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 $\mathcal{L}_{p} = \mathcal{L}_{p,p}$ ?

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    $\begingroup$ 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. $\endgroup$ – Giorgio Metafune Mar 6 at 10:00
  • $\begingroup$ @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 $\endgroup$ – user69642 Mar 6 at 10:11
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    $\begingroup$ 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. $\endgroup$ – Giorgio Metafune Mar 6 at 10:24

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