# Existence of a Lyapunov function for a log-concave measure

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 $$K=\left\{|x|\le R\right\}$$.)

• Why does (5) follow from (4)? – Nawaf Bou-Rabee Jan 27 '19 at 16:24
• @NawafBou-Rabee By definition, $\lambda=\lim_{r\to\infty}\inf_{|x|\ge r}\frac{\langle\nabla f(x),x\rangle}{|x|}$. The only problematic thing could be if $\lambda=\infty$ (for example, when $f(x)=\frac12 x^2+\text{constant}$). Am I missing something? – 0xbadf00d Jan 27 '19 at 18:23
• Still don't see how (5) follows from the definition of $\lambda$. Suppose $R \ge r$. Then $\inf_{|x| \ge R} \nabla f(x) \cdot x / |x| \ge \inf_{|x| \ge r} \nabla f(x) \cdot x / |x|$, since the infimum in the RHS of the inequality is taken over a larger set because $\{ |x| \ge R \} \subseteq \{ |x| \ge r \}$, and so $\lambda \ge \inf_{|x| \ge r} \nabla f(x) \cdot x / |x|$. This conclusion seems different from (5). – Nawaf Bou-Rabee Jan 27 '19 at 19:03
• @NawafBou-Rabee By definition of $\lambda$ (assuming $\lambda<\infty$), $$\forall\varepsilon>0:\exists R>0:\forall r\ge R:\left|\lambda-\inf_{|x|\ge r}\frac{\langle\nabla f(x),x\rangle}{|x|}\right|<\varepsilon.$$ And $\lambda\ge\inf_{|x|\ge r}\frac{\langle\nabla f(x),x\rangle}{|x|}$ is trivial for all $r>0$, since $\lambda$ is the supremum of the values on the right-hand side over all $r>0$. As you noted, the right-hand side is increasing in $r$. – 0xbadf00d Jan 27 '19 at 19:34
• How does (8) follow from (7)? The sign of the RHS of (8) looks off. Seems like (7) only implies that $A J / (\lambda J) \le 1$. – Nawaf Bou-Rabee Jan 30 '19 at 17:27

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 $$LJ\le-\lambda J+b1_{\left\{\:|x|\:\le\:\tilde r\:\right\}}\tag{15}.$$
• In deriving a Foster-Lyapunov drift condition, typically $c$ in (4) is not given, but is chosen small enough such that something like (14) holds. It also seems natural to choose $\alpha = \min(1, \lim_{r \to \infty} \inf_{|x| \ge r x \cdot \nabla f(x)/|x|$. However, these are minor points, and overall this is nice. – Nawaf Bou-Rabee Feb 3 '19 at 15:19