What is the gelfand transform of an operator in the algebra generated by a bounded normal operator and it's adjoint?
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2026-03-30 14:08:10.1774879690
Gelfand Transform in a specific case
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The algebra is commutative, and isomorphic to $C(X)$ where $X$ is the maximal ideal space. Under the Gelfand Transform, the operator itself goes to the function $f(z)=z$.
Have a look at continuos functional calculus in your book- they should prove this statement over there.