# If $L$ is a diffusion operator with corresponding carré du champ operator $\Gamma$, then $Lf^2=2fLf+2\Gamma(f)$

Let

• $$(E,\mathcal E,\mu)$$ be a measure space and $$\mu f:=\int f\:{\rm d}\mu\;\;\;\text{for }f\in L^1(\mu)$$
• $$\mathcal A_0$$ be a subspace of $$\left\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\right\}$$ stable under multiplication and dense in $$L^p(\mu)$$ for all $$p\in[1,\infty)$$
• $$\Gamma$$ be a bilinear symmetric operator on $$\mathcal A_0$$ and $$\Gamma(f):=\Gamma(f,f)\;\;\;\text{for }f\in\mathcal A_0$$

Assume $$\forall f\in\mathcal A_0:\exists c\ge0:\forall g\in\mathcal A_0:|\mu\Gamma(f,g)|\le c\left\|g\right\|_{L^2(\mu)}\tag1.$$ By $$(1)$$ and Riesz' representation theorem, there is a unique linear operator $$L:\mathcal A_0\to L^2(\mu)$$ with $$\langle Lf,g\rangle_{L^2(\mu)}=-\mu\Gamma(f,g)\;\;\;\text{for all }f,g\in\mathcal A_0.\tag2$$

Now, assume $$\mu\Gamma(f^2,g)+2\langle\Gamma(f),g\rangle_{L^2(\mu)}=2\mu\Gamma(fg,f)\;\;\;\text{for all }f,g\in\mathcal A_0\tag3$$ and $$L\mathcal A_0\subseteq\mathcal A_0\tag4.$$

Why do we need $$(4)$$ in order to conclude that $$Lf^2=2fLf+2\Gamma(f)\tag5$$ for all $$f\in\mathcal A_0$$?

From $$(2)$$, we obtain $$\langle Lf^2,g\rangle_{L^2(\mu)}=-\mu\Gamma(f^2,g)\;\;\;\text{for all }g\in\mathcal A_0\tag6.$$ By $$(3)$$, $$-\mu\Gamma(f^2,g)=2\langle\Gamma(f),g\rangle_{L^2(\mu)}-2\mu\Gamma(fg,f)\;\;\;\text{for all }g\in\mathcal A_0\tag7.$$ Again By $$(2)$$, $$-\mu\Gamma(fg,f)=-\mu\Gamma(f,fg)=\langle Lf,fg\rangle_{L^2(\mu)}\;\;\;\text{for all }g\in\mathcal A_0\tag8.$$ Now, the crucial point might be to write $$\langle Lf,fg\rangle_{L^2(\mu)}=\langle fLf,g\rangle_{L^2(\mu)}\;\;\;\text{for all }g\in\mathcal A_0\tag9.$$ In order to do this, we need $$fLf\in L^2(\mu)$$. But since $$f$$ is bounded, we clearly have $$\left\|fLf\right\|_{L^2(\mu)}\le\left\|f\right\|_\infty\left\|Lf\right\|_{L^2(\mu)}<\infty\tag{10}$$ and hence there should be no problem. Thus, we should be able to conclude $$\langle Lf^2,g\rangle_{L^2(\mu)}=2\langle\Gamma(f),g\rangle_{L^2(\mu)}+2\langle fLf,g\rangle_{L^2(\mu)}\;\;\;\text{for all }g\in\mathcal A_0\tag{11}.$$ By density of $$\mathcal A_0$$ in $$L^2(\mu)$$, this should prove $$(5)$$.

What am I missing? The necessity of $$(4)$$ is claimed in the book Analysis and Geometry of Markov Diffusion Operators, but I don't see where we need $$(5)$$.

• $(4)$ seems to be a very restrictive assumption. In the application, I've got in mind, $\mathcal A_0=C_c^\infty(\mathbb R)$ and $Lf=\frac12(\ln\varrho)'f'+\frac12f''$. So, $(4)$ would force me to assume $\varrho\in C^\infty(\mathbb R)$, while I'm only willing to assume $\varrho\in C^1(\mathbb R)$ ... – 0xbadf00d Feb 10 at 14:02
• I can't see anything at all wrong with your argument since it essentially amounts to writing down $(3.1.1)$ in the text in terms of $L$. I also can't see how $(4)$ is being used elsewhere. I don't think you miss anything. – Rhys Steele Feb 10 at 18:20
• @RhysSteele Thank you for your comment. Do you know the mentioned book? I would like to know how one would typically choose the objects in the context of this question (I'm willing to assume a curvature dimension condition). It seems to be an unreasonable strict assumption to assume that $L$ (or even $\kappa_t$) preserves $\mathcal A_0$. On the other hand, they consider an extended algebra $\mathcal A$ later on ... – 0xbadf00d Feb 11 at 10:34
• Unfortunately, I'm not familiar with the book, I only briefly read through the relevant section to see if that assumption was implicitly used elsewhere so I don't think I can be of much more help, sorry. – Rhys Steele Feb 11 at 11:54
• @RhysSteele Thank you, anyway. – 0xbadf00d Feb 11 at 19:43