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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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up vote 1 down vote accepted

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.

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