Measure, Integration & Real Analysis, 2021a

Section 4A Hardy–Littlewood Maximal Function 105
4.8 Hardy–Littlewood maximal inequality
Suppose h ∈L 1 (R). Then
for every c > 0.
|{b ∈ R : h ∗ (b) > c}| ≤ 3 c ‖h‖ 1
Proof Suppose F is a closed bounded subset of {b ∈ R : h ∗ (b) > c}. We will
show that |F| ≤ 3 ∫ ∞
c −∞
|h|, which implies our desired result [see Exercise 24(a) in
Section 2D].
For each b ∈ F, there exists t b > 0 such that
4.9
Clearly
1
2t b
∫ b+tb
b−t b
|h| > c.
F ⊂ ⋃
(b − t b , b + t b ).
b∈F
The Heine–Borel Theorem (2.12) tells us that this open cover of a closed bounded set
has a finite subcover. In other words, there exist b 1 ,...,b n ∈ F such that
F ⊂ (b 1 − t b1 , b 1 + t b1 ) ∪···∪(b n − t bn , b n + t bn ).
To make the notation cleaner, relabel the open intervals above as I 1 ,...,I n .
Now apply the Vitali Covering Lemma (4.4) to the list I 1 ,...,I n , producing a
disjoint sublist I k1 ,...,I km such that
Thus
I 1 ∪···∪I n ⊂ (3 ∗ I k1 ) ∪···∪(3 ∗ I km ).
|F| ≤|I 1 ∪···∪I n |
≤|(3 ∗ I k1 ) ∪···∪(3 ∗ I km )|
≤|3 ∗ I k1 | + ···+ |3 ∗ I km |
= 3(|I k1 | + ···+ |I km |)
< 3 (∫
∫
|h| + ···+
c I k1
≤ 3 c
∫ ∞
−∞
|h|,
I k m
)
|h|
where the second-to-last inequality above comes from 4.9 (note that |I kj | = 2t b for
the choice of b corresponding to I kj ) and the last inequality holds because I k1 ,...,I km
are disjoint.
The last inequality completes the proof.
Measure, Integration & Real Analysis, by Sheldon Axler