Talk:Continuous functional calculus

The article invokes the Gelfand representation of the C* algebra, but that only applies for commutative algebras. Seems to me that maybe what is meant is the algebra generated by x, which, since generated by a normal element, will be commutative. Is it so? -lethe ^{talk +} 17:44, 6 April 2006 (UTC) [reply ]

I thought this was clear. The article states:

The proof of this fact is almost immediate from the Gelfand representation: it suffices to assume A is the C*-algebra of continuous functions on some compact space X and define

${\displaystyle \pi (f)=f\circ x.}${\displaystyle \pi (f)=f\circ x.}

Uniqueness follows from application of the Stone-Weierstrass theorem.

However, to reduce risk of confusion, maybe we should put in "without loss of generality, we may assume A is commutative since the C*-algebra generated by a single normal element is commutative.--CSTAR 18:22, 6 April 2006 (UTC) [reply ]

I'm still trying to decide whether I find it confusing or not. The article doesn't make clear what algebra A is. Is it the algebra generated by x? What if I take A to be some larger commutative algebra? Then X will not be the spectrum of x. Also, this composition fx is bothering me. A priori, it's not clear that the composition makes sense, since, as they're defined, the codomain of x (as a function on X) is all of C, while the domain of f is only sp(x). I guess the range of x turns out to coincide with the spectrum so that the composition does actually make sense. But maybe this needs to be made clear to the reader? -lethe ^{talk +} 18:46, 6 April 2006 (UTC) [reply ]

One way to deal with this technically (altough not in the article) is to note that the functional calculus does not depend on the choice of commutative C*-algebra A containing x.--CSTAR 19:03, 6 April 2006 (UTC) [reply ]

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