Proof of Out of an Absolutely Continuous Measure the Optimal Map and the Optimal Displacement are Tangent

Throughout, each result cited is universally quantified over the data appearing in its own statement and is applied to the data named here. Write $\Pi(\mu,\nu)$ for the set of couplings of $\mu$ and $\nu$.

Step 1 (Claim 1). By Existence of an Optimal Coupling of Two Probability Measures with Finite Second Moment there is an optimal $\pi\in\Pi(\mu,\nu)$. Since $\mu$ is absolutely continuous, Brenier's Theorem: Optimal Couplings out of an Absolutely Continuous Measure are Induced by a Unique Map §map provides a Borel map $S:\mathbb{R}^{d}\to\mathbb{R}^{d}$ with $\pi=(\mathrm{id},S)_{\#}\mu$ and $S_{\#}\mu=\nu$. The coupling $(\mathrm{id},S)_{\#}\mu$ is $\pi$, hence optimal, so $S$ is an optimal map from $\mu$ to $\nu$. By Brenier's Theorem: Optimal Couplings out of an Absolutely Continuous Measure are Induced by a Unique Map §unique-coupling every optimal coupling of $\mu$ and $\nu$ equals $\pi=(\mathrm{id},S)_{\#}\mu$. Therefore $(\mu,\nu)$ is uniquely mapped, and in particular an optimal map exists.

Step 2 (The second moment). Let $T$ be an optimal map from $\mu$ to $\nu$, so $T_{\#}\mu=\nu$ and $(\mathrm{id},T)_{\#}\mu$ is an optimal coupling. Applying Brenier's Theorem: Optimal Couplings out of an Absolutely Continuous Measure are Induced by a Unique Map §map to that optimal coupling and to the Borel map $T$ itself gives

so the class of $T$ belongs to $L^{2}(\mu;\mathbb{R}^{d})$ by Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation §l2mu.

Step 3 (Tangency of the optimal map). Apply Brenier's Theorem: Optimal Couplings out of an Absolutely Continuous Measure are Induced by a Unique Map §potential to the optimal coupling $(\mathrm{id},T)_{\#}\mu$: there are a Borel map $T'$ with $(\mathrm{id},T')_{\#}\mu=(\mathrm{id},T)_{\#}\mu$, an open convex set $G\subseteq\mathbb{R}^{d}$ with $\mu(G)=1$, a function $\phi:G\to\mathbb{R}$ convex on $G$, and a set $D\in\mathcal{B}(\mathbb{R}^{d})$ with $D\subseteq G$ and $\mu(D)=1$, such that $\partial_{G}\phi(x)=\{T'(x)\}$ for every $x\in D$. Applying Brenier's Theorem: Optimal Couplings out of an Absolutely Continuous Measure are Induced by a Unique Map §map to the optimal coupling $(\mathrm{id},T)_{\#}\mu$, which equals $(\mathrm{id},T')_{\#}\mu$, and to the Borel map $T'$, gives $T'_{\#}\mu=\nu$ and $\int\lVert T'\rVert^{2}\,d\mu=M_{2}(\nu)<\infty$.

The hypotheses of A Square-Integrable Selection of the Subdifferential of a Convex Potential Belongs to the Tangent Space §tangent are therefore met by $G$, $\phi$, $D$ and $T'$, and it gives $T'\in T_{\mu}$.

By Brenier's Theorem: Optimal Couplings out of an Absolutely Continuous Measure are Induced by a Unique Map §unique-map, applied to $T$ and $T'$, both of which induce optimal couplings of $\mu$ and $\nu$, the set $\{x:T(x)\ne T'(x)\}$ is $\mu$-null. Hence $T$ and $T'$ determine the same class in $L^{2}(\mu;\mathbb{R}^{d})$, by Square-Integrable Vector Fields Against a Probability Measure on Euclidean Space, and Test Functions: Standing Notation §l2mu, and $T\in T_{\mu}$.

Step 4 (Tangency of the optimal displacement). By Basic Properties of the Tangent Space: Closed Subspace, the Identity Map Belongs to It, Second-Moment Limits, and Representation of Bounded Functionals on Gradients §identity the class of $\mathrm{id}$ belongs to $T_{\mu}$, and by Basic Properties of the Tangent Space: Closed Subspace, the Identity Map Belongs to It, Second-Moment Limits, and Representation of Bounded Functionals on Gradients §closed the set $T_{\mu}$ is a closed linear subspace of $L^{2}(\mu;\mathbb{R}^{d})$, hence closed under differences by Linear Subspace. Therefore $\mathrm{id}-T\in T_{\mu}$, which together with Step 3 proves claim 2.