# Clarification about extensions of Ornstein-Uhlenbeck operator

I am reading stuffs regarding the Ornstein-Uhlenbeck operator and its various extensions to $$L^p(\gamma)$$, with $$p \in (1,+\infty)$$ and with $$\gamma$$ the standard Gaussian measure on $$\mathbb{R}^d$$. On $$\mathcal{C}_c^\infty(\mathbb{R}^d)$$, this second order differential operator reads as \begin{align*} \mathcal{L}(f)(x) = \Delta(f)(x) - \langle x; \nabla(f)(x) \rangle. \end{align*} Now, I can extend it to a closed densely defined linear operator on $$L^p(\gamma)$$ at least in two ways: through the integration by parts formula; for all $$f,g \in \mathcal{C}_c^{\infty}(\mathbb{R}^d)$$, \begin{align*} \int_{\mathbb{R}^d}(-\mathcal{L}(f)(x))g(x)\gamma(dx) = \int_{\mathbb{R}^d} \langle \nabla(f)(x) ; \nabla(g)(x) \rangle \gamma(dx). \end{align*} Indeed, thanks to it, the operator is closable on $$L^p(\gamma)$$ so that it admits a minimal closed extension denoted by $$(\mathcal{L}_{p,p},\mathcal{D}(\mathcal{L}_{p,p}))$$. Now, I can see the operator $$\mathcal{L}$$ as the restriction on $$\mathcal{C}_c^\infty(\mathbb{R}^d)$$ of the $$L^p(\gamma)$$-generator of the Ornstein-Uhlenbeck semigroup which is a contraction Markovian $$C_0$$-semigroup of operators on $$L^p(\gamma)$$. Let's denote it $$(\mathcal{L}_{p},\mathcal{D}(\mathcal{L}_{p}))$$.

Question: Do we have $$\mathcal{L}_{p} = \mathcal{L}_{p,p}$$ ?

• Yes, they are the same. The reason is that $C_c^\infty(R^d)$ is a core for ${\cal L}_p$. I can give a reference if you don't find it. – Giorgio Metafune Mar 6 at 10:00
• @Giorgio: thank you for your comment. Indeed, once you have identified the domain of $\mathcal{L}_p$ as the Sobolev space $W^{2,p}(\gamma)$ and prove that $\mathcal{C}_c^{\infty}(\mathbb{R})$ is dense in $W^{2,p}(\gamma)$, you are done. I guess it is contained in your papers '02 regarding OU. Thanks again – user69642 Mar 6 at 10:11
• Yes true, but you can check that is a core directly: the Schwartz class is a core, since it is invariant under the semigroup and approximating a function in the Schwartz class with functions with compact support, in the graph norm, is not a problem. – Giorgio Metafune Mar 6 at 10:24