# why is this a sufficient condition for a domain to be a core of an unbounded operator?

Let $\alpha:\mathbb R\to U(H)$ be a strongly continuous action of the reals on some Hilbert space, and let $A=-i\frac d{dt}\alpha(t)|_{t=0}$ be its infinitesimal generator, so that $\alpha(t)=e^{itA}$. Finally, let $D\subset H$ be the domain of $A$.

A subset $D_0\subset D$ is called a core of $A$ if the closure of $A|_{D_0}$ is $A$.

I've read the following statement in some paper, without reference or explanation, and I'd like to know why it's true:

If $D_0$ is dense in $H$ and invariant under $\alpha$, then it is a core of $A$.

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## 2 Answers

This is stated and proved as Proposition 2.20 here: https://isem-mathematik.uibk.ac.at/isemwiki/images/9/94/ISEM15_Lecture2.pdf

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Great! Having read the proof, I'd like to make the following two remarks to help other readers. (1) The fact that $H$ was a Hilbert space is irrelevant, the proof works equally well for Banach spaces (2) The fact that the operators $\alpha(t)$ are invertible is also not used, it is enough for $\alpha$ to be defined on the semigroup of positive reals $\mathbb R_{\ge 0}$. –  André Henriques Sep 9 '13 at 10:29
I also found this reference useful: fuw.edu.pl/~derezins/mat-u.pdf –  André Henriques Oct 5 '13 at 23:48

See also Theorem VIII.11 in Reed-Simon (p.269 vol. 1).

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