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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 ...
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 $\eta_\mathscr I^2=\eta_\mathscr ... 1 Yes, your argument is the standard way of proving it. 1 Because you are trying to prove (first) that$\phi$is unbounded on$A^+$. You do this because you want to use that your map is positive, so it makes sense to work on the positive part of$A$. The summands are positive. Because$\|\phi(p)\|$would be an upper bound for the natural numbers. 1 In a$C^\ast$algebra, this is easy. You have$\Vert a \Vert = \Vert a^\ast \Vert$and (by definition of a$C^\ast$algebra) you have $$\Vert a^\ast \Vert^2 = \Vert (a^\ast)^\ast a^\ast \Vert = \Vert a \cdot a^\ast\Vert,$$ so that taking$b = \frac{a^\ast}{\Vert a\Vert}$yields your claim, since the estimate$\Vert ab \Vert \leq \Vert a \Vert \Vert b ...