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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.

In Euclidean space, push an optimal transport plan $\pi$ forward by $(x,y)\mapsto (1-t)x+ty$. The resulting probability measures form a constant speed curve for the p-Wasserstein distance:If the plan is induced by a transport map $T$, this becomes the displayed title formula.

For a proper convex function $\Phi$ and its convex conjugate ${\Phi }^{\ast }$, this gap is nonnegative by the Fenchel–Young inequality. It vanishes exactly when $y\in \partial \Phi (x)$. Its integral against a transport plan is the difference between the half-squared-distance cost and the value of the integrable Kantorovich potentials $u=|x{|}^2/2-\Phi$, $v=|y{|}^2/2-{\Phi }^{\ast }$.

For probability measures defined as Borel measures on Polish spaces and a nonnegative sequentially lower semicontinuous cost, the minimum cost over transport plans equals the supremum of $\int ud\mu +\int vd\nu$ over integrable Kantorovich potentials satisfying $u(x)+v(y)\le c(x,y)$. The primal minimum is attained; a dual maximum needs additional assumptions. Compact metric spaces and a finite continuous cost suffice for attainment of both extrema.

The Kantorovich problem minimizes $\int c(x,y)d\pi (x,y)$ over transport plans $\pi$ with marginal distributions $\mu ,\nu$. It relaxes the Monge optimal transport problem by allowing source mass to split among destinations.

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.

For probability measures on ${\mathbb{R}}^d$ with finite second moments, a transport plan minimizes the quadratic cost exactly when it is concentrated on the graph of the subdifferential of a sequentially lower semicontinuous proper convex function.

For an atomless measure $\mu$ on $\mathbb{R}$ with cumulative distribution function $F$, the monotone transport to a target with quantile function $G^{-1}$ is $T=G^{-1}\circ F$, defined $\mu$-almost everywhere. It minimizes well-defined costs $d(x-y)$ for convex continuous $d$. With atoms in the source, the common-quantile transport plan $(F^{-1},G^{-1}{)}_{\#}\mathcal{U}(0,1)$ remains available but need not be induced by a map.

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.

The intended monotone rearrangement isHere $G^{-1}$ is the quantile function of $\nu$, agreeing with the ordinary inverse when $G$ is continuous and strictly increasing. With an atomless measure $\mu$, its cumulative distribution function $F$ is continuous, and the probability integral transform makes $F(X)$ uniform on $(0,1)$ for $X\sim \mu$. Thus $G^{-1}(F(X))\sim \nu$. The one-dimensional monotone rearrangement theorem says this transport map minimizes the cost $d(x-y)$ for convex continuous $d$, whenever the cost integrals are well defined. Values at exceptional endpoints may be chosen arbitrarily.

The printed assumptions omit an essential source condition. Invertibility of $G$ alone does not ensure an admissible transport map. For example, $\mu ={\delta }_0$ and $\nu$ a standard normal distribution satisfy the stated condition on $G$, but $T_{\#}{\delta }_0$ is always a Dirac measure. There is no solution to the Monge optimal transport problem in this example. The boxed answer therefore requires the additional assumption that $\mu$ is an atomless measure, or an equivalent condition making the displayed map admissible. For arbitrary sources the always admissible monotone transport plan is $(F^{-1},G^{-1}{)}_{\#}\mathcal{U}(0,1)$, which need not be induced by a map.

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.

As in part (c), every admissible transport map has nonnegative displacement $D(x)=T(x)-x$ with ${\int }_0^{1}D(x)dx=1$. The square root is a strictly concave function, so the reversed Jensen inequality givesThe map $T^{†}(x)=x+1$ is admissible and attains one. ThereforeThus “worst” means largest cost among admissible transport maps. Equality in the Jensen inequality for a strictly concave function forces $D(x)$ to be constant almost everywhere, so this maximizer is unique up to a $\mu$-null set. For comparison, the admissible reflection $T(x)=2-x$ has smaller costThe upper bound also holds for every transport plan.

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.

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 $\Phi (x)=\lambda |x{|}^2/2$.