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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?

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  • $\begingroup$ What is $l^0$? Do you mean $l^\infty$ or $c_0$? $\endgroup$ – daw Jul 17 at 6:41
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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 $Se_n \not\to 0$.

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  • $\begingroup$ @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. $\endgroup$ – mechanodroid Jul 16 at 17:52
  • $\begingroup$ @SABOY $S(e_n)$ are scalars in $\mathbb{R}$. Weak$^*$ convergence is equivalent to strong convergence in $\mathbb{R}$. $\endgroup$ – mechanodroid Jul 18 at 18:56
  • $\begingroup$ @SABOY Precisely. Notice that finite-dimensional spaces are self-dual so weak$^*$ convergence can be considered alongside weak and strong. $\endgroup$ – mechanodroid Jul 18 at 19:20

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