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Definition: Let $\ell_2$ be the linear space that consists of all sequences $x=(x_n)$ in $\mathbb{F}$ for which $\sum\limits_{n=1}^{\infty}|x_n|^2<\infty$. Then $\|x\|_2=(\sum\limits_{n=1}^{\infty}|x_n|^2)^{\frac{1}{2}}$ is a norm on $\ell_2$, and $\ell_2$ is a Banach space with respect to this norm. Let $c_{00}$ be the linear space of all sequences in $\mathbb{F}$ that are eventually zero. Then $c_{00}$ is a linear subspace of $\ell_2$.

Equip $c_{00}$ with $\|\cdot\|_2$. Let $B$ be the norm closed unit ball of $c_{00}$. I want to show $B$ is not weakly compact. It suffices to show that there every sequence $(x_n)$ in $B$ such that has no weakly cluster point $x$ in $B$.

In fact, let $y=(y_n)\in\ell_2$. Then $f_y(x)=\sum\limits_{n=1}^{\infty}x_ny_n$ ($x=(x_n)\in c_{00}$) defines a bounded linear functional on $c_{00}$; that is $f_y\in {c_{00}}^*$.

I saw this. And I am trying to use it to find an example. But I am stuck. Can anyone help me? Thank you!

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  • $\begingroup$ @uniquesolution But the sequence in $c_0$ does work for $B$. $\endgroup$
    – Answer Lee
    Jul 6, 2018 at 18:49
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    $\begingroup$ @uniquesolution I am sorry the sequence in $c_0$ does not work for $B$. As $x_n=(\underbrace{1,1,\ldots,1}_{n\text{-terms}},0,0,\ldots)$ is not in $B$. $\endgroup$
    – Answer Lee
    Jul 6, 2018 at 18:56
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    $\begingroup$ @uniquesolution Equip $c_{00}$ with $\|\cdot\|_2$. And $B$ is a norm closed unit ball of $c_{00}$. $\endgroup$
    – Answer Lee
    Jul 6, 2018 at 19:08
  • $\begingroup$ Sorry, I missed that part. $\endgroup$
    – uniquesolution
    Jul 6, 2018 at 19:09
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    $\begingroup$ @uniquesolution $c_{00}$ is not reflexive as it is not a Banach space. $\endgroup$
    – Answer Lee
    Jul 6, 2018 at 19:21

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In this answer to your previous question we considered the sequence $(x_n)_n$ defined as $$x_n = \left(\frac12, \frac14, \frac18, \ldots, \frac1{2^n}, 0, 0, \ldots\right) \in B$$

We then showed that the sets $E_n = \{x_k : k \ge n\}$ are weakly closed and satisfy $\bigcap_{n=1}^\infty E_n = \emptyset$.

By taking complements we get $\bigcup_{n=1}^\infty E_n^c = c_{00}$.

Assume that $(x_n)_n$ has a weak limit point $y \in c_{00}$. Then there exists $n \in \mathbb{N}$ such that $y \in E_n^c$. Therefore $E_n^c$ is a weakly open neighbourhood of $y$ so $E_n^c$ contains infinitely many terms of the sequence $(x_n)_n$.

But this is a direct contradiction with $E_n = \{x_k : k \ge n\}$.

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  • $\begingroup$ Let $x=(\frac{1}{2^n})_{n\in\mathbb{N}}$. Then $x\in\ell_2-c_{00}$. The sequence $(x_i)=\{\frac{1}{2},\cdots,\frac{1}{2^i},0\cdots\}$ $(i\in\mathbb{N})$ is a sequence in $B$. We have $\|x-x_i\|_2=(\sum\limits_{n=i+1}^{\infty}|\frac{1}{2^n}|^2)^{\frac{1}{2}}\rightarrow0$. Thus $x_i\overset{\|\cdot\|_2}{\longrightarrow} x$ and hence $x_i\overset{wk}{\longrightarrow} x$. So, every subsequence of $(x_i)$ converges to $x$. Now assume that $B$ is weakly compact. Then there is a subsequence $(y_i)$ of $(x_i)$ such that $y_i\overset{wk}{\longrightarrow} y=x$, which is a contradiction. $\endgroup$
    – Answer Lee
    Jul 6, 2018 at 21:04
  • $\begingroup$ Is this proof ok? Thank you! $\endgroup$
    – Answer Lee
    Jul 6, 2018 at 21:04
  • $\begingroup$ @AnswerLee Yes, it is a correct proof that $B$ is not weakly compact. However, it doesn't show that the sequence $(x_n)_n$ has no limit points. This is because in a non-metrizable space there can be limit points which are not limits of subsequences of $(x_n)_n$. $\endgroup$
    – mechanodroid
    Jul 6, 2018 at 21:07
  • $\begingroup$ I am not familiar with the theorem you mentioned. But do you mean there may exist a subsequence of $(x_n)$ that converges in $B$. So we can not get $B$ is not weakly compact? $\endgroup$
    – Answer Lee
    Jul 6, 2018 at 21:13
  • $\begingroup$ @AnswerLee You have proved that $(x_n)_n$ has no weakly convergent subsequence. Hence $B$ is not weakly sequentially compact, and therefore it is not weakly compact, by Eberlein–Šmulian. That part is perfectly fine. However, you didn't show that $(x_n)_n$ doesn't have a limit point (which is what you asked for in this question). If $y$ is a weak limit point of $(x_n)_n$, there need not exist a subsequence $(x_{p(n)})_n$ such that $x_{p(n)} \xrightarrow{w} y$. $\endgroup$
    – mechanodroid
    Jul 6, 2018 at 21:17

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