For every integer $n \geq 1$ let $$e^n=(0,0,\dots,\underset{(n)}{1},0,\dots).$$
- Prove that $e^n \underset{n \rightarrow \infty}{\rightharpoonup} 0$ in $\ell^p$ weakly $\sigma(\ell^p,\ell^{p'})$ with $1 < p \leq \infty$.
My attempt: For $1<p<\infty$ and $f \in (\ell^p)'$, there exists $g=(g_1,g_2,\dots) \in \ell^{p'}$ such that $$\langle f, x \rangle_{(\ell^p)',\ell^p}=\sum_{n=1}^{\infty}g_n x_n \hbox{ for all } x=(x_1,x_2,\dots) \in \ell^p.$$ So that, $$\langle f, e^n \rangle_{(\ell^p)',\ell^p}=g_n. \tag{*}$$
Since $g \in \ell^{p'}$, then $\sum_{n=1}^{\infty}|g_n|^{p'}<\infty$ which implies that $|g_n|^{p'} \rightarrow 0$ as $n \rightarrow \infty.$ Hence, $|g_n|\rightarrow 0$ as $n \rightarrow \infty$. Therefore, from $(*)$ we obtain that $$\langle f, e^n \rangle_{(\ell^p)',\ell^p} \rightarrow 0 \hbox{ as } n \rightarrow \infty.$$ Is this correct?
My question: How to prove that $e^n \underset{n \rightarrow \infty}{\rightharpoonup} 0$ in $\ell^\infty$ weakly $\sigma(\ell^\infty,(\ell^{\infty})')$?
