Gaussian log-Sobolev bound for exponentials of Lipschitz functions: $\mathrm{Ent}_\gamma(e^{sf})\le\frac{s^2L^2}{2}\,\mathbb{E}_\gamma e^{sf}$

Let $\iota$ be a finite index set and $\gamma_\iota = \bigotimes_{i\in\iota} N(0,1)$ the standard Gaussian measure on $\mathbb{R}^\iota$. Let $L \in \mathbb{R}$, and let $f : \mathbb{R}^\iota \to \mathbb{R}$ satisfy

so that $f$ is $L$-Lipschitz for the Euclidean distance. Then for every $s \in \mathbb{R}$,

that is, $\mathrm{Ent}_{\gamma_\iota}(e^{sf}) \le \tfrac{s^2L^2}{2}\,\mathbb{E}_{\gamma_\iota} e^{sf}$.

This is the Gaussian logarithmic Sobolev inequality $\mathrm{Ent}_\gamma(g^2) \le 2\int \|\nabla g\|^2\, d\gamma$ (Gross, 1975) applied to $g = e^{sf/2}$. Here $\|\nabla g\|^2 \le \tfrac{s^2L^2}{4} e^{sf}$ almost everywhere, by Rademacher's theorem, or after approximating $f$ by smooth Lipschitz functions. It is the analytic input of Herbst's argument, which together with the Chernoff bound gives the sharp Gaussian concentration inequality.

Formalization Note. Every integral in the statement is finite, because Lipschitz functions have Gaussian exponential moments of all orders. The hypothesis allows any real $L$. If $L < 0$ and $\iota \neq \emptyset$, the hypothesis cannot hold. If $\iota = \emptyset$, then $f$ is constant and the left-hand side is $0$. The case $s \le 0$ is included and is equally true (replace $f$ by $-f$).