Instructions for use
A uthor(s ) C ho,Y onggeun; K im,Y oungcheol; L ee,S anghyuk; S him,Y ongsun
C itation Hokkaido University Preprint S eries in Mathematics, 684: 1-17
Is s ue D ate 2005
D O I 10.14943/83835
D oc UR L http://hdl.handle.net/2115/69489
T ype bulletin (article)
F ile Information pre684.pdf
KERNELS HAVING SINGULARITIES ON THE LIGHT CONE
YONGGEUN CHO, YOUNGCHEOL KIM, SANGHYUK LEE, AND YONGSUN SHIM
Abstract. We study the convolution operatorTzwith the distribution kernel
given by analytic continuation from the function
e
Kz(y, s, t) =
{
(t2−s2− |y|2)z
+/Γ(z+ 1) if t >0
0 if t≤0
}
, Re(z)>−1
where (y, s, t)∈Rn−1×R×R. We obtain some improvement upon the previous known estimates forTz. Then we extend the result of the cone multiplier of
negative order onR3 [8] to the case of general
Rn+1, n≥2.
1. Introduction and statement of results
LetKez be the family of distributions onRn+1, n≥2 defined by analytic
con-tinuation from the equation
e
Kz(y, s, t) =
{
(t2−s2− |y|2)z
+/Γ(z+ 1) ift >0
0 ift≤0
}
, Re(z)>−1
where (y, s, t) ∈ Rn−1×R×R, and r
+ = r if r ≥ 0 and r+ = 0 if r < 0. The
Lp-Lq estimates for the convolution operatorf →Kez∗f were studied by Oberlin
[12]. Oberlin [12] showed that for−n
2 ≤Re(z)≤0,f →Ke
z∗f is bounded from
Lp(Rn+1) toLq(Rn+1) provided
(1.1) 1
p−
1
q = 1 +
2Re(z)
n+ 1 , 1 +
Re(z)
n <
1
p <1 +
Re(z)(n−1)
n2+n .
He also showed when z is non-integer, for−n+12 < z <0,f →Kez∗f is bounded
fromLp(Rn+1) toLq(Rn+1) only if (1.1) is satisfied (by simple modification of his
argument one can see the necessary condition is also valid for integerz). Using the
L2boundedness off →Kez∗f whenRe(z) =−(n+ 1)/2, by interpolation it can be
shown thatf →Kez∗f is bounded formLp(Rn+1) toLq(Rn+1) if 1≤p <2< q≤
∞and (1.1) is satisfied. The same results were also obtained by Harmse [7] using the method based on the interpolation along analytic family of operators. These and the previously known results are obtained by some variants of T∗T method
2000Mathematics Subject Classification.Primary 46F10; Secondary 26D10 .
Key words and phrases.cone, convolution estimates.
The first author was supported by Japan Society for the Promotion of Science under JSPS Postdoctoral Fellowship for Foreign Researchers and the third author was partially supported by the Post-doctoral Fellowship Program of Korea Science & Engineering Foundation (KOSEF).
which is basically anL2-theory. The aim of this paper is to extend the range ofp, q
for whichTz is bounded beyond the region mentioned above.
By a simple linear transformation, the distributionKez is essentially equivalent
toKz given by analytic continuation from the function
Kz(y, s, t) =
{
(ts−|y4|2)z
+/Γ(z+ 1) ift >0
0 ift≤0
}
, Re(z)>−1.
It is easy to see that theLp-Lq boundedness properties off →Kz∗f andf →Kez∗f
are equivalent. It is also possible to handle f →Kez∗f directly but computations
involved can be simplified by consideringf →Kz∗f. Thus from now on, we study
f →Kz∗f instead off →Kez∗f.
Our results are stated in terms of Besov spaces. Let ϕ∈C∞
0 (1/2,2) satisfying ∑
kϕ(| · |/2k) = 1. We use a kind of homogeneous Besov space Besp,r, which is
equipped with the norm:
(1.2) ∥f∥Bes p,r :=
( ∑
k
2srk∥∆kf∥rp )1
r
, 1≤p, r≤ ∞, s∈R,
where ∆dkf(η, ρ, τ) = ϕ(|ρ|/2k)fb(η, ρ, τ), (η, ρ, τ)∈ Rn−1×R×R, b stands for
the Fourier transform defined byfb(ξ) =∫ e−2πix·ξf(x)dxand∥ · ∥
p=∥ · ∥Lp(Rn+1).
The following is our main result.
Theorem 1.1. For−2(nn(n+1)2+2n−21) ≤Re(z)<0, there is a constantC such that
(1.3) ∥Kz∗f∥Bes
q,r ≤C∥f∥Besp,r
for allf ∈ S(Rn+1), providedp, q satisfies(1.1).
For realz, the conditions (1.1) are necessary for (1.3). Indeed, the first is a simple consequence of homogeneity. For the second, consider a positive functionf, whose Fourier transform is supported in {(η, ρ, τ) :ρ ∼1}. Here for A, B > 0,A ∼ B
means thatA/2 ≤B ≤2A. Then choosing a suitable ϕ, we have ∆0Tzf =Tzf.
Employing the argument of Oberlin [12], we seeTzf ∈Lq only if −z/n <1/q.
If |s| ≤ 1p, then by virtue of the density of Schwartz class in Bes
p,r [13], one
can replace the condition thatf ∈ S(Rn+1) withf ∈Bes
p,r in Theorem 1.1. On the
other hand, from Littlewood-Paley theory, we have∥f∥q ≤C∥f∥Be0
q,2 and∥f∥Be0p,2≤ C2∥f∥p when 1 ≤p≤2 ≤q ≤ ∞(see [3], p.152). Setting s = 0 and r = 2, the
estimates (1.3) implies the previously knownLp-Lq estimates forf →Kz∗f when 1≤p≤2≤q≤ ∞. However, when 2≤p, q≤ ∞, the estimates (1.3) seems to be slightly weaker thanLp-Lq boundedness.
Unlike the approaches in [12], [7], our study of f → Kz∗f is carried out by
analyzing Fourier transform of Kz ∗f in a more direct way. To obtain (1.3),
supported in the sets {(η, ρ, τ) ∈ Rn−1 ×R×R : ρ ∼ 2l}. Then a simple
re-scaling argument reduces the problem to evaluating the sharpLp-Lq norm of the
Fourier multiplier operator Tδ whose Fourier multiplier is essentially supported in
aδ-neighborhood of the light cone. More precisely, for 0< δ≪1, let us define
(1.4) Tδf(y, s, t) = ∫
Rn+1 ϕ(ρ)ψ
(
τ− |η|2/ρ
δ
)
b(η)fb(η, ρ, τ)dηdρdτ
whereb∈C∞
0 (B(0,1)),ϕ∈C0∞(1/2,2) andψ∈ S(R).
Using the L2(n+1)/(n+3) → L2 restriction estimate for the light cone in Rn+1
(see [15], p. 365-367) and its dual estimate, it follows that ∥Tδf∥ L
2(n+1)
n−1 )(Rn+1)≤ Cδ1/2∥f∥
L2(Rn+1). Interpolating this with false estimate ∥Tδf∥ 2n
n−1 ≤ C∥f∥n2−n1
corresponding to the cone multiplier conjecture, one might expect that if (n−
1)(1−1
p) = (n+ 1) 1 q and
2(n+1) n−1 ≥q >
2n
n−1, then there is a constantC such that
(1.5) ∥Tδf∥q ≤Cδ
n+1 2 (
1
p−
1
q)∥f∥
p.
Using a smooth bump function adapted to a rectangle of size (δ1/2)n−1×δ×1
con-tained inδ-neighborhood of the light cone, the sharpness of the boundCδn+12 ( 1
p−
1
q)
can easily seen.
In order to obtain Theorem 1.1, it is important to obtain the sharpLp-Lq bound
for Tδ in terms ofδ. The estimate (1.3) is actually a consequence of establishing
(1.5) forq > 2(n2n+2n2−1−1) (see Proposition 2.5). This will be done using the bilinear
restriction estimates due to Wolff [21] and Tao [16], for which we refer the readers to [9] and [6]. Combining the known local smoothing estimates ([10], [20]) with the argument in this note (also see [8]), one can extend the range ofq for which (1.5) holds with bound Cδn+12 (
1
p−
1
q)−ϵ. But these estimates can not be used to obtain (1.3) because of the homogeneity condition 1/p−1/q= 1 +2Re(z)n+1 .
The method used in this paper is similar to the argument used by the third author to get some sharp estimates for the cone multiplier of negative order inR3
([8]). However, exploiting the parabolic structure of the cone in a more direct way, the arguments here are simpler than those in [8]. The convolution estimate forKz
is essentially equivalent to the cone multiplier problem of negative order, since for
−n+1
2 < Re(z)<− n−1
2 , Fourier transform ofKe z is
C(z){(τ2−ρ2− |η|2)−z− n+1
2
− −sinπ(z+
n
2)(τ
2−ρ2− |η|2)−z−n+12
+ }
where C(z) = π−2z−n+32 Γ(z+ n+1
2 ) (see [4], p.284). Consequently, our estimates
forTδ provide some new sharp bounds for the cone multiplier of negative order in
Rn+1,n≥3. Some remarks on the cone multiplier operator are in the last section.
Throughout this paper,C is a positive constant which may vary line to line. In addition to the symbolb, we useF(·),F−1(·) to denote Fourier transform, inverse
2. Proof of Theroem 1.1
2.1. Dyadic decomposition of the distribution Kz. Let’s denote by Dz the
distribution analytically continued from the function
Dz=tz+/Γ(z+ 1), Re(z)>−1.
We decomposeKz so that the Fourier transforms of the decomposed distributions
have dyadically disjoint supports.
Lemma 2.1. Let z be a complex number which is not integer and −(n+ 1)/2 < Re(z) < 0. Then there are smooth functions ψz,ϕz with ψz ∈ C0∞(1/2,2), ϕbz ∈
C∞
0 ((−2,−1/2)∪(1/2,2)) such that for allh∈ S(Rn+1),
(2.1) ⟨Kz, h⟩=∑
l ∑
k
2z(k+l)
∫∫∫
ψz(t/2k)ϕz(
s− |y|2/4t
2l )h(y, s, t)ds dy dt.
Proof. First we prove Lemma 2.1 for−1< Re(z)<0. SinceKz is a function for
−1< Re(z)<0, we have
⟨Kz, h⟩=
∫∫
tz
∫ (s− |y|2/4t)z +
Γ(z+ 1) h(y, s, t)ds dy dt.
Let us choose ψz ∈ C0∞(1/2,2) such that tz = ∑
k2kzψ(t/2k) and we use the
following fact from [4](p.173). Ifzis not integer
b
Dz(ρ) =i(2π)−z−1(eizπ/2ρ−−z−1−e−izπ/2ρ−+z−1).
Let g ∈ C∞
0 ((−2,−1/2)∪(1/2,2)) satisfying ∑
g(2l·) = 1. Applying Parseval’s
formula to the inner integral (ins) and dominated convergence theorem inρ(note
b
Dz(ρ) is locally integrable), we get
⟨Kz, h⟩=∑
k ∑
l
2zk
∫∫
ψz(t/2k) ∫
e−i2πρ|y|2/4tDbz(ρ)g(2lρ)Fs−1(h)(y, ρ, t)dρ dy dt
whereF−1
s (h)(y,·, t) is the inverse Fourier transform ofh(y,·, t) insvariable.
Now set
ϕz(s) =i(2π)−z−1F−1(g(ρ)(eizπ/2ρ−−z−1−e−izπ/2ρ−+z−1))(s).
By Parseval’s formula one can see
∫
e−i2πρ|y|2/4tDbz(ρ)g(2lρ)Fs−1(h)(y, ρ, t)dρ= 2lz ∫
ϕz(s− |y| 2/4t
2l )h(y, s, t)ds.
Therefore we have
⟨Kz, h⟩=∑
k ∑
l
2z(k+l)
∫
ψz(t/2k)ϕz(s− |y| 2/4t
2l )h(y, s, t)dy ds dt.
Now we prove Lemma 2.1 for−(n+ 1)/2< Re(z)<−1. Let us set a differential operator L= ∂2
∂s∂t − ∑n−1
i=1 ∂
2
∂yi2. Then LK
z+1= 4(z+ (n+ 1)/2)Kz. Using this,
we seeKz is analytically continued by the formula
(2.2) ⟨Kz, h⟩= 1
C(z, k, n)⟨K
z+k, Lkh⟩
whereC(z, k, n) = 4k(z+ (n+ 1)/2)(z+ 1 + (n+ 1)/2)· · ·(z+k−1 + (n+ 1)/2).
Suppose−2< Re(z)<−1. Then, using above formula, we have
⟨Kz, h⟩= 1
4(z+ (n+ 1)/2)⟨K
z+1, Lh⟩
SinceRe(z+ 1)>−1, from the above results we haveϕz+1 andψz+1 such that
⟨Kz+1, Lh⟩=
∑
l, k
2(z+1)(k+l)
∫∫∫
ψz+1(t/2k)ϕz+1
(s− |y|2/4t
2l )
Lh(y, s, t)dy ds dt.
(2.3)
Set
ψz=
ψ′
z+1(t) + (n−1)
2t ψz+1(t)
4(z+ (n+ 1)/2) , ϕz=
ϕ′ z+1
4(z+ (n+ 1)/2) and note that thatψz∈C0∞(1/2,2), ϕbz∈C0∞((−2,−1/2)∪(1/2,2)). Since
L
[
ψz+1 (
t
2k )
ϕz+1 (
s− |y|2/4t
2l )]
=
2−k−l
[
ψz+1′ (
t
2k )
+(n−1)2
k
2t ψz+1
(
t
2k )]
ϕ′z+1 (
s− |y|2/4t
2l )
,
by integration by parts in (2.3) one can easily see (2.1) holds. One can repeat this argument using (2.2) forRe(z)<−n+1
2 . This completed the proof. □
Let us set
Kl,kz (y, s, t) = 2zk+zlψz(t/2k)ϕz((s− |y|2/t)/2l)
with ψz, ϕz given in Lemma 2.1. Then by Lemma 2.1 for complexz which is not
integer we can writeKz∗f as
(2.4) Kz∗f =∑
k ∑
l
Kl,kz ∗f
provided−(n+ 1)/2< Re(z)<0. By the presence of quadratic term|y|2/t, the
Fourier transform ofKz
l,k can be computed in an exact form. We state this in the
following:
Lemma 2.2. Let ψ ∈ C∞
0 (1/2,2) and ϕ∈ S(R) with ϕb= 0 on (−ϵ, ϵ) for some
ϵ >0. Then
(2.5) F(ψ(t)ϕ(s− |y|2/4t))(η, ρ, τ) =ϕe(ρ)ψe(τ− |η|2/ρ)
where ϕe(ρ) = (sgn(−ρ))n−21|ρ| 1−n
2 ϕb(ρ), ψe(τ) = cnF(t
n−1
2 ψ(t))(τ), where cn =
Sinceϕbz is supported in (−2,−1/2)∪(1/2,2), using this and re-scaling, we see
(2.6) Kdz
l,k(η, ρ, τ) = 2(z+
n+1 2 )k2(z+
n+1
2 )lϕez(2lρ)ψez(2k(τ− |η|2/ρ)).
Note that ϕez(2l·) is supported on the set{|ρ| ∼2−l}.
Proof of Lemma 2.2. Note that
F(ψ(t)ϕ(s− |y|2/4t))(η, ρ, τ)
=
∫∫∫
ψ(t)
∫ b
ϕ(a)e2πia(s−|y|2/4t)da e−2πi(tτ+sρ+y·η)dydsdt.
Integrating the right hand side of the above iny, we see
F(ψ(t)ϕ(s− |y|2/4t))(η, ρ, τ)
=cn ∫∫
ψ(t)tn−21ϕb(a)(sgn(−a))
n−1 2 |a|
1−n
2 e2πit(|η| 2/a−τ)∫
e2πis(a−ρ)dsdtda.
Since the inner integral is the delta function, one can easily see (2.5). In fact, these calculations are not rigorous. But it can be made so in sense of tempered
distribution. □
2.2. A summation method. Now we prove Theorem 1.1. To do this we need to establish the sharpLp-Lq estimates forf →Kz
l,k∗f. The following will be shown
in the next Subsection 2.3.
Proposition 2.3. If(1/p,1/q)is contained in the open quadrangle Qwith vertices
(1/2,1/2),(1/p0,1/q0),(1,0),(1−1/q0,1−1/p0)where
(2.7) p0=
2(n2+ 2n−1)
n2+ 2n−3 , q0=
2(n2+ 2n−1)
n2−1 ,
then there is a constantC such that
(2.8) ∥Kl,kz ∗f∥q≤C2(Re(z)+
n+1
2 (1−p1+1q))(k+l)∥f∥
p.
Note that the point (1/p0,1/q0) is on the line (n−1)(1−1p) = (n+ 1)1q. Once
Proposition 2.3 has been established, the proof of Theorem 1.1 is rather straight-forward. Define pointsQz,Q′z∈[0,1]×[0,1] by
Qz= (1 +Re(z)
n ,
(1−n)Re(z)
n2+n ), Q ′
z= (1 +
(n−1)Re(z)
n2+n ,−
Re(z)
n )
which are on the line 1 p−
1 q = 1 +
2Re(z)
n+1 . By a simple computation, one can see
that the conditions (1.1) are equivalent to (1/p,1/q)∈(Qz, Q′z). Set
Kz l ∗f :=
∑
k
Kz l,k∗f.
We claim that if (1/p,1/q)∈(Qz, Q′z)∩ Q, then
(2.9) ∥Kz
l ∗f∥q≤C∥f∥p.
One can verify that if Re(z) ≤ −2(nn(n+1)2+2n−21), then (Qz, Q′z)∩ Q = (Qz, Q′z). To
throughout this paper. We use a simple extension of the Lemma in [8] which is implicit in [2](also see [5]).
Lemma 2.4 (Interpolation lemma). Let ε1, ε2 > 0. Suppose that {Tj}j∈Z be l-linear operators satisfying that
∥Tj(f1, . . . , fl)∥q1,∞≤M12
ε1j
l ∏
i=1
∥fi∥pi
1,1 and
∥Tj(f1, . . . , fl)∥q2,∞≤M22 −ε2j
l ∏
i=1
∥fi∥pi
2,1 for1 ≤pi
1, pi2, < ∞, i= 1, . . . l (here, the superscript i is not an exponent, but an
index) and1< q1, q2<∞. Then T :=∑Tj is bounded from Lp
1,1
× · · · ×Lpl,1
to Lq,∞ with
∥T(f1, . . . , fl)∥Lq,∞ ≤CMθ 1M21−θ
l ∏
i=1
∥fi∥pi,1
where θ = ε2/(ε1+ε2), 1/q = θ/q1+ (1−θ)/q2, 1/pi = θ/pi1+ (1−θ)/pi2, for
i= 1, . . . , l.
Proof of Lemma 2.4. LetN ∈Z, which will be chosen later. Set TN =∑N−∞Tj,
TN =∑∞
N+1Tj. Sinceq1, q2 >1, the spaces Lq1,∞,Lq2,∞ are Banach spaces. By
summation we have
∥TN∥Lp1
1,1×···×Lpl1,1→Lq1,∞≤CM12
N ε1, ∥T
N∥Lp1
2,1×···×Lpl2,1→Lq2,∞ ≤CM22 −N ε2.
Let E1, . . . , El be measurable sets and let λ >0. By Chebyshev’s inequality, the
measure of the set{x:|T(χE1, . . . , χEl)(x)|> λ} is bounded above by
|{x:|TN(χE1, . . . , χEl)(x)|> 1
2λ}|+|{x:|T
N(χ
E1, . . . , χEl)(x)|> 1 2λ}|
≤C(Mq1
1 2ε1N q1 l ∏
i=1
|Ei|q1/p
i
1λ−q1+Mq2
2 2−ε2N q2 l ∏
i=1
|Ei|q2/p
i
2λ−q2).
ChoosingN that optimizes the above, we have
|{x∈Rn:|T(χE1, . . . , χEl)(x)|> λ}| ≤C(M
θ 1M21−θ
l ∏
i=1
|Ei|1/p
i λ−1)q.
This completes the proof. □
Fix z, 0 > Re(z) > −n+1
2 . Let (1/p,1/q) ∈ (Qz, Q ′
z)∩ Q. Then we can
find two points (1/p1,1/q1), (1/p2,1/q2) contained inQ such that the line joining
(1/p1,1/q1) and (1/p2,1/q2) passes through (1/p,1/q), and 1/p1−1/q1<1+2Re(z)n+1
and 1/p2−1/q2>1 +2Re(z)n+1 . Then from Proposition 2.3, we have two estimates
∥Kl,kz ∗f∥qi ≤C2 (
Re(z)+n+12 (1−pi1+qi1)
)
(k+l)
Using this and applying Lemma 2.4 toKz l ∗f =
∑
kKl,kz ∗f, we see∥Klz∗f∥q,∞≤
C∥f∥p,1 if (1/p,1/q)∈(Qz, Q′z)∩ Q. By real interpolation among these, (for fixed
z) we obtain (2.9).
From (2.6), it follows that Kbl is supported in the set{(η, ρ, τ) : 2−1−l ≤ |ρ| ≤
21−l}. Therefore, invoking the definition of Besov spaces Bes
p,r and Kz = ∑Kz
l,
and using (2.9), we see that
∥Kz∗f∥
e
Bs q,r≤C
( ∑
l
2−srl∥Kz
l ∗∆(−l)f∥rq )1/r
≤C
( ∑
l
2−srl∥∆ (−l)f∥rp
)1/r
provided (1/p,1/q)∈(Qz, Q′z)∩ Q. This completes the proof of Theorem 1.1.
2.3. Convolution estimates forKz
l,k; proof of Proposition 2.3. LetTeδ be the
multiplier operator given by
F(Teδf)(η, ρ, τ) =ϕez(ρ)ψez
(τ− |η|2/ρ
δ
) b
f(η, ρ, τ).
Since the Fourier transform ofKz
l,kis given as (2.6), re-scaling (η, ρ)→(2−l/2η,2−lρ)
in the Fourier transform side, for (2.8) it suffices to show that∥Teδ∥p→q ≤Cδ
n+1 2 (
1
p−
1
q)
for (1 p,
1
q)∈ Q. Now considerTe R
δ defined by
F(TeδRf)(η, ρ, τ) =F(Teδ)(η, ρ, τ)b(
η
R)fb(η, ρ, τ)
where b is a smooth function supported in B(0,1) ∈Rn−1 with b(0) = 1. Then,
it suffices to show that if (1p,1q) ∈ Q, then ∥TeR
δ ∥p→q ≤ Cδ
n+1
2 (1p−1q), uniformly
in R. Again by re-scaling (η, ρ) → (Rη, R2ρ), to prove (2.8), it is sufficient to
show that if (1p,1q) ∈ Q, then ∥Te1
δ∥p→q ≤ Cδ
n+1
2 (1p−1q). Since L2 boundedness
for Teδ is trivial, by duality, we need only to show ∥Teδ1∥p→q ≤ Cδ
n+1 2 (
1
p−
1
q) for
(1p,1q)∈((1/p0,1/q0),(∞,∞)]. Therefore Proposition 2.3 follows from
Proposition 2.5. Let δ >0 and letTδ be a multiplier operator given by d
Tδf(η, ρ, τ) =ϕ(ρ)ψ(
(τ− |η|2/ρ)
δ )b(η)fb(η, ρ, τ) whereϕ∈C∞
0 (1/2,2),ψ∈ S(R)andbis a smooth function supported inC0∞(B(0,1)),
B(0,1)⊂Rn−1. Then if(n−1)(1−1
p) = (n+ 1) 1
q andq > q0=
2(n2+2n−1)
n2−1 , then there is a constant C such that
(2.10) ∥Tδf∥q ≤Cδ
n+1 2 (
1
p−
1
q)∥f∥
p.
And when(p, q) = (p0, q0),Tδf is of restricted weak type(p0, q0). Namely,
(2.11) ∥Tδf∥q0,∞≤Cδ
n+1
2 (p10−q10)∥f∥
p0,1.
When δ ≥1, (2.10) follows from direct kernel estimate. Indeed, by differenti-ation, ∂α(ϕe(ρ)ψe((τ−|η|2/ρ)
kernel ofTδ is contained in anyLp spaces, 1≤p≤ ∞with its norm O(1).
There-fore∥Tδ∥p→q ≤Cfor any 1≤p≤q. Hence it is sufficient to show Proposition 2.5
for the case 0< δ <1.
2.3.1. Proof of Proposition 2.5. Since from Lemma 2.2 we have
F(δn+12 ψ(tδ)ϕ((s− |y|2/4t))) =ϕe(τ)ψe(τ− |η|
2/ρ
δ ),
it follows that
(2.12) ∥Tδf∥∞≤Cδ
n+1 2 ∥f∥
1.
Interpolation (real) between this and (2.11) gives (2.10). Therefore it is sufficient to show (2.11).
Even though ψe is not compactly supported, by the rapid decay of ψe, it can be replaced by a function with compact support. In fact, Proposition 2.5 can be deduced from:
Proposition 2.6. Let 0 < δ ≤ 1 and let ψδ be a smooth function supported on
[−δ, δ] with|dn
dtnψδ| ≤Cδ−n for anyn and define a multiplier operatorSδ by
(2.13) Sdδf(η, ρ, τ) =ϕ(ρ)ψδ(τ− |η|2/ρ)b(η)fb(η, ρ, τ)
whereϕ∈C0∞(1/2,2)andbis a smooth function supported inC0∞(B(0,1)),B(0,1)⊂
Rn−1. Then there is a constantC such that for0< δ≤1,
(2.14) ∥Sδf∥q0,∞≤Cδ
n+1
2 (p10−q10)∥f∥
p0,1.
Assuming Proposition 2.6 for a moment, we prove Proposition 2.5. Let χ ∈ C∞
0 ((−2,−1/2)∪(1/2,2)) with ∑kχ(·/2k) = 1. Let us set ψk := χ(·/2k)ψe(·/δ)
and
m1:=ϕ(ρ)b(η) ∑
2k≤δ
ψk(τ− |η|2/ρ),
m2:=ϕ(ρ)b(η) ∑
δ<2k≤1
ψk(τ− |η|2/ρ),
m3:=ϕ(ρ)b(η) ∑
1<2k
ψk(τ− |η|2/ρ).
Also letT1, T2, T3 be the multiplier operators given byTdif =mifb,i = 1,2,3, so
thatSδ =T1+T2+T3. Since∑2k≤δψk =ψ0(·/δ) for someψ0∈C0∞(−1,1), one can
easily seeT1 falls into the category ofSδ. By Proposition 2.6, ∥T1∥Lp0,1→Lq0,∞ ≤
Cδn+12 (p10−q10). Next, from the rapid decay ofψwe see that∥T
3∥p→q ≤CδM∥f∥pfor
allp≤qandM. LetTk be the multiplier operator with multiplierϕ(ρ)b(η)ψ k(τ−
|η|2/ρ) so thatT2=∑δ<2k≤1Tk. Observe that ifδ≤2−k, then
d
n
dτnψk(τ)
for any M >0. Since ψk is supported in [−2k,2k], using Proposition 2.6, we see
that
∥Tkf∥
q0,∞≤Cδ
M2−kM2kn+1 2 (
1
p0− 1
q0)∥f∥p 0,1.
Choosing largeM, we get the required estimates forT2 after summing ink. From
this Proposition 2.5 follows.
2.3.2. A bilinear type estimates for Sδ. For the proof of 2.6 we use the bilinear
restriction estimates for the cone [16], [21]. For the reader’s convenience, we give a statement which is slightly different from those in [16], [21]. To exploit the parabolic structure of the cone in effective way, we reformulate it in co-normal coordinates. LetQ= [−1,1]n−1and define
(2.15) R∗(F)(y, s, t) =
∫ 2
1/2 ∫
Q
e2πi(y·η+sρ+t|η|2/ρ)F(η, ρ)dηdρ.
And let Θ(F) ={(η/ρ)∈Q: (η, ρ)∈supp (F) for someρ∈[1/2,2]}. The follow-ing is the bilinear restriction theorem for the cone:
Suppose dist (Θ(F),Θ(G))∼1. Then for q≥ n+3n+1, there is a constant C such that
(2.16) ∥R∗(F)R∗(G)∥Lq(Rn+1)≤C∥F∥L2(Q×[1,2])∥G∥L2(Q×[1,2]).
Using this we obtain a bilinear type estimates for Tδ. Let us define ∆(f) =
{(η/ρ)∈Rn−1: (η, ρ, τ)∈suppf, for someρ, τ}.
Lemma 2.7. Let0< δ≪1. If dist(∆(fb),∆(bg))∼1, then for n+3
n+1 ≤p≤2,there
is a constant C such that∥Sδf Sδg∥p≤Cδ∥f∥2∥g∥2.
Proof. Note that the Fourier transforms of Sδf and Sδg are supported in a δ
-neighborhoodC(δ) of the cone{(η, ρ, τ) :τ=|η|2/ρ, ρ∼1}. C(δ) can be written as
a union of the surfacesCs={(η, ρ, τ)∈:τ =|η|2/ρ+s, ρ∼1, η∈Q},s∈(−δ, δ).
Namely,
C(δ)⊂ ∪
s∈(−δ,δ)
Cs.
LetQ1= ∆(fb),Q2= ∆(bg). Set
Ci
s:={(η, ρ, τ)∈ Cs:η/ρ∈Qi}
fori,= 1,2. Letdµi
sbe the surface measure ofCis. Then using the bilinear restriction
estimates (2.16), we can see that forp≥ n+3
n+1, there is constantC, independent of
s, t∈(−δ, δ), such that
(2.17) \F dµ1
s\Gdµ2t
(Here we discard harmless factors which come from changing surface measures.) Now we set
e
f(η, ρ, τ) :=ϕ(ρ)ψδ(τ− |η|2/ρ)b(η)fb(η, ρ, τ), e
g(η, ρ, τ) :=ϕ(ρ)ψδ(τ− |η|2/ρ)b(η)bg(η, ρ, τ)
and also letfes=fe|C1
s,egt=eg|Ct2. Sincef ,eeg are supported inC(δ), it follows that
∥Sδf Sδg∥pp= ∫ ∫ δ −δ ∫ δ −δ \e
fsdµ1seg\tdµ2tdtds p dx.
Since dist (Θ(fes),Θ(get)) ∼ 1 for all s, t ∈ (−δ, δ), using H¨older’s inequality and
(2.17), we see
∥Sδf Sδg∥pp≤Cδ2p−2 ∫ δ
−δ ∫ δ
−δ ∫
f\esdµ1s\egtdµ2t p dxdtds
≤Cδ2p−2
∫ δ
−δ ∫ δ
−δ
∥fes∥p2∥egt∥p2dsdt.
Since p≤2, H¨older’s inequality yields ∫−δδ∫−δδ∥fes∥2p∥egt∥p2dsdt ≤Cδ2−p∥fe∥ p 2∥eg∥
p 2.
Using Plancherel’s theorem, we get Lemma 2.7. □
2.3.3. Proof of Proposition 2.6. We use a decomposition technique introduced in [19], which is useful in exploiting bilinear estimate. For eachj ≥1 we dyadically decompose Q ⊂ Rn−1 into ∼ 2(n−1)j dyadic cubes Qj
k with side length 2−(j−1).
We sayQjk∼Q j
k′ to mean thatQ j k,Q
j
k′ are not adjacent but have adjacent parent
cubes of diameter 2−j+2. So ifQj k∼Q
j
k′, then dist (Q j k, Q
j k′)∼2
−j.By a Whitney
decomposition of Q×Q away from the diagonal D of Q×Q (e.g. [14], p.16), ignoring some harmless measure zero sets, we have
(2.18) Q×Q\D=∪j≥1
∪
Qjk∼Qjk′Q j k×Q
j k′.
Letfkj be defined by
(2.19) fbkj(η, ρ, τ) =χQj
k(η/ρ)fb(η, ρ, τ).
We may assume that the Fourier transform off is supported inQ×[1/2,2]×[−1,1]. Since∑j≥1∑Qj
k∼Q j k′χQ
j kχQ
j
k′ = 1 almost everywhere inQ×Q, we see that
(2.20) Sδf(x)·Sδf(x) = ∑
j≥1 ∑
Qjk∼Qjk′Sδf j
k(x)·Sδf j k′(x).
Fixingj, we define a bilinear operator by
Bj(f, g)(x) = ∑
Qj k∼Q
j k′Sδf
j
k(x)·Sδgkj′(x).
Then, it is easy to see that
(2.21) (Sδf(x))2=
∑
j≥1
Bj(f, f).
Lemma 2.8. Suppose for some p, q satisfying 2 ≤ p < q and 1/2 < 1/q+ 1/p, there is a constant B such that for Qjk∼Qjk′,
(2.22) ∥Sδfkj·Sδgkj′∥Lq/2(Rn+1)≤B∥f j
k∥Lp(Rn+1)∥gjk′∥Lp(Rn+1).
Then there is a constant C, independent ofj andδ, such that
(2.23) ∥Bj(f, g)∥Lq/2(Rn+1)≤CB∥f∥Lp(Rn+1)∥g∥Lp(Rn+1). Proof. For a fixed j, if Qjk ∼ Qjk′, then the Fourier support of Sδf
j k ·Sδg
j k′ is
contained in the set {(η, ρ, τ) :|η/ρ−cjk| ≤23−j,|η| ≤ C,2 ≤ρ≤4} where cj k is
the center ofQjk. So the Fourier transforms of{Sδfkj·Sδgkj′}
Qjk∼Qjk′ are supported in
boundedly (at most 8n) overlapping (infinite) rectangles. By Plancherel’s theorem
and a standard argument (see, [TVV, Lemma 6.1]), we have for 1≤q/2≤ ∞,
(2.24) ∥Bj(f, g)∥q/2≤C (∑
Qjk∼Qjk′∥Sδf j k ·Sδf
j k′∥
(q/2)∗ q/2
)1/(q/2)∗
wherep∗= min(p, p′). Now by the assumption (2.22), we have
∥Bj(f, g)∥q/2≤CB (∑
Qjk∼Qjk′∥f j k∥
(q/2)∗ p ∥g
j k′∥
(q/2)∗ p
)1/(q/2)∗
.
Since the number ofQjk satisfyingQjk∼Qjk′ is at most 4 n, ∑
Qjk∼Qjk′∥f j k∥
(q
2)
∗ p ∥gkj′∥
(q
2)
∗
p ≤C
( ∑
k
∥fkj∥2(
q
2)
∗ p
)1 2(∑
k
∥gjk∥2(
q
2)
∗ p
)1 2 .
Since 1/2<1/q+ 1/p and 2≤p < q, it is easy to see that p <2(q/2)∗. We see
that the right hand side of the above is bounded above byC∥f∥(
q
2)
∗ p ∥g∥(
q
2)
∗
p using
the following: If 2≤r < p <∞, then there is a constant C=C(p, r) such that
(2.25)
( ∑
k
∥fkj∥pr )1/p
≤C∥f∥r.
The inequality (2.25) follows from the observations: (i) supj,k∥f j
k∥p ≤C∥f∥p for
any 1 < p < ∞ by the singular integral theory, especially Lp boundedness of
Hilbert transform ([St1, p.100]), (ii) (∑k∥fkj∥2
2)1/2≤ ∥f∥2by Plancherel theorem,
and (iii) interpolation between the above estimates (i) and (ii). □
Now we state a simple lemma which is to be used for kernel estimates ofSδ.
Lemma 2.9. For0< r, δ≤1, letQ(a, r)⊂Q= [−1,1]n−1 be a cube centered ata
with side length2rand letχQ(a,r)be a smooth function adapted to the cubeQ(a, r),
that is,χQ(a,r)=χ(·−ra)whereχ be a smooth function supported inQ. Set
ma,rδ =ϕe(ρ)ψeδ(τ− |η|2/ρ)χQ(a,r)(η/ρ).
Then, there is a constantC, independent ofa, δ, r, such that
∥F−1[ma,rδ ]∥∞≤Cδr(n−1), ∥F−1[ma,rδ ]∥1≤Cmax(1, rn−1δ−
Proof. The first inequality is trivial, because ∫ |ma,rδ (η, ρ, τ)|dηdρdτ ≤ Cδr(n−1).
For the second, we consider the casesδ≥r2,δ < r2, separately. By the linear map
La,r(η, ρ, τ) = (rη+aρ, ρ, r2τ+ 2a·rη+a2ρ),
we may assume a = 0, using the facts ∥F−1[ma,r
δ ]∥1 = ∥F−1[ma,rδ (La,1)]∥1 and
ma,rδ (La,1(η, ρ, τ)) =m0,rδ (η, ρ, τ).
First we consider the caseδ≥r2. It suffices to show that∥F−1[m0,r
δ (L0,r)]∥1≤
C. Since∥(∂ ∂τ)
α1(∂
∂η) α2(∂
∂ρ) α3m0,r
δ (L0,r)∥1≤C (
r2 δ
)α1−1
for any index (α1, α2, α3),
by integration by parts one can see that
|F−1[m0,rδ (L0,r)](y, s, t)| ≤C
δ r2
(
1 +|y|+|s|+| δ r2t|
)−N
for some largeN. By integration, we get∥F−1[m0,r
δ (L0,r)]∥1≤C.
Now for the case δ ≤ r2, it suffices to show that ∥F−1[m0,r
δ (L0,δ1/2)]∥1 ≤ Crn−1δ−n−1
2 by re-scaling. Note thatrδ−1/2≥1 and
m0,rδ (L0,δ1/2(η, ρ, τ)) =ϕe(ρ)ψeδ(δ(τ− |η|2/ρ))χQ(0,rδ−1/2)(η/ρ).
Letχe0be smooth function supported inQwith∑j∈Zn−1χe0(· −j). We decompose
χQ(0,rδ−1/2)intoO((rδ−1/2)n−1) smooth functionsχej=χe0(·−j)χQ(0,rδ−1/2), which
are uniformly contained inC∞. Since
(∂τ∂ )α1(∂η∂ )α2(∂ρ∂ )α3m0,rδ (L0,δ1/2)χej 1
≤C for any (α1, α2, α3),
the integration by parts yields that for any largeN,
|F−1[m0,rδ (L0,δ1/2)χej](y, s, t)| ≤C(1 +|y|+|t|+|s|)−N.
Thus we have∥F−1[m0,r
δ (L0,δ1/2)χej]∥1≤Cand hence ∥F−1[ma,rδ (L0,δ1/2)]∥1≤Crn−1δ−
n−1 2 ,
because the number ofjareO(rn−1δ−n−1
2 ). This completes the proof of lemma. □
In view of Lemma 2.8, we only need to estimate∥Sδfkj·Sδfkj′∥q/2withQ j k ∼Q
j k′.
This is to be done by treating the cases 22jδ≥1 and 22jδ <1 separately.
I. Case 2−2j≥δ. We claim that if 2−2j≥δ, then for (1/p,1/q) contained in the
line segment [(0,0),(1/2,(n+ 1)/2(n+ 3))], we have
(2.26) ∥Sδfkj·Sδgjk′∥q/2≤C2 2(n+1
q −(n−1)(1−
1
p))jδ1−n+
2n p∥fj
k∥p∥gjk′∥p.
By virtue of interpolation, it is sufficient to show (2.26) for (p, q) = (∞,∞) and (p, q) = (2,2(n+ 3)/(n+ 1)). The estimate (2.26) for (p, q) = (∞,∞) follows from Lemma 2.9, becauseSδfkj=κ∗fkj for someκwith∥κ∥1≤C2−(n−1)jδ−
n−1 2 .
Now we show (2.26) when (p, q) = (2,2(n+ 3)/(n+ 1)). The cubesQjk andQjk′
η→η+aρon the Fourier transform side, we may assumeQjkandQjk′are contained
in the ballB(0, C2−j). By re-scaling (η, ρ, τ)→(2−jη, ρ,2−2jτ),
Sδfkj(y, s, t)
=
∫
ei(2−jy·η+sρ+2−2jtτ)ϕe(ρ)ψeδ (
τ− |η|2/ρ
22j )
χQ1(η/ρ)fbj(η, ρ, τ)dηdρdτ
where∥fj∥2≤C2−(n+1)j/2∥fkj∥2 andQ1 is a cube contained inB(0, C). Applying
the same re-scaling toSδgjk′, we see that
(2.27) (Sδfkj·Sδgkj′)(y, s, t) = (Se22jδfj·Se22jδgj)(2
−jy, s,2−2jt)
where ∥gj∥2 ≤C2−(n+1)j/2∥gkj′∥2 and dist (∆(fbj),∆(gbj))∼1. Here, Se22jδ is
de-fined by replacingψδ byψδ(2·2j) in (2.13). Sinceδ22j≤1 andψδ(22·j) is supported in [−δ22j, δ22j], applying Lemma 2.7 to (2.27) with δ replaced by δ22j and
re-scaling, we have the required estimate (2.26).
In case that 22jδ <1, using (2.26) and Lemma 2.8, we have
∥Bj(f, g)∥q/2≤C22(
n+1
q −(n−1)(1−
1
p))jδ1−n+
2n p∥f∥
p∥g∥p
for p, q satisfying 1 2 <
1 p +
1 q and (
1 p,
1
q) ∈ [(0,0),(1/2,(n+ 1)/2(n+ 3))]. Now
applying Lemma 2.4 to the bilinear operators {Bj(f, g)}2−2j≥δ, using the above, we obtain
(2.28)
∑
2−2j≥δ
Bj(f, g) q
0/2,∞
≤Cδ1−n+2pn0∥f∥p
0,1∥g∥p0,1.
Note that the line n+1
q = (n−1)(1− 1
p) intersects the line segment [(0,0),(1/2,(n+
1)/2(n+ 3))] at (1/p0,1/q0).
II. Case 2−2j≤δ. In this case, the condition Qj k ∼ Q
j
k′ is unnecessary. From
Lemma 2.9, we see thatSδfkj=κ∗f j
k with∥κ∥1≤Cand∥κ∥∞≤Cδ2−j(n−1). So ∥Sδfkj∥p≤C∥fkj∥pfor 1≤p≤ ∞and∥Sδfkj∥∞≤Cδ2−j(n−1)∥fkj∥1. Interpolation
between these two estimates gives us
∥Sδfkj∥q ≤(δ2−j(n−1))1/p−1/q∥fkj∥p,
provided p ≤ q. Observe that if n+1q = (n−1)(1− p1), (δ2−j(n−1))2/p−2/q =
2−j4n(n
−1)
n+1 ( 1
p− n−1
2n)δ
4n n+1(
1
p− n−1
2n). Thus we deduce from H¨older inequality that there
is a constantC, independent ofQjk,Q j
k′, such that
(2.29) ∥Sδfkj·Sδgkj′∥q/2≤C2
−j4n(n−1)
n+1 (1p− n−1
2n)δn4+1n(p1− n−1
2n)∥fj
k∥p∥g j k′∥p
forp, q satisfying n2n−1 ≤q≤4, 2≤p≤q and n+1q = (n−1)(1−1p). And hence from the application of Lemma 2.8 to (2.29) we also deduce that there exists a constantC, independent ofj, such that
∥Bj(f, g)∥q/2≤C2−j
4n(n−1)
n+1 (1p− n−1
2n )δn4+1n(1p− n−1
2n )∥f∥
for the same exponent p, q as in (2.29). Sincep < n2n−1, we conclude from direct summation onj that if n+1
q = (n−1)(1− 1
p) andp < 2n n−1,
(2.30)
∑
2−2j≤δ
Bj(f, g)
q/2≤Cδ 1−n+2n
p∥f∥
p∥g∥p.
Therefore, combining (2.28) and (2.30), we finally get (2.11). This completes the proof of Proposition 2.6.
3. Remarks on the cone multiplier of negative order
The cone multiplier operator Sµ of order µ in Rn+1, n ≥ 2, is a multiplier
operator defined by
(3.1) Sdµf(ξ, τ) = ϕ(τ)
Γ(µ+ 1)(1− |ξ|
2/τ2)µ
+fb(ξ, τ),(ξ, τ)∈Rn×R
whereϕ∈C∞ 0 (1,2).
We are concerned with Lp-Lq estimates for S−α of negative order. For easier
description, we introduce some notations. For 0< α <(n+ 1)/2, define vertices in square [0,1]×[0,1] by
Aα(n) = (n−1
2n + α n,0
)
, Bα(n) =
(n−1
2n + α n,
n−1 2n −
α(n−1)
n2+n )
,
A′α(n) = (
0,n+ 1
2n − α n
)
, B′α(n) = (
n+ 1 2n +
α(n−1)
n2+n ,
n+ 1 2n −
α n
)
.
And let us denote ∆α(n) be the closed pentagon with verticesAα(n), Bα(n), B′α(n),
A′
α(n), (1,0) from which closed line segments [Aα(n), Bα(n)],[A′α(n), Bα′(n)] are
removed. In [8] it was shown thatS−αis bounded fromLptoLq only if (1/p,1/q)∈
∆α(n) and that inR3 (n= 2) the necessary condition is sufficient for 3/14< α <
3/2.
Our estimates for Tδ in the previous section give some new bounds for the
cone multiplier of negative order in Rn+1, n ≥ 3. By a linear transform, we
may replace the multiplier Γ(µ+1)ϕ(τ) (1− |ξ|2/τ2)µ
+ byeb(η)Kz(2η, ρ, τ) for some fixed eb∈C∞
0 (B(0,2)\B(0,12)), sinceL
p−Lq boundedness of both multiplier operators
defined by these two distrubutions are equivalent to each other. Using Lemma 2.1, one can have the decomposition
b(η)Kz(2η, ρ, τ) =∑
l,k
b(η)Kl,kz (2η, ρ, τ),
where Kz
l,k(2η, ρ, τ) = 2z(k+l)ψz(ρ/2k)ϕz(τ−|η|
2/ρ
2l ), and ϕz and ψz is defined in Lemma 2.1. Then using Lemma 2.2, one can obtain
F−1(Kz)(y, s, t) =∑
l, k
2(z+n+12 )(l+k)ϕe
Denote byTz
l the convolution operator with kernel
Klz= ∑
k
2(z+n+12 )(l+k)ϕe
z(2ls)ψez(2k(t− |η|2/s)).
Using re-scaling and Proposition 2.5 and interpolation, the resulting estimates with simple estimates from H¨older’s inequality, one can see that forp, qsatisfying n+1q ≤
(n−1)(1−1
p) andq≥
2(n2+2n−1)
n2−1 ,
(3.2) ∥Tlzf∥q≤C2(z+
n+1 2 −
n(n+1) (n−1)q)l∥f∥
p.
Using Lemma 2.4 and some summation method in [8], [9] or [6], the following can be obtained from (3.2).
Theorem 3.1. Let S−α be defined by (3.1), n ≥ 2. Then, the followings hold
whenever 2(n2n+2n2−1−1)≤α <(n+ 1)/2.
a) If(1/p,1/q) =Bα(n),orBα′(n), thenS−αis of restricted weak type(p, q).
b) If (1/p,1/q)∈(B′
α(n), A′α(n)], thenS−α is of weak type(p, q).
c) If (1/p,1/q)∈∆α(n), then there is a constantC such that
∥S−αf∥Lq(Rn+1)≤C∥f∥Lp(Rn+1).
Apart from sharp estimates, it is possible to get some bound for S−α beyond
Theorem 3.1 if one use the recently proven local smoothing estimates due to Wolff and Laba [10] and the argument in [8] based on some scaling method. For the related results to the cone multiplier, one may refer to [1], [11], [17], [18] and [21].
References
[1] J. Bourgain, Estimates for cone multiplier, Operator Theory: Advances and Application79
(1995), 41-60.
[2] J. Bourgain, Estimations de certaines functions maximales, C.R. Acad. Sci. Paris 310
(1985), 499-502.
[3] J. Bergh and J. L¨ofstr¨om,Interpolation spaces: An introduction, Springer-Verlag, New York, 1976.
[4] I.M. Gel’fand and G.E. Shilov, Generalized functions: Properites and operations, vol. 1, Academic Press, New York and London (1964).
[5] A. Carbery, A. Seeger, S. Wainger, and J. Wright,Class of singular integral operators along variable lines, Journal of Geometric Analysis9(1999), 583-605.
[6] Y. Cho, Y. Kim, S. Lee and Y. Shim,SharpLp−Lq estimates for Bochner-Riesz operators
of negative index inRn, n≥3,Journ. Func. Anal.218(2005), 150-167.
[7] J. Harmse, On Lebesgue space estimates for the wave equation, Indiana Univ. Math. J.39
(1990), 229-248
[8] S. Lee, Some sharp bounds for the cone multiplier of negative order in R3, Bull. London Math. Soc.35(2003), 373-390.
[9] S. Lee, Improved bounds for Bochner-Riesz and maximal Bochner-Riesz operators, Duke Math. Journ.122(2004), 205-232.
[10] I. Laba and T. Wolff,A local smoothing estimate in higher dimensions, J. d’Analyse Math.
88(2002), 149-171.
[11] G. Mochenhaupt,A note on the cone multiplier, Proc. Amer. Math. Soc.177(1993), 145-151.
[13] J. Peetre,New thoughts on Besov spaces,Duke University Mathematical Series 1, Durham N.C., (1976).
[14] E. M. Stein,Singular integrals and differentiability properties of functions, Princeton Uni-versity Press, Princeton, (1970).
[15] E. M. Stein,Real variable methods, othogonality, and oscillatory integrals, Princeton Univ. Press, Princeton, (1993).
[16] T. Tao,Endpoint bilinear restriction theorems for the cone and some sharp null from esti-mates, Math. Z.238(2001), 215-268.
[17] T. Tao and A. Vargas,A bilinear approach to cone multipliers I. Resriction estimates, Geom. Funct. Anal.10(2000), 185-215.
[18] T. Tao and A. Vargas,A bilinear approach to cone multipliers. II. Application, Geom. Funct. Anal.10(2000), 216-258.
[19] T. Tao, A. Vargas and L. VegaA bilinear approach to the restriction and Kakeya conjecture, J. Amer. Math. Soc.11(1998), 967-1000.
[20] T. Wolff, Local smoothing type estimates onLpfor largep, Geom. Funct. Anal.10(2000),
1237-1288.
[21] T. Wolff, A sharp cone restriction estimate, Annals of Math.153(2001), 661-698.
Yonggeun Cho:
Department of Mathematics, Hokkaido University, Sapporo 060-0810, Japan E-mail address:[email protected]
Youngchoel Kim:
Department of Mathematics, Pohang University of Science and Technology, Pohang 790-784, Korea
E-mail address:[email protected]
Sanghyuk Lee:
Department of Mathematics, University of Wisconsin-Madison, WI 53706-1388, USA E-mail address:[email protected]
Yongsun Shim:
Department of Mathematics, Pohang University of Science and Technology, Pohang 790-784, Korea
