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I'm currently studying Stone's theorem on one-parameter unitary groups (see https://en.wikipedia.org/wiki/Stone%27s_theorem_on_one-parameter_unitary_groups). I was asking myself if I drop the hypothesis of the one parameter group $\{U_t\}_{t \in \mathbb{R}}$ of unitaries to be strongly continuous, can I find an explicit counterexample that $\{U_t\}_t$ is not of the form $e^{itA}$ for some self-adjoint operator $A$?

Stefani Joanne
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  • Yes, you can take any non-linear solution of the functional equation $f(s+t)=f(s)+f(t)$ (see here: https://math.stackexchange.com/questions/385586/do-there-exist-functions-satisfying-fxy-fxfy-that-arent-linear) and define $U_t=e^{if(t)} I$, where $I$ is the identity operator. – MaoWao Feb 07 '25 at 15:59
  • I'm pretty sure this question has been asked and answered before on this site, but I could not find a suitable duplicate target. Maybe you can search the site yourself to see if something turns up. – MaoWao Feb 07 '25 at 16:07
  • @MaoWao, I tried to search some old posts here before post the question, but I also did not find anything. – Stefani Joanne Feb 07 '25 at 16:39
  • Yes, the site search does not work particularly well. Anyway, do you understand the answer from my first comment? – MaoWao Feb 07 '25 at 17:08
  • Yes, I appreciate it! Although I think the construction of such a function f is very artificial. – Stefani Joanne Feb 07 '25 at 17:31
  • It is, and necessarily so. If you only assume that $t\mapsto \langle\xi,U_t\eta\rangle$ is measurable for all $\xi,\eta\in H$, then $(U_t)$ is already strongly continuous. So any counterexample will be somewhat artificial (and require the axiom of choice in a way). – MaoWao Feb 07 '25 at 17:34

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