13.2. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 383

It follows that for a.e. x,∫Rn | f (x−y)g(y)|dy < ∞ and for each of these values of x, it

follows that∫Rn f (x−y)g(y)dy exists and equals a function of x which is in L1 (Rn) , f ∗

g(x). Now

F( f ∗g)(t)≡ (2π)−n/2∫Rn

e−it·x f ∗g(x)dx

= (2π)−n/2∫Rn

e−it·x∫Rn

f (x−y)g(y)dydx

= (2π)−n/2∫Rn

e−it·yg(y)∫Rn

e−it·(x−y) f (x−y)dxdy

= (2π)n/2 F f (t)Fg(t) . ■

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

13.2.2 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 13.2.12 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). (13.6)

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,∫

Rnφ(F−1

ψ)dx =∫Rn(F−1

φ)ψdt. (13.7)

Now, 13.6 - 13.7 imply∫Rn|φ |2dx =

∫Rn

φφdx =∫Rn

φ(F−1(Fφ))dx =∫Rn

φF(Fφ)dx

=∫Rn

Fφ(Fφ)dx =∫Rn|Fφ |2dx.

Similarly ∥φ∥2 = ∥F−1φ∥2. ■

Lemma 13.2.13 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).) 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 ) .