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A very simple example on a $2\times 2$ image $$I_0=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ with (very crude Gaussian) low-pass: $$g=1/4\begin{bmatrix}1&1\\1&1\end{bmatrix}$$ yields a downsampled $I_1$ after filtering, with only one pixel (out of 4): $$I_1=\begin{bmatrix}(a+b+c+d/4)\end{bmatrix}$$ It can be upsampled as: $$U(I_1) = I_1^\... 2 Actually the down sampling has no role here. It is all based on a real simple equation:$$ I = A + B $$It is always enough to keep 2 terms of the 3 to restore completely and perfectly the information. So let's look on this:$$ {I}_{0} = \left( \left( {I}_{0} \downarrow \right) \uparrow \right) + {R}_{0}  So if we keep ${R}_{0}$ and we have \$ \left( ...