1098 CHAPTER 32. FOURIER TRANSFORMS

Proof: Let f ∈ Lp (Rn) . Then there exists g ∈Cc (Rn) such that || f −g||p < ε . Now letb > 0 be large enough that ∫

Rn

(e−b|x|2

)pdx < ε

p.

Then x→ g(x)eb|x|2 is in Cc (Rn) ⊆C0 (Rn) . Therefore, from Lemma 32.1.3 there existsψ ∈ G such that ∣∣∣∣∣∣geb|·|2 −ψ

∣∣∣∣∣∣∞

< 1

Therefore, letting φ (x)≡ e−b|x|2ψ (x) it follows that φ ∈ G and for all x ∈ Rn,

|g(x)−φ (x)|< e−b|x|2

Therefore, (∫Rn|g(x)−φ (x)|p dx

)1/p

≤(∫

Rn

(e−b|x|2

)pdx)1/p

< ε .

It follows|| f −φ ||p ≤ || f −g||p + ||g−φ ||p < 2ε.

Since ε > 0 is arbitrary, this proves the theorem.The following lemma is also interesting even if it is obvious.

Lemma 32.1.5 For ψ ∈ G , p a polynomial, and α,β multiindices, Dα ψ ∈ G and pψ ∈ G .Also

sup{|xβ Dαψ(x)| : x ∈ Rn}< ∞

32.2 Fourier Transforms Of Functions In G

Definition 32.2.1 For ψ ∈ G Define the Fourier transform, F and the inverse Fouriertransform, F−1 by

Fψ(t)≡ (2π)−n/2∫Rn

e−it·xψ(x)dx,

F−1ψ(t)≡ (2π)−n/2

∫Rn

eit·xψ(x)dx.

where t ·x≡∑ni=1 tixi.Note there is no problem with this definition because ψ is in L1 (Rn)

and therefore, ∣∣eit·xψ(x)

∣∣≤ |ψ(x)| ,

an integrable function.

One reason for using the functions, G is that it is very easy to compute the Fouriertransform of these functions. The first thing to do is to verify F and F−1 map G to G andthat F−1 ◦F (ψ) = ψ.