Proof of McCann's Tangent Inequality: the Entropy Lies Above its Tangent Along Optimal Maps

Each result cited is universally quantified over the data in its own statement. 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 write $\phi(s)=s\log s$ as in The Function $s\log s$: Continuity, Young's Inequality and Lower Bounds, with the Elementary Bounds for the Exponential and the Logarithm, and use that $\log(ab)=\log a+\log b$ and $\log(a^{-1})=-\log a$ for positive $a,b$, since $\log$ is the inverse of $\exp$ (The Natural Logarithm) and $\exp$ turns sums into products (Basic Properties of the Exponential Function).

Step 0: densities. By The Entropy of a Probability Measure on Euclidean Space §entropy, $\mu$ and $\nu$ have densities $\rho$ and $\rho_{1}$ with respect to $\lambda_{d}$ (Borel, real, nonnegative), with $\phi\circ\rho$ and $\phi\circ\rho_{1}$ integrable, $\mathrm{Ent}(\mu)=\int\phi\circ\rho\,d\lambda_{d}$ and $\mathrm{Ent}(\nu)=\int\phi\circ\rho_{1}\,d\lambda_{d}$. By claim 3 of Image Measures, Measures with Densities, and Change of Variables, $\int u\,d\mu=\int u\rho\,d\lambda_{d}$ for Borel $u\ge0$, and for Borel real $u$ with $u\rho$ integrable; likewise for $\nu$ and $\rho_{1}$. Both measures are absolutely continuous by Basic Properties of the Entropy on the Wasserstein Space: Comparison with the Gaussian Relative Entropy, Lower Bound, Translation Invariance, Absolute Continuity, Closed Sublevel Sets and Lower Semicontinuity §absolutely-continuous. The sets $P=\{\rho>0\}$ and $P_{1}=\{\rho_{1}>0\}$ are Borel, and $\mu(\mathbb{R}^{d}\setminus P)=\int\mathbf{1}_{\{\rho=0\}}\rho\,d\lambda_{d}=0$, so $P$ is $\mu$-full; likewise $P_{1}$ is $\nu$-full. Let $\ell,\ell_{1}:\mathbb{R}^{d}\to\mathbb{R}$ equal $\log\rho$ on $P$, respectively $\log\rho_{1}$ on $P_{1}$, and $0$ elsewhere; they are Borel. Since $|\ell|\rho=|\phi\circ\rho|$ everywhere (both vanish off $P$), $\ell$ is $\mu$-integrable with $\int\ell\,d\mu=\int\phi\circ\rho\,d\lambda_{d}=\mathrm{Ent}(\mu)$; likewise $\ell_{1}$ is $\nu$-integrable with $\int\ell_{1}\,d\nu=\mathrm{Ent}(\nu)$.

Step 1: the potentials. The coupling $(\mathrm{id},T)_{\#}\mu$ is optimal. 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, $(T_{0})_{\#}\mu=\nu$, 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, so $T$ and $T_{0}$ have the same class in $L^{2}(\mu;\mathbb{R}^{d})$. By Existence of an Optimal Coupling of Two Probability Measures with Finite Second Moment there is an optimal coupling of $\nu$ and $\mu$, and Brenier's Theorem: Optimal Couplings out of an Absolutely Continuous Measure are Induced by a Unique Map §potential, applied to it with $\nu$ absolutely continuous, gives a Borel $S$, 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 $y\in D'$, where $(\mathrm{id},S)_{\#}\nu$ is that optimal coupling and $S_{\#}\nu=\mu$ by Brenier's Theorem: Optimal Couplings out of an Absolutely Continuous Measure are Induced by a Unique Map §map, so that $S$ is an optimal map from $\nu$ to $\mu$.

Step 2: inverse Hessians. By Along an Optimal Map between Absolutely Continuous Measures the Hessians of the Two Convex Potentials are Inverse Matrices §hessians, applied to $\mu,\nu,T_{0},S,G,G',\varphi,\psi,D,D'$, there is a $\mu$-full Borel $X\subseteq D$ such that every $x\in X$ satisfies: $T_{0}(x)\in D'$; $\varphi$ is twice differentiable at $x$ with first-order coefficient $T_{0}(x)$; $\psi$ is twice differentiable at $T_{0}(x)$ with first-order coefficient $x$; $D^{2}\varphi(x)$ is positive definite; and $\det D^{2}\psi(T_{0}(x))=(\det D^{2}\varphi(x))^{-1}$, the determinant $\det D^{2}\varphi(x)$ being positive 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.

Step 3: the area inequality. By 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 §full for $\psi$ on $G'$ there is a Borel $A_{\psi}\subseteq G'$ with $\lambda_{d}(G'\setminus A_{\psi})=0$ at whose points $\psi$ is twice differentiable; put $A'=D'\cap A_{\psi}$, which is $\nu$-full ($\nu(G'\setminus A_{\psi})=0$ by absolute continuity). Let $k$ be the function of The Area Inequality for the Gradient of a Convex Function §area for $f=\psi$, $U=G'$, $A=A'$ and $h=\rho$: $k(y)=\rho(D\psi(y))\det D^{2}\psi(y)$ for $y\in A'$ and $k(y)=0$ otherwise. That theorem gives $\int k\,d\lambda_{d}\le\int\rho\,d\lambda_{d}=\mu(\mathbb{R}^{d})=1$. Let $u:\mathbb{R}^{d}\to\mathbb{R}$ be $u=k\rho_{1}^{-1}$ on $P_{1}$ and $u=0$ elsewhere, a nonnegative Borel function. Then $\int u\,d\nu=\int\mathbf{1}_{P_{1}}k\,d\lambda_{d}\le1$, and since $(T_{0})_{\#}\mu=\nu$, Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward gives

Step 4: the pointwise inequality. Let $X'=X\cap P\cap T_{0}^{-1}(P_{1}\cap A')$, a Borel set which is $\mu$-full because $\mu(T_{0}^{-1}(P_{1}\cap A'))=\nu(P_{1}\cap A')=1$. Let $\Delta$ be the Borel function 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 $X'$: $\Delta(x)=\mathrm{tr}\,D^{2}\varphi(x)$ on $X'$ and $0$ elsewhere, with $0\le\Delta$ by 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 §nonnegative. Let $x\in X'$ and $y=T_{0}(x)$. Then $y\in A'\cap P_{1}$ and $D\psi(y)=x$ by Step 2, so $k(y)=\rho(x)\,(\det D^{2}\varphi(x))^{-1}$ and

By The Function $s\log s$: Continuity, Young's Inequality and Lower Bounds, with the Elementary Bounds for the Exponential and the Logarithm §log, $\log u(y)\le u(y)-1$, that is, $\ell(x)-\ell_{1}(y)-\log\det D^{2}\varphi(x)\le u(y)-1$. By Determinants of Positive Definite Matrices: Positivity, the Bound $\log\det A\le\mathrm{tr}\,A-d$, Bounds under Pinching, and the Expansion of $\det(I+tB)$ §log-det, $\log\det D^{2}\varphi(x)\le\Delta(x)-d$. Hence, for every $x\in X'$,

Step 5: the score. Apply The Score Paired with the Identity, and the Laplacian of a Convex Potential Bounded by its Pairing with the Score §comparison with $G$, $\varphi$ and $A=X'$ (a $\mu$-full Borel subset of $G$ at whose points $\varphi$ is twice differentiable); $\mu$ is absolutely continuous with finite Fisher information. Its map $g$ equals $D\varphi=T_{0}$ on $X'$, so $g=T_{0}=T$ $\mu$-almost everywhere; hence $\int\lVert g\rVert^{2}\,d\mu=\int\lVert T\rVert^{2}\,d\mu<\infty$ and $g$ and $T$ have the same class in $L^{2}(\mu;\mathbb{R}^{d})$. Its function $\Delta$ is the one of Step 4. The lemma gives that $\Delta$ is $\mu$-integrable and

and The Score Paired with the Identity, and the Laplacian of a Convex Potential Bounded by its Pairing with the Score §identity gives $\langle\xi_{\mu},\mathrm{id}\rangle_{\mu}=-d$.

Step 6: integration. By Step 0 and Probability Measures on Euclidean Space and Random Vectors: Standing Notation §pushforward, $\ell_{1}\circ T_{0}$ is $\mu$-integrable with $\int\ell_{1}\circ T_{0}\,d\mu=\int\ell_{1}\,d\nu=\mathrm{Ent}(\nu)$; $\ell$ is $\mu$-integrable with $\int\ell\,d\mu=\mathrm{Ent}(\mu)$ by Step 0; $u\circ T_{0}$ is nonnegative and $\mu$-integrable by (1); and $\Delta$ is $\mu$-integrable by Step 5. So every term of (2) is $\mu$-integrable, (2) holds on the $\mu$-full set $X'$, and integrating it (linearity and monotonicity of the integral of integrable functions, claim 2 of Linearity and Monotonicity of the Lebesgue Integral, with The Lebesgue Integral and Null Sets: Almost-Everywhere Comparison, Markov's Inequality, and Dominated Convergence Almost Everywhere §comparison for the $\mu$-null set $\mathbb{R}^{d}\setminus X'$) and using (1) gives

By bilinearity of the inner product, the identity of Step 5, (3) and (4),

which is the tangent inequality.