11.6. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 289

Proof: Note that 11.19 follows from Definition 11.6.25 and both assertions hold forf ∈ G . Consider 11.20. Here is a simple formula involving a pair of functions in G .(

ψ ∗F−1F−1φ)(x)

=

(∫ ∫ ∫ψ (x−y)eiy·y1eiy1·zφ (z)dzdy1dy

)(2π)n

=

(∫ ∫ ∫ψ (x−y)e−iy·ỹ1e−iỹ1·zφ (z)dzdỹ1dy

)(2π)n

= (ψ ∗FFφ)(x) .

Now for ψ ∈ G ,

(2π)n/2 F(F−1

φF−1 f)(ψ)≡ (2π)n/2 (F−1

φF−1 f)(Fψ)≡

(2π)n/2 F−1 f(F−1

φFψ)≡ (2π)n/2 f

(F−1 (F−1

φFψ))

=

f((2π)n/2 F−1 ((FF−1F−1

φ)(Fψ)

))≡

f(ψ ∗F−1F−1

φ)= f (ψ ∗FFφ) (11.21)

Also

(2π)n/2 F−1 (FφF f )(ψ)≡ (2π)n/2 (FφF f )(F−1

ψ)≡

(2π)n/2 F f(FφF−1

ψ)≡ (2π)n/2 f

(F(FφF−1

ψ))

=

= f(

F((2π)n/2 (FφF−1

ψ)))

= f(

F((2π)n/2 (F−1FFφF−1

ψ)))

= f(F(F−1 (FFφ ∗ψ)

))f (FFφ ∗ψ) = f (ψ ∗FFφ) . (11.22)

The last line follows from the following.∫FFφ (x−y)ψ (y)dy =

∫Fφ (x−y)Fψ (y)dy =

∫Fψ (x−y)Fφ (y)dy

=∫

ψ (x−y)FFφ (y)dy.

From 11.22 and 11.21 , since ψ was arbitrary,

(2π)n/2 F(F−1

φF−1 f)= (2π)n/2 F−1 (FφF f )≡ f ∗φ

which shows 11.20. ■