text:
\section{Introduction}\label{s:intro}
Path-interaction processes have been introduced by Norris, Rogers and Williams during the late 80s in~\cite{NRW}. Since this period, they have been an intensive research area. Under the name of Brownian Polymers, Durrett and Rogers~\cite{DR} studied a family of self-interacting diffusions, as a model for the shape of a growing polymer.  Denoting by $X_t$ the location of the end of the growing polymer at time $t$, the process $X$ satisfies a stochastic differential equation driven by a Brownian motion,  with a drift term depending on its own occupation measure.   One is then interested in finding the scale for which the process converges to a non trivial limit. Later, another model of growing polymer has been introduced by Bena\"im, Ledoux and Raimond~\cite{BLR}, for which the drift term depends on its own empirical measure. Namely, they have studied  the following process living in a compact
smooth connected Riemannian manifold $M$ without boundary:
\begin{equation*}
\mathrm{d}X_t = \sum_{i=1}^N F_i(X_t)\circ\mathrm{d}B_t^i - \int_M \nabla_x W(X_t,y)\mu_t(\mathrm{d}y)\mathrm{d}t,
\end{equation*}
where $W$ is a (smooth) interaction potential, $(B^1,\cdots,B^N)$
is a standard Brownian motion on $\mathbb{R}^N$, $\mu_t =\frac{1}{t} \int_0^t \delta_{X_s} \mathrm{d} s$ and the symbol
$\circ$ stands for the Stratonovich stochastic integration.
In the compact setting, they have shown that the asymptotic behaviour of the empirical measure of the process can be related to the analysis of some deterministic dynamical flow. Later, Bena\"im and Raimond~\cite{BR} gave sufficient conditions for the almost sure convergence of the empirical measure (again in the compact setting). More recently, Raimond~\cite{Ra} has generalized the previous study and has proved that for the solution of the SDE living on a compact manifold $$\mathrm{d} X_t = \mathrm{d} B_t - \frac{g(t)}{t} \int_0^t \nabla_x V(X_t, X_s) \mathrm{d} s \ \mathrm{d} t$$ unless $g$ is constant, the approximation of the empirical measure by a deterministic flow is no longer valid.

Similar questions have also been answered in the non-compact setting, that is $\bbR^d$. Chambeu and Kurtzmann~\cite{CK} have studied the ergodic behaviour of the self-interacting diffusion depending on the empirical mean of the process. They have proved, under some convexity assumptions (ensuring the non-explosion in finite time of the process), a convergence criterion for the diffusion solution to the SDE $$\mathrm{d} X_t = \mathrm{d} B_t - g(t) \nabla V\left(X_t - \frac{1}{t}\int_0^t X_s \mathrm{d} s \right)  \ \mathrm{d} t$$
where $g$ is a positive function.
This model could represent for instance the behaviour of some social insects, as ants who are marking their paths with the trails' pheromones. This paper shows in particular how difficult is the study of general self-interacting diffusions in non-compact spaces as in~\cite{AK}, driven by the generic equation $$\mathrm{d} X_t = \mathrm{d} B_t - \frac{1}{t} \int_0^t \nabla_x V(X_t, X_s) \mathrm{d} s \ \mathrm{d} t.$$ Nevertheless, if the interaction function $V$ is symmetric and  uniformly convex, then Kleptsyn and Kurtzmann~\cite{kk-ejp} obtained the limit-quotient ergodic theorem for the self-attracting diffusion. Moreover, they managed to obtain a speed of convergence. As the results of this former paper are essential for the present work, we will explain them more precisely in \S\ref{ss:speed}.
\medskip

Another problem related to this paper is the diffusion corresponding to McKean-Vlasov's PDE. This corresponds to the Markov process governed by the SDE
\begin{equation}
\mathrm{d} X_t = \mathrm{d} B_t - \nabla W * \nu_t (X_t)\mathrm{d} t
\end{equation}
where $\nu_t = \mathcal{L}(X_t)$, $W$ is a smooth convex potential and * stands for the convolution. The asymptotic behaviour of $X$ has been studied by various authors these last years, see for instance Cattiaux, Guillin and Malrieu~\cite{CGM}. It turns out that under some assumptions, the law $\nu_t$ converges to the (unique if $W$ is strictly convex) probability measure solution to the equation $\nu = \frac{1}{Z} e^{-2W*\nu}$ where $Z=Z(\nu)$ is the normalisation constant. In the latter paper, the authors use a particle system to prove both a convergence result (with convergence rate) and a deviation inequality for solutions of granular media equation when the interaction potential is uniformly convex at infinity. To this end, they use a uniform propagation of chaos property and a control in Wasserstein distance of solutions starting from different initial conditions.

A related question to this problem concerns the exit-times from domains of attraction for the following motion
\begin{equation}\label{eq:MKV}
\mathrm{d} X_t^\sigma = \sigma \mathrm{d} B_t - \nabla V(X_t^\sigma) \mathrm{d} t - \nabla W*\nu_t (X_t^\sigma) \mathrm{d} t
\end{equation}
where $V$ is a potential, * stands for the convolution, $\nu_t = \mathcal{L}(X_t^\sigma)$ and $\sigma >0$. This was addressed by Herrmann, Imkeller and Peithmann~\cite{HIP}, who exhibited a Kramers' type law for the particle's exit from the potential's domains of attraction and a large deviations principle for the self-stabilizing (also named McKean-Vlasov) diffusion. To get this, they reconstructed the Freidlin-Wentzell theory for the self-stabilizing
diffusion. More precisely, they established a large deviations principle with a good rate function. The exit-problem for the McKean-Vlasov diffusion has also been already studied recently, without using the Freidlin-Wentzell method. In~\cite{T2011f}, Tugaut has analysed the exit-problem (time and location) in convex landscapes, showing the same result as Herrmann, Imkeller and Peithmann, but without reconstructing the proofs of Freidlin and Wentzell. He has then generalised very recently his results in the case of double-wells landscape in~\cite{JOTP}. In~\cite{Tug14c,PTRF}, Tugaut did not use large deviations principle but a coupling method
between the time-homogeneous diffusion $$\mathrm{d} X_t = \sigma \mathrm{d} B_t - \nabla V (X_t) \mathrm{d} t - \nabla W (X_t - m) \mathrm{d} t$$ (where $m$ is the unique point at which the vector field $\nabla V$ equals 0) and the McKean-Vlasov diffusion so that the results on the exit-time of $X$ can be used for the exit-time of the self-stabilizing diffusion~\eqref{eq:MKV}.

 The present paper also deals with the exit-time problem of a specific diffusion, driven by the SDE
\begin{equation}\label{eq:sid}
\mathrm{d}X_t = \sigma \mathrm{d} B_t - \left( \nabla V (X_t) + \frac{1}{t}\int_0^t \nabla W( X_t -  X_s) \mathrm{d} s \right) \mathrm{d} t, \quad X_0 = x\in \bbR^d
\end{equation}
where $V,W$ are two potentials and $\sigma  >0$. We could adapt the techniques introduced by Herrmann, Imkeller and Peithmann but only in the case of a convex gradient $V$. Our aim is to generalize the study also to non-convex potentials. In the present work, we will solve the exit-problem (time and location) for the diffusion~\eqref{eq:sid}. Indeed, the exit-location will be easily obtained once we know the exit-time. The motivation for the study of such a diffusion is twofold. First, we wish to obtain the basin of attraction  when the diffusion converges (if we know the speed of convergence, or at least a nice upper bound of it). And more challenging, our main aim consists in improving the simulated annealing method (even if we need a concave interaction in that case). This paper is a first  step in this direction.

In the following, for readability issue, we will omit the $sigma-$exponent for the process $X$ as well as the empirical measure $\mu_t$. Nevertheless, the reader has to keep in mind that the process and $\mu_t$ depend on $\sigma$.

\subsection{Some useful notations}
As usual, we denote by $\mathcal{M}(\bbR^d)$ the space of signed
(bounded) Borel measures on $\bbR^d$ and by $\mathcal{P}(\bbR^d)$ its
subspace of probability measures. We will need the following measure
space:
\begin{equation}
\mathcal{M}(\bbR^d;P):= \{\mu \in \mathcal{M}(\bbR^d); \int_{\bbR^d} P(|y|) \,
\vert\mu\vert(\mathrm{d}y)< +\infty\},
\end{equation}
where $\vert\mu\vert$ is the variation of $\mu$ (that is
$\vert\mu\vert := \mu^{+} + \mu^{-}$ with $(\mu^{+},\mu^{-})$ the
Hahn-Jordan decomposition of $\mu$: $\mu = \mu^+ -\mu^-$) and $P$ is some polynomial. Belonging to this space will enable us to
always check the integrability of $P$ (and therefore of $V,W$ and their
derivatives thanks to the domination
condition~\eqref{domination}) with respect to the (random)
measures to be considered. We endow this space with the dual weighted supremum norm (or dual $P$-norm) defined for $\mu \in \mathcal{M}(\bbR^d;P)$ by
\begin{equation}
\vert \vert \mu\vert \vert_P := \sup_{\varphi\in \mathcal{C}(\bbR^d); \vert \varphi\vert
\leq P} \left\vert \int_{\bbR^d} \varphi\, \mathrm{d}\mu\right\vert = \int_{\bbR^d} P(|y|) \, |\mu|(\mathrm{d}y),
\end{equation}
where $\mathcal{C}(\bbR^d)$ is the set of continuous functions $\bbR^d \to \mathbb{R}$. Without any loss of generality, we suppose that $P(|x|)\ge 1$, so that $\|\mu\|_P \ge |\mu(\bbR^d)|$.
This norm makes $\mathcal{M}(\bbR^d;P)$ a Banach space. Next, we consider $\mathcal{P}(\bbR^d;P) =
\mathcal{M}(\bbR^d;P) \cap \mathcal{P}(\bbR^d)$. In the sequel, $(\cdot,\cdot)$ stands for the Euclidean
scalar product.

\begin{definition}
\label{stasta}
Let $d$ be any positive integer. Let $\mathcal{G}$ be a subset of $\mathbb{R}^{d}$ and let $U: \mathbb{R}^{d} \to \mathbb{R}^d$ be a vector field  satisfying some ``good assumptions''. For all $x\in\mathbb{R}^d$, we consider the dynamical system $\rho_t(x)=x+\int_0^tU\left(\rho_s(x)\right) \mathrm{d}{s}$. We say that the domain $\mathcal{G}$ is positively invariant for the flow generated by $U$ if the orbit $\left\{\rho_t(x)\,;\,t\in\mathbb{R}_+\right\}$ is included in $\mathcal{G}$ for all $x\in\mathcal{G}$.
\end{definition}

\subsection{Main results}

The precise assumptions on the potentials will be given later in Section~\ref{s:diffusion-linear}.

The goal of this paper consists in finding some precise upper and lower bounds for the exit-time of some positively invariant domain.
\begin{theorem}
\label{dell}
Let $\mathcal{D}$ be a domain that is positively invariant for the flow $x\mapsto -\nabla V(x) - \nabla W(x-m)$ and denote by $\tau$ the first time the process $X$ exits the domain $\mathcal{D}$.
Let  $H:=\inf_{x\in\partial\mathcal{D}}\left(V(x)+W(x-m)-V(m)\right)$ be the exit cost from $\mathcal{D}$. Then $\displaystyle\mathbb{P}-\lim_{\sigma\to0}\frac{\sigma^2}{2}\log(\tau)=H$ that is for any $\delta>0$, we have
\begin{equation}
\label{eq:dell}
\lim_{\sigma\to0}\mathbb{P}\left(\exp\left\{\frac{2}{\sigma^2}\left(H-\delta\right)\right\}\leq\tau\leq\exp\left\{\frac{2}{\sigma^2}\left(H+\delta\right)\right\}\right)=1\,.
\end{equation}
\end{theorem}

From Theorem \ref{dell}, we immediately obtain the classical statement on the exit-location.
\begin{cor}
\label{hphp}
Under the same assumptions as the ones of Theorem \ref{dell}, if $\mathcal{N}\subset\partial\mathcal{D}$ is such that $\inf_{z\in\mathcal{N}}\left(V(z)+W(z-m)-V(m)\right)>H$, then
\begin{equation}
\label{eq:hphp}
\lim_{\sigma\to0}\mathbb{P}\left(X_{\tau}\in\mathcal{N}\right)=0\,.
\end{equation}

\end{cor}

\subsection{Outline}
Our paper is divided in two parts. First, Section~\ref{s:diffusion-linear} is devoted to the study of the self-interacting diffusion and more specifically, we will explain the former results of Kleptsyn and Kurtzmann~\cite{kk-ejp} and, more precisely, how we adapt them in our context. In particular, we will show that the empirical measure of the studied process converges almost surely with a rate  upper bounded independently of $\sigma$. After that, we will prove our main result, that is Theorem~\ref{dell}, in Section~\ref{s:proof}. To this aim, we will show that our process $X$ is close to the solution of a given deterministic flow in~\S\ref{s:majoration}. We then prove in~\S\ref{s:leaving} that the probability of leaving a positively invariant domain before the empirical mean remains stuck in the ball of center $m$ and radius $\kappa$ vanishes as $\sigma$ goes to zero. Finally, a coupling permits to conclude the proof of the main theorem in~\S\ref{ss:conclusion} and we give the proof of the corollary in~\S\ref{s:hphp}.

\section{The self-interacting diffusion}\label{s:diffusion-linear}

 We assume that $V$ and $W$  satisfy the following set of hypotheses denoted by~\textbf{(H)}:
\begin{enumerate}
    \item[i)] (\textit{regularity and positivity}) $V \in \mathcal{C}^2(\mathbb{R}^d)$, $W\in
    \mathcal{C}^2(\mathbb{R}^d)$ and $V \geq 0, \; W \geq 0$;
    \item[ii)] (\textit{growth}) $V$ and $W$ have at most a polynomial growth: for some polynomial $P$, such that $P(|x|)\geq1$ for any $x\in\bbR^d$, we have $\forall x\in\bbR^d$:
    \begin{equation}\label{domination}
    \begin{aligned}
    |W(x)| +|\nabla W(x)| +  \|\nabla^2 W(x)\|
    + |V(x)| +|\nabla V(x)| + \|\nabla^2 V(x)\|\leq P(|x|)
    \end{aligned}
    \end{equation}
    and $ \inf  \nabla^2 V\ge \rho>0$, $\inf \nabla^2 W\ge \alpha>0$,
    \begin{equation}\label{eq:growth}
    \Delta V(x)\leq aV(x) \,\, \text{and} \,\, \lim_{\vert x\vert\rightarrow \infty} \frac{\vert\nabla V(x)\vert^2}{V(x)} = \infty.
   \end{equation}

	\item[iv)] (\textit{curvature}) $V$ and $W$ are uniformly strictly convex functions. $V$ has a unique minimum in $m$;
	\item[v)] (\textit{spherical symmetry}) $W(x)=G(|x|)$ for some function $G$ from $\bbR_+$ to $\bbR$.
\end{enumerate}
\begin{rem}
By the growth condition \eqref{eq:growth}, $\vert\nabla V\vert^2 - \Delta V$ is bounded by below.
\end{rem}
\begin{rem}
    Note, that without loss of generality we can choose polynomials $P$ to be of the form $P(|x|) = C(1 + |x|^k)$. Then the following property holds: there exists a constant $\gamma > 0$ such that $P(|x + y|) \leq \gamma(P(|x|) + P(|y|))$.
\end{rem}

Let us first show existence and uniqueness of the solution to the latter equation.
\begin{proposition}\label{p:existence-W} For any $x\in \mathbb{R}^d$, there exists a
unique global strong solution $(X_t,t\geq 0)$.
\end{proposition}
\begin{proof}
Local existence and uniqueness of the solution to \eqref{eq:sid} is standard (see for instance~\cite{DR}).
We only need to prove here that $X$
does not explode in a finite time. Let us introduce the increasing sequence of stopping times $\tau_0 = 0$ and $$\tau_n :=
\inf\left\{t\geq \tau_{n-1}; \mathcal{E}_{t}(X_t) + \int_0^t \left\vert\nabla
\mathcal{E}_{s}(X_s)\right\vert^2 \mathrm{d}s > n\right\}$$ where $\mathcal{E}_{t}(X_t) := V(X_t) + \frac{1}{t} \int_0^t W(X_t-  X_s) \mathrm{d} s$.
In order to show that the solution never explodes, we use the Lyapunov
functional $(x,t)\mapsto \mathcal{E}_t(x)$. As the process $(t,x)\mapsto\mathcal{E}_{\mu_t}(x)$ is of class $\mathcal{C}^2$ (in the space variable) and is a $\mathcal{C}^1$-semi-martingale
(in the time variable),
It\^o-Ventzell formula applied to
$(x,t)\mapsto\mathcal{E}_{t\wedge \tau_n}(x)$ implies
\begin{eqnarray}\label{eq:ito}
\mathcal{E}_{t\wedge \tau_n}(X_{t\wedge \tau_n}) = V(x)  + \int_0^{t\wedge \tau_n}
(\nabla \mathcal{E}_{s}(X_s), \mathrm{d}B_s) - \int_0^{t\wedge \tau_n} \left\vert\nabla \mathcal{E}_{s}(X_s)\right\vert^2 \mathrm{d}s\\
+ \frac{\sigma^2}{2}\int_0^{t\wedge \tau_n} \Delta \mathcal{E}_{s}(X_s)\mathrm{d}s
- \int_0^{t\wedge \tau_n}  \int_0^s  W(X_s-  X_u)\mathrm{d} u \frac{\mathrm{d}s}{s^2}.\notag
\end{eqnarray}
We note that $\int_0^{t\wedge \tau_n} (\nabla \mathcal{E}_{s}(X_s),
\mathrm{d}B_s)$ is a true martingale.
We then get
\begin{equation*}
\mathbb{E}\mathcal{E}_{t\wedge \tau_n}(X_{t\wedge \tau_n})
\leq V(x) + a \int_0^{t}
\mathbb{E}\mathcal{E}_{s\wedge \tau_n}(X_{s\wedge \tau_n})
\mathrm{d}s.
\end{equation*}
So, Gronwall's lemma leads to: $$\mathbb{E}V(X_{t\wedge \tau_n})\le \mathbb{E}\mathcal{E}_{t\wedge
\tau_n}(X_{t\wedge \tau_n}) \leq V(x)e^{at}.$$
As $\underset{\vert x\vert\rightarrow \infty}{\lim}V(x) =\infty$, the process $(X_t,t\geq 0)$ does not explode in a finite time and there exists a global strong solution.
\end{proof}

\begin{rem}
A large family of path-dependent process has been studied by Saporito, see for instance~\cite{yuri}. He proves, with his co-authors, existence and uniqueness of such processes. The difference with our process is that we normalize the occupation measure.
\end{rem}

\subsection{Speed of convergence}\label{ss:speed}
Our results are based on the paper of Kleptsyn and Kurtzmann~\cite{kk-ejp}. More precisely, they have proved the following
\begin{theorem}\cite[Theorem 1.6]{kk-ejp}\label{t:main-2}
Let $X$ be the solution to the equation \eqref{eq:sid} with $\sigma=\sqrt{2}$.

Suppose, that $V\in \mathcal{C}^2(\bbR^d)$ and $W\in \mathcal{C}^2(\bbR^d)$, satisfy \textbf{(H)} and either $V$ or $W$ is uniformly strictly convex that is there exists $C>0$ such that
$$
\forall x\in \bbR^d, \forall v\in \bbS^{d-1} , \, \left.\frac{\partial^2 V}{\partial v^2}\right|_{x}\ge C\quad \text{ or } \quad  \forall x\in \bbR^d, \forall v\in \bbS^{d-1},\ \left.\frac{\partial^2 W}{\partial v^2}\right|_{x}\ge C.
$$
Then there exists a unique density $\rho_{\infty}:\bbR^d\to\mathbb{R}_+$, such that almost surely
$$
\mu_t = \frac{1}{t}\int_0^t \delta_{X_s}\mathrm{d}s \xrightarrow[t\to+\infty]{*-weakly}
\rho_{\infty}(x) \, \mathrm{d}x.
$$
\end{theorem}
Moreover, if $V$ is symmetric with respect to some point $q$, then the corresponding density $\rho_\infty$ is also symmetric with respect to the same point $q$. Remark that the density $\rho_{\infty}$ is the same limit density as in the result of~\cite{CMV}, uniquely defined by the following property: $\rho_{\infty}$ is a positive function, proportional to~$e^{-(V+W*\rho_{\infty})}$.

 And what is more important, Kleptsyn and Kurtzmann obtained a speed of convergence in the following way. First, let us recall the definition of the Wasserstein distance.
\begin{definition}
  For $\mu_1,\mu_2 \in \mathcal{P}(\bbR^d;P)$, the quadratic Wasserstein distance is defined as $$\mathbb{W}_2(\mu_1,\mu_2) := \left(\inf \{ \mathbb{E}(|\xi_1-\xi_2|^2)\}\right)^{1/2},$$ where the infimum is taken over all the random variables such that $\{$law of $\xi_1\}= \mu_1$ and $\{$law of $\xi_2\}= \mu_2$. This corresponds to the minimal $L^2$-distance taken over all the couplings between $\mu_1$ and $\mu_2$.

  Similarly, the Wasserstein distance $\mathbb{W}_{2k}$ is defined as
  $$\mathbb{W}_{2k}(\mu_1,\mu_2) := \left(\inf \{ \mathbb{E}(|\xi_1-\xi_2|^{2k})\}\right)^{1/(2k)}.$$
\end{definition}

More precisely, in \cite[Theorem 1.12]{kk-ejp}, it is proved the existence of a constant $a>0$ such that almost surely, for $t$ large enough one has
\begin{equation*}
\mathbb{W}_2(\mu_t^c, \rho_\infty) = O(\exp\{-a\sqrt[2k+1]{\log t}\})\,,
\end{equation*}
where $2k$ is the degree of the polynomial $P$, $\mu_t^c$ is the translation of the empirical measure $\mu_t$ such that $\mathbb{E} (\mu_t^c)=0$ and $\mathbb{W}_2$ is the quadratic Wasserstein distance.

Of course, in the case $W(x) = \alpha \frac{|x|^2}{2}$, the polynomial $P$ corresponds to the growth of $V$, and we have to replace the Brownian motion by the rescaled Brownian motion $\sigma B_t$. So that the density $\rho_{\infty}$ is  uniquely defined by the following property: $\rho_{\infty}$ is a positive function, proportional to~$e^{-2(V+W*\rho_{\infty})/\sigma^2}$. Let us sketch the proof of these results and explain how $\sigma$ appears  here.

First, note that the empirical measure $\mu_t=\frac{1}{t}\int_0^t \delta_{X_s} \rm ds$ evolves very slowly. Indeed,  choose a deterministic sequence of times $T_n\to+\infty$, with
$T_n\gg T_{n+1}-T_{n}\gg 1$, and consider the behaviour of the
measures~$\mu_{T_n}$. As $T_n\gg T_{n+1}-T_n$, it is natural to
expect  that the
empirical measures $\mu_t$ on the interval $[T_n,T_{n+1}]$ almost
do not change and thus stay close to $\mu_{T_n}$. So we can approximate, on this
interval, the solution $X_t$ of~\eqref{eq:sid} with $\sigma=\sqrt{2}$ by the solution of the same equation
with $\mu_t\equiv \mu_{T_n}$:
$$
\mathrm{d}Y_t = \sigma\, \mathrm{d}B_t - (\nabla V+\nabla W*\mu_{T_n})(Y_t) \, \mathrm{d}t,\quad t\in [T_n,T_{n+1}],
$$
in other words, by a Brownian motion in a potential~$V+W*\mu_{T_n}$ that does not depend on time.

On the other hand, the series of general term $T_{n+1}-T_n$ increases. So, using Birkhoff Ergodic Theorem,
we see that the (normalized)
distribution $\mu_{[T_n,T_{n+1}]}$ of values of $X_t$ on these
intervals becomes (as $n$ increases) close to the equilibrium measures
$\Pi(\mu_{T_n})$ for a Brownian motion in the potential
$V+W*\mu_{T_n}$, where
$$
\Pi(\mu)(\mathrm{d}x) := \frac{1}{Z(\mu,\sigma)} e^{-2(V+W*\mu)(x) /\sigma^2} \,
\mathrm{d}x,\quad Z(\mu,\sigma):=\int_{\bbR^d} e^{-2(V+W*\mu)(x) /\sigma^2} \,
\mathrm{d}x.$$
As
$$
\mu_{T_{n+1}} = \frac{T_n}{T_{n+1}} \,\mu_{T_n} +
\frac{T_{n+1}-T_n}{T_{n+1}} \,\mu_{[T_n,T_{n+1}]},
$$
 we then have
$$
\mu_{T_{n+1}} \approx \frac{T_n}{T_{n+1}} \,\mu_{T_n} +
\frac{T_{n+1}-T_n}{T_{n+1}} \,\Pi(\mu_{T_n}) = \mu_{T_n} +
\frac{T_{n+1}-T_n}{T_{n+1}} (\Pi(\mu_{T_n})-\mu_{T_n}),
$$
and
$$
\frac{\mu_{T_{n+1}}-\mu_{T_n}}{T_{n+1}-T_n} \approx
\frac{1}{T_{n+1}}(\Pi(\mu_{T_n})-\mu_{T_n}).
$$
This motivates Kleptsyn and Kurtzmann to approximate the behaviour of the measures $\mu_t$
by trajectories of the flow (on the infinite-dimensional space of
measures)
\begin{equation}\label{eq:9-3/4}
\dot{\mu}=\frac{1}{t}(\Pi(\mu)-\mu),
\end{equation}
or after a logarithmic change of variable $\theta=\log t$,
\begin{equation}\label{eq:flow}
\mu'=\Pi(\mu)-\mu.
\end{equation}

Indeed, choose an appropriate interval $[T_n, T_{n+1})$. On this interval, fix the empirical measure $\mu_t$ at $\mu_{T_n}$. Then construct a new process $Y$, coupled with $X$ (the coupling is such that $X$ and $Y$ are driven by the same Brownian motion), such that for all $t\in [T_n, T_{n+1})$, we have $$\mathrm{d}Y_t = \sigma \mathrm{d}B_t - (\nabla V(Y_t) + \nabla W*\mu_{T_n}(Y_t)) \mathrm{d}t.$$
This new process has  two advantages. First, it is Markovian (and its invariant probability measure is $\Pi(\mu_{T_n})(\mathrm{d}x) = \frac{1}{Z} e^{-2(V+W*\mu_{T_n})(x)/\sigma^2} \mathrm{d}x$), and so is easier than $X$ to study. Second, its evolution is very close to the evolution of the desired $X$. Indeed, we will use $Y$ to prove that the transport distance between the empirical measure on $[T_n, T_{n+1}]$, denoted by $\mu_{[T_n, T_{n+1}]}$ in~\cite[Proposition 3.2]{kk-ejp}, and the probability measure $\Pi(\mu_{T_n})$ (both measures being centered in $c_{T_n}$) is controlled by $T_n^{-\frac{1}{3}\min (8C_W,1/5d)}$ and so, this distance vanishes as $n\to +\infty$. This has been done in \cite[\S 3.1.1]{kk-ejp}.\\
After that, remark that if a.s. the empirical measure $\mu_t$ converges weakly* to $\mu_\infty$, then for $t$ large enough, the process $X$ shall be very close to $Z$, defined by $$\mathrm{d}Z_t = \sigma \mathrm{d}B_t - (\nabla V+\nabla W*\mu_{\infty})(Z_t) \mathrm{d}t.$$
The process $Z$ is obviously Markovian and the ergodic theorem can be applied:
$$\frac{1}{t} \int_0^t \delta_{Z_s} \mathrm{d}s \underset{t\to +\infty}{\longrightarrow} \Pi(\mu_\infty)\quad \text{a.s.}$$ for the weak* convergence of measures. So when the limit $\mu_\infty$ exists, it satisfies
\begin{equation}\label{eq:mu_infty_and_Pi}
\mu_\infty = \Pi(\mu_\infty)\,.
\end{equation}

This explains the idea of introducing the dynamical system \mbox{$\dot{\mu} = \Pi(\mu)-\mu$} (after the time-shift $t\mapsto e^t$ in order to work with a time-homogeneous system) defined on the set of probability measures that are integrable for the polynomial $P$. Note that, instead of considering the latter dynamical system, Kleptsyn and Kurtzmann work with its discretized version, with the knots chosen at the moments $T_n$. They then prove, in~\cite[Proposition 3.5]{kk-ejp}, that the transport distance between the deterministic trajectory induced by the smoothened (discrete) dynamical system and the (centered) random trajectory $\mu_{T_n}$ is controlled and decreases to 0. This has been done in~\cite[\S 3.1.2]{kk-ejp}.\\
Next, it remains to show that the free energy  between this (centered) deterministic trajectory and the set of translates of $\rho_\infty$ goes to 0. We recall that  for the dynamics in presence of an exterior potential $V$, the free energy function is
$$
\mathcal{F}_{V,W}(\mu) := -\frac{\sigma^2}{2} \mathcal{H}(\mu) + \int_{\bbR^d} V(x) \mu(x)\, \mathrm{d}x \\
+ \frac{1}{2} \iint_{\bbR^d\times\bbR^d} \mu(x) W(x-y) \mu(y)\, \mathrm{d}x\, \mathrm{d}y.
$$
and consider $\mathcal{F}_{V+W*\mu}= -\frac{\sigma^2}{2} \mathcal{H}(\mu) + \int_{\bbR^d} (V(x)+W*\mu (x)) \mu(x)\, \mathrm{d}x$ for the energy of ``small parts'', where the entropy of the measure $\mu$ is
\begin{equation}\label{eq:entropy}
\mathcal{H}(\mu) := -\int_{\bbR^d} \mu(x) \log\mu(x) \mathrm{d}x.
\end{equation}

As the free energy is controlled by the quadratic Wasserstein distance $\mathbb{W}_2$, this implies that the transport distance between the two previous quantities decreases, as asserted in \cite[Proposition 3.6]{kk-ejp}.\\
To conclude, it remains to put all the pieces together and use the triangle inequality: $\mathbb{W}_2 (\mu_t^c,\rho_\infty)$ is upper bounded by the sum of three distances, involving the flow  $\Phi_n$ induced by the discretization of the dynamical system $\dot{\mu} = \Pi(\mu) - \mu$ on the interval $[T_n, T_{n+1})$, for $n$ large enough. The first term of the summation bound will be $\mathbb{W}_2 (\mu_t^c, \Pi(\mu_{T_n}^c))$, the second one $\mathbb{W}_2(\Pi(\mu_{T_n}^c), \Phi_n^n(\mu_{T_n}^c))$ and the third one $\mathbb{W}_2 (\Phi_n^n(\mu_{T_n}^c),\rho_\infty)$.

Finally, the previous decrease estimates will allow Kleptsyn and Kurtzmann to show the convergence of the center, after having made the appropriate choice $T_n = n^{3/2}$.

\subsection{The speed of convergence for the solution of \texorpdfstring{\eqref{eq:sid}}{(\ref{eq:sid})}}
For this paper, the corresponding result of \cite[Theorem 1.12]{kk-ejp} is the following
\begin{theorem}
\label{thm:KK12}
Let $X$ be the solution to the equation \eqref{eq:sid}. There exists a constant $a>0$ such that almost surely, we have for $t$ large enough $\mathbb{W}_{2k}(\mu_t, \rho_\infty) = O(\exp\{-a\sqrt[2k+1]{\log t}\})$, where $2k$ is the degree of the polynomial $P$, $\mu_t$ is  the empirical measure and $\mathbb{W}_{2k}$ is the $2k$-Wasserstein distance, that is the minimal $L^{2k}$-distance taken over all the couplings between $\mu_t$ and $\rho_\infty$ that is the unique density proportional to $e^{-\frac{2}{\sigma^2}(V+W*\rho_\infty)}$.
\end{theorem}
To prove this result, we mimic the proof of \cite[Theorem 1.12]{kk-ejp}, and show that the speed of convergence is less than the one of the case corresponding to $\sigma =1$. We will not reproduce it here. We will only show how we handle this inequality for the following particular result, as it is representative of the difficulty and shows how $\sigma$ appears in the calculation.

Let us prove for instance the exponential decrease for the centered measure $\Pi(\mu)$ and show that, as $\sigma^2 \ll 1$, we can obtain a lower bound of the speed of convergence that does not depend on $\sigma$. We have seen previously that the centered $\mu^c$ has the same asymptotic behaviour as $\Pi(\mu)(\cdot + c_\mu)$, up to a time-scale. An important step to prove the latter result consists in estimating the behaviour of the centered measures $\mu_t^c$. More precisely, one has to prove that the tail of these measures are exponentially decreasing. This is why we have to introduce the following sets:
 \begin{definition}\label{eq:K-alpha-C}
Let $\alpha,C>0$ be given. Define
\begin{eqnarray}
K_{\alpha,C}^0 &:=& \{\mu \in \mathcal{P}(\bbR^d); \quad \forall R>0, \, \mu(\{y; |y|>R\})<Ce^{-\alpha R}\},\\
K_{\alpha,C} &:=& \{\mu \in \mathcal{P}(\bbR^d); \quad \mu^c \in K_{\alpha,C}^0 \}.
\end{eqnarray}
\end{definition}
We can now prove the exponential decrease of $\Pi(\mu)(\cdot + c_\mu)$.
\begin{proposition} \label{prop:exp-decrease}
There exist $C_W,C_\Pi>0$, two constants  independent of $\sigma$, such that for all $\mu \in \mathcal{P}(\mathbb{R};P)$, we have $\Pi(\mu)(\cdot +c_\mu) \in K^0_{C_W,C_\Pi}$ where $c_\mu$ is defined by the equation $(\nabla V + \nabla W *\mu )(c_\mu) = 0$.
\end{proposition}
\begin{proof}
Let us fix $R>0$. Note first that, imposing a condition $C_\Pi\ge e^{2C_W}$, we can restrict ourselves only to $R\ge 2$: for $R<2$, the estimate is obvious.

The measure $\Pi(\mu)$ has the density $\frac{1}{Z(\mu,\sigma)} e^{-2(V+W*\mu)(x)/\sigma^2}$. To avoid working with the normalization constant $Z(\mu,\sigma)$, we will prove a stronger inequality, that is
\begin{equation}\label{eq:exp-dec}
\Pi(\mu)(|x-c_\mu|\ge R)\le
C_\Pi e^{-C R}\cdot \Pi(\mu)(|x-c_\mu|\le 2),
\end{equation}
which is equivalent to $$\int_{|x-c_\mu|\ge R} e^{-2(V+W*\mu)(x) /\sigma^2} \mathrm{d}x \le C_\Pi e^{-C R} \int_{|x-c_\mu|\le 2} e^{-2(V+W*\mu)(x)/\sigma^2} \mathrm{d}x.$$
We use the polar coordinates, centered at the center $c_\mu$, and so we want to prove that
\begin{eqnarray*}
& & \int_{\mathbb{S}^{d-1}} \int_R^{+\infty} e^{-2(V+W*\mu) (c_\mu + \lambda v) / \sigma^2}\lambda^{d-1}\mathrm{d}\lambda \mathrm{d}v \\
& & \quad \le   C_\Pi e^{-C R} \int_{\mathbb{S}^{d-1}} \int_0^2 e^{-2(V+W*\mu) (c_\mu + \lambda v)/\sigma^2}\lambda^{d-1}\mathrm{d}\lambda \mathrm{d}v.
\end{eqnarray*}
It suffices to prove such an inequality ``directionwise'': for all $v\in \mathbb{S}^{d-1}$, for all $R\ge 2$
$$\int_R^{+\infty} e^{-2(V+W*\mu) (c_\mu + \lambda v) /\sigma^2}\lambda^{d-1}\mathrm{d}\lambda \le C_\Pi e^{-C R} \int_0^2 e^{-2(V+W*\mu) (c_\mu + \lambda v)/\sigma^2}\lambda^{d-1}\mathrm{d}\lambda.$$ But from the uniform convexity of $V$ and $W$ and the definition of the center, the function $f(\lambda) = 2(V+ W*\mu) (c_\mu + \lambda v)$ satisfies $f'(0) =0$ and $\forall r>0$, $f''(r)\ge C$. Hence, $f$ is monotone increasing on $[0,+\infty)$, and in particular,
\begin{equation}\label{eq:coord-polaires}
\int_0^2 e^{-f(\lambda)/\sigma^2} \lambda^{d-1} \mathrm{d}\lambda \ge e^{-f(2)/\sigma^2} \int_0^2 \lambda^{d-1}\mathrm{d}\lambda =: C_1 e^{-f(2)/\sigma^2}.
\end{equation}
On the other hand, for all $\lambda \ge 2$, $f'(\lambda) \ge f'(2)\ge 2C$, and thus $f(\lambda)\ge 2C (\lambda -2) + f(2)$. Hence, as $\sigma^2\ll 1$, we have
\begin{eqnarray}
\int_R^{+\infty} e^{-f(\lambda)/\sigma^2} \lambda^{d-1} \mathrm{d}\lambda & \le & e^{-f(2)/\sigma^2} \int_R^{+\infty} \lambda^{d-1} e^{-2C(\lambda-2)/\sigma^2} \mathrm{d}\lambda \notag\\
& \le & e^{-\frac{f(2)}{\sigma^2}} \int_R^{+\infty} \lambda^{d-1} e^{-2C(\lambda-2)} \mathrm{d}\lambda.
\end{eqnarray}
So finally, using that $\sigma\ll 1$
\begin{eqnarray}\label{eq:coord-polaires-2}
\int_R^{+\infty} e^{-f(\lambda)/\sigma^2} \lambda^{d-1} \mathrm{d}\lambda \le C_2 R^{d-1} e^{-2C R} \cdot e^{-\frac{f(2)}{\sigma^2}} \le C_3 e^{-CR} \cdot e^{-\frac{f(2)}{\sigma^2}}.
\end{eqnarray}
Comparing~\eqref{eq:coord-polaires} and~\eqref{eq:coord-polaires-2}, we obtain the desired exponential decrease. Indeed, as $\sigma$ only appears in the exponential under the form $\exp (-const /\sigma^2)$, we can upper bound the previous quantity by $e^{-const}$, so that the constant $C_W$ is upper bounded by a constant that is independent of $\sigma$. By an abuse of notation, we call again this constant $C_W$.
\end{proof}
\begin{rem}
This result corresponds to \cite[Proposition 2.9]{kk-ejp} for the case $\sigma =1$.
\end{rem}

\subsection{Small-noise limit}
In the following, we remind the reader that we don't emphasize the dependence on $\sigma$, but it will appear everywhere in the computations.

As was mentioned above, the invariant probability measure of self-interacting diffusion is the unique solution to the equation
\begin{equation*}
    \rho_\infty = \Pi(\rho_\infty),
\end{equation*}
where $\Pi(\mu)(x) =  e^{-2(V+W*\mu)(x) /\sigma^2} / \int e^{-2(V+W*\mu)(z) /\sigma^2} \mathrm{d} z$. The same invariant probability measure appears in the self-stabilizing diffusion, small-noise limit of which was studied in \cite{HT-EJP}. There, authors studied the case of double-wells potentials which is more general then our diffusion. In this paper the result, that can be transformed in our context as following, was proved. If the moments of invariant probability measures $\rho_\infty$ are uniformly bounded with respect to $\sigma$, then $\delta_m$ is the weak-* limit of $\rho_\infty$ with $\sigma \to 0$ a.s. Note, that indeed, moments of $\rho_t$ are uniformly bounded, since $\mu_t \in K_{\alpha, C}$ for any $t > 0$ and some $\alpha, C$ that do not depend on $\sigma$.

Thus, consider the following deterministic time, representing the time of stabilization of the occupation measure, if it occurs, around its supposed limit $\delta_m$:
\begin{equation}
\label{def:Tkappa}
T_\kappa(\sigma):=\inf\left\{t_0\geq0\,\,:\,\,\forall t\geq t_0,\,\mathbb{E}\left(\mathbb{W}_{2k}\left(\mu_t; \delta_m\right)\right)\leq\kappa\right\}\,.
\end{equation}

First, let us discuss why expectation $\mathbb{E} \left(\mathbb{W}_{2k}\left(\mu_t; \delta_m\right)\right)$ exists in the first place. To show that, we use the fact that $\mu_t \in K_{\alpha, C}$ almost surely and get
\begin{equation*}
    \mathbb{W}_{2k}\left(\mu_t; \delta_m\right) \leq \left( 2^{2k - 1} \int |x|^{2k} \mu_t(\mathrm{d} x) + 2^{2k - 1} |m|^{2k}\right)^{1/(2k)} \leq \text{Const},
\end{equation*}
where the last constant depends only on $\alpha, C, m$ and $k$. Therefore, since the random variable is bounded by a constant almost surely, expectation exists.

Second, let us show that the definition of the time $T_\kappa(\sigma)$ makes sense. Indeed,
\begin{equation*}
    \mathbb{E}\mathbb{W}_{2k}\left(\mu_t; \delta_m\right) \leq \mathbb{E}\mathbb{W}_{2k}\left(\mu_t; \rho_\infty\right) + \mathbb{E}\mathbb{W}_{2k}\left(\rho_\infty; \delta_m \right) \xrightarrow[\substack{t \to \infty \\ \sigma \to 0}]{} 0,
\end{equation*}
where the limit is not just iterated, but holds for the pair $(t, \sigma)$, since the speed of convergence of $\mu_t$ towards $\rho_{\infty}$ in time does not depend on $\sigma$, which was shown in Theorem \ref{thm:KK12}. Therefore, for any $\kappa > 0$ we can find $\sigma_0$ small enough and $t_0$ big enough such that $T_\kappa(\sigma) < T_\kappa < \infty$ for any $\sigma < \sigma_0$, which does not only prove existence and finiteness of $T_\kappa(\sigma)$, but also its uniformness with respect to $\sigma$.

\section{Proof of Theorem \ref{dell}} \label{s:proof}
In this section, we prove our main result. First, we show that the process $X$ solution to \eqref{eq:sid} is close to the solution of a deterministic flow $(\psi_t)_{t\ge 0}$ in~\S\ref{s:majoration}. Using that, we prove in~\S\ref{s:leaving} that the probability of leaving a positively invariant domain before the empirical mean remains stuck in the ball of center $m$ and radius $\kappa$ vanishes as $\sigma$ goes to zero. Then, we consider the coupling between the studied diffusion and the one where the empirical measure is frozen to $\delta_m$ and we show that these diffusions are close in~\S\ref{ss:coupling}. We conclude the proof in~\S\ref{ss:conclusion}.

\subsection{Upper bound}\label{s:majoration}
We remind the reader that in this work, the noise vanishes. Consequently, it is natural to introduce the deterministic flow $(\psi_t)_t$ defined by the following
\begin{equation}
\dot{\psi}_t = -\nabla V(\psi_t) -\frac{1}{t}\int_0^t \nabla W(\psi_t-\psi_s ) \mathrm{d} s, \quad \quad \psi_0 = x_0.
\end{equation}
We will show that $X_t$ and $\psi_t$ are uniformly close while the noise goes to zero. Namely
\begin{proposition}
For any $\xi>0$ and for any $T>0$, we have:
\begin{equation}
\label{apple}
\lim_{\sigma\to 0}\mathbb{P}\left(\sup_{t\in [0;T]}\left|\left|X_t-\psi_t(x_0)\right|\right|^2>\xi\right)=0\,.
\end{equation}
\end{proposition}

 \begin{proof}

First of all, we fix some $\xi$ and introduce the following stopping time $\mathcal{T} := \inf\{t: |X_t^\sigma - \psi_t|^2 \geq \xi\}$. We apply Itô formula and get the following result, for  $\omega \in \{\mathcal{T} > t\}$ (the choice of this event will be clear further) :
\begin{equation*}\label{eq:X_t-psi_t}
\begin{aligned}
    |X_t - \psi_t|^2 &= 2\int_{0}^t (X_s - \psi_s, \mathrm{d}{X_s}-\mathrm{d}{\psi_s}) + d \sigma^2 t \\
    & \leq d\sigma^2t -2C_V \int_0^t |X_s - \psi_s|^2 \mathrm{d} s + \sigma \int_0^t (X_s - \psi_s , \mathrm{d} B_s) \\
    & \quad - \int_0^t\frac{1}{s} \int_0^s (X_s - \psi_s, \nabla W(X_s - X_z) - \nabla W(\psi_s - \psi_z))\mathrm{d} z\mathrm{d} s.
 \end{aligned}
\end{equation*}
Let $\text{Lip}_{\nabla W}^{K^\prime}$ be a Lipschitz constant of $\nabla W$ inside the following compact\\ $K^\prime:= \{x: |x - \psi_t|^2 \leq \xi \text{, for some } t > 0\}$ (due to our assumptions, this set is indeed a compact at least for small $\xi$, which we can decrease without loss of generality), and $C_V$ is the convexity constant of $V$. We thus have
\begin{eqnarray}\label{eq:X_t-psi_tbis}
    |X_t - \psi_t|^2
    & \leq & d\sigma^2 t - 2C_V\int_0^t |X_s - \psi_s|^2\mathrm{d} s + \sigma \int_0^t(X_s - \psi_s , \mathrm{d} B_s)\notag \\
    &+& \text{Lip}_{\nabla W}^{K^\prime}\int_0^t\frac{1}{s}\int_0^s \big(|X_s - \psi_s|^2 + |X_s - \psi_s|\cdot|X_z - \psi_z|\big) \mathrm{d} z\notag \\
    & \leq & d\sigma^2 t - 2C_V\int_0^t |X_s - \psi_s|^2\mathrm{d} s + \sigma \int_0^t(X_s - \psi_s , \mathrm{d} B_s) \\
    &+ & \frac{\text{Lip}_{\nabla W}^{K^\prime}}{2}\int_0^t\frac{1}{s}\int_0^s \big(3|X_s - \psi_s|^2 + |X_z - \psi_z|^2\big) \mathrm{d} z.\notag
\end{eqnarray}
Note then, that by the BDG inequality we get for some constant $C > 0$:
\begin{equation*}
\begin{aligned}
    \mathbb{E} \left(\sup_{[0,t \wedge \mathcal{T}]} \left|\sigma \int_0^s (X_z-\psi_z, \mathrm{d} B_z )\right| \right) &\le C \sigma^2 \mathbb{E} \sqrt{\int_0^{t\wedge \mathcal{T}} |X_s - \psi_s|^2 \mathrm{d} s}\\
    & \le C \sigma^2 \sqrt{\int_0^t \mathbb{E} \left(\sup_{z \in [0, s \wedge \mathcal{T}]}(|X_z - \psi_z|^2)\right) \mathrm{d} s}.
\end{aligned}
\end{equation*}

Let us consider the following random variable $\sup_{s \in [0; t \wedge \mathcal{T}]} |X_s - \psi_s|^2$. The fact that we consider the supremum before time $t \wedge \mathcal{T}$ gives us that for any $\omega$ we consider only such $s$, that $s \leq \mathcal{T}(\omega)$, which in turn means that we can apply estimation \eqref{eq:X_t-psi_tbis} for any $s \in [0, t\wedge \mathcal{T}]$. We also remind that $t \leq T$ and derive:
\begin{equation*}
\begin{aligned}
    \mathbb{E} \left(\sup_{s \in [0; t \wedge \mathcal{T}]} |X_s - \psi_s|^2\right) &\leq d\sigma^2T+ C \sigma^2 \sqrt{\int_0^t \mathbb{E} \left(\sup_{z \in [0 , s \wedge \mathcal{T}]} (|X_z - \psi_z|^2)\right) \mathrm{d} z} \\
    &\quad + 2 \text{Lip}_{\nabla W}^{K^\prime} \int_0^t \mathbb{E} \left(\sup_{z \in [0, s \wedge \mathcal{T}]} |X_z - \psi_z|^2\right) \mathrm{d} s \\
    &\leq  d\sigma^2T + \frac{C \sigma^2}{2} \left[1 +  T\mathbb{E}\left( \sup_{s \in [0 , t \wedge \mathcal{T}]} (|X_s - \psi_s|^2)\right)  \right] \\
    & \quad + 2 \text{Lip}_{\nabla W}^{K^\prime} \int_0^t \mathbb{E} \left(\sup_{z \in [0, s \wedge \mathcal{T}]} |X_z - \psi_z|^2\right) \mathrm{d} s,
\end{aligned}
\end{equation*}
where in the last inequality we used $\sqrt{x} \leq (1 + x)/2$. Now, if we denote\\ $u_t := \mathbb{E} \left(\sup_{s \in [0; t \wedge \mathcal{T}]} |X_s - \psi_s|^2\right)$, we have
\begin{equation*}
    u_t \leq \frac{1}{1 - C T \sigma^2/2} \left(\frac{2Td + C}{2}\sigma^2 + 2 \text{Lip}_{\nabla W}^{K^\prime} \int_0^t u_s \mathrm{d} s \right),
\end{equation*}
for small enough $\sigma$ (such that $1 - C T \sigma^2/2 > 0$). Thus, using Gr\"onwall lemma, we get
\begin{equation}\label{eq:ut_bound}
    u_t \leq \frac{(2Td + C) \sigma^2}{2(1 - C T \sigma^2/2)}\exp\left\{\frac{\text{Lip}_{\nabla W}^{K^\prime} }{1 - C T \sigma^2/2} T\right\} = O(\sigma^2).
\end{equation}
This in particular means, that $\mathbb{E}\left( \sup_{s \in [0; T \wedge \mathcal{T}]} |X_s - \psi_s|^2 \right) \leq O(\sigma^2)$. Nevertheless, to show the necessary result, we have to get rid of the stopping time $\mathcal{T}$ in the previous equation. It is sufficient to show, that $\mathbb{P} (\mathcal{T} \leq T) \xrightarrow[\sigma \to 0]{} 0$.

Indeed, by its definition, $\mathcal{T}$ is the first time when the difference $|X_t - \psi_t|^2$ reaches $\xi$. But under the assumption $\mathcal{T} \leq T$ and due to \eqref{eq:ut_bound}, by decreasing $\sigma$ we can control $|X_t - \psi_t|^2$ and make it small enough, such that $|X_{\mathcal{T}} - \psi_{\mathcal{T}}|^2 < \xi$ (in some sense), which contradicts the definition of $\mathcal{T}$. Rigorously,
\begin{equation*}
    \mathcal{T} < T \Rightarrow \sup_{[0, \mathcal{T}\wedge T]} |X_s - \psi_s|^2 = \sup_{[0, \mathcal{T}]} |X_s - \psi_s|^2 \ge \xi.
\end{equation*}
Thereby,
\begin{equation*}
    \mathbb{P} (\mathcal{T} < T) \leq \mathbb{P}(\sup_{[0, \mathcal{T}\wedge T]} |X_s - \psi_s|^2 \ge \xi) \leq O(\sigma^2),
\end{equation*}
by Markov inequality.

To conclude the proof of the Proposition, we consider
\begin{equation*}
\begin{aligned}
    \mathbb{P}\left(\sup_{t\in [0;T]}|X_t-\psi_t(x_0)|^2>\xi\right) & \leq \mathbb{P}\left(\sup_{t\in [0;T]}|X_t-\psi_t(x_0)|^2>\xi, \mathcal{T} > T\right) \\
    & \quad\quad + \mathbb{P} \left( \mathcal{T} \leq T \right) \\
    & \leq \mathbb{P}\left(\sup_{t\in [0;T \wedge \mathcal{T}]}|X_t-\psi_t(x_0)|^2>\xi \right) + O(\sigma^2) \\
    & \leq O(\sigma^2),
\end{aligned}
\end{equation*}
by Markov inequality and \eqref{eq:ut_bound}, which completes the proof.
\end{proof}

\subsection{Probability of leaving before \texorpdfstring{$T_{\kappa}(\sigma)$}{Tkappasigma}}\label{s:leaving}

Remind that we denoted in~\eqref{def:Tkappa} by $T_{\kappa}(\sigma)$ the first time at which the expectation of the $2k$-Wasserstein distance between the occupation measure of the process and $\delta_m$ is smaller than $\kappa$. By $\mathbb{B}(m;\kappa)$, we denote the ball of center $\delta_m$ and radius $\kappa$ for $\mathbb{W}_{2k}$.
\begin{proposition}
\label{craie}
We put $\tau:=\inf\{t\geq0\,\,:\,\,X_t\notin\mathcal{D}\}$ where $\mathcal{D}$ is a domain that is positively invariant for the flow $x\mapsto-\nabla V(x)-\nabla W(x-m)$. For any $\kappa>0$,
\begin{equation}
\label{apple2}
\lim_{\sigma\to0}\mathbb{P}\left(\tau\leq T_\kappa(\sigma)\right)=0\,.
\end{equation}
\end{proposition}

\begin{proof}
First, we remind that if $\sigma$ is small enough, $T_\kappa(\sigma)\leq T_\kappa$ where $T_\kappa$ does not depend on $\sigma$. Moreover, $T_\kappa$ and $T_\kappa(\sigma)$ are deterministic. As a consequence, we have
\begin{equation*}
\mathbb{P}\left(\tau\leq T_\kappa(\sigma)\right)\leq\mathbb{P}\left(\tau\leq T_\kappa\right)\,.
\end{equation*}

We now prove that for any $T>0$, $\mathbb{P}\left(\tau\leq T\right)\longrightarrow0$ as $\sigma$ goes to $0$.
According to the assumption $\left\{\psi_t(x_0)\,;\,0\leq t\leq T\right\}\subset\mathcal{D}$, we know (since $\mathcal{D}$ is an open set) that there exists $\epsilon>0$ such that $\mathbb{B}\left(0;\epsilon\right)+\left\{\psi_t(x_0)\,;\,0\leq t\leq T\right\}\subset\mathcal{D}$.

Now, on the event $\left\{\tau\leq T\right\}$, we deduce that $\sup_{t\in[0;T]}\left|\left|X_t-\psi_t(x_0)\right|\right|^2>\epsilon^2$. As a consequence:
\begin{equation*}
\mathbb{P}\left(\tau\leq T\right)\leq\mathbb{P}\left(\sup_{t\in[0;T]}\left|\left|X_t-\psi_t(x_0)\right|\right|^2>\epsilon^2\right)\,,
\end{equation*}
which goes to $0$ as $\sigma$ goes to $0$, thanks to \eqref{apple}.
This concludes the proof.
\end{proof}

\subsection{Coupling for \texorpdfstring{$t\geq T_k(\sigma)$}{t>=Tkappasigma}}\label{ss:coupling}

In \cite{Tug14c,JOTP}, Tugaut has proved the Kramers' type law for the exit-time. He has used a coupling between the diffusion of interest ($X$ here) and another diffusion that is expected to be close from $X$ if the time is sufficiently large. The main difficulty with the considered self-stabilizing diffusion is in fact that we do not have a uniform (with respect to the time) control of the law.

Here, we have proved that the nonlinear quantity appearing in the equation (that is $\frac{1}{t}\int_0^t \delta_{X_s} \mathrm{d} s$) remains stuck - with high probability - in a small ball (for $\mathbb{W}_{2k}$) of center $\delta_m$ and radius $\kappa$ for any $t\geq T_\kappa(\sigma)$. The idea is thus to replace $\frac{1}{t}\int_0^t \delta_{X_s} \mathrm{d} s$ by $\delta_m$ and to compare the new diffusion with the self-interacting one.

In other words, we consider the diffusion
\begin{equation}
\label{sandra}
Y_t=X_{T_\kappa(\sigma)}+\sigma\left(B_t-B_{T_{\kappa}(\sigma)}\right)-\int_{T_\kappa(\sigma)}^t\nabla V\left(Y_s\right)\mathrm{d} s-\int_{T_\kappa(\sigma)}^t \nabla W(Y_s-m)\mathrm{d} s\,,
\end{equation}
for any $t\geq T_\kappa(\sigma)$ and $Y_t=X_t$ if $t\leq T_\kappa(\sigma)$.

\begin{proposition}
\label{prop:coupling}
For any $\xi>0$, if $\kappa$ is small enough, we have
\begin{equation}
\label{eq:prop:coupling}
\limsup_{\sigma\to0}\mathbb{P}\left(\sup_{T_\kappa(\sigma)\leq t\leq\exp\left[\frac{2H+10}{\sigma^2}\right]}\left|X_t-Y_t\right|\geq\xi\right)\leq\sqrt{\kappa}\,.
\end{equation}
\end{proposition}

\begin{proof}
For any $t\geq T_\kappa(\sigma)$, we have
\begin{align*}
&\mathrm{d}\left|X_t-Y_t\right|^2\\
&=-2\left( X_t-Y_t\,;\,\left(\nabla V\left(X_t\right)+\nabla W\ast\mu_t(X_t) \right)-\left(\nabla V\left(Y_t\right)+\nabla W(Y_t-m)\right)\right)\mathrm{d} t\,,
\end{align*}
with the empirical measure $\mu_t:=\frac{1}{t}\int_0^t\delta_{X_s}\mathrm{d} s$. Let us define $W_m(x):=V(x)+ W(x-m)$ and $W_{\mu_t}(x):=V(x)+W\ast\mu_t(x)$.
We thus have
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d} t}\left|X_t-Y_t\right|^2=&-2\left( X_t-Y_t\,;\,\nabla W_{\mu_t}\left(X_t\right)-\nabla W_{\mu_t}\left(Y_t\right)\right)\\
&+2\left( X_t-Y_t\,;\,\nabla W(Y_t-m)-\nabla W\ast\mu_t(Y_t)\right)\,.
\end{align*}
However, $\nabla^2W_{\mu_t}=\nabla^2V+\nabla^2 W\ast\mu_t\geq\rho+\alpha>0$. So, putting $\gamma(t):=\left|X_t-Y_t\right|^2$, Cauchy-Schwarz inequality yields to
\begin{equation*}
\gamma'(t)\leq-2\left(\alpha+\rho\right)\gamma(t)+2\sqrt{\gamma(t)}\left|\nabla W(Y_t-m)-\nabla W\ast\mu_t(Y_t)\right|\,.
\end{equation*}
But by the growth assumption \eqref{domination} on $W$, we have for any probability measures $\mu,\nu$ the following control $$\left|\nabla W\ast\mu(x)-\nabla W\ast\nu(x)\right|\leq C\left(1+|x|^{2k}\right)\mathbb{W}_{2k}^{2k}\left(\mu;\nu\right)$$ where $2k$ is the degree of the polynomial $P$. We introduce the set
\begin{equation*}
\mathcal{A}_\kappa:=\left\{\omega\in\Omega\,\,:\,\,\mathbb{W}_{2k}^{2k}\left(\mu_t;\delta_m\right)\leq\kappa^k\right\}\,.
\end{equation*}

By Markov inequality, we have $\mathbb{P}\left(\mathcal{A}_\kappa\right)\geq1-\kappa^k$. This implies for any $t\geq T_\kappa(\sigma)$ and for any $\omega\in\mathcal{A}_\kappa$:
\begin{equation*}
\gamma'(t)\leq-2\left(\alpha+\rho\right)\gamma(t)+2C\kappa^k\sqrt{\gamma(t)}\left(1+\left|Y_t\right|^{2k}\right)\,.
\end{equation*}
 However, $\gamma(t)=0$ for any $t\leq T_\kappa(\sigma)$. This means that $$\left\{t\geq0\,\,:\,\,\gamma(t)>\frac{C^2\kappa^{2k}\left(1+\left|Y_t\right|^{2k}\right)^2}{(\alpha+\rho)^2}\right\}\subset\left\{t\geq0\,\,:\,\,\gamma'(t)<0\right\}.$$ We deduce that
\begin{equation*}
\sup_{T_\kappa(\sigma)\leq t\leq\exp\left[\frac{2H+10}{\sigma^2}\right]}\gamma(t)\leq\frac{C^2\kappa^{2k}\left(1+\sup_{T_\kappa(\sigma)\leq t\leq\exp\left[\frac{2H+10}{\sigma^2}\right]}\left|Y_t\right|^{2k}\right)^2}{(\alpha+\rho)^2}\,,
\end{equation*}
if $\omega\in\mathcal{A}_\kappa$. We now consider $R>0$ such that the exit cost of the diffusion $Y$ from the ball of center $m$ and radius $R$ is at least $H+6$, meaning that $\inf \{V(x) + W(x~\!-~\!m) - V(m): x \in B(m, R)\} \geq H + 6$.

Then, by Freidlin-Wentzell theory, we deduce that\\ $\lim_{\sigma\to 0}\mathbb{P}\left(\sup_{T_\kappa(\sigma)\leq t\leq\exp\left[\frac{2H+10}{\sigma^2}\right]}\left|Y_t - m\right|\geq R\right)=0$. However, we have
\begin{eqnarray*}
&\mathbb{P}& \left(\sup_{T_\kappa(\sigma)\leq t\leq\exp\left[\frac{2H+10}{\sigma^2}\right]} \left|X_t-Y_t\right|\geq\xi\right)\\
& &\quad \leq \mathbb{P}\left(\sup_{T_\kappa(\sigma)\leq t\leq\exp\left[\frac{2H+10}{\sigma^2}\right]}\left|Y_t - m\right|\geq R\right)\\
& & \quad + \mathbb{P}\left(\sup_{T_\kappa(\sigma)\leq t\leq\exp\left[\frac{2H+10}{\sigma^2}\right]}\gamma(t)\geq\xi^2,\sup_{T_\kappa(\sigma)\leq t\leq\exp\left[\frac{2H+10}{\sigma^2}\right]}\left|Y_t - m\right|<R\right)\\
&&\quad+ \mathbb{P}\left(\mathcal{A}_\kappa^c\right)\,.
\end{eqnarray*}
The first term tends to $0$ as $\sigma$ goes to $0$. The second term is equal to $0$ provided that $\xi>\frac{C\kappa^{k} \big(1+ 2^{2k - 1}(R + |m|^{2k}) \big)}{\alpha+\rho}$. In other words, if $\kappa$ is small enough, the second term is equal to $0$ uniformly with respect to $\sigma$. The third term is less than $\sqrt{\kappa}$. This concludes the proof.
\end{proof}

\subsection{Proof of Theorem~\ref{dell}}\label{ss:conclusion}

Now we will prove the main Theorem~\ref{dell}. The idea of the proof is to use the fact that diffusions $Y$ and $X$ are close to each other after the deterministic stabilization time $T_{\kappa}(\sigma)$ and until some fixed deterministic time $\exp{\frac{2(H + 5)}{\sigma^2}}$, which we chose to be sufficiently big for our line of reasoning. We can control the proximity of these two diffusions by parameter $\kappa$, which represents how close time-limit of our occupation measure and $\delta_m$ are. It was already shown in Proposition~\ref{craie}, that with $\sigma \to 0$ probability of exiting before time $T_{\kappa}(\sigma)$ tends to zero, which means that we can only focus on our dynamics after the stabilization of occupation measure happens. After that, for the upper bound we show, that the event $\tau > \exp\{\frac{2(H + \delta)}{\sigma^2}\}$ is not very likely due to the fact that, for small $\sigma$, diffusion $Y$ can even leave some bigger domain in a smaller time, which contradicts the closeness of $X$ and $Y$. Same type of reasoning takes place for the lower bound. Let us now provide the rigorous proof.

Fix some $\delta, \kappa > 0$, decrease it if necessary to be $\delta < 5$. For the upper bound, consider the following inequality:
\begin{equation}\label{eq:tau_lowbound1}
    \mathbb{P}(\tau > e^{\frac{2(H + \delta)}{\sigma^2}}) \leq \mathbb{P}(\tau > e^{\frac{2(H + \delta)}{\sigma^2}}, \tau^Y_{\mathcal{D}^e} \leq e^{\frac{2(H + \delta)}{\sigma^2}}) + \mathbb{P}(\tau^Y_{\mathcal{D}^e} > e^{\frac{2(H + \delta)}{\sigma^2}}),
\end{equation}
where $\mathcal{D}^e$ is some enlargement of domain $\mathcal{D}$ such that its exit cost is equal to $H + \frac{\delta}{2}$, i.e.:
\begin{equation*}
    \mathcal{D}^e:= \{x \in \mathbb{R}^d: V(x) + W(x - m) - V(m) < H + \frac{\delta}{2} \};
\end{equation*}
and $\tau^Y_{\mathcal{D}^e}$ is exit time of diffusion $Y$ from this domain, i.e.:
\begin{equation*}
    \tau^Y_{\mathcal{D}^e} := \inf \{t: Y_t \notin \mathcal{D}^e\}.
\end{equation*}
Note, that domain $\mathcal{D}^e$ (since both $V$ and $W$ are continuous and convex) satisfies the usual assumptions (see \cite{DZ}) and $d_e:= d(\mathcal{D}, \mathcal{D}^e) > 0$. By classical result of Freidlin-Wentzell theory,
\begin{equation*}
    \mathbb{P}(\tau^Y_{\mathcal{D}^e} > e^{\frac{2((H + \delta/2)  + \delta/2)}{\sigma^2}}) \xrightarrow[\sigma \to 0]{} 0.
\end{equation*}
Let us decrease $\sigma_\kappa$ if necessary, such that the quantity above will be less then $\sqrt{\kappa}$ for any $\sigma < \sigma_\kappa$. Moreover, the first probability in \eqref{eq:tau_lowbound1} can be bounded by:
\begin{equation*}
    \mathbb{P}(\tau^Y_{\mathcal{D}^e} \leq e^{\frac{2(H + \delta)}{\sigma^2}} < \tau) \leq \mathbb{P}(|X_{\tau^Y_{\mathcal{D}^e}} - Y_{\tau^Y_{\mathcal{D}^e}}| \geq d_e) \leq 2\kappa^k,
\end{equation*}
where we use Proposition \ref{prop:coupling} and decrease $\kappa$ and $\sigma_\kappa$ if necessary.

We approach the lower bound similarly and introduce the contraction of the domain $\mathcal{D}$:
\begin{equation*}
    \mathcal{D}^c := \inf \{x \in \mathbb{R}^d: V(x) + W(x - m) - V(m) < H - \frac{\delta}{2} \}.
\end{equation*}
If $\mathcal{D}^c$ turns out to be empty, decrease $\delta$. As previously, the domain $\mathcal{D}^c$ satisfies usual properties and has positive distance with the initial domain, that is $d_c := d(\mathcal{D}^c, \mathcal{D}) > 0$. We introduce exit-time from the contracted domain for diffusion $Y$:
\begin{equation*}
    \tau_{\mathcal{D}^c}^Y := \inf \{t: Y_t \notin \mathcal{D}^c\},
\end{equation*}
and have the following estimate:
\begin{equation*}
    \begin{aligned}
    \mathbb{P}( \tau < e^{\frac{2(H - \delta)}{\sigma^2}}) & \leq \mathbb{P}(T_{\kappa}(\sigma) < \tau < e^{\frac{2(H - \delta)}{\sigma^2}} \leq \tau_{\mathcal{D}^c}^Y) \\
    &\quad\quad + \mathbb{P}(\tau \leq T_{\kappa}(\sigma))  + \mathbb{P}(\tau_{\mathcal{D}^c}^Y \geq e^{\frac{2((H - \delta/2) - \delta/2)}{\sigma^2}}) \\
    & \leq \mathbb{P}(|X_{\tau} - Y_{\tau}| \geq d_c) ++ \mathbb{P}(\tau \leq T_{\kappa}(\sigma))+  2\kappa^k \\
    & \leq 3\kappa^k + \mathbb{P}(\tau \leq T_{\kappa}(\sigma))\leq 4\kappa^k,
    \end{aligned}
\end{equation*}
by Proposition~\ref{craie} \eqref{apple2}, with $\kappa$ and $\sigma_\kappa$ small enough. That leads to:
\begin{equation*}
    \mathbb{P}(e^{\frac{2(H - \delta)}{\sigma^2}} \leq \tau \leq e^{\frac{2(H + \delta)}{\sigma^2}}) \geq 1 - 7\kappa^k,
\end{equation*}
which proves the theorem if we consider $\kappa \to 0$, parameter that uniformly controls the convergence of $\sigma$ towards $0$.

\subsection{Proof of Corollary \ref{hphp}}
\label{s:hphp}

We can apply Theorem~\ref{dell} to the level sets of the potential $V+W\ast\delta_m$.

By definition of $\mathcal{N}$ in Corollary~\ref{hphp}, there exists a constant $\xi>0$ such that
$\displaystyle\inf_{z\in\mathcal{N}}\left(V(z)+W(z-m)-V(m)\right)=H+3\xi$. We introduce the set
\begin{eqnarray*}
\mathcal{K}_{H+2\xi}:=\left\{x\in\mathbb{R}^d\,\,:\,\,V(x)+W(x-m)-V(m)<H+2\xi\right\}\,.
\end{eqnarray*}
If we denote by $\tau_\xi$ the first exit time of $X$ from $\mathcal{K}_{H+2\xi}$, then we obtain
\begin{eqnarray}
\label{oz}
\lim_{\sigma \to0}\mathbb{P}\left\{\exp\left[\frac{2}{\sigma^2}\left(H+2\xi-\rho\right)\right]<\tau_\xi<\exp\left[\frac{2}{\sigma^2}\left(H+2\xi+\rho\right)\right]\right\}=1
\end{eqnarray}
for any $\rho>0$. By construction of $\mathcal{K}_{H+2\xi}$, $\mathcal{N}\subset\mathcal{K}_{H+2\xi}^c$, which implies
\begin{align*}
\mathbb{P}\left\{X_{\tau}\in\mathcal{N}\right\}\leq&\mathbb{P}\left\{X_{\tau}\notin\mathcal{K}_{H+2\xi}\right\}\\
\leq&\mathbb{P}\left\{\tau_{\xi}\leq\tau\right\}\\
\leq&\mathbb{P}\left\{\tau_{\xi}\leq\exp\left[\frac{2(H+3\xi)}{\sigma^2}\right]\right\}+\mathbb{P}\left\{\exp\left[\frac{2H+\xi}{\sigma^2}\right]\leq\tau\right\}\,.
\end{align*}
Applying \eqref{oz} with $\rho:=\xi$ to the first term and Theorem \ref{dell} to the second one, we obtain the result.

\begin{small}
\def$'${$'$}

redpajama_set_name: RedPajamaArXiv


text:
#pragma once
#include <aws/fsx/FSx_EXPORTS.h>
#include <aws/fsx/model/LustreAccessAuditLogLevel.h>
#include <aws/core/utils/memory/stl/AWSString.h>
#include <utility>

namespace Aws
{
namespace Utils
{
namespace Json
{
  class JsonValue;
  class JsonView;
} // namespace Json
} // namespace Utils
namespace FSx
{
namespace Model
{

  /**
   *
The configuration for Lustre logging used to write the enabled logging events
   * for your file system to Amazon CloudWatch Logs.

When logging is enabled,
   * Lustre logs error and warning events from data repository operations such as
   * automatic export and data repository tasks. To learn more about Lustre logging,
   * see Logging
   * with Amazon CloudWatch Logs.

See Also:
   AWS
   * API Reference

   */
  class AWS_FSX_API LustreLogConfiguration
  {
  public:
    LustreLogConfiguration();
    LustreLogConfiguration(Aws::Utils::Json::JsonView jsonValue);
    LustreLogConfiguration& operator=(Aws::Utils::Json::JsonView jsonValue);
    Aws::Utils::Json::JsonValue Jsonize() const;

    /**
     *
The data repository events that are logged by Amazon FSx.

     * WARN_ONLY - only warning events are logged.

     * ERROR_ONLY - only error events are logged.

     * WARN_ERROR - both warning events and error events are logged.

     *

 DISABLED - logging of data repository events is
     * turned off.

     */
    inline const LustreAccessAuditLogLevel& GetLevel() const{ return m_level; }

    /**
     *
The data repository events that are logged by Amazon FSx.

     * WARN_ONLY - only warning events are logged.

     * ERROR_ONLY - only error events are logged.

     * WARN_ERROR - both warning events and error events are logged.

     *

 DISABLED - logging of data repository events is
     * turned off.

     */
    inline bool LevelHasBeenSet() const { return m_levelHasBeenSet; }

    /**
     *
The data repository events that are logged by Amazon FSx.

     * WARN_ONLY - only warning events are logged.

     * ERROR_ONLY - only error events are logged.

     * WARN_ERROR - both warning events and error events are logged.

     *

 DISABLED - logging of data repository events is
     * turned off.

     */
    inline void SetLevel(const LustreAccessAuditLogLevel& value) { m_levelHasBeenSet = true; m_level = value; }

    /**
     *
The data repository events that are logged by Amazon FSx.

     * WARN_ONLY - only warning events are logged.

     * ERROR_ONLY - only error events are logged.

     * WARN_ERROR - both warning events and error events are logged.

     *

 DISABLED - logging of data repository events is
     * turned off.

     */
    inline void SetLevel(LustreAccessAuditLogLevel&& value) { m_levelHasBeenSet = true; m_level = std::move(value); }

    /**
     *
The data repository events that are logged by Amazon FSx.

     * WARN_ONLY - only warning events are logged.

     * ERROR_ONLY - only error events are logged.

     * WARN_ERROR - both warning events and error events are logged.

     *

 DISABLED - logging of data repository events is
     * turned off.

     */
    inline LustreLogConfiguration& WithLevel(const LustreAccessAuditLogLevel& value) { SetLevel(value); return *this;}

    /**
     *
The data repository events that are logged by Amazon FSx.

     * WARN_ONLY - only warning events are logged.

     * ERROR_ONLY - only error events are logged.

     * WARN_ERROR - both warning events and error events are logged.

     *

 DISABLED - logging of data repository events is
     * turned off.

     */
    inline LustreLogConfiguration& WithLevel(LustreAccessAuditLogLevel&& value) { SetLevel(std::move(value)); return *this;}

    /**
     *
The Amazon Resource Name (ARN) that specifies the destination of the logs.
     * The destination can be any Amazon CloudWatch Logs log group ARN. The destination
     * ARN must be in the same Amazon Web Services partition, Amazon Web Services
     * Region, and Amazon Web Services account as your Amazon FSx file system.

     */
    inline const Aws::String& GetDestination() const{ return m_destination; }

    /**
     *
The Amazon Resource Name (ARN) that specifies the destination of the logs.
     * The destination can be any Amazon CloudWatch Logs log group ARN. The destination
     * ARN must be in the same Amazon Web Services partition, Amazon Web Services
     * Region, and Amazon Web Services account as your Amazon FSx file system.

     */
    inline bool DestinationHasBeenSet() const { return m_destinationHasBeenSet; }

    /**
     *
The Amazon Resource Name (ARN) that specifies the destination of the logs.
     * The destination can be any Amazon CloudWatch Logs log group ARN. The destination
     * ARN must be in the same Amazon Web Services partition, Amazon Web Services
     * Region, and Amazon Web Services account as your Amazon FSx file system.

     */
    inline void SetDestination(const Aws::String& value) { m_destinationHasBeenSet = true; m_destination = value; }

    /**
     *
The Amazon Resource Name (ARN) that specifies the destination of the logs.
     * The destination can be any Amazon CloudWatch Logs log group ARN. The destination
     * ARN must be in the same Amazon Web Services partition, Amazon Web Services
     * Region, and Amazon Web Services account as your Amazon FSx file system.

     */
    inline void SetDestination(Aws::String&& value) { m_destinationHasBeenSet = true; m_destination = std::move(value); }

    /**
     *
The Amazon Resource Name (ARN) that specifies the destination of the logs.
     * The destination can be any Amazon CloudWatch Logs log group ARN. The destination
     * ARN must be in the same Amazon Web Services partition, Amazon Web Services
     * Region, and Amazon Web Services account as your Amazon FSx file system.

     */
    inline void SetDestination(const char* value) { m_destinationHasBeenSet = true; m_destination.assign(value); }

    /**
     *
The Amazon Resource Name (ARN) that specifies the destination of the logs.
     * The destination can be any Amazon CloudWatch Logs log group ARN. The destination
     * ARN must be in the same Amazon Web Services partition, Amazon Web Services
     * Region, and Amazon Web Services account as your Amazon FSx file system.

     */
    inline LustreLogConfiguration& WithDestination(const Aws::String& value) { SetDestination(value); return *this;}

    /**
     *
The Amazon Resource Name (ARN) that specifies the destination of the logs.
     * The destination can be any Amazon CloudWatch Logs log group ARN. The destination
     * ARN must be in the same Amazon Web Services partition, Amazon Web Services
     * Region, and Amazon Web Services account as your Amazon FSx file system.

     */
    inline LustreLogConfiguration& WithDestination(Aws::String&& value) { SetDestination(std::move(value)); return *this;}

    /**
     *
The Amazon Resource Name (ARN) that specifies the destination of the logs.
     * The destination can be any Amazon CloudWatch Logs log group ARN. The destination
     * ARN must be in the same Amazon Web Services partition, Amazon Web Services
     * Region, and Amazon Web Services account as your Amazon FSx file system.

     */
    inline LustreLogConfiguration& WithDestination(const char* value) { SetDestination(value); return *this;}

  private:

    LustreAccessAuditLogLevel m_level;
    bool m_levelHasBeenSet;

    Aws::String m_destination;
    bool m_destinationHasBeenSet;
  };

} // namespace Model
} // namespace FSx
} // namespace Aws

redpajama_set_name: RedPajamaGithub


Items where Author is "Abraham, M J"
Paul, M and Abraham, M J and Menon, N R (1998) Settlement pattern of the larvae of Hydroides elegans (Polychaeta) in the presence of petroleum hydrocarbon compounds. Indian Journal of Marine Sciences, 27. pp. 489-491.
This list was generated on Sat Apr 20 05:54:58 2019 IST.


text:
Q: Does playing non-shareable games prevent sharing? According to the Steam Family Sharing FAQ:

Can all Steam games be shared with friends and family?
No, due to technical limitations, some Steam games may be unavailable for sharing. For example, titles that require an additional third-party key, account, or subscription in order to play cannot be shared between accounts.

I understand that if I am playing any game in my shared library other people cannot use my games.  However, when I am playing my own non-shareable game will other people be able to play my shared games?
Sorry if this seems an odd question.  I am trying to understand the feature limitations.  Basically, if there is no such limitation I could start up one of my non-shareable games whilst a family member continues playing a shared game.  I just want to know if that is actually an option.

A: Short answer is no. You can't play any Steam games if you want a family member to play any of the shared games. Even games that are not shared lock your library to you only.
If for example one plays GTA V (which is not possible to be shared as far as I know) which is using Rockstar Social Club too, a friend or family member won't be able to play any of the shared games. When you stop playing he/she will get a notification from Steam saying "Hooray, 's library is free to play" or something like that.
To clarify - whenever Steam turns your status to green (in-game) and it's not a non-steam shortcut the library is locked for you. Basically all games activated on Steam apply here.

A: This simply means that you can't share all games for example everything on UPlay (Assassins Creed, Splinter Cell and so on). If you play your own non-shareable game, it doesn't matter, everyone of your shared accounts can access the other games which are shareable. The only limitation within the family sharing is that you can't play the same game at the same time.

A: Steam Games which are tightly combined with your Steam ID belong into this category as well.
The most famous example for this is CS:GO, you can launch it though, but you can't connect any game since the game is connected to the buyer's Steam ID.

redpajama_set_name: RedPajamaStackExchange


For the last decade U.S. Corporations have rushed to move Information Technology. and Software Development operations offshore in an effort to presumably "shore up" their organization's bottom line and save dollars through cheaper costs via overseas development centers in India, China, Vietnam and the Philippines.
These Corporations placed their bets that the savings gained via the off-shoring model would outweigh the inherent challenges faced by managing enterprise level development half a world away. And the challenges were many as was soon realized.
Aside from the obvious language barriers and time zone differences, there were also geopolitical risks, business process knowledge deficiencies, security risks, productivity factors, inflation risks via a declining global dollar, not to mention travel and expense costs.
Project Managers and PMO's faced these challenges and more on a daily basis. What they learned quickly was that sending work offshore doesn't work for all classes of development. Not only is offshore getting more expensive, but industry analysts understand that one developer in the United States is generally more productive than three developers in India when you take into account 'the context' of what they know about American business and business processes. Overseas developers understand JAVA, HTML, and PHP but they don't understand how American banks or health care systems work and that lack of understanding oftentimes requires more development for the off shore team to get right.
How many of these companies are employing agile development practices across these far flung development centers? How are sprint cycles handled by U.S. Project Managers and development leaders while their teams of programmers are 8 time zones away? It clearly goes against the iterative process that is agile. It doesn't make sense.
In 2014 corporations started to take a hard look at the benefits of bringing development back to United States and outsourcing to on-shore software development centers. Rural sourcing has gained traction again as the rates are typically lower than on either coast and the development is as good or better.
Corporations are beginning to understand that in today's evaluation of sourcing options. It is imperative to evaluate the full picture and total costs of software development and software maintenance projects, not just the hourly rate. Mature organizations must take all factors into account and assign a value to get a crystal clear picture of the cost savings that can be achieved. After all, if you're saving 50% in hourly rate overseas but the project takes 3 times as long to complete, what are you really saving?


The mercy he asks for and receives also is the direct result of the Lord Jesus Christ's atoning work upon the Cross and here in the Old Testament, the gift was given looking ahead to the Redeemer's death, burial and resurrection, the Fountainhead of God's matchless grace!
David's focus quickly turns to his adversaries: 'O ye sons of men, how long will ye turn my glory into shame? How long will ye love vanity and seek after leasing. Selah.'(v.2) He questions the actions of these lost men as they are turning his honorable testimony into a reproach through worthless and empty lies. Selah, here denotes his deep contemplation of the situation as this assault upon his reputation is serious.
His response to this fierce and intense character assassination is key: 'But know that the LORD has set apart him that is godly for Himself: the LORD will hear when I call unto Him.' (v.3) His understanding includes the knowledge that He has been separated unto God for a high and holy purpose, therefore, he knows this attack by the enemy will be taken care of by the LORD and His wisdom.
The first calls for an attitude of awareness of the actual presence of the holy God of heaven, His presence is dependent upon the sinless condition of the soul, then heart to heart communication takes place, in the quietness of meditation. Selah, again calls for focused consideration on what has just been revealed.
The righteous sacrifices of joy (Psalm 27:6), praise (Hebrews 13:15), thanksgiving (Psalm 50:14), a broken spirit (Psalm 51:17)and uplifted hands (Psalm 141:2) are emphasized here in this private and personal time of fellowship and close communion with the LORD as we have entered into David's closet, alone with him and his LORD.
David's one on one communion continues as the thoughts of his mind reveal the taunting statement of the adversary. His immediate response is to request additional illumination from the LORD which causes him to remember the paling contrast of their temporal joy for their 'good', which is earthly abundance, with the powerful presence of his eternal 'good' which is the heavenly abundance of joy in his heart when the LORD manifests His presence.
This intimate and private look into David's prayer life ends with him falling to sleep, safely in the arms of Jesus! 'I will never leave thee nor forsake thee' (Hebrews 13:5) guarantees the presence of our loving Savior, Who is with us in each and every situation we face.


No unknown third-parties/sub-contractors. Professionally trained crews, masters in their trade and the phenomenal leadership of our owner/production manager Frank Martinez. Watching homeowners check-off items on their home improvement to-do lists.


text:
The Supreme Court of Ohio
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Drug Court Graduation Illustrates Many Ties to Addiction
Ninth District Judge Hears Supreme Court Case
Statewide Symposiums Share Best Practices on Using Medication to Fight Opioid Addiction
Change to Out-of-State Attorney Admission Rule Adopted
Roger Bacon's Challenge to OHSAA Competitive-Balance Rules Back to Local Courts – Cincinnati Enquirer
County Drug Court Earns State Certification – Portsmouth Daily Times
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redpajama_set_name: RedPajamaCommonCrawl


Mila Kunis and Ashton Kutcher—who first met on the set of That '70s Show and welcomed their daughter, Wyatt, in October 2014—reportedly married over the Fourth of July weekend in a private ceremony at the Secret Garden at Parrish Ranch in Oak Glen, California. Kunis has spoken in the past about wanting a low-key wedding: "My theory on weddings is: Don't invite anyone. Do it privately and secretly. My parents are OK with that." And if she took her own advice, she's following a string of chic brides, from Olivia Palermo to Scarlett Johansson—who decided to forgo an over-the-top ceremony.
The five-acre property is tucked away in the ranch's apple orchards, features scenic views, spring-fed ponds, and most importantly: privacy. A secluded venue can be one of the smartest ways to keep paparazzi at bay. "Privacy, security, and peace of mind are everything for a high-profile couple on their most special day," says master event planner Bronson van Wyck. "It can be more important than flowers, catering, or entertainment." But even if prying eyes are not a concern of the non-celebrity bride, a remote location is a wonderful way to bring guests together for a shared experience. Rather than being distracted by, say, the bright lights of New York City, guests are more focused on the event itself. From South America's Patagonia to Norfolk, Connecticut, we're taking a look back at some of _Vogue'_s favorite en plein air locales.


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In politics, a mafia state is a state system where the government is tied with organized crime to the degree when government officials, the police, and/or military became a part of the criminal enterprise. According to US diplomats, the expression "mafia state" was coined by Alexander Litvinenko.

Particular applications of the concept

Mafia in Italy and Yakuza in Japan

In a critical review of Moisés Naím's essay in Foreign Affairs, Peter Andreas pointed to the long existence of Italian mafia and Japanese Yakuza, writing that there were close relationships between those illicit organisations and respective governments. According to Andreas, these examples speak against incidences of mafia states as a historically new threat.

In Italy, there are three main mafia organisations that originated in the 19th century: the Cosa Nostra originating from the region of Sicily, the Camorra originating from the region of Campania, and the 'Ndrangheta originating from the region of Calabria.

Former Prime Minister of Italy, Giulio Andreotti, had legal action against him, with a trial for mafia association on 27 March 1993 in the city of Palermo. The prosecution accused the former prime minister of "[making] available to the mafia association named Cosa Nostra for the defense of its interests and attainment of its criminal goals, the influence and power coming from his position as the leader of a political faction".  Prosecutors said in return for electoral support of Salvo Lima and assassination of Andreotti's enemies, he had agreed to protect the Mafia, which had expected him to fix the Maxi Trial. Andreotti's defense was predicated on character attacks against the prosecution's key witnesses who were themselves involved with the mafia. Andreotti was eventually acquitted on 23 October 1999. However, together with the greater series of corruption cases of Mani pulite, Andreotti's trials marked the purging and renewal of Italy's political system.

The Camorra Casalesi clan rose in the 1980s, gaining control of large areas of the local economy "partly by manipulating politicians and intimidating judges". The clan was heavily involved in the Naples waste management crisis that dumped toxic waste around Campania in the 1990s and 2000s; the boss of the clan, Gaetano Vassallo, admitted to systematically working for 20 years to bribe local politicians and officials to gain their acquiescence to dumping toxic waste.

Countries described as Mafia states

Republics, mobs and territories of the former Yugoslavia

Kosovo, a partially recognised independent state formerly part of Serbia, was called a "mafia state" by Italian MEP Pino Arlacchi in 2011, and also by Moisés Naím in his 2012 essay "Mafia States" in the Foreign Affairs. Naím pointed out that Prime Minister of Kosovo Hashim Thaçi is allegedly connected to the heroin trade. Many other crime allegations have been made, and investigated by several countries, against Thaçi.

Naím also labeled Montenegro as a "mafia state" in the same essay, describing it as a hub for cigarette smuggling.

Transnistria

Transnistria, an unrecognised break-away state from Moldova, has long been described by journalists, researchers, politicians and diplomats as a quasi-state whose economy is dependent on contraband and gunrunning, with some having directly referred to Transnistria as a mafia state.

For instance, in 2002, Moldova's president, Vladimir Voronin, called Transnistria a "residence of international mafia", "smuggling stronghold" and "outpost of Islamic combatants". The allegations were followed by attempts of customs blockade. Reacting to the allegations, Russian state-run RTR aired an investigative program revealing that Transnistrian firms were conducting industrial-level manufacturing of small arms purposely for subsequent illegal trafficking via the Ukrainian port of Odessa. According to the program, the trade was controlled by and benefited from Transnistria's founder and then-ruler Igor Smirnov.

However, more recent investigations and monitoring missions did not prove continuity in arms trafficking concerns. According to regular reports of the European Border Assistance Mission to Moldova and Ukraine (EUBAM), there have been no signs of significant weapons smuggling from Transnistria. During the press-conference on 30 November 2006 head of EUBAM Ferenc Banfi officially stated that organised smuggling of weapons in Transnistria did not exist. In 2013, Ukrainian Foreign Minister and acting chairman of the Organization for Security and Cooperation in Europe (OSCE) Leonid Kozhara gave an interview to El País, commenting on situation in Transnistria and results of work of the EUBAM mission. According to Kozhara, there have been no cases of arms traffic found.

Some experts from Russia and Transnistria state that allegations of Transnistria being a "mafia state", "black hole of Europe", "heaven for arms trafficking", etc. are a carefully planned defamation campaign paid by the Moldovan government and aimed at producing negative image of Transnistria. Officials from the European Union and the OSCE, say they have no evidence that the Tiraspol regime has ever trafficked arms or nuclear material. Much of the alarm is attributed to efforts by the Moldovan government to increase pressure on Transnistria.

Russia

The term has been used by slain Russian dissident Alexander Litvinenko and media to describe the political system in Russia under Vladimir Putin's rule. This characterization came to prominence following the United States diplomatic cables leak, which revealed that US diplomats viewed Russia as "a corrupt, autocratic kleptocracy centred on the leadership of Vladimir Putin, in which officials, oligarchs and organised crime are bound together to create a 'virtual mafia state.'" In his book titled Mafia State, journalist and author Luke Harding argues that Putin has "created a state peopled by ex-KGB and FSB officers, like himself, [who are] bent on making money above all." In the estimation of American diplomats, "the government [of Russia] effectively [is] the mafia." Russian-American journalist Masha Gessen has also described Russia as a "mafia state".

According to the New Statesman, "the term had entered the lexicon of expert discussion" several years before the cables leak, "and not as a frivolous metaphor. Those most familiar with the country had come to see it as a kleptocracy with Vladimir Putin in the role of capo di tutti capi, dividing the spoils and preventing turf wars between rival clans of an essentially criminal elite." In 2008, Stephen Blank noted that Russia under Putin is "a state that European officials privately call a Mafia state" that "naturally gravitates toward Mafia-like behavior."

Nikolay Petrov, an analyst at the Carnegie Moscow Centre, said "it's pretty hard to damage the Russian image in the world because it's already not very good".

Syria

For more than half a century, Syria has been under the tight grip of Assad dynasty and its ruling system has been described as a mafia state. Due to the Syrian economy being dominated by businessmen tied to Assad family, control of various services like electricity by loyalist warlords, etc. more than 90% of Syrians are stricken by poverty.  The regime also profits from a multi-billion dollar captagon industry, made of a network consisting of over 35 manufacturing centres and factories, known as the "Syndicate" (Niqabeh). Assad's wife Asma stripped the assets of pro-Assad billionaire Rami Makhlouf and subsequently took control of all foreign charities and UN aid money; enabling the Assads to directly own revenue money. The fall of Rami Makhlouf led to widespread characterisations of Assad acting as a crime boss presiding over "a crime syndicate simply designed to enrich himself and his family", deploying violence to advance the interests of his mafia. Businessman Ribal al-Assad, a cousin of Bashar al-Assad who lives in exile outlines the situation: "When it's all boiled down, the family is still in charge.. They are very sensitive to internal issues and they know how to manage them. They have a saying: 'You may not have to listen to your cousins, but you do need to listen to their mothers."

Various academics compare Assad's reign to tribal rule; with Assad as the tribal chieftain glorified by the system and extolled as its saviour. Syrian-French Professor of Sociology Burhan Ghalioun asserts that Bashar al-Assad doesnt see Syrians as an independent people, but rather as guests in his family property. Hence, Assad's dictatorship is akin to the colonial project of an invading empire, according to Ghalioun. Syrian historian Rana Kabbani designates Assad's dynastic rule as "internal colonialism" owing to the deep disconnect of the regime's post-colonial ruling classes from its native population and their blatant disregard for Syrian wealth and lives. Describing how neo-Ba'athist ideologues and Assadist security apparatus constructed a totalitarian state built around cult of personality to the Assad dynasty, Rana Kabbani writes: "When Bashar al-Assad's father, Hafez, came to power through yet another violent army squabble leading to his coup of 1970, an alarming cult of the leader was systematically formed around him, modelled on Ceausescu. The Romanian dictator was Assad's political ally, strategic adviser in matters of popular repression, and close personal and family friend...In the government school I attended in Damascus between 1971 and 1974, a process of wholesale brainwashing had begun. It was designed to create a population with no political personality or affiliation – other than to the head of what would become, in my children's generation, a vindictive family mafia, monopolising business and power with the crudest of propaganda machines and the most lethal of security services."

Mexico

The scholar of Law and Economics Edgardo Buscaglia describes the political system of Mexico as a "Mafiacracy". Buscaglia characterises the condition between the state, the economy and organized crime in Mexico as a mutual interweaving, Mexico has also been labeled as a Narco-state (a political and economic term applied to countries where all legitimate institutions become penetrated by the power and wealth of the illegal drug trade).

Other
Venezuela under President Hugo Chávez has also been evaluated as creating the conditions for President Nicolás Maduro to consolidate a "mafia state", according to political scientists in Policy Studies in 2021. Moisés Naím, Venezuelan writer and the author of Illicit: How Smugglers, Traffickers, and Copycats Are Hijacking the Global Economy, wrote in an article for the American magazine Foreign Policy: "In mafia states such as Bulgaria, Guinea-Bissau, Montenegro, Myanmar, Ukraine, and Venezuela, the national interest and the interests of organised crime are now inextricably intertwined."

During the 1990s, Aruba, an overseas constituent country of the Kingdom of the Netherlands, was alleged to be a mafia state under then Prime Minister Henny Eman. The Sicilian Cuntrera-Caruana Mafia clan was said to "own" over 60% of the island through investments in hotels, casinos, and donations towards Eman's election-campaign. These claims later turned out to be exaggerated, however.

According to journalist Masha Gessen in 2016, sociologist Bálint Magyar had popularized the term "mafia state" to describe the administration of Viktor Orbán in Hungary. Magyar argued that Orbán's method of taking power in Hungary, removing rivals and politicizing institutions for personal power created a kind of regime known as a "mafia state". Magyar said that besides Hungary, other such states included Azerbaijan, Russia, and Kazakhstan as of 2016.

In 2010, Libyan leader Muammar Gaddafi had labelled Switzerland as a mafia state, stating that "Switzerland is behaving like a criminal organisation and it is involved in money laundering, assassinations and terrorism."

Jonathan Benton, the former head of a United Kingdom anti-corruption agency, described Malta as a "mafia state" where money laundering transactions of hundreds of millions of euros are made every year without any problem. He made this statement while speaking on BBC radio following the murder of investigative journalist Daphne Caruana Galizia.

See also

Related concepts
 Mafia Raj
 Counterintelligence state
 Police state
 Narco-state
 Failed state
 Terrorist state
 North Korea's illicit activities
 Kleptocracy

References

Further reading

 Naím, M. (2012). "Mafia states: Organized crime takes office." Foreign Affairs, 91, 100.
 Wang, P. & Blancke, S. (2014). "Mafia State: The Evolving Threat of North Korean Narcotics Trafficking." The RUSI Journal. 159 (5). 52–59.

Authoritarianism
Mafia
Oligarchy
Political corruption
Political terminology

redpajama_set_name: RedPajamaWikipedia


text:
Nomansland er en film instrueret af Karsten Geisnæs.

Handling
Christian må med ét tage sit liv op til revision, da hans ekskæreste begår selvmord, og han efterfølgende selv får konstateret HIV. Ramt af kendsgerningerne ryger han ud på en seksuel flugt, indtil han endelig ser sig selv i øjnene.

Referencer

Eksterne henvisninger

Danske film fra 2013

Film fra Danmark (administrationskategori)
Kortfilm fra Danmark