[Math] Weakly sequentially compact sets

Question

From Peter Lax Functional analysis page 104:

Show that a weakly sequentially compact set is bounded.

Definition. A subset $C$ of a Banach space $X$ is called weakly sequentially compact
if any sequence of points in $C$ has a subsequence weakly convergent to a point of $C$.

Answer 1

The key is the following result:

If $X$ is a normed space, and $\{x_n\}\subset X$ a sequence which converges weakly to $X$, then $\sup_{n\in\Bbb N}\lVert x_n\rVert<\infty$.

It's a consequence of Baire's categories theorem applied to $X^*$, the topological dual of $X$.

We show that an unbounded set cannot be weakly compact. If $S$ is not bounded, let $\{x_n\}\subset S$ such that $\lVert x_n\rVert\geqslant n$. If $\{x_{n_k}\}$ is a subsequence of $\{x_n\}$, then for each $k$, $\lVert x_{n_k}\rVert\geqslant n_k$. In particular, $\{x_{n_k}\}$ is not bounded hence cannot be weakly convergent.