11.6. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 283

There are many other considerations involving Fourier transforms of functions whichare in L1 (Rn). Some others are in the exercises.

11.6.4 Fourier Transforms of Functions In L2 (Rn)

Consider F f and F−1 f for f ∈ L2(Rn). First note that the formula given for F f and F−1 fwhen f ∈ L1 (Rn) will not work for f ∈ L2(Rn) unless f is also in L1(Rn). However, thereis no problem for functions in G .

Theorem 11.6.13 For φ ∈ G , ∥Fφ∥2 = ∥F−1φ∥2 = ∥φ∥2.

Proof: First note that for ψ ∈ G ,

F(ψ) = F−1(ψ) , F−1(ψ) = F(ψ). (11.10)

This follows from the definition. For example,

Fψ (t) = (2π)−n/2∫Rn

e−it·xψ (x)dx = (2π)−n/2

∫Rn

eit·xψ (x)dx = F(ψ)(t)

Let φ ,ψ ∈ G . It was shown above that∫Rn(Fφ)ψ(t)dt =

∫Rn φ(Fψ)dx. Similarly,∫

Rnφ(F−1

ψ)dx =∫Rn(F−1

φ)ψdt. (11.11)

Now, 11.10 - 11.11 imply∫Rn|φ |2dx =

∫Rn

φφdx =∫Rn

φF−1 (Fφ) dx =∫Rn

φF(Fφ)dx

=∫Rn

Fφ(Fφ)dx =∫Rn|Fφ |2dx.

Similarly ∥φ∥2 = ∥F−1φ∥2. ■

Lemma 11.6.14 Let f ∈ L2 (Rn) and let φ k → f in L2 (Rn) where φ k ∈ G . (Such asequence exists because of density of G in L2 (Rn). Theorem 11.5.4) Then F f and F−1 fare both in L2 (Rn) and the following limits take place in L2.

limk→∞

F (φ k) = F ( f ) , limk→∞

F−1 (φ k) = F−1 ( f ) .

Proof: Let φ k→ f in L2 (Rn) . Then if ψ ∈ G , by the Cauchy Schwarz inequality,∣∣∣∣∫Rnφ kFψdx−

∫Rn

φ mFψdx∣∣∣∣≤ ∥φ k−φ m∥L2 ∥Fψ∥L2

and so limk→∞

∫Rn φ k (x)Fψ (x)dx exists. Now

F f (ψ) ≡ f (Fψ)≡∫Rn

f (x)Fψ (x)dx

= limk→∞

∫Rn

φ k (x)Fψ (x)dx = limk→∞

∫Rn

Fφ k (x)ψ (x)dx.