Proof of Basic Properties of the Lift: Law Invariance, the Correspondence on a Rich Space, and Transfer of Boundedness, Lipschitz Constants and Continuity

Each result cited is universally quantified over the data in its own statement. Throughout, $U(X)=u(\mathcal{L}(X))$ for every $X\in L^{2}(\Omega;\mathbb{R}^{d})$ by The Lift of a Function on the Wasserstein Space to the Space of Square-Integrable Random Vectors (i), and $\mathcal{L}(X)\in\mathcal{P}_{2}(\mathbb{R}^{d})$ by The Wasserstein Space and Its Lift to Square-Integrable Random Vectors: Standing Notation (v). The metric on $\mathbb{R}$ is $d_{\mathbb{R}}(s,t)=|s-t|$ (Real Hilbert Spaces: Standing Notation and Background (i)), so that for real-valued functions the distance between two values is the absolute value of their difference; the metric on $L^{2}(\Omega;\mathbb{R}^{d})$ is $d_{L^{2}}(X,Y)=\lVert X-Y\rVert_{L^{2}}$ (The Wasserstein Space and Its Lift to Square-Integrable Random Vectors: Standing Notation (iii)) and that on $\mathcal{P}_{2}(\mathbb{R}^{d})$ is $W_{2}$ (The Wasserstein Space and Its Lift to Square-Integrable Random Vectors: Standing Notation (iv)). Two facts about the law map are used repeatedly, both from The Wasserstein Space and Its Lift to Square-Integrable Random Vectors: Standing Notation (v): the domination inequality of The Wasserstein Distance and the Mean-Square Distance of Random Vectors (ii), namely, for all $X,Y\in L^{2}(\Omega;\mathbb{R}^{d})$,

$$W_{2}(\mathcal{L}(X),\mathcal{L}(Y))\le\lVert X-Y\rVert_{L^{2}} ,$$

and, when $(\Omega,\mathcal{F},P)$ is rich, every $\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$ is the law of some $X\in L^{2}(\Omega;\mathbb{R}^{d})$ (The Wasserstein Distance and the Mean-Square Distance of Random Vectors (iii)), and for all $\mu,\nu\in\mathcal{P}_{2}(\mathbb{R}^{d})$ the set $D(\mu,\nu)=\{\lVert X-Y\rVert_{L^{2}}:\mathcal{L}(X)=\mu,\ \mathcal{L}(Y)=\nu\}$ (the letter $D$ denoting this set and not a gradient throughout this proof) is nonempty and bounded below with $\inf D(\mu,\nu)=W_{2}(\mu,\nu)$ (The Wasserstein Distance and the Mean-Square Distance of Random Vectors (iv)), the infimum being the greatest lower bound.

Claim 1. Let $X,Y\in L^{2}(\Omega;\mathbb{R}^{d})$ with $\mathcal{L}(X)=\mathcal{L}(Y)$. Then $U(X)=u(\mathcal{L}(X))=u(\mathcal{L}(Y))=U(Y)$, so $U$ is law-invariant in the sense of Law-Invariant Function on the Space of Square-Integrable Random Vectors (i).

Claim 2. Let $(\Omega,\mathcal{F},P)$ be rich and let $\Phi$ be law-invariant. Existence. For $\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$ the set $\{\Phi(X):X\in L^{2}(\Omega;\mathbb{R}^{d}),\ \mathcal{L}(X)=\mu\}$ is nonempty by The Wasserstein Distance and the Mean-Square Distance of Random Vectors (iii) and has exactly one element, since $\Phi(X)=\Phi(X')$ whenever $\mathcal{L}(X)=\mu=\mathcal{L}(X')$ by Law-Invariant Function on the Space of Square-Integrable Random Vectors (i); let $v(\mu)$ be that element. This defines $v:\mathcal{P}_{2}(\mathbb{R}^{d})\to\mathbb{R}$, and for every $X\in L^{2}(\Omega;\mathbb{R}^{d})$ the element $X$ itself has law $\mathcal{L}(X)$, so $v(\mathcal{L}(X))=\Phi(X)$; that is, the lift of $v$ is $\Phi$. Uniqueness. Let $v,v':\mathcal{P}_{2}(\mathbb{R}^{d})\to\mathbb{R}$ both have lift $\Phi$ and let $\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$. Choose $X$ with $\mathcal{L}(X)=\mu$ by The Wasserstein Distance and the Mean-Square Distance of Random Vectors (iii); then $v(\mu)=v(\mathcal{L}(X))=\Phi(X)=v'(\mathcal{L}(X))=v'(\mu)$. Hence $v=v'$.

Claim 3. Let $u$ be bounded with bound $M$, so that $|u(\mu)|\le M$ for every $\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$ by Bounded Real-Valued Function on a Set. For $X\in L^{2}(\Omega;\mathbb{R}^{d})$, $|U(X)|=|u(\mathcal{L}(X))|\le M$; so $U$ is bounded with bound $M$.

Claim 4. Let $(\Omega,\mathcal{F},P)$ be rich and $U$ bounded with bound $M$. For $\mu\in\mathcal{P}_{2}(\mathbb{R}^{d})$ choose $X$ with $\mathcal{L}(X)=\mu$ by The Wasserstein Distance and the Mean-Square Distance of Random Vectors (iii); then $|u(\mu)|=|U(X)|\le M$. So $u$ is bounded with bound $M$.

Claim 5. Let $u$ be Lipschitz with constant $\kappa$, so that $|u(\mu)-u(\nu)|\le\kappa\,W_{2}(\mu,\nu)$ for all $\mu,\nu\in\mathcal{P}_{2}(\mathbb{R}^{d})$ by Lipschitz Map Between Metric Spaces. For $X,Y\in L^{2}(\Omega;\mathbb{R}^{d})$,

$$|U(X)-U(Y)|=|u(\mathcal{L}(X))-u(\mathcal{L}(Y))|\le\kappa\,W_{2}(\mathcal{L}(X),\mathcal{L}(Y))\le\kappa\,\lVert X-Y\rVert_{L^{2}},$$

the last step by the domination inequality and claim 5 of Elementary Arithmetic in an Ordered Field (multiplication by the nonnegative $\kappa$), and the two inequalities combine by the transitivity of $\le$, the order of $\mathbb{R}$ being that of an ordered field (Real Hilbert Spaces: Standing Notation and Background (i)). So $U$ is Lipschitz with constant $\kappa$.

Claim 6. Let $(\Omega,\mathcal{F},P)$ be rich and $U$ Lipschitz with constant $\kappa$, and let $\mu,\nu\in\mathcal{P}_{2}(\mathbb{R}^{d})$; put $c=|u(\mu)-u(\nu)|$. For every $s\in D(\mu,\nu)$, say $s=\lVert X-Y\rVert_{L^{2}}$ with $\mathcal{L}(X)=\mu$ and $\mathcal{L}(Y)=\nu$, we have $c=|U(X)-U(Y)|\le\kappa\,s$ by Lipschitz Map Between Metric Spaces. If $\kappa=0$, choose $s\in D(\mu,\nu)$, which is nonempty; then $c\le\kappa\,s=0=\kappa\,W_{2}(\mu,\nu)$ by claim 1 of Zero Products and Elementary Identities in a Field. If $\kappa\ne0$, then $\kappa^{-1}$ exists, and $0\le\kappa^{-1}$ by claim 4 of Elementary Arithmetic in an Ordered Field; multiplying $c\le\kappa\,s$ by $\kappa^{-1}$ (claim 5 there) gives $\kappa^{-1}c\le\kappa^{-1}(\kappa\,s)=s$ for every $s\in D(\mu,\nu)$, so $\kappa^{-1}c$ is a lower bound of $D(\mu,\nu)$ and hence $\kappa^{-1}c\le\inf D(\mu,\nu)=W_{2}(\mu,\nu)$ by Lower Bound and Greatest Lower Bound; multiplying by the nonnegative $\kappa$ (claim 5 again) gives $c=\kappa(\kappa^{-1}c)\le\kappa\,W_{2}(\mu,\nu)$. In both cases $|u(\mu)-u(\nu)|\le\kappa\,W_{2}(\mu,\nu)$, so $u$ is Lipschitz with constant $\kappa$.

Claim 7. Let $u$ be uniformly continuous and let $\varepsilon>0$. By Uniformly Continuous Map Between Metric Spaces there is $\delta>0$ such that $|u(\mu)-u(\nu)|<\varepsilon$ whenever $W_{2}(\mu,\nu)<\delta$. Let $X,Y\in L^{2}(\Omega;\mathbb{R}^{d})$ with $\lVert X-Y\rVert_{L^{2}}<\delta$. Then $W_{2}(\mathcal{L}(X),\mathcal{L}(Y))<\delta$ by the domination inequality and claim 2 of Elementary Order Arithmetic in an Ordered Field, hence $|U(X)-U(Y)|=|u(\mathcal{L}(X))-u(\mathcal{L}(Y))|<\varepsilon$. So $U$ is uniformly continuous.

Claim 8. Let $(\Omega,\mathcal{F},P)$ be rich and $U$ uniformly continuous, and let $\varepsilon>0$; choose $\delta>0$ with $|U(X)-U(Y)|<\varepsilon$ whenever $\lVert X-Y\rVert_{L^{2}}<\delta$. Let $\mu,\nu\in\mathcal{P}_{2}(\mathbb{R}^{d})$ with $W_{2}(\mu,\nu)<\delta$. Since $D(\mu,\nu)$ is nonempty and bounded below with $\inf D(\mu,\nu)=W_{2}(\mu,\nu)<\delta$, claim 2 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$ gives $s\in D(\mu,\nu)$ with $s<\delta$, say $s=\lVert X-Y\rVert_{L^{2}}$ with $\mathcal{L}(X)=\mu$ and $\mathcal{L}(Y)=\nu$. Then $|u(\mu)-u(\nu)|=|U(X)-U(Y)|<\varepsilon$. So $u$ is uniformly continuous.

Claim 9. Let $u$ be continuous on $\mathcal{P}_{2}(\mathbb{R}^{d})$, let $X\in L^{2}(\Omega;\mathbb{R}^{d})$ and let $\varepsilon>0$. By Continuous Map Between Metric Spaces applied at $\mathcal{L}(X)$ there is $\delta>0$ such that every $\nu\in\mathcal{P}_{2}(\mathbb{R}^{d})$ with $W_{2}(\mathcal{L}(X),\nu)<\delta$ satisfies $|u(\nu)-u(\mathcal{L}(X))|<\varepsilon$. Let $Y\in L^{2}(\Omega;\mathbb{R}^{d})$ with $\lVert X-Y\rVert_{L^{2}}<\delta$. Then $W_{2}(\mathcal{L}(X),\mathcal{L}(Y))<\delta$ by the domination inequality and claim 2 of Elementary Order Arithmetic in an Ordered Field, so $|U(Y)-U(X)|=|u(\mathcal{L}(Y))-u(\mathcal{L}(X))|<\varepsilon$. Hence $U$ is continuous at $X$ relative to $L^{2}(\Omega;\mathbb{R}^{d})$, and, $X$ being arbitrary, continuous on $L^{2}(\Omega;\mathbb{R}^{d})$.