Theorem 6 part (i) of Lax's Functional Analysis book (Chpt 17) states (paraphrased)
Suppose that the spectrum of $M$ can be decomposed as the union of $n$ pairwise disjoint closed components:
$\sigma(M) = \sigma_1 \cup ... \cup\sigma_N, \quad \sigma_j \cap\sigma_k = \emptyset$
For each $j$, denote $C_j$ a contour in $\rho(M)$ that winds once around $\sigma_j$ but not $\sigma_k$, $k\neq j$. We define
$P_j = \frac{1}{2\pi i}\oint_{C_j} (\zeta I-M)^-1 d\zeta$
Then $P_j$ are disjoint projections, that is $P_j^2 = P_j$ and $P_jP_k = 0$ for all $j\neq k$.
For the proof, Lax cites part (ii) of Theorem 5 of that same chapter, which states:
The mappings
$f(M) = \oint_C(\zeta - M)^{-1} f(\zeta) d\zeta$
from the algebra of functions analytic on an open set containing $\sigma(M)$ into $\mathcal L$ is a homomorphism.
(here $C$ surrounds $\sigma(M)$).
My question is I really don't see how that statement implies that $P_j$ are disjoint projections. My limited understanding is: up until now in the chapter, we always start with a linear operator $M$ with spectra $\sigma(M)$ and define $f(M)$ as given. Is this second theorem saying that we can start with some open region $O$, some function analytic on $O$, and generate an $M$ a linear operator with spectra $\sigma(M)\subset O$? Then I can see how the projections are disjoint, since taking that integral around a region with no spectral is like taking a contour integral around an analytic function, which gives 0. But Lax's proof of Theorem 5 (ii) is simply showing that the mapping is multiplicative so I'm not sure how this logic works. Any filling in of the blanks would be appreciated!
Thank you for any insight!
You only need to assume that the functions you're using are analytic on an open neighborhood of the spectrum. That's not as constraining as you would think when the spectrum is not connected. In the case that the spectrum is not connected, you're allowed to define $f$ to be one analytic function on a neighborhood of one component and an entirely different analytic function on a neighborhood of the other component. When you do that, you still get a homomorphism: $(fg)(M)=f(M)g(M)$.
For your case, you are allowed to take $f_{j}$ to be $1$ on a neighborhood of $\sigma_{j}$ and $0$ on neighborhoods of all the other components. Then $f_{j}^{2}=f_{j}$ is also analytic, and you see that $P_{j}=f_{j}(M)$ is a projection. Furthermore $f_{j}f_{k}=0$ for $j\ne k$, which gives $P_{j}P_{k}=P_{k}P_{j}=0$. Finally $P_{1}+\cdots+P_{n}=I$ because $1$ is always mapped to $I$. If $f$ is any function which is analytic on an open neighborhood of the spectrum, then $f(M)P_{j}$ reduces to an integral of $f$ around a single component.
I'm not saying any of this is obvious, but normally the full-blown analytic functional calculus is stated in these terms. The proofs require looking at integrals around components. Multiplying integrals over disjoint components does give you zero because of the resolvent equation for $R(\lambda) = (M-\lambda I)^{-1}$: $$ (\lambda-\lambda')R(\lambda)R(\lambda')=R(\lambda)-R(\lambda'). $$ To see how this works, rewrite the following using the resolvent equation, assuming that $C_{1}$ and $C_{2}$ are disjoint simple closed rectifiable curves with non-overlapping interiors: $$ \oint_{C_{1}}f(\lambda)R(\lambda)\,d\lambda \oint_{C_{2}}f(\lambda')R(\lambda')\,d\lambda'. $$ You'll get $0$.