This question comes from Villani's book, Optimal Transport: Old and New. Consider the cost function $c(x, y) = |x - y|^2$ on $X \times Y$, where $X$ is the right half of the unit ball, and $Y$ is the right half of the unit ball, shifted one unit to the right. Let $\mu$ be the uniform distribution on $X$ and $\nu$ the uniform distribution on $Y$.
The optimal transport map $T$ is given by $(x, y) \mapsto (x + 1, y)$, and by a theorem from chapter 10 we know that $T(x, y) = (x,y) + \nabla \psi(x, y)$ for some $c$-convex function $\psi$ on $X$. Since this means that $\psi(x, y) = (1, 0)$ we have $\psi(x, y) = x$.
My question: Is this $\psi$ equal to the optimal $\psi$ in the Kantorovich dual problem? That is, is the following inequality true: $$ \int_X c(x, Tx)\, d\mu(x) = \int_Y \psi^c \, d\nu - \int_X \psi\, d\mu, $$ where $\psi^c(y) := \inf_x \psi(x) + c(x, y)$. I've tried computing the $c$-transform and as far as I can tell the answer to my question is negative, since I am getting that $\partial_c \psi(x)$ is not the translate by 1, which it should be to guarantee optimality by Kantorovich duality.
Any help is much appreciated!
I've since figured out the answer, which is somewhat annoyingly trivial. The transport map is given by $x + \frac{1}{2}\nabla \psi(x)$, not by $x + \nabla \psi (x)$; the latter occurs for $c = d^2/2$. Then we get in fact that $\psi(x) = 2|x|$, which can also be checked is optimal for the dual problem.