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Let $h\in C^1(\mathbb R)$ such that $h'$ is Lipschitz continuous and $$L\varphi:=-h'\varphi'+\varphi''\;\;\;\text{for }\varphi\in C^2(\mathbb R).$$ The formal adjoint of $L$ is $$L^\ast\psi:=\psi''+(h'\psi)'\;\;\;\text{for }\psi'\in C^2(\mathbb R).$$ Note that $L^\ast e^{-h}=0$.

Are we able to show that there is a Borel measurable (hopefully continuous) function $v:\mathbb R\to[0,\infty)$ such that $$Lv\le c-\lambda v\tag1$$ for some $c\ge0$ and $\lambda>0$?

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  • $\begingroup$ One needs more assumptions on $h$, e.g., a dissipativity condition is sufficient. $\endgroup$
    – Nawaf Bou-Rabee
    Commented Jan 18, 2019 at 12:41
  • $\begingroup$ @NawafBou-Rabee Can you provide details? $\endgroup$
    – 0xbadf00d
    Commented Jan 18, 2019 at 13:05

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Suppose that $h'(x)$ satisfies a dissipativity condition: there exists $K>0$ and $A \ge 0$ such that for all $x \in \mathbb{R}$ we have $x h'(x) \ge K x^2 - A$. Consider as a candidate function $v(x) = x^2$. Then, $$ L v(x) = - x h'(x) + 1 \le - K x^2 + A + 1 \;. $$ So, (1) is satisfied with $\lambda=K$ and $c=A+1$.

Other conditions are possible. See, e.g., Assumption 2.1 (B) and Lemma 2.5 of Non-Asymptotic Mixing of the MALA Algorithm.

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  • $\begingroup$ Thank you very much for your answer. This question arose in me while I've tried to find a solution for an other question. Is this the right track or is there an easier solution available which doesn't need furhter assumptions on $h$? $\endgroup$
    – 0xbadf00d
    Commented Jan 18, 2019 at 13:22
  • $\begingroup$ Adding further assumptions is the right track. Otherwise you could be in a situation where $e^{-h}$ might not be integrable or the infinitesimally invariant measure with density $e^{-h}$ might not be an invariant measure for the process. $\endgroup$
    – Nawaf Bou-Rabee
    Commented Jan 18, 2019 at 13:29
  • $\begingroup$ Actually, in the application I've got in mind $h=-\ln f$ for some positive $f\in C^2(\mathbb R)$ such that $f'/f$ is Lipschitz and $\lambda f=1$. Does that change the situation? $\endgroup$
    – 0xbadf00d
    Commented Jan 18, 2019 at 13:33
  • $\begingroup$ What do you mean by $\lambda f = 1$? $\endgroup$
    – Nawaf Bou-Rabee
    Commented Jan 18, 2019 at 13:35
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    $\begingroup$ Not sure if the existence of a Foster-Lyapunov function implies (9). The easiest and most quantitative approach is via couplings, which on unbounded spaces rely on Foster-Lyapunov functions or something comparable to confine most of the Langevin dynamics to a compact set. You can read more about this here: arxiv.org/abs/1305.1233 $\endgroup$
    – Nawaf Bou-Rabee
    Commented Jan 18, 2019 at 14:01

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