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