Maximal averaging operators: from geometry to boundedness ...Maximal averaging operators: from...

Maximal averaging operators: from geometry

to boundedness through duality

Arraz Almirall, Alexis (Universitat de Barcelona)

Cobollo Gomez, Christian (Universitat de Valencia)

Jardon Sanchez, Hector (Universida d’Uvieu)

Martın Murillo, Cristina (Universidad de Extremadura)

Quilis Sandemetrio, Andres (Universitat Politecnica de Valencia)

Ribera Baraut, Pol (Universitat de Barcelona)

March 8, 2019

Introduction

Motivation Question

Lebesgue Differentiation Theorem

|Q(x , r)|

∫Q(x,r)

f −−−→r→0

f (x) a.e x ∈ Rn, for all f ∈ Lp(Rn)

Q(x , r) is a cube centered at x and sidelength r .

Some Notation

Definition

B := {Collection of bounded open sets on Rn}Rn := {Open rectangles on Rn}Qn := {Open cubes on Rn}

Remark

They are all homothecy invariant.

Definition

Maximal operator associated to B:

MBf (x) := supB∈Bx∈B

|f |, f ∈ Lp(Rn), x ∈ Rn.

Some Notation

Definition

Remark

Definition

|f |, f ∈ Lp(Rn), x ∈ Rn.

Some Notation

Definition

Remark

Definition

|f |, f ∈ Lp(Rn), x ∈ Rn.

Introduction

Definition

• Lp(Rn) :=

{f measurable :

(∫Rn

|f |p) 1

=: ||f ||Lp(Rn) <∞

• ||f ||Lp,∞ := {The smallest C0 that verifies (*)}• Lp,∞(Rn) := {f measurable : ||f ||Lp,∞(Rn) <∞}

For all f ∈ Lp(Rn) (1 ≤ p <∞) we have

|{x ∈ Rn : |f (x)| > λ}| ≤ C p0

λp(*)

Introduction

Definition

• Lp(Rn) :=

{f measurable :

(∫Rn

|f |p) 1

=: ||f ||Lp(Rn) <∞

}• ||f ||Lp,∞ := {The smallest C0 that verifies (*)}

• Lp,∞(Rn) := {f measurable : ||f ||Lp,∞(Rn) <∞}

|{x ∈ Rn : |f (x)| > λ}| ≤ C p0

λp(*)

Introduction

Definition

• Lp(Rn) :=

{f measurable :

(∫Rn

|f |p) 1

=: ||f ||Lp(Rn) <∞

}• ||f ||Lp,∞ := {The smallest C0 that verifies (*)}• Lp,∞(Rn) := {f measurable : ||f ||Lp,∞(Rn) <∞}

|{x ∈ Rn : |f (x)| > λ}| ≤ C p0

λp(*)

Introduction

Definition

We say that T is strong-type (p, p) if T : Lp −→ Lp is bounded.

We say that T is weak-type (p, p) if T : Lp −→ Lp,∞ is bounded,

|{x ∈ Rn : |Tf (x)| > λ}| ≤ C p

|f (x)|pdx , λ > 0,

for some C > 0.

T is of weak-type (p, p) (1 < p <∞) if and only if for every E with

0 < |E | <∞

∣∣∣∣ ∫E

∣∣∣∣ ≤ Cp‖f ‖Lp(Rn)|E |1p′ .

Introduction

Definition

We say that T is strong-type (p, p) if T : Lp −→ Lp is bounded.

We say that T is weak-type (p, p) if T : Lp −→ Lp,∞ is bounded,

|{x ∈ Rn : |Tf (x)| > λ}| ≤ C p

|f (x)|pdx , λ > 0,

for some C > 0.

T is of weak-type (p, p) (1 < p <∞) if and only if for every E with

0 < |E | <∞

∣∣∣∣ ∫E

∣∣∣∣ ≤ Cp‖f ‖Lp(Rn)|E |1p′ .

Strong Maximal Theorem

Strong Maximal Theorem (Jessen, Marcinkiewicz, Zygmund 1935)

The estimate

|{x ∈ Rn : MRn f (x) > λ}| ≤ Cn

|f |λ

(1 + log+ |f |

)n−1

holds for every λ > 0, with log+ t = max{0, log t}. It follows that Rn

differentiates functions f for which∫K

|f |(1 + log+ |f |)n−1 <∞,

for every compact set K ⊂ Rn.

Duality link between analysis and

geometry

Duality approach

Duality approach for general B (A. Cordoba - R. Fefferman).

Definition (Covering property Vq)

Let 1 ≤ q ≤ ∞. We say that B has the covering property Vq if there

exist c1, c2 > 0 such that for every finite collection {Bj}Nj=1 ⊂ B there

exists a finite subcollection {Bk}Mk=1 such that

∣∣∣∣ N⋃j=1

∣∣∣∣ ≤ c1

∣∣∣∣ M⋃k=1

∣∣∣∣ (We don’t lose much of the measure)

∥∥∥∥ M∑k=1

∥∥∥∥Lq(Rn)

≤ c2

∣∣∣∣ N⋃j=1

∣∣∣∣ 1q

(Control of the overlap)

Proposition

Let 1 < p <∞ and 1p + 1

p′ = 1. The maximal operator MB is of

weak-type (p, p) if and only if B has the covering property Vp′ .

Duality approach

∣∣∣∣ N⋃j=1

∣∣∣∣ ≤ c1

∣∣∣∣ M⋃k=1

∥∥∥∥ M∑k=1

∥∥∥∥Lq(Rn)

≤ c2

∣∣∣∣ N⋃j=1

∣∣∣∣ 1q

Proposition

Let 1 < p <∞ and 1p + 1

Duality approach

∣∣∣∣ N⋃j=1

∣∣∣∣ ≤ c1

∣∣∣∣ M⋃k=1

∥∥∥∥ M∑k=1

∥∥∥∥Lq(Rn)

≤ c2

∣∣∣∣ N⋃j=1

∣∣∣∣ 1q

Proposition

Let 1 < p <∞ and 1p + 1

Proof: Vp′ ⇒ weak-type (p, p)

Consider

Eλ = {x ∈ Rn : MBf (x) > λ}

Consider

Eλ = {x ∈ Rn : MBf (x) > λ} =⋃

x∈Eλ

such that

|f (y)|dy > λ, x ∈ Bx .

Consider

K ⊂ Eλ = {x ∈ Rn : MBf (x) > λ} =⋃

x∈Eλ

such that

|f (y)|dy > λ, x ∈ Bx .

Consider

K ⊂N⋃j=1

such that

|f (y)|dy > λ.

Consider

K ⊂N⋃j=1

such that

|f (y)|dy > λ.

By the property (i) of Vp′

|K | ≤∣∣∣∣ N⋃j=1

∣∣∣∣ ≤ c1

∣∣∣∣ M⋃k=1

∣∣∣∣.

Consider

K ⊂N⋃j=1

such that

|f (y)|dy > λ.

By the property (i) of Vp′

|K | ≤∣∣∣∣ N⋃j=1

∣∣∣∣ ≤ c1

∣∣∣∣ M⋃k=1

∣∣∣∣ ≤ c1

M∑k=1

1Bk(y)|f (y)|dy

Consider

K ⊂N⋃j=1

such that

|f (y)|dy > λ.

By Holder’s inequality and (ii) of Vp′ :

|K | ≤∣∣∣∣ N⋃j=1

∣∣∣∣ ≤ c1

∣∣∣∣ M⋃k=1

∣∣∣∣ ≤ c1

M∑k=1

1Bk(y)|f (y)|dy

Holder≤ c1

∥∥∥∥∥M∑k=1

∥∥∥∥∥Lp′ (Rn)

‖f ‖Lp(Rn)

≤ c1c2

∣∣∣∣ N⋃j=1

∣∣∣∣ 1p′

‖f ‖Lp(Rn).

Consider

K ⊂N⋃j=1

such that

|f (y)|dy > λ

With a few more simple computations:

|K | ≤ cp1 cp2

λp‖f ‖pp

and let K ↗ Eλ.

Proof: weak-type (p, p)⇒ Vp′

Let {Bj}Nj=1 ⊂ B, and suppose without loss of generality that

|B1| ≥ · · · ≥ |BN |.

We extract a subcollection as follows:

- B1 = B1

|B1| ≥ · · · ≥ |BN |.

We extract a subcollection as follows:

- B1 = B1

- Selected {Bk}mk=1, m < N, choose the first B ∈ {Bj}Nj=1 \ {Bk}mk=1

verifying: ∣∣∣∣B ∩ ⋃k≤m

∣∣∣∣ ≤ 1

The overlapping is less than 50% in measure.

We have selected {Bk}Mk=1 such that, defining

Ek = Bk \⋃j<k

it holds

|Ek | ≥1

2|Bk | &

M⋃k=1

Ek =M⋃k=1

Ek = Bk \⋃j<k

it holds

|Ek | ≥1

2|Bk | &

M⋃k=1

Ek =M⋃k=1

To prove (i):

⋃B not

selected

B ⊂⋃{

B :|B ∩

⋃Bk |

}⊂{MB(1⋃

Ek = Bk \⋃j<k

it holds

|Ek | ≥1

2|Bk | &

M⋃k=1

Ek =M⋃k=1

To prove (i): ∣∣∣∣ ⋃B not

selected

∣∣∣∣ ≤ ∣∣∣∣{MB(1⋃Bk

}∣∣∣∣.9

Ek = Bk \⋃j<k

it holds

|Ek | ≥1

2|Bk | &

M⋃k=1

Ek =M⋃k=1

To prove (i):

∣∣∣∣ ⋃B not

selected

∣∣∣∣ ≤ ∣∣∣∣{MB(1⋃Bk

}∣∣∣∣ ≤ C p2p

∣∣∣∣ M⋃k=1

∣∣∣∣.9

To prove (ii) we define the linear and weak-type (p, p) operator

Tf (x) :=M∑k=1

f (y)dy

(x) ≤ MBf (x).

To prove (ii) we define the linear and weak-type (p, p) operator

Tf (x) :=M∑k=1

f (y)dy

(x) ≤ MBf (x).

Computing the adjoint of T and evaluating at 1⋃k Bk

we get

T ∗(1⋃k Bk

) =M∑k=1

|Ek ||Bk |

1Bk≥ 1

M∑k=1

Therefore

∣∣∣∣ ∫Rn

M∑k=1

∣∣∣∣ ≤ 2

∣∣∣∣ ∫Rn

T ∗(1⋃k Bk

∣∣∣∣ = 2

∣∣∣∣ ∫⋃k Bk

∣∣∣∣ ≤ 2Cp‖f ‖Lp(Rn)

∣∣∣∣ M⋃k=1

∣∣∣∣ 1p′

Therefore

∣∣∣∣ ∫Rn

M∑k=1

∣∣∣∣ ≤ 2

∣∣∣∣ ∫Rn

T ∗(1⋃k Bk

∣∣∣∣ = 2

∣∣∣∣ ∫⋃k Bk

∣∣∣∣ ≤ 2Cp‖f ‖Lp(Rn)

∣∣∣∣ M⋃k=1

∣∣∣∣ 1p′

Taking the supremum over all f ∈ Lp(Rn) with ‖f ‖Lp(Rn) ≤ 1 we obtain

∥∥∥∥ M∑k=1

∥∥∥∥Lp′ (Rn)

≤ 2Cp

∣∣∣∣ M⋃k=1

∣∣∣∣ 1p′

Some remarks

Remark 1

The covering property V∞ implies weak-type (1, 1).

Remark 2

The differentiation basis given by all cubes Qn verifies the property V∞(Vitali’s covering lemma). We conclude that Qn differentiates L1(Rn).

Some remarks

Remark 1

The covering property V∞ implies weak-type (1, 1).

Remark 2

The differentiation basis given by all cubes Qn verifies the property V∞(Vitali’s covering lemma). We conclude that Qn differentiates L1(Rn).

• Setting: rectangles in Rn. For simplicity, we will restrict to R2. Let

R2 be the set of all rectangles in R2 with sides parallel to the axes.

• MR2 is not weak-type (1, 1). The following does not hold for any c

|{x ∈ R2 : MR2 f (x) > λ}| ≤ c

|f (y)|dy .

• Setting: rectangles in Rn. For simplicity, we will restrict to R2. Let

R2 be the set of all rectangles in R2 with sides parallel to the axes.

• MR2 is not weak-type (1, 1). The following does not hold for any c

|{x ∈ R2 : MR2 f (x) > λ}| ≤ c

|f (y)|dy .

New covering property Vexp: we say that B has the covering property

Vexp if there exist c1, c2 > 0 such that for every finite collection

{Rj}Nj=1 ⊂ B there exists a finite subcollection {Rk}Mk=1 such that

∣∣∣∣∣ N⋃j=1

∣∣∣∣∣ ≤ c1

∣∣∣∣ M⋃k=1

∣∣∣∣ ,ii)

M∑k=1

)− 1

]≤ θc2

∣∣∣∣∣M⋃k=1

∣∣∣∣∣ for θ small.

Strong Maximal Theorem: Vexp ⇒ boundedness

Vexp implies the strong maximal theorem in R2.

Theorem

If B has the covering property Vexp, then

∣∣{x ∈ R2 : MBf (x) > λ}∣∣ ≤ C

|f (x)|λ

(1 + log+ |f (x)|

where log+ t := max{0, log t}.

Strong Maximal Theorem: Vexp ⇒ boundedness

Sketch of proof:

• Instead of Holder’s, use Young’s inequality (Holder’s is a particular

case):

st ≤ cθs(1 + log+ s

)+ exp (θt)− 1,

where s, t ≥ 0 and θ > 0.

|K | ≤ c1

∣∣∣∣∣M⋃k=1

∣∣∣∣∣ ≤ c1

(M∑k=1

1Rk(y)

)|f (y)|λ

Young≤ c1

[cθ|f (y)|λ

(1 + log+ |f (y)|

)+ exp

M∑k=1

1Rk(y)

)− 1

≤ . . . + c1c2θ

∣∣∣∣∣M⋃k=1

∣∣∣∣∣ .16

Strong Maximal Theorem: R2 has Vexp

Theorem

The differentiation basis of rectangles R2 satisfies the Vexp covering

property.

Sketch of proof (inductive selection scheme): Let Π1(R),Π2(R) denote

the projections of R ∈ R2 on the x and y axis (resp.).

• Let {Rj}Nj=1 ⊂ R2 ordered by |Π2(R1)| ≥ · · · ≥ |Π2(RN)|.• Fix R1 := R1.

• Having selected {Rk}mk=1, m < N, choose the first

R ∈ {Rj}Nj=1\{Rk}mk=1 verifying either∣∣∣R ∩ (∪j<k R∗k

)∣∣∣ ≤ 1

or R ∩(∪j<k Rk

)= ∅.

Theorem

property.

• Let {Rj}Nj=1 ⊂ R2 ordered by |Π2(R1)| ≥ · · · ≥ |Π2(RN)|.

• Fix R1 := R1.

)∣∣∣ ≤ 1

or R ∩(∪j<k Rk

)= ∅.

Theorem

property.

)∣∣∣ ≤ 1

or R ∩(∪j<k Rk

)= ∅.

Theorem

property.

)∣∣∣ ≤ 1

or R ∩(∪j<k Rk

)= ∅.

Sketch of proof: claim {Rk}mk=1 satisfies the Vexp estimates.

• The selection scheme allows us to show that the 1D slices of the

{Rk}mk=1 satisfy similar sparseness properties.

• Fix y ∈ Π2(R), R not selected. Then,

(2D)∣∣∣R ∩ (∪j<k R

)∣∣∣ > 1

2|R| ⇒

⇒ (1D)∣∣∣Πy

1(R) ∩(∪j<kΠy

1(Rk))∣∣∣ > 1

1(R)|,

and these are 1D averages.

⋃R not

selected

Πy1(R) ⊂

{x ∈ R : MR1 1 ⋃

Πy1 (Rk )(x) >

(2D)∣∣∣R ∩ (∪j<k R

)∣∣∣ > 1

2|R| ⇒

⇒ (1D)∣∣∣Πy

1(R) ∩(∪j<kΠy

1(Rk))∣∣∣ > 1

1(R)|,

⋃R not

selected

Πy1(R) ⊂

{x ∈ R : MR1 1 ⋃

Πy1 (Rk )(x) >

(2D)∣∣∣R ∩ (∪j<k R

)∣∣∣ > 1

2|R| ⇒

⇒ (1D)∣∣∣Πy

1(R) ∩(∪j<kΠy

1(Rk))∣∣∣ > 1

1(R)|,

⋃R not

selected

Πy1(R) ⊂

{x ∈ R : MR1 1 ⋃

Πy1 (Rk )(x) >

(2D)∣∣∣R ∩ (∪j<k R

)∣∣∣ > 1

2|R| ⇒

⇒ (1D)∣∣∣Πy

1(R) ∩(∪j<kΠy

1(Rk))∣∣∣ > 1

1(R)|,

⋃R not

selected

Πy1(R) ⊂

{x ∈ R : MR1 1 ⋃

Πy1 (Rk )(x) >

Sketch of proof: i)

∣∣∣∣∣ N⋃j=1

∣∣∣∣∣ ≤ C1

∣∣∣∣ M⋃k=1

∣∣∣∣ .

• Maximal operator MR1 is Hardy-Littlewood maximal operator M1.

• weak-type properties of M1 prove the estimates∣∣∣∣∣∣∣⋃

R notselected

Πy1(R)

∣∣∣∣∣∣∣ ≤ C1

∣∣∣∣∣M⋃k=1

Πy1(Rk)

∣∣∣∣∣ .

• Integrate in y .

Sketch of proof: i)

∣∣∣∣∣ N⋃j=1

∣∣∣∣∣ ≤ C1

∣∣∣∣ M⋃k=1

∣∣∣∣ .• Maximal operator MR1 is Hardy-Littlewood maximal operator M1.

R notselected

Πy1(R)

∣∣∣∣∣∣∣ ≤ C1

∣∣∣∣∣M⋃k=1

Πy1(Rk)

∣∣∣∣∣ .

Sketch of proof: i)

∣∣∣∣∣ N⋃j=1

∣∣∣∣∣ ≤ C1

∣∣∣∣ M⋃k=1

R notselected

Πy1(R)

∣∣∣∣∣∣∣ ≤ C1

∣∣∣∣∣M⋃k=1

Πy1(Rk)

∣∣∣∣∣ .

Sketch of proof: i)

∣∣∣∣∣ N⋃j=1

∣∣∣∣∣ ≤ C1

∣∣∣∣ M⋃k=1

R notselected

Πy1(R)

∣∣∣∣∣∣∣ ≤ C1

∣∣∣∣∣M⋃k=1

Πy1(Rk)

∣∣∣∣∣ .

Sketch of proof: ii)

M∑k=1

)− 1

]≤ θC2

∣∣∣∣∣M⋃k=1

∣∣∣∣∣ .

• Take the exponential series expansion∫R2

∞∑p=1

(M∑k=1

=∞∑p=1

∥∥∥∥∥M∑k=1

∥∥∥∥∥p

Lp(R2)

• Estimate the Lp(R2) norms of the overlaps from the slices∥∥∥∥∥M∑k=1

1Πy1 (Rk )

∥∥∥∥∥p

Lp(R2)

≤ cp

∣∣∣∣∣M⋃k=1

Πy1(Rk)

∣∣∣∣∣ ..

M∑k=1

)− 1

]≤ θC2

∣∣∣∣∣M⋃k=1

∣∣∣∣∣ .• Take the exponential series expansion∫

∞∑p=1

(M∑k=1

=∞∑p=1

∥∥∥∥∥M∑k=1

∥∥∥∥∥p

Lp(R2)

1Πy1 (Rk )

∥∥∥∥∥p

Lp(R2)

≤ cp

∣∣∣∣∣M⋃k=1

Πy1(Rk)

∣∣∣∣∣ ..

M∑k=1

)− 1

]≤ θC2

∣∣∣∣∣M⋃k=1

∣∣∣∣∣ .• Take the exponential series expansion∫

∞∑p=1

(M∑k=1

=∞∑p=1

∥∥∥∥∥M∑k=1

∥∥∥∥∥p

Lp(R2)

1Πy1 (Rk )

∥∥∥∥∥p

Lp(R2)

≤ cp

∣∣∣∣∣M⋃k=1

Πy1(Rk)

∣∣∣∣∣ ..

Conclusion

• Boundedness of maximal operator equivalent to covering properties

of the differentiation basis. Equivalence is provided by duality.

• Rectangles 6= cubes in covering terms, but boundedness properties

still hold: strong maximal theorem.

Conclusion

Thanks for your attention!

Maximal averaging operators: from geometry to boundedness ...Maximal averaging operators: from...

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