Hardy-Littlewood maximal function

1. Hardy-Littlewood maximal function

1.1 Introduction What Carleson proved in 1966 was Luzin’s conjecture of 1913, and this proof depended on many results obtained in the fifty years since the conjecture was stated. In this chapter we make a rapid exposition of one of these prerequisites. We can also see one of the best ideas, that is, taking a maximal operator when one wants to prove pointwise convergence. The convergence result obtained is simple: the differentiability of the definite integral. This permits one to observe one of the pieces of Carleson’s proof without any technical problems. Given a function f ∈ L1 (R) we ask about the differentiability properties of the definite integral  x

F (x) =

f (t) dt. −∞

This is equivalent to the question of whether there exists  F (x + h) − F (x) 1 x+h = lim lim f (t) dt. h→0 h→0 h x h When we are confronted with questions of convergence it is advisable to study the corresponding maximal function. Here,  1  x+h    sup f (t) dt. h h x An analogous result in dimension n will be  1 f (x) = lim f (t) dt, Qx |Q| Q

(1.1)

where Q denotes a cube of center x and side h and we write Q  x to express that the side h → 0+ . In the one-dimensional case we have Q = [x − h, x + h], this difference ([x − h, x + h] instead of [x, x + h]) has no consequence, as we will see. For every locally integrable function f : Rn → C, we put

J.A. de Reyna: LNM 1785, pp. 3–10, 2002. c Springer-Verlag Berlin Heidelberg 2002


4

1. Hardy-Littlewood maximal function

Mf (x) = sup Q

1 |Q|

 |f (t)| dt, Q

where the supremum is taken over all cubes Q ⊂ Rn with center x. Mf is the Hardy-Littlewood maximal function.

1.2 Weak inequality First observe that given f locally integrable, the function Mf : Rn → [0, +∞] is measurable. In fact for every positive real number α the set {Mf (x) > α} is open, because given x ∈ Rn with Mf (x) > α there exists a cube Q with center at x and such that  1 |f (t)| dt > α. |Q| Q We only have to observe that the function  1 |f (t)| dt y → |Q| y+Q is continuous. If f ∈ Lp (Rn ), with 1 < p < +∞, we shall show that Mf ∈ Lp (Rn ). However, for p = 1 this is no longer true. What we can say is only that f belongs to weak-L1 . That is to say m{Mf (x) > α} ≤ cn


f
1 . α

The proof is really wonderful. The set where Mf (x) > α is covered by cubes where the mean of |f | is greater than α. If this set has a big measure, we shall have plenty of these cubes. Then we can select a big pairwise disjoint subfamily and this implies that the norm of f is big. The most delicate point of this proof is that at which we select the disjoint cubes. This is accomplished by the following covering lemma Lemma 1.1 (Covering lemma) Let Rd be endowed with some norm, and let cd = 2 · 3d . If A ⊂ Rd is a non-empty set of finite exterior measure, and U is a covering of A by open balls, then there is a finite subfamily of disjoint balls B1 ,. . . ,Bn of U such that cd

n 

m(Bj ) ≥ m∗ (A).

j=1

Proof. We can assume that A is measurable, because if it were not, there would exist open set G ⊃ A with m(G) finite and such that U would be a

1.2 Weak inequality

5

covering of G. Now, assuming that A is measurable, there exists a compact set K ⊂ A with m(K) ≥ m(A)/2. Now select a finite subcovering of K, say that with the balls U1 , U2 , . . . , Um . Assume that these balls are ordered with decreasing radii. Then we select the balls Bj in the following way. First B1 = U1 is the greatest of them all. Then B2 is the first ball in the sequence of Uj that is disjoint from B1 , if there is one, in the other case we put n = 1. Then B3 will be the first ball from the Uj that is disjoint from B1 ∪ B2 . We continue in this way, until every ball from the sequence Uj has non-empty intersection with some Bj .  m n Now we claim that K ⊂ j=1 3Bj . In fact we know that K ⊂ j=1 Uj . Hence for every x ∈ K, there is a first j such that x ∈ Uj . If this Uj is equal to some Bk obviously we have x ∈ Bk ⊂ 3Bk . In other case Uj intersects some Bk = Us . Selecting the minimum k, it must be that s < j, for otherwise we would have selected Uj instead of Bk in our process. So the radius of the ball Bk is greater than or equal to that of Uj . It follows that Uj ⊂ 3Bk . Therefore n  1 3d m(Bj ), m(A) ≤ m(K) ≤ 2 j=1 and the construction implies that these balls are disjoint.




Lemma 1.2 (Hardy and Littlewood) If f ∈ L1 (Rd ) then Mf satisfies, for each α > 0, the weak inequality m{x ∈ Rd | Mf (x) > α} ≤ cd


f
1 . α

Proof. Let A = {x ∈ Rd | Mf (x) > α}, it is an open set. We do not know yet that it has finite measure, so we consider An = A ∩ Bn , where Bn is a ball of radius n and center 0. Now each x ∈ An has Mf (x) > α; hence there exists an open cube Q, with center at x and such that  1 |f (t)| dt > α. (1.2) |Q| Q Now cubes are balls for the norm
·
∞ on Rd . So we can apply the covering lemma to obtain a finite set of disjoint cubes (Qj )m j=1 such that every one of them satisfies (1.2), and m(An ) ≤ cd

m  j=1

Therefore we have

m(Qj ).

6

1. Hardy-Littlewood maximal function

 m  1 m(An ) ≤ cd |f (t)| dt. α Q j=1 Since the cubes are disjoint m(An ) ≤ cd


f
1 . α

Taking limits when n → ∞, we obtain our desired bound.




1.3 Differentiability As an application we desire to obtain (1.1). In fact we can prove something more. It is not only that at almost every point x ∈ Rd we have 

 1  f (t) − f (x) dt = 0, lim  Qx |Q| Q but that we have

1 Qx |Q|



lim

  f (t) − f (x) dt = 0.

Q

A point where this is true is called a Lebesgue point of f . Theorem 1.3 (Differentiability Theorem) Let f : Rd → C be a locally integrable function. There exists a subset Z ⊂ Rd of null measure and such that every x ∈ / Z is a Lebesgue point of f . That is    1 f (t) − f (x) dt = 0. lim Qx |Q| Q Proof. Whether x is a Lebesgue point of f or not, depends only on the values of f in a neighborhood of x. So we can reduce to the case of f integrable. Also the results are true for a dense set on L1 (Rd ). In fact if f is continuous, given x and ε > 0, there is a neighborhood of x such that |f (t)−f (x)| < ε. Hence if Q denotes a cube with a sufficiently small radius we have    1 f (t) − f (x) dt ≤ ε. |Q| Q Hence, for a continuous function f , every point is a Lebesgue point. Now we can observe for the first time how the maximal function intervenes in pointwise convergence matters. We are going to define the operator Ω. If f ∈ L1 (Rd ),    1 f (t) − f (x) dt Ωf (x) = lim sup |Q| Qx Q

1.3 Differentiability

7

Note that Ωf (x) ≤ Mf (x) + |f (x)|. Now our objective is to prove that Ωf (x) = 0 almost everywhere. Fix ε > 0. Since the continuous functions are dense on L1 (Rd ), we obtain a continuous ϕ ∈ L1 (Rd ), such that
f − ϕ
1 < ε. By the triangle inequality Ωf (x) ≤ Ωϕ(x) + Ω(f − ϕ)(x) = Ω(f − ϕ)(x) ≤ M(f − ϕ)(x) + |f (x) − ϕ(x)|. Hence for every α > 0 we have {Ωf (x) > α} ⊂ {M(f − ϕ)(x) > α/2} ∪ {|f (x) − ϕ(x)| > α/2}. Now we use the weak inequality for the Hardy-Littlewood maximal function and the Chebyshev inequality for |f − ϕ| m{Ωf (x) > α} ≤ 2cd


f − ϕ
1 ε
f − ϕ
1 +2 ≤ Cd . α α α

Since this inequality is true for every ε > 0, we deduce m{Ωf (x) > α} = 0. And this is true for every α > 0, hence Ωf (x) = 0 almost everywhere.
As an example we prove that 

x

F (x) =

f (t) dt −∞

is differentiable at every Lebesgue point of f . We assume that f is integrable. For h > 0 

1 x+h F (x + h) − F (x) − f (x) = f (t) − f (x) dt. h h x Hence

 F (x + h) − F (x)  1  x+h     f (t) − f (x) dt − f (x) ≤  h h x  x+h   2 f (t) − f (x) dt. ≤ 2h x−h

If x is a Lebesgue point of f we know that the limit when h → 0 is equal to zero. An analogous procedure proves the existence of the left-hand limit at x.

8

1. Hardy-Littlewood maximal function

1.4 Interpolation At one extreme, with p = 1, the maximal function Mf satisfies a weak inequality. At the other extreme p = +∞, it is obvious from the definition that if f ∈ L∞ (Rd )
Mf
∞ ≤
f
∞ . An idea of Marcinkiewicz permits us to interpolate between these two extremes. Theorem 1.4 For every f ∈ Lp (Rd ), 1 < p < +∞ we have
Mf
p ≤ Cd

p
f
p . p−1

Proof. For every α > 0 we decompose f , f = f χA +f χRd
A , where A = {|f | > α}. Then Mf ≤ α + M(f χA ). Consequently  cd m{Mf > 2α} ≤ m{M(f χA ) > α} ≤ |f | χ{|f |>α} dm. α Rd The proof depends on a judicious use of this inequality. In particular observe that we have used a different decomposition of f for every α. We have the following chain of inequalities  +∞ p
Mf
p = p tp−1 m{Mf > t} dt ≤ 0



+∞

tp−1

p 0

2cd t

Applying Fubini’s theorem   |f (x)| 2cd p Rd

 2cd p

Rd

0

 Rd

|f | χ{|f |>t/2} dm dt ≤

+∞

tp−2 χ{|f (x)|>t/2} dt dx =

2p cd p (2|f (x)|)p−1 |f (x)| dx =
f
pp p−1 p−1

It is easy to see that (p/(p − 1))1/p is equivalent to p/(p − 1). Hence we obtain our claim about the norm.
In the case of p = 1 the best we can say is the weak inequality. For example if
f
1 > 0, then Mf is not integrable. In spite of this we shall need in the proof of Carleson theorem a bound of the integral of the maximal function on a set of finite measure; that is a consequence of the weak inequality

1.5 A general inequality

9

Proposition 1.5 For every function f ∈ L1 (Rd ) and B ⊂ Rd a measurable   set Mf (x) dx ≤ m(B) + 2cd |f (x)| log+ |f (x)| dx. Rd

B

Proof. Let mB be the measure mB (M ) = m(B ∩ M ). We have   +∞ Mf (x) dx = mB {Mf (x) > t} dt. 0

B

Now we have two inequalities mB {Mf (x) > t} ≤ m(B), and the weak inequality. The point of the proof is to use adequately the weak inequality. For every α we have f = f χA +f χRd
A where A = {f (x) > α}. Therefore Mf ≤ α + M(f χA ), and {Mf (x) > 2α} ⊂ {M(f χA )(x) > α}. It follows that cd m{Mf (x) > 2α} ≤ α Hence



 Mf (x) dx ≤ m(B) + 2

B

1



+∞

{|f (x)|>α}

cd  t

Therefore by Fubini’s theorem   Mf (x) dx ≤ m(B) + 2cd

Rd

B

|f (x)| dx.

 {|f (x)|>t}

 |f (x)| dx dt.

|f (x)| log+ |f (x)| dx.


1.5 A general inequality The Hardy-Littlewood maximal function can be used to prove many theorems of pointwise convergence. This and many other applications of these functions derive from the following inequality. Theorem 1.6 Let ϕ: Rd → R be a positive, radial, decreasing, and integrable function. Then for every f ∈ Lp (Rd ) and x ∈ Rd we have |ϕ ∗ f (x)| ≤ Cd
ϕ
1 Mf (x), where Cd is a constant depending only on the dimension, and equal to 1 for d = 1.

10

1. Hardy-Littlewood maximal function

Proof. We say that ϕ is radial if there is a function u: [0, +∞) → R such that ϕ(x) = u(|x|) for every x ∈ Rd . Also we say that a radial function ϕ is decreasing if u is decreasing. The function u is measurable, hence there is an increasing sequence of simple functions (un ) such that un (t) converges to u(t) for every t ≥ 0. In this case, since u is decreasing, it is possible to choose each un un (t) =

N 

hj χ[0,tj ] (t),

j=1

where 0 < t1 < t2 < · · · < tN and hj > 0 and the natural number N depends on n. Now the proof is straightforward. Let ϕn (x) = un (|x|). By the monotone convergence theorem |ϕ ∗ f (x)| ≤ ϕ ∗ |f |(x) = lim ϕn ∗ |f |(x). n

Therefore ϕn ∗ |f |(x) =

N  j=1

 hj

B(x,tj )

|f (y)| dy.

We can replace the ball B(x, tj ) by the cube with center x and side 2tj . The quotient between the volume of the ball and the cube is bounded by a constant. Thus ϕn ∗ |f |(x) ≤

N 



hj m Q(xj , tj ) · Mf (x) ≤ Cd
ϕ
1 Mf (x).

j=1