Is this map (from the space of probability densities to the Wasserstein space) Lipschitz?

Question

Let $p \in [1, \infty)$. Let $\mathcal P_p(\mathbb R^d)$ be the space of all Borel probability measures on $\mathbb R^d$ with finite $p$-th moments. Let $D_p$ be the collection of all Borel measurable functions $f:\mathbb R^d \to \mathbb R_{\ge 0}$ such that $\int_{\mathbb R^d} f (x) \, \mathrm d x = 1$ and $\int_{\mathbb R^d} |x|^p f (x) \, \mathrm d x < \infty$. So $f \in D_p$ if and only if $f$ is a density of some $\mu \in \mathcal P_p(\mathbb R^d)$.

Let $F:D_p \to \mathcal P_p(\mathbb R^d)$ that sends a density $f$ to its corresponding $\mu \in \mathcal P_p(\mathbb R^d)$. We endow $\mathcal P_p(\mathbb R^d)$ with the $L_p$-Wasserstein metric $W_p$. We endow $D_p$ with the norm $[\cdot]$ defined by $$ [f-g]_p := \int_{\mathbb R^d} |f(x)-g(x)| \cdot |x|^p \, \mathrm d x \quad \forall f,g \in D_p. $$

Is $F$ Lipschitz? Are there some special properties about $F$?

Thank you so much for your elaboration?

Answer 1

$\newcommand{\vpi}{\varphi}\newcommand\R{\mathbb R}$

Claim 1: The map $F$ is not Lipschitz if $p>1$.

Claim 2: The map $F$ is $1$-Lipschitz if $p=1$: For all $f,g$ in $D_p$, \begin{equation*} W_1(F(f),F(g))\le[f-g]_1. \tag{1}\label{1} \end{equation*}

Proof of Claim 1: Take any real $a>0$ and a small $h>0$. Let $f$ and $g$ be the densities of the uniform distributions over the cubes $[0,h]^d$ and $[a,a+h]\times[0,h]^{d-1}$. Then $[f-g]_p\to a^p$ whereas $W_p(F(f),F(g))\to a$ as $h\downarrow0$. Letting now $a\downarrow0$, we complete the proof of Claim 1. $\quad\Box$

Proof of Claim 2: By (say) Theorem 1.17 (Duality) in A user's guide to optimal transport (and the last paragraph on p. 8 there), \begin{equation*} W_1(F(f),F(g))=\int\vpi\,d\mu+\int\psi\,d\nu, \tag{2}\label{2} \end{equation*} for some $\vpi\in L^1(\mu)$ and $\psi\in L^1(\nu)$ such that $\vpi=\psi^*$ and $\psi=\vpi^*$, where \begin{equation*} \theta^*(y):=\inf_{x\in\R^d}(|x-y|-\theta(x)) \tag{3}\label{3} \end{equation*} for $y\in\R^d$; note that the function $\theta^*$ is $1$-Lipschitz, as the infimum of $1$-Lipschitz functions $y\mapsto|x-y|-\theta(x)$.

So, $\vpi=\psi^*$ is $1$-Lipschitz. Using \eqref{3} again, we see that $\psi(y)=\vpi^*(y)\le|y-y|-\vpi(y)$ for all $y$, so that $\psi\le-\vpi$. So, by \eqref{2}, \begin{equation*} \begin{gathered} W_1(F(f),F(g))\le\int\vpi\,d\mu-\int\vpi\,d\nu =\int\vpi\,d(\mu-\nu)=\int(\vpi-\vpi(0))\,d(\mu-\nu) \\ \le\int|\vpi-\vpi(0)|\,d|\mu-\nu| =\int|\vpi(x)-\vpi(0)|\,|f(x)-g(x)|\,dx \\ \le\int|x|\,|f(x)-g(x)|\,dx=[f-g]_1, \end{gathered} \end{equation*} which completes the proof of Claim 2 (the latter inequality follows because $\vpi$ is $1$-Lipschitz). $\quad\Box$