Let $\mathcal H = L^2 [0,1],$ let $T$ be the operator $Tf(x) = x f(x).$ Let $\{e_n\ |\ n \in \mathbb N \}$ be an orthonormal basis for $\mathcal H.$ Let $D = \{r_n\ |\ n \in \mathbb N \}$ be the set of all rationals in $(0,1),$ and let $S$ be the operator on $\mathcal H$ given by $S e_n = r_n e_n.$ Show that
$(a)$ $\sigma (T) = \sigma (S).$
$(b)$ Let $\phi$ and $\psi$ be the $C^{\ast}$-homomorphism from $C \left (\sigma (T) \right ) = C \left (\sigma (S) \right )$ to $\mathcal L (\mathcal H)$ given by the continuous functional calculus for $T$ and $S$ respectively. Show that $\phi$ and $\psi$ are not equivalent i.e. there doesn't exist any unitary operator $U \in \mathcal L (\mathcal H)$ such that $\phi (f) = U \psi (f) U^{\ast}$ for all $f.$
I have shown $(a)$ but I have no idea how to approach $(b).$ Could anyone please give me some hint in this regard?
Thanks for your time.
It is enough for the equality to fail at one $f$. You can take $f$ to be the identity, and that's the point of part a), as the problem becomes to show that $T$ and $S$ are not unitarily equivalent. For this, take any rational $\lambda$ and show that it is an eigenvalue for $S$ but not for $T$.