I have the following question. I guess it's quite simple for experts.
Unfortunately, I could not come up with an answer yet.


Let $X$ be a Banach space which contains no copy of $c\_0$.
Does it impply that $X''$ (the bidual of $X$) contains no copy of $c\_0$?



Containment of $c_0$
No. For complicated and important examples, consider any $\mathcal{L}\_\infty$ space that does not contain a subspace isomorphic to $c\_0$. The first such examples were constructed by Bourgain and Delbaen in the early 1980s. Some had their duals isomorphic to $\ell\_1$. The constructions there were subsequently used by Argyros and Haydon to produce a space $X$ s.t. $X^\*$ is isomorphic to $\ell\_1$ but every operator on $X$ is of the form $\lambda I +K$ with $K$ compact. Since then much more has been done.


For a simple example, consider $(\sum\_{n=1}^\infty \ell\_\infty^n)\_1$. This space obviously does not contain a copy of $c\_0$, and it is just an exercise to to point this prove that its dual contains a norm one complemented subspace isometric to $\ell\_1$. Charles Stegall was the first to point this out. I think the right reference is lemma 1 in:


Stegall, C. Banach spaces whose duals contain l1(Γ) with applications to the study of dual L1(μ) spaces. Trans. Amer. Math. Soc. 176 (1973), 463–477. 


