Timeline for property of convex functions
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Sep 19, 2017 at 18:16 | history | edited | Hammerhead | CC BY-SA 3.0 |
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| Sep 7, 2017 at 4:59 | vote | accept | Hammerhead | ||
| S Sep 7, 2017 at 4:56 | history | bounty ended | Hammerhead | ||
| S Sep 7, 2017 at 4:56 | history | notice removed | Hammerhead | ||
| Sep 7, 2017 at 4:55 | vote | accept | Hammerhead | ||
| Sep 7, 2017 at 4:59 | |||||
| Sep 6, 2017 at 3:16 | answer | added | fedja | timeline score: 10 | |
| S Aug 31, 2017 at 16:48 | history | bounty started | Hammerhead | ||
| S Aug 31, 2017 at 16:48 | history | notice added | Hammerhead | Draw attention | |
| Aug 25, 2017 at 19:27 | history | edited | Hammerhead | CC BY-SA 3.0 |
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| Aug 23, 2017 at 15:53 | history | edited | Hammerhead | CC BY-SA 3.0 |
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| Aug 23, 2017 at 5:23 | history | edited | Hammerhead | CC BY-SA 3.0 |
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| Aug 23, 2017 at 5:23 | comment | added | Hammerhead | Indeed, Fan. It is better to assume that (edited). Though otherwise, $P$ would have to be contained in a lower dimensional hyperplane, in which case we could just take a lower dimensional integral. | |
| Aug 23, 2017 at 5:14 | comment | added | Fan Zheng | Probably you want $P$ to be open (or at least has positive measure). Otherwise when $P$ is a segment in $\mathbb R^2$ the left bound cannot hold. | |
| Aug 23, 2017 at 2:48 | history | edited | Hammerhead | CC BY-SA 3.0 |
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| Aug 23, 2017 at 1:55 | comment | added | Hammerhead | Fedja, indeed $A$ and $B$ can be take to be zero. Thanks for pointing that out, I edited the question. I was blinded with the particular application of this inequality that interests me, and that naturally involves the constants $A,B$. Anyway, If you know a 5 line proof for the first inequality please share it with us. My argument is more convoluted, that is why I am (still) interested in a reference. | |
| Aug 23, 2017 at 1:47 | history | edited | Hammerhead | CC BY-SA 3.0 |
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| Aug 22, 2017 at 22:32 | comment | added | fedja | It is one of those situations where sending the reader to the library wastes more of his time than writing a 5-line proof, so I would advise not to bother about any references here though the choice is yours. Still I'm unable to understand what $A$ and $B$ are doing in a completely homogeneous problem. | |
| Aug 22, 2017 at 7:43 | comment | added | Kavi Rama Murthy | The right-hand inequality does not even require convexity of P or f provided P is bounded and f is continuous.: let g=max{f,0} and h=-min{f.0}. Then the right-hand inequality follows from the fact that 2$\int g$ $\leq$ 2(sup g) m(P)=2(sup f) m(P) by changing f to -f. Note that sup f $\geq$ 0 by the hypothesis (unless f is 0). | |
| Aug 21, 2017 at 22:15 | comment | added | Yoav Kallus | @GeraldEdgar I think you missed that $\alpha,\beta,A,B>0$. | |
| Aug 21, 2017 at 21:41 | history | asked | Hammerhead | CC BY-SA 3.0 |