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An atom of a measure $\mu$ is a measurable set $A$ of positive measure for which every measurable $B\subseteq A$ has either $\mu (B)=0$ or $\mu (B)=\mu (A)$. A singleton of positive mass is an example. A transport map cannot split the mass of a singleton among several destinations.

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.

Given a cost $c$ and probability measures $\mu ,\nu$, the Monge problem minimizes $\int c(x,T(x))d\mu (x)$ over measurable transport maps with pushforward measure $T_{\#}\mu =\nu$. The feasible set may be empty because a map cannot split an atom of a measure.

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.

Interpret the invertibility assumption on $G$ as continuity and strict increase on the relevant range, so that $\nu$ is an atomless measure. Suppose a non-decreasing transport map $T$ with $T_{\#}\mu =\nu$ exists. Then $\mu$ is also an atomless measure: if $\mu (\{x\})>0$, the pushforward measure would give $\nu (\{T(x)\})\ge \mu (\{x\})>0$. Thus $F$ is continuous.

At any point $x$ where the non-decreasing representative is defined, the definition of a monotone function givesUsing the pushforward measure identity and continuity of the two cumulative distribution functions, we obtainConsequentlyThe monotone rearrangement in part (c) is optimal for the convex difference cost, so $T$ has the same cost and solves the Monge optimal transport problem. The meaningful uniqueness is up to a $\mu$-null set; arbitrary values away from the source do not affect transport or cost. Existence of the non-decreasing transport map supplies the source condition missing in part (c).

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.

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

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.

Define $T^{†}(0)=0$; the printed quotient is undefined at the origin, but this continuous extension changes no transport cost because $\mu (\{0\})=0$. In polar coordinates, the two probability density functions giveBoth angular distributions are uniform, with independence of angle and radius. The proposed transport map preserves the angle and sends $r$ to $s=\sqrt{r}$. Thereforeand preservation of the angle proves $(T^{†}{)}_{\#}\mu =\nu$. As a local check using the Jacobian determinant, its radial and tangential derivatives for $r>0$ are $1/(2\sqrt{r})$ and $1/\sqrt{r}$, respectively, so $\det D{T}^{†}=1/(2r)$ and $g(T^{†}(x))\det D{T}^{†}(x)=(2r/\pi )/(2r)=f(x)$.

Now use the convex functionIt is convex because the Euclidean norm is convex and $s\mapsto (2/3)s^{3/2}$ is increasing and convex on $[0,\infty )$. It is differentiable, including at zero, andIts graph transport plan lies in the graph of $\partial \Phi$, so the Knott–Smith optimality criterion proves quadratic optimality. Since that plan is induced by a map, part 1(b) proves optimality for the Monge optimal transport problem as well. The minimum provides a useful independent check:

For $1\le p<\infty$, define the probability measures with finite $p$th absolute moment bywhere $x_0\in X$ is any fixed reference point. The condition is independent of the choice of $x_0$. The p-Wasserstein distance isThe infimum is over all transport plans, rather than only transport maps. The product measure gives a finite upper bound usingFor $p=1$ this is the Wasserstein distance with the metric of Euclidean space; for $p=2$ it is the second Wasserstein distance. “Bounded moment” here means a finite integral, and does not require $X$ to be bounded.

A transport plan is a coupling of probability distributions $\mu ,\nu$, that is, a probability measure on the product space with those marginal distributions. Unlike a transport map, it need not be concentrated on the graph of a function.