What I have here is a sequence $c\in\ell^2(\mathbb Z^2)$ for which each convolution power $c*\ldots*c$ exists and is contained in $\ell^2(\mathbb Z^2)$, too. It is clear from $c\in\ell^2(\mathbb Z^2)$ that $c^{[2]} = c*c$ is contained in $\ell^\infty(\mathbb Z^2)$, but I actually know that $c^{[2]}\in\ell^2(\mathbb Z^2)$. Same for $c^{[3]} = (c*c)*c$ and so on...
Clearly the class of these sequences $c$ contains $\ell^1(\mathbb Z^2)$. But is it larger? In addition, I know that there exists some $n$ such that $c^{[n]}$ is a basis vector $e_m$.
What I would like to have is that $\widehat{c^{[n]}} = \hat c^n$ (convolution theorem), where $\hat d$ is the DTFT of $d\in\ell^2(\mathbb Z^2)$, i.e., $$ \hat d(x) = \sum_{k\in\mathbb Z^2}d_ke^{-2\pi i\langle k,x\rangle}, $$ and that $\hat c$ is continuous (which both would be true if $c\in\ell^1(\mathbb Z^2)$).
Any idea anyone?