Proof of A Log-Sobolev Inequality Survives a Bounded Perturbation of the Measure (Holley-Stroock)

Each result cited is universally quantified over the data in its own statement.

Throughout, $\mu$, $C$, $U$, $u_{-}$ and $u_{+}$ are as in the statement, $Z$ and $\nu$ are as introduced in claim 1, and $w:X\to\mathbb{R}$ denotes the function $w(x)=\exp(-U(x))$. A Borel function $X\to\mathbb{R}$ that is bounded is integrable with respect to every Borel probability measure on $X$ by Relative Entropy on a Measurable Space: the Gibbs Inequality, the Variational Criterion and Formula, the Entropy Inequality, Small Sets and Data Processing §functional, applied on the measurable space $(X,\mathcal{B}(X))$; this is used without further mention for the bounded Borel functions below. For a real constant $r$ and a Borel probability measure $\vartheta$ on $X$, $\int_{X}r\,d\vartheta=r\,\vartheta(X)=r$ by The Integral of an Indicator Function is the Measure of the Set (applied with $A=X$) and Linearity and Monotonicity of the Lebesgue Integral §integrable.

Step 1 (Claim 1). The function $w$ is Borel. Indeed, by the criterion of Measure Spaces and the Lebesgue Integral: Standing Notation §measurable it suffices that $\{x\in X:w(x)>r\}$ be Borel for every real $r$. By claim 2 of Basic Properties of the Exponential Function, $w(x)>0$ for every $x$, so this set is $X$ when $r\le0$. When $r>0$, write $r=\exp(\log r)$ with $\log$ the natural logarithm; since $\exp$ is strictly increasing by claim 4 of Basic Properties of the Exponential Function, $\exp(-U(x))>\exp(\log r)$ holds exactly when $-U(x)>\log r$, so the set is $\{x\in X:-U(x)>\log r\}$, which is Borel because $-U$ is Borel by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. Moreover $-u_{+}\le-U(x)\le-u_{-}$ for every $x$, so, $\exp$ being increasing,

$$0<\exp(-u_{+})\le w(x)\le\exp(-u_{-})\qquad(x\in X).$$

In particular $w$ is bounded and Borel, hence integrable with respect to $\mu$, and $Z=\int_{X}w\,d\mu$ is a real number. By the monotonicity of Linearity and Monotonicity of the Lebesgue Integral §integrable, comparing $w$ with the constants $\exp(-u_{+})$ and $\exp(-u_{-})$ and using $\mu(X)=1$, we obtain $\exp(-u_{+})\le Z\le\exp(-u_{-})$; in particular $Z>0$.

Let $h=Z^{-1}w$, a Borel function $X\to[0,\infty)$ by claim 2 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions. By claim 3 of Image Measures, Measures with Densities, and Change of Variables, the measure $\nu_{h}$ with density $h$ with respect to $\mu$, $\nu_{h}(A)=\int_{X}\mathbf{1}_{A}h\,d\mu$ for $A\in\mathcal{B}(X)$, is a measure on $\mathcal{B}(X)$. For $A\in\mathcal{B}(X)$ the function $\mathbf{1}_{A}w$ is Borel (claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions), bounded and nonnegative, as is $\mathbf{1}_{A}h$, so the $[0,\infty]$-valued integral defining $\nu_{h}(A)$ is real and equals the integral of $\mathbf{1}_{A}h$ as an integrable function (claim 6(c) of Borel Measurability and Bounded Integration on a Metric Space, with $\mu(X)=1$); hence, by Linearity and Monotonicity of the Lebesgue Integral §integrable,

$$\nu_{h}(A)=\int_{X}\mathbf{1}_{A}\,Z^{-1}w\,d\mu=Z^{-1}\int_{X}\mathbf{1}_{A}\exp(-U)\,d\mu=\nu(A).$$

Hence $\nu=\nu_{h}$ is a measure on $\mathcal{B}(X)$, and $\nu(X)=Z^{-1}\int_{X}w\,d\mu=Z^{-1}Z=1$, so $\nu\in\mathcal{P}(X)$. Now let $f:X\to\mathbb{R}$ be bounded and Borel. Then $fh$ is Borel by claim 3 of Arithmetic, Absolute Values, and Pointwise Limits of Measurable Real-Valued Functions and bounded, hence integrable with respect to $\mu$, and claim 3 of Image Measures, Measures with Densities, and Change of Variables shows that $f$ is integrable with respect to $\nu_{h}=\nu$ with $\int_{X}f\,d\nu=\int_{X}fh\,d\mu=Z^{-1}\int_{X}f\exp(-U)\,d\mu$, the last step by Linearity and Monotonicity of the Lebesgue Integral §integrable. This completes the proof of claim 1.

Step 2 (A variational bound for the entropy). For a positive real $t$ let $\chi_{t}:[0,\infty)\to\mathbb{R}$ be $\chi_{t}(r)=\phi(r)-r\log t-r+t$. We show $\chi_{t}(r)\ge0$ for $r\ge0$. For $r>0$, the identity $\log r=\log(r/t)+\log t$ of The Natural Logarithm gives $t\,\phi(r/t)=r\log(r/t)=\phi(r)-r\log t$; for $r=0$ both sides are $0$, as $\phi(0)=0$. Hence $\chi_{t}(r)=t\bigl(\phi(r/t)-(r/t)+1\bigr)$, which is nonnegative by the lower bound $s-1\le\phi(s)$ of The Function $s\log s$: Continuity, Young's Inequality and Lower Bounds, with the Elementary Bounds for the Exponential and the Logarithm §lower, applied with $s=r/t$, and $t>0$. Moreover, if $r>0$ then $\chi_{r}(r)=r\log r-r\log r-r+r=0$, while $\chi_{t}(0)=t$.

Now let $\vartheta$ be a Borel probability measure on $X$, let $G:X\to\mathbb{R}$ be Borel with values in an interval $[0,b]$, and put $m=\int_{X}G\,d\vartheta$, a real number with $m\ge0$. The function $\phi\circ G$ is Borel by The Function $s\log s$: Continuity, Young's Inequality and Lower Bounds, with the Elementary Bounds for the Exponential and the Logarithm §continuous, and bounded since $\phi$ is continuous on $[0,b]$ by that clause, hence bounded there by Extreme Value Theorem on a Closed Real Interval; so $\chi_{t}\circ G=\phi\circ G-(\log t+1)G+t$ is bounded and Borel. By Linearity and Monotonicity of the Lebesgue Integral §integrable and The Entropy of a Nonnegative Function with Respect to a Probability Measure §entropy,

$$\int_{X}\chi_{t}\circ G\,d\vartheta=\int_{X}\phi\circ G\,d\vartheta-m\log t-m+t=\operatorname{Ent}_{\vartheta}(G)+\chi_{t}(m).$$

Since $\chi_{t}(m)\ge0$, this gives $\operatorname{Ent}_{\vartheta}(G)\le\int_{X}\chi_{t}\circ G\,d\vartheta$ for every $t>0$. Also, for every positive real $\tau$ there is $t>0$ with $\chi_{t}(m)\le\tau$: take $t=m$ if $m>0$, so that $\chi_{t}(m)=0$, and $t=\tau$ if $m=0$, so that $\chi_{t}(m)=\tau$.

Step 3 (Comparison of the gradient terms). Let $F\in\mathcal{F}C^{1}_{b}(X)$ and let $g=|\nabla_{a}F|_{a}^{2}$, a bounded nonnegative Borel function by The Noise Gradient of a Cylindrical Function and the Noise Tangent Space at a Probability Measure on a Hilbert Space §gradient. By Step 1, $w(x)\ge\exp(-u_{+})>0$, so $1\le\exp(u_{+})\,w(x)$, using $\exp(u_{+})\exp(-u_{+})=\exp(0)=1$ from claim 1 of Basic Properties of the Exponential Function; hence $g\le\exp(u_{+})\,g\,w$ pointwise, $g$ being nonnegative. By the monotonicity of Linearity and Monotonicity of the Lebesgue Integral §integrable and claim 1 (Step 1) applied with $f=g$,

$$\int_{X}g\,d\mu\le\exp(u_{+})\int_{X}g\,w\,d\mu=\exp(u_{+})\,Z\int_{X}g\,d\nu .$$

Step 4 (Claim 2). Let $F\in\mathcal{F}C^{1}_{b}(X)$, let $G=F^{2}$, which is Borel with values in an interval $[0,b]$ as recorded in the statement, and put $m=\int_{X}G\,d\mu$. Let $t>0$. By Step 2 with $\vartheta=\nu$, then claim 1 applied with the bounded Borel function $f=\chi_{t}\circ G$, then $0\le\chi_{t}\circ G$ (Step 2) and $w\le\exp(-u_{-})$ (Step 1) with the monotonicity of Linearity and Monotonicity of the Lebesgue Integral §integrable, and finally Step 2 with $\vartheta=\mu$,

$$\operatorname{Ent}_{\nu}(G)\le\int_{X}\chi_{t}\circ G\,d\nu=Z^{-1}\int_{X}(\chi_{t}\circ G)\,w\,d\mu\le Z^{-1}\exp(-u_{-})\int_{X}\chi_{t}\circ G\,d\mu=Z^{-1}\exp(-u_{-})\bigl(\operatorname{Ent}_{\mu}(G)+\chi_{t}(m)\bigr).$$

Given a positive real $\varepsilon$, put $\tau=\varepsilon Z\exp(u_{-})$, a positive real, and choose $t$ by Step 2 with $\chi_{t}(m)\le\tau$. Since $\exp(-u_{-})\exp(u_{-})=\exp(0)=1$ by claim 1 of Basic Properties of the Exponential Function, $Z^{-1}\exp(-u_{-})\,\tau=\varepsilon$, so $\operatorname{Ent}_{\nu}(G)\le Z^{-1}\exp(-u_{-})\operatorname{Ent}_{\mu}(G)+\varepsilon$. As $\varepsilon>0$ is arbitrary, Comparison of Real Numbers with Arbitrary Positive Slack §slack-above gives $\operatorname{Ent}_{\nu}(G)\le Z^{-1}\exp(-u_{-})\operatorname{Ent}_{\mu}(G)$. By the hypothesis on $\mu$, then Step 3 together with $C\ge0$,

$$\operatorname{Ent}_{\nu}(F^{2})\le Z^{-1}\exp(-u_{-})\cdot2C\int_{X}|\nabla_{a}F|_{a}^{2}\,d\mu\le Z^{-1}\exp(-u_{-})\cdot2C\exp(u_{+})\,Z\int_{X}|\nabla_{a}F|_{a}^{2}\,d\nu .$$

Finally $Z^{-1}Z=1$ and $\exp(-u_{-})\exp(u_{+})=\exp(u_{+}-u_{-})$ by claim 1 of Basic Properties of the Exponential Function, which gives claim 2.