Let $A$ be a unital Banach algebra and $S\subset A$ a subset consisting of commuting elements. Let $C(S)$ be the centralizer of $S$, that is, $y \in C(S)$ iff $ys=sy$ for all $s \in S$. I have already proved that $C(C(S))$ is a commutative Banach subalgebra and I want to prove that $C(C(S))$ is actually the smallest Banach subalgebra containing $S$. Let's denote by $M$ such smallest subalgebra. Now, if $\langle D \rangle$ denotes the subspace generated by a set $D$ then it's clear that if $B$ is any subalgebra containing $S$ then $E:=\langle s_1 s_2 \ldots s_n : s_i \in S \textrm{ and } n \in \mathbb{N} \rangle \subset B$ and therefore $\overline{E} \subset B$. Since $\overline{E}$ is clearly a subalgebra it follows that $M=\overline{E}$ so I just have to prove that $C(C(S))=\overline{E}$.
In order to do this, I thought the following: we have that $E \subset C(S)$ and thus, since $C(S)$ is a closed subset it follows that $\overline{E} \subset C(S)$. This last contention entails that $C(C(S))\subset C(\overline{E})$. Hence, if I could prove that $C(\overline{E})=\overline{E}$ it'd be done. However, I am not sure this last claim it's true. Actually, it doesn't seem true and therefore I'm probably following the wrong path.
Any hint in order to prove this statement?
In advance thank you.