If $A$ is a non-unital $C^*$ algebra,$a$ is a normal element of $A$, there is a $*$ isometric isomorphism from $C_0(\sigma_{A}(a))$ to $C^*(a)$.
I have a question:what is the set of $C_0(\sigma_{A}(a))$?Is it the set of all continuous functions $f$ defined on $\sigma_{A}(a) $ which vanish at infinity? or besides, we need $f(0)=0$? see continuous functional calculus for nonunital $c^*$-algebras
Best Answer
No, it's just bad notation. When $A$ is not unital and you consider the C$^*$-algebra generated by $a$ in $A$, it cannot contain the unit of $A$. So when $a$ is normal C$^*(a)$ is the closure of the span of the words in $a$ and $a^*$, without constants.
The spectrum $\sigma(A)$ is compact, so $C(\sigma(A))$ is unital. For the identification $C^*(a)\simeq C_0(\sigma(a))$, one is taking the continuous functions where $f(0)=0$. That's what the subscript means. It's bad notation because it is not the set of functions that vanish at infinity.