Square Root of the shift operator indexed by $\mathbb{Z}$ My question is very similar to this question, but instead of indexing by $\mathbb N$ I am indexing by $\mathbb Z$.
Consider the shift operator $T : \ell^1(\mathbb Z) \to \ell^1(\mathbb Z) $ given by $$T(...,x_{-1},x_0,x_{1},... )=(...,x_{-2},x_{-1},x_0,...)$$
that is, the operator that shifts elements one to the right.
Does there exist $R : \ell^1(\mathbb Z) \to \ell^1(\mathbb Z) $ such that $R^2=T$? 
Since $\ker T=0$ the proof from the post I linked to no longer works, but the intuition of the poster still remains.
 A: 
Consider the definition of $f(U)$ given by
$$
    f(U)x = \frac{1}{2\pi}\int_{0}^{2\pi}f(e^{i\theta})\left(\sum_{n=-\infty}^{\infty}e^{-in\theta}U^{n}x\right)d\theta
$$
If $f(z)=z^{n}$, then $f(U)=U^{n}$ for $n=0,\pm 1,\pm 2,\pm 3,\cdots$. The inner sum is a vector function of $\theta$:
$$
    S(\theta)x=\sum_{n=-\infty}^{\infty}e^{-in\theta}U^nx
$$
The $k$ element of the sequence is
$
          (S(\theta)x)_{k} = \sum_{n=-\infty}^{\infty}e^{-in\theta}x_{k-n}.
$
This is defined for every $k$, but there is some question of whether or not the resulting sequence is in $\ell^1$.
So it makes sense to try
$$
        (\sqrt{U}x)_k = \frac{1}{2\pi}\int_{0}^{2\pi}e^{i\theta/2}\sum_{n=-\infty}^{\infty}e^{in\theta}x_{k-n}d\theta \\
       = \sum_{n=-\infty}^{\infty}\left(\frac{1}{2\pi}\int_{0}^{2\pi}e^{i(\theta/2-n\theta)}d\theta\right)x_{k-n} \\
       = \sum_{n=-\infty}^{\infty}\left.\frac{e^{i(1/2-n)\theta}}{2\pi i(1/2-n)}x_{k-n}\right|_{\theta=0}^{2\pi} \\
       = \sum_{n=-\infty}^{\infty}\frac{1}{\pi i(n-1/2)}x_{k-n}.
$$
Different branches of $\sqrt{z}$ will give different expressions because different branches will translate to differ intervals of integration. I would start by applying this construction to the simple sequences $e_n$ where $(e_n)_k = \delta_{n,k}$ is the Kronecker delta.
