• $C_0(\mathbb R)$ denote the space of continuous functions vanishing at infinity equipped with the supremum norm
  • $b,\sigma:\mathbb R\to\mathbb R$ be Lipschitz continuous, $$Lf:=bf'+\frac12\sigma^2f''\;\;\;\text{for }f\in C^2(\mathbb R)$$ and $$L^\ast g:=\frac12(\sigma^2g)''-(bg)'\;\;\;\text{for }g\in C^2(\mathbb R)$$
  • $A$ be a closed linear operator on $C_0(\mathbb R)$ with $$A\phi=L\phi\;\;\;\text{for all }\phi\in C_c^2(\mathbb R)\tag1$$
  • $\lambda^1$ denote the Lebesgue measure
  • $g\in C^2(\mathbb R)$ with $g\ge0$, $$\int g\;{\rm d}\lambda^1=1$$ and $$\int\phi L^\ast g\:{\rm d}\lambda^1=0\;\;\;\text{for all }\phi\in C_c^\infty(\mathbb R)\tag2$$
  • $\mu$ denote the measure with density $g$ with respect to $\lambda^1$

I want to show that $$\int Af\:{\rm d}\mu=0\tag3\;\;\;\text{for all }f\in C_0(\mathbb R).$$

By $(1)$ and $(2)$, we should obtain $$\int A\phi\:{\rm d}\mu=\int (L\phi)g\:{\rm d}\lambda^1\stackrel{\text{(4)}}=\int\phi(L^\ast g)\:{\rm d}\lambda^1\tag5=0\;\;\;\text{for all }\phi\in C_c^\infty(\mathbb R)$$ (Could anyone point me to a reference for $(4)$?). Now, let $f\in C_0(\mathbb R)$. Since, $C_c^\infty(\mathbb R)$ is dense in $C_0(\mathbb R)$ (again, any reference for that?), there is a $(\phi_n)_{n\in\mathbb N}\subseteq C_c^\infty(\mathbb R)$ with $$\left\|\phi_n-f\right\|_\infty\xrightarrow{n\to\infty}0\tag6.$$

How can we conclude?

Remark: Any idea for a better title? The problem is occuring in the infinitesimal characterization of invariant distributions for diffusion processes on $\mathbb R$.

  • $\begingroup$ Reference for density: Corollary 1, page 159, in Treves book. $\endgroup$ – Pedro Dec 26 '18 at 17:52
  • $\begingroup$ I could be wrong here but I believe you might be missing the hypothesis: $\mathcal{C}_c^\infty(\mathbb{R})$ is a core for the operator $A$ which would allow you to conclude. $\endgroup$ – Michh Dec 28 '18 at 3:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.