Let $e_n = (0,0,\dots,1,0,0,\dots)$,(i.e. the n-th component is 1). Show that $e_n\rightharpoonup 0$ in $\sigma(\ell^{\infty},(\ell^{\infty})')$. I'm having trouble because the dual of $\ell^{\infty}$ is not $\ell^1$. Could someone give me some ideas? Thanks!
Weak convergence in $\ell^{\infty}$
functional-analysis
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Best Answer
You're right that you can't use the direct argument for $1 \le p < \infty$. Have you tried contradiction? This gives you an $x' \in (\ell^\infty)'$ and $\varepsilon > 0$ with $| x'(e_{n_k}) | \ge \varepsilon$ for some subsequence. Can you use this to construct $x \in \ell^\infty$ that exploits the $e_n$?