1100 CHAPTER 32. FOURIER TRANSFORMS

It remains to verify the other assertion. As in the first case, it suffices to considerψ (x) = xα e−b|x|2 .

F−1 ◦F (ψ)(s)≡(

12π

)n/2 ∫Rn

eis·tF (ψ)(t)dt

=

(1

2π

)n/2 ∫Rn

eis·t(

12π

)n/2

(i)−|α|Dαt

(e−|t|24b

(√π√b

)n)dt

=

(1

2π

)n

(i)−|α|∫Rn

eis·tDαt

(e−|t|24b

(√π√b

)n)dt

=

(1

2π

)n

(i)−|α|∫Rn

(i)|α| sα eis·t(

e−|t|24b

(√π√b

)n)dt

and by Lemma 32.2.2,

=

(1

2π

)n(√π√b

)n ∫Rn

sα eis·t(

e−|t|24b

)dt

=

=1︷ ︸︸ ︷(1

2π

)n(√π√b

)n( √

π√1/4b

)n

sα e−b|s|2 = ψ (s) .

32.3 Fourier Transforms Of Just About Anything32.3.1 Fourier Transforms Of G ∗

Definition 32.3.1 Let G ∗ denote the vector space of linear functions defined on G whichhave values in C. Thus T ∈ G ∗ means T : G →C and T is linear,

T (aψ +bφ) = aT (ψ)+bT (φ) for all a,b ∈ C, ψ,φ ∈ G

Let ψ ∈ G . Then we can regard ψ as an element of G ∗ by defining

ψ (φ)≡∫Rn

ψ (x)φ (x)dx.

Then we have the following important lemma.

Lemma 32.3.2 The following is obtained for all φ ,ψ ∈ G .

Fψ (φ) = ψ (Fφ) ,F−1ψ (φ) = ψ

(F−1

φ)

Also if ψ ∈ G and ψ = 0 in G ∗ so that ψ (φ) = 0 for all φ ∈ G , then ψ = 0 as a function.