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\|$ ?

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    $\begingroup$ Because we even have $\lVert T_n\rVert = \lVert x_n\rVert$ for every $n$. The canonical map $X \to X^{\ast\ast}$ is an isometry. $\endgroup$ – Daniel Fischer Jun 8 '14 at 11:53
  • $\begingroup$ Is it difficult to prove it? $\endgroup$ – luka5z Jun 8 '14 at 12:03
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    $\begingroup$ No. It's the definition of the norms plus a small application of Hahn-Banach. $\endgroup$ – Daniel Fischer Jun 8 '14 at 12:04
  • $\begingroup$ Is the answer in this topic what you mean: math.stackexchange.com/questions/793083/… ? Thanks $\endgroup$ – luka5z Jun 8 '14 at 12:09
  • $\begingroup$ Yes, that's the argument. $\endgroup$ – Daniel Fischer Jun 8 '14 at 12:17

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.


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