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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.

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    $\begingroup$ 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? $\endgroup$ Sep 19, 2021 at 17:34

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