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?



Local Lipschitzness of parameterization of Gaussians in Wasserstein space
$\newcommand{\R}{\mathbb R}\newcommand{\tr}{\operatorname{tr}}$The answer is yes.


Indeed, it is easy to see (cf. e.g. [Proposition 7](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-31/issue-2/A-class-of-Wasserstein-metrics-for-probability-distributions/10.1307/mmj/1029003026.full) 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](https://projecteuclid.org/journals/michigan-mathematical-journal/volume-31/issue-2/A-class-of-Wasserstein-metrics-for-probability-distributions/10.1307/mmj/1029003026.full)) 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](https://en.wikipedia.org/wiki/Wasserstein_metric#Definition), 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](https://mathoverflow.net/questions/449265/local-lipschitzness-of-parameterization-of-gaussians-in-wasserstein-space/449287#comment1161164_449287): "@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. $\quad\Box$


