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? 



Are weak and strong convergence of sequences not equivalent?
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".


