Consider $L_{\infty}(\Omega,\Sigma,\mu)$, where $(\Omega,\Sigma,\mu)$ is any measure space. Does it it have the Grothendieck property? If the measure space is localizable, then it is true. The Bourgain-Delbaen space is a separable $\mathcal{L}_\infty$ space with the Schur property and since its bidual is complemented in $L_{\infty}(\Omega,\Sigma,\mu)$ for some measure space, this could be an example of an $L_{\infty}(\Omega,\Sigma,\mu)$ without the Grothendieck-property.
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1$\begingroup$ The Encyclopedia of Mathematics lists $L_\infty(\mu)$ as examples of Grothendieck spaces, without any conditions. I guess you should check the references they cite, see encyclopediaofmath.org/…. $\endgroup$– UwFCommented Mar 26, 2014 at 17:34
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2$\begingroup$ Since the Encyclopedia of Mathematics is not explicit in the references for $L^\infty$, consider this: a generic $L^\infty$ as vector lattice of self-adjoint elements is Dedekind $\sigma$-complete (and complete iff localizable); its spectrum as $C^*$-algebra is then $\sigma$-stonean (see Berberian, Baer$^*$ rings, page 45 or Meyer-Nieberg, Banach lattices, page 54) and now see Ando or Seever cited by the Encyclopedia. $\endgroup$– user46855Commented Mar 31, 2014 at 20:48
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$\begingroup$ user46855, you should post that as the answer. $\endgroup$– Tomasz KaniaCommented Apr 14, 2014 at 3:10
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