# Are weak and strong convergence of sequences not equivalent?

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?

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".

• See also Wikipedia: en.wikipedia.org/wiki/Schur%27s_property I can also recommend Albiac and Kalton's book. Commented Mar 1, 2012 at 15:39
• As to $L^1[0,1]$ and more generally $L^1(X,\mu)$ the complete statement is: $f_n$ converges strongly if and only if it converges weakly and in measure. (Of course in a discrete measure space such as $\mathbb{N}$, for a bounded sequence, weak convergence, convergence in measure and pointwise convergence all coincide). Check e.g. Dunford&Schwartz. Commented Mar 1, 2012 at 17:20
• @Pietro Majer: In $\ell^1(\mathbb N_0)$, the sequence of standard unit vectors converges to zero pointwise but neither weakly, nor (globally) in measure. Anyway, thanks for the additional reference. en.wikipedia.org/wiki/Convergence_in_measure
– TaQ
Commented Mar 1, 2012 at 19:35
• Now that Schur has been mentioned, I found from Jarchow's Locally Convex Spaces, Theorem 10.5.2, proof pp. 205−206, which is the result of Schur Jochen Wengenroth referred to, however without the phrase "Schur property".
– TaQ
Commented Mar 1, 2012 at 20:38
• I have now also observed that Schur's result is contained in Corollary 4.21.6 (p. 276) in R. E. Edwards' Functional Analysis. An example of a weakly but not in norm to zero convergent sequence in $L^p([0,1])$ for $1\le p<+\infty$ is $\langle[\langle \sin(2\pi(i+1)t):0\le t\le 1\rangle]:i\in\mathbb N_0\rangle$. Thanks again to Pietro Majer for providing refreshment to my memory.
– TaQ
Commented Mar 3, 2012 at 0:52

A remark in H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Chapter 3:

In any infinite dimensional space the weak topology is strictly coarser than the strong topology.

In the same place there are two examples:

1. The unit sphere $S=\{x \in E : \|x \|=1\}$, with $E$ infinite dimensional, is never closed in the weak topology $\sigma(E,E^*)$

2. The unit ball $U=\{x \in E : \|x\|<1\}$, with $E$ infinite dimensional is never open in the weak topology $\sigma(E,E^*)$

The proofs of these two facts can be found in the book.

You can find some useful facts and examples in the following documents: http://www.uio.no/studier/emner/matnat/math/MAT4380/v06/Weakconvergence.pdf

http://people.sissa.it/~bianchin/Courses/Functionanal/lecture06.weaktopologies.pdf

• This says the topologies are different... It doesn't mean that convergence of sequences is different. Commented Mar 1, 2012 at 15:38
• @Gerald Edgar: Yes. I realize that now. Thank you for your comment. Commented Mar 1, 2012 at 17:23