32.3. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 1113

Theorem 32.3.22 Let ψ ∈S. Then (F ◦F−1)(ψ) = ψ and (F−1 ◦F)(ψ) = ψ wheneverψ ∈S. Also F and F−1 map S one to one and onto S.

Proof: The first claim follows from the fact that F and F−1 are inverses of each otheron G ∗ which was established above. For the second, let ψ ∈ S. Then ψ = F

(F−1ψ

).

Thus F maps S onto S. If Fψ = 0, then do F−1 to both sides to conclude ψ = 0. Thus Fis one to one and onto. Similarly, F−1 is one to one and onto.

32.3.5 ConvolutionTo begin with it is necessary to discuss the meaning of φ f where f ∈ G ∗ and φ ∈ G . Whatshould it mean? First suppose f ∈ Lp (Rn) or measurable with polynomial growth. Thenφ f also has these properties. Hence, it should be the case that φ f (ψ) =

∫Rn φ f ψdx =∫

Rn f (φψ)dx. This motivates the following definition.

Definition 32.3.23 Let T ∈ G ∗ and let φ ∈ G . Then φT ≡ T φ ∈ G ∗ will be defined by

φT (ψ)≡ T (φψ) .

The next topic is that of convolution. It was just shown that

F ( f ∗φ) = (2π)n/2 FφF f , F−1 ( f ∗φ) = (2π)n/2 F−1φF−1 f

whenever f ∈ L2 (Rn) and φ ∈ G so the same definition is retained in the general casebecause it makes perfect sense and agrees with the earlier definition.

Definition 32.3.24 Let f ∈ G ∗ and let φ ∈ G . Then define the convolution of f with anelement of G as follows.

f ∗φ ≡ (2π)n/2 F−1 (FφF f ) ∈ G ∗

There is an obvious question. With this definition, is it true that

F−1 ( f ∗φ) = (2π)n/2 F−1φF−1 f

as it was earlier?

Theorem 32.3.25 Let f ∈ G ∗ and let φ ∈ G .

F ( f ∗φ) = (2π)n/2 FφF f , (32.3.16)

F−1 ( f ∗φ) = (2π)n/2 F−1φF−1 f . (32.3.17)

Proof: Note that 32.3.16 follows from Definition 32.3.24 and both assertions hold forf ∈ G . Consider 32.3.17. Here is a simple formula involving a pair of functions in G .(

ψ ∗F−1F−1φ)(x)