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From this problem， I see that

Theorem 1:(Eberlein-Smul'yan Theorem) A Banach space $E$ is reflexive if and only if every (norm) bounded sequence in $E$ has a subsequence which converges weakly to an element of $E$.

But in a Chinese book, I seemly find that

Theorem 2: $E$ is normed space. If $E$ is reflexive, then, every bounded sequence in $E$ has a subsequence which converges weakly.

The proof of Theorem 2 really does not use completeness. But I always felt a little uneasy, so I wanted to ask, is theorem 2 right?

Enhao Lan
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1 Answers1

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Dual of a normed linear space is complete, and so is the second dual. So reflexive spaces (being isomorphic to their second duals) are complete.

Kavi Rama Murthy
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