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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 2

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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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    $\begingroup$ 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 $\mathbb R_{\ge 0}$. $\endgroup$
    – André Henriques
    Commented Sep 9, 2013 at 10:29
  • $\begingroup$ I also found this reference useful: fuw.edu.pl/~derezins/mat-u.pdf $\endgroup$
    – André Henriques
    Commented Oct 5, 2013 at 23:48
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See also Theorem VIII.11 in Reed-Simon (p.269 vol. 1).

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