# Spectrum included in spectral measure support

Let $\mathcal{H}$ be an Hilbert space and $A=\displaystyle{\int} \lambda \,dE(\lambda)$ a spectral operator on $\mathcal{H}$ associated to the spectral measure $E$. I am interested in the very basic property: $$\operatorname{supp}(E)=\sigma(A)$$ where $\operatorname{supp}(E)$ is the support of $E$ and $\sigma(A)$ the spectrum of $A$. In this discussion it has been shown by Mateus Sampaio that $\operatorname{supp}(E) \subset \sigma(A)$.

In the case of a bounded operator the other inclusion can be shown as follow:

For $\zeta \in \mathbb{C} \setminus \operatorname{supp}(E), \lambda \mapsto \frac{1}{\lambda-\zeta}$ is bounded on $\operatorname{supp}(E)$ and we can define the resolvent: $R(\zeta)=\displaystyle{\int} \frac{1}{\lambda-\zeta} \,dE(\lambda)$ which is a bounded operator. Hence $\mathbb{C} \setminus \operatorname{supp}(E) \subset \rho(A)$ and $\sigma(A) \subset \operatorname{supp}(E)$. $\blacksquare$

Does anyone knows how to generalize to unbounded operator ?

## 1 Answer

Note that every spectral measure $E$ defines a resolution of the identity $E: \mathbb{R} \to L(H)$ by $E_\lambda := E((-\infty, \lambda])$. Then we have $A = \int_\mathbb{R} \lambda dE_\lambda$. Let $\mu \in \sigma(A)$. For every $\varepsilon > 0$ we have $E_{\mu + \varepsilon} - E_{\mu - \varepsilon} \neq 0$:
Assume there is an $\varepsilon > 0$ which satisfies $E_{\mu + \varepsilon} - E_{\mu - \varepsilon} = 0$. This yields $\Vert (A - \mu)x\Vert^2 = \int (t - \mu)^2 d \rho_x(t)$, where $\rho_x$ is the induced Stieltjes-meausure of $t \mapsto \langle x, E(t)x \rangle = \Vert E_tx\Vert^2$.
Then we have $\rho_x(\{ t \in \mathbb{R} \colon (\mu - t)^2 < \varepsilon^2 \}) \leq \rho_x((\mu - \varepsilon, \mu + \varepsilon]) = \Vert E_{\mu + \varepsilon}x \Vert ^2 - \Vert E_{\mu - \varepsilon}x \Vert^2 = 0$. In other words $(t - \mu)^2 \geq \varepsilon^2$ is valid for $\rho_x$-a.e. $t \in \mathbb{R}$. This means
$\Vert (A - \mu)x\Vert^2 = \int (t - \mu)^2 d \rho_x(t) \geq \varepsilon^2 \Vert x \Vert ^2$ for every $x \in D(A)$ which implies $(A - \mu)$ is continously invertable, a contradiction to $\mu \in \sigma(A)$.

Why is $\mu \in \rm{supp}(E)$? Let $U \subset \mathbb{C}$ open with $\mu \in U$. Then there is an $\varepsilon > 0$ with $(\mu - \varepsilon, \mu + \varepsilon] \subset U$. Now $E(U) = E\left((\mu - \varepsilon, \mu + \varepsilon] \dot\cup (U \setminus (\mu - \varepsilon, \mu + \varepsilon])\right) = E((\mu - \varepsilon, \mu + \varepsilon]) + E\left( U \setminus (\mu - \varepsilon, \mu + \varepsilon] \right) \neq 0$ and therefore $\mu \in \rm{supp}(E)$.