I am working on an example from the spectral theory of linear operators on Banach spaces. This is example 6.10 from chapter VII of Conway's A Course in Functional Analysis.
Let $\{\alpha_n \}_{n=1}^\infty \subset l^\infty$. For $1 \le p \le \infty$, define the bounded linear operator $A: l^p \to l^p$ by $Ax(n) = \alpha_n x(n)$ for all $x \in l^p$. It's not too difficult to show that the spectrum of $A$, denoted by $\sigma(A)$, equals $\text{cl} \left(\{\alpha_n\}_{n=1}^\infty\right)$.
Suppose now that, for some index $k \in \mathbb{N}$, $\alpha_k$ is an isolated point with respect to the subspace topology on $\sigma(A)$. Then the singleton $\{\alpha_k\}$ is a clopen set in $\sigma(A)$. And we can find a smooth contour $\Gamma$ that winds once around $\{\alpha_k\}$ but not around $\sigma(A)\setminus \{\alpha_k\}$. According to the Riesz functional calculus, we may define a new bounded operator on $l^p$ by:
$$E( \alpha_k; A) = \int_\Gamma (z - A)^{-1} dz.$$
And by proposition 4.11 earlier in the chapter, $E(\alpha_k ; A)$ is a idempotent. That is $E(\alpha_k ; A) \circ E(\alpha_k ; A) = E(\alpha_k ; A)$.
Now, I think I understand this example so far. But then, Conway goes on to claim that $E(\alpha_k ; A)$ is equal the operator $P: l^p \to l^p $ defined by $Px(n) = \chi_{N_k}(n)x(n)$ where $\chi_{N_k}$ is the characteristic function (defined on $\mathbb{N}$) of $N_k = \{n \in \mathbb{N} : \alpha_n = \alpha_k \}$.
I'm not able to show why $E(\alpha_k ; A) = P$ as claimed.
This is equality is puzzling me because I am not sure how to get started with the operator $\int_\Gamma (z - A)^{-1} dz$. Is there a trick that I could use to explicitly compute $\left( \int_\Gamma (z - A)^{-1} dz \right) x$ for arbitrary $x \in l^p$? I think the only useful observation I have made to this point is that it's clear $P$ is an idempotent (as it must be if we are to have $P = E(\alpha_k ; A)$).
Hints or solutions are greatly appreciated.
You left out the factors $1/(2\pi i)$.
For the standard unit vector $e_k$ (i.e. $e_k(k) = 1$, $e_k(n) = 0$ otherwise) we have $A e_k = \alpha_k e_k$ so $(z - A)^{-1} e_k = (z - \alpha_k)^{-1} e_k$, and $$\dfrac{1}{2\pi i}\int_\Gamma (z - A)^{-1} e_k \; dz = \left(\dfrac{1}{2\pi i}\int_\Gamma (z - \alpha_k)^{-1} \; dz \right) e_k$$ which is $e_k$ if $\alpha_k$ is inside $\Gamma$ and $0$ if it is outside $\Gamma$.