In a book about $ C^* $-algebra, in the section of continuous functional calculus says that:

Suppose $ x $ is a normal element of $ C^*$-algebra $ A $, then the continuous functional calculus has this property that If $ \Phi: A \to B $ is a $ C^*$-homorphism ($B$ is an arbitrary $C^*$-algebra) then for every $ f\in C(\sigma(x))$ we have $ \Phi (f(x))= f(\Phi(x))$. Please help me about the proof of this statement. thanks

  • $\begingroup$ @julien: Does this have something to do with the question? $\endgroup$ – Martin Brandenburg Nov 17 '13 at 16:37
  • $\begingroup$ @MartinBrandenburg Absolutely not... I read too fast... $\endgroup$ – Julien Nov 17 '13 at 16:52

This is clear if $f$ is a polynomial in $t,\overline{t}$. But these polynomials are dense in $C(\sigma(x))$ (and actually this is how one constructs the functional calculus) and both sides of the equation are continuous in $f$. This proves the equation in general.


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