# convolution of non-zero functions

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.

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

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.

• 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". – Sergio Nov 16 '14 at 23:29
• 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. – Sergio Nov 16 '14 at 23:29
• 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 $f \ast g \equiv 0$. – PhoemueX Nov 17 '14 at 20:58