When two multiplication operators $M_{f}$ and $M_{g}$ acting on $L^2(X,\mu) $and $L^2(Y,\nu)$ are unitary equivalent? How multiplicity function look like here? What is the spectral multiplicity in this case and moreover is algebraic multiplicity replaced by cyclic vectors in the Hilbert space?. I know for bounded self-adjoint operators acting on $\mathcal{H}$ unitary equivalent when these have same spectrum and same multiplicity function. For an unbounded positive self-adjoint operator, can be possible to say it is unitary equivalent to unbounded function corresponding multiplication operator?
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2$\begingroup$ This is fairly basic material and not very well suited for this site. You could read about these topics in any standard textbook on the subject (such as Reed-Simon 1). This should answer your questions and clear up some of your confusion (same spectrum and multiplicity is not enough for unitary equivalence). $\endgroup$– Christian RemlingCommented May 24, 2018 at 20:16
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$\begingroup$ No problem. Actually, RS may not be the gentlest introduction to this (it was just the first book that came to my mind), for example my own lecture notes (on my homepage) might be friendlier. $\endgroup$– Christian RemlingCommented May 24, 2018 at 20:24
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$\begingroup$ Actually, I want to see spectral theorem in terms of von Neumann algebra point of view. Experiencing chapter 9 from Kadison I understand spectral theorem has something to do with commutant, strong connection with multiplicity, as spectral theorem talks about abelian von Neumann algebra, somewhat it decomposed the algebra in masa"s, can I have the better point of view from you. Please have a look at my comment. $\endgroup$– mathloverCommented May 24, 2018 at 20:24
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2$\begingroup$ Well, there's the highbrow proof of the spectral theorem through the realization of commutative $C^*$ or vN algebras (I actually do it that way in my notes), but that doesn't seem very relevant to the questions you actually asked. $\endgroup$– Christian RemlingCommented May 24, 2018 at 20:27
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