# Spectral Measures: Concentration

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

Consider spectral measures: $$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad E(\mathbb{C})=1$$

Define its support: $$\operatorname{supp}(E):=\bigg(\bigcup_{U=\mathring{U}:E(U)=0}U\bigg)^\complement=\bigcap_{C=\overline{C}:E(C)=1}C$$

By second countability: $$E\bigg(\operatorname{supp}E^\complement\bigg)\varphi=E\left(\bigcup_{k=1}^\infty B_k'\right)\varphi=\sum_{k=1}^\infty E(B_k')\varphi=0$$

But it may happen: $$\Omega\subsetneq\operatorname{supp}E:\quad E(\Omega)=E(\operatorname{supp}E)=1$$

What is an example?

• If the spectral measure is defined on $\mathbb{C}$, then the spectrum of $\int \lambda dE(\lambda)$ is the the support of $E$. For $\mu \notin\mbox{supp} E$ you can write down the resolvent $(\int \lambda dE(\lambda)-\mu I)^{-1}=\int(\lambda-\mu)^{-1}dE(\lambda)$ Commented Sep 22, 2014 at 22:00
• @T.A.E.: Yepp... and $\lambda$ is an eigenvalue iff $E(\{\lambda\})\neq 0$. So can one construct an example with $\mathrm{supp}E=[0,1]$ but also $E((0,1])=1$. For nonseparable Hilbert spaces I have this construction in mind by just "filling up" with eigenvalues for all $z\in(0,1]$. But I cannot imagine yet what else would work... Commented Sep 22, 2014 at 22:35
• @Freeze_S Why does the example I give below not work? Commented Sep 23, 2014 at 7:29
• @user161825: Ah right yes that answers my question. Commented Sep 23, 2014 at 8:23

Consider the operator $(Af)(x)=xf(x)$ on the Hilbert space $L^2([0,1])$ with Lebesgue measure. Then $E_A(M)=1_M$ is the spectral measure associated with $A$. Note that $\mbox{supp }E_A=\sigma(A)=[0,1]$, but $E_A(\{\lambda\})=0$ for any singleton $\lambda$, since $A$ does not have any eigenvalues. Now consider, say, the set $(0,1]$, for which $I=E([0,1])=E(\{0\})+E((0,1])=E((0,1])$.

I'm assuming that your spectral measure is defined on Borel subsets of $\mathbb{C}$, which is the standard definition of the spectral measure which is the spectral resolution for a normal operator $T$.

You have defined the support $\mbox{supp}E$ to be the intersection of all closed subsets $C$ for which $E(C)=I$. If $E(A)=I$ for some closed subset $A\subsetneq \mbox{supp}E$, then $A$ is one of the sets in your intersection, yielding an obvious contradiction.

If $\mbox{supp}E$ is contained in some closed disk of finite radius $R$, then $T_{E}=\int \lambda dE(\lambda)$ satisfies \begin{align} \|T_{E}x\|^{2} & = \int_{\mbox{supp}E} |\lambda|^{2}d\|E(\lambda)x\|^{2} \\ & \le \int_{\mbox{supp}E}d\|E(\lambda)x\|^{2}R^{2} = \|E(\mbox{supp}E)x\|^{2}R^{2} = R^{2}\|x\|^{2}. \end{align} So $\|T_{E}\| \le \|R\|$.

• The point was rather to have a not necessarily closed $A\in\Sigma$ with $E(A)=1$ but $A\subsetneq\mathrm{supp}E$. Commented Sep 23, 2014 at 8:27
• By the way thanks for the nice quick proof of boundedness :) Commented Sep 23, 2014 at 8:29
• Yes, it's obvious that $E(\mbox{supp}E\setminus\{\lambda\})=E(\mbox{supp}E)$ for some $\lambda \in \mbox{supp}E$ iff $E(\{\lambda\})= 0$. I thought you surely must be asking something else. Commented Sep 23, 2014 at 9:01
• Yes I edited my question what I looked for is a nonclosed subset Commented Sep 23, 2014 at 9:04
• Same remark applies. $E(\mbox{supp}E\setminus F)=E(\mbox{supp}E)$ for some $F \subseteq\mbox{supp}E$ iff $E(F)=0$. Commented Sep 23, 2014 at 9:08