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.




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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 $X$.



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> 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
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> 
> $$ W\_1(\mu,\nu) = \max\left\{ \int f \, d\mu - \int f \, d\nu \, : \, f \in \mathcal{K}\_r\right\} \text{?} $$
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Is there a "smooth Kantorovich-Rubinstein duality" for Wasserstein distances on smooth/Euclidean space?
$\newcommand{\K}{\mathcal K}\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$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. $\quad\Box$


