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I have this question also in MSE (see: https://math.stackexchange.com/questions/666053/centralizers-and-containment-of-c-0), but I have not got an answer there. So I thought I try my luck here.


Let $X$ be a Banach space over $\mathbb{R}$ or $\mathbb{C}$.

By a multiplier on $X$ we mean a bounded linear operator $T$ on $X$ such that every extreme point of $B_{X^*}$ becomes an eigenvector for $T^*.$ Given a multiplier $T$ on $X$, and an extreme point $p$ of $B_{X^*}$, there exists a unique number $a_T(p)$ satisfying $p\circ T^* = a_T(p)p$.

The centralizer $Z(X)$ of $X$ is the set of those multipliers $T$ on $X$ such that there exists a multiplier $S$ on $X$ satisfying $a_S(p)=\overline{a_T(p)}$ for every extreme point $p$ of $B_{X^*}$. (For more one can find in Harmand, Werner, Werner, "M-ideals in Banach spaces and Banach algebras.")

I wonder if the answer to the following question is known:

If $Z(X)$ is infinite-dimensional does then $X$ contain an isomporphic copy of $c_0$?

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Cameron's answer indicates why $Z(X)$ rather than $X$ contains an isomorphic copy of $c_0$. Using E. Behrends's function module representation theory, one actually obtains an isometric copy of $c_0$ in $X$. This can be found explicitly in E. Behrends, $M$-Structure and the Banach-Stone Theorem; LNM 736 (1979), Prop. 4.22.

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The discussion in your cited reference around Theorem 3.6 / Definition 3.7 indicates that $Z(X)$ is always isomorphic to some $C(K)$ space, so if it is infinite-dimensional it will contain an isomorphic copy of $c_0$.

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