Timeline for Precise decay of density through Fourier transform
Current License: CC BY-SA 4.0
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| Nov 19, 2021 at 18:06 | comment | added | Christian Remling | The general principle is that decay of the FT corresponds to smoothness of the function and vice versa. | |
| Nov 19, 2021 at 17:30 | comment | added | Mateusz Kwaśnicki | One way to relate the decay of $f$ at infinity with regularity of $\varphi$ is provided by the theory of regular variation — if your $f$ is regularly varying at infinity, you may find this interesting. This definitely applies to symmetric densities $f$; I can give some references if you are interested. I never really needed that for non-symmetric $f$. | |
| Nov 19, 2021 at 13:24 | history | edited | Uchiha | CC BY-SA 4.0 |
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| Nov 19, 2021 at 12:44 | history | edited | Uchiha | CC BY-SA 4.0 |
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| Nov 19, 2021 at 12:13 | comment | added | Uchiha | Thanks for the comment. Yes I am aware that in general the estimate may not be improved. However like I illustrated for function of power decay, there is a gap. So maybe an alternative question is, how is the power decay precisely reflected by Fourier transform. | |
| Nov 19, 2021 at 2:12 | comment | added | Willie Wong | Isn't the loss coming just from chaining together non-sharp inequalities? You certainly do not need $f = o(|x|^{-p-1})$ for it to have finite moment $\int |x|^p f(x) ~dx < \infty$. And there are also $f\in O(|x|^{-p})$ which do not have Fourier transform with integrable $p$th derivatives. This is to say nothing of changing between $L^\infty$ and $L^1$ scales. It is unclear to me what you are hoping for as an answer here. | |
| Nov 19, 2021 at 1:11 | history | asked | Uchiha | CC BY-SA 4.0 |