Given $\mathbf{x} \in \mathbb{R}^n$ that the norm is fixed $\left\Vert \mathbf{x}\right\Vert^2 = n\times \rho$.
Define circular shift operator $\mathrm{circshift}([x_0, x_1, ..., x_n-1], \tau) = [x_{n-\tau}, x_{n-\tau +1}, ..., x_{n-1}, x_0, ..., x_{n-\tau-1}]$ with $\tau \in\{1,2,...,n-1\}$
I want to find the bounds of $\left\Vert \mathbf{x} - \mathrm{circshift(}\mathbf{x},\tau)\right\Vert^2$ by $\tau$, i.e. $G$ and $H$ that
$$\left\Vert \mathbf{x} - \mathrm{circshift(}\mathbf{x},\tau)\right\Vert^2 < G \quad\textrm{ for } 1 \leq \tau \leq n-1$$
and $$\left\Vert \mathbf{x} - \mathrm{circshift(}\mathbf{x},\tau)\right\Vert^2 > H \quad\textrm{ for } 1 \leq \tau \leq n-1$$
and I expect that the bounds depend on the norm $\rho$ and the length $n$.
Call $\mathbf{X}$ the DFT of $\mathbf{x}$.
Thus by Parseval theorem, $$\left\Vert \mathbf{x}\right\Vert^2 = \frac{1}{n}\sum_{k=0}^{m-1} |X(k)|^2 = P$$
and
$$\left\Vert \mathbf{x} - \mathrm{circshift(}\mathbf{x},\tau)\right\Vert^2 = \frac{1}{n}\sum_{k=0}^{m-1} |X(k)|^2 |1 - e^{j2\pi\frac{k}{m}\tau}|^2$$
The first thing that cross my mind is that $|1 - e^{j2\pi\frac{k}{m}\tau}| < 2$, thus $\left\Vert \mathbf{x} - \mathrm{circshift(}\mathbf{x},\tau)\right\Vert^2 = \frac{1}{n}\sum_{k=0}^{m-1} |X(k)|^2 |1 - e^{j2\pi\frac{k}{m}\tau}|^2 < 4\times\frac{1}{n}\sum_{k=0}^{m-1} |X(k)|^2 = 4n\rho$ which seems quite loose.
Any hints or inequalites that I can use to derive $G$ and $H$ would be appreciated.