TITLE: Weak convergence in reflexive Banach space
QUESTION [10 upvotes]: Let $X$ be a reflexive Banach space. Let $T: X \to Y$ a linear operator. I want to show that:
$$T \in \mathcal{L}(X, Y) \iff ((x_n \stackrel{w}{\rightharpoonup} x) \implies (T(x_n) \stackrel{w}{\rightharpoonup} T(x))) $$
Any help, thoughts, hints, solutions will be greatly appreciated. I am lots of trouble regarding weak convergence.

REPLY [2 votes]: For the backward direction (⟸), here's the proof given by Norbert in a simplified form:
Let $x_n \to x$ and $Tx_n \to y$, then we have $x_n \to x$ weakly and $Tx_n \to y$ weakly, now using the assumption we have also $Tx_n \to Tx$ weakly. By the uniqueness of the weak limit we have $y=Tx$. T is then closed and therefore bounded (continuous) by the closed graph theorem.