380 CHAPTER 13. FOURIER TRANSFORMS

Proof: First note that if f ∈ Lp (Rn) or has polynomial growth, then it makes senseto write the integral

∫f ψdx described above. This is obvious in the case of polynomial

growth. In the case where f ∈ Lp (Rn) it also makes sense because

∫| f | |ψ|dx≤

(∫| f |p dx

)1/p(∫|ψ|p

′dx)1/p′

< ∞

due to the fact mentioned above that all these functions in G are in Lp (Rn) for every p≥ 1.Suppose now that f ∈ Lp, p ≥ 1. The case where f ∈ L1 (Rn) was dealt with in Corollary13.2.6. Suppose f ∈ Lp (Rn) for p > 1. Then

| f |p−2 f ∈ Lp′ (Rn) ,

(p′ = q,

1p+

1q= 1)

and by density of G in Lp′ (Rn) (Theorem 13.1.4), there exists a sequence {gk} ⊆ G suchthat

∥∥∥gk−| f |p−2 f∥∥∥

p′→ 0. Then

∫Rn| f |p dx =

∫Rn

f(| f |p−2 f −gk

)dx+

∫Rn

f gkdx

=∫Rn

f(| f |p−2 f −gk

)dx≤ ∥ f∥Lp

∥∥∥gk−| f |p−2 f∥∥∥

p′

which converges to 0. Hence f = 0.It remains to consider the case where f has polynomial growth. Thusx→ 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 13.2.8 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. The theoretical basis for thisassertion will be given a little later.