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For probability measures $\mu ,\nu$ on ${\mathbb{R}}^d$ with finite second moments, the Knott–Smith optimality criterion states that a transport plan $\pi \in \Pi (\mu ,\nu )$ minimizes the quadratic cost $|x-y{|}^2$ if and only if there is a proper convex function $\Phi :{\mathbb{R}}^d\to \mathbb{R}\cup \{+\infty \}$ that is sequentially lower semicontinuous and satisfiesThe subdifferential is characterized bywith $\Phi (x)<\infty$. Thus the transport plan is concentrated on the graph of the subdifferential. No absolute continuity of measures assumption on $\mu$ is needed. Multiplying the cost by $1/2$ leaves the criterion unchanged. The original quadratic optimal mapping result is Knott and Smith, On the optimal mapping of distributions.

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.

Let $\Phi ={\phi }_{\epsilon }$ and let ${\Phi }^{\ast }$ denote its convex conjugate. Its Fenchel–Young gapis nonnegative by the Fenchel–Young inequality, and the hypothesis says $\int Hd{\pi }_{\epsilon }\le \epsilon$.

First justify the integrability needed for the certificate. Finite second moments and the Cauchy-Schwarz inequality imply, for every transport plan $\pi$,Since $\Phi \in L^1(\mu )$ and $H\in L^1({\pi }_{\epsilon })$, the identity ${\Phi }^{\ast }(y)=H(x,y)-\Phi (x)+x\cdot y$ proves ${\Phi }^{\ast }\in L^1(\nu )$ by the marginal distribution property. In particular, no subtraction of infinite integrals is being used.

SetThese are integrable Kantorovich potentials. The Fenchel–Young inequality gives $u(x)+v(y)\le \frac{1}{2}|x-y{|}^2$. More precisely, for every transport plan $\pi$,Taking the infimum in the first inequality therefore provesThe factor $1/2$ in the quadratic cost is essential for this exact gap identity.

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.

Let $D=\mathrm{diam}(X)<\infty$, and first suppose $1\le p\le q<\infty$. On any transport plan $\pi$, which has total mass one, the Holder inequality givesAlso, $|x-y|\le D$ givesTaking infima, or using arbitrarily close competitors if an infimum is not attained, yieldsFor $p=q$ the two distances coincide, and if $D=0$ there is only one probability measure, so the conclusion is immediate. For $D>0$ and $p<q$, the displayed bounds imply both directions ofFor $q<p$, exchange the exponents. Thus all finite-order p-Wasserstein distances induce the same convergence on a bounded subset of ${\mathbb{R}}^d$. Closedness of $X$ is not required for these estimates.

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.

Let $\Delta =\{(x,y):x=y\}$. For any transport plan $\pi$ and any Borel set $A$,Both rectangles are contained in ${\Delta }^c$, soThus $\inf \mathbb{K}\ge {\sup }_A|\mu (A)-\nu (A)|$.

To attain this bound, use the dominating measure $\rho =\mu +\nu$ and Radon-Nikodym derivatives $f=d\mu /d\rho$, $g=d\nu /d\rho$. Define the common measure and residual mass bySince $\int (f-g)d\rho =0$, the positive and negative parts have equal mass. Taking $A=\{f>g\}$ givesIf $r=0$, then $\mu =\nu$ and the diagonal transport plan has zero cost. If $r>0$, put $\beta =\mu -\alpha$, $\gamma =\nu -\alpha$ and defineThe residual measures each have mass $r$, so the marginal distributions of ${\pi }^{†}$ are $\alpha +\beta =\mu$ and $\alpha +\gamma =\nu$, and its total mass is $1-r+r=1$. They are mutually singular measures, concentrated respectively on $\{f>g\}$ and $\{g>f\}$. Their product measure therefore gives no mass to $\Delta$, and $\mathbb{K}({\pi }^{†})=r$. This is a maximal coupling.

In the paper's convention for the zero-mass signed measure $\mu -\nu$, the answer isThe repository's total variation distance uses the supremum itself. The paper's formula $2{\sup }_A|\sigma (A)|$ agrees with the usual total variation norm when $\sigma$ has total mass zero, as here; it is not the usual norm for an arbitrary positive measure. All suprema above are over Borel sets.

For $1\le p<\infty$ and probability measures with finite $p$th absolute moments,where $\rho$ is the ground metric and $\pi$ ranges over transport plans. For $p=2$ this is the second Wasserstein distance.

A transport map from $\mu$ to $\nu$ is a measurable function $T$ satisfying $T_{\#}\mu =\nu$. Its graph defines the transport plan $(\mathrm{Id},T{)}_{\#}\mu$.