1130 CHAPTER 33. FOURIER ANALYSIS IN Rn

Proof: From Lemma 33.3.2, F−1ρ∗ is weak (1,1), weak (2,2), and maps

L1 (Rn)+L2 (Rn)

to measurable functions. Therefore, by the Marcinkiewicz interpolation theorem, thereexists a constant Ap depending only on p,C1,n, and ||ρ||

∞for p ∈ (1,2], such that for

f ∈ Lp (Rn), and p ∈ (1,2], ∣∣∣∣F−1ρ ∗ f

∣∣∣∣p ≤ Ap || f ||p .

Thus the theorem is proved for these values of p. Now suppose p > 2. Then p′ < 2 where

1p+

1p′

= 1.

By Plancherel’s theorem and Theorem 32.3.25,∫F−1

ρ ∗φ (x)ψ (x)dx = (2π)n/2∫

ρ (x)Fφ (x)Fψ (x)dx

=∫

F(F−1

ρ ∗ψ)

Fφdx

=∫ (

F−1ρ ∗ψ

)(φ)dx.

Thus by the case for p ∈ (1,2) and Holder’s inequality,∣∣∣∣∫ F−1ρ ∗φ (x)ψ (x)dx

∣∣∣∣ =

∣∣∣∣∫ (F−1ρ ∗ψ

)(φ)dx

∣∣∣∣≤

∣∣∣∣F−1ρ ∗ψ

∣∣∣∣p′ ||φ ||p

≤ Ap′ ||ψ||p′ ||φ ||p .

Letting Lψ ≡∫

F−1ρ ∗φ (x)ψ (x)dx, this shows that L∈Lp′ (Rn)′ and also that ||L||(Lp′)

′ ≤

Ap′ ||φ ||pwhich implies by the Riesz representation theorem that F−1ρ ∗φ represents L and

||L||(Lp′)

′ =∣∣∣∣F−1

ρ ∗φ∣∣∣∣

Lp ≤ Ap′ ||φ ||p

Since p′ = p/(p−1), this proves the theorem.It is possible to give verifiable conditions on ρ which imply 33.3.20. The condition on

ρ which is presented here is the existence of a constant, C0 such that

C0 ≥ sup{|x||α| |Dαρ (x)| : |α| ≤ L, x ∈ Rn \{0}}, L > n/2. (33.3.25)

ρ ∈CL (Rn \{0}) where L is an integer.

Here α is a multi-index and |α| = ∑ni=1 α i. The condition says roughly that ρ is pretty

smooth away from 0 and all the partial derivatives vanish pretty fast as |x| →∞. Also recallthe notation

xα ≡ xα11 · · ·x

αnn

where α = (α1 · · ·αn). For more general conditions, see [69].