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?



Is this map (from the space of probability densities to the Wasserstein space) Lipschitz?
$\newcommand{\vpi}{\varphi}\newcommand\R{\mathbb R}$



> 
> **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](https://www.math.umd.edu/%7Eyanir/OT/AmbrosioGigliDec2011.pdf) (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). $\quad\Box$


