284 CHAPTER 11. FUNDAMENTAL TRANSFORMS
Also by Theorem 11.6.13 {Fφ k}∞
k=1 is Cauchy in L2 (Rn) and so it converges to someh ∈ L2 (Rn). Therefore, from the above,
F f (ψ) =∫Rn
h(x)ψ (x)
which shows that F ( f ) ∈ L2 (Rn) and h = F ( f ) . The case of F−1 is entirely similar. ■Since F f and F−1 f are in L2 (Rn) , this also proves the following theorem.
Theorem 11.6.15 If f ∈ L2(Rn), F f and F−1 f are the unique elements of L2 (Rn)such that for all φ ∈ G , ∫
RnF f (x)φ(x)dx =
∫Rn
f (x)Fφ(x)dx, (11.12)
∫Rn
F−1 f (x)φ(x)dx =∫Rn
f (x)F−1φ(x)dx. (11.13)
Theorem 11.6.16 (Plancherel)
∥ f∥2 = ∥F f∥2 = ∥F−1 f∥2. (11.14)
Proof: Use the density of G in L2 (Rn) to obtain a sequence, {φ k} converging to f inL2 (Rn). Then by Lemma 11.6.14
∥F f∥2 = limk→∞
∥Fφ k∥2 = limk→∞
∥φ k∥2 = ∥ f∥2 .
Similarly, ∥ f∥2 = ∥F−1 f∥2. ■The following corollary is a simple generalization of this. To prove this corollary,
use the following simple lemma which comes as a consequence of the Cauchy Schwarzinequality.
Lemma 11.6.17 Suppose fk→ f in L2 (Rn) and gk→ g in L2 (Rn). Then
limk→∞
∫Rn
fkgkdx =∫Rn
f gdx.
Proof: ∣∣∣∣∫Rnfkgkdx−
∫Rn
f gdx∣∣∣∣≤ ∣∣∣∣∫Rn
fkgkdx−∫Rn
fkgdx∣∣∣∣+∣∣∣∣∫Rn
fkgdx−∫Rn
f gdx∣∣∣∣
≤ ∥ fk∥2 ∥g−gk∥2 +∥g∥2 ∥ fk− f∥2 .
Now ∥ fk∥2 is a Cauchy sequence and so it is bounded independent of k. Therefore, theabove expression is smaller than ε whenever k is large enough. ■
Corollary 11.6.18 For f ,g ∈ L2(Rn),∫Rn
f gdx =∫Rn
F f Fgdx =∫Rn
F−1 f F−1gdx.