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

Each result cited is universally quantified over the data in its own statement. Put $\mu=\mathcal{L}(X_{0})$ and $\nu=\mathcal{L}(Y_{0})$, both in $\mathcal{P}_{2}(\mathbb{R}^{d})$ by The Wasserstein Space and Its Lift to Square-Integrable Random Vectors: Standing Notation (v), and let $D(\mu,\nu)=\{\lVert X-Y\rVert_{L^{2}}:X\sim\mu,\ Y\sim\nu\}$ be the set of The Wasserstein Distance and the Mean-Square Distance of Random Vectors (iv), which is in force through The Wasserstein Space and Its Lift to Square-Integrable Random Vectors: Standing Notation (v); since $(\Omega,\mathcal{F},P)$ is rich, that clause gives that $D(\mu,\nu)$ is nonempty, bounded below, and has greatest lower bound $\inf D(\mu,\nu)=W_{2}(\mu,\nu)$. Norms in $L^{2}(\Omega;\mathbb{R}^{d})$ are nonnegative by Real Inner Product Spaces (ii).

Claim 1. Put $r_{0}=\lVert X_{0}-Y_{0}\rVert_{L^{2}}$; then $r_{0}\in D(\mu,\nu)$. Let $s\in D(\mu,\nu)$, say $s=\lVert X-Y\rVert_{L^{2}}$ with $\mathcal{L}(X)=\mu$ and $\mathcal{L}(Y)=\nu$. By Law-Invariant Function on the Space of Square-Integrable Random Vectors (i), $U(X)=U(X_{0})$ and $V(Y)=V(Y_{0})$, so the hypothesis of the lemma for this pair reads

$$U(X_{0})-V(Y_{0})-\alpha\,s^{2}\le U(X_{0})-V(Y_{0})-\alpha\,r_{0}^{2}.$$

By claim 3 of Elementary Arithmetic in an Ordered Field this is equivalent to $0\le\alpha\,s^{2}-\alpha\,r_{0}^{2}$, that is, to $\alpha\,r_{0}^{2}\le\alpha\,s^{2}$ by the same claim. Since $\alpha$ is positive, $\alpha^{-1}$ exists and $0\le\alpha^{-1}$ by claim 4 of Elementary Arithmetic in an Ordered Field, and multiplying by $\alpha^{-1}$ (claim 5 there) gives $r_{0}^{2}\le s^{2}$. As $0\le r_{0}$ and $0\le s$, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field yields $r_{0}\le s$. Thus $r_{0}$ is a lower bound of $D(\mu,\nu)$, so $r_{0}\le\inf D(\mu,\nu)$ by Lower Bound and Greatest Lower Bound; and $\inf D(\mu,\nu)\le r_{0}$ because $r_{0}\in D(\mu,\nu)$ and the infimum is a lower bound. By antisymmetry of the order of $\mathbb{R}$ (Real Hilbert Spaces: Standing Notation and Background (i)), $r_{0}=\inf D(\mu,\nu)=W_{2}(\mu,\nu)$.

Claim 2. Fix representatives of $X_{0}$ and $Y_{0}$, again written $X_{0},Y_{0}$, and let $\pi$ be the law of their pairing $(X_{0},Y_{0})$. By Random Vectors: Coordinates, Functions, Pairs and Almost-Sure Equality (v), $\pi$ is a coupling of the laws of the two representatives, which are $\mu$ and $\nu$ by The Space of Square-Integrable Random Vectors (iv), with quadratic cost $I(\pi)=\mathbb{E}[\lVert X_{0}-Y_{0}\rVert^{2}]$. The class $X_{0}-Y_{0}$ has the pointwise difference of the chosen representatives as a representative, the additive inverse of the class of $Y_{0}$ being the class of $(-1)Y_{0}$ by Square-Integrable Random Vectors: Coordinates, Operations, Almost Sure Equality and the Mean-Square Form (v) and sums of classes being classes of pointwise sums by The Space of Square-Integrable Random Vectors (ii), so by The Space of Square-Integrable Random Vectors (iii) the norm $\lVert X_{0}-Y_{0}\rVert_{L^{2}}$ is the nonnegative square root of $\mathbb{E}[\lVert X_{0}-Y_{0}\rVert^{2}]$, whence $\lVert X_{0}-Y_{0}\rVert_{L^{2}}^{2}=\mathbb{E}[\lVert X_{0}-Y_{0}\rVert^{2}]$ by Existence and Uniqueness of the Nonnegative Square Root. Together with claim 1,

$$I(\pi)=\lVert X_{0}-Y_{0}\rVert_{L^{2}}^{2}=W_{2}(\mu,\nu)^{2},$$

which is the defining identity of an optimal coupling, Optimal Coupling of Two Probability Measures with Finite Second Moment (i).