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?

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.