32.3. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 1107

32.3.3 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). Recall thata+ ib = a− ib.

Theorem 32.3.12 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). (32.3.6)

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

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

∫Rn

φ(Fψ)dx.

Similarly, ∫Rn

φ(F−1ψ)dx =

∫Rn(F−1

φ)ψdt. (32.3.7)

Now, 32.3.6 - 32.3.7 imply∫Rn|φ |2dx =

∫Rn

φF−1(Fφ)dx =∫Rn

φF(Fφ)dx

=∫Rn

Fφ(Fφ)dx =∫Rn|Fφ |2dx.

Similarly||φ ||2 = ||F−1

φ ||2.

Lemma 32.3.13 Let f ∈ L2 (Rn) and let φ k → f in L2 (Rn) where φ k ∈ G . (Such a se-quence exists because of density of G in L2 (Rn).) Then F f and F−1 f are 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 ψ ∈ G be given. Then

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.