I have some problems in understanding the convolution theorem in the case of a Fourier transform (FT) of a discrete convolution. Suppose I have the discrete convolution $$ r(x) = \sum_j f(x) g(x-z_j) $$ where the functions $f(x)$ and $g(x)$ have FT $F(k)$ and $G(k)$ respectively. What would be the FT $$ R(k) = \int dx r(x) e^{i k x} $$
In the case of a continuous convolution I would expect $R(k)=F(k)G(k)$, but according to some numerical tests I made it looks more something like $R(k)\approx\sum_j F(q_j) G(k-q_j)$.
I do not have clear if at some point I need to introduce a discrete FT.