For $f : [1,\infty) \to \mathbb C$, it is known that $g(x)= \sum_{1\le n \le x, n \in \mathbb N} f(x/n), \forall x \in [1,\infty)$ iff $f(x)=\sum_{1\le n \le x, n \in \mathbb N} \mu(n) g(x/n), \forall x \in [1,\infty)$ .
So if we define $\widehat f(x):=\sum_{1\le n \le x, n \in \mathbb N} f(x/n)$ and $\check f(x):=\sum_{1\le n \le x, n \in \mathbb N} \mu(n) f(x/n)$, then $f=\widehat{\check f}=\check {\widehat f}$ .
My question is : Let $f : [1,\infty) \to \mathbb C$ be a not identically zero function; then can both $f$ and $\widehat f$ have compact support ? If this is possible, my next question would be , is it possible that one of $f, \widehat f$ has finite support and the other compact support ?
If $supp (f) \subseteq \mathbb N$ then I can show that both $supp (f)$ and $supp(\widehat f)$ cannot be compact i.e. cannot be finite (compact subsets of the subspace $\mathbb N$ are finite) ; but I am unable to see what happens in general . Please help. Thanks in advance.