# Measurable functional calculus

I am struggeling with this exercise:

Let $T \in L(H)$ be a self-adjoint operator and $\Psi$ be a measurable (Borel) functional calculus on the spectrum of $T$. For a Borel set $\Delta \subset \sigma (T)$, we have an orthogonal projection $E_{\Delta} := \Psi(1_{\Delta})$, where $1_A$ is the indicator function on set $A$. Now the restriction of $T$ given by $T|_{ran(E_{\Delta})}$ is a self-adjoint operator too. Now I want to show that $\sigma(T|_{ran(E_{\Delta})}) \subset \overline{\Delta}$.

There is also a hint: I shall assume that $\lambda \notin \overline{\Delta}$ and show that $\lambda \in \rho(T|_{ran(E_{\Delta})})$. To do this, I shall consider the function $f(t) = t 1_{\Delta} (t) + \lambda_0 1_{\Delta^C}(t)$ first for some $\lambda_0 \in \Delta$ and show $\lambda \in \rho(f(T))$. From this, I should conclude that $\lambda \in \rho(T)$.

My ideas and questions:

Is it correct that $\Psi(f) = T E_{\Delta} + \lambda_0 E_{\Delta^C} = T|_{ran(E_{\Delta})} + \lambda_0 E_{\Delta^C}$?

I guess $E_{\Delta^C}$ is the projection on the complement of $ran(E_{\Delta})$. This should follow from $\Psi(1) = \Psi( 1_{\Delta} + 1_{\Delta^C}) = E_\Delta + E_{\Delta^C}. What kind of$\lambda_0$should I consider and where can I use that we are looking at the closure of$\Delta$? Some hints or a solution to this excercise would be highly appreciated. • For$x$in the range of$E(\Delta)$, you have that$\|(T-\lambda)x\|^2 = \int_{\Delta} |t-\lambda|^2\, d\|E(t)x\|^2$, which shows you that$\lambda\notin\overline{\Delta}$is not in the spectrum. – user138530 Oct 26 '14 at 2:10 • @ChristianRemling: "element of the spectrum"$\ne$eigenvalue. – Martin Argerami Oct 26 '14 at 3:00 • @MartinArgerami: I am well aware of that. – user138530 Oct 26 '14 at 3:04 • Then maybe I don't understand how your equality would prove that if$\lambda\not\in\bar\Delta$then$\lambda $is not in the spectrum. – Martin Argerami Oct 26 '14 at 3:34 • @MartinArgerami: If$\lambda\notin\overline{\Delta}$, then$|t-\lambda|\ge\delta>0$for all$t\in\Delta$, so$\|(T-\lambda)x\|\ge \delta \|x\|$and$T-\lambda$on$R(E(\Delta))$is boundedly invertible. – user138530 Oct 26 '14 at 5:21 ## 1 Answer You probably do not mean the operator$TE_{\Delta}$because this operator has a spectrum which now includes$0$unless$E_{\Delta} = I$. I assume that you want to deal with the restriction$T_{\Delta}$of$T$to the range of$E_{\Delta}$as an operator on the Hilbert space$\mathcal{H}_{\Delta}=E_{\Delta}\mathcal{H}$under the induced norm; this makes sense because$\mathcal{H}_{\Delta}$is invariant under$T$. And it also makes sense because you don't pick up anything unexpected in the spectrum with this interpretation. If$\lambda\notin\Delta^{c}$then$r_{\lambda}(x)=\frac{1}{x-\lambda}1_{\Delta}$is a bounded Borel function on$\mathbb{R}$and, therefore,$\Phi(r_{\lambda})$is a bounded linear operator with$\mathcal{H}_{\Delta}$as an invariant subspace. Using the functional calculus gives $$\Psi(1_{\Delta})=\Psi(r_{\lambda})\Psi((x-\lambda)1_{\Delta}) =\Psi((x-\lambda)1_{\Delta})\Psi(r_{\lambda}).$$ When considered on$\mathcal{H}_{\Delta}$, the above becomes $$I=R_{\lambda}(T_{\Delta}-\lambda I)=(T_{\Delta}-\lambda I)R_{\lambda},$$ where$R_{\lambda}$is the restriction of$\Psi(r_{\lambda})$to$\mathcal{H}_{\Delta}$. This proves that$\lambda\in\rho(T_{\Delta})$whenever$\lambda\notin\Delta^{c}$, which is what you wanted to prove. In other words$\mathbb{C}\setminus\Delta^{c}\subseteq\rho(T_{\Delta})$or, equivalently,$\sigma(T_{\Delta})\subseteq\Delta^{c}\$.

• Comments are not for extended discussion; this conversation has been moved to chat. – user642796 Oct 27 '14 at 5:32