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This is a nice problem i came across yesterday and an attempt of a solution.

Let $X$ be a Banach space with $X=\bigcup_{n=1}^{\infty}A_n$ and each $A_n$ weakly compact then $X$ is reflexive.

Ok so if each $A_n$ is w-compact this means it is norm closed and bounded (Principle of uniform Boundeness).So the complete space $X$ is a union of closed sets hence from Baires theorem one(let $A_L$) has nonempty interior .

Does this mean that it contains the unit ball $(B_X,w)$ which is compact as closed subset of w compact set?

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  • $\begingroup$ I've modified your question title to remove the all caps - that came across as shouting. $\endgroup$ – Sean Roberson Jun 4 '17 at 15:55
  • $\begingroup$ I think your proof is correct. The set $A_L$ contains an open ball $B_r(x)$. Being $A_L$ weakly compact, it contains also its weak closure (which by convexity coincided with its strong closure) $\overline{B}_r(x)$, and this ball il weakly compact. But then, by Kakutani's theorem, $X$ is reflexive. $\endgroup$ – Rigel Jun 4 '17 at 16:04
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Your argument is almost correct. We just do not know that your set $A_L$ contains the unit ball $B_X$. However, we do know that it must contain some closed ball $B(a,r)$, for some $a\in A_L$ and $r>0$. However, scaling and translation are homeomorphisms, which will give you what you need.

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  • $\begingroup$ It could be weakened to $ B_X= \bigcup A_n $ i think $\endgroup$ – Pmorphy Jun 4 '17 at 20:10

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