I need to take the Fourier transform of a product of three terms. The statements I have of the convolution theorem are thus: [math]\frac 1 {2\pi} \int dk F(k)G(k)e^{ik x} = \int du f(u)g(x-u) . [/math] How do we express the inverse, namely: [math] \int dx f(x)g(x)e^{-ik x} [/math]? Can I just follow my nose and write a similar convolution integral in k-space?

I've found Wikipedia is a good reference for many general equations of the Fourier transform (although you have to be a little careful of the normalization that you're using). Anyway, yes, the convolution identity you've got up there will work in reverse and I believe the proof will work in exactly the same way.

I don't think there is any real difference between x and k, so the formula will work "both ways". Just be careful of normalisation and any minus sign conventions.

Thanks, Dave and ajb, I hope to lose my fear of convolution.