State and prove Fourier convolution theorem

Definitions

Fourier transform: F{f}(s) = F(s) = ∫_{−∞}^{∞} f(x) e^{−isx} dx.

Convolution: (f * g)(x) = ∫_{−∞}^{∞} f(u) g(x − u) du.

Statement

If F(s) and G(s) are the Fourier transforms of f(x) and g(x), then the Fourier transform of their convolution is the product of the transforms:

F{f * g} = F(s) · G(s)

Equivalently, f * g = F⁻¹{F(s)G(s)}. (With the symmetric convention F(s) = (1/√(2π)) ∫ f e^{−isx} dx, the result reads F{f * g} = √(2π) F(s)G(s).)

Proof

F{f * g} = ∫ e^{−isx} [ ∫ f(u) g(x − u) du ] dx

Interchange the order of integration (f and g are absolutely integrable):

F{f * g} = ∫ f(u) [ ∫ g(x − u) e^{−isx} dx ] du

In the inner integral put x − u = w, so x = w + u and dx = dw:

∫ g(w) e^{−is(w + u)} dw = e^{−isu} ∫ g(w) e^{−isw} dw = e^{−isu} G(s)

Hence

F{f * g} = ∫ f(u) e^{−isu} G(s) du
         = G(s) ∫ f(u) e^{−isu} du
         = F(s) · G(s)      ∎

Consequence

The inverse form is ∫_{−∞}^{∞} F(s)G(s) e^{isx} ds = 2π (f * g)(x). It lets us compute a convolution from a simple product.