Exponential decay of Fisher information along the OU semigroup

Question

I read from a paper that there is a "well-known" exponential decay of Fisher information along the OU semigroup, that is $$J(\nu^t\mid\gamma)\leq e^{-2t}J(\nu\mid\gamma),$$

where $\gamma$ is the standard gaussian distribution and $\nu\ll\gamma$ is any probability measure with density $h$, and $d\nu^t=P_thd\gamma$ where $P_t$ is the OU semigroup corresponding to the SDE $dX_t=-X_tdt+\sqrt{2}dW_t$.

I found this blog as a reference for the proof(I appreciate it if you can direct me to another source!) which is as in the screenshot below.

And I got stuck on the red-step: how is that a Jensen's inequality? What is the convex function here? What I have so far is that if we view the numerator as a constant(for fixed x), and apply Jensen's inequality with $1/x$, then we get something like $$\frac{P_t|\nabla h|^2}{P_th}\leq P_t|\nabla h|^2P_t\frac{1}{h},$$ which is still different from what we want: $P_t\frac{|\nabla h|^2}{h}$.

Answer 1

This is an answer to the specific question about the type of Jensen inequality.

For this, you can use the function $\alpha : [0,\infty)\times [0,\infty) \to [0,\infty]$ defined by $$ \alpha(a,b) = \begin{cases} \frac{a^2}{b}, & b>0 \\ +\infty , & a>0, b=0\\ 0 , & a=0, b=0 \end{cases} $$ The function $\alpha$ is jointly convex, jointly positively one-homogeneous, i.e. $\alpha(\lambda a, \lambda b) = \lambda \alpha(a,b)$ for $\lambda>0$.

Hence, we can apply Jensen and get $$ \frac{|\nabla f|_t^2}{f_t} = \alpha(P_t |\nabla f|^2, P_t f) \leq P_t \alpha(|\nabla f|^2,f) = \left( \frac{|\nabla f|^2}{f}\right)_t. $$

Answer 2

I guess that the 2014 book "Analysis and Geometry of Markov Diffusion Operators" by Bakry-Gentil-Ledoux will answer all the questions (see in particular Chapter 5).