I am wondering about what can be said about the spectral theorem for unbounded, self-adjoint operators in a non-separable Hilbert space. There is a comment in this sense to the question "Does spectral theory assume separability", but for compact normal operators. It says that "the space splits into a direct sum of two spaces, one on which the operator vanishes and a separable one which is spanned by the eigenvectors". Does something similar happen for SA unbounded operators? I would greatly appreciate indications for references on this topic.
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7$\begingroup$ 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 $\ell^2(\mathbf R)$). $\endgroup$– Francois ZieglerCommented Sep 27, 2019 at 3:58
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1$\begingroup$ 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. $\endgroup$– Robert FurberCommented Sep 27, 2019 at 16:56
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$\begingroup$ 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. $\endgroup$– sbisafCommented Sep 28, 2019 at 22:55
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2 Answers
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As mentioned above, it‘s not entirely clear what you want. Here are two suggestions
1) the space can be split into a direct sum of separable closed subspaces, each of which is invariant under the given operator.
2) the space can be split as the direct sum of closed invariant subspaces on each of which the operator is bounded.
There are many ways of stating the spectral theorem and these results allow one to deduce them for the non-separable case from the separable or bounded one. The non-separability isn‘t really an issue.
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$\begingroup$ Thanks @user131781. I agree I didn't express precisely my question. I actually had in mind what you suggest as point 1). The main point was to understand what can be said about the behavior of the operator in the subspace to which the eigenvectors do not belong. I was wondering for which class of operators the property of vanishing on that subspace holds, if it is wider than the class of compact normal operators of not. Your answer was valuable to me. $\endgroup$– sbisafCommented Sep 28, 2019 at 23:02
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I agree with user131781, but wanted to add that there is a strong form of the spectral theorem which requires separability. Namely: if $A \in B(H)$ is self-adjoint then there is a probability measure $\mu$ on ${\rm sp}(A)$ and a measurable bundle $\mathcal{H}$ of Hilbert spaces over ${\rm sp}(A)$, such that $H$ can be identified with $L^2({\rm sp}(A), \mathcal{H})$ in a way that takes $A$ to multiplication by $x$.
I like this version because it not only "diagonalizes" $A$, it tells you very explicitly how $A$ sits inside of $B(H)$. This can be useful.
It utterly fails in the nonseparable setting because you can have things like multiplication by $x$ acting on $L^2[0,1]\oplus l^2[0,1]$.
answered Sep 27, 2019 at 13:07
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$\begingroup$ A reference: Theorem 3.5.1 of my book Mathematical Quantization. $\endgroup$ Commented Sep 27, 2019 at 13:08
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$\begingroup$ Thanks @Nik. Your comments was helpful to me. Thanks also for the reference. $\endgroup$– sbisafCommented Sep 28, 2019 at 23:05