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 §law-map, 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 §infimum, which is in force through The Wasserstein Space and Its Lift to Square-Integrable Random Vectors: Standing Notation §law-map; 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 Space §norm.

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 §invariant, $U(X)=U(X_{0})$ and $V(Y)=V(Y_{0})$, so the hypothesis of the lemma for this pair reads

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 §numbers), $r_{0}=\inf D(\mu,\nu)=W_{2}(\mu,\nu)$.

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 Basic Properties of Random Vectors: Coordinates, Borel Images and Arithmetic, Change of Variables, Almost Sure Equality and Pairs §pair, $\pi$ is a coupling of the laws of the two representatives, which are $\mu$ and $\nu$ by The Space of Square-Integrable Random Vectors §law, 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 §vector-space and sums of classes being classes of pointwise sums by The Space of Square-Integrable Random Vectors §classes, so by The Space of Square-Integrable Random Vectors §inner-product 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,

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