Timeline for Discontinuous convolutions
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Aug 17, 2010 at 18:18 | history | edited | Vince | CC BY-SA 2.5 |
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| Aug 17, 2010 at 15:38 | comment | added | Vince | My mistake, you are right - I have been working with $\mathbb{T}$ for so long I forgot I require $f$ to be compactly supported. | |
| Aug 17, 2010 at 4:49 | comment | added | Zen Harper | Also, the space of Fourier transforms of continuous $L^1$ functions (as well as the Fourier transforms of $L^1$ itself) doesn't have any nice characterisation as far as I know, so even if we could take the Fourier transform, it wouldn't necessarily be able to help us...consider $e^{-x^2} sin(\exp(\exp(\exp(x))))$, which is $C^\infty$ and $L^1$, but with very nasty derivatives!! | |
| Aug 17, 2010 at 1:36 | comment | added | Zen Harper | I think this problem is more subtle than it looks, and this answer is incorrect as written (without a lot of extra explanation). For example, I'm fairly sure that $f'$ need not make sense as a tempered distribution in general, so how can you take the Fourier transform? $f'$ can be very badly behaved! | |
| Aug 17, 2010 at 1:19 | comment | added | Nate Eldredge | Please double-check the hypotheses of the theorems you're using. The proofs I'm familiar with require $f'$ to be integrable, which is not assumed here. | |
| Aug 17, 2010 at 1:14 | comment | added | senti_today | What if f' is not integrable? | |
| Aug 17, 2010 at 1:13 | history | edited | Vince | CC BY-SA 2.5 |
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| Aug 17, 2010 at 1:02 | history | answered | Vince | CC BY-SA 2.5 |