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Equip $c_{00}$ with $\|\cdot\|_2$. Let $B$ be the norm closed unit ball of $c_{00}$. Then how to show $B$ is not weakly compact?

I know I can use the fact that $c_{00}$ is not reflexive to show it. But is that possible we can show $B$ is not weakly compact directly? Thank you in advance!!

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  • $\begingroup$ The title and the question disagree - do you want to show $B$ is weakly compact or not weakly compact? $\endgroup$ – David C. Ullrich Jul 5 '18 at 17:57
  • $\begingroup$ @DavidC.Ullrich Thanks for pointing out. I have changed the title. $\endgroup$ – Answer Lee Jul 5 '18 at 17:58
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    $\begingroup$ I'm guessing that $c_{00}$ is the space of sequences with only finitely many non-zero terms. If so then no, $c_{00}$ is _not reflexive! It can't be because it's not even a Banach space. $\endgroup$ – David C. Ullrich Jul 5 '18 at 17:59
  • $\begingroup$ @DavidC.Ullrich Yes. And $c_{00}$ is not reflexive. My question is: is that possible we can show $B$ is not weakly compact directly without using reflexivity? $\endgroup$ – Answer Lee Jul 5 '18 at 18:01
  • $\begingroup$ Of course you can show it directly - this is trivial from the definitions. No, you cannot say the unit ball of $c_{00}$ is not weakly compact because $c_{00}$ is not reflexive, because $c_{00}$ is not a Banach space! So the theorem ""the closed unit ball of a Banach space $X$ is weakkly compact if and only if $X$ is reflexive" simply does not apply. $\endgroup$ – David C. Ullrich Jul 5 '18 at 18:05
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Inspired by this, consider $x_n = \left(\frac12, \frac14, \frac18, \ldots, \frac1{2^n}, 0, 0, \ldots\right) \in c_{00}$ and the sets $E_{n} = \{x_k : k \ge n\}$ contained in the closed unit ball of $c_{00}$.

We claim that the sets $E_n$ are weakly closed. Recall that a basis for the weak topology on $c_{00}$ is given by the sets of the form

$$V(y, a_1, \ldots, a_n, \varepsilon) = \{x \in c_{00} : \left|\langle x - y, a_i\rangle\right| < \varepsilon, 1 \le i \le n\}$$ for some $y\in c_{00}$, $a_1, \ldots, a_n \in \ell^2$ and $\varepsilon > 0$.

Notice that for $y \in c_{00}$ holds $y \in E_n$ if and only if

  • $y_k \in \left\{0, \frac1{2^k}\right\}, \forall k \in \mathbb{N}$
  • $y_k = \frac1{2^k}, \forall k \le n$
  • $y_k = 0 \implies y_j = 0, \forall j \ge k$

Assume $y \notin E_n$. Verify the following:

  • if $y_k \notin \left\{0, \frac1{2^k}\right\}$ for some $k \in \mathbb{N}$ then $$V\left(y, e_k, \min\left\{|y_k|, \left|y_k - \frac1{2^k}\right|\right\}\right) \subseteq E_n^c$$
  • if $y_k \ne \frac1{2^k}$ for some $k \le n$ then $$V\left(y, e_k, \left|y_k - \frac1{2^k}\right|\right) \subseteq E_n^c$$
  • if $y_k = 0$ but $y_j \ne 0$ for some $j > k$ then $$V\left(y, e_k,e_j ,\min\left\{\frac1{2^{k+1}}, |y_j|\right\}\right) \subseteq E_n^c$$

where $(e_n)_n$ are the canonical vectors in $c_{00}$.

Therefore $E_n^c$ is weakly open so $E_n$ is weakly closed.

Now notice that $\bigcap_{i=1}^n E_i = E_n \ne \emptyset, \forall n\in\mathbb{N}$ but $\bigcap_{i=1}^\infty E_i = \emptyset$, so the closed unit ball in $c_{00}$ cannot be weakly compact.

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  • $\begingroup$ Thanks for your answer! Is that possible we use [this] (math.stackexchange.com/questions/146727/…) to find a sequence $(x_n)$ in $B$ and show that $(x_n)$ has no weakly cluster point? $\endgroup$ – Answer Lee Jul 6 '18 at 1:34
  • $\begingroup$ @AnswerLee I don't think that is enough to show that $B$ is not weakly compact. The equivalence between weak compactness, weak sequential compactness and weak limit point compactness only holds for Banach spaces: Eberlein–Šmulian theorem $\endgroup$ – mechanodroid Jul 6 '18 at 1:35
  • $\begingroup$ I think it also holds for normed space. See Google Books. $\endgroup$ – Answer Lee Jul 6 '18 at 1:48
  • $\begingroup$ But I tried for a long time it is so closed but I couldn't prove it. $\endgroup$ – Answer Lee Jul 6 '18 at 1:49
  • $\begingroup$ @AnswerLee It seems you're right. Then just notice that the sequence $(x_n)_n$ we defined above converges to $\left(\frac1{2^n}\right)_n$ in $\ell^2$, so it is the only possible limit point of $(x_n)_n$. However, it is not in $c_{00}$ so $(x_n)_n$ has no limit points in $c_{00}$. Still, the argument above is completely elementary, it doesn't rely on Eberlein–Šmulian. $\endgroup$ – mechanodroid Jul 6 '18 at 9:22

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