text:
\section{Introduction}

  We are interested in the stability properties, in the sense of Lyapounov, of linearly stable elliptic fixed points (equilibrium) denoted
$x_*$ on a symplectic manifold $(M,\Omega)$ for a symplectic vector field $X$ (i.e. : $i_X\Omega$ is closed), that is $X(x_*)=0$.

As the problem is local, it is enough to consider a Hamiltonian $H$, defined and analytic on an open neighbourhood of $0 \in\R^n\times\R^n$ equipped with
the standard symplectic form, having the origin as a fixed point. Up to an irrelevant additive constant, expanding the Hamiltonian as a power series at the origin we can write
\[ H(z)=H_2(z)+ V(z) \]
where $z=(x,y)\in\R^n\times\R^n$ is sufficiently close to the origin in $\R^n\times\R^n$, $H_2$ is the Hessian of $H$ at $0$ and $V(z)=O(|z|^3)$.

 Recall that the fixed point is said to be elliptic if the spectrum of the linearized system is purely imaginary. Due to the symplectic character of the equations, these are the only linearly stable fixed points and the spectrum has the form $\{\pm i\omega_1, \dots, \pm i\omega_n\}$ for some vector $\omega=(\omega_1, \dots, \omega_n)$ which is called the normal (or characteristic) frequency. Now we assume that the components of $\omega$ are all distinct so that we can make a symplectic linear change of variables that diagonalize the quadratic part, hence
\[ H(z)=\omega.\mathcal{I}+V(z) \]
where $\mathcal{I}=\mathcal{I}(z)$ are the ``formal actions", that is
\[ \mathcal{I}(z)=(\mathcal{I}_1,\dots ,\mathcal{I}_n)=\frac{1}{2}(x_1^2+y_1^2, \dots, x_n^2+y_n^2)\in\R^n\ \text{
\rm and}\ \omega\!.\mathcal{I}=\displaystyle{\sum_{j=1}^{n}}\omega_j\mathcal{I}_j.
\]

 If all the components of $\omega$ have all the same sign, then $H$ is a
Lyapounov function and the fixed point is stable. But in a general setting,
one has to take into account the higher order terms $V(z)=O(|z|^3)$.

\medskip

 Given a solution $z(t)$, if we denote $\mathcal{I}(t)=\mathcal{I}(z(t))$, then $||\mathcal{I}(t)||$ is essentially the distance of $z(t)$ to the origin and $||\mathcal{I}(t)-\mathcal{I}(0)||$ measures the deviation of $z(t)$ from its original position.

 Without loss of generality, we can assume that we consider a Hamiltonian defined on the ball $B_R$ of radius $R<1$ in $\R^{2n}$. Then by analyticity, we extend the resulting Hamiltonian as a holomorphic function on some closed complex ball $D_s$ of radius $s<1$ in $\C^{2n}$. If we define $\mathcal{A}_s$ as the space of holomorphic Hamiltonians on $D_s$ which are real valued for real arguments, and $||.||_s$ its usual supremum norm, then we are led to consider
\begin{equation}\label{Ham1}
\begin{cases}
H(z)=\omega.\mathcal{I}+f(z) \\
H \in \mathcal{A}_s\ ;\ f(z)=O(z^3).
\end{cases}\tag{$A$}
\end{equation}

\medskip

To obtain results of stability, one of the main tool is to construct normal forms via classical averagings, and in this case these are the so-called {\it Birkhoff normal forms}.

 For an integer $m\geq 1$, assuming $\omega$ is non-resonant up to order $2m$, that~is
\[ k.\omega \neq 0, \quad k\in\Z^n, \; 0<||k||_1 =\sum_{j=1}^n | k_j |\leq 2m  \]
then there exists an analytic symplectic transformation $\Phi_m$ close to identity such that $H\circ\Phi_m$ is in Birkhoff normal form up to order $2m$, that is
\[ H\circ\Phi_m(z)=h_m(\mathcal{I})+f_m(z)\ \text{\rm where}\ h_m(\mathcal{I})=
\displaystyle{\sum_{k=1}^m}\mathcal{B}^{(k)}(\mathcal{I})\]
with homogenous polynomial $\mathcal{B}^{(k)}$ of degree $k$ in the $\mathcal{I}$ variables and the remainder $f_m$ of order $z^{2m}$ (see \cite{Bir66} or \cite{Dou88}, \cite{HZ94} for a more recent exposition). The polynomials $\left(\mathcal{B}^{(k)}\right)_{k\in\N^*}$ are uniquely defined (even though the transformations are not) and are usually called the {\it Birkhoff invariants}. The transformed Hamiltonian is therefore the sum of an integrable part $h_m$, for which the origin is trivially stable in the sense that $\mathcal{I}(t)$ is constant for all times, and a much smaller perturbation $f_m$ for a small enough neighbourhood of the origin. Moreover, if $\omega$ is non-resonant up to any order, we can even define a formal symplectic transformation $\Phi_\infty$ and a formal power series $h_\infty=\displaystyle{\sum_{k\geq 1}}\mathcal{B}^{(k)}$ such that

\begin{equation}
\label{hinfini}H\circ\Phi_\infty(z)=h_\infty(\mathcal{I}).
\end{equation}

However, for a certain topology on the coefficients, the transformation $\Phi_\infty$ is generically divergent as it was proved by Siegel \cite{Si41} and the convergence properties of  the series $h_\infty$ are even more subtle. Actually, Perez-Marco (\cite{PM03}) has
showed that if $h_\infty$ is divergent for some $H\in\mathcal{A}_s$, then the complete Birkhoff normal form $h_\infty$ is divergent for a "typical"
$H\in\mathcal{A}_s$. But nevertheless, we cannot ensure that $h_\infty$ is generically convergent or divergent.

\medskip

The first kind of stability result is given by an application of KAM theory. Assume that $\omega$ is non-resonant up to order $4$ so that our system reduce to
\[ H(z)=h_2(z)+f_2(z)=\omega.\mathcal{I}+\mathcal{B}^{(2)}(\mathcal{I})+f_2(z). \]
where $\mathcal{B}^{(2)}$ is a quadratic form.

 Under the so-called Kolmogorov non-degeneracy condition (or the twist condition in the context of an elliptic fixed point) : the quadratic form $\mathcal{B}^{(2)}$
is {\it non-degenerate}, KAM theory (see \cite{AKN97} and \cite{DG96} for the specific case of an elliptic fixed point) ensures stability for all times for most of the initial conditions. That is, for a large set of solution $z(t)$ :
if $|\mathcal{I}(0)|\leq\rho$ then the variations $|\mathcal{I}(t)-\mathcal{I}(0)|$ are of order $\rho$ for all times, hence we have Lyapounov stability. More precisely, KAM theory yields the existence of invariant Lagrangian tori which form a set of positive measure with a density going to one at the origin. However because of the dimensions, it is only for $n=2$ that we can deduce perpetual stability results. Actually, for $n=2$ the invariant tori are 2-dimensional and divide the 3-dimensional energy level, therefore the solutions of the perturbed system are global and bounded over infinite times. On the other
hand, an arbitrary large drift of the orbits can still occur for $n\geq
3$ in the complement of the invariant tori which is a dense set with a complicated
topology. In fact it is believed that, except for two degrees of freedom systems (and of course one degree), ``generic" elliptic fixed points are unstable (see \cite{DLC83} and \cite{Dou88} for examples and \cite{KMV04} for an announcement in a generic setting).

 Thus, for $n>2$, KAM theory provides results of stability only in the sense
of measure and theorems of stability which are valid for an open set of initial condition can only be proved over {\it finite} times.

\begin{definition}
The origin is stable over {\it finite times} if $|\mathcal{I}(0)|\leq\rho$ small enough implies
\[ |\mathcal{I}(t)-\mathcal{I}(0)|=R(\rho)\ \text{\rm for}\ |t|<T(\rho)\]
where $R(\rho )$ is of order $\rho$ and $T(\rho )$ is at least of order
$\rho^{-1}$ (otherwise we have a trivial bound).

\medskip

 We have {\bf polynomial stability} if $T(\rho )$ is at least of order
$\rho^{-m}$ for some fixed integer $m\in\N^*$.

 We have {\bf exponential stability} if $T(\rho )$ is at least of order
$\exp\left( C\rho^{-\alpha}\right)$ for some fixed constants $C>0$ and $\alpha
>0$.

 We have {\bf superexponential stability} if $\exp\left( C\rho^{-\alpha}\right)$
 is negligible with respect to $T(\rho )$ for {\it any} constants $C>0$ and $\alpha >0$.
\end{definition}

\medskip

 Using the mean value theorem, a polynomial time of stability can be easily obtained once a Birkhoff normal form up to a finite order has been built. The first result of this kind around an elliptic fixed point was obtained by Littlewood in 1959 (\cite{Lit59}).

 To obtain results of exponential stability, there is basically two distinct methods.

In the first approach, we call \text{\it Birkhoff type estimates}, one assumes that the frequency $\omega$ is a $(\gamma ,\tau )${\it -Diophantine vector} for some positive constants $\gamma$ and $\tau$, hence $\omega\in\Omega_{\gamma ,\tau}$ where :
\begin{equation}
\Omega_{\gamma ,\tau}=\left\{\omega\in\R^n\ {\rm such\ that}\ \vert
k.\omega\vert\geq{{\gamma}\over{\vert\vert k\vert \vert_\infty^\tau}}\ {\rm for\ all}\ k\in\Z^n\backslash\{ (0,\ldots ,0\}\right\}
\end{equation}

We recall that the measure of the complementary set of $\Omega_{\gamma ,\tau}$ is of order ${\cal O}(\gamma )$ for $\tau >n-1$.

 In particular $\omega$ is completely non-resonant, hence one can perform any finite number $m$ of Birkhoff normalizations, and since we have a control on the small divisors, one can estimate the size of the remainder $f_m$.

 Under these assumptions, one can prove (\cite{BG86}, \cite{GDFGS89}) that for a real-analytic system, the ``action variables'' $\mathcal{I}$ become quasi integrals over
exponentially long times, more specifically :

\begin{theoreme}

{\it Consider an elliptic fixed point for a real analytic Hamiltonian system of type $(A)$.

 There exists positive constants $C_1$, $C_2$, $C_3$, $C_4$ which depend
only of the analyticity width $s$, the size $\vert\vert H\vert\vert_s$ of
the Hamiltonian $H$, the number of degree of freedom $n$ and $\tau$ the exponent
in the Diophantine condition such that if $\rho <C_1\gamma$, an arbitrary solution $(x(t),y(t))$ with an initial action $\vert\vert\mathcal{I}(0)\vert\vert
\leq\rho$ is defined at least over an exponentially long time and satisfies~:
\begin{equation}
\left\vert\left\vert \mathcal{I}(t)-\mathcal{I}(0)\right\vert\right\vert\leq C_2\rho\ {\rm if}\ \vert t\vert\leq C_3\exp\left( C_4\left(\frac{\gamma}{\rho}
\right)^{{{1}\over{1+\tau}}}\right) .
\end{equation}
}
\end{theoreme}

\medskip

 The proof rely on the existence of a normalizing transformation up
to an exponentially large order $m$ comparable with the inverse of the distance
to the origin. This is possible since we have lower bounds on the small denominators and the growth of the coefficients in the normalizing expansion is reduced to a combinatorial problem (the "great multipliers" along Poincar\'e terminology).
Actually, under the assumptions of linear diophantine frequencies and the
real analyticity of considered Hamiltonian, there exists a decreasing sequence
of radii $s_m\sim \gamma /m^{1+\tau}$ for $m\in\N^*$ such that the normalizing transformation and the Birkhoff normal up to order $m$ can be build on the
ball of radius $s_m$ with a remainder of size comparable with $s_m^m.(m!)^{1+\tau}\!\sim\! (m!/m^m)^{1+\tau}$ which is dominated by $\exp\left( -(1+\tau )m\right)$ or $\exp\!\left(\!\! -C\left(\frac{\gamma}{s_m}\right)^{{{1}\over
{1+\tau}}}\!\!\right)$.

 Finally, since the averaged Hamiltonian is integrable, the speed of drift of the action variables is very slow.

 More specifically, there exists a sequence of annuli of decreasing radii $(s_m)_{m\in\N^*}$ such that :
$$s_{m+1}\!\leq\! \vert\vert\mathcal{I}(0)\vert\vert\! <\! s_m\!\!\Longrightarrow
\!\!\left\vert\left\vert\mathcal{I}(t)-\mathcal{I}(0)\right\vert\right\vert\!\leq\! C_2 s_m\ {\rm if}\ \vert t\vert\!\leq\! C_3\exp\left(\!\! C_4\left(\!\frac{\gamma}
{s_m}\!\right)^{{{1}\over{1+\tau}}}\!\!\right)$$

 In order to obtain uniform estimates with respect to the initial radius
 $\rho$, as in our theorem 1.2, it only remains to slightly lower the coefficients $C_2, C_3, C_4$. Treschev continuous averaging method (\cite{Tr02}) would allow to obtain directly a result with respect to the initial radius $\rho$ and to avoid this sequence $(s_m)_{m\in\N^*}$ but we will stay in the setting of Giorgilli et al (\cite{GDFGS89}, see also \cite{DG96}).

\vskip0,5truecm

The second approach, we call \text{\it Nekhoroshev type estimates}, is fundamentally different. One just assume that $\omega$ is non-resonant up to order $4$
so that the Hamiltonian reduces as for KAM theory to
\[ H(z)=h_2(z)+f_2(z)=\omega.\mathcal{I}+\mathcal{B}^{(2)}(\mathcal{I})+f_2(z). \]
where $\mathcal{B}^{(2)}$ is a quadratic form. But this time we require that the non-linear part $h_2(z)=\omega.\mathcal{I}+\mathcal{B}^{(2)}(\mathcal{I})$ is {\it convex} in actions $\mathcal{I}$, which is equivalent to $\mathcal{B}^{(2)}$ being sign definite. Under those assumptions, it was predicted by Lochak (\cite{Loc92}) and proved independently by Niederman (\cite{Nie98}) and Benettin, Fasso and Guzzo (\cite{BFG98}) that exponential stability holds. Their proofs are based on the implementation of Nekhoroshev estimates in cartesian coordinates, but they are radically different : the first one uses Lochak method of periodic averagings and simultaneous Diophantine approximations while the second is based on Nekhoroshev original mechanism. The proof of Niederman was later clarified by Poschel (\cite{Pos99a}).

Actually, Birkhoff estimates rely on the existence of an accurate integrable normal form while Nekhoroshev theory (see \cite{Nek77} or \cite{Nie11} for
a survey) gives exponential stability for all the solutions even if resonances of {\it low order} are present. This problem is overcome by a suitable transversality condition called {\it steepness} on the geometry of the integrable system. Steepness is a generic condition infinitely differentiable functions which is implied by (quasi)-convexity.

\bigskip

 A remarkable result of Morbidelli and Giorgilli (\cite{MG95}) clearly shows that the two previous results which come respectively from Hamiltonian perturbation
theory and Nekhoroshev's theorem are {\it independent} and can be {\it superimposed}.

\medskip

\begin{theoreme}

{\it Consider an elliptic fixed point for a real analytic Hamiltonian system of type $(A)$ with a Diophantine normal frequency $\omega$ and a sign definite
torsion, that is : $\mathcal{B}^{(2)}$ is a sign definite quadratic form.

 There exists positive constants $C_1$, $C_2$, $C_3$, $C_4$ which depend
only of the analyticity width $s$, the size $\vert\vert H\vert\vert_s$ of
the Hamiltonian $H$, the number of degree of freedom $n$ and $\tau$ the exponent
in the Diophantine condition such that if $\rho <C_1\gamma$, an arbitrary solution $(x(t),y(t))$ with an initial action $\vert\vert\mathcal{I}(0)\vert\vert
\leq\rho$ is defined at least over a superexponentially long time and satisfies~:
\begin{equation}
\left\vert\left\vert \mathcal{I}(t)-\mathcal{I}(0)\right\vert\right\vert\leq C_2\rho\ {\rm if}\ \vert t\vert\leq C_3\exp\left(\exp\left( C_4\left(\frac{\gamma}
{\rho}\right)^{{{1}\over{1+\tau}}}\right)\right)
\end{equation}
}
\end{theoreme}

 The proof starts with a Birkhoff's normal form with an exponentially small remainder like in the theorem 1.2. Then, Nekhoroshev's theory is applied with a perturbation which is already exponentially small, hence we obtain a superexponential time of stability. More specifically, we need the
(quasi-)convexity of the Birkhoff normal forms at all order. Since (quasi-)
convexity
is an open condition for the $\mathcal{C}^{(2)}-$topology, the convexity
of the second Birkhoff invariant $\mathcal{B}^{(2)}$ implies the same property
for $h_m$ ($\forall m\in\mathbb{N}^*$).

\medskip

 One of the main problem to extend the later kind of results in a generic setting is that we need a property on the Birkhoff invariants at all order
which is {\it not} an open condition.

 In this direction, Bounemoura (\cite{Bo11a}) has proved that these reasonings
remain valid in the case of a non resonant elliptic fixed point under a generic
condition on the associated Birkhoff normal form $h_\infty$, more specifically :

\begin{theoreme}

{\it Consider an elliptic fixed point for a real analytic Hamiltonian system of type $(A)$ with a non-resonant normal frequency $\omega$.

 If its Birkhoff normal form $h_\infty$ belongs to a generic set in the space of the formal power expansions $\mathbb{R}[[X]]$ (genericity is in the sense
of prevalence, see the next section) the estimates of superexponential stability given in the previous theorem remain valid.
}\end{theoreme}

\section{Setting and statement of our results :}

 Bounemoura gave only a partial answer to the problem of superexponential
stability of elliptic fixed position. Actually, we cannot ensure that generically the Birkhoff normal form associated to a non resonant elliptic fixed point belongs to the set introduced by Bounemoura even if it is a big set in the sense of measure. This problem is not straightforward since we have very few general information on the Birkhoff normal forms. Actually, in the context
of KAM theory, we are only aware of one result of Eliasson, Fayad and Krikorian
(so far unpublished) which involves {\it all} the Birkhoff invariants and ensure the existence of invariant tori which accumulate an elliptic fixed point in a Hamiltonian system with a Diophantine linear spectrum but without any torsion condition.

 Here, we provide a complete answer to our problem of stability.

\medskip

 We first specify our space of Hamiltonians
$$\mathcal{H}_s=\left\{ H(x,y)=\omega.\mathcal{I} +f(x,y)\ \text{\rm where}\
\left\vert\begin{split}
 H &\in \mathcal{A}_s,\\
 \omega &\in \R^n\ \text{\rm and}\ f(x,y)=O_3 (x,y) .
\end{split}\right.\right\}$$
we recall that $\mathcal{A}_s$ is the space of holomorphic Hamiltonians on $D_s\subset\mathbb{C}^{2n}$ the ball of radius $s$ around the origin which are real valued for real arguments equipped with the supremum norm $\vert\vert .\vert\vert_s$ on $D_s$, hence it is a {\it Banach} space.

\medskip

 Without loss of generality, we can assume that $s<1$.

\bigskip

 For an homogenous polynomial $P(\mathcal{I})=\!\!\!\displaystyle{\sum_{\vert l\vert =k, l\in\N^{2n}}}\!\! p_k \mathcal{I}^k$, as in \cite{GHK06} we consider the
following Euclidean norm (also called Bombieri norm)
$$\vert\vert P\vert\vert_k =\sum_{\vert l\vert =k}\frac{\vert p_l\vert^2}{C_k^l}$$
with the multinomial coefficient $C_k^l$.

 This norm is invariant by isometry in~$\C^{2n}$ (see \cite{GHK06}).

\bigskip

  There exists a Borelian measure canonically associated to the norm $\vert\vert P\vert\vert_k$ on the space $\mathcal{P}_k$ of homogenous polynomials of degree $k$.

 \medskip

 We consider the {Hilbert brick}
$$HB^\N =\left\{ h(\mathcal{I})=\sum_{k\in\N^*}h_k(\mathcal{I})\ \text{with}\ \vert\vert h_k\vert\vert_k\leq 1\ (\forall k)\right\}$$
it is a {\bf compact} set for the product topology when the space of polynomial
is identified as the infinite product $\displaystyle{\prod_{k\in\N^*}}\mathcal{P}_k$.

This compact can be continuously embedded in our Banach space of Hamiltonian
$\mathcal{H}_s$ with the assumption that $s<1$, hence it is a compact set in $(\mathcal{H}_s,\vert\vert .\vert\vert_s)$.

\medskip

 If we denote by $\mu_k$ the Borelian probability measure on the unit ball of $\mathcal{P}_k$
for all $k\in\N^*$, the embedding of $HB^\N$ in $(\mathcal{H}_s,\vert\vert .\vert\vert_s)$ can be equipped with the image of the product measure $\mu =\displaystyle{\prod_{k\in\N^*}}
\mu_k$ which define a {\it compactly supported Borelian probability measure} in $(\mathcal{H}_s,\vert\vert .\vert\vert_s)$.

 This setting was previously considered by Gorodetski, Hunt and Kaloshin
 (\cite{GHK06}).

\medskip

\begin{theoreme}{\bf (Main theorem)}
For an arbitrary $H\in\mathcal{H}_s$, the modified Hamiltonian $H+h$ is {super-exponentially stable} for almost all $h\in HB^\N$ with respect to the probability $\mu$.

More precisely, there exist positive constants $a$, $C_1$, $C_2$, $C_3$ such that for a small enough $\rho$, every solution of $H+h$ with initial condition $|\mathcal{I}(0)|<\rho$ satisfies
\[ |\mathcal{I}(t)-\mathcal{I}(0)|< C_1\rho\ f\! or\ |t| < \exp\left( C_2\rho^{-C_3\vert
\ln(\rho )\vert^a}\right) .  \]
\end{theoreme}

 In other words, the set of super exponentially stable Hamiltonians is prevalent in our Banach space since $\mu$ defines a compactly supported Borelian measure which is transverse to the set of non super-exponentially stable Hamiltonian (see \cite{OY05} and \cite{HK10}).

 We do not see obstructions to obtain similar results for Lagrangian tori
 (see the remarks at the end of this section).

\medskip

\section{Scheme of the proof :}

\medskip

 Either in this paper or in Bounemoura work (\cite{Bo11a}), the proofs rely
crucially on the extension of Nekhoroshev theory obtained in \cite{Nie07}
for a much wider class of unperturbed Hamiltonian than the steep integrable Hamiltonian considered originally in \cite{Nek77}.

 For instance, it can be proved (\cite{Bo11a}) that almost all quadratic
form in action variables fall into this class of Diophantine Morse
functions while only sign definite quadratic forms are steep according to
Nekhoroshev definition.

 Actually, the definition of Diophantine Morse functions involves countable quantitative transversality conditions which are stated in adapted coordinates. It is inspired on one hand by the steepness condition introduced by Nekhoroshev (\cite{Nek77}) and on the other hand by the quantitative Morse-Sard theory of Yomdin (\cite{Yom83} and \cite{YC}) where we consider "nearly-critical" points which are quantitatively non degenerate.

 The application of Yomdin quantitative Morse Sard theory allows to prove
(\cite{Nie07}) that for an arbitrary real analytic integrable Hamiltonian $h$ defined in a neighbourhood of the the origin in the action space $\R^n$, then for almost any parameter $\omega\in\R^n$, the modified integrable Hamiltonian $h_\omega (x)=h(I)-\omega .I$ is exponentially stable (i.e. : Nekhoroshev estimates are valid).

 In the case of an elliptic equilibrium, we have to deal with cartesian coordinates instead of action-angles ones and we cannot use the latter since they become singular at the origin, this yields specific problems detailed in \cite{Loc92}
and overcomes in \cite{BFG98}, \cite{Nie98} for an elliptic point with a
sign definite torsion. In \cite{BN12}, Nekhoroshev estimates are
proved in a generic setting by means of successive periodic averagings and
this method can be implemented in cartesian coordinates. These observations
allowed Bounemoura \cite{Bo11a} to prove that a Morse Diophantine condition is satisfied at all order on a full Lebesgue measure set of Birkhoff invariants
which is strong enough to prove superexponential results of stability over
times of order $\exp (\exp (C/r^a ))$.

 As it will be specified in the sequel, the set considered by Bounemoura
 is not big enough to ensure that generically the Birkhoff normal form of
 a non resonant elliptic fixed point falls into this good set. We will
 have to consider more refined estimates on the set of bad parameter $\omega\in\R^n$
 according to the following proposition which is a direct extension of (\cite{Nie07}, theorem 3.2.6) or (\cite{BN12}, theorem 2.2) :
 \begin{proposition} \label{estbadpara}
Consider an integrable Hamiltonian
$$\mathrm{H}(x,y)=h(\mathcal{I})=h\left(\frac{
x^2+y^2}{2}\right)$$
holomorphic on the ball of radius $s$ around the origin
in $\mathbb{C}^{2n}$.

 There exists an open set $\mathcal{C}^{(\rho )}_\varepsilon\subset
\mathcal{B}_\rho$ where $\mathcal{B}_\rho$ is the ball of radius $\rho$ around the origin in $\mathbb{R}^n$ such that
\begin{equation*}
{\rm Vol}(\mathcal{C}^{(\rho )}_\varepsilon )< C_\rho \varepsilon^\alpha
\end{equation*}
where $C_\rho >0$ depends only of $n$, $s$, $\rho$, $\vert\vert\mathrm{H}\vert\vert_s$ and $\alpha >0$ depends only of $n$.

 For $\omega\in\mathcal{B}_\rho\backslash\mathcal{C}^{(\rho )}_\varepsilon$, Nekhoroshev estimates of stability can be proved for the modified integrable Hamiltonian $\mathrm{H}_\omega (x,y)=h_\omega (\mathcal{I})=h(\mathcal{I})-\omega .\mathcal{I}$ with an holomorphic perturbation $f(x,y)$ of size $\vert\vert f\vert\vert_s <\varepsilon$.

\end{proposition}

\begin{remarque}\label{remsdm}{\rm
Note also that $h$ does not need to be polynomial, we only require that the
Hamiltonian is smooth enough.}
\end{remarque}

\medskip

 Now, following the strategy of Morbidelli and Giorgilli (\cite{MG95}), the estimates of superexponential stability given by Bounemoura rely
on the existence for all $m\in\mathbb{N}^*$ of a Birkhoff normalization $\Phi_m$
specified in theorem 1.2. which integrate $H$ up to order $2m$ on the ball of radius $s_m >0$ around the origin in $\mathbb{C}^n$, hence $H\circ\Phi_m(z)\!
=\! h_m(\mathcal{I})+f_m(z)$ where $h_m\in\mathbb{R}_m[X]$ and the remainder $f_m$ is of order $z^{2m}$. Moreover, the size of the perturbation $\vert\vert f_m\vert\vert_{s_m}$ is comparable with $\exp\left( -(1+\tau )m\right)\vert\vert f_1\vert\vert_{s_1}$ and $\vert\vert f_1\vert\vert_{s_1}\!\sim\! s_1$ (see \cite{GDFGS89}, \cite{DG96}).

 Then, in order to obtain superexponential estimates of stability, we should be able to apply Nekhoroshev theory (with the proof given in \cite{BN12}) on the Birkhoff normal forms $\left( h_m\right)_{m\in\mathbb{N}^*}$ at all order $m$ with a perturbation $f_m$ which decrease geometrically.

 Going back to the genericity problem, according to the previous proposition for an arbitrary $m\in\N^*$, Nekhoroshev estimates can be proved on the ball of radius $s_m >0$ for the perturbation $f_m$ of $h_m(\mathcal{I}) +\omega .\mathcal{I}$ if $\omega\in\mathcal{B}_\rho\backslash\mathcal{C}^{(\rho )}_m$ where $\mathcal{C}^{(\rho )}_m$ is an open set which contain the critical parameters in the ball $\mathcal{B}_\rho
\subset\mathbb{R}^n$ with a volume which decrease geometrically with $m$ hence ${\rm Vol}\left(\displaystyle{\bigcup_{m\in\mathbb{N}^*}}\mathcal{C}^{(\rho )}_m\right) <\displaystyle{\sum_{m\in \mathbb{N}^*}}{\rm Vol}\left(\mathcal{C}^{(\rho )}_m\right) \sim {\rm Vol}\left(\mathcal{C}^{(\rho )}_1\right) <\infty$.

 In our infinite-dimensional dimensional setting, we will say that a formal
 expansion $h_\infty\in\mathbb{R}[[X]]$ is {\it superexponentially stable}
 if there exists a sequence of radii $\left( s_m\right)_{m\in\N^*}$ such
 that Nekhoroshev estimates can be proved for all $m\in\N^*$ with the integrable
 part given by the truncation ${\rm Proj}_{\mathbb{R}_m[X]}(h_\infty )$ at
 order $m$ and a perturbation $f_m$ of size $\vert\vert f_m\vert\vert_{s_m}\sim
 s_1/e^{(1+\tau )m}$\!\! .

  We note that we have ${\rm Proj}_{\mathbb{R}_m[X]}(h_\infty +\ell_\omega )={\rm Proj}_{\mathbb{R}_m [X]}(h_\infty )+\ell_\omega$ where $\ell_\omega (X)=\omega .X$ for all $\omega\in\mathbb{R}^n$ hence, for an arbitrary formal expansion $h_\infty$ in $\mathbb{R}[[X]]$, the modified expansion $h_\infty (\mathcal{I})+\omega .\mathcal{I}$ is superexponentially stable if  $\omega\in\mathcal{B}_\rho\backslash{\displaystyle{\bigcup_{m\in\mathbb{N}^*}}}
 \mathcal{C}^{(\rho )}_m$.

 From these estimates Bounemoura (\cite{Bo11a}) proves that, for an arbitrary
formal expansion $h_\infty\in\mathbb{R}[[X]]$, superexponential estimates
of stability are valid for $h_\infty (\mathcal{I})-\omega .\mathcal{I}$
on a ball of small enough radius $\varepsilon$ around the origin if $\omega\in\mathcal{B}_\rho
\backslash\mathcal{C}^{(\rho )}_\varepsilon$ where $\mathcal{C}^{(\rho )}_\varepsilon
\!\! =\!\!\!\displaystyle{\bigcup_{m\in \mathbb{N}^*}}\mathcal{C}^{(\rho )}_m$ with ${\rm Vol}\left(\mathcal{C}^{(\rho )}_\varepsilon\right)$ of the order of $\varepsilon^\alpha$ where $\alpha$ given in the previous proposition depends only of $n$.

 This implies the generic property stated in Theorem 1.4.

 Especially, Bounemoura (\cite{Bo11a}) shows the existence of a foliation
of the space of formal series $\mathbb{R}[[X]]$ in $n$-dimensional probe
spaces directed by the space of linear forms where the properties required
to prove superexponential estimates are satisfied at almost every point on the leaves for the $n$-dimensional Lebesgue measure. The issue is that when we want to come back in the initial (non-normalized) coordinates, a change of the linear frequencies implies a change of all the Birkhoff invariants and all the normalizing transformations. Hence it is impossible to recover a foliation of the initial space of Hamiltonian $\mathcal{H}_s$ by $n$-parameter
families where superexponential stability can be proved for almost all parameters
with the $n$-dimensional Lebesgue measure. We must use more global estimates
with an {\it infinite} number of parameters and the important observations in this direction are the following.

\medskip

 To state our results we have considered the embedding of $HB^\mathbb{N}$
in the Banach space $\mathcal{H}_s$. Here, we first consider the Hilbert
brick $HB^\mathbb{N}$ embedded in the space of formal expansion $\mathbb{R}[[X]]$ identified with the product space ${\prod_{\N^*}}\mathcal{P}_k$
equipped with the product topology (we recall that $\mathcal{P}_k$ is the space of homogenous
polynomial of degree $k\in\N^*$).

 Actually, despite we are in an infinite dimensional space,
the central estimates concern polynomial integrable Hamiltonian (the truncated
Birkhoff normal forms) and now we revisit the study of Bounemoura for polynomial
Hamiltonian.  For all $m\in\mathbb{N}^*$, we identify the space $\mathbb{R}_m[X]$
of polynomial of degree bounded by $m$ with the product $\displaystyle{\prod_{1\leq k\leq m}}\mathcal{P}_k$ of the spaces $(\mathcal{P}_k)_{k\in\mathbb{N}^*}$ of homogenous polynomial of degree $k\leq m$ equipped with the product topology.
Moreover, for all $m\in\N^*$, we denote the projection ${\rm Proj}_{\mathbb{R}_m[X]}
\left( HB^\mathbb{N}\right)$ by $HB^{\leq m}\subset\mathbb{R}_m[X]$ equipped
with the product measure $\mu_{\leq m}=\!\!\displaystyle{\prod_{1\leq k\leq m}}\!\!\mu_k$.

 We note that, for all $m\in\N^*$, we have $\mu =\mu_{\leq
m}\times\displaystyle{\!\!\prod_{k\geq m+1}}\!\!\mu_k$ and especially $\mu =\mu_1\times
\displaystyle{\prod_{k\geq 2}}\mu_k$ where $\mu_1$ is the classical Euclidean volume up to a scaling (to define a probability) on the unit ball in the space of linear forms.

\medskip

 As in the previous reasonings, for all $m\in\mathbb{N}^*$, we consider
an open set $X_\varepsilon^{(m)}$ in $HB^{\leq m}\subset\mathbb{R}_m[X]$ given by Yomdin Morse-Sard theory which include the expansions in $HB^{\leq
m}$ (considered as integrable Hamiltonians) where Nekhoroshev theory cannot be applied to prove superexponential results of stability on a ball of radius
$\varepsilon$. Using Bounemoura construction, we can prove that $X_\varepsilon^{(m)}$ intersects any affine subspace in $\mathbb{R}_m[X]$ directed by the vectorial space of linear forms along a set of volume dominated by $\left(\varepsilon /\xi^m\right)^\alpha$ with $\alpha >0$, given in proposition 3.1, which depends only of the number of degree of freedom $n$ and a constant $\xi >1$.

 With the fact that $\mu_{\leq m}=\mu_1\times\displaystyle{\prod_{k=2}^m}\mu_k$
where $\mu_1$ is the Euclidean volume and $\displaystyle{\prod_{k=2}^m}\mu_k$
is a probability measure, the application of Fubini theorem implies that the measure $\mu_{\leq m}\left( X_\varepsilon^{(m)}\right)$ remains is of the order of $\left(\displaystyle{\frac{\varepsilon}{\xi^m}}\right)^\alpha$.

 Now we come back to our initial functional space $\mathcal{H}_s$, for an arbitrary Hamiltonian $H\in\mathcal{H}_s$ we consider the application :
$$\begin{array}{ccll}
\mathcal{B}\mathcal{N}\!\mathcal{F}_m : &\mathcal{M}_m &\longrightarrow &\ \ \ \ \mathbb{R}_m [X]\ \ \ \ \ \ \ {\mathrm w}{\mathrm i}{\mathrm t}{\mathrm h}\ \ \ \mathbb{R}_m[X]\simeq\!\!\!\!
\displaystyle{\prod_{1\leq k\leq m}}\mathcal{P}_k\\
& \left( P_1, \ldots , P_m\right) &\longmapsto &\left( Q_1, \ldots , Q_m\right)
\end{array}$$
where $Q_k$ is the Birkhoff invariant of order $k\in\{ 1,\ldots ,m\}$ of $H+P$ for $P={\displaystyle\sum_{k=1}^m} P_k$ and $\mathcal{M}_m$ is an open set in $\mathbb{R}_m [X]$ where the Birkhoff normal form at order $m$ can be build.

 We have :

\medskip

\begin{proposition}

 For $k=1$, the transformation is $Q_1=P_1+\mathcal{B}^{(1)}(H)$ where $\mathcal{B}^{(1)}(H)$ is the first Birkhoff invariant of $H$ hence it is a translation.

 For $k\in\{ 2,\ldots ,m\}$ :
$$Q_k=P_k +\mathcal{F}_k (P_1,\ldots ,P_{k-1})$$
where $\mathcal{F}_k$ is a differentiable function over $\mathcal{M}_m$, hence the mapping $\mathcal{B}\mathcal{N}\!\mathcal{F}_m$ is one-to-one with a Jacobian equal to 1.

\end{proposition}

 A similar property on the Newton interpolation polynomials was used by
Gorodetski-Hunt-Kaloshin (\cite{GHK06}).

\medskip

 Despite our transformation has a unit Jacobian, it does not preserve the
measure $\mu_{\leq m}$ since it does not leave invariant the projection $HB^{\leq m}$ of the Hilbert Brick~$HB^\mathbb{N}$.

 Actually, for an analytic Hamiltonian, the Birkhoff normal form should admit a Gevrey regularity and not remain analytic (we recall that we have no results on the generic divergence or convergence of the Birkhoff normal forms). Hence, the Birkhoff invariants should growth much more rapidly then the coefficient of the initial Hamiltonian and the set $HB^{\leq m}\cap\mathcal{M}_m$ is sent to a much bigger domain by the mapping $\mathcal{B}\mathcal{N}\!\mathcal{F}_m$.
Consequently, we cannot compute the measure $\mu_{\leq m}$ of an image set in $\mathcal{B}\mathcal{N}\!\mathcal{F}_m\left( HB^{\leq m}\cap\mathcal{M}_m\right)$ by the change of variables formula.

\medskip

 In order to overcome this problem, we rescale the normalized variables in $\mathbb{C}^{2n}$ on the ball of radius $s_m$ around the origin where the Birkhoff transformation $\Phi_m$ at order $m$ is defined. Denoting the initial
variables by $z\in\mathbb{C}^{2n}$, we consider $z'\in\mathbb{C}^{2n}$ defined by $z=\Phi_m (s_m z')$ which correspond to a conformal symplectic transformation giving the Hamiltonian~:
\begin{equation}
K_m(z')=\frac{1}{s_m^2}H\circ\Phi_m (s_m z')\ \text{\rm or}\ K_m(x',y')=\frac{1}
{s_m^2}H\circ\Phi_m (s_m x',s_m y').
\end{equation}

 In these rescaled variables, the integrable part $k_m(z')=\frac{1}{s_m^2}h_m(s_m^2 \mathcal{I}(x',y'))$ belongs to $HB^{\leq m}$.

 Our initial issue is settled but the price to pay is a huge deformation of the measure $\mu_{\leq m}$ since the Jacobian of the transformation $\widetilde
{\mathcal{B}\mathcal{N}\!\mathcal{F}}_m$ which maps to the rescaled Birkhoff invariants (denoted $\left({\tilde Q}_1,\ldots ,{\tilde Q}_m\right)$ in the
sequel) is now $s_m^{-{\rm D}(m)}$ where ${\rm D}(m)$ growth like dimension of the space of polynomial $\mathbb{R}_m[X]$ of degree $m$ in $n$ variables hence ${\rm D}(m)$ is of the order of~$m^n$.

\medskip

 Let ${\widetilde X}_\varepsilon^{(m)}$ be a measurable set given by Yomdin quantitative Morse Sard theory of polynomial integrable Hamiltonians in $HB^{\leq m}$ which include the set of critical Hamiltonians at order $m$ where Nekhoroshev
theory cannot be applied in the rescaled variables to prove superexponential
estimates of stability.

 We must bound the measure of the set of Hamiltonian in $HB^{\leq m}\cap\mathcal{M}_m$ which are sent in ${\widetilde X}_\varepsilon^{(m)}$ by $\widetilde{\mathcal{B}\mathcal
{N}\!\mathcal{F}}_m$, hence :
$${\displaystyle\int\limits_{\mathcal{M}_m}}\!\!\!\! 1_{{\widetilde X}_\varepsilon^{(m)}}\!\!
\left(\widetilde{\mathcal{B}\mathcal{N}\!\mathcal{F}}_m (P_1,\ldots ,P_m)\right) \!\! d\mu_{\leq m}(P)\! =\!\frac{1}{s_m^{{\rm D}(m)}}\!\displaystyle{\int\limits_{\widetilde
{\mathcal{B}\mathcal{N}\!\mathcal{F}}_m (\mathcal{M}_m)}}\!\!\!\!\!\!\!\!\!\!\!\! \!\! 1_{{\widetilde X}_\varepsilon^{(m)}}\!\!\left({\tilde Q}_1,\ldots ,{\tilde Q}_m\right)\!\! d\mu_{\leq m} ({\tilde Q})$$
and this is bounded by ${\displaystyle\frac{\mu_{\leq m}\left({\widetilde X}_\varepsilon^{(m)}\right)}{s_m^{{\rm D}(m)}}}$.

\medskip

 With the growth of ${\rm D}(m)$, the divisor $s_m^{-{\rm D}(m)}$ decreases
 in a {\it superexponential} way with respect to the order of normalization
 $m$. At this step, the set introduced by Bounemoura is not big enough to
 ensure that its preimage in the construction of the Birkhoff normal forms
 is a generic set in $\mathcal{H}_s$. More specifically, we cannot afford a {\it geometric} decrease of the volume the critical parameters as in the convex case or in Bounemoura construction with the set $X_\varepsilon^{(m)}$.

 Here, we need a superexponential decrease of the volume of ${\widetilde X}_\varepsilon^{(m)}$\!\!\! .

 In order to settle this issue, we will not use Nekhoroshev theory on the
 whole domain where the Birkhoff normalization at order $m$ is defined (hence
 on the ball of radius $s_m$) but on a much smaller domain which is the ball
 ${\widetilde{\mathcal{B}}}_{r_m}$ of radius $r_m$ around the origin in the
 rescaled variables (hence $r_m s_m$ in the normalized variables).

 After a normalization at order $m\in\N^*$ and our rescaling, the size of the perturbation is of the order of $(m!)^{1+\tau}(s_m r_m)^m\sim \exp
(-(1+\tau )m)r_m^m$ (the prefactor comes from the growth of the coefficients).
Hence the size of the perturbation is dominated by $r_m^m$ since the expansion
of the perturbation starts at order $m$ (this is another important point),
with the proposition 3.1 and the previous reasonings we have $\mu_{\leq m}
\left({\widetilde X}_\varepsilon^{(m)}\right)$ of the order of $(r_m)^{m\alpha}$\!\!\!
.

\medskip

 Now we come back in our infinite-dimensional setting and, for all $m\in\N^*$,
 we consider the set ${\mathbf{{\widetilde X}_{\mathbf\varepsilon}^{\mathbf(m)}}}\!\!\!
\subset\mathbb{R}[[X]]$ such that $h_\infty\in{\mathbf{{\widetilde X}_{\mathbf
\varepsilon}^{\mathbf(m)}}}$ if and only if ${\rm Proj}_{\mathbb{R}_m[X]}\left( h_\infty\right)$ is in $\left(\widetilde{\mathcal{B}\mathcal{N}\!\mathcal{F}}_m\right)^{-1}\!\! \left({\widetilde X}_\varepsilon^{(m)}\right)$ or $\widetilde{\mathcal{B}\mathcal{N}\!
\mathcal{F}}_m\!\!\left({\rm Proj}_{\mathbb{R}_m[X]}\left( h_\infty\right)\right)\!\!\in
\!\!{\widetilde X}_\varepsilon^{(m)}$.

\medskip

 For all $m\in\N^*$, the product $\displaystyle{\prod_{k\geq m+1}}\mu_k$ is a probability measure and we have $\mu =\mu_{\leq m}\times\!\!\!\displaystyle
{\prod_{k\geq m+1}}\mu_k$, hence : $\mu\left({\mathbf{{\widetilde X}_{\mathbf\varepsilon}^{\mathbf(m)}}}
\right)\!\! =\!\mu_{\leq m}\left(\left(\widetilde{\mathcal{B}
\mathcal{N}\!\mathcal{F}}_m\right)^{-1}\!\! \left({\widetilde X}_\varepsilon^{(m)}
\right)\right)$.

 With the previous reasonings for all $m\in\N^*$, we have the upper bounds~:
$$\mu\left({\mathbf{{\widetilde X}_{\mathbf\varepsilon}^{\mathbf(m)}}}\right)\!\!
\leq\frac{\mu_{\leq m}\left({\widetilde X}_\varepsilon^{(m)}\right)}{s_m^{{\rm D}(m)}}\sim\frac{(r_m)^{m\alpha}}{s_m^{{\rm D}(m)}}.$$

 Finally, the set of critical Hamiltonian $\mathbf{C}$ where Nekhoroshev
theory cannot be applied at some order $m\in\mathbb{N}^*$ to prove superexponential
results of stability is included in $\displaystyle{\bigcup_{m\in\mathbb{N}^*}}{\mathbf
{{\widetilde X}_{\mathbf\varepsilon}^{\mathbf(m)}}}$ which is of small measure if the sum
$${\displaystyle\sum_{m\in\mathbb{N}^*}\frac
{(r_m)^{m\alpha}}{s_m^{{\rm D}(m)}}}<\infty.$$

 The second radius $r_m$ is a {\it free} parameter which has to be chosen small enough to ensure the convergence of ${\displaystyle\sum_{m\in\mathbb{N}^*}\frac
{(r_m)^{m\alpha}}{s_m^{{\rm D}(m)}}}$.

 This imposes to choose a radius $r_m$ which is {\it exponentially small} with respect to the order of normalization $m$, here $m$ is of the order of $\log (r_m)^a$ where $a>0$ depends only of $n$. We cannot recover an exponentially small remainder with $m$ of the order of $1/r_m$ as in the convex case but only a remainder which is smaller than any polynomially small remainder with respect to the radius. This is an important point which comes exclusively from the expected divergence of the Birkhoff normal form (actually, as previously said, we know that this normal form is always obtained as a Gevrey divergent expansion in the Diophantine case but we don't know if it diverges or converges generically). Another misleading fact is that on the ball of radius $s_m r_m$, we could obtain directly a Birkhoff normal form which integrate the system up to a superexponentially small remainder with respect to $m$. But in this case, the Birkhoff normal form would give only a remainder which is exponentially small with respect to the radius while here we start with a perturbation which is smaller than any polynomially small remainder with respect to the radius before going to the exponential with Nekhoroshev estimates.

 Summarizing we have proved that there exists positive constants $a$, $C_1$,
 $C_2$, $C_3$ and a sequence of annuli of decreasing radii $(r_m)_{m\in\N^*}$ such that, for an arbitrary function $H\in\mathcal{H}_s$ and for almost all $h\in HB^\N$ with respect to the probability $\mu$, every solution of the Hamiltonian system governed by $H+h$ with an initial condition $r_{m+1}\leq
\vert\vert{\widetilde{\mathcal{I}}}(0)\vert\vert <r_m$ in the rescaled variables satisfies :
\[ \left\vert\left\vert{\widetilde{\mathcal{I}}}(t)-{\widetilde{\mathcal{I}}}(0)\right\vert\right\vert< C_1 r_m\ f\! or\ |t| < \exp\left( C_2 r_m^{-C_3\vert\ln( r_m )\vert^a}\right)\!\! .  \]

 In order to obtain uniform estimates with respect to the initial radius
 $\rho$, as in our theorem 1.4, it only remains to slightly lower the coefficients $C_1, C_2, C_3$ and the exponent $a$.

\medskip

\section{Comments and prospects :}

\medskip

 A natural extension of this work would be for the case of a Lagrangian invariant
torus for an Hamiltonian system. In this setting, we consider a system governed
by the Hamiltonian $\omega .I +f(I,\varphi )I^2$ where $f$ is analytic in
action-angle variables $(I,\varphi )\in\mathbb{R}^n\times\mathbb{T}^n$. The only missing ingredient to make the previous reasonings is a quantitative
construction of the Birkhoff normal form to a given order in action. Up to the author knowledge, the available constructions (\cite{Fas90}, \cite{PW94},
\cite{Pos93}) in the literature yield an integrable Birkhoff normal form with a perturbation which is small but with terms of all order in action
including low order terms. Here, we need a Birkhoff normal form of the kind
$h_m(I)+f_m(I,\varphi )I^m$ for $m\in\mathbb{N}^*$ to follow the scheme of
our proof. This is a purely technical issue and there is certainly no obstruction to overcome this problem.

  The second extension would be for the case of an Hamiltonian with a lower regularity, that is for Gevrey or differentiable Hamiltonians. With the classical
papers on Birkhoff normal forms in the differentiable case or the work of Popov-Mitev \cite{MP10} in the Gevrey case together with the article of Bounemoura \cite{Bo11b} on generic Nekhoroshev estimates for Gevrey or differentiable Hamiltonian we have all the required technical tools.

 A final interesting extension would be for the case of invariant tori provided
by KAM theory. A model statement would be : for an analytic integrable hamiltonian which satisfy Kolmogorov (or Russman) non-degeneracy condition (\cite{AKN97}), KAM invariant tori are superexponentially stable for a small enough generic perturbation. The papers of Popov (\cite{Pop00}, \cite{Pop04}) on the existence of a global Birkhoff normal form for all the tori given by KAM theory should be useful in this setting.

\bigskip

\addcontentsline{toc}{section}{References}
\bibliographystyle{amsalpha}

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\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }
\providecommand{\MRhref}[2]
  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
}
\providecommand{\href}[2]{#2}

redpajama_set_name: RedPajamaArXiv


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text:
FILE - Bosnian-Serb General Ratko Mladic is seen in this 1995 file photo during a visit to troops in the east Bosnian town of Vlasenica. Belgrade's B-92 radio reports Thursday May 26, 2011 that a man suspected to be Europe's most wanted war crimes fugitive Ratko Mladic has been arrested in Serbia. Serbia's war crimes prosecutors refused to confirm or deny the report. (AP Photo/Oleg Stjepanovic) (AP)
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May 26, 2011 4:19PM (UTC)
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redpajama_set_name: RedPajamaCommonCrawl


text:
Q: How to use Java annotations to guide Android's Proguard? When using Proguard with Android, methods that are only invoked via reflection (e.g., callbacks defined in onClick XML attributes) are erroneously stripped out.
One solution for this issue is to add each affected class and method to your proguard.cfg.
How can I use Java annotations to achieve the same effect?
I feel that would make the code self-documenting and it would avoid code and proguard.cfg drifting out of sync. However, Android's Proguard doesn't seem to ship with the annotations.jar mentioned in its documentation.

A: You can retrieve annotations.jar and annotations.pro from an official ProGuard release. You should then be able to use annotations as discussed here
All the necessary options can go in proguard.cfg.

A: I ran in to this problem recently. Here is what you need to do:
To fix the onClick events add this to Proguard settings
-keepclassmembers class * extends android.app.Activity{
   public void *(android.view.View);
}

To keep annotation add
-keepattributes **  or -keepattributes  *Annotation*
More information is available here http://www.simpligility.com/2010/12/hints-for-using-proguard-on-your-android-app/ I use maven-android-plugin to compile android app and this article pretty much sums up what I do for the android to get it working. Hope this helps!

redpajama_set_name: RedPajamaStackExchange


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Squash shoes are an essential piece of squash equipment. they help you move around the court, protect your feet, and are nonmarking, meaning they will not leave marks on the squash court. . D .


A few days after launching Modi app, the government introduced the Swasthya Samiksha mobile application on Thursday. Through this application, the Members of Parliament can track the women's maternal health related issues in their constituencies.
The Swasthya Samiksha app was collaboratively launched by White Ribbon Alliance, Centre for Catalysing Change & Swaniti Initiative and all not-for-profit organisations to help MP's oversee the women's reproductive health services.
Initially, the Swasthya Samiksha app will be provided in the states such as Bihar, Rajasthan, Jharkhand and Madhya Pradesh. Currently, the application is still in experimental phase which will provide district-wise data for the respective state.
"Swasthya Samiksha aims to engage and encourage the Members of Parliament to take significant steps to ensure better health for women in their constituencies. It is an attempt to provide data on health notices in an easy and convenient way," Aparajita Gogoi, Executive Director of Centre for Catalysing Change and National Coordinator of White Ribbon Alliance India, said in a statement.
According to the reports, the Swasthya Samiksha app will address the issue of quality of maternal health services and helps to generate demand for institutional deliveries. Through this app, Members of Parliament can track their respective constituency's reproductive and maternal health-related data and the indicators.
This app also enables the comparison of districts in the state and subsequent improvement in the quality of care. It is an attempt to provide information on health indicators in a comfortable and more convenient way.


A stay that you will not forget! Lulu has received us with great sympathy, we have lacked nothing.
At the right distance from everything and the beautiful beach of the Orient Bay just in front. On the advice of Lulu we spent a day at the island of Pinel ..... magic !!! another world !! Thank you again for these wonderful moments.
Thank you so much for your kind comments, not only about me, but mostly about your stay with us.
It was a real pleasure to help you with some advice and I am really delighted that you took the time to go to the Pinel Island and that you enjoyed your ballad.
Thank you once again for your kindness and we look forward to welcoming you again soon.
We spent a stay, in family with the children, it was awesome. The studio is big and quiet behind the house. It's a small apartment because to sleep, there is a separate bedroom and a mezzanine. It is well ventilated and well equipped, even against the mosquitos. And we had the most beautiful beach of the island just two minutes walk away.
Thank you for posting this comment and to learn that you liked the space of this studio with its mezzanine and its courtyard, and not to be bothered by the mosquitoes.
Great family holidays, beach next door and good hospitality!
We stayed as a couple accompanied by our 6 year old daughter in April 2016. We were greeted by the owner of the place in a very warm manner. "Lulu" the owner of the place gave us right away the right places to go: everything is close by! The beach of Orient Bay is simply magnificent, the surroundings of the residence are very secure and the attendance is of quality, it is all the interest of the "private" domains.
The swimming pool is big, warm and well maintained and our daughter loved it! Thanks for everything and see you next year.
'The 5 stars of orient bay' called the ST TROPEZ of the Caribbean, restaurants on the beach combined with ST TROPEZ. A place at 200 meters called 'the village' includes shops (bakery, grocery store, hairdresser, press, tobacco, snack bars, restaurants) with evening entertainment.


Despite advances in DNA testing, forensics, and the investigative skills used by police, hundreds of crimes remain unsolved across Canada. With every passing day trails grow colder and decades can pass before a new lead or witness comes forward if one comes forward.
"As Hoshowsky shows in his sensitive, non-sensationalist survey of 11 Canadian cases from 1968 to 2006, what progress has been made in solving [cold cases] has been a matter of technological wizardry."
Hoshowsky is a good writer. His prose, which is of a very high caliber, flows smoothly, almost in poetic fashion, with the result that Unsolved is very hard to put down.
"Peggy Dymond Leavey has written an engaging and moving story of a likable, imaginative girl. I kept reading late into the night, empathizing with Ivy and wanting to know what happened next. The realities of the Great Depression are brought to life with authentic, well-researched details"


April 30, 2015 • Fresno native Mark Arax has written about the war over water in his state for decades. "It used to be the farmers against the delta smelt fish, and now it's the urbanite against the almond," he says.
April 28, 2015 • Pope Francis wants world leaders to work together to enact laws that will slow or stop emissions that fuel climate change. He is expected to deliver a papal encyclical in June highlighting environmental degradation and the effect of climate change on the poor. The Vatican held a one-day conference on climate change Tuesday.
April 28, 2015 • More and more city dwellers are trying their hand at urban gardening. Most know to be wary of lead in their soil, but fewer are aware of how to avoid other types of contaminants.
April 27, 2015 • New mandates in California will limit how much water graveyards can use to keep their grass green. The people who run cemeteries don't want crispy, parched grass to feel like an insult on top of loss.
April 26, 2015 • The state is accustomed to tornadoes and severe weather. But since 2008, there's been a surge of quakes — linked to oil and gas drilling.
April 26, 2015 • NPR's Arun Rath speaks with a California landscaper who says demand for her business is booming.
April 24, 2015 • Once embraced by cities for its beautiful white flowers, disease resistance and ability to grow just about anywhere, the Callery pear is now considered a nuisance due to its smell and invasive nature.
April 24, 2015 • Early next month, California plans to finalize its emergency water conservation plan. Cities are under the gun to cut their water usage from anywhere between 15 and 40 percent.
April 24, 2015 • The world's largest ant colony stretches over 3,700 miles. It succeeds, biologist Deborah Gordon says, because no ant is in charge. They communicate with algorithmic patterns to survive and thrive.
April 23, 2015 • In Michigan's orchard country, extreme heat and cold can mean disaster for fruit growers. Now some are using a new twist on old technology to fool trees when sudden, unexpected weather changes occur.
April 22, 2015 • "Climate change can no longer be denied," Obama said. "It can't be edited out. It can't be omitted from the conversation. And action can no longer be delayed."


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Under the framework of this new program, Israel Innovation Authority (Formely OCS and MATIMOP) and NEDO are pleased to present the second editions of the Japan-Israel Partnerships Catalogs – a collection of profiles of Israeli companies and Japanese companies seeking partnerships. The Israeli Company profiles are available in both English and Japanese!
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Israeli and Japanese companies interested in finding partners for R&D cooperation and other business ties are invited to join the catalog. These profiles will contribute to the next edition of the Japan-Israel Partnerships Catalog, which will be distributed widely among Japanese and Israeli Industry and relevant stakeholders. Inclusion in the catalog will provide a great deal of exposure and potential business opportunities for participating companies.


text:
That'd Be A No
RP Logs » Archives: Logs » That'd Be A No
November 02 2014: May and Triplett visit Trent after the meeting with Fury. Hunter shows up later.
Unit 24F
An apartment with a view of the Hudson
Agent Triplett
K'nert
Since the meeting at the cafe Jericho's been keeping a lower profile than normal. Mostly he's been at home. Well, the place he currently thinks of as home anyway. He's making some stir fry in case the other two people who presently live with him show up. At the moment its just him and K'nert. He has been waiting to hear back from May on how things went at SHIELD but being a government organization of a sort. So he's not holding his breath.
Melinda May didn't actually get to meet with Director Fury herself, but she knows that Triplett did. So rather than play the rumor game or make the taller agent repeat himself, she takes him to meet with Jericho in person. Again. At least she warns the hacker ahead of time by sending him a text just before they get underway. She's driving, so no texting while driving.
"You know I'mma' get you back, right?" That's all Trip has to say to Agent May whilst they are en route to her contact. After all, he knows she set him up for that. But at least he got to meet Fury. He's also got a small smile on his face so May doesn't take it as more than a teasing threat. He's not trying to get punched out of a moving vehicle or anything like that. But then he's back to silence because May likes the quiet. Safer that way.
Jericho has bugs in the hallway, and very good recognition software. He recognizes the sound of May's gait when she comes up the stairs and just as she and Trip reach the door, it opens. "Hey May. You're here a lot faster than I thought you would be. Agent Triplett… why don't you two come in. Don't mind K'nert… do you two want some stir fry?"
Melinda May nods to Trent when he opens the door, apparently not at all surprised that he knew they were approaching the apartment, then steps inside proper. She visually locates K'Nert first thing, and opens the insulated bag she'd refused to let Trip inquire about on the way here. Pulling a Klondike Bar from the bag, she holds it up for the odd little creature to see for a second while she levels one of her more dangerous warning glares. She then lets the little demon take the ice cream bar from her and answers Jericho while heading into the kitchen, presumably to put the bag into the freezer. "Stir fry sounds good… as long as you leave out the water chestnuts."
"I'm good. Thanks." Triplett just stands on the inside of the place of residence and semi-gawks. Just a little bit. Honestly, when there's random demons just lurking around in a place like this and they ain't freakin' May out and it's just a whole bunch of something that nobody else seems to be weirded out by, he's kind of got to just kind of shrug it off. But he still can't not go a little wide-eyed. He's seen some stuff but this is kind of pushing it. What the hell.
Jericho spoons some out into a bowl. No water chestnuts, no but there are bamboo shoots. "Here you go, May."
The hacker's eyes gleam in amusement at the imp and Trip. The little lizard like being eyes the agent and then goes off to eat his klondike bar in a corner.
"Come on out to the terrace." Jericho motions. It's a nice place, overlooking the river, particularly at night when the city lights up. "So, how did it go?"
Maybe this will help toward explaining why May keeps Trent as an off-the-books asset/contact/informant. There's just some shit that most of SHIELD can't handle. Case in point, the cat-sized demon currently munching on a Klondike Bar in the corner of the kitchen's dining nook. At least the little Jim Henson reject remembered to open the wrapper this time.
May accepts the bowl from Jericho, snags a fork unless the hacker actually keeps chopsticks on hand, then nods and steps out onto the terrace. "Figured it was best to make Triplett go through this only once." She suspects she knows what Fury's response was, but she knows better than to assume anything.
Trip heads out towards the Terrace while still trying to keep an eye on the Muppet in the Corner. He's ready for action if he needs to be, but still manages to make it out to the Terrace without getting attacked. So that's a good thing. "Right." It takes him a moment to remember what was said to him. "After he tore me a couple of new ones, one of them about you" Trip eyes Jericho. " he moved right along to giving me, or us, what I like to call a Soft Yes." Trip shrugs a little bit. "But he wants to meet with us before we head out, guns blazing. /All/ of us." Another look at Trent.
Jericho's eyebrows quirk. "Meet and greet mmm?" By the way he folds his arms and leans back against the wall he's not exactly amused. "Interesting. He say way?"
Whatever's going on behind Jericho's eyes, it's clearly generating smoke. Which is to say, Trip can almost see him thinking, considering.
Melinda May just shakes her head slowly, waiting until she's swallowed a bite of the stir fry before speaking up. "So he didn't share any intel about Black Valley with you, just demanded a full briefing?" Typical. She takes another small forkful of the stir fry then nods to Trent slightly. Good stuff. And she's PICKY about her Asian food.
"No idea." Trip responds to Trent with a shrug of his shoulders. "Probably has something to do with you playing as many sides of the fence as you can." Trip gives an almost accusing look in Trent's direction. "Or could have something to do with your girlfriend." This time Trip's look is actually accusing. But then he's shrugging it off and giving a wave of his hand. "It doesn't even matter. If we want to do this Op, we're gonna' need Fury's signature. And the only way we're gettin' that is if he debriefs us himself."
Jericho straightens up and his eyes narrow. "Two things. Firstly I had no idea SHIELD took such an interest in my personal life, but I think I object. A lot. Second, I don't need Fury's signature to do anything. I came to SHIELD because I thought you guys might like to see whatever intel I dig up first hand rather than having me relate it after the fact. If I decide to share it at all, that is. I'm not Fury's dog to call and send as he pleases. I've got other resources and allies, so if SHIELD wants to play this way they can damn well do it without me." Beat. "That's a 'no' by the way."
Melinda May sighs inwardly. That's pretty much what she expected was going to happen. Though, Trent just confirmed her suspicions in the worst possible way, and that's clear by the way her eyebrows draw together briefly. How in the HELL did Fury find out? That's what she wants to know. Because she sure as hell never so much as breathed Illyana's name in conjunction with Trent's. Ever.
"Well. I guess that's that then." Trip offers a shrug of his shoulders and a dismissive wave of his hand. He's not about to get in the middle of all this nonsense. There are plenty more assignments that he can pick and choose from. He doesn't want to go to Russia anyway. It's cold. "May? I'll be downstairs." Trip gives a nod to Trent as he turns to make with the getting the heck out of here before the demon thing decides he's the enemy. "Good luck." And that's all he's got to say to him.
"You can stay and have a beer if you'd like, Agent. K'nert won't bite. He knows better." Jericho waves as the other guy turns to leave and then turns his flickering gaze on May. The fact that he's a bit annoyed is pretty clear.
"So, May, any ideas why a multi-billion dollar UN spy Agency thinks it's any of its business what I do or don't do personally?"
"Trip. Hang on." May shifts her bowl of stir fry to her other hand, then fishes out the keys to the car they arrived in and tosses them at the younger agent. "I'll find my own way back, and then we'll talk some more."
"Yeah, yeah. I ain't see nothin'. I ain't hear nothin'. Next time?" Trip catches the keys, before giving a little salute in the direction of both May and Trent. Because this is definitely not up his alley. "Don't send me to get my ass chewed out by Fury for nothing, huh?" He frowns just a bit before he goes forth to take his leave. Still being cautiously wary of the demon thing on the inside. And he's muttering something or other. Probably just about looking like a fool in front of the Director.
Jericho leaves May on the terrace momentarily to get his own bowl of stir fry and then returns. He eats in silence, watching may his question as yet unanswered. "And just when I thought I might actually start getting along with the Division, mmm?"
Melinda May shakes her head again. "That's unlikely to happen. You refuse to drink the KoolAid." She's quiet for a moment longer before adding, "Back when Coulson and I first started working as partners, the water cooler talk could not get enough of speculating. I'm pretty sure there was even a betting pool at some point." She doesn't elaborate, but it's pretty easy to guess how she reacted to finding out about the rumors and betting pools and such.
"Maybe I'll just keep working with you, then." Jericho makes a face. "When I can. Sorry, May, I honestly thought this'd be juicy enough to warrant dispensing with the usual BS. Apparently I was wrong. Also, apparently, I'm going to need to make plans to ditch this safe house at some point. I know for a fact SHIELD doesn't have surveillance devices in here. I'm a bit peeved that apparently you have a detail out there on me somewhere. Count on me making their lives miserable tomorrow."
Melinda May nods. "Can't say I blame you. And I honestly would have expected this kind of crap from Hill, not Fury." She takes a deep breath as she finishes her stir fry. "If you need help moving anything… Well, maybe it's best if I don't know where you're going." And when Trent says that SHE has a detail out there, she frowns. "I don't have anything out there. Not even my car right now. And if you don't give them grief, I will." With that, she stands and heads inside to rinse out her bowl.
Jericho comes back in with her, shutting the door. He'll see about some more active countermeasures soon. The nice thing about dealing with spies is that they don't often want to admit they're spying on you so if you make their lives difficult, they'll often try to figure something else out. And Jericho has enough experience playing this game with Hydra to have some good ways to do just that.
"Sorry May." He murmurs. "Didn't mean to make it personal. I'll keep in touch but yeah, if I burn this place I won't let on where the next one is. I have some options out there." Which is all he'll say on that. Plausible deniability and all that. "You going to stick around a bit? Or do you need a lift somewhere?"
"I can stick around if you want me to." May would totally understand if he would rather she move along now. "But no matter how this shakes out, if you need someone to drive while you're in Russia, I'll tell Fury that I'm using vacation time and he can just deal with it." It's not like she doesn't have a ton of personal time banked.
"I don't mind you staying a bit. Help me eat some of this Stir Fry if nothing else. I know you probably have a few questions for me that are… unrelated to this present mess, mmm?" May often does and generally speaking, offline, Jericho doesn't mind answering them.
Melinda May says, "Nothing that can't wait," May says as she shrugs at the suggestion that she might have other questions. Instead of getting seconds of the stir fry, she opens the freezer to claim one of the Klondike bars she brought. And of course, K'Nert gets a warning stare. This one's hers, punk. "I've lived by the need-to-know for long enough that I know when I don't need to know.""
"Yeah, but you've been enough of a friend that I think you've earned some right to know. So long as it doesn't get back into SHIELD's files, but I trust you." Jericho chuckles wryly. "I trust you enough not to go try hacking into SHIELD servers and erasing my data out of paranoia and spite. Though the way their survieling me it makes me wonder if I might not have a real reason to worry about having to do that soon."
Melinda May scoffs faintly. "If your file has anything more than bare bones in it, they're getting their intel from elsewhere." And if she finds out who IS spilling to SHIELD, she might well have to have a TALK with them.
Jericho grins wryly and his amber eyes gleam in amusement. "What? SHIELD poach info from other sources? Would they do that?"
The hacker shakes his head as K'nert, the imp, comes to perch atop the 'fridge and peers over at May. He doesn't even try for the klondike bar though. Jericho himself dishes out another bowl of stir fry. "Honestly though, that'd be somewhat worrisome. The only other group with anything resembling a file on me would be Hydra. And why HYDRA would be feeding SHIELD information is…" Beat. "Well, you get the idea."
A creak of a floorboard and then they have a new arrival, quickly through one of the internal doors, shutting it behind him. He shrugs out of his leather jacket, giving them a brief grin, lopsided, charming, lazy, as he heads to the fridge, predictably seeking out a beer. "Careless of you both. Did you know you are being followed?" The casual statement comes as he opens the beer, leaning against the side, Hunter's grin cocky, amused, "If you wanted to leave now, they can find themselves immobilized, too busy to follow you." The offer is spoken in the sarcastic, British accent, with one eyebrow cocked. He takes a swig, adding, "No, May, they didn't see me."
Melinda May levels a VERY unimpressed glare at Hunter. "Don't make me kick your ass, Hunter. Again. Or better. I think I'll let this apartment's owner do the ass kicking." She considers the ice cream sandwich in her hand, then holds it up for K'Nert to take. You'd better remember this, you little punk. "Why are you here, anyway?"
Jericho turns, flicking the leather case behind him. In an eyeblink the device inside unfolds into a wicked looking blade. The man's eyes glow brightly in surprise and then anger. Any move to attack though is held up when May seems to know the man. "May, who is this?" He says carefully, his traces starting to glow blue, a sure sign that he was just a second ago preparing for metahuman level combat.
Hunter gives her that grin, untouched by the glare, mock innocent leveling in his features. "Me? Well, I saw people tailing my good friend, my companion in arms? How could I not offer my help?" He spreads his hands, taking a swig of the beer before he makes the gesture. He toasts Jericho, "Lance Hunter, at your service, mate. Why are they following you? Why are our guys following you?"
Melinda May breathes out a long sigh, dusting her hands off after relinquishing the ice cream sandwich to the cat-sized demon on top of the fridge. "Hunter, you are a bloody git sometimes. And this is Trent, the person whose home you just broke into. Trent, Hunter is former MI5 with a Matt Smith is 007 complex."
The way May's reacting makes Jericho think that perhaps it's not necessary to try slicing this guy. The blade folds away and goes back with practiced ease into the case slung on the hacker's back. "Alright Hunter. May hasn't tried to throw you out, so I'm not going to either. Yet." Beat. "Before I answer your question let me pose one of my own. Why… are you in my house?"
"Hey!" The protest would sound more genuine without the grin that follows, "007 had nothing on me. The chap drank weak martinis for fucks sake." The insult does bring a grin, and he shifts the beer to his other hand, offering his free one to Trent. "Someone followed you earlier. Consider me the back up, something I can't do from outside so easily. My pleasure, no need to thank me. So who are we kicking the fuck out of tonight?"
"How about Nick Fury. Up for that?" It's in jest, well, half jest. Jericho's not crazy enough to actually do that to the head of one of the largest spy agencies on the planet. But he's not happy at all with SHIELD right now both for the way they've apparently been watching him - something he intends to fix - and the way they expected him to come when called.
"Depends. How much are you paying?" Half in jest in reply, and he takes a sip of the beer, his gaze steadily on Jericho, watching him closely. "Did you resent the dog whistling then? SHIELD isn't subtle, mate." The glance at the door is thoughtful, and his gaze moves over the room, taking in the set up, examining exits, entry points, defenses.
Melinda May resists the urge to roll her eyes at Hunter then tosses a sidelong glare at Trent. "No one. You'll have to get your jollies elsewhere." She crosses her arms and watches Hunter casing the room. "He's not an official contact, Hunter. SHIELD has no rights to boss him around."
Jericho shrugs at Hunter's question though he does chuckle at the joke. "I'd pay quite a lot to watch, at any rate. I like working with SHIELD when that's an option. And when I have contacts I can trust. So far I've got…" He eyes May meaningfully. "One."
K'nert, the imp, hops down off the fridge and onto the countertop to peer over at Hunter, hissing a bit, though it's not clear what, if anything he's saying.
Hunter gives May a look of injured innocence, "She never lets me have any fun." The mock-complaint is done with a barely repressed grin, and he simply offers the beer to the imp, lifting his eyebrows, "Want some of mine, mate? I suppose I should be going any road."
"Yes, you should be going, Hunter." May tosses a look toward Trent before adding to the Brit, "I'll go with you, so you point out those people that tailed me. I want to have a talk with them."
Jericho waves. "Have fun you two. May, my apologies for anything that comes across your desk in the future." She'll know what he's talking about. He's about to make keeping tabs on him unpleasant, expensive and in as many cases as he can manage, actually impossible. Yes, he's being vindictive. No, it's not exactly petty. "Talk to you offline later. Have a good evening."
_201411antoine_triplettaspecthunterk'nertshieldthe_cavalry


text:
Интерлингви́стика — раздел языкознания, изучающий различные аспекты межъязыкового общения и, в частности, международные языки как средство такого общения. Основное внимание обращается на процессы создания и функционирования международных искусственных языков, которые исследуются в связи с вопросами многоязычия, взаимовлияния языков, образования интернационализмов и тому подобным.

Интерлингвистика сформировалась на базе теории лингвопроектирования, заложенной работами Рене Декарта (1629 г.) и развитой Готтфридом Лейбницем и другими. Термин «интерлингвистика» появился в 1911 г. и был определён его автором, бельгийским учёным Жюлем Мейсмансом, как наука о «естественных законах формирования общих вспомогательных языков», под которыми он подразумевал любые языки, способные выступать в роли посредников межъязыкового (и даже междиалектного) общения. Со второй половины XX века интерлингвистика концентрируется на изучении международных искусственных языков (рассматриваемых как одно из средств преодоления языкового барьера), и обычно рассматривается как раздел языкознания, изучающий международные языки в контексте общей теории межъязыкового общения.

Применительно к кругу проблем, касающихся искусственных языков, интерлингвистика подразделяется на два основных раздела: теорию лингвопроектирования, которая распространяется на искусственные языки любого типа — вне зависимости от того, получили ли они коммуникативную реализацию, и теорию функционирования планового языка, которая распространяется только на системы, реализовавшиеся в практике человеческого общения.

Примечания

Литература

 (eo) Vĕra Barandovská-Frank, (2020), Interlingvistiko. Enkonduko en la sciencon pri planlingvoj, 333 pp., Poznań, Univ. Adam Mickiewicz, ISBN 9788365483539
(en) Gobbo, Federico, 2020, Introduction to Interlinguistics, München: GRIN Publishing

  — С. 196—197.
  — С. 25—40.

Ссылки
 (ru) Интерлингвистика, Lingvo.info
 Университет Адама Мицкевича, Познань, Польша
 Заметка об интерлингвистических курсах при Познанском Университете, по-русски
 Германское интерлингвистическое общество, Берлин, Германия

Языковые контакты

Интерлингвистика
Переводоведение

redpajama_set_name: RedPajamaWikipedia


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DEEP POLITICS FORUM > Activities and Strategies > Iceland's Efforts To Become an International Transparency Haven and Free Speech Zone.
View Full Version : Iceland's Efforts To Become an International Transparency Haven and Free Speech Zone.
"I am proud to advise the Icelandic Modern Media Initiative's proposal to create a global safe haven for investigative journalism. I believe this proposal is a strong way of encouraging integrity and responsive government around the world, including in Iceland. In my work investigating corruption I have seen how important it is to have have robust mechanisms to get information out to the public. Iceland, with its fresh perspectives and courageous, independent people seems to be the perfect place to initiate such an effort towards global transparency and justice."
On June 16th the Icelandic Parliament unanimously passed a proposal tasking the government to introduce a new legislative regime to protect and strengthen modern freedom of expression, and the free flow of information in Iceland and around the world. The unanimous vote included all government members.
According to the RUV, the Icelandic National Broadcasting Service, FBI agents landed in Reykjavík in August 2011 without prior notification in an attempt to investigate WikiLeaks operations within the country. However, their plan was interupted when Home Secretary Ögmundur Jónasson learned about the FBI's visit and sent them packing. The Icelandic government then issued a formal protest to United States authorities.
On several different occasions, the U.S. government's hunt for WikiLeaks has involved private individuals and companies in Iceland. For example, the U.S. has been successful in obtaining account information from Twitter on parliamentarian Birgitta Jónsdóttir, who now refuses to travel outside of Iceland for fear of being arrested for her connections with WikiLeaks.
Iceland's government had already protested against the United States' activities in Reykjavík. It has already been learned that the U.S. embassy located there is monitoring people who enter the immediate area, and try to stop people from taking pictures of their building, despite the fact that it is perfectly legal to do so.
According to the report from RUV, a private plane landed at Reykjavík airport in August 2011 and onboard were FBI agents who had flown directly from the U.S. to Iceland with the mission to investigate WikiLeaks operations in the country as a part of a larger investigation into the organization. The FBI agents reportedly contacted the head of the national Icelandic police, as well as the head prosecutor in an attempt to gain access to all available information on WikiLeaks.
When Home Secretary Ögmundur Jónasson found out about the FBI's visit, he summoned the FBI agents, who were then told that the Icelandic government wouldn't permit a foreign power running their own investigations in the country. Jónasson then ordered the FBI agents to return to the U.S. After a special meeting of the cabinet, Foreign minister Össur Skarphéðinsson was then charged with formally protesting against the United States' behavior.
"The FBI arrived in private planes and landed at the Reykjavík airport. According to my sources, which are highly reliable and which I have been able to corroborate, news of the visit reached Home Secretary Ögmundur Jónasson who reacted sharply, as it was unbelievably presumptuous to come to Iceland that way. According to my sources, Jónasson demanded that the FBI agents pack their bags, get back onboard, and leave the country. The matter was then brought before the cabinet and a formal protest was issued to US authorities."
"I can corroborate this took place in August 2011. Agents from the FBI arrived in Iceland. They'll have to answer for what their plans were. I can also corroborate that they wanted to get the cooperation of the national police and the national prosecutor's office."
The Federal Bureau of Intimidation got what they deserved. Few countries would have the bravery to kick them out. Bravo Iceland!
Yeah - it wasn't Mulder and Scully.
And expect the Wrath of J Edgar.