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 ∗.
378 CHAPTER 13. FOURIER TRANSFORMSSimilarly, for y (a) = ate“tray — del _qylal (1)? og 6pFU (yw) =i" (-1) 5 se % (13.2)CcFor y (a) = ateciel,ja|-+n/ ;FloF(y)=F"! ((-a" (-1)'“' (=) . tebe)|o|+n/2_ lal (=) E>! (steel)2cFrom 13.2 with c > 1/ (4c),1 \ laltn/2 1 Vian? ;= jel ( — jloel (—y Il ( —__ Oe aay! = gt —-<ls|(=) eC" aya) ee eeSince Y consists of sums of multiples of such y, this has shown that F~! oF (y) = y.13.2 Fourier Transforms of Just about AnythingI will define Fourier Transforms of the linear functions acting on Y and then show that thisincludes virtually anything which could possibly be of any interest.Definition 13.2.1 Let Y* denote the vector space of linear functions defined on GYwhich have values in C. Thus T € Y* means T :Y — C and T is linear,T (aw+bo) =aT (w)+5bT (@) foralla,bE C,w,d EYLet y € Y. Then y is an element of Y* by defining W(¢) = Jan W(x) (x) dx.Then we have the following important lemma.Lemma 13.2.2 The following is obtained for all ¢,y € G.Fy(¢)=V(Fo), F'w()=w(F '9)Also if we FY and y =0 in Y* so that y(o) =0 for all @ € Y, then y = 0 as a function.Proof:1rvi)= [,rumo@a=[ (2) [em veo naR”- R” onThe other claim is similar.Suppose now y(¢) = 0 for all @ € Y. Then fen wodx = 0 for all 6 € Y. Therefore,this is true for @ = W and so y=0. HfThis lemma suggests a way to define the Fourier transform of something in Y*.= [.wa(5 y" [eto tatdx= [ w(a)Fo(a)dx= (Fo)