continuous functional calculus for nonunital $c^*$-algebras

In lecture we had the continuous functional calculus for unital $c^*$-algebras: Let $A$ be an unital $c^*$-algebra, $a\in A$ normal and let $$alg(a,a^*)=\overline{ \{ \sum\limits_{k,l=0}^n\lambda_{k,l}a^k\overline{a}^l; \lambda_{k,l}\in\mathbb{C}, n\in\mathbb{N}\} \\}$$ be the smallest $C^*$-subalgebra of A which contains $a$ (it is a commutative, unital $c^*$-algebra). Then there exists an unique (continuous) *-homomorphism of $c^*$-algebras $$\phi:C(\sigma_A(a))\to A,\; f\mapsto f(a)$$ with $\phi(x\mapsto x)=a$, $\phi(x\mapsto 1)=e$ and $\phi:C(\sigma_A(a))\to alg(a,a^*),\; f\mapsto f(a)$ is an isometric *-isomorphism of $c^*$-algebras.

Now, I want to understand the proof of the non-unital version. Continuous functional calculus theorem- non-unital version:

Let $A$ be a $c^*$-algebra, $a\in A$ normal and let $$C^*(a,a^*)=\overline{ \{ \sum\limits_{k,l=0}^n\lambda_{k,l}a^k\overline{a}^l; \lambda_{k,l}\in\mathbb{C}, n\in\mathbb{N}, \lambda_{0,0}=0\} \\}.$$ Then there exists an unique (continuous) *-homomorphism of $c^*$-algebras $$\phi:C_0(\sigma_A(a))\to A,\; f\mapsto f(a)$$ with $\phi(x\mapsto x)=a$ and $\phi:C_0(\sigma_A(a))\to C^*(a,a^*),\; f\mapsto f(a)$ is an isometric *-isomorphism of $c^*$-algebras.

1)Here is $C_0(\sigma_A(a))$ the set of continuous functions $f:\sigma_A(a)\to \mathbb{C}$ such that f(0)=0.
2)If $A$ is unital and $0\notin \sigma_A(a)$, it is $C_0(\sigma_A(a))=C(\sigma_A(a))$.

1. My first question is about $C_0(\sigma_A(a))$ (and based on this question here unitalization of the $c^*$-algebra of complex polynoms without constant term / compactness of the spectrum of elements in non-unital $c^*$-algebras ). For locally compact Hausdorff spaces X, I know $C_0(X)$ as the set of continuous functions vanishing at infinity. If $X$ is compact, it is $C_0(X)=C(X)$, or is it false? Clearly $C_0(\sigma_A(a))$ as stated in remark 1) is non-unital. And $\sigma_A(a)$ is a compact subset of $\mathbb{C}$. What is the relation of the definitions of $C_0(\sigma_A(a))$ and $C_0(X)$, where $C_0(X)$ is again the set of continuous functions vanishing at infinity with $X$ is a noncompact, locally compact Hausdorff space?

Now, the proof of this theorem (nonunital version):
existence: If $A$ is unital, $\phi$ is the restriction of the unital functional calculus $\phi^1:C(\sigma_{A_1}(a))\to alg(a,a^*)\subseteq A_1,\; f\mapsto f(a)$ ($A_1$ is the unitarization of A) on $C_0(\sigma_A(a)\setminus\{0\})$. If A is nonunital, let $\phi^1:C(\sigma_{A_1}(a))\to alg(a,a^*)$ the functional calculus for $A_1$. Now consider the restriction $\phi^1|_{C_0(\sigma_A(a)\setminus\{0\})}:C_0(\sigma_A(a)\setminus\{0\})\to A_1$. It is $\phi^1(C_0(\sigma_A(a)\setminus\{0\}))\subseteq A$ and therefore $\phi^1|_{C_0(\sigma_A(a)\setminus\{0\})}=\phi$ and $\phi$ an isometric $*$-homomorphism. Then the proof continues with the uniqueness of $\phi$.

My questions are:

1. How is $C_0(\sigma_A(a)\setminus\{0\})$ defined and what is the relation between this set and $C_0(\sigma_A(a))$ and $C_0(X)$ ($C_0(\sigma_A(a))$ and $C_0(X)$ are defined as in question 1)?

2. Why is $\phi:C_0(\sigma_A(a))\to C^*(a,a^*),\; f\mapsto f(a)$ surjective?

Regards

If $A$ is non-unital, then by definition $$\sigma_A(a) = \sigma_{A_1}(a)$$ where $A_1$ is the unitization of $A$, and necessarily one has that $0\in \sigma_A(a)$. One can then define $$I = \{f \in C(\sigma_A(a)) : f(0) = 0\}$$ and obtain a homomorphism $\varphi : I \to A$ as you have mentioned. However, $I \neq C_0(\sigma_A(a))$ in the sense of continuous functions vanishing at infinity.
For instance, if $A:= K(H)$ is the non-unital $C^{\ast}$-algebra of compact operators on an infinite dimensional Hilbert space and $a \in K(H)$ is any finite rank projection, then $$\sigma_A(a) = \{0,1\}$$ Hence, $$C_0(\sigma_A(a)) = C(\sigma_A(a)), \text{ but } I \neq C(\sigma_A(a))$$ The notation $C_0(\sigma_A(a))$ instead of $I$ is just one of convenience. It does not coincide with the usual definition of $C_0(X)$ for a non-compact locally compact space $X$.
For your third question, note that the image of $\phi$ contains all polynomials $p(a,a^{\ast})$ which do not have a constant term. But the image of a $\ast$-homomorphism is necessarily closed (first prove this in the injective case, and then use the first isomorphism theorem in the general case), and so it must be all of $C^{\ast}(a,a^{\ast})$
• It's probably useful to mention that $I=C_0(\sigma_(A)\setminus\{0\})$ (in the sense of "vanishing at infinity"). – Martin Argerami May 14 '15 at 14:32
• @MartinArgerami thank you. Why is $I=C_0(\sigma(A)\setminus \{0\})$ in the sense of "vanishing at infinity"? – banach-c May 14 '15 at 15:16