I am trying to find two non-zero functions $f,g\in L^1(\mathbb{R}^n;\mathbb{C})$ such that $f\star g=0$. For that I applied the Fourier transform and got that $f$ and $g$ must satisfy $\hat{f}\hat{g}=0$. My idea was to choose $\hat{f},\hat{g}$ continuous with disjoint (compact) support (We could define $h_1,h_2:\mathbb{R}\to\mathbb{C}$ to be two disjoint triangles in $\mathbb{R}^+$ and then make $\hat{f}(\xi)=h_1(||\xi||)$, $\hat{g}(\xi)=h_2(||\xi||)$).
However the resolution of this exercice says that we have to chose $\hat{f},\hat{g}$ infinitely differentiable and with disjoint support. It says that this is needed so that we will have $\lim_{|x|\to\infty}|x|^n f(x)=0$, which would imply that $f\in L^1(\mathbb{R}^n;\mathbb{C})$.
Why is it true and why my first attempt fails?
Suppose $\lim_{n\to \infty}|x|^nf(x)=0$ for all $n\in \mathbb{N}$, and $f\in L^{1}_{loc}(\mathbb{R}^d)$ (i.e. $f\in L^1(K)$ for all $K\subset \mathbb{R^d}$ compact). Here you just need $n=d+1$: if $\lim_{n\to \infty}|x|^{d+1}f(x)=0$ then we have
$$|f(x)|\leq C|x|^{-(d+1)},\qquad \forall |x|\geq r $$ for some $r>0$ and $C>0$. But then $$\int_{\mathbb{R}^d} |f|\leq \int_{|x|\leq r}|f|+C\int_{|x|>r}\frac{1}{|x|^{d+1}}dx<\infty$$ and so $f\in L^1(\mathbb{R}^d)$. The assumption $f\in L^1_{loc}(\mathbb{R}^d)$ is necessary, otherwise $f$ might have a non-integrable singularity, which is obviously not affected by the condition $\lim_{x\to \infty}|x|^nf(x)=0$.
Either way, if $\hat{f}$, $\hat{g}$ are smooth and compactly supported then they are in $L^2(\mathbb{R}^d)$, and so since the Fourier transform maps $L^2$ onto itself bijectively, $f$ and $g$ are also in $L^2$, and in particular in $L^1_{loc}$ (here you might also replace $L^2$ with the Schwartz class).