About convolution and Fourier transform

Question

I have some doubts with this question:

I we have $f,g\in\cal{S}$ (where $\cal{S}$ is the Schartz space) with $f\ast g=0$,

Can we deduce that $f=0$ or $g=0$?

What I did is apply Fourier transform, so $\cal{TF}(f\ast g) =0$. This means $\cal{TF}(f)\cdot\cal{TF}{(g)}=0$.

So $\cal{TF}(f)=0$ or $\cal{TF}(g)=0$. And as $\cal{TF}:\cal{S}\to\cal{S}$ is injective, we can deduce that $f=0$ or $g=0$.

Is that right? I'm not sure about my reasoning...

Thanks a lot in advance.

Answer 1

You were on the right track. Let's denote $\hat f$ and $\hat g $ the respective Fourier transforms. You correctly concluded that $\hat f(\xi) \hat g(\xi)=0$ for all $\xi$ (i.e. pointwise).

Now take two functions $h_1,\,h_2$ in $\mathcal C^\infty_c(\Bbb R)\subset \mathcal S(\Bbb R)$ (i.e. compactly supported). Suppose also that their supports are disjoint, then their product is always zero. Now take $f,\,g$ such that $\hat f=h_1$, $\hat g=h_2$.

Hence, we can have two Schwartz functions with their convolution being zero everywhere.