we denote $\int (d\omega/d\psi) d\psi = \omega(1)$. For an arbitrary $1 \leq p<\infty$, $L^p({\mathcal{L}}G)$ is defined as the set of closed densily defined operators $T$ with polar decomposition $T=u|T|$ satisfying $u \in {\mathcal{L}}G$, $|T|^p \in L^1({\mathcal{L}}G)$, and one denotes $\|T\|_{L^p({\mathcal{L}}G)} = (\int |T|^p d\psi)^{1/p}$. In general the non-zero elements of $L^p({\mathcal{L}}G)$ are not bounded operators. Hilsum [@Hilsum] proved that the sum of two elements of $L^p({\mathcal{L}}G)$ is densely defined, closable, that its closure belongs to $L^p({\mathcal{L}}G)$, and that for this linear structure if $p\geq 1$, $L^p({\mathcal{L}}G)$ is a Banach space for the norm $\| \cdot \|_{L^p({\mathcal{L}}G)}$. Moreover, for $0\leq p,q,r \leq \infty$ with $1/r=1/p+1/q$ and $a \in L^p({\mathcal{L}}G)$ and $b \in L^q({\mathcal{L}}G)$, $ab$ is closable and its closure (still denoted by $ab$) belongs to $L^r({\mathcal{L}}G)$ and this product is associative. Lastly, if $r=1$, $\int a b d\psi=\int b a d\psi$, and if $p\neq \infty$ the pairing $\langle a,b\rangle=\int a b d\psi$ realizes $L^q({\mathcal{L}}G)$ isometrically as the dual of $L^p({\mathcal{L}}G)$.

Apart from what we just recalled, we will use some facts from [@Hilsum Proposition 11] that we collect in the following proposition.

\[prop=reminders\_hilsum\] We have the following properties.

1.  We have $u D(L^2G,\psi)\subset D(L^2G,\psi)$ for every $u \in {\mathcal{L}}G$.

2.  If 