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Let $X$ be a normed space. If $X$ is weakly sequentially complete, show that $X$ is reflexive.

And my effort is to show the closed unit ball of X is weakly compact. But I am stuck. Any help will be appreciate!

Answer Lee
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1 Answers1

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It's not true. $L^1$ is weakly sequentially complete but not reflexive.

For the special case of $\ell^1$ this can be proved as a corollary of Schur's theorem that a sequence in $\ell^1$ converges weakly iff it converges in norm. See Proposition 2.3.12 of Albiac, Fernando; Kalton, Nigel J., Topics in Banach space theory, Graduate Texts in Mathematics 233. Cham: Springer (ISBN 978-3-319-31555-3/hbk; 978-3-319-31557-7/ebook). xx, 508 p. (2016). ZBL1352.46002. Google Books

Nate Eldredge
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  • This is very helpful!! Thank you!! By the way, what if we add a condition to make $X$ reflexive what condition can be added here? – Answer Lee Feb 09 '18 at 06:58