11.6. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 279

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 Corollary 7.7.6, there exists {sk}, a sequence of simple functions converging pointwiseto sgn f XB(0,R) such that |sk| ≤ 1. Then by the dominated convergence theorem again,∫

| f |XB(0,R)dmn = limk→∞

∫f XB(0,R)skdmn = 0.

Since R is arbitrary, | f |= 0 a.e. ■

Corollary 11.6.7 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 11.6.6 f = 0. ■

The next theorem is the main result of this sort.

Theorem 11.6.8 Let f ∈ Lp (Rn) , p≥ 1, or suppose f is measurable and has poly-

nomial growth, defined as | f (x)| ≤ K(

1+ |x|2)m

for some K and m. Then if∫

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

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 Corollary11.6.7. 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 11.5.4), 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.