Proof of A Displacement Convex Penalty Pair Has a Monotone Score Along Optimal Couplings

Each result cited is universally quantified over the data in its own statement. Write $\sigma:\mathbb{R}^{d+d}\to\mathbb{R}^{d+d}$ for the swap, a Borel map with $\mathrm{pr}_{1}(\sigma(z))=y$ and $\mathrm{pr}_{2}(\sigma(z))=x$ for $z\in\mathbb{R}^{d+d}$, by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections.

The swapped coupling is optimal. By Couplings on Euclidean Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, and the Lipschitz Bound §swap one has $\sigma_{\#}\pi\in\Pi(\nu,\mu)$ and $I(\sigma_{\#}\pi)=I(\pi)$. Since $\pi$ is optimal, $I(\pi)=W_{2}(\mu,\nu)^{2}$ by Optimal Coupling of Two Probability Measures with Finite Second Moment §optimal, and $W_{2}(\mu,\nu)=W_{2}(\nu,\mu)$ by The Quadratic Wasserstein Distance is a Metric on the Wasserstein Space §symmetry. Hence $I(\sigma_{\#}\pi)=W_{2}(\nu,\mu)^{2}$, so $\sigma_{\#}\pi$ is an optimal coupling of $\nu$ and $\mu$.

The two integrals. By Displacement Convexity of a Penalty Pair on the Wasserstein Space §displacement, applied to $\mu$, $\nu$ and $\pi$, the displacement plan $\pi^{-}=(\mathrm{pr}_{1},\mathrm{pr}_{2}-\mathrm{pr}_{1})_{\#}\pi$ is a plan with first marginal $\mu$ and

is a real number, the variables $x,p$ being those of Plans, Marginals, Vector Fields and Symmetric Matrices on the Wasserstein Space: Standing Notation §plans for the plan $\pi^{-}$. The function $z\mapsto\Sigma(\mu)(x)\cdot(y-x)$ on $\mathbb{R}^{d+d}$ is the composition with $(\mathrm{pr}_{1},\mathrm{pr}_{2}-\mathrm{pr}_{1})$ of the integrand defining $A$, by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §projections and Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets §pairing; hence it is integrable with respect to $\pi$ with integral $A$, by the change-of-variables formula of Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward.

Applying Displacement Convexity of a Penalty Pair on the Wasserstein Space §displacement to $\nu$, $\mu$ and the optimal coupling $\sigma_{\#}\pi$ in the same way, the real number

is the integral with respect to $\sigma_{\#}\pi$ of the function $w\mapsto\Sigma(\nu)(\mathrm{pr}_{1}(w))\cdot(\mathrm{pr}_{2}(w)-\mathrm{pr}_{1}(w))$, which is integrable with respect to it; composing with $\sigma$ and using the change-of-variables formula of Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward once more, now for the push-forward $\sigma_{\#}\pi$, the function $z\mapsto\Sigma(\nu)(y)\cdot(x-y)$ is integrable with respect to $\pi$ with integral $B$.

The two inequalities. Both $\mu$ and $\nu$ lie in $\mathcal{D}_{\Sigma}$, hence in $\mathcal{D}$, so Displacement Convexity of a Penalty Pair on the Wasserstein Space §convex applies in both directions and gives

Adding these two inequalities, by claim 3 of Elementary Arithmetic in an Ordered Field and the compatibility of the order with addition (an axiom of Ordered Field), and cancelling the real numbers $\mathcal{E}(\mu)$ and $\mathcal{E}(\nu)$ by the associativity, commutativity, inverse and identity axioms of a field, gives $A+B\le0$.

Conclusion. By the previous paragraphs the two functions $z\mapsto\Sigma(\mu)(x)\cdot(y-x)$ and $z\mapsto\Sigma(\nu)(y)\cdot(x-y)$ are integrable with respect to $\pi$, so their sum is, and its integral is $A+B$ by claim 2 of Linearity and Monotonicity of the Lebesgue Integral. At each $z$ that sum equals

by claim 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$ one has $w\cdot(y-x)=w\cdot y-w\cdot x=-(w\cdot x-w\cdot y)=-\bigl(w\cdot(x-y)\bigr)$ for every $w\in\mathbb{R}^{d}$, so the left-hand side is $\Sigma(\nu)(y)\cdot(x-y)-\Sigma(\mu)(x)\cdot(x-y)$, which by claim 3 of that lemma is $\bigl(\Sigma(\nu)(y)-\Sigma(\mu)(x)\bigr)\cdot(x-y)$, and this is the right-hand side by claim 3 once more together with the field identity $-(s-t)=t-s$ for real numbers. Hence the function of the statement is integrable with respect to $\pi$, being $(-1)$ times this sum, with integral $-(A+B)$ by claim 2 of Linearity and Monotonicity of the Lebesgue Integral; and $0\le-(A+B)$ follows from $A+B\le0$ by claim 4 of Elementary Order Arithmetic in an Ordered Field.