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Aug 24, 2020 at 11:33 history edited Asaf Shachar CC BY-SA 4.0
deleted 192 characters in body
Aug 21, 2020 at 5:25 history edited Asaf Shachar CC BY-SA 4.0
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Aug 20, 2020 at 17:24 comment added Ron P @AsafShachar. Got it. See another proof below. Thanks!
Aug 20, 2020 at 11:53 answer added Ron P timeline score: 2
Aug 20, 2020 at 11:38 comment added Asaf Shachar @RonP No, I don't assume that, why? Do you have a simpler proof for the case where $F$ is increasing?
Aug 20, 2020 at 11:32 comment added Ron P Is assumed that $F$ is increasing?
Aug 19, 2020 at 6:27 comment added Pietro Majer Yes! I realized it with the tentative counterexample in the answer then deleted. The fact it does not work may be turned into a positive proof
Aug 18, 2020 at 18:31 comment added Asaf Shachar Dear Pietro, it seems that your suggestion cannot work. You may see Iosif Pinelis's answer below which proves that the answer is in fact positive.
Aug 18, 2020 at 18:29 vote accept Asaf Shachar
Aug 18, 2020 at 15:25 answer added Iosif Pinelis timeline score: 4
Aug 18, 2020 at 7:21 comment added Pietro Majer One can also just do it for powers of 2, b_n=2^n
Aug 18, 2020 at 6:39 comment added Pietro Majer It should be easy to make a strictly convex function with $F(2n)/2+F(0)/2 -F(n)=1/n$, as envelope of a family of lines. (that is a counterexample with $a_n=0, c_n=n, b_n=2n, \lambda_n=1/2$)
Aug 18, 2020 at 6:20 history edited Asaf Shachar CC BY-SA 4.0
added 50 characters in body
Aug 18, 2020 at 5:20 history asked Asaf Shachar CC BY-SA 4.0