Let $\mathcal{A}$ be a unital commutative C* algebra. Its Gelfand transform $$ \hat{a}: \mathcal{A} \rightarrow C(\hat{\mathcal{A}}) $$ is an isometric *-isomorphism. (partially because its range is norm-closed and dense in $C(\hat{\mathcal{A}})$).
This theorem was proved by using several other theorems.
First, I know that for a unital commutative Banach algebra, $||\hat{a}||_\infty = r(a)$, where $r(a)$ is a spectrum of $a \in \mathcal{A}$ (commutative is important)
Second, I know that for (unital?) C* algebras, since $ ||a^*a|| = ||a||^2, r(a) = ||a||$ (do we even require unital here?).
These two give me isometric.
For any unital Banach algebra (complete is the key), Gelfand transform is a unital algebra homomorphism. (homomorphism is an isomorphism if and only if it is bijective).
- For unital symmetric Banach* algebra $\mathcal{A}$, the range of Gelfand is dense in the sup norm (note C* algebra is Banach* algebra, and every unital C* algebra is symmetric)
I know that if a subset is a norm-closed and dense, then it is equal to the whole set. So the last missing piece is to show the range of $\hat{a}$ is norm-closed. Does it trivially follow from one of the assumptions? And last but not least, what does it mean to be * isomorphism? Any other condition to check?