I am trying to prove that for all $x,y \in l^{2}(\mathbb{R})$, it holds that $$\left\langle\left(\downarrow M\right)\left[x\right],y \right\rangle = \left\langle x,\left(\uparrow M\right)\left[y\right] \right\rangle \text{.}$$ How should I proceed to prove this?
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$\begingroup$ Attention! This demands that the down- and upsampling operators are defined to be "compatible" in the sense that if you assigned indices to $x$ and $\tilde y= (\uparrow M) [y]$, the non-zero entries of $\tilde y$ have the same indices as the downsampling operator would keep! $\endgroup$– Marcus MüllerCommented Jan 6, 2017 at 11:05
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1 Answer
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The left-hand side is
$$ \sum_{i \% M=0}\sum_{j\%M=0} x(i/M,j/M)y(i,j) $$
where the sum can only be done over all i,j which are dividable by M (otherwise x(i/M)=0). Hence, by substitution $i=Ma, j=Mb$ we get
$$ =\sum_{a\in\mathbb{Z}}\sum_{b\in\mathbb{Z}}x(a,b)y(aM,bM) $$
which equals the right-hand side.
answered Jan 6, 2017 at 7:00