Timeline for About optimizing decay rate of Fourier transforms?
Current License: CC BY-SA 3.0
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| Nov 16, 2017 at 15:10 | comment | added | Willie Wong | why do you think the expression you wrote decay to 0 "infinite times when $t\to \infty$"? (In fact, by the Fourier inversion formula + Riemann-Lebesgue, if you have an integrable function whose Fourier transform decays "infinite times when $t\to\infty$", then the original function must be $C^\infty$.) | |
| Nov 16, 2017 at 15:03 | history | edited | Willie Wong |
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| Nov 16, 2017 at 12:11 | comment | added | CC95 | Thank you for your comments. Suppose $g(x) = |x|^p, x \in [-1,1]$ and is smooth and compactly supported otherwise. Since $\int_{-1}^{1}|x|^p cos(tx)dx =2*( \frac{sin(t)}{t} - \frac{\int_{0}^{t}px^{p-1}sin(x)dx}{t^{p+1}})$ and it will decay to $0$ infinite times when $t \rightarrow \infty$, the tail is still not as slow as $O(x^{-\beta})$. So it confuses me wether we can find a $\phi_{f}(x)$ that has a polynomial tail. | |
| Nov 15, 2017 at 20:47 | comment | added | Willie Wong | The condition $|f'| < \epsilon |f|$ is not that restrictive. Let $g$ be any compactly supported $C^1$ function. Then for $\delta$ sufficiently small the function $$ f = \exp (- \frac12 \epsilon \sqrt{1 + x^2}) + \delta g(x) $$ would satisfy your constraint. The part of the Fourier transform coming from the $\exp$ would decay exponentially fast. You just need to make $g$ very poorly differentiable at some points (say something like $|x-a|^p$ where $p > 1$ is very close to 1) to guarantee a slow decay of the Fourier transform. | |
| Nov 15, 2017 at 20:13 | review | Close votes | |||
| Nov 16, 2017 at 12:10 | |||||
| Nov 15, 2017 at 19:32 | history | edited | CC95 | CC BY-SA 3.0 |
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| Nov 15, 2017 at 19:22 | review | First posts | |||
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| Nov 15, 2017 at 19:20 | history | asked | CC95 | CC BY-SA 3.0 |