Timeline for Square root of dirac delta function
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 16, 2021 at 5:27 | review | Suggested edits | |||
| Apr 16, 2021 at 8:32 | |||||
| Apr 11, 2016 at 4:46 | comment | added | BigM | Very interesting. If I remember it correctly in probability they call this cross-correlation. So according to T.Tao's statement there are measurable functions of constant cross-correlation. | |
| Apr 10, 2016 at 18:25 | history | edited | GH from MO |
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| Apr 10, 2016 at 17:35 | vote | accept | DLN | ||
| Apr 10, 2016 at 17:25 | answer | added | Terry Tao | timeline score: 49 | |
| Apr 10, 2016 at 16:40 | comment | added | Terry Tao | Nice question! There is an a priori bound: for any interval $I$, one has $(\int_I f)^2 \leq \int_{I+I} f*f = |I+I| = 2|I|$, so $\int_I f \leq \sqrt{2} |I|^{1/2}$. In particular $\hat f$ exists as a tempered distribution at least. Unfortunately there is not yet enough regularity to justify the equation $\hat f^2 = \delta$. Nevertheless the answer should still be negative, I think. | |
| Apr 10, 2016 at 16:00 | comment | added | DLN | @Carlo Beenakker, in chapter 5, it has been proved that there is no real valued function whose square is the dirac delta function. But I can't exactly see how that solves this problem. I will read the chapter thoroughly and see if I can find something that helps. Thanks for the reference. | |
| Apr 10, 2016 at 15:48 | comment | added | DLN | Yeah. But in this case its a positive function. So the integrations make sense. | |
| Apr 10, 2016 at 15:46 | comment | added | Alexandre Eremenko | "Measurable" is not enough for convolution to exist. You should describe precisely what is your class of functions. | |
| Apr 10, 2016 at 15:37 | comment | added | Carlo Beenakker | euclid.ucc.ie/pages/staff/hanzon/Report2.pdf (in particular chapter 5) | |
| Apr 10, 2016 at 14:49 | history | asked | DLN | CC BY-SA 3.0 |