ANALYSIS
1. Covering Theorems
Here out reference is Heinonen
Theorem 1.1 (5r-Covering Theorem). Every family F of balls of uniformly bounded radius
in a metric Space X contains a disjointed subfamily G such that
[
[
B⊆
5B.
B∈F
B∈G
In fact, every ball B from F meets a ball from G with radius at elast half that of B.
Proof. Let Ω denote the partially ordered (by inclusion) set consisting of all disjointed
subfamilies ω of F withe the following property: If a ball B from F meets some ball from
ω, then it meets one whose radius is at least half the radius of B. Then if C ⊆ Ω is a chain,
we see that
[
ω0 =
ω
C
belongs to Ω. We see that Ω is nonempty because let B be a ball from F such that the
radius of B is close to the supremal radius is in Ω. So by Zorn’s Lemma there is a maximal
element G in Ω. By construction B is disjointed.
If there is a ball B in F that does not meet any ball from G, then pick a ball B0 from F
such that the radius of B0 is larger than half the radius of any other ball that does not meet
the balls from G. Then, if a ball B from F meets a ball form the collection G 0 = G ∪ {B0 },
by construction it meets one whose radius is at least half that of B, showing that G 0 belongs
to Ω contradicting maximality of G.
Thus, every ball B = B(x, r) from F meets ball B 0 = B(x0 , r0 ) from G so that r ≤ 2r0
and so by the triangle inequality we have that B ⊆ 5B 0 .
t
u
Theorem 1.2 (Vitali Covering Theorem). Let A be a subset in a doubling metric measure
space (X, µ), and let F be a collection of closed balls centered on A such that
inf{r > 0 : B(a, r) ∈ F} = 0
for each a ∈ A. Then there is a countable disjointed subfamily G of F such that the balls in
G cover µ almost all of A, namely
!
[
µ A \ B = 0.
G
Proof. Assume that A is bounded. Next, we may assume that the balls in F have uniformly
bounded radii. Thus,
S by the 5r-Covering theorem,
S that is a disjointed subcollection G of
F such that A ⊆ G 5B. Moreover, the union G B is contianed in some fixed ball, and
1
2
ANALYSIS
hence, because µ is doubling, G is countable (because uncountable sums cannot be finite).
By doubling we have that
X
X
µ(5Bi ) ≤ C
µ(Bi ) < ∞
i≥1
since Bi disjoint and are all in some big ball. Thus,
X
lim
µ(Bi ) = 0.
N →∞
i>N
Therefore, it suffices to show that
A\
N
[
Bi ⊆
i=1
[
5Bi .
i>N
S
Take a ∈ A \ N
i=1 Bi . As the balls in F are closed, we can find a small ball B(a, r) ∈ F
that does not meet any of the balls Bi for i ≤ N . On the other hand, by the 5r-covering
theorem, the family G can be chosen so that B(a, r) meets
S some ball Bj from G with radius
atleast 2r . Thus, j > N and B(a, r) ⊆ 5Bj . So a ∈ i>N 5Bi . For unbounded case we
can take larger and larger balls and prove the result on each of them or use dyadic cubes
(Christ Cubes).
t
u
2. Maximal Functions
Here our reference is Heinonen and Mattila and Duoandikoetxea
Definition 2.1. Let (X, µ) be a doubling metric measure space. And let f ∈ L1loc then we
define the maximal function of f as
Z
M (f )(x) = sup −
|f | dµ.
r>0 B(x,r)
Theorem 2.2 (Maximal Function Theorem). If f ∈ L1 then for all t > 0 we have
Z
C1
|f | dµ.
µ({M (f ) > t}) ≤
t X
And if f ∈ Lp for p > 1 we have
Z
Z
p
|M (f )| dµ ≤ Cp
X
|f |p dµ.
X
The constants depend only on p and the doubling constant.
R
Proof. Let MR (f )(x) = supR>r>0 −B(x,r) |f | dµ. For each now fix f positive and fix t > 0.
Now for each x ∈ {MR (f ) > t} pick a ball B(x, r) such that
Z
|f | dµ > tµ(B(x, r)).
B(x,r)
From this collection of balls we extract a disjoint subcollection G by the 5r-covering theorem.
And we see that
Z
Z
X
X
CX
C
µ({MR (f ) > t}) ≤
µ(5B) ≤ C
µ(B) ≤
|f | dµ ≤
|f | dµ.
t
t X
B
G
G
G
ANALYSIS
3
Thus as our bound is independent of R we can take limits to get
Z
C
|f | dµ.
µ({M (f ) > t}) ≤
t X
Now we note that ||M (f )||∞ ≤ ||f ||∞ so by an interpolation result (Marcinkiewicz) we have
that ||M (f )||p ≤ C 0 ||f ||p for p > 1.
More explicitly let f ∈ Lp . And we note that
f = f · χ{f ≤ t } + f · χ{f > t } : g + b
2
2
with fixed t > 0. Thus we see that
t
+ M (b)
2
M (f ) ≤ M (g) + M (B) ≤
which implies that
t
{M (f ) > t} ⊆ {M (b) > }.
2
We note that b ∈ L1 because
||b||1 = ||f ||1 χ{f > t }
2
p−1
2
≤2
||f ||p < ∞.
t
And so
t
2C1
µ({M (f ) > t}) ≤ µ({M (b) > }) ≤
2
t
Z
2C1
b dµ ≤
t
X
Z
f dµ.
{f (x)> 2t }
Therefore we see
Z
p
Z
∞
tp−1 µ ({M (f ) > t}) dt
Z
Z ∞
p−2
≤ 2C1 p
t
f dµ dt
|M (f )| dµ = p
0
X
{f (x)> 2t }
0
Z
= 2C1 p
f (x)
Z X
= Cp
|f |p dµ
Z
2f (x)
tp−2 dt dµ
0
X
t
u
Theorem 2.3 (Lebesgue Differentiation Theorem). Let (X, µ) be a locally compact doubling
metric measure space. Assume f ∈ L1loc , then
Z
|f (y) − f (x)| dµ(y) = 0
lim sup −
r→0
for almost every x ∈ X.
B(x,r)
4
ANALYSIS
R
Proof. Let Λ(f )(x) = lim supr→0 −B(x,r) |f (y) − f (x)| dµ(y). Clearly Λ is sublinear and
vanishes on continuous functions. Note
Z
|f (y) − f (x)| dµ(y)
Λ(f )(x) = lim sup −
r→0
B(x,r)
Z
≤ lim sup −
|f (y)| + |f (x)| dµ(y)
r→0
B(x,r)
= M (f ) + |f |.
We note that continuous functions are dense in L1 (since X is locally compact). Now let
f ∈ L1loc and take g continuous such that ||f − g||1 < and let t > 0. Then we see
µ({Λ(f ) > t}) ≤ µ({Λ(f − g) > t})
t
t
≤ µ({M (f − g) > }) + µ{(|f − g| > })
2
2
Z
Z
2C
2
≤
|f − g| dµ +
|f − g| dµ
t X
t X
2C − 2
.
≤
t
This we have that Λ(f ) = 0 a.e. in X.
t
u
Now will discuss the dyadic maximal function and a decomposition of a space that respects a given function well (Calderon-Zygmund decomposition).
Definition 2.4. In Rn we define the unit cube to be the set [0, 1), let Q0 be the collection
of cubes in Rn which are congruent to [0, 1) and whose vertices lie on the lattice Zn . Let
−k
Q
Sk , k ∈ Z be the collection of cubes which are the dilation of Q0 by. 2 . The cubes in
k Qk are called dyadic cubes.
Dyadic Cubes have following properties:
(1) Given x ∈ Rn there is a unique cube in each family Qk which contains x
(2) Any two dyadic cubes are either disjoint of one is wholly contained in the other.
(3) A dyadic cube in Qk is contained in a unique cube of each family Qj , j < k, and
contains 2n dyadic cubes of Qk + 1.
Definition 2.5. Let f ∈ L1loc (Rn ) define
X Z
Ek f (x) =
− f dm χQ (x).
Q
Q∈Qk
This is the conditional expectation of f with respect to the σ-algebra generated by Qk .
Also we define the dyadic maxiaml function as
Md f (x) = sup |Ek f (x)|.
k
WeSnote that Ek f is a discrete analogue of an approximation of the identity. Also if
Ω = Qk then
Z
Z
Ek f dm =
f dm.
Ω
Ω
Theorem 2.6 (Dyadic Maximal Function Theorem).
ANALYSIS
5
(1) The dyadic maximal function is weak (1, 1)
(2) If f ∈ L1loc ,
lim Ek f (x) = f (x)a.e.
k→∞
Proof. Let f ∈ L1 . Define a set Ωk = {x ∈ Rn : Ek fS
(x) > λ and Ej f (x) ≤ λ for all j < k}.
We see that Ωk is a union of cubes from Qk . Also k Ωk = {Md f (x) > λ}. We note that
the Ωk are disjoint. Thus,
!
X
[
m(Ωk )
m({Md f (x) > λ}) = m
Ωk =
k
k
≤
XZ
Ωk
k
1
≤
λ
X1
1
Ek f (x) =
λ
λ
k
Z
f (x)
Ωk
Z
f.
Rn
For the second part we note that its clearly true continuous functions. Let g be a
continuous function such that ||f − g||1 < . Fix x ∈ Rn let Qxk be the unique cube in Qk
such that x ∈ Qxk .
m({lim sup |Ek f (x) − f (x)| > λ})
k→∞
(
Z
=m
)!
lim sup − f − f (x)dm > λ
k→∞
Qx
k
(
=m
)!
Z
Z
lim sup − f − gdm + − g − g(x)dm + g(x) − f (x) > λ
k→∞
Qx
k
Qx
k
λ
λ
+m
(f − g) >
≤m
Md (f − g) >
2
2
Z n
C
≤
f − gdm
λ R
C
≤ t
t
u
Theorem 2.7 (Calderon-Zygmund Decomposition). Given a f ∈ L1 such f ≥ 0 and a
number λ, there exists a sequence {Qj } of disjoint dyadic cubes such that
S
1. f (x) ≤ λ for almost every x ∈
/ j Qj
S
2. m
Qj ≤ λ1 ||f ||1
Rj
3. λ < −Qj f dm ≤ 2n λ.
Proof. Define a set Ωk = {x ∈ Rn : Ek f (x) > λ and Ej f (x) ≤ λ for all j < k}. We see
S
that Ωk is a union of cubes Ck := {Qkj } from Qk . Also k Ωk = {Md f (x) > λ}. We let
S
{Qj } = SCk .
If x ∈
/ j Qj then for all k we have that Ek f (x) ≤ λ and so by part 2. of the Dyadic
Maximal function theorem we have that f (x) ≤ λ.
6
ANALYSIS
S
We note that j Qj = {Md f > λ} so by part 1. of the Dyadic Maximal function theorem
S
1
we have that m
Q
j j ≤ λ ||f ||1 .
Fix
R a Qj from {Qj }. Then Qj ⊆ Ωk for some k. By definition of Ωk we see that
λ < −Qj f dm. Now let Q̃j be the parent of Qj thus
Z
− f dm ≤ λ
Q̃j
by definition of Ωk . Thus,
Z
Z
m(Q̃j )
− f dm ≤ 2n λ.
− f dm ≤
m(Q
)
j Q̃j
Qj
t
u
3. Singular Integrals
Here our reference is Duoandikoetxea, Javier. Fourier Analysis. 2001.
3.1. Fourier Series and Integrals. We wish to study the representation of a function f
by
∞
X
ck e2πikx
f (x) =
k=−∞
As the right hand side is periodic with period it 1. It will suffice to consider f on an interval
of length 1.
Assuming the series converges uniformly, then mulitplying both sides by e2πimx we see
Z 1
cm =
f (x)e−2πimx dx
0
since
Z
1
e
−2πikx −2πimx
e
0
For each function in
defined as
L1 (T)
(
0
dx =
1
if k 6= m
if k = m.
we can associate a sequence {fˆ(k)} of Fourier coefficient of f
fˆ(k) =
Z
1
f (x)e−2πikx dx.
0
We call
∞
X
fˆ(k)e2πikx
k=−∞
the Fourier series of f .
We note that we immediate see that
|fˆ(k)| ≤ ||f ||1
Now define
SN f (x) =
N
X
fˆ(k)e2πikx .
k=−N
We want to know if limN →∞ SN f (x) exists for each x and whether it is equal to f (x).
ANALYSIS
7
Now we rewrite SN f (x) as
N
X
SN f (x) =
fˆ(k)e2πikx
k=−N
N Z
X
=
k=−N
Z 1
1
f (x)e−2πikt dt · e2πikx
0
f (t)DN (x − t) dt
=
0
1
Z
f (x − t)DN (t) dt,
=
0
where
DN (t) =
N
X
e2πikt =
k=−N
sin(π(2N + 1)t)
.
sin(πt)
is called the Dirichlet kernel. We note that
Z 1
DN (t) = 1
and
|DN (t)| ≤
0
1
1
, δ ≤ |t| ≤
sin(πδ)
2
Theorem 3.1 (Riemann-Lebesgue). If f ∈ L1 (T) then
lim fˆ(k) = 0.
|k|→∞
Proof. As e2πix has period 1,
fˆ(k) =
1
Z
f (x)e−2πikx dx
0
Z
1
1
f (x)e−2πik(x+ 2k ) dx
0
Z 1 1
f x−
e−2πikx dx.
=−
2k
0
=−
Thus,
1
fˆ(x) =
2
Z
1
f (x) − f
0
1
x−
e−2πikx dx.
2k
If f continuous then its immediate that
lim fˆ(k) = 0.
|k|→∞
If f ∈ L( T), given > 0, we can choose a g continuous such that ||f − g||1 < and choose
|k| sufficfiently large so that |ĝ(k)| < . Then
|fˆ(k)| ≤ |(f\
− g)(k)| + |ĝ(k)| ≤ ||f − g||1 + |ĝ(k)| < 2.
t
u
8
ANALYSIS
Theorem 3.2 (Dini’s Criterion). If for some x there exists δ such that
Z
f (x + t) − f (x)
<∞
t
|t|<δ
then
lim SN f (x) = f (x).
N →∞
Proof. As
R1
DN (t) dt = 1
Z 1
sin(π(2N + 1)t)
|SN f (x) − f (x)| =
(f (x − t) − f (x))
dt
sin(πt)
0
Z
sin(π(2N + 1)t)
=
(f (x − t) − f (x))
dt
sin(πt)
|t|<δ
Z
sin(π(2N + 1)t)
(f (x − t) − f (x))
dt
+
sin(πt)
δ≤|t|≤ 12
1 |ĝ(2N + 1)| + |ĥ(2N + 1)| + |ĝ(−2N − 1)| + |ĥ(−2N − 1)|
≤
2i
0
(x)
(x)
We let g(t) = f (x−t)−f
χ|t|<δ and let h(t) = f (x−t)−f
χδ≤|t|≤ 1 . We note that by assumpsin(πt)
sin(πt)
2
tion that g is integrable and it is clear that h is integrable so by Riemann-Lebesgue we are
done.
t
u
Remark 3.3. We note that if f is Hölder continuous then it satisfies the Dini Criterion
for some δ. However, there exists continuous functions whose Fourier series diverges at a
point.
Theorem 3.4. The mapping f 7→ {fˆ(k)} is an isometry from L2 to `2 , that is
∞
X
2
||f ||2 =
|fˆ(k)|2 .
k=−∞
This follows from the fact that {e2πikx } from an orthonormal basis of L2 . Also we have
the following
Lemma 3.5. SN f converges to f in the Lp 1 ≤ p < ∞, if and only if there exists Cp
independent of N such that
||SN f ||p ≤ Cp ||f ||p .
This proof uses the uniform boundedness principle to show that converges implies there
is a Cp . And the other directions use the fact that trigonometric polynomials are dense,
SN g = g for deg g ≥ N for g a trigonometric polynomial.
Theorem 3.6 (Uniform Boundedness Principle). Let X be a Banach Space, Y a normed
vector space, and let {Ta }a∈A be a family of bounded linear operators from X to Y . Then
either
sup ||Ta || < ∞
a
or there exists x ∈ X such that
sup ||Ta x||Y = ∞.
a
ANALYSIS
9
Remark 3.7. There are other kernels besides the Dirichlet kernel. There is the Fejér kernel:
N
sin(π(N + 1)t) 2
1
1 X
Dk (t) =
FN =
N +1
N +l
sin(πt)
k=0
this is analogous to Césaro summability. And has the properties
FN (t) ≥ 0
||FN ||1 = 1
Z
lim
N →∞ δ<|t|< 1
2
FN (t) dt = 0 if δ > 0
There is the Poisson kernel
Pr (t) =
∞
X
r|k| e2πikt =
k=−∞
1 − r2
1 − 2r cos(2πt) + r2
this treats the Fourier transform as a formal limit on the unit circle in the complex plane.
And has the properties
Pr (t) ≥ 0
||Pr ||1 = 1
Z
lim
r→1−
δ<|t|< 12
Pr (t) dt = 0 if δ > 0
Theorem 3.8. If f ∈ Lp , 1 ≤ p < ∞ or if f is continuous and p = ∞, then
lim ||FN ∗ f − f ||p = 0
N →∞
Theorem 3.9. If f ∈ Lp , 1 ≤ p < ∞ or if f is continuous and p = ∞, then
lim ||Pr ∗ f − f ||p = 0
r→1−
Now take f ∈ L1 (Rn ), define the Fourier transform by
Z n
ˆ
f (ξ) =
f (x)e−2πixξ dx
R
Here are some properties of the the Fourier Transform
1. (αf + βg)b = αfˆ + βĝ
2. ||fˆ||∞ ||f ||1 and fˆ is continuous
3. lim|ξ|→∞ fˆ = 0 (Riemann-Lebesgue)
4. f[
∗ g = fˆĝ
ˆ 2πih·ξ , where τh f (x) = f (x + h). And (f e2πih·x )b (ξ) = fˆ(ξ − h).
5. τd
h f (ξ) = f e
6. if ρ ∈ On (an orthogonal transformation), then (f ◦ ρ)b (ξ) = fˆ(ξ − h)
7. if g(x) = λ−n f (λ−1 x), then ĝ(ξ) = fˆ(λξ)
∂f
)b (ξ) = 2πiξj fˆ(ξ)
8. ( ∂x
j
9. (−2πixj f )b (ξ) =
∂ fˆ
∂ξj (ξ)
We define the Schwartz class as
S(Rn ) = f ∈ C ∞ (Rn ) : sup |xα Dβ f (x)| =: pα,β (f ) < ∞ for all α, β ∈ Nn
x
10
ANALYSIS
Theorem 3.10. The Fourier transform is a continuous map form S to S such that
Z
Z
f ĝ dx =
fˆg dx
Rn
Rn
and
Z
fˆ(ξ)e2πix·ξ dξ.
f (x) =
Rn
The Fourier series has period 4.
Theorem 3.11. The Fourier transform is an isometry on L2 , that is fˆ ∈ L2 and ||fˆ||2 =
||f ||2 . Furthermore,
Z
ˆ
f (x)e−2πix·ξ dx
f (ξ) = lim
R→∞ |x|<R
and
Z
f (x) = lim
R→∞ |ξ|<R
fˆ(x)e2πix·ξ dξ
where the limits are in L2 .
Theorem 3.12 (Riez-Thorin Interpolation). Let 1 ≤ p0 , p1 , q0 , q1 ≤ ∞, and for 0 < θ < 1
define p and q by
1−θ
1−θ
1
θ
1
θ
=
=
+ ,
+ .
p
p0
p1
q
q0
q1
If T is a linear operator from Lp0 + Lp1 to Lq0 + Lq1 such that
||T f ||q0 ≤ M0 ||f ||p0 for f ∈ Lp0
and
||T f ||q1 ≤ M1 ||f ||p1 for f ∈ Lp1
then
||T f ||q ≤ M01−θ M1θ ||f ||p for f ∈ Lp .
Theorem 3.13 (Marcinkiewicz Interpolation). Let (X, µ) and (Y, ν) be measure spaces,
1 ≤ p0 < p1 ≤ ∞, and let T be a sublinear operator from Lp0 (X, µ)+Lp1 (X, µ) to measurable
functions on Y that is weak (p0 , p0 ) and weak (p1 , p1 ). Then T is strong (p, p) for p0 < p <
p1 .
Interpolation results like the ones above let us prove Lp inequalities for only certain values
of p and then you get everything in between.
Similar issues of convergence and summability arise in this case leading to the definition
of more kernels that are suited to this situation.
References
[1] Duoandikoetxea, Javier. Fourier Analysis. 2001.
[2] Heinonen, Juha. Lectures on Analysis on Metric Spaces. 2001.
[3] Mattila, Pertti. Geometry of Sets and Measures in Euclidean Spaces: Fractals and Rectifiability. 1995