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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**).2012-11-20
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    Ok. How do you conclude the argument?2012-11-20
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    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.2012-11-20
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    I don't know whether your argument will work, but I posted an answer below.2012-11-21
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    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.2012-11-22

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