Timeline for Poincare inequality on the hemisphere
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Feb 8 at 23:20 | comment | added | Student | Oh wait $\int \sin(r)^2\neq 0,$ maybe I am missing something here.... | |
| Feb 8 at 23:13 | comment | added | Student | @GiuseppeNegro I am guessing the extremizers are eigenfunctions on the hemisphere, which are the coordinate functions. Writing them co-ordinates, $x_1=\sin(r)cos(\theta),x_2=\sin(r)\sin(\theta).$ Then $\int \sin(r)^3 = 2/3$ while $\int \cos(r)^2 \sin(r) = 1/3$ so I am guessing that the constant has to be at least $2$? | |
| Feb 8 at 23:02 | comment | added | Student | @IosifPinelis (i) mathoverflow.net/questions/295683/… might be relevant (ii) you are right, thanks for pointing that out! | |
| Feb 8 at 22:59 | history | edited | Student | CC BY-SA 4.0 |
deleted 30 characters in body
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| Feb 8 at 19:14 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Feb 8 at 18:58 | comment | added | Giuseppe Negro | I agree with Iosif that $r$ should run on $[0, \pi/2]$. Also, if you show some details on the proof of the first statement, it will probably shed light on the second. It may well be that a maximizer to the first inequality is radial (i.e. it depends on $r$ only), in which case the second follows. | |
| Feb 8 at 18:52 | comment | added | Iosif Pinelis | (i) "we know that [...]" -- How/from where do we know that? (ii) Shouldn't $\int_{-\pi/2}^{\pi/2}$ be replaced everywhere by $\int_0^{\pi/2}$? | |
| Feb 8 at 18:42 | history | asked | Student | CC BY-SA 4.0 |