Spectral mapping theorem of the measurable functional calculus

Question

Let $H$ be a hilbert space and $T\in L(H)$ a self adjoint operator.

Show that we have in general $\sigma(f(T))\neq f(\sigma(T))$

Any tips?

If I choose a self adjoint operator how the measurable functional calculus looks like?

Answer 1

Just choose the simplest setting, i.e. the setting of a multiplication operator.

Choose $H=L^2((0,1))$ (with Lebesgue measure),

$$ Tg(x)= x \cdot g(x) $$

and $f(x) = 1$ for $x\neq 1/2$ as well as $f(1/2)=42$.

It is not hard to see that

$$ f(T)g( x) = f(x) g( x) = g( x), $$ where the last equality holds a.e.

Hence, $\sigma(T)=[0,1]$, $\sigma(f(T))={1}$, but $f(\sigma(T))=\{1,42\}$.

EDIT: Further explanation regarding the measurable functional calculus: If $T$ is a multiplication operator on some $L^2$ space as above (multiply with $h$), then $f(T)$ is multiplication with $f\circ h$.

In the general case, the spectral theorem yields a unitary operator $U:H \to L^2(\mu)$ with $T = U^\ast M U$, where $M$ is a multiplication operator as above. Then $f(T) = U^\ast N U$, where $N$ corresponds to the modified multiplication operator (as above).