I'm working on the following problem:
Suppose that $X$ and $Y$ are Banach spaces. If $\{T_n\} \subset L(X,Y)$ ($L(X,Y)$ is the space of bounded linear ops from $X \rightarrow Y$) and $T_n \rightarrow T$ weakly then $\sup_n \|T_n\| < \infty$.
A solution is here: https://math24.files.wordpress.com/2013/02/ch5-folland.pdf (47a). I came up with a different solution though, and I'm wondering if its correct:
$T_n \rightarrow T$ weakly so that $\|f(T_nx)-f(Tx)\| \rightarrow 0 \, \forall f \in Y^*, x\in X$. In particular, choose $f = 1$ so that $\|T_nx-Tx\| \rightarrow 0 \, \forall x\in X$. Then $\sup_n\|T_nx\|<\infty$ pointwise in $X$. So the result follows from uniform boundedness principle.
Observe that the dual $X^*$ contains bounded linear functionals, i.e. maps of $X$ into $\mathbb{C}$, whereas the identity map $f = 1$ maps $X$ into $X$, so it is not a functional, and therefore does not belong to the dual.