A kind of Tauberian theorem for $L^2(\mathbb R^n)$?

Question

For $f \in L^2(\mathbb R^n)$, let $\hat f(y)=\int_{\mathbb R^n} f(x) e^{-2\pi ixy}dx , \forall y \in \mathbb R^n $ . Let $M$ be a closed translation invariant (i.e. $f \in M \implies f_a\in M , \forall a \in \mathbb R^n$, where $f_a(x)=f(x+a),\forall x \in \mathbb R^n$) linear subspace of $L^2(\mathbb R^n)$ ; then is it true that there is a measurable subset $E \subseteq \mathbb R^n$ such that $M=\{f \in L^2(\mathbb R^n) : \hat f$ vanishes a.e. on $E \}$ ?

Answer 1

Yes. This follows from the usual $L^2$ Wiener Tauberian theorem.

I will work entirely in Fourier space, with $\hat M=\{\hat f\vert f\in M\}.$ Let $G$ be a Gaussian (or another strictly positive $L^2$ function of your choice).

Let $K$ be an arbitrary compact set. Given any $f$ in $\hat M,$ the usual theorem says we can approximate $1_{K}$ by linear combinations of modulations of $f+G1_{f=0}.$ But taking the same linear combinations of modulations of $f,$ we get approximations to $1_K1_{f\neq 0}$ that are at least as good - we’re just removing part of the integral from the calculation of the $L^2$ distance. So $1_K1_{f\neq 0}\in M.$

The set of indicator functions of subsets of $K$ in $\hat M$ is closed under finite unions, and hence under taking countable unions. We can take a countable union $E’$ achieving the supremal measure of all such indicator functions. The earlier trick using $G$ shows that $\hat M$ contains every $L^2$ function that is zero on the complement of $E’$ plus a null set. So $K\setminus E’$ does the job of the desired $E,$ for functions with Fourier transform restricted to $K.$

By a standard argument expressing the space as a countable union of compact sets, and using approximation by compactly supported functions, we can build a set $E$ such that a function is in $\hat M$ if and only if it is zero on $E$ up to a null set.