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.