Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 3, Article 89, 2006
CONVOLUTION OPERATORS WITH HOMOGENEOUS SINGULAR MEASURES
ON R3 OF POLYNOMIAL TYPE. THE REMAINDER CASE.
MARTA URCIUOLO
FAMAF -C IEM , U NIVERSIDAD NACIONAL DE C ÓRDOBA -C ONICET.
M EDINA A LLENDE S / N C IUDAD U NIVERSITARIA
5000, C ÓRDOBA , A RGENTINA .
urciuolo@mate.uncor.edu
Received 22 December, 2005; accepted 23 September, 2006
Communicated by A. Fiorenza
A BSTRACT. Let ϕ (y1 , y2 ) = y2l P (y1 , y2 ) where P is a polynomial function
R of degree l such
that P (1, 0) 6= 0. Let µδ be the Borel measure on R3 defined by µδ (E) = Vδ χE (x, ϕ (x)) dx
where
Vδ = x = (x1 , x2 ) ∈ R2 : |x1 | ≤ 1, and |x1 | ≤ δ |x2 |
and let Tµδ be the convolution operator with the measure µδ . In this paper we explicitely describe
the type set
1 1
Eµδ :=
,
∈ [0, 1] × [0, 1] : kTµδ kp,q < ∞ ,
p q
for δ small enough.
Key words and phrases: Convolution operators, Singular measures.
2000 Mathematics Subject Classification. 42B20, 26B10.
1. I NTRODUCTION
Let ϕ : R2 → R be a homogeneous polynomial function of degree m ≥ 2 and let D =
{y ∈ R2 : |y| ≤ 1} . Let µ be the Borel measure on R3 given by
Z
(1.1)
µ (E) =
χE (y, ϕ (y)) dy
D
3
and
Tµ be the operator defined, for f ∈ S (R ) , by Tµ f = µ∗f. Let Eµ be the set of the pairs
let
1 1
,
∈ [0, 1] × [0, 1] such that there exists a positive constant c satisfying kT f kq ≤ c kf kp
p q
for all f ∈ S (R3 ) , where the Lp spaces are taken with respect to the Lebesgue measure on R3 .
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
Partially supported by Conicet, Agencia Córdoba Ciencia, Agencia Nacional de Promoción Científica y Tecnológica y Secyt-UNC.
The author is deeply indebted to Prof. F. Ricci for his useful suggestions.
374-05
2
M ARTA U RCIUOLO
For
1 1
,
p q
q
3
∈ Eµ , T can be extended to a bounded operator, still denoted by T, from Lp (R3 )
into L (R ) .
Let ϕ = ϕe11 ...ϕenn be a decomposition of ϕ in irreducible factors with ϕi - ϕj for i 6= j. In [3]
we could give a complete description of the set Eµ under the assumption that ei 6= m2 for each
ϕi of degree 1. If det ϕ00 (y) is not identically zero and if it vanishes somewhere on R2 − {0},
the set of the points y where det ϕ00 (y) vanishes is a finite union of lines L1 , ..., Lk through the
origin. So, after a possibly linear change of variables, we localized the problem to the x axes
and we studied the type set corresponding to measures µδ defined by
Z
µδ (E) =
χE (y, ϕ (y)) dy,
Vδ
where Vδ = D ∩ {(y1 , y2 ) ∈ R2 : |y2 | ≤ δ |y1 |} and δ is small enough such that det ϕ00 (y) only
vanishes, on Vδ , along the x axes. The only case left was the one corresponding to functions ϕ
of the form ϕ (y1 , y2 ) = y2l P (y1 , y2 ) with l = m2 , P being a homogeneous polynomial function
of degree l such that P (1, 0) 6= 0 .
In this paper we characterize Eµδ in this remainder case.
Lp improving properties of convolution operators with singular measures supported on hypersurfaces in Rn have been widely studied in [2], [5], [6]. In particular, in [5], the type set was
studied under our actual hypothesis, but the endpoint problem was left open there. Our proof
of the main result involves a biparametric family of dilations and will be based on a suitable
adaptation of arguments due to M. Christ, developed in [1], where the author studied the type
set associated to the two dimensional measure supported on the parabola.
Also, oscillatory integral estimates are involved. A very careful study of this kind of estimate
can be found in [4] where the authors study the boundedness of maximal operators associated
to mixed homogeneous hypersurfaces.
Throughout this paper c will denote a positive constant, not the same at each occurrence.
2. T HE M AIN R ESULT
We assume ϕ (y1 , y2 ) = y2l P (y1 , y2 ) , where l = m2 and P is a homogeneous polynomial
function of degree l such that P (1, 0) 6= 0. We take δ1 > 0 such that, for y ∈ Vδ1 such that y2 6=
0, det ϕ00 (y) 6= 0. Moreover, since P (1, 0) 6= 0 we can assume that P (y) 6= 0 and P1 (y) 6=0
for all y ∈ Vδ1 . Now, if maxVδ1 |P2 (y1 , y2 )| =
6 0, we choose δ < min
l minVδ |P (y1 ,y2 )|
1
2 maxVδ |P2 (y1 ,y2 )|
, δ1 .
1
In the other case we take δ = δ1 .
The main result we prove is the following.
Theorem 2.1. Let ϕ (y1 , y2 ) = y2l P (y1 , y2 ) where l = m2 and P is a homogeneous polynomial
function of degree l such that P (1, 0) 6= 0 and y2 - P (y1 , y2 ) . Let Vδ be defined as above and
let EVδ be the corresponding
type set. Then
EVδ is the closed polygonal region with vertices
2l+1 2l−1
3
1
(0, 0) , (1, 1) , 2l+2 , 2l+2 and 2l+2 , 2l+2 .
Standard arguments (see, for example Lemma 2 and Lemma 3 in [3]) imply the following
result.
1 1
1
Lemma 2.2. If p , q ∈ Eµδ then 1q ≤ p1 , 1q ≥ p3 − 2 and 1q ≥ p1 − l+1
.
So, since kTµδ k1,1 < ∞, by duality arguments it only remains to prove that
(2.1)
J. Inequal. Pure and Appl. Math., 7(3) Art. 89, 2006
kTµδ k 2l+2
, 2l+2
2l+1 2l−1
< ∞.
http://jipam.vu.edu.au/
C ONVOLUTION O PERATORS WITH H OMOGENEOUS S INGULAR M EASURES
3
δ δ
, 8 . We
function θ ∈ C ∞ (R2 ), θ (y1 , y2 ) ≥ 0,
We set Q0 = 41 , 2 × 64
1 take
atruncation
δ
δ
supp θ ⊂ Q0 and θ (y1 , y2 ) = 1 on 2 , 1 × 32
, 16
. We define, for ε, γ > 0, the biparametric
2
3
family of dilations
on R and R given by (ε, γ)◦(y1 , y2 ) = (εy1 , γy2 ) and (ε, γ)◦(y1 , y2 , y3 ) =
εy1 , γy2 , εl γ l y3 repectively. Also, for j, k ≥ 0, we set Qj,k = 2−j , 2−k ◦ Q0 .
For f ∈ S (R3 ) , we define
Z
(2.2)
Tj,k f (x1 , x2 , x3 ) = f (x1 − y1 , x2 − y2 , x3 − ϕ (y1 , y2 )) θ 2j y1 , 2k y2 dy1 dy2
so for f ≥ 0,
Tµ δ f ≤ c
(2.3)
8
To study
P
X
Tj,k f.
0≤j≤k
Tj,k f, we will adapt the argument developed by M. Christ (see [1]) to the setting
0≤j≤k
of biparametric dilations. First of all, we prove the following
Proposition 2.3. There exists a positive constant c > 0 such that for 0 ≤ j ≤ k,
kTj,k k 2l+2 , 2l+2 ≤ c.
2l+1 2l−1
Proof.
Tj,k f (x1 , x2 , x3 )
Z
= f (x1 − y1 , x2 − y2 , x3 − ϕ (y1 , y2 )) θ 2j y1 , 2k y2 dy1 dy2
Z
−(j+k)
=2
f x1 − 2−j y1 , x2 − 2−k y2 , x3 − ϕ 2−j y1 , 2−k y2 θ (y1 , y2 ) dy1 dy2
= 2−(j+k) T (j−k) fj,k 2j x1 , 2k x2 , 2(j+k)l x3 ,
where we denote
T
(j)
Z
f (x1 , x2 , x3 ) =
f x1 − y1 , x2 − y2 , x3 − y2l P y1 , 2j y2
θ (y1 , y2 ) dy1 dy2
and
fj,k (x1 , x2 , x3 ) = f
2−j , 2−k ◦ (x1 , x2 , x3 ) .
So
(2.4)
kTj,k f (x1 , x2 , x3 )kq = 2(j+k)(
Now,
1+l
− 1+l
−1
p
q
)
T (j−k) p,q
kf kp .
00
= 2(2−2l)(j−k) det (ϕ)00 y1 , 2j−k y2 .
so as in the proof of Lemma 4 in [3] we obtain that there exists c > 0 such that T (j−k) 2l+2
det y2l P y1 , 2j−k y2
, 2l+2
2l+1 2l−1
c for 0 ≤ j ≤ k, and the proposition follows.
≤
We take 0 ≤ j ≤ k, and denote by µj,k and µ(j) the measures associated to Tj,k and T (j)
respectively. For ξ = (ξ1, ξ2 , ξ3 ) ,
Z
l
j−k
\
(j−k)
µ
(ξ) = e−i(ξ1 y1 +ξ2 y2 +ξ3 y2 P (y1 ,2 y2 )) θ (y1 , y2 ) dy1 dy2 .
(j−k)
If for some ξ on the unit sphere, Ωξ
point belonging to the supp θ, then
(y1 , y2 ) = ξ1 y1 + ξ2 y2 + ξ3 y2l P y1 , 2j−k y2 has a critical
ξ1 + ξ3 y2l P1 y1 , 2j−k y2 = 0
J. Inequal. Pure and Appl. Math., 7(3) Art. 89, 2006
http://jipam.vu.edu.au/
4
M ARTA U RCIUOLO
and
ξ2 + ξ3 2j−k y2l P2 y1 , 2j−k y2 + ly2l−1 P y1 , 2j−k y2 = 0,
but then, since P1 (y) 6= 0 for y ∈ Vδ1 , from the first equation we obtain that there exist constants
a, b ∈ Z with a < b such 2a |ξ3 | ≤ |ξ1 | ≤ 2b |ξ3 | , and, from the second one and the choice of δ
we obtain constants c, d ∈ Z2 with c < d such that 2c |ξ3 | ≤ |ξ2 | ≤ 2d |ξ3 |. So ξ belongs to the
cone
C0 = ξ ∈ R3 : 2a |ξ3 | < |ξ1 | < 2b |ξ3 | , 2c |ξ3 | < |ξ2 | < 2d |ξ3 | .
Lemma 2.4. Suppose C0 is as above. Then the family of cones 2j , 2k ◦ C0 j,k∈Z has finite
overlapping (i.e., # (j, k) ∈ Z2 : C0 ∩ 2j , 2k ◦ C0 6= ∅ < ∞).
Proof. We suppose ξ ∈ C0 and 2j , 2k ◦ ξ ∈ C0 , then
2a |ξ3 | < |ξ1 | < 2b |ξ3 | ,
2c |ξ3 | < |ξ2 | < 2d |ξ3 |
and
2(j+k)l+a |ξ3 | < 2j |ξ1 | < 2(j+k)l+b |ξ3 | ,
2(j+k)l+c |ξ3 | < 2k |ξ2 | < 2(j+k)l+d |ξ3 |
so
2j |ξ1 | < 2(j+k)l+b |ξ3 | < 2(j+k)l+b−a |ξ1 |
and
2b |ξ3 | > |ξ1 | > 2−j 2(j+k)l+a |ξ3 | ,
so
a − b − kl < j (l − 1) < b − a − kl,
analogously we obtain
c − d − jl < k (l − 1) < d − c − jl,
thus
(c − d) (l − 1) + (a − b) l
(d − c) (l − 1) + (b − a) l
<k<
2
l2 − (l − 1)
l2 − (l − 1)2
and so
a−b
(b − a)
(d − c) (l − 1) + (b − a) l
(d − c) (l − 1) + (b − a) l
−l
<j<
+l
.
2
l−1
l−1
l2 − (l − 1) (l − 1)
l2 − (l − 1)2 (l − 1)
We define m0 (ξ) = n (ξ1 , ξ3 ) r (ξ2 , ξ3 ) where n and r belong to C ∞ (R2 − {0}) , are homogeneous of degree zero with respect to the isotropic dilations,
supp n ⊂ (ξ1 , ξ3 ) : 2a−1 |ξ3 | < |ξ1 | < 2b+1 |ξ3 |
n ≥ 0 and n ≡ 1 on (ξ1 , ξ3 ) : 2a |ξ3 | < |ξ1 | < 2b |ξ3 | ,
supp r ⊂ (ξ2 , ξ3 ) : 2c−1 |ξ3 | < |ξ2 | < 2d+1 |ξ3 | ,
r ≥ 0 and r ≡ 1 on (ξ2 , ξ3 ) : 2c |ξ3 | < |ξ2 | < 2d |ξ3 | , so m0 is homogeneous of degree zero
with respect to the isotropic dilations, it belongs to C ∞ on each octant of R3 , m0 ≥ 0, m0 ≡ 1
on C0 and
f0 = ξ ∈ R3 : 2a−1 |ξ3 | < |ξ1 | < 2b+1 |ξ3 | , 2c−1 |ξ3 | < |ξ2 | < 2d+1 |ξ3 | .
supp m0 ⊂ C
For (j, k) ∈ Z2 , we define mj,k (ξ) = m0 2−j , 2−k ◦ ξ and Qj,k the operator with multiplier
mj,k . If ξ belongs to an open octant of R3 then ξ belongs to 2j , 2k ◦ C0 for some (j, k) ∈ Z2
1|
2|
(indeed 2−k ∼ |ξ
and 2−j ∼ |ξ
) and from the previous lemma, it belongs to a finite number of
|ξ3 |
|ξ3 |
J. Inequal. Pure and Appl. Math., 7(3) Art. 89, 2006
http://jipam.vu.edu.au/
C ONVOLUTION O PERATORS WITH H OMOGENEOUS S INGULAR M EASURES
P
them (independent of ξ). So
5
mj,k (ξ) ≤ c. Now it is easy to check that, for 1 < p < ∞,
(j,k)∈Z2
there exists Ap > 0 such that for f ∈ L2 ∩ Lp and any choice of εj,k = ±1,
X
≤ Ap kf k .
(2.5)
ε
Q
f
j,k
j,k
p
(j,k)∈Z2
p
Indeed, we now show that
X
m (ξ) =
εj,k mj,k (ξ)
(j,k)∈Z2
satisfies the hypothesis of the Marcinkiewicz Theorem, as stated in Theorem 6’ in [7].
We have just observed that
X
|m (ξ)| ≤
mj,k (ξ) ≤ c.
(j,k)∈Z2
Now we want to estimate ∂ξ∂1 m (ξ) . We recall that ∂ξ∂1 m0 is homogeneous of degree −1.
We pick ξ in an open octant. In a small neighborhood of ξ only finitely many (j, k) ∈ Z2
(independent of ξ) are involved. For each one of them,
∂
∂
mj,k (ξ) = 2−j
m0 2−j ξ1 , 2−k ξ2 , 2−(j+k)l ξ3
∂ξ1
∂ξ1
−1
−1
≤ c2−j 2−j ξ1 , 2−k ξ2 , 2−(j+k)l ξ3 ≤ c2−j 2−j ξ1 ,
so
Z
2s+1
sup
ξ2 ,ξ3
2s
∂
∂ξ1 m (ξ) dξ1 ≤ c,
and in a similar way (using the homogeneity of the derivatives of mj,k ) we obtain that for each
0 < k ≤ 3,
Z ∂k
dξ1 ≤ c,
sup
m
(ξ)
ξk+1 ,...,ξ3 ρ ∂ξ1 ...∂ξk
as ρ ranges over dyadic rectangles of Rk and that this inequality holds for every one of the six
pemutations of the variables ξ1 , ξ2 , ξ3 .
We now define h (ξ) ∈ C ∞ (R3 ) , h ≥ 0, h ≡ 1 on the unit ball of R3 , hj,k (ξ) = h 2−j , 2−k ◦ ξ
and Rj,k the operators with multipliers hj,k .
Lemma 2.5. There exists a constant C > 0, independent of K, such that
X
Tj,k Rj,k ≤ C.
0≤j≤k≤K
2l+2 , 2l+2
2l+1 2l−1
Proof. Let Kj,k be the kernel of Tj,k Rj,k . A computation shows that,
Kj,k (x) = 2(j+k)l µ(j−k) ∗ hˆ∨
2j , 2k ◦ x .
Thus
X
|Kj,k (ξ)| ≤
0≤j≤k≤K
J. Inequal. Pure and Appl. Math., 7(3) Art. 89, 2006
X
2(j+k)l G(j,k)
2j , 2k ◦ ξ 0≤j≤k
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6
M ARTA U RCIUOLO
∧
∧
with G(j,k) defined by G(j,k) = µ(j−k) h. Since j − k ≤ 0, as in Lemma 7 in [3] we obtain
∧
that G(j,k) ∈ S (R3 ) with each seminorm bounded on j, k, it follows that the same holds for
G(j,k) . Now
X
X
2ja+ka+ha G(j,k,h) 2j ξ1 , 2k ξ2 , 2h ξ3 2(j+k)l G(j,k) 2j , 2k ◦ ξ ≤
0≤j≤k
j,k,h≥0
l
with a = l+1
, G(j,k,h) = G(j,k) for h = l (j + k) and G(j,k,h) = 0 otherwise. It is well known
that from the uniform boundedness properties of G(j,k,h) it follows that
X
c
2ja+ka+ha G(j,k,h) 2j ξ1 , 2k ξ2 , 2h ξ3 ≤
,
a
|ξ1 | |ξ2 |a |ξ3 |a
j,k,h≥0
so
X
|Kj,k (ξ)| ≤
0≤j≤k≤K
P
so
c
|ξ1 |
l
l+1
l
l
,
|ξ2 | l+1 |ξ3 | l+1
Tj,k Rj,k convolves Lp (R3 ) into Lq (R3 ) for
0≤j≤k≤K
1
q
=
1
p
−
1
l+1
with bounds independent
of K.
Lemma 2.6. There exists a constant C > 0, independent of K, such that
X
Tj,k (I − Pj,k ) (I − Qj,k )
≤ C.
1≤j≤k≤K
2l+2 , 2l+2
2l+1 2l−1
Proof. The kernel Hj,k of
X
Tj,k (I − Pj,k ) (I − Qj,k )
1≤j≤k≤K
satisfies
X
|Hj,k (ξ)| ≤
X
2(j+k)l g (j,k)
2j , 2k ◦ ξ 0≤j≤k
1≤j≤k≤K
(j,k) ∧
∧
with g (j,k) defined by g
= µ(j−k) (1 − h) (1 − m0 ) .
∧
Observe that, from Lemma 7 in [3], we have µ(j−k) (1 − h) (1 − m0 ) ∈ S(R3 ) with each
seminorm bounded on j, k. From this fact the proof follows as in the previous lemma.
Proof of the theorem. We have just observed that it is enough to prove (2.1). Since we can
suppose f ≥ 0, by (2.3), we need only check that there exists C > 0, independent of K such
that
X
≤ C,
Tj,k 0≤j≤k≤K
2l+2 , 2l+2
2l+1 2l−1
where Tj,k are defined by (2.2). For a constant c0 > 0, we define Q0j,k =
P
Qi,k . So Q0j,k
|i−j|≤c0
have the same properties as Qj,k and Q0j,k ◦ Qj,k = Qj,k thus we have that (2.5) holds for Q0j,k .
Then, for 1 < p < ∞ and
X
Q0j,k fj,k ≤ cp kF kLp (l2 ) .
F = {fj,k }j,k≥0 ∈ Lp l2 ,
j,k≥0
J. Inequal. Pure and Appl. Math., 7(3) Art. 89, 2006
p
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C ONVOLUTION O PERATORS WITH H OMOGENEOUS S INGULAR M EASURES
We decompose
X
Tj,k f =
0≤j≤k≤K
X
Tj,k (I − Pj,k ) I − Q0j,k f +
0≤j≤k≤K
X
7
Tj,k Pj,k f
0≤j≤k≤K
+
X
Tj,k Q0j,k (I − Pj,k ) f.
0≤j≤k≤K
Now, proceeding as in [1], the theorem follows from Proposition 2.3, Lemmas 2.5 and 2.6 and
the remarks in [8, p. 85] concerning the multiparameter maximal function.
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J. Inequal. Pure and Appl. Math., 7(3) Art. 89, 2006
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