Prove Operator is a Projector Let $\mathscr{H}$ be a complex Hilbert space. A projector is a linear map $P:\mathscr{H}\to\mathscr{H}$ such that $P\circ P = P$.
I'm trying to prove the following claim, from the information given in the following picture (a snapshot of page 2 in Friedrich's book "Pertrubation of Spectra in Hilbert Space"): 
Claim: For any interval $\mathscr{I}\subseteq\mathbb{R}$, the map $\eta_{\mathscr{I}}(A)$ is a projector.
I have the following problems:
1) The author defined the $f$ map $\mathscr{H}^\mathscr{H}\to\mathscr{H}^\mathscr{H}$ only for polynomials $f$ (not even power series). I am not sure how to define $\eta_{\mathscr{I}}(A)$, the characteristic function, from the definitions given. I have looked on Wikipedia which helped but I'm still not sure how it relates to the definitions given so far.
2) Suppose you somehow "swallow" that $\eta_{\mathscr{I}}(A)$ is the identity mapping if $\mathscr{I}$ contains the spectrum of $A$ (which hasn't been defined yet) and is the zero mapping otherwise. How do you prove the claim, strictly from the fact that $\left[\eta_{\mathscr{I}}(\alpha)\right]^2=\eta_{\mathscr{I}}(\alpha)$?
 A: 1) In that page the author is not claiming that $\eta_{\mathscr{I}}(A)$ exists (yet), but is rather discussing what properties it should have before constructing it. 
2) Note that $\eta_{\mathscr{I}}(A)$ is an operator, not a function. The idea of functional calculus is that the map $f\longmapsto f(A)$ should be a $*$-homomorphism, i.e. it should preserve sums and products, and send conjugates to adjoints:
$$
(f+g)(A)=f(A)+g(A),\ \ (fg)(A)=f(A)g(A),\ \ \bar f(A)=f(A)^*.
$$
Since the function $\alpha\longmapsto\eta_{\mathscr{I}}(\alpha)$ is a characteristic function, it is real valued and satisfies $\eta_{\mathscr{I}}(\alpha)^2=\eta_{\mathscr{I}}(\alpha)$, so you expect to have
$$
\eta_{\mathscr{I}}(A)^2=\eta_{\mathscr{I}}(A),\ \ \eta_{\mathscr{I}}(A)^*=\eta_{\mathscr{I}}(A),
$$
which is exactly to say that $\eta_{\mathscr{I}}(A)$ is a projection. 
A: You only need to use that

The sum and product of two such operators correspond to the sum and product of the corresponding functions.

That said, for any operator $A$, the square of the operator $\eta_\mathscr I(A)$ equals to $\eta_\mathscr I^2(A)$ -- whatever it will mean --, but as a real (or complex) function, we have $\eta_\mathscr I^2=\eta_\mathscr I$, so
$$[\eta_\mathscr I(A)]^2=\eta_\mathscr I^2(A)=\eta_\mathscr I(A)\,.$$
