33.3. MIHLIN’S THEOREM 1129

Thus, 33.3.23 holds with A = ||ρ||∞(2π)n/2. Consequently,

A || f ||2 ≥

(∫[|F−1ρ∗ f |>α]

∣∣F−1ρ ∗ f (x)

∣∣2 dx

)1/2

≥ αm([∣∣F−1

ρ ∗ f∣∣> α

])1/2

and so 33.3.22 follows.It remains to prove 33.3.21 which holds for all f ∈G by Lemma 33.3.1. Let f ∈ L1 (Rn)

and let φ k→ f in L1 (Rn) ,φ k ∈G . Without loss of generality, assume that both f and F−1ρ

are Borel measurable. Therefore, by Minkowski’s inequality, and Plancherel’s theorem,∣∣∣∣F−1ρ ∗φ k−F−1

ρ ∗ f∣∣∣∣

2

≤

(∫ ∣∣∣∣∫ F−1ρ (x−y)(φ k (y)− f (y))dy

∣∣∣∣2 dx

)1/2

≤ ||φ k− f ||1 ||ρ||2

which shows that F−1ρ ∗ φ k converges to F−1ρ ∗ f in L2 (Rn). Therefore, there exists asubsequence such that the convergence is pointwise a.e. Then, denoting the subsequenceby k,

X[|F−1ρ∗ f |>α] (x)≤ lim infk→∞

X[|F−1ρ∗φ k|>α] (x) a.e. x.

Thus by Lemma 33.3.1 and Fatou’s lemma, there exists a constant, A, depending on C1,n,and ||ρ||

∞such that

m([∣∣F−1

ρ ∗ f∣∣> α

])≤ lim inf

k→∞m([∣∣F−1

ρ ∗φ k

∣∣> α])

≤ lim infk→∞

A||φ k||1

α= A|| f ||1

α.

This shows 33.3.21 and proves the lemma.

Theorem 33.3.3 Let ρ ∈ L2 (Rn)∩L∞ (Rn) and suppose∫|x|≥2|y|

∣∣F−1ρ (x−y)−F−1

ρ (x)∣∣dx≤C1.

Then for each p ∈ (1,∞), there exists a constant, Ap, depending only on

p,n, ||ρ||∞,

and C1 such that for all φ ∈ G , ∣∣∣∣F−1ρ ∗φ

∣∣∣∣p ≤ Ap ||φ ||p .