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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 $u$ ?

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This is the classic "double subsequence trick".

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 $u$.

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  • $\begingroup$ How do you know that such a neighborhood exitsts? It is clear while considering a sequance convergent with respect to the norm. But now, it is only weakly convergent. $\endgroup$ – zorro47 Apr 24 '17 at 15:50
  • $\begingroup$ @Vessemir: It's the definition of sequence convergence in any topology. I should have said "weakly open", but everything here is with respect to the weak topology. $\endgroup$ – Nate Eldredge Apr 24 '17 at 17:25
  • $\begingroup$ Now it's perfectly clear. Thank you! $\endgroup$ – zorro47 Apr 24 '17 at 18:05

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