32.3. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 1109

Corollary 32.3.17 For f ,g ∈ L2(Rn),∫Rn

f gdx =∫Rn

F f Fgdx =∫Rn

F−1 f F−1gdx.

Proof: First note the above formula is obvious if f ,g ∈ G . To see this, note∫Rn

F f Fgdx =∫Rn

F f (x)1

(2π)n/2

∫Rn

e−ix·tg(t)dtdx

=∫Rn

1

(2π)n/2

∫Rn

eix·tF f (x)dxg(t)dt =∫Rn

(F−1 ◦F

)f (t)g(t)dt

=∫Rn

f (t)g(t)dt.

The formula with F−1 is exactly similar.Now to verify the corollary, let φ k→ f in L2 (Rn) and let ψk→ g in L2 (Rn). Then by

Lemma 32.3.13∫Rn

F f Fgdx = limk→∞

∫Rn

Fφ k Fψkdx = limk→∞

∫Rn

φ kψkdx =∫Rn

f gdx

A similar argument holds for F−1.How does one compute F f and F−1 f ?

Theorem 32.3.18 For f ∈ L2(Rn), let fr = f XEr where Er is a bounded measurable setwith Er ↑ Rn. Then the following limits hold in L2 (Rn) .

F f = limr→∞

F fr , F−1 f = limr→∞

F−1 fr.

Proof: || f − fr||2→ 0 and so ||F f −F fr||2→ 0 and ||F−1 f −F−1 fr||2→ 0 by Plan-cherel’s Theorem.

What are F fr and F−1 fr? Let φ ∈ G∫Rn

F frφdx =∫Rn

frFφdx

= (2π)−n2

∫Rn

∫Rn

fr(x)e−ix·yφ(y)dydx

=∫Rn[(2π)−

n2

∫Rn

fr(x)e−ix·ydx]φ(y)dy.

Since this holds for all φ ∈ G , a dense subset of L2(Rn), it follows that

F fr(y) = (2π)−n2

∫Rn

fr(x)e−ix·ydx.

Similarly

F−1 fr(y) = (2π)−n2

∫Rn

fr(x)eix·ydx.