# why for every $f\in C(\sigma(x))$ we have $\Phi (f(x))= f(\Phi(x))$?

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

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