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I am studying the spectral theory of operators on Banach and Hilbert spaces, making use of Conway's A Course in Functional Analysis. In section VII.4, Conway states Proposition 4.11 as a consequence of the spectral mapping theorem.

I am having trouble understanding a crucial step in this argument. I have summarized Conway's proof below.

Proposition 4.11

Let $\mathcal{A}$ be a Banach algebra with identity. Suppose that $a \in \mathcal{A}$ with $\sigma(a) = F_1 \cup F_2$, where $F_1,F_2$ are disjoint, non-empty closed subsets of $\mathbb{C}$. Then there exists a nontrivial idempotent $e \in \mathcal{A}$ such that

(i) If $b \in \mathcal{A}$ commutes with $a$, then $b$ commutes with $e$.

(ii) If $a_1 = ae$ and $a_2 = a(1-e)$, then $a = a_1 + a_2$ and $a_1a_2 = a_2a_1 = 0$.

(iii) $\sigma(a_1) = F_1 \cup \{0\}$, $\sigma(a_2) = F_2 \cup \{0\}$.

This theorem is achieved by the following argument, with utilizes the Riesz functional calculus. We let $G_1, G_2$ be disjoint open subsets of $\mathbb{C}$ such that $G_1 \supset F_1, G_2 \supset F_2$. Then we choose $\Gamma$, a positively oriented system of closed curves such that $F_2$ lies in the inside of $\Gamma$ and $F_2$ lies on the outside of $\Gamma$. We then let $\chi_1$ be the characteristic function of $G_1$ an observe that $\chi_1$ is a holomorphic function on $G_1 \cup G_2$, hence holomorphic on a neighborhood of $\sigma(a) = F_1 \cup F_2 \subset G_1 \cup G_2$. We then use the Riesz functional calculus to define:

$$\rho(\chi_1) = \int_{\Gamma} \chi_1(z) (z - a)^{-1} dz,$$

which is an element of $\mathcal{A}$. The claim is that, since $\rho$ is a homomorphism and $\chi_1^2 = \chi_1$, $\rho(\chi_1)$ is the idempotent we are looking for. The spectral mapping theorem allows us to prove (iii).

My concern with the above proof is about the selection of the contour $\Gamma$. According to the definition of the functional calculus for an element $a \in \mathcal{A}$ (see, for instance (4.5) in the same section of Conway), we must choose $\Gamma$ so that the entire spectrum $\sigma(a)$ lies on the inside of $\Gamma$. But here it is specifically stated that $F_2$ (which is part of $\sigma(a)$) lies on the outside of $\Gamma$. How, then, can the contour integral above be considered as an instance of the functional calculus?

Any clarification is greatly appreciated.

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The holomorphic functional calculus allows holomorphic functions on an open neighborhood of the spectrum. Such functions can be defined independently on disjoint neighborhoods of the closed components. In particular, the holomorphic function $f$ used for defining $f(A)$ may be $0$ on an open neighborhood of one component and non-zero on another. And that's how the idempotents associated with different components may be viewed as coming from the holomorphic functional calculus.

The system of contours to enclose the various components may not be simple, which is the case if $\sigma =\{ 0\} \cup \{ \lambda : |\lambda| = 1 \}$. In this case, it takes two disjoint curves to enclose the unit circle. But this is allowed for the general holomorphic functional calculus. Then you can define $f(z)=1$ on a neighborhood of the unit circle, and define $f=0$ on a neighborhood of $0$; in this case $f(A)$ is the idempotent associated with the unit circle. Such holomorphic functions still define an algebra on which $f(A)$ is defined.

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