Some doubts concerning spectral theory.

Question

Probably I'm saying something wrong (that's why the conclusions are strange) so please correct me!

There is the continuous functional calculus for a normal element $N$ of a C*-Algebra. This means that there is a $*$-isomorphism between $C^\ast(N)$, the C*-Algebra generated by N, and $C(\sigma(N))$, the continuous functions on the spectrum of $N$.

There is also the Borel functional calculus for a normal element N that extends the continuous one. And the range of $\phi(N)$ for $\phi$ Borel function on $\sigma(N)$ is also $C^\ast(N)$ and there is a *-isomorphism between $B(\sigma(N))$ and $C^\ast(N)$. (Probably the error is here somewhere but I don't realize it totally).

Then if what is written up here is true that means that there is a *-isomorphism between $C(\sigma(N))$ and $B(\sigma(N))$ and this is quite absurd.

Furthermore I would ask if it's true or not that $W^\ast(N)$, the Von neumann algebra generated by N, contains $C^\ast(N)$, the C*-algebra generated by N. Intuitively I would say yes, because the C*-algebra is closed only in the norm topology and the Von Neuman algebra is closed in the WOT topology.

Thank you!

Answer 1

No, the range for Borel functional calculus is not $C^*(N)$, in fact in general it takes you to the strong operator closure of $C^*(N)$.

Answer 2

Robert already answered your question but let me give an extra illustration.

Think of a normal operator with spectrum $[0,1]$. So you are in $C[0,1]$. Now you can compose functions from $C[0,1]$ with Borel functions on $C[0,1]$. For example, take your favourite discontinuous Borel function $f$. Then $ f = f\circ {\rm id}_{[0,1]}$. This takes you out of $C[0,1]$.

What I am saying is rather trivial but my point is that you should always look at the commutative situation first.

By the way, note that $C(\sigma(N))$ and $B(\sigma(N))$ are not even isomorphic as Banach spaces unless $N$ is finite. Indeed, the former space is separable, whereas the latter one is not.