Volume 10 (2009), Issue 2, Article 35, 11 pp.
FOURIER RESTRICTION ESTIMATES TO MIXED HOMOGENEOUS SURFACES
E. FERREYRA AND M. URCIUOLO
FAMAF - CIEM (U NIVERSIDAD NACIONAL DE C ÓRDOBA - C ONICET ).
M EDINA A LLENDE S / N , C IUDAD U NIVERSITARIA ,
5000 C ÓRDOBA .
eferrey@mate.uncor.edu
urciuolo@gmail.com
Received 17 September, 2008; accepted 13 February, 2009
Communicated by L. Pick
A BSTRACT. Let a, b be real numbers such that 2 ≤ a < b, and let ϕ : R2 → R a mixed
homogeneous function. We consider polynomial functions ϕ and also functions of the type
a
b
ϕ (x1 , x2 ) = A |x1 | + B |x2 | . Let Σ = {(x, ϕ (x)) : x ∈ B} with the Lebesgue induced
measure. For f ∈ S R3 and x ∈ B, let (Rf ) (x, ϕ (x)) = fb(x, ϕ (x)) , where fb denotes the
usual Fourier transform.
For a large class of functions ϕ and for 1 ≤ p < 34 we characterize, up to endpoints, the pairs
(p, q) such that R is a bounded operator from Lp R3 on Lq (Σ) . We also give some sharp
Lp → L2 estimates.
Key words and phrases: Restriction theorems, Fourier transform.
2000 Mathematics Subject Classification. Primary 42B10, 26D10.
1. I NTRODUCTION
Let a, b be real numbers such that 2 ≤ a < b, let ϕ : R2 → R be a mixed homogeneous
1
1
a
b
function of degree one with respect to the non isotropic dilations r · (x1 , x2 ) = r x1 , r x2 ,
i.e.
1
1
(1.1)
ϕ r a x1 , r b x2 = rϕ (x1 , x2 ) , r > 0.
We also suppose ϕ to be smooth enough. We denote by B the closed unit ball of R2 , by
Σ = {(x, ϕ (x)) : x ∈ B}
and by σ the induced Lebesgue measure. For f ∈ S (R3 ) , let Rf : Σ → C be defined by
(1.2)
(Rf ) (x, ϕ (x)) = fb(x, ϕ (x)) ,
x ∈ B,
Research partially supported by Secyt-UNC, Agencia Nacional de Promoción Científica y Tecnológica.
The authors wish to thank Professor Fulvio Ricci for fruitful conversations about this subject.
254-08
2
E. F ERREYRA AND M. U RCIUOLO
where fb denotes the usual Fourier transform of f. We denote by E the type set associated to R,
given by
1 1
E=
,
∈ [0, 1] × [0, 1] : kRkLp (R3 ),Lq (Σ) < ∞ .
p q
Our aim in this paper is to obtain as much information as possible about the set E, for certain
surfaces Σ of the type above described.
In the general n-dimensional case, the Lp (Rn+1 ) − Lq (Σ) boundedness properties of the
restriction operator R have been studied by different authors. A very interesting survey about
recent progress in this research area can be found in [11]. The Lp (Rn+1 ) − L2 (Σ) restriction
n+4
theorems for the sphere were proved by E. Stein in 1967, for 3n+4
< p1 ≤ 1; for 2n+4
<
4n+4
1
n+4
1
≤ 1 by P. Tomas in [12] and then in the same year by Stein for 2n+4 ≤ p ≤ 1. The last
p
argument has been used in several related contexts by R. Strichartz in [9] and by A. Greenleaf
in [6]. This method provides a general tool to obtain, from suitable estimates for σ
b, Lp (Rn+1 )−
L2 (Σ) estimates for R. Moreover, a general theorem, due to Stein, holds for smooth enough
hypersurfaces
never vanishing Gaussian curvature ([8], pp.386). There it is shown that in
with
n+4
1
≤ p1 ≤ 1 and − n+2
+ n+2
≤ 1q ≤ 1, also that this last relation is
this case, p1 , 1q ∈ E if 2n+4
n p
n
the best possible and that no restriction theorem of any kind can hold for f ∈ Lp (Rn+1 ) when
1
n+2
n+2
n+4
≤ 2n+2
([8, pp.388]). The cases 2n+2
< p1 < 2n+4
are not completely solved. The best results
p
for surfaces with non vanishing curvature like the paraboloid and the sphere are due to T. Tao
[10]. Restriction theorems for the Fourier transform to homogeneous polynomial surfaces in
R3 are obtained in [4]. Also, in [1] the authors obtain sharp Lp Rn+l − L2 (Σ) estimates for
certain homogeneous surfaces Σ of codimension l in Rn+l .
In Section 2 we give some preliminary results.
a
b
In Section 3 we consider ϕ (x1 , x2 ) = A
|x1 | + B |x2 | , A 6= 0, B 6= 0. We describe
completely, up to endpoints, the pairs p1 , 1q ∈ E with p1 > 34 . A fundamental tool we use is
Theorem 2.1 of [2].
In Section 4 we deal with polynomial
ϕ. Under certain hypothesis about
functions
ϕ we can
3
1
1 1
prove that if 4 < p ≤ 1 and the pair p , q satisfies some sharp conditions, then p1 , 1q ∈ E.
4
Finally we obtain some L 3 − Lq estimates and also some sharp Lp − L2 estimates.
2. P RELIMINARIES
We take ϕ to be a mixed homogeneous and smooth enough function that satisfies (1.1). If
V is a measurable set in R2 , we denote ΣV = {(x, ϕ (x)) : x ∈ V } and σ V as the associated
surface measure. Also, for f ∈ S (R3 ) , we define RV f : ΣV → C by
RV f (x, ϕ (x)) = fb(x, ϕ (x))
x∈V;
we note that RB = R, σ B = σ and ΣB = Σ.
For x = (x1 , x2 ) letting kxk = |x1 |a + |x2 |b , we define
2 1
A0 = x ∈ R : ≤ kxk ≤ 1
2
and for j ∈ N,
Aj = 2−j · A0 .
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F OURIER R ESTRICTION E STIMATES TO M IXED H OMOGENEOUS S URFACES
S
Thus B ⊆
3
Aj . A standard homogeneity argument (see, e.g. [5]) gives, for 1 ≤ p, q ≤ ∞,
j∈N∪{0}
A
1
− a+b+ab
+ p1 a+b+ab
R j p 3 q Aj = 2−j a+b
ab ( q
a+b
a+b ) RA0 .
L (R ),L (Σ )
Lp (R3 ),Lq (ΣA0 )
(2.1)
From this we obtain the following remarks.
1
1 1
Remark 1. If p , q ∈ E then 1q ≥ − a+b+ab
+
a+b p
1
Remark 2. If − a+b+ab
+
a+b p
(2.2)
then
1 1
,
p q
a+b+ab
.
a+b
a+b+ab
a+b
< 1q ≤ 1 and
A
R 0 p 3 q A < ∞,
L (R ),L (Σ 0 )
∈ E.
We will use a theorem due to Strichartz (see [9]),whose proof relies on the Stein complex
interpolation theorem, which gives Lp (R3 )−L2 ΣV estimates for the operator RV depending
V . In [4] we obtained information about the size of the constants.
on the behavior at infinity of σc
There we found the following:
Remark 3. If V is a measurable set in R2 of positive measure and if
c
−τ
V
σ (ξ) ≤ A (1 + |ξ3 |)
for some τ > 0 and for all ξ = (ξ1 , ξ2 , ξ3 ) ∈ R3 , then there exists a positive constant cτ such
that
V
1
R p 3 2 V ≤ cτ A 2(1+τ )
L (R ),L (Σ )
for p =
2+2τ
.
2+τ
In [2] the authors obtain a result (Theorem 2.1, p.155) from which they also obtain the following consequence
Remark 4 ([2, Corollary 2.2]). Let I, J be two real intervals, and let
M = {(x1 , x2 , ψ (x1 , x2 )) : (x1 , x2 ) ∈ I × J} ,
2
where ψ : I × J → R is a smooth function such that either ∂∂xψ2 (x1 , x2 ) ≥ c > 0 or
1
2
∂ ψ
∂x2 (x1 , x2 ) ≥ c > 0, uniformly on I × J. If M has the Lebesgue surface measure, 1q =
2
3 1 − p1 and 34 < p1 ≤ 1 then there exists a positive constant c such that
b (2.3)
≤ c kf kLp (R3 )
f |M Lq (M )
for f ∈ S(R3 ).
Following the proof of Theorem 2.1 in [2]
that if in the last remark we take
2 we can check
−k −k+1 ∂ ψ
J = 2 ,2
, k ∈ N in the case that ∂x2 (x1 , x2 ) ≥ c > 0 uniformly on I × J with c
1
independent of k, or I = 2−k , 2−k+1 , k ∈ N in the other case, then we can replace (2.3) by
1
1
b (2.4)
≤ c0 2−k( p + q −1) kf kLp (R3 )
f |M Lq (M )
with c0 independent of k.
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 35, 11 pp.
http://jipam.vu.edu.au/
4
E. F ERREYRA AND M. U RCIUOLO
3. T HE C ASES ϕ (x1 , x2 ) = A |x1 |a + B |x2 |b
1 1
In this cases we characterize, up to endpoints, the pairs p , q ∈ E with 34 < p1 ≤ 1. We also
obtain some border segments. If either A = 0 or B = 0, ϕ becomes homogeneous and these
cases are treated in [4]. For the remainder situation we obtain the following
Theorem 3.1. Let a, b, A, B ∈ R with 2 ≤ a ≤ b, A 6= 0, B 6= 0, let ϕ (x1 , x2 ) = A |x1 |a +
1
B |x2 |b and let E be the type set associated to ϕ. If 43 < p1 ≤ 1 and − a+b+ab
+ a+b+ab
< 1q ≤ 1
a+b p
a+b
then p1 , 1q ∈ E.
1
Proof. Suppose 34 < p1 ≤ 1 and − a+b+ab
+ a+b+ab
< 1q ≤ 1. By Remark 2 it is enough
a+b p
a+b
1 to prove
(2.2).
Now,
A
is
contained
in
the
union
of
the
rectangles
Q
=
[−1,
1]
×
,1 ,
0
2
1 0
Q = 2 , 1 × [−1, 1] , and its symmetrics with respect to the x1 and x2 axes. Now we will
S
Qk with
study RQ Lp (R3 ),Lq (ΣQ ) . We decompose Q =
k∈N
Qk =
−k+1
−2
−k
, −2
−k −k+1 1
∪ 2 ,2
× ,1 .
2
Now, as in Theorem 1, (3.2), in [3] we have
a−2
d
−1
σ Qk (ξ) ≤ A2k 2 (1 + |ξ3 |)
and then Remark 3 implies
Q R k 4
L 3 (R3 ),L2
(3.1)
(
Σ Qk
)
≤ c2k
a−2
8
.
2
Also, since ∂∂xϕ2 (x1 , x2 ) ≥ c > 0 uniformly on Qk , from (2.4) we obtain
2
Q R k p 3 q Q ≤ c0 2−k( p1 + 1q −1)
L (R ),L (Σ k )
for 1q = 3 1 − p1 and 43 < p1 ≤ 1. Applying the Riesz interpolation theorem and then performing the sum on k ∈ N we obtain
Q
R p 3 q Q < ∞,
L (R ),L (Σ )
for 2+3a
1 − p1 < 1q ≤ 1 and 34 < p1 ≤ 1. In a similar way we get that
2+a
Q0 R
p 3 q Q0 < ∞,
L (R ),L (Σ )
for 2+3b
1 − p1 < 1q ≤ 1 and 34 < p1 ≤ 1. The study for the symmetric rectangles is analogous.
2+b
Thus
A
R 0 p 3 q A < ∞
L (R ),L (Σ 0 )
for
3
4
<
1
p
1
≤ 1 and − a+b+ab
+
a+b p
a+b+ab
a+b
<
1
q
≤ 1 and the theorem follows.
Remark 5.
i) If
b+2
8
<
1
q
ii) The point
≤ 1 then
3 1
,
4 q
a+b+2ab 1
,
2a+2b+2ab 2
∈ E.
∈ E.
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F OURIER R ESTRICTION E STIMATES TO M IXED H OMOGENEOUS S URFACES
5
From (3.1) and the Hölder inequality we obtain that
Q k( a−2
− 2−q
R k 4
8
2q )
≤
c2
Q
3
q
L 3 (R ),L (Σ k )
for
1
2
≤
1
q
≤ 1. Then if
a+2
8
<
for these q’s. Analogously, if
thus since a ≤ b, if
b+2
8
<
1
q
1
q
≤ 1 we perform the sum over k ∈ N to get
Q
R 4
< ∞,
L 3 (R3 ),Lq (ΣQ )
b+2
8
< 1q ≤ 1 we get
Q0 R 4 3 q Q0 < ∞,
L 3 (R ),L (Σ )
≤ 1,
A
R 0 4
L 3 (R3 ),Lq
(ΣA0 )
< ∞,
and i) follows from Remark 2.
Assertion ii) follows from Remark 3, since from Lemma 3 in [3] we have that
1
1
|b
σ (ξ)| ≤ c (1 + |ξ3 |)− a − b .
4. T HE P OLYNOMIAL C ASES
In this section we deal with mixed homogeneous polynomial functions ϕ satisfying (1.1).
The following result is sharp (up to the endpoints) for 34 < p1 ≤ 1, as a consequence of Remark
1.
Theorem 4.1. Let ϕ be a mixed homogeneous polynomial function satisfying (1.1). Suppose
that the gaussian curvature of Σ does not vanish identically and that at each point of ΣB−{0}
with vanishing curvature, at least one principal curvature
is different from zero. If (a, b) 6=
3
1
a+b+ab 1
a+b+ab
1
(2, 4) , 4 < p ≤ 1 and − a+b p + a+b < q ≤ 1 then p1 , 1q ∈ E.
Proof. We first study the operator RA0 . Let (x01 , x02 ) ∈ A0 . If Hessϕ (x01 , x02 ) 6= 0 there exists a neighborhood U of (x01 , x02 ) such that Hessϕ (x1 , x2 ) 6= 0 for (x1 , x2 ) ∈ U. From the
proposition in [8, pp. 386], it follows that
U
R p 3 q U < ∞
(4.1)
L (R ),L (Σ )
for 1q = 2 1 − p1 and 34 ≤ p1 ≤ 1. Suppose now that Hessϕ (x01 , x02 ) = 0 and that either
∂2ϕ
∂x21
2
(x01 , x02 ) 6= 0 or ∂∂xϕ2 (x01 , x02 ) 6= 0. Then there exists a neighborhood V = I × J of (x01 , x02 )
2
2 2
such that either ∂∂xϕ2 (x1 , x2 ) ≥ c > 0 or ∂∂xϕ2 (x1 , x2 ) ≥ c > 0 uniformly on V. So from
1
2
Remark 4 we obtain that
V
R p 3 q V < ∞
(4.2)
L (R ),L (Σ )
for 1q = 3 1 − p1 and 34 < p1 ≤ 1. From (4.1), (4.2) and Hölder´s inequality, it follows that
A
R 0 p 3 q A < ∞
(4.3)
L (R ),L (Σ 0 )
for 1q ≥ 3 1 − p1 and 34 < p1 ≤ 1. So, if a+b+ab
≥ 3, the theorem follows from Remark 2. The
a+b
only cases left are (a, b) = (3, 4) , (a, b) = (3, 5) , (a, b) = (4, 5) and (a, b) = (2, b) , b > 2.
If (a, b) = (3, 4) and ϕ has a monomial of the form ai,j xi y j , with aij 6= 0, then 3i + 4j = 1 so
4i + 3j = 12 and so either (i, j) = (0, 4) or (i, j) = (3, 0). So ϕ (x1 , x2 ) = a3,0 x31 + a0,4 x42 .
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6
E. F ERREYRA AND M. U RCIUOLO
The hypothesis about the derivatives of ϕ imply that a3,0 6= 0 and a0,4 6= 0 and the theorem
follows using Theorem 3.1 in each quadrant. The cases (a, b) = (3, 5) , or (a, b) = (4, 5) are
completely analogous.
Now we deal with the cases (a, b) = (2, b) , b > 2. We note that
b
ϕ (x1 , x2 ) = Ax21 + Bx1 x22 + Cxb2
(4.4)
where B = 0 for b odd. The hypothesis about ϕ implies A 6= 0. For b odd, ϕ (x1 , x2 ) =
Ax21 + Cxb2 and since C 6= 0 (on the contrary Hessϕ (x1 , x2 ) ≡ 0), the theorem follows using
Theorem 3.1 as before. Now we consider b even and ϕ given by (4.4). If B = 0 the theorem
follows as above, so we suppose B 6= 0.
b
−2
b
x2 2 2
(4.5)
Hessϕ (x1 , x2 ) = − 2
B b + 8ACb − 8ACb2 x22 − 2(b − 2)ABbx1 .
4
0
0
So if Hessϕ (x1 , x2 ) = 0 then either x02 = 0 or
b
B 2 b2 + 8ACb − 8ACb2 x02 2 − 2(b − 2)ABbx01 = 0.
In the first case we have b > 4. We take a neighborhood W1 = I × −2−k0 , 2−k0 ⊂ A0 , k0 ∈ N,
of the point (x01 , 0) such that Hessϕ vanishes, on W1 , only along the x1 axes. For k ∈ N,
k > k0 , we take Uk = I × Jk where Jk = [−2−k+1 , −2−k ] ∪ [2−k , 2−k+1 ] . So W1 = ∪Uk . For
(x1 , x2 ) ∈ Uk , it follows from (4.5) that
b
|Hessϕ (x1 , x2 )| ≥ c2−k( 2 −2) ,
so for ξ = (ξ1 , ξ2 , ξ3 ) ∈ R3 ,
b−4
d
Uk (ξ) ≤ c2k 4 (1 + |ξ |)−1
σ
3
and from Remark 3 we get
U R k 4
L 3 (R3 ),L2
(4.6)
(ΣUk )
≤ c2k
b−4
16
.
2
∂ ϕ
Also, since ∂x2 (x1 , x2 ) ≥ c > 0 uniformly on Uk , as in (2.4) we obtain
1
(4.7)
for
3
4
<
(4.8)
for
1
qt
1
p
≤ 1 and
1
q
U R k p 3 q U ≤ c2−k(2− p2 )
L (R ),L (Σ k )
= 3 1 − p1 . From (4.6), (4.7) and the Riesz Thorin theorem we obtain
U −(1−t)(2− p2 ))
R k p 3 q U ≤ c2k(t b−4
16
L t (R ),L t (Σ k )
1
1
= t 2 + (1 − t) 3 1 − p and p1t = t 34 + (1 − t) p1 .
A simple computation shows that if
8
and that
4+b
1
p
=
3
4
then the exponent in (4.8) is negative for t < t0 =
1
2 + 3b
−
< 0,
qt0
4 (2 + b)
so for
1
p
>
3
4
and t < t0 , both near enough, the exponent is still negative and
1
2 + 3b
1
−
1−
< 0,
qt
2+b
pt
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F OURIER R ESTRICTION E STIMATES TO M IXED H OMOGENEOUS S URFACES
7
thus
W
R 1 p 3 q W < ∞
L (R ),L (Σ 1 )
1
2+3b
= 2+b 1 − p . Finally, if
(4.9)
for
3
4
<
1
p
near enough and
1
q
B 2 b2 + 8ACb − 8ACb2
x02
2b
− 2(b − 2)ABbx01 = 0
then we study the order of Hessϕ (x1 , x02 ) for 2−k−1 ≤ |x1 − x01 | ≤ 2−k , k ∈ N.
(x0 ) 2b −2 b
2
(4.10) B 2 b2 + 8ACb − 8ACb2 x02 2 − 2(b − 2)ABbx1 4
(x0 ) 2b −2
= 2
(b − 2)ABb x1 − x01 ≥ c2−k .
2
We take the following neighborhood of (x01 , x02 ) , W2 = ∪k∈N Vk , with
1
1
1
0
−k−1
0
−k
r 2 x1 , r b x2 : 2
Vk =
≤ x1 − x1 ≤ 2 , ≤ r ≤ 2 .
2
From the homogeneity of ϕ and (4.10) we obtain
1
1
2 Hessϕ r 2 x1 , r b x02 = r1− b Hessϕ x1 , x02 ≥ c2−k ,
then from Proposition 6 in [8, p. 344], we get for ξ = (ξ1 , ξ2 , ξ3 ) ∈ R3
k
c
−1
σ Vk (ξ) ≤ c2 2 (1 + |ξ3 |) ,
so from Remark 3
V R k 4
L 3 (R3 ),L2
k
(ΣVk )
≤ c2 8
and by Hölder’s inequality, for q < 2 we have
V 2−q
1
R k 4
≤ c2k( 8 − 2q ) .
L 3 (R3 ),Lq (ΣVk )
This exponent is negative for
1
q
>
(4.11)
for
5
8
<
1
q
≤ 1. Since b ≥ 6,
5
8
≤
5
8
and so we sum on k to obtain
W
R 2 4
<∞
L 3 (R3 ),Lq (ΣW2 )
2+3b
4(2+b)
and then from (4.1), (4.9) and (4.11), we get
A
R 0 p 3 q A < ∞,
L (R ),L (Σ 0 )
3
1
1
2+3b
1
for 4 < p near enough and q > 2+b 1 − p and the theorem follows from standard considerations involving Hölder’s inequality, the Riesz Thorin theorem and from Remark 2.
Remark 6. In the case (a, b) = (2, b) , b > 2, we have (4.11). In a similar way we get, from
(4.6) and Hölder’s inequality,
W
R 1 4
<∞
L 3 (R3 ),Lq (ΣW1 )
for
b+4
16
<
1
q
≤ 1. So
kRkL 34 (R3 ),Lq (Σ) < ∞
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8
E. F ERREYRA AND M. U RCIUOLO
for max 58 , b+4
, 2+3b < 1q ≤ 1. We observe that if b = 6 then 58 = b+4
= 2+3b
, thus from
16 8+4b
16
8+4b
1
Remark 1 we see that, in this case, this condition for q is sharp, up to the end point.
Now we will show some examples of functions ϕ not satisfying the hypothesis of the previous
theorem, for which we obtain that the portion of the type set E in the region 34 < p1 ≤ 1 is smaller
than the region
1 1
3
1
a + b + ab
1
1
Ea,b =
,
: < ≤ 1,
1−
< ≤1
p q
4
p
a+b
p
q
stated in Theorem 4.1.
We consider ϕ (x1 , x2 ) = x21 , which is a mixed homogeneous function satisfying (1.1) for
any b > 2. In this case ϕx1 x1 ≡ 2 but Hessϕ ≡ 0. FromRemark
2.8 in [4] and Remark 4 we
1
1
obtain that the corresponding type set is the region q ≥ 3 1 − p , 43 < p1 ≤ 1 which is smaller
than the region Ea,b .
We consider now a mixed homogeneous function ϕ satisfying (1.1), of the form
ϕ (x1 , x2 ) = xl2 P (x1 , x2 ) ,
(4.12)
with P (x1 , 0) 6= 0 for x1 6= 0. Since a < b it can be checked that l ≥ 2 and that for l > 2,
ϕx1 x1 (x1 , 0) = ϕx2 x2 (x1 , 0) = 0. Moreover
(4.13)
Hessϕ = x2l−2
Px1 x1 l (l − 1) P + 2lx2 Px2 + x22 Px2 x2 − (lPx1 + x2 Px1 x2 )2 ,
2
which vanishes at (x1 , 0) . A computation shows that the second factor is different from zero at
a point of the form (x1 , 0) . So Hessϕ does not vanish identically.
Proposition 4.2. Let ϕ be a mixed homogeneous function satisfying (1.1) and (4.12). If p1 , 1q ∈
1
1
E then q ≥ (l + 1) 1 − p .
h −1 i h ε−l i
Proof. Let f ε = χKε the characteristic function of the set Kε = 0, 13 × 0, ε 3 × 0, 3M
,
with M =
max
P (x1 , x2 ) . If p1 , 1q ∈ E then
(x1 ,x2 )∈[0,1]×[0,1]
kRfε kLq (Σ) ≤ c kfε kLp (R3 ) = cε−
(4.14)
1+l
p
.
By the other side,
Z
kRfε kLq (Σ) ≥
Wε
1q
q
b
fε (x1 , x2 , ϕ (x1 , x2 )) dx1 dx2
1
where Wε = 2 , 1 × [0, ε] . Now, for (x1 , x2 ) ∈ Wε and (y1 , y2 , y3 ) ∈ Kε ,
|x1 y1 + x2 y2 + ϕ (x1 , x2 ) y3 | ≤ 1
so
b
fε (x1 , x2 , ϕ (x1 , x2 ))
Z
−i(x1 y1 +x2 y2 +ϕ(x1 ,x2 )y3 )
=
e
dy1 dy2 dy3 Z Kε
≥
cos (x1 y1 + x2 y2 + ϕ (x1 , x2 ) y3 ) dy1 dy2 dy3 ≥ cε−1−l .
Kε
Thus
(4.15)
1
kRfε kLq (Σ) ≥ cε−1−l+ q .
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 35, 11 pp.
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F OURIER R ESTRICTION E STIMATES TO M IXED H OMOGENEOUS S URFACES
9
The proposition follows from (4.14) and (4.15).
We note that in the case that (a + b) l > ab (for example ϕ (x1 , x2 ) = x42 (x21 + x42 )) the
portion of the type set corresponding to 43 < p1 ≤ 1 will be smaller than the region Ea,b .
Also, ϕ (x1 , x2 ) = x22 (x1 + x22 ) is an example where a = 2, b = 4, Hessϕ (x1 , x2 ) = −4x22
and if x2 = 0 and x1 6= 0, ϕx2 x2 (x1 , x2 ) = 2x1 6= 0. Again, since 12 = (a + b) l > ab = 8, we
get that the portion of the type set corresponding to 34 < p1 ≤ 1 will be smaller than the region
Ea,b .
Proposition 4.3. Let ϕ be a mixed
function satisfying (1.1) and (4.12) with l ≥ 2b .
homogeneous
If
3
4
≤
1
p
≤ 1 and
1
q
> (l + 1) 1 − p1 , then
A
R 0 p 3 q A ≤ c.
L (R ),L (Σ 0 )
Proof. Let (x01 , x02 ) ∈ A0 , if Hessϕ (x01 , x02 ) 6= 0, as in the proof of Theorem 4.1 we find a
neighborhood U of (x01 , x02 ) such that (4.1) holds. If Hessϕ (x01 , x02 ) = 0, by (4.13), either x02 =
0 or the polynomial Q given by Px1 x1 (l (l − 1) P + 2lx2 Px2 + x22 Px2 x2 ) − (lPx1 + x2 Px1 x2 )2
vanishes at (x01 , x02 ) . In the first case, using the fact that P (x1 , 0) 6= 0 for x1 6= 0, we get that
Px1 x1 l (l − 1) P − l2 Px21 x01 , 0 6= 0.
We take a neighborhood W1 of the point (x01 , 0) and Uk as in the proof of Theorem 4.1. So for
(x1 , x2 ) ∈ Uk ,
|Hessϕ (x1 , x2 )| ≥ c2−k(2l−2)
and so
2k(l−1)
d
U
k
.
σ (ξ1 , ξ2 , ξ3 ) ≤
1 + |ξ3 |
By the other side,
d
σ Uk (ξ1 , ξ2 , ξ3 ) ≤ 2−k
so for 0 ≤ τ ≤ 1,
d
U
k
σ (ξ1 , ξ2 , ξ3 ) ≤
2k(τ l−1)
(1 + |ξ3 |)τ
and by Remark 3
U k(τ l−1)
R k p 3 2 U ≤ cτ 2 2(1+τ )
k
L (R ),L (Σ )
for p =
2(1+τ )
2+τ
and so Hölder’s inequality implies, for 1 ≤ q < 2,
U 2−q
τ l−1
R k p 3 q U ≤ cτ 2k( 2(1+τ ) − 2q )
L (R ),L (Σ k )
and a computation shows that this exponent is negative for 1q > (l + 1) 1 − p1 . Thus
W
R 1 p 3 q W < ∞
(4.16)
L (R ),L (Σ 1 )
for 43 ≤ p1 ≤ 1 and (l + 1) 1 − p1 < 1q ≤ 1. Now we suppose Q (x01 , x02 ) = 0. We observe that
deg Q ≤ 2 deg P − 2 ≤ 2 (b − l) − 2 ≤ 2l − 2
and so Hessϕ (x1 , x02 ) vanishes at x01 with order at most 2l − 2. Then defining W2 and Vk as in
the proof of Theorem 4.1, we have
Hessϕ x1 , x02 ≥ 2−k(2l−2)
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 35, 11 pp.
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10
E. F ERREYRA AND M. U RCIUOLO
and as in the previous case we obtain
W
R 2 (4.17)
Lp (R3 ),Lq (ΣW2 )
for 34 ≤
(4.1).
1
p
≤ 1 and
1
q
<∞
> (l + 1) 1 − p1 . The proposition follows from (4.16), (4.17) and
From Proposition 4.3 and Remark 2 we obtain the following result, sharp up to the end points,
for 43 ≤ p1 ≤ 1.
b
Theorem 4.4. Let ϕ be a mixed homogeneous function
(1.1)
satisfying
and
(4.12) with l ≥ 2 .
If m = max l + 1, a+b+ab
, 34 ≤ p1 ≤ 1 and 1q > m 1 − p1 , then p1 , 1q ∈ E.
a+b
4.1. Sharp Lp − L2 Estimates. In [4] we obtain sharp Lp − L2 estimates for the restriction
of the Fourier transform to homogeneous polynomial surfaces in R3 . The principal tools we
used there were two Littlewood Paley decompositions. Adapting this proof to the setting of non
isotropic dilations we obtain the following results.
Lemma 4.5. Let
then
1 1
,
p 2
a+b+2ab
2a+2b+2ab
≤
1
p
≤ 1. If
A
R 0 p 3 2 A < ∞
L (R ),L (Σ 0 )
∈ E.
Proof. From (2.1), the lemma follows from a process analogous to the proof of Lemma 4.3 in
[4].
Theorem 4.6.
i) If ϕ is a mixed homogeneous
polynomial function satisfying the hypothesis of Theorem
a+b+2ab 1
4.1 then 2a+2b+2ab , 2 ∈ E.
a+b+2ab 2l+1 ii) Let p10 = max 2a+2b+2ab
, 2l+2 . If ϕ is a mixed homogeneous polynomial function
1 1
satisfying the hypothesis of Theorem 4.4 then p0 , 2 ∈ E.
Proof. i) If a+b+ab
≥ 3, i) follows from (4.3) and Lemma 4.5. The cases (a, b) = (3, 4) ,
a+b
(a, b) = (3, 5) and (a, b) = (4, 5) are solved in Remark 5, part ii). The cases (a, b) = (2, b)
with b odd or B = 0 are also included in Remark 5, part ii). For the remainder cases (2, b), we
observe that, if b > 6, from the proof of Theorem 4.1 we obtain
A
R 0 p 3 2 A < ∞,
(4.18)
L (R ),L (Σ 0 )
for
1
p
=
a+b+2ab
,
2a+2b+2ab
so i) follows from Lemma 4.5. For b = 6, as before we get
W
R 1 p 3 2 W < ∞,
L (R ),L (Σ 1 )
and
V R k p 3 2 V < ∞
L (R ),L (Σ k )
a+b+2ab
for k ∈ N, p1 = 2a+2b+2ab
. In a similar way to Lemma 4.3 of [4], we use a uni-dimensional
Littlewood Paley decomposition to obtain
W
R 2 p 3 2 W < ∞
L (R ),L (Σ 2 )
and then we have (4.18) for
1
p
=
a+b+2ab
.
2a+2b+2ab
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 35, 11 pp.
So i) follows from Lemma 4.5.
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F OURIER R ESTRICTION E STIMATES TO M IXED H OMOGENEOUS S URFACES
11
ii) From the proof of Proposition
4.3, we use auni-dimensional Littlewood Paley decomposition
a+b+2ab
1
to obtain (4.18) for p = max 2a+2b+2ab , 2l+1
, and ii) follows from Lemma 4.5.
2l+2
Remark 7. In [7] the authors obtain sharp estimates for the Fourier transform of measures σ
associated to surfaces Σ like ours, when ϕ is a polynomial function satisfiyng (1.1) and the
condition that ϕ and Hessϕ do not vanish simultaneously on B − {(0, 0)} . In these cases,
part i) of the above theorem follows from Remark 3. We observe that our hypotheses are less
restrictive, for example ϕ (x1 , x2 ) = x41 x22 + x10
2 satisfies the hypothesis of part i) of the above
theorem but ϕ and Hessϕ vanish at any (x1 , x2 ) with x2 = 0.
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J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 35, 11 pp.
http://jipam.vu.edu.au/