TITLE: Elliptic Regularity with Gibbs Measure Satisfying Bakry-Emery Condition
QUESTION [6 upvotes]: Consider $\mathbb{R}^d$ with Gibbs measure $d\mu=Z^{-1}\exp(-V(x))dx$, where the potential $V(x)$ is strongly convex ($\nabla^2 V(x) \ge \lambda Id $). We can assume the regularity of $V$ is as good as we need, and the Hessian is also bounded above. 
Now, for “Laplace equation” in $(\mathbb{R}^d, d\mu)$: $$\nabla^* \nabla u=f,$$ Here $\nabla^*$ is the $L^2(d\mu)$-adjoint of $\nabla$. Assume $f\in L^2(d\mu)$ and solvability condition $\mathbb{E}_{\mu} f=0$, do we always have well-posedness of the equation in the space of mean-zero functions? If yes, do we have $u \in H^2(d\mu)$, and $\| D^2 u\|_{L^2(d\mu)} \le C \|f\|_{L^2(d\mu)}$? Finally, if yes, what is the constant $C$? 
Thanks!

REPLY [2 votes]: For the estimate of the Hessian you might use Bochner's formula: on such weighted $R^d$ for any smooth compactly supported function $u$ it holds
$$
\nabla^*\nabla\frac{|\nabla u|^2}2=|D^2u|^2+\langle\nabla u,\nabla\nabla^*\nabla u\rangle+D^2V(\nabla u,\nabla u)
$$ 
Now using the assumption $D^2V\geq \lambda Id$, $\nabla^*\nabla u=f$ and integrating (the left hand side integrates to 0), after rearrangement we get
$$
\int |D^2u|^2\leq \int |f|^2-\lambda |\nabla u|^2
$$
In particular, if $\lambda\geq 0$ then your desired inequality holds with $C=1$. Once you got this for smooth and compactly supported $u$'s, the general case follows by approximation.