· Denegerate Poincare [email protected] Sobolev inequalities Sobolev inequalities...

DEGENERATE POINCARE INEQUALITIES

Carlos Perez

University of the Basque Country&

BCAM, Basque Center for Applied Math.

Atelier d’Analyse Harmonique 2018CNRS-Paul Langevin Center

Aussois, March 30, 2018

Denegerate Poincare inequalities [email protected]

collaborators

1


collaborators

most of the lecture will be on an almost finished work with:

1


collaborators


Ezequiel Rela

1


collaborators


Ezequiel Rela

And if I have time few results from a work in progress with

1


collaborators


Ezequiel Rela

And if I have time few results from a work in progress with

Sheldy Ombrosi, Ezequiel Rela and Israel Rios-Rivera

1


THE CLASSICAL POINCARE INEQUALITY

2



The simplest context: Rn with the metric associated to the cubes with thelebesgue measure.

2



The simplest context: Rn with the metric associated to the cubes with thelebesgue measure.• (1,1) Poincare inequality:

1

|Q|

∫Q|f − fQ| ≤ cn

`(Q)

|Q|

∫Q|∇f |

fQ = 1|Q|

∫Q f = average of f over the cube Q &

2




1

|Q|


`(Q)

|Q|

∫Q|∇f |

fQ = 1|Q|

∫Q f = average of f over the cube Q & `(Q) = sidelength ofQ.

2




1

|Q|


`(Q)

|Q|

∫Q|∇f |

fQ = 1|Q|


• (p, p), p ≥ 1 Poincare inequality:(1

|Q|

∫Q|f − fQ|p dx

)1/p

≤ c `(Q)

(1

|Q|

∫Q|∇f |pdx

)1/p

2




1

|Q|


`(Q)

|Q|

∫Q|∇f |

fQ = 1|Q|



|Q|

∫Q|f − fQ|p dx

)1/p

≤ c `(Q)

(1

|Q|

∫Q|∇f |pdx

)1/p

• Higher order case. Somewhat less-known result.

2




1

|Q|


`(Q)

|Q|

∫Q|∇f |

fQ = 1|Q|



|Q|

∫Q|f − fQ|p dx

)1/p

≤ c `(Q)

(1

|Q|

∫Q|∇f |pdx

)1/p

• Higher order case. Somewhat less-known result.

1

|Q|

∫Q|f − P (Q, f)| ≤ cn

`(Q)m

|Q|

∫Q|∇mf |

2


INTEGRAL REPRESENTATION FORMULA

3



The proof is based on the following classical formula:

|f(x)− fQ| ≤c

|Q|

∫Q

|∇f(z)||x− z|n−1

dz = c I1(|∇f |χQ

)(x).

3




|f(x)− fQ| ≤c

|Q|

∫Q

|∇f(z)||x− z|n−1


)(x).

• They are equivalent (Franchi-Lu-Wheeden)

3




|f(x)− fQ| ≤c

|Q|

∫Q

|∇f(z)||x− z|n−1


)(x).


Here

Iαf(x) = c∫Rn

f(y)

|x− y|n−αdy

0 < α < n.

3




|f(x)− fQ| ≤c

|Q|

∫Q

|∇f(z)||x− z|n−1


)(x).


Here

Iαf(x) = c∫Rn

f(y)

|x− y|n−αdy

0 < α < n.

These are called fractional integrals of order α or Riesz potentials.

3




|f(x)− fQ| ≤c

|Q|

∫Q

|∇f(z)||x− z|n−1


)(x).


Here

Iαf(x) = c∫Rn

f(y)

|x− y|n−αdy

0 < α < n.


The classical well–known estimates use these type of representation.

3




|f(x)− fQ| ≤c

|Q|

∫Q

|∇f(z)||x− z|n−1


)(x).


Here

Iαf(x) = c∫Rn

f(y)

|x− y|n−αdy

0 < α < n.


The classical well–known estimates use these type of representation.

It goes back to Sobolev.

3


Sobolev inequalities

Sobolev inequalities can be seen as sharp versions of the (p, p) Poincareinequalities:

4




Let p∗ = pnn−p when 1 ≤ p < n.

4




Let p∗ = pnn−p when 1 ≤ p < n. Observe that 1∗ = n

n−1 = n′

4





n−1 = n′

(1

|Q|

∫Q|f − fQ|p

∗)1/p∗

≤ c `(Q)

(1

|Q|

∫Q|∇f |p

)1/p

4





n−1 = n′

(1

|Q|

∫Q|f − fQ|p

∗)1/p∗

≤ c `(Q)

(1

|Q|

∫Q|∇f |p

)1/p

The SELF-IMPROVING property

4





n−1 = n′

(1

|Q|

∫Q|f − fQ|p

∗)1/p∗

≤ c `(Q)

(1

|Q|

∫Q|∇f |p

)1/p


Observations:

4





n−1 = n′

(1

|Q|

∫Q|f − fQ|p

∗)1/p∗

≤ c `(Q)

(1

|Q|

∫Q|∇f |p

)1/p


Observations:• These estimates are sharper than the corresponding (p, p) Poincare inequal-ities since p∗ > p by Jensen’s inequality.

4





n−1 = n′

(1

|Q|

∫Q|f − fQ|p

∗)1/p∗

≤ c `(Q)

(1

|Q|

∫Q|∇f |p

)1/p


Observations:• These estimates are sharper than the corresponding (p, p) Poincare inequal-ities since p∗ > p by Jensen’s inequality.• p∗ is optimal, that is, we cannot replace p∗ by a larger exponent.

4





n−1 = n′

(1

|Q|

∫Q|f − fQ|p

∗)1/p∗

≤ c `(Q)

(1

|Q|

∫Q|∇f |p

)1/p


Observations:• These estimates are sharper than the corresponding (p, p) Poincare inequal-ities since p∗ > p by Jensen’s inequality.• p∗ is optimal, that is, we cannot replace p∗ by a larger exponent.• p∗ is usually called the Sobolev exponent.

4





n−1 = n′

(1

|Q|

∫Q|f − fQ|p

∗)1/p∗

≤ c `(Q)

(1

|Q|

∫Q|∇f |p

)1/p


Observations:• These estimates are sharper than the corresponding (p, p) Poincare inequal-ities since p∗ > p by Jensen’s inequality.• p∗ is optimal, that is, we cannot replace p∗ by a larger exponent.• p∗ is usually called the Sobolev exponent.• One of the points of this talk is to show how to avoid such representationformulae.

4





n−1 = n′

(1

|Q|

∫Q|f − fQ|p

∗)1/p∗

≤ c `(Q)

(1

|Q|

∫Q|∇f |p

)1/p


Observations:• These estimates are sharper than the corresponding (p, p) Poincare inequal-ities since p∗ > p by Jensen’s inequality.• p∗ is optimal, that is, we cannot replace p∗ by a larger exponent.• p∗ is usually called the Sobolev exponent.• One of the points of this talk is to show how to avoid such representationformulae.•We will use Calderon-Zygmund theory instead.

4


Elliptic Theory

5


Elliptic TheoryThe case p > 1 is important in the theory of elliptic P.D.E. .

5



•The elliptic Operator:

5



•The elliptic Operator: Lu = div(A(x).∇u) = 0

5




where λ|ξ|2 ≤ A(x)ξ.ξ ≤ Λ|ξ|2

5





• Goal: to prove local Holder continuity of the (weak) solutions of the equation.

5






• Classical theory: De Giorgi, Nash (local Holder continuity theory of solutions)

5







• Moser (Harnack inequality from which Holder continuity of solutions can bederived). This became the standard machinery for these questions.

5








Key point besides the (2,2) PI,

5








Key point besides the (2,2) PI, the PS inequality:

(1

|Q|

∫Q|f − fQ|2

∗dx

)1/2∗

≤ c `(Q)

(1

|Q|

∫Q|∇f |2dx

)1/2

5


“Degenerate” elliptic equations

• “Degenerate” elliptic equations

6




λ |ξ|2w(x) ≤ A(x)ξ.ξ ≤ Λ |ξ|2w(x)

6




λ |ξ|2w(x) ≤ A(x)ξ.ξ ≤ Λ |ξ|2w(x)

where w is a weight with some sort of singularity.

6




λ |ξ|2w(x) ≤ A(x)ξ.ξ ≤ Λ |ξ|2w(x)


Late 60’s and early 70’s:

6




λ |ξ|2w(x) ≤ A(x)ξ.ξ ≤ Λ |ξ|2w(x)


Late 60’s and early 70’s: Kruzkov, Murthy, Stampacchia, Trudinger.

6




λ |ξ|2w(x) ≤ A(x)ξ.ξ ≤ Λ |ξ|2w(x)



supQ

(1

|Q|

∫Qws dx

)1/s ( 1

|Q|

∫Qw−t dx

)1/t

<∞,1

s+

1

t<

2

n

6




λ |ξ|2w(x) ≤ A(x)ξ.ξ ≤ Λ |ξ|2w(x)



supQ

(1

|Q|

∫Qws dx

)1/s ( 1

|Q|

∫Qw−t dx

)1/t

<∞,1

s+

1

t<

2

n

• The relevant work is due Fabes-Kenig-Serapioni (1982),

6




λ |ξ|2w(x) ≤ A(x)ξ.ξ ≤ Λ |ξ|2w(x)



supQ

(1

|Q|

∫Qws dx

)1/s ( 1

|Q|

∫Qw−t dx

)1/t

<∞,1

s+

1

t<

2

n

• The relevant work is due Fabes-Kenig-Serapioni (1982),they removed the restriction in s, t,

6




λ |ξ|2w(x) ≤ A(x)ξ.ξ ≤ Λ |ξ|2w(x)



supQ

(1

|Q|

∫Qws dx

)1/s ( 1

|Q|

∫Qw−t dx

)1/t

<∞,1

s+

1

t<

2

n

• The relevant work is due Fabes-Kenig-Serapioni (1982),they removed the restriction in s, t,and consider the A2 condition instead:

6




λ |ξ|2w(x) ≤ A(x)ξ.ξ ≤ Λ |ξ|2w(x)



supQ

(1

|Q|

∫Qws dx

)1/s ( 1

|Q|

∫Qw−t dx

)1/t

<∞,1

s+

1

t<

2

n


[w]A2

= supQ

(1

|Q|

∫Qw dx

) (1

|Q|

∫Qw−1 dx

)

6




λ |ξ|2w(x) ≤ A(x)ξ.ξ ≤ Λ |ξ|2w(x)



supQ

(1

|Q|

∫Qws dx

)1/s ( 1

|Q|

∫Qw−t dx

)1/t

<∞,1

s+

1

t<

2

n


[w]A2

= supQ

(1

|Q|

∫Qw dx

) (1

|Q|

∫Qw−1 dx

)

• method of proof is based on the Moser iteration technique

6




λ |ξ|2w(x) ≤ A(x)ξ.ξ ≤ Λ |ξ|2w(x)



supQ

(1

|Q|

∫Qws dx

)1/s ( 1

|Q|

∫Qw−t dx

)1/t

<∞,1

s+

1

t<

2

n


[w]A2

= supQ

(1

|Q|

∫Qw dx

) (1

|Q|

∫Qw−1 dx

)

• method of proof is based on the Moser iteration technique

6


Weighted Poincare and Poincare-Sobolev inequalties

7



If w ∈ A2

7



If w ∈ A2

(1

w(Q)

∫Q|f − fQ|2w

)12≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12,

7



If w ∈ A2

(1

w(Q)

∫Q|f − fQ|2w

)12≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12,

and there is gain for some δ:

7



If w ∈ A2

(1

w(Q)

∫Q|f − fQ|2w

)12≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12,


(1

w(Q)

∫Q|f − fQ|2+δw

) 12+δ

≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12

7



If w ∈ A2

(1

w(Q)

∫Q|f − fQ|2w

)12≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12,


(1

w(Q)

∫Q|f − fQ|2+δw

) 12+δ

≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12

• Due to Fabes-Kenig-Serapioni.

7



If w ∈ A2

(1

w(Q)

∫Q|f − fQ|2w

)12≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12,


(1

w(Q)

∫Q|f − fQ|2+δw

) 12+δ

≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12

• Due to Fabes-Kenig-Serapioni.• Method of proof is by fractional integrals.

7



If w ∈ A2

(1

w(Q)

∫Q|f − fQ|2w

)12≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12,


(1

w(Q)

∫Q|f − fQ|2+δw

) 12+δ

≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12

• Due to Fabes-Kenig-Serapioni.• Method of proof is by fractional integrals.Since there have been a lot of variants of these results:

7



If w ∈ A2

(1

w(Q)

∫Q|f − fQ|2w

)12≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12,


(1

w(Q)

∫Q|f − fQ|2+δw

) 12+δ

≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12

• Due to Fabes-Kenig-Serapioni.• Method of proof is by fractional integrals.Since there have been a lot of variants of these results:• Chanillo-Wheeden, Franchi-Lu-Wheeden, Chua-Wheeden

7



If w ∈ A2

(1

w(Q)

∫Q|f − fQ|2w

)12≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12,


(1

w(Q)

∫Q|f − fQ|2+δw

) 12+δ

≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12

• Due to Fabes-Kenig-Serapioni.• Method of proof is by fractional integrals.Since there have been a lot of variants of these results:• Chanillo-Wheeden, Franchi-Lu-Wheeden, Chua-WheedenWe will see again the gain again but it is not that precise anymore.

7



If w ∈ A2

(1

w(Q)

∫Q|f − fQ|2w

)12≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12,


(1

w(Q)

∫Q|f − fQ|2+δw

) 12+δ

≤ C(w)`(Q)

(1

w(Q)

∫Q|∇f |2w

)12

• Due to Fabes-Kenig-Serapioni.• Method of proof is by fractional integrals.Since there have been a lot of variants of these results:• Chanillo-Wheeden, Franchi-Lu-Wheeden, Chua-WheedenWe will see again the gain again but it is not that precise anymore.

• First example of this property is due to L. Saloff-Coste.

7


GENERAL SITUATION

STARTING POINT:

1

|Q|

∫Q|f(y)− fQ| dy ≤ a(Q)

where a is a “functional”

8


GENERAL SITUATION

STARTING POINT:

1

|Q|


where a is a “functional” a : Q → (0,∞)

8


GENERAL SITUATION

STARTING POINT:

1

|Q|



where Q denotes the family of all cubes from Rn.

8


GENERAL SITUATION

STARTING POINT:

1

|Q|




Question: What kind of condition can we impose on a to get the self-improvingproperty?

8


GENERAL SITUATION

STARTING POINT:

1

|Q|





• There is the Lp self-improving (model example: Sobolev inequalities)

8


GENERAL SITUATION

STARTING POINT:

1

|Q|






• There is also exponential self-improving:

8


GENERAL SITUATION

STARTING POINT:

1

|Q|







a lo John-Nirenberg or

8


GENERAL SITUATION

STARTING POINT:

1

|Q|







a lo John-Nirenberg or

of Trudinger type

8


model example

Our model example is associated to the fractional average

9


model example


a(Q) = `(Q)α(ν(Q)

|Q|

)1/p

9


model example


a(Q) = `(Q)α(ν(Q)

|Q|

)1/p

• They enjoy a Lp self-improving

9


model example


a(Q) = `(Q)α(ν(Q)

|Q|

)1/p

• They enjoy a Lp self-improving• Motivated by the theory developed in the papers by Hajlasz, Heinonen andKoskela.

9


model example


a(Q) = `(Q)α(ν(Q)

|Q|

)1/p


• Other type of examples are given by

9


model example


a(Q) = `(Q)α(ν(Q)

|Q|

)1/p



a(Q) = ν(Q)1p

9


model example


a(Q) = `(Q)α(ν(Q)

|Q|

)1/p



a(Q) = ν(Q)1p

• related to the exponential self-improving property.

9


the Dr condition

10


the Dr conditionGOAL: find some conditions on a such if f satisfies

10



1

|Q|

∫Q|f − fQ| ≤ a(Q) Q ⊂ Rn

10



1

|Q|

∫Q|f − fQ| ≤ a(Q) Q ⊂ Rn

implies a Lr self–improving property of the form for some r > 1,(1

|Q|

∫Q|f(y)− fQ|r dy

)1/r

≤ c a(Q)

10



1

|Q|

∫Q|f − fQ| ≤ a(Q) Q ⊂ Rn


|Q|


)1/r

≤ c a(Q)

We impose a geometrical type condition which will be key in what follows:

10



1

|Q|

∫Q|f − fQ| ≤ a(Q) Q ⊂ Rn


|Q|


)1/r

≤ c a(Q)


Let 0 < r <∞. We say that the functional a satisfies the Dr condition if thereexists a finite constant c such that for each cube Q and any family {Qi} ofpairwise disjoint dyadic subcubes of Q,∑

i

a(Qi)r|Qi| ≤ cr a(Q)r|Q|

10



1

|Q|

∫Q|f − fQ| ≤ a(Q) Q ⊂ Rn


|Q|


)1/r

≤ c a(Q)



i


• Resembles a little bit the Carleson condition.

10



1

|Q|

∫Q|f − fQ| ≤ a(Q) Q ⊂ Rn


|Q|


)1/r

≤ c a(Q)



i


• Resembles a little bit the Carleson condition.• r < s =⇒ Ds ⊂ Dr.

10



1

|Q|

∫Q|f − fQ| ≤ a(Q) Q ⊂ Rn


|Q|


)1/r

≤ c a(Q)



i


• Resembles a little bit the Carleson condition.• r < s =⇒ Ds ⊂ Dr.• Then we can define for a given a the optimal exponent

ra = sup{r : a ∈ Dr}.

10


EXAMPLES

11


EXAMPLESRecall the fractional functional given by

a(Q) = `(Q)α(ν(Q)

|Q|

)1/p

,

11



a(Q) = `(Q)α(ν(Q)

|Q|

)1/p

,

Observe that with r = npn−αp we have

a(Qi)r|Qi| = ν(Qi)

r/p

11



a(Q) = `(Q)α(ν(Q)

|Q|

)1/p

,


a(Qi)r|Qi| = ν(Qi)

r/p

and then if {Qi} is a family of disjoint dyadic subcubes of Q

11



a(Q) = `(Q)α(ν(Q)

|Q|

)1/p

,


a(Qi)r|Qi| = ν(Qi)

r/p


∑i

a(Qi)r|Qi| =

∑i

ν(Qi)r/p ≤

∑i

ν(Qi)

r/p

11



a(Q) = `(Q)α(ν(Q)

|Q|

)1/p

,


a(Qi)r|Qi| = ν(Qi)

r/p


∑i

a(Qi)r|Qi| =

∑i

ν(Qi)r/p ≤

∑i

ν(Qi)

r/p

≤ ν(Q)r/p = a(Q)r|Q|

which means that a ∈ Dr.

11



a(Q) = `(Q)α(ν(Q)

|Q|

)1/p

,


a(Qi)r|Qi| = ν(Qi)

r/p


∑i

a(Qi)r|Qi| =

∑i

ν(Qi)r/p ≤

∑i

ν(Qi)

r/p


which means that a ∈ Dr.Some observations:• If α = 1, r = p∗, the Sobolev exponent.

11



a(Q) = `(Q)α(ν(Q)

|Q|

)1/p

,


a(Qi)r|Qi| = ν(Qi)

r/p


∑i

a(Qi)r|Qi| =

∑i

ν(Qi)r/p ≤

∑i

ν(Qi)

r/p



• If α = m = 1,2 · · · , r = mpn−pm, the Sobolev exponent related to higher

order PI.

11



a(Q) = `(Q)α(ν(Q)

|Q|

)1/p

,


a(Qi)r|Qi| = ν(Qi)

r/p


∑i

a(Qi)r|Qi| =

∑i

ν(Qi)r/p ≤

∑i

ν(Qi)

r/p



• If α = m = 1,2 · · · , r = mpn−pm, the Sobolev exponent related to higher

order PI.• If ν ≡ 1, a satisfies the Dr condition for every r > 1.

11


Variants of the Dr condition

12



The (weighted) Dr(w) condition for some 0 < r < ∞: for each cube Q andfor any family {Qi} of pairwise disjoint cubes contained in Q,∑

i

a(Qi)rw(Qi) ≤ cr a(Q)rw(Q)

12




i


Theorem (Franchi, P, Wheeden, 1998)Let a ∈ Dr(w) for some r > 0 and let w ∈ A∞. Let f such that

1

|Q|

∫Q|f − fQ| ≤ a(Q),

12




i



1

|Q|

∫Q|f − fQ| ≤ a(Q),

Then there exists a constant c such that

‖f − fQ‖Lr,∞(Q,w) ≤ c a(Q)

12




i



1

|Q|

∫Q|f − fQ| ≤ a(Q),


‖f − fQ‖Lr,∞(Q,w) ≤ c a(Q)

• The proof combines Calderon–Zygmund theory with an appropriate variantof the good–λ inequality of Burkholder–Gundy.

12




i



1

|Q|

∫Q|f − fQ| ≤ a(Q),


‖f − fQ‖Lr,∞(Q,w) ≤ c a(Q)

• The proof combines Calderon–Zygmund theory with an appropriate variantof the good–λ inequality of Burkholder–Gundy.• Can be extended to the context of a space of homogeneous type.

12


How to recover the weighted Poincare-Sobolev inequality

13



We start by using the L1 unweighted Poincare inequality

1

|Q||f(x)− fQ|dx ≤ cn`(Q)−

∫Q|∇f(x)|dx.

13




1


∫Q|∇f(x)|dx.

By the Ap condition, we obtain that

1

|Q||f(x)− fQ|dx ≤ cn [w]

1pAp`(Q)

(1

w(Q)

∫Q|∇f(x)|pw dx

)1/p

.

13




1


∫Q|∇f(x)|dx.


1


1pAp`(Q)

(1

w(Q)

∫Q|∇f(x)|pw dx

)1/p

.

Hence appears naturally the weighted fractional integral

af(Q) := cn [w]1pAp`(Q)

(1

w(Q)

∫Q|∇f(x)|pw dx

)1/p

13




1


∫Q|∇f(x)|dx.


1


1pAp`(Q)

(1

w(Q)

∫Q|∇f(x)|pw dx

)1/p

.



(1

w(Q)

∫Q|∇f(x)|pw dx

)1/p

which satisfies trivially

13




1


∫Q|∇f(x)|dx.


1


1pAp`(Q)

(1

w(Q)

∫Q|∇f(x)|pw dx

)1/p

.



(1

w(Q)

∫Q|∇f(x)|pw dx

)1/p

which satisfies trivially a ∈ Dp(w)

13




1


∫Q|∇f(x)|dx.


1


1pAp`(Q)

(1

w(Q)

∫Q|∇f(x)|pw dx

)1/p

.



(1

w(Q)

∫Q|∇f(x)|pw dx

)1/p

which satisfies trivially a ∈ Dp(w)

and hence(1

w(Q)

∫Q|f(x)− fQ|pwdx

)1/p

≤ c `(Q)

(1

w(Q)

∫Q|∇f(x)|pwdx

)1/p

13



14



It is more interesting to get for some p∗ > p(1

w(Q)

∫Q|f(y)− fQ|p

∗w(y)dy

) 1p∗≤ c `(Q)

(1

w(Q)

∫Q|∇f(x)|pw dx

)1p

14




w(Q)

∫Q|f(y)− fQ|p

∗w(y)dy

) 1p∗≤ c `(Q)

(1

w(Q)

∫Q|∇f(x)|pw dx

)1p

Hence the question reduces to understand the Dp∗(w) for a larger exponent.

Lemma Let w ∈ Ap, a ∈ Dp(n′+δ)(w) where δ depends on w.

14




w(Q)

∫Q|f(y)− fQ|p

∗w(y)dy

) 1p∗≤ c `(Q)

(1

w(Q)

∫Q|∇f(x)|pw dx

)1p



and hence

14




w(Q)

∫Q|f(y)− fQ|p

∗w(y)dy

) 1p∗≤ c `(Q)

(1

w(Q)

∫Q|∇f(x)|pw dx

)1p



and hence(1

w(Q)

∫Q|f(y)− fQ|p(n′+δ)wdx

) 1p(n′+δ)

≤ c `(Q)

(1

w(Q)

∫Q|∇f(x)|pw dx

)1p

14


Other Interesting Poincare Inequalities

15



Subelliptic operators,

15



Subelliptic operators, from the theory of several complex variables.

15



Subelliptic operators, from the theory of several complex variables.In R2:

15




(X1, X2) = (∂

∂x, x

∂

∂y)

15




(X1, X2) = (∂

∂x, x

∂

∂y)

associated to the Grushin operator:

15




(X1, X2) = (∂

∂x, x

∂

∂y)


X21 +X2

2 =∂2

∂x2+ x2 ∂

2

∂y2.

15




(X1, X2) = (∂

∂x, x

∂

∂y)


X21 +X2

2 =∂2

∂x2+ x2 ∂

2

∂y2.

Nagel-Stein-Wainger proved that there is a metric dX

, the Carnot-Caratheodorymetric,

15




(X1, X2) = (∂

∂x, x

∂

∂y)


X21 +X2

2 =∂2

∂x2+ x2 ∂

2

∂y2.


, the Carnot-Caratheodorymetric, which is doubling with respect to the Lebesgue measure.

15




(X1, X2) = (∂

∂x, x

∂

∂y)


X21 +X2

2 =∂2

∂x2+ x2 ∂

2

∂y2.


, the Carnot-Caratheodorymetric, which is doubling with respect to the Lebesgue measure.Hence

15




(X1, X2) = (∂

∂x, x

∂

∂y)


X21 +X2

2 =∂2

∂x2+ x2 ∂

2

∂y2.



(Rn, dX, dx)

15




(X1, X2) = (∂

∂x, x

∂

∂y)


X21 +X2

2 =∂2

∂x2+ x2 ∂

2

∂y2.



(Rn, dX, dx)

becomes a space of homogeneous type.

15




(X1, X2) = (∂

∂x, x

∂

∂y)


X21 +X2

2 =∂2

∂x2+ x2 ∂

2

∂y2.



(Rn, dX, dx)

becomes a space of homogeneous type.•. There is a key PI:

15




(X1, X2) = (∂

∂x, x

∂

∂y)


X21 +X2

2 =∂2

∂x2+ x2 ∂

2

∂y2.



(Rn, dX, dx)


1

|B|

∫B|f − fB| ≤ c

rB|B|

∫B|Xf |

15




(X1, X2) = (∂

∂x, x

∂

∂y)


X21 +X2

2 =∂2

∂x2+ x2 ∂

2

∂y2.



(Rn, dX, dx)


1

|B|

∫B|f − fB| ≤ c

rB|B|

∫B|Xf |

• Jerison (1986)

15


A NON–SMOOTH EXAMPLE

Consider the vector field

16




Xα = (X1, X2) = (∂

∂x, |x|α

∂

∂y)

16




Xα = (X1, X2) = (∂

∂x, |x|α

∂

∂y)

associated to the operator:

16




Xα = (X1, X2) = (∂

∂x, |x|α

∂

∂y)


∂2

∂x2+ |x|2α

∂2

∂y2.

16




Xα = (X1, X2) = (∂

∂x, |x|α

∂

∂y)


∂2

∂x2+ |x|2α

∂2

∂y2.

These type of non–smooth examples were considered by Franchi-Lanconelliin the mid 80’s.

16




Xα = (X1, X2) = (∂

∂x, |x|α

∂

∂y)


∂2

∂x2+ |x|2α

∂2

∂y2.

These type of non–smooth examples were considered by Franchi-Lanconelliin the mid 80’s.As above there is a corresponding Carnot-Caratheodory metric d

Xsuch

(Rn, dX, dx)

becomes space of homogeneous type.

16




Xα = (X1, X2) = (∂

∂x, |x|α

∂

∂y)


∂2

∂x2+ |x|2α

∂2

∂y2.


Xsuch

(Rn, dX, dx)

becomes space of homogeneous type.• There is another key PI:

16




Xα = (X1, X2) = (∂

∂x, |x|α

∂

∂y)


∂2

∂x2+ |x|2α

∂2

∂y2.


Xsuch

(Rn, dX, dx)


1

|B|

∫B|f − fB| ≤ c

rB|B|

∫B|Xαf |

16




Xα = (X1, X2) = (∂

∂x, |x|α

∂

∂y)


∂2

∂x2+ |x|2α

∂2

∂y2.


Xsuch

(Rn, dX, dx)


1

|B|

∫B|f − fB| ≤ c

rB|B|

∫B|Xαf |

• Franchi–Gutierrez–Wheeden (1994).

16


Poincare-Sobolev

17


Poincare-Sobolev

Let

17


Poincare-Sobolev

Let p∗ = pDD−p where

17


Poincare-Sobolev

Let p∗ = pDD−p where D = doubling order

17


Poincare-Sobolev


Then

17


Poincare-Sobolev


Then• the smooth case:

17


Poincare-Sobolev


Then• the smooth case:(

1

|B|

∫B|f − fB|p

∗) 1p∗≤ c rB

(1

|B|

∫B|Xf |p

)1/p

17


Poincare-Sobolev



1

|B|

∫B|f − fB|p

∗) 1p∗≤ c rB

(1

|B|

∫B|Xf |p

)1/p

• Lu p > 1

17


Poincare-Sobolev



1

|B|

∫B|f − fB|p

∗) 1p∗≤ c rB

(1

|B|

∫B|Xf |p

)1/p

• Lu p > 1

• Franchi–Lu–Wheeden p = 1.

17


Poincare-Sobolev



1

|B|

∫B|f − fB|p

∗) 1p∗≤ c rB

(1

|B|

∫B|Xf |p

)1/p

• Lu p > 1


• the non–smooth case:

17


Poincare-Sobolev



1

|B|

∫B|f − fB|p

∗) 1p∗≤ c rB

(1

|B|

∫B|Xf |p

)1/p

• Lu p > 1



(1

|B|

∫B|f − fB|p

∗)1/p∗

≤ c rB

(1

|B|

∫B|Xαf |p

)1/p

17


Poincare-Sobolev



1

|B|

∫B|f − fB|p

∗) 1p∗≤ c rB

(1

|B|

∫B|Xf |p

)1/p

• Lu p > 1



(1

|B|

∫B|f − fB|p

∗)1/p∗

≤ c rB

(1

|B|

∫B|Xαf |p

)1/p

Franchi–Gutierrez–Wheeden, p ≥ 1

17


Poincare-Sobolev



1

|B|

∫B|f − fB|p

∗) 1p∗≤ c rB

(1

|B|

∫B|Xf |p

)1/p

• Lu p > 1



(1

|B|

∫B|f − fB|p

∗)1/p∗

≤ c rB

(1

|B|

∫B|Xαf |p

)1/p


• Each case has its own proof all of them based on a representation formula.

17


Poincare-Sobolev



1

|B|

∫B|f − fB|p

∗) 1p∗≤ c rB

(1

|B|

∫B|Xf |p

)1/p

• Lu p > 1



(1

|B|

∫B|f − fB|p

∗)1/p∗

≤ c rB

(1

|B|

∫B|Xαf |p

)1/p


• Each case has its own proof all of them based on a representation formula.•We avoid all these.

17


THE SHARPEST RESULT IN THE GENERAL CONTEXT

18



Let (X, d, µ) be a metric space with a doubling measure µ

18




Dr(w) condition: for some 0 < r < ∞ namely for each ball B and for anyfamily {Bi} of pairwise disjoint balls contained in B,

18




Dr(w) condition: for some 0 < r < ∞ namely for each ball B and for anyfamily {Bi} of pairwise disjoint balls contained in B,∑

i

a(Bi)rwµ(Bi) ≤ cr a(B)rwµ(B)

18





i


Theorem (MacManus, P.) Let w ∈ A∞(µ) and suppose that

1

µ(B)

∫B|f − fB| dµ ≤ a(B)

18





i



1

µ(B)


Then, if δ > 0, there is a constant C independent of f and B such that

‖f − fB‖Lr,∞(B,w) ≤ C a((1 + δ)B)

18





i



1

µ(B)



‖f − fB‖Lr,∞(B,w) ≤ C a((1 + δ)B)

• It is not so clean because of the factor (1 + δ). (lack of dyadic structure)

18





i



1

µ(B)



‖f − fB‖Lr,∞(B,w) ≤ C a((1 + δ)B)

• It is not so clean because of the factor (1 + δ). (lack of dyadic structure)• Other situations: non homogeneous spaces

18





i



1

µ(B)



‖f − fB‖Lr,∞(B,w) ≤ C a((1 + δ)B)

• It is not so clean because of the factor (1 + δ). (lack of dyadic structure)• Other situations: non homogeneous spaces(joint work with J. Orobitg)

18





i



1

µ(B)



‖f − fB‖Lr,∞(B,w) ≤ C a((1 + δ)B)

• It is not so clean because of the factor (1 + δ). (lack of dyadic structure)• Other situations: non homogeneous spaces(joint work with J. Orobitg)•Weaker hypothesis: replace L1 norm by much weaker norms.

18





i



1

µ(B)



‖f − fB‖Lr,∞(B,w) ≤ C a((1 + δ)B)

• It is not so clean because of the factor (1 + δ). (lack of dyadic structure)• Other situations: non homogeneous spaces(joint work with J. Orobitg)•Weaker hypothesis: replace L1 norm by much weaker norms.(joint work with A. Lerner)

18


The small Dp condition

19



Definition 1 Let L > 1 and let Q be a cube. We will say that a family ofpairwise disjoint subcubes {Qi} of Q is L-small if

∑i

|Qi| ≤|Q|L

We will say {Qi} ∈ S(L)

19




∑i

|Qi| ≤|Q|L


Now, the correct notion of Dp condition in this context is the following.

19




∑i

|Qi| ≤|Q|L


Now, the correct notion of Dp condition in this context is the following.

Definition 1Let w be any weight and let s > 1. We say that the functional a satisfies theweighted SDs

p(w) condition for 0 < p < ∞ if there is a constant c such thatfor any cube Q and any family {Qi} of pairwise disjoint subcubes of Q suchthat {Qi} ∈ S(L), the following inequality holds:

∑i

a(Qi)pw(Qi) ≤ cp

(1

L

)psa(Q)pw(Q)

19


main example

20


main example

Let µ be any Radon measure and define

a(Q) = `(Q)

(1

w(Q)µ(Q)

)1/p

.

20


main example


a(Q) = `(Q)

(1

w(Q)µ(Q)

)1/p

.

Let w be a weight, L > 1,1 ≤ p < n and let a ∈ SDnp (w).

20


main example


a(Q) = `(Q)

(1

w(Q)µ(Q)

)1/p

.


The proof is an easy consequence of Holder’s inequality. Let {Qi} ∈ S(L),then

20


main example


a(Q) = `(Q)

(1

w(Q)µ(Q)

)1/p

.


The proof is an easy consequence of Holder’s inequality. Let {Qi} ∈ S(L),then ∑

i

a(Qi)pw(Qi) =

∑i

`(Qi)pµ(Qi) =

∑i

|Qi|p/nµ(Qi)

20


main example


a(Q) = `(Q)

(1

w(Q)µ(Q)

)1/p

.



i

a(Qi)pw(Qi) =

∑i

`(Qi)pµ(Qi) =

∑i

|Qi|p/nµ(Qi)

≤

∑i

|Qi|

p/n∑i

µ(Qi)(n/p)′

1(n/p)′

20


main example


a(Q) = `(Q)

(1

w(Q)µ(Q)

)1/p

.



i

a(Qi)pw(Qi) =

∑i

`(Qi)pµ(Qi) =

∑i

|Qi|p/nµ(Qi)

≤

∑i

|Qi|

p/n∑i

µ(Qi)(n/p)′

1(n/p)′

≤(|Q|L

)p/nµ(Q) =

(1

L

)p/na(Q)pw(Q)

20


A first result

21


A first result

Theorem. Let w be any weight. Consider a functional a satisfiyngSDs

p(w) with s > 1 and p ≥ 1. Suppose that

1

|Q|

∫Q|f − fQ| ≤ a(Q) (H)

for every cube Q. Then, there exists a dimensional constant cn such that forany cube Q

21


A first result



1

|Q|

∫Q|f − fQ| ≤ a(Q) (H)

for every cube Q. Then, there exists a dimensional constant cn such that forany cube QThen (

1

w(Q)

∫Q|f − fQ|pwdx

)1p

≤ s cn a(Q)

21


A first result



1

|Q|

∫Q|f − fQ| ≤ a(Q) (H)

for every cube Q. Then, there exists a dimensional constant cn such that forany cube QThen (

1

w(Q)

∫Q|f − fQ|pwdx

)1p

≤ s cn a(Q)

CorollaryLet (u, v) ∈ Ap. The the following Poincare (p, p) inequality holds(

1

u(Q)

∫Q|f − fQ|p u dx

)1/p

≤ cn[u, v]1pAp`(Q)

(1

u(Q)

∫Q|∇f |p v dx

)1/p

,

where cn is a dimensional constant.

21


Two more corollaries

22



Corollary (A generalized John-Nirenberg)Let a be an increasing functional and suppose that f satisfies (H).

22



Corollary (A generalized John-Nirenberg)Let a be an increasing functional and suppose that f satisfies (H).Then, ∥∥∥f − fQ∥∥∥expL(Q,w)

≤ cn[w]A∞ a(Q)

22




≤ cn[w]A∞ a(Q)

Corollary (The Keith-Zhong phenomenon)Let 1 < p0 and let (f, g) be a couple of functions satisfying

1

|Q|

∫Q|f − fQ| dx ≤ C[w]Ap0

`(Q)

(1

w(Q)

∫Qgp0 wdx

) 1p0

w ∈ Ap0

22




≤ cn[w]A∞ a(Q)

Corollary (The Keith-Zhong phenomenon)Let 1 < p0 and let (f, g) be a couple of functions satisfying

1

|Q|

∫Q|f − fQ| dx ≤ C[w]Ap0

`(Q)

(1

w(Q)

∫Qgp0 wdx

) 1p0

w ∈ Ap0

Then, for any 1 ≤ p < p0, the following estimate holds for any w ∈ Ap(1

w(Q)

∫Q|f − fQ|wdx

)1/p

≤ cC[w]Ap`(Q)

(1

w(Q)

∫Qgpwdx

)1/p

22


Improving Poincare-Sobolev

23



As before let

23



As before let

a(Q) = `(Q)

(1

w(Q)µ(Q)

)1/p

23



As before let

a(Q) = `(Q)

(1

w(Q)µ(Q)

)1/p

Lemma Let 1 ≤ q ≤ p < n, and let w ∈ Aq. If E > 1 we let p∗ be

1

p−

1

p∗=

1

nqE.

23



As before let

a(Q) = `(Q)

(1

w(Q)µ(Q)

)1/p


1

p−

1

p∗=

1

nqE.

Then, if {Qi} ∈ S(L), L > 1, the following inequality holds:

∑i

a(Qi)p∗w(Qi) ≤ [w]

p∗nqEAq

(1

L

) p∗nE′

a(Q)p∗w(Q)

23



As before let

a(Q) = `(Q)

(1

w(Q)µ(Q)

)1/p


1

p−

1

p∗=

1

nqE.


∑i


p∗nqEAq

(1

L

) p∗nE′

a(Q)p∗w(Q)

• The functional a “preserves smallness” with index nE′ and constant [w]1nqEAq

23



As before let

a(Q) = `(Q)

(1

w(Q)µ(Q)

)1/p


1

p−

1

p∗=

1

nqE.


∑i


p∗nqEAq

(1

L

) p∗nE′

a(Q)p∗w(Q)

• The functional a “preserves smallness” with index nE′ and constant [w]1nqEAq

• E can be seen as “error” it is the made.

23


Consequence

24


Consequence

Corollary Let 1 ≤ q ≤ p < n, and let w ∈ Aq. Let p∗ be defined by

1

p−

1

p∗=

1

n(q + log[w]Aq)

and suppose that f satisfies (H). Then

24


Consequence


1

p−

1

p∗=

1

n(q + log[w]Aq)


(1

w(Q)

∫Q|f − fQ|p

∗wdx

) 1p∗≤ cn a(Q)

24


Consequence


1

p−

1

p∗=

1

n(q + log[w]Aq)


(1

w(Q)

∫Q|f − fQ|p

∗wdx

) 1p∗≤ cn a(Q)

In particular(1

w(Q)

∫Q|f − fQ|p

∗wdx

) 1p∗≤ cn [w]

1pAp`(Q)

(1

w(Q)

∫Q|∇f(x)|pw dx

)1/p

24


Consequence


1

p−

1

p∗=

1

n(q + log[w]Aq)


(1

w(Q)

∫Q|f − fQ|p

∗wdx

) 1p∗≤ cn a(Q)

In particular(1

w(Q)

∫Q|f − fQ|p

∗wdx

) 1p∗≤ cn [w]

1pAp`(Q)

(1

w(Q)

∫Q|∇f(x)|pw dx

)1/p

• Again, this is a very “clean” inequality.

24


Poincare-Sobolev via Good-λ

25



TheoremLet 1 ≤ q ≤ p < n an let w ∈ Aq. Let p∗ be defined by

1

p−

1

p∗=

1

nq


25



TheoremLet 1 ≤ q ≤ p < n an let w ∈ Aq. Let p∗ be defined by

1

p−

1

p∗=

1

nq


(1

w(Q)

∫Q|f − fQ|p

∗wdx

) 1p∗≤ c[w]

1nqAq

[w]2pAp`(Q)

(1

w(Q)

∫Q|∇f |pwdx

)1p

,

25


Bloom BMO and Muckenhoupt-Wheeden

Let f be a locally integrable function and let w be a weight such that

‖f‖BMOw

= supQ

1

w(Q)

∫Q|f − fQ| <∞,

26




‖f‖BMOw

= supQ

1

w(Q)

∫Q|f − fQ| <∞,

Theorema) A1 case: If w ∈ A1, there exists a constant c such that for any cube Qand any q > 1(

1

w(Q)

∫Q

(|f − fQ|

w

)qwdx

)1q

≤ c q [w]A1‖f‖BMOw

and hence for any cube Q∥∥∥∥f − fQw

∥∥∥∥expL(Q,w)

≤ c [w]A1‖f‖

BMOw

26




‖f‖BMOw

= supQ

1

w(Q)

∫Q|f − fQ| <∞,

Theorema) A1 case: If w ∈ A1, there exists a constant c such that for any cube Qand any q > 1(

1

w(Q)

∫Q

(|f − fQ|

w

)qwdx

)1q

≤ c q [w]A1‖f‖BMOw

and hence for any cube Q∥∥∥∥f − fQw

∥∥∥∥expL(Q,w)

≤ c [w]A1‖f‖

BMOw

b) Ap case: If w ∈ Ap, 1 < p < ∞, there exists a constant c such that forany cube Q 1

w(Q)

∫Q

(|f − fQ|

w

)p′wdx

1p′

≤ c2np p′ [w]Ap‖f‖BMOw

26


Using the sparse method

27



Theorema) A1 case: If w ∈ A1, there exists a constant c such that for any cube Q andq > 1(

1

w(Q)

∫Q

∣∣∣∣∣f(x)− fQw

∣∣∣∣∣q

w(x)dx

)1q

≤ cn‖f‖BMOw

qq′[w]1q′A1

[w]1qA∞

27




1

w(Q)

∫Q

∣∣∣∣∣f(x)− fQw

∣∣∣∣∣q

w(x)dx

)1q

≤ cn‖f‖BMOw

qq′[w]1q′A1

[w]1qA∞

b) Ap case: then, if 1 ≤ q ≤ p′

(1

w(Q)

∫Q

∣∣∣∣∣f(x)− fQw

∣∣∣∣∣q

w(x)dx

)1q

≤ cnp‖f‖BMOw

[w]1p′A∞[w]

1pAp

27




1

w(Q)

∫Q

∣∣∣∣∣f(x)− fQw

∣∣∣∣∣q

w(x)dx

)1q

≤ cn‖f‖BMOw

qq′[w]1q′A1

[w]1qA∞


(1

w(Q)

∫Q

∣∣∣∣∣f(x)− fQw

∣∣∣∣∣q

w(x)dx

)1q

≤ cnp‖f‖BMOw

[w]1p′A∞[w]

1pAp

27




1

w(Q)

∫Q

∣∣∣∣∣f(x)− fQw

∣∣∣∣∣q

w(x)dx

)1q

≤ cn‖f‖BMOw

qq′[w]1q′A1

[w]1qA∞


(1

w(Q)

∫Q

∣∣∣∣∣f(x)− fQw

∣∣∣∣∣q

w(x)dx

)1q

≤ cnp‖f‖BMOw

[w]1p′A∞[w]

1pAp

27




1

w(Q)

∫Q

∣∣∣∣∣f(x)− fQw

∣∣∣∣∣q

w(x)dx

)1q

≤ cn‖f‖BMOw

qq′[w]1q′A1

[w]1qA∞


(1

w(Q)

∫Q

∣∣∣∣∣f(x)− fQw

∣∣∣∣∣q

w(x)dx

)1q

≤ cnp‖f‖BMOw

[w]1p′A∞[w]

1pAp

27


mercibeaucoup

28


mercibeaucoup

thank you very much

28

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