1150 CHAPTER 33. FOURIER ANALYSIS IN Rn

Lemma 33.5.5 Let f ∈ Lp (U) and let

wi (x)≡∫

UΦ,i (x−y) f (y)dy.

Then wi, j ∈ Lp (U) for each j = 1 · · ·n and the map f → wi, j is continuous and linear onLp (U).

Proof: First let f ∈C∞c (U). For such f ,

wi (x) =∫

UΦ,i (x−y) f (y)dy =

∫Rn

Φ,i (x−y) f (y)dy

=∫Rn

Φ,i (y) f (x−y)dy =∫

BΦ,i (y) f (x−y)dy

and

wi, j (x) =∫

BΦ,i (y) f, j (x−y)dy

=∫

B\B(0,ε)Φ,i (y) f, j (x−y)dy+

∫B(0,ε)

Φ,i (y) f, j (x−y)dy.

The second term converges to 0 because f, j is bounded and by 33.5.65, Φ,i ∈ L1loc. Thus

wi, j (x) =∫

B\B(0,ε)Φ,i (y) f, j (x−y)dy+ e(ε)

=∫

B\B(0,ε)−(Φ,i (y) f (x−y)), j +Φ,i j (y) f (x−y)dy+ e(ε)

where e(ε)→ 0 as ε → 0. Using the divergence theorem, this yields

wi, j (x) =∫

∂B(0,ε)Φ,i (y) f (x−y)n jdσ +

∫B\B(0,ε)

Φ,i j (y) f (x−y)dy+ e(ε).

Consider the first term on the right. This term equals, after letting y = εz,

εn−1

∫∂B(0,1)

Φ,i (εz) f (x−εz)n jdσ = Cnεn−1

∫∂B(0,1)

ε1−nziz j f (x−εz)dσ (z)

= Cn

∫∂B(0,1)

ziz j f (x−εz)dσ (z)

and this converges to 0 if i ̸= j and it converges to

Cn f (x)∫

∂B(0,1)z2

i dσ (z)

if i = j. Thus

wi, j (x) =Cnδ i j f (x)+∫

B\B(0,ε)Φ,i j (y) f (x−y)dy+ e(ε).