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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!!

Answer Lee
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  • The title and the question disagree - do you want to show $B$ is weakly compact or not weakly compact? – David C. Ullrich Jul 05 '18 at 17:57
  • @DavidC.Ullrich Thanks for pointing out. I have changed the title. – Answer Lee Jul 05 '18 at 17:58
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    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. – David C. Ullrich Jul 05 '18 at 17:59
  • @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? – Answer Lee Jul 05 '18 at 18:01
  • 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. – David C. Ullrich Jul 05 '18 at 18:05
  • @DavidC.Ullrich The theorem you mentioned works for a normed space. Can you tell me how to show it is not weakly compact. Thank you! – Answer Lee Jul 05 '18 at 18:16
  • Ok, if you have that result for normed spaces you're done, because $c_{00}$ is not reflexive. Note the reason $c_{00}$ is not reflexive is that $X^{}$ is a Banach space and $c_{00}$ is not a Banach space. Note** it has nothing to do with the fact that $c_0$ is not reflexive, because $c_0$ is not the completion of $(c_{00},||.||2)$; in fact the completion of $c{00}$ in the $\ell_2$ norm is reflexive! – David C. Ullrich Jul 05 '18 at 18:24
  • @DavidC.Ullrich Thanks for your answer. But I wondering how to show $B$ is not weakly compact directly without using reflexivity? – Answer Lee Jul 05 '18 at 18:27
  • @RhysSteele Thanks for your answer! I read it before and basically understand the idea. But we can pick $x_n=(\underbrace{1,1,\ldots,1}_{n\text{-terms}},0,0,\ldots)$ in $B$ as it is not in $B$. What sequence should me pick then? – Answer Lee Jul 05 '18 at 20:06

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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.

mechanodroid
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  • 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? – Answer Lee Jul 06 '18 at 01:34
  • @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 – mechanodroid Jul 06 '18 at 01:35
  • I think it also holds for normed space. See Google Books. – Answer Lee Jul 06 '18 at 01:48
  • But I tried for a long time it is so closed but I couldn't prove it. – Answer Lee Jul 06 '18 at 01:49
  • @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. – mechanodroid Jul 06 '18 at 09:22
  • Thank for your answer! Can you be more specific why $(x_n)$ has no limit points? – Answer Lee Jul 06 '18 at 18:41
  • @AnswerLee We can show that $(x_n)n$ has no weakly convergent subsequences. Namely, every subsequence $(x{p(n)})n$ converges strongly to $\left(\frac1{2^n}\right)_n$ in $\ell^2$, so it also converges weakly to $\left(\frac1{2^n}\right)_n$. Since $\left(\frac1{2^n}\right)_n \notin c{00}$, we conclude that $(x_{p(n)})n$ doesn't converge weakly in $c{00}$. Note that if it converges weakly to something in $c_{00}$, then it also converges weakly to the same thing in $\ell^2$. – mechanodroid Jul 06 '18 at 18:46
  • I'm not entirely sure that the above argument shows that $(x_n)_n$ has no weak limit points, but it certainly shows that $B$ is not weakly sequentially compact, so it suffices. – mechanodroid Jul 06 '18 at 18:49
  • I see what you mean. But how to show every subsequence of $(x_n)$ converges strongly to $(\frac{1}{2^n})$? – Answer Lee Jul 06 '18 at 19:27
  • @AnswerLee Because $(x_n)_n$ converges strongly to $\left(\frac1{2^n}\right)_n$. If a sequence converges to a limit $L$, every subsequence also converges to the same limit $L$. This holds in any topological space. – mechanodroid Jul 06 '18 at 19:28
  • Let us continue this discussion in chat. – Answer Lee Jul 06 '18 at 21:00