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Claim:In a reflexive Banach space, the weak compactness of a subset is equivalent to the boundedness of the subset.

But there is no guarantee that the bounded subset would even have its sequences converge in the bounded subset. 1) Is this statement true how it is? 2) Is it true with the additional condition of the subset being closed?

Note: I understand that weakly compact implies bounded direction works from the definition of weak compactness having these subsequences converge in the subset.

Thank you

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  1. No, the statement is not true as it is: Take the closed unit ball in Hilbert space, and throw away the zero element. Obviously the punctured ball is bounded, but it is not weakly compact because any orthonormal basis converges weakly to zero, which we intentionally left out.

  2. No, because you can take just the unit sphere in Hilbert space - i.e. all vectors whose norm is precisely $1$, and again you have that zero is in the weak closure of this set, so it is not weakly closed, let alone weakly compact.

  3. What is true is that in a reflexive Banach space, every convex, bounded and closed set is weakly compact.

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  • $\begingroup$ Actually while I am here, could you possibly tell me a reference in which $3)$ is proved? $\endgroup$ – I'm mostly just an idiot Oct 28 '15 at 11:19
  • $\begingroup$ It follows quite immediately from Banach-Alauglo's theorem, which states that a convex, closed and bounded set is $w^*$-compact in the dual space $X^*$ of a Banach space $X$. Now, if $X$ is reflexive, then the weak topology on $X$ is homeomorphic to the $w^*$ topology on $X^{**}$, viewed as the dual of $X^*$, so the result follows. $\endgroup$ – uniquesolution Oct 28 '15 at 11:22

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