Timeline for Are Fourier transforms of L^p stable under diffeomorphisms?
Current License: CC BY-SA 3.0
8 events
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| Aug 20, 2016 at 15:36 | comment | added | Christian Remling | @PabloShmerkin: Yes, you're right of course, that was obvious in fact. | |
| Aug 20, 2016 at 13:16 | comment | added | Pablo Shmerkin | @ChristianRemling - $\widehat{\mu}$ is Lipschitz, so in this case it is true that if it doesn't go to $0$ then it is in no $L^p$, added edit explaining this. | |
| Aug 20, 2016 at 13:15 | history | edited | Pablo Shmerkin | CC BY-SA 3.0 |
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| Aug 19, 2016 at 15:43 | history | edited | Pablo Shmerkin | CC BY-SA 3.0 |
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| Aug 19, 2016 at 15:37 | comment | added | Pablo Shmerkin | @Denis Serre - thanks, it was a typo, $\phi$ has to be strictly convex, not strictly increasing. | |
| Aug 19, 2016 at 15:37 | history | edited | Pablo Shmerkin | CC BY-SA 3.0 |
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| Aug 19, 2016 at 15:35 | comment | added | Denis Serre | I don't understand: one can take $\phi$ the identity, and then $\widehat{\phi^*\mu}=\hat\mu$ does not decay. | |
| Aug 19, 2016 at 15:24 | history | answered | Pablo Shmerkin | CC BY-SA 3.0 |