Spectral measures 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?
 A: 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 $\varepsilon$.
A: 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, $E(\sigma(A))=I$.
