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Sep 28, 2019 at 22:55 comment added sbisaf I acknowledge your comments. I agree my question was not clearly expressed. The example of Francois is very fitting and both comments were clarifying to me. Although I didn't express it precisely, my concern was with the behavior of the operator outside the the subspace spanned by the eigenvectors. I understand that there is no universal behavior under the only assumption of self-adjointness.
Sep 27, 2019 at 16:56 comment added Robert Furber In the formulation of the spectral theorem using projection-valued measures, there's no difference in the statement between separable and inseparable Hilbert spaces, for example.
Sep 27, 2019 at 13:07 answer added Nik Weaver timeline score: 2
Sep 27, 2019 at 9:25 answer added user131781 timeline score: 2
Sep 27, 2019 at 4:15 review Close votes
Oct 1, 2019 at 2:37
Sep 27, 2019 at 3:58 comment added Francois Ziegler Clarification needed on what “something similar” means. The spectral theorem doesn’t require separability, and the property in quotes fails in general (consider multiplication by $x\mapsto x$ in $\ell^2(\mathbf R)$).
Sep 27, 2019 at 3:05 review First posts
Sep 27, 2019 at 3:08
Sep 27, 2019 at 3:02 history asked sbisaf CC BY-SA 4.0