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I'm reading about Gelfand's representation theory for Banach algebras and I have a small problem with one of the proofs.

Let A be a commutative Banach algebra with unit, define

[itex]

\Gamma_A=\left\{\varphi:A\to\mathbb{C}:\varphi\neq0, \varphi \text{ an algebra homomorphism }\right\}

[/itex]

to be the Gelfand space and let

[tex]

\hat{}:A\to C(\Gamma_A),\quad x\mapsto\hat x= \{\varphi\mapsto\varphi(x)\}

[/tex]

be the Gelfand transform.

Then the statement is that ^ is a (continuous) algebra homomorphism between the Banach algebra A and [itex]C(\Gamma_A)[/itex], the space of continuous functions on the Gelfand space.

My problem is: Why is ^ surjective? I agree that for any x in A, [itex]\hat x[/itex] is a continuous function on the Gelfand space but why should ANY function in the Gelfand space come as an evaluation functional at some point x...is this not equivalent to assuming that the A as an Banach space is reflexive? If so, does this follow from the fact that A is commutative and has a unit?

Thanks

Pere

oooh, I just see, it is not claimed that ^ is surjective. So is it right that A is isomorphic to [itex]\hat{A}[/itex] which is a closed subspace of [itex]C(\Gamma_A)[/itex] or is ^ not even necessarily injective...?

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# Gelfand representation theory

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