Singular Integrals Along N Directions in R^2

0.14 MB·English

by  Ciprian Demeter

SINGULAR INTEGRALS ALONG N DIRECTIONS IN R2
CIPRIAN DEMETER
0 Abstract. We prove optimal bounds in L2(R2) for the maximal oper-
1 ator obtained by taking a singular integral along N arbitrary directions
0
in the plane. We also give a new proof for the optimal L2 bound for the
2
single scale Kakeya maximal function.
n
a
J

2
1 1. Introduction

] Our main result is the following:
A
Theorem 1.1. Let Σ a setof N unit vectors in R2. Letm be a H¨ormander-
C N
Mikhlin multiplier, that is
.
h
t 1
a ∂αm(ξ) . , ξ = 0
m | | α ξ α 6
| |
[ for each α 0. Define

≥
1
v T∗f(x,y) := sup fˆ(ξ,η)m(v ξ +v η)ei(xξ+yη)dξdη .
8 N v∈ΣN |Z 1 2 |
9
Then
9
1 T∗ . logN.
. k Nk2→2
1
In Section 3 we also prove that the bound logN is optimal. It is well
0
0 known (see for example Proposition 2, page 245 in [8]) that the distribution
1 mˆ coincides with a function K outside the origin, which means that if f is
:
v compactly supported and (x,y) is outside the support of f, then we also
i
X have the following kernel representation
r
a
T∗f(x,y) = sup f((x,y)+tv)K(t)dt
N | |
v∈ΣN Z
Thus, Theorem 1.1isthesingularintegral analogueofthefollowing bound
for the Kakeya maximal function due to Nets Katz:
Theorem 1.2 ([5]). Let Σ a set of N unit vectors in R2. Define
N
1
M∗f(x,y) := sup sup f((x,y)+tv)dt .
N |2ǫ |
v∈ΣN ǫ>0 Z|t|<ǫ
Key words and phrases. Kakeya maximal function, singular integrals.
The author is supported by a Sloan Research Fellowship and by NSF Grants DMS-
0742740and 0901208.
1
2 CIPRIANDEMETER
Then
M∗ . logN.
k Nk2→2
In section 4 we give a new proof for the following fact proved in [6]
Theorem 1.3. Let Σ a set of N unit vectors in R2. Define the single scale
N
Kakeya operator
M0f(x,y) := sup f((x,y)+tv)dt .
N | |
v∈ΣN Z|t|<1
Then
M0 . logN.
k Nk2→2
The proofs in both [5] and [6] rely opn stopping time arguments similar
to John Nirenberg’s inequality in the context of product BMO. Perhaps
surprisingly, our arguments in this paper avoid product theory, and are
essentially L2 based.
The proof of the Theorem 1.1 is very similar to the proof of Theorem 1.1.
in [4], in particular it relies on the deep inequality of Chang, Wilson and
Wolff. The second major ingredient we use is a result of Lacey and Li [7]
for single annulus operators, which is essentially equivalent with Carleson’s
theorem [2].
In the case K(t) = p.v.1, the bound
t
H∗ . logN.
k Nk2→2
for the operator
H∗f(x,y) :=
N
dt
sup p.v. f((x,y)+tv) = sup fˆ(ξ,η)sgn(v ξ +v η)ei(xξ+yη)dξdη
1 2
| t | | |
v∈ΣN Z v∈ΣN Z
immediately follows from the Rademacher-Menshov theorem. We do not see
a way to extend this type of approach to the general case.
Wewouldlike to thankZubinGautam, NetsKatzandFrancesco DiPlinio
for stimulating discussions on the subject.
2. Discrete versus continuous square function
To simplify notation, we will often replace (x,y) with x.
Recall the conditional expectation with respect to the σ- algebra consist-
ing of dyadic squares of side length 2−j,
1
Q
E f(x) := f, 1 (x)
j Q
h Q i
Q:|Q|=2−2j | |
X
and let ∆ = E E be the martingale difference. Denote by
j j+1 j
−
∆(f)(x) = ( ∆ f(x) 2)1/2
k
| |
k≥0
X
SINGULAR INTEGRALS ALONG N DIRECTIONS IN R2 3
the discrete square function. We will need the following fundamental in-
equality.
Lemma 2.1 (The Chang-Wilson-Wolff inequality, [3]). There exist con-
stants c ,c such that for all λ > 0 and 0 < ǫ < 1 one has
1 2
x R2 : f(x) E0f(x) > 2λ,∆(f)(x) < ǫλ c2e−cǫ21 x R2 : sup Ekf(x) > ǫλ .
|{ ∈ | − | }| ≤ |{ ∈ | | }|
k≥0
Define for each v = (v ,v ) Σ
1 2 N
∈
T f(x,y) := fˆ(ξ,η)m(v ξ +v η)ei(xξ+yη)dξdη.
v 1 2
Z
Let φ C∞(R2) be supported in 1/2 < ξ < 2 so that
∈ | |
φ(2−kξ) = 0, ξ = 0.
6
k∈Z
X
ˆ
Choose β a Schwartz function such that β is supported in x 1/4 and β is
| | ≤
nonzero in 1/4 ξ 4 and β(0) = 0. Finally, choose a Schwartz function
≤ | | ≤
ψ such that ψ(ξ)β2(ξ) = 1 for ξ in the support of φ.
For each linear bounded operator T : L2 L2 denote by T f = T(S f) :
k k
→
L2 L2, where S f(ξ) = fˆφ(2−kξ).
k
→
We need the following variant of Lemma 3.1 from [4].
d
Lemma 2.2. Let T be a linear bounded multiplier operator T : L2 L2,
→
that is
ˆ
Tf = mf,
for some m L∞(R2). Then there exists c > 0 independent of N, T and x
3
such that for∈almost every x R2cand each f L2(R2)
∈ ∈
∆(Tf)(x) c ( M (M(T f)(x)) 2)1/2,
3 2 k
≤ | |
k∈Z
X
where
M g(x) = (M(g2)(x))1/2,
2
Mg being the standard Hardy-Littlewood maximal function.
Proof Define
B f(ξ) = β(2−kξ)fˆ(ξ)
k
L f(ξ) = ψ(2−kξ)fˆ(ξ).
k
d
Note that
∆ Tf = d(∆ B )(B L )T f.
k k k+n k+n k+n k+n
n∈Z
X
Sublemma 4.2 from [4] implies that (uniformly) for each x
∆ B g(x) . 2−|n|/2M g(x).
k k+n 2
| |
4 CIPRIANDEMETER
Since also
B L g(x) . Mg(x),
k+n k+n
| |
uniformly over k,n, the result follows from the triangle inequality.
Lemma 2.3. Let T be a linear bounded multiplier operator T : L2(R2)
L2(R2) associated with multiplier m. Assume in addition that m is zero o→n
the ball with radius 22N. Then there exists c > 0 independent of N, T and
4
x such that for almost every x R2 and each f L2(R2)
∈ ∈
E (Tf)(x) c 2−N( M (M(T f)(x)) 2)1/2
0 4 2 k
| | ≤ | |
k∈Z
X
Proof The argument follows as in the previous lemma, by using the fact
that (part of Sublemma 4.2 in [4])
E B g(x) . 2−|n|/2M g(x),
0 n 2
| |
for each n 0.
≥
3. Proof of the Theorem 1.1
We have proved in the previous section that
sup ∆(T )f(x) G(f)(x) := c ( M (M(sup T (S f) )(x)) 2)1/2.
v 3 2 v k
| | ≤ | | | |
v∈ΣN k∈Z v∈ΣN
X
The following Lemma is a variant of similar lemmas from [1] and [5].
Lemma 3.1. Let T be a sublinear operator on some measure space (X,m).
Assume
T . N
L1→L1,∞
k k
T . 1
L2→L2,∞
k k
T . N,
Lp→Lp
k k
for some 2 < p < . Then
∞
T . logN
L2→L2
k k
Proof We need to prove that p
∞
λ Tf > 3λ dλ . logN f 2.
|{ }| k k2
Z0
Let
ν (α) := f > α .
f
|{| | }|
Split
f = fλ +fλ +fλ,
1 2 3
where
f, : if f > Nλ
fλ = | | ,
1 0 : otherwise
(
SINGULAR INTEGRALS ALONG N DIRECTIONS IN R2 5
f, : if Nλ f > N−qλ
fλ = ≥ | | ,
2 0 : otherwise
(
f, : if N−qλ f
fλ = ≥ | | .
3 0 : otherwise
(
where q(p 2) = p. Now,
−
∞ ∞ ∞ ∞
λ Tfλ > λ dλ . N fλ dλ = N ν (α)dαdλ =
|{ 1 }| k 1k1 f
Z0 Z0 Z0 ZNλ
∞ α/N
= N ν (α) dλdα = f 2,
f k k2
Z0 Z0
∞ ∞ ∞ Nλ
λ Tfλ > λ dλ . λ−1 fλ 2dλ = λ−1 αν (α)dαdλ =
|{ 2 }| k 2k2 f
Z0 Z0 Z0 ZN−qλ
∞ Nqα
= αν (α) λ−1dλdα . (logN) f 2
f k k2
Z0 Zα/N
∞ ∞ ∞ N−qλ
λ Tfλ > λ dλ . Np λ1−p fλ pdλ = Np λ1−p αp−1ν (α)dαdλ =
|{ 3 }| k 3kp f
Z0 Z0 Z0 Z0
∞ ∞
= Np αp−1ν (α) λ1−pdλdα . f 2
f k k2
Z0 ZNqα
Proposition 3.2.
G(f) C logN f ,
2 2
k k ≤ k k
with C independent of N.
p
Proof It suffices to prove that for each k
A (f) := sup T (S f) C logN S f .
k v k 2 k 2
k | |k ≤ k k
v∈ΣN
p
It was proved in [7] that A maps L2 to L2,∞ and Lp to itself for 2 < p < ,
k
∞
with implicit bounds independent of N. That is a deep result, essentially
equivalent with Carleson’s theorem [2].
Since each T maps L1 to L1,∞,
v
A (f) . N f .
k 1,∞ 1
k k k k
The result now follows from Lemma 3.1 above.
Proof[of Theorem 1.1]Byusing alimiting argument, we canassume that m
is supported in a finite union of dyadic annuli 2j ξ < 2j+1. By rescaling,
≤ | |
we can assume m is actually supported away from the ball of radius 22N (for
example). Let ǫ = √c logN. For each λ > 0,
N 1
x : sup T f(x) > 4λ E E E ,
v λ,1 λ,2 λ,3
{ | | } ⊂ ∪ ∪
v∈ΣN
6 CIPRIANDEMETER
where
E = x : sup T f(x) E T f(x) > 2λ,Gf(x) ǫ λ
λ,1 v 0 v N
{ | − | ≤ }
v∈ΣN
E = x : Gf(x) > ǫ λ
λ,2 N
{ }
E = x : sup E T f(x) > 2λ .
λ,3 0 v
{ | | }
v∈ΣN
By Lemma 2.1,
E x : T f(x) E T f(x) > 2λ,∆(T f)(x) ǫ λ .
λ,1 v 0 v v N
| | ≤ |{ | − | ≤ }|
vX∈ΣN
1
. x : M(T f)(x) > ǫ λ .
v N
N |{ }|
vX∈ΣN
Using this, Lemma 2.3 and Proposition 3.2, we get that
∞ 3
λ E dλ . (logN)2 f 2.
| λ,i| k k2
Z0 i=1
X
Theorem 1.1 now follows.
The following result shows that the bound in Theorem 1.1 is optimal.
Proposition 3.3. Let Σ a set of N equidistributed unit vectors in R2.
N
Define
dt
T∗f(x,y) := sup lim f((x,y)+tv) .
N v∈ΣN |ǫ→∞Z|t|>ǫ t |
Then,
T∗ & (logN).
k Nk2→2
Proof This is a standard example, we only briefly sketch the details. Define
f(x,y) = 1 1 .
k(x,y)k 1≤k(x,y)k≤N/100
Let 100 (x,y) N/100. If v := (x,y) , then
≤ k k ≤ k(x,y)k
dt 1 N2 (x,y) 2
lim f((x,y)+tv) = log −k k ln( (x,y) 2 1)
|ǫ→∞ t | (x,y) | N2 − k k − |
Z|t|>ǫ k k
log (x,y)
k k.
≥ 2 (x,y)
k k
Let w Σ be such that v w 10. It easily follows that
∈ N k − k ≤ N
dt dt log x
lim f((x,y)+tv) lim f((x,y)+tw) < k k,
|ǫ→∞ t −ǫ→∞ t | 10 x
Z|t|>ǫ Z|t|>ǫ k k
SINGULAR INTEGRALS ALONG N DIRECTIONS IN R2 7
simply from the fact that tv tw 1/5 if t < N/50 and the fact that
f(x,y) f(x′,y′) 2k(x,yk)−(x−′,y′)kkif≤1 (x|,y|) , (x′,y′) N/100 and
| − | ≤ k(x,y)k2 ≤ k k k k ≤
(x,y) (x′,y′) 1. Thus,
k − k ≤
dt log (x,y)
T∗f(x,y) lim f((x,y)+tw) k k,
N ≥ |ǫ→∞ t | ≥ 4 (x,y)
Z|t|>ǫ k k
and hence
T∗f & (logN)3/2.
k N k2
Since f . (logN)1/2, the Proposition follows.
2
k k
4. Proof of Theorem 1.3
First, letusremarkthatthereisaproofalongthelinesoftheargumentfor
Theorem1.1. Wehoweverchoosetogiveadifferent, selfcontainedargument,
one that in particular does not appeal to the Chang-Wilson-Wolff inequality.
It suffices to prove the bound for
M F(x,y) := sup F(x+tv ,y +tv )ψ(t)dt ,
0 1 2
| |
v∈ΣN Z
ˆ
where ψ is a positive function such that ψ is supported in [0,1].
Decompose
F = F + F ,
0 n
n≥1
X
such that
F 2 = F 2 + F 2.
k k2 k 0k2 k nk2
n≥1
X
Here F is the Fourier restriction to the unit ball B (that is, F = Fˆ1 ),
0 0 B
while F the Fourier restriction to the annulus 2n−1 (ξ,η) 2n. Note
n
≤ | | ≤
that c
M F (x,y) = sup Fˆ (ξ,η)φ(ξ,η)ψˆ(v ξ+v η)ei(xξ+yη)dξdη . MF (x,y),
0 0 0 1 2 0
| |
v∈ΣN Z
where M is the Hardy-Littlewood maximal function, and
1 φ 1
B 2B
≤ ≤
ˆ ˇ
issmooth. This is since K (x,y) := (φ(ξ,η)ψ(v ξ+v η))(x,y)easily satisfies
v 1 2
K (x,y) . (1+ (x,y) )−3,
v
| | k k
with bound independent of v, and M F (x,y) = sup F K (x,y) .
0 0 v∈ΣN | 0 ∗ v |
Next, we analyze the case n 1.
≥
Definition 4.1 (Grids). A collection of intervals on the unit circle S1 is
called a grid if for any two intervals from the collection that intersect, one
of them should contain the other one.
8 CIPRIANDEMETER
The standard dyadic grid 0 is the collection dyadic intervals on S1, ob-
G
tained by identifying S1 with [0,1). In addition, for i 1/3,2/3 , define
∈ { }
i to consist of the projection on S1 of the intervals
G
[2jk,2j(k +( 1)ji)] : 2j 1,k Z .
{ − ≤ ∈ }
It is easy to see that each i is a grid. Moreover, each interval I on the
G
circle is contained inside an interval of similar length from one of these grids.
Denoteby Ω anypartitionoftheannulus 2n−1 (ξ,η) 2n innsectors
n
≤ | | ≤
with equal area. For each ω Ω , call d(ω) the direction of the radius that
n
∈
splits ω in two sectors with equal area. For each ω Ω , let ω˜ be the sector
n
∈
inthe larger annulus 2n−2 (ξ,η) 2n+1, withtwice theaperture ofω and
≤ | | ≤
having the same bisector d(ω). Let also A(ω) be the projection of the sector
ω on S1. Denote by B(ω) some interval from some grid i that contains
G
10A(ω)+π1, and whose length is comparable to that of A(ω), which in turn
2
is comparable to 2−n.
Decompose further
F = F ,
n ω
ωX∈Ωn
ˆ
where F = F1 . Note that
ω ω
F (x+tv ,y +tv )ψ(t)dt = Fˆ (ξ,η)ψˆ(v ξ +v η)ei(xξ+yη)dξdη
cω 1 2 ω 1 2
Z Z
is nonzero only if there exists (ξ,η) ω such that 0 v ξ +v η 1. This
1 2
∈ ≤ ≤
implies that v 10A(ω)+ π B(ω).
∈ 2 ⊂
Let θ be a bump function adapted to and supported on ω˜ and which
ω
equals 1 on ω. Note that
F (x+tv ,y +tv )ψ(t)dt = Fˆ (ξ,η)θ (ξ,η)ψˆ(v ξ +v η)ei(xξ+yη)dξdη.
ω 1 2 ω ω 1 2
Z Z
Lemma 4.2. If v,v′ B(ω) and ω Ω , then
n
∈ ∈
F (x+tv ,y +tv )ψ(t)dt F (x+tv′,y +tv′)ψ(t)dt .
| ω 1 2 − ω 1 2 |
Z Z
. 2n v v′ M∗F(x,y), (1)
k − k
where
1
M∗F(x,y) = sup F (x+t,y +s)dtds.
ǫδ | |
ǫ,δ>0 Z|t|<ǫZ|s|<δ
Proof Indeed, by rotation invariance it suffices to assume that d(ω) =
(0, 1). Then, B(ω) is an interval of length roughly 2−n containing (1,0).
−
110A(ω) is the interval on S1 with the same center as A(ω) and 10 times its length
SINGULAR INTEGRALS ALONG N DIRECTIONS IN R2 9
Thus, if v,v′ B(ω), then v , v′ . 2−n. This immediately shows that for
∈ | 2| | 2|
each α,β 0
≥
∂α∂β[θ (ξ,η)(ψˆ(v ξ +v η) ψˆ(v′ξ +v′η))] . 2n(1−β) v v′ .
| ξ η ω 1 2 − 1 2 | k − k
By using this, integration by parts and the fact that θ is supported in ω˜,
ω
it follows that
K (x,y) := θ (ξ,η)[ψˆ(v ξ +v η) ψˆ(v′ξ +v′η)]ei(xξ+yη)dξdη
| ω ω 1 2 − 1 2 |
Z
. 22n v v′ min 1, x −2, y2n −2 . 22n v v′ (1+ x +2n y )−2.
k − k { | | | | } k − k | | | |
The lemma follows.
Define
F˜ (x,y,v) := F (x+tv ,y +tv )ψ(t)dt,
ω ω 1 2
Z
and Ω = Ω . Thus,
n≥1 n
∪
˜
M ( F )(x,y) = sup F (x,y,v) .
0 n ω
| |
n≥1 v∈ΣN n≥1ω∈Ω
X XX
By splitting the sum above in three parts, we can assume without loss of
generality that all intervals B(ω) belong to a fixed grid.
For 1 κ logN, let Ωκ consist of those ω Ω such that
≤ ≤ ∈
2κ−1 < B(ω) Σ 2κ. (2)
N
| ∩ | ≤
Let
T F(x,y) = sup F˜ (x,y,v) .
κ ω
| |
v∈ΣN n≥1ω∈Ωκ
X X
It suffices to prove that T . 1 for each κ, by orthogonality and the fact
κ 2
k k
that
T F = T ( F ).
κ κ ω
ω∈Ωκ
X
Definition 4.3. A cluster is a subset C of Ωκ such that B(ω) = .
Let t(C) be the center of the smallest B(ω) in the cluster. ω∈C 6 ∅
T
The grid property and (2) imply that Ωκ can be split in clusters such that
the intervals B(ω) and B(ω′) are pairwise disjoint if ω, ω′ belong to distinct
clusters. Thus,
sup F˜ (x,y,v) = sup sup F˜ (x,y,v) .
ω ω
v∈ΣN |ω∈Ωκ | C v∈ΣN |ω∈C |
X X
Note also that F and F are orthogonal, for two distinct clus-
ω∈C ω ω∈C′ ω
ters C and C′. Using these two facts, it suffices to prove that for each cluster
P P
C,
sup F˜ (x,y,v) . F
ω 2 2
k | |k k k
v∈ΣN ω∈C
X
10 CIPRIANDEMETER
Note that for each v Σ
N
∈
F˜ (x,y,v) = F˜ (x,y,v) =
ω ω
ω∈C ω∈C
X v∈XB(ω)
= F˜ (x,y,t(C))+O(M∗F(x,y)) (by Lemma 4.2) =
ω
ω∈C
v∈XB(ω)
= P ( F˜ (x,y,t(C)))+O(M∗F(x,y))
m ω
ω∈C
X
for some appropriate m = m(v,C), where P denotes the Fourier projection
m
ontheballwithradius2m. Thelastequalityabovefollowssince F˜ (x,y,v)is
ω
supported in frequency in ω. Recall that the maximal operator sup P F
m| m |
is bounded on L2, by invoking standard maximal function arguments. Also,
the operator F˜ (x,y,t(C)) has a bounded multiplier and thus it is
ω∈C ω
bounded on L2. This ends the proof.
P
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DepartmentofMathematics,IndianaUniversity,831East3rdSt.,Bloom-
ington IN 47405
E-mail address: [email protected]