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 ∗.