2

If $T\in B(H)$ is normal,and if $f\in C(S_p(T))$ is never zero,how to prove that the functional calculus $\theta$ for T satisfies

$$\theta(1/f)=\theta(f)^{-1}$$

89085731
  • 7,614
  • I think if you have the correct definition of $\theta$, then this question should be very easy to answer. So, what is $\theta$, formally? – Paul McKenney Feb 26 '13 at 15:33
  • You'll crucially need a normal operator. (There's also a functional calculus in the realm of closed operators on Banach space but I guess that was not your intention, or?) – C-star-W-star Jan 26 '15 at 21:58

1 Answers1

1

The functional calculus is a homomorphism which sends $1$ to $Id$, so $$ \theta(f)\theta(1/f)=\theta(1/f)\theta(f)=\theta(1)=Id. $$

Julien
  • 44,791