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.
 A: 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.
