Proof of The First Variation of the Entropy Along a Gradient Perturbation of the Identity

Each result cited is universally quantified over the data in its own statement. Write $B(x)=D^{2}\psi(x)$ for $x\in\mathbb{R}^{d}$ and, for $t\in\mathbb{R}$, $\mu_{t}=(\mathrm{id}+t\nabla\psi)_{\#}\mu$.

Step 1 (the potential). Fix $t\in\mathbb{R}$ and let $\Phi_{t}:\mathbb{R}^{d}\to\mathbb{R}$ be $\Phi_{t}(x)=\tfrac12\lVert x\rVert^{2}+t\psi(x)=\tfrac12\sum_{i=1}^{d}x_{i}^{2}+t\psi(x)$. Coordinate functions, products, sums and scalar multiples of $C^{2}$ functions are $C^{2}$ by Constants, Coordinate Functions, Sums and Products of $C^k$ Functions on a Euclidean Open Set, and $\psi$ is smooth, hence $C^{2}$ (Test Functions on Euclidean Space, Their Gradient Maps and Laplacians §gradient); so $\Phi_{t}$ is of class $C^{2}$ on $\mathbb{R}^{d}$, with

where $\delta_{ij}$ is the entry of $I_{d}$ in row $i$ and column $j$. Thus the gradient map of $\Phi_{t}$ is $\mathrm{id}+t\nabla\psi$ and $D^{2}\Phi_{t}(x)=I_{d}+tB(x)$.

Step 2 (pinching). Let $x,v\in\mathbb{R}^{d}$. Since $|B_{ij}(x)|\le b$,

the first inequality by the triangle inequality and claim 4 of Properties of the Absolute Value in an Ordered Field, the second by Cauchy-Schwarz Inequality for the Euclidean Dot Product applied to $(|v_{1}|,\dots,|v_{d}|)$ and $(1,\dots,1)$. Hence, if $2d|t|b\le1$, then $v^{\mathsf T}(I_{d}+tB(x))v=\lVert v\rVert^{2}+t\,v^{\mathsf T}B(x)v$ lies between $\tfrac12\lVert v\rVert^{2}$ and $\tfrac32\lVert v\rVert^{2}$; that is, $\tfrac12I_{d}\preceq D^{2}\Phi_{t}(x)\preceq\tfrac32I_{d}$ in the positive semidefinite ordering.

Claim 1. Let $2d|t|b\le1$. By Step 2 and Determinants of Positive Definite Matrices: Positivity, the Bound $\log\det A\le\mathrm{tr}\,A-d$, Bounds under Pinching, and the Expansion of $\det(I+tB)$ §pinching (with $\varepsilon=\tfrac12$, $L=\tfrac32$) the matrix $I_{d}+tB(x)$ is positive definite, so $0<\det(I_{d}+tB(x))$ 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. By Steps 1 and 2, The Entropy of the Push-Forward of a Measure by the Gradient of a Twice Continuously Differentiable Function with Pinched Hessian applies to $\Phi=\Phi_{t}$ with $\varepsilon=\tfrac12$ and $L=\tfrac32$: its preamble gives that $\nabla\Phi_{t}=\mathrm{id}+t\nabla\psi$ is Borel, and The Entropy of the Push-Forward of a Measure by the Gradient of a Twice Continuously Differentiable Function with Pinched Hessian §entropy gives that $x\mapsto\log\det(I_{d}+tB(x))$ is Borel and bounded, that $\mu_{t}\in\mathcal{P}_{2}^{\mathrm{Ent}}(\mathbb{R}^{d})$, and the stated formula.

Claim 2. Let $2d|t|b\le1$ and $|t|b\le c_{d}$. The trace of $B(x)$ is $\sum_{i}\partial_{i}\partial_{i}\psi(x)=\Delta\psi(x)$ by The Laplacian of a Twice Continuously Differentiable Function §laplacian. Since $c_{d}\le1$, Determinants of Positive Definite Matrices: Positivity, the Bound $\log\det A\le\mathrm{tr}\,A-d$, Bounds under Pinching, and the Expansion of $\det(I+tB)$ §expansion applies with the matrix $B(x)$, its entry bound taken to be $b$, and this $t$, and gives for every $x$

Both $\log\det(I_{d}+tB)$ (Claim 1) and $\Delta\psi$ (The Gradient of a Test Function is Bounded and Square-Integrable, and Its Laplacian Bounded and Integrable, Against Every Probability Measure §laplacian) are bounded Borel functions, hence $\mu$-integrable, so by Claim 1 and claim 2 of Linearity and Monotonicity of the Lebesgue Integral,

whose absolute value is at most $\int K_{d}t^{2}b^{2}\,d\mu=K_{d}t^{2}b^{2}$ by monotonicity of the integral (the same theorem) and $\mu(\mathbb{R}^{d})=1$.

Claim 3. Let $t_{0}=1$ if $b=0$, and otherwise let $t_{0}$ be the smaller of $(2db)^{-1}$ and $c_{d}b^{-1}$; in either case $t_{0}>0$, and every $t\in(-t_{0},t_{0})$ satisfies $2d|t|b\le1$ and $|t|b\le c_{d}$. By Claim 1, $\mu_{t}\in\mathcal{P}_{2}^{\mathrm{Ent}}(\mathbb{R}^{d})$ for all such $t$; let $F(t)=\mathrm{Ent}(\mu_{t})$. Since $\mu_{0}=\mathrm{id}_{\#}\mu=\mu$, Claim 2 gives, for $0<|t|<t_{0}$,

Given $\varepsilon>0$, the right side is below $\varepsilon$ whenever $0<|t|<\delta$, where $\delta$ is the smaller of $t_{0}$ and $\varepsilon(K_{d}b^{2}+1)^{-1}$. The point $0$ is an interior point of $(-t_{0},t_{0})$ by Basic Facts about Intervals of the Real Line and Their Interior Points §open-interval, so by Derivative at an Interior Point the function $F$ is differentiable at $0$ with derivative $-\int_{\mathbb{R}^{d}}\Delta\psi\,d\mu$. $\blacksquare$