11.6. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 285

Proof: First note the above formula is obvious if f ,g ∈ G . To see this, use 11.10 towrite ∫

RnF f Fgdx =

∫Rn

F f F−1 (ḡ)dx =∫Rn

F−1 (F f )(ḡ)dx =∫Rn

f ḡdx

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 11.6.14 Fφ k→ F f and Fψk→ Fg and so∫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 11.6.19 For f ∈ L2(Rn), let fr = f XEr where Er is a bounded measur-able set with 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.

This shows that to take the Fourier transform of a function in L2 (Rn), it suffices to take thelimit as r→ ∞ in L2 (Rn) of (2π)−

n2∫Rn fr(x)e−ix·ydx. A similar procedure works for the

inverse Fourier transform.Note this reduces to the earlier definition in case f ∈ L1 (Rn). Now consider the convo-

lution of a function in L2 with one in L1.

Theorem 11.6.20 Let h ∈ L2 (Rn) and let f ∈ L1 (Rn). Then h∗ f ∈ L2 (Rn),

F−1 (h∗ f ) = (2π)n/2 F−1hF−1 f ,

F (h∗ f ) = (2π)n/2 FhF f ,

and∥h∗ f∥2 ≤ ∥h∥2 ∥ f∥1 . (11.15)