Timeline for Fourier transform of a function of bounded variation
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Jan 11 at 19:30 | comment | converted from answer | Juan H. Arredondo | When f is of bounded variation the Fourier sine transform F^s(f) and the Fourier cosine transform F^c(f) -which form the Fourier transform of a function f := F^f(f)= F^c(f) -i F^s(f) have different behavior. Respectively, F^s(f) is not (HK-)integrable while F^c(f) is. | |
| Jun 28, 2020 at 5:20 | answer | added | PhoemueX | timeline score: 1 | |
| Jun 27, 2020 at 20:20 | comment | added | Christian Remling | No, for example the FT of step function has decay $1/x$. (I assume that a hat is missing and you're asking if a BV function has integrable FT.) | |
| Jun 27, 2020 at 20:15 | history | asked | A beginner mathmatician | CC BY-SA 4.0 |