The goal of this section is to establish the convergence and to 
find an explicit expression for 
$\displaystyle \IE(\t) := \lim_{N \ra \infty} 
\frac 1 N \log I_N(\t, E_N)$ 
as far as $E_N$ satisfies Hypothesis \ref{hypsurEN}
but $\t$ do not necessarily satisfy the hypotheses of Theorem \ref{sh}.
This corresponds to show
Theorem \ref{expreel} (we again restrict to the case $\b = 1$ to
avoid heavy notations).\\
We recall that 
$$ I_N(\t, E_N) = \EE\left[ \exp\left(N\t \frac{\sum_{i=1}^N \l_i g_i^2}{\sum_{i=1}^N g_i^2}\right)\right],$$
therefore one main step of the proof will be to get a large deviation principle for $\displaystyle z_N:=\frac{\sum_{i=1}^N \l_i g_i^2}{\sum_{i=1}^N g_i^2}$.

\subsection{Large deviation
bounds for $z_N$}

We denote by $\displaystyle u_N := \frac 1 N \sum_{i=1}^N g_i^2$
and  $\displaystyle v_N := \frac 1 N \sum_{i=1}^N \l_i g_i^2.$
We intend to get the following result
\begin{prop}
\label{gdz}
If the empirical measure $\mun_{E_N} = \frac 1 N \sum_{i=1}^N \d_{\l_i}$ satisfies
Hypothesis  \ref{hypsurEN}, 
the law $\hat \pi_N$ of 
$\left( u_N^{-1} v_N \right)$ 
under the standard $N$-di\-men\-sio\-nal Gaussian measure satisfies a
large deviation principle in the scale $N$ with good rate function
$$T (\a)=\left\lbc
\begin{array}{ll}
\frac 1 2 h_\a(\KE(\QE(\a))) & \mbox{ if }
\a \in [\a_{\rm min}, \a_{\rm max}],\\
\frac 1 2  h_\a^{\rm max} & 
\mbox{ if }\a \in ]\a_{\rm max}, \lM[,  \\
\frac 1 2  h_\a^{\rm min} &
\mbox{ if }\a \in ]\lm, \a_{\rm min}[ , \\
 +\infty
\mbox{ if } \a \notin ]\lm, \lM[\\
\end{array}
\right. $$
with $$ \a_{\rm max} = \lM - \frac 1 \HM \quad \textrm{ and }\quad  \a_{\rm min} = \lm - \frac 1 \Hm,$$
where  we recall that $\HM = \lim_{z \downarrow \lM} \int \frac{1}{z-\l} d\mu_E(\l)$ and 
$\Hm = \lim_{z \uparrow \lm} \int \frac{1}{z-\l} d\mu_E(\l)$; \linebreak 
we denote also,
for $\k \in [\lm, \lM]^c$,
$$ h_\a(\k) = \int \log\left( \frac{\k - \l}{\k-\a}\right) d\mu_E(\l),$$ 
$h_\a^{\rm min} = \lim_{\k \uparrow \lm} h_\a(\k)$ and 
$h_\a^{\rm max} = \lim_{\k \downarrow \lM} h_\a(\k)$.
Finally, the functions $\KE$ and $\QE$ were defined respectively in Definition 
\ref{KRreel} and Property \ref{Rreel}. \\
Note that $\HM$ and $\Hm$ can be infinite (respectively $+\infty$
and $-\infty$); in this case,  we adopt the convention that $\frac{1}{\infty} =0$.
\end{prop}


The \textbf{proof of Proposition \ref{gdz}} decomposes mainly in four steps,
expressed in the following four lemmata :
\begin{lem}
\label{lem1}
  For any $\a \in [\lm, \lM]$,
\begin{align*}
 \lim_{\e\ra 0}\varliminf_{N\ra\infty} &\frac 1N \log
\hat\pi_N\left( \bigl|{v_N}  -\a {u_N}\bigr|<\sqrt{\e}\right)  \\
 & \ppq \lim_{\e\ra 0}\varliminf_{N\ra\infty}\frac 1N \log
\hat\pi_N\left( \left|\frac{v_N}{u_N} -\a\right|<\e\right)   \\
& \ppq \lim_{\e\ra 0}\varlimsup_{N\ra\infty}\frac 1N \log
\hat\pi_N\left(  \left|\frac{v_N}{u_N} -\a \right|<\e\right)  \\
& \qquad \qquad \ppq \lim_{\e\ra 0}\varlimsup_{N\ra\infty}\frac 1N \log
\hat\pi_N\left( \bigl|{v_N} -\a {u_N}\bigr|<\sqrt{\e}\right) \label{encadr} 
\end{align*}
\end{lem}

\begin{lem}
\label{lem2}
We denote by $v_N(\g) :=N^{-1}\sum_{i=1}^N \g_i g_i^2$ and we assume that
the $\g_i$'s are such that
\begin{enumerate}
\item $\g_{\rm max}^N:=\max_{1\le i\le N} \g_i$
(resp. $\g_{\rm min}^N=\min_{1\le i\le N} \g_i$)
converges towards $\g_{\rm max}<\infty$ (resp. $\g_{\rm
min}>-\infty$).
\item The empirical measure $N^{-1}\sum_{i=1}^N \d_{\g_i}$ converges to 
a compactly supported measure $\mu$; we denote by $\g^+$ and $\g^-$
the edges of the support of $\mu$.
\end{enumerate}
Then, the law of $v_N(\g)$ satisfies a
large deviation principle in the
scale $N$ with rate function
$$J_{\mu,\g_{\rm min}, \g_{\rm max}} (x)=\left\lbc
\begin{array}{ll}
L(x)& \mbox{ if }
x\in [x_1, x_2]\\
L(x_1) +\frac{1}{2 \g_{\rm min}}(x-x_1)& 
\mbox{ if } x<x_1\\
L(x_2) +\frac{1}{2 \g_{\rm max}}(x-x_2) &
\mbox{ if } x>x_2\\
\end{array}
\right.$$
with
$$L(x)=\sup\left\{ ux+\frac 12 \int\log(1-2\l u) d\mu(\l)\right\}$$
where the supremum is taken over $u$ such that
$1-2\l u>0$ for every 
$\l \in [\g_{\rm min},\g_{\rm max}]$,
\[
x_1 = \left\{ \begin{array}{ll}
 \g_{\rm min}( \g_{\rm min} \Hm^\g - 1), & \textrm{ if } \g_{\rm min} < 0\\
-\infty & \textrm{ otherwise, }
\end{array}\right.
\]
whereas
\[
x_2 = \left\{ \begin{array}{ll}
 \g_{\rm max}( \g_{\rm max} \HM^\g - 1), & \textrm{ if } \g_{\rm max} > 0\\
\infty & \textrm{ otherwise,}
\end{array}\right.
\]
with the obvious notations $\HM^\g = \lim_{z\downarrow \g_{\rm max}} H_\mu(z)$
and $\Hm^\g = \lim_{z\uparrow \g_{\rm min}} H_\mu(z)$.

\end{lem}

\begin{lem}
\label{lem3}
If we denote $\g_i^\a := \l_i - \a$, $\mu^\a$ the weak limit of the
empirical measure $\frac 1 N  \sum_{i=1}^N \d_{\g_i^\a}$ (note that
$\mu^\a$ is just $\tau_{-\a} \sharp \mu$, where $\tau_{-\a} $ is the shift
given by $\tau_{-\a}(x) = x-\a$), $\g_{\rm max}^\a$ and $\g_{\rm min}^\a$
are respectively the limits of ${\rm max} \g_i^\a$ and ${\rm min} \g_i^\a$,
then
$$ J_{\mu^\a, \g_{\rm max}^\a, \g_{\rm min}^\a}(0) = T(\a),$$
with $T $ as defined in Proposition \ref{gdz}.
\end{lem}

\begin{lem}
\label{lem4}
$T$ is a good rate function.
\end{lem}

Then,  Proposition \ref{gdz} follows easily from these lemmata.
 Indeed,by definition of $u_N$ and $v_N$,
 we have that, for all $\e>0$ and $N$
large enough  $z_N \in [\lm-\e, \lM +\e]$ 
so that,
$$ \limsup_{N\ra\infty}\frac 1N \log 
\hat\pi_N\left( z_N \in [\lm -\e,\lM +\e]^c\right) = -\infty. $$ 
Thus,
from Theorem 4.1.11 in \cite{DZ}, it is enough to consider small balls
ie to show that, for any $\a\in [\lm,\lM]$,
$$ \limsup_{\e\ra 0}\limsup_{N\ra\infty}\frac 1N \log 
\hat\pi_N\left( \bigl|z_N -\a \bigr|\le \e\right) \ppq -T(\a),$$
and 
$$ \liminf_{\e\ra 0}\liminf_{N\ra\infty}\frac 1N \log 
\hat\pi_N\left( \bigl|z_N -\a \bigr|<\e\right) \pgq -T(\a).$$
Now, if $\g_i^\a = \l_i - \a$ and the $\l_i$'s satisfy Hypothesis 
\ref{hypsurEN}, $v_N(\g^\a):= \frac 1 N \sum(\l_i-\a) g_i^2 = v_N-\a u_N$
satisfy the hypotheses $(1)$ and $(2)$ of Lemma \ref{lem2}.
Therefore it satisfies a large deviation principle with rate function
$ J_{\mu^\a, \g_{\rm max}^\a, \g_{\rm min}^\a}$.
In particular this gives that in  Lemma \ref{lem1},
the rightmost and leftmost members coincide, so that
\begin{multline*}
\lim_{\e\ra 0}\liminf_{N\ra\infty}\frac 1N \log
\hat\pi_N\left( \bigl|{v_N} -\a {u_N}\bigr|<\sqrt{\e}\right) \\
= \lim_{\e\ra 0}\limsup_{N\ra\infty}\frac 1N \log
\hat\pi_N\left( \bigl|{v_N} -\a {u_N}\bigr|<\sqrt{\e}\right) 
= -J_{\mu^\a,\g_{\rm min}^\a, \g_{\rm max}^\a}(0) = -T(\a)
\end{multline*}
where the last equality comes from Lemma \ref{lem3}.\\
The study of the function $T$, that will give Lemma \ref{lem4},
allows to conclude the proof.

\subsection{Proofs of the lemmata}

\textbf{Proof of Lemma \ref{lem1}:}\\
For any $\a \in\RR$ and $\e >0$, we have 
\begin{multline*}
\hat\pi_N\left( |{v_N} -\a {u_N}|<\sqrt{\e}\right)-
\hat\pi_N\left( |u_N|\pgq \sqrt{\e}^{-1}\right)
\ppq \hat\pi_N\left( \left|\frac{v_N}{u_N} -\a\right|<\e\right) \\
\ppq
\hat\pi_N\left( |{v_N} -\a{u_N}|<\sqrt{\e}\right)+
\hat\pi_N\left( |u_N|\pgq \sqrt{\e}^{-1}\right).
\end{multline*}
Now, by Chebychev's inequality,
$$\hat\pi_N\left( |u_N|\pgq \sqrt{\e}^{-1}\right)
\ppq e^{-\frac{1}{4\sqrt{\e} }N} \hat\pi_N\left(e^{\frac{1}{4}
u_N}\right)
\ppq 2^N e^{-\frac{1}{4\sqrt{\e} }N},$$
so that
$$ \lim_{\e\ra 0}\limsup_{N\ra\infty}\frac 1N \log 
\hat\pi_N\left( |u_N|\pgq \sqrt{\e}^{-1}\right) = -\infty,
$$
what gives immediately Lemma \ref{lem1}.\\


Lemma \ref{lem2} is proved  in \cite{BGR},  Theorem 1;
we omit it here.

\textbf{Proof of Lemma \ref{lem3}:}\\   
Our goal is to identify $T(\a) = J_{\mu^\a,\g_{\rm min}^\a, \g_{\rm max}^\a}(0)$. As we said above, it is enough to restrict to 
$\a \in [\l_{\rm min}, \l_{\rm max}]$.\\
We have of course $\g_{\rm min}^\a = \l_{\rm min}-\a$ and 
$\g_{\rm max}^\a = \l_{\rm max}-\a$ and it is easy to check that
$$\HM^\a := \lim_{z \downarrow \lM -\a} \int \frac{1}{z-\l} d\mu^\a(\l) = \HM$$
(and respectively for $\Hm$).\\
Therefore, if we denote by $x_1^\a$ and $x_2^\a$ the bounds corresponding
to $\mu^\a$, we have that :
$$ x_1^\a = (\lm -\a) ((\lm -\a)\Hm -1)$$ 
(as the inequality $\g_{\rm min}^\a = \lm-\a <0$ is always satisfied for the
$\a$'s we are interested in) and similarly
$ x_2^\a = (\lM -\a) ((\lM -\a)\HM -1).$
We now have to determine the sign of $x_1^\a$ and $x_2^\a$ with respect to
$\a$. It is easy to check that
\begin{itemize}
\item $x_1^\a \ppq 0$ and $x_2^\a \pgq 0$ if $\displaystyle \a \in \left[\a_{\rm min}:= \lm - \frac  1 \Hm, \a_{\rm max}:= \lM - \frac  1 \HM \right]$ 
\item $x_1^\a \ppq 0$ and $x_2^\a \ppq 0$ if $\displaystyle \a \in [\a_{\rm max}, \l_{\rm max}]$
\item  $x_1^\a \pgq 0$ and $x_2^\a \pgq 0$ if $\displaystyle \a \in [\a_{\rm min}, \l_{\rm min}]$
\end{itemize}
Therefore, we deduce
$$J_{\mu^\a,\g_{\rm min}^\a, \g_{\rm max}^\a}(0)=\left\lbc
\begin{array}{ll}
L^\a(0) & \mbox{ if }
\a \in [\a_{\rm min}, \a_{\rm max}]\\
L^\a(x_2^\a) -\frac{1}{2} \HM(\a_{\rm max}-\a) &
\mbox{ if } \a_{\rm max} \ppq \a \ppq \lM\\
L^\a(x_1^\a) -\frac{1}{2} \Hm(\a_{\rm min}-\a) &
\mbox{ if } \lm \ppq \a \ppq \a_{\rm min},\\
\end{array}
\right.$$
where we recall that 
$$ L^\a(x) = \sup \left\{ ux + \frac 1 2 \int \log(1+2\a u - 2\l u) d\mu(\l)\right\},$$
with the supremum on $u$ such that $1+2\a u - 2\l u>0$ 
for all $\l \in [\l_{\rm min}, \l_{\rm max}]$.\\

We now get interested in the expression of $L^\a $ on $[x_1^\a, x_2^\a]$.\\
Obviously, the supremum is not reached at $u=0$.\\
For $u \neq 0$, we denote $\k := \a + \frac{1}{2u}$,
then we have that $1+2\a u - 2\l u = \frac{\k-\l}{\k - \a}$.
Moreover, if for all $\l \in [\l_{\rm min}, \l_{\rm max}]$,
$1+2\a u - 2\l u>0$ then ($\k > \lM$ and $u>0$) or ($\k < \lm$ and $u<0$) 
and conversely, so that

$$ L^\a(x)  =  \frac 1 2 \sup_{\k \in [\l_{\rm min}, \l_{\rm max}]^c}
\left\{  \frac{x}{\k - \a} + h_\a(\k)\right\},$$
with the notations of Proposition \ref{gdz}.\\

$\bullet$ If $\a \in \Ipp := [\a_{\rm min}, \a_{\rm max}]$,
$$ J_{\mu_\a,\g_{\rm min}^\a, \g_{\rm max}^\a}(0) = L^\a(0)
= \frac 1 2 \sup_{\k \in [\l_{\rm min}, \l_{\rm max}]^c} h_\a(\k).$$
We now want to check that in this case, the supremum of $h_\a$
is reached at $\k_0 = \KE(\QE(\a))$.\\

The first point is to show that in this case, there is a unique 
$\k_0$ where $h_\a^\prime$ cancels. Indeed :
$$ h^\prime_\a(\k_0)=0 \Longleftrightarrow \HE(\k_0) = \frac{1}{\k_0 - \a}
\Longleftrightarrow \k_0 = \KE(\QE(\a))$$
We now check that the maximum of $h_\a$ is reached at 
$\k_0$; 
\begin{itemize}
\item if $\k_0 > \lM$, $h_\a$ is decreasing from $0$ to $h^\a_{min}$ 
on $]-\infty, \lm[$, it is increasing from $h^\a_{max}$ to $h_\a(\k_0)$
on $]\lM,\k_0 ]$ and then decreasing from $h_\a(\k_0)$ to $0$ 
on $]\k_0, +\infty ]$,
\item if $\k_0 < \lm$, $h_\a$ is increasing from $0$ to $h_\a(\k_0)$
on $]-\infty, \k_0]$ then decreasing from $h_\a(\k_0)$ to $h^\a_{min}$ 
on $]\k_0, \lm[$, it is increasing from $h^\a_{max}$ to $0$ 
on $]\lM, +\infty [$.
\end{itemize}
We treat in details the proof of the first point, when $\k_0 > \lM$, 
the other one being very similar.
We recall from Property \ref{Rreel} that $\Ipp$ is the image of $\RE$.\\
If $\k_0 > \lM$, $h_\a^\prime$ does not cancel on $]-\infty, \lm[$.
It is negative since, when $\a \in \Ipp$, $ \lm-\frac 1 \Hm$
and so  $\lim_{\k \ra \lm} h_\a^\prime(\k)  < 0$.
On the other side, we want to find the sign of $h_\a^\prime$
on $]\lM, +\infty [$ knowing that it cancels at $\k_0$. 
As above, we show that 
$ \lim_{\k \ra \lM} h_\a^\prime(\k) > 0$ and we deduce from that 
and the continuity of $h_\a^\prime$,
that  it is positive till $\k_0$. 
Furthermore, $h_\a$ is also twice differentiable at $\k_0$ and
\begin{eqnarray*}
h_\a^{\prime\prime}(\k_0) & = & - \int \frac{1}{(\k_0 -\l)^2} d\mu_E(\l) 
+ \left(\frac{1}{\k_0-\a}\right)^2 \\
& < & - \left(\int \frac{1}{\k_0 -\l} d\mu_E(\l) \right)^2 
+(\HE(\k_0))^2 <0, 
\end{eqnarray*}
where we used Cauchy-Schwarz inequality and the definition of $\k_0$.
Therefore $h_\a^\prime$
is negative for $\k > \k_0$
and the fact that $ \lim_{\k \ra +\infty} h_\a(\k) =0$
concludes the proof of the first point.\\


Finally, we got that if $\a \in  [\a_{\rm min}, \a_{\rm max}]$,
$$ J_{\mu^\a,\g_{\rm min}^\a, \g_{\rm max}^\a}(0) = \frac 1 2 h_\a (\KE(\QE(\a)))$$

$\bullet$ If $\a > \a_{\rm max}$, our starting point is
$$ J_{\mu^\a,\g_{\rm min}^\a, \g_{\rm max}^\a}(0) = \frac 1 2
\sup_{\k \in [\lm, \lM]^c} \left\{ x_2^\a \left( \frac{1}{\k-\a}
- \frac{1}{\lM-\a} \right) + h_\a(\k)\right\}$$
Using arguments as above, we show that the function
$$ g_\a(\k) = \frac{x_2^\a}{\k-\a}  + h_\a(\k)$$
on $ [\lm, \lM]^c$ takes its supremum as $\k$ goes to $\lM$
by showing that its derivative is negative on $[\lm, \lM]^c$.
Hence, $ J_{\mu^\a,\g_{\rm min}^\a, \g_{\rm max}^\a}(0) = 
\frac 1 2 h_{\rm max}^\a$. 

$\bullet$ The case $\a< \a_{\rm min} $ is treated similarly,
which concludes the proof of Lemma \ref{lem3}.\\


The \textbf{proof of Lemma \ref{lem4}} is easy : $T$
is in fact continuous on $]\lm, \lM[$. 
Indeed, it is continuous on each interval $]\lm, \a_{\rm min}[$,
$]\a_{\rm min}, \a_{\rm max}[$ and $]\lM, \a_{\rm max}[$  so that it is enough to check
that $\KE(\QE(\a)) \xrightarrow[\a \ra \a_{\rm max}]{} \lM$ (see Property
\ref{Rreel}) so that $T(\a) \xrightarrow[\a \ra \a_{\rm max}]{} \frac 1 2
h^\a_{\rm max}$; and similarly at $\a_{\rm min}$.\\

\subsection{Proof of Theorem \ref{expreel}}

By Varadhan's lemma, we have 
\begin{lem}
\label{lem5}
For any $\t \in \RR$, if $T$ is the function introduced
in Proposition \ref{gdz}, we have
$$ \lim_{N \ra \infty} \frac 1 N \log I_N(\t, E_N)
= \sup_\a \{\t \a - T(\a)\}.$$
\end{lem}


Lemma \ref{lem5} therefore gives the existence of the limit,
the last step 
to conclude the proof of Theorem \ref{expreel} is to check that it coincides with the function $\IE$
introduced in Theorem \ref{expreel}. \\

We denote by 
$$ G(\t):= \sup_{\a \in \Ipp} \left[ \t \a - \frac 1 2 h_\a(\KE(\QE(\a)))\right],$$
$$ G_1(\t):= \sup_{\a \in I_1} \left[ \t \a - \frac 1 2 h^\a_{\rm max} \right], \quad
G_2(\t):= \sup_{\a \in I_2} \left[ \t \a - \frac 1 2 h^\a_{\rm min} \right],$$
where we recall that $\Ipp = [\a_{\rm min}, \a_{\rm max}]$
and we denote by $I_1 = ]\a_{\rm max}, \lM]$ and $I_2 = [\lm, \a_{\rm min}[$.\\ 

The main part of the work for this last step will rely on proving
\begin{lem}
\label{exprG}
With the notations introduced above, we have\footnote{$\sharp = - \infty$ if $\Hm = - \infty$ and otherwise these expressions 
are well defined in virtue of the fact that
$\displaystyle \int_0^1 \frac 1 \l d\mu(\l) < +\infty \Rightarrow  -\int_0^1 \log \l d\mu(\l) < +\infty$,\\
$* = - \infty$ if $\HM = + \infty$ and otherwise these expressions 
are well defined for the same reason.}
\[
 G(\t) = \left\{
\begin{array}[c]{ll}
\frac 1 2 \int_0^{2\t} \RE(u)du, & \textrm{ if } 2\t \in \Ip\cup \{0\} = ]\Hm, \HM[ \\
\t \a_{\rm min} - \frac 1 2 \int \log (\Hm(\lm-\l))d\mu_E(\l)^\sharp, & \! \!\textrm{ if } 2\t  \!\ppq  \!\Hm \\
\t \a_{\rm max} - \frac 1 2 \int \log (\HM(\lM-\l))d\mu_E(\l)^*, &   \! \! \!\textrm{ if } 2\t  \! \pgq  \!\HM ,
\end{array} 
\right.
\]

\[
 G_1(\t) = \left\{
\begin{array}[c]{ll}
\t\left( \lM - \frac{1}{2 \t} \right) - \frac 1 2 \int \log (2\t (\lM-\l))d\mu_E(\l)^*, &   \!\!\textrm{ if } 2\t  \! >  \!\HM \\
\t \a_{\rm max} - \frac 1 2 \int \log (\HM(\lM-\l))d\mu_E(\l)^*, &  \! \!\textrm{ if } 2\t  \!<  \!\HM ,
\end{array} 
\right.
\]

\[
 G_2(\t) = \left\{
\begin{array}[c]{ll}
\t\left( \lm - \frac{1}{2 \t} \right) - \frac 1 2 \int \log (2\t (\lm-\l))d\mu_E(\l)^\sharp, &   \!\! \!\textrm{ if } 2\t  \! <  \!\Hm \\
\t\left( \lm - \frac 1 \Hm \right) - \frac 1 2 \int \log (\Hm(\lm-\l))d\mu_E(\l)^\sharp, &   \!\! \!\textrm{ if } 2\t  \!>  \!\Hm .
\end{array} 
\right.
\]
\end{lem}

\noindent \textbf{Proof of Lemma \ref{exprG} :}\\
$\bullet$ We first study $G$. \\
This is finding the supremum of
$j_\t(\a) := \t \a - \frac 1 2 h_\a(\KE(\QE(\a))) $ on $\Ipp$.
From Definition \ref{KRreel} and Property \ref{Rreel}, we have that 
$j_\t$ is differentiable  on $\Ipp$ and an easy computation gives
$$ j_\t^\prime(\a) = \frac 1 2 (2 \t - \QE(\a)).$$ 
\begin{itemize}
\item If $2 \t \in \Ip$, $j_\t$ is maximized at
$\a_0 = \RE(2 \t)$ 
and so,  if $2\t \in ]\Hm, \HM[ \setminus \{0\}$,
\begin{eqnarray*}
 G(\t) &=& \frac 1 2 \left(2\t \RE(2\t) - \log(2\t)
- \int \log (\KE(2\t) - \l)d\mu_E(\l)
\right)\\
&=&= \frac 1 2 \int_0^{2\t} \RE(u) du. \\
\end{eqnarray*}


\item If $\Hm > - \infty$ and $2 \t < \Hm$, the 
equation $  j_\t^\prime(\a_0) = 0$ has no solution
and actually  $  j_\t^\prime$ is negativeso that  the supremum is reached at the left boundary 
$\displaystyle \a_{\rm min}$ of $\Ipp$
and is equal to 
$$ \t \a_{\rm min} - \frac 1 2 \int \log (\Hm(\lm-\l))d\mu_E(\l).$$

\item If $\HM < + \infty$, a similar treatment in the case $2\t > \HM$ concludes the proof for $G$.
\end{itemize}

$\bullet$ The formulas for $G_1$ and $G_2$
are derived similarly.


By virtue of Lemmata \ref{lem5} and \ref{exprG},
 to finish the proof of Theorem \ref{expreel}, 
we have now 
\begin{enumerate}
\item to compare $G_{|\Ip}$,  $G_{1|\Ip}$ and  $G_{2|\Ip}$ to get  $I_{\mu_E |\Ip}$.\\
Since 
$\displaystyle \lim_{\a \uparrow \HM} j_\t(\a) = G_1(\t)$ and
$\displaystyle \lim_{\a \downarrow \Hm} j_\t(\a) = G_2(\t)$
whereas $G(\t) = \sup_{\a \in \Ip} [j_\t(\a)]$, we get that
 $I_{\mu_E |\Ip} = G_{|\Ip}$.\\
\item if $\HM < +\infty$, to compare $G_{|\{ 2\t > \HM\}}$, $G_{1|\{ 2\t > \HM\}}$ 
and $G_{2|\{ 2\t > \Hm\}}$ to get $I_{\mu_E ||\{ 2\t > \HM\}}$.\\
By studying the function 
$\displaystyle x \mapsto - \frac{\t}{x} - \frac 1 2 \log x $, 
which reaches its maximum at $\t$, we can easily deduce that
 $G_{|\{ 2\t > \HM\}} < G_{1|\{ 2\t > \HM\}}$.\\
Moreover $G_{1|\{ 2\t > \HM\}}$ and  $G_{2|\{ 2\t > \HM\}}$ 
are the limits of $j_\t$ respectively at 
$\a_{\rm max}$ and  $\a_{\rm min}$
and we know that in the case $2\t > \HM$, $j_\t$ is
increasing. This gives $G_{2|\{ 2\t > \HM\}} < G_{1|\{ 2\t > \HM\}}$.\\
In this case we conclude that the maximum is given by 
$G_{1|\{ 2\t > \HM\}}$.
\item Arguing similarly, we can see that in the case where 
$ 2\t < \Hm$ the maximum is given by  $G_{2|\{ 2\t < \Hm\}}$. 
\end{enumerate} 

To conclude the proof of Theorem \ref{expreel},
we use the continuity of $\IE$ with respect to $\t$ given by
the first point of Lemma \ref{cont} 
to specify its value at $\lm$, $\a_{\rm min}$,
$\a_{\rm max}$ and $ \lM $. \hfill \xx
{"config": "arxiv", "file": "math0406121/newproof0312.tex"}