I noticed that although $C_c^{\infty}$functions are dense in some quite large spaces and well understood (especially their Fourier transform) I have never encountered an explicit example of a Fourier transform $F(f)$ for $f \in C_c^{\infty}$ in my life (there are good reasons for this, as most explicit expression for $C_C^{\infty}$ look rather difficult to integrate). Does anybody know an example, where the Fourier transform can be explicitly calculated or are there just no examples?

$\begingroup$ First how about a Fourier transform that is not realanalytic? $\endgroup$– Gerald EdgarApr 24, 2016 at 1:08

2$\begingroup$ @GeraldEdgar: We need the opposite here: $\widehat{f}$ is entire, of exponential type, and a Schwartz function on $\mathbb R^d$. (And I don't think there will be a very explicit example.) $\endgroup$– Christian RemlingApr 24, 2016 at 1:47

$\begingroup$ ... and conversely, any such function will have a FT in $C_0^{\infty}$. $\endgroup$– Christian RemlingApr 24, 2016 at 1:48

$\begingroup$ Also posted on MSE: math.stackexchange.com/questions/1755999/… $\endgroup$– Christian RemlingApr 24, 2016 at 2:02
2 Answers
$$ \hat f(\xi) = \prod_{j=1}^\infty \operatorname{sinc}(\xi/2^j)$$ will work (it is rapidly decreasing, and obeys the hypotheses of the PaleyWiener theorem; alternatively, its inverse Fourier transform can be written explicitly as an infinite convolution that is manifestly compactly supported).


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4$\begingroup$ For the lazy, $\operatorname{sinc} x=\sin(\pi x)/(\pi x)$. $\endgroup$ Apr 24, 2016 at 9:19

1$\begingroup$ Nice. This is done in great generality in Theorem 1.3.5 in the first volume of Hormander's book. By the same trick one can construct compactly supported functions with higher degree of smoothness $\endgroup$ Apr 24, 2016 at 20:30
The simplest infinitely differentiable function with compact support is $f(x)=\exp\left(1/x(1x)\right), \; 0<x<1$. This is the first example given in textbooks. Its Fourier transform is $$F(z)=\int_0^1e^{izx}f(x)dx.$$ Is this an "explicit" formula? In not, why is the infinite product in Terence Tao's answer "explicit"?
(I think the word "explicit" (and "analytic expression", which some people think is synonimous) are very much abused on this site. These words always require a further explanation of their meaning).
Back to $C^\infty$ functions with compact support. The inverse Fourier transform of $\mathrm{sinc}$ is the characteristic function of an interval. Thus the example in Tao's answer is an infinite convolution of these characteristic functions. (Is this explicit or not?) H/"ormander, Analysis of partial differential operators, vol. I, Chap I, sect 1.3 discusses at great length how to construct $C^\infty$ functions with compact support by taking infinite convolutions of characteristic functions, and surprisingly, obtains a proof of the DenjoyCarleman quasianalyticity theorem by elementary methods:)

2$\begingroup$ In this context, "explicit formula" could be taken to mean "formula not visibly the same as the definition of the Fourier transform". $\endgroup$ Apr 29, 2016 at 19:56