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 ∈S

2 Thus, to have the L -estimate, it suffices to verify

  Z k ψ(u, .) k du < ∞.

  

R

  But 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 −1

k

  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+ ǫ

  4

  r

  −1−iu

  dr =

  Z ∞ b ν (r) e

  • Z ∞ b ν
  • Z ∞ b ν (r) e
For ψ (u), we have

  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|

−2

  dr < ∞ (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

j

  ϕ (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−1

  Z

  dr ≤ C

  2

  |Φ(r)|

  ≤|r|≤2 j+1

  

2

j−1

  Z

  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
Then, by Plancherel’s theorem, Z Z

  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

R

  Applying 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.

  

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  R [1] M. Cowling, S. Disney, G. Mauceri and D. M¨ uller, Damping oscillatory integrals, Invent.

  Math. 101 (1990), 237-260. [2] M. Cowling and G. Mauceri, On maximal functions, Rend. Sem. Mat. Fis. Mil. XLIX (1979), 79-87.

  [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.

  [7] J. Marcinkiewicz and A. Zygmund, On the summability of double Fourier series, Fund. Math. 32 (1939), 122-132.