Suppose $f$ is a smooth function with $\mathbb{supp}{(\mathcal{F}{f})} \subset B(0,1)$. In addition, assume $f$ is non-negative.
We can observe that the function $|f|^2$ has a nice property : $$\mathbb{supp{(\mathcal{F}(|f|^2))}}=\mathbb{supp{(\mathcal{F}(f)\ast\mathcal{F}(f)})} \subset B(0,2)$$ Let $1 < p < 2$. Does there exist a constant $C$ independent of $p$ such that $$\mathbb{supp(\mathcal{F}(|f|^p))} \subset B(0,C)$$ for all $f$ satisfying the above conditions? This is not homework or exercise in some book but my pure question.
What I did :
Let $p$ be a rational number. Put $p=\frac{n}{m}$, $n$, $m$ $\in \mathbb{N}$. First, we have $$\mathbb{supp{(\mathcal{F}(|f|^n))}} \subset B(0,n)$$ Put $g(x)=|f(x)|^{\frac{n}{m}}$. Then $$\mathcal{F}(g) \; \ast \; ... \; \ast \; \mathcal{F}(g) (m-times) = \mathcal{F}(|f|^n)$$ If $\mathcal{F}(g)$ is always non-negative, then $$\mathcal{F}(g) \subset B(0,\frac{n}{m})$$ But in general we cannot deduce that because $\mathcal{F}g$ can be negative. I don't believe that my claim is correct, but finding counterexample is extremely hard. Any comments or hints are welcomed.