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In a normed space $X$ is there an equivalence between these two proposition?

$1)$ $X$ is reflexive;

$2)$ $B$, the unit ball of $X$, is weakly compact.

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4… – Nate Eldredge Sep 20 '12 at 4:08
@Nate: I would suggest yours is an answer and not a comment. – Martin Argerami Sep 20 '12 at 14:23
Dear Maria, Since the double dual of $X$ will necessarily be complete, you will need $X$ to be Banach (i.e. normed and complete). Regards, – Matt E Sep 21 '12 at 1:11
Related question:… – Nate Eldredge Sep 21 '12 at 2:01
Dear Maria, Regarding my previous comment: in fact weak compactness implies complete, as Nate noted in the edit to his answer. Best wishes, – Matt E Sep 21 '12 at 3:36
up vote 6 down vote accepted


A proof of this theorem can be found in:

Marian Fabian, Petr Habala, Petr Hajek, Vicente Montesinos Santalucia, Jan Pelant, Vaclav Zizler. Functional Analysis and Infinite-Dimensional Geometry.

See Theorem 3.31.

Google Books link

Edit: The referenced theorem assumes that $X$ is Banach; however, this automatically follows from either of conditions (1) and (2):

  1. Since $X^{**}$ is always complete, if $X$ is reflexive then it is complete (as noted in Matt E's comment).

  2. Suppose $B$ is weakly compact. Let $\{x_n\}$ be Cauchy in $X$. Cauchy sequences are bounded so by rescaling we may assume $\{x_n\} \subset B$. By weak compactness, $\{x_n\}$ has a weak cluster point $x$. Fix $\epsilon > 0$ and choose $N$ so large that $\|x_n - x_m\| < \epsilon$ for $n,m \ge N$. Let $n \ge N$. Now choose an arbitrary $f \in X^*$ with $\| f \| \le 1$. As $x$ is a weak cluster point, there exists $m \ge N$ with $|f(x_m) - f(x)| < \epsilon$. We also have $|f(x_m) - f(x_n)| \le \|x_m - x_n\| < \epsilon$. Hence $|f(x_n) - f(x)| < 2 \epsilon$. Taking the supremum over $f$ and using the Hahn-Banach theorem, we have $\|x_n - x\| < 2 \epsilon$. Thus $x_n \to x$ in norm, and we have shown that $X$ is complete.

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Dear Nate, As you know, you need $X$ to be complete (i.e. a Banach space), not merely normed. Regards, – Matt E Sep 21 '12 at 1:12
@MattE: You actually don't; completeness follows from either of the given conditions. I added a proof. – Nate Eldredge Sep 21 '12 at 1:46
Dear Nate, Thanks for this; I hadn't realized that weak compactness of the unit ball implies completeness. Best wishes, Matt – Matt E Sep 21 '12 at 3:35
Hi, the following link says that it could be the case that a closed and convex subset of unit ball in L^1 is weakly compact?So the theorem does not apply to this case? – user91360 Jun 16 '15 at 17:09
@user91360: In that question, it is shown that a particular closed convex subset $S$ of the unit ball (defined in the question) is weakly compact. It is not claimed, and not true, that every closed convex subset of the unit ball of $L^1$ is weakly compact. In particular, the unit ball itself is not weakly compact in $L^1$. There is no contradiction here. – Nate Eldredge Jun 16 '15 at 17:18

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