32.3. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 1103

Theorem 32.3.7 Let f ∈ Lp (Rn) , p≥ 1, or suppose f is measurable and has polynomialgrowth,

| 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.

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 Corollary32.3.6. 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 32.1.4), 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. 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.

The following theorem shows that you can consider most functions you are likely toencounter as elements of G ∗.