Proof of A Displacement Convex Penalty Pair Has a Monotone Score Along Optimally Coupled Pairs

Each result cited is universally quantified over the data in its own statement and is applied here to the data named in the statement of the lemma. Write $\mu=\mathcal{L}(X)$ and $\nu=\mathcal{L}(Y)$, both in $\mathcal{D}_{\Sigma}$, hence in $\mathcal{D}$ by Penalty Pairs on the Wasserstein Space: the Penalty, Its Score, and Their Domains §pair, so that $\mathcal{E}(\mu)$ and $\mathcal{E}(\nu)$ are real numbers; write $\sigma_{X}=\Sigma(\mu)\circ X$ and $\sigma_{Y}=\Sigma(\nu)\circ Y$, elements of $L^{2}(\Omega;\mathbb{R}^{d})$ by Composition of a Square-Integrable Vector Field with a Random Vector, and the Lifted Score: Isometry, Norm, Weak Identity and Second-Moment Identity §composition. The space $L^{2}(\Omega;\mathbb{R}^{d})$ is a real Hilbert space by Wasserstein Spaces, Random Vectors, Vector Fields and Symmetric Matrices in Every Dimension: Standing Notation §dimensions, so Elementary Identities in a Real Inner Product Space and The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity apply to it with the norm $\lVert\cdot\rVert_{L^{2}}$; real arithmetic is that of the field axioms of Field and of Elementary Arithmetic in an Ordered Field.

The reversed pair is optimally coupled. By Optimally Coupled Pairs of Square-Integrable Random Vectors §optimal, $\lVert X-Y\rVert_{L^{2}}^{2}=W_{2}(\mu,\nu)^{2}$. By The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §metric the function $d_{L^{2}}(Z,Z')=\lVert Z-Z'\rVert_{L^{2}}$ is a metric, hence symmetric, so $\lVert Y-X\rVert_{L^{2}}=\lVert X-Y\rVert_{L^{2}}$; and $W_{2}(\nu,\mu)=W_{2}(\mu,\nu)$ by The Quadratic Wasserstein Distance is a Metric on the Wasserstein Space §symmetry. Hence $\lVert Y-X\rVert_{L^{2}}^{2}=W_{2}(\nu,\mu)^{2}$, that is, $(Y,X)$ is optimally coupled by Optimally Coupled Pairs of Square-Integrable Random Vectors §optimal.

Two applications of displacement convexity. Since $(X,Y)$ is optimally coupled with $\mu\in\mathcal{D}_{\Sigma}$ and $\nu\in\mathcal{D}$, and $(Y,X)$ is optimally coupled with $\nu\in\mathcal{D}_{\Sigma}$ and $\mu\in\mathcal{D}$, Displacement Convexity of a Penalty Pair on the Wasserstein Space §convex gives

By claim 3 of Elementary Arithmetic in an Ordered Field, $0\le\mathcal{E}(\nu)-\mathcal{E}(\mu)-\langle\sigma_{X},Y-X\rangle_{L^{2}}$ and $0\le\mathcal{E}(\mu)-\mathcal{E}(\nu)-\langle\sigma_{Y},X-Y\rangle_{L^{2}}$, and by claim 2 of that lemma the sum of these two nonnegative numbers is nonnegative; the terms $\mathcal{E}(\mu)$ and $\mathcal{E}(\nu)$ cancel in the sum by the field axioms, so

Rewriting the right-hand side. By Elementary Identities in a Real Inner Product Space §bilinear, $\langle\sigma_{X},Y-X\rangle_{L^{2}}=\langle\sigma_{X},Y\rangle_{L^{2}}-\langle\sigma_{X},X\rangle_{L^{2}}$ and $\langle\sigma_{X},X-Y\rangle_{L^{2}}=\langle\sigma_{X},X\rangle_{L^{2}}-\langle\sigma_{X},Y\rangle_{L^{2}}$, so $-\langle\sigma_{X},Y-X\rangle_{L^{2}}=\langle\sigma_{X},X-Y\rangle_{L^{2}}$ by the field axioms. Therefore, again by Elementary Identities in a Real Inner Product Space §bilinear in the first argument,

which is the claim.