does anybody know an example of a real function $f\in C_c^\infty(\mathbb R)\setminus\{0\}$ (non-zero, smooth with compact support) such that for its fourier transform $\hat f(\xi)=\int_\mathbb R f(x)e^{-2\pi i x\xi}~dx$ a closed/explicit form (without integral) is known?

Or the other way around: Does somebody know an explicitly represented schwartz function $g\neq 0$ on $\mathbb R$ such that $\hat g\in C_c^\infty(\mathbb R)$?

I've tried some constructions with $e^{-1/x^2}$, but the occurring integrals aren't solvable (at least for me and maple ;)).

Thank you.

  • $\begingroup$ There are some (entire) power series representations, but no closed-form as far as I know. $\endgroup$ – reuns Oct 5 '17 at 17:01
  • $\begingroup$ Okay, I feared that.. but can you give me such a series, please? $\endgroup$ – tofurind Oct 5 '17 at 21:45
  • $\begingroup$ Are you sure that there is a analytic functions g with this property after all? Assume yes. Since $\hat g\in C_c^\infty$, we may assume that it is symmetric and vanishing in a neighborhood of zero (otherwise translate, add copies of that, etc). Then, $\hat g$ has a root in zero of infinite order which means that $0=((i\partial)^k \hat g)(0) = \widehat{x^kg}(0) = \int x^k g(x)dx$ for all $k$. So g is orthogonal on all polynomials. Moreover, g should be orthogonal on analytic functions, especially on itself. But this means $g=0$. (?) $\endgroup$ – tofurind Oct 5 '17 at 23:03
  • 1
    $\begingroup$ Although I disclaim serious expertise on this precise issue, it is my impression that it is essentially impossible to write a completely elementary expression for the Fourier transform of a test function, or for its power series expansion. (@tofurind, the existence, in fact, classification of such things is the enhanced form of the Paley-Wiener theorem. Google-able.) $\endgroup$ – paul garrett Oct 5 '17 at 23:05
  • 1
    $\begingroup$ @tofurind I essentially meant if $f \in C^\infty_c$ then $F(z) = \int_{-\infty}^\infty f(x) e^{-xz}dx$ is entire. $\endgroup$ – reuns Oct 5 '17 at 23:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.