upsampling and downsampling operator
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.