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Let $X$ be a compact metric space, and fix an arbitrary point $x_\ast \in X$. By the Kantorovich-Rubinstein duality theorem, the $1$-Wasserstein metric $W_1$ on the set of Borel probability measures on $X$ can be expressed as $$ W_1(\mu,\nu) = \max\left\{ \int f \, d\mu - \int f \, d\nu \, : \, f \in \mathcal{K}_0\right\}\text{,} $$ where $\mathcal{K}_0$ is the compact set (in the topology of uniform convergence) consisting of all $1$-Lipschitz functions $f \colon X \to \mathbb{R}$ with $f(x_\ast)=0$.

I now want to move from the purely topological setting to a "smooth" setting.


Suppose $X$ is the closure of a bounded convex open subset of $\mathbb{R}^N$, with the boundary being an $(N-1)$-dimensional $C^\infty$ embedded submanifold of $\mathbb{R}^N$. For each $r \in \mathbb{N}$, let $C^r(X,\mathbb{R})$ be the set of functions $f \colon X \to \mathbb{R}$ admitting a $C^r$ extension to an $\mathbb{R}^N$-neighbourhood of $X$.

For each $r \in \mathbb{N}$, does there exist a set $\mathcal{K}_r \subset C^r(X,\mathbb{R})$ that is compact in the $C^0$ topology on $X$ (i.e. topology of uniform convergence) such that $W_1$ can be expressed as

$$ W_1(\mu,\nu) = \max\left\{ \int f \, d\mu - \int f \, d\nu \, : \, f \in \mathcal{K}_r\right\} \text{?} $$

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  • $\begingroup$ Do you insist on the $\max$ (rather than $\sup$ in your last formula? $\endgroup$ Commented Jul 11, 2022 at 11:32
  • $\begingroup$ @IosifPinelis I'm confused as to how it can make a difference, since I require $\mathcal{K}_r$ to be compact in the topology of uniform convergence, and the map $f \mapsto \int f \, d\mu$ is continuous on $C(X,\mathbb{R})$ equipped with the topology of uniform convergence. $\endgroup$ Commented Jul 11, 2022 at 15:46

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$\newcommand{\K}{\mathcal K}\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$No, the formula \begin{equation*} W_1(\mu,\nu) = \max\Big\{ \int f \, d\mu - \int f \, d\nu \, : \, f \in \K_r\Big\}, \tag{1}\label{1} \end{equation*} with $\max$ rather than $\sup$, does not hold in general, for whatever choices of an integer $r\ge1$ and a set $\K_r \subset C^r(X,\R)$ (the $\sup$ version of \eqref{1} holds by approximation).

Indeed, let $N=1$ and $X:=[-1,1]$.

Note that all functions in $\K_r$ must be $1$-Lipschitz. Otherwise, there would exist a function $f\in\K_r$ and points $x,y$ in $X$ such that $f(x)-f(y)>|x-y|$. But then we would have \begin{equation*} |x-y|=W_1(\de_x,\de_y)\ge\int f \, d\de_x - \int f \, d\de_y=f(x)-f(y)>|x-y|, \end{equation*} a contradiction. (Here, of course, $\de_z$ is the Dirac probability measure supported on $\{z\}$.)

Suppose now that \eqref{1} holds for \begin{equation*} \mu:=\tfrac12\,(\de_{-1}+\de_1)\quad\text{and}\quad\nu:=\de_0. \end{equation*} Then there exists a $1$-Lipschitz function $f\colon[-1,1]\to\R$ in $C^1(X,\R)$ such that \begin{equation*} \begin{aligned} 1&= W_1(\mu,\nu)=\int f \, d\mu- \int f \, d\nu=\tfrac12\,(f(-1)+f(1))-f(0) \\ &=\tfrac12\,(f(-1)-f(0)+f(1)-f(0))\le\tfrac12\,(|-1-0|+|1-0|)=1. \end{aligned} \end{equation*} It follows that \begin{equation*} f(-1)-f(0)=f(1)-f(0)=1. \end{equation*} Since $f$ is $1$-Lipschitz, it now follows that $f(x)-f(0)=|x|$ for all $x\in[-1,1]$. So, $f\notin C^1(X,\R)$, a contradiction. $\quad\Box$

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  • $\begingroup$ Beautiful! (albeit highly inconvenient for the problem I'm working on). I don't think weakening $\max$ to $\sup$ will help, as I need my $\mathcal{K}_r$ to be compact. Or am I making a mistake? $\endgroup$ Commented Jul 11, 2022 at 15:59
  • $\begingroup$ @JulianNewman : Thank you for your appreciation. I don't know what the goals of your research project are to say anything meaningful about $\mathcal K_r$, except that in this counterexample no properties of $\mathcal K_r$ are used except that it consists of continuously differentiable functions. $\endgroup$ Commented Jul 11, 2022 at 16:59

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