B
where the supremeum is taken all balls containing the point x.
Theorem 1 (The main properties of f ∗) Suppose f is integrable on Rd. Then: (i)f ∗ is measurable.
(i)Almost every x ∈ E is a point of density of E. (ii)Almost every x ∈/ E is not a point of density of E.
3
Next we consider the pointwise continuity. If f is locally integrable on Rd, the Lebesgue set of f consists of all points x¯ ∈ Rd for which f (x¯) is finite and
1
limm(B)→0 m(B)
f (y)dy = f (x), f or a.e. x
B
for all x ∈ B.
¶Note that as an immediate consequence of the theorem applied to |f |, we see that f ∗(x) ≥ |f (x)|
then
1
limm(B)→0 m(B)
f (y)dy = f (x¯)
B
Corollary 6 If f is locally integrable on Rd, then almost every point belongs to the Lebesgue set of f .
A collection of sets {Uα} is said to shrink regularly to x¯ (or has bounded eccentricity at x) if there is a constant c > 0 such that for each {Uα} ther is a ball B with
for every point x¯ in the Lebesgue ste of f .
• Application: Good kernels and approximations to the indentity.
Defination 8 The function is ”Good kernel” if it is integrable and satisfy the following conditions for δ > 0:
m({x : f ∗(x) > α}) ≥ c α
4
for some c > 0 and all sufficiently small α. 2.Let R denote the set of all rectangles in R2 that contain the origin, and with sides parallel to the
B
for all x ∈ B. ¶It is clearly true when f is continuous at x. We need to make some quantitative estimates bearing on the overall behavior of the averages of f .
for a.e. x.
Actually, we only focus on the ”local”. Indeed, the limit in Lebesgue’s theorem is taken over balls
that shrink to the point x, so the behavior of f far from x is irrelevant. To make this precise, we say that a measurable function f on Rd is locally integrable, if for every ball B the function f (x)χB(x) is integrable.We shall denote by L1loc(Rd) the space of all locally integrable functions.
1
limm(B)→0 m(B)
|f (y) − f (x¯)| dy = 0
B
for all x¯ ∈ B.
¶x¯ belongs to the Lebesgue set of f whenever f is continuous at x¯. If x¯ is in the Lebesgue set of f ,
¶The functions e|x| and |x|−1/2 are both locally integrable, but not integrable on Rd.
Theorem 4 If f ∈ Lຫໍສະໝຸດ loc(Rd), then1
limm(B)→0 m(B)
f (y)dy = f (x), f or a.e. x
1
2
(ii) f ∗ < ∞ for a.e. x.
(iii) f ∗ stisfies
m({x
∈
R
:
f∗
>
α})
≤
A α
|f (y)| dy
Rd
(1)
for all α > 0, where A = 3d.
¶We can not conclude that f ∗ is integrable on R (see Exercises 1 ) An inequality of the type (1) is called a weak − type inequality because it is weaker than the corresponding inequality for the L1-norms.Indeed, this can be seen from the Tchebychev inequality. The proof of inequality (1) relies on an elementary version of a Vitali covering argument.
f∗ ≥ c |xd|
for some c > 0 and all |x| ≥ 1. Conclude that f ∗ is not integrable on Rd.Then, show that the weak type estimate m({x : f ∗(x) > α}) ≤ c α
for all α > 0 whenever |f | = 1, then
x a
f
(y)dy.
Does
this
imply
that F is differentiable (at least for almost everywhere x), and that F = f ?
To deal with F (x), we recall the difination of the derivative as the limit of the quotient
k
m(∪Nl=1Bl) ≤ 3d m(Bij )
j=1
.
The esimate obtained for the maximal function now leads to a solution of the averaging problem.
Theorem 3 If f is integrable on Rd, then
This will be done in terms of the maximal averages of |f |, to which we now turn. We define the maximal function f ∗ by
f ∗ = sup 1 x∈B m(B)
|f (y)| dy, x ∈ Rd
In high dimensions, we can pose a similar question, where the anerages of f are taken over appropriate sets that generalize the intervals in one dimension. Initially we shall study this problem where the sets involved are the balls B containing x, with their volume m(B) (the Lebesgue measure) replacing the length |I| of I.
We restate our first problem in the context Rd, for all d ≥ 1. •(Averaging Problem) Suppose f is integrable on Rd. Is it true that
1
limm(B)→0 m(B)
f (y)dy = f (x), f or a.e. x
x¯ ∈ B, Uα ⊂ B, and m(Uα) ≥ cm(B) Corollary 7 Suppose f is locally integrable on Rd. If {Uα} shrinks regularly to x¯, then
lim
f (y)dy = f (x¯)
m(Uα)→0,x∈Uα Uα
B
for all x ∈ B.