I'm currently looking at a subspace of $A \subset \ell^p(\mathbb{Z}^n)$ which is generated by some finitely supported elements and their translations. My question is an old one (but the answer is apparently behind a paywall): is $A$ complemented in $\ell^p$?
If I remember correctly, this reduces [result of Rudin] to the question whether there is a bounded projection $\ell^p(\mathbb{Z}^n) \to A$ which is translation invariant. To do this one first make some Fourier analysis. The dual of $\mathbb{Z}^n$ is the torus $\mathbb{T}^n$ (seen as $\subset \mathbb{C}^n$). Translation-invariance becomes multiplication in the dual. One then looks at the dual of $A$: this will be generated by a polynomial (the variables take on unit-valued complex numbers) together with multiplication with $x_i$ and $x_i^{-1}$.
Since polynomials are fairly dense in $\ell^p$, this reduces to studying the place where they all vanish. More precisely, the annihilator of $A$ in the dual $\mathbb{T}^n$ is the intersection $Z$ of the zero loci of those polynomials... it has measure 0. The projection [in the dual] is then given by the multiplication with the characteristic function $\chi_{Z^\mathsf{c}}$ of the complement of $Z^\mathsf{c}$. I believe that if $p >1$, $A$ is just equal to $\ell^p(\mathbb{Z}^n)$. In the case $p=2$, this follows from the fact the von Neumann dimension of $A$ is 1, which means that the complement has 0-dimension and hence is trivial. Does this also hold for $p \in ]1;2[$?
The case $p=1$ is particularly interesting: I'm pretty sure, that $A$ is not always dense. Is there a [translation invariant] projection $\ell^1 \to A$?
