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Fix a positive integer $n$ and consider the $2$-Wasserstein space $\mathcal{P}_2(\mathbb{R}^n)$. Let $X$ be the cone of $n\times n$ symmetric positive semidefinite matrices with Frobenius norm and consider the map $f:\mathbb{R}^n\times X \rightarrow \mathcal{P}_2(\mathbb{R}^n)$ given by $$ f(a,B) = N(a,B) $$ where $N(a,B)$ is the $n$-dimensional Gaussian measure with mean $a$ and covariance $B$.

Clearly the map $f$ is locally-Lipschitz and injective. However, is its inverse on its image in $\mathcal{P}_2(\mathbb{R}^n)$ also locally-Lipschitz?

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$\newcommand{\R}{\mathbb R}\newcommand{\tr}{\operatorname{tr}}$The answer is yes.

Indeed, it is easy to see (cf. e.g. Proposition 7 or the beginning of its proof) that the Wasserstein distance between $N(a,A)$ and $N(b,B)$ is \begin{equation*} W_2(N(a,A),N(b,B))=\sqrt{\|a-b\|^2+W_2(N(0,A),N(0,B))^2}, \tag{10}\label{10} \end{equation*} where $\|\cdot\|$ is the Euclidean norm. So, \begin{equation*} \|a-b\|\le W_2(N(a,A),N(b,B)). \tag{20}\label{20} \end{equation*}

Let now $X\sim N(0,A)$ and $Y\sim N(0,B)$. Then for any unit vector $u\in\R^n=\R^{n\times1}$ \begin{equation*} E\|X-Y\|^2\ge E(u^\top X-u^\top Y)^2\ge(\sqrt{u^\top A u}-\sqrt{u^\top B u})^2, \end{equation*} where the last inequality holds (say, by mentioned Proposition 7) because $u^\top X$ and $u^\top Y$ are real-valued Gaussian zero-mean random variables with variances $u^\top A u$ and $u^\top B u$. So, by \eqref{10} and the definition of the $W_2$-distance, for any unit vector $u\in\R^n=\R^{n\times1}$, \begin{equation*} \begin{aligned} W_2(N(a,A),N(b,B))&\ge W_2(N(0,A),N(0,B)) \\ &\ge|\sqrt{u^\top A u}-\sqrt{u^\top B u}| \\ &=\frac{|u^\top A u-u^\top B u|}{\sqrt{u^\top A u}+\sqrt{u^\top B u}} \\ &\ge\frac{|u^\top(A-B)u|}{\sqrt{\|A\|}+\sqrt{\|B\|}} \\ &=\frac{\|A-B\|}{\sqrt{\|A\|}+\sqrt{\|B\|}} \end{aligned} \end{equation*} for some unit vector $u\in\R^n=\R^{n\times1}$, where $\|M\|$ is the spectral norm of a matrix $M$. So, \begin{equation*} \|A-B\|\le(\sqrt{\|A\|}+\sqrt{\|B\|\,})\, W_2(N(a,A),N(b,B)). \tag{30}\label{30} \end{equation*}

Thus, by \eqref{20} and \eqref{30}, we have the desired local Lipschitz property. $\quad\Box$

Remark: Inequality \eqref{20} is "exact in the limit", when $A$ and $B$ are each (close to) the zero matrix.

Inequality \eqref{30} turns into the equality when $n=1$ to, more generally, "exact in the limit" when $A$ and $B$ are (close to) commuting matrices of rank $1$ each. $\quad\Box$


User ABIM wrote in a comment: "@Justin_other_PhD OP claimed that $f$ is bi-Lipschitz but why is the upper-bound true?"

Let me answer this question as well. First of all, it is not true that $f$ is Lipschitz (and the OP did not claim that). Indeed, even when $n=1$, we have $W_2(N(0,A),N(0,B))=|\sqrt A-\sqrt B|$, so that there is no real $L>0$ such that $W_2(N(0,A),N(0,B))\le L|A-B|$ for all real $A,B>0$.

What the OP said, and what is true, is that $f$ is locally Lipschitz. Indeed, let $X\sim N(0,A)$ and $Y:=B^{1/2}A^{-1/2}X$. Then $Y\sim N(0,B)$ and hence \begin{equation*} \begin{aligned} W_2(N(0,A),N(0,B))^2&\le E\|X-Y\|^2 \\ &=\tr(A+B-B^{1/2}A^{1/2}-A^{1/2}B^{1/2}) \\ &=\tr[(A^{1/2}-B^{1/2})^2]=\|A^{1/2}-B^{1/2}\|_F^2 \\ &\le n\,\|A^{1/2}-B^{1/2}\|^2, \end{aligned} \tag{40}\label{40} \end{equation*} where $\tr$ denotes the trace and $\|\cdot\|_F$ is the Frobenius norm.

Next, for real $a\ge0$, \begin{equation*} a^{1/2}=\frac1\pi\int_0^\infty dt\,t^{-1/2}\Big(1-\frac t{t+a}\Big) \end{equation*} and hence the value of the derivative of the function $M\mapsto M^{1/2}$ at $A$ at a symmetric matrix $H$ is \begin{equation*} \begin{aligned} (A^{1/2})'(H)&=\frac1\pi\int_0^\infty dt\,t^{1/2}(tI+A)^{-1}H(tI+A)^{-1} \\ & \le\frac1\pi\int_0^\infty dt\,t^{1/2}(t+c)^{-2}|H| =\frac1{2c^{1/2}}\,|H| \end{aligned} \end{equation*} provided that $A\ge cI$ for some $c\in(0,\infty)$ (where $I$ is the identity matrix), so that the operator norm of $(A^{1/2})'$ is $\le\dfrac1{2c^{1/2}}$. So, by \eqref{10} and \eqref{40}, \begin{equation*} W_2(N(a,A),N(b,B))\le\|a-b\|+W_2(N(0,A),N(0,B)) \\ \le \|a-b\|+\dfrac{n^{1/2}}{2c^{1/2}}\,\|A-B\| \end{equation*} provided that $A\ge cI$ and $B\ge cI$ for some $c\in(0,\infty)$. Thus, $f[=N]$ is locally Lipschitz. $\quad\Box$

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    $\begingroup$ Or just note that, more generally, taking the mean and variance of a probability distribution are continuous operations in 2-Wasserstein space... $\endgroup$ Commented Jun 21, 2023 at 7:48
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    $\begingroup$ @MartinHairer : I am not sure how to understand your comment. The question here was about Lipschitz continuity, not just continuity. $\endgroup$ Commented Jun 21, 2023 at 11:23
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    $\begingroup$ You're right, I wasn't careful in reading the question and the variance is in general only 1/2-Hölder in Wasserstein-2 space. $\endgroup$ Commented Jun 21, 2023 at 18:23
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    $\begingroup$ @MartinHairer : Thank you for your further comment. Do you have a readily available reference to the 1/2-Hölder condition in general? $\endgroup$ Commented Jun 21, 2023 at 19:39
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    $\begingroup$ @IosifPinelis I wanted to ask the same question; to me it is not obvious $\endgroup$ Commented Jun 21, 2023 at 22:38

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