2
$\begingroup$

Let $f,g$ be two continuous functions with compact support. Show that if $f$ and $g$ are not identically $0$, then neither is $f\ast g$.

This statement seems rather elementary, and I would prefer if the proof was also so, i.e avoiding reference to Titchsmarsh convolution theorem, and if possible not using Fourier transforms.

$\endgroup$
  • $\begingroup$ Laplace transforms ok? I'm guessing no $\endgroup$ – Simon S Nov 16 '14 at 15:21
  • $\begingroup$ Well if you want to write a solution it's always welcome ! $\endgroup$ – Sergio Nov 16 '14 at 17:43
2
$\begingroup$

A basic property of the Fourier transform is that it is not possible for $f$ and $\hat{f}$ to be both compactor supported (if $f\not \equiv 0$). To see this, note that

$$ \hat{f}(\xi) = \int_{[-R,R]^n} f(x) \cdot e^{2\pi i \langle x,\xi \rangle} \, dx $$

defines a holomorphic function on $\Bbb{C}^n$, as can be seen by differentiation under the integral sign (here $\mathrm{supp} (f) \subset [-R,R]^n$ for suitable $R>0$), so that the identity theorem for holomorphic functions implies that the support of $\hat{f}$ can not be compact. Even more, $\hat{f}$ can not vanish on any nondegenerate cube.

The convolution theorem implies

$$ \widehat{f\ast g} = \hat{f} \cdot \hat{g}, $$

where none of the two factors on the right can vanish on a nondegenerate cube. By continuity, $\hat{f} \cdot \hat{g}$ can not vanish on any nondegenerate cube.

Hence, $\widehat{f\ast g} \not\equiv 0$, so that $f\ast g \not\equiv 0$.

I am aware that this solution does not satisfy your requirement of avoiding the Fourier transform, but maybe it is better than having no proof at all.

$\endgroup$
  • $\begingroup$ Ok thanks a lot ! Indeed this exercise comes from a complex analysis book, and the previous exercise was "Prove that the Fourier transform of a continuous compact supported function cannot also be continuous compact supported". $\endgroup$ – Sergio Nov 16 '14 at 23:29
  • $\begingroup$ It is though a lot of heavy machinery for a result so easily formulated, and I'm trying to see deeper into this, perhaps find counter-examples if we don't suppose $f$ and $g$ compact supported. $\endgroup$ – Sergio Nov 16 '14 at 23:29
  • 1
    $\begingroup$ If you do not assume compact support, the Fourier transform can give you an easy counterexample. Take e.g. $\varphi, \psi \in C_c^\infty$ with disjoint supports and let $f := \mathcal{F}^{-1} \varphi$ and $g := \mathcal{F}^{-1} \psi$. Then $\widehat{f \ast g} = \varphi \cdot \psi \equiv 0$, so that also $f \ast g \equiv 0$. $\endgroup$ – PhoemueX Nov 17 '14 at 20:58

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.