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

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?

• Do you mean $e_n\xrightarrow{n\rightarrow\infty}0$ weakly as an element of $\ell_\infty=(\ell_1)^*$? Commented Nov 15, 2023 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? Commented Nov 15, 2023 at 23:46
• In such case, this my be helpful. Commented Nov 16, 2023 at 4:33
• 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 $c_0$. Commented Nov 18, 2023 at 2:32
• @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? Commented Nov 19, 2023 at 0:28

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

• Thank you very much! By the way, can I ask if my proof by contradiction work for this question? Commented Nov 17, 2023 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. Commented Nov 17, 2023 at 23:25

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 $$f(e_k) \to 0$$

• 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 $|f(e_k)|\to 0$. Commented Nov 17, 2023 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
Commented Nov 18, 2023 at 17:41