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Dec 20, 2012 at 15:26 comment added Robert Rauch Although I have already looked over your reference a couple of days ago, I did not realize that a nice answer to my question is hidden therein. Thanks a lot pointing that out!
Dec 20, 2012 at 14:15 comment added Branimir Ćaćić Yes, you're absolutely right. I imagine that the result you want is precisely the theorem on Page 7 of math.uchicago.edu/~may/VIGRE/VIGRE2006/PAPERS/RaskinSam.pdf, namely, that the support of a compactly-supported projection-valued measure is one and the same as the spectrum of the associated normal operator.
Dec 20, 2012 at 13:23 comment added Robert Rauch Yes, I mean a projection valued measure. By definition, $E(\mathbb{C})=id_H$, and also if $E$ is defined as the spectral measure associated to a given normal operator $A$, this is essentially true by definition. However, in our situation we start with a generic spectral measure $E$ from wich it is only known that there is some compact $K\subset\mathbb{C}$ such that $E(K)=id_H$ and define $A$ by the above formula. One can then show that the spectral measure associated to this $A$ is exactly $E$, but this is not obvious and actually the crucial point is to show that $E(spec A)=id_H$
Dec 20, 2012 at 12:32 comment added Branimir Ćaćić By "spectral" do you mean a projection-valued measure? Because if so, then doesn't this hold just by definition?
Dec 20, 2012 at 12:27 history asked Robert Rauch CC BY-SA 3.0