1 Introduction

Let R n , n2, be the n-dimensional Euclidean space and S n 1 denote the unit sphere in R n equipped with the induced Lebesgue measure . Let α j 1 (j=1,,n) be fixed real numbers. Define the function F: R n ×(0,)R by F(x,ρ)= j = 1 n x j 2 ρ 2 α j , x=( x 1 , x 2 ,, x n ). It is clear that, for each fixed x R n , the function F(x,ρ) is a decreasing function in ρ>0. We let ρ(x) denote the unique solution of the equation F(x,ρ)=1. Fabes and Rivière [1] showed that ( R n ,ρ) is a metric space, which is often called the mixed homogeneity space related to { α j } j = 1 n . For λ>0, we let A λ be the diagonal n×n matrix A λ =diag{ λ α 1 ,, λ α n }. Let R + :=(0,) and φ: R + R + , we denote A φ ( ρ ( y ) ) y by A φ (y) for y R n , where y = A ρ ( y ) 1 y S n 1 .

The change of variables related to the spaces ( R n ,ρ) is given by the transformation

x 1 = ρ α 1 cos θ 1 cos θ n 2 cos θ n 1 , x 2 = ρ α 2 cos θ 1 cos θ n 2 sin θ n 1 , , x n 1 = ρ α n 1 cos θ 1 sin θ 2 , x n = ρ α n sin θ 1 .

Thus dx= ρ α 1 J( x )dρdσ( x ), where ρ α 1 J( x ) is the Jacobian of the above transform and α= j = 1 n α j , J( x )= j = 1 n α j ( x j ) 2 . Obviously, J( x ) C ( S n 1 ) and there exists M>0 such that

1J ( x ) M, x S n 1 .

It is easy to see that

ρ(x)=|x|,if  α 1 = α 2 == α n =1.

Let Ω be integrable on S n 1 and satisfy

S n 1 Ω(u)J(u)dσ(u)=0,
(1.1)
Ω( A s x)=Ω(x),s>0 and x R n .
(1.2)

For d2 and a suitable function Φ: R n R d , we define the parabolic Marcinkiewicz integral operator M Ω Φ on R d by

M Ω Φ (f)(x)= ( 0 | 1 t ρ ( y ) t Ω ( y ) ρ ( y ) α 1 f ( x Φ ( y ) ) d y | 2 d t t ) 1 / 2 .
(1.3)

When α 1 == α n =1, we denote M Ω Φ by μ Ω Φ . Clearly, if n=d and Φ(y)=y, the operator μ Ω Φ reduces to the classical Marcinkiewicz integral operator denoted by μ Ω , which was introduced by Stein [2] and investigated by many authors (see [39] for example). In particular, Ding et al. [5] proved that if Ω H 1 ( S n 1 ), then μ Ω is bounded on L p ( R n ) for 1<p<. Subsequently, Chen et al. [4] showed that μ Ω is bounded on L p ( R n ) for 2β/(2β1)<p<2β if Ω F β ( S n 1 ) for some β>1. Here

F β ( S n 1 ) : = { Ω L 1 ( S n 1 ) : sup ξ S n 1 S n 1 | Ω ( y ) | ( log 1 | ξ y | ) β d σ ( y ) < } , β > 0 .
(1.4)

The functions class F β ( S n 1 ) was introduced by Grafakos and Stefanov [10] in the study of L p boundedness of singular integral operator with rough kernels. It follows from [10] that F β 1 ( S n 1 )⊆̷ F β 2 ( S n 1 ) for 0< β 2 < β 1 , and q > 1 L q ( S n 1 )⊆̷ F β ( S n 1 ) for any β>0. Moreover,

β > 1 F β ( S n 1 ) H 1 ( S n 1 ) β > 1 F β ( S n 1 )

and

β > 1 F β ( S n 1 ) L log + L ( S n 1 ) .

Later on, Al-Salman et al. [11] proved that μ Ω is bounded on L p ( R n ) for 1<p< provided that ΩL ( log + L ) 1 / 2 ( S n 1 ). It is well known that L ( log + L ) 1 / 2 ( S n 1 ) and H 1 ( S n 1 ) do not contain each other. When n=d and Φ(y)=P(|y|) y with P(y) being a real polynomial on ℝ satisfying P(0)=0, Wu [12] proved that μ Ω Φ is bounded on L p ( R n ) for 1+1/(2β)<p<1+2β provided that Ω F β ( S n 1 ) for some β>1/2. The L p boundedness for the Marcinkiewicz integral operator associated to polynomial mappings has also been obtained (see [6, 13]).

When α j 1 (j=1,,n), n=d and Φ(y)=y, we denote M Ω Φ by M Ω . In 2008, Ding et al. [14] proved that M Ω is bounded on L p ( R n ) for 1<p<, provided that Ω L q ( S n 1 ) for fixed q>1. Chen and Ding [15] extended the above result to the case ΩL ( log L ) 1 / 2 ( S n 1 ). Later on, Chen and Lu [16] proved that M Ω is bounded on L p ( R n ) for 2β/(2β1)<p<2β, provided that Ω F β ( S n 1 ) for some β>1. This result was recently refined by Liu and Wu [17], who extended the range of β to the case β>1/2 and the range of p to the case 1+1/(2β)<p<1+2β. When n=d and Φ(y)= A φ (y), Al-Salman [18] obtained the following result.

Theorem A Let n=d and Φ(y)= A φ (y). Suppose that Ω F β ( S n 1 ) for some β>1 with satisfying (1.1)-(1.2).

  1. (i)

    If φ(t)=P(t) with P being a real polynomial on ℝ, then M Ω Φ are bounded on L p ( R n ) for 2β/(2β1)<p<2β. The bounds are independent of the coefficients of P.

  2. (ii)

    If φF, then M Ω Φ are bounded on L p ( R n ) for 2β/(2β1)<p<2β. Here F is the set of all functions ϕ which satisfy:

    1. (a)

      ϕ: R + R + is continuous increasing C 1 function satisfying that ϕ is monotonous;

    2. (b)

      there exist constant C ϕ and c ϕ such that t ϕ (t) C ϕ ϕ(t) and ϕ(2t) c ϕ ϕ(t) for all t>0.

Remark 1.1 There are some model examples in the class F, such as t α (α>0), t α ( ln ( 1 + t ) ) β (α,β>0), tlnln(e+t), real-valued polynomials P on ℝ with positive coefficients and P(0)=0 and so on. For φF, there exists a constant B φ >1 such that φ(2t) B φ φ(t) (see [19]).

It is natural to ask whether Theorem A also holds if the range of β is relaxed to β>1/2 and the range of p is relaxed to 1+1/(2β)<p<1+2β. In this paper, we will give an affirmative answer to this question. Our main results can be stated as follows.

Theorem 1.1 Let n=d and Φ(y)=( P 1 (φ(ρ(y))) y 1 ,, P n (φ(ρ(y))) y n ) with P i (t) being real valued polynomials onsatisfying P i (0)=0 and φF. Suppose that Ω F β ( S n 1 ) for some β>1/2 satisfying (1.1)-(1.2). Then M Ω Φ are bounded on L p ( R n ) for 1+1/(2β)<p<1+2β. The bounds are independent of the coefficients of P j for all 1jn but depend on max 1 j d deg( P j ) and φ.

Theorem 1.2 Let n=d and Φ(y)= A P N ( φ ) (y) with φF and P N (t)= i = 1 N a i t i and P N (t)>0 if t0. Suppose that Ω F β ( S n 1 ) for some β>1/2 satisfying (1.1)-(1.2). Then M Ω Φ are bounded on L p ( R n ) for 1+1/(2β)<p<1+2β. The bounds are independent of the coefficients of P N but depend on N and φ.

Remark 1.2 It is clear that Theorem 1.1 implies Theorem 1.2. When α 1 == α n =1, φ(t)=t and P 1 (t)== P n (t)= i = 1 N a i t i , Theorem 1.1 implies the result of [12]. In fact, Theorem 1.2 with φ(t)=t extends the result of [12] to the mixed case. Comparing Theorem A with Theorem 1.2, the range of β is extended to the case β>1/2 and the range of p is enlarged to the case 1+1/(2β)<p<1+2β. Thus Theorem 1.2 essentially improves and generalizes the corresponding results in Theorem A. In addition, Theorem 1.2 implies the result [[17], Theorem 1.3] when P N (t)=φ(t)=t.

When n=2, we have the following result.

Theorem 1.3 Let ϕ=( ϕ 1 ,, ϕ d ) be real analytic on S 1 . Let Φ(y)= P N (φ(ρ(y)))ϕ( y )=( P N (φ(ρ(y))) ϕ 1 ( y ),, P N (φ(ρ(y))) ϕ d ( y )) with P N (t)= i = 1 N a i t i and φF. Suppose that Ω F β ( S 1 ) for some β>1 satisfying (1.1)-(1.2). Then M Ω Φ are bounded on L p ( R d ) for 1+1/(2β)<p<1+2β. The bounds are independent of the coefficients of P N but depend on φ and N.

We remark that when α 1 == α n =1 and φ(t)=t, the surface {Φ(y):y R n } given as in Theorem 1.3 recovers {( P N (|y|) ϕ 1 ( y ),, P N (|y|) ϕ d ( y ));y R n }, which was originally introduced by Al-Balushi and Al-Salman [20] in the study of L p bounds of singular integrals associated to certain surfaces.

The third type of surfaces we consider are polynomial compound subvarieties. To state the rest of our result, we need to recall some notations. Let A(n,m) be the set of polynomials on R n which have real coefficients and degrees not exceeding m, and let V(n,m) be the collection of polynomials in A(n,m) which are homogeneous of degree m. For PA(n,m), we set

P= | λ | m a λ y λ = ( | λ | m | a λ | 2 ) 1 / 2 .

Definition 1.1 ([21])

Let n2, mN and β>0. An integrable function Ω on S n 1 is said to be in the space F(n,m,β) if

sup P V ( n , m ) , P = 1 S n 1 |Ω(y)| ( log + 1 | P ( y ) | ) β dσ(y)<.
(1.5)

It should be pointed out that the condition (1.5) was introduced by Al-Salman and Pan [21] (also see [22]) in a study of the L p boundedness of singular integrals with rough kernels. It is easy to check that F(n,1,β)= F β ( S n 1 ). Moreover, it was shown in [21] that

F β ( S 1 ) = m = 1 F(2,m,β).
(1.6)

The rest of the results can be stated as follows.

Theorem 1.4 Let P=( P 1 ,, P d ) with P j : R n R being a polynomial for 1jd. Let Φ(y)=P(φ(ρ(y)) y ) and φF. Suppose that Ω satisfies (1.1)-(1.2) and Ω s = 1 F(n,s,β) for some β>1/2. Then M Ω Φ are bounded on L p ( R d ) for 1+1/(2β)<p<1+2β. The bounds are independent of the coefficients of P j for all 1jd but depend on max 1 j d deg( P j ) and φ.

Theorem 1.5 Let P=( P 1 ,, P d ) with P j : R 2 R being a polynomial for 1jd. Let Φ(y)=P(φ(ρ(y)) y ) and φF. Suppose that Ω satisfies (1.1)-(1.2) and Ω F β ( S 1 ) for some β>1/2. Then M Ω Φ are bounded on L p ( R d ) for 1+1/(2β)<p<1+2β. The bounds are independent of the coefficients of P j for all 1jd but depend on max 1 j d deg( P j ) and φ.

Remark 1.3 When α j =1 (j=1,,n), Theorem 1.4 implies Theorem 1.2 with ρ=1 in [13]. Obviously, Theorem 1.5 follows from Theorem 1.4 because of (1.6).

The rest of this paper is organized as follows. After recalling some preliminary notations and lemmas in Section 2, we will prove our results in Section 3. We would like to remark that the main methods employed in this paper is a combination of ideas and arguments from [12, 21, 23]. The main ingredient in our proofs is to give a systematic treatment with these operators mentioned above.

Throughout this paper, we let p satisfy 1/p+1/ p =1. The letter C, sometimes with additional parameters, will stand for positive constants, not necessarily the same one at each occurrence, but independent of the essential variables.

2 Preliminaries

Lemma 2.1 Let { σ j , t } be a family of measures. Suppose that

sup j Z sup t > 0 | | σ j , t | g | p C g p

holds for some p>1 and g L p ( R n ). Then there exists a constant C>0 such that

( 1 2 j Z | σ j , t g j | 2 d t ) 1 / 2 p C ( j Z | g j | 2 ) 1 / 2 p

for arbitrary functions { g j } j Z L p ( 2 , R n ).

Proof By the assumption, we have

sup j Z sup t [ 1 , 2 ] | σ j , t g j | p sup j Z sup t > 0 | σ j , t | sup j Z | g j | p C sup j Z | g j | p .

On the other hand, by the dual argument, there exists a function h L p ( R n ) satisfying h p =1 such that

1 2 j Z | σ j , t g j | d t p = R n j Z 1 2 | σ j , t g j ( x ) | d t h ( x ) d x R n j Z 1 2 | σ j , t | | g j ( x ) | d t | h ( x ) | d x R n j Z | g j ( x ) | sup j Z sup t [ 1 , 2 ] | σ j , t | | h ¯ | ( x ) d x j Z | g j | p sup j Z sup t > 0 | σ j , t | | h ¯ | ( x ) p C j Z | g j | p ,

where h ¯ (x)=h(x). Thus, Lemma 2.1 follows from the standard interpolation arguments. □

Let { a k } k Z be a sequence of real positive numbers with satisfying inf k Z a k + 1 / a k =a>1. Let { λ k } k Z be a collection of C (0,) functions satisfying the following conditions:

supp ( λ k ) [ a k + 1 1 , a k 1 1 ] ; 0 λ k 1 ; k Z λ k 2 ( t ) = 1 ; | d λ k ( t ) / d t | C / t ,

where C is independent of t and k. Let MN{0} and L: R n R M be a linear transformation. For each kZ, we define the multiplier operators S k in R n by

S k f ˆ (ξ)= λ k ( | L ( ξ ) | ) f ˆ (ξ).
(2.1)

By an argument which is similar to those used in [[8], Proposition 3.1], one can easily get the following lemma. The details are omitted here.

Lemma 2.2 Let S k be as in (2.1) and { g j , k , t } arbitrary functions on R n . Then

  1. (i)

    for each fixed 1<p<2 and 1<q<p,

    ( j Z 1 2 | k Z S j + k g j , k , t | 2 d t ) 1 / 2 p q C k Z ( j Z 1 2 | g j , k , t | 2 d t ) 1 / 2 p q ;
    (2.2)
  2. (ii)

    for each fixed 2<p< and 1<q< p ,

    ( j Z 1 2 | k Z S j + k g j , k , t | 2 d t ) 1 / 2 p q C k Z ( 1 2 ( j Z | g j , k , t | 2 ) 1 / 2 p 2 d t ) q / 2 .
    (2.3)

The following lemma is our main ingredient in the proof of our main results.

Lemma 2.3 Let { τ k , t :kZ,t R + } be a family of uniformly bounded Borel measures on R n . Let { a k } k Z be a sequence of real numbers with satisfying inf k Z a k + 1 / a k =a>1. Let MN{0} and L: R n R M be a linear transformation. Suppose that

(i)| τ k , t ˆ (ξ)|Cmin { 1 , a k | L ( ξ ) | } ;
(2.4)
(ii)| τ k , t ˆ (ξ)|C ( log | a k L ( ξ ) | ) β for some β>0, if  a k |L(ξ)|>1;
(2.5)
(iii) sup k Z sup t > 0 | | τ k , t | f | q C f q
(2.6)

for all 1<q<. Then for p(1+1/(2β),1+2β) and β>1/2, there exists a constant C(a)>0 such that

( 1 2 k Z | τ k , t f | 2 d t ) 1 / 2 p C(a) f p .

Proof Let S k be as in (2.1). Then we can write

G ( f ) ( x ) : = ( 1 2 k Z | τ k , t f ( x ) | 2 d t ) 1 / 2 = ( 1 2 k Z | τ k , t ( j Z S j + k S j + k f ) ( x ) | 2 d t ) 1 / 2 = ( k Z 1 2 | j Z S j + k ( τ k , t S j + k f ) ( x ) | 2 d t ) 1 / 2 .
(2.7)

Case 1. 1+1/(2β)<p<2. It follows from (2.2) and (2.7) that

G ( f ) p q C j Z ( k Z 1 2 | τ k , t S j + k f | 2 d t ) 1 / 2 p q ,1<q<p.
(2.8)

For each fixed kZ, we set

I j f(x):= ( k Z 1 2 | τ k , t S j + k f ( x ) | 2 d t ) 1 / 2 .

Invoking Lemma 2.1 and the Littlewood-Paley theory imply

I j f p C ( k Z | S j + k f | 2 ) 1 / 2 p C f p ,1<p<.
(2.9)

On the other hand, by Plancherel’s theorem and (2.4)-(2.5), we have

I j f 2 2 = 1 2 k Z R n | f ˆ ( ξ ) | 2 | λ j + k ( | L ( ξ ) | ) | 2 | τ k , t ˆ ( ξ ) | 2 d ξ d t C 1 2 k Z { a j + k + 1 1 | L ( ξ ) | a j + k 1 1 } | f ˆ ( ξ ) | 2 | τ k , t ˆ ( ξ ) | 2 d ξ d t C B j 2 f 2 2 ,

where B j = a j + 1 χ { j 1 } + ( | j + 1 | log a ) β χ { j < 1 } . That is,

I j f 2 C B j f 2 .
(2.10)

Interpolating between (2.9) and (2.10), there exists ϵ(2/(2β+1),1) such that

I j f p C B j ϵ f p ,1+1/(2β)<p<2.

For fixed p(1+1/(2β),2) and β>1/2, we can choose q(1,p) such that qϵβ>1. Thus

j Z I j f p q C ( j 1 a q ϵ ( j 1 ) + j < 1 ( | j + 1 | log a ) q ϵ β ) f p q C(a) f p q ,

which, together with (2.8), implies

G ( f ) p C(a) f p ,for 1+1/(2β)<p<2.
(2.11)

Case 2. 2<p<1+2β. By (2.3) and (2.7), we have for 2<p< and 1<q< p ,

G ( f ) p q C j Z ( 1 2 ( k Z | τ k , t S j + k f | 2 ) 1 / 2 p 2 d t ) q / 2 .
(2.12)

Let

J j , t f(x):= ( k Z | τ k , t S j + k f ( x ) | 2 ) 1 / 2 .

By (2.6), [[23], p.544, Lemma] and the Littlewood-Paley theory, we have, for kZ and t[1,2],

J j , t f p 0 C ( k Z | S j + k f | 2 ) 1 / 2 p 0 C f p 0 ,1< p 0 <.
(2.13)

On the other hand, by the same arguments as in (2.10), we have

J j , t f 2 C B j f 2 ,
(2.14)

where B j is as in (2.10). On interpolation between (2.13) and (2.14), for fixed p(2,1+2β) and β>1/2, we can choose q(1, p ) and δ(2/(2β+1),1) such that qδβ>1 and

J j , t f p C B j δ f p ,for 2<p<1+2β.

This, combined with (2.12), implies

G ( f ) p q C ( j 1 a q δ ( j 1 ) + j < 1 ( | j + 1 | log a ) q δ β ) f p q C(a) f p q ,

which, together with (2.11), completes the proof of Lemma 2.3. □

Lemma 2.4 ([[13], Lemma 2.2])

Suppose Φ(t)= t α 1 + μ 2 t α 2 ++ μ n t α n and φF, where μ 2 ,, μ n are real parameters, and α 1 ,, α n are distinct positive (not necessarily integer) exponents. Then for any r>0 and λR{0},

| r / 2 r exp ( i λ Φ ( φ ( t ) ) ) d t t |C(φ)|λφ ( r ) α 1 | ϵ ,

where ϵ=min{1/ α 1 ,1/n} and C(φ) does not depend on μ 2 ,, μ n .

Lemma 2.5 ([[24], Lemma 2.2])

Let P(t)=( P 1 (t),, P d (t)) with P j being real polynomials defined on R + . Suppose that φF. Then the operator M P , φ defined by

M P , φ (f)(x)= sup r > 0 r 2 r |f ( x P ( φ ( t ) ) ) | d t t

is bounded on L p ( R d ) for 1<p<. The bound is independent of the coefficients of P j for all 1jd and f but depends on φ.

Lemma 2.6 ([25])

Let Φ: S 1 R d , Φ=( Φ 1 ,, Φ d ) be real analytic on S 1 . Suppose that { Φ 1 ,, Φ d } is linearly independent set. If Ω F β ( S 1 ) for some β>1, then

sup ξ S d 1 S 1 |Ω ( y ) | ( log + 1 | ξ Φ ( y ) | ) β dσ ( y ) <.

3 Proofs of main theorems

Proof of Theorem 1.1 Let N= max 1 j n deg( P j ). For 1ln, let P l (t)= i = 1 N a i , l t i . For 1sN, and 1ln, let P l ( s ) (t)= i = 1 s a i , l t i and P ( s ) (t)=( P 1 ( s ) (t),, P n ( s ) (t)). Set P ( 0 ) (t)=0 and

Φ s (y)= ( P 1 ( s ) ( φ ( ρ ( y ) ) ) y 1 , , P n ( s ) ( φ ( ρ ( y ) ) ) y n ) .

Then we can write

Φ s ( y ) ξ = l = 1 n ξ l y l P l ( s ) ( φ ( ρ ( y ) ) ) = l = 1 n i = 1 s ξ l y l a i , l φ ( ρ ( y ) ) i = i = 1 s ( L i ( ξ ) y ) φ ( ρ ( y ) ) i ,
(3.1)

where L i : R n R n is the linear transformation given by

L i (ξ)=( a i , 1 ξ 1 ,, a i , n ξ n ).

For each jZ, t R + and 1sN, we define the measures { σ j , t s } and {| σ j , t s |} by

σ j , t s ˆ ( ξ ) = 1 2 j t 2 j 1 t < ρ ( y ) 2 j t exp ( 2 π i Φ s ( y ) ξ ) Ω ( y ) ρ ( y ) α 1 d y , | σ j , t s | ˆ ( ξ ) = 1 2 j t 2 j 1 t < ρ ( y ) 2 j t exp ( 2 π i Φ s ( y ) ξ ) | Ω ( y ) | ρ ( y ) α 1 d y .

We get from (3.1)

| σ j , t s ˆ ( ξ ) σ j , t s 1 ˆ ( ξ ) | 1 2 j t 2 j 1 t < ρ ( y ) 2 j t | exp ( 2 π i Φ s ( y ) ξ ) exp ( 2 π i Φ s 1 ( y ) ξ ) | | Ω ( y ) | ρ ( y ) α 1 d y C ( φ ( 2 j t ) s | L s ( ξ ) | ) .
(3.2)

On the other hand, by a change of variable, we have

| σ j , t s ˆ ( ξ ) | = 1 2 j t | 2 j 1 t 2 j t S n 1 exp ( 2 π i k = 1 s L k ( ξ ) θ φ ( r ) k ) Ω ( θ ) J ( θ ) d σ ( θ ) d r | C S n 1 | Ω ( θ ) | | I j , t , s , ξ ( θ ) | d σ ( θ ) ,
(3.3)

where

I j , t , s , ξ (θ):= 1 2 j t 2 j 1 t 2 j t exp ( 2 π i k = 1 s L k ( ξ ) θ φ ( r ) k ) dr.

By Lemma 2.4, we have

| I j , t , s , ξ (θ)|C|φ ( 2 j t ) s L s (ξ)θ | 1 / s .

Combining the trivial inequality | I j , t , s , ξ (θ)|C with the fact that t/ ( log t ) β is increasing in ( e β ,), we have

| I j , t , s , ξ (θ)|C ( log e β s | η θ | 1 ) β ( log | φ ( 2 j t ) s L s ( ξ ) | ) β ,if |φ ( 2 j t ) s L s (ξ)|>1,
(3.4)

where η= L s (ξ)/| L s (ξ)|. This, together (3.3) with the fact that Ω F β ( S n 1 ), implies

| σ j , t s ˆ (ξ)|C ( log | φ ( 2 j t ) s L s ( ξ ) | ) β ,if |φ ( 2 j t ) s L s (ξ)|>1.
(3.5)

Now we can choose a function ψ C 0 (R) such that ψ(t)1 for |t|1/2 and ψ(t)0 for |t|>1. For 1sN, jZ and t R + , we define the measures { τ j , t s } by

τ j , t s ˆ ( ξ ) = σ j , t s ˆ ( ξ ) k = s + 1 N ψ ( | φ ( 2 j t ) k L k ( ξ ) | ) σ j , t s 1 ˆ ( ξ ) k = s N ψ ( | φ ( 2 j t ) k L k ( ξ ) | ) .

Here we use the convention j a j =1. It is easy to see that

σ j , t N = s = 1 N τ j , t s .
(3.6)

It follows from (3.2), (3.5), and the trivial estimate | σ j , t s ˆ (ξ)|C that, for 1sN,

| τ j , t s ˆ (ξ)|C(φ)min { 1 , φ ( 2 j t ) s | L s ( ξ ) | } ,
(3.7)
| τ j , t s ˆ (ξ)|C(φ) ( log | φ ( 2 j t ) s L s ( ξ ) | ) β ,if |φ ( 2 j t ) s L s (ξ)|>1.
(3.8)

By the definition of σ j , t s and (3.6), we can write

M Ω Φ ( f ) ( x ) = ( 0 | j = 0 2 j σ j , t N f ( x ) | 2 d t t ) 1 / 2 j = 0 2 j ( 0 | σ j , t N f ( x ) | 2 d t t ) 1 / 2 2 ( j Z 2 j 2 j + 1 | σ 0 , t N f ( x ) | 2 d t t ) 1 / 2 2 ( j Z 1 2 | σ j , t N f ( x ) | 2 d t t ) 1 / 2 2 s = 1 N ( 1 2 j Z | τ j , t s f ( x ) | 2 d t t ) 1 / 2 .
(3.9)

On the other hand, by a change of variable we have

| | σ j , t s | f ( x ) | 1 2 j t 2 j 1 t < ρ ( y ) 2 j t | f ( x Φ s ( y ) ) | | Ω ( y ) | ρ ( y ) α 1 d y C S n 1 | Ω ( y ) | ( 1 2 j t 2 j 1 t 2 j t | f ( x Φ s ( A r y ) ) | d r ) d σ ( y ) C S n 1 | Ω ( y ) | M P , φ ( f ) ( x ) d σ ( y ) ,

where M P , φ is as in Lemma 2.5 and P(t)=( P 1 ( s ) (t) y 1 ,, P n ( s ) (t) y n ). By Lemma 2.5 and Minkowski’s inequality, we have

sup j Z sup t > 0 | | σ j , t s | f | p C f p .

This inequality, together with the definition of τ j , t s , yields

sup j Z sup t > 0 | | τ j , t s | f | p C f p .
(3.10)

Then Theorem 1.1 follows from (3.7)-(3.10) and Lemma 2.3. □

Proof of Theorem 1.3 Let Φ, P N , φ, ϕ be as in Theorem 1.3. For 1sN, we set P s (t)= m = 1 s a m t m . Define the measures { σ j , t s } and {| σ j , t s |} by

σ j , t s ˆ (ξ)= 1 2 j t 2 j 1 t < ρ ( y ) 2 j t exp ( 2 π i P s ( φ ( ρ ( y ) ) ) ϕ ( y ) ξ ) Ω ( y ) ρ ( y ) α 1 dy;
(3.11)
| σ j , t s | ˆ (ξ)= 1 2 j t 2 j 1 t < ρ ( y ) 2 j t exp ( 2 π i P s ( φ ( ρ ( y ) ) ) ϕ ( y ) ξ ) | Ω ( y ) | ρ ( y ) α 1 dy.
(3.12)

Following the notation in [20], let { ϕ i 1 ,, ϕ i l } be a maximal linearly independent subset of { ϕ 1 ,, ϕ d }, where 1ld, 1 i r d and r=1,,l. Thus, for j{ i 1 ,, i l }, there exist a ( j ) =( a j , 1 ,, a j , l ) R l such that

ϕ j ( y ) = a ( j ) ( ϕ i 1 ( y ) , , ϕ i l ( y ) ) = k = 1 l a j , k ϕ i k ( y ) .

This implies that there exists a linear transformation L: R d R l such that

ϕ ( y ) ξ=L(ξ) ϕ ˜ ( y ) ,ξ R d ,

where ϕ ˜ ( y )=( ϕ i 1 ( y ),, ϕ i l ( y )). Thus

| σ j , t s ˆ (ξ)|C S 1 |Ω ( z ) || I j , t , s , ξ ( z ) |dσ ( z ) ,

where

I j , t , s , ξ ( z ) := 1 2 j t 2 j 1 t 2 j t exp [ 2 π i P s ( φ ( u ) ) ( L ( ξ ) ϕ ˜ ( z ) ) ] du.

By Lemma 2.4, we have

| I j , t , s , ξ ( z ) |C|φ ( 2 j t ) s a s L(ξ) ϕ ˜ ( z ) | 1 / s .
(3.13)

Since t/ ( log t ) β is increasing in ( e β ,) for any β>0, and | ϕ ˜ ( z )|B with B>1 for any z S n 1 . We can deduce from (3.13) and the trivial estimate | I j , t , s , ξ ( z )|C that

| I j , t , s , ξ ( z ) |C ( log ( B e β s | ζ ϕ ˜ ( z ) | 1 ) ) β ( log | φ ( 2 j t ) s a s L ( ξ ) | ) β ,if |φ ( 2 j t ) a s L(ξ)|>1,
(3.14)

where ζ=L(ξ)/|L(ξ)|. Invoking Lemma 2.6 and (3.14), we obtain, for β>1,

| σ j , t s ˆ (ξ)|C ( log | φ ( 2 j t ) s a s L ( ξ ) | ) β ,if |φ ( 2 j t ) s a s L(ξ)|>1.
(3.15)

On the other hand, we have

| σ j , t s ˆ ( ξ ) σ j , t s 1 ˆ ( ξ ) | 1 2 j t 2 j 1 t < ρ ( y ) 2 j t | exp ( 2 π i ( a s φ ( ρ ( y ) ) s | L ( ξ ) | ϕ ˜ ( y ) ) ) 1 | | Ω ( y ) | ρ ( y ) α 1 d y C | φ ( 2 j t ) s a s L ( ξ ) | S 1 | Ω ( z ) | | ϕ ˜ ( z ) | d σ ( z ) C | φ ( 2 j t ) s a s L ( ξ ) | sup z S 1 | ϕ ˜ ( z ) | C | φ ( 2 j t ) s a s L ( ξ ) | .
(3.16)

Notice that

| | σ j , t s | f ( x ) | 1 2 j t 2 j 1 t < ρ ( y ) 2 j t | f ( x P s ( φ ( ρ ( y ) ) ) ϕ ( y ) ) | | Ω ( y ) | ρ ( y ) α 1 d y C S 1 | Ω ( y ) | ( 0 2 j t | f ( x P s ( φ ( u ) ) ϕ ( y ) ) | d u u ) d σ ( y ) .

This combining with Lemma 2.5 and Minkowski’s inequality, implies

sup j Z sup t > 0 | | σ j , t s | f | p C f p .
(3.17)

Then the rest of the proof of Theorem 1.3 follows from an argument which is similar to those in the proof of Theorem 1.1 and (3.15)-(3.17). We omit the details. □

Proof of Theorem 1.4 Let n2, P=( P 1 ,, P d ), where P j : R n R is a polynomial for 1jd. Let

M=max { deg ( P 1 ) , , deg ( P d ) } ,

and

P i (y)= | γ | M a i γ y γ for i=1,,d.

For 1sM, we let

P ( s ) =( P 1 , s ,, P d , s ),

where

P i , s (y)= | γ | s a i γ y γ .

Set P ( 0 ) =0 and Φ s (y)= P ( s ) (φ(ρ(y)) y ).

For each jZ, t R + and 0sM, we define the measures { σ j , t s } and {| σ j , t s |} by

σ j , t s ˆ ( ξ ) = 1 2 j t 2 j 1 t < ρ ( y ) 2 j t exp ( 2 π i Φ s ( y ) ξ ) Ω ( y ) ρ ( y ) α 1 d y ; | σ j , t s | ˆ ( ξ ) = 1 2 j t 2 j 1 t < ρ ( y ) 2 j t exp ( 2 π i Φ s ( y ) ξ ) | Ω ( y ) | ρ ( y ) α 1 d y .

For 1sM, let l s denote the number of multi-indices γ=( γ 1 ,, γ n ) satisfying |γ|=s, and define the linear transformation L s : R d R l s by

L s (ξ)= ( ( L s ( ξ ) ) γ ) | γ | = s = ( i = 1 d a i γ ξ i ) | γ | = s .

By the change of variables, we have

| σ j , t s ˆ ( ξ ) | = | S n 1 1 2 j t 2 j 1 t 2 j t exp ( 2 π i ξ P ( s ) ( φ ( u ) θ ) ) d u Ω ( θ ) J ( θ ) d σ ( θ ) | C S n 1 | Ω ( θ ) | | 1 2 j t 2 j 1 t 2 j t exp ( 2 π i ξ P ( s ) ( φ ( u ) θ ) ) d u | d σ ( θ ) C S n 1 | Ω ( θ ) | | 1 2 j t 2 j 1 t 2 j t exp ( 2 π i | γ | s i = 1 d ξ j a j γ θ γ φ ( u ) | γ | ) d u | d σ ( θ ) C S n 1 | Ω ( θ ) | | J j , t , s , ξ ( θ ) | d σ ( θ ) ,

where

J j , t , s , ξ (θ):= 1 2 j t 2 j 1 t 2 j t exp ( 2 π i | γ | = s ( L s ( ξ ) ) γ θ γ φ ( u ) s + lower powers in  u ) du.

Let

Q s , ξ (θ):=| L s (ξ) | 1 | γ | = s ( L s ( ξ ) ) γ θ γ .

By Lemma 2.4, we have

| J j , t , s , ξ (θ)|C ( φ ( 2 j t ) s | L s ( ξ ) | | Q s , ξ ( θ ) | ) 1 / s .

By this inequality, together with the trivial estimate | J j , t , s , ξ (θ)|C, we get

| J j , t , s , θ (ξ)|C ( log ( e β s l s 1 / 2 | Q s , ξ ( θ ) | 1 ) ) β ( log | φ ( 2 j t ) s L s ( ξ ) | ) β ,if |φ ( 2 j t ) s L s (ξ)|>1.

Since Ω s = 1 F(n,s,β), Q s , ξ V(n,s) and Q s , ξ =1, we immediately obtain

| σ j , t s ˆ (ξ)|C ( log | φ ( 2 j t ) s L s ( ξ ) | ) β ,if |φ ( 2 j t ) s L s (ξ)|>1.
(3.18)

On the other hand, we have

| σ j , t s ˆ ( ξ ) σ j , t s 1 ˆ ( ξ ) | 1 2 j t 2 j 1 t < ρ ( y ) 2 j t | exp ( 2 π i j = 1 d | ι | = s ξ j a j ι φ ( ρ ( y ) ) s ( y ) ι ) 1 | | Ω ( y ) | ρ ( y ) α 1 d y C | φ ( 2 j t ) s L s ( ξ ) | .
(3.19)

In addition, using Lemma 2.5, one can easily check that

sup j Z sup t > 0 | | σ j , t s | f | p C f p .
(3.20)

Then the rest proof of Theorem 1.4 follows from similar arguments to the proof of Theorem 1.1 and (3.18)-(3.20). Details will be omitted. □