# Injectivity of the convolution operator $f \mapsto (k*f(s))_{s \in S}$ via sampling at $S=\alpha \mathbb Z$

Let $$S \subset \mathbb R$$ be a set of sampling points, say $$S = \alpha \mathbb Z, \alpha >0$$. Let $$k$$ be some convolution kernel and $$A$$ the operator which maps some $$f$$ to the sequence $$Af = (k*f(s))_{s \in S}$$ I.e. $$A$$ maps $$f$$ to the samples of the convolution of $$f$$ with $$k$$.

Problem:

Restrict $$A$$ to the space of compactly supported $$L^1$$-functions on $$[0,1]$$, i.e. $$L^1[0,1]$$. For which $$k$$ does there exist a set $$S=\alpha \mathbb Z$$ such that the resulting operator $$A$$ is injective? In formulas: $$\forall f,g \in L^1[0,1] : (k*f(x)=k*g(x) \ \forall x \in S \implies f=g).$$ There are obvious candidates for $$k$$ where this is not true, e.g. compactly supported $$k$$.

I was wondering if anyone of you came accross such problems, knows if there's a characterization of such functions $$k$$ or can point towards papers regarding this problem. The papers I found so far mainly deal with situation where the domain of $$A$$ is equal to $$L^p(\mathbb R)$$. Here, I'm intersted in functions with compact support.

• Since $(K*f)(n)=\langle K_n, f\rangle$, with $K_n(x)=\overline{K(n-x)}$, an obvious reformulation is: For what $K$ do the translates $K_n$ span the whole space? Sep 19, 2021 at 17:34