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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
...
We can also see one of the best ideas, that is, taking a maximal
operator when one wants to prove pointwise convergence
...
This
permits one to observe one of the pieces of Carleson’s proof without any
technical problems
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−∞

This is equivalent to the question of whether there exists

F (x + h) − F (x)
1 x+h
= lim
lim
f (t) dt
...
Here,
 1  x+h



sup
f (t) dt
...
1)

where Q denotes a cube of center x and side h and we write Q  x to express
that the side h → 0+
...

For every locally integrable function f : Rn → C, we put

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



4

1
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Mf is
the Hardy-Littlewood maximal function
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2 Weak inequality
First observe that given f locally integrable, the function Mf : Rn → [0, +∞]
is measurable
...

|Q| Q
We only have to observe that the function

1
|f (t)| dt
y
→
|Q| y+Q
is continuous
...

However, for p = 1 this is no longer true
...
That is to say
m{Mf (x) > α} ≤ cn


f
1

...
The set where Mf (x) > α is covered by
cubes where the mean of |f | is greater than α
...
Then we can select a big pairwise disjoint
subfamily and this implies that the norm of f is big
...
This is accomplished by the following covering lemma
Lemma 1
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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 ,
...


j=1

Proof
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2 Weak inequality

5

covering of G
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Now select a finite subcovering of K, say
that with the balls U1 , U2 ,
...
Assume that these balls are ordered
with decreasing radii
...
First
B1 = U1 is the greatest of them all
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Then B3 will be the first ball from the Uj that is disjoint from B1 ∪ B2
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m
n
Now we claim that K ⊂ j=1 3Bj
...

Hence for every x ∈ K, there is a first j such that x ∈ Uj
...
In other case Uj intersects
some Bk = Us
...
So the radius of the
ball Bk is greater than or equal to that of Uj
...

Therefore
n

1
3d m(Bj ),
m(A) ≤ m(K) ≤
2
j=1
and the construction implies that these balls are disjoint
...
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

...
Let A = {x ∈ Rd | Mf (x) > α}, it is an open set
...
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
...
2)
|Q| Q
Now cubes are balls for the norm
·
∞ on Rd
...
2), and
m(An ) ≤ cd

m

j=1

Therefore we have

m(Qj )
...
Hardy-Littlewood maximal function


m

1
m(An ) ≤ cd
|f (t)| dt
...

α

Taking limits when n → ∞, we obtain our desired bound
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3 Differentiability
As an application we desire to obtain (1
...
In fact we can prove something
more
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Q

A point where this is true is called a Lebesgue point of f
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3 (Differentiability Theorem) Let f : Rd → C be a locally
integrable function
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That is



1
f (t) − f (x) dt = 0
...
Whether x is a Lebesgue point of f or not, depends only on the values
of f in a neighborhood of x
...

Also the results are true for a dense set on L1 (Rd )
...

Hence if Q denotes a cube with a sufficiently small radius we have



1
f (t) − f (x) dt ≤ ε
...

Now we can observe for the first time how the maximal function intervenes
in pointwise convergence matters
...
If f ∈ L1 (Rd ),



1
f (t) − f (x) dt
Ωf (x) = lim sup
|Q|
Qx
Q

1
...

Now our objective is to prove that Ωf (x) = 0 almost everywhere
...
Since the continuous functions are dense on L1 (Rd ), we obtain
a continuous ϕ ∈ L1 (Rd ), such that
f − ϕ
1 < ε
...

Hence for every α > 0 we have
{Ωf (x) > α} ⊂ {M(f − ϕ)(x) > α/2} ∪ {|f (x) − ϕ(x)| > α/2}
...

α
α
α

Since this inequality is true for every ε > 0, we deduce m{Ωf (x) > α} = 0
...



As an example we prove that


x

F (x) =

f (t) dt
−∞

is differentiable at every Lebesgue point of f
...

For h > 0



1 x+h
F (x + h) − F (x)
− f (x) =
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
...


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4 Interpolation
At one extreme, with p = 1, the maximal function Mf satisfies a weak
inequality
...

An idea of Marcinkiewicz permits us to interpolate between these two
extremes
...
4 For every f ∈ Lp (Rd ), 1 < p < +∞ we have

Mf
p ≤ Cd

p

f
p
...
For every α > 0 we decompose f , f = f χA +f χRd
A , where A =
{|f | > α}
...
Consequently

cd
m{Mf > 2α} ≤ m{M(f χA ) > α} ≤
|f | χ{|f |>α} dm
...
In particular
observe that we have used a different decomposition of f for every α
...
Hence we
obtain our claim about the norm
...
For example
if
f
1 > 0, then Mf is not integrable
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5 A general inequality

9

Proposition 1
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Rd

B

Proof
...
We have

 +∞
Mf (x) dx =
mB {Mf (x) > t} dt
...

The point of the proof is to use adequately the weak inequality
...
Therefore Mf ≤
α + M(f χA ), and
{Mf (x) > 2α} ⊂ {M(f χA )(x) > α}
...



{|f (x)|>t}


|f (x)| dx dt
...




1
...
This and many other applications of these functions
derive from the following inequality
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6 Let ϕ: Rd → R be a positive, radial, decreasing, and integrable
function
...


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We say that ϕ is radial if there is a function u: [0, +∞) → R such
that ϕ(x) = u(|x|) for every x ∈ Rd
...

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
...

Now the proof is straightforward
...
By the monotone
convergence theorem
|ϕ ∗ f (x)| ≤ ϕ ∗ |f |(x) = lim ϕn ∗ |f |(x)
...


We can replace the ball B(x, tj ) by the cube with center x and side 2tj
...
Thus
ϕn ∗ |f |(x) ≤

N





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

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