I came across this question while exploring the growth rates of abelian groups. (It's tangentially related.)
Suppose $G$ and $K$ are abelian groups, and $G$ is generated by some finite subset $S$. Now let $\hat G$ be the abelian group (under pointwise addition) of functions $G \to K$, i.e. $\hat G = \{f : G \to K\}$. Finally, suppose $c : S \to \mathbb Z$ is any function, and let $\phi : \hat G \to \hat G$ be the endomorphism given by $$\phi(f) = g \mapsto \sum_{s \in S} c(s) \cdot f(g - s).$$ Thus $\phi$ is a kind of discrete convolution, and $c$ is the "kernel".
I'm trying to determine when it is that solutions to $\phi(f) = 0$ have finite support, i.e. what some necessary and sufficient conditions on $c$ such that if $\phi(f) = 0$ then $f = 0$ or $f$ has infinite support.
For example, if $c$ has singleton support, say $c(s_0) = n$ and $c(s) = 0$ for all other $s$, and no torsion element of $K$ (except for the identity) has order dividing $n$, then $\ker \phi$ is trivial, so if $\phi(f) = 0$ we necessarily have $f = 0$.
Another example: if $G = K = \mathbb Z$, and we define $c : \{-1, 0, 1\} \to \mathbb Z$ by $c(\pm 1) = 1$ and $c(0) = 2$, then it turns out that all solutions to $\phi(f) = 0$ are of the form $$f(n) = (-1)^n ((n-1)a + nb)$$ with $a, b \in \mathbb Z$, and it is easy to see that if $f \neq 0$ then $f$ necessarily has infinite support.
Examples like this lead me to expect the following result:
Conjecture: Suppose $\ker \phi \subsetneq \hat G$ and $f \in \ker \phi$. Then $f = 0$ or $f$ has infinite support.
This is trivially true in the singleton-support example above since $\ker \phi = \{0\}$, but it less obviously true in the second example, since there $\ker f$ is not all of $\hat G$, but every non-zero solution to $\phi(f) = 0$ has infinite support.
Hence my question: is my conjecture correct? If not, what conditions are necessary on $c$ and/or $\phi$ instead of requiring $\ker \phi \subsetneq \hat G$?