Timeline for An inequality inspired by the isoperimetric inequality
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 27, 2017 at 5:05 | comment | added | Fan Zheng | @user50396 I guess the fact that the extremal $\rho$ can describe a translated circle means that any proof of the isoperimetric inequality must deal with this "gauge freedom", one way or another, but polar coordinates is not best suited to deal with translations. | |
| Jul 27, 2017 at 4:12 | answer | added | Paata Ivanishvili | timeline score: 3 | |
| Jul 24, 2017 at 13:08 | comment | added | user50396 | @PaulBryan For the time being I just want to understand this simple case. Since the inequality looks simple enough, there must be a way to avoid the isoperimetric inequality. But let me remark that for many curves (even non-star-shaped ones), both area and length can be expressed as a signed integral over the circle, possibly by partitioning the circle and regard $\rho$ as multi-valued functions (and also add suitable signs for the area integral). Not sure how useful it is though. | |
| Jul 24, 2017 at 12:57 | comment | added | user50396 | @PaataIvanisvili I am not familiar with this approach. Can you elaborate more? | |
| Jul 24, 2017 at 12:53 | history | edited | user50396 | CC BY-SA 3.0 |
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| Jul 23, 2017 at 1:17 | comment | added | Paul Bryan | I think your problem is that an arbitrary (even convex) simple, closed curve doesn't have a polar representation $r(\theta) e^{i\theta}$ the points on the curve are one to one with $\theta \in [0, 2\pi)$. For that you need to do something like translate the curve so it encloses the origin in such a way that the resulting curve becomes star shaped around the origin (so convex if fine). Then I suspect you are just working backwards from Wirtinger $\Rightarrow$ Isoperimetric inequality modulo translation. | |
| Jul 22, 2017 at 22:46 | comment | added | Paata Ivanishvili | If you are interested I can elaborate little bit more: it reduces the problem to finding a certain Bellman function of 4 variables with a very simple obstacle condition. | |
| Jul 22, 2017 at 22:44 | comment | added | Paata Ivanishvili | @user50396 One dimensional inequality: you can always try the standard Bellman function approach like I discussed here before mathoverflow.net/questions/199418/… using Hamilton--Jacobi--Bellman PDE | |
| Jul 22, 2017 at 3:17 | comment | added | fedja | but the equality in our inequality can only hold when $\rho$ is constant Not really: it holds for circles containing the origin but not all such circles are centered at the origin. You have actually noticed it yourself in the next sentence. | |
| Jul 21, 2017 at 20:28 | history | asked | user50396 | CC BY-SA 3.0 |