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The unit ball is compact in the weak topology iff the space is reflexive. Is there an analogous topology under which the unit ball is compact iff the space is super-reflexive?

(I know a space is super-reflexive iff the unit ball is super weakly compact, but I'm not aware of a topology which makes super weak compactness equivalent to compactness in that topology.)

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If the supposed topology is compatible with the norm topology, then it must be the weak topology on the ball of a superreflexive space but not the weak topology on the ball of a reflexive, non superreflexive space. If the topology is also locally convex, then it lies between the weak and norm topologies. I guess the real question is: is their a (presumably functorial) assignment of a locally convex topology between the weak and norm topology which gives the weak topology on the ball of a superreflexive space but not on other reflexive spaces? –  Bill Johnson Feb 21 '11 at 0:25
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I am guessing the answer is No. Because if you take, for instance, the product of a family of $\ell_p$ spaces for $p>1$ and $p$'s tends to 1, then it won't be superreflexive anymore but the unit ball of the product under the sought topology would be compact.

Edit: This was a quick late night answer and it doesn't work. By product I had $\ell_2$ direct sum in mind but either way it doesn't solve the problem (see the comments below).

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I don't understand this answer. What do you mean by product? The $\ell_2$ direct sum? Some ultraproduct of the spaces? If you mean the usual product, the generated Banach space would be the $\ell_\infty$ direct sum of the spaces. –  Bill Johnson Feb 20 '11 at 14:45
    
Actually, now that I think about it, I'm less convinced that this disposes of the problem. The norm topology is compact in finite spaces, but not in an infinite product. Couldn't the definition in this topology be such that in the infinite product of spaces, it's finer than the product of the topologies of its component spaces? –  Henry Towsner Feb 20 '11 at 18:54
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