Timeline for (Quasi) convexity of separately convex homogeneous functions
Current License: CC BY-SA 3.0
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| Oct 29, 2016 at 21:05 | history | edited | user_lambda | CC BY-SA 3.0 |
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| Oct 18, 2016 at 13:47 | history | edited | user_lambda | CC BY-SA 3.0 |
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| Oct 12, 2016 at 18:01 | history | edited | user_lambda | CC BY-SA 3.0 |
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| Oct 9, 2016 at 15:33 | history | edited | user_lambda | CC BY-SA 3.0 |
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| Oct 8, 2016 at 20:07 | history | edited | user_lambda | CC BY-SA 3.0 |
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| Oct 8, 2016 at 15:36 | comment | added | user_lambda | Thanks, that's useful to know. I've edited the question more time (sorry about the back and forth) to impose that $f$ is decreasing in each of its arguments. The examples I've found are all convex. | |
| Oct 8, 2016 at 15:34 | history | edited | user_lambda | CC BY-SA 3.0 |
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| Oct 8, 2016 at 15:20 | comment | added | Paata Ivanishvili | this example $B(x,y,z)=\frac{xy}{z}$ is 1-homogeneous, separately convex, and increasing only in x and y variables but not in z, is not quasicovex. I understand that in general counterexample cannot be made by product functions like $B(x,y,z,..)=f_{1}(x) f_{2}(y) f_{3}(z)..$. This will never work. So joint structure should be involved. | |
| Oct 8, 2016 at 14:58 | comment | added | user_lambda | I haven't been able to find a counterexample. I have also relaxed some of the assumptions on $f$ since I couldn't come up with an example that was not linear. | |
| Oct 8, 2016 at 14:46 | history | edited | user_lambda | CC BY-SA 3.0 |
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| Oct 8, 2016 at 2:45 | comment | added | Paata Ivanishvili | I doubt that this is true. First I would try to construct any function which is 1 homogeneous, separately convex, increasing in each variable but not jointly convex. Do you have such example? | |
| Oct 8, 2016 at 0:39 | history | edited | user_lambda | CC BY-SA 3.0 |
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| Oct 7, 2016 at 19:33 | history | edited | user_lambda | CC BY-SA 3.0 |
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| Oct 7, 2016 at 19:09 | history | asked | user_lambda | CC BY-SA 3.0 |