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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 ...


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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 ...


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Yes, your argument is the standard way of proving it.


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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.


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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 ...



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