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Apr 27, 2023 at 21:11 history edited Medo CC BY-SA 4.0
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Apr 27, 2023 at 21:06 history edited Medo CC BY-SA 4.0
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Apr 27, 2023 at 21:02 comment added Fedor Petrov Ah, only at 0. Sorry, I misread
Apr 27, 2023 at 20:35 comment added Medo Would you like to elaborate a bit? Condition (3) says the Hessian is positive definite at the origin, so by smoothness, the Hessian is positive definite in some neighborhood of the origin. On that neighborhood the function is strictly convex. What do we infer outside that neighborhood given conditions (1)-(3) ?
Apr 27, 2023 at 8:10 comment added Medo This is brilliant. But the inequality $f(x)>\delta^{-1} f(\delta x)$ applies to $x$ that lies in convex set that contains $0$ on which $f$ is convex. The question remains : Is the claim ''$f$ is convex '' true and what is the best way to argue that?
Apr 27, 2023 at 8:00 history edited Medo CC BY-SA 4.0
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Apr 27, 2023 at 7:31 history edited Medo CC BY-SA 4.0
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Apr 27, 2023 at 7:22 history edited Medo CC BY-SA 4.0
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Apr 26, 2023 at 20:26 answer added Bazin timeline score: 1
Apr 26, 2023 at 19:29 history edited Medo
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Apr 26, 2023 at 19:17 history undeleted Medo
Apr 26, 2023 at 19:15 history deleted Medo via Vote
Apr 26, 2023 at 19:14 history asked Medo CC BY-SA 4.0