Timeline for Projection exists ⇒ Uniformly convex?
Current License: CC BY-SA 4.0
13 events
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Oct 13, 2022 at 18:37 | history | edited | LSpice | CC BY-SA 4.0 |
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S Oct 13, 2022 at 14:49 | history | suggested | ViktorStein | CC BY-SA 4.0 |
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Oct 13, 2022 at 11:54 | review | Suggested edits | |||
S Oct 13, 2022 at 14:49 | |||||
Nov 29, 2011 at 0:12 | history | edited | Bill Johnson |
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Nov 27, 2011 at 18:40 | comment | added | Matthew Daws | @Mark: Ah, yes! Very silly... | |
Nov 27, 2011 at 17:52 | vote | accept | Thomas Kuhn | ||
Nov 27, 2011 at 17:35 | answer | added | Bill Johnson | timeline score: 8 | |
Nov 27, 2011 at 17:28 | history | edited | Thomas Kuhn | CC BY-SA 3.0 |
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Nov 27, 2011 at 17:23 | comment | added | Thomas Kuhn | @Hsueh-Yung Lin: interesting paper. M. Day provides a example for a reflexive strictly convex space, which is not isomorphic to a uniformly convex space. So we get the unique best approximation, because every bounded sequence admits a weakly-convergent subsequence, so I have to modify my question. Is it true, that, if every closed convex set admits a best approximation, then every bounded sequence admits a weakly-convergent subsequence. | |
Nov 27, 2011 at 12:32 | comment | added | Mark Meckes | @Matthew: a compactness argument actually shows that a finite dimensional strictly convex Banach space is uniformly convex. | |
Nov 27, 2011 at 10:27 | comment | added | HYL | The following article might help you: ams.org/journals/bull/1941-47-04/S0002-9904-1941-07451-3/… | |
Nov 27, 2011 at 9:44 | comment | added | Matthew Daws | Well, take a finite-dimensional strictly convex (but not uniformly convex Banach space). Then compactness will ensure the existence of a point of best approximation. For an infinite-dimensional example, use a reflexive strictly (but not uniformly) convex space and argue via weak compactness. I'm not convinced this is a great question for this site: math.stackexchange.com might be better. | |
Nov 27, 2011 at 2:24 | history | asked | Thomas Kuhn | CC BY-SA 3.0 |