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Let us consider the finite group algebra $\ell^1(\mathbb{Z}_n)$. Let $x=(x_0,\cdots,x_{n-1})$ in $\ell^1(\mathbb{Z}_n)$ and define $$M_x: \ell^1(\mathbb{Z}_n)\to \ell^1(\mathbb{Z}_n) : M_x(a)=a*x$$ where $*$ is just the convolution product.

Q. Any characterization (formula) for linear maps $T: \ell^1(\mathbb{Z}_n)\to \ell^1(\mathbb{Z}_n)$ satisfying $T^2=M_x$?

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    $\begingroup$ I suppose $\mathbb{Z}_n$ means $\mathbb{Z}/n\mathbb{Z}$, not $n$-adics. But then why do you write $\ell^1(\mathbb{Z}_n)$ for what I suppose is just $\mathbb{C}^{\mathbb{Z}/n\mathbb{Z}}$ (the $1$ is purely irrelevant, right?)? Doesn't Fourier analysis immediately show the answer to be “convolution by the inverse Fourier transform of any square root of the Fourier transform of $x$”? Is there some subtlety that I missed? $\endgroup$
    – Gro-Tsen
    Commented Jun 16, 2022 at 12:47

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