Shift invariant subspaces of $l^1$ 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 $l^1$?
 A: 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.)
