Let $\mathcal{H}$ be a Hilbert space, and let $\Delta$ be a positive, self-adjoint, unbounded operator. I think I can prove the following statement, but I find it quite surprising, and I can't find it written down anywhere:
The set of vectors {$x$ | $x$ is in the domain of $\Delta^{s}$ for every $s > 0$} is dense in $\mathcal{H}.$
Obviously each $\Delta^s$ is densely defined, but the domains shrink as you increase $s,$ and I wouldn't expect the intersection of all the domains to still be dense. Is this true? My suggested proof is as follows, but it gets at the result from an unexpected path, so I want to make sure I'm not missing something.
One can prove, as in the analytic proof of Stone's theorem (cf. Stratila-Zsido "Lecture on von Neumann Algebras" chapter 9), that $x$ is in the domain of $\Delta$ if and only if the map $i t \mapsto \Delta^{it} x$ has an analytic continuation to the strip $0 \leq \operatorname{Re}(z) \leq 1.$ One can also show (see e.g. Stratila-Zsido 9.17) that if $U(t)$ is a strongly continuous unitary group, then there is a dense set of vectors $x$ such that $i t \mapsto U(t) x$ admits an entire analytic continuation. So I should be able to conclude that:
- There is a dense set of vectors $x$ such that the map $i t \mapsto \Delta^{it} x$ has an entire analytic continuation.
- Any such vector $x$ is in the domain of $\Delta^{s}$ for any $s > 0.$
- QED.