How to find the weak limit of $e_n$ in $\ell_{\infty}$ directly?

Question

Consider the normed space $\ell_\infty$ with sup norm. I try to verify if the sequence $e_n$ converges weakly to $0$ as $n\to \infty$ in $\ell_\infty$, where $e_n$ is $n$-th coordinate 1 and others 0 vector. Here we assume that linear functional is bounded.

I think $e_n\to 0$ weakly. In other words, for $f\in \ell_{\infty}^*$, as $n\to \infty$, $$ f(e_n)\to 0 $$

I have the following proof by contradiction. $f\in \ell_{\infty}^*$ and $x\in \ell_{\infty}$, $$ f(x)=\sum_{j\ge 1}^nx_j f(e_j)=\sum_{j\ge 1} x_j y_i\tag{0} $$ where $y_j:=f(e_j)$.

Note that $$ |f(x)|\le \sup|x_i| \sum|f(e_j)|=\|x\|_\infty \|y\|_1 $$ Then $\|f\|\le \|y\|_1$

If $f(e_n)$ does not converge to $0$, then $\sum_j y_j=\infty$. So $f$ cannot be bounded.

Question: how to find the weak limit directly?

Answer 1

$\ell^*_\infty$ is the space of all charges (finitely additive functions $\nu$ on $\mathbb{2}^{\mathbb{N}}$ with $\nu(\emptyset)=0$ and with finite variation). This space has also a simple representation: $$\ell^*_\infty\cong \ell_1\oplus \mathcal{c}^\perp_0$$ where $\mathcal{c}^\perp_0=\{h\in \ell^*_\infty: \text{if $x\in\mathcal{c}_0$, then $h(x)=0$}\}$. That is, if $f\in\ell^*_\infty$, then there is $b\in\ell_1$ and $a\in\mathcal{c}^\perp_0$ such that $$f(x)=\sum_nb(n)x(n) + a(x)$$ where $a(y)=0$ for all $y\in \mathcal{c}_0$. See the posting for a short simple proof and the comments accompanying it.

From this, it follows that

$$f(e_n)=b(e_n)+a(e_n)=b(n)\xrightarrow{n\rightarrow\infty}0$$ since $e_n\in\mathcal{c}_0$ for each $n\in\mathbb{N}$.

A shorter proof is to consider the action of any $f\in \ell^*_\infty$ on the subspace $\mathcal{c}_0$. Recall that $\mathcal{c}^*_0\cong \ell_1$.

Answer 2

I don't know how to find weak limit in general case, but in this particular case, it can be shown like this: Consider the sequence $\left\lbrace c_k\right\rbrace_{k=1}^\infty$ defined by $$c_k = \begin{cases} -1; &f(e_k)<0\\ 1,& f(e_k)>0\\ \end{cases}.$$ Then the finite sum $$S_n=\sum_{k=1}^n |f(c_ke_k)|= \left|f\left(\sum_{k=1}^n c_ke_k\right)\right| \le \|f\|$$ Then the sequence of partial sum $\left\lbrace S_n\right\rbrace_{n=1}^\infty$ is bounded above, which implies that the series $\sum_{k=1}^\infty |f(c_ke_k)|$ is convergent. A consequence is that the $k$-term $ |f(c_ke_k)|=|f(e_k)| \to 0$, which is equivalent to $f(e_k) \to 0$