Sequential version of the Eberlein-Shmul'yan theorem

257 Views Asked by At

Theorem: A Banach space is $(i)$ reflexive iff $(ii)$ every bounded sequence possesses a weakly convergent subsequence; see e.g. Thm 3.18 and 3.19 in Brezis' 2010 book.

The implication $(i) \implies (ii)$ is essentially due to the Hahn-Banach Theorems.

Can $(ii) \implies (i)$ be shown without the use of Banach-Alaoglu and the Eberlein-Shmul'yan theorems, and by a more direct proof (i.e. one that possibly avoids topological compactness, for it is not mentioned in the statement of the theorem)?

References. H. Brezis, Functional analysis, Sobolev spaces and partial differential equations, Springer-Verlag (2010)