I have seen that, for $1 < p < \infty$, a sequence in $\ell^p$ is weakly convergent if and only if it is componentwise convergent and bounded.
Is there a counterexample for $p = 1$? That is, what would be an example of a sequence in $\ell^1$ that is componentwise convergent and bounded but not weakly convergent?
Best Answer
By a theorem of Schur, in $\ell^1$ weak convergence is equivalent to strong convergence, so any bounded sequence which is not strongly convergent serves as a counterexample.
As David pointed out in the comments, the sequence of standard unit vectors, $(e_n)_{n \in \mathbb{N}}$, is one such example, as $\lVert e_n \rVert = 1$ for all $n$ and the sequence doesn't (strongly) converge to anything.
A more direct way to see that this sequence is not weakly convergent is to consider the functional $\phi \in (\ell^1)^*$ corresponding to $((-1)^k)_{k \in \mathbb{N}} \in \ell^\infty \cong (\ell^1)^*$: $$ \phi(e_n) = \sum_{k=1}^\infty (-1)^k e_n(k) = (-1)^n$$ which doesn't converge to anything.