Timeline for Fourier Transform, for entire function
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Dec 3, 2012 at 2:34 | comment | added | Michael Renardy | I don't understand the last comment. The Fourier transform of any distribution of compact support, in particular any test function in D, is an entire function. It is holomorphic everywhere, not just outside a strip. | |
| Dec 2, 2012 at 21:47 | comment | added | Peter Michor | Attention here: The Fourier transform of a test function (on $\mathbb R$, say) is rapidly decreasing on dual $\mathbb R$. If you insist on extending the Fourier transform to the complexification $\mathbb C$, then you look at the Laplace transform in the imaginary direction, which has exponential growth. The extension to $\mathbb C$ is holomorphic outside a strip parallel to $\mathbb R$ whose thickness depends on the support of the function you started with. | |
| Dec 2, 2012 at 17:13 | comment | added | Michael Renardy | The Fourier transforms of functions in D lie in a certain space of analytic functions, which is called Z. So the Fourier transform maps D to Z and vice versa. By duality, the Fourier transform then maps D' to Z' and Z' to D'. Your function f is in D', and its Fourier transform is in Z'. If you want more details, read Gelfand and Shilov as I suggested in the first place. Then you can tell me if I am making sense. | |
| Dec 2, 2012 at 15:12 | comment | added | Matthias Ludewig | Your statement about defining the Fourier transform by duality does not make sense... As you pointed out yourself, the fourier transform of a function in D is not in D again, but in S. Hence taking the dual operator gives again only an operator from S' to D'. | |
| Dec 2, 2012 at 15:02 | comment | added | Michael Renardy | Any distribution f (and therefore any locally integrable function defined on the reals) has a Fourier transform F, which is defined as an analytic functional. In general F is not a distribution. Analytic functionals act on test functions which are entire, for instance in your example $\delta(\xi-is)$ can be defined only for analytic test functions. | |
| Dec 2, 2012 at 11:33 | comment | added | Lwins | Mmm.. Then, is a entire function $f$ have a Fourier transform $F$ (perhaps $F$ is a distribution) for sure? If not, then the answer for [THIS][1] is incomplete. [1]: mathoverflow.net/questions/114875 | |
| Dec 2, 2012 at 11:24 | history | answered | Michael Renardy | CC BY-SA 3.0 |