An Lp theory for outer measures. Application to singular integrals.II

Preview:

DESCRIPTION

An Lp theory for outer measures. Application to singular integrals.II. Christoph Thiele Santander, September 2014. Recall Tents (or Carleson boxes). X is the open upper half plane, generating sets are tents T(x,s ) : Define outer measure on X by. Sizes. - PowerPoint PPT Presentation

Citation preview

An Lp theory for outer measures.Application to singular integrals.II

Christoph ThieleSantander, September 2014

Recall Tents (or Carleson boxes)

X is the open upper half plane, generating sets are tents T(x,s) :

Define outer measure on X by

σ(T(x,s)) := s (= μ(T(x,s)))

Sizes

Sp size of function on a tent

Alos Sinfty

S p ( f )(T(x,s)) = (1

sf (y, t)p dy

dt

tT (x,s)

∫∫ )1/ p

S∞( f )(T(x,s)) = sup(y,t )∈T (x,s) f (y, t)

Outer essential supremum

Space of functions with finite out.ess.supremum

Outer essential supremum on a subset F:€

outess( f ) = sup{S( f )(E) : E ∈Σ}

L∞(X,μ,S)

outessF ( f ) = sup{S( f 1F )(E) : E ∈Σ}

Outer Lp spaces

Define super level measures

Define Lp norms

Also weak Lp (Lorentz space) €

μ( f > λ ) := inf{μ(F) : outessF c f ≤ λ}

fLp (X ,μ ,S )

:= ( pλ p

0

∫ μ( f > λ )dλ /λ )1/ p

fLp ,∞ (X ,μ ,S )

:= supλ λ pμ( f > λ )

Embedding theorems

Thm: Define for fixed Schwartz function ϕ

Then we have:

If φ has integral zero:

Fφ ( f )(y, t) := f (z)t−1φ(t−1(z − y))dz∫

Fφ ( f )Lp (X ,μ ,S ∞ )

≤ Cφ fp

Fφ ( f )Lp (X ,μ ,S 2 )

≤ Cφ fp

Proof of embedding thm for Sinfty size

By Marcinkiewicz interpolation between

and

Fφ ( f )L∞ (X ,μ ,S ∞ )

≤ Cφ f∞

Fφ ( f )L1,∞ (X ,μ ,S ∞ )

≤ Cφ f1

Linfty-Sinfty estimate

To show for all tents T(x,s)

But this follows from

S∞(Fφ )(T(x,s)) ≤ C f∞

sup(y,t )∈T (x,s) Fφ (y, t) ≤ sup(y,t ) Fφ (y, t)

=sup(y,t ) f (z)t−1φ(t−1(z − y))dzR

∫ ≤ φ 1 f∞

Weak L1-Sinfty estimate

Ned for all lambda

Need to find a collection of tents T(xi,si) with union E such that for all x,s

S∞(Fφ1E c )(T(x,s)) ≤ Cλ

sii

∑ ≤ Cλ−1 f1

λ μ(Fφ > λ ) ≤ C f1

Weak L1-Sinfty estimate

Hardy Littlewood maximal function

There is an open set where Mf is larger than lambda. Let (xi-si,xi+si) be the collection of connected components of this set. These are the tents. By Hardy Littlewood maximal thm

Mf (x) = supε

1

2εf (x + y)dy

−ε

ε

sii

∑ = x : Mf (x) > λ{ } ≤ Cλ−1 f1

More on weak L1-Sinfty estimate

We haveSince F(y,t) is testing f against bump function at y of width t, may estimate by Hardy Littlewood

S∞(Fφ1E c )(T(x,s)) ≤ sup

(y,t )∈E c Fφ (y, t)

≤C supy −t ≤z≤y +t Mf (z) ≤ Cλ

Proof of embedding thm for S2 size

By Marcinkiewicz interpolation between

and

Fφ ( f )L∞ (X ,μ ,S 2 )

≤ Cφ f∞

Fφ ( f )L1,∞ (X ,μ ,S 2 )

≤ Cφ f1

Use Calderon reproducing formula

Fφ (y, t)2

∫∫ dydt / t = f *ϕ t (y)2∫∫ dydt / t

= ˆ f ˆ ϕ t (η)2

∫∫ dη dt / t = ˆ f (η)2

ˆ ϕ (η / t)dt / t∫ dη ≤ C∫ ˆ f 2

dη∫

Fφ (y, t)2

T (x,s)

∫∫ dydt / t ≤ f 1[x −Cs,x +Cs]

2dη∫ ≤ Cs f

2

Proof of Linfty-S2 estimate

Apply Calderon reproducing formula.If we integrate over arbitrary tent T(x,s), only

restriction of f to X-Cs,x+Cs matters if phi has compact support

Dividing by s yields the desired

Fφ (y, t)2

T (x,s)

∫∫ dydt / t ≤ f 1[x −Cs,x +Cs]

2dη∫ ≤ Cs f

2

S(Fφ )(T(x,s)) ≤ C f∞

2

BMO estimate

Note that we have in fact proven for any m

Thus we have the stronger embedding theorem

For the space BMO, which is defined by

Fφ ( f )L∞ (X ,μ ,S 2 )

≤ Cφ fBMO

fBMO

2:= supI

1

I| f (x) − f I |2 dx

I

S(Fφ )(T(x,s) ≤1

sf (y) − m

2dy

x −Cs

x +Cs

Weak L1-S2 estimate

Let (xi-si,x-+si) be the connected components of the set where Mf(x) is larger than lambda/2.

Let E be the union of tents T(xi,3si).Do Calderon Zygmund decomposition of f

Where g is bounded by lambda, and bi is supported on interval xi-si,xi+si and has integral zero.

f = g + bii

Weak L1-S2 estimate

For the good function use Linfty estimate.

For bi, do a careful estimate and accounting using1) Partial integration of bi to use mean zero when

paired with a phi-t of large support2) Support considerations of bi and phi when paired

with phi-t of small support. .

Summary of proof of embedding thm

Encodes much of singular integral theory:Hardy Littlewood maximal theorem, Vitali covering argumentsCalderon reproducing formula, BMO estimates, square function techniques, Calderon Zygmund decomposition

Use to prove boundedness of operators

Suppose we have operator T mapping functions on real line to functions on real line.

Want to prove

(If T is commutes with dilations, it is forced thatBoth exponents are the same)

Tfp

≤ C fp

Use to prove boundedness of operators

Duality implies

Hence it suffices to prove

Tfp

= supg p ' =1

Tf ,g

Tf ,g ≤ C fp

gp'

Use to prove boundedness

Express <Tf,g> by Fphi and Gphi and prove

Where either S is S2 or Sinfty, as the case may beBy outer Hoelder suffices to prove

Which itself my be result of outer triangle ineq.

Tf ,g ≤ C Fφ Lp (X ,μ ,S )Gϕ Lp ' (X ,μ ,S )

Tf ,g ≤ C FφGϕ L1 (X ,μ ,S )

Example identity operator

By polarization of Calderon reproducing f.:

Provided phi has mean zero and

Triangle ineq. and outer triangle ineq. imply€

f ,g = Fφ ,Gφ dydt / t

ˆ ϕ (η / t)2dt / t∫ =1

f ,g ≤ FφGφ L1 (X ,μ ,S1 )

Cauchy projection operator

By polarization of Calderon reproducing f.:

Provided phi has mean zero and

Boundedness of identity operator and of Cauchy projection imply that of Hilbert transform

Πf ,g = Fφ ,Gφ

ˆ ϕ (η / t)2dt / t∫ =1(0,∞)(η)

Hf (x) = p.v. f (x − t)dt / t∫

Paraproduct estimates

For three Schwartz functions

If two Schwartz functions have integral zero

Λ( f1, f2, f3) := ( Fφ i( f i)(y, t))dy

dt

ti=1

3

∏∫∫

Λ( f1, f2, f3) ≤ Fφ i( f i)

i=1

3

∏L1 (X ,μ ,S1 )

≤ C Fφ i( f i)

i=1

3

∏Lpi (X ,μ ,Sqi

)

≤ C f ii=1

3

∏Lpi

Special paraproducts

Paraproducts ard trilinear forms which are dual to bilinear operators

If phi-2 has mean 1 and f2 is 1, then F2 is 1 And we are redued to previous case. In

particular there is a paraproduct withFixing h and considering as operator in f we haveAn operator with

Λ( f1, f2,.)

Λ(h,1,.) = h

T( f ) = Λ(h, f ,.)

T(1) = h

Basic T(1) Theorem

Let T be a bounded operator in L2 with

whereFor some nonzero test function phi with mean

zero. Then for 1<p<infty we have for all f,

with universal constant Cp independent of T

Tφx,s,φy,t ≤min(t,s)

max(t,s, x − y )

Tfp

≤ Cp fp

φx,s(z) = s−1φ(s−1(z − x))

Why T(1) theorem?

Usually have a different set of assumptions.If s<t<|x-y| then we write

and demand suitable pointwise estimates on the partial derivative of K (Calderon Zygmund kernel).

Symmetrically if t<s<|x-y|€

Tφx,s,φy,t = K(u,v)φx,s(u)φy,t (v)dudv∫∫

=− ∂1K(u,v)Φx,s(u)φy,t (v)dudv∫∫

Why T(1) theorem?

We further demand T(1)=0If |x-y|<s and t<s, then we write

And again demand suitable bounds on K. Similarly we ask T*(1)=0 to address the case|x-y|<t, s<t

= K(u,v)(u − v)(φx,s(u) − φx,s(v))

u − vφy,t (v)dudv∫∫

K(u,v)(φx,s(u) − φx,s(v))φy,t (v)dudv∫∫

More general T(1) theorem

It suffices to demand T(1)=h, T*(1)=0, for some h in BMO. One reduces to the previous case by subtracting a suitable paraproduct:

Similarly one can relax the condition on T* by subtracting a dual paraproduct.

′ T f ,g = Tf ,g − P(h, f ,g)

Proof of T(1) theorem

By Calderons reproducing formula

At which time we use the assumption. To apply outer Hoelder, we break up the region, e.g

|x-y|<s, t<s where we obtain€

F(x,s) Tφx,s,φy,t∫∫ G(y, t)dxdyds /sdt / t∫∫€

Tf ,g =

≤ F(x,s)x −s

x +s

∫0

s

∫ G(y, t)dydt / tR

∫ dx0

∫ ds /s3

Proof of T(1) theorem

Applying Fubini

Setting Gab(x,s)=G(x+as,bs) and using outer triangle inequality

≤ F(x,s)R

∫0

∫ G(x + as,bs)dxds /s−1

1

∫ da0

1

∫ db

≤ FGa,b L1 (X ,S1 )−1

1

∫ da0

1

∫ db

FGa,b L1 (X ,S 2 )

Proof of T(1) theorem

Applying outer Hoelder

Integrating trivially:

≤ F Lp (X ,S 2 ) Ga,b Lp ' (X ,S 2 )−1

1

∫ da0

1

∫ db

≤C F Lp (X ,S 2 ) Ga,b Lp ' (X ,S 2 )

Proof of T(1) theorem

Using embedding theorem (a modified one with tilted triangles for Gab)

Similarly one proves this estimate for the other regions other than|x-y|<s, t<s. €

≤C fp

gp'

Recommended