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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?

Mittens
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Hermi
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    Do you mean $e_n\xrightarrow{n\rightarrow\infty}0$ weakly as an element of $\ell_\infty=(\ell_1)^*$? – Mittens Nov 15 '23 at 23:13
  • @Mittens We say $x_n$ converges weakly to $x$ in $X$ if $f(x_n)\to f(x)$ for $f\in X^*$. So in our question $X=\ell_\infty$. Does it make sense? – Hermi Nov 15 '23 at 23:46
  • In such case, this my be helpful. – Mittens Nov 16 '23 at 04:33
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    Can you prove this property or the Banach space $c_0$ ? Next $c_0$ is a closed linear subspace of $l^\infty$. Any element of $(l^\infty)^$, when restricted to the subspace $c_0$, is an element of $(c_0)^$. The dual of $c_0$ is well-understood. Our vectors $e_n$ lie in $c_0$. – GEdgar Nov 18 '23 at 02:32
  • @GEdgar Thank you! So it is enough to show that $e_n\to 0$ weakly in $c_0$. Note that for every $f\in c_0^*=\ell_1$, $x\in c_0$, then $f(x)=\sum x_i f(e_i)$ where $f(e_i)\in \ell_1$. Since $\sum|f(e_i)|<\infty$, then $f(e_i)\to 0$. Thus, $e_i$ converges to $0$ weakly in $c_0$. Does it make sense? – Hermi Nov 19 '23 at 00:28
  • You are right. So note that solving this problem does not require a description of the dual $(l^\infty)^*$. – GEdgar Nov 19 '23 at 02:23

2 Answers2

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$\ell^*_\infty$ 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 $\mathcal{c}^*_0\cong \ell_1$.

Mittens
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  • Thank you very much! By the way, can I ask if my proof by contradiction work for this question? – Hermi Nov 17 '23 at 23:12
  • @Hermi: Your identity (0) (I made an edit to label that identity) requires some arguments. The full description of $\ell^*_\infty$ given in my posting (see Martin's posting for a simple proof) gives the justification. – Mittens Nov 17 '23 at 23:25
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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 $f(e_k) \to 0$

Tri
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  • Thank you! But I am confused why do you set $c_k$? It seems that just consider $\sum_{k=1}^n |f(e_k)|$ is enough? Because $S_n\le |f|$. Then $|f(e_k)|\to 0$. – Hermi Nov 17 '23 at 23:30
  • It is true that $S_n$ is indeed just the sum $\sum_{k=1}^n |f(e_k)|$, but how could you compare $S_n$ with $|f|_\infty$ directly? I defined the $c_k$ so that I could compare these two quantities. – Tri Nov 18 '23 at 17:41