12.2 The size of the maximal function
Let f be a function on [0, 1), and let M(f ) be the corresponding dyadic maximal function, as in the preceding section.
What can we say about M(f )? How does it compare to f , in terms of overall size? In other words, if f is not too large, can we say that M(f ) is not too large?
Lemma 12.13 (Supremum bound for M(f)) Suppose that A is a non-negative real number such that |f (x)| ≤ A for all x ∈ [0, 1). Then M(f )(x) ≤ A for all x ∈ [0, 1).
This is an easy consequence of the definitions. If |f (x)| ≤ A for all x ∈ [0, 1), then all averages of f have absolute value less than or equal to A.
This implies that M(f ) ≤ A at all points in [0, 1), since M(f ) is defined in terms of suprema of absolute values of averages of f .
This would also work if we assumed that |f (x)| ≤ A holds on [0, 1) except for a very small set, like a finite set, a countable set, or, more generally, a set of measure 0. That is, the behavior of f on a small set like this would not affect any of the integrals of f , and so one could just as well replace the values of f with 0 on such a set.
Proposition 12.14 (Weak-type estimate for M(f)) For each positive real number λ,
|{x ∈ [0, 1) : M(f )(x) > λ}| ≤ 1 λ
Z
[0,1)|f (w)| dw.
(12.15)
The left-hand side of (12.15) refers to the measure of the set in question, and the proof of the lemma will give a very simple meaning for this.
Let λ > 0 be given. Define F to be the collection of dyadic intervals L in
[0, 1) such that
1
|L|
Z
Lf (y) dy
> λ, (12.16)
Let us check that
{x ∈ [0, 1) : M(f )(x) > λ} = [
L∈F
L.
(12.17)
If x ∈ [0, 1) and M(f )(x) > λ, then there is a dyadic interval L in [0, 1) such that x ∈ L and L satisfies (12.16), because of (12.2). This shows that the left
100 CHAPTER 12. MAXIMAL FUNCTIONS side of (12.17) is contained in the right side of (12.17). Conversely, if L ∈ F , then
for all x ∈ L. This shows that L is contained in the left side of (12.17). Hence the right side of (12.17) is contained in the left side. This proves (12.17).
We may as well assume that the left side of (12.17) is not empty, since otherwise (12.15) is automatic. This implies that F is nonempty as well.
Now we apply Lemma 2.5 to obtain a subcollection F0 of F such that
[
and the intervals in F′are pairwise disjoint. Combining these properties with (12.17), we obtain that
|{x ∈ [0, 1) : M(f )(x) > λ}| = X
L∈F0
|L|.
(12.20)
For the proof of (12.15), we do not need equality in (12.20), but only the inequality ≤. This inequality does not require the disjointness of the intervals in F0, but this will be needed in a moment.
Remark 12.21 If f is a dyadic step function, then the set {x ∈ [0, 1) : M(f )(x) > λ} can be given as a union of finitely many dyadic intervals.
More precisely, one could use only intervals L with size |L| ≥ 2−j for some j, because of Corollary 12.12.
Now let us look at the right side of (12.20). Each interval L in the sum satisfies (12.16), which we can rewrite as
|L| < 1
12.3. SOME VARIATIONS 101 The last step uses the disjointness of the intervals L in F0.
The combination of (12.20) and (12.23) implies (12.15). In fact, we get the slightly more precise inequality that
|{x ∈ [0, 1) : M(f )(x) > λ}| ≤ 1 λ
Z
{x∈[0,1):M (f )(x)>λ}|f (y)| dy, (12.24)
using (12.17) and (12.19).
12.3 Some variations
Instead of dyadic intervals (and averages of functions over them, etc.), one can also look at arbitrary intervals in the real line. For these, it is not as easy to have disjointness, or reduce to that case. However, there is a simple substitute, indicated by the following lemmas.
Lemma 12.25 (3 intervals to 2) Suppose that I, J, and K are intervals in the real line such that there is a point x ∈ R which lies in all three of them. Then one of the intervals is contained in the union of the other two.
This is not hard to check. One of the intervals will go as far as possible to the left of x, another of the intervals (perhaps the same one) will go as far as possible to the right of x, and the union of these two intervals (which may be the same interval) will contain all of I ∪ J ∪ K.
Lemma 12.26 (No point counted more than twice) Let A be an arbi-trary finite collection of intervals in R. There is a subcollection A1 of A with the following properties:
[
J∈A1
J = [
J∈A
J;
(12.27)
no point in R belongs to more than two intervals in A1. (12.28)
This can be derived from Lemma 12.25. Specifically, one starts with A itself, and one leaves it alone if it already satisfies (12.28). If not, there is a point in R which lies in 3 intervals in A, and one can throw away one of the intervals without changing the total union, by Lemma 12.25. One repeats this process until there are no points in R which lie in 3 intervals remaining in the collection (i.e., that have not been thrown away). This has to happen
102 CHAPTER 12. MAXIMAL FUNCTIONS eventually, since we assumed that A is finite. The resulting collection can be used for A1.
One can rephrase (12.28) in terms of indicator functions, as follows:
X
J∈A1
1J(x) ≤ 2 · 1(SJ ∈A1J)(x) for all x ∈ R.
(12.29)
If f is a nonnegative function on R, one can integrate (12.29) with f to get
X
J∈A1
Z
Jf (x) dx ≤ 2
Z S
J ∈A1Jf (x) dx.
(12.30)
This could be used in a setting like that of the last step in (12.23). Of course
X
J∈A1
Z
Jf (x) dx ≥
Z S
J ∈A1Jf (x) dx (12.31)
holds automatically (so that the equality in the last step in (12.23) corre-sponds to these two inequalities, going in opposite directions).
These observations are special to dimension 1, although there are well-known substitutes for other settings (like Rn, n > 1).
In Rn, n > 1, one can also consider dyadic cubes, which behave in much the same manner as dyadic intervals in R. A dyadic cube in Rnis a Cartesian product of n dyadic intervals in R of the same size. As above, one can focus on dyadic subcubes of the unit cube, which is the Cartesian product of n copies of the unit interval [0, 1). (This is not a serious restriction, however.) With dyadic cubes, one has much the same kind of disjointness and partitioning properties as for dyadic intervals.