32.3. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 1101

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)(

12π

)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∫

Rnψφdx = 0

for all φ ∈ G . Therefore, this is true for φ = ψ and so ψ = 0.This lemma suggests a way to define the Fourier transform of something in G ∗.

Definition 32.3.3 For T ∈ G ∗, define FT,F−1T ∈ G ∗ by

FT (φ)≡ T (Fφ) , F−1T (φ)≡ T(F−1

φ)

Lemma 32.3.4 F and F−1 are both one to one, onto, and are inverses of each other.

Proof: First note F and F−1 are both linear. This follows directly from the definition.Suppose now FT = 0. Then FT (φ) = T (Fφ) = 0 for all φ ∈ G . But F and F−1 map Gonto G because if ψ ∈ G , then as shown above, ψ = F

(F−1 (ψ)

). Therefore, T = 0 and

so F is one to one. Similarly F−1 is one to one. Now

F−1 (FT )(φ)≡ (FT )(F−1

φ)≡ T

(F(F−1 (φ)

))= T φ .

Therefore, F−1 ◦F (T ) = T. Similarly, F ◦F−1 (T ) = T. Thus both F and F−1 are one toone and onto and are inverses of each other as suggested by the notation.

Probably the most interesting things in G ∗ are functions of various kinds. The followinglemma will be useful in considering this situation.

Lemma 32.3.5 If f ∈ L1loc (Rn) and

∫Rn f φdx = 0 for all φ ∈Cc (Rn), then f = 0 a.e.

Proof: For r > 0, let

E ≡ {x : f (x)≥ r}, ER ≡ E ∩B(0,R).

Let Km be an increasing sequence of compact sets, and let Vm be a decreasing sequence ofopen sets satisfying

Km ⊆ ER ⊆Vm, mn (Vm)≤ mn (Km)+2−m,V1 ⊆ B(0,R) .