L²-estimates for some maximal functions
2 L -ESTIMATES FOR SOME MAXIMAL FUNCTIONS
By Hendra Gunawan
UNTO
S . Si presentono alcuni risultati sugli operatori massimali associati alle mis-
n
ure superficiali in R . E. M. Stein [9] ha iniziato lo studio di quest’argomento, dimostrando una disuguaglianza a priori per la funzione massimale sferica, medi- ante l’utilizzo delle funzioni “g”. In [3], M. Cowling e G. Mauceri hanno generaliz- zato il risultato di Stein. Qui si d`a una dimostrazione diversa e forse pi` u semplice della stima a priori, utilizzando la transformata di Mellin, come suggerito in un altro lavoro di Cowling e Mauceri [2].
NTRODUCTION
I
n
For a locally integrable function f on R , one can define Z
1
n
x Mf (x) = sup |f (x − ry)| dy, ∈ R
n
- n
m (B )
∈R r B n n n
where B is the unit ball in R and m(B ) is its measure. This is known as the maximal
p
function of f . One basic property of Mf is that it satisfies the L -inequality k Mf k ≤ C k f k
p p p p n whenever f ∈ L (R ), 1 < p ≤ ∞. This was first studied by G. H. Hardy and J. E.
Littlewood [4] for n = 1 and, for the general case, by N. Wiener [12] and by J. Marcinkiewicz AMS Classification: 42B25. and A. Zygmund [7]. Covering lemmas were used to prove the result. Marcinkiewicz [6] later proved this via an interpolation theorem. For a brief introduction to the theory of maximal functions we refer the reader to Stein [8, Ch. 1].
n
If f is smooth and rapidly decreasing, then one can replace B in the definition of
n −1
Mf by the unit sphere S and thus obtain the spherical maximal function of f . Stein [9] developed the theory and proved that, for n ≥ 3, the a priori inequality still holds
p n n
provided f ∈ L (R ), < p ≤ ∞. The detailed proof of this can be found in Stein
n −1
and S. Wainger [10]; it involves the theory of the Fourier transform and an argument using g -functions.
Cowling and Mauceri [2] reproved Stein’s result and gave a new approach to the study of maximal functions. Here the maximal function of f , denoted now by M f , is defined
φ
by
n
M f (x) = sup |φ ∗ f (x)|, x ∈ R
φ r r ∈R
- n
where φ is a distribution on R and φ is its dilate; Mellin transform techniques were used
r to tackle the problems.
A generalisation of Stein’s treatment of spherical maximal functions was also found
2
by Cowling and Mauceri [3]. One of their results, namely the L -estimate, asserts that for
n
any smooth function f on R , the inequality k M f k ≤ C k f k
φ
2
2
- n −α
holds whenever φ is compactly supported in R and | b φ (rξ)| ≤ C (1 + r) , r ∈ R , ξ ∈
n −1
1 S
, for some α > . This was proved via a study of Riesz operators, the theory
2
of fractional differentiation, some properties of Bessel functions, and an argument of g- functions.
The aim of this note is to prove a similar result for surface measures, using Mellin transform techniques. Throughout this note we shall use the following notation. The Fourier transform of a
n
function f on R is given by Z
iξx n f b f dx, ξ .
(ξ) = (x) e ∈ R
n R S(R ) denotes the space of all smooth and rapidly decreasing functions on R . By ex- , C pressions C, C k k,l etc. we mean various constants which may vary from line to line. These constants usually depend on n — the dimension of the ambient space.
- n
Let φ be a distribution on R . For r ∈ R , we define the dilate φ by duality
r
Z Z
n
φ φ f (x) f (x) dx = (x) f (rx) dx, ∈ S(R ).
r n n
R R
One may observe that
- n −1 φ b φ r , ξ .
(ξ) = b (rξ), ∈ R ∈ S
r n
f Now, for each f ∈ S(R ), define the associated maximal function M by
φ n
M f (x) = sup |φ ∗ f (x)|, x ∈ R .
φ r
- r ∈R
2 The focus of our interest is the L -inequality n
k M f k ≤ C k f k , f ∈ S(R ).
φ
2
2 Using the Mellin transform, we write
Z
- iu
b φ (ξ) = ψ (u, ξ) r du, r ∈ R ,
r R
where
∞
Z
1
−1−iu
ψ φ b dr, u (u, ξ) = (ξ) r ∈ R.
r
2π (See [2] for more about the techniques.) For s > 0, one can observe that
iu n ψ ψ u .
(u, sξ) = s (u, ξ), ∈ R, ξ ∈ R
n
Moreover, from the theory of the Fourier transform, for all f in S(R ), b (φ ∗ f ) b = b φ f
r r
whence b φ ∗ f = ( b φ f ) ˇ
r r
¾ ½Z
ˇ
iu
= ψ (u, .) b f r du
R
Z ª
iu
© ˇ ψ r du. = (u, .) ∗ f
R
(Here ˇ denotes the inverse Fourier transform.) Hence we find that Z
n f ψ x .
M φ (x) ≤ | ˇ (u, .) ∗ f (x)| du, ∈ R
R
Applying Minkowski’s inequality and Plancherel’s theorem, we obtain Z f ψ du k M k
2 ≤ k ˇ (u, .) ∗ f k
2 φ R
Z f du = k ψ(u, .) b k
2 R
Z f du ≤ k ψ(u, .) k ∞ k b k
2 R
Z
1
2
du = (2π) k f k
2 k ψ(u, .) k ∞ R
where u k ψ(u, .) k ∞ = sup |ψ(u, ξ)|, ∈ R.
n−1
ξ ∈S2 Thus, to have the L -estimate, it suffices to verify
Z k ψ(u, .) k du < ∞.
∞
RBut this would be satisfied when
−1−δ n −1
|ψ(u, ξ)| ≤ C(1 + |u|) , u ∈ R, ξ ∈ S , for some δ > 0. With this approach, therefore, we attempt to find conditions on φ such that this inequality holds.
We are grateful to Prof. M. Cowling for his suggestions during the preparation and the writing of this paper.
I In this part, we are concerned with some maximal operators associated to distributions
n
on R . Our results here are based on a study of Riesz operators, due to Cowling and Mauceri [3].
For a, b in C, with Re(a) > 0, and a distribution φ on R , we define the Riesz operator R
a,b via
Z
1
2
−1 2 −1 a b n
b (R φ )b(ξ) = s (1 − s ) φ (sξ) ds, ξ ∈ R .
a,b
Γ(b) This clearly makes sense for Re(b) > 0. Moreover, by analytic continuation, it extends to
- Re(b) > –N for any N ∈ Z (see [3] for justification).
The following lemma generalises [3, Lemma 1.2].
n Lemma 1.1 Suppose φ is a compactly supported distribution on R and for k = 0, 1, 2, . . .
there exist α > 0 such that
k k
¯ ¯ ∂
- −α k n −1
¯ ¯ b φ , r , ξ .
(rξ) (1 + r) ∈ R ∈ S
k
¯ ¯ ≤ C
k
∂r Then for Re(a ) > α − k, k = 0, 1, 2, . . . , and any δ > 0, we have
k k k
¯ ¯ ∂
−α +m+δ −1 k n
¯ ¯
- (R φ )b(rξ) (1 + r) , r ∈ R , ξ ∈ S ,
a k ,b k
¯ ¯ ≤ C
k
∂r where m = max(0, –Re(b)).
- n −1
Proof. For each k = 0, 1, 2, . . . , we have, whenever r ∈ R , ξ ∈ S , Z
1 k k
∂ 2 ∂
a −1
2 b −1k
b (R φ )b(rξ) = s (1 − s ) φ (srξ) ds
a k ,b k k
∂r ∂r
Γ(b)
1 Z k
∂
2
a k +k−1 2 b −1
s φ b = (1 − s ) (srξ) ds.
k
Γ(b) ∂ (sr) As in the proof of [3, Lemma 1.2], we find that for Re(a ) > α − k, and δ > 0,
k k k
¯ ¯ ∂
−α +m+δ −1 + k n
¯ ¯ φ , r , ξ ,
(R a ,b )b(rξ) k (1 + r) ∈ R ∈ S
k
¯ ¯ ≤ C
k
∂r where m = max(0, –Re(b)). ⊔ ⊓ Now we have the following theorem. Theorem 1.2 Suppose φ is a compactly supported distribution on R with b φ (0) = 0 such that ¯ ¯
−ǫ n −1
- φ (rξ) ¯ ≤ C (1 + r) , r ∈ R , ξ ∈ S and
¯b
¯ ¯ ∂
−1−ǫ n −1
¯ ¯
- φ b , r , ξ
(rξ) ∈ R ∈ S ¯ ¯ ≤ C (1 + r)
∂r
2
for some ǫ > 0. Then we have the L -estimate
n
f , f k M k
2 ≤ C k f k 2 ∈ S(R ). φ n −1
Proof.
For each ξ in S , write Z
- iu
b φ (rξ) = ψ (u, ξ) r du, r ∈ R ,
R
where
∞
Z
1
−1−iu
ψ φ b dr, u (u, ξ) = (rξ) r ∈ R.
2π
−1−δ We need to show that |ψ(u, ξ)| ≤ C (1 + |u|) , u ∈ R, for some δ > 0.
φ φ First, since b (0) = 0 and b is differentiable, we have
| b φ (rξ)| ≤ C r, 0 ≤ r ≤ 1, and thus, whenever u ∈ R, we obtain
∞
Z
−1
φ dr |ψ(u, ξ)| ≤ | b (rξ)| r
Z
1 Z ∞ −1−ǫ
- ≤ C dr C r dr
1 ≤ C.
Next, we invoke the identity φ = R R φ
a +2b,−b a,b
where Re(a) > 0, Re(a + 2b) > 0 (see [3, Lemma 1.3]), to have Z
1
∂ ∂
2
- 2b−1 2 −b−1 + a
b φ s φ r .
(rξ) = (1 − s ) (R a,b )b(srξ) ds, ∈ R ∂r
∂r Γ(−b) Applying Fubini’s theorem, we obtain
∞
Z
−1−iu
φ b dr 2π iu ψ(u, ξ) = − (rξ) (−iu) r
Z ∞ ∂
−iu
b φ r dr
= − (rξ) ∂r
Z ∞ ∂
−iu
b = φ (rξ) r dr
∂r Z ∞ Z
1
∂
2
- 2b−1 2 −b−1 −iu a
s φ dr = (1 − s ) (R a,b )b(srξ) ds r
∂r Γ(−b)
Z
1 Z ∞
2 ∂
−iu a +2b−1+iu 2 −b−1
= (R φ )b(srξ) (sr) d (sr) s (1 − s ) ds
a,b
Γ(−b) ∂ (sr) Z ∞ Z
1
∂
2
−iu a +2b−1+iu 2 −b−1
φ dt s ds = (R )b(tξ) t (1 − s )
a,b
Γ(−b) ∂t
- 2b+iu a Z ∞
Γ( ) ∂
2 −iu = (R φ )b(tξ) t dt. a,b a +iu
∂t Γ( )
2 ǫ
For Re(a) > ǫ and b = − , we have
4
¯ ¯
ǫ a +iu Z ∞
¯ ¯
ǫ
¯ + Γ(− ) ¯ ∂
−
4
2
¯ ¯
ǫ
4
¯ ¯ φ ,
|u ψ(u, ξ)| ≤ (R
a, − )b(tξ) |u| > 1,
¯
4 ¯ dt ≤ C |u| a +iu
¯ ¯ ∂t
Γ( )
2 π
1
¯ ¯
− |d| − ∂ c ǫ
2
2
¯ φ since |Γ(c + id)| ∼ C e |d| as |d| → ∞ (see [11, p. 151]) and (R )b(tξ) ¯ ≤
a, − ∂t 4 ǫ
- −1−
ǫ
(1 + t) for t ∈ R (by Lemma 1.1, with δ = ). We therefore find that
2 C
4 ǫ
−1−
4
, |ψ(u, ξ)| ≤ C |u| |u| > 1. Combining this with the previous inequality, we obtain
ǫ −1− −1 n
4
|ψ(u, ξ)| ≤ C (1 + |u|) , u ∈ R, ξ ∈ S , as desired. ⊔ ⊓ Remark. The condition b φ (0) = 0 in the theorem can in fact be removed. When φ does not
n
ϕ φ satisfy this condition, we can set g = φ − ϕ, where ϕ ∈ S(R (0) = b (0). Thus ) with b g satisfies the hypothesis, and hence the conclusion holds for M f . But this implies that
g
f f the conclusion also holds for M since M is known to be dominated by the standard
φ ϕ maximal function of f . Moreover, the theorem is also true for a distribution φ which is
1
φ not compactly supported but such that b is C .
II
n Here we deal with some maximal operators associated to surface measures on R .
We first outline the result of Cowling et al [1]. Let H be a smooth compact convex
∞ n
hypersurface of class C in R whose tangent lines have order of contact at most m < ∞. Suppose µ is a smooth surface measure on H. Then, it is shown in [1] that for all
- n −1
r ∈ R , ξ ∈ S , we have
−irp(ξ)·ξ −irp(−ξ)·ξ
µ (rξ) = F (r, ξ) e + F (r, −ξ) e + E(r, ξ) b where
|p(±ξ) · ξ| ≤ C,
k
¯ ¯ ∂
−k
¯ ¯ F r K k (r, ±ξ) (r), = 0, 1, 2, . . .
k
¯ ¯ ≤ C
k
∂r
−α
with K(r) = O(r ) as r → ∞, for some α > 0,
k
¯ ¯ ∂
−l
¯ ¯ and E (r, ξ) r , k, l = 0, 1, 2, . . . .
k,l
¯ ¯ ≤ C
k
∂r By rearranging F and E for 0 ≤ r ≤ 1, we may assume that F (0, ±ξ) = 0 (or even
- n −1
F (r, ±ξ) = 0, for 0 ≤ r < s < 1). If, in addition, we have for all r ∈ R , ξ ∈ S ,
k
¯ ¯ ∂
−k−α
¯ ¯ F r , k
(r, ±ξ) = 0, 1, 2, . . . ,
k
¯ ¯ ≤ C
k
∂r
1 for some α > , we then say that µ is ‘of good decay type’.
2 Our results are the following.
n
Theorem 2.1 Suppose that H is a hypersurface in R , and that µ is a surface measure
2
of good decay type on H. Then we have the L -estimate
n
f , f k M k ≤ C k f k ∈ S(R ).
µ
2
2 Theorem 2.2 Let µ be a measure on R , p be a bounded function on S , and 0 < ǫ < 3.
n −1
If, for each ξ in S , there exists a measure ν = ν ξ on R such that
1+ ǫ irp (ξ)·ξ 2 +
4
ν µ , r , (i) (rξ) e (1 + r ) ∈ R b (r) = b
Z
2
(ii) (1 + s ) |dν(s)| < C,
R
2
with C being independent of ξ, then we have the L -estimate
n
f , f k M µ k
2 ≤ C k f k 2 ∈ S(R ).
We shall prove later that Theorem 2.1 is in fact a special case of Theorem 2.2. Now, Theorem 2.2 follows from the lemma below. We are indebted to Prof. G. Mauceri for suggesting its proof.
n
Lemma 2.3 Let µ be a measure on R µ µ (0) = 0, p be a such that b (0) = 0 and ∇b
n −1 n −1
bounded function on S , and 0 < ǫ < 3. Suppose that for each ξ in S there exists a measure ν = ν on R such that
ξ 1+ ǫ
(ξ)·ξ +
2 irp
4
ν µ , r , (i) (rξ) e (1 + r ) ∈ R b
(r) = b Z
2
(ii) (1 + s ) |dν(s)| < C,
R
with C being independent of ξ. Define ψ = ψ by
ξ ∞
Z
1
−1−iu ψ (u) = µ (rξ) r dr, u ∈ R.
b 2π
Then we have
ǫ −1−
2
|ψ(u)| ≤ C (1 + |u|) , u ∈ R, R
2 with C = C(ǫ, p) (1 + s ) |dν(s)|. R
dr ≤
|b ν
)
− 1+ ǫ
4
, r ∈ R
|ψ(u)| ≤ Z ∞
|b µ
(rξ)| r
−1
dr ≤
Z
1
(r)| r
(1 + r
−1
dr
1
|b ν
(r)| (1 + r
2
)
− 1+ ǫ
4
r
−1
2
−irq
Z
∂b ν
Z ∞
Proof.
We note first that b ν is C from (ii). We also have b
ν (0) = 0 and b
ν (0) = 0. And, for 0 ≤ r ≤ 1, we have |b
ν (r)| = |b
ν (r) − b
ν (0)| (as b
ν (0) = 0) =
¯ ¯ ¯r
∂r (ρ)
(rξ) = b ν (r) e
¯ ¯ ¯
(for some 0 < ρ < r) ≤ r
Z
R
|s| |dν(s)| ≤ r
Z
R
(1 + s
2
) |dν(s)| ≤ C r.
Now, from (i), we have b µ
- , with q = p(ξ) · ξ. Thus, whenever u ∈ R,
- Z ∞
1 C dr +
∞
- such that ϕ(r) = 0 if 0 ≤ r ≤
4
− 1+ ǫ
4
r
−1−iu
dr
(r) e
−irq
ϕ (r) n
(1 + r
2
)
− 1+ ǫ
− 1+ ǫ
− r
2
2
o r
−1−iu
dr
−irq
ϕ (r) r
− 1+ ǫ
2
r
−1−iu
dr = ψ
1 (u) + ψ 2 (u) + ψ
3 (u), say.
)
{1 − ϕ(r)} (1 + r
1
3
kb ν k
r
−
3
2
dr ≤ C
Z
R
(1 + s
2
) |dν(s)| < ∞.
From here on we assume |u| > 1. Take a smooth function ϕ on R
2
−irq
and ϕ(r) = 1 if r ≥ 2. We then write 2π ψ(u) =
Z ∞ b ν
(r) e
−irq
(1 + r
2
)
−
1+ ǫ4
r
−1−iu
dr =
Z ∞ b ν (r) e
- Z ∞ b ν
- Z ∞ b ν (r) e
1 ∞
Z
1+ ǫ
∂
−irq 2 − −iu
4
ν r dr −iu ψ (u) = (r) e {1 − ϕ(r)} (1 + r )
1
b ∂r
Z ∞ h i
1+ ǫ
∂
−irq 2 − −iu
4
ν r dr, = − (r) e {1 − ϕ(r)} (1 + r ) b
∂r and similarly Z ∞
2
h 1+ ǫ i ∂
−irq 2 − 1−iu
1
b
4 −iu(1 − iu) ψ (u) = ν (r) e {1 − ϕ(r)} (1 + r ) r dr.
2
∂r
2 1+ ǫ
£ ¤
∂ −irq 2 −
b
∂r
4 But 1 − ϕ(r) = 0 for r ≥ 2, and 2 ν (r) e {1 − ϕ(r)} (1 + r ) is in fact a
linear combination of products of derivatives of the four factors, which are all bounded for R
2
ν are dominated by (1 + s ) |dν(s)|). Thus 0 ≤ r ≤ 2 (in particular, the derivatives of b R we obtain
2
¯ h 1+ ǫ i¯ ∂
2 Z
−irq
2 −¯ ¯
4
ν |iu(1 − iu) ψ (u)| ≤ (r) e {1 − ϕ(r)} (1 + r )
1
¯ b ¯ r dr ≤ C,
2
∂r which gives
−2 |ψ (u)| ≤ C |u| .
1 For ψ (u), we also have
2 Z ∞
2
h n oi
1+ ǫ 1+ ǫ
∂
−irq 2 − − 1−iu
4
2
b
2 −iu(1 − iu) ψ (u) = ν (r) e ϕ (r) (1 + r ) − r r dr.
2
∂r Put
1+ ǫ 1+ ǫ 1+ ǫ n 1+ ǫ o 2 − − − −2 − +
4
2
2
(r) = (1 + r ) − r = r (1 + r ) − 1 ∈ R
4 R , r .
1+ ǫ
3 −2 −
2 −4
4 For r ≥ , we can expand (1 + r ) into its convergent Taylor series to get
ǫ 1+
1 + ǫ 1 + ǫ 5 + ǫ r
−2 − −2
- 4 (1 + r ) = 1 − r − · · · .
4
4 4 2! Hence
5+ ǫ
1 + ǫ
3
−
(r) = − + terms of a lower power of r, ≥
2 R r r .
4
2
2 5+ ǫ
¯ ¯
3
2
¯ We therefore find that |R(r)|, | R (r)| and
2 R (r) ¯ ≤ C r for r ≥ . And thus, ∂r ∂r
2 −irq
ν , e and ϕ are bounded, we obtain since the derivatives of b
Z ∞
5+ ǫ −
2
|iu(1 − iu) ψ (u)| ≤ C r dr ≤ C,
2
3
2
which gives
−2 |ψ (u)| ≤ C |u| .
2 For ψ (u), we write
3 Z ∞ ǫ 3+
−irq − −iu
2
ψ (u) = ν (r) e ϕ (r) r dr
3
b Z ∞
1 + ǫ
−irq z −1
ν ϕ dr = (r) e (r) r (with z = − − iu) b
2 Z Z ∞
3
−1 −ir(s+q) z
ϕ e dr dν = (r) r (s) (as ϕ(r) = 0 for 0 ≤ r ≤ ).
R
2 Then, for fixed s ∈ R, consider Z ∞
−1 −ir(s+q) z
I ϕ e dr, z (z) = (r) r ∈ C. We note that I(z) continues analytically into Re(z) ≤ 1. For 0 < Re(z) < 1, we write
Z ∞ Z ∞
z −1 −ir(s+q) z −1 −ir(s+q)
I (z) = + {ϕ(r) − 1} r e dr r e dr = I
4 (z) + I 5 (z), say.
Corresponding to I
4 (z), we have
Z Z ∞
−irq −1 z
ψ (z) = I (z) dν(s) = ν (r) e {ϕ(r) − 1} r dr,
4
4
b
R ′
ν ν (0) = 0. Hence which continues analytically into –2 < Re(z) < 1, since b (0) = 0 and b
Z ∞
3+ ǫ −irq − −iu
2
ψ (u) = ν (r) e {ϕ(r) − 1} r dr
4
b Z ∞ h i
1+ ǫ −irq − −1−iu
= (r) e {ϕ(r) − 1} r b Integrating by parts twice, we obtain
2 ν r dr.
Z ∞
2
h i
1+ ǫ
∂
−irq − 1−iu
−iu(1 − iu) ψ
4 (u) = (r) e {ϕ(r) − 1} r
b
2 ν r dr.
2
∂r But ϕ(r)−1 = 0 for r ≥ 2, and ¯ ¯
2 ∂r
2
£ b ν (r) e {ϕ(r)−1} r
1+ ǫ
2
¤¯ ¯ ≤ C r
1+ ǫ
2
for 0 ≤ r ≤ 2 since again the derivatives of b ν
, e
−irq
and ϕ are bounded. Hence we find that |iu(1 − iu) ψ
2 C r
(u)| ≤ Z
, z = − 1 + ǫ 2 − iu, t ∈ R.
The contour γ
Fig. 1 The contour γ
Γ(z), t 6= 0.
−z
dr = (it)
−irt
e
z −1
(see Fig. 2) for t < 0, one can observe that, as R → ∞, Z ∞ r
2
1 (see Fig. 1) for t > 0, or around the contour γ
By integrating h around the contour γ
−wt
1−ǫ
e
z −1
− iu. Let h be the function on C defined by h (w) = w
2
5 (z) dν(s), where z = − 1+ǫ
I
4
R
5 (u) =
. It remains to estimate ψ
4 (u)| ≤ C |u|
−2dr < ∞ (as 0 < ǫ < 3), whence |ψ
2
R
1 Fig. 2
5 (u)| =
{1 + (s + q)
|s + q|
1+ ǫ
2
|dν(s)| ≤ C |u|
−1− ǫ
2 Z R
|s + q|
1+ ǫ
2
|dν(s)| (as |u| > 1) ≤ C |u|
−1− ǫ
2 Z R
2
− iu)| Z
} |dν(s)| (as 0 < ǫ < 3) ≤ C |u|
−1− ǫ
2
(1 + q
2
) Z
R
(1 + s
2
) |dν(s)| ≤ C |u|
−1− ǫ
2 .
2 Hence
R
2
¯ ¯ ¯
Z ∞ r
Z
R
I
5 (z) dν(s)
¯ ¯ ¯
(with z = −
|ψ
2
− iu) =
¯ ¯ ¯
Z
R
z −1
1+ǫ
e
−ir(s+q)
dr dν (s) ¯ ¯ ¯
= ¯ ¯ ¯
Z
R
{i(s + q)}
−z
Γ(z) dν(s) ¯ ¯ ¯
≤ e
− π
2 u
|Γ(−
1+ǫ Combining this with the previous results, we obtain
ǫ −1−
2
|ψ(u)| ≤ C |u| , |u| > 1, R
2
with C = C(ǫ, p) (1 + s ) |dν(s)|. This completes the proof. ⊔ ⊓
R To prove Theorem 2.1, we adapt a result of H¨ormander [5, pp. 121-122].
Proposition 2.4 Suppose that Φ is a smooth function on R, and that for k = 0, 1, 2, . . .
k
¯ ¯ ∂
−k−ǫ
¯ ¯ Φ(r) (1 + |r|) , r ∈ R,
k
¯ ¯ ≤ C
k
∂r
1 k
1
ˇ for some ǫ > 0. It then follows that ˇ Φ ∈ L (R). In general, s Φ(s) ∈ L (R), for k = 0, 1, 2, . . . .
∞
Proof. First of all (see [5, p. 121]), there exists a function ϕ ∈ C (R) supported in
1
< {r : |r| < 2} such that
2 ∞
X
−j ϕ (2 r ) = 1, r 6= 0. j =−∞
For this ϕ, we have
∞
X
1
−j
ϕ r (2 ) = 0, 0 < |r| ≤
2
j =1
and
∞
X
−j ϕ (2 r ) = 1, |r| > 2. j =1
Let us put ½
P ∞
−j
ϕ r 1 − (2 ), if r 6= 0,
=1
jϕ (r) = 1, if r = 0.
∞
It is clear that ϕ ∈ C (R) and that
∞
X
−j
ϕ ϕ r r (r) + (2 ) = 1, ∈ R.
j =1
(1−2ǫ)j .
∂ϕ ∂r
2
Φ(r) ¯ ¯ ¯
∂ ∂r
r )
−j
ϕ (2
j
r ) Φ(r) + 2
−j
(2
¯ ¯ ¯
Z
R
Z
dr =
2
(r) ¯ ¯ ¯
j
Φ
∂ ∂r
j
¯ ¯ ¯2
R
dr ≤ C
2 j−1
(1−2ǫ)j
Z
o dr ≤ C 2
−2−2ǫ
|r|
2j
−1−2ǫ
|r|
j
−2ǫ
n |r|
≤|r|≤2 j+1
2 j−1
o dr ≤ C
≤|r|≤2 j+1
2
Φ(r) ¯ ¯
∂ ∂r
¯ ¯
2j
Φ(r) ¯ ¯ + 2
∂ ∂r
|Φ(r)| ¯ ¯
j
2
n |Φ(r)|
, and similarly Z
dr ≤ C 2
This enables us to decompose Φ into Φ = Φ +
R
2
∂r
2
∂
¯ ¯ ¯
R
Z
1 2π
ˇ Φ (s)| ≤
2
|Φ (r)| dr ≤ C and |s
Z
Hence | ˇ Φ (s)| ≤ C (1 + |s|)
1 2π
We see that Φ = ϕ Φ is smooth and compactly supported on R. Thus for all s ∈ R | ˇ Φ (s)| ≤
−j r ) Φ(r), r ∈ R (j = 0, 1, 2, . . . ).
(r) = ϕ(2
j
where Φ = ϕ Φ and Φ
j
Φ
j =1
X
∞
Φ (r) ¯ ¯ ¯ dr ≤ C.
−2
−2ǫ
2
|r|
≤|r|≤2 j+1
2
j−1Z
dr ≤ C
2
|Φ(r)|
≤|r|≤2 j+1
2
j−1Z
dr ≤ C
r ) Φ(r)|
, s ∈ R, yielding Z
−j
|ϕ(2
R
dr = Z
2
(r)|
j
|Φ
R
Now, for j = 1, 2, 3, . . . , we have Z
R | ˇ Φ (s)| ds < ∞.
- 2
- 2
- 2
2 2 (1−2ǫ)j
| ˇ Φ (s)| ds = |Φ (r)| dr ≤ C 2
j j R R
and Z Z
∂ ¯ ¯
2 ¯ ¯
2 j j (1−2ǫ)j
¯ ¯ ¯2 s ˇ Φ (s) ds = ¯2 Φ (r) dr ≤ C 2 .
j j R R ∂r
These give Z
¡ ¢
−j 2j
2 2 −2ǫj s ds .
2 1 + 2 | ˇ Φ (s)| ≤ C 2
j
RApplying the Cauchy-Schwartz inequality, we obtain
1
¾ ¾ ½Z
2 ½Z
2 −1
1 Z
¡ ¢ ¡ ¢
−j 2j
2 2 j 2j
2
s ds s ds | ˇ Φ (s)| ds ≤ 2 1 + 2 | ˇ Φ (s)| 2 1 + 2
j j R R R
nZ o
2 −ǫj
1 Z Z
−j −2 j
≤ C 2 −j −j
2 2 +
- ds ds s ds
|s|≤2 2 ≤|s|≤1 |s|≥1 −ǫj
. ≤ C 2
P ∞ ˇ
Since ˇ Φ = ˇ Φ Φ , we therefore have +
j j =1 ∞
X
1 ≤ k ˇ Φ k
1 k ˇ Φ j k
- k ˇ Φ k
1 ≤ C(ǫ) < ∞, =1 j
1 meaning that ˇ Φ ∈ L (R). k
∂ k
1
ˇ In general, the proof also applies to k Φ(r), and so we have s Φ(s) ∈ L (R), for all
∂r
k = 0, 1, 2, . . . . ⊔ ⊓ We now come to the proof of Theorem 2.1.
- n −1
Proof (of Theorem 2.1). For r ∈ R , ξ ∈ S , we have
−irp(ξ)·ξ −irp(−ξ)·ξ
µ (rξ) = F (r, ξ) e + F (r, −ξ) e + E(r, ξ), b with p, F , and E as prescribed. We assume here that F (r, ±ξ) = 0, for 0 ≤ r < s < 1, to
k ∂
F have k (r, ξ)| = 0, for all k = 0, 1, 2, . . . .
r =0 ∂r
- 1
- 1
, if r ≥ 0, F (−r, −ξ) (1 + r
(1 + r
2
1+ ǫ
F (r, ξ) r
(r) = ½
Writing b ν
4 , if r < 0.
1+ ǫ
)
2
4
)
1+ ǫ
)
2
F (r, ξ) (1 + r
(r) = ½
). Thus b ν
2
1
, if r < 0, for some 0 < ǫ < min(3, α −
4
1+ ǫ
−2
1+ ǫ
2
k
2
−k− ǫ
(r) ¯ ¯ ¯ ≤ C |r|
b ν
k
∂r
k
∂
|r| ≤ 1 and ¯ ¯ ¯
(r) ¯ ¯ ¯ ≤ C,
b ν
∂r
4
k
∂
, if r < 0, and applying Leibnitz’s formula, we obtain that for k = 0, 1, 2, . . . , ¯ ¯ ¯
4
1+ ǫ
)
−2
(1 + r
2
1+ ǫ
, if r ≥ 0, F (−r, −ξ) (−r)
)
(1 + r
For u ∈ R, ξ ∈ S , consider ψ
2π Z ∞
Let us fix ξ ∈ S
1 (u, ξ) + ψ 2 (u, ξ) + ψ 3 (u, ξ), say.
dr = ψ
−1−iu
E (r, ξ) r
2π Z ∞
dr
−1−iu
r
−irp(−ξ)·ξ
F (r, −ξ) e
dr
hereafter. By Theorem 1.2, we have |ψ
−1−iu
r
−irp(ξ)·ξ
Z ∞ F (r, ξ) e
1 2π
dr =
−1−iu
(rξ) r
Z ∞ b µ
1 2π
(u, ξ) =
n −1
3
irp (−ξ)·ξ
, and define b ν on R by b ν (r) =
((−r)(−ξ)) e
1
µ
, if r ≥ 0, b
4
1+ ǫ
)
2
(1 + r
1 (rξ) e irp (ξ)·ξ
½ b µ
−irp(ξ)·ξ
(u, ξ)| ≤ C (1 + |u|)
(rξ) = F (r, ξ) e
1
b µ
1 say. We put
. But since they are similar, it suffices to work with one of them, ψ
2
and ψ
1
It then remains to tackle ψ
, u ∈ R, for some δ > 0, assuming that E(0, ξ) = 0.
−1−δ
, |r| > 1. Hence, for all k = 0, 1, 2, . . . , we have
k
¯ ¯
ǫ
∂
−k−
¯ ¯
2
ν , r (r) ∈ R. ¯ b ¯ ≤ C (1 + |r|)
k
∂r
1
ν )ˇ ∈ L (R) (which assures us that ν defines
It follows from Proposition 2.4 that ν = (b
2
1
a measure on R), and further s ν (s) ∈ L (R). So we find that there exists a measure ν on R which satisfies the hypothesis of Lemma 2.3. The proof is therefore complete. ⊔ ⊓
K Remark.
The result extends to surface measures of class C for some finite K; K is sufficiently large so that the hypothesis
k
¯ ¯ ∂
−k−α n −1
¯ ¯
- F r , r , ξ ,
(r, ±ξ) ∈ R ∈ S
k
¯ ¯ ≤ C
k
∂r is satisfied for k = 0, 1, 2, 3 and 4.
EFERENCES
R [1] M. Cowling, S. Disney, G. Mauceri and D. M¨ uller, Damping oscillatory integrals, Invent.
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[3] M. Cowling and G. Mauceri, Inequalities for some maximal functions. II, Trans. Amer. Math. Soc. 296 (1986), 341-365. [4] G. H. Hardy and J. E. Littlewood, A maximal theorem with function-theoretic applica- tions
, Acta Math. 54 (1930), 81-116.
p [5] L. H¨ormander, Estimates for translation invariant operators in L -spaces , Acta Math.
104 (1960), 93-139. [6] J. Marcinkiewicz, Sur l’interpolation d’operations, C.R. Acad. des Sciences, Paris 208 (1939), 1272-1273.
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