Convolution identically $0$ implies $f =0$?

Question

Let $f, g \in L^2(\mathbb R^n)$. Is it true that if $g > 0$ and $$(f * g)(z) = 0$$ for all $z \in \mathbb R^n $, then $$f=0$$ almost everywhere?

Answer 1

Hint: Define $$ \phi(\xi)=\left\{\begin{array}{} e^{\frac1{|\xi|^2-1}}&\text{if $|\xi|\lt1$}\\ 0&\text{if $|\xi|\ge1$} \end{array}\right. $$ Since $\phi$ is real and even, $\widehat{\phi}$ is real.

Define $g(x)=\widehat{\phi}(x)^2$ and $f_\alpha(x)=g(x)\cos(\alpha\cdot x)$.

What can you say about the supports of $\widehat{g}$ and $\widehat{f_\alpha}$?

What can you say about $g\ast f_\alpha$?