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

  • $\begingroup$ Why does (5) follow from (4)? $\endgroup$ – Nawaf Bou-Rabee Jan 27 '19 at 16:24
  • $\begingroup$ @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? $\endgroup$ – 0xbadf00d Jan 27 '19 at 18:23
  • $\begingroup$ 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). $\endgroup$ – Nawaf Bou-Rabee Jan 27 '19 at 19:03
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    $\begingroup$ @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 $r$. $\endgroup$ – 0xbadf00d Jan 27 '19 at 19:34
  • $\begingroup$ How does (8) follow from (7)? The sign of the RHS of (8) looks off. Seems like (7) only implies that $A J / (\lambda J) \le 1$. $\endgroup$ – 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 $$LJ\le-\lambda J+b1_{\left\{\:|x|\:\le\:\tilde r\:\right\}}\tag{15}.$$

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  • $\begingroup$ 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. $\endgroup$ – Nawaf Bou-Rabee Feb 3 '19 at 15:19

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