  Suppose that ${\cal L}$ is a divergence form differential operator of the
form ${\cal L}f:=(1/2) e^{U}\nabla_x\cdot\big[e^{-U}(I+H)\nabla_x f\big]$,
where $U$ is scalar valued, $I$ identity matrix and $H$ an anti-symmetric
matrix valued function. The coefficients are not assumed to be bounded, but are
$C^2$ regular. We show that if $Z=\int_{\mathbb{R}^d}e^{-U(x) }dx<+\infty$ and
the supremum of the numerical range of matrix $-\frac12\nabla^2_x
U+\frac12\nabla_x\left\{\nabla_x\cdot H-[\nabla_x U]^TH\right\}$ satisfies some
exponential integrability condition with respect to measure
$d\mu=Z^{-1}e^{-U}dx$, then for any $1 \le p<q<+\infty$ there exists a constant
$C>0$ such that $\left\| f\right\|_{W^{2,p}(\mu)}\le C\Big(\left\|{\cal
L}f\right\|_{L^q(\mu)}+\left\|f\right\|_{L^q(\mu)}\Big)$ for $f\in
C_0^\infty(\mathbb{R}^d)$. Here $W^{2,p}(\mu)$ is the Sobolev space of
functions that are $L^p(\mu)$ integrable with two derivatives. Our proof is
probabilistic and relies on an application of the Malliavin calculus.
