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?
Best Answer
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".