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There is a simple characterization of shift-invariant closed subspaces of $l^2$: for any measurable subset $S$ of $\mathbb{T} = \mathbb{R}/2\pi\mathbb{Z}$, the set of elements of $l^2$ whose Fourier transform is supported on $S$ is a shift-invariant closed subspace, and every such subspace has this form. This is pretty easy.

Is the analogous statement true of shift-invariant closed subspaces of $l^1$? I.e., for any such subspace $E$ there is a measurable subset $S \subseteq \mathbb{T}$ such that $E = \{f \in l^1: \hat{f}$ is supported on $S\}$?

If so, my next question would be which measurable subsets of $\mathbb{T}$ arise in this way, as the "support" of a shift-invariant closed subspace of $l^1$?

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    $\begingroup$ To be picky, to have a complete description in the $\ell^2$ case, one has to say for which pairs $S_1,S_2$ of measurable subsets of the circle we have $F_{S_1}=F_{S_2}$, where $F_S$ is the set of elements in $L^2$ of the circle that are zero outside $S$; I guess it holds iff $S\triangle T$ has measure zero. In the $\ell^1$ case, Fourier transforms are continuous and hence the discussion will be different: letting $G_S$ be the set of $\ell^1$ elements whose Fourier transform is zero outside $S$, I guess $G_{S_1}=G_{S_2}$ iff $S_1$ and $S_2$ have equal interiors. $\endgroup$
    – YCor
    Commented Jun 30, 2017 at 8:16
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    $\begingroup$ Shift-invariant closed subspaces of $\ell^1$ are ideals wrto convolution, so the question equivalently asks for a characterization of the closed ideals of the Wiener algebra $A(\mathbb{T})$, doesn't it? $\endgroup$
    – Pietro Majer
    Commented Jun 30, 2017 at 10:14
  • $\begingroup$ These are both great insights ... surely the ideals of $A(\mathbb{T})$ must be well understood? If they all have the desired form, I think these two comments solve the problem completely. $\endgroup$
    – Nik Weaver
    Commented Jun 30, 2017 at 13:23
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    $\begingroup$ Nik: as Matt's answer indicates below, the ideals of A(T) are not well understood in full generality, which is why "Banach algebras bad, operator algebras trendy" or some such. Pietro: the closed ideals of C^1(T) are well understood but they are NOT merely the ones given by the zero-set construction; in the language of Matt's answer, points of T are not sets of synthesis for C^1(T). It's a nice exercise to work out exactly which ideals have "hull" equal to a given singleton... $\endgroup$
    – Yemon Choi
    Commented Jun 30, 2017 at 16:21
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    $\begingroup$ @Yemon Choi: The idea was that the function $f_E:=\mathrm{dist}(\cdot,E)$ is lipschitz, hence it is in $A(\mathbb{T})$; and the hull of the ideal generated by $f_E$ is $E$ $\endgroup$
    – Pietro Majer
    Commented Jun 30, 2017 at 21:12

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This is not a full answer, as my memory / personal bookshelf is not good enough. But it should give some hints.

As the comments suggest, we can reduce this to a question about the Fourier algebra $A(\mathbb T)$. Indeed, the steps are:

  • We can turn $\ell^1(\mathbb Z)$ into a commutative Banach algebra for the convolution product.
  • Closed, shift-invariant subspaces of $\ell^1(\mathbb Z)$ are the same as closed ideals of $\ell^1(\mathbb Z)$.
  • The Fourier transform $\ell^1(\mathbb Z) \rightarrow \mathbb A(\mathbb T)$ is an isometric algebra homomorphism.
  • So we wish to classify closed ideals of $A(\mathbb T)$.
  • All this works for a general locally compact abelian group.
  • Under a suitable identification, $\mathbb T$ is the spectrum of $\ell^1(\mathbb Z)$ and the Fourier transform is nothing but the Gelfand transform.

Let $I$ be a closed ideal of $A(\mathbb T)$. The hull of $I$ is $$ \nu(I) = \{ h\in \mathbb T : f(h)=0 \ (f\in I) \} $$ a closed subset of $\mathbb T$. Conversely, if $N\subseteq\mathbb T$ is a closed subspace then the kernel of $N$ is $$ \iota(N) = \{ f\in A(\mathbb T) : f(h)=0 \ (h\in N) \} $$ a closed ideal in $A(\mathbb T)$. Clearly $I \subseteq \iota(\nu(I))$ but sadly we do not always have equality.

We say that an ideal $I$ has spectral synthesis if we have $I = \iota(\nu(I))$. If $G$ is a discrete abelian group then every closed ideal in $A(G)$ has spectral synthesis. If $G$ is a non-discete abelian group, then a theorem of Malliavin shows that there is a closed ideal without spectral synthesis.

In conclusion, closed subsets of $\mathbb T$ do not classify closed ideals of $\ell^1(\mathbb Z)$.

Theorem: If the boundary of $\nu(I)$ does not contain a non-empty perfect set then $\iota(\nu(I))=I$ so $I$ is of spectral synthesis.

Theorem: If $f\in A(G)$ vanishes on a neighbourhood of $\nu(I)$ then $f\in I$.

This 2nd result hints at the connection between spectral synthesis and the ability (or not) to be able to approximate elements in a closed ideals by elements which vanish on a slightly large set than the hull.

(This is all in Folland's book "A course in Abstract Harmonic Analysis". You should find a lot more in Hewitt and Ross Vol 2, or Rudin's book, etc.)

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  • $\begingroup$ Apologies if you mentioned this already in your answer, but if not: you can address Nik's second question by using the fact $A({\bf T})$ is regular, I think? (Regular in the sense of function algebras, that is) $\endgroup$
    – Yemon Choi
    Commented Jun 30, 2017 at 16:35
  • $\begingroup$ Thank you! That's funny, I was somehow under the impression that if $G$ is compact abelian then every closed set is a set of synthesis. Not sure why I thought that ... $\endgroup$
    – Nik Weaver
    Commented Jun 30, 2017 at 16:42
  • $\begingroup$ @Nik I think the confusion comes from whether you are studying the algebra $L^1(G) = A(\hat G)$ or $A(G) = L^1(\hat G)$. I gave my answer for the Fourier algebra, but it's more common in the (classical) literature to work with the L^1 algebra, hence the compact / discrete confusion. $\endgroup$
    – Matthew Daws
    Commented Jun 30, 2017 at 16:45

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