13.2. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 379

Definition 13.2.3 For T ∈ G ∗, define FT,F−1T ∈ G ∗ by

FT (φ)≡ T (Fφ) , F−1T (φ)≡ T(F−1

φ)

Lemma 13.2.4 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 13.2.5 If f ∈ L1loc (Rn) and

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

Proof: Let E be bounded and Lebesgue measurable. By regularity, there exists a com-pact set Kk ⊆ E and an open set Vk ⊇ E such that mn (Vk \Kk)< 2−k. Let hk equal 1 on Kk,vanish on VC

k , and take values between 0 and 1. Let Ek ≡[∣∣hk−XEk

∣∣> ( 23

)k]. Then

mn (Ek)≤(

32

)k ∫Ek

∣∣hk−XEk

∣∣dmn <

(32

)k ∫Vk\Kk

2dmn = 2(

32

)k 12k = 2

(34

)k

and so ∑k mn (Ek)< ∞ so there is a set of measure zero such that off this set, hk→XEk .Hence, by the dominated convergence theorem,

∫f XEdmn = limk→∞

∫f hkdmn = 0. It

follows that for E an arbitrary Lebesgue measurable set,∫

f XB(0,R)XEdmn = 0. Let

sgn f =

{f| f | if | f | ̸= 00 if | f |= 0

By Theorem 9.1.6 applied to positive and negative parts, there exists {sk}, a sequence ofsimple functions converging pointwise to sgn f such that |sk| ≤ 1. Then by the dominatedconvergence theorem again,

∫| f |XB(0,R)dmn = limk→∞

∫f XB(0,R)skdmn = 0. Since R is

arbitrary, | f |= 0 a.e. ■

Corollary 13.2.6 Let f ∈ L1 (Rn) and suppose∫Rn f (x)φ (x)dx = 0 for all φ ∈ G .

Then f = 0 a.e.

Proof: Let ψ ∈Cc (Rn) . Then by the Stone Weierstrass approximation theorem, thereexists a sequence of functions, {φ k} ⊆ G such that φ k→ ψ uniformly. Then by the domi-nated convergence theorem,

∫f ψdx = limk→∞

∫f φ kdx = 0. By Lemma 13.2.5 f = 0. ■

The next theorem is the main result of this sort.

Theorem 13.2.7 Let f ∈ Lp (Rn) , p≥ 1, or suppose f is measurable and has poly-

nomial growth, | f (x)| ≤K(

1+ |x|2)m

for some m∈N. Then if∫

f ψdx= 0, for all ψ ∈ G ,then it follows f = 0.