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## 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: FT

^{2}f(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)

FT

^{2}(f*g) = FT[FT(f)FT(g)]

LHS = FT

^{2}(f*g)(x) = (2pi)(f*g)(-x) (by duality) =(2pi)∫f(u)g(-x-u)du (by convolution) = (2pi)∫{[FT

^{2}f(-u)][FT

^{2}g(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!