Spectral Measures: Support vs. Spectrum

Question

Given a Hilbert space $\mathcal{H}$.

Consider a normal operator: $$N:\mathcal{D}(N)\to\mathcal{H}:\quad N^*N=NN^*$$

And its spectral measure: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad N=\int\lambda\mathrm{d}E(\lambda)$$

Then one has: $$\sigma(T)=\operatorname{supp}E$$

How can I prove this?

Answer 1

By measurable calculus: $$\lambda\in\rho(N)\iff(\lambda-N)^{-1}\in\mathcal{B}(\mathcal{H})\iff\|\frac{1}{\lambda-\mathrm{id}}\|_\infty<\infty$$ $$\|\frac{1}{\lambda-\mathrm{id}}\|_\infty<\infty\iff E(\lambda-B_{R_0})=0\iff\lambda\in\operatorname{supp}E$$

Concluding the assertion.

Answer 2

Suppose that $t_0\in \operatorname{supp}E$, that is $E(B(t_0,\varepsilon))\neq0, \forall\varepsilon>0$. Then, there exists a normalized sequence $(x_j)$, with $$x_j\in E(B(t_0,1/j))\mathcal{H}, \qquad \forall j\in\mathbb{N}.$$ Hence $E(B(t_0,1/j))x_j=x_j$, and it follows that $\mu_{x_j}^T\left(\Bbb{C}\backslash B(t_0,1/j)\right)=0$. Thus $$\|(T-t_0)x_j\|^2=\int(t-t_0)^2d\mu_{x_j}^T(t)\leq\frac{1}{j^2}\|x_j\|^2=\frac{1}{j^2}\to0,$$ so $(x_j)$ is a Weyl sequence at $t_0$ for $T$, and $t_0\in \sigma(T)$.

Answer 3

I'm going to assume the facts that I proved in a previous answer to a question of yours: Spectral Measures: Integration

For any Borel function $f$, I defined an operator $T_{f}$ on $\mathcal{D}(T_{f})$ consisting of all $x \in \mathcal{H}$ for which $\int |f(\lambda)|^{2}\,d\|E(\lambda)x\|^{2}<\infty$. I showed that $T_{f}$ is a closed densely-defined linear operator. For $x \in \mathcal{D}(T_{f})$, $T_{f}x$ is defined as the unique vector such that $$ \int f(\lambda)d(E(\lambda)x,y)=(T_{f}x,y),\;\;\; y \in \mathcal{H}. $$ It was found that $$ \|T_{f}x\|^{2} = \int |f(\lambda)|^{2}\,d\|E(\lambda)x\|^{2},\;\;\; x\in\mathcal{D}(T_{f}). $$ Therefore, if $g$ is a bounded Borel function on $\mbox{supp}E$, then $T_{g}\in\mathcal{B}(\mathcal{H})$ with $\|T_{g}\| \le \sup_{\lambda\in\mbox{supp}E}|g(\lambda)|$.

Algebraic homomorphism: Suppose $f$ is a Borel function and $x \in \mathcal{D}(T_{f})$. If $S$ is a Borel subset, then $E(S)x \in \mathcal{D}(T_{f})$ because $$ \int |f(\lambda)|^{2}d\|E(\lambda)E(S)x\|^{2}=\int_{S}|f(\lambda)|^{2}d\|E(\lambda)x\|^{2} \le \|T_{f}x\|^{2} < \infty. $$ Furthermore, for $x \in \mathcal{D}(T_{f})$ and $y\in\mathcal{H}$, $$ (T_{f}E(S)x,y) = \int f(\lambda)d(E(\lambda)E(S)x,y) \\ = \int f(\lambda)\chi_{S}(\lambda)d(E(\lambda)x,y) = (T_{\chi_{S}f}x,y) $$ Thus, $T_{f}E(S)=T_{\chi_{S}f}$ holds on $\mathcal{D}(T_{f})$. In fact, $$ E(S)T_{f}x=T_{f}E(S)x = T_{\chi_{S}f}x,\;\;\; x \in \mathcal{D}(T_{f}). $$ If $g$ is a bounded Borel function then $T_{g}$ can be approximated in the norm of $\mathcal{B}(\mathcal{H})$ by functions of the form $\sum_{j=1}^{n}\lambda_{j}E(S_{j})$. Using the above and the fact that $T_{f}$ is closed then implies $T_{g} : \mathcal{D}(T_{f})\subseteq \mathcal{D}(T_{f})$ and $$ T_{f}T_{g}x = T_{g}T_{f}x= T_{fg}x,\;\;\; x\in\mathcal{D}(T_{f}). $$ This is enough to get what you want.

Spectrum of $T_{z} = \mbox{supp}E$: If $\lambda \notin\mbox{supp}E$, then there exists an closed disk $D_{r}[\lambda]$ with $r > 0$ such that $D_{r}[\lambda]\cap \mbox{supp}E=\emptyset$. Then $$ T_{1/(z-\lambda)}=\int \frac{1}{z-\lambda}dE(z) $$ is in $\mathcal{B}(\mathcal{H})$ with $\|T_{1/(z-\lambda)}\| \le 1/r$. Furthermore, the following holds on $\mathcal{D}(T_{f})$: $$ (T_{z}-\lambda I)T_{1/(z-\lambda)}=T_{1/(z-\lambda I)}(T_{z}-\lambda)=I. $$ Using the fact that $T_{z}$ is closed leads to the conclusion that $\lambda$ is in the resolvent set $\rho(T_{z})$ and $(T_{z}-\lambda I)=T_{1/(z-\lambda)}$. On the other hand, suppose $\lambda \in \mbox{supp}(T)$. Define $\chi_{n}$ to be the characteristic function of $D_{1/n}(\lambda)$. Then there exists $x_{n} \ne 0$ such that $T_{\chi_{n}}x_{n}=x_{n}$, which leads to $$ \|(T_{z}-\lambda I)x_{n}\|=\|T_{(z-\lambda)\chi_{n}}x_{n}\|\le \frac{1}{n}\|x_{n}\|. $$ This last equality guarantees that $T_{z}-\lambda I$ cannot have a bounded inverse. Hence, $\mbox{supp}E \subseteq \sigma(T_{z})$. Finally, $$ \sigma(T_{z}) = \mbox{supp}E. $$