In my research in operator theory, specifically in C\* algebras and enveloping, I came across this strange footnote in a text (locally published in non English where I study) which states the following:



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> Suppose we have X a compact topological space, now suppose we have A, a sub-algebra of the algebra of continuous functions $ C(X) $ and we know A contains 1 (the identity) and that A separates points, in this case we know A is an operator algebra, and we also know the C\*-enveloping $ C\_{e}^{\*}(A) $ is commutative and that there exists a topological space F such that $ C\_{e}^{\*}(A) \simeq C(F) $.
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OK, I'll admit I am not an expert on enveloping of C\* algebras, but this seems really strange to me non the less as it is said as if it is an elementary result but still it is beyond my reach, why is $ C\_{e}^{\*}(A) $ necessarily commutative and why do we must have $ C\_{e}^{\*}(A) \simeq C(F) $, I do not even know what the topological space F is nor do I have any idea how to begin to construct it. Can someone please help me by explaining this strange remark I came across? It seems like valuable information to know but I cannot find it anywhere or understand it. I thank all helpers.



The C*-envelope of the algebra of continuous functions on a compact topological space is commutative
What they are aiming at is the following result:


Let $A \subset C(X)$ be a uniform algebra. Then there exists a unique compact set $F \subset X$, known as the Shilov boundary w.r.t. to $A$, such that every function in $A$ achieves its maximum modulus on $F$. Moreover,
$$
C\_e^\*(A) \cong C(F).
$$
See chapter 16 in Completely Bounded Maps and Operator Algebras by Vern Paulsen for a proof of this result.


