Timeline for Showing that a Gaussian achieves equality in a logarithmic Sobolev inequality
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 29, 2021 at 19:38 | answer | added | Iosif Pinelis | timeline score: 2 | |
| Apr 29, 2021 at 17:37 | comment | added | Deane Yang | I also am puzzled by why they included the second term. There is a virtually identical sharp log-Sobolev inequality that holds without the second term. | |
| Apr 29, 2021 at 17:36 | comment | added | Deane Yang | Sorry. I misread what you wrote. I don't know the exact constants for this form of the log-Sobolev inequality. Their proof should also show that equality holds only if $f$ is an appropriate Gaussian. You should go through that part of the proof carefully and verify that it is consistent with the constants shown. | |
| Apr 29, 2021 at 17:12 | history | edited | Ma Joad | CC BY-SA 4.0 |
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| Apr 29, 2021 at 17:11 | comment | added | Ma Joad | @DeaneYang Yes I have tried. I know where $\ln a$ comes from. It is in the norm $\|f\|_2$ when we take logarithm of it. But I did not get $n.$ So I do not get the $1$ in $n(1+\ln a).$ | |
| Apr 29, 2021 at 17:05 | comment | added | Deane Yang | Did you try computing the RHS? The $\ln a$ is hidden inside the $C$. | |
| Apr 29, 2021 at 16:42 | comment | added | Ma Joad | Thank you for pointing out. I have missed a square at the very end. $a$ is a free parameter. This expression does come from the proof in some way, but I'd like to see it more directly. | |
| Apr 29, 2021 at 16:40 | history | edited | Ma Joad | CC BY-SA 4.0 |
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| Apr 29, 2021 at 15:41 | comment | added | Iosif Pinelis | This inequality cannot be true in general, as it is not homogeneous in $f$. Did you copy it correctly? Also, what is $a$ here? Plus, I guess, the equality, if true, follows from the proof of the inequality. | |
| Apr 29, 2021 at 15:00 | history | asked | Ma Joad | CC BY-SA 4.0 |