280 CHAPTER 11. FUNDAMENTAL TRANSFORMS

It remains to consider the case where f has polynomial growth. Thus x→ f (x)e−|x|2∈

L1 (Rn) . Therefore, for all ψ ∈ G ,

0 =∫

f (x)e−|x|2ψ (x)dx

because e−|x|2ψ (x) ∈ G . Therefore, by the first part, f (x)e−|x|

2= 0 a.e. ■

Note that “polynomial growth” could be replaced with a condition of the form

| f (x)| ≤ K(

1+ |x|2)m

ek|x|α , α < 2

and the same proof would yield that these functions are in G ∗. The main thing to observeis that almost all functions of interest are in G ∗.

Theorem 11.6.9 Let f be a measurable function with polynomial growth,

| f (x)| ≤C(

1+ |x|2)N

for some N,

or let f ∈ Lp (Rn) for some p ∈ [1,∞]. Then f ∈ G ∗ if f (φ)≡∫

f φdx.

Proof: Let f have polynomial growth first. Then the above integral is clearly welldefined and so in this case, f ∈ G ∗.

Next suppose f ∈ Lp (Rn) with ∞ > p≥ 1. Then it is clear again that the above integralis well defined because of the fact that φ is a sum of polynomials times exponentials of theform e−c|x|2 and these are in Lp′ (Rn). Also φ → f (φ) is clearly linear in both cases. ■

This has shown that for nearly any reasonable function, you can define its Fourier trans-form as described above. You could also define the Fourier transform of a finite Borel mea-sure µ because for such a measure ψ→

∫Rn ψdµ is a linear functional on G . This includes

the very important case of probability distribution measures.

11.6.2 Fourier Transforms of Functions In L1 (Rn)

First suppose f ∈ L1 (Rn) . As mentioned, you can think of it as being in G ∗ and so onecan take its Fourier transform as described above. However, since it is in L1 (Rn) , there isa natural way to define its Fourier transform. Do the two give the same thing?

Theorem 11.6.10 Let f ∈ L1 (Rn) . Then F f (φ) =∫Rn gφdt where

g(t) =(

12π

)n/2 ∫Rn

e−it·x f (x)dx

and F−1 f (φ) =∫Rn gφdt where g(t) =

( 12π

)n/2 ∫Rn eit·x f (x)dx. In short,

F f (t)≡ (2π)−n/2∫Rn

e−it·x f (x)dx,

F−1 f (t)≡ (2π)−n/2∫Rn

eit·x f (x)dx.