I am looking for a proof, other than the original article by Grothendieck which is in French, that the space $C(K)$ is Grothendieck when $K$ is extremely disconnected.
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I'm not sure, but you may want to use the fact that (for extremely disconnected $K$) $C(K)$ is $1$-injective, i.e. there is a norm 1 projection $\ell^\infty(K)\to C(K)$. – tomasz Jun 01 '16 at 18:25
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Would you mind considering accepting the answer or asking if anything is unclear? – Tomasz Kania Aug 08 '16 at 20:51
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You will find one proof here. Here's a possible different route.
Show that for every compact Hausdorff space $K$, the Banach space $C(K)$ has Pełczyński's property (V):
A. Pełczyński, Banach spaces in which every unconditionally converging operator is weakly compact, Bull. Acad. Polon. Sci. 10 (1962), 641-648.
Show that dual spaces with property (V) are Grothendieck. Conclude that $\ell_\infty(\Gamma)$ is Grothendieck for any index set $\Gamma$.
Use the fact that $C(K)$ are injective for $K$ extremely disconnected, hence complemented in $\ell_\infty(\Gamma)$ for some $\Gamma$. Conclude the Grothendieck property.
You may also apply
T. Andô, Convergent sequences of finitely additive measures, Pacific J. Math. 11 (1961), 395-404.
where it is proved that $C(K)$ is Grothendieck for a compact $\sigma$-Stonean space $K$. Extremely disconnected spaces are certainly $\sigma$-Stonean spaces. You may also want to see the proof for compact F-spaces:
G.L. Seever, Measures on F-spaces, Trans. Amer. Math. Soc. 133 (1968), 267-280.
Tomasz Kania
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