Every weakly convergent sequence is bounded

Solution 1:

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.

Solution 2:

One way to think about it is to recall the following 'duality property' for norms, which holds for any normed space $(X,\lVert{\cdot}\rVert)$:

$$ \forall x \in X: \lVert x \rVert = \sup_{\lVert f\rVert_*=1} \lvert f(x) \rvert $$

$$ \forall f \in X^*: \lVert f \rVert_* = \sup_{\lVert x\rVert =1} \lvert f(x) \rvert $$

where the second equality is almost a tautology, depending on your definition of the dual norm $\lVert \cdot \rVert_*$ and the first equality is a consequence of the Hahn-Banach Theorem and is a pretty easy exercise. From this, the fact that $\lVert T_x \rVert_{**} \equiv \lVert x \rVert$ is pretty obvious: $$ \lVert T_x \rVert_{**} = \sup_{\Vert f \Vert_* = 1} \vert T_x(f) \vert = \sup_{\Vert f \Vert_* = 1} \vert x(f) \vert = \lVert x \rVert$$