I am interested in the Radon transformation for separable functions $F(x,y) = f(x)g(y)$. Why is it in tomography that a separable function is determined completely by two of its projections ?
And first step towards the Radon transform is solving an integral $$I(a,b,t) = \int_{\mathbb{R}^2} f(x)g(y) \delta(ax + by - t)\;dA$$
where $\delta$ denotes the Dirac delta function. How is the delta function simplified ?
Why is it that a separable function is thus determined? Consider any real-valued separable function $F(x,y) = f(x)g(y)$. Then we have the integrals
$$I(1,0,t) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x)g(y)\delta(x-t)\;\mathrm dy \mathrm dx = f(t) \int_{-\infty}^{\infty} g(y) \mathrm dy$$
$$I(0,1,t) = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x)g(y)\delta(y-t)\;\mathrm dy \mathrm dx = g(t) \int_{-\infty}^{\infty} f(x) \mathrm dx$$
Thus considering the product $$H(x,y) = I(1,0,x)I(0,1,y) = f(x)g(y) \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x')g(y') \; \mathrm dx' \mathrm dy' = F(x,y) \int_{\mathbb{R}^2} F \;\mathrm dA$$
which satisfies
$$\int_{\mathbb{R}^2} H \;\mathrm dA = \left(\int_{\mathbb{R}^2} F \;\mathrm dA\right)^2$$
and thus we have that
$$F(x,y) = \frac{H(x,y)}{\sqrt{\int_{\mathbb{R}^2} H \; \mathrm dA}}$$
Since $H$ is simply the product of the two integrals $I(1,0,x)$ and $I(0,1,y)$, we have that $F$ is determined by those two integrals as well.