I am looking forward to having an example of a sequence in the closed unit ball of $\ell^1$ which does not have any weakly convergent subsequence. I need this to prove that the closed unit ball in $\ell^1$ is not weakly compact and hence in turn it will imply that $\ell^1$ is non-reflexive.
My teacher hinted to take the unit basis elements $(e_n)_{n \geq 1}$ of $\ell^1.$ I know that this sequence does not converge in the weak topology but I can't argue whether it has any subsequence which is weakly convergent. Any help in this regard would be warmly appreciated.
Thanks!
Best Answer
A stronger conclusion can be proved. The sequence $e_n$ does not admit any weak accumulation point. Assume by contradiction that $u\in \ell^1$ is such an accumulation point. There is $n_0$ such that $\sum_{n=n_0}^\infty |u_n|<{1\over 2}.$ Let $$v_n=\begin{cases} 0 & n<n_0\\ 1 &n\ge n_0\end{cases}$$ Then the set $$U=\left \{x\in \ell^1\,:\, |v(x)-v(u)|<{1\over 2}\right \}$$ is open in the weak topology, where $v(x):=\sum v_nx_n.$ Moreover $u\in U,$ but $e_n\notin U$ for $n\ge n_0.$