A Pair Maximising a Quadratically Penalised Difference of Law-Invariant Functions Within Its Laws Realises an Optimal Coupling

In the setting of The Wasserstein Space and Its Lift to Square-Integrable Random Vectors: Standing Notation, assume that $(\Omega,\mathcal{F},P)$ is rich. Let $L^{2}(\Omega;\mathbb{R}^{d})$ be the space of classes of square-integrable random vectors, and $(\mathcal{P}_{2}(\mathbb{R}^{d}),W_{2})$ the Wasserstein space, so that $\mathcal{L}(X)\in\mathcal{P}_{2}(\mathbb{R}^{d})$ for every $X\in L^{2}(\Omega;\mathbb{R}^{d})$ by The Wasserstein Space and Its Lift to Square-Integrable Random Vectors: Standing Notation (v). Let $U,V:L^{2}(\Omega;\mathbb{R}^{d})\to\mathbb{R}$ be law-invariant, let $\alpha\in\mathbb{R}$ be positive, and let $X_{0},Y_{0}\in L^{2}(\Omega;\mathbb{R}^{d})$ satisfy

$$U(X)-V(Y)-\alpha\,\lVert X-Y\rVert_{L^{2}}^{2}\le U(X_{0})-V(Y_{0})-\alpha\,\lVert X_{0}-Y_{0}\rVert_{L^{2}}^{2}$$

for all $X,Y\in L^{2}(\Omega;\mathbb{R}^{d})$ with $\mathcal{L}(X)=\mathcal{L}(X_{0})$ and $\mathcal{L}(Y)=\mathcal{L}(Y_{0})$. Then the following hold.

1. (The penalty is the Wasserstein distance)(i)

$$\lVert X_{0}-Y_{0}\rVert_{L^{2}}=W_{2}\bigl(\mathcal{L}(X_{0}),\mathcal{L}(Y_{0})\bigr).$$

2. (The pair realises an optimal coupling)(ii) The law of the pairing $(X_{0},Y_{0})$ of any representatives of $X_{0}$ and $Y_{0}$, a random vector in $\mathbb{R}^{d+d}$ by Random Vectors: Coordinates, Functions, Pairs and Almost-Sure Equality (v), is an optimal coupling of $\mathcal{L}(X_{0})$ and $\mathcal{L}(Y_{0})$.