Every Bounded Sequence in a Ref. Space Has a W. Conv. Subseq.

Every Bounded Sequence in a Reflexive Space Has a Weak Convergent Subsequence

Theorem 1: Let $X$ be a normed linear space. If $X$ is reflexive then every bounded sequence in $X$ has a weak convergent subsequence.
  • Proof: Let $(x_n)_{n=1}^{\infty}$ be a bounded sequence in $X$. Let $Y \subseteq X$ be the subspace defined by:
\begin{align} \quad Y = \overline{\mathrm{span}} (x_1, x_2, ...) \end{align}
  • Let $J_Y : Y \to Y^{**}$ be the canonical embedding and consider the sequence of functionals in $Y^{**}$
\begin{align} \quad (J_Y(x_n))_{n=1}^{\infty} \end{align}
  • Then this is a sequence of bounded linear functionals. By Helly's Theorem there exists a subsequence $(J_Y(x_{n_k}))_{k=1}^{\infty}$ that weak* converges to some $\xi \in X^{**}$. Since $Y$ is reflexive there exists a $y \in Y$ such that:
\begin{align} \quad J_Y(y) = \xi \end{align}
  • We aim to show that the subsequence $(x_{n_k})_{k=1}^{\infty}$ of $X$ weak converges to $y$ Let $\varphi \in Y^*$. Then:
\begin{align} \quad \lim_{k \to \infty} \varphi(x_{n_k}) = \lim_{k \to \infty} J_Y(x_{n_k})(\varphi) = \xi(\varphi) = (J_Y(y))(\varphi) = \varphi(y) \end{align}
  • So $(x_{n_k})_{k=1}^{\infty}$ weak converges to $y$. $\blacksquare$
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