Timeline for Fourier transform of $\exp(-\|x\|_p)$: more general question
Current License: CC BY-SA 4.0
7 events
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| May 1, 2021 at 21:30 | history | edited | Mark Lewko | CC BY-SA 4.0 |
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| Apr 16, 2021 at 18:32 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Math Jaxed
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| Oct 24, 2009 at 18:09 | history | edited | Mark Lewko | CC BY-SA 2.5 |
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| Oct 20, 2009 at 14:21 | comment | added | Tom Leinster | Just a note on terminology, because it's somewhat treacherous. What many analysts call "positive definite" is what one might these days prefer to call "positive semidefinite": the assertion is that some quantity is \geq 0, with no conditions on when equality is attained. What they call "strictly positive definite" is what one might these days call "positive definite". I'm using the analysts' convention. | |
| Oct 20, 2009 at 14:19 | comment | added | Tom Leinster | Thanks again. But I'm still not sold. A version of Bochner's Theorem is that a continuous L^1 function is strictly positive definite iff it is bounded and its Fourier transform is non-negative and not identically zero. So, my question is equivalent to asking whether e^{-||x||_p} is strictly positive definite - and in fact, that's the question I really want answered. The question about Fourier transforms was a reformulation that I thought people would find easier to answer. The results of Schoenberg show that e^{-||x||_p} is pos def. But my difficulty is: why is it strictly pos def? | |
| Oct 20, 2009 at 3:30 | history | edited | Ben Webster♦ | CC BY-SA 2.5 |
added 8 characters in body
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| Oct 18, 2009 at 19:38 | history | answered | Mark Lewko | CC BY-SA 2.5 |