Existence of a Lyapunov function for a log-concave measure

Question

Let $d\in\mathbb N$, $f:\mathbb R^d\to\mathbb R$ be convex with $$\int e^{-f(x)}\:{\rm d}x<\infty\tag1$$ and $\mu$ denote the measure with density $e^{-f}$ with respect to the Lebesgue measure on $\mathcal B(\mathbb R^d)$. Moreover, let $$\Gamma(\varphi,\psi):=\langle\nabla\varphi,\nabla\psi\rangle\;\;\;\text{for }\varphi,\psi\in\mathcal A_0:=C_c^\infty(\mathbb R)$$ and $$A\varphi:=\Delta\varphi-\langle\nabla f,\nabla\varphi\rangle\;\;\;\text{for }\varphi,\psi\in\mathcal A_0.$$

I want to show that we can find the following objects:

1. $K\in\mathcal B(\mathbb R^d)$ with $\mu(K)\in(0,\infty)$
2. $L\in\mathcal B(\mathbb R^d)$ with $L\supseteq K$ and $$\int_K\left|\varphi-m_K(\varphi)\right|^2\:{\rm d}\mu\le C_{K,\:L}\int_L\Gamma(\varphi)\:{\rm d}\mu\tag2\;\;\;\text{for all }\varphi\in\mathcal A_0,$$ where $$m_K(\varphi):=\frac1{\mu(K)}\int_K\varphi\:{\rm d}\mu$$ and $\Gamma(\varphi):=\Gamma(\varphi,\varphi)$
3. $J:\mathbb R^d\to[1,\infty)$ with $J\in\mathcal A:=C^\infty(\mathbb R^d)$ and $$1\le-\frac{AJ}{\lambda J}+b1_K\tag3$$ for some $\lambda,b>0$

My idea is as follows: Let $c,R>0$ and $J:\mathbb R^d\to\mathbb R$ with $$J(x)=e^{c|x|}\;\;\;\text{for all }|x|\ge R\tag4$$ and $$J(x)\ge1\;\;\;\text{for all }|x|\le R\tag5.$$ Note that $$(LJ)(x)=\left(c+\frac{d-1}{|x|}\right)cJ(x)-\frac{cJ(x)}{|x|}\langle\nabla f(x),x\rangle\tag6$$ for all $|x|>R$ and hence $$1\le-\frac{(LJ)(x)}{\lambda|x|}\Leftrightarrow\frac\lambda c+c+\frac{d-1}{|x|}\le\frac{\langle\nabla f(x),x\rangle}{|x|}\tag7$$ for all $|x|>R$. Now, we somehow need to use that by convexity of $f$ and $(1)$ $$\lim_{r\to\infty}\inf_{|x|\:\ge\:r}\frac{\langle\nabla f(x),x\rangle}{|x|}=\liminf_{r\to\infty}\frac{\langle\nabla f(x),x\rangle}{|x|}\in(0,\infty]\tag8.$$ By $(8)$ we may choose $R\ge d-1$ with $$l:=\inf_{|x|\:\ge\:R}\frac{\langle\nabla f(x),x\rangle}{|x|},$$ $\lambda=l^2/4$ and $c=l/2$ to obtain $$\frac\lambda c+c+\frac{d-1}{|x|}\le l\le\frac{\langle\nabla f(x),x\rangle}{|x|}\;\;\;\text{for all }|x|\ge R\tag9.$$

This ensures at least that $(3)$ is satisfied for $|x|\ge R$. How do we need to choose $b$ and how do we need to choose $L$ such that $(2)$ is satisfied? (Clearly, with the argumentation above, we would choose $K=\left\{|x|\le R\right\}$.)

Answer 1

By $(8)$, $$\langle\nabla f(x),x\rangle\ge\alpha|x|\;\;\;\text{for all }|x|\ge r\tag{10}$$ for some $\alpha>0$ and $r\ge0$. Let $c\in(0,\alpha)$, $\tilde r\ge r$ with $$\tilde r>\frac{d-1}{\alpha-c}\tag{11}$$ and $J\in C^\infty\left(\mathbb R^d\right)$ with $$J(x)=e^{c|x|}\;\;\;\text{for all }|x|\ge\tilde r\tag{12}$$ and $$J(x)\ge1\;\;\;\text{for all }|x|\le\tilde r\tag{13}.$$ Note that $$(LJ)(x)=-c\left(\frac{\langle\nabla f(x),x\rangle}{|x|}-c-\frac{d-1}{|x|}\right)J(x)\le-\underbrace{c\left(\alpha-c-\frac{d-1}{\tilde r}\right)}_{=:\:\lambda\:>\:0}J(x)\tag{14}$$ for all $|x|>\tilde r$. Now, $J$ and $LJ$ are continuous and hence locally bounded (above). Thus, $b_1:=\sup_{|x|\:\le\:\tilde r}J(x)<\infty$ and $b_2:=\sup_{|x|\:\le\:\tilde r}(LJ)(x)<\infty$. Letting $b:=\lambda b_1+b_2$, we obtain $$LJ\le-\lambda J+b1_{\left\{\:|x|\:\le\:\tilde r\:\right\}}\tag{15}.$$