Elliptic Regularity with Gibbs Measure Satisfying Bakry-Emery Condition 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!
 A: 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.
A: As you observed, by Lax-Milgram it is easy to show well-posedenss in $H^1(d\mu)$ with a right-side in $H^{-1}(d\mu)$. The constant in the estimate should be given by whatever the optimal constant in the Poincar\'e inequality for the measure is. Now let's try to differentiate the equation. First, we can write $\nabla^* = -\nabla + \nabla V$ and thus
$$
\nabla^*\nabla u = -\Delta u + \nabla V\cdot \nabla u.
$$
Now: formally differentiate the equation
$$
-\Delta u + \nabla V\cdot \nabla u = f
$$
with respect to $x_i$ and put $v := \partial_{x_i}u$:
\begin{equation}
-\Delta v + \nabla V\cdot \nabla v = \partial_{x_i} f-\nabla (\partial_{x_i} V)\cdot \nabla u.
\end{equation}
We now should ask ourselves whether the right side belongs to $H^{-1}(d\mu)$. The answer is yes. First, the term $\nabla (\partial_{x_i} V)$ is bounded because of assumptions on $V$, and the $\nabla u$ term belongs to $L^2(d\mu)$, so their product belongs to $L^2(d\mu)$ and hence $H^{-1}(d\mu)$. Now we have to establish that $\partial_{x_i}f$ belongs to $H^{-1}(d\mu)$. So pick a test function $\phi\in H^1(d\mu)$ with unit norm and test:
$$
\left| \int \partial_{x_i} f \phi \,d\mu \right| 
=
\left| \int  f \left( \partial_{x_i}\phi - \phi \partial_{x_i} V \right) \,d\mu \right|.
$$
Obviously the term $f\partial_{x_i}\phi$ is ok by the assumption that $\phi\in H^1(d\mu)$ and $f\in L^2(d\mu)$. The second term is also ok because $\partial_{x_i} V$ grows at most like $C(|x|+1)$ and $(1+|x|)\phi(x)$ belongs to $L^2(d\mu)$ by the logarithmic Sobolev inequality. 
This is so far purely formal: the argument only works as an a priori estimate. I do not believe it is a good idea to use difference quotients in making this rigorous, since these don't play well with the spaces due of the Gaussian-like decay of the measure. I think you can work backwards: Since we have established that  $\partial_{x_i} f-\nabla (\partial_{x_i} V)\cdot \nabla u$ belongs to $H^{-1}(d\mu)$, solve the equation for it and find a solution $v_i$. Then try to argue that $v_i = \partial_{x_i} u$. (I consider this last step more of a formality so I won't go on, but I can think harder about the right way to justify it if needed.)
