Proof of The Quadratic Wasserstein Distance is a Metric on the Probability Measures with Finite Second Moment on a Hilbert Space

Each result cited below is universally quantified over the data in its own statement and is applied with the data named at the point of citation.

Step 0 (Preliminaries). Couplings $\Pi(\cdot,\cdot)$ and the quadratic cost $I$ are those of Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §coupling and Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §cost. Let $\alpha,\beta\in\mathcal{P}_{2}(X)$ and put $J(\alpha,\beta)=\{I(\pi):\pi\in\Pi(\alpha,\beta)\}$. By The Quadratic Wasserstein Distance on a Hilbert Space §distance, $J(\alpha,\beta)$ is a nonempty set of nonnegative real numbers, $W_{2}(\alpha,\beta)$ is the nonnegative square root of its greatest lower bound, so that $W_{2}(\alpha,\beta)^{2}=\inf J(\alpha,\beta)$ by Existence and Uniqueness of the Nonnegative Square Root, and $W_{2}(\alpha,\beta)^{2}\le I(\pi)$ for every $\pi\in\Pi(\alpha,\beta)$. Since $\sqrt{I(\pi)}^{2}=I(\pi)$, claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field turns the last inequality into

$$W_{2}(\alpha,\beta)\le\sqrt{I(\pi)}\qquad\text{for every }\pi\in\Pi(\alpha,\beta).$$

We also use the following approximation (A): for every real $\varepsilon>0$ there is $\pi\in\Pi(\alpha,\beta)$ with $\sqrt{I(\pi)}<W_{2}(\alpha,\beta)+\varepsilon$. Indeed, with $W=W_{2}(\alpha,\beta)\ge0$ the number $\eta=2W\varepsilon+\varepsilon^{2}$ is positive, so claim 4 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$, applied to the set $J(\alpha,\beta)$, bounded below by $0$, and to $\eta$, gives $\pi\in\Pi(\alpha,\beta)$ with $I(\pi)<W^{2}+\eta=(W+\varepsilon)^{2}$; as $\sqrt{I(\pi)}$ and $W+\varepsilon$ are nonnegative, claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $\sqrt{I(\pi)}<W+\varepsilon$.

Step 1 (Claim 1, symmetry). Let $\sigma$ be the swap of Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §swap. By that clause, applied to $\mu,\nu$, every $\pi\in\Pi(\mu,\nu)$ gives $\sigma_{\#}\pi\in\Pi(\nu,\mu)$ with $I(\sigma_{\#}\pi)=I(\pi)$, so $J(\mu,\nu)\subseteq J(\nu,\mu)$; applied to $\nu,\mu$ it gives $J(\nu,\mu)\subseteq J(\mu,\nu)$. Hence $J(\mu,\nu)=J(\nu,\mu)$, the two sets have the same greatest lower bound, and its nonnegative square root is unique by Existence and Uniqueness of the Nonnegative Square Root; so $W_{2}(\mu,\nu)=W_{2}(\nu,\mu)$ by The Quadratic Wasserstein Distance on a Hilbert Space §distance.

Step 2 (Claim 2, separation: equal measures). Suppose $\mu=\nu$. By Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §pushforward, $(\mathrm{id}_{X},\mathrm{id}_{X})_{\#}\mu\in\Pi(\mu,\mu)$ has quadratic cost $0$, so $W_{2}(\mu,\mu)^{2}\le0$ by Step 0. Being the square of a real number, $W_{2}(\mu,\mu)^{2}\ge0$, so $W_{2}(\mu,\mu)^{2}=0$ and therefore $W_{2}(\mu,\mu)=0$.

Step 3 (Claim 2, separation: zero distance). Suppose $W_{2}(\mu,\nu)=0$. By Existence of an Optimal Coupling of Two Borel Probability Measures with Finite Second Moment on a Hilbert Space §existence there is $\pi\in\Pi(\mu,\nu)$ with $I(\pi)=W_{2}(\mu,\nu)^{2}=0$. Let $g:X\times X\to[0,\infty)$ be $g(z)=|\pi_{1}(z)-\pi_{2}(z)|^{2}$, which is nonnegative and Borel as recorded in Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §cost, and whose integral against $\pi$ is $I(\pi)=0$. By The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §vanishing, applied to the measure space $(X\times X,\mathcal{B}(X\times X),\pi)$ and to $g$, we have $g(z)=0$ for $\pi$-almost every $z$; that is, by A Property Holding Almost Everywhere and Null Set of a Measure, there is $B\in\mathcal{B}(X\times X)$ with $\pi(B)=0$ such that $g(z)=0$ for every $z\in(X\times X)\setminus B$. For such $z$ the nonnegative real number $|\pi_{1}(z)-\pi_{2}(z)|$ has square $0$ and hence is $0$, so $d(\pi_{1}(z),\pi_{2}(z))=0$ and $\pi_{1}(z)=\pi_{2}(z)$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §metric and condition 2 of Metric Space.

Now let $A\in\mathcal{B}(X)$. The sets $A\times X=\pi_{1}^{-1}(A)$ and $X\times A=\pi_{2}^{-1}(A)$ belong to $\mathcal{B}(X\times X)$, since $\pi_{1},\pi_{2}$ are Borel by Borel Sets of a Hilbert Space with an Orthonormal Basis: Coordinates, Determination by Finite-Dimensional Projections, and Pairs §product-sigma. For $z\notin B$ we have $\pi_{1}(z)=\pi_{2}(z)$, so $z\in A\times X$ if and only if $z\in X\times A$; hence the Borel set $D=(A\times X)\setminus B$ equals $(X\times A)\setminus B$. The set $A\times X$ is the disjoint union of $D$ and $(A\times X)\cap B$, and $\pi((A\times X)\cap B)\le\pi(B)=0$, so by Basic Properties of a Measure §additivity and Basic Properties of a Measure §monotone, applied to the measure $\pi$, $\pi(A\times X)=\pi(D)$; in the same way $\pi(X\times A)=\pi(D)$. By Couplings of Two Borel Probability Measures on a Hilbert Space and Their Quadratic Cost §coupling and the definition of the push-forward in Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §pushforward,

$$\mu(A)=(\pi_{1})_{\#}\pi(A)=\pi(A\times X)=\pi(D)=\pi(X\times A)=(\pi_{2})_{\#}\pi(A)=\nu(A).$$

As $A\in\mathcal{B}(X)$ was arbitrary and $\mu,\nu$ are both defined on $\mathcal{B}(X)$, $\mu=\nu$. Together with Step 2 this proves claim 2.

Step 4 (Claim 3, triangle inequality). Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$ be given; the following choices are made in this order. First, by the approximation (A) of Step 0 choose $\alpha\in\Pi(\mu,\nu)$ and $\beta\in\Pi(\nu,\lambda)$ with

$$\sqrt{I(\alpha)}<W_{2}(\mu,\nu)+\varepsilon,\qquad\sqrt{I(\beta)}<W_{2}(\nu,\lambda)+\varepsilon ;$$

both costs are finite by Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §cost-finite. Second, by Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §quantisation, applied to $\nu\in\mathcal{P}_{2}(X)$ and $\varepsilon$, choose a Borel map $T:X\to X$ whose image $F=T(X)$ is a finite set and with $e=\int_{X}|T(y)-y|^{2}\,\nu(dy)\le\varepsilon^{2}$; then $e$ is a nonnegative real number and $\sqrt{e}\le\varepsilon$ by claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field.

Put $\rho=T_{\#}\nu$, a Borel measure on $X$ with $\rho(X)=\nu(T^{-1}(X))=1$, so $\rho\in\mathcal{P}(X)$ by Borel Probability Measures on a Real Hilbert Space with an Orthonormal Basis: Standing Notation §pushforward; by the first sentence of Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §quantisation, $\rho(X\setminus F)=0$. By the first assertion of Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §modification, applied to the coupling $\alpha\in\Pi(\mu,\nu)$ and the map $T$, the measure $\alpha'=(\pi_{1},T\circ\pi_{2})_{\#}\alpha$ belongs to $\Pi(\mu,\rho)$, and since $I(\alpha)<\infty$ and $e<\infty$, $I(\alpha')<\infty$ and $\sqrt{I(\alpha')}\le\sqrt{I(\alpha)}+\sqrt{e}$. By the last assertion of the same clause, applied to the coupling $\beta\in\Pi(\nu,\lambda)$ and the map $S=T$, the measure $\beta'=(T\circ\pi_{1},\pi_{2})_{\#}\beta$ belongs to $\Pi(\rho,\lambda)$, and since $I(\beta)<\infty$ and $e<\infty$, $I(\beta')<\infty$ and $\sqrt{I(\beta')}\le\sqrt{I(\beta)}+\sqrt{e}$. By Couplings on a Hilbert Space: Product Coupling, Swap, Finiteness of the Cost, Push-Forward Couplings, Modifying One Marginal, Quantisation, Gluing over a Finitely Supported Measure, the Lipschitz Bound and the Moment Bound §gluing, applied to the measures $\mu$, $\lambda$, $\rho$, the finite set $F$ and the couplings $\pi_{12}=\alpha'$ and $\pi_{23}=\beta'$, there is $\pi_{13}\in\Pi(\mu,\lambda)$ with $\sqrt{I(\pi_{13})}\le\sqrt{I(\alpha')}+\sqrt{I(\beta')}$. Since $\mu,\lambda\in\mathcal{P}_{2}(X)$, Step 0 applies to $\pi_{13}$, and combining the inequalities above gives

$$W_{2}(\mu,\lambda)\le\sqrt{I(\pi_{13})}\le\sqrt{I(\alpha)}+\sqrt{I(\beta)}+2\sqrt{e}<W_{2}(\mu,\nu)+W_{2}(\nu,\lambda)+4\varepsilon .$$

Write $s=W_{2}(\mu,\nu)+W_{2}(\nu,\lambda)$. The display holds for every real $\varepsilon>0$; if $s<W_{2}(\mu,\lambda)$ held, the choice $\varepsilon=(W_{2}(\mu,\lambda)-s)/8$ would give $W_{2}(\mu,\lambda)<s+(W_{2}(\mu,\lambda)-s)/2<W_{2}(\mu,\lambda)$, which is impossible. Hence $W_{2}(\mu,\lambda)\le s$, which is claim 3.

Step 5 (Claim 4, the metric). By Metric Space, $W_{2}$ is a metric on $\mathcal{P}_{2}(X)$ if it is a real-valued function on pairs of elements of $\mathcal{P}_{2}(X)$ satisfying, for all $\mu,\nu,\lambda\in\mathcal{P}_{2}(X)$, conditions 1 to 4 there. $W_{2}$ is real-valued and nonnegative by The Quadratic Wasserstein Distance on a Hilbert Space §distance, which is condition 1; condition 2 is claim 2 (Steps 2 and 3); condition 3 is claim 1 (Step 1); and condition 4, with $\mu,\nu,\lambda$ in the roles of $x,y,z$, is claim 3 (Step 4). Hence $(\mathcal{P}_{2}(X),W_{2})$ is a metric space.