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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?

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Errh... Why do you want to avoid the spectral theorem? –  Berci Dec 14 '12 at 10:28
    
Because 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. –  Robert Rauch Dec 14 '12 at 12:12

2 Answers 2

I'm assuming that $E$ has all of the properties of a spectral measure: values commute, values are selfadjoint projections, $E(S)E(T)=E(S\cap T)$, etc.. Then it's not hard to show that the support $K$ of $E$ is $\sigma(A)$, when one defines $A=\int \lambda dE(\lambda)$. Automatically $A^{\star}=\int \overline{\lambda}dE(\lambda)$.

To show $\sigma(A) \subseteq K$: Suppose $\lambda \notin K$ with $d=\mbox{dist}(\lambda,K)$, and show $\lambda \notin \sigma(A)$. To do this, notice that, for each $x\in\mathscr{H}$, $$ \|(A-\lambda I)x\|^{2}=\int_{K}|\lambda-\mu|^{2}d\|E(\mu)x\|^{2} \ge d^{2}\|E(\mu)x\|^{2} \ge d^{2}\|x\|^{2}. $$ Likewise $\|(A^{\star}-\overline{\lambda} I)x\|\ge d\|x\|$. This is true for all $x\in\mathscr{H}$. Because $A$ is normal, then $(A-\lambda I)$ is invertible. Thus $\sigma(A)\subseteq K$.

To show $K\subseteq \sigma(A)$: Suppose $\lambda \in K$, and show $\lambda \in \sigma(A)$. Let $B_{r}(\lambda)$ be the open ball of radius $r$ centered at $\lambda$. By assumption, for each $r > 0$, $E(B_{r}(\lambda))\ne 0$, which gives the existence of unit vectors $\{ x_{n}\}_{n=1}^{\infty}$ such that $E(B_{1/n}(\lambda))x_{n}=x_{n}$. Then $$ \|(A-\lambda I)x_{n}\|^{2}=\int_{|\mu-\lambda| < 1/n}|\mu-\lambda|^{2}d\|E(\mu)x_{n}\|^{2}\le \frac{1}{n^{2}}\int d\|E(\mu)x_{n}\|^{2}=\|x_{n}\|^{2}/n^{2}. $$ $\|(A-\lambda I)x_{n}\| \le \frac{1}{n}\|x_{n}\|$. So $K\subseteq \sigma(A)$ because $A-\lambda I$ cannot have a continuous inverse. Therefore, $E(\sigma(A))=I$.

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Hint. It suffices to show that if $\lambda$ is in the resolvent set of $A$, then for some $\varepsilon>0$, $$ E\big(D(\lambda,\varepsilon)\big)=0, $$ where $D(\lambda,\varepsilon)$ is a disk centered at $\lambda$ with radius $\varepsilon$.

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