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4
votes
Integrating the resolvent of a self-adjoint operator across a continuous part of the spectrum
Sure, why not? I think it's a neat idea!
Probably there are lots of ways to do this. In the simplest case, where $C$ intersects the real line transversally at $a$ and $b$, I guess you could just let …
3
votes
Accepted
Does Borel functional calculus commute with *-isomorphism?
Any $*$-isomorphism between von Neumann algebra is also an order isomorphism and hence is normal. So it is automatically weak* continuous.
1
vote
An inverse to functional calculus
I think you want functions from $\mathbb{R}$ to $\mathbb{R}$, not to $\mathbb{R} \cup \{\infty\}$. Because if $f$ takes the value $\infty$ at $\lambda$ then $f(A)$ won't be self adjoint --- unless $\l …