Fourier transform convolution proof

[mcheung4]

Homework Statement

Let FT(f) = Fourier transform of f, (f*g)(x) = convolution of f and g.
Given FT(f*g) = FT(f)FT(g), the first part of the convolution theorem, show that FT[fg] = [FT(f)*FT(g)]/2pi.

Homework Equations

Duality: FT²f(x) = (2pi)f(-x)

Convolution: (f*g)(x) = ∫f(u)g(x-u)du

The Attempt at a Solution

FT(f*g) = FT(f)FT(g)
FT²(f*g) = FT[FT(f)FT(g)]
LHS = FT²(f*g)(x) = (2pi)(f*g)(-x) (by duality) =(2pi)∫f(u)g(-x-u)du (by convolution) = (2pi)∫{[FT²f(-u)][FT²g(x+u)]}/(2pi)² du
implying that RHS = FT[FT(f)FT(g)] = (1/2pi)∫FT[FT(f(-u))]FT[FT(g(x+u))] du ≠ FT[fg] = [FT(f)*FT(g)]/2pi

Where did I go wrong?

Thanks!

 

[vela]

Try letting f = FT(F) and g=FT(G). Then evaluate FT²(f*g) and FT[FT(f)FT(g)] separately and show they are equal.