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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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    http://books.google.com/books?id=Hcqm4_lW4EkC&pg=PA75&lpg=PA75&dq=weakly+compact+reflexive&source=bl&ots=_YW7BRi8tl&sig=5_bVXWyr2u_7nApt6vCQ3SWHB14&hl=en&sa=X&ei=UpZaUK_kG6q00AGY-4DABw&ved=0CDwQ6AEwBA#v=onepage&q=weakly%20compact%20reflexive&f=false2012-09-20
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    @Nate: I would suggest yours is an answer and not a comment.2012-09-20
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    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,2012-09-21
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    Related question: http://math.stackexchange.com/questions/143394/is-unit-ball-weakly-compact?rq=12012-09-21
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    Dear Maria, Regarding my previous comment: in fact weak compactness implies complete, as Nate noted in the edit to his answer. Best wishes,2012-09-21

1 Answers 1

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Yes.

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,2012-09-21
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    @MattE: You actually don't; completeness follows from either of the given conditions. I added a proof.2012-09-21
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    Dear Nate, Thanks for this; I hadn't realized that weak compactness of the unit ball implies completeness. Best wishes, Matt2012-09-21
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    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? http://math.stackexchange.com/questions/420778/l1-weak-topology2015-06-16
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    @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.2015-06-16
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    Thanks. If in that question, the functions are bounded by 1 everywhere is replaced by almost everywhere, that claim is still true? What is the most important element for that particular set to be weakly compact?2015-06-16
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    @user91360: Yes, almost everywhere would be fine. The key point is that the set is bounded in $L^p$ norm for some $p > 1$ (here we can take $p=\infty$). More generally, you can look at the [Dunford-Pettis theorem](http://math.stackexchange.com/questions/134376/dunford-pettis-theorem) which says that the essential condition for weak compactness is uniform integrability; when working with a finite measure, an $L^p$ bounded set ($p > 1$) is always uniformly integrable.2015-06-16
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    Thanks. This would be very helpful.2015-06-16