Theorem: Every weakly convergent sequence in X is bounded.
Let $\{x_n\}$ be a weakly convergent sequence in X. Let $T_n \in X^{**}$ be defined by $T_n(\ell) = \ell(x_n)$ for all $\ell \in X^*$. Fix an $\ell \in X^*$. For any $n \in \mathbb{N}$, since the sequence $\{\ell(x_n)\}$ is convergent, $\{T_n(\ell)\}$ is a bounded set. By Uniform Boundedness Principle $ \sup_{n \in \mathbb{N}} \|x_n\| = \sup_{n \in \mathbb{N}} \|T_n\| < \infty,$ i.e. $\{x_n\}$ is bounded.
My question is: why $ \sup_{n \in \mathbb{N}} \|x_n\| = \sup_{n \in \mathbb{N}} \|T_n\|$ ?
Best Answer
The equality $\|x_n\|=\|T_n\|$ is an instance of the fact that the canonical embedding into the second dual is an isometry.
See also Weak convergence implies uniform boundedness which is stated for $L^p$ but the proof works for all Banach spaces.