Timeline for Minimization problem for convolution
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 2, 2019 at 23:04 | answer | added | Stefan Steinerberger | timeline score: 5 | |
| Oct 1, 2019 at 8:49 | comment | added | Ilya Bogdanov | I would rather say that the Laplace transform of the uniform distribution is not a square of an entire function, due to the behaviour at zeroes. | |
| Oct 1, 2019 at 8:45 | comment | added | Ilya Bogdanov | Sorry for being silly, but which consequences you make from squaring a complex-valued function? | |
| Oct 1, 2019 at 8:06 | comment | added | Kurisuto Asutora | The Fourier transform of an indicator function is not non-negative real. In contrast, a convolution of a function with itself leads to a non-negative real Fourier transform (since we are squaring). So 1 cannot be reached exactly. However, I don't have an estimate how close to 1 one can actually get. | |
| Sep 30, 2019 at 21:47 | comment | added | kodlu | Silly question. Is the reason that the L_1 norm lower bound of 1 cannot be achieved the normalization of the Fourier transform on [0,1]? | |
| Sep 30, 2019 at 11:08 | answer | added | Mateusz Kwaśnicki | timeline score: 4 | |
| Sep 30, 2019 at 7:22 | comment | added | Mateusz Kwaśnicki | Interestingly, exactly the same norm $\|g\|_1 = 2/\sqrt{\pi} \approx 1.13$ is attained both by $g(x) = (\pi x)^{-1/2}$ and by $g(x) = \max\{(\pi x)^{-1/2},(\pi (1/2-x))^{-1/2}\} \mathbb{1}_{(0,1/2)}(x)$. | |
| Sep 30, 2019 at 7:03 | history | asked | Kurisuto Asutora | CC BY-SA 4.0 |