First of all I want to thank you for the help you provide on this website! Whenever I had a hard time understanding things in math I visited this website and (nearly) allways found a hint or a solution to a problem I had. But this time it seems that I have to ask a question myself, because I haven't found anything to solve a problem I found in a book I'm currently reading. Although it seems like a pretty basic question in functional calculus.
Setting of the Problem: Let $T$ be a self adjoint (unbounded) linear operator in an Hilbert space $X$ with spectral family (resolution of the identity) $E$. Let $f : \mathbb{R} \rightarrow \mathbb{C}$ be a measurable function, then $f(T)$ can be defined by the spectral theorem:
$f(T)x := \int\limits_{\mathbb{R}} f(t) dE(t)x$ for all $x \in D(f(T)) := \left\{ y \in X \Big | \int\limits_{\mathbb{R}} |f(t)|^2 d \langle y, E(t) y \rangle < \infty \right\}$
Problem: Assume $f$ is continous and real-valued, ie: $f : \mathbb{R} \rightarrow \mathbb{R}$, then there holds: $\sigma(f(T)) = \overline{f(\sigma(T))}$, where $f(A) := \left\{f(t) \Big | t \in A \right\}$ for $A \subseteq \mathbb{R}$. Moreover if $|f(t)| \rightarrow 0$ for $|t| \rightarrow \infty$ then the upper equation holds without the closure.
My attempted solution: I know from functional calculus that $\sigma(f(T)) = \left\{ \lambda \in \mathbb{C} \Big | \forall \varepsilon > 0 \text{ there holds } E \left( \left\{ t \in \mathbb{R} \Big | |f(t) - \lambda| < \varepsilon \right\} \right) \neq 0 \right\}$ and I have the feeling that this might be the solution but I can't see the relation between this and the stated problem.
I hope somebody has a solution or hint to this problem and is able to help out a poor little fellow who is relatively new to functional calculus. Thanks in advance, GordonFreeman
Ps.: I apologize for any mistakes in my english. If somethings unclear because of my phrasing, please let me know.