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

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?

• What metric do you take on $D_p$? Couldn't you take $W_p$ as well, rendering the problem trivial?
– Dirk
Feb 13 at 14:16
• @Dirk Thank you so much for your comment. I have edited the question. Feb 13 at 14:26
• Thanks! Unfortunately, I do not know the answer…
– Dirk
Feb 13 at 14:30

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

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). $$\quad\Box$$

• Thank you so much for your help! Feb 14 at 8:54