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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 $(5)$.

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  • $\begingroup$ $(4)$ 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 $\varrho\in C^1(\mathbb R)$ ... $\endgroup$ – 0xbadf00d Feb 10 at 14:02
  • $\begingroup$ 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. $\endgroup$ – Rhys Steele Feb 10 at 18:20
  • $\begingroup$ @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 ... $\endgroup$ – 0xbadf00d Feb 11 at 10:34
  • $\begingroup$ 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. $\endgroup$ – Rhys Steele Feb 11 at 11:54
  • $\begingroup$ @RhysSteele Thank you, anyway. $\endgroup$ – 0xbadf00d Feb 11 at 19:43

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