Proof of A $\lambda$-Displacement Convex Penalty Pair Has a $\lambda$-Monotone Score Along Optimal Couplings

Each result cited is universally quantified over the data in its own statement. Write $W=W_{2}(\mu,\nu)^{2}$.

Step 1 (the swap of an optimal coupling is optimal). Let $\pi\in\Pi(\mu,\nu)$ be optimal, so $I(\pi)=W$ by Optimal Coupling of Two Probability Measures with Finite Second Moment §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, $\pi^{\top}\in\Pi(\nu,\mu)$ and $I(\pi^{\top})=I(\pi)=W$, and $W=W_{2}(\nu,\mu)^{2}$ by The Quadratic Wasserstein Distance is a Metric on the Wasserstein Space §symmetry. Hence $\pi^{\top}$ is an optimal coupling of $\nu$ and $\mu$, again by Optimal Coupling of Two Probability Measures with Finite Second Moment §optimal.

Step 2 (claim 1). Let $\pi$ be as in Step 1. Since $\mu,\nu\in\mathcal{D}_{\Sigma}\subseteq\mathcal{D}$ by Penalty Pairs on the Wasserstein Space: the Penalty, Its Score, and Their Domains §pair, the $\lambda$-displacement convexity of $\lambda$-Displacement Convexity of a Penalty Pair on the Wasserstein Space §convex applies to the measure $\mu\in\mathcal{D}_{\Sigma}$, the measure $\nu\in\mathcal{D}$ and the optimal coupling $\pi$, and also, by Step 1, to the measure $\nu\in\mathcal{D}_{\Sigma}$, the measure $\mu\in\mathcal{D}$ and the optimal coupling $\pi^{\top}$:

By claim 3 of Elementary Arithmetic in an Ordered Field the differences of the right and left sides are nonnegative, so their sum is nonnegative by claim 2 there. In that sum $\mathcal{E}(\mu)$ and $\mathcal{E}(\nu)$ cancel, and $\tfrac{\lambda}{2}W+\tfrac{\lambda}{2}W=\lambda W$ because $2$ times the multiplicative inverse of $2$ is $1$; so

and claim 3 of Elementary Arithmetic in an Ordered Field turns this into claim 1.

Step 3 (claim 2). Let $S$ and $S'$ be as in claim 2. The coupling $\pi_{S}=(\mathrm{id},S)_{\#}\mu$ belongs to $\Pi(\mu,\nu)$ and is optimal, and $(\mathrm{id},S')_{\#}\nu$ belongs to $\Pi(\nu,\mu)$ and is optimal, both by Optimal Transport Maps and Uniquely Mapped Pairs of Probability Measures §map. By Step 1, $\pi_{S}^{\top}$ is an optimal coupling of $\nu$ and $\mu$. Since $(\nu,\mu)$ is uniquely mapped, there is an optimal map $T$ from $\nu$ to $\mu$ such that every optimal coupling of $\nu$ and $\mu$ equals $(\mathrm{id},T)_{\#}\nu$ (Optimal Transport Maps and Uniquely Mapped Pairs of Probability Measures §uniquely-mapped); applying this to both optimal couplings just named,

Next, $\lVert\mathrm{id}-S\rVert_{\mu}^{2}=W<\infty$ by The Optimal Map as a Square-Integrable Vector Field: Integrability, Transport Cost and Uniqueness of the Class §cost, and $\lVert S(x)-x\rVert=\lVert x-S(x)\rVert$ for every $x$ by claim 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ (with the scalar $-1$), so $\int_{\mathbb{R}^{d}}\lVert S-\mathrm{id}\rVert^{2}\,d\mu=\lVert\mathrm{id}-S\rVert_{\mu}^{2}<\infty$. Hence The Displacement Pairing of a Square-Integrable Vector Field Along a Coupling §displacement applies to $S$ and gives $\mathcal{J}(\Sigma(\mu),\pi_{S})=\langle\Sigma(\mu),S-\mathrm{id}\rangle_{\mu}$. The class $S-\mathrm{id}$ is $(-1)(\mathrm{id}-S)$, the pointwise relation $S(x)-x=(-1)(x-S(x))$ passing to classes by Square-Integrable Random Vectors: Coordinates, Operations, Almost Sure Equality and the Mean-Square Form §vector-space; by linearity of the inner product of $L^{2}(\mu;\mathbb{R}^{d})$ in its second argument,

The same argument with $\nu$, $\mu$ and $S'$ in place of $\mu$, $\nu$ and $S$ gives $-\mathcal{J}(\Sigma(\nu),(\mathrm{id},S')_{\#}\nu)=\langle\Sigma(\nu),\mathrm{id}-S'\rangle_{\nu}$. Claim 1, applied to the optimal coupling $\pi_{S}$ with $\pi_{S}^{\top}=(\mathrm{id},S')_{\#}\nu$, now reads

which is claim 2. $\blacksquare$