Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 348 / 3

The dual of the Kantorovich optimal transport problem iswhere the Kantorovich potentials are measurable representatives satisfyingOne standard form of the Kantorovich duality theorem assumes that $X,Y$ are Polish spaces, $\mu ,\nu$ are probability measures defined as Borel measures, and $c:X\times Y\to [0,+\infty ]$ is sequentially lower semicontinuous. ThenThe primal infimum is attained; its value may be $+\infty$. Nonnegativity can be replaced by a constant lower bound by shifting the cost. This duality for lower semicontinuous costs is also discussed in Beiglboeck, Leonard and Schachermayer's duality paper.

Equality of values does not by itself assert a dual maximum. A sufficient stronger setting for attainment on both sides is compact metric spaces $X,Y$ and a finite continuous cost $c$; then continuous Kantorovich potentials attain the dual supremum. The general statement above correctly uses a supremum.

Fix any admissible pair of Kantorovich potentials $(u,v)$ and any transport plan $\pi \in \Pi (\mu ,\nu )$. Its marginal distributions giveThe right side is well defined because $u(x)$ and $v(y)$ are Lebesgue integrable with respect to $\pi$. Integrating their pointwise feasibility inequality givesSince this holds for every feasible pair and every transport plan,This proves the required inequality directly from the transport constraints, without any convex optimization duality theorem. Whenever the extrema are attained, the supremum and infimum can respectively be written as a maximum and minimum. Even the Kantorovich duality theorem is unnecessary for this direction.

The Kantorovich–Rubinstein theorem concerns a Polish space $(X,\rho )$ and probability measures $\mu ,\nu$ with finite first moments, meaning $\int \rho (x,x_0)d\mu (x)+\int \rho (x,x_0)d\nu (x)<\infty$ for one, hence every, $x_0\in X$. It statesHere $\mathrm{Lip}(f)\le 1$ means $|f(x)-f(y)|\le \rho (x,y)$ for every $x,y$. The Wasserstein distance with ground cost $\rho$ therefore equals a supremum over functions with Lipschitz constant at most one. One may normalize $f(x_0)=0$, since adding a constant does not change the difference of integrals. This normalization bounds $|f(x)|$ by $\rho (x,x_0)$, so finite first moments ensure integrability. The supremum is unchanged if an absolute value is placed around the difference, because $-f$ is admissible whenever $f$ is.

Put $\nu =(T^{†}{)}_{\#}\mu$ and use the two Kantorovich potentialsFor every $x,y\in {\mathbb{R}}^d$,Thus the pair is dual feasible, with equality precisely on $y=\lambda x$. Finite second moments make both potentials integrable; compactness of $X$ is more than sufficient. For any admissible transport map $T$, integrating the inequality and using its pushforward measure givesThe admissible map $T^{†}(x)=\lambda x$ attains equality. ConsequentlyThe same certificate proves optimality of its graph transport plan among all transport plans. Notice that $u$ need not be a convex function when $\lambda >1$: it is a cost dual potential. The associated convex gradient potential is instead $\Phi (x)=\lambda |x{|}^2/2$.