2
$\begingroup$

Let $c:\mathbb{R}^d\times \mathbb{R}^d\to [0,\infty)$ be a Lipschitz cost function and consider the optimal transport problem $$ C(\mu,\nu):=\inf_{\pi}\, \int c(x,y)\,\pi(dxdy) $$ where, as usual, the infimum is taken over all couplings between two given probability measures $\mu,\nu\in \mathcal{P}(\mathbb{R}^d)$.

Then, the Kantorovich duality implies that $$ \tag{1} \label{1} C(\mu,\nu) = \sup_{\phi \in L^1(\nu)} \int \phi(u)\,\nu(du) - \int \phi^c(x)\,\mu(dx) $$ where, $\phi^c$ is given by the infimal convolution $\phi^c:= \inf_x [c(x,y)-\phi(x)]$.


Are the settings where there is some $\phi^{\star}$ realizing \eqref{1} exists and is Lipschitz? That is, some $L$-Lipschitz $\phi^{\star}:\mathbb{R}^d\to \mathbb{R}$ such that $$ C(\mu,\nu) = \int \phi^{\star}(u)\nu(du) - \int (\phi^{\star})^c(v)\mu(dv)? $$

Is there any relationship between the Lipschitz constant of $c$ and that of $\phi^{\star}$?

$\endgroup$
3
  • 2
    $\begingroup$ Once you have the existence of a solution $\phi$ of (1), you can take $\phi^{cc}$. The objective value of $\phi^{cc}$ is at least that of $\phi$, hence, it is again a solution and further, the $L$-Lipschitz continuity of $c$ implies that $c$-conjugates are $L$-Lipschitz. $\endgroup$
    – gerw
    Commented Aug 7 at 12:12
  • $\begingroup$ @gerw But is it clear when $\phi^{\star}$ is $L$-Lipschitz? $\endgroup$ Commented Aug 7 at 18:02
  • 2
    $\begingroup$ No, but you can replace $\phi$ by $\phi^{cc}$. $\endgroup$
    – gerw
    Commented Aug 8 at 9:39

0

You must log in to answer this question.

Browse other questions tagged .