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A Banach space $X$ is called weakly sequentially complete if all weakly Cauchy sequences are weakly convergent.

Question: If $Y$ is a subspace of a Banach space $X$, must $Y$ be weakly sequentially complete?

My guess is yes, but I don't know how to prove it.

Any hint is much appreciated.

Martin Sleziak
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Idonknow
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  • Yes. See Norbert's answer here. – David Mitra Feb 03 '16 at 08:59
  • To transport a quick summary: subspaces are strongly closed and convex, thus weakly closed. It is a general fact that a sequentially closed subset of a sequentially complete space is sequentially complete, in whatever topology. – Eric Thoma Feb 03 '16 at 23:41

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