I am interested in examples of dual Banach spaces $X$ with the Schur property (weakly convergent sequences in $X$ are norm convergent) like $\ell_1$. The Lorentz spaces $d(w,1)$ [Lindenstrauss and Tzafriri. Classical Banach spaces I. Sequence spaces. Section 4.3] are candidates because they admit a predual and are hereditarily-$\ell_1$ (Proposition 4.e.3 in the cited reference).
Do the spaces $d(w,1)$ have the Schur property?
No. The unit vector basis $(e_n)$ is unconditional and symmetric but not equivalent to the unit vector basis for $\ell_1$, hence $(e_n)$ converges weakly to zero.
By Rosenthal's $\ell_1$ Theorem, every normalized basic sequence in $X$ either admits a subsequence equivalent to the canonical basis of $\ell_1$, or else admits a normalized 2-block basic sequence which is weakly null. If $X$ has the Schur property, it follows that every normalized basic sequence admits a subsequence equivalent to the canonical basis of $\ell_1$. What sort of spaces might have this property?
Two interesting candidates come to mind. First of all, you could try $X=(\oplus\ell_2^n)_{\ell_1}$. Second, you could let $X=S^*$, where $S$ is the Schreier space, i.e. the completion of $c_{00}$ under the norm $\|(a_n)\|_S=\|(a_n)\|_\infty\vee\sup_{F\in\mathcal{S}_1}\|(a_n)_{n\in F}\|_{\ell_1}$. (Here, $\mathcal{S}_1$ denotes the first Schreier family.) I don't know whether these spaces have the Schur property, but I strongly suspect $(\ell_2^n)_{\ell_1}$ has it. I'm less optimistic about $S^*$, but it's still worth looking at.
EDIT: The reason I suspect $S*$ is because $S$ is known to be $c_0$-saturated. Since it is also (uniformly) subprojective, that means $S^*$ is $\ell_1$-saturated. Does this imply the Schur property? Even if not, it makes it plausible.
