Timeline for Is a function of several variables convex near a local minimum when the derivatives are non-degenerate?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 29, 2020 at 21:18 | history | became hot network question | |||
| Jul 29, 2020 at 17:29 | comment | added | Willie Wong | So something like $d^k Hess(f)$ being positive would probably work. (So if $k = 4$ is the smallest $k$ for which $d^k f \neq 0$ you want $\partial_v \partial_v Hess(f)$ to be a PD matrix.) This can be written as for any $v, w\neq 0$ that $d^kf(v,v,\ldots, v, w,w) > 0$. | |
| Jul 29, 2020 at 17:22 | comment | added | Willie Wong | @Mateusz already gave you a counter example, but I think that $d^kf > 0$ is definitely too weak. Morally you want $Hess(f)$ to be positive semi definite in a neighborhood, and this suggests that since you are in the case $Hess(f)(0) = 0$, you want a suitable number of higher derivatives of the matrix valued function $Hess(f)$ to take values in the symmetric positive semidefinite matrices. Assuming $d^3f > 0$ does not control $\partial_x \partial^2_{yy} f$ (which is essentially what Mateusz used in his example). | |
| Jul 29, 2020 at 16:40 | vote | accept | Asaf Shachar | ||
| Jul 29, 2020 at 15:18 | answer | added | Mateusz Kwaśnicki | timeline score: 12 | |
| Jul 29, 2020 at 14:25 | answer | added | Iosif Pinelis | timeline score: 5 | |
| Jul 29, 2020 at 13:17 | history | asked | Asaf Shachar | CC BY-SA 4.0 |