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!
Best Answer
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 $E_n = \{x_k : k \ge n\}$.