Are dual spaces with unconditional bases weakly sequentially complete?

Question

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?

Answer 1

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.)