The Hilbert transform along a one variable vector field

Christoph Thiele

(joint work with M. Bateman)

Conference in honor of Eli Stein,

Princeton, 2011

Partial list of work by Eli Stein that had impact on this research

- Stein: Problems in harmonic analysis related to curvature and oscillatory integrals, Proc ICM 1986

- Phong, Stein: Hilbert integrals, singular integrals, and Radon transforms II, Invent. Math, 1986

- Christ, Nagel, Stein, Wainger: Singular and maximal Radon transforms. Annals of Math, 1999

- Stein, Wainger: Oscillatory integrals related to Carleson’s theorem, Math Research Lett, 2001

- Stein, Street: Multi-parameter singular Radon transforms, preprint, 2011

Outline of lecture

1) Hilbert transform along vector fields with

a) regularity (analytic, Lipshitz) condition

b) one variable condition (main topic here)

2) Connection with Carleson’s theorem

3) Reduction to covering lemmas

4) Three different covering lemmas

Vector Field in the Plane

22: RRv

/

Hilbert Transform/Maximal Operator along Vector Field

Question: - bounds

tdttyxvyxfvpyxfHv /)),(),((..),(

pL

|/)),(),((|sup),(0

dttyxvyxfyxfM v

First observations

Bounded by 1D result if vector field constant.

Value independent of length of v(x,y). May assume unit length vector field.

Alternatively, may assume v(x,y)=(1,u(x,y)) for scalar slope field u.

Nikodym set example

Set E of null measure containing for each

(x,y) a line punctured at (x,y). If vector field

points in direction of this line then averages

of characteristic fct of set along vf are one.

Gravitation vector field

HT/MO of bump function asymptotically , only for p>2, weak type 2

x

c

pL

Propose modification/conditions

Truncate integral at (normalize )

Demand slow rotation (Lipshitz: )1v

1vt

Zygmund/Stein conjectures

Assume , , and truncate .

Conj.1:Truncated MO bounded .

Conj.2:Truncated HT bounded .

No bounds known except 1) if p is infinity.

1v 1

v

,22 LL

,22 LL

pL

t

Analytic vector fields

If v is real analytic, then on a bounded domain:

Bourgain (1989):

MO is bounded in , p>1.

Christ, Nagel, Stein, Wainger (1999):

HT bounded in (assume no straight integral curves. Stein,Street announce without assumption)

pL

pL

One variable (meas.) vector field

)0,(),( xvyxv

Theorem (M. Bateman, C.T.)

Measurable, one variable vector field

(HT not truncated;

Related earlier work: Bateman; Lacey/Li)

pppv fCfH

p2/3

Linear symmetry group

Isotropic dilations

Dilation of second variable

Shearing

TfHTfH vTvT

)()(1

Constant along Lipschitz

Angle of to x axis less than

Angle of to x-axis less than or equal to

Conjecture:

Same bounds for

as in Bateman,CT

2

v v

vH

Relation with Carleson’s theorem and time-frequency analysis

Carleson’s operator

Carleson 1966, Hunt 1968:

Carleson’s operator is bounded in ,pL p1

)(

2)(ˆ)(x

ix defxfC

tdtetxfvp xit /)(.. )(

Coifman’s argument

),(2

/))(,(yxLR

tdttxuytxf

),(

)(

2

/),(ˆ

yxLR

txiu

R

iy dtdtetxfe

),(

)(

2

/),(ˆ

xLR

txiu tdtetxf 2),(2

),(ˆ~ fxfxL

The argument visualized

A Littlewood Paley band

Bound for supported on Littlewood Paley band

Lacey and Li: Bound on for

arbitrary two variable measurable vector field

Bateman: Bound on for

one variable vector field.

p2

21 p

f̂

fHv

fHv

Vector valued inequality, reduction to covering arguments

Littlewood Paley decomp.

Vector valued inequality

Since LP projection commutes with HT

(vector field constant in vertical direction)

,

Enough to prove for any sequence

p

kpff

2/12 p

kvpv fHfH2/12

kf

p

kp

kkv fCfH2/122/12

Weak type interpolation

Enough to prove for

Whenever

p3

2

pp

pGkkv GHCfH/11/12/12

1,

Hkf 12

Cauchy Schwarz

Enough to prove

Which follows from

pp

pGkkv GHCfH/21/22

1,

2/21

21,

k

p

pGkkv fH

GCfH

Single band operator estimate

By interpolation enough for

Lacey-Li (p>2) /Bateman (p<2) proved

Note: F,E depend on k, while G,H do not

2

/12/1

2/)1(1 fHGCfH

p

HkvG

2/12/1/12/1/1,1 EFHGCH

p

pEFkv

qq

qEFkv EFCH/11/1

1,1

HFGE ,

Induction on .

If p<2 and

Find such that and

the desired estimate holds (prove!) for .

Apply induction hypothesis on (gain )

GH GGexc 2/GGexc

)/(log2 HG

excG

excGG \2

Induction on .

If p>2 and

Find such that and

And the desired estimate holds for

Apply induction hypothesis on

HG HH exc 2/HH exc

)/(log2 GH

excH

excHH \

Finding exceptional sets. Covering arguments

Parallelograms

Kakeya example

Try union of all parallelograms R

with for appropriate .

Bad control on , example of Kakeya set.

RGR

excH

excH

Vector field comes to aid

Restrict attention to U( R), set of points in R where the direction of the vector field is in angular interval E( R) of uncertainty of R

1st covering lemma

The union of all parallelograms with

has measure bounded by for q>1

(vector field measurable, no other assumption)

RRURG )(

GC qq

excH

Outline of argument

Find good subset of set of

parallelograms with large density, such that

1.

2.

Then:

'

'RR

RCR

'

'

')(1

R

q

RRU RC

qq

RRR

GRGRUR/1

'/1

'

1

'

1

'

)(

Greedy selection

Iteratively select R for with maximal

shadow such that for previously selected R’

Here 7R means stretch in vertical direction.

'

100/'770)'()(','

RRRRURUR

Vertical maximal function

The non-selected parallelograms are all inside

which has acceptable size.

10000/1),)(1(:),('

yxMyxR

RV

Additional property

R’ selected prior to R; U(R’) intersect U(R).

Then

RSR R 7'7

argument

Assume in order of selection and

' ,...,

1)(

21

)(...)()1(R RRR

nn

RU

n

RURUC

0)()( 1 ii RURU

nR RRSC 7...7 11

11 7...71 nR RRSC

17... RC

nL

2nd covering lemma (Lacey-Li)

The union of all parallelograms with

and

Has measure bounded by .

Use vector field Lipshitz in vertical direction. -power responsible for .

RRH

HC 21

RRUR )(

excG

2

3p

Outline of argument

From set of parallelograms with large

densities select as before. Using as before

obtain:

To prove:

Power of 2 here responsible for -power

'

'RR

RCR

'

1

2

'

1RR

R RC

VM

Expansion of .

Sum over all pairs with R selected before R’.

Case 1

Case 2

Second case has aligned directions, as before.

'RR

)'(100)( RERE

)'(100)( RERE

Second selection in Case 1

Fix R, consider R’ selected later with Case 1.

Prove

Select so that each selected R’ has

Projection of U(R’) onto x-axis disjoint from

Projection of U(R’’) for previously selected R’’

RCRR 1'

'''

Removing δ

Use disjointness of projections of U(R’) and

density δ to reduce to showing for fixed R’

Summing over those R’’ in which where

not selected for because of prior selection

of R’ . All R’’ have similar angle as R’.

(use vector field Lipschitz in vertical direction.)

''' RSRCRR '

''

Picture of situation

Back to maximal function

1) Hard case is when all rectangles are thin.

2) Intersection with R is only fraction α (depending on angle) of R

3) If too much overlap, then vertical maximal function becomes too large in extended rectangle (1/α) R.

4) Two effects of α cancel.

3rd covering lemma (Bateman)

One variable vf, parallelograms of fixed

height h. Union of all parallelograms R with

has measure bounded by

,RRH RRUR )(

HC )1(1

Difference to previous situation

Single height and constant vf in vertical

direction causes approximately constant slope

for all R’’ and thus avoids overlap.