1
$\begingroup$

Let $\mathcal{P}(\mathbb{R}^n)$ be the set of Borel probability measures on the Euclidean space $\mathbb{R}^n$ and consider thereof consisting of all probability measures $\mathbb{P}$ satisfying $\int\, \|x\|\mathbb{P}(dx)<\infty$. One can identify this as a subspace of the pre-dual of $\operatorname{Lip}(\mathbb{R}^n)$ (via $\mathbb{P}\mapsto \mathbb{P}-\delta_0$) normed by $\|f\|_{\operatorname{Lip}}:=\sup_{x\in \mathbb{R}^n}\, \|f(x)\| + \sup_{x_1,x_2\in \mathbb{R}^n}\, \frac{|f(x)-f(x_2)|}{\|x_1-x_2\|}$ which we metrize using the restriction of the dual-norm $$ d(\mathbb{P},\mathbb{Q}) := \sup_{\|f\|_{Lip}\leq 1}\,\int f d\mathbb{P} - \int f d\mathbb{Q}. $$ What's rather nice is that the Kantorovich duality (essentially) identifies this norm with the Wasserstein 1 distance from optimal transport on the set $\mathcal{P}(\mathbb{R}^n)$.


Suppose that $k,p\geq 1, k\in \mathbb{N}$ and consider the Sobolev space $W^{k,p}(\mathbb{R}^n)$. Define the norm $$ d_{k,p}(\mathbb{P},\mathbb{Q}) := \sup_{\|f\|_{W^{k,p}}\leq 1}\, \int f d\mathbb{P} - \int f d\mathbb{Q}, $$ Is there a (somewhat well-studied or classical) metric $\tilde{d}_{k,p}$ on a (reasonable) subset of $\mathcal{P}(\mathbb{R}^n)$ for which there is a similar duality; or at-least $$ d_{k,p}\leq \tilde{d}_{k,p}? $$

$\endgroup$
6
  • 2
    $\begingroup$ When $k > n/p$ you can use Sobolev embedding to bound it from above by the Wasserstein distance for the metric $(x,y) \mapsto |x-y|^\gamma$ for suitable $\gamma \in (0,1]$. $\endgroup$ May 5, 2022 at 19:36
  • $\begingroup$ By duality $d_{k,p}(\mathbb{P},\mathbb{Q}) = \Vert\mathbb{P}-\mathbb{Q}\Vert_{W^{-k,q}}$ with $p^{-1}+q^{-1}=1$ is the norm of the according dual negative Sobolev space of the difference of the measures. $\endgroup$ May 5, 2022 at 21:40
  • $\begingroup$ @MartinHairer I don't fully see why? $\endgroup$
    – ABIM
    May 6, 2022 at 7:51
  • $\begingroup$ @AndréSchlichting Here you're assuming that $\mathbb{P}$ and $\mathbb{Q}$ are dominated by Lebesgue measure no? This I can't do since I eventually need one of the measures to be singular with respect to Lebesgue measure. $\endgroup$
    – ABIM
    May 6, 2022 at 7:53
  • $\begingroup$ @TomTheQuant Since $|f|_\gamma \le C \|f\|_{W^{k,p}}$, the set $|f|_\gamma \le C$ is larger than the set $\|f\|_{W^{k,p}} \le 1$, so that distance is larger than (some multiple of) $d_{k,p}$. $\endgroup$ May 6, 2022 at 12:10

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.