Let $E:\Sigma\to\mathcal{L}(\mathcal{H})$ be a spectral measure on the Borel $\sigma$-algebra $\Sigma$ of $\mathbb{C}$. Assume also that $E$ is compactly supported in the sense that $E(K)=\operatorname{id}_\mathcal{H}$ for some compact $K\subset\mathbb{C}$, so that $$A:=\int\lambda dE$$ is a well-defined, bounded normal operator on $\mathcal{H}$. Do you know a nice proof for the fact, that $E(\operatorname{spec}A)=\operatorname{id}_\mathcal{H}$, which (of course) does not use the spectral theorem?
Spectral measures
5
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operator-algebras
spectral-theory
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0Errh... Why do you want to avoid the spectral theorem? – 2012-12-14
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0Because I am actually working on a proof for the spectral theorem. Maybe I should be more precise here. What we can use is the continuous functional calculus (including the spectral mapping theorem) as well as the bounded Borel functional calculus for bounded normal operators. – 2012-12-14