Kantorovich Dual Problem (for c(x,y) = d(x,y)^2 where d is a metric)

Kantorovich Dual Problem (for $c(x,y)=d(x,y)^{2}$ where $d$ is a metric).

The case where the cost is given by $c(x,y)={\frac {1}{2}}|x-y|^{2}$ is treated as a special case due to its relationship to c-concavity and convexity.

Relationship to c-concavity and convexity

Proposition

Given a function $f:\mathbb {R} ^{d}\to \mathbb {R} \cup \{+\infty \}$, define $u_{f}:\mathbb {R} ^{d}\to \mathbb {R} \cup \{+\infty \}$ by $u_{f}(x)={\frac {1}{2}}|x|^{2}-f(x)$. Then $u_{f^{c}}=(u_{f})^{*}$. Then a function $\zeta$ is c-concave if and only if $x\mapsto {\frac {1}{2}}|x|^{2}-\zeta (x)$ is convex and lower semicontinuous.

Theorem

Theorem

Let $\mu ,\nu$ be probabilities over $\mathbb {R} ^{d}$ and $c(x,y)={\frac {1}{2}}|x-y|^{2}$. Suppose $\int |x|^{2}dx,\int |y|^{2}dy<+\infty$, which implies min(KP) $<+\infty$ and suppose that $\mu$ gives no mass to $(d-1)$ surfaces of class $C^{2}$. Then there exists a unique optimal transport map $T$ from $\mu$ to $\nu$, and it is of the form $T=\nabla u$ for a convex function $u$.