Downsampling, shifting, high pass and low pass filter commutativity strong textI have been reading "The Stationary Wavelet Transform and some Statistical Applications" by Nason and Silverman, and there is a claim in the their paper of which I cannot convince myself.

Defininitions: 


$\mathcal{H}$: A low-pass filter.  


The action of the low pass filter on a doubly infinite sequence $\{..., x_{-1}, x_0, x_1, x2, ... \}$ is defined $(\mathcal{H}x)_k = \sum_n h_{n-k} x_n$ 
The filter satisfied internal orthogonality, i.e. $\sum_n h_{n} h_{n + 2j} = 0$

$\mathcal{G}$: A high-pass filter, defined by the filter $g_n = (-1)^n h_{1-n}$ 


Clearly $\mathcal{G}$ satisfied the same internal orthogonality, and is mutually orthogonal to $\mathcal{H}$, i.e. $\sum_n h_{n} g_{n + 2j} = 0$ $\forall j$

$\mathcal{S}$: A shift operator defined by $(\mathcal{S} x)_j = x_{j+1}$
$\mathcal{D}_0$:  A binary decimation operator that chooses every even member of a sequence.


$(\mathcal{D}_0 x)_j = x_{2j}$
$\mathcal{R}_0$ the inverse operation.

$\mathcal{D}_1$: A binary decimation operator that chooses every odd member of a sequence, similar to $\mathcal{D}_0$ above.
It is not even necessary for the same choice of "odd" or "even" to be used
throughout. 


Suppose $\epsilon_{J-1}, \epsilon_{J-2}, ..., \epsilon_{0}$ are a sequence of 0's and 1's.
One can use the binary operator $\mathcal{D}_{\epsilon_j}$ at level $j$, and the original signal can be recovered by applying the corresponding sequence $\mathcal{R}_{\epsilon_j}$



On page 4, the paper claims, 


  "If $x$ is a finite sequence, define the shift periodically at the boundary. It then immediate from the definitions that $\mathcal{D}_1$ = $\mathcal{D}_0 \mathcal{S}$ and hence that $\mathcal{R}_1$ = $\mathcal{S}^{-1} \mathcal{R}_0$.




NOW comes the portion I can't seem to grasp...


  It is also easy to see that $\mathcal{S} \mathcal{D}_0 = \mathcal{D}_0 \mathcal{S}^2$.
  and that the operator $\mathcal{S}$ commutes with $\mathcal{H}$ and $\mathcal{G}$.


How is this obvious? I have tried recreating these properties in R, but I cannot convince myself that $\mathcal{S} \mathcal{D}_0 = \mathcal{D}_0 \mathcal{S}^2$, or that S and the filters commute.  Can anyone help?
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.