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Does this argument work? Let $X$ be a normed vector space, with $A$ weakly compact in $X$. The collection of sets of the form

$$\{x \in X: |f(x) - f(a)| < 1 \},f \in X^*,a \in A$$

forms a cover of $A$ consisting of weakly open sets in $X$, and so should has a finite subcover.

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Showing a map T:X to Y weakly compact iff T** takes values in Y (identified with its image under the canonical isometric embedding in Y**). – Frank Nov 20 '12 at 21:58
Ok. How do you conclude the argument? – Davide Giraudo Nov 20 '12 at 22:12
Haven't managed to do it yet. For the forward direction, I've reduced the problem to showing: There exists some finite set {y1,...,yn} in Y such that mod(f(yi))< d (for all i = 1,...,n) implies that mod(f(T(x)) < e (for given e > 0, for all x in the unit ball of X). This should follow from the weak compactness of the closure of T(unit ball of X), I think. But I have not shown this. – Frank Nov 20 '12 at 23:37
I don't know whether your argument will work, but I posted an answer below. – Davide Giraudo Nov 21 '12 at 10:34
The "not necessarily complete" remark can be eliminated. Indeed, $A$ is weakly compact and/or bounded in the completion if and only if in $X$ itself. – GEdgar Nov 22 '12 at 14:56

For $x\in A$ define the map $T_x\colon X^*\to \Bbb K$ by $T_x(f)=f(x)$. Fix $f\in X^*$. Let for $n$ integer $O_n:=\{x\in X,|f(x)|<n\}$. These ones are weak open sets and form an open cover of $A$. So there is $N_f$ integer such that for all $x\in A$, $|T_x(f)|=|f(x)|<N_f$. As $X^*$ is complete, by the Principle of Uniform Boundedness, $\sup_{x\in A}\lVert T_x\rVert$ is bounded. By Hahn-Banach theorem, $\lVert T_x\rVert=\lVert x\rVert$, proving that $A$ is bounded.

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