Is the dual of a reflexive Banach space strictly convex? Why?
This is a question that arouse trying to understand the theory behind approximation by finite element methods.
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At best, it would have a strictly convex equivalent norm. Otherwise, for a counterexample take any finite dimensional, non-strictly convex space. (E.g., $\mathbb{R^2}$ with $L^\infty$ norm.) – Harald Hanche-Olsen Jan 13 '14 at 17:51
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1Also, “dual of a reflexive Banach space” is just a difficult way of saying “reflexive Banach space” … – Harald Hanche-Olsen Jan 13 '14 at 17:55
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These are valid remarks. I would settle for a strictly convex equivalent norm. – user66081 Jan 13 '14 at 18:13
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See this. – David Mitra Jan 13 '14 at 18:32
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Very interesting, thank you. – user66081 Jan 13 '14 at 18:35
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Joram Lindenstrauss has shown that a reflexive Banach space admits a strictly convex equivalent norm. The paper in which this result appears (as Corollary 1) can be freely downloaded from this page. (This paper also seems relevant; in fact, the results mentioned in the first paragraph show how Corollary 1 of the Lindenstrauss paper follows from Theorem 1).
It is worth pointing out that it's relatively easy to show that a separable Banach space admits a strictly convex norm (and smooth as well). The proof of this can be found on page 33 of W. B. Johnson and J. Lindenstrauss' article in the Handbook of the Geometry of Banach Spaces, Vol 1. (The paper mentioned in the first paragraph in part aimed to show that this result holds for reflexive non-separable spaces.)
David Mitra
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Any hope for a canonical choice of the equivalent strictly convex norm? – user66081 Jan 13 '14 at 19:34