378 CHAPTER 13. FOURIER TRANSFORMS

Similarly, for ψ (x) = xα e−c|x|2 ,

F−1 (ψ) = i|α| (−1)|α|(

12c

)|α|+n/2

sα e−14c |s|

2(13.2)

For ψ (x) = xα e−c|x|2 ,

F−1 ◦F (ψ) = F−1

((−i)|α| (−1)|α|

(12c

)|α|+n/2

sα e−14c |s|

2

)

= i|α|(

12c

)|α|+n/2

F−1(sα e−

14c |s|

2)

From 13.2 with c→ 1/(4c) ,

= i|α|(

12c

)|α|+n/2

i|α| (−1)|α|(

12(1/4c)

)|α|+n/2

sα e−1

4(1/(4c)) |s|2= sα e−c|s|2

Since G consists of sums of multiples of such ψ, this has shown that F−1 ◦F (ψ) = ψ . ■

13.2 Fourier Transforms of Just about AnythingI will define Fourier Transforms of the linear functions acting on G and then show that thisincludes virtually anything which could possibly be of any interest.

Definition 13.2.1 Let G ∗ denote the vector space of linear functions defined on Gwhich have 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 ψ is an element of G ∗ by defining ψ (φ)≡∫Rn ψ (x)φ (x)dx.

Then we have the following important lemma.

Lemma 13.2.2 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.

Proof:

Fψ (φ)≡∫Rn

Fψ (t)φ (t)dt =∫Rn

(1

2π

)n/2 ∫Rn

e−it·xψ(x)dxφ (t)dt

=∫Rn

ψ(x)

(1

2π

)n/2 ∫Rn

e−it·xφ (t)dtdx =

∫Rn

ψ(x)Fφ (x)dx≡ ψ (Fφ)

The other claim is similar.Suppose now ψ (φ) = 0 for all φ ∈ G . Then

∫Rn ψφdx = 0 for all φ ∈ G . Therefore,

this is true for φ = ψ̄ and so ψ = 0. ■This lemma suggests a way to define the Fourier transform of something in G ∗.