Downsampling, shifting, high pass and low pass filter commutativity
I am suspicious due to the claim that $\mathcal{R}_0$ is the inverse of $\mathcal{D}_0$. Decimation is not invertible. Once samples/entries are deleted, those values are forgotten. They cannot be reconstructed from what remains.



Let's say the input finite sequences are in $\mathbb{R}^{2N}$, numbered as $x = (x[0],x[1],\ldots,x[2N-1])$. Then $\mathcal{D}_0:\mathbb{R}^{2N}\to\mathbb{R}^N$, and $\mathcal{S}$ is overloaded, so that it maps from $\mathbb{R}^{2N}$ to $\mathbb{R}^{2N}$ and from $\mathbb{R}^{N}$ to $\mathbb{R}^{N}$:
\begin{eqnarray}
(\mathcal{S}x)[n] &amp;=&amp; x[n+1~\textrm{mod}~2N]~~\textrm{for}~~x\in\mathbb{R}^{2N},\\
(\mathcal{S}y)[n] &amp;=&amp; y[n+1~\textrm{mod}~N]~~\textrm{for}~~y\in\mathbb{R}^{N}.
\end{eqnarray}


Let $u = \mathcal{D}_0x\in\mathbb{R}^{N}$:

\begin{equation}
u[n] ~=~ (\mathcal{D}_0x)[n] ~=~ x[2n],
\end{equation}
where $n$ runs from 0 to $N-1$.



Then $\mathcal{S}u = \mathcal{S}\mathcal{D}_0x\in\mathbb{R}^{N}$:

\begin{equation}
\begin{split}
(\mathcal{S}u)[n] &amp;=~ u[n+1 ~\textrm{mod}~N]\\
 &amp;=~ x[2\times((n+1)~\textrm{mod}~N)]\\
 &amp;=~ x[2n+2~\textrm{mod}~2N]
\end{split}
\end{equation}


Now let $v = \mathcal{S}^2x\in\mathbb{R}^{2N}$:

\begin{equation}
v[n] ~=~ (\mathcal{S}^2x)[n] ~=~ x[n+2~\textrm{mod}~2N]
\end{equation}

Then $\mathcal{D}_0v = \mathcal{D}_0\mathcal{S}^2x\in\mathbb{R}^{N}$:

\begin{equation}
\begin{split}
(\mathcal{D}_0v)[n] &amp;=~ v[2n]\\
 &amp;=~ x[(2n)+2~\textrm{mod}~2N]\\
&amp;=~ x[2n+2~\textrm{mod}~2N]
\end{split}
\end{equation}

Hence
\begin{equation}
(\mathcal{D}_0\mathcal{S}^2x)[n] ~=~ x[2n+2~\textrm{mod}~2N] ~=~ (\mathcal{S}\mathcal{D}_0x)[n]
\end{equation}
for $0\leq n &lt; N$ (because all results are in $\mathbb{R}^{N}$), so $\mathcal{D}_0\mathcal{S}^2$ and $\mathcal{S}\mathcal{D}_0$ are the same.