How do you prove that the Fourier transform has the distributive property over addition:
but not over multiplication:
By using the 1D DFT (Discrete Fourier Transform) definition (no need to use the 2D equation with more variables):
$F(u) = \displaystyle\frac{1}{N}\sum_{x = 0}^{N-1} f(x) exp(\frac{-j 2 \pi u x}{N})$
or $F(u) = \displaystyle\frac{1}{N}\sum_{x = 0}^{N-1} f(x) cos(\frac{2 \pi u x}{N}) - j sin(\frac{2 \pi u x}{N})$ by Euler's formula
Of course it may happen that for certain special $f_1$ and $f_2$ you have ${\cal F}(f_1\cdot f_2)={\cal F}(f_1)\cdot{\cal F}(f_2)$. But this is not true in general. In order to prove this it suffices to give a single counterexample. To this end let $f_1$ and $f_2$ be the characteristic functions of two disjoint open squares. Then $f_1\cdot f_2=0$, hence ${\cal F}(f_1\cdot f_2)\equiv0$. But both Fourier transforms ${\cal F}(f_i)$ are spread out over the whole frequency domain, hence do not have disjoint supports. It follows that ${\cal F}(f_1)\cdot{\cal F}(f_2)\not \equiv 0$.