# Proving the $x \mapsto \sum_{n}x_{n}$ is not weak-* sequentially continuous on $\ell^{1}$

Let $$S: \ell^{1} \to \mathbb R$$ where $$x \mapsto \sum_{n}x_{n}$$. Show that $$S$$ is not weak-* sequentially continuous (when identifying $$\ell^{1}$$ and $$(\ell^{0})^{*}$$. We are given as a hint to use the standard basis.

Defining the standard basis $$(e^{n})_{n}$$ where $$(e^{j})_{i}=\delta_{ij}$$ we can identify a corresponding set of linear operators $$(T_{n})_{n}\in (\ell_{0})^{*}$$ where for a particular $$m \in \mathbb N$$:

$$T_{m}(y)=\sum\limits_{n \in \mathbb N}y_{n}e_{n}^{m}$$. We now note that $$T_{m}(y)=y_{m} \xrightarrow{m \to \infty} 0$$ for all $$y \in \ell_{0}$$ and hence (I am not sure about this) $$T_{m}\xrightarrow{w-*} 0$$. But now how does this help me in showing that $$S$$ is not weak-* sequentially continuous?

• What is $l^0$? Do you mean $l^\infty$ or $c_0$? – daw Jul 17 at 6:41

The canonical vectors $$(e_n)_n$$ in $$\ell^1$$ satisfy $$e_n(x) = x_n \xrightarrow{n\to\infty} 0, \quad\text{ for all }x \in c_0$$
so $$e_n \to 0$$ weakly-$$*$$. However $$Se_n = 1, \forall n \in \mathbb{N}$$ so $$Se_n \not\to 0$$.
• @SABOY $e_n\to 0$ weakly-$^*$ by definition means that $T_n(x) = e_n(x) \to 0$ for all $x \in c_0$, which is precisely $T_n \to 0$ pointwise. – mechanodroid Jul 16 at 17:52
• @SABOY $S(e_n)$ are scalars in $\mathbb{R}$. Weak$^*$ convergence is equivalent to strong convergence in $\mathbb{R}$. – mechanodroid Jul 18 at 18:56
• @SABOY Precisely. Notice that finite-dimensional spaces are self-dual so weak$^*$ convergence can be considered alongside weak and strong. – mechanodroid Jul 18 at 19:20