Proof of The Density Cost Along Optimal Maps is Controlled by the Monotonicity of the Score

Each result cited is universally quantified over the data in its own statement. Elementary order and field arithmetic in $\mathbb{R}$ is used without mention (Elementary Order Arithmetic in an Ordered Field, Elementary Arithmetic in an Ordered Field). Integrals against $\lambda_{d}$ are written $\int\cdots d\lambda_{d}$; a Borel set is $\mu$-full if its complement is $\mu$-null, and finitely many $\mu$-full sets have a $\mu$-full intersection (claim 4 of Basic Properties of a Measure); likewise for $\nu$. We use that $\exp(u+v)=\exp(u)\exp(v)$, $\exp(0)=1$, $0<\exp(u)$ and $\exp(-u)=\exp(u)^{-1}$ (claims 1 and 2 of Basic Properties of the Exponential Function), and that $\log$ is the inverse of $\exp$ (The Natural Logarithm).

Step 0: densities. By Finite Fisher Information, the Score and the Fisher Information of a Probability Measure §finite, $\mu,\nu\in\mathcal{P}_{2}(\mathbb{R}^{d})$. By The Density Cost of a Convex Lipschitz Integrand §cost, $\mu$ and $\nu$ have densities $\rho$ and $\rho_{1}$ with respect to $\lambda_{d}$ (Borel, real, nonnegative), the functions $\Phi\circ\rho$ and $\Phi\circ\rho_{1}$ are Borel and $\lambda_{d}$-integrable, and $\mathcal{G}_{\Phi}(\mu)=\int\Phi\circ\rho\,d\lambda_{d}$, $\mathcal{G}_{\Phi}(\nu)=\int\Phi\circ\rho_{1}\,d\lambda_{d}$. Since $\mu$ is the measure with density $\rho$ with respect to $\lambda_{d}$, claim 3 of Image Measures, Measures with Densities, and Change of Variables gives $\int u\,d\mu=\int u\rho\,d\lambda_{d}$ for Borel $u\ge0$, and, for Borel real $u$, that $u$ is $\mu$-integrable exactly when $u\rho$ is $\lambda_{d}$-integrable, the same identity then holding. By the Lipschitz bound of Convex Lipschitz Integrands §integrand with $a=0$ and $\Phi(0)=0$, $0\le\Phi(s)$ for every $s\ge0$.

Step 1: the potentials. The coupling $(\mathrm{id},T)_{\#}\mu$ is optimal (Optimal Transport Maps and Uniquely Mapped Pairs of Probability Measures §map). By Brenier's Theorem: Optimal Couplings out of an Absolutely Continuous Measure are Induced by a Unique Map §potential there are a Borel $T_{0}$ with $(\mathrm{id},T_{0})_{\#}\mu=(\mathrm{id},T)_{\#}\mu$, an open convex $G$ with $\mu(G)=1$, a convex $\varphi:G\to\mathbb{R}$ and a Borel $D\subseteq G$ with $\mu(D)=1$ and $\partial_{G}\varphi(x)=\{T_{0}(x)\}$ for $x\in D$. By Brenier's Theorem: Optimal Couplings out of an Absolutely Continuous Measure are Induced by a Unique Map §map, $(T_{0})_{\#}\mu=\nu$ and $\int\lVert T_{0}\rVert^{2}\,d\mu<\infty$, so $T_{0}$ is an optimal map from $\mu$ to $\nu$, and by Brenier's Theorem: Optimal Couplings out of an Absolutely Continuous Measure are Induced by a Unique Map §unique-map, $T=T_{0}$ $\mu$-almost everywhere. Likewise $(\mathrm{id},T')_{\#}\nu$ is an optimal coupling of $\nu$ and $\mu$, and the same three clauses, applied with $\nu$ (absolutely continuous) in place of $\mu$, give a Borel $S$ with $(\mathrm{id},S)_{\#}\nu=(\mathrm{id},T')_{\#}\nu$, an open convex $G'$ with $\nu(G')=1$, a convex $\psi:G'\to\mathbb{R}$ and a Borel $D'\subseteq G'$ with $\nu(D')=1$ and $\partial_{G'}\psi(y)=\{S(y)\}$ for $y\in D'$, such that $S_{\#}\nu=\mu$, $\int\lVert S\rVert^{2}\,d\nu<\infty$, $S$ is an optimal map from $\nu$ to $\mu$, and $S=T'$ $\nu$-almost everywhere.

Step 2: the good set. By McCann's Jacobian Equation Along an Optimal Map Between Absolutely Continuous Measures §jacobian, applied to $\mu,\nu,\rho,\rho_{1},T_{0},S,G,G',\varphi,\psi,D,D'$, there is a $\mu$-full Borel $X\subseteq D$ such that every $x\in X$ satisfies: $\varphi$ is twice differentiable at $x$ with first-order coefficient $T_{0}(x)$; $\psi$ is twice differentiable at $T_{0}(x)$; $D^{2}\varphi(x)$ and $D^{2}\psi(T_{0}(x))$ are positive definite; $\det D^{2}\varphi(x)\cdot\det D^{2}\psi(T_{0}(x))=1$ (these from Along an Optimal Map between Absolutely Continuous Measures the Hessians of the Two Convex Potentials are Inverse Matrices §hessians); and $\rho(x)=\rho_{1}(T_{0}(x))\det D^{2}\varphi(x)$ with $\rho(x)>0$. By Along an Optimal Map between Absolutely Continuous Measures the Hessians of the Two Convex Potentials are Inverse Matrices §hessians, applied to $\nu,\mu,S,T_{0},G',G,\psi,\varphi,D',D$ (its hypotheses are symmetric in the two measures), there is a $\nu$-full Borel $Y\subseteq D'$ such that at every $y\in Y$ the function $\psi$ is twice differentiable with first-order coefficient $S(y)$. Let $X'=X\cap T_{0}^{-1}(Y)$, a Borel set which is $\mu$-full because $\mu(T_{0}^{-1}(Y))=\nu(Y)=1$. Let $\Delta_{\varphi}$ be the function $\Delta$ of The Points of Twice Differentiability of a Convex Function: a Borel Set of Full Measure, and Borel Measurability of the Gradient and Hessian on It §borel for $\varphi$, $G$ and the set $X'$, and $\Delta_{\psi}$ the one for $\psi$, $G'$ and the set $Y$: $\Delta_{\varphi}(x)=\mathrm{tr}\,D^{2}\varphi(x)$ on $X'$, $\Delta_{\psi}(y)=\mathrm{tr}\,D^{2}\psi(y)$ on $Y$, both $0$ elsewhere, and both Borel. Let $H:\mathbb{R}^{d}\to\mathbb{R}$ be the Borel function

Step 3: the pointwise inequality. Fix $x\in X'$ and write $r=\rho(x)>0$, $M=D^{2}\varphi(x)$, $N=D^{2}\psi(T_{0}(x))$ and $\delta=\det M$, so that $0<\delta$ by Determinants of Positive Definite Matrices: Positivity, the Bound $\log\det A\le\mathrm{tr}\,A-d$, Bounds under Pinching, and the Expansion of $\det(I+tB)$ §positive, $\det N=\delta^{-1}$, $\rho_{1}(T_{0}(x))=r\delta^{-1}$ by Step 2, and $\mathrm{tr}\,M=\Delta_{\varphi}(x)$, $\mathrm{tr}\,N=\Delta_{\psi}(T_{0}(x))$ since $T_{0}(x)\in Y$, so that $H(x)=A(\mathrm{tr}\,M-d)+A(\mathrm{tr}\,N-d)+dL^{2}A^{-1}$. We show

Trace bounds. Let $g=d^{-1}\log\delta$, $t=\exp(g/2)$ and $s=t^{2}=\exp(g)$; by the exponential rules and induction on $d$, $s^{d}=\exp(dg)=\exp(\log\delta)=\delta$. For a positive real $c$ and a positive definite $B\in\mathcal{S}(d)$, the matrix $cB$ is symmetric and positive definite ($z\cdot(cBz)=c\,(z\cdot Bz)$ for $z\in\mathbb{R}^{d}$, Symmetric, Positive Semidefinite, and Positive Definite Real Matrices), $\mathrm{tr}(cB)=c\,\mathrm{tr}\,B$ (Trace of a Real Square Matrix), and $\det(cB)=\det(cI_{d})\det B=c^{d}\det B$ by The Determinant is Multiplicative and claim 1 of The Determinant of a Triangular Matrix is the Product of its Diagonal Entries, $cI_{d}$ being lower triangular with all diagonal entries $c$. With $c=s^{-1}$ and $B=M$ this gives $\det(s^{-1}M)=s^{-d}\delta=1$, so Determinants of Positive Definite Matrices: Positivity, the Bound $\log\det A\le\mathrm{tr}\,A-d$, Bounds under Pinching, and the Expansion of $\det(I+tB)$ §log-det yields $0=\log1\le s^{-1}\mathrm{tr}\,M-d$, that is $\mathrm{tr}\,M\ge d\,s$; with $c=s$ and $B=N$ it gives $\det(sN)=\delta\delta^{-1}=1$ and likewise $\mathrm{tr}\,N\ge d\,s^{-1}$. Hence

Case $\delta\le1$. Convexity in Convex Lipschitz Integrands §integrand, with $a=r\delta^{-1}$, $b=0$ and weight $\delta$, together with $\Phi(0)=0$, gives $\Phi(r)=\Phi(\delta\cdot r\delta^{-1}+(1-\delta)\cdot0)\le\delta\,\Phi(r\delta^{-1})$, so the left side of (1) is at most $0$, while $rH(x)\ge0$ by (2), since $r>0$, $A>0$ and $dL^{2}A^{-1}\ge0$.

Case $1<\delta$. By The Function $s\log s$: Continuity, Young's Inequality and Lower Bounds, with the Elementary Bounds for the Exponential and the Logarithm §log, $1-\delta^{-1}\le\log\delta$, so $0<\log\delta$ and $0<g$. By Step 0, $0\le(\delta-1)\Phi(r\delta^{-1})$, i.e. $\Phi(r\delta^{-1})\le\delta\Phi(r\delta^{-1})$; and $r\delta^{-1}\le r$, so the Lipschitz bound of Convex Lipschitz Integrands §integrand gives

By The Function $s\log s$: Continuity, Young's Inequality and Lower Bounds, with the Elementary Bounds for the Exponential and the Logarithm §exp, $t\ge p$ with $p=1+g/2\ge1$, so $t^{-1}\le p^{-1}$ and $t-t^{-1}\ge p-p^{-1}=(p-1)(p+1)p^{-1}\ge p-1=g/2\ge0$; hence $(t-t^{-1})^{2}\ge g^{2}/4$. From $0\le A\,(g/2-L/A)^{2}=A\,g^{2}/4-L\,g+L^{2}/A$ we get $L\,g\le A\,g^{2}/4+L^{2}/A\le A\,(t-t^{-1})^{2}+L^{2}/A$, and multiplying by $r\,d>0$ and using (2),

This proves (1) in both cases. By Step 2, $\varphi$ has first-order coefficient $T_{0}(x)$ at $x$ and $\delta\,\Phi(r\delta^{-1})=\Phi(\rho_{1}(T_{0}(x)))\det D^{2}\varphi(x)$.

Step 4: the scores. Apply The Score Paired with the Identity, and the Laplacian of a Convex Potential Bounded by its Pairing with the Score §comparison to $\mu$ (absolutely continuous, with finite Fisher information) with $G$, $\varphi$ and, as the set called $A$ there (not our constant $A$), the set $X'$, a $\mu$-full Borel subset of $G$ at whose points $\varphi$ is twice differentiable; its function $\Delta$ is $\Delta_{\varphi}$, and its map $g_{\mu}$ equals $D\varphi=T_{0}$ on $X'$, hence $g_{\mu}=T_{0}=T$ $\mu$-almost everywhere by Step 1. So $\int\lVert g_{\mu}\rVert^{2}\,d\mu=\int\lVert T_{0}\rVert^{2}\,d\mu<\infty$ by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison, and $g_{\mu}$ and $T$ have the same class in $L^{2}(\mu;\mathbb{R}^{d})$. That clause gives that $\Delta_{\varphi}$ is $\mu$-integrable with $\int\Delta_{\varphi}\,d\mu\le-\langle\xi_{\mu},T\rangle_{\mu}$, and The Score Paired with the Identity, and the Laplacian of a Convex Potential Bounded by its Pairing with the Score §identity gives $\langle\xi_{\mu},\mathrm{id}\rangle_{\mu}=-d$; by bilinearity of the inner product,

In the same way, The Score Paired with the Identity, and the Laplacian of a Convex Potential Bounded by its Pairing with the Score §comparison applied to $\nu$ with $G'$, $\psi$ and the set $Y$, whose map equals $S$ on $Y$ and hence $T'$ $\nu$-almost everywhere, with $\int\lVert S\rVert^{2}\,d\nu<\infty$ by Step 1, together with The Score Paired with the Identity, and the Laplacian of a Convex Potential Bounded by its Pairing with the Score §identity for $\nu$, shows that $\Delta_{\psi}$ is $\nu$-integrable and

Since $(T_{0})_{\#}\mu=\nu$, Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward shows that $\Delta_{\psi}\circ T_{0}$ is $\mu$-integrable with $\int\Delta_{\psi}\circ T_{0}\,d\mu=\int\Delta_{\psi}\,d\nu$. Constants being $\mu$-integrable with integral equal to themselves ($\mu(\mathbb{R}^{d})=1$), $H$ is $\mu$-integrable and, by claim 2 of Linearity and Monotonicity of the Lebesgue Integral, (3) and (4),

Step 5: integration. Let $k$ be the function of The Area Inequality for the Gradient of a Convex Function §area for $f=\varphi$, $U=G$, the set $X'$ in place of its set $A$, and $h=\Phi\circ\rho_{1}$ (Borel and nonnegative by Step 0): by Step 3, $k(x)=\Phi(\rho_{1}(T_{0}(x)))\det D^{2}\varphi(x)=\delta\,\Phi(r\delta^{-1})$ in the notation there for $x\in X'$, and $k(x)=0$ otherwise. That theorem gives that $k$ is Borel and nonnegative with $\int k\,d\lambda_{d}\le\int\Phi\circ\rho_{1}\,d\lambda_{d}=\mathcal{G}_{\Phi}(\nu)$; so $k$ is $\lambda_{d}$-integrable. Next, $\int\mathbf{1}_{\mathbb{R}^{d}\setminus X'}\rho\,d\lambda_{d}=\mu(\mathbb{R}^{d}\setminus X')=0$, so by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §vanishing there is a $\lambda_{d}$-null set $Z$ with $\rho(x)=0$ for every $x\notin X'\cup Z$. Outside $Z$, $\Phi\circ\rho=\mathbf{1}_{X'}\,\Phi\circ\rho$ (as $\Phi(0)=0$) and $H\rho=\mathbf{1}_{X'}H\rho$. By Step 0, $H\rho$ is $\lambda_{d}$-integrable with $\int H\rho\,d\lambda_{d}=\int H\,d\mu$, so by The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison the functions $\mathbf{1}_{X'}\,\Phi\circ\rho$ and $\mathbf{1}_{X'}H\rho$ are $\lambda_{d}$-integrable with integrals $\mathcal{G}_{\Phi}(\mu)$ and $\int H\,d\mu$. By (1), $\mathbf{1}_{X'}\,\Phi\circ\rho-k\le\mathbf{1}_{X'}H\rho$ on $X'$, and also off $X'$, where both sides vanish. Integrating (claim 2 of Linearity and Monotonicity of the Lebesgue Integral) and using (5),

which is the displacement bound.