Journal of Fourier Analysis and Applications https://doi.org/10.1007/s00041-018-9618-2

Characterization of H 1 Sobolev Spaces by Square Functions of Marcinkiewicz Type Shuichi Sato1 Received: 5 April 2017 © Springer Science+Business Media, LLC, part of Springer Nature 2018

Abstract We establish characterization of H 1 Sobolev spaces by certain square functions of Marcinkiewicz type. The square functions are related to the Lusin area integrals. Also, in the one dimensional case, the non-periodic version of the function of Marcinkiewicz is applied to characterize weighted H 1 Sobolev spaces. Keywords Sobolev space · Littlewood–Paley function · Marcinkiewicz function · Lusin area integral Mathematics Subject Classification Primary 46E35 · Secondary 42B25

1 Introduction We recall that a function  belongs to Mα , α > 0, if  is a bounded, compactly supported function on Rn satisfying Rn (x) d x = 1; if α ≥ 1, we further require that  (x)x γ d x = 0 for all γ with 1 ≤ |γ | ≤ [α], Rn

Communicated by Dachun Yang. The author is partly supported by Grant-in-Aid for Scientific Research (C) No. 16K05195, Japan Society for the Promotion of Science.

B 1

Shuichi Sato [email protected] Department of Mathematics, Faculty of Education, Kanazawa University, Kanazawa 920-1192, Japan

Journal of Fourier Analysis and Applications

where γ is a multi-index, γ = (γ1 , γ2 , . . . , γn ), |γ | = γ1 + γ2 + · · · + γn , x γ = γ γ x1 1 . . . xn n and [α] denotes the greatest integer not exceeding α. Let  G α ( f )(x) =

∞

| f (x) − t ∗ f (x)|2

0

dt

1/2 , α > 0,

t 1+2α

(1.1)

where  ∈ Mα and t (x) = t −n (t −1 x). p Let L w , 0 < p < ∞, be the weighted Lebesgue space with the norm  f  p,w  1/ p defined as  f  p,w = Rn | f (x)| p w(x) d x . When w = 1 (the unweighted case),  f  p,w is written simply as  f  p . Let w ∈ A p , 1 < p < ∞, where A p denotes the α, p weight class of Muckenhoupt (see [7]), and let α > 0. The Sobolev space Ww (Rn ) p n is defined to be the collection of functions f ∈ L w (R ) which can be expressed as p f = Jα (g) with g ∈ L w (Rn ); the norm is given by  f  p,α,w = g p,w , where Jα is the Bessel potential operator defined as Jα (g) = K α ∗ g with Kˆ α (ξ ) = (1 + 4π 2 |ξ |2 )−α/2 . The Fourier transform is defined as fˆ(ξ ) =

 Rn

f (x)e−2πix,ξ d x, x, ξ =

n 

xjξj.

j=1 α, p

It was proved in [17] that G α can characterize the weighted Sobolev spaces Ww , 1 < p < ∞. Theorem A Let 1 < p < ∞, 0 < α < n and w ∈ A p ,  ∈ Mα . Let G α be as in α, p p p (1.1). Then f ∈ Ww (Rn ) if and only if f ∈ L w and G α ( f ) ∈ L w ; furthermore,  f  p,α,w  f  p,w + G α ( f ) p,w , that is, c1  f  p,α,w ≤  f  p,w + G α ( f ) p,w ≤ c2  f  p,α,w for positive constants c1 , c2 independent of f . (See [17, Corollary 5.2].) A square function characterization of the Sobolev spaces of this type was established by [1]. It has been developed by [8,15]. Special cases of α, p Theorem A can be found in [15]. In [16] a square function characterization of Ww different from Theorem A was given when α = 2. Also, an alternative proof of a result of [8] using a pointwise relation between a square function of Marcinkiewicz type and one arising from the Bochner–Riesz operators can be found in [18]. We now state an application of Theorem A. Let B(x, t) be the ball centered at x with radius t: B(x, t) = {y ∈ Rn : |x − y| < t} and B0 = B(0, 1). Put χ0 = |B0 |−1 χ B0 ,

Journal of Fourier Analysis and Applications

where |B0 | denotes the Lebesgue measure of B0 and χ B0 is the characteristic function of B0 . Then χ0 ∈ Mα for α ∈ (0, 2). If  = χ0 in (1.1), G α ( f ) can be expressed as 



f (x) − −

∞

0

B(x,t)

1/2

2

dt

f (y) dy 1+2α , t

  where −B(x,t) f (y) dy = |B(x, t)|−1 B(x,t) f (y) dy. Theorem A applies to this square function for 0 < α < min(n, 2), as shown in [1] in the unweighted case. In this note we consider characterization of H 1 Sobolev spaces by square functions of Marcinkiewicz type. Let S(Rn ) be the Schwartz  class of rapidly decreasing smooth functions. We choose ϕ ∈ S(Rn ) satisfying Rn ϕ d x = 1, ϕ ≥ 0 and supp(ϕ) ⊂ B(0, 1). The function ϕ will be fixed in what follows. Let H p (Rn ), 0 < p < ∞, be the Hardy space of tempered distributions on Rn such that f ∗ ∈ L p (Rn ), where f ∗ (x) = supt>0 |ϕt ∗ f (x)|. The norm of f in H p (Rn ) is defined to be  f  H p =  f ∗  p . It is known that a different choice of such ϕ gives an equivalent norm. The space H p coincides with L p when 1 < p < ∞ and H 1 is contained in L 1 . We denote by S0 (Rn ) a dense subspace of H p (Rn ) consisting of those functions f in S(Rn ) such that fˆ = 0 outside a compact set not containing the origin (see [26, Chapters V and VII]). 1 and Let f ∈ L loc 

∞

ν( f )(x) = 0

dt | f (x + t) + f (x − t) − 2 f (x)| 3 t 2

Put μ( f ) = ν(I( f )), where I( f )(x) = Marcinkiewicz and we have

x 0

1/2 .

(1.2)

f (y) dy. Then μ( f ) is the function of

μ( f ) p  f  H p for f ∈ S0 (R) when 2/3 < p < ∞ (see [14, Theorem 4]). It follows that ν( f ) p  f  H p

(1.3)

if f ∈ S0 (R) when 2/3 < p < ∞. Let gψ be the Littlewood-Paley function on Rn defined as  gψ ( f )(x) =

0

∞

| f ∗ ψt (x)|2

dt t

1/2 ,

 with ψ ∈ L 1 (Rn ) satisfying Rn ψ(x) d x = 0. The Marcinkiewicz function μ was introduced by the author in 1938 (see [11]) in the setting of periodic functions. The non-periodic version μ( f ) can be realized as a Littlewood–Paley function gψ ( f ) on R with ψ(x) = χ[0,1] (x) − χ[−1,0] (x).

Journal of Fourier Analysis and Applications

We define the H 1 Sobolev space by W Hα 1 (Rn ) = { f ∈ H 1 (Rn ) : f = Jα (h) for some h ∈ H 1 (Rn )}. For f ∈ W Hα 1 , define  f W α 1 = h H 1 with f = Jα (h). This is well-defined since H

Jα is an injection from H 1 to H 1 . To see that Jα is injective on H 1 , suppose that Jα ( f ) = Jα (g) for f , g ∈ H 1 . Then by part (2) of Lemma 2.5 below, we have 

 (K α ∗ h)(y) f (y) dy =



=

(K α ∗ f )(y)h(y) dy



(K α ∗ g)(y)h(y) dy =

(K α ∗ h)(y)g(y) dy

for all h ∈ S(Rn ), which implies that f = g, since Jα is a surjection from S(Rn ) to S(Rn ). The estimates (1.3) indicates that the square function ν can characterize W H1 1 (R). Indeed, we have Theorem 1.2 below. In general dimensions, we can give a characterization of the space W Hα 1 (Rn ) in terms of Lusin area integral functions of Marcinkiewicz type. Let Iα be the Riesz potential operator defined by  Iα ( f )(ξ ) = (2π |ξ |)−α fˆ(ξ ),

f ∈ S(Rn ).

L α (ξ ) = (2π |ξ |)−α , 0 < α < n, where If L α (x) = τ (α)|x|α−n , then τ (α) =

(n/2 − α/2) π n/2 2α (α/2)

(see [23, Chapter V.1]). Let ψ (α) (x) = L α (x) − L α ∗ (x),

(1.4)

where  ∈ Mα , 0 < α < n. We note that ψ (α) ∈ L 1 (Rn ) (see [17, p. 37]). Define  Sψ (α) ( f )(x) =

0

 =

∞ ∞

0

B0

|ψt(α) ∗ f (x − t z)|2 dz

B(x,t)

(α) |ψt

∗ f (z)| dz t 2

dt t

−n

1/2

dt t

1/2 .

Then, by homogeneity of the Riesz potential,  Sψ (α) ( f )(x) =

0

 =

0

∞ ∞

B0

|Iα f (x − t z) − t ∗ Iα f (x − t z)|2 dz t −2α |Iα f (z) − t ∗ Iα f (z)| dz t 2

B(x,t)

−2α−n dt

t

dt t

1/2

1/2

.

Journal of Fourier Analysis and Applications

Also, let  Uα ( f )(x) = 

∞ 0

=

∞

| f (x − t z) − t ∗ f (x − t z)| dz t 2

B0

| f (z) − t ∗ f (z)| dz t 2

−2α−n dt

B(x,t)

0

−2α dt

1/2 (1.5)

t

1/2

t

.

Then Uα is available to characterize the Sobolev space W Hα 1 (Rn ). Theorem 1.1 Let Uα be as in (1.5). Suppose that n/2 < α < n,  ∈ Mα and that there exists β > 0 such that ˆ )| ≤ C(1 + |ξ |)−β , α + β > n. |(ξ Then the following two statements are equivalent: (1) f ∈ W Hα 1 (Rn ), (2) f ∈ H 1 (Rn ) and Uα ( f ) ∈ L 1 (Rn ). Further, we have  f W α 1  f  H 1 + Uα ( f )1 . H

We refer to [25] for a related previous work. We note that |χˆ0 (ξ )| ≤ C(1 + |ξ |)−(n+1)/2 . When n = 1, this can be seen by a direct computation. If n ≥ 2, this follows by applications of Theorem 3.3 and Lemmas 3.11, 4.13 of [24, Chapter IV]. Further, χ0 ∈ Mα , 0 < α < 2. Thus Theorem 1.1 is available for n = 1, 2, 3 when  = χ0 . ˆ )| ≤ C(1 + Also, if  is a bounded radial function with compact support, then |(ξ |ξ |)−(n−1)/2 , which easily follows from a formula in [24, Theorem 3.3, Chapter IV]. If  = χ0 in (1.5), then Uα ( f ) can be written, up to a constant factor, as 

∞

−

0

B(x,t)



f (z) − −

B(z,t)

1/2

2

−2α dt

f (y) dy dz t . t

 This may extend to the case of metric measure spaces, since the expression −B(x,t) f makes sense in general metric measure spaces. Thus, Theorem 1.1 suggests that we may define H 1 Sobolev spaces in metric measure spaces by considering H 1,∞ on metric measure spaces as the definition of Hardy spaces, where H 1,∞ denotes the atomic H 1 (see [10, Chap. 3] for H 1,∞ ). The square function Uα ( f ) with  = χ0 was used in [9] to characterize the classical Sobolev space W α, p (Rn ) with α ∈ (0, 2) and p ∈ (1, ∞). We can also consider weighted H 1 Sobolev spaces. Recall the weight class A1 of Muckenhoupt consisting of weights w such that M(w) ≤ Cw almost everywhere. We

Journal of Fourier Analysis and Applications

have denoted by M the Hardy–Littlewood maximal operator defined as M( f )(x) = sup |B| B

−1

 | f (y)| dy, B

where the supremum is taken over all balls B containing x. Let Hw1 , w ∈ A1 , be the subspace of L 1w such that f ∗ ∈ L 1w with the norm  f  Hw1 =  f ∗ 1,w . Similarly, p we can define Hw (Rn ) for p ∈ (0, 1) with w ∈ A1 to be the space of tempered p p distributions f such that f ∗ ∈ L w . Then S0 (Rn ) is dense in Hw (Rn ), 0 < p ≤ 1 (see [26]). Define W Hα 1 (Rn ) = { f ∈ Hw1 (Rn ) : f = Jα (h) for some h ∈ Hw1 (Rn )}. w

For f ∈ W Hα 1 , let  f W α 1 = h Hw1 with f = Jα (h), which is also well-defined w

Hw

since Jα is injective on Hw1 (see Lemma 2.5 (2) below and recall that Jα : S → S is onto). Here we note that Jα (h) ∈ Hw1 whenever h ∈ Hw1 . To see this, we first observe that Jα (h)∗ ≤ K α ∗ h ∗ , where we recall that K α is the kernel of Jα and K α > 0 (see [23, Chapter V, identity (26)]). Also, we have K α ∗ w ≤ C M(w) by applying suitably [23, Chapter III, Theorem 2], since the least decreasing radial majorant of K α is integrable (see [23, Chapter V, identities (29), (30)]). Thus Jα (h)∗ 1,w ≤ Ch ∗ 1,M(w) ≤ Ch ∗ 1,w = Ch Hw1 . In the one dimensional case, we have the following result. Theorem 1.2 Let ν be as in (1.2) and w ∈ A1 . Then we have the equivalence of the following two statements: (1) f ∈ W H1 1 (R), w

(2) f ∈ Hw1 (R) and ν( f ) ∈ L 1w (R). Also, we have  f W 1  f  Hw1 + ν( f )1,w . 1 Hw

Remark 1.3 The square function Sψ (α) ( f ) can be treated in some respects similarly to  Dα ( f )(x) =

dy |Iα ( f )(x − y) − Iα ( f )(x)| n+2α n |y| R 2

1/2 .

(See [3,21] and also [5,22] for Dα .) Remark 1.4 Let 0 < α < n/2 and 1 ≤ p < 2n/(2α + n) (we note that p < 2). Then, Sψ (α) is not bounded on L p , where ψ (α) is as in (1.4). (See Sect. 5 for a proof, which is based on [3]; see also [21] for the weak type estimates at p = 2n/(2α + n).)

Journal of Fourier Analysis and Applications

Remark 1.5 In Theorem 1.1, the condition α > n/2 is optimal in the sense that if 0 < α < n/2, the estimate Uα ( f )1 ≤ C f W α 1 does not hold, where Uα is as in H (1.5) with  ∈ Mα . (A proof can be found in Sect. 5.) Remark 1.6 The proof of Theorem A is based on the estimates gψ (α) ( f ) p,w

 f  p,w , w ∈ A p , 1 < p < ∞, where ψ (α) is as in (1.4). If gψ (α) ( f )1  f  H 1 , then we would be able to characterize W Hα 1 by G α . We shall prove Theorem 1.1 in Sect. 2 and Theorem 1.2 in Sect. 4. The proof of Theorem 1.1 is based on the equivalence of Sψ (α) ( f )1 and  f  H 1 , f ∈ S0 (Rn ) (Theorem 2.3). In proving Theorem 1.2, we shall show in Sect. 3 the estimates of the kind  f  Hw1 ≤ Cgψ ( f )1,w for f ∈ S0 (Rn ) (see Theorem 3.2), which generalizes a result of [27] to the case of weighted Hardy spaces.

2 Proof of Theorem 1.1 We will show Theorem 1.1 by means of the equivalence c f  H 1 ≤ Sψ (α) ( f )1 ≤ C f  H 1 which will be established in Theorem 2.3. The first inequality is obtained via a duality argument based on Lemma 2.2 below, while the converse inequality will come from the application of a Calderón–Zygmund estimate for vector valued kernels in [7, Chapter V, Corollary 3.10], whose crucial step is checking the Hörmander condition in Lemma 2.1 below. Lemma 2.1 Let ψ (α) be as in (1.4) with α and  as in Theorem 1.1. Then



 |x|>2|y|



2 dt 1/2

(α)

(α) ψ (x − y − t z) − ψ (x − t z) dz d x ≤ C,

t

t t B0 ×(0,∞)

with a constant C independent of y ∈ Rn . Proof To prove the lemma, by a change of variables we may assume that supp() ⊂ B0 . This can be seen as follows. Suppose that supp() ⊂ B(0, M) with M > 1. Let c = 1/M. Then by a change of variables, we see that 



2 dt

(α)

(α)

ψt (x − y − t z) − ψt (x − t z) dz t B0 ×(0,∞) 

2

dt

(α) (α) = c−n

ψct (x − y − t z) − ψct (x − t z) dz t B(0,c)×(0,∞) 

2 dt

(α)

(α) ≤ c−n

ψct (x − y − t z) − ψct (x − t z) dz . t B0 ×(0,∞)

We note that ψc(α) (x) = c−α (L α (x) − L α ∗ c (x))

Journal of Fourier Analysis and Applications

and that supp(c ) ⊂ B0 . This implies what we need. Fix x, y ∈ Rn with |x| > 2|y|. Let 

2



(α) (α)

ψt (x − y − t z) − ψt (x − t z) dz.

I (x, y, t) = B0

We write I (x, y, t) = I1 (x, y, t) + I2 (x, y, t), where 



2

(α)

(α)

ψt (x − y − t z) − ψt (x − t z) dz,

I1 (x, y, t) = z∈B0 ,|x/t−z|<6





2

(α)

(α)

ψt (x − y − t z) − ψt (x − t z) dz.

I2 (x, y, t) = z∈B0 ,|x/t−z|>6

We first estimate I1 (x, y, t). Decompose I1 (x, y, t) = I1,1 (x, y, t) + I1,2 (x, y, t), where 

2



(α) (α)

ψt (x − y − t z) − ψt (x − t z) dz,

I1,1 (x, y, t) = z∈B0 ,2|y|/t<|x/t−z|<6





2

(α)

(α)

ψt (x − y − t z) − ψt (x − t z) dz.

I1,2 (x, y, t) = z∈B0 ,|x/t−z|<6,|x/t−z|<2|y|/t

Since L α ∗  is bounded and n/2 < α < n, we easily see that I1,2 (x, y, t) ≤ Ct



−2n

|z|

−2n+2α

dz ≤ Ct

|z|<3|y|/t

−2n



|y| t

−n+2α

.

If |x/t − z| < 6 and |z| < 1, then t > |x|/7. Thus 

∞ 0

dt = I1,2 (x, y, t) t



∞

|x|/7 ∞

≤C

t |x|/7

dt t −n+2α

I1,2 (x, y, t) −2n



|y| t

dt ≤ C|y|−n+2α |x|−n−2α .(2.1) t

(x, y, t) and I (x, y, t) be defined in the same manner as I (x, y, t) with Let I1,1 1,1 1,1 + 2I . By the mean L α and L α ∗  in place of ψ (α) , respectively. Then I1,1 ≤ 2I1,1 1,1

Journal of Fourier Analysis and Applications

value theorem we see that I1,1 (x, y, t) ≤ Ct −2n

= Ct

−2n



|y| t

2



2(−n−1+α)

x



dz

− z

t

2|y|/t<|x/t−z|<6



|y| t

2



|z|2(−n−1+α) dz. 2|y|/t<|z|<6

Thus, if n/2 < α < n/2 + 1, (x, I1,1

y, t) ≤ Ct

−2n



|y| t

2

−n−2+2α

(2|y|/t)

≤ Ct

−2n



|y| t

−n+2α

;

(2.2)

if n/2 + 1 < α, (x, I1,1

y, t) ≤ Ct

−2n



|y| t

2 ;

(2.3)

and if α = n/2 + 1, (x, y, t) ≤ Ct −2n I1,1



|y| t



2 log+

 3t , |y|

(2.4)

where log+ s = max(log s, 0). By (2.2), (2.3) and (2.4), we have 

∞ 0

I1,1 (x, y, t)

dt t

 =

∞

|x|/7

I1,1 (x, y, t)

dt t

⎧ −n+2α |x|−n−2α if n/2 < α < n/2 + 1, ⎪ ⎨ C|y| 2 |x|−2n−2 C|y| (2.5) ≤   if α > n/2 + 1, ⎪ ⎩ C|y|2 |x|−2n−2 log |x| if α = n/2 + 1. |y|

We note that the case α ≥ n/2 + 1 may occur only when n ≥ 3, if n/2 < α < n. , we note that To estimate I1,1



|L α ∗ (x − y) − L α ∗ (x)| =

−α

 

2πix,ξ −2πiy,ξ ˆ e − 1 dξ

(ξ )e

|2π ξ |  ≤C |ξ |−α (1 + |ξ |)−β |y| |ξ | dξ Rn

Rn 

≤ C|y| , where 0 <  < min(α + β − n, 1). Thus I1,1 (x, y, t) ≤ Ct −2n



|y| t

2 ,

Journal of Fourier Analysis and Applications

and hence 

∞

0

I1,1 (x,

dt = y, t) t



∞

|x|/7

I1,1 (x, y, t)

dt ≤ C|y|2 |x|−2n−2 . t

(2.6)

We recall that it is assumed that |x| > 2|y|. To estimate I2 (x, y, t), we observe that if |x/t −z| > 6 and |z| < 1, then |x|/t ≥ 5 and |x/t −y/t −z| ≥ |x|/(2t)−|z| > 3/2. So, further if |x/t − z| < c0 |x|/t with 0 < c0 < 2/3, we would have |z| ≥ (1−c0 )|x|/t > 1. Thus 



2

(α)

(α)

ψt (x − y − t z) − ψt (x − t z) dz.

I2 (x, y, t) =

(2.7)

z∈B0 ,|x/t−z|>6,|x/t−z|≥c0 |x|/t

Take c0 ∈ (1/2, 2/3). Since  ∈ Mα , by Taylor’s formula, we see that (α) ψt (x

− y − t z) = t

−n

 

  x y y − z − Lα − − z − w (w) dw t t t t  [α] + 1 γ!

Lα

= −t −n

x

−

|γ |=[α]+1

  ×

0

1

(1 − s)[α] (∂ γ L α )

x t

−

 y − z − sw ds (−w)γ (w) dw t

and (α)

(α)

ψt (x − y − t z) − ψt (x − t z)  [α] + 1   1 (1 − s)[α] Nγ (x, y, z, w, t, s) ds (−w)γ (w) dw, = −t −n γ! 0 |γ |=[α]+1

where   x y − − z − sw − (∂ γ L α ) − z − sw t t t  x  −y β  1 uy − − z − sw du. = (∂ γ +β L α ) t t t 0

Nγ (x, y, z, w, t, s) = (∂ γ L α )

x

|β|=1

γ

γ

Here ∂ γ = ∂1 1 . . . ∂n n , ∂ j = ∂/∂ x j , 1 ≤ j ≤ n, and γ ! = γ1 ! . . . γn !. We note that |x/t − uy/t − z − sw| ≥ c|x/t − z| for some c > 0 if |x/t − z| > 6, 2|y| < |x|, z, w ∈ B0 , u, s ∈ [0, 1] and |x/t − z| ≥ c0 |x|/t. This can be seen as follows. First, as in the case u = 1 above we have |x/t − uy/t − z| ≥ 3/2. Thus 

1 x

1

x uy uy 1

x







− z − sw ≥ − − z ≥ 1−

−

− z , t t 3 t t 3 2c0 t

Journal of Fourier Analysis and Applications

as claimed. Using these results in (2.7), we have 



t −2n

I2 (x, y, t) ≤ C z∈B0 ,|x/t−z|>6,|x/t−z|≥c0 |x|/t

≤ Ct

−2n



|y| t

2 

|x| t

|y| t

2

2(−n+α−[α]−2)

x dz

− z

t

2(−n+α−[α]−2)

≤ Ct −2α+2[α]+2 |y|2 |x|2(−n+α−[α]−2) and hence 

∞ 0

dt = I2 (x, y, t) t



|x|/5

I2 (x, y, t)

0

dt ≤ C|y|2 |x|−2n−2 . t

(2.8)

Since 

 ∞

2

dt dt

(α) (α) ≤C I1,2 (x, y, t)

ψt (x − y − t z) − ψt (x − t z) dz t t B0 ×(0,∞) 0  ∞  ∞  ∞ dt dt dt +C I1,1 (x, y, t) I1,1 (x, y, t) I2 (x, y, t) , +C +C t t t 0 0 0

using (2.1), (2.5), (2.6), (2.8) and recalling that α > n/2, we can get the conclusion.   Also, we need the next result for the proof of Theorem 2.3. Lemma 2.2 Let ψ (α) be as in (1.4). Then, if f ∈ S0 (Rn ) and g ∈ BMO(Rn ), we have







Rn



f (x)g(x) d x

≤ CgBMO

Rn

Sψ (α) ( f )(x) d x.

This can be shown by the methods of part (a) of Remarks on pp. 148–149 of [6]. Proof of Lemma 2.2 There exists η ∈ S0 (Rn ) such that  0

∞

dt  = 1 for all ξ = 0. ˆ ) ψ (α) (tξ )η(−tξ t

We can find such η since  ˆ )) ψ (α) (ξ ) = (2π |ξ |)−α (1 − (ξ

(2.9)

ψ (α) (tξ )| > 0 for ξ = 0. (See [2, satisfies a non-degeneracy condition supt>0 | Lemma 4.1] and its proof.) Since g ∈ BMO, we have the Carleson measure estimate

Journal of Fourier Analysis and Applications

(see [6, p. 145]) sup

y∈Rn ,h>0

h −n



h

 |g ∗ ηt (x)|2 d x B(y,h)

0

dt ≤ Cg2BMO . t

(2.10)

Let Sη(h) (g)(x)

 =

h



 Rn

0

χ[0,1]

 1/2 |x − y| 2 −n dt |g ∗ ηt (y)| dy t . t t

Then (2.10) implies that sup

z∈Rn ,h>0

|B(z, h)|−1

 B(z,h)

Sη(h) (g)(x)2 d x ≤ C02 g2BMO .

(2.11)

This can be seen as follows.  B(z,h)

Sη(h) (g)(x)2 d x =



h

 Rn

0

|B(z, h) ∩ B(y, t)||g ∗ ηt (y)|2 dy t −n



≤C

h



0

|g ∗ ηt (y)|2 dy B(z,2h)

dt t

dt t

≤ Cg2BMO h n . Let h(x) = sup{h : Sη(h) (g)(x) ≤ 21/2 C0 gBMO }. Then |{x ∈ B(z, h 0 ) : h(x) ≥ h 0 }| ≥ |B(z, h 0 )|/2

(2.12)

for all z ∈ Rn and h 0 > 0. To see this, by (2.11) we observe that  C02 g2BMO |B(z, h 0 )|

Sη(h 0 ) (g)(x)2 d x

≥ B(z,h 0 )\{x:h(x)≥h 0 }

≥ 2C02 g2BMO (|B(z, h 0 )| − |{x ∈ B(z, h 0 ) : h(x) ≥ h 0 }|) , from which (2.12) follows. Now we can show  ∞ 0

Rn

(α)

| f ∗ ψt

(y)||g ∗ ηt (y)| dy

 dt Sψ (α) ( f )(x) d x. ≤ CgBMO t Rn

(2.13)

Journal of Fourier Analysis and Applications

To see this, we first note that by (2.12) the left hand side is bounded up to a constant factor by 

∞

dt (α) |{x ∈ B(y, t) : h(x) ≥ t}|t −n | f ∗ ψt (y)||g ∗ ηt (y)| dy t Rn 0 ⎛ ⎞    ∞  |x − y| dt ⎜ ⎟ (α) χ[0,1] = d x ⎠ | f ∗ ψt (y)||g ∗ ηt (y)| dy t −n . ⎝ t t Rn 0 h(x)≥t

Fubini’s theorem implies that this quantity is equal to 

 Rn



h(x)  Rn

0

χ[0,1]

  |x − y| dt (α) | f ∗ ψt (y)||g ∗ ηt (y)| dy t −n d x. t t

Via Schwarz’s inequality, this is bounded by  Rn

Sψ (α) ( f )(x)Sη(h(x)) (g)(x) d x, (h(x))

from which (2.13) follows, since Sη Finally we prove 

 Rn

(g)(x) ≤ 21/2 C0 gBMO .

∞

f (x)g(x) d x = 0

(α)

Rn

f ∗ ψt (y)g ∗ ηt (y) dy

dt , t

(2.14)

for f ∈ S0 (Rn ) and g ∈ B M O(Rn ), assuming that Sψ (α) ( f ) ∈ L 1 (Rn ). Combining we get the conclusion of the lemma. To see (2.14), we first note that (2.13) and (2.14),−n−1 dz < ∞, since g ∈ BMO(Rn ), and hence |g(z)|(1 + |z|) 

|g(z)|(1 + |y − z|)−n−1 dz ≤ C(1 + |y|)n+1 .

From this we have 



(α)

f ∗ ψt (y)g ∗ ηt (y) dy = lim

m→∞ Rn

Rn

(α)

f ∗ ψt (y)g(m) ∗ ηt (y) dy

for each t > 0, where g(m) (x) = g(x)χ[0,m] (|x|), m = 1, 2, . . . , which can be seen by the dominated convergence theorem, since f ∗ψt(α) ∈ S, |g(m) ∗ηt (y)| ≤ Ct (1+|y|)n+1 and g(m) ∗ ηt → g ∗ ηt pointwise. We notice that g(m) ∈ L 1 and  Im (t) :=

Rn

f ∗ ψt(α) (y)g(m) ∗ ηt (y) dy =

 Rn

g(m) (−ξ )η(−tξ ˆ ) dξ. fˆ(ξ ) ψ (α) (tξ )

Journal of Fourier Analysis and Applications

Using this and the fact f , η ∈ S0 , we can see that there exists  ∈ (0, 1) such that / (,  −1 ) and Im (t) = 0 if t ∈  1=

∞

dt  = ψ (α) (tξ )η(−tξ ˆ ) t

0

 

 −1

dt  for ξ ∈ supp( fˆ). ψ (α) (tξ )η(−tξ ˆ ) t

Also, we note that  sup

m≥1,t∈[, −1 ] Rn





(α)

f ∗ ψt (y)g(m) ∗ ηt (y) dy ≤ C.

Applying these results, we have 

∞

dt (α) f ∗ ψt (y)g ∗ ηt (y) dy t 0   ∞  dt (α) lim = f ∗ ψt (y)g(m) ∗ ηt (y) dy m→∞ n t R 0    −1   dt (α) lim = f ∗ ψt (y)g(m) ∗ ηt (y) dy m→∞ Rn t     −1  dt = lim f ∗ ψt(α) (y)g(m) ∗ ηt (y) dy m→∞  n t R    −1  dt = lim g(m) (−ξ )η(−tξ ˆ ) dξ fˆ(ξ ) ψ (α) (tξ ) m→∞  n t R  −1    dt  = lim ψ (α) (tξ )η(−tξ ˆ ) fˆ(ξ ) g(m) (−ξ ) dξ m→∞ Rn t    ˆ f (x)g(m) (x) d x f (ξ ) g(m) (−ξ ) dξ = lim = lim m→∞ Rn m→∞ Rn  = f (x)g(x) d x. Rn

Rn

This completes the proof of (2.14) and hence that of the lemma.

 

Now we can prove the following result, which will be applied in the proof of Theorem 1.1. Theorem 2.3 Suppose that ψ (α) is as in (1.4) with α and  as in Theorem 1.1. Then Sψ (α) ( f )1  f  H 1 ,

f ∈ S0 (Rn ).

Proof The inequality  f  H 1 ≤ CSψ (α) ( f )1 follows from the estimate of Lemma 2.2 and duality of H 1 and BMO by taking supremum over g ∈ BMO with gBMO ≤ 1.

Journal of Fourier Analysis and Applications

The reverse inequality will be proved by applying a result of [7]. We first note that Sψ (α) is bounded on L 2 (Rn ). To see this, by the Plancherel theorem we observe that 

 Sψ (α) ( f ) (x) d x = |B0 | 2

Rn

Rn

| fˆ(ξ )|2



∞

0

dt dξ | ψ (α) (tξ )|2 t

From this we can deduce the L 2 boundedness, since | ψ (α) (ξ )| ≤ C|ξ |−α and | ψ (α) (ξ )| ≤ C|ξ |[α]+1−α , which is a consequence of the hypothesis  ∈ Mα and (2.9). In Lemma 2.1 we have checked the Hörmander condition which is required for [7, Chapter V, Corollary 3.10] to apply. Thus the reverse inequality follows from Corollary 3.10 of [7, Chapter V] for H-valued singular integrals, where H is the   1/2 ∞ .   Hilbert space with the norm gH = 0 B0 |g(z, t)|2 dz dt/t We also need the next result for the proof of Theorem 1.1. Lemma 2.4 Let α > 0 and w ∈ A1 . There exist Fourier multipliers  and m for Hw1 (Rn ) such that (2π |ξ |)α = (ξ )(1 + 4π 2 |ξ |2 )α/2 , (1 + 4π |ξ |2 )α/2 = m(ξ )(1 + (2π |ξ |)α ). 2

(2.15) (2.16)

This follows from [26, Chapter XI, Theorem 14]. Proof of Theorem 1.1 Let n/2 < α < n and let Uα be as in Theorem 1.1. We show that Uα (Jα (g))1 + Jα (g) H 1 g H 1

(2.17)

for g ∈ H 1 . First we assume that g ∈ S0 (Rn ). Then Uα (Jα (g)) = Sψ (α) (I−α Jα (g)) and by Theorem 2.3 we have Uα (Jα (g))1 I−α Jα (g)) H 1 .

(2.18)

Thus by (2.15) with w = 1 Uα (Jα (g))1 ≤ Cg H 1 .

(2.19)

Jα (g) H 1 ≤ CK α 1 g ∗ 1 ≤ Cg H 1 .

(2.20)

Also, we easily see that

Next, by (2.16) with w = 1 and (2.18) we see that g H 1 = J−α Jα (g) H 1 ≤ CJα (g) H 1 + CI−α Jα (g) H 1 ≤ CJα (g) H 1 + CUα (Jα (g))1 .

(2.21)

Journal of Fourier Analysis and Applications

By (2.19), (2.20) and (2.21), we have (2.17) for g ∈ S0 (Rn ). For g ∈ H 1 , take a sequence {gk } in S0 (Rn ) satisfying gk → g, Jα (gk ) → Jα (g) in H 1 and almost everywhere as k → ∞. Fix x ∈ Rn and t > 0. Then |Jα (gk )(z) − t ∗ Jα (gk )(z)| → |Jα (g)(z) − t ∗ Jα (g)(z)| for almost every z ∈ B(x, t).

Thus Fatou’s lemma implies that 

 B(x,t)

|Jα (g)(z) − t ∗ Jα (g)(z)|2 dz ≤ lim inf k→∞

B(x,t)

|Jα (gk )(z) − t ∗ Jα (gk )(z)|2 dz,

and hence by Fatou’s lemma it follows that Uα (Jα (g))(x) ≤ lim inf Uα (Jα (gk ))(x) k→∞

for all x. By (2.17) for g ∈ S0 (Rn ) we have Uα (Jα (gk ))1 ≤ Cgk  H 1 . Using this and Fatou’s lemma, we have Uα (Jα (g))1 ≤ lim inf Uα (Jα (gk ))1 k→∞

≤ C lim inf gk  H 1 k→∞

≤ Cg H 1 . Applying this we can deduce that lim Uα (Jα (g)) − Uα (Jα (gk ))1 ≤ lim Uα (Jα (g − gk ))1

k→∞

k→∞

≤ C lim g − gk  H 1 = 0. k→∞

Therefore, letting k → ∞ in Uα (Jα (gk ))1 + Jα (gk ) H 1 gk  H 1 , which follows from (2.17) on S0 (Rn ), we have (2.17) on the whole H 1 . Since we have shown (2.17), to complete the proof of Theorem 1.1, it suffices to prove that f ∈ W Hα 1 if f ∈ H 1 and Uα ( f ) ∈ L 1 . For this we need the next two lemmas. Lemma 2.5 Let f ∈ L 1w , w ∈ A1 , g ∈ S(Rn ) and α > 0. Then (1) we have K α ∗ ( f ∗ g)(x) = (K α ∗ f ) ∗ g(x) = (K α ∗ g) ∗ f (x) for every x ∈ Rn ;

Journal of Fourier Analysis and Applications

(2) also, 

 Rn

(K α ∗ f )(y)g(y) dy =

Rn

(K α ∗ g)(y) f (y) dy.

Proof By Fubini’s theorem, the result is a consequence of the estimate  K α (x − z − y)| f (y)||g(z)| dy dz < ∞, x ∈ Rn . This follows from the inequality  K α (x − z − y)|g(z)| dz ≤ C x,M (1 + |y|)−M for any M > 0, which is a consequence of the fact that g ∈ S(Rn ) and K α is rapidly 1 decreasing as |x|−n→ ∞ (see [23, p.132]), since the assumption f ∈ L w implies   | f (y)|(1 + |y|) dy < ∞ (see [13, Section 4]). 1 Lemma 2.6 Let w ∈ A1 and let {gm }∞ m=1 be a sequence of functions in Hw satisfying ∞ supm≥1 gm  Hw1 < ∞. Then there exist a subsequence {gm k }k=1 and g ∈ Hw1 such that   gm k (x)v(x) d x → g(x)v(x) d x as k → ∞ for v ∈ Cc (Rn ), Rn

Rn

where Cc (Rn ) is the space of all continuous functions with compact support on Rn ; also this is valid for all v ∈ S(Rn ). Proof Let Ol = B(0, l), l = 1, 2, 3, . . . . We note that w(x) ≥ Cw (1 + |x|)−n for ≤ Cw(x) valid for A1 weights w ∈ A1 , which follows from the property M(w)(x)  (see [13, Section 4]). Thus we have supm≥1 Ol |gm | d x < ∞ for every l. It is well-known that there exist a subsequence {gm k }∞ k=1 and a regular Borel measure μl such that   χ Ol+1 (y)gm k (y)v(y) dy → v(y) dμl (y) as k → ∞ for v ∈ Cc (Rn ) (2.22) Rn

Rn

(see [4, 1.9, Theorem 2]). We note that the subsequence can be chosen independent of l by the diagonal process. Taking ϕ (x − y) in (2.22) in place of v(y) for x ∈ Ol ,  ∈ (0, 1), we easily see that











∗



ϕ (x − y) dμl (y) ≤ lim inf

ϕ (x − y)gm k (y) dy

≤ lim inf gm (x),

k k→∞

Rn

Rn

since y ∈ Ol+1 if ϕ (x − y) = 0, and hence  ∗ μl† (x) d x ≤ lim inf gm  ≤ C, k 1 Ol

k→∞

k→∞

Journal of Fourier Analysis and Applications

where μl† (x)



= sup

R

∈(0,1)





∗

ϕ (x − y) dμl (y) , gm k (x) = sup

n >0

R



ϕ (x − y)gm k (y) dy

. n

Also, for v ∈ Cc (Rn ), 

 Rn

Rn





ϕ (x − y) dμl (y) v(x) d x →

Rn

v(y) dμl (y) as  → 0.

Thus







Rn



v(y) dμl (y)

≤

Ol

|v(x)|μl† (x) d x

for all v ∈ Cc (Rn ) with support in Ol , which implies  |μl |(O) ≤ O

μl† (x) d x

 ≤ Ol

μl† (x) d x

for any open set O in Ol , where |μl | is the total variation of μl . Thus μl is absolutely continuous when restricted to Ol and there exists gl ∈ L 1 (Rn ) such that 

 v(x) dμl (x) =

v(x)gl (x) d x for v ∈ Cc (Rn ) with support in Ol . (2.23)

By (2.22) and (2.23), we can see that there is a locally integrable function g on Rn such that g = gl on Ol and  Rn

 gm k (y)v(y) dy →

Rn

g(y)v(y) dy as k → ∞ for v ∈ Cc (Rn ).

∗ (x), Applying this with v(y) = ϕ (x − y), as above, we have g ∗ (x) ≤ lim inf k→∞ gm k which combined with the assumption that {gk∗ } is L 1w bounded implies g ∗ ∈ L 1w and   hence g ∈ Hw1 . The result for v ∈ S(Rn ) follows from that for v ∈ Cc (Rn ).

Now we can finish the proof of Theorem 1.1. Suppose that f ∈ H 1 and Uα ( f ) ∈ L 1 . Let f () (x) = ϕ ∗ f (x) and g () (x) = J−α (ϕ ) ∗ f (x). Then g () ∈ H 1 and by part (1) of Lemma 2.5 with w = 1, f () = Jα (g () ) and (2.17) implies Uα ( f () )1 +  f ()  H 1 g ()  H 1 .

(2.24)

sup  f ()  H 1 ≤ C f  H 1 .

(2.25)

We easily see that >0

Journal of Fourier Analysis and Applications

Further, using Minkowski’s inequality, we have Uα ( f () )(x) =

 0

 ≤

Rn

 =

Rn

∞ B(x,t)

|ϕ ∗ f (z) − t ∗ ϕ ∗ f (z)|2 dz t −2α−n



ϕ (y)

∞ 0

dt t

1/2

| f (z − y) − t ∗ f (z − y)|2 dz t −2α−n B(x,t)

dt t

1/2 dy

ϕ (y)Uα ( f )(x − y) dy,

which implies sup Uα ( f () )1 ≤ Uα ( f )1 . >0

(2.26)

Combining (2.24), (2.25) and (2.26), we have sup>0 g ()  H 1 < ∞. Thus by Lemma 2.6 for w = 1 we have a sequence {g (k ) } with k → 0 and g ∈ H 1 such that   (k ) g (x)v(x) d x → g(x)v(x) d x as k → ∞ for v ∈ S(Rn ). Rn

Rn

Also, { f (k ) } converges to f in L 1 . Thus, we can see that f = Jα (g). We show this in more detail as follows. Let v ∈ S(Rn ). Then, using part (2) of Lemma 2.5 with w = 1 and the fact that Jα (v) ∈ S(Rn ), we see that    f (x)v(x) d x = lim f (k ) (x)v(x) d x = lim Jα (g (k ) )(x)v(x) d x k→∞ k→∞   = lim g (k ) (x)Jα (v)(x) d x = g(x)Jα (v)(x) d x k→∞  (2.27) = Jα (g)(x)v(x) d x. It follows that f = Jα (g). Thus f ∈ W Hα 1 (Rn ).

 

3 Estimates for Littlewood–Paley Functions on the Weighted Hardy Spaces To prove Theorem 1.2, we need Theorem 3.2 below with p = 1 and n = 1 (see [19] and [20] for the unweighted case). Definition 3.1 Let ψ ∈ L 1 (Rn ). We say ψ ∈ B if (1) (2) (3) (4)

ψˆ ∈ C ∞ (Rn \ {0}); ˆ )| > 0 for all ξ = 0; supt>0 |ψ(tξ ψ ∈ C 1 (Rn ), ∂k ψ ∈ L 1 (Rn ), 1 ≤ k ≤ n; ˆ )| ≤ C|ξ | for some  > 0; |ψ(ξ

Journal of Fourier Analysis and Applications

ˆ )| ≤ Cγ ,τ |ξ |−τ outside a neighborhood of the origin for all multi-indices (5) |∂ γ ψ(ξ γ and τ > 0. Theorem 3.2 Let 0 < p ≤ 1, w ∈ A1 and ψ ∈ B. Then we have  f  Hwp ≤ C p gψ ( f ) p,w for f ∈ S0 (Rn ) with a positive constant C p independent of f , where we recall that  f  Hwp =  f ∗  p,w . Let H be the Hilbert space of functions u(t) on (0, ∞) such that uH =  ∞ 1/2 2 < ∞. Let w ∈ A1 . We consider the weighted Lebesgue space 0 |u(t)| dt/t q n L H,w (R ) of functions h(y, t) with the norm 

1/q

q

hq,H,w =

Rn

h y H w(y) dy

, q

q

where h y (t) = h(y, t). When w = 1, we write simply L H,w (Rn ) = L H (Rn ). Let 0 < p ≤ 1. We consider the weighted Hardy space of functions on Rn with p p values in H, which is denoted by HH,w (Rn ). We say that h ∈ HH,w (Rn ) if h ∈ L 2H (Rn ) and h H p = h ∗  p,w < ∞ with H,w

h ∗ (x) = sup



s>0

∞

|ϕs ∗ h t (x)|2

0

dt t

1/2 ,

where we write h t (x) = h(x, t) and we recall that ϕ is the function in S(Rn ) fixed in Sect. 1. In proving Theorem 3.2, we need the next result. ˆ ⊂ {1 ≤ |ξ | ≤ 2} Lemma 3.3 Let ψˆ ∈ S(Rn ) be a radial function such that supp(ψ) and  ∞ dt ˆ )|2 = 1 for all ξ = 0. |ψ(tξ t 0 p

Let F(y, t) = f ∗ ψt (y) with f ∈ S0 (Rn ). Let w ∈ A1 . Then F ∈ HH,w (Rn ), 0 < p ≤ 1, and  f  Hwp ≤ CF H p . H,w

Let ψ be as in Lemma 3.3 and  ∞ E ψ (h)(x) = 0

Rn

ψt (x − y)h () (y, t) dy

where h ∈ L 2H and h () (y, t) = h(y, t)χ(, −1 ) (t), 0 <  < 1. To prove Lemma 3.3 we apply the following.

dt , t

Journal of Fourier Analysis and Applications

Lemma 3.4 Suppose that w ∈ A1 . Then sup E ψ (h) Hwp ≤ Ch H p , 0 < p ≤ 1. H,w

∈(0,1)

To prove Lemma 3.4 we use the atomic decomposition. p

Definition 3.5 We say that a is a ( p, ∞) atom in HH,w (Rn ), w ∈ A1 , 0 < p ≤ 1, if 1/2  ∞ ≤ w(B)−1/ p , where B is a ball in Rn and w(B) = (i)  0 |a(x, t)|2 dt/t B w(x) d x; (ii) supp(a(·, t)) ⊂ B for all t > 0, where B is the same as in (i);  (iii) Rn a(x, t)x γ d x = 0 for all t > 0 and multi-indices γ such that |γ | ≤ [n(1/ p − 1)]. p

Lemma 3.6 Let w ∈ A1 , 0 < p ≤ 1. Suppose that h ∈ HH,w (Rn ). Then there p exist a sequence {ak } of ( p, ∞) atoms in HH,w (Rn ) and a sequence {λk } of positive ∞ p p numbers such that k=1 λk ≤ Ch H p , where C is a constant independent of h, H,w  p 2 n n and h = ∞ k=1 λk ak in HH,w (R ) and in L H (R ). p

A proof of the atomic decomposition for Hw (Rn ) can be found in [26, Chapter VIII] and the proof for the vector valued case is similar; we can apply the same arguments as in the case of the scalar valued case by replacing absolute values with H-norms in appropriate places. Also, we need the following result in proving Lemma 3.4. Lemma 3.7 Let w ∈ A2 and h ∈ L 2H,w (Rn ) ∩ L 2H (Rn ). Let ψ be as in Lemma 3.3. Then sup E ψ (h)2,w ≤ Ch2,H,w .

∈(0,1)

Proof Let  ∈ C0∞ (Rn ) satisfy supp() ⊂ {1/2 ≤ |ξ | ≤ 2} and 1 for ξ = 0. Define  ˆ t)e2πiξ,x dξ, m (h)(x, t) = (2−m ξ )h(ξ,



m∈Z (2

where the Fourier transform hˆ is with respect to x variable, and  Am (h () )(x) =

∞ 0

Rn

ψt (x − y)m (h () )(y, t) dy

dt . t

Then by taking the Fourier transform we see that  Am (h () )(x) =

2−m+2 2−m−1

 Rn

ψt (x − y)m (h () )(y, t) dy

dt . t

−m ξ )

=

Journal of Fourier Analysis and Applications

We note that







Rn



ψt (x − y)m (h () )(y, t) dy

≤ C M(m (h () )(·, t))(x),

since the least decreasing radial majorant of ψ is integrable (see [23, Chapter III, Theorem 2]). Similarly,



m (h () )(x, t) ≤ C M(h () (·, t))(x). It follows that  |Am (h () )(x)| ≤ C 2

2−m+2

M 2 (h () (·, t))(x)2

2−m−1

dt . t

Thus, since the maximal operator M is bounded on L 2w for w ∈ A2 , 

 |Am (h () )(x)| w(x) d x ≤ C 2

2−m+2 2−m−1

 |h(x, t)|2 w(x) d x

dt . t

Using this and applying Littlewood–Paley inequality with w ∈ A2 , we have 



Am (h () )22,w ≤ C

m∈Z



Am (h () )22,w

m∈Z

≤ Ch22,H,w . This completes the proof, since E ψ (h) =

 m∈Z

Am (h () ).

 

Proof of Lemma 3.4 The proof is analogous to the one for Lemma 3.5 of [19]. So p we put it briefly. Let a be a ( p, ∞) atom in HH,w (Rn ) supported on the ball B of Definition 3.5. If  B is a concentric enlargement of B such that 2|y − y | < |x − y | B. Then, as in the proof of Lemma 3.5 of [19], using for y, y ∈ B and x ∈ Rn \  B and y ∈ B we have properties of an atom, for x ∈ Rn \  





|y − y | M+1 dy

ϕs ∗ E ψ (a)(x) ≤ Cw(B)−1/ p |x − y |−n−M−1

(3.1)

B

where M = [n(1/ p − 1)]. To prove (3.1), let s,t = ϕs ∗ ψt , s, t > 0. Let Px (y, y ) be the Taylor polynomial in y of order M = [n(1/ p − 1)] at y for ϕs/t ∗ ψ(x − y). Then, if |x − y| > 2|y − y|, we see that |s,t (x − y) − t −n Px/t (y/t, y /t)| ≤ Ct −n−M−1 |y − y | M+1 (1 + |x − y |/t)−L ,

Journal of Fourier Analysis and Applications

where L > n + M + 1 and the constant C is independent of s, t, x, y , y. Thus, using the properties of an atom and the Schwarz inequality, for x ∈ Rn \  B we have





ϕs ∗ E ψ (a)(x)



 

  dt

s,t (x − y) − t −n Px/t (y/t, y /t) a() (y, t) dy

=

t Rn ×(0,∞) 1/2  ∞    ∞



dt 1/2

s,t (x − y) − t −n Px/t (y/t, y /t) 2 dt ≤ |a(y, t)|2 dy t t B 0 0 1/2   ∞



s,t (x − y) − t −n Px/t (y/t, y /t) 2 dt ≤ Cw(B)−1/ p dy t 0 B  ≤ Cw(B)−1/ p |y − y | M+1 |x − y |−n−M−1 dy, B

which proves (3.1). Since p > n/(n + M + 1), by a straightforward computation, using (3.1), we see that 

p



sup ϕs ∗ E ψ (a)(x) w(x) d x ≤ Cw(B)−1 |B| inf M(w)(y ) y ∈B Rn \  B s>0  ≤ Cw(B)−1 w(y) dy ≤ C. (3.2) B

By Hölder’s inequality, the L 2w -boundedness of M, Lemma 3.7 and the properties (i), (ii) of Definition 3.5, we get 

 p/2 

p



sup ϕs ∗ E ψ (a)(x) w(x) d x ≤ Cw(B)1− p/2 |M(E ψ (a))(x)|2 w(x) d x

 B s>0

 ≤ Cw(B)1− p/2

 B

 B

|E ψ (a)(x)|2 w(x) d x

  ≤ Cw(B)1− p/2 B

∞

|a(y, t)|2 w(y)

0

≤ C,

 p/2

dt dy t

 p/2

(3.3)





where we have used the estimate ϕs ∗ E ψ (a) ≤ C M(E ψ (a)). Combining (3.2) and (3.3), we have 



p



sup ϕs ∗ E ψ (a)(x) w(x) d x ≤ C.

(3.4)

Rn s>0

By Lemma 3.6 and (3.4) we can prove 



p



p sup ϕs ∗ E ψ (h)(x) w(x) d x ≤ Ch H p .

Rn s>0

H,w

Journal of Fourier Analysis and Applications

 

This completes the proof. p HH,w (Rn )

Proof of Lemma 3.3 It can be shown that F ∈ similarly to the proof of p Lemma 3.4 by using the atomic decomposition for f ∈ Hw (Rn ); recall that f ∈ S0 p and that S0 is a subspace of Hw (Rn ). We give a sketch of the proof. First we can prove an estimate analogous to (3.1): 

∞

sup s>0

0

dt |ϕs ∗ ψt ∗ a(x)| t

1/2

2

≤ Cw(B)

−1/ p

−n−M−1



|y − y | M+1 dy,

|x − y |

(3.5)

B p

where a is a ( p, ∞) atom for Hw (Rn ) supported on the ball B with properties analB with  B denoting a ogous to those for the atom in (3.1) and y ∈ B and x ∈ Rn \  concentric enlargement of B as in the case of (3.1); further M = [n(1/ p − 1)]. Also, we have the following L 2w -boundedness: 

 Rn

∞

sup

|ϕs ∗ ψt ∗ f (x)|2

s>0 0

dt w(x) d x ≤ C f 2L 2 . w t

(3.6)

This can be shown by using the L 2w -boundedness of M and gψ as follows. 



∞

sup

Rn s>0 0

|ϕs ∗ ψt ∗ f (x)|2

dt w(x) d x ≤ C t



∞

0



≤C

Rn

∞

Rn

0 =C

Rn

|M(ψt ∗ f )(x)|2 w(x) d x |ψt ∗ f (x)|2 w(x) d x

dt t

dt t

gψ ( f )(x)2 w(x) d x

≤ C f 2L 2 . w

p

Using (3.5) and (3.6), we can show F ∈ HH,w (Rn ) as in the proof of Lemma 3.4. Let ψ, F be as in Lemma 3.3 and let ψ¯ denote the complex conjugate. Then E ψ¯ (F)(x)

 =

 −1 

 Rn

ψt ∗ f (y)ψ¯ t (y − x) dy

dt = t

 Rn

 () (x − z) f (z) dz,

where  () (x) =



 −1



 Rn

ψt (x + y)ψ¯ t (y) dy

dt . t

There exists 0 ∈ (0, 1) such that  (0 ) (ξ ) = 



0−1 0

dt ¯ ˆ )ψ(−tξ ψ(tξ ) = t



0−1

0

ˆ )|2 |ψ(tξ

dt = 1 for ξ ∈ supp( fˆ). t

Journal of Fourier Analysis and Applications

Thus E ψ¯0 (F)(x)

 =

 (0 ) (ξ ) fˆ(ξ )e2πix,ξ dξ = 



fˆ(ξ )e2πix,ξ dξ = f (x)

and hence by Lemma 3.4  f  Hwp = E ψ¯0 (F) Hwp ≤ CF H p . H,w

  Along with Lemma 3.3, the next two results (Lemmas 3.8, 3.9) are used in proving Theorem 3.2. Lemma 3.8 Let w ∈ A∞ = ∪ p>1 A p . Suppose that η ∈ S(Rn ), supp(η) ˆ ⊂ {1/2 ≤ |ξ | ≤ 4} and η(ξ ˆ ) = 1 on {1 ≤ |ξ | ≤ 2}. Let ψ be as in Lemma 3.3. Then for p, q > 0 and f ∈ S0 (Rn ) we have   1/q  1/q      ∞ ∞ dt dt     sup |ϕs ∗ ψt ∗ f |q ≤C |ηt ∗ f |q .     0 s>0    0 t t p,w

p,w

This can be established by the proof of Lemma 3.3 of [19], where only the unweighted version of Lemma 3.8 is explicitly treated but the proof is exactly the same in the weighted cases. Lemma 3.9 Let  ∈ B and w ∈ A∞ . Suppose that 0 < p, q < ∞. Let η ∈ S(Rn ) satisfy ηˆ = 0 in a neighborhood of the origin. Then   1/q  1/q      ∞ ∞     q dt q dt | f ∗ ηt | ≤C | f ∗ t |     0    0 t t p,w

p,w

for f ∈ S0 (Rn ) with a positive constant C independent of f . This is a particular case of Theorem 2.4 of [20] (also, results of [19] imply Lemma 3.9). Now we can complete the proof of Theorem 3.2. Proof of Theorem 3.2 Let η and ψ be as in Lemma 3.8 and w ∈ A1 . Applying successively Lemmas 3.3 and 3.8 with q = 2, we have   ∞    dt 1/2    2  f  Hwp ≤ C sup |ϕs ∗ ψt ∗ f |    s>0 0 t p,w   ≤ C gη ( f ) p,w

for f ∈ S0 (Rn ). By this and Lemma 3.9 with q = 2 we can finish the proof of Theorem 3.2.  

Journal of Fourier Analysis and Applications

4 Proof of Theorem 1.2 We first note the following. Lemma 4.1 Let W H1 1 (R), w ∈ A1 , be as in Sect. 1. Then w

 f W 1  f  Hw1 +  f  Hw1 ,

f ∈ S0 (R).

1 Hw

Proof Let f ∈ S0 (R). Then f ∈ Hw1 and f = J1 (g) for g ∈ S0 (R). Applying integration by parts, we have 

ˆ dx = − f (x)η(x)



f (x)(η) ˆ (x) d x =



1 (ξ )g(ξ K ˆ )2πiξ η(ξ ) dξ

for η ∈ S(R). Since ξ = (sgnξ )|ξ |, by Lemma 2.4 (2.15) and the fact that the Hilbert ˆ ) for h ∈ S0 (R) with 1 (ξ )g(ξ transform is bounded on Hw1 , we see that 2πiξ K ˆ ) = h(ξ h Hw1 ≤ Cg Hw1 (see [26, Chapter XI, Theorem 14]). Thus we have 





ˆ dx = f (x)η(x)

ˆ )η(ξ ) dξ = h(ξ

 h(x)η(x) ˆ d x,

which implies that f = h and hence  f  Hw1 ≤ C f W 1 . Also, a straightforward 1 Hw

computation implies that

 f  Hw1 = (J1 (g))∗ 1,w ≤ Cg ∗ 1,M(w) ≤ Cg ∗ 1,w = C f W 1 . 1 Hw

Here we give a proof of the first inequality for completeness. As in Sect. 1, we have (J1 (g))∗ ≤ K 1 ∗ g ∗ and K 1 ∗ w ≤ C M(w). Thus (J1 (g))∗ 1,w ≤



K 1 ∗ g ∗ (x)w(x) d x =  ≤ C g ∗ (y)M(w)(y) dy,



g ∗ (y)K 1 ∗ w(y) dy

where we have used the fact that K 1 is even. On the other hand, let g = J−1 ( f ) ∈ S0 (R). Then, by (2.16) −1 (ξ ) = m(ξ ) fˆ(ξ ) + m(ξ )(−isgn(ξ )) f (ξ ). g(ξ ˆ ) = fˆ(ξ ) K Using again the boundedness of the Hilbert transform together with Lemma 2.4, we   get that  f W 1 = g Hw1 ≤ C f  Hw1 + C f  Hw1 . 1 Hw

Also, we require the next result to prove Theorem 1.2.

Journal of Fourier Analysis and Applications

Lemma 4.2 Let ν be as in (1.2) and w ∈ A1 . Then ν( f )1,w  f  Hw1 ,

f ∈ S0 (R).

Proof It suffices to show that μ( f )1,w  f  Hw1 for f ∈ S0 (R). We recall a Littlewood–Paley function g0 defined as 

∞

g0 ( f )(x) =

1/2 |(∂/∂ x)u(x, t)|2 t dt

,

0

ˆ ) = e−2π |ξ | . where u(x, t) is the Poisson integral of f : u(x, t) = Pt ∗ f (x), P(ξ −2π |ξ | ˆ . We note that R = P and that R ∈ B; the condition (3) Let R(ξ ) = 2πiξ e of Definition 3.1 is obvious from the explicit forms: −2x , π(1 + x 2 )2

P (x) =

P (x) =

2(3x 2 − 1) . π(1 + x 2 )3

We see that g0 ( f ) = g R ( f ) and, since R ∈ B, by Theorem 3.2 with p = 1 we have  f  Hw1 ≤ Cg R ( f )1,w for f ∈ S0 (R) and w ∈ A1 (see [6,27] for the unweighted case). Also we have the pointwise relation g0 ( f ) ≤ Cμ( f ) for f ∈ S0 (R) (see [14, Theorem 5]). Combining results, we see that  f  Hw1 ≤ Cμ( f )1,w . In proving the reverse inequality, we apply the pointwise equivalence between g3∗ and μ to get μ( f )1,w ≤ Cg3∗ ( f 1 )1,w + Cg3∗ ( f 2 )1,w , where gλ∗ ( f )(x)

  =

R×(0,∞)



t t + |x − y|

λ

1/2 |∇u(y, t)| dy dt 2

is another Littlewood–Paley function and fˆ1 = fˆχ[0,∞) , fˆ2 = fˆχ(−∞,0] (see [14, Theorems 1, 2 and Remark 1] for the pointwise equivalence). It follows that μ( f )1,w ≤ Cg3∗ ( f )1,w + Cg3∗ (H f )1,w , where H denotes the Hilbert transform; this can be seen by noting χ[0,∞) (ξ ) =

  1 1 χ(−∞,∞) (ξ ) + sgn(ξ ) , χ(−∞,0] (ξ ) = χ(−∞,∞) (ξ ) − sgn(ξ ) 2 2

for ξ = 0. Then, we apply the Hw1 − L 1w boundedness with w ∈ A1 of g3∗ due to [12] and the boundedness of H on Hw1 . Here we would like to recall the following. In [12]

Journal of Fourier Analysis and Applications

Hw1 norm is defined as N (u)1,w for u(x, t) = f ∗ Pt (x), where N (u) denotes the non-tangential maximal function and we have N (u)1,w ≤ C f ∗ 1,w , which can be shown, for example, by applying the atomic decomposition for f ∈ Hw1 .   To prove μ( f )1,w ≤ C f  Hw1 , alternatively, we can apply an argument similar to the one in the proof of Lemma 3.4, using an atomic decomposition for Hw1 and an estimate from (4.7) of [17]: 

∞

|ψt (x − y) − ψt (x)|2

0

dt t

1/2 ≤C

|y|1/2 for 2|y| < |x|, |x|3/2

where ψ(x) = χ[0,1] (x) − χ[−1,0] (x) (see the proof of Theorem 4.5 of [17]). Proof of Theorem 1.2 The proof is analogous to that of Theorem 1.1. Lemmas 4.1 and 4.2 imply that J1 (g) Hw1 + ν(J1 (g))1,w g Hw1

(4.1)

for g ∈ S0 (R). We prove (4.1) for all g ∈ Hw1 . Let g ∈ Hw1 and take a sequence {gk } in S0 (R) such that gk → g, J1 (gk ) → J1 (g) in Hw1 and almost everywhere as k → ∞. We note that, if x is fixed, |J1 (gk )(x + t) + J1 (gk )(x − t) − 2J1 (gk )(x)| → |J1 (g)(x + t) + J1 (g)(x − t) − 2J1 (g)(x)| for a.e. t ∈ (0, ∞), which implies that ν(J1 (g))(x) ≤ lim inf ν(J1 (gk ))(x) k→∞

for every x by Fatou’s lemma. By (4.1) established with g ∈ S0 (Rn ), for gk ∈ S0 (Rn ) we have ν(J1 (gk ))1,w ≤ Cgk  Hw1 , from which and Fatou’s lemma, it follows that ν(J1 (g))1,w ≤ lim inf ν(J1 (gk ))1,w k→∞

≤ C lim inf gk  Hw1 k→∞

≤ Cg Hw1 .

Journal of Fourier Analysis and Applications

Therefore lim ν(J1 (g)) − ν(J1 (gk ))1,w ≤ lim ν(J1 (g − gk ))1,w

k→∞

k→∞

≤ C lim g − gk  Hw1 = 0. k→∞

Thus, letting k → ∞ in the relation J1 (gk ) Hw1 + ν(J1 (gk ))1,w gk  Hw1 , gk ∈ S0 (R), which is already shown, we have (4.1) for g ∈ Hw1 . Thus, to complete the proof of Theorem 1.2, it suffices to prove that if f ∈ Hw1 and ν( f ) ∈ L 1w , then f ∈ W H1 1 . Suppose that f ∈ Hw1 and ν( f ) ∈ L 1w . Let f () (x) = w

ϕ ∗ f (x) and g () (x) = J−1 (ϕ ) ∗ f (x). Then g () ∈ Hw1 and f () = J1 (g () ) (see Lemma 2.5). The relation (4.1) implies  f ()  Hw1 + ν( f () )1,w g ()  Hw1 .

(4.2)

sup  f ()  Hw1 ≤ C f ∗ 1,M(w) ≤ C f ∗ 1,w = C f  Hw1 ,

(4.3)

We see that >0

since w ∈ A1 . By Minkowski’s inequality, ν( f () )(x) =



∞ 0

|ϕ ∗ f (x + t) + ϕ ∗ f (x − t) − 2ϕ ∗ f (x)|2 



≤  =

R

R

ϕ (y)

∞

dt t3

1/2

| f (x + t − y) + f (x − t − y) − 2 f (x − y)|2

0

dt t3

1/2 dy

ϕ (y)ν( f )(x − y) dy.

Thus sup ν( f () )1,w ≤ Cν( f )1,M(w) ≤ Cν( f )1,w . >0

(4.4)

Consequently, it follows that sup>0 g ()  Hw1 < ∞ from (4.2), (4.3) and (4.4). Applying Lemma 2.6, we choose a sequence {g (k ) } in Hw1 with k → 0 and g ∈ Hw1 such that   g (k ) (x)v(x) d x → g(x)v(x) d x as k → ∞ for v ∈ S(R). R

R

Further, { f (k ) } converges to f in L 1w . Thus, (2.27) applies and f = J1 (g), so f ∈   W H1 1 (R). w

Journal of Fourier Analysis and Applications

Remark 4.3 The function of Marcinkiewicz μ( f ) is generalized. Let μβ ( f ) = gη(β) ( f ), β > 0, where η(β) (x) = β|1 − |x||β−1 sgn(x)χ(−1,1) (x). Then μβ generalizes μ in the sense that μ1 = μ. See [14] for properties of μβ .

5 Proofs of Results in Remarks 1.4 and 1.5 Here we give proofs of Remarks 1.4 and 1.5 for completeness. Proof of Remark 1.4 We prove that if 0 < α < n/2, 1 ≤ p ≤ 2 and Sψ (α) is bounded on L p (Rn ), then p ≥ 2n/(n + 2α). Let f ∈ S0 (Rn ), f = 0. We estimate Sψ (α) ( f ) as 2α+(n+1)/2 Sψ (α) ( f )(x)  2  1/2 2 ≥ |Iα ( f )(z) − t ∗ Iα ( f )(z)| dz dt 

1

≥

2

B(x,1)

B(x,1)

1



1/2 |Iα ( f )(z)| dz dt

=

1/2



|Iα ( f )(z)| dz

B(x,1)

1

1/2

2 1

2

−

2

 −

2

|t ∗ Iα ( f )(z)| dz dt 2

B(x,1)

1/2

|t ∗ Iα ( f )(z)| dz dt 2

B(x,1)

.

We easily see that 2

 1

B(x,1)

 2 |t ∗ Iα ( f )(z)|2 dz dt ≤ C2∞ χ B(0,C1 ) ∗ |Iα ( f )|(x) .

Thus 

1/2 B(x,1)

|Iα ( f )(z)|2 dz

≤ C Sψ (α) ( f )(x) + Cχ B(0,C1 ) ∗ |Iα ( f )|(x). (5.1)

On the other hand we will show that   p/2  2 |Iα ( f )(x)| d x ≤C Rn



 p/2 |Iα ( f )(z)| dz 2

Rn

B(x,1)

d x.

(5.2)

n To see this, we consider a covering of Rn : ∪∞ j=1 B(c j , 1) = R . We assume that there ∞ n exists τ > 0 such that ∪ j=1 B(x j (y), 1) = R for all y ∈ B(0, τ ) with x j (y) = c j + y and B(c j , τ ) ∩ B(ck , τ ) = ∅ if j = k. Then, since p/2 ≤ 1, we have



 p/2 |Iα ( f )(x)| d x 2

Rn

≤

 p/2

 ∞  j=1

|Iα ( f )(x)| d x 2

B(x j (y),1)

Journal of Fourier Analysis and Applications

for all y ∈ B(0, τ ). Thus  Rn

|Iα ( f )(x)|2 d x

≤

inf

y∈B(0,τ )

= Cτ = Cτ

B(0,τ ) j=1 ∞  

 p/2 |Iα ( f )(x)| d x 2

B(x j (y),1)

|Iα ( f )(x)| d x 2

B(x j (y),1)



B(c j ,τ )

dy

 p/2 |Iα ( f )(x)| d x 2

B(y,1)



dy

 p/2 |Iα ( f )(x)| d x 2

Rn

dy

 p/2



j=1 B(0,τ ) ∞  



|Iα ( f )(x)|2 d x



∞ 

j=1

≤ Cτ

B(x j (y),1)

j=1

 ≤ Cτ

 p/2

 ∞ 

 p/2

B(y,1)

dy,

which proves (5.2). From (5.1) and (5.2), it follows that Iα ( f )2 ≤ CSψ (α) ( f ) p + Cχ B(0,C1 ) ∗ |Iα ( f )| p . Thus if Sψ (α) ( f ) p ≤ C f  p , since χ B(0,C1 ) ∗ |Iα ( f )| p ≤ CIα ( f ) p , we have Iα ( f )2 ≤ C f  p + CIα ( f ) p . From this with f ρ in place of f , by homogeneity, it readily follows that ρ α−n/2 ≤ Cρ −n+n/ p + Cρ α+n(1/ p−1) ≤ Cρ −n+n/ p for all ρ ∈ (0, 1), which implies that p ≥ 2n/(n + 2α) as claimed.

 

Proof of Remark 1.5 Suppose that 0 < α < n/2, 1 < p < 2n/(2α + n). We see that Sψ (α) is not bounded from H 1 to L 1 ; otherwise Sψ (α) would be bounded on L p by interpolation between the H 1 − L 1 and L 2 boundedness of Sψ (α) (for the L 2 boundedness see the proof of Theorem 2.3 in Sect. 2). This contradicts Remark 1.4. However, if Uα was bounded from W Hα 1 to L 1 , then Uα (Jα (g))1 ≤ Cg H 1 for g ∈ S0 (Rn ). Since Uα ( f ) = Sψ (α) (I−α ( f )), it follows that Sψ (α) (g)1 ≤ CIα J−α (g) H 1 . Thus by Lemma 2.4 with w = 1 Sψ (α) (g)1 ≤ CIα (g) H 1 + Cg H 1 , g ∈ S0 (Rn ).

(5.3)

Since Sψ (α) (gρ ) = (Sψ (α) (g))ρ , Iα (gρ ) H 1 = ρ α Iα (g) H 1 and gρ  H 1 = g H 1 , by (5.3) with gρ in place of g we have Sψ (α) (g)1 ≤ Cρ α Iα (g) H 1 + Cg H 1

Journal of Fourier Analysis and Applications

for all ρ > 0. Thus, letting ρ → 0, we see that Sψ (α) (g)1 ≤ Cg H 1 , from which the H 1 − L 1 boundedness of Sψ (α) follows. This contradicts what we have already observed.  

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