Gross logarithmic Sobolev inequality for standard Gaussian measure

For $t>0$, put $a=e^{-t}$, $b=\sqrt{1-e^{-2t}}$, and $P_{t}g(x)=\int g(ax+by)d{\gamma }_d(y)$. By [F1], $aX+bY$ is standard Gaussian for independent $X,Y\sim {\gamma }_d$; Fubini [F4] gives $\int P_{t}gd{\gamma }_d=\int gd{\gamma }_d$. Differentiation under the integral [F2] gives $\mathrm{\nabla }{P}_{t}g=a{P}_t(\mathrm{\nabla }g)$ and $\mathrm{\Delta }{P}_{t}g=a^2{P}_t(\mathrm{\Delta }g)$. It also gives ${\mathrm{\partial }}_t{P}_{t}g=-ax\cdot P_t(\mathrm{\nabla }g)+(a^2\mathrm{/}b)\int y\cdot \mathrm{\nabla }g(ax+by)d{\gamma }_d(y)$. For each coordinate, [F3] applied against the Gaussian density gives $\int y_i{\mathrm{\partial }}_{i}g(ax+by)d{\gamma }_d=b\int {\mathrm{\partial }}_{ii}g(ax+by)d{\gamma }_d$; the boundary terms vanish because ${\mathrm{\partial }}_{i}g$ is bounded and the Gaussian density tends to zero at both ends. Thus ${\mathrm{\partial }}_t{P}_{t}g=\mathrm{\Delta }{P}_{t}g-x\cdot \mathrm{\nabla }{P}_{t}g$.

Put $u_t=P_{t}g$. Differentiate the entropy using step 1.1 and [F2]: $\frac{d}{dt}{\mathrm{Ent}}_{{\gamma }_d}(u_t)=\int (1+\log u_t)(\mathrm{\Delta }{u}_t-x\cdot \mathrm{\nabla }{u}_t)d{\gamma }_d$. For each coordinate, [F3] applied to $v=1+\log u_t$ and ${\mathrm{\partial }}_i{u}_t$ against the Gaussian density [F1] gives $\int v({\mathrm{\partial }}_{ii}{u}_t-x_i{\mathrm{\partial }}_i{u}_t)d{\gamma }_d=-\int ({\mathrm{\partial }}_i{u}_t{)}^2\mathrm{/}{u}_{t}d{\gamma }_d$; the boundary term vanishes because $v$ and $\mathrm{\nabla }{u}_t$ are bounded and the density tends to zero. Summing coordinates yields $\frac{d}{dt}{\mathrm{Ent}}_{{\gamma }_d}(u_t)=-\int \frac{\mathrm{\mid }\mathrm{\nabla }{u}_t{\mathrm{\mid }}^2}{u_t}d{\gamma }_d$. The differentiated integrands are dominated on each compact time interval because $u_t$ stays between the positive bounds for $g$ and its first two derivatives are bounded. As $t\downarrow 0$, continuity of $g$ and bounded convergence [F4] give $P_{t}g(x)\to g(x)$ and ${\mathrm{Ent}}_{{\gamma }_d}(u_t)\to {\mathrm{Ent}}_{{\gamma }_d}(g)$. As $t\to \mathrm{\infty }$, $u_t(x)\to \int gd{\gamma }_d$, so [F4] gives entropy limit zero. Integrate the derivative on $[\delta ,T]$ by [F6], then let $\delta \downarrow 0$ and $T\uparrow \mathrm{\infty }$ using [F5] on the nonnegative dissipation. Therefore ${\mathrm{Ent}}_{{\gamma }_d}(g)={\int }_0^{\mathrm{\infty }}\int \mathrm{\mid }\mathrm{\nabla }{P}_{t}g{\mathrm{\mid }}^2\mathrm{/}(P_{t}g)d{\gamma }_{d}dt$.

For each $x,t$, apply [F7] to each coordinate after writing $\mathrm{\nabla }g=(\mathrm{\nabla }g\mathrm{/}\sqrt{g})\sqrt{g}$ inside the probability integral defining $P_t$. Summing the coordinate inequalities gives $\frac{\mathrm{\mid }{P}_t(\mathrm{\nabla }g)(x){\mathrm{\mid }}^2}{P_{t}g(x)}\le P_t\text{\hspace{0.17em}⁣}\left(\frac{\mathrm{\mid }\mathrm{\nabla }g{\mathrm{\mid }}^2}{g}\right)(x)\mathrm{.}$

By step 1.1, $\mathrm{\mid }\mathrm{\nabla }{P}_{t}g{\mathrm{\mid }}^2=e^{-2t}\mathrm{\mid }{P}_t(\mathrm{\nabla }g){\mathrm{\mid }}^2$; apply step 2.2, integrate using invariance of ${\gamma }_d$, and use ${\int }_0^{\mathrm{\infty }}{e}^{-2t}dt=1\mathrm{/}2$ in the entropy representation of step 2.1. This gives ${\mathrm{Ent}}_{{\gamma }_d}(g)\le \frac{1}{2}\int \frac{\mathrm{\mid }\mathrm{\nabla }g{\mathrm{\mid }}^2}{g}d{\gamma }_d\mathrm{.}$