This is a special case of a previous question I asked, which I hope will receive more attention due to the specificity.
Suppose we have a function space $F = \{f : \mathbb Z^2 \to \mathbb R\}$ and a discrete convolution $\phi : F \to F$ given by $$\phi(f) = p \mapsto \sum_{\delta \in \mathbb Z^2}\gamma_\delta f(p - \delta)$$ where $\gamma \in \mathbb R[\mathbb Z^2]$ is an element of the group ring of reals indexed by the 2d integer lattice. (We may treat $\gamma$ simply as a finitely-supported map $\gamma : \mathbb Z^2 \to \mathbb R$, so that the sum defining $\phi(f)$ above makes sense: the sum becomes a finite sum since only finitely many $\gamma_\delta$ are nonzero.) We place a condition on $\gamma$: the support of $\gamma$ must generate $\mathbb Z^2$ (as a group), so that the kernel of $\phi$ is not made of infinitely many equivalent "slices".
I am trying to prove the following conjecture:
Conjecture: If $\phi(f) = 0$ then $f$ is either the zero function or has infinite support.
For convenience let $F_{fin} \leq F$ be the subspace of finitely-supported functions in $F$.
One plan I have is to show that $\ker \phi$ is finite-dimensional, since if $\ker \phi$ is finite-dimensional then $\ker \phi \cap F_{fin}$ must also be finite-dimensional, and it is not hard to see that $\ker \phi \cap F_{fin}$ must have dimension either 0 or $\infty$ (since if $\phi(f) = 0$ with $f$ finitely supported, then for any translate $f'$ of $f$, still $\phi(f') = 0$ but for infinitely many $f'$ we have $f'$ linearly independent from $f$).
However, I am not yet able to show that $\ker \phi$ is finite-dimensional.
Can anyone prove the finite-dimensionality of $\ker \phi$? Or perhaps solve the conjecture a different way?
I'll give it a try:
If $\gamma:\>{\mathbb Z}^2\to{\mathbb C}$ has finite support then $$\hat\gamma(x,y):=\sum_{k, l}\gamma(k,l)e^{2\pi i(kx+ly)}$$ is a trigonometric polynomial in the variables $x$ and $y$. Furthermore, if $f:\>{\mathbb Z}^2\to{\mathbb C}$ has finite support then $\hat f(x,y)$ is a trigonometric polynomial as well. Now one computes $$\eqalign{\widehat{\phi(f)}(x,y)&=\sum_{k,l}\phi(f)(k,l)e^{2\pi i(kx+ly)}\cr &=\sum_{k,l}\left(\sum_{r,s}\gamma(r,s)f(k-r,l-s)\right)e^{2\pi i(kx+ly)}\cr &=\sum_{r,s}\sum_{k',l'}\gamma(r,s)f(k',l')e^{2\pi i((k'+r)x+(l'+s)y)}\cr &=\hat\gamma(x,y)\>\hat f(x,y)\ .\cr}$$ If we now had $\phi(f)=0$ then this would imply $\widehat{\phi(f)}(x,y)\equiv0$. On the other hand, the product of two nonzero trigonometric polynomials cannot be identically $0$.
Note that I did not use your technical assumption on $\gamma$.