Discrete convolution and supports: can we have $\operatorname{supp}(f * g) \subsetneq \operatorname{supp}(f) + \operatorname{supp}(g)$?

Question

Let $G$ be a finite abelian group. For $f,g:G\rightarrow \mathbb{C}$ functions. Define their convolution in a point $x\in G$ as $$(f*g)(x)=\sum_{y\in G}f(x-y)g(y)$$

From the definition of convolution is easy show that $$\operatorname{supp}(f*g)\subseteq \operatorname{supp}(f)+\operatorname{supp}(g)$$

When $\operatorname{supp}(f)$ denotes the set of points in $x\in G$ for which $f(x)\neq 0$, and with similar definitions with $\operatorname{supp}(g)$ and $\operatorname{supp}(f*g)$.

But, I spent a pair of hours trying to find functions $f$ and $g$ for which the equality does not hold in general. Since I do a lot of examples in finite abelian groups of the form $\mathbb{Z}/n\mathbb{Z}$ and I always get the equality (is important remark the fact that I always use indicator functions), I'm interested in know examples in finite abelian groups for which the equality does not hold.

Any suggestion or example is welcome.

Answer 1

Consider $G := \mathbb{Z}/2n\mathbb{Z}$ for some $n \in \mathbb{N}$, $n \neq 0$. Let $A := \left\{\bar{0}, \bar{2}, \bar{4}, \cdots, \overline{2n-2}\right\}$ and $B := G \setminus A$.
Then, we can observe, using $1_C$ to denote the indicator function for any $C \subset G$ and $1$ to denote the constant function equal to $1$ on $G$, that: $$\left(\left(1_A - 1_B\right) * 1\right)(x) = \sum_{y \in G} \left(1_A(x - y) - 1_B(x-y)\right) = \left|\left\{y \in G \mid x - y \in A\right\}\right| - \left|\left\{y \in G \mid x - y \in B\right\}\right|$$ Now, $G$, $A$ and $B$ were chosen in this particular manner in order to have the two cardinals be equal no matter the $x$ taken, hence the previous convolution is actually the zero function. Indeed, we can see that:

• $G = A \cup B$ with $A \cap B = \emptyset$ and $|A| = |B| = n$;
• If $x \in A$, then $x - y \in A$ iff $y \in A$;
• If $x \in B$, then $x - y \in A$ iff $y \in B$.

Hence: $$\operatorname{supp}\left(\left(1_A - 1_B\right) * 1\right) = \emptyset \neq \operatorname{supp}\left(1_A - 1_B\right) + \operatorname{supp}(1) = G + G = G$$ and we have the desired counterexample.