \e
$$
Let
\begin{equation}\label{E:Etas}
\eta_n := \ga_{\frac 1n , -n , n}
\qquad\qquad , \qquad\qquad
T_n := T\bigl(\ga_{\frac 1n , -n , n}\bigr)
\end{equation}
We can assume that $\lv \dot{\eta}(t)\rv < A$
for all $0\le t\le T_n$.

The rest of the proof is similar to item (a), but now the corresponding
${\Bbb M}\subseteq \widehat{\Sigma}(L)$.
\qed












\stepcounter{section}


\bigskip
\bigskip
\newpage

\centerline{\large \bf  \thesection. Graph, covering and coboundary properties. }

\bigskip

\noindent{\bf Theorem VI. (Graph Properties)}

{\it
\begin{itemize}
\item[(a)] If $\ga(t)$, $t\ge 0$ is an orbit in $\Sigma^+(L)$, then, denoting
$\pi:TM\to M$ the canonical projection, the map $\pi\vert_{\{\dot{\ga}\vert
t>0\}}$ is {\it injective with Lipschitz inverse}.

\item[(b)] Denoting $\Sigma_0(L)\subset M$, the projection of $\SH $, for every
$p\in\Sigma_0(L)$ there exists a {\it unique} $\xi(p)\in T_pM$ such that
$$
( p,\xi(p))\in \Sigma^+(L)\, .
$$
Moreover
$$
(p,\xi(p))\in \SH \, ,
$$
and the vector field $\xi$ {\it is Lipschitz}. Obviously
$$
\SH = \text{\rm graph}(\xi)\, .
$$
\end{itemize}
}

\medskip


For a proof of the following lemma see Mather~\cite{Mather} 
or Ma\~n\'e \cite{Ma0}.

\begin{Lemma}{\bf (Mather)}\label{L:MATHER2}
 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 $x_i(a)=p_i$,
$\dot{x}_i(p_i)=v_i$, 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) \, , \\
S_L(x_1\vert_{[a-\e,a+\e]}&)+S_L(x_2\vert_{[a-\e,a+\e]})>S_L(\ga_1)+S_L(\ga_2)
\end{align*}
\end{Lemma}


\noindent{\bf Proof of theorem VI:}

{\bf (a)}
 Since the curve $\ga$ is semistatic,  corollary \ref{corvel2} implies that
there exists $A>0$ such that $\lv\dot{\ga}(t)\rv<A$.
Let $K>0$, $\e_1=1$, $\de>0$ be from lemma \ref{L:MATHER2}.
We prove that if $(p,v)$, $(q,w)\in \{\,\dot{\ga}(t)\,\vert\, t>0\,\}$ and
$d_M(p,q)<\de$, then
$$
d_{TM}\bigl((p,v),(q,w)\bigr)<K\, d_M(p,q) \, .
$$
This implies item(a). Suppose it is false. Then there exist $0<t_1<t_2$ such
that
\begin{align*}
d_M\bigl(\ga(t_1),\ga(t_2)\bigr) &< \de \\
d_{TM}\bigl(\dot{\ga}(t_1),\dot{\ga}(t_2)\bigr) &> K\,
d_M\bigl(\ga(t_1),\ga(t_2)\bigr)
\end{align*}
By lemma \ref{L:MATHER2} there exist $0<\e<t_1$, and solutions
 $\a$, $\be:[-\e,\e]\to M$,
of (E-L) such that
\begin{alignat*}{2}
\a(-\e) &= \ga(t_1-\e)=:p \qquad\qquad &\a(\e)  &=\ga(t_2+\e)=:q  \\
\be(-\e) &=\ga(t_2-\e)=:s\qquad\qquad &\be(\e)  &=\ga(t_1+\e)=:r
\end{alignat*}
$$
S_L(\a)+S_L(\be) < S_L(\ga\vert_{[t_1-\e\, , \, t_1+\e]})
+S_L\bigl(\ga\vert_{[t_2-\e\, , \,t_2+\e]}\bigr)
$$
Since $\ga\vert_{]0,+\infty[}$ is semistatic, we have that
\begin{align*}
\Phi_c(p,q)+\Phi_c(s,r) &\le S_{L+c}(\a) +S_{L+c}(\be)  \\
   &< S_{L+c}(\ga\vert_{[t_1-\e\, ,\, t_1+\e]})
      + S_{L+c}(\ga\vert_{[t_2-\e\, ,\, t_2+\e]}) \\
   &< \Phi_c(p,r) +\Phi_c(s,q)
\end{align*}
Since $0\le \Phi_c(s,r)+\Phi_c(r,s)$,
\begin{align}
\Phi_c(p,q)&< \Phi_c(p,r)-\Phi_c(s,r)+\Phi_c(s,q) \notag\\
\Phi_c(p,q)&< \Phi_c(p,r) +\Phi_c(r,s) +\Phi_c(s,q) \label{E:VI.1}
\end{align}
Since $\ga\vert_{]0,+\infty[}$ is semistatic, we have that
\begin{align*}
\Phi_c(p,q) &= S_{L+c}(\ga\vert_{[t_1 -\e\, ,\, t_1+\e]})
    + S_{L+c}(\ga\vert_{[t_1+\e\, ,\, t_2-\e]})
    + S_{L+c}(\ga\vert_{[t_2-\e\, ,\, t_2+\e]}) \, ,\\
\Phi_c(p,q) &= \Phi_c(p,r)+\Phi_c(r,s)+\Phi_c(s,q) \, .
\end{align*}
This contradicts \eqref{E:VI.1}.

{\bf (b)} Now we prove item (b). We prove that if $p$, 
$q\in\Sigma_0(L)$, $(p,v)\in\SH $, $(q,w) \in \Sigma^+(L)$, 
and $d(p,q)<\de$, then
$$
d_{TM}\bigl((p,v),(q,w)\bigr)<K\, d_M(p,q)\, .
$$
Observe that this implies item (b). 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) &= f_{-\e}(q,w) &&=:q_{-\e }\qquad \qquad &\a(\e)  
              &= f_\e (p,v)&&=:p_{-\e} \\
\be(-\e)&= f_{-\e}(p,v) &&=:p_\e \qquad \qquad &\be(\e) &= f_\e(q,w) 
             &&=:q_\e
\end{alignat*}
and
$$
S_L(\a) +S_L(\be) <S_L(x_w\vert_{[-\e,\e]}) +S_L(x_v\vert_{[-\e,\e]})
$$
So
\begin{align*}
\Phi_c(q_{-\e},p_\e)+\Phi_c(p_{-\e},q_\e) &<\Phi_c(q_{-\e},q_\e)+
\Phi_c(p_{-\e},p_\e) \\
  & =\Phi_c(q_{-\e},q_\e)-\Phi_c(p_\e,p_{-\e})
\end{align*}

\centerline{\epsfxsize=6cm\epsfbox{6b.eps}}
\centerline{\small \sc theorem vi b.}


Thus
$$
\Phi_c(q_{-\e},q_\e)\le\Phi_c(q_{-\e},p_\e)+\Phi_c(p_\e,p_{-\e})+
\Phi_c(p_{-\e},q_\e) < \Phi_c(q_{-\e},q_\e)
$$
which is a contradiction.

\qed
\bigskip



Denote by ${\cal L}$ the set of equivalence classes of the equivalence relation
$d_c(a,b)=0$ lifted to $\SH$ by the vectorfield $\xi$ of  theorem VI.
Denote by $\om(v)$ the $\om$-limit of $v\in TM$ by the flow $f_t$.
If $\Ga\in{\cal L}$ set
$$
\Ga^+ :=\bigl\{\, w\in\Sigma^+(L)\, \big\vert\, \om(w)\in\Ga\, \bigr\} \, .
$$
Clearly $\Ga^+$ is foward invariant. Let
$$
\Ga^+_0:=\bigcup_{t>0}\pi\, f_t\,\Ga^+\,\, .
$$

\noindent{\bf Theorem VII. (Covering Property)}

{\it
\begin{itemize}
\item[(a)] $\pi\Sigma^+(L) = M$.

\item[(b)] For all $p\in\Ga^+_0$, there exists a {\it unique}
$\xi_\Ga(p)\in T_pM$ such that
$$
(p,\xi_\Ga(p))\in\Ga^+\, .
$$
Moreover,  $\xi_\Ga$ is
Lipschitz.
\end{itemize}
}


Observe that $\pi:\Sigma^+(L)\to M$ is not necessarily injective.
We recall that Ma\~n\'e stated item (a) in a stronger form: 
$\pi \Gamma =M$ for every
 equivalence class $\Gamma $. This may not be true as 
 the following example shows. Let
 $M=S^1=\re/{\Bbb Z}$ be the unit circle and $L=\frac{1}{2} v^2 -\cos 6\pi x$.
Then the three maximums of the potential $A=0=2\pi$, $B=\frac{2\pi}3$ and
$C=\frac{4\pi}3$ are singular (hyperbolic saddle) points of the Lagrangian
flow.
For {\it mechanical Lagrangians} $L=\frac 12 v^2 - \phi(x)$, we have that
$$
c(L) = e_0 = - \min_{x\in M} L(x,0)\,,
$$
and the static points are the critical points $(x,0)$ of the Lagrangian flow
such that $L(x,0)=-e_0$. In this example $c(L)= 1$ and
$\widehat{\Sigma}(L)=\{(A,0),(B,0),(C,0)\}$.
We shall prove below that $d_1(A,B)=\Phi_1(A,B)+\Phi_1(B,A)>0$. Hence
$(A,0)$ and $(B,0)$ are not in the same equivalence class. By \eqref{$*$},
$\Sigma^+(L)\subseteq E^{-1}\{c(L)\}$, hence if $\Ga$ is the equivalence class
of $(A,0)$, then $\pi\, \Ga^+ \subseteq [0,\frac{2\pi}{3}[\cup
]\frac{2\pi}{3},2\pi]\neq S^1$. This example can easily generalized
to higher dimensions.

\centerline{\epsfxsize=8cm\epsfbox{example.eps}}
%\centerline{Theorem 6b.}


We show now that in the example above $d_1(A,B)>0$.
By theorem V, the set $\Sigma(L)$ is chain transitive and contains
$\widehat{\Sigma}(L)=\{(A,0),(B,0),(C,0)\}$.
Moreover, by \eqref{$*$}, $\Sigma^+(L)\subseteq E^{-1}\{c(L)\}$.
In our example this is only possible if $\Sigma(L)$ contains 
complete components
of the saddle connections (so that $\pi\Sigma^+(L)=M$).
\, From the symmetry of this Lagrangian we get that
$\Sigma(L)$ contains all the saddle connections.
Hence $\Sigma(L)=E^{-1}\{ c(L)\}$.
Let $(x,\dot{x})$ be the orbit on $E^{-1}\{c(L)\}$ with $\a$-limit
$(A,0)$ and $\om$-limit $(B,0)$. Then it is semistatic and hence
\begin{gather*}
L(x,\dot{x})+c(L) = \dot{x}\, L_v(x,\dot{x}) = \lv \dot{x}\rv^2,\\
\Phi_1(A,B)=\lim_{T\to +\infty} S_{L+c}(x\vert_{[-T,+T]})
=\int_{-\infty}^{+\infty} \lv\dot{x}(t)\rv^2 dt > 0 \,.
\end{gather*}
The same argument gives that $\Phi_1(B,A)>0$ and thus $d_1(A,B)>0$.


\bigskip


\noindent{\bf Proof of theorem VII:}

We prove (a).
We may assume that $\pi \SH\ne M$ otherwise the proof is trivial.
Let $p\in M\setminus \pi \SH$ and  $q\in \pi \SH \ne \o$.
For each $n\in{\Bbb N}$ let $\gamma_n :[0,T_n] \to M$ be such that
\begin{enumerate}
\item $\gamma_n(0)=p$, $\gamma_n (T_n) =q$.
\item $S_{L+c}(\ga_n) \le \Phi_c(p,q) +\frac{1}{n}$.
\item $\gamma_n$ minimizes $L+c$ in $AC(p,q,T_n)$.
\end{enumerate}
Then
\begin{itemize}
\item[(i)] $\ga_n$ is a solution of (E-L).
\item[(ii)] $\lv\dot{\ga}_n\rv < A$ for all $t\in [0,T_n]$.
\end{itemize}
Since $\SH$ is invariant under the Lagrangian flow, item (b) of theorem VI
implies that
\begin{itemize}
\item[(iii)] $T_n\to +\infty$.
\end{itemize}

Let $\eta(t):=\pi f_t(p,v)$. By (i) and (iii), for any fixed $T>0$
we have that $\ga_n\vert_{[0,T]}\to \eta\vert_{[0,T]}$ in the $C^1$
topology, and hence
$$
S_{L+c}(\eta\vert_{[0,T]}) = \lim
S_{L+c}(\ga_n\vert_{[0,T]})\,.
$$
Clearly $\eta(0)=p$. It is enough to show that $\eta\vert_{[0,+\infty[}$
is semistatic. For, we have that
\begin{align*}
S_{L+c}(\ga_n) &=
S_{L+c}(\ga_n\vert_{[0,T]}) +
S_{L+c}(\ga_n\vert_{[T,T_n]})   \\
       &\le \Phi_c(p,q) + \tfrac 1n          \\
       &\le \Phi_c(p,\ga_n(T)) + \Phi_c(\ga_n(T),q) + \tfrac 1n \\
       &\le \Phi_c(p,\ga_n(T))
           + S_{L+c}(\ga_n\vert_{[T,T_n]})+\tfrac 1n\,.
\end{align*}
Hence
$$
S_{L+c}(\ga_n\vert_{[0,T]}) \le \Phi_c(p,\ga_n(T)) + \tfrac 1n\,.
$$
Taking the limit when $n\to +\infty$ we obtain that $\eta$ is semistatic.


We now prove  item (b).  Let $A>0$ be from
corollary  \ref{corvel2} and $K=K(A)>0$, $\e_1=\e_1(A)>0$,
$\de=\de(A)>0$ be from
Mather's lemma \ref{L:MATHER2}.
It is enough to prove that if $p_1,\,p_2\in\Ga^+_0$, $v_1,\, v_2\in\Ga^+$,
$\pi(v_i)=p_i$, $i=1,2$ and $d_M(p_1,p_2)<\de$, then
$d_{TM}(v_1,v_2)\le K\,d_M(p_1,p_2)$.

Suppose it is false. Then there exists $p_i\in\Ga^+_0$, $v_i\in\Ga^+$,
$\pi(v_i)=p_i$ such that $d_M(p_1,p_2)<\de$ and
$d_{TM}(v_1,v_2)> K\, d_M(p_1,p_2)$.
Since $p_i\in\Ga^+_0$, there exists $0<\e<\e_1$ such that
$f_t(v_i)\in\Ga^+$ for $t>-\e$. Let $q_i:=\pi f_{-\e}(v_i)$,
$r_i:=f_{+\e}(v_i)$. By Mather's lemma \ref{L:MATHER2}, there exist
$\eta_i:[-\e,\e]\to M$ such that $\eta_i(-\e)=q_i$, $\eta_i(+\e)=r_i$
and
\begin{alignat}{5}
\SLC(\eta_1)\, &+& \SLC(\eta_2)\, &<&
        \,\SLC(x_{v_1}\vert_{[-\e,+\e]})\, &+&\, \SLC(x_{v_2}\vert_{[-\e,+\e]}).
\notag \\
\intertext{Thus}
\Phi_c(q_1,r_2)\, &+&\, \Phi_c(q_2,r_1) &<&\, \Phi_c(q_1,r_1)\quad &+&
\Phi_c(q_2,r_2).\quad
\label{E:S3.A}
\end{alignat}

Let $u_i\in\om(v_i)$ and $z_i:=\pi(u_i)$. If
$T^i_n\overset{n}{\longrightarrow}+\infty$
is such that $f_{T^i_n}(v_i)\overset{n}{\longrightarrow}u_i$,
we have that
\begin{align*}
\Phi_c(q_i,z_i) &= \lim_n \Phi_c(q_i,x_{v_i}(T^i_n))
                    = \lim_n \SLC(x_{v_i}\vert_{[-\e,T^i_n]}) \\
                &= \SLC(x_{v_i}\vert_{[-\e,+\e]})
                    + \lim_n \SLC(x_{v_i}\vert_{[+\e,T^i_n]})         \\
                &\ge  \Phi_c(q_i,r_i) + \Phi_c(r_i,z_i)\,.
\end{align*}
By the triangle inequality for $\Phi_c$ we get that
\begin{equation}\label{E:S3.B}
\Phi_c(q_i,z_i) = \Phi_c(q_i,r_i) + \Phi_c(r_i,z_i)
\end{equation}
 \, From \eqref{E:S3.B} and \eqref{E:S3.A} we have that
\begin{align}
\Phi_c(q_1,z_1) + \Phi_c(q_2,z_2)
              &= \Phi_c(q_1,r_1) +  \Phi_c(r_1,z_1)
                    + \Phi_c(q_2,r_2)  + \Phi_c(r_2,z_2)
\notag \\
              &> \Phi_c(q_1,r_2) + \Phi_c(q_2,r_1)
                    + \Phi_c(r_1,z_1) + \Phi_c(r_2,z_2)
\notag \\
              &\ge \Phi_c(q_1,z_2) + \Phi_c(q_2,z_1)\,.
\label{E:S3.C}
\end{align}

Since $u_i\in \Ga$, then $d_c(z_1,z_2)=\Phi_c(z_1,z_2)+\Phi_c(z_2,z_1)=0$.
Adding $d_c(z_1,z_2)=0$ to the right of inequality \eqref{E:S3.C}, we obtain
that
\begin{align*}
\Phi_c(q_1,z_1) + \Phi_c(q_2,z_2)
          &> \Phi_c(q_1,z_2) + \Phi_c(z_2,z_1)
                        + \Phi_c(z_1,z_2) + \Phi_c(q_2,z_1)  \\
          &\ge \Phi_c(q_1,z_1) + \Phi_c(q_2,z_2)\, .
\end{align*}
This is a contradiction.

\qed







\bigskip

\noindent
{\bf Theorem VIII. (Generic Structure of $\SH $)}

{\it
For a generic Lagrangian $L$, $\SH $ is a uniquely ergodic set.
If it is a periodic orbit then it is a hyperbolic periodic orbit.
}
\medskip

\noindent {\bf Proof:}

This should be thought as a corollary of theorems III and IV. 
Take the Generic set given by theorem III of Lagrangians $L$ 
that satisfy $ \# \widehat{\cal M}(L)=1$ and
call this unique minimizing measure $\mu (L)$.
Then if $\mu$ is an invariant measure of $L$ and it is
supported in $ \SH $ then by theorem IV
it is minimizing. Thus  $\mu =\mu (L)$. This proves the theorem.

\qed


\bigskip

\newpage
\noindent
{\bf Theorem IX. (Coboundary Property)}

{\it
If $c=c(L)$, then $(L+c)\big\vert_\SH$ is a Lipschitz coboundary. More
precisely, taking any $p\in M$ and defining $G:\SH\to\re$ by
$$
G(w)=\Phi_c\bigl(p,\pi(w)\bigr)
$$
Then
$$
(L+c)\big\vert_\SH = \frac{dG}{df} \, ,
$$
where
$$
\frac{dG}{df}(w):= \lim_{h\to 0}\frac 1h
\bigl[\, G\bigl(f_h(w)\bigr)-G(w)\, \bigr]\, .
$$
}

\medskip



%\newpage
\noindent{\bf Proof:}

Let $w\in\SH$ and define $F_w(v):=\Phi_c\bigl(\pi(w),\pi(v)\bigr)$. We have
that
\begin{align*}
\left.\frac{dF_w}{df}\right\vert_w
    &= \lim_{h\to 0}\frac 1h\, \bigl[\, F_w(f_h\, w) - F_w(w)\,\bigr] \\
    &= \lim_{h\to 0}\frac 1h\, \bigl[\, \Phi_c(\pi w, \pi f_h\, w)
          - \Phi_c(\pi w, \pi w)\, \bigr] \\
    &= \lim_{h\to 0}\frac 1h\,
          \left[\, S_{L+c}\left(x_w\vert_{[0,h]}\right)\,\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 $w\in\SH$, $h\in\re$,
\begin{align}
G(f_h\, w) &= \Phi_c\bigl(p,\pi(f_h\, w) \bigr)
               = \Phi_c\bigl( p,\pi(w)\bigr)
               + \Phi_c\bigl(\pi(w),\pi(f_h \,w)\bigr) \notag  \\
G(f_h\, w) &= \Phi_c\bigl( p,\pi(w)\bigr) + F_w\bigl( f_h(w)\bigr) \, .
\label{E:IX.1}
\end{align}
This is enough to prove the theorem because then
$$
\left.\frac{dG}{df}\right\vert_w
     = \left.\frac{d\,}{dh} F_h(f_h\, w)\right\vert_{h=0}
     = \left.\frac{F_w}{df}\right\vert_w
     = L(w)\, +c,
$$
and $G$ is Lipschitz by theorem I.

We now prove \eqref{E:IX.1}. Let $q:=\pi(w)$, $x:=\pi(f_h\, w)$. 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
$$
\Phi_c(p,q)\le \Phi_c(p,x)+\Phi_c(x,q)=
\Phi_c(p,x)-\Phi_c(q,x)\le r\Phi_c(p,q)\,
.
$$
This implies \eqref{E:IX.2}.

\qed










\bigskip
\bigskip

%\newpage                                     
\stepcounter{section}

\centerline{\large \bf  \thesection. Connecting orbits 
inside fixed energy levels.}

\bigskip


We quote a paragraph from Ma\~n\'e \cite{Ma3}:

``Exploiting that the energy, $E:TM\to\re$, defined as usual by
$E(x,v)=\frac{\partial L}{\partial v}\, v - L$, is a first integral
of the flow generated by $L$, leads to {\it information on the position of
$\Sigma^+(L)$}. First observe that it is easy to check that a semistatic
curve $x:[a,b]\to M$ satisfies:
\begin{equation}\label{$*$}
E(x(t),\dot{x}(t)) = c(L)\,.
\end{equation}
This follows from calculating the derivative at $\la=1$ of
the function $F:\re\to\re$ given by:
$$
F(\la)=\int_a^{\frac{b}{\la}} (L+c)(x_\la(t),\dot{x}_\la (t))\, dt\,,
$$
where $x_\la:[a,\frac b{\la}]\to M$ is given by $x_\la(t)=x(\la t)$.
 \, From \eqref{$*$} follows that:
$$
\Sigma^+(L)\subset E^{-1}(c)\,,
$$
that together with $\pi\,\Sigma^+(L)=M$ implies:
$$
\pi\, E^{-1}(c)=M\,.
$$
Hence,
$$
c\ge \max_q E(q,0)\,.
$$
Moreover $\widehat{\Sigma}(L)\subset E^{-1}(c)$ implies:''

\begin{Corollary}
$\mu\in {\cal M}(L)$ is minimizing if and only if
\begin{gather*}
\int\left(\frac{\partial L}{\partial v}\right)v\, d\mu = 0\,, \\
\text{supp}(\mu)\subset E^{-1}(c(L))\,.
\end{gather*}
\end{Corollary}






\bigskip

\noindent{\bf Theorem X.}

{\it
If $k>c(L)$ then for all  $p,\,q\in M$, $p\ne q$ there exists a solution $x(t)$
of (E-L) such that $x\in AC(p,q)$  and
$$
\Phi_k(p,q)=S_{L+k}(x)
$$
Moreover the solution $(x(t)\dot{x}(t))$ is contained in the energy level
$E^{-1}\{k\}$.
}


\begin{remark}
If $p=q$ then the infimum
$$
\Phi_k(p,p) = \inf\{S_{L+k}(\ga)\,\vert\, \ga\in AC(p,q)\,\} =0
$$ can not be realized by a path defined on an interval with nonzero length.
For otherwise if $\ga:[0,T]\to M$, $T>0$ is a  minimum, then
$$
\Phi_c(p,p)\le S_{L+c}(\ga) = \Phi_k(p,p) - (k-c(L))\, T
< \Phi_k(p,p) =0 \,.
$$
Contradicting theorem I.
\end{remark}

\medskip

\begin{remark}      This theorem, with the same proof, holds for coverings
$\pi: \widehat{M} \to M$ of a compact manifold $M$, with the lifted
Lagrangian $\widehat{L}=L\circ \pi$.
\end{remark}




\medskip

%\newpage
\noindent {\bf Proof:}

Suppose that $p,\, q\in M$ and $p\ne q$.
For each $T>0$  there exists a minimizer $x_T$ of $S_L$ on
$AC(p,q,T)$. Then
$$
\Phi_k(p,q) = \inf_{T>0} S_L(x_T)+k\, T\, .
$$
Observe that
\begin{align*}
\lim_{T\to +\infty}\bigl(S_L(x_T)+k\,T\bigr)
  &= \lim_{T\to +\infty} \bigl( S_{L+c}(x_T)+(k-c)\,T \bigr) \\
  &\ge \lim_{T\to +\infty}\bigl(\Phi_c(p,q) + (k-c)\, T \bigr)\\
  &= +\infty\, .
\end{align*}



Choose a sequence $\{T_i\}$ such that $\lim_i S_L(x_{T_i}+k\,
T_i)=\Phi_k(p,q)$. Then $\{T_i\}$ must be bounded and we may assume that it has
a limit $T_0$. Since $\lV\dot{x}_{T_i}\rV<A$ and $p\ne q$, then $T_0>0$.
We shall prove that we actually have a minimal solution
with time interval $[0,T_0]$. We proceed as in the proof of Tonelli's theorem
(see Mather \cite{Mather} or Ma\~n\'e \cite{Ma0}, lemma 6.1). Observe that
by corollary \ref{lemmavel} the sequence $\{x_{T_i}\}$ is absolutely
equicontinuous because $\{ S_L(x_{T_i}) \}$ is bounded. Since
$T_i\overset{i}{\longrightarrow} T_0$ there is a subsequence of $\{x_{T_i}\}$
which converges to a curve $x_{T_0}\in AC(p,q,T_0)$. It remains to prove that
\begin{equation}\label{E:10.1}
S_{L+k}(x_{T_0})=\Phi_k(p,q)\,.
\end{equation}

The following lemma uses the superlinearity of the Lagrangian. A proof
can be found in Ma\`n\'e \cite{Ma0}. We may assume that $M=\re^n$.

\begin{Lemma} Given $C>0$ and $\e >0$ and a
compact subset $S\subset M=\re^n$ there exist  $\delta>0$ such that
if $(x,v),\,(y,w)\in TM = \re^n\times\re^n$ satisfy 
$x\in S$, $d(x,y)<\delta $ and $|v|<C$ then
$$
L(y,w)\ge L(x,v)+\frac{\partial L}{\partial v}(x,v)(w-v)-\e
$$
\end{Lemma}

Let $\Delta:=\Phi _k(p\,,\,q)-k\,T_0$. Given $C>0$ define
$$
S(C):= \{\, t\in [0,T_0]\,\vert\, \lv \dot{x}(t)\rv \le C\, \} \,.
$$
Since $L$ is bounded below, we can assume that $L$ is positive.
We will prove that for all $h>0$ and $C>0$,
\begin{equation}\label{E:10.2}
\int_{S(C)}L(x(t),\dot x(t))dt\le \Delta + h\,.
\end{equation}
Taking the limits when $h\to 0$ and $C\to \infty$ we get that
$S_L(x)\le \Delta$, completing the proof of \eqref{E:10.1}.
If $i$ is big enough
\begin{align*}
\Delta + h &> \int_{S(C)} L(x_{T_i}(t), \dot{x}_{T_i}(t))\, dt \\
&\ge \int_{S(C)} L(x(t),\dot{x}(t))\, dt - \e\, T_0
+ \int_{S(C)}\frac{\partial L}{\partial v}(\dot{x}_{T_i}(t) -\dot{x}(t))\,
dt\,.
\end{align*}
Then
\begin{equation}\label{E:10.3}
\int_{S(C)}L(x(t),\dot x(t))\, dt\le \Delta + h
+ \e\, T-\lim\inf_i \int_{S(C)}
\frac{\partial L}{\partial v}(\dot x_{T_i}(t)-\dot x(t))\,dt\,.
\end{equation}
We claim that
\begin{equation}\label{E:10.4}
\lim \inf \int_{S(C)}\frac{\partial L}{\partial v}(\dot x_{T_i}(t)-\dot
x(t))\,dt=0.
\end{equation}
This follows from the fact that $x$ is absolutely continuous,
$\frac{\partial L}{\partial v}$ is bounded in $S(C)$ and that for
any interval $(a,b)\subset[0,T_0]$
$$
\lim _{i\to\infty}\int_a^b(\dot x_{T_i}(t)-\dot x(t))\,dt
=\lim_{i\to\infty}(x_{T_i}(b)-x(b))-(x_{T_i}(a)-x(a))=0.
$$
Since $\e>0$ is arbitrary \eqref{E:10.4} together with \eqref{E:10.3} prove
\eqref{E:10.2}.

We now prove that minimizers of $S_{L+k}$ are in the energy
level $E=k$. Suppose that $x\in AC(p,q,T)$ is such that
$$
S_{L+k}(x) = \min_{y\in AC(p,q)} S_{L+k}(y)\,.
$$

Define
$$
F(\lambda )=\int_0^{\lambda T}(L+k)(x_\lambda ,\dot x_\lambda )
$$
where $x_\lambda (t):[0,\lambda T]\rightarrow M$ is defined as $x_\lambda
(t)=x(\frac t\lambda ).$ By the minimizing condition $F^{\prime }(1)=0$. On
the other hand
\begin{align*}
F^{\prime }(\lambda )
  &=T\,L\bigl(x_\lambda (\lambda T),\dot x_\lambda (\lambda T)\bigr)
      +\int_0^{\lambda T}\frac{\partial L}{\partial \lambda }\, dt+ kT  \\
\intertext{now}
\frac{\partial L}{\partial \lambda }
  &=\frac{\partial L}{\partial x}\frac{
     \partial x_\lambda }{\partial \lambda }+
     \frac{\partial L}{\partial v}\frac{
     \partial \dot x_\lambda }{\partial \lambda }     \\
  &=-\frac{\partial L}{\partial x}\dot x_\lambda \left(\frac t\lambda \right)
     \frac t{\lambda ^2}-\frac{\partial L}{\partial v}
     \left(\frac 1{\lambda ^3}\ddot x_\lambda \left(\frac t\lambda \right)t
     +\frac{\dot x_\lambda (\frac t\lambda )}{\lambda^2}\right)
\end{align*}
So
\begin{align*}
0&=T\,L\bigl(x(T),\dot x(T)\bigr)+T\, k-\int_0^T
 \left(\frac{\partial L}{\partial x}
\dot x+\frac{\partial L}{\partial v}\ddot x\right)t\, dt  -\int_0^T
\frac{\partial L}{\partial v}\dot x\,dt \\
&=T\,L\bigl(x(T),\dot x(T)\bigr)+T\,k
-\int_0^T\frac{\partial L}{\partial v}\dot x\, dt
-\int_0^T\frac d{dt}(L)t\, dt \\
&=T\,L\bigl(x(T),\dot x(T)\bigr)+T\,k-\int_0^T
\frac{\partial L}{\partial v}\dot x\,dt+\int_0^TL\,dt -L\,t\big\vert_0^T\\
&=Tk-\int_0^T E\,dt \\
&=T(k-E).
\end{align*}
This proves that the energy level of the solution $x$ is $k.$

\qed


\newpage
Observe that
\begin{equation}\label{E:10.5.0}
L + k = (\partial L /\partial v) v \quad \text{ on } E^{-1}\{k\}\,.
\end{equation}






\begin{Corollary}\label{C:10.5}
\end{Corollary}
{\it \begin{itemize}
\item[(a)] If $k>c(L)$ and $a,b\in M$, there exists a solution $x(t)$ of (E-L)
such that $x(0)=a$, $x(T)=b$ for some $T>0$, $E(x(t),\dot{x}(t))=k$ for all
$t\in\re$, and:
\begin{equation}\label{E:10.5.1}
\int_0^T\frac{\partial L}{\partial v}(x,\dot{x})\, \dot{x}\, dt
=\min \int_0^{T_1} \frac{\partial L}{\partial v}(y,\dot{y})\, \dot{y}\, dt\,,
\end{equation}
where the minimum is taken over all the absolutely continuous $y:[0,T_1]\to M$,
$T_1\ge 0$, with $y(0)=a$, $y(T_1)=b$ and $E(y(t),\dot{y}(t))=k$ for a.e. $t\in
[0,T_1]$.

\item[(b)] Conversely, if given $k>c(L)$ and $a,\, b\in M$, there exists an
absolutely continuous $x:[0,T]\to M$ with $x(0)=a$, $x(T)=b$,
$E(x(t),\dot{x}(t))=k$ for a.e. $t\in [0,T]$ and satisfying the minimization
property \eqref{E:10.5.1}, then $x(t)$ is a solution of (E-L).
\end{itemize}
}

If $p,q\in \pi E^{-1}\{k\}$, define
$$
ACE(p,q;k):=\{ x\in AC(p,q) \, \vert\, E(x,\dot{x}) = k \text{ a.e.}\,\}\,.
$$


Item (a) of corollary \ref{C:10.5} follows from \eqref{E:10.5.0} and item (b)
follows from the fact that since minimizers of $S_{L+k}$ have energy $k$, then
minimizing $S_{L+k}$ on $AC(p,q)$ is equivalent to minimize it on $ACE(p,q;k)$.


\bigskip

Given $\mu\in{\cal M}(L)$ define its {\it homology} or {\it asymptotic cycle}
(cf. Schwartzman \cite{S}), by $\rho(\mu)\in H_1(M,\re)=H^1(M,\re)^*$ such that
$$
\rho(\mu)\cdot [\theta] = \int_{TM}\theta\, d\mu
\quad , \quad
\forall\, [\theta]\in H^1(M,\re)\,,
$$
where $\theta$ is a closed 1-form and $[\theta]$ its homology class.
Define the {\it Mather's betha function} $\beta:H_1(M,\re)\to\re$, by
$$
\beta(h):= \min\{\, S_L(\mu)\,\vert\, \mu\in{\cal M}(L),\; \rho(\mu)=h\,\}\,.
$$
Since, for any $h\in H_1(M,\re)$ the set $K(h):=\{\,\mu\in{\cal
M}(L)\,\vert\,\rho(\mu)=h\,\}$ is convex, it follows that $\beta$ is a convex
function. Let $\beta^*$ be its Legendre transform: $\beta^*:H^1(M,\re)\to\re$,
$$
\beta^*(w):=\max_{h\in H_1(M,\re)} \{\,w(h)-\beta(h)\,\}\,.
$$
The reader can check that
$$
\beta^*([\theta])= -c(L-\theta),
\quad \forall\, [\theta]\in H^1(M,\re)\,.
$$
Define the {\it strict critical value} $c_0(L)$ by
$$
c_0(L) = \min_\theta c(L-\theta)\,.
$$
Then
$$
c_0(L) = -\beta^*(0) = - \min \{\, S_L(\mu)\,\vert\, \mu\in{\cal
M}(L),\;\rho(\mu)=0\,\}\,.
$$

Observe that
\begin{equation}\label{E:S4.D}
\frac{d\,}{dt} \, E(x,tv)\big\vert_{t=1} = v\cdot L_{vv}(x,v)\cdot v > 0\; ,
\end{equation}
Therefore if $(x,v)$ is a critical point of the energy function $E$ then $v=0$
and $\frac{\partial L}{\partial x}(x,0)=0$.
Let
$$
e_0 := \max_{p\in M} E(p,0).
$$
By \eqref{E:S4.D}, we have that
\begin{equation}\label{E:S4.E}
E(p,0)= \min_{v\in T_pM} E(p,v)\, .
\end{equation}
In particular
$$
e_0 = \min\{\,k\in\re\,\vert\, \pi(E^{-1}\{k\}) = M\, \}\,.
$$
Let $\theta_0$ be a closed 1-form such that $c_0(L)=c(L-\theta_0)$.
Then the energy function and the Euler-Lagrange equations for $L-\theta_0$ and
$L$ are the same. Theorem X implies that $\pi(E^{-1}\{k\})=M$ for all
$k>c(L-\theta_0)=c_0(L)$. Hence
$$
e_0 \le c_0(L)\,.
$$

As observed in Ma\~n\'e \cite{Ma3}, for mechanical Lagrangians $L(x,v) =\frac
12 \langle v ,v\rangle_x - \phi(x)$, with $\langle\;,\,\rangle_x$ a riemannian
metric, we have that
$d_0=e_0=c_0(L)=c(L)$. There is an example in \cite{Ma3} of a Lagrangian $L$
with $e_0<c_0(L)$. It uses the following corollary to the proof of theorem X:

\begin{Corollary}

If $k>c_0(L)$, for every free homotopy class $H\ne 0$ of $M$, there exists a
periodic orbit in $E^{-1}\{k\}$ such that its projection on $M$ belongs to that
free homotopy class.
\end{Corollary}

\noindent{\bf Proof:}

Fix $k>c_0(L)$. By adding a closed 1-form we can assume that $c(L)=c_0(L)$.
Let $AC(H)$ be the set of absolutely continuous closed curves in $M$ with free
homotopy class $H$. Let $x_n\in AC(H)$ with $x_n:[0,T_n]\to M$ and
$$
\lim_{n\to \infty}S_{L+k}(x_n) = \inf_{x\in AC(H)}S_{L+k}(x).
$$
Let $\widetilde{x}_n$ be a lift of $x_n$ to the universal cover $\widetilde{M}$
of $M$. We can assume that $\tx_n$ is a minimizer of $_{\widetilde{L}+k}$ on
$AC(\tx_n(0),\tx_n(T_n);T_n)$ and in particular, that it is a solution of
(E-L). Then the same arguments as in the proof theorem X yield that $\{ T_n\}$
must be bounded and $\lV\dot{x}_n\rV<A$. We can assume thet $\lim_n T_n=T_0$.
Moreover
$$
T_0>\frac 1A \,\min\{\,\text{length}(\ga)\,\vert\, \ga\in AC(H)\,\}>0.
$$
The same arguments as in theorem X give a closed curve $\ga:[0,T_0]\to M$ wich
is a uniform limit of a subsequence of $x_n$ and hence $\ga\in AC(H)$. Moreover
\begin{equation}\label{E:C.X.1}
S_{L+k}(\ga)=\min_{x\in AC(H)} S_{L+k}(x)\,,
\end{equation}
and $(\ga,\dot{\ga})$ is in the energy level $E^{-1}\{k\}$.

It remains to prove that $\dot{\ga}(0)=\dot{\ga}(T_n)$. Suppose that
$\dot{\ga}(0)\ne \dot{\ga}(T_n)$. Let $\widetilde{M}$ be the universal cover of
$M$ and $\widetilde{L}$ the lift of $L$. Let $\widetilde{\ga}$ be a lift of
$\ga$. Consider the
path $\eta\vert_{[-\e,\e]}=\widetilde{\ga}\vert_{[T_0-\e,T_0]} *
\widetilde{\ga}\vert_{[0,\e]}$. We have that $\eta$ is not $C^1$ and hence it
is not a solution of (E-L). Since $k>c_0(L)\ge e_0$, then $k$ is a regular
value of the energy $\widetilde{E}$ of $\widetilde{L}$. By the Maupertius
principle (see theorem 3.8.5 of Abraham \& Marsden \cite{A-M}),
$\eta\vert_{[-\e,\e]}$ is not a minimizer of the $(\widetilde{L}+k)$-action on
$\widetilde{E}^{-1}\{k\}$. Then there exists $\xi\in(\eta(-\e),\eta(\e))$, with
energy $\widetilde{E}(\xi,\dot{\xi})\equiv k$ and $S_{\widetilde{L}+k}(\xi)<
S_{L+k}(\eta)$. Moreover, since $\widetilde{M}$ is simply connected, the paths
$\xi$ and $\eta$ are homotopic by a homotopy which fixes their endpoints. Hence
$\pi(\xi * \widetilde{\ga}\vert_{[-\e, T_0-\e]})$ and
$\dot{\ga}\vert_{[0,T_0]}$ are in the same free homotopy class of $M$. We have
that
\begin{align*}
S_{L+k}\left(\pi(\xi *\widetilde{\ga}\vert_{[\e,T_0-\e]})\right)
    &= S_{\widetilde{L}+k}(\xi * \widetilde{\ga}\vert_{[\e,T_0-\e]})    \\
    &= S_{\widetilde{L}+k}(\xi) + 
             S_{L+k}(\widetilde{\ga}\vert_{[\e,T_0-\e]}) \\
    &< S_{L+k}(\ga)\,.
\end{align*}
This contradicts the minimizing property \eqref{E:C.X.1} of $\ga$.

\qed

\medskip

\newpage

``An interesting characterization of the critical value $c(L)$, in terms
of an analogous to Tonelli's theorem (Mather \cite{Mather}) in a prescribed
energy level is given by the following result:''
\hfill {\it --- Ma\~n\'e \cite{Ma3}}

\medskip

\noindent{\bf Theorem XI. }

{\it
Assume that $k$ is a regular value of $E$ and  $\dim M >1$.
 Suppose that $k$ has the following property: for all $a,b$ in
$\pi E^{-1}(k)$ there exists an absolutely continuous 
curve $x:[0,T]\to M$ such that:
\begin{itemize}
\item[(i)] $x(0)=a$ and $x(T)=b$.
\item[(ii)] $E(x(t),\dot x(t))=k$ a.e. in $[0,T]$.
\item[(iii)]{$\displaystyle{
            \int_0^T \frac{\partial L}{\partial v} (x(t),\dot x(t))\,
            \dot x(t)\, dt
            =\min
            \int_0^{T_1} \frac{\partial L}{\partial v} (y(t),\dot y(t))\,
            \dot y(t)\, dt}$,}\newline
            where the minimum is taken over all absolutely continuous
            $y:[0,T_1]\to M$, with $y(0)=a,y(T_1)=b$ and
             $E(x(t),\dot x(t))=k$ a.e. in $[0,T_1]$.
\end{itemize}

Then $k>c(L)$ and $x(t)$ is a solution of the Euler Lagrange equation.
}

\medskip

The hypothesis $\dim M>1$ is necessary as the example of a simple pendulum
shows. Indeed, for $L(x,v) = \frac 12 \lv v\rv^2 - \cos x$ and any regular
value $k < e_0 = -\min_{p\in S^1}L(p,0)=c(L)=1$ there are such
minimizers. This is because a non-empty energy level $E^{-1}(k)$ with
$k<e_0$ is a topological circle and given $a,b\in\pi\bigl(E^{-1}(k)\bigr)$
there are two injective paths on $E^{-1}(k)$ from $\pi^{-1}\{a\}$ to
$\pi^{-1}\{b\}$. One of them must minimize the $(L+k)$-action
because $L+k=v\cdot L_v = \frac 12\lv v\rv^2 \ge 0$ on $E^{-1}(k)$.
In fact, the minimizer is the one whose projection on $S^1$ is injective.


%\centerline{\epsfysize=3cm\epsfbox{pendulum.eps}}


Another example in $M=S^1$ is ${\Bbb L}=L + \theta$ where
$\theta_x(v)$ is a (closed) 1-form such that $c({\Bbb L}) =
c(L+\theta)>c(L)=e_0$. The energy functions and the Lagrangian
flows for ${\Bbb L}$ and $L$ are the same. For $e_0<k<c({\Bbb L})$ the energy
levels $E^{-1}(k)$ support two periodic orbits. Lemma~\ref{L:11.1}
below shows that these orbits have negative $(L+k)$-action, and hence
there are no minimizers on these levels. On the other hand,
a regular energy level $k<e_0$ consists of one periodic orbit $(\ga,\dot{\ga})$.
By the symmetry of $E^{-1}(k)$, we have that $\int_\ga v\;{\Bbb L}_v
= \int_\ga \frac 12 \lv v\rv^2 + 0 >0$. Then the same arguments as for $L$ show
that there exists minimizers for $k<c(L)$.



We comment now the hypothesis of the regularity of the energy value $k$.
Recall that all the critical points of $E$ are in the zero section of $TM$.
The following lemma shows that if
$(p_0,0)$ is a critical point of $E$, then it is a singularity of the
Lagrangian flow and that the point $p_0$ can not be joined to other 
points by a path with energy $k$ which is differentiable at $p_0$. 
Thus completing the picture of theorem XI: If we require the 
differentiability of the minimizers $x(t)$ at the endpoints, 
then the same statement holds for critical values $k$
of $E$ if $E^{-1}\{k\}$ consists of more than one point, and if 
$\# E^{-1}\{k\}=1$ the (unique) minimizing curve is a 
(singular) solution of (E-L). Conversely, if 
$(q_0,0)\in E^{-1}\{k\}$ is not a critical point, then it is 
necessarily reached on finite time by curves on $E^{-1}\{k\}$.

Define
$$
d_0 := \min_{p\in M}E(p,0) = - \max_{p\in M} L(p,0)
$$
If $k<d_0$ then $E^{-1}(k)=\o$. Given $d_0\le k\le e_0$ define
$$
\theta(k):=\{\, p\in M \,\vert\, E(p,0)=k\,\}\, ,
$$
then $\theta(k)\ne \o$. The proof of the following lemma is delayed to the end
of the section.




\def \dy{{\dot{y}}}

\begin{Lemma}\label{L:11.2}
\end{Lemma}
{\it
\begin{itemize}
\item[(a)]
If $p_0\in\theta(k)$ and $D_pE(p_0,0)=0$, then the only curve $p:[0,\de]\to M$
with $E(p(t),\dot{p}(t))=k$ and $p(0)=p_0$ is the constant curve $p(t)=p_0$.
\item[(b)]
If $p:[0,t_0[\to M$ is such that $E(p,\dot{p})\equiv k$,
$\lim_{t\to t_0^-}p(t) = q_0\in\theta(k)$ and
$D_pE(q_0,0)\ne 0$, then $t_0$ is finite.
\end{itemize}
}




\bigskip




\noindent
{\bf Proof of theorem XI:}

\medskip

For $k\le c(L)$ we have to show that there are points in $E^{-1}\{k\}$ which
can not be joined by a curve  which minimizes the action  of $S_{L+k}$ on
$ACE(p,q;k)$.

Observe that $d_0$ is necessarily a critical value of $E$.
Suppose first that $d_0<k<e_0$. Since $k$ is a regular value of $E$, by the
Maupertius principle (see theorem 3.8.5 of Abraham \& Marsden \cite{A-M}),
the critical points of the functional $S_{L+k}$ on $ACE(p,q,k)$ are solutions
of (E-L). Let $p\in\theta(k)\ne \o$. Then $\{(p,0)\}=T_pM\cap E^{-1}\{k\}$.
Hence there is only one solution $x$ of (E-L) with $E(x,\dot{x})\equiv k$ and
$x(0)=p$. The set $\{\, x(t)\,\vert\, \dot{x}=0\,\} = \{\, x(t)\,\vert\,
x(t)\in\theta(k)\,\}$ has at most countably many points. But since $k$ is a
regular value of $E$ then $\theta(k)$ is a submanifold of $M$ of codimension 1
and $\dim \theta(k)>0$. Hence there are points in $\theta(k)=\partial \pi
E^{-1}(k)$ which can not be joined to $p$ by $(L+k)$-minimizers with
$E(x,\dot{x})\equiv k$.


Suppose now that $k=c(L)=:c$.  The arguments of equation \eqref{$*$} show that
minimizers of $S_{L+c}$ in $AC(p,q)$ are in the energy level $E=c(L)$. Thus
minimizers of $S_{L+c}$ on $ACE(p,q;c(L))$ are also minimizers on $AC(p,q)$
and hence semistatic curves. Let $(q,\xi(q))\in\widehat{\Sigma}\subseteq
E^{-1}\{k\}$. Let $p\notin \pi\{f_t(q,\xi(q))\,\vert\, t\in\re\,\}$.
Suppose that there exists a semistatic curve $x\in AC(p,q,T)$.
Then by theorem VI (b), we have that $\dot{x}(T)=\xi(q)$. This
contradicts the choice of $p$.

Now suppose that $e_0<k<c(L)$. We shall prove in lemma \ref{L:11.1}
that there exists a closed curve
$\ga(t)$ such that $E(\ga,\dot{\ga})\equiv k$ and
$$
\int_\ga (L+k) < 0\, .
$$
By making a large number of loops along $\ga$ one can produce a curve
with arbitrarily negative action. By adding two connecting segments
this implies that for any $p,q\in M$ there
are not minimizers of $S_{L+k}$ on $ACE(p,q;k)$.

This completes the proof of theorem XI.
\qed


Given an absolutely continuous closed curve $\Ga:[0,T]\to M$,
define the probability measure $\mu_\Ga$ on $TM$ by
$$
\int_{TM} \vr\; d\mu_\Ga :=
\frac 1T \int_0^T \vr\bigl( \Ga(t),\dot{\Ga}(t)\bigr)\; dt\, .
$$
Let ${\cal C}(k)$ be the set of measures $\mu_\Ga$ supported on the energy
level $E^{-1}(k)$ and let $\ov{{\cal C}(k)}$ be its closure in the weak*
topology. Define
$$
\ga(k):= \min_{\mu\in \ov{\cal C}(k)} \int v\cdot L_v\; d\mu\, .
$$

\medskip


Measures realizing this minimum may not be invariant.
For example if $L(x,v) = \frac 12 \lv v\rv^2 -\phi(x)$
is a mechanical lagrangian and $k<e_0$ is a regular
energy level, then any such measure will be supported
on $\partial E^{-1}\{k\}$, where $v=0$ and $v L_v = \frac 12 \lv v\rv^2 =0$.
Nevertheless,  $\partial E^{-1}\{k\}$ has no invariant subsets.



In the following lemma we use ideas of Dias Carneiro \cite{Ca}.

\begin{Lemma}\label{L:11.1}
For all $e_0<k<c(L)$ we have that $\ga(k)<0$.
\end{Lemma}



\noindent{\bf Proof:}

\, From \eqref{E:S4.E} we get that given any $k>e_0$ and $(p,v)\in TM$,
there exists a unique
$\la>0$ such that $E(p,\la\, v)=k$. Moreover,
$\la = \la(p,v,k):TM\times ]e_0,+\infty[\to]0,+\infty[$ is a smooth function.


Let $\mu$ be an invariant measure $\mu\in {\cal M}(L)$ such that
\begin{equation}\label{E:11.2}
\int \bigl( L + c(L) \bigr) \; d\mu =0\; .
\end{equation}
By theorem 4 and the fact that static curves have energy level 
$c(L)$ we have that $\text{supp}(\mu)\subseteq E^{-1}(c(L))$, and by the
Poincar\'e recurrence theorem we have that $\mu\in \ov{{\cal C}(c(L))}$.

For $k>e_0$ define the
measure $\nu_k$ on $E^{-1}(k)$ by
\begin{equation}\label{E:11.3}
\int_{E^{-1}(k)} f\, d\nu_k :=
A(k) \int_{E^{-1}(k)}\frac{f(p,\la(p,v,k)v)}{\la(p,v,k)}\,
d\mu (p,v)
\end{equation}
for any continuous function $f:E^{-1}(k)\to\re$, where
$$
A(k) := \left( \int \frac 1{\la(p,v,k)}\, d\mu (p,v)\right)^{-1} .
$$
Then $\nu_k\in\ov{{\cal C}(k)}$ and $\nu_{c(L)}=\mu$.
This measure $\nu_k$ is just the (probability) measure obtained by
reparametrizing the solutions of (E-L) on $\pi(\text{supp}(\mu))$ so
as to have energy $k$. This process is reversible, i.e. we can recover
$\mu$ by reparametrizing $\nu_k$.
Let
$$
g(k):= \int v\cdot L_v\; d\nu_k = k + \int L \; d\nu_k\; .
$$
Then $g$ is a differentiable function with derivative
$$
g'(k) = 1 + A(k) \int \frac{\partial\,}{\partial k}
\left(\frac{L(p,\la v)}{\la}\right)\, d\mu
+ A'(k) \int \frac{L(p,\la v)}{\la}\, d\mu \, .
$$

If we change the reference energy level $c(L)$ to $k_1>e_0$,
we can use $\nu_{k_1}$ instead of $\mu$ on formula \eqref{E:11.3}.
The function $g(k)$ does not change but now $\la(k_1)\equiv 1$,
$A(k_1)=1$ and
$$
g'(k_1) = 1
+ \int\frac{\partial \,}{\partial k} 
\left(\frac{\Bbb L}{\la}\right)\; d\nu_{k_1}
+ A'(k_1) \int L\; d\nu_{k_1} \; ,
$$
where ${\Bbb L}(p,v):= L(p,\la\, v)$. We compute this derivative:
\begin{align*}
\frac{\partial\,}{\partial k}\left(\frac{{\Bbb L}} 
{\la}\right)\Big\vert_{k=k_1}
             &= \frac{{\Bbb L}_k \, \la - {\Bbb L}\, \la_k}{\la^2}
                   = {\Bbb L}_k - \la_k\, {\Bbb L}\vert_{k=k_1}\; ,  \\
{\Bbb L}_k   &= v \cdot  L_{v} \; \la_k\; .
\end{align*}
Since $E(p,\la\, v)=k$, we have that
\begin{gather*}
\frac{\partial E}{\partial k}
  = E_v \cdot (\la_k \, v)
  = (v\cdot L_{vv}\cdot v)\ \; \la_k
  = 1 \; ,  \\
\la_k = \frac 1{v\cdot L_{vv} \cdot v}
  \quad \text{and} \quad
  {\Bbb L}_k = \frac{v\cdot L_v}{v\cdot L_{vv}\cdot v}\; .
\end{gather*}
Moreover
\begin{align*}
\frac{\partial \,}{\partial k} \left( \frac 1{\la}\right)\Big\vert_{k=k_1}
      &= -\frac{\la_k}{\la^2}\Big\vert_{k=k_1}
          = - \frac 1{v\cdot L_{vv}\cdot v} \\
A'(k_1)
      &= - \frac 1{A(k_1)} \int \frac{\partial \,}{\partial k}
             \left( \frac 1{\la}\right)\; d\mu
          = \int \frac 1{v\cdot L_{vv}\cdot v}\; d\nu_{k_1} \; .
\end{align*}
Therefore
\begin{gather}
\begin{aligned}
g'(k) &= 1 + \int \frac{v\cdot L_v}{v\cdot L_{vv}\cdot v}\; d\nu_k
         - \int \frac L{v\cdot L_{vv}\cdot v}\; d\nu_k
         + \left( \int \frac 1{v\cdot L_{vv}\cdot v}\; d\nu_k\right)
           \left(\int L \; d\nu_k\right)
\notag\\
g'(k) &= 1 + \int \frac k{v\cdot L_{vv}\cdot v}\; d\nu_k
         + \left(\int \frac 1{v\cdot L_{vv}\cdot v}\; d\nu_k\right)
           \left(\int L\; d\nu_k\right)
\notag\\
g'(k) &= \left( \int (L+k)\; d\nu_k\right)                  
         \left( \int \frac 1{v\cdot L_{vv}\cdot v}\; d\nu_k\right)
         + 1
\notag
\end{aligned}
\\
g'(k) = b(k)\; g(k) + 1 \; ,
\label{E:11.4}
\end{gather}
where $b(k) := \int (v\cdot L_{vv}\cdot v)^{-1}\, d\nu_k >0$. Let
$$
B(k) := \int_{e_0}^k b(s)\; ds
$$
and $h(k) := e^{-B(k)}\, g(k)$. From \eqref{E:11.4} we get that
$$
h'(k) = e^{-B(k)} > 0\; .
$$
By \eqref{E:11.2}, $h(c(L))=0$, therefore $h(k)<0$
for all $e_0<k<c(L)$. And thus
$$
\ga(k)\le g(k) = e^{B(k)}\, h(k) < 0
$$
for all $e_0<k<c(L)$. \qed


\bigskip



\noindent{\bf Proof of lemma \ref{L:11.2}:}

We first prove (a).
Using local coordinates we can assume that $M=\re^n$ and $TM=
\re^n\times\re^n$. For $v$ small and $p$ near $p_0$ such that $E(p,v)=k$, we
have that
\begin{align*}
E &= v\cdot L_v - L \; , \\
k &= \left( v\cdot L_v (p,0) + v\cdot L_{vv}(p,0) \cdot v + 
{\cal O}(v^3) \right)
\\
&\qquad\qquad - \left( L(p,0) + v\cdot L_v(p,0) + \tfrac 12 \, v\cdot 
   L_{vv}(p,0) \cdot v
+ {\cal O}(v^3) \right) \; ,
\\
\tfrac 12 \; &v \cdot L_{vv}(p,0)\cdot v + {\cal O}(v^3)
= k + L(p,0) \, .
\end{align*}
Since $k=-L(p_0,0)$,  there exists a function
$F(p,v)$ defined on a neighbourhood of $(p_0,0)$ on the energy level
$E(p,v)=k$ by
\begin{align}
F(p,v) \, \lv v\rv^2
       &:= k + L(p,0)
\notag \\
       &= L_p(p_0,0)\, \Delta p
           + \tfrac 12 \,\Delta p\, L_{pp}(p_0,0)\, \Delta p
           + {\cal O}(\Delta p^3)\; ,
\label{E:11.2.7}
\end{align}
where $\Delta p = p - p_0$, $F(p,v)$ is smooth and $F(p_0,v)>0$. We have that
\begin{equation}\label{E:11.2.8}
D_p E(p_0,0) = - L_p(p_0,0) = 0 \; .
\end{equation}
Since the left hand side of \eqref{E:11.2.7} is positive, using
\eqref{E:11.2.8} we have that $\Delta p \, L_{pp}(p_0,0)\, \Delta p > 0$.
Then there exists a function $G(p,v)>A>0$ such that for $v$ small and $p$ near
$p_0$ such that $E(p,v)=k$, we have
\begin{align}
\lv v \rv^2 &= G(p,v)^2\, \lv \Delta p \rv^2 \; ,
\notag \\
\lv v \rv &\ge A \, \lv \Delta p \rv \; .
\label{E:11.2.9}
\end{align}
Suppose that there exists a differentiable curve $p:[0,\de]\to M$ such that
$p(0)=p_0$ and $E(p(t),\dot{p}(t))=k$. For simplicity suppose that $\de =2$.
Let $x(t):=\lv p(t)-p_0\rv$. Writing $v(t)=\dot{p}(t)$, we have that
\begin{align*}
\frac{d\,}{dt} x(t)^2
    &= 2\, x\, \dot{x} = 2\, \langle p(t)-p_0, v(t) \rangle \, ,
\\
\dot{x}(t)
    &= \left\langle \frac{p(t)-p_0}{\lv p(t)-p_0\rv} , v(t)\right\rangle
         \ge -B \, \lv v(t)\rv \; .
\end{align*}
for some $B>0$, because $p(t)$ is differentiable at $t=0$. From
\eqref{E:11.2.9}, we have that
\begin{align*}
\dot{x}(t) &\ge -B \, \lv v(t) \rv \ge -A B\, x(t) \; ,  \\
x(t) &\ge x(1) \, \exp\bigl( - A B\, (t-1)\bigr) \; .
\end{align*}
In particular $x(0)\ne 0$. This contradicts the choice of $p(t)$.

Now we prove (b).
Since $D_pE(q_0,0)=L_p(q_0,0)\ne 0$, then $\theta(k)$ is locally a codimension
1 submanifold near $q_0$. Let $q(t)\in\theta(k)$ be given by the condition
$$
L_p(q(t),0)\, \bigl( p(t)-q(t)\bigr) = 0 \; .
$$
Using formula \eqref{E:11.2.7} with $\Delta p = p(t) - q(t)$, we have that
$$
F(p,v)\, \lv v \rv^2 = L_p(q(t),0)\, \Delta p
+ {\cal O}\left( \lv \Delta p \rv^2\right)
$$
and then
$$
\lv v \rv \ge A \; \sqrt{\lv \Delta p \rv} \; ,
$$
for some $A>0$. Let $y(t):= \lv p(t) -q(t)\rv$. Computing
$\frac{d\,}{dt}\, y(t)^2$, we obtain
$$
\dy (t)
= \left\langle \frac{p(t)-q(t)}{\lv p(t)-q(t)\rv} \; ,\; v(t)\right\rangle
\le - B \; \lv v(t)\rv
\le - A\, B \,\sqrt{y(t)} \; ,
$$
for some $B>0$. Therefore
$$
\frac{\dy}{2\, y^{\frac 12}} \le - \frac{A \, B} 2
\quad \text{ and } \quad
y(t) \le \frac{A^2\, B^2}{4} \; (t-t_0)^2\; .
$$
For future reference, we also note that
\begin{align}
\lv \dy(t) \rv &\ge b\; \lv v(t)\rv \ge b_1\; \sqrt{y(t)} \; ,
\label{E:11.2.I} \\
y(t) &\ge b_2 \; (t-t_0)^2 \; ,
\label{E:11.2.II}
\end{align}
for some  $b,\; b_1,\; b_2 >0$.

\qed









\stepcounter{section}


\bigskip
\bigskip



\centerline{\large \bf \thesection. Properties of weaker global minimizers.}

\bigskip

\noindent{\bf Definition.}
We say that a solution $x(t)$ of (E-L) is a {\it minimizer} (resp.
{\it forward minimizer}) if
$$
S_L(x\vert_{[t_0,t_1]})\le S_L(y)
$$
for every $t_0\le t_1$ (resp. $0<t_0\le t_1$) and every absolutely continuous
$y:[t_0,t_1]\to M$, with $y(t_i)=x(t_i)$, $i=1,2$.

Denote by $\La(L)$ (resp. $\La^+(L)$) the set of $(p,v)\in TM$
such that the solution $x(t)$ of (E-L) with initial condition
$(x(0),\dot{x}(0))=(p,v)$ is a minimizer (resp. forward minimizer).

Let $w\in\La^+(L)$ for $0\le s \le t$ define $\delta_w (s,t)$ by
\begin{equation}
S_{L+c}(x_w\vert_{[s,t]}) =\Phi_c(x_w(s),x_w(t)) + \de_w(s,t).
\label{defdelta}
\end{equation}
It is clear that
\begin{equation}\label{E:S5.2}
\delta_w (s,t) \ge 0.
\end{equation}
The triangle inequality for $\Phi$ implies that
\begin{equation}\label{destrianginv}
\delta_w (s,t)+\delta_w (t,r)\le \delta_w (s,r)
\end{equation}
for any $w\in\La^+(L)$ and $0\le s \le t\le r$.

\noindent{\bf Claim.}
{\it $\delta_w (s,t)$ is uniformly bounded on $w\in \La^+(L)$, and $0\le s\le
t<\infty$.}

We shall  prove first theorems XIV, XII and XIII and then the claim.
\medskip

\noindent {\bf Theorem XIV.}

{\it
\begin{itemize}
\item[(a)] There exists $C>0$ such that setting $c=c(L)$
$$
|S_{L+c}(x_w\vert_{[s,t]})|\le C
$$
for every $w\in\La^+(L)$ and all $0\le s \le t$.

\item[(b)] If $w\in\La^+(L)$ and $p\in M$ is such that
$p=\lim_{n\to \infty} x_w(t_n)$ for some sequence $t_n \to \infty$ 
then the limit
$$
\lim_{n\to \infty}S_{L+c}(x\vert_{[0,t_n]})
$$
exists and does not depend in the sequence $\{ t_n\}$.
\end{itemize}
}

\medskip

\noindent{\bf Proof:}

Item (a) is straightforward consequence of the claim and the definition of
$\delta_w$. From equation (\ref{destrianginv}) it follows that the function
$t\to\delta_w(0,t)$ is  increasing. Then the claim implies that
$D:=\lim_{t\to +\infty} \de_w(0,t)$ exists. Thus
\begin{align*}
\lim_{n\to \infty}S_{L+c}(x_w\vert_{[0,t_n]})
          &= \lim_{n\to \infty}\Phi_c(x_w(0),x_w(t_n))
                   +  \lim_{n\to \infty}\delta_w(0,t_n) \\
          &= \Phi_c(x_w(0),p)+ D\,.
\end{align*}

\qed

\bigskip

\noindent{\bf Theorem XII.}

{\it
\begin{itemize}
\item[(a)] The $\om$-limit set of an orbit in $\La^+(L)$ is contained in $\SH$.

\item[(b)] The $\a$ and $\om$-limit sets of an orbit in $\La(L)$ are contained
in $\SH$.
\end{itemize}
}

\medskip

\noindent{\bf Proof:}

We only prove (a). Let $w\in\La^+(L)$. Let $(p,v)\in\om(w)$, 
$T>0$ and let $\langle n_k\rangle$ be a sequence in $\re^+$ such that
$$
n_{k+1}>n_k + T\,,
$$
$$
(p,v) = \lim_k \bigl( x_w(n_k),\dot{x}_w(n_k)\bigr)\,.
$$

Let
$$
p_k := x_w(n_k) \qquad ,\qquad q_k :=x_w(n_k +T)\,,
$$
$$
(q,u) :=
\bigl(x_v(T),\dot{x}_v(T)\bigr)=\lim_k\,\bigl(q_k\,,\,\dot{x}_w(n_k+T)\bigr)\,.
$$
we have that
\begin{align*}
S_{L+c}(x_w\vert_{[n_k,n_k +T]}) &=\Phi_c(p_k,q_k) + \de_w(n_k,n_k+T) \,,\\
S_{L+c}(x_w\vert_{[n_k+T,n_{k+1}]}) &= \Phi_c(q_k,p_k)+\de_w(n_k+T,n_{k+1})\,,\\
S_{L+c}(x_w\vert_{[n_k,n_{k+1}]})&=\Phi_c(p_k,p_{k+1})+\de_w(n_k,n_{k+1})\,.
\end{align*}

By the claim and equation \eqref{destrianginv}
there is a constant $Q$ such that
\begin{align*}
   \sum_{k=1}^\infty\bigl( \de_w(n_k,n_k+T)+\de_w(n_k+T,n_{k+1})\bigr)
& \le\sum_{k=1}^{\infty} \de_w(n_k,n_{k+1}) \\
& \le \lim_{t \to \infty} \delta_w(0,t)
  \le Q\,.
\end{align*}
Therefore
\begin{equation}\label{E:S5.zero}
\lim_{k\to\infty}\de_w(n_k,n_{k+1})
=\lim_{k\to\infty}\de_w(n_k,n_k+T)
=\lim_{k\to\infty}\de_w(n_k+T,n_{k+1})
=0\,.
\end{equation}
Hence
\begin{align*}
S_{L+c}(x_w\vert_{[0,T]})
   &= \lim_k S_{L+c}(x_w\vert_{[n_k,n_k+T]})\\
   &=\lim_k \Phi_c(p_k,q_k) =  \Phi_c(p,q)\,.
\end{align*}
and from \eqref{E:S5.zero},
\begin{align*}
   \Phi_c(p,q)+ \Phi_c(q,p)
         & = \lim_k \Phi_c(p_k,q_k) + \lim_k\Phi_c(q_k,p_k) \\
   &= \lim_k\,\bigl[\, S_{L+c}(x_w\vert_{[n_k,n_k+T]})
                  +S_{L+c}(x_w\vert_{[n_k+T,n_{k+1}]})\,\bigr] \\
   &= \lim_k \,\bigl[\, \Phi_c(p_k,p_k) + \de_w(n_k,n_{k+1})\,\bigr]  \\
   &= \Phi_c(p,p) \\
  &= 0 \, .
\end{align*}
This implies that $\om(w)\subseteq\SH$.

\qed


\bigskip

\noindent{\bf Theorem XIII.}
\quad
{\it $f_t\vert_{\La(L)}$ is chain transitive.}

\medskip

\noindent{\bf Proof:}
Let $v$, $w\in\La(L)$ and $\e>0$. It is enough to prove that there exists an
$\e$-chain in $\La(L)$ joining $\om(w)$ to $\a(v)$. By theorem XIII,
$\a(v)\cup\om(w)\subseteq\SH\subseteq\Sigma(L)$.
By theorem V there exists such $\e$-chain contained
in $\Sigma(L)\subseteq\La(L)$. This completes the proof.

\qed

\medskip
\noindent {\bf Proof of the claim:}


Let
\begin{align*}
A &:= 2\, \max\,\bigl\{\, L(p,v)+ c(L)\,\big\vert\, 
\lV v\rV\le\text{ diam}M + 1 \,\bigr\}\\
B &:= \max\,\bigl\{\,\lv\Phi_c(p,q)\rv\,\big\vert\, p,q\in M\,\bigr\} \\
Q &:= 3\, \max\,\{ \, A , B\,\} + 2 \, .
\end{align*}
Suppose that there exist $\om\in\La^+(L)$, $0\le a\le b$ such that
$\de_w(a,b)>Q$.


Let $\ga:[a,T_b]\to M$ be in $AC(x_w(a),x_w(b))$ and such that
$$
S_{L+c}(\ga) < \Phi_c(x_w(a),x_w(b))+1\,.
$$
We have that
\begin{align}
S_{L+c}(\ga) &< S_{L+c}(x_w\vert_{[a,b]}) -\de_w(a,b) +1\,,   \notag\\
S_{L+c}(\ga) &< S_{L+c}(x_w\vert_{[a,b]}) - Q + 1\,, \label{E:XII.2}  \\
S_{L+c}(\ga) &< S_{L+c}(x_w\vert_{[a,b]}) - (A+B)\,. \label{E:XII.3}
\end{align}
Suppose that $T_b\ge b-a$. Let $\eta:[0,1]\to M$ be a geodesic on $M$ such that
$$
\eta(0)=x_w(b)   \quad ,\quad
\eta(1)=x_w(T_b+1) \quad ,\quad
\lV\dot{\eta}\rV\le \text{ diam}(M)\, .
$$
Then
$$
S_{L+c}(\eta) < A\, .
$$
Since
$$
-B\le \Phi_c(x_w(b),x_w(T_b+1))
\le S_{L+c}(x_w\vert_{[b,T_b+1]})\, ,
$$
using \eqref{E:XII.3} (or \eqref{E:XII.4}), we have that
\begin{align*}
S_{L+c}(\ga *\eta)
      &< S_{L+c}(x_w\vert_{[a,b]}) - (A+B) +A  \\
      &< S_{L+c}(x_w\vert_{[a,b]}) + S_{L+c}(x_w\vert_{[b,T_b +1]})  \\
S_{L+c}(\ga *\eta)
      &< S_{L+c}(x_w\vert_{[a,T_b+1]}) \, .
\end{align*}
This contradicts the hypothesis $w\in\La^+(L)$.

Now suppose that $T_b<b-a$. Let $u\in\SH$ and let $\la:[0,1]\to M$
be a geodesic on $M$ such that
$$
\la(0)=x_w(b)=\ga(T_b)
\quad ,\quad \la(1) =\pi\, u
\quad ,\quad \Vert\dot{\la}\Vert \le
\text{diam }M\, ,
$$
and let $\ov{\la}:[0,1]\to M$ be $\ov{\la}(t):=\la(1-t)$. Then
$$
S_{L+c}(\la)+S_{L+c}(\ov{\la})\le \frac A 2 + \frac A 2 = A\, .
$$
Let $\tau>(b-a)-T_b$ and let $\sigma:[0,T_\sigma]\to M$ be a curve in
$AC(x_u(\tau),\pi\, u)$ such that
$$
S_{L+c}(\sigma) < -\Phi_c\bigl(\pi\, u, x_u(\tau)\bigr)+1\, .
$$
Let $\ov{\ga}:= \ga * \la * x_u\vert_{[0,\tau]}*\sigma *\ov{\la}$. This curve
is in $AC(x_w(a),x_w(b))$, it is defined on a time interval of length
$$
T_b + 1 +\tau + T_\sigma + 1 > T_b +\tau > b-a\, ,
$$
and has $(L+c)$-action
\begin{align}
S_{L+c}(\ov{\ga})
   &\le S_{L+c}(\ga) + \bigl( S_{L+c}(\la) + S_{L+c}(\ov{\la})\bigr)
       + \bigl( S_{L+c}(x_u\vert_{[0,\tau]}) + S_{L+c}(\sigma)\bigr) \notag \\
   &\le S_{L+c}(\ga) + A\, +
\bigl(  \Phi_c(\pi\, u, x_u(\tau))
       - \Phi_c(\pi\, u, x_u(\tau)) + 1 \bigr) \notag \\
   &< \bigl( S_{L+c}(x_w\vert_{[a,b]}) - Q + 1 \bigr) + A + 1\,, \notag \\
S_{L+c}(\ov{\ga})
   &< S_{L+c}(x_w\vert_{[a,b]}) - (A+B)\,. \label{E:XII.4}
\end{align}
Now the same argument as in the case $T_b\ge b$, using $\ov{\ga}$ instead of
$\ga$, gives a contradiction.

\qed








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\end{document}

\date{September, 1996}


