Timeline for Why isn't the theorem of approximation applicable in Banach spaces?
Current License: CC BY-SA 2.5
9 events
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| Aug 24, 2011 at 18:49 | comment | added | Margaret Friedland | In fact, a normed vector space $X$ is strictly convex if and only if for every $x$ in $X$ and every affinely convex closed set $A$ there is a unique projection of $x$ onto $A$. (see e.g. "Metric spaces, convexity and nonpositive curvature" by Athanase Papadopoulos, IRMA lectures in mathematics and theoretical physics). All Hilbert spaces are strictly convex; the space $l_\infty$ which appears in Andrew Stacey's answer is not. | |
| Aug 13, 2010 at 10:59 | answer | added | Andrew Stacey | timeline score: 14 | |
| Aug 13, 2010 at 9:52 | history | edited | Ian Morris |
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| Aug 13, 2010 at 9:11 | answer | added | Zen Harper | timeline score: 2 | |
| Aug 13, 2010 at 8:47 | answer | added | Gjergji Zaimi | timeline score: 11 | |
| Aug 13, 2010 at 8:41 | history | edited | Yemon Choi | CC BY-SA 2.5 |
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| Aug 13, 2010 at 8:34 | comment | added | Ian Morris | Non-uniqueness of the best approximation doesn't require a complicated or unusual Banach space: take X to be \mathbb{R}^2 equipped with a norm whose unit ball is not strictly convex, and play around with some examples. (I assume you meant to require that y should be unique, otherwise the two statements which you give are not equivalent.) | |
| Aug 13, 2010 at 8:33 | comment | added | Martijn | Can you give a counterexample? | |
| Aug 13, 2010 at 8:25 | history | asked | Linda Raabe | CC BY-SA 2.5 |