Show that the norm-closed unit ball of $c_0$ is not weakly compact; recall that $c_0^*=\ell_1$.
Functional Analysis – Norm-Closed Unit Ball of $c_0$ Not Weakly Compact
functional-analysis
Related Solutions
Let $e_n$ be $0$ except for a $1$ in the $n$th coordinate. Clearly $e_n \in c_0$. Then $\|e_n\|_\infty = 1$, but $\|e_n -e_m \|_\infty = 1$ whenever $n\neq m$. Hence $e_n$ can have no convergent subsequence, hence the set $\{ x \in c_0 | \|x\|_\infty \leq 1 \}$ cannot be compact.
And yes, all subsequences of $c_0$ must converge, since any subsequence of a convergent sequence must converge to the same limit.
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.
Best Answer
Hint: Let $x_n=(\underbrace{1,1,\ldots,1}_{n\text{-terms}},0,0,\ldots)$. Suppose $z\in c_0$ is a weak cluster point of $(x_n)$. By considering the action of the standard unit vectors of $\ell_1$ on the $x_n$, obtain a contradiction by showing that we must have $z=(1,1,\ldots)$.