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
v:R R
2
2
/
Hilbert Transform/Maximal
Operator along Vector Field

H v f ( x, y )  p.v.  f (( x, y )  v( x, y )t )dt / t


M v f ( x, y )  sup |  f (( x, y )  v( x, y )t )dt /  |
0

p
L
Question:
- bounds
First observations
Bounded by 1D result if vector field constant.
Value independent of length of v(x,y).
assume unit length vector field.
May
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
Lp only for p>2, weak type 2
c
x
,
Propose modification/conditions
Truncate integral at t   (normalize v   1 )
Demand slow rotation (Lipshitz: v   1 )
Zygmund/Stein conjectures
Assume v   1, v   1 , and truncate t   .
2 ,
Conj.1:Truncated MO bounded L  L
2
Conj.2:Truncated HT bounded L2  L2,
p
.
.
No L bounds known except 1) if p is infinity.
Analytic vector fields
If v is real analytic, then on a bounded domain:
Bourgain (1989):
MO is bounded in Lp , p>1.
Christ, Nagel, Stein, Wainger (1999):
HT bounded in Lp (assume no straight integral
curves. Stein,Street announce without assumption)
One variable (meas.) vector field
v( x, y )  v( x,0)
Theorem (M. Bateman, C.T.)
Measurable, one variable vector field
3/ 2  p  
Hv f
p
 Cp f
p
(HT not truncated;
Related earlier work: Bateman; Lacey/Li)
Linear symmetry group
HT 1vT ( f  T )  ( Hv f )  T
Isotropic dilations
Dilation of second variable
Shearing
Constant along Lipschitz
Angle of
Angle of
v

to x axis less than 2  
 v to x-axis less than or equal to 
Conjecture:
Same bounds for H v
as in Bateman,CT
Relation with Carleson’s theorem
and time-frequency analysis
Carleson’s operator

C f ( x ) 


ˆf ( ) e 2ix d
( x)

p.v.  f ( x  t )e
it ( x )
dt / t

Carleson 1966, Hunt 1968:
Carleson’s operator is bounded in
p
L
, 1 p  
Coifman’s argument
 f ( x  t, y  u( x)t )dt / t
R

iy
ˆ ( x  t , )eiu( x ) t dt / t d
e
f
 
R


R
L2 ( x , y )
R
ˆf ( x  t , )eiu( x ) t dt / t
~

L2 ( x , )
fˆ ( x, )
L2 ( x , y )
L ( x , )
2
 f
2
The argument visualized
A Littlewood Paley band
fˆ
Bound for supported on
Littlewood Paley band
Lacey and Li: Bound on H v f for 2  p  
arbitrary two variable measurable vector field
Bateman: Bound on H v f for
one variable vector field.
1 p  2
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)
Hv f
p

 H f 
2 1/ 2
v k
p
f
,
p

 f 
2 1/ 2
k
Enough to prove for any sequence f k
 H  f 
2 1/ 2
v
k
C
k
p
 f 
2 1/ 2
k
p
p
Weak type interpolation
Enough to prove for
 H  f 
2 1/ 2
v
Whenever
k
k

2
 p
3
,1G  C p H
2
f k  1H
1/ p
G
11 / p
Cauchy Schwarz
Enough to prove
H 
v
2
k
f k ,1G  C p H
Which follows from
H 
v
2
k
f k ,1G
2/ p
G
12 / p
12 / p
G
 C p  
H 
 f
2
k
Single band operator estimate
1G H v  k (1H f ) 2  C  G / H

1 / 2 1 / p
f
2
By interpolation enough for E  G, F  H
H v  k 1F ,1E  C p  G / H 
1 / 2 1 / p
F
1/ 2
E
1/ 2
Lacey-Li (p>2) /Bateman (p<2) proved
H v  k 1F ,1E  Cq F
1/ q
E
11 / q
Note: F,E depend on k, while G,H do not
Induction on log ( G / H ).
2
If p<2 and H  G
exc
exc
G
 G / 2 and
Find G  G such that
the desired estimate holds (prove!) for G \ G exc
.

exc
2
Apply induction hypothesis on G (gain )
Induction on log ( H / G ).
2
If p>2 and G  H
exc
exc
Find H  H such that H  H / 2 and
And the desired estimate holds for H \ H exc
Apply induction hypothesis on H exc
Finding exceptional sets.
Covering arguments
Parallelograms
Kakeya example
Try H exc union of all parallelograms R
with R  G   R for appropriate  .
exc
H
Bad control on
, example of Kakeya set.
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 H
exc
of all parallelograms with
G  R U ( R)   R
q
C

G
has measure bounded by
q
for q>1
(vector field measurable, no other assumption)
Outline of argument
Find good subset ' of set  of
parallelograms with large density, such that
1.
R C
R


R'
R
q'
2.


1U ( R )   C  R
  R
 '
R '

1/ q'
Then:


R    U ( R)  G     R 

R'
R'
 R' 
1
1
G
1/ q
Greedy selection
Iteratively select R for ' with maximal
shadow such that for previously selected R’
 7R  7R'  R / 100
R '',U ( R )U ( R ')0
Here 7R means stretch in vertical direction.
Vertical maximal function
The non-selected parallelograms are all inside


( x, y ) : MV ( 1R )( x, y )  1 / 10000
R'


which has acceptable size.
Additional property
R’ selected prior to R; U(R’) intersect U(R).
Then
7 R'SR  7 R
n
L argument
n
(
1
)
  U (R)  C
R '
 U ( R )  ...  U ( R )
1
n
R1 , R2 ,...Rn
Assume in order of selection and
U ( Ri ) U ( Ri1 )  0
 C  S R1  7 R1  ...  7 Rn
 C  S R1  7 R1  ...  7 Rn 1
 ...  C 7R1
2nd covering lemma (Lacey-Li)
The union G exc of all parallelograms with
H R  R
and
R U ( R)   R
Has measure bounded by C  H .
Use vector field Lipshitz in vertical direction.
3
p

.
 -power responsible for
2
1 2
Outline of argument
From set  of parallelograms with large
densities select ' as before. Using M V as before
obtain:
R  C  R
R
To prove:
R'
2


1
1

C

R



R
  R
'
R'

Power of 2 here responsible for  -power
Expansion of  R  R' .
Sum over all pairs with R selected before R’.
Case 1
E ( R)  100E ( R' )
Case 2
E ( R)  100E ( R' )
Second case has aligned directions, as before.
Second selection in Case 1
Fix R, consider R’ selected later with Case 1.
Prove
 R  R'  C
1
R
Select ' '  ' so that each selected R’ has
Projection of U(R’) onto x-axis disjoint from
Projection of U(R’’) for previously selected R’’
Removing δ
Use disjointness of projections of U(R’) and
density δ to reduce to showing for fixed R’
 R' 'R  C R  S
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.)
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
H R   R ,
R U ( R)   R
has measure bounded by
1 (1 )
C 
H
Difference to previous situation
Single height and constant vf in vertical
direction causes approximately constant slope
for all R’’ and thus avoids overlap.