spectral thoerem question

Aug 2009
122
19
Pretoria
Suppose T is a normal operator on a Hilbert space H, show that there exists a self-adjoint operator S on H such that T=f(S), where f is a continuous function form the spectrum of S into C
 

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
Suppose T is a normal operator on a Hilbert space H, show that there exists a self-adjoint operator S on H such that T=f(S), where f is a continuous function form the spectrum of S into C
This is a strange result, and I'm not convinced that it is true.

Suppose for example that \(\displaystyle H = L^2(\overline{\mathbb{D}})\), the Hilbert space of all square-integrable functions on the closed unit disk \(\displaystyle \overline{\mathbb{D}}\) in the complex plane (with respect to 2-dimensional Lebesgue measure). Let \(\displaystyle M_z\) be the operator on H defined by \(\displaystyle M_z\xi(z) = z\xi(z)\ (\xi\in H)\). In other words, the operator \(\displaystyle M_z\) consists of multiplication by the coordinate function. Then \(\displaystyle M_z\) is a normal operator on H, and its spectrum is equal to \(\displaystyle \overline{\mathbb{D}}\). If there is a selfadjoint operator S on H such that T=f(S), where f is a continuous function from the spectrum of S into \(\displaystyle \mathbb{C}\) then the spectral mapping theorem states that f maps the spectrum of S onto the spectrum of T. So f would have to be something like a space-filling curve, mapping a compact subset of the real line onto the unit disk in the plane.

I suppose it's possible that every normal operator could come from a selfadjoint operator in this sort of way, but it somehow seems unlikely to me.
 
Aug 2009
122
19
Pretoria
I see, so when the operator T is in fact unitary the result is trivial, since then we can find a self-adjoint operator, say S such that \(\displaystyle T=e^{iS}\)