In my engineering class, I was told that
$(FF(f))(x)=2\pi f(-x) $, where $F$ is the Fourier transform ----(1)
and $F(f(x-a))(k)=\exp(-ika) X(k)$ where $X(k)=F(f(x))$ ----(2)
implies $F(\exp(iax)f(x))(k)=X(k-a)$. ----(3)
Reworded: Perhaps my question would be clearer if I said why does the the duality property of the FT ----(1) allow us to obtain the modulation property ----(2) from the translation property ----(3)?
But I don't see how that is done... I am quite happy with getting $F^{-1}X(k-a)=\exp(iax)f(x)$ by brute force calculation. I would like to see how to use duality though. (It was said that the duality should simplify stuff...)