Kenneth Kuttler

11.6. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 277

From Lemma 11.5.7,

F−1 (F ( fb))(s) = F−1((2b)−n/2 f

(4b)−1

)(s) = (2b)−n/2 F−1

(f(4b)−1

)(s)

= (2b)−n/2(

2(4b)−1)−n/2

f(4(4b)−1)

−1 (s) = fb (s)

Hence F−1 ◦F (ψ)(s) = sα fb (s) = ψ (s).Another way to see this is to use Observation 11.3.7. If n = 1 then F−1 ◦F (ψ)(s) =

ψ (s) from Observation 11.3.7 and Theorem 11.3.6. So suppose F−1 ◦F (ψ)(s) = ψ (s) onRn−1. From the definition and the observation and Fubini’s theorem, if ψ ∈ G on Rn, thenfrom the special form of ψ and neglecting the (1/2π)n/2 to make it simpler to write,

F−1 ◦F (ψ)(s)≡∫R

eisntn∫R

e−itnxn

∫Rn−1

eiŝn·t̂n

∫Rn−1

e−it̂n·x̂nψ (x̂n,xn)dx̂ndt̂ndxndtn

Now by induction and Theorem 11.3.6, this is∫R

eisntn∫R

e−itnxnψ (ŝn,xn)dxndtn = ψ (ŝn,sn) = ψ (s) ■

11.6 Fourier Transforms of Just About Anything11.6.1 Fourier Transforms in G ∗

It turns out you can make sense of the Fourier transform of any linear map defined on G .This is a very abstract way to look at things but if you want ultimate generality, you mustdo something like this. Part of the problem is that it is desired to take Fourier transforms offunctions which are not in L1 (Rn). Thus the integral which defines the Fourier transformin the above will not make sense. You run into this problem as soon as you try to take theFourier transform of a function in L2 because such functions might not be in L1 if they aredefined on R or Rn. However, it was realized long ago that if a function is in L1∩L2, thenthe L2 norm of the function is equal to the L2 norm of the Fourier transform of the function.Thus there is an obvious question about whether you can get a definition which will allowyou to directly deal with the Fourier transform on L2. If you solve this, perhaps by usingdensity of L1∩L2 in L2, you are still faced with the problem of taking the Fourier transformof an arbitrary function in Lp. The method developed here removes all these difficulties atonce.

Definition 11.6.1 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 we can regard ψ as an element of G ∗ by defining

ψ (φ)≡∫Rn

ψ (x)φ (x)dx.

This implies the following important lemma.

11.6. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 277From Lemma 11.5.7,F-'(F (fy))(s)F* (26) "? fugyy-1) (8) = (26)? F* (Fugyy-1) (8)= (26) (240) Figg) 11 (8) =o)(401)!Hence F~!o F (w) (s) =s% fy (s) = w(s).Another way to see this is to use Observation 11.3.7. If n = 1 then F~! oF (w)(s) =y(s) from Observation 11.3.7 and Theorem 11.3.6. So suppose F~! 0 F (yw) (s) = w(s) onR"—!. From the definition and the observation and Fubini’s theorem, if y € on R", thenfrom the special form of y and neglecting the (1/ 2n)"/ * to make it simpler to write,Fl oF (w)(s)=/ eldntn / etinxn / elsntn | e ttn Xn y (Rn,Xn) di, dt,dx,dtyR R Rl RilNow by induction and Theorem 11.3.6, this is[ cv | ey 8 ,,Xn) dxndtn = YW (8n5n) = W(s)R R11.6 Fourier Transforms of Just About Anything11.6.1 Fourier Transforms in 4*It turns out you can make sense of the Fourier transform of any linear map defined on Y.This is a very abstract way to look at things but if you want ultimate generality, you mustdo something like this. Part of the problem is that it is desired to take Fourier transforms offunctions which are not in L! (IR”). Thus the integral which defines the Fourier transformin the above will not make sense. You run into this problem as soon as you try to take theFourier transform of a function in L* because such functions might not be in L! if they aredefined on R or R”. However, it was realized long ago that if a function is in L'L’, thenthe L? norm of the function is equal to the L* norm of the Fourier transform of the function.Thus there is an obvious question about whether you can get a definition which will allowyou to directly deal with the Fourier transform on L”. If you solve this, perhaps by usingdensity of L'ML? in L’, you are still faced with the problem of taking the Fourier transformof an arbitrary function in L’. The method developed here removes all these difficulties atonce.Definition 11.6.1 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)+bT (6) forallabeC, w,gegGLet w € Y. Then we can regard w as an element of Y* by definingv(0)= |. w(0)0(x)dx.This implies the following important lemma.