3
$\begingroup$

Let $\mu$ and $\nu$ be probability measures on $\mathbb{R}^n$ with first moment and suppose that both $\mu$ and $\nu$ have a densities with respect to the $n$-dimensional Lebesgue measure. Fix some positive integer $k$.

Are the "simple and broad conditions" guaranteeing that exist a class $C^k$ Monge map $T:\mathbb{R}^n\rightarrow \mathbb{R}^n$ i.e.: $$ T_{\#}\mu=\nu \mbox{, } \int \|T(x)-x\|\mu(dx) = \mathcal{W}_1(\mu,\nu) \mbox{, and } T\in C^k(\mathbb{R}^n,\mathbb{R}^n)? $$

$\endgroup$

1 Answer 1

4
$\begingroup$

This is an open question. The difficulty is that Monge's cost function is very degenerate, so doesn't satisfy the assumptions of the standard regularity theory of optimal transport.

The first difficulty is that the solutions to this problem need not be unique. However, the transport will occur along disjoint line segments (called transport rays) and in order for the transport to be continuous along each line segment, it is generally necessary to assume that the transport is monotonic along transport rays. ​With this additional assumption, you do obtain a unique solution to the Monge problem (for reasonable measures). In order to obtain any sort of regularity for this transport, you then need to assume that the support of the target measure is convex (which was originally shown by Caffarelli when the cost is the squared distance).

​However, even among smooth measures with convex supports, it is possible to find examples where the ray monotone solution fails to be Lipschitz continuous [1]. On the other hand, in two dimensions, there is a result which shows that the monotone optimal mapping is continuous in the interior of the transfer set (i.e., the union of all transfer rays), under the assumptions that the densities of $\mu$ and $\nu$ are positive, continuous, and have compact, convex and disjoint supports [2]. It seems that the optimal estimate, even for smooth measures, might be a $C^{1/2}$ estimate [3], but this is still an open problem.

[1] Li, Qi-Rui; Santambrogio, Filippo; Wang, Xu-Jia, Regularity in Monge’s mass transfer problem, J. Math. Pures Appl. (9) 102, No. 6, 1015-1040 (2014). ZBL1304.49094.

[2] Fragalà, Ilaria; Gelli, Maria Stella; Pratelli, Aldo, Continuity of an optimal transport in Monge problem, J. Math. Pures Appl., IX. Sér. 84, No. 9, 1261-1294 (2005). ZBL1075.49018.

[3] Colombo, Maria; Indrei, Emanuel, Obstructions to regularity in the classical Monge problem, Math. Res. Lett. 21, No. 4, 697-712 (2014). ZBL1305.49068.

$\endgroup$
6
  • 1
    $\begingroup$ Also, I guess this means that for every $\epsilon>0$ every Monge map is $\epsilon$-close to a $C^k$ function; by density of $C^k$ functions in $L^1$ no? So at-least all almost-optimal maps are $C^k$? $\endgroup$
    – ABIM
    Commented Nov 18, 2021 at 22:18
  • $\begingroup$ Yes, that's correct. One elegant way to see this is to consider the cost function $c_\epsilon(x,y) = \sqrt{\epsilon + \|x-y\|^2}$, which approximates the Monge's cost in $C^\alpha$. It turns out that this cost is weakly regular (i.e., has non-negative MTW tensor), so you can show that the solutions to the associated Monge problem are smooth when the measures are smooth and the supports have the right type of convexity. $\endgroup$
    – Gabe K
    Commented Nov 18, 2021 at 23:10
  • $\begingroup$ Furthermore, these solutions will converge to the ray monotone solution with respect to the Monge cost in the weak-* topology (and perhaps in a stronger topology as well, but that's essentially equivalent to deriving regularity for the original Monge problem). $\endgroup$
    – Gabe K
    Commented Nov 18, 2021 at 23:14
  • $\begingroup$ Very interesting! Do you have a reference to these things? $\endgroup$
    – ABIM
    Commented Nov 18, 2021 at 23:20
  • 1
    $\begingroup$ A good starting point might be the survey paper by De Phillipis and Figalli “The Monge-Ampere equation and its link to optimal transport.” The proof of the regularity of the cost $ c_\epsilon$ comes from the Ma-Trudinger-Wang paper. $\endgroup$
    – Gabe K
    Commented Nov 18, 2021 at 23:41

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.