TITLE: Are weak and strong convergence of sequences not equivalent?
QUESTION [12 upvotes]: For some infinite-dimensional Banach spaces $E$, it is easy to find sequences $\langle x_i:i\in\mathbb N_0\rangle$ which converge to zero weakly but not in the norm topology, i.e. we have $\lim_{i\to\infty}y(x_i)=0$ for all $y\in E^*$ but $\lim_{i\to\infty}\|x_i\|\not=0$. For example, taking $E=c_0(\mathbb N_0)$ or $E=\ell^p(\mathbb N_0)$ with $ 1 < p < +\infty $, the sequence of standard unit vectors provides such an example. For $C([0,1])$, the sequence $\langle\langle (i+1)(1-t)t^{i+1}:0\le t\le 1\rangle:i\in\mathbb N_0\rangle$ is an example. However, I cannot find such an example for $\ell^1(\mathbb N_0)$. So I ask
 Are weak and strong convergence of sequences equivalent in $\ell^1(\mathbb N_0)$? What about $L^1([0,1])$? Is there possibly a general result saying that in any infinite-dimensional Banach space weak and strong convergence of sequences are not equivalent?

REPLY [21 votes]: Banach spaces where all weakly convergent sequences are norm convergent are said to have the Schur property. A classical Theorem of Schur says that $\ell^1(I)$ has the Schur property for every set $I$. $L^1([0,1])$ does not have it. I guess that you can find this in
P. Wojtaszczyk's "Banach Spaces for Analysts".