It is well known that a weakly sequentially complete Banach space with an unconditional basis is isomorphic to a conjugate space. Is the converse to this statement true?
If a Banach space is a conjugate space and has an unconditional basis, must it be weakly sequentially complete?
Yes. More generally this holds for separable Banach lattices that are dual spaces (spaces with unconditional bases are Banach lattices with the positive cone consisting of vectors having non-negative coefficients with respect to a given unconditional basis).
Indeed, we have the following generalisation of a theorem of James (This is Theorem 1.4.c in the second volume of Lindenstrauss and Tzafriri):
Let $X$ be a Banach lattice. Then $X$ is weakly sequentially complete if and only if it does not contain subspaces isomorphic to $c_0$.
Now, a separable dual cannot contain subspaces isomorphic to $c_0$ as every such subspace would be complemented (Sobczyk's theorem) and this would contradict the Phillips-Sobczyk theorem. (You will find more details here.)
