text:
\begin{document}
\title{Limit behavior of the invariant measure for Langevin dynamics}
\author{Gerardo Barrera}
\address{University of Helsinki, Department of Mathematical and Statistical Sciences. PL 68, Pietari Kalmin katu 5.
Postal Code: 00560. Helsinki, Finland.}
\thanks{The author  has been supported by the Academy of Finland via
the Matter and Materials Profi4 university profiling action.}
\email{gerardo.barreravargas@helsinki.fi}
\keywords{Coupling; Invariant distribution; Langevin dynamics; Ornstein-Uhlenbeck process; Perturbations of dynamical systems; Wasserstein distance}
\subjclass[2000]{60H10; 34D10; 37M25; 60F05; 49Q22}
\begin{abstract}
In this manuscript, we consider the Langevin dynamics on $\mathbb{R}^d$ with an overdamped vector field and driven by multiplicative Brownian noise  of small amplitude $\sqrt{\epsilon}$, $\epsilon>0$. Under suitable assumptions on the vector field and the diffusion coefficient, it is well-known that it possesses a unique invariant probability measure $\mu^{\epsilon}$. As $\epsilon$ tends to zero, we prove that the probability measure
$\epsilon^{\nicefrac{d}{2}} \mu^{\epsilon}(\sqrt{\epsilon}\ud x)$ converges in the $p$-Wasserstein  distance for $p\in [1,2]$ to a Gaussian measure with zero-mean vector and non-degenerate covariance matrix which solves a Lyapunov matrix equation.
Moreover, the error term is estimated.
We emphasize that generically no explicit formula for $\mu^{\epsilon}$ can be found.
\end{abstract}
\maketitle
\markboth{Invariant probability measure}{Gaussian approximation}
\section{\textbf{Introduction}}
\subsection{\textbf{The overdamped Langevin dynamics}}
Random dynamical systems  arise in the modeling of a (realistic) physical system
subject to noise perturbations from its surrounding environments or
from intrinsic uncertainties associated with the system.
The Langevin dynamics was introduced by P. Langevin in 1908 in his celebrated article
``Sur la th\'eorie du mouvement brownien'',
\textit{C. R. Acad. Sci. Paris} \textbf{146},
530--533.
It is perhaps one of the most
popular models in molecular systems. For details about the history of the Langevin equation, see \cite{POPI}.
For a phenomenological treatment, we recommend the monography \cite{COK}.

In the last decades, there have been many applications
of Markov chain Monte Carlo methods to complex  systems in
 Computer Science and Statistical Physics.
Since
sampling
high-dimensional distributions is typically a difficult task,
the use of stochastic equations  for sampling has become important in many applications as Artificial intelligence and  Bayesian algorithms.
Stochastic algorithms based on Langevin equations have been proposed to simulate and improve the rate of convergence  to limiting distributions.
For further details we refer to
\cite{Dalalyan2017}, \cite{Duncan2017},
\cite{Eberle2019},
 \cite{Hwang193}, \cite{Levi2013}, \cite{Wu2014} and the references therein.

Differential equations subject to small noise perturbations are one of the classic directions of modern mathematical physics.
Let $\epsilon\in (0,1]$ be a parameter that measures the perturbation strength
and let $(B_t)_{t \geq 0}$ be a standard Brownian motion on $\mathbb{R}^d$.
For any (deterministic) $x\in \mathbb{R}^d$ we consider the unique strong solution $(X^{\epsilon}_t(x))_{t \geq 0}$ of the following stochastic differential equation (SDE for short) on $\mathbb{R}^d$
\begin{equation}\label{over}
\left\{
\begin{array}{r@{\;=\;}l}
\ud X^{\epsilon}_t(x)&-F(X^{\epsilon}_t(x))\ud t +
\sqrt{\epsilon}\sigma(X^\epsilon_t(x))\ud B_t\quad
\textrm{  for any } t\geq 0,\\
X^{\epsilon}_0(x) & x,
\end{array}
\right.
\end{equation}
where the vector field $F\in\mathcal{C}^2(\mathbb{R}^d,\mathbb{R}^d)$ and the diffusion coefficient  $\sigma\in
\mathcal{C}^2(\mathbb{R}^d,\mathbb{R}^{d\times d})$ satisfy the following assumptions.
We assume that $0_d$ is a fixed point for $F$, i.e., $F(0_d)=0_d$ and the
following hypotheses for $F$.
\begin{itemize}
\item[a)] \textbf{
Bakry-\'Emery condition:}
there exists a positive constant $\delta$  such that
\begin{equation}\label{H}
\tag{{\bf{A}}}
\<F(x_1)-F(x_2),x_1-x_2\>\geq \delta \|x_1-x_2\|^2\quad \textrm{ for any } x_1,x_2\in \mathbb{R}^d,
\end{equation}
where
$\<\cdot,\cdot\>$ denotes the standard inner product on $\mathbb{R}^d$ and
$\|\cdot\|$ denotes the standard Euclidean  norm on $\mathbb{R}^d$.
\item[b)] \textbf{Exponential growth condition:}
there exist positive constants $c_0$ and $c_1$ satisfying
\begin{equation}
\tag{\textbf{B}}
\label{C3}
\|F(x)\| \leq c_0e^{c_1\|x\|^2}\quad \textrm{ for any } x \in \mathbb{R}^d.
\end{equation}
\end{itemize}
Observe that $F(x)=0$ if and only if $x=0_d$.
For the diffusion coefficient
$\sigma$ we assume the following standard hypotheses.
\begin{enumerate}
\item[c)]
\textbf{Lipschitz continuity:} there exists a positive constant $\ell$ such that
\begin{equation}\label{eq:lipt}
\tag{{\bf{C}}}
\|\sigma(x)-\sigma(x_0)\|_{\mathrm{F}}\leq \ell \|x-x_0\| \quad\textrm{ for all } x, x_0\in \mathbb{R}^d,
\end{equation}
where $\|\cdot\|_{\mathrm{F}}$ denotes the Frobenius norm.
\item[d)]
\textbf{Ellipticity:}
there is a positive constant $\kappa$ such
that
\begin{equation}\label{eq:elip}
\tag{{\bf{D}}}
  \<\sigma(x_0) \sigma^*(x_0)x, x\>\geq \kappa\|x\|^2
\quad\textrm{ for all } x, x_0\in \mathbb{R}^d,
\end{equation}
where $*$ denotes the transpose operator.
\end{enumerate}

Replacing the classic linear growth condition for  $F$ by $\eqref{H}$ and using similar reasoning as in the proof of Theorem 2.2, p.~104 in \cite{Friedman}, one can deduce the existence and uniqueness of the unique strong solution of \eqref{over}.
We denote by $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\geq 0},\mathbb{P})$ the filtered probability space where
$\eqref{over}$ is defined and by $\mathbb{E}$ the expectation with respect to $\mathbb{P}$.

\subsection{\textbf{Invariant distribution}}
Existence of invariant measures for stochastic processes are an important feature in probability
theory and mathematical physics; and typically they are not so easy to describe explicitly.
By Theorem~3.3.4 p.~91 in \cite{KUL}, it is not hard to verify that  Hypotheses \eqref{H}, \eqref{C3}, \eqref{eq:lipt} and \eqref{eq:elip} yield the existence and uniqueness of an invariant probability measure $\mu^{\epsilon}$ for the stochastic dynamics \eqref{over}.
If in addition,
$F(x)=\nabla V(x) + b(x)$ and $\sigma(x)=I_d$ for any $x\in \mathbb{R}^d$,
where $V:\mathbb{R}^d\rightarrow \mathbb{R}$ is a scalar function and $b:\mathbb{R}^d\rightarrow \mathbb{R}^d$
is a vector field
which satisfies the divergence-free condition
\begin{equation}\label{eq:dvfree}
\sum_{j=1}^d \frac{\partial}{\partial x_j}\left(\exp(-(\nicefrac{2}{\epsilon})V(x))b(x)\right)=0\quad \textrm{ for any } x=(x_1,\ldots,x_d)\in \mathbb{R}^d,
\end{equation}
one can verify that
$\exp(-(\nicefrac{2}{\epsilon})V(x))\ud x$
is a stationary measure for the random dynamics \eqref{over}. However,  it might not be a probability measure.
Under some appropriate assumptions on $V$ for $\|x\|\gg 1$,
the unique invariant probability measure $\mu^{\epsilon}$ of \eqref{over} is of the Gibbs type
\begin{equation}\label{eq:gibbs}
\mu^{\epsilon}(\ud x) =
\frac{\exp(-(\nicefrac{2}{\epsilon})V(x))}{\mathcal{Z}^{\epsilon}}\ud x,
\end{equation}
where $\mathcal{Z}^\epsilon$ is
the so-called partition function (normalized constant).
See for instance Chapter 2, Convergence of the Langevin process, p.~21-23 in \cite{VI}.
Using the Laplace Method (Saddle-point Method), asymptotics as
$\epsilon\to 0^+$ for the density of $\mu^{\epsilon}$
can be carried out, see for instance \cite{AHWANG} and \cite{Hwang1980}.

If we drop the free-divergence condition \eqref{eq:dvfree} and replace it by the
transversal condition
$\<\nabla V(x),b(x)\>=0$ for all $x\in \mathbb{R}^d$, in \cite{Sheu1986}
 it is shown a beautiful expansion on $\epsilon$ for the density of  $\mu^{\epsilon}$.
However, this expansion requires smoothness of the so-called  Freidlin-Wentzell quasipotential.
The latter is a nontrivial mathematical problem since it is expressed by a variational principle.
For additive noise, and bounded and dissipative vector field $F$, in \cite{Mikami1988},
by way of Watanabe's theory and Malliavin calculus, it is proved an asymptotic expansion of $\mu^\epsilon$.
Later, in \cite{Mikami} it is shown that $\mu^\epsilon$ can be expanded in Wentzel-Kramers-Brillouin (W.K.B.) type, as $\epsilon \to 0^+$, in the set in which the quasipotential is of $\mathcal{C}^\infty$-class and each coefficient which appears in the expansion is of $\mathcal{C}^\infty$-class. More recently, in \cite{Biswas2009}, using control theoretic methods, it is proved that $\mu^\epsilon(\ud x)\approx \exp(-\nicefrac{V_*(x)}{\epsilon})$, $\epsilon\ll 1$, where  $V_*$ is characterized as the optimal cost of a deterministic control problem. Nevertheless, the control problem is not easy to solve explicitly.
In \eqref{over} we consider  multiplicative noise, and no transverse condition on the vector field $F$ is assumed. Moreover, we do not
need that the Gibbs measure \eqref{eq:gibbs} remains stationary, and no smoothness on
$\mu^\epsilon$ and the Freidlin-Wentzell quasipotential are needed.
We remark that \textit{generically} it is not possible to compute an explicit formula for $\mu^{\epsilon}$.

\subsection{\textbf{Informal result}}
Our goal is to prove that the probability  $\epsilon^{\nicefrac{d}{2}}
\mu^{\epsilon}(\sqrt{\epsilon} \ud x)$ has a Gaussian shape in
the small noise limit. To be more precise, under Hypotheses  \eqref{H}, \eqref{C3}, \eqref{eq:lipt} and \eqref{eq:elip},
it follows that the probability measure
\[
\epsilon^{\nicefrac{d}{2}}
\mu^{\epsilon}(\sqrt{\epsilon} \ud x)
\]
converges in the $p$-Wasserstein ($p\in [1,2]$) to a Gaussian $\mathcal{N}$ distribution with zero-mean vector and covariance matrix given by the unique solution of the Lyapunov matrix equation
\begin{equation*}\label{eq:mlyaeq}
DF(0)\mathbb{X}+\mathbb{X}(DF(0))^*=\sigma(0_d)(\sigma(0_d))^*.
\end{equation*}
Generically, it is hard to find an explicit formula for the solution of
the preceding matrix equation. Nevertheless, it can be estimated via numerical algorithms, see for instance \cite{Benner}, \cite{Hodel} and the references therein.
Roughly speaking, it is shown that
\begin{equation}\label{eq:cuantif}
\mu^{\epsilon}(\ud x) \approx \epsilon^{-d/2}\mathcal{N}\left(\nicefrac{\ud x}{\sqrt{\epsilon}}\right)\quad \textrm{ for } \epsilon \ll 1.
\end{equation}
We anticipate that the proof does not rely on explicit computations of the distribution $\mu^\epsilon$. It is based
 on the linearization of the nonlinear dynamics around the stationary
point $0_d$. It is not hard to see that the resulting linear process has the target Gaussian as invariant distribution. It is then necessary to control the
difference between this linear process and the nonlinear dynamics. This is done using
the so-called synchronous coupling techniques.
In this manuscript, we provide the rigorous mathematical
proof.
The proof is purely dynamic and
it does not require techniques as
Malliavin calculus, large deviation theory for SDEs as in \cite{Freidlin}, smoothness of the quasipotential, smoothness of the density $\mu^\epsilon$,
analysis of the infinitesimal generator or the  W.K.B. expansion.
We also quantify in the Wasserstein distance  the implicit error term given in \eqref{eq:cuantif}.

\subsection{\textbf{Wasserstein distance}}
Let $\mathcal{P}$ be the set of probability measures in $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$, where $\mathcal{B}(\mathbb{R}^d)$ denotes the Borel $\sigma$-algebra of $\mathbb{R}^d$.
For $p\geq 1$ we define
\[
\mathcal{P}_p:=\left\{\mu\in \mathcal{P}: \int_{\mathbb{R}^d}\|x\|^p\mu(\ud x)<\infty \right\},
\]
the space of probability measures with finite $p$-moment. For any $\mu,\nu\in \mathcal{P}$ we say that a probability measure ${\pi}_*$ in the measurable space
$(\mathbb{R}^d \times \mathbb{R}^d,\mathcal{B}(\mathbb{R}^d\times \mathbb{R}^d))$ is a {\em coupling} between $\mu$ and $\nu$ if the marginals of ${\pi}_*$ are $\mu$ and $\nu$, that is,
for any $B\in \mathcal{B}{(\mathbb{R}^d)}$ it follows that
\[
{\pi}_*(B \times \mathbb{R}^d) = \mu(B)\quad \text{ and } \quad {\pi}_*(\mathbb{R}^d \times B) = \nu(B).
\]
Let $\Pi(\mu,\nu)$ the set of all coupling between $\mu$ and $\nu$.
For any $\mu,\nu\in \mathcal{P}_p$, the Wasserstein distance of order $p$ between $\mu$ and $\nu$, $\mathcal{W}_p(\mu,\nu)$, is defined by
\[
\mathcal{W}_p(\mu,\nu):=\inf\limits
\left\{\left(\int_{\mathbb{R}^d \times \mathbb{R}^d}\|x-y\|^p {\pi}_*(\ud x, \ud y)\right)^{\nicefrac{1}{p}}:
{\pi}_* \in \Pi(\mu,\nu)
\right\}.
\]
Let $X$ and $Y$ be two random vectors on $\mathbb{R}^d$ defined on the probability space $(\Omega,\mathcal{F},\mathbb{P})$ with finite $p$-moment. The Wasserstein distance of order $p$ between $X$ and $Y$, $\mathfrak{W}_p(X,Y)$, is defined by
$\mathfrak{W}_p(X,Y):=\mathcal{W}_p(\mathbb{P}_X,\mathbb{P}_Y)$,
where $\mathbb{P}_X$ and $\mathbb{P}_Y$ are  the push-forward probability measures
$\mathbb{P}_X(B):=\mathbb{P}(X\in B)$
and
$\mathbb{P}_Y(B):=\mathbb{P}(Y\in B)$
for any $B\in\mathcal{B}(\mathbb{R}^d)$.
For short, we write
$\mathcal{W}_p(X,Y)$ in place of $\mathfrak{W}_p(X,Y)$.
A remarkable property that we use along this manuscript is the following scaling property
\begin{equation}\label{eq:scaling}
\mathcal{W}_p(\mathfrak{c}X,\mathfrak{c}Y)=|\mathfrak{c}|\mathcal{W}_p(X,Y)
\quad \textrm{ for any } \mathfrak{c}\in \mathbb{R}.
\end{equation}
The Wasserstein distance metrizes the weak convergence in the space of probabilities with finite $p$-moment.
It is convenient to use in optimal transport problems.
The Wasserstein distance is a natural way  to compare (even for degenerate) the law  of two random variables  $X$ and $Y$, where one variable is derived from the other by a small perturbation.
For further details and properties of the Wassertein distance, we refer to the monographies \cite{PZ} and \cite{VI}.

\subsection{\textbf{Results}}
We denote by  $\mathcal{N}{\left(v,\Xi \right)}$ the Gaussian distribution in $\mathbb{R}^d$ with vector mean $v$ and positive definite covariance matrix $\Xi$. Let $I_d$ be the identity $d\times d$-matrix. Given a matrix $A\in \mathbb{R}^{d\times d}$, denote by $A^*$ the transpose matrix of $A$ and denote by $\mathrm{Tr}(A)$ the trace of $A$.

The main result of this manuscript is the following.
\begin{theorem}\label{main}
Assume Hypotheses \eqref{H}, \eqref{C3}, \eqref{eq:lipt} and \eqref{eq:elip} are valid.
Let $\mathcal{J}^{\epsilon}$ be a random vector on $\mathbb{R}^d$ with distribution $\mu^{\epsilon}$. Then there exists a positive  constant $K:=K(\delta,\ell,d,c_0,\sigma(0_d))$
such that for any $\epsilon\in (0,\epsilon_*)$ with
\[
\epsilon_*=\min\left\{\frac{\delta}{8c_1\|\sigma(0_d)(\sigma(0_d))^*\|_{\mathrm{F}}\cdot d^2},\frac{ \delta}{\ell^2}\right\}
\]
 it follows that
\begin{equation}\label{eq:resultado}
\mathcal{W}_2\left(\frac{\mathcal{J}^{\epsilon}}{\sqrt{\epsilon}},\mathcal{N}\right)\leq K\sqrt{\epsilon},
\end{equation}
where $\mathcal{N}$ denotes
the Gaussian distribution on $\mathbb{R}^d$ with zero-mean vector and covariance matrix $\Sigma$ which is the unique solution of the Lyapunov matrix equation
\begin{equation}\label{eq:mleq}
DF(0)\mathbb{X}+\mathbb{X}(DF(0))^*=\sigma(0_d)(\sigma(0_d))^*.
\end{equation}
\end{theorem}
Using the coupling approach, rates of
convergence of the time evolution to equilibrium in the Wasserstein distance for Langevin processes are given in \cite{Eberle2019}. However, the authors in \cite{Eberle2019} do not study small random perturbations of dynamical systems, and hence,  an asymptotic analysis for the invariant measure is not needed there.

The proof of Theorem~\ref{main}
does not rely on explicit computations of  $\mu^\epsilon$
neither on explicit formula of the Wasserstein distance of order $2$ between Gaussian distributions. The It\^o formula with the help of \eqref{H} and \eqref{eq:lipt} implies that
 the $p$-moments are bounded recursively as a function of moments of order $p$ and $p-2$.
Consequently, by an analogous reasoning (but more involved) one can see that the proof of Theorem~\ref{main} can be adapted for any $p\in \{4,6,\ldots,\}$ and hence for any $p\geq 1$.

\begin{remark}
The constant $K$ can be taken as
\[
K=\frac{96c_0}{\|\sigma(0_d)(\sigma(0_d))^*\|_{\mathrm{F}}\cdot d^2}+
\frac{2\ell C^{\nicefrac{1}{2}}_0}{\delta}\quad \textrm{ with } C_0=\mathrm{Tr}((\sigma(0_d))^*\sigma(0_d)).
\]
We emphasize that the error term $K\sqrt{\epsilon}$ may not be optimal
\end{remark}
\begin{remark}
By \eqref{H} and \eqref{eq:elip}, Theorem~1, p. ~443 of \cite{LANTI} implies that \eqref{eq:mleq} possesses a unique solution. Moreover, Theorem~3, p.~414 of \cite{LANTI} yields the integral representation
\[
\mathbb{X}=\int_{0}^\infty
e^{-DF(0)s}
\sigma(0_d)(\sigma(0_d))^*
e^{-(DF(0))^*s}
\ud s.
\]
\end{remark}

As a consequence of Theorem~\ref{main} we have the following corollaries.
\begin{corollary}[$\mathcal{W}_p$ convergence for \textrm{$p\in [1,2]$}]\label{cor:p}
Assume Hypotheses \eqref{H}, \eqref{C3}, \eqref{eq:lipt} and \eqref{eq:elip} are valid.
Let $\mathcal{J}^\epsilon $, $\mathcal{N}$, $K$ and $\epsilon_*$ be as in Theorem~\ref{main}.
For any $p\in [1,2]$ it follows that
\[
\mathcal{W}_p\left(\frac{\mathcal{J}^{\epsilon}}{\sqrt{\epsilon}},\mathcal{N}\right)\leq K\sqrt{\epsilon}\quad \textrm{ for all } \epsilon\in (0,\epsilon_*).
\]
\end{corollary}
\begin{proof}
The proof follows immediately by the H\"older inequality and \eqref{eq:resultado}.
\end{proof}
\begin{corollary}[Concentration]\label{cor:conce}
Assume Hypotheses \eqref{H}, \eqref{C3}, \eqref{eq:lipt} and \eqref{eq:elip} are valid.
Let $\mathcal{J}^\epsilon $ be as in Theorem~\ref{main}. Denote by $\delta_{0_d}$ the Dirac measure at $0_d$.
For any $p\in [1,2]$ and $\beta<\nicefrac{1}{2}$ it follows that
\[
\lim\limits_{\epsilon \to 0^+}\frac{1}{\epsilon^{\beta}}\mathcal{W}_p(\mathcal{J}^{\epsilon},\delta_{0_d})=0.
\]
\end{corollary}
Theorem~1 in \cite{Biswas2009} implies that $\mathcal{J}^\epsilon\to \delta_{0_d}$ as $\epsilon \to 0^+$ in distribution sense. However, it does not say anything about its rate of convergence as Corollary~\ref{cor:conce}.
Results about quantitative
concentration of stationary measures on
attractors and repellers for multiplicative noise are given in \cite{Yingfei} and\cite{YingfeiShen}.
\begin{proof}
By the triangle inequality for the $\mathcal{W}_p$ distance we have
\begin{align}\label{eq:cor1}
\mathcal{W}_p(\mathcal{J}^{\epsilon},\delta_{0_d})&\leq
\mathcal{W}_p(\mathcal{J}^{\epsilon},\sqrt{\epsilon}\mathcal{N})+
\mathcal{W}_p(\sqrt{\epsilon}\mathcal{N},\delta_{0_d}),
\end{align}
where $\mathcal{N}$ denotes the Gaussian distribution defined in Theorem~\ref{main}.
Property \eqref{eq:scaling} yields
\begin{equation}\label{eq:cor2}
\mathcal{W}_p(\mathcal{J}^{\epsilon},\sqrt{\epsilon}\mathcal{N})=\sqrt{\epsilon}\mathcal{W}_p\left(\nicefrac{\mathcal{J}^{\epsilon}}{\sqrt{\epsilon}},\mathcal{N}\right).
\end{equation}
Note that
\begin{equation}\label{eq:cor3}
\mathcal{W}_p(\sqrt{\epsilon}\mathcal{N},\delta_{0_d})=\sqrt{\epsilon}(\mathbb{E}[\|\mathcal{N}\|^p])^{\nicefrac{1}{p}}.
\end{equation}
Combining \eqref{eq:cor1}, \eqref{eq:cor2} and \eqref{eq:cor3} with the help of Corollary~\ref{cor:p} implies the statement.
\end{proof}

The rest of the manuscript is organized as follows. Section \ref{sec:outline} describes the outline of the proof for the main Theorem~\ref{main}.
Section \ref{sec:proof} is
devoted to the proofs of the results skipped in Section \ref{sec:outline}.
Finally, in Appendix \ref{Ap:moments} we provide polynomials and exponential moments estimates for the Ornstein-Uhlenbeck process that we use  in Section \ref{sec:proof}.

\section{\textbf{Outline of the proof}}\label{sec:outline}
\subsection{\textbf{Linear diffusion approximation}}\label{sub:slda}
Due to the dissipativity condition \eqref{H},
the nonlinear random dynamics $(X^{\epsilon}_t(x))_{t\geq 0}$ is pushed-back to the origin with high probability.
In a neighbourhood of the origin, it is reasonable
that an Ornstein-Uhlenbeck process helps us to understand $(X^{\epsilon}_t(x))_{t\geq 0}$ for large times.
Let $(Y_t(x))_{t\geq 0}$ be the unique strong solution of the following linear SDE
\begin{equation}
\label{SEDONOU}
\left\{
\begin{array}{r@{\;=\;}l}
\ud Y^\epsilon_t(x) & - DF(0_d)Y^\epsilon_t(x) \ud t + \sqrt{\epsilon}\sigma(0_d)\ud B_t\quad
\textrm{ for any } t\geq 0,\\
Y^\epsilon_0(x) & x,
\end{array}
\right.
\end{equation}
where $(B_t)_{t\geq 0}$ is a standard Brownian motion on $\mathbb{R}^d$ and
$DF(0_d)$ denotes the Jacobian matrix at the point $0_d$.
The variation of constant formula yields
\begin{equation*}
Y^{\epsilon}_t(x)=e^{-DF(0_d)t}x+\sqrt{\epsilon}\, e^{-DF(0_d)t}\int_{0}^{t}e^{DF(0_d)s}\sigma(0_d)\ud B_s\quad \textrm{ for any } t\geq 0.
\end{equation*}
The preceding formula implies that
for any $t>0$, $Y^{\epsilon}_t(x)$ possesses Gaussian distribution with vector mean $m^{\epsilon}_t(x):=e^{-DF(0_d)t}x$ and covariance matrix
$\Sigma^{\epsilon}_t:=\epsilon\Sigma_t$ for any $t\geq 0$,
where $(\Sigma_t)_{t\geq 0}$ solves the following matrix differential equation
\begin{equation}
\label{EDO111}
\left\{
\begin{array}{r@{\;=\;}l}
\frac{\ud}{ \ud t}\Sigma_t & -DF(0_d)\Sigma_t-\Sigma_t(DF(0_d))^*+\sigma(0_d)(\sigma(0_d))^*\quad \textrm{ for any } t\geq 0,\\
\Sigma_0 & 0_{d\times d},
\end{array}
\right.
\end{equation}
where $0_{d\times d}$ is the $d$-squared zero matrix. We refer to Section 3.7 in \cite{pavliotisbook} for further details.
By \eqref{H} one can easily see that
the eigenvalues of $DF(0_d)$ are contained in the set
$\{z\in \mathbb{C}: \Re{(z)}\geq \delta\}$.
As a consequence, we have
\[
\|m^{\epsilon}_t(x)\|\leq e^{-\delta t}\|x\|\to 0 \quad \textrm{ as } t\to \infty.
\]
Lemma~\ref{lem:covariance} in Appendix \ref{Ap:moments} implies
\begin{equation*}
\|\Sigma_t-\Sigma\|_{\mathrm{F}}\leq
\|\Sigma\|^2_{\mathrm{F}} \, e^{-2\delta t}\to 0 \quad \textrm{ as } t\to \infty,
\end{equation*}
where
$\Sigma$ is the unique solution of the matrix Lyapunov equation \eqref{eq:mleq}.
Therefore, the limiting distribution of $Y^{\epsilon}_t(x)$ is a Gaussian law with zero-mean vector and covariance matrix
$\epsilon\Sigma$.
Moreover, Proposition~3.5 in \cite{pavliotisbook} implies that $\mathcal{N}(0,\epsilon\Sigma)$
is the unique invariant probability measure for the dynamics given by \eqref{SEDONOU}.
\subsection{\textbf{Disintegration}}
For short we write $\mathcal{N}$
in a place of $\mathcal{N}(0,\Sigma)$.
Recall that $\mathcal{J}^{\epsilon}$ denotes a random vector on $\mathbb{R}^d$ with distribution $\mu^{\epsilon}$.
Let $t\geq 0$ and $x_0\in \mathbb{R}^d$. The triangle inequality for the distance $\mathcal{W}_2$ yields
\begin{align}\label{ine0}
\mathcal{W}_2\left(\mathcal{J}^{\epsilon},\sqrt{\epsilon}\mathcal{N}\right)&\leq
\mathcal{W}_2\left(\mathcal{J}^{\epsilon},X^{\epsilon}_t(x_0)\right)+\mathcal{W}_2\left(X^{\epsilon}_t(x_0),Y^{\epsilon}_t(x_0)\right)+\mathcal{W}_2\left(Y^{\epsilon}_t(x_0),\sqrt{\epsilon}\mathcal{N}\right).
\end{align}
Since $\mu^{\epsilon}$ is invariant for the dynamics \eqref{over}, for any $t\geq 0$,
$X^{\epsilon}_t(\mathcal{J}^{\epsilon})$ has distribution $\mu^{\epsilon}$.
By disintegration, the first-term of the right-hand side of \eqref{ine0} can be estimated as follows
\begin{align}\label{ine200}
\mathcal{W}_2\left(\mathcal{J}^{\epsilon},X^{\epsilon}_t(x_0)\right)
\leq \int_{\mathbb{R}^d}\mathcal{W}_2\left(X^{\epsilon}_t(x),X^{\epsilon}_t(x_0)\right)
\mu^{\epsilon}(\ud x).
\end{align}
Analogously,
\begin{align}\label{ine600}
\mathcal{W}_2\left(\sqrt{\epsilon}\mathcal{N},Y^{\epsilon}_t(x_0)\right)
\leq \int_{\mathbb{R}^d}\mathcal{W}_2\left(Y^{\epsilon}_t(x),Y^{\epsilon}_t(x_0)\right)
\mathcal{N}(0_d,\epsilon \Sigma)(\ud x),
\end{align}
where $\mathcal{N}(0_d,\epsilon \Sigma)(\ud x)$ denotes the density of $\sqrt{\epsilon}\mathcal{N}$.
Combining \eqref{ine0},
\eqref{ine200} and \eqref{ine600} we obtain
\begin{align*}
\mathcal{W}_2\left(\mathcal{J}^{\epsilon},\sqrt{\epsilon}\mathcal{N}\right)\leq & \int_{\mathbb{R}^d}\mathcal{W}_2\left(X^{\epsilon}_t(x),X^{\epsilon}_t(x_0)\right)
\mu^{\epsilon}(\ud x)
+\mathcal{W}_2\left(X^{\epsilon}_t(x_0),Y^{\epsilon}_t(x_0)\right)\\
&\quad+\int_{\mathbb{R}^d}\mathcal{W}_2\left(Y^{\epsilon}_t(x),Y^{\epsilon}_t(x_0)\right)
\mathcal{N}(0_d,\epsilon \Sigma)(\ud x)
\end{align*}
for any $t\geq 0$ and $x_0\in \mathbb{R}^d$.
In particular,  for any $t\geq 0$ we have
\begin{equation}\label{formulainequality}
\begin{split}
\mathcal{W}_2\left(\mathcal{J}^{\epsilon},\sqrt{\epsilon}\mathcal{N}\right)\leq & \int_{\mathbb{R}^d}\mathcal{W}_2\left(X^{\epsilon}_t(x),X^{\epsilon}_t(0_d)\right)
\mu^{\epsilon}(\ud x)
+\mathcal{W}_2\left(X^{\epsilon}_t(0_d),Y^{\epsilon}_t(0_d)\right)\\
&\quad+\int_{\mathbb{R}^d}\mathcal{W}_2\left(Y^{\epsilon}_t(x),Y^{\epsilon}_t(0_d)\right)
\mathcal{N}(0_d,\epsilon \Sigma)(\ud x).
\end{split}
\end{equation}
In what follows, we provide the tools for estimating the right-hand side of the preceding inequality.
The following lemma allows us to couple two solutions of \eqref{over} starting in  different initial conditions.
\begin{lemma}[Synchronous coupling I]\label{lemmaA}
Assume Hypotheses \eqref{H}, \eqref{C3}, \eqref{eq:lipt} and \eqref{eq:elip} are valid.
Let $x,x_0\in \mathbb{R}^d$. Then
\[
\mathcal{W}_2\left(X^{\epsilon}_t(x),X^{\epsilon}_t(x_0)\right)\leq e^{-(\nicefrac{\delta}{2}) t}\|x-x_0\|\quad \textrm{ for all } t\geq 0,\,\epsilon \in (0,\nicefrac{\delta}{\ell^2}],
\]
where $\delta>0$ is the dissipativity constant that appears in \eqref{H} and $\ell$ is the Lipschitz constant that appears in \eqref{eq:lipt}.
In particular,
\[
\mathcal{W}_2\left(X^{\epsilon}_t(x),X^{\epsilon}_t(0_d)\right)\leq e^{-(\nicefrac{\delta}{2}) t}\|x\|\quad \textrm{ for all } t\geq 0,\,\epsilon \in (0,\nicefrac{\delta}{\ell^2}].
\]
\end{lemma}
The following lemma provides second moment estimates for the marginals of the process \eqref{over} and also for its invariant probability measure $\mu^\epsilon$.
\begin{lemma}[Second moment estimates]\label{lemmaB}
Assume Hypotheses \eqref{H}, \eqref{C3}, \eqref{eq:lipt} and \eqref{eq:elip} are valid.
For any $x\in \mathbb{R}^d$ we have
\[
\mathbb{E}[\|X^{\epsilon}_t(x)\|^2]\leq
\|x\|^2 e^{-\delta t}+\frac{ \epsilon C_0}{\delta}\quad \textrm{ for all } t\geq 0,\,\epsilon \in (0,\nicefrac{\delta}{\ell^2}],
\]
where $\delta>0$ is the dissipativity constant that appears in \eqref{H}, $\ell$ is the Lipschitz constant that appears in \eqref{eq:lipt}
and $C_0=\mathrm{Tr}((\sigma(0_d))^*\sigma(0_d))$.
In addition,
\[
\int_{\mathbb{R}^d}\|x\|^2 \mu^{\epsilon}(\ud x)\leq \frac{\epsilon C_0}{\delta}
\quad \textrm{ for all }\epsilon \in (0,\nicefrac{\delta}{\ell^2}].
\]
\end{lemma}
We stress that $C_0\geq \kappa d$
where $\kappa$ is the positive constant that appears in \eqref{eq:elip}.
The next lemma is crucial in our argument. Due to the contracting nature of the dynamics, the random dynamics around zero,
$(X^\epsilon_t(0))_{t\geq 0}$, can be approximated from its linearization $(Y^\epsilon_t(0))_{t\geq 0}$.
\begin{lemma}[Synchronous coupling II]\label{lemmaC}
Assume Hypotheses \eqref{H}, \eqref{C3}, \eqref{eq:lipt} and \eqref{eq:elip} are valid.
Then there exists a positive constant $C:=C(\delta,\ell, d,c_0,\sigma(0_d))$ such that for any $\epsilon\in (0,\epsilon_*)$ with
$\epsilon_*:=\min\big\{\frac{\delta}{8c_1\|\sigma(0_d)(\sigma(0_d))^*\|_{\mathrm{F}}\cdot d^2},\frac{ \delta}{\ell^2}\big\}$,
 and for all $t\geq 0$ we have
\[
\mathcal{W}_2\left(X^{\epsilon}_t(0_d),Y^{\epsilon}_t(0_d)\right)\leq
\epsilon C,
\]
where $\delta>0$ is the dissipativity constant that appears in \eqref{H},
$\ell$ is the Lipschitz constant that appears in \eqref{eq:lipt},
 and $c_0$, $c_1$ are the positive constants that appear in \eqref{C3}.
\end{lemma}
We point out that the constant $C$ can be taken as
\[
C=\frac{48c_0}{\|\sigma(0_d)(\sigma(0_d))^*\|_{\mathrm{F}}\cdot d^2}+\frac{\ell C^{\nicefrac{1}{2}}_0}{\delta}.
\]
The latter is deducted from \eqref{eq:contc}.
Recall that $\Sigma$ is the solution of \eqref{EDO111}.
Since for any $t\geq 0$, $Y^{\epsilon}_t(\mathcal{N}(0_d,\epsilon\Sigma_t))$ has distribution  $\mathcal{N}(0_d,\epsilon\Sigma)$, an analogous reasoning using in the proofs of Lemma~\ref{lemmaA} and Lemma~\ref{lemmaB} implies the following lemma.
\begin{lemma}[Synchronous coupling III]\label{lemmaD}
Assume Hypotheses \eqref{H}, \eqref{C3}, \eqref{eq:lipt} and \eqref{eq:elip} are valid.
For any $x\in \mathbb{R}^d$ it follows that
\[
\mathcal{W}_2\left(Y^{\epsilon}_t(x),Y^{\epsilon}_t(0_d)\right)\leq e^{-(\nicefrac{\delta}{2}) t}\|x\|\quad \textrm{ for all } t\geq 0,\,\epsilon \in (0,\nicefrac{\delta}{\ell^2}],
\]
where $\delta>0$ is the dissipativity constant that appears in \eqref{H} and $\ell$ is the Lipschitz constant that appears in \eqref{eq:lipt}.
In addition,
\[
\int_{\mathbb{R}^d}\|x\|^2 \mathcal{N}(0_d,\epsilon \Sigma)(\ud x)\leq  \epsilon d \|\Sigma^{\nicefrac{1}{2}}\|^2_{\mathrm{F}}.
\]
\end{lemma}
In the sequel, we stress the fact that
Theorem~\ref{main} is just a consequence of what we have already stated up to here.
Indeed, by  \eqref{formulainequality}, Lemma~\ref{lemmaA}, Lemma~\ref{lemmaB}, Lemma~\ref{lemmaC} and Lemma~\ref{lemmaD} we have
\begin{align*}
\mathcal{W}_2\left(\mathcal{J}^{\epsilon},\sqrt{\epsilon}\mathcal{N}\right)\leq &
\sqrt{\frac{\epsilon C_0}{\delta}}e^{-(\nicefrac{\delta}{2}) t}
+C\epsilon
+
\sqrt{\epsilon d \|\Sigma^{\nicefrac{1}{2}}\|^2_{\mathrm{F}}}e^{-(\nicefrac{\delta}{2}) t}
\end{align*}
for any $t\geq 0$ and $\epsilon\in (0,\epsilon_*]$.
Due to \eqref{eq:scaling}, the preceding inequality implies
\begin{align*}
\mathcal{W}_2\left(\frac{\mathcal{J}^{\epsilon}}{\sqrt{\epsilon}},\mathcal{N}\right)\leq & \sqrt{\frac{ C_0}{\delta}}e^{-(\nicefrac{\delta}{2}) t}
+C\sqrt{\epsilon}+
\sqrt{ d \|\Sigma^{\nicefrac{1}{2}}\|^2_{\mathrm{F}}}e^{-(\nicefrac{\delta}{2}) t}
\end{align*}
for any $t \geq 0$ and $\epsilon\in (0,\epsilon_*]$.
The cunning choice
\[
t_\epsilon=\max\left\{\frac{1}{\delta}\ln\Big(\frac{4C_0}{\delta C^2\epsilon}\Big),\frac{1}{\delta}\ln\Big(\frac{4d \|\Sigma^{\nicefrac{1}{2}}\|^2_{\mathrm{F}}}{ C^2\epsilon}\Big)\right\}
\]
 yields
\[
\mathcal{W}_2\left(\frac{\mathcal{J}^{\epsilon}}{\sqrt{\epsilon}},\mathcal{N}\right)\leq 2C\sqrt{\epsilon},
\]
which concludes Theorem~\ref{main}.
\section{\textbf{Proofs}}\label{sec:proof}
In this section, we give the proofs of the Lemma~\ref{lemmaA}, Lemma~\ref{lemmaB}, Lemma~\ref{lemmaC} and Lemma~\ref{lemmaD}.
Along their proofs, we use several times
the celebrated Gr\"onwall inequality.
We state it here as a lemma for the sake of completeness.
\begin{lemma}[Gr\"onwall inequality]\label{lem:gron} Let $T>0$ be fixed, $\mathfrak{g}:[0,T]\rightarrow \mathbb{R}$
be a $\mathcal{C}^1$-function and
$\mathfrak{h}:[0,T]\rightarrow \mathbb{R}$ be a $\mathcal{C}^0$-function. Assume that
\[
\frac{\ud}{\ud t} \mathfrak{g}(t)\leq -a
\mathfrak{g}(t) + \mathfrak{h}(t)\quad
\textrm{ for any } t \in [0, T],
\]
where $a\in\mathbb{R}$, and the derivative at $0$ and $T$ are understood as the right
and left derivatives, respectively.
Then
\[
\mathfrak{g}(t)\leq e^{-at}\mathfrak{g}(0) + e^{-at}\int_{0}^{t}
e^{as}\mathfrak{h}(s)\ud s\quad \textrm{ for any } t\in [0, T].
\]
\end{lemma}

\subsection{\textbf{The synchronous coupling I}}
For any $x,x_0\in \mathbb{R}^d$, let
$(X^{\epsilon}_t(x))_{t\geq 0}$ and $(X^{\epsilon}_t(x_0))_{t\geq 0}$ be the solutions of \eqref{over} with initial conditions $x$ and $x_0$, respectively.
In the sequel, we consider the so-called synchronous coupling, i.e., both processes $(X^{\epsilon}_t(x))_{t\geq 0}$ and $(X^{\epsilon}_t(x_0))_{t\geq 0}$ have  the same  driving noise $(B_t)_{t\geq 0}$.

\noindent
\textbf{\textit{Proof of Lemma~\ref{lemmaA}}.}
By the It\^o formula we have
\begin{align*}
\ud \|X^{\epsilon}_t(x)-X^{\epsilon}_t(x_0)\|^2&=-2\<X^{\epsilon}_t(x)-X^{\epsilon}_t(x_0),F(X^{\epsilon}_t(x)) -F(X^{\epsilon}_t(x_0))\>\ud t\\
&\quad+
\epsilon\,\mathrm{Tr}[(\sigma(X^{\epsilon}_t(x))-\sigma(X^{\epsilon}_t(x_0)))^* (\sigma(X^{\epsilon}_t(x))-\sigma(X^{\epsilon}_t(x_0)))]\ud t
\\
&\quad+2\sqrt{\epsilon}\<X^{\epsilon}_t(x)-X^{\epsilon}_t(x_0),(\sigma(X^{\epsilon}_t(x))-\sigma(X^{\epsilon}_t(x_0)))\ud B_t\>.
\end{align*}
By \eqref{eq:lipt} we have
\begin{equation}\label{eq:traze1}
\mathrm{Tr}[(\sigma(X^{\epsilon}_t(x))-\sigma(X^{\epsilon}_t(x_0)))^* (\sigma(X^{\epsilon}_t(x))-\sigma(X^{\epsilon}_t(x_0)))]\leq \ell^2 \|X^{\epsilon}_t(x)-X^{\epsilon}_t(x_0)\|^2.
\end{equation}
A localization argument with the help of \eqref{H} and
\eqref{eq:traze1}
implies
\begin{align*}
\frac{\ud}{\ud t} \mathbb{E}[\|X^{\epsilon}_t(x)-X^{\epsilon}_t(x_0)\|^2] &\leq -2\delta \mathbb{E}\left[\|X^{\epsilon}_t(x)-X^{\epsilon}_t(x_0)\|^2\right]
+ \epsilon  \ell^2  \, \mathbb{E}\left[\|X^{\epsilon}_t(x)-X^{\epsilon}_t(x_0)\|^2\right]\\
&\leq -(2\delta-\epsilon  \ell^2)\mathbb{E}\left[\|X^{\epsilon}_t(x)-X^{\epsilon}_t(x_0)\|^2\right]
\end{align*}
for all $t\geq 0$.
Since $\mathbb{E}[\|X^{\epsilon}_0(x)-X^{\epsilon}_0(x_0)\|^2]=\|x-x_0\|^2$,
Lemma~\ref{lem:gron} yields
\[
\mathbb{E}[\|X^{\epsilon}_t(x)-X^{\epsilon}_t(x_0)\|^2]\leq e^{-(2\delta-\epsilon  \ell^2) t}\|x-x_0\|^2\quad \textrm{  for  any } t\geq 0.
\]
Therefore, for any
$\epsilon\in (0,\nicefrac{\delta}{\ell^2}]$
we have
\[
\mathcal{W}_2(X^{\epsilon}_t(x),X^{\epsilon}_t(x_0))
\leq e^{-(\nicefrac{\delta}{2}) t}\|x-x_0\|
\quad \textrm{ for any }
x,x_0\in \mathbb{R}^d,\; t\geq 0.
\]
\subsection{\textbf{Second moment estimates}}
For any $x\in \mathbb{R}^d$, let
$(X^{\epsilon}_t(x))_{t\geq 0}$ be the solution of \eqref{over} with initial condition $x$.

\noindent
\textbf{\textit{Proof of Lemma~\ref{lemmaB}}.}
In the sequel, we  estimate
$\mathbb{E}[\|X^\epsilon_t(x)\|^2]$.
The It\^o formula
and \eqref{H} yield
\begin{align*}
\ud \|X^\epsilon_t(x)\|^2
		&= -2 \<X^\epsilon_t(x), F(X^\epsilon_t(x))\>\ud t+  \epsilon\,\mathrm{Tr}[(\sigma(X^{\epsilon}_t(x)))^*
\sigma(X^{\epsilon}_t(x))]+
		\<M^\epsilon_t(x), \ud B_t\> \nonumber\\
		&\leq -2\delta \|X^\epsilon_t(x)\|^2 \ud t +
\epsilon\,\mathrm{Tr}[(\sigma(X^{\epsilon}_t(x)))^*
\sigma(X^{\epsilon}_t(x))]
		 \ud t
		+\<M^\epsilon_t(x), \ud B_t\>,
\end{align*}
where ${M}^\epsilon_t(x):=2\sqrt{\epsilon}X^\epsilon_t(x)$ for every $t\geq 0$.
By \eqref{eq:lipt} we have
\begin{equation}\label{eq:traza2}
\begin{split}
\mathrm{Tr}[(\sigma(X^{\epsilon}_t(x)))^*
\sigma(X^{\epsilon}_t(x))]&=
\mathrm{Tr}[(\sigma(X^{\epsilon}_t(x))-\sigma(0_d))^*(
\sigma(X^{\epsilon}_t(x))-\sigma(0)_d)]
\\
&\quad+\mathrm{Tr}((\sigma(0_d))^*\sigma(0_d))\\
&\leq
\ell^2 \|X^{\epsilon}_t(x) \|^2+C_0,
\end{split}
\end{equation}
where $C_0:=\mathrm{Tr}((\sigma(0_d))^*\sigma(0_d))$.
 A localization argument with the help of \eqref{H} and
\eqref{eq:traza2}
implies
\begin{align*}
\frac{\ud}{\ud t}\mathbb{E}[\|X^\epsilon_t(x)\|^2]
	\leq -(2\delta-\epsilon \ell^2 ) \mathbb{E}[\|X^\epsilon_t(x)\|^2] + \epsilon C_0 \quad \textrm{ for any } t\geq 0.
\end{align*}
Since $\mathbb{E}[\|X^\epsilon_0(x)\|^2]=\|x\|^2$,
for any
$\epsilon\in (0,\nicefrac{\delta}{\ell^2}]$
Lemma~\ref{lem:gron}
yields
\[
\mathbb{E}[\|X^\epsilon_t(x)\|^2]\leq e^{-\delta t}\|x\|^2 +
\frac{\epsilon C_0}{\delta}(1-e^{-\delta t})\leq e^{-\delta t}\|x\|^2 +\frac{\epsilon C_0}{\delta}\quad
\textrm{
for any } t\geq 0, \, x\in \mathbb{R}^d.
\]
Following the same reasoning using in  p.~39 in \cite{BJ1},
it is not hard to see that  the preceding inequality implies
\begin{equation*}
\int_{\mathbb{R}^d} \|x\|^2\mu^{\epsilon}(\ud x)\leq \frac{\epsilon C_0}{\delta}\quad \textrm{ for all } \epsilon\in (0,\nicefrac{\delta}{\ell^2}].
\end{equation*}
\subsection{\textbf{The synchronous coupling II}}\label{sec: natcouII}
We consider the
solution of \eqref{over} with initial condition $x=0_d$,
$(X^{\epsilon}_t(0_d))_{t\geq 0}$.
Let $(Y^{\epsilon}_t(0_d))_{t\geq 0}$ being as \eqref{SEDONOU}. In this section, we use the synchronous coupling between
$X^{\epsilon}_t(0_d)$ and $Y^{\epsilon}_t(0_d)$, i.e., both processes $(X^{\epsilon}_t(0_d))_{t\geq 0}$ and $(Y^{\epsilon}_t(0_d))_{t\geq 0}$ have  the same  driving noise $(B_t)_{t\geq 0}$.

\noindent
\textbf{\textit{Proof of Lemma~\ref{lemmaC}}.}
In the sequel, we estimate
$\mathbb{E}\left[\|X^{\epsilon}_t(0_d)-Y^{\epsilon}_t(0_d)\|^2\right]$.
Note that $X^{\epsilon}_0(0_d)=Y^{\epsilon}_0(0_d)=0_d$,
\[
\ud X^{\epsilon}_t(0_d)=-F(X^{\epsilon}_t(0_d))\ud t+\sqrt{\epsilon} \sigma(X^{\epsilon}_t(0_d))\ud B_t
\]
and
\[
\ud Y^{\epsilon}_t(0_d)=-DF(0_d)Y^{\epsilon}_t(0_d)\ud t+\sqrt{\epsilon} \sigma(0_d) \ud B_t.
\]
Let $\Delta^{\epsilon}_t(0_d):=X^{\epsilon}_t(0_d)-Y^{\epsilon}_t(0_d)$, $t\geq 0$. Then
\[
\begin{split}
\ud \Delta^{\epsilon}_t(0_d)
&=-\left[F(X^{\epsilon}_t(0_d))-F(Y^{\epsilon}_t(0_d))\right]\ud t
+\left[DF(0_d)Y^{\epsilon}_t(0_d)-F(Y^{\epsilon}_t(0_d))\right]\ud t\\
&\quad+
\sqrt{\epsilon}( \sigma(X^{\epsilon}_t(0_d))
-\sigma(0_d)
)
\ud B_t.
\end{split}
\]
Hence, the It\^o formula reads
\[
\begin{split}
\ud \|\Delta^{\epsilon}_t(0_d)\|^2
&=-2\<\Delta^{\epsilon}_t(0_d),F(X^{\epsilon}_t(0_d))-F(Y^{\epsilon}_t(0_d))\>\ud t\\
&\quad+2\<\Delta^{\epsilon}_t(0_d),DF(0_d)Y^{\epsilon}_t(0_d)-F(Y^{\epsilon}_t(0_d))\>\ud t
\\
&\quad +\epsilon\, \mathrm{Tr}
[
(\sigma(X^{\epsilon}_t(0_d))
-\sigma(0_d))^*( \sigma(X^{\epsilon}_t(0_d))
-\sigma(0_d))
]\ud t \\
&\quad +2\sqrt{\epsilon}
\<\Delta^{\epsilon}_t(0_d),( \sigma(X^{\epsilon}_t(0_d))
-\sigma(0_d)
)
\ud B_t\>.
\end{split}
\]
By \eqref{eq:lipt} we have
\begin{equation}\label{eq:trazita}
\mathrm{Tr}
[
(\sigma(X^{\epsilon}_t(0_d))
-\sigma(0_d))^*( \sigma(X^{\epsilon}_t(0_d))
-\sigma(0_d)]\leq \ell^2 \|X^{\epsilon}_t(0_d)\|^2.
\end{equation}
A localization argument with the help of \eqref{H},
the Cauchy-Schwarz inequality
and \eqref{eq:trazita}
implies
\[
\begin{split}
\frac{\ud}{\ud t}
\mathbb{E}[\|\Delta^{\epsilon}_t(0_d)\|^2] &\leq -2\delta \mathbb{E}[\|\Delta^{\epsilon}_t(0_d)\|^2]\\
&\hspace{-0.5cm}+2
\mathbb{E}[\|\Delta^{\epsilon}_t(0_d)\|\cdot
\|F(Y^{\epsilon}_t(0_d))-DF(0_d)Y^{\epsilon}_t(0_d)\|]+\epsilon \ell^2 \mathbb{E}[\|X^{\epsilon}_t(0_d)\|^2].
\end{split}
\]
The preceding differential inequality and the Young inequality (for $p=2$) yield
\[
\begin{split}
\frac{\ud}{\ud t}
\mathbb{E}[\|\Delta^{\epsilon}_t(0_d)\|^2]&\leq -\delta \mathbb{E}[\|\Delta^{\epsilon}_t(0_d)\|^2]\\
&\quad+\frac{1}{\delta}
\mathbb{E}[\|F(Y^{\epsilon}_t(0_d))-DF(0_d)Y^{\epsilon}_t(0_d)\|^2]+\epsilon \ell^2 \mathbb{E}[\|X^{\epsilon}_t(0_d)\|^2].
\end{split}
\]
By Lemma~\ref{lemmaB} we have
\[
\mathbb{E}[\|X^{\epsilon}_t(0_d)\|^2]\leq
\frac{ \epsilon C_0}{\delta}\quad \textrm{ for all } t\geq 0,\,\epsilon \in (0,\nicefrac{\delta}{\ell^2}],
\]
where $C_0=\mathrm{Tr}((\sigma(0_d))^*\sigma(0_d))$.
Since $\Delta^{\epsilon}_t(0_d)=0$, Lemma~\ref{lem:gron} implies
\begin{equation}
\begin{split}
\mathbb{E}[\|\Delta^{\epsilon}_t(0_d)\|^2]&\leq \frac{1}{\delta}e^{-\delta t}\int_{0}^{t}
e^{\delta s}\mathbb{E}[\|F(Y^{\epsilon}_s(0_d))-DF(0_d)Y^{\epsilon}_s(0_d)\|^2]\ud s
+
\frac{ \epsilon^2 \ell^2 C_0}{\delta^2}
\\
& \leq \frac{1}{\delta^2}\sup\limits_{0\leq s\leq t}\mathbb{E}[\|F(Y^{\epsilon}_s(0_d))-DF(0_d)Y^{\epsilon}_s(0_d)\|^2]
+
\frac{ \epsilon^2 \ell^2 C_0}{\delta^2}
\label{eq: es0}
\end{split}
\end{equation}
for all  $t\geq 0$ and $\epsilon \in (0,\nicefrac{\delta}{\ell^2}]$.
Next, we estimate
\[
\sup\limits_{0\leq s\leq t}\mathbb{E}[\|F(Y^{\epsilon}_s(0_d))-DF(0_d)Y^{\epsilon}_s(0_d)\|^2].
\]
Let $s\in [0,t]$.
Recall that $F\in \mathcal{C}^2(\mathbb{R}^d,\mathbb{R}^d)$. Since $F(0_d)=0_d$, the mean value theorem yields
\[
\begin{split}
F(Y^{\epsilon}_s(0_d))-F(0_d)
=
\int_{0}^{1} DF(\theta_1 Y^{\epsilon}_s(0_d))\ud \theta_1 \cdot Y^{\epsilon}_s(0_d).
\end{split}
\]
Then we have
\[
\begin{split}
&F(Y^{\epsilon}_s(0_d))-
DF(0_d) Y^{\epsilon}_s(0_d)=
\int_{0}^{1} \left[DF(\theta_1 Y^{\epsilon}_s(0_d))-DF(0_d)\right]\ud \theta_1\cdot Y^{\epsilon}_s(0_d).
\end{split}
\]
Applying the mean value theorem
to the preceding integral we deduce
\begin{equation}\label{ine6}
\begin{split}
\|F(Y^{\epsilon}_s(0_d))-
DF(0_d)Y^{\epsilon}_s(0_d))\|
\leq C^{\epsilon}_{s} \|Y^{\epsilon}_s(0_d)\|^2,
\end{split}
\end{equation}
where
\[
C^{\epsilon}_{s}:=\int_{0}^{1}\int_{0}^{1}\|D^2 F(\theta_1\theta_2 Y^{\epsilon}_s(0_d))\| \ud \theta_1 \ud \theta_2
\]
and $D^2 F$ denotes the second order derivative of $F$.
Note that
\begin{equation}\label{eq:resc}
Y^{\epsilon}_t(0_d)=\sqrt{\epsilon}\, Y_t\quad \textrm{ for any } t\geq 0,
\end{equation}
where $(Y_t)_{t\geq 0}$ is the unique strong solution of
\begin{equation}\label{eq:Ynw}
\left\{
\begin{array}{r@{\;=\;}l}
\ud Y_t&-DF(0)Y_t\ud t+\sigma(0_d)\ud B_t\quad
\textrm{  for any } t\geq 0,\\
Y_0 & 0_d.
\end{array}
\right.
\end{equation}
By \eqref{eq:resc} and \eqref{C3} we have
\[
\begin{split}
\|D^2 F(\theta_1\theta_2 Y^{\epsilon}_s(0_d))\|=
\|D^2 F(\theta_1\theta_2 \sqrt{\epsilon}Y_s)\|&\leq c_0e^{c_1\theta^2_1\theta^2_2 \epsilon \|Y_s\|^2}.
\end{split}
\]
Since $\theta_1,\theta_2\in [0,1]$, we obtain
\[
\|D^2 F(\theta_1\theta_2 Y^{\epsilon}_s(0_d))\|\leq c_0e^{c_1\epsilon \|Y_s\|^2}.
\]
The preceding inequality  with the help of  inequality \eqref{ine6} and equality \eqref{eq:resc} yields
\begin{equation}\label{eq:nomet}
\begin{split}
&\|F(Y^{\epsilon}_s(0_d))-
DF(0_d) Y^{\epsilon}_s(0_d)\|^2
\leq c^2_0e^{2c_1\epsilon \|Y_s\|^2}\epsilon^2\|Y_s\|^4
\end{split}
\end{equation}
for any $s\geq 0$, where $(Y_t)_{t\geq 0}$ is the solution of \eqref{eq:Ynw}.
By item i) of Lemma~\ref{A1} in  Appendix \ref{Ap:moments}
it follows that
\begin{equation}\label{eq:notmet1}
\mathbb{E}[\|Y_s\|^8]\leq 24 C^4_*\quad \textrm{ for any } s\geq 0,
\end{equation}
where $C_*=\frac{\|VV^*\|_{\mathrm{F}}\cdot d^2}{\delta}$ and $\|\cdot\|_{\mathrm{F}}$ denotes the Frobenius norm.
Moreover, by item ii) Lemma~\ref{A1} in  Appendix \ref{Ap:moments} we have
for $\epsilon\in (0,\frac{1}{4c_1C_*})$ we have
\begin{equation}\label{eq:nomet2}
\mathbb{E}[e^{4 c_1\epsilon\|Y_s\|^2}]\leq \frac{1}{1-4\epsilon c_1 C_*}\quad \textrm{ for any } s\geq 0.
\end{equation}
Estimate
\eqref{eq:nomet} with the help of
the Cauchy-Schwarz inequality,
 \eqref{eq:notmet1} and \eqref{eq:nomet2} implies
\[
\begin{split}
  \mathbb{E}[\|F(Y^{\epsilon}_s(0_d))-
DF(0_d)Y^{\epsilon}_s(0_d)\|^2]&\leq \epsilon^2 c^2_0
\left(\mathbb{E}[e^{4 c_1\epsilon\|Y_s\|^2}]
 \mathbb{E}[\|Y_s\|^8]\right)^{\nicefrac{1}{2}}\\
 &
\leq \tilde{C}(\delta,d,c_0)\epsilon^2\Big(\frac{1}{1-4\epsilon c_1 C_*} \Big)^{\nicefrac{1}{2}}
\end{split}
\]
for any $s\geq 0$,
$\epsilon\in (0,\frac{1}{4c_1C_*})$,
 where $\tilde{C}(\delta, d,c_0)=\sqrt{24}c^2_0C^2_*$ is a positive constant.
Consequently, for $\epsilon\in (0,\frac{1}{4c_1C})$ we obtain
\begin{align}\label{eq:utu}
\sup\limits_{0\leq s\leq t} \mathbb{E}[\|F(Y^{\epsilon}_s(0_d))-
DF(0_d)Y^{\epsilon}_s(0_d)\|^2]
\leq \tilde{C}(\delta,d,c_0)\epsilon^2\Big(\frac{1}{1-4\epsilon c_1 C_*} \Big)^{\nicefrac{1}{2}}.
\end{align}
Note that if $\epsilon\in(0,\frac{1}{8c_1C_*})$, then
$(1-4\epsilon c_1 C_*)\leq 2$.
Let $\epsilon_*:=\min\big\{\frac{1}{8c_1C_*},\frac{ \delta}{\ell^2}\big\}$.
By  \eqref{eq: es0} and \eqref{eq:utu} we have for all
$\epsilon\in (0,\epsilon_*]$ and all $t\geq 0$
\[
\mathbb{E}\left[\|X^{\epsilon}_t(0_d)-Y^{\epsilon}_t(0_d)\|^2\right]\leq
\frac{\sqrt{2}}{\delta^2}\tilde{C}(\delta,d,c_0)\epsilon^2+\frac{ \epsilon^2 \ell^2 C_0}{\delta^2}.
\]
As a consequence, for any  $\epsilon\in (0,\epsilon_*]$ and  $t\geq 0$
we have
\begin{align}\label{eq:contc}
\mathcal{W}_2\left(X^{\epsilon}_t(0_d),Y^{\epsilon}_t(0_d)\right)&\leq
\frac{\epsilon}{\delta}(\sqrt{2}\tilde{C}(\delta,d,c_0)+\ell^2 C_0 )^{\nicefrac{1}{2}}\leq
\frac{\epsilon}{\delta}(48c_0 C_*+\ell C^{\nicefrac{1}{2}}_0 ),
\end{align}
where in the last inequality we use
the subadditivity property of the root-map.

\appendix
\section{\textbf{Moment estimates}}\label{Ap:moments}
In this section, we compute the even moments and exponential moments of the Ornstein-Uhlenbeck process starting at zero.
Let$(Z_t)_{t\geq 0}$ be the unique strong solution of the linear SDE
\begin{equation}\label{eq:Z990}
\left\{
\begin{array}{r@{\;=\;}l}
\ud Z_t&-UZ_t\ud t+V\ud B_t\quad
\textrm{  for any } t\geq 0,\\
Z_0 & 0_d,
\end{array}
\right.
\end{equation}
where $U,V\in \mathbb{R}^{d\times d}$
are given matrices. The drift matrix $U$ satisfies the following condition:
 there exists a positive $\delta>0$ such that
\begin{equation}\label{eq:matrix}
\<Ux,x\>\geq \delta \|x\|^2\quad \textrm{ for all } x\in \mathbb{R}^d.
\end{equation}
We recall the definitions and properties of 1-norm $\|\cdot\|_{1}$ and the Frobenius norm $\|\cdot\|_{\mathrm{F}}$. For a matrix $A=(a^{i,j})_{i,j\in \{1,\ldots,d\}}$  the norms are defined as follows
\[
\|A\|_{1}:=\sum\limits_{i,j=1}^{d}|a^{i,j}|\quad \textrm{ and } \quad \|A\|_{\mathrm{F}}:=\sqrt{\sum\limits_{j=1}^{d}|a^{i,j}|^2}.
\]
We  use several times in the proof of Lemma~\ref{A1} the following
classical inequalities.
For any matrices $A,B\in \mathbb{R}^{d\times d}$ it follows that
\begin{enumerate}
\item[i)] \textbf{Equivalence of norms:}
\begin{equation}\label{eq:equinorm}
\|A\|_{\mathrm{F}}\leq \|A\|_{1}\leq d \,\|A\|_{\mathrm{F}}.
\end{equation}
\item[ii)] \textbf{Sub-multiplicative property:}
\begin{equation}\label{eq:multipli}
\|AB\|_{\mathrm{F}}\leq \|A\|_{\mathrm{F}} \|B\|_{\mathrm{F}}.
\end{equation}
\item[iii)] \textbf{Trace bound:}
\begin{equation}\label{eq:traza}
|\mathrm{Tr}(AB)|\leq \|AB\|_{1}\leq d\|A\|_{1}\|B\|_{\mathrm{F}}.
\end{equation}
\end{enumerate}
\begin{lemma}[Polynomial and exponential moments]\label{A1}
Assume that \eqref{eq:matrix} is valid and let $(Z_t)_{t\geq 0}$ be the unique strong solution of the SDE \eqref{eq:Z990}. Then the following holds.
\begin{itemize}
\item[i)]
For each $j\in \mathbb{N}$ it follows that
\begin{equation}\label{eq:supremum}
\mathbb{E}[\|Z_t\|^{2j}]\leq C^j_* j!\quad \textrm{ for all } t\geq 0, \quad
\textrm{ where }
C_*:=\frac{\|VV^*\|_{\mathrm{F}}\cdot d^2}{\delta}.
\end{equation}
\item[ii)]
Let $C_*$ be as in item i).
For any $\lambda \in (0,1/C_*)$ and all $t\geq 0$ it follows that
\begin{equation*}
\mathbb{E}[e^{\lambda \|Z_t\|^2}]\leq \frac{1}{1-\lambda C_*}.
\end{equation*}
\end{itemize}
\end{lemma}
\begin{proof}
We start with the proof of item i).
The proof is done by the induction method.
We start the induction basis, $j=1$.
The
It\^o formula yields
\begin{equation}\label{eq:Z32}
\ud \|Z_t\|^2=-2\<Z_t,UZ_t\>\ud t+\mathrm{Tr}[V^*V]\ud t+2\<Z_t,V \ud B_t\>.
\end{equation}
A localization argument in \eqref{eq:Z32} with the help of \eqref{eq:matrix} implies
\[
\frac{\ud }{\ud t} \mathbb{E}[\|Z_t\|^2]\leq -2\delta \mathbb{E}[\|Z_t\|^2]+\mathrm{Tr}[V^*V].
\]
Since $Z_t=0_d$,
Lemma~\ref{lem:gron} yields
\begin{equation}\label{eq:Z2}
 \mathbb{E}[\|Z_t\|^2]\leq \frac{\mathrm{Tr}[V^*V]}{2\delta}\quad \textrm{ for all } t\geq 0.
\end{equation}
By \eqref{eq:equinorm} and \eqref{eq:multipli} we deduce
\begin{align}\label{eq:hest}
\frac{\mathrm{Tr}[V^*V]}{2\delta}\leq \frac{\|VV^*\|_{1}}{2\delta}\leq
\frac{d\, \|VV^*\|_{\mathrm{F}}}{2\delta}\leq
\frac{d^2\, \|VV^*\|_{\mathrm{F}}}{\delta}.
\end{align}
Combining \eqref{eq:Z2} and \eqref{eq:hest} we prove the induction basis.

We assume that \eqref{eq:supremum} holds for $j=n$ and we  prove that it remains valid for $j=n+1$.
The It\^o formula for the function $f(x)=\|x\|^{2(n+1)}$, $x\in \mathbb{R}$,
reads
\begin{equation}\label{eq:Zn}
\begin{split}
\ud \|Z_t\|^{2(n+1)}&=-2(n+1)\|Z_t\|^{2n}\<Z_t,UZ_t\>\ud t+(\nicefrac{1}{2})\mathrm{Tr}[V^*H(Z_t)V]\ud t\\
&\quad+2(n+1)\|Z_t\|^{2n}\<Z_t,V \ud B_t\>,
\end{split}
\end{equation}
where the matrix valued function $\mathbb{R}^d\ni x\mapsto H(x):=(H_{i,j}(x))_{i,j\in \{1,\ldots,d\}}\in \mathbb{R}^{d\times d}$ is given by
\begin{equation*}
\begin{split}
H_{i,j}(x):=
\begin{cases}
4(n+1)n \|x\|^{2(n-1)}x^2_i+2(n+1)\|x\|^{2n} & \textrm{ for } i=j,\\
4(n+1)n \|x\|^{2(n-1)} x_ix_j &\textrm{ for } i\neq j.
\end{cases}
\end{split}
\end{equation*}
By definition of $\|\cdot \|_{1}$  it follows that
\begin{align}\label{eq:H11}
\|H(x)\|_{1}&\leq 2d(n+1)\|x\|^{2n}+
4d(n+1)n \|x\|^{2n}=2d(n+1)(1+2n)\|x\|^{2n}
\end{align}
for all $x\in \mathbb{R}^d$.
Note that $\mathrm{Tr}[V^*H(Z_t)V]=\mathrm{Tr}[H(Z_t)VV^*]$.
By \eqref{eq:traza} and \eqref{eq:H11} we deduce
\begin{equation}
\label{eq:trace101}
\begin{split}
|\mathrm{Tr}[H(Z_t)VV^*]|
&\leq
d \|H(Z_t)\|_{1}\|VV^*\|_{\mathrm{F}}\leq 2d^2(n+1)(1+2n)\|VV^*\|_{\mathrm{F}} \|Z_t\|^{2n}.
\end{split}
\end{equation}
Using a localization argument in
\eqref{eq:Zn}
with the help of \eqref{eq:matrix} and \eqref{eq:trace101} yields
\begin{align*}
\frac{\ud}{\ud t} \mathbb{E}[\|Z_t\|^{2(n+1)}]&\leq -2(n+1)\delta \mathbb{E}[\|Z_t\|^{2(n+1)}]+
d^2(n+1)(1+2n)\|VV^*\|_{\mathrm{F}}\mathbb{E}[\|Z_t\|^{2n}].
\end{align*}
By induction hypothesis we have
$
\mathbb{E}[\|Z_t\|^{2n}]\leq C^n_* n!$  for all  $t\geq 0$.
Since $Z_0=0_d$, Lemma~\ref{lem:gron} yields for all $t\geq 0$
\begin{equation*}
\mathbb{E}[\|Z_t\|^{2(n+1)}]\leq \frac{d^2(n+1)(1+2n)\|VV^*\|_{\mathrm{F}}C^n\, n!}{2(n+1)\delta}
\leq C^{n+1}_*(n+1)!,
\end{equation*}
which finishes the induction step.
This concludes the proof of item i).

We continue with the proof of item ii).
By the Monotone Convergence Theorem we have
\begin{equation*}
\mathbb{E}[e^{\lambda \|Z_t\|^2}]=\sum\limits_{j=0}^{\infty} \frac{\lambda^j \mathbb{E}[\|Z_t\|^{2j}]}{j!}\quad \textrm{ for all } \lambda\geq 0.
\end{equation*}
By item i)
for all $\lambda \in (0,1/C_*)$ and $t\geq 0$ it follows that
\begin{equation*}
\mathbb{E}[e^{\lambda \|Z_t\|^2}]\leq \sum\limits_{j=0}^{\infty} (\lambda C_*)^j=\frac{1}{1-\lambda C_*}.
\end{equation*}
\end{proof}
\begin{lemma}[Covariance]\label{lem:covariance}
Assume that \eqref{eq:matrix} holds and that the diffusion matrix $V$ is invertible.
Let $(Z_t)_{t\geq 0}$ be the unique strong solution of the SDE \eqref{eq:Z990}
 and let $\Theta_t:=\mathbb{E}[ Z_t Z^*_t]$ for any $t\geq 0$. Then
\[
\| \Theta_t-\Theta \|_{\mathrm{F}}\leq \|\Theta \|^2_{\mathrm{F}}\, e^{-2\delta t} \quad  \textrm{ for all } t\geq 0,
\]
where  $\Theta\in \mathbb{R}^{d\times d}$
is the unique symmetric and positive definite solution of
the Lyapunov matrix equation
$U\Theta+\Theta U^*=VV^*$.
\end{lemma}
\begin{proof}
The proof follows by analogous reasoning using in the proof of Lemma~C.4 in Appendix C of \cite{BJ1}.
\end{proof}

\section*{\textbf{Acknowledgments}}
The research of G. Barrera  has been supported by the Academy of Finland via
the Matter and Materials Profi4 university profiling action. He gratefully acknowledges support from a post-doctorate grant (2020-2023)  held
at the Department of Mathematical and Statistical Sciences at University of Helsinki
 and expresses his gratitude for all the facilities used along with the realization of this work.


TITLE: Can I solve this integral problem?
QUESTION [0 upvotes]: I want to know how can i solve this function.
$\int{(1-y^d)^n}dy$
Is it possible to solve it? If you know the method, please teach me.

REPLY [2 votes]: Assuming that $d$ and $n$ are positive integers, you can just use the binomial theorem to expand the integrand, and integrate term by term, since it's a polynomial in $y$.
$$
(1 - u)^n
= \sum_{i = 0}^n (1)^{n-i}{n \choose i} (-u)^i
= \sum_{i = 0}^n (-1)^i{n \choose i} u^i
$$
In your case, $u = y^d$, so this becomes
$$
(1 - y^d)^n = \sum_{i = 0}^n (-1)^i{n \choose i} (y^d)^i
 = \sum_{i = 0}^n (-1)^i{n \choose i} y^{di}
$$
and
\begin{align}
\int (1 - y^d)^n ~dy
&=  \int \left( \sum_{i = 0}^n (-1)^i{n \choose i} y^{di} \right) ~dy\\
&=  \sum_{i = 0}^n (-1)^i{n \choose i} \int y^{di}  ~dy \\
&=  \sum_{i = 0}^n (-1)^i {n \choose i} \frac{y^{di+1}}{di + 1}.
\end{align}


TITLE: Are $ \tau _{1}$ and $\tau _{2}$ same?
QUESTION [1 upvotes]: Let $ \tau _{1}$be the product (standard) topology on $\mathbb{R}^{2}$ and $ \tau _{2}$ be the topology generated by the base $B_{2}=\{ C(a,r,R): a \in \mathbb{R}^{2}, r,R \in \mathbb{R}, 0<r<R \} $ where
  $C(a,r,R)=\{(x,y) \in \mathbb{R}^{2}:r^{2}<(x-a_{1})^{2}+(y-a_{2})^{2} < R^{2}\}$ where $a=(a_{1},a_{2})$.
  Then is $ \tau _{1} =  \tau _{2}?$

I feel that both are equal.
 Open sets in $ \tau _{1}$ are open rectangles and in $ \tau _{2} $ they are annuli. Given a rectangle we can find an annulus containing that rectangle and given an annulus we can always find a rectangle which contains that annulus. Hence the two are equal. Am I correct?

REPLY [0 votes]: All basic open subsets of $\tau_2$ are open in $\tau_1$. This implies that $\tau_2 \subseteq \tau_1$ right away. (unions of $\tau_2$-basic sets are also $\tau_1$-open.)
For every basic open ball $B(a,R)$ for $\tau_1$ we have that $C(a,r,R) \subseteq B(a,R)$ and this implies $\tau_1 \subseteq \tau_2$. ($O$ open in $\tau_1$ iff for every $a \in O$ we have some $R>0$ with $B(a,R) \subseteq O$; we can now also take $C(a,r,R)$ and get $O$ as a union of $\tau_2$-basic sets as well).
Hence we have equality, as you suggested.


TITLE: Closed subset of a manifold is a submanifold if there exists a smooth retract.
QUESTION [6 upvotes]: My question is regarding a specific exercise that I'm not really sure how to approach. Suppose I have a smooth manifold $M$ and let $A$ be a closed subset. Now suppose that there exists a smooth function $f:M \to A$ such that $f|_A = id$ (smooth retract).
I want to show that $A$ is a submanifold, i.e, it is Hausdorff, second countable and locally euclidean. Hausdorff and second countable are both hereditary properties, so only the latter is left.
My attempt was to prove that if $(U, \varphi)$ is a chart around $x \in A \subset M$ for $M$, then $\left(f(U), \varphi|_{f(U)}\right)$ is the desired chart. However, I'm not sure if $f(U)$ is open. I probably should use that $A$ is closed and $f$ is a retract, but I couldn't connect both facts.
There is no need for a solution, just a hint is enough.
PS: John Lee suggested that I should ask a question instead of leaving the exercise for the community, so here it is. I rewrote my previous post here. :)

REPLY [1 votes]: Assume $M$ connected. Note that $A=Im(f)$ is therefore connected. Let $a\in A$, $f'(a)$ is a linear projection on $T_aM$. As the rank of a projection is a contious function on the set of projection (it is the trace), along $A$ the rank of $f'$ is constant $r$. Now, let us check that it is constant in a neigbourhood of $A$ . As $f\circ f= f$, $f'(f(p))\circ f'(p)= f'(p)$, and $f'(f(p))$ induces a isomorphism on $Im ( f'(p))$, so that the rank of $f'(p)$ is less than or equal to $r$. But the rank can only increase in the neigbourhood of $A$, so it is constant. If you know the "constant rank theorem" you are done. If not let us argue as follows.
Now choose a point $a\in A$. In order to study the image of $f$ near this point, let us choose  coordinates $x_1,...,x_n$ near $a$ such that $f'(a)$ is the projection onto $x_{r+1}==x_n=0$, and let us consider the new system of coordiantes $y_i=x_i\circ f$ for $i\leq r$, $y_i=x_i$  if $i> r$. There are functions $F_j, j>r$ such that $f(y_1,..y_n)=(y_1,,..,y_r, F_{r+1},...F_{n})$ Writing that the rank of $f'$ is $r$ we see that ${\partial F_j \over \partial y_k }=0$ if $k>r$. Or $f(y_1,...y_n)= (y_1,..,y_r, F_1(y_1,..) ..)$. So near $a$ the image of $F$ is just the graph of the function $F=(F_{r+1},...F_n$), a submanifold.


TITLE: Differential identity and wedge products
QUESTION [0 upvotes]: Apparently
$dx^{i_1} \wedge ... \wedge dx^{i_k}=d(x^{i_1}dx^{i_2}\wedge ... \wedge dx^{i_k})$
which I cannot see proved anywhere in my notes. It just stated as if it is obvious which I don't believe it is.
I cannot see how you could prove this identity. Firstly, I am not sure as to what $x^{i_1}dx^{i_2}\wedge ... \wedge dx^{i_k}$ is and secondly how you could apply $d$ to it.

REPLY [1 votes]: It is an application of the following fundamental property of the exterior derivative: $d(\alpha \wedge \beta) = d\alpha \wedge \beta + (-1)^{\mathrm{deg} \alpha}(\alpha \wedge d\beta)$.
When $\alpha = f$ is a $0$-form (a function), this identity says that $d(f \beta) = df \wedge \beta + f d\beta$.
In your situation, $f = x^{i_1}$ and $\beta = dx^{i_2}\wedge ... \wedge dx^{i_k}$. You can show that $d \beta = 0$, so you get $d(x^{i_1}\, dx^{i_2}\wedge ... \wedge dx^{i_k}) = dx^{i_1}\wedge dx^{i_2}\wedge ... \wedge dx^{i_k}$.
Oops, I just realized this answer is just what Franck Science wrote in the comments to your question (with a bit more details).


TITLE: Positivity and Interchange of Summation
QUESTION [17 upvotes]: Suppose I have $\sum\limits_{n = 1}^{\infty}\sum\limits_{m = 1}^{\infty} a_{m, n}$ where $a_{m, n} \geq 0$ for all $m$ and $n$. Can I interchange the two summations? If so why?

REPLY [4 votes]: Here's an elementary proof fleshing out @JonasMeyer's answer.
Claim: If $a_{m,n}\ge 0$ for all $m,n$ then
$$
\sum_{m=1}^\infty\sum_{n=1}^\infty a_{m,n}= \sup_{M,N}\sum_{m=1}^M\sum_{n=1}^N a_{m,n}.\tag1
$$
Proof. Let $S$ denote the RHS of (1). Write $$S_{M,N}:=\sum_{m=1}^M\sum_{n=1}^Na_{m,n}\quad\text{and}\quad S_M:=\lim_N S_{M,N}=\sum_{m=1}^M\sum_{n=1}^\infty a_{m,n}.$$ The limit $S_M$ exists, possibly with value infinity, since $S_{M,N}$ is monotonically increasing in $N$. By definition of sup, we have $S_{M,N}\le S\ \forall M,N$, hence $S_M\le S\,\forall M$, and $\color{green}{\lim_M S_M\le  S}$. But $S_{M,N}\le S_M$ by non-negativity of $a_{m,n}$, so $S:=\sup S_{M,N}\le\sup S_M$. Since $S_M$ is nondecreasing (non-negativity again), this means $\color{green}{S\le\lim_MS_M}$. To finish, check that $\lim_M S_M$ equals the LHS of (1).
Corollary:  If $a_{m,n}\ge 0$ for all $m,n$ then
$$
\sum_{m=1}^\infty\sum_{n=1}^\infty a_{m,n}= \sum_{n=1}^\infty\sum_{m=1}^\infty a_{m,n}.
$$
Proof. First apply the Claim to $a_{m,n}':=a_{n,m}$, then swap indices to obtain
$$
\sum_{n=1}^\infty\sum_{m=1}^\infty a_{m,n}= \sup_{M,N}\sum_{n=1}^N\sum_{m=1}^M a_{m,n}.\tag2
$$
Finally note that the RHS of (1) equals the RHS of (2).


TITLE: When the subgroup generated by all elements of some fixed finite order equals the group
QUESTION [3 upvotes]: Let $G$ be an infinite non-abelian group.
Let $H$ be the subgroup of $G$ generated by all elements of order $n$,
for some fixed $n \geq 2$.
My question: Is it possible to tell when $H=G$? Namely, are there some known groups $G$ that are generated by all elements of some fixed finite order?
(It is easy to see that $H$ is normal in $G$, see this; I do not know if this fact should help in answering my question).
Edit: The motivation for my question is Proposition 5(b), which seems to me too good to be true (perhaps I am missing some implicit assumptions in that proposition).
A relevant question is this question.

REPLY [1 votes]: You can modify your question a bit: which groups $G$ admit surjection from a free product $\mathbb Z / n\mathbb Z \,* \dots \mathbb * \, Z / n\mathbb Z \twoheadrightarrow^f G$? If you restrict it a bit and assume that $f$ induces isomorphism on abelianisations and $H_2(G, \mathbb Z) = 0$, then (by beautiful Stallings theorem) $f$ induces isomorphisms on all lower central quotients, and this is rather restrictive property.
In other direction, if you know that $G$ is nilpotent, then torsion elements form a normal subgroup, so if your group is simultaneously nilpotent, finitely generated and infinite, it cannot be generated by torsion even normally.
Third aspect is that if your group is finitely generated and equals its $n$-torsion (is subject to relation $x^n = 1)$, then it's finite for $n =$ 2 (obvious), 3 (easy lemma), 4 or 6 (not easy, but can be proven by elementary methods), and it's not known if it must be finite for $n$ = 5, 7, 8 $\dots$, 664 wiki link
Lastly, very plentiful source of natural examples are reflection groups in homogeneous spaces with most well-known case of plane Coxeter groups