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}

- Then $Y$ is separable. Furthermore, by the theorem on the Closed Subspaces of Reflexive Spaces are Reflexive, $Y$ is reflexive.

- 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 $y$. $\blacksquare$