3

Let $A$ be a closed subspace of $\ell^2(\Bbb Z)$. Suppose for every $\{a_n\}_n \in A$, we have $\{a_{n+m}\}_n \in A$ for each $m \in \Bbb Z$. Show that there exists a measurable set $E\subset \Bbb T$ such that $$A = \{\{\hat f(n)\}_{n\in \Bbb Z}: f\in L^2(\Bbb T), \operatorname{supp} f \subset E\}$$

Let $\mathcal F: L^2(\Bbb T) \to \ell^2(\Bbb Z)$ be the usual map $f\mapsto \{\hat f(n)\}_{n\in \Bbb Z}$. This is a surjective isometry, and so $\mathcal F^{-1}$ makes sense. My first guess for $E$ is $\bigcup_{f\in \mathcal F^{-1} (A)} \operatorname{supp} f$ but this may not even be a measurable set. I'd like some hints or ideas to complete this proof!

stoic-santiago
  • 13,705
  • I think you should be able to take the borel closure of the set you were proposing (see borel regular measure) – Eric Sep 05 '23 at 09:42
  • 1
    One way to think of the operator on $\ell^2$ given by $(a_n) \mapsto (a_{n-1})$ (a unitary often called the bilateral shift) is as "multiplication by $z$" in the usual Fourier basis (often written $M_z$). If $A$ is invariant under $M_z$ and $P$ is orthogonal projection onto $A$, then $P$ commutes with $M_z$, and in fact with any multiplication operator, and thus (by some work) is itself a multiplication operator (it's $\chi_E$!). See the answer to https://math.stackexchange.com/questions/2541462/showing-that-if-an-operator-commutes-with-all-multiplication-operator-it-must-b for several details – leslie townes Sep 05 '23 at 09:44

0 Answers0