2

Let $\mathcal{H}$ be a Hilbert space and $E:\Sigma\to\mathcal{L}(\mathcal{H})$ be a compactly supported spectral on the Borel $\sigma$-algebra $\Sigma$ of $\mathbb{C}$. Then we can form the bounded, normal operator $$A=\int \operatorname{id}_\mathbb{C}\;dE\in\mathcal{L}(\mathcal{H})$$ Do you know a proof for the fact, that $E(\operatorname{spec} A)=\operatorname{id}_H$?

flag
By "spectral" do you mean a projection-valued measure? Because if so, then doesn't this hold just by definition? – Branimir Ćaćić Dec 20 at 12:32
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 $E(spec A)=id_H$ – Robert Rauch Dec 20 at 13:23
1 
 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/…, namely, that the support of a compactly-supported projection-valued measure is one and the same as the spectrum of the associated normal operator. – Branimir Ćaćić Dec 20 at 14:15
1 
 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! – Robert Rauch Dec 20 at 15:26

Your Answer

Get an OpenID
or

Browse other questions tagged or ask your own question.