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For probability measures on ${\mathbb{R}}^d$ with finite second moments and a source satisfying absolute continuity of measures with respect to Lebesgue measure, the quadratic Kantorovich optimal transport problem has a unique optimal transport plan. It is induced by the gradient of a convex function, which also uniquely solves the Monge optimal transport problem up to a source-null set.

Kantorovich potentials are the functions in the dual of the Kantorovich optimal transport problem. A feasible pair satisfies $u(x)+v(y)\le c(x,y)$ and provides a lower bound $\int ud\mu +\int vd\nu$ for every transport plan. An optimal pair attaining the dual value provides an optimality certificate.

Optimal transport minimizes the cost of moving one probability measure to another. The Monge optimal transport problem uses a transport map; the Kantorovich optimal transport problem allows a transport plan that can split mass.

For a cost $c:X\times Y\to [0,+\infty ]$ that is a Borel measurable function, a transport map is a measurable $T:X\to Y$ whose pushforward measure satisfiesThe Monge optimal transport problem moves every source point to one destination:The Kantorovich optimal transport problem permits mass to split. Its admissible transport plans are the probability measures on $X\times Y$ with prescribed marginal distributions:Thus a transport plan is a coupling of probability distributions. The set $\Pi (\mu ,\nu )$ is never empty: it contains the product measure $\mu \otimes \nu$. Signed costs can also be used when their integrals are well defined, for example with an integrable lower bound of the form $a(x)+b(y)$.

On the Polish space $\mathbb{R}$, take the Dirac measuresEvery measurable map satisfies $T_{\#}{\delta }_0={\delta }_{T(0)}$, which cannot equal $\nu$. A transport map cannot split an atom of a measure, whereas the transport plan $\frac{1}{2}{\delta }_{(0,-1)}+\frac{1}{2}{\delta }_{(0,1)}$ can.

Given any admissible transport map $T$, form its graph transport planFor Borel sets $A\subseteq X$ and $B\subseteq Y$, the definition of a pushforward measure givesHence ${\pi }_T\in \Pi (\mu ,\nu )$. Integration against a pushforward measure also givesThe Kantorovich optimal transport problem therefore has at least all the competitors of the Monge optimal transport problem, with exactly the same costs. ConsequentlyIf there is no admissible transport map, the left side is $+\infty$ by the convention $\inf \varnothing =+\infty$, so the conclusion still holds. No existence of an optimizer is needed.

For every admissible transport map $T$, the pushforward measure condition gives $T(x)\in [1,2]$ for $\mu$-almost every $x\in [0,1]$. Hence $|T(x)-x|=T(x)-x$, andApply the Jensen inequality to the convex function $h$ and the probability measure $\mu$:Translation by one sends the source uniform distribution to the target uniform distribution, so $T^{†}(x)=x+1$ is admissible. Its displacement is constantly one, givingNo monotonicity assumption on $h$ is needed, because all displacements have the same sign. The same argument with a transport plan also gives the identical minimum for the Kantorovich optimal transport problem.

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.

A standard sufficient form of Brenier theorem assumes $\mu ,\nu \in \mathcal{P}({\mathbb{R}}^d)$ have finite second moments and $\mu \ll {\mathcal{L}}^d$, that is, absolute continuity of measures with respect to Lebesgue measure. For the cost $|x-y{|}^2$, there exists a unique optimal transport plan, and it is induced by a transport map:Here $\Phi$ is a proper convex function, which may be chosen sequentially lower semicontinuous, and its gradient exists $\mu$-almost everywhere. The map $\nabla \Phi$ is unique $\mu$-almost everywhere and is the unique minimizer of the Monge optimal transport problem; its cost equals the Kantorovich optimal transport problem minimum. Equivalently, it is the unique gradient of a convex function transporting $\mu$ to $\nu$.

The uniqueness claim concerns the map and the transport plan, not a globally unique potential. The potential may be shifted by a constant, and additional nonuniqueness away from the source can occur. No density assumption is required on $\nu$. A primary reference is Brenier's Polar factorization and monotone rearrangement of vector-valued functions.

Write $L=W_p(\mu ,\nu )$. The assumed equality of the Monge optimal transport problem and Kantorovich optimal transport problem values givesEach interpolated pushforward measure ${\mu }_t=(P_t{)}_{\#}\mu$ has a finite $p$th absolute moment, since $|P_t(x){|}^p\le 2^{p-1}(|x{|}^p+|T^{†}(x){|}^p)$ and $(T^{†}{)}_{\#}\mu =\nu$.

For any $s,t\in [0,1]$, the common-source transport planprovides the upper boundFor the reverse bound assume $0\le s\le t\le 1$. Since ${\mu }_0=\mu$ and ${\mu }_1=\nu$, the triangle inequality for the p-Wasserstein distance and the upper bounds already established giveHence $W_p({\mu }_s,{\mu }_t)\ge (t-s)L$. Combining the bounds and using symmetry provesThis is the constant speed property of displacement interpolation. The given invertibility of $P_s$ also lets one realize the competitor as the map $P_t\circ P_s^{-1}$, assuming its inverse is measurable, but the transport plan argument proves the result without invertibility.