33.4. SINGULAR INTEGRALS 1143

Proof:Kε ∗g(x)≡

∫Kε (y)g(x−y)dy.

Let 0 < η < ε. Then since g ∈ C1c (Rn) , there exists a constant, K such that K |u−v| ≥

|g(u)−g(v)| for all u,v ∈ Rn.∣∣Kε ∗g(x)−Kη ∗g(x)∣∣ ≤ BK

∫η<|y|<ε

1|y|n|y|dy

= BK∫

Sn−1

∫ε

η

dρdσ =Cn |ε−η | .

This proves the corollary.

Theorem 33.4.5 Suppose 33.4.40 - 33.4.42. Then for Kε given by 33.4.43 and p > 1, thereexists a constant A(p,n,B) such that for all f ∈ Lp (Rn),

||Kε ∗ f ||p ≤ A(p,n,B) || f ||p . (33.4.52)

Also, for each f ∈ Lp (Rn),T f ≡ lim

ε→0Kε ∗ f (33.4.53)

exists in Lp (Rn) and for all f ∈ Lp (Rn),

||T f ||p ≤ A(p,n,B) || f ||p . (33.4.54)

Thus T is a linear and continuous map defined on Lp (Rn) for each p > 1.

Proof: From 33.4.40 it follows Kε ∈ Lp′ (Rn)∩L2 (Rn) where, as usual, 1/p+1/p′= 1.By continuity of translation in Lp′ (Rn), x→ Kε ∗ f (x) is a continuous function.By Lemma33.4.3, ||FKε ||∞ ≤C (n)B for all ε . Therefore, by Lemma 33.4.2,

||Kε ∗g||p ≤ A(p,n,B) ||g||p

for all g ∈ G . Now let f ∈ Lp (Rn) and gk→ f in Lp (Rn) where gk ∈ G . Then

|Kε ∗ f (x)−Kε ∗gk (x)| ≤∫|Kε (x−y)| |gk (y)− f (y)|dy

≤ ||Kε ||p′ ||gk− f ||p

which shows that Kε ∗gk (x)→ Kε ∗ f (x) pointwise and so by Fatou’s lemma,

||Kε ∗ f ||p ≤ lim infk→∞

||Kε ∗gk||p ≤ lim infk→∞

A(p,n,B) ||gk||p= A(p,n,B) || f ||p .

This verifies 33.4.52.To verify 33.4.53, let δ > 0 be given and let

f ∈ Lp (Rn) ,g ∈C∞c (Rn) .