Show that a certain sequence in $l^\infty(\mathbb N) $ does not converge weakly.

Question

Let $V = l^\infty(\mathbb N) $ (space of bounded sequences with sup norm), I want to show that the sequence $ \{\sum_{ m =n }^\infty e_m\}_{n \ge 1} $ does not converge weakly. Here $ e_i = ( 0,..., 0 ,1 ,0,0,...) $ with $1$ in the $i$th entry.

---

The hint says use a generalised limit functional. But for $ f \in V^*$ be the generalised limit functional, $ f( \sum_{m=n}^\infty e_m) = 1$ for all $n$. So how does this prevent $\sum_{m=n} e_m$ from converging weakly to some sequence with limit 1, say (1,1,1,1,...) ?

Any help is appreciated.

Answer 1

If your sequence $x_n=\sum_{m\ge n} e_m = (0,\ldots,0,1,1,1,\ldots)$ converges weakly then the limit is necessarily $0$ (because all "evaluations" $\delta_k((z_1,z_2,\ldots))=z_k$ are continuous linear functionals). For a "generalized limit functional" $f$ you thus have $1= f(x_n) \not\to f(0)=0$.