# Corollary from Eberlein - Smulyan theorem

Theorem. If $X$ is reflexive Banach space, then each bounded sequance in $X$ has a weakly convergent subsequance.

It follows that if $X$ is a reflexive Banach space and the sequnace $(u_n)$ is bounded, then we can find a subsequance $(u_{n_{k}})\subset (u_{n})$ and an element $u\in X$ such that $u_{n_{k}}\rightarrow u$ weakly in $X$. How to prove that if the limit is independent of the subsequance extracted, then the whole sequance $(u_{n})$ converges weakly to $u$ ?

Prove the contrapositive. Suppose $u_n$ does not converge weakly to $u$. Then there is a weakly open neighborhood $V$ of $u$ such that infinitely many of the $u_n$, call them $u_{n_1}, u_{n_2}, \dots$, are outside $V$. But this subsequence $u_{n_k}$ is again bounded, so it has a weakly convergent subsequence $u_{n_{k_j}}$, whose limit cannot equal $u$.