Every Bounded Seq. in a Reflexive Space X has a Weak Con. Subseq.
Every Bounded Sequence in a Reflexive Space X has a Weakly Convergent Subsequence
Recall from the Helley's Theorem page that if $X$ is a normed linear space and $X$ is separable then every bounded sequence in $X^*$ has a weak-* convergent subsequence.
We will use Helley's theorem to show that if $X$ is a normed linear space and $X$ is reflexive then every bounded sequence in $X$ has a weakly convergent subsequence.
| Proposition 1: Let $X$ be a normed linear space. If $X$ is reflexive then every bounded sequence in $X$ has a weakly-convergent subsequence. |
- Proof: Let $(x_n)$ be a bounded sequence in $X$. Let:
\begin{align} \quad X_0 = \mathrm{cl} (\mathrm{span} (x_1, x_2, ...)) \end{align}
- Note that $X_0$ is separable. It is clear that all finite linear combinations of points in $(x_n)$ with rational coefficients is a countable dense subset of $X_0$.
- Since $X_0$ is a closed subspace of the reflexive space $X$ we have by the Closed Subspaces of Reflexive Spaces are Reflexive theorem that $X_0$ is reflexive.
- Let $J_0 : X_0 \to X_0^{**}$ be the natural embedding of $X_0$ into $X_0^{**}$. Consider the sequence $(J_0(x_n))$. This sequence is a bounded sequence in $X^{**}$. By Helley's theorem, there exists a weak-* convergent subsequence, $(J_0(x_{n_k})$. Say, $(J_0(x_{n_k}))$ weak-* converges to $\varphi \in X_0^{**}$.
- Since $X_0$ is reflexive there exists an $x_0 \in X_0$ such that $J(x_0) = \varphi$.
- Since $(J_0(x_{n_k}))$ weak-* converges to $J_0(x_0)$ we have that for every $f \in X^*$:
\begin{align} \quad \lim_{k \to \infty} \hat{x_{n_k}} (f) = \hat{x_0} (f) \end{align}
- Or equivalently, for all $f \in X^*$:
\begin{align} \quad \lim_{k \to \infty} f(x_{n_k}) = f(x_0) \end{align}
- So $(x_{n_k})$ weakly converges to $x_0$. $\blacksquare$