Lemma 2.4.2 (Kato Inequality, [Lemma 4.24, Bau14]).label Let
\[Lu = \dpn{A, D^2u}{\real^{d \times d}}+ \dpn{b, Du}{\real^d}\]
be a diffusion operator on $\real^{d}$ with symmetrising measure $\mu$, then for any $u \in C_{c}^{\infty}(\real^{d}; \real)$,
\[L(|\mu|) \ge \sgn(u) \cdot Lu\]
in the sense of distributions.
Proof. First suppose that $u \in C_{c}^{\infty}(\real^{d}; \real)$. For any $\phi \in C^{2}(\real; \real)$,
\begin{align*}D(\phi \circ u)&= (\phi' \circ u) \cdot Du \\ D^{2}(\phi \circ u)&= (\phi' \circ u) \cdot Du \otimes Du + (\phi' \circ u) \cdot D^{2}u\end{align*}
so
\[L(\phi \circ u) = (\phi' \circ u) Lu + (\phi'' \circ u)\dpn{Du, ADu}{\real^d}\]
In particular, if $\phi$ is also convex, then
\[L(\phi \circ u) \ge (\phi' \circ u)Lu\]
Now, for each $\eps > 0$, let $\phi_{\eps}(x) = \sqrt{x^{2} + \eps^{2}}$, then
\[L(\phi_{\eps} \circ u)(x) \ge \frac{u(x)}{\sqrt{x^{2} + \eps^{2}}}Lu(x)\]
by the Dominated Convergence Theorem,
\[L|u| \ge \sgn(u) Lu\]
$\square$