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.

  • $\begingroup$ 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. $\endgroup$ – user138530 Oct 26 '14 at 2:10
  • $\begingroup$ @ChristianRemling: "element of the spectrum" $\ne$ eigenvalue. $\endgroup$ – Martin Argerami Oct 26 '14 at 3:00
  • $\begingroup$ @MartinArgerami: I am well aware of that. $\endgroup$ – user138530 Oct 26 '14 at 3:04
  • $\begingroup$ 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. $\endgroup$ – Martin Argerami Oct 26 '14 at 3:34
  • $\begingroup$ @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. $\endgroup$ – user138530 Oct 26 '14 at 5:21

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, $\sigma(T_{\Delta})\subseteq\Delta^{c}$.

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

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