# 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 $$\mathbb{R}^{N}$$: $$\begin{eqnarray} (\mathcal{S}x)[n] &=& x[n+1~\textrm{mod}~2N]~~\textrm{for}~~x\in\mathbb{R}^{2N},\\ (\mathcal{S}y)[n] &=& y[n+1~\textrm{mod}~N]~~\textrm{for}~~y\in\mathbb{R}^{N}. \end{eqnarray}$$

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

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

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

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

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

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

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

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

Hence $$$$(\mathcal{D}_0\mathcal{S}^2x)[n] ~=~ x[2n+2~\textrm{mod}~2N] ~=~ (\mathcal{S}\mathcal{D}_0x)[n]$$$$ for $$0\leq n < 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.