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Oct 13, 2022 at 18:37 history edited LSpice CC BY-SA 4.0
While this is on the front page, ASCII -> Unicode arrow in title, and link to comment
S Oct 13, 2022 at 14:49 history suggested ViktorStein CC BY-SA 4.0
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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