# Functional calculus - Typo in Blackadar's Operator Algebras?

I was reviewing some C*-algebra theory in Bruce Blackadar's Operator Algebras - Theory of C*-Algebras and von Neumann Algebras, when I came upon what seems to be a typo. On page 61, the following corollary is stated:

Let $$A$$ be a C*-algebra, and $$x$$ a normal element of $$A$$. Then $$C^*(x)$$ is isometrically isomorphic to $$C_0(\sigma_A(x))$$ under an isomorphism which sends $$x$$ to the function $$f(t) = t$$.

Here, $$C^*(x)$$ is the C*-subalgebra generated by $$x$$. My confusion is about the notation $$C_0(\sigma_A(x))$$. The index of the book refers to page 52 for this notation, where it says $$C_0(X)$$ is the space of continuous (complex-valued) functions vanishing at infinity. However, in the context of the above result, it seems to me that this should instead be the space of continuous functions with $$f(0)=0$$.

Question 1: Is this use of notation a typo in Blackadar's book?

I realize my question is similar to this question about the same bad notation, but I thought it was worth drawing attention to this specific typo in this book for others who might be confused.

I would also like to ask a follow-up question to ensure that I have understood this material correctly. As far as I understand, there are two closely related isomorphism theorems for the functional calculus of C*-algebras:

1. If $$A$$ is a C*-algebra and $$x \in A$$ is normal, then $$C^*(x) \cong \{f : \sigma_A(x) \rightarrow \mathbb{C}$$ continuous, with $$f(0)=0$$ if $$0 \in \sigma_A(x)$$ $$\}$$ via an isomorphism sending $$x$$ to the function $$f(t)=t$$.

2. If $$A$$ is a unitary C*-algebra (with unit $$1$$) and $$x \in A$$ is normal, then $$C^*(1,x) \cong \{f : \sigma_A(x) \rightarrow \mathbb{C}$$ continuous $$\}$$ via an isomorphism sending $$x$$ to the function $$f(t)=t$$ and sending $$1$$ to the function $$f(t)=1$$.

Question 2: Are the above statements correct? If so, is there a reference that states these results explicitly?

In the unital case, the natural isomorphism that the Gelfand transform gives is not onto $$C(\sigma(x))$$ but onto $$C(\Sigma)$$, where $$\Sigma$$ is the locally compact space of the characters with the weak$$^*$$-topology. Afterwards, one identifies the characters with the point evaluations.
When $$A$$ is not unital, the Gelfand transform gives you $$A\simeq C_0^{\phantom{0}}(\Sigma)$$, where here one indeed means the continuous functions vanishing at infinity. When one identifies the characters with the point evaluations at elements of $$\sigma(x)$$, $$C_0(\Sigma)$$ is mapped to $$C_0(\sigma(x)\setminus\{0\})$$; this in turn can be identified with the functions such that $$f(0)=0$$. Hence the notation, while not great, kind of makes sense.
You can find this stuff for instance in pages 7 and 8 in Davidson's C$$^*$$-Algebra by Example.