Spectral Measures: Support vs. Spectrum

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?

• This is true in general. What proof of this fact do you know for bounded self-adjoint operators? Probably it's possible to adapt such a proof for the general case. – Mateus Sampaio Oct 15 '14 at 22:36
• I only know the one via the Neumann series for normal elements of a Banach algebra which applies to bounded normal operators... – C-Star-W-Star Oct 15 '14 at 23:23
• Check my answer for a general proof then. – Mateus Sampaio Oct 15 '14 at 23:25

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, $$\sigma(T_{z}) = \mbox{supp}E.$$

• I posted an answer myself - I really hope you don't mind - especially as you put so much effort in yours. (+1) – C-Star-W-Star Oct 21 '14 at 17:11
• @Freeze_S : I don't mind. I wanted to post something where all of the i's are dotted and t's are crossed because of the many subtle issues concerning unbounded operators and their dense domains. It helps to see these things if you're relatively new to the subject. – DisintegratingByParts Oct 21 '14 at 17:20
• Yeah also for next visitors. I'm pretty sure you will earn more thumbs up. ;) – C-Star-W-Star Oct 21 '14 at 17:21

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 $t_0\in \sigma(T)$.

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.

• does $||\frac{1}{1-id}||_\infty$ mean $\frac{1}{dist(\lambda,\sigma(N))}$? What does $E(\lambda - B_{R_0})$ mean and why is equality with 0 equivalent to the previous function being bounded? thanks! – abe.nong Mar 7 '16 at 7:32