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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 Müller Jan 6 '17 at 11:05
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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.

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