Lemma 2.2.3.label Let $L$ be a diffusion operator, then for any $f, g \in C_{c}^{\infty}(\real^{d}; \real)$,
\[\dpn{Df, ADg}{\real^d}= \frac{1}{2}[L(fg) - fLg - gLf]\]
Proof. Firstly,
\begin{align*}D(fg)&= fDg + gDf \\ D^{2}(fg)&= fD^{2}g + 2Df \otimes Dg + gD^{2}f \\\end{align*}
so
\[L(fg) = fLg + gLf + 2\dpn{Df, ADg}{\real^d}\]
$\square$