Timeline for Quadratic Farkas' Lemma?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Nov 7, 2012 at 8:25 | vote | accept | Seva | ||
| Nov 6, 2012 at 22:42 | history | edited | Markus Schweighofer | CC BY-SA 3.0 |
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| Nov 6, 2012 at 21:17 | history | edited | Seva | CC BY-SA 3.0 |
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| Nov 6, 2012 at 21:16 | comment | added | Seva | It does seem to work for a pentagon - thanks for the nice example! | |
| Nov 6, 2012 at 20:14 | history | edited | Markus Schweighofer | CC BY-SA 3.0 |
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| Nov 6, 2012 at 20:12 | comment | added | Markus Schweighofer | I am so sorry, you are again right. But if you take a pentagon instead of a square, then it works. Not all of the $L_i L_j$ were strictly positive on at least one of the vertices of the square but for the pentagon this works. | |
| Nov 6, 2012 at 20:06 | history | edited | Markus Schweighofer | CC BY-SA 3.0 |
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| Nov 6, 2012 at 17:59 | comment | added | Seva | $L_1=1-x$, $L_2=1+x$, $L_3=1-y$, $L_4=1+y$, and $P=2-x^2-y^2$ is not a counterexample in view of $P=L_1L_2+L_3L_4$. | |
| Nov 6, 2012 at 14:50 | comment | added | Markus Schweighofer | You are right, I am sorry for suggesting this counterexample. However, I think that I have now a valid counterexample, see above. | |
| Nov 6, 2012 at 14:48 | history | edited | Markus Schweighofer | CC BY-SA 3.0 |
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| Nov 6, 2012 at 7:29 | history | edited | Markus Schweighofer | CC BY-SA 3.0 |
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| Nov 6, 2012 at 7:10 | comment | added | Seva | $L_1=1+x$, $L_2=1-x$, and $P=x^2$ is not a counterexample: take all $c_i$ and $c_{ij}$ to be equal to $0$. Many thanks for the references! | |
| Nov 6, 2012 at 7:00 | history | edited | Markus Schweighofer | CC BY-SA 3.0 |
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| Nov 6, 2012 at 4:16 | history | edited | Markus Schweighofer | CC BY-SA 3.0 |
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| Nov 6, 2012 at 4:11 | history | answered | Markus Schweighofer | CC BY-SA 3.0 |