Show the norm closed unit ball of $c_{00}$ is not weakly compact

functional-analysislp-spacesnormed-spaces

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!!

Best Answer

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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