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 , the Hilbert space of all square-integrable functions on the closed unit disk in the complex plane (with respect to 2-dimensional Lebesgue measure). Let be the operator on H defined by . In other words, the operator consists of multiplication by the coordinate function. Then is a normal operator on H, and its spectrum is equal to . 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 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.