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https://doi.org/10.1007/s11425-017-9193-1

Sharp Lp decay of oscillatory integral operators with certain homogeneous polynomial hases in several variables Shaozhen Xu∗ & Dunyan Yan School of Mathematical Sciences, University of Chinese Academy of Sciences, Beijing 100049, China Email: [email protected], [email protected] Received August 6, 2017; accepted October 6, 2017

Abstract

In this paper, we obtain the Lp decay of oscillatory integral operators Tλ with certain homogeneous

polynomial phase functions of degree d in (n + n)-dimensions; we require that d > 2n. If d/(d − n) < p < d/n, the decay is sharp and the decay rate is related to the Newton distance. For p = d/n or d/(d − n), we obtain the almost sharp decay, where “almost” means that the decay contains a log(λ) term. For otherwise, the Lp decay of Tλ is also obtained but not sharp. Finally, we provide a counterexample to show that d/(d − n) 6 p 6 d/n is not necessary to guarantee the sharp decay. Keywords MSC(2010)

oscillatory integral operators, sharp Lp decay, several variables, Newton distance 42B20, 47G10

Citation: Xu S Z, Yan D Y. Sharp Lp decay of oscillatory integral operators with certain homogeneous polynomial phases in several variables. Sci China Math, 2018, 61, https://doi.org/10.1007/s11425-017-9193-1

1

Introduction

We consider the following oscillatory operator: ∫ Tλ (f )(x) = eiλS(x,y) ψ(x, y)f (y)dy, Rn

n > 2,

(1.1)

where x ∈ Rn , ψ(x, y) is a smooth function supported in a compact neighborhood of the origin and ∑ S(x, y) = |α|+|β|=d aα,β xα y β is a real-valued homogeneous polynomial in higher dimensions with degree d, where α and β are multi-indices. Research on this operator focuses on the decay of the Lp bound as the parameter λ tends to infinity. In one-dimensional case, Phong and Stein [7–10] made significant contributions to the research on this subject. In a series of articles [7–10], they developed an almostorthogonality method to obtain the sharp L2 decay of oscillatory integral operators with phase functions varying from homogeneous polynomials to real-valued analytic functions. They also clarified the relation between decay rate and Newton distance proposed by Arnold et al. [1]. Later, the sharp L2 estimate was extended to C ∞ phases by Rychkov [12] and Greenblatt [2]. Greenleaf and Seeger [4] obtained the endpoint estimates for the Lp decay rate of Tλ when S is smooth and Tλ has a two-sided Whitney fold. * Corresponding author c Science China Press and Springer-Verlag GmbH Germany 2018 ⃝

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Yang [18] obtained a sharp endpoint estimate with the assumption a1,d−1 ad−1,1 ̸= 0, where aα,β are coefficients of the homogeneous polynomial phase function S(x, y) in R × R. Shi and Yan [13] established the sharp endpoint Lp decay for arbitrary homogeneous polynomial phase functions. Later, Xiao [17] extended this result to arbitrary analytic phases and presented a very specific review for this subject. Higher-dimensional cases, even those of the L2 estimate, have not been well understood. The onedimensional result of L2 decay has been partially extended to (2+1)-dimensions by Tang [16]. However, a remarkable study on a higher-dimension case has been reported [3]. Greenleaf et al. [3] obtained the L2 estimate for oscillatory integral operators with homogeneous polynomials that satisfy various genericity assumptions. Inspired by the method used in [13, 18], we prove our main result by embedding Tλ into a family of analytic operators and using complex interpolation. This method requires us to establish the L2 -L2 decay estimate as well as the H 1 -L1 boundedness of operators with different amplitude functions. Before we state our main theorem, some definitions should be illustrated. ∑ Definition 1.1 (See [3]). If S(x, y) ∈ C ω (RnX × RnY ) with Taylor series α,β aα,β xα y β that have no pure x- or z-terms, we denote the reduced Newton polyhedron by ( ∪ ) nX +nY N0 (S) = convex hull (α, β) + R+ . aα,β ̸=0

Then the Newton polytope of S(x, y) (at (0, 0)) is N (S) := ∂(N0 (S)), and the Newton distance δ(S) of S is then δ(S) := inf{δ −1 > 0 : (δ −1 , . . . , δ −1 ) ∈ N (S)}. These definitions correspond to the one-dimension definitions provided in [9]. In our main theorem, the next definition is necessary. Definition 1.2.

Denote the Hilbert-Schmidt norm of a matrix A = (aij ) by ∥A∥HS = (tr(A · A )) T

1/2

=

(∑

)1/2 |aij |

2

.

(1.2)

i,j

Denote by S d (Rn × Rn ) the space of homogeneous polynomials of degree d on Rn × Rn . In fact, for oscillatory operators with homogeneous polynomial phases, we are only interested in polynomial phase functions not containing pure x- or y-terms since these leave the operator norm unchanged. Thus, we denote the space consisting of such polynomials by Od (Rn × Rn ). Now, we formulate our main result. Theorem A. then it follows

′′ Suppose S(x, y) ∈ Od (Rn × Rn ) and d > 2n > 4. If ∥Sxy ∥HS

1/(d−2)

  λ−δ/2 ,    λ−δ/2 (log(λ))δ , ∥Tλ ∥p . ′  λ−1/p ,     −1/p λ ,

is a norm of Rn × Rn ,

d/(d − n) < p < d/n, p = d/n

or

p = d/(d − n),

1 < p < d/(d − n),

(1.3)

d/n < p < ∞,

where δ is the Newton distance. If d/(d − n) < p < d/n, the decay is sharp. If p = d/n or d/(d − n), the decay is sharp except possibly for a log(λ) term. In addition, d/(d − n) 6 p 6 d/n is not necessary to guarantee the sharp decay. To clarify the relation between Newton distance and the Lp decay rate, a proposition in [3] should be mentioned.

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Proposition 1.3.

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3

If S(x, y) ∈ Od (Rn × Rn ) satisfies the rank one condition ′′ rank(Sxy ) > 1,

for all

(x, y) ̸= (0, 0),

then δ(S) = 2n/d. Obviously, S(x, y) satisfies the rank one condition because of the assumptions in Theorem A; thus, the Newton distance in Theorem A is actually 2n/d. The main tool in our proof is the interpolation of analytic families of operators, which was due to Stein and Weiss [15]. Here, the analytic families of operators are ∫ ′′ z z Tλ (f )(x) = eiλS(x,y) ∥Sxy ∥HS ψ(x, y)f (y)dy, z = σ + it ∈ C. (1.4) Rn

Especially, Tλ0 = Tλ . Theorem A naturally follows from the interpolation between L2 -L2 decay of Tλz and the H 1 -L1 mapping property of Tλz as well as dual arguments. The layout of this paper is as follows. In Section 2, we obtain the L2 -L2 decay estimate of Tλz . In Section 3, we prove the H 1 -L1 boundedness of Tλz . In Section 4, we obtain the optimality of decay and give an example to demonstrate our main theorem and a counterexample to show that d/(d − n) 6 p 6 d/n is not necessary to guarantee the sharp decay. Notation. In this paper, the notation C denotes a positive constant used in the usual way, and it may vary according to different conditions.

2

L2 decay of the damped oscillatory integral operators

In this section, we desire the following result. Theorem 2.1. Set σ1 = −n/(d − 2), σ2 = (d − 2n)/(2(d − 2)); moreover, we consider the operators defined in (1.4). If the Hessian of its phase function satisfies the rank one condition, then the following estimates hold:   σ > σ2 , Cz |λ|−1/2 ,  ∥Tλz ∥2 .

Cz |λ|−1/2 log(λ),   C |λ|−[(d−2)σ+n]/d , z

σ = σ2 ,

(2.1)

σ1 < σ < σ2 .

Our proof roughly follows the pattern in [3, 16], in which the authors offered a good perspective of higher-dimensional oscillatory integral operators. They combined dyadic decompositions of the entire space and the local H¨ormander lemma (see [5]) on a dyadic shell to obtain the next lemma. Lemma 2.2 (See [3]). rank one condition

′′ satisfies the For a homogeneous phase function S(x, y) of degree d, where Sxy ′′ rank(Sxy ) > 1,

for all

on RnX × RnY (nX > nY > 2), there holds  −(nX +nY )/(2d)  ,  Cλ ∥Tλ ∥2 6

−1/2

Cλ log λ,   Cλ−1/2 ,

(x, y) ̸= (0, 0)

if d > nX + nY , if d = nX + nY ,

(2.2)

if 2 6 d < nX + nY .

Our proof is different in that we combine the dyadic decomposition and local oscillatory estimate (see [8, Lemma 1.1]). Now, we address our proof of Theorem 2.1. Proof of Theorem 2.1. Since the support of ψ(x, y) is compact, we may assume that supp (ψ) is contained in {(x, y) : |(x, y)| 6 1}. Considering the compactness of the sphere |(x, y)| = 1, we can make a partition of unity over the unit sphere and then extend it to a partition of unity on R2n \ {0}, homogeneous of degree 0. Thus, to conclude the result of (2.1), it suffices to show that for each point on the unit

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sphere of Rn × Rn , an operator supported in one of its sufficiently small convex conic neighborhood has ∑∞ the desired decay rate. Decompose the unit ball by a dyadic partition of unity {ak }, k=0 ak (x, y) ≡ 1, and supp (ak ) ⊆ {2−k−1 < |(x, y)| 6 2−k+1 }. Set ψk = ψak and

∫ z Tλ,k f (x) =

Rn

eiλS(x,y) ψk (x, y)f (y)dy.

′′ Since the Hessian Sxy satisfies the rank one condition, for each (x0 , y0 ) ∈ S2n−1 , there exists at least a pair of indices (i0 , j0 ) such that Sx′′i0 yj0 (x0 , y0 ) ̸= 0. Set

C0 = max{|Sx′′i yj (x0 , y0 )| : 1 6 i, j 6 n}. Thus, C0 > 0 for each (x0 , y0 ) ∈ S2n−1 . Without confusion, we may assume |Sx′′1 y1 (x0 , y0 )| = C0 , and there exists a sufficiently small neighborhood U of (x0 , y0 ) on the unit sphere such that C0 /2 < |Sx′′1 y1 (x, y)| < 2C0 ,

|Sx′′i yj (x, y)| < 2C0 ,

∀ (i, j) ̸= (1, 1),

∀ (x, y) ∈ U.

We denote the conic convex hull of origin and U by Uc . A finite number of such Uc cover the unit ball. Thus, ψ(x, y) can be assumed to be supported in Uc . Obviously, on the support of ψk , we have |Sx′′1 y1 (x, y)| ≈ 2−(d−2)k C0 . ′′ z ∥HS ψk (x, y), we have Setting x = (x1 , x′ ), y = (y1 , y ′ ), ϕzk (x, y) = ∥Sxy ∫ ′′ z z ∥HS ψk (x, y)f (y)dy (f )(x) = eiλS(x,y) ∥Sxy Tλ,k n R ∫ ∫ ′ ′ = eiλS(x1 ,x ,y1 ,y ) ϕk (x1 , x′ , y1 , y ′ )f (y1 , y ′ )dy1 dy ′ . Rn−1

R

Set Sx′ ,y′ (x1 , y1 ) = S(x1 , x′ , y1 , y ′ ), ϕzk,x′ ,y′ (x1 , y1 ) = ϕzk (x1 , x′ , y1 , y ′ ), and fy′ (y1 ) = f (y1 , y ′ ); then, ∫ ∫ z Tλ,k (f )(x1 , x′ ) = eiλSx′ ,y′ (x1 ,y1 ) ϕzk,x′ ,y′ (x1 , y1 )fy′ (y1 )dy1 dy ′ Rn−1 R ∫ ′ z T˜λ,k,x := ′ ,y ′ fy1 (x1 )dy , Rn−1

z where T˜λ,k,x ′ ,y ′ fy1 (x1 ) are the one-dimensional oscillatory integral operators investigated in [8]. Repeating the proof of Lemma 1.1 in [8] and assuming that Sx′ ,y′ (x1 , y1 ) is uniformly polynomial-like in y1 , we obtain −(d−2)kσ z ∥T˜λ,k,x |λ2−(d−2)k |−1/2 ∥f (·, y ′ )∥L2 (R) . ′ ,y ′ fy1 (x1 )∥L2 (R) . |z(z − 1)| 2

Combining this with the size of the support in x′ yields z ∥Tλ,k ∥2 . |z(z − 1)| 2−(d−2)kσ |λ2−(d−2)k |−1/2 2−(2n−2)k/2

= Cz 2−(d−2)kσ 2(d−2n)k/2 |λ|

−1/2

,

(2.3)

where Cz = |z(z − 1)|. On the other hand, from the size estimate, it is easy to verify that z ∥Tλ,k ∥2 . 2−(d−2)kσ 2−nk .

(2.4)

The estimates in (2.3) and (2.4) are comparable if and only if 2−(d−2)kσ 2k(d−2n)/2 |λ|

−1/2

∼ 2−(d−2)kσ 2−nk ,

or

Thus, ∥Tλz ∥2

. Cz

+∞ ∑ k=0

min{2−(d−2)kσ 2k(d−2n)/2 |λ|

−1/2

1/d

2k ∼ |λ|

, 2−(d−2)kσ 2−nk }

.

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= Cz

[ d1 log 2 |λ| ∑

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2−(d−2)kσ 2(d−2n)k/2 |λ|

−1/2

+∞ ∑

+

2−(d−2)kσ 2−nk

]

1 k= d log2 |λ|

k=0

= Cz

5

[ d1 log 2 |λ| ∑

((d−2n)−2(d−2)σ)k/2

2

|λ|

−1/2

+∞ ∑

+

−k((d−2)σ+n)

2

] .

(2.5)

1 k= d log2 |λ|

k=0

−1/2

If σ > σ2 , then (d − 2n) − 2(d − 2)σ < 0; therefore, the first sum in (2.5) is less than Cz |λ| and −1/2 the second one is less than Cz |λ| . If σ = σ2 , then (d − 2n) − 2(d − 2)σ = 0; therefore, the first term is less than Cz |λ|−1/2 log2 |λ| and the second term is less than Cz |λ|−1/2 . If σ1 < σ < σ2 , then (d − 2n) − 2(d − 2)σ > 0 and (d − 2)σ + n > 0; therefore, the first sum is less than −((d−2)σ+n)/d Cz |λ| and so is the second sum. Summing up the three cases above, we complete the proof of Theorem 2.1.

3

H 1 -L1 mapping property of the damped oscillatory integral operators

1 Using the result in [11], Pan [6] established the HE -L1 boundedness for oscillatory singular integral 1 operators, where HE is a modified Hardy space. Later, Yang [18] and Shi [13] developed the method used 1 by Pan [6] to get their corresponding H 1 -L1 and HE -L1 boundedness results for the oscillatory operators with a homogeneous polynomial phase function. Based on these studies, the next result can be obtained.

Theorem 3.1.

Define an operator ∫ P

T f (x) = Rn

′′ σ1 +it eiP (x,y) ∥Sxy ∥HS ψ(x, y)f (y)dy,

∑ ′′ 1/(d−2) ∥HS where P (x, y) = α,β cα,β xα y β is a higher-dimensional polynomial with c0,β = 0 for any β. If ∥Sxy n n P 1 n 1 n is a norm of R × R , then T maps H (R ) to L (R ) with operator norm less than C(1 + |t|), in which C is a constant independent of the coefficients of P (x, y). The inductive argument in [6] starts with the Lp boundedness of the oscillatory singular integral operator obtained in [11]. This method requires us to consider the following operator: ∫ ′′ σ1 +it T0 (f )(x) = ∥Sxy ∥HS ψ(x, y)f (y)dy. Rn

′′ σ1 +it If we set K(x, y) = ∥Sxy ∥HS ψ(x, y), then

∫ T0 (f )(x) = 3.1

K(x, y)f (y)dy.

(3.1)

Rn

Mapping property of T0

′′ Theorem 3.2. Considering the operator T0 defined in (3.1), if ∥Sxy ∥HS then it follows (i) T0 is of type (p, p) whenever 1 < p < +∞; (ii) T0 is of weak-type (1, 1); (iii) T0 maps H 1 (Rn ) to L1 (Rn ) with operator norm less than C(1 + |t|).

1/(d−2)

Proof.

′′ (i) By the assumption that ∥Sxy ∥HS

1/(d−2)

is a norm of Rn × Rn and the fact that the norms in

′′ finite-dimensional linear normed space are equivalent, we have ∥Sxy ∥HS σ +it ′′ 1 Since K(x, y) = ∥Sxy ∥HS ψ(x, y) and ψ ∈ C0∞ (Rn × Rn ), we have

1/(d−2)

|K(x, y)| .

is a norm of Rn × Rn ,

1 1 ≈ n n. (|x| + |y|)n |x| + |y|

≈ (|x|2 + |y|2 )1/2 ≈ |x| + |y|.

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For any f ∈ Lp (Rn ), 1 < p < +∞, (∫ ∥T0 f (x)∥p =

)1/p p

|T0 f (x)| dx

p )1/p (∫ ∫ |K(x, y)||f (y)|dy dx (∫ ∫ )1/p p |f (y)| . dy dx |x|n + |y|n (∫ ∫ )1/p |f (|x|y)| p = dy dx . 1 + |y|n 6

Using the polar coordinates x = Rθ and y = rω, the last term equals p )1/p ( ∫ +∞ ∫ +∞ ∫ n−1 |f (Rrω)| n−1 1/p ωn−1 r dωdr R dR n 0 0 Sn−1 1 + |r| p )1/p n−1 ∫ +∞ ( ∫ +∞ ∫ n−1 r 1/p R 6 ωn−1 |f (Rrω)|dω dR dr n−1 1 + |r|n 0 0 S )1/p n−1 ∫ +∞ ( ∫ +∞ ∫ r 1/p p/q p n−1 6 ωn−1 |f (Rrω)| R dr dω · ωn−1 dR 1 + |r|n 0 0 Sn−1 )1/p n−1 ∫ +∞ ( ∫ +∞ ∫ r p n−1 = ωn−1 |f (Rrω)| R dωdR dr + |r|n 1 n−1 0 0 S )1/p −n/p n−1 ∫ +∞ ( ∫ +∞ ∫ r ·r = ωn−1 |f (Rω)|p Rn−1 dωdR dr n + |r| 1 n−1 0 0 S ∫ +∞ −n/p n−1 r ·r = ωn−1 ∥f ∥p dr. n 1 + |r| 0 Since the integral in the last term is finite, we obviously have ∥T0 f ∥p 6 C∥f ∥p . (ii) For any f ∈ L1 (Rn ) and λ > 0, we can decompose it into f (x) = g(x) + b(x) by Calder`on-Zygmund decomposition. Here, ∑ b= bj , ( ) ∫ 1 bj = f (x) − f (y)dy χQj (x), |Qj | Qj where Qj is a cube centered at xQj with side-length dQj . Let Q∗j denote the cube centered at xQj with side-length M dQj , where M is a sufficiently large constant. Thus, |{x : |T0 f (x)| > λ}| 6 |{x : |T0 g(x)| > λ/2}| + |{x : |T0 b(x)| > λ/2}| { ( ∪ )c } ∑ ∗ ∗ 6 2/λ∥g∥1 + |Qj | + x ∈ Qj : |T0 b(x)| > λ/2 j

j

{ ( ∪ )c } . ∥g∥1 /λ + ∥f1 ∥1 /λ + x ∈ Q∗j : |T0 b(x)| > λ/2 j

{ ( ∪ )c } . ∥f ∥1 /λ + x ∈ Q∗j : |T0 b(x)| > λ/2 . j

From (3.1), it follows that { ( ∪ )c } ∫ 2 ∗ x∈ Q : |T b(x)| > λ/2 6 0 j λ ∪ ∗ c |T0 b(x)|dx ( j Qj ) j

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∫ K(x, y)b(y)dy dx ∪ c ( j Q∗ Rn j) ∫ ∫ ∑ 2 dx = K(x, y)b (y)dy j λ (∪j Q∗j )c j Qj ∫ ∑2∫ 6 K(x, y)bj (y)dy dx. ∗ λ c (Qj ) Qj j 2 = λ

By the vanishing property of bj , we have ∫ ∫ ∫ dx = K(x, y)b (y)dy j c (Q∗ j)

∫

∫ c

(Q∗ j)

Qj

Qj

(K(x, y) − K(x, xQ ))bj (y)dy dx.

Obviously, ∫ c

∫

(Q∗ j)

(K(x, y) − K(x, xQj ))bj (y)dy dx

Qj

∫

6 sup

∫

c (Q∗ j)

y∈Qj

If we can prove that

|bj (y)|dy.

|K(x, y) − K(x, xQj )|dx · Qj

∫ sup y∈Qj

c (Q∗ j)

|K(x, y) − K(x, xQj )|dx 6 C,

(3.2)

∑ where C is a constant independent of Qj , on account of j ∥bj ∥1 6 C∥f ∥1 , we would conclude (ii). However, analysis of this supremum should be split into two cases as follows. ′′ Case I. |xQj | < 2dQj . In this case, |y − xQj | < dQj yields |y| < 3dQj . Note that each entry in Sxy is a homogeneous polynomial of degree d − 2, and |x| ≈ |x − xQj | > M dQj ≫ |y|. Thus, provided that |σ1 | 6 1/2, we have C(1 + |t|) |∇y K(x, y)| 6 . |x|n+1

Therefore, ∫

∫ sup y∈Qj

c (Q∗ j)

|K(x, y) − K(x, xQ )|dx .

|x−xQj |>M dQj

∫ 6

|x|>(M −2)dQj

dQj dx |x|n+1 dQj dx |x|n+1

6 C. Case II. |xQj | > 2dQj . In this case, since y ∈ Qj , then |y − xQj | < dQj ; i.e., Hence, ∫ sup |K(x, y) − K(x, xQ )|dx y∈Qj

.

c (Q∗ j)

∫

|x−xQj |>M dQj

|K(x, y) − K(x, xQ )|dx

∫ =

|x−xQj |>M |xQj |

|K(x, y) − K(x, xQ )|dx

∫

+ M dQj <|x−xQj |6M |xQj |

=: A + B.

|K(x, y) − K(x, xQ )|dx

1 2 |xQj |

< |y| < 32 |xQj |.

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Observe that |y| ≈ |xQj |, the estimate of A, is the same as that in Case I; thus, it is omitted here. For B, we have ∫ B= |K(x, y) − K(x, xQ )|dx M dQj <|x−xQj |6M |xQj |

∫ 6

M dQj <|x−xQj |6M |xQj |

∫ 6

M dQj <|x−xQj |6M |xQj |

∫

|K(x, y)| + |K(x, xQ )|dx 1 1 + n dx |x|n + |y|n |x| + |xQj |n

1 1 + dx n n |y| |x Q |x−xQj |6M |xQj | j| ( ) 1 1 6 |xQj |n + |y|n |xQj |n

6

6 C. Thus, the proof of (ii) is completed. (iii) Let a denote an H 1 atom associated with a cube Q centered at xQ with side-length dQ and supp a ⊂ Q, 1 ∥a∥∞ 6 , |Q| ∫ adx = 0.

(3.3) (3.4) (3.5)

Q

Our goal is to prove ∥T0 a∥1 6 C, where C is independent of Q. Analogous to the argument of (ii), the proof should be divided into two cases. Case I.

|xQ | < 2dQ . It holds that ∫ ∫ ∥T0 a∥1 = |T0 a| dx =

∫

|x−xQ |6M |dQ |

|T0 a| dx +

|x−xQ |>M |dQ |

|T0 a| dx =: I1 + I2 .

From the Lp (1 < p < +∞) boundedness of T0 in (i), we have ∫ I1 = |T0 a| dx 6 (M |dQ |)n/2 ∥T0 a∥2 6 (M |dQ |)n/2 ∥a∥2 6 C. |x−xQ |6M |dQ |

By employing the argument of Case I in (ii) to I2 , we obtain ∫ I2 = |T0 a| dx (Q∗ )c ∫ ∫ dx = K(x, y)a(y)dy (Q∗ )c Q ∫ ∫ = (K(x, y) − K(x, xQ ))a(y)dy dx (Q∗ )c Q ∫ ∫ 6 sup |K(x, y) − K(x, xQ )|dx · |a(y)|dy. y∈Q

(Q∗ )c

Q

Thus, (3.2) and (3.4) imply I2 6 C. Case II.

|xQ | > 2dQ . It holds that ∫ ∫ ∥T0 a∥1 = |T0 a| dx =

|x−xQ |6M |xQ |

∫ |T0 a| dx +

|x−xQ |>M |xQ |

|T0 a| dx =: I3 + I4 .

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9

Since |xQ | > 2dQ and y ∈ Q, we have 12 |xQ | 6 |y| 6 32 |xQ |. It is easy to verify that ∫ ∫ ∫ I3 = |T0 a| dx = K(x, y)a(y)dy dx |x−xQ |6M |xQ |

|x−xQ |6M |xQ |

∫

6

∫

|x|6(M +1)|xQ |

∫ ∫

Q

Q

|x|n

= Q

∫ ∫

|x|6(M +1)|xQ |/|y|

= Q

|x|62(M +1)

1 |a(y)|dydx + |y|n

1 dx|a(y)|dy |x|n + 1

1 dx|a(y)|dy |x|n + 1

6 C. Observe that

∫ I4 6

∫

|x−xQ |>M |dQ |

|T0 a| dx =

(Q∗ )c

|T0 a| dx

∫ = K(x, y)a(y)dy dx ∗ c (Q ) Q ∫ ∫ = (K(x, y) − K(x, xQ ))a(y)dy dx (Q∗ )c Q ∫ ∫ 6 sup |K(x, y) − K(x, xQ )|dx · |a(y)|dy. ∫

y∈Q

(Q∗ )c

Q

On account of (3.2), I4 6 C obviously. 3.2

Some useful lemmas

Before we prove Theorem 3.1, some useful lemmas should be stated. ∑ α n Lemma 3.3. Let ϕ(x) = |α|6d aα x be a real-valued polynomial in R of degree d, and φ(x) ∈ C0∞ (Rn ). If aα0 ̸= 0 for α0 = d, we have ∫ iϕ(x) 6 C |aα0 |−1/d (∥φ∥∞ + ∥∇φ∥1 ). e φ(x)dx Rn

More details about this lemma can be found in [14]. The following lemma about polynomials was due to Ricci and Stein [11]. ∑ Lemma 3.4. Let P (x) = |α|6d aα xα denote a polynomial in Rn of degree d. Suppose ϵ < 1/d. Then, ∫ |x|61

|P (x)|

−ϵ

dx 6 Aϵ

( ∑

)−ϵ |aα |

.

|α|6d

The bound Aϵ depends on ϵ and dimension n but not on the coefficients {aα }. 3.3

Proof of Theorem 3.1

Proof of Theorem 3.1. For the atoms in Hardy space, we use the same notation as that used in the proof of (iii). To prove this theorem, we shall use induction on the degree l of y in P (x, y). If l = 0, P (x, y) contains only the pure x-term, then from (iii) we know that ∥T P a∥1 = ∥T0 a∥1 6 C. We suppose that ∥T P a∥1 6 C holds if the degree of P in y is less than l. As the proof of (iii), we consider the following two cases:

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Case I.

|xQ | 6 2dQ . It holds that ∫ ∫ |T P a(x)|dx =

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∫

|x−xQ |6M dQ

|T P a(x)|dx +

|x−xQ |>M dQ

|T P a(x)|dx

=: I5 + I6 . Taking an absolute value in I5 and recalling the argument of (i), I5 6 C obviously. ∑ Set P (x, y) = |α|>1,|β|=l cα,β xα y β + Q(x, y), where Q(x, y) is a polynomial with degree in y less than or equal to l − 1. We split I6 into two parts, ∫ ∫ iP (x,y) I6 = K(x, y)a(y)dy dx e M dQ <|x−xQ |
∫

+ |x−xQ |>max{M dQ ,r}

∫ eiP (x,y) K(x, y)a(y)dy dx

=: I7 + I8 . Then, for I7 , we have ∫ I7 =

M dQ <|x−xQ |
|T P a|dx

∫ iP (x,y) = K(x, y)a(y)dy dx e M dQ <|x−xQ |
M dQ <|x−xQ |
=: I9 + I10 . For I9 , we have ∫ I9 = M dQ <|x−xQ |
(e −e )K(x, y)a(y)dx dx ∑ 1 α β |a(y)|dydx cα,β x y n |x| + |y|n M dQ <|x−xQ |1,|β|=l ∫ ∫ ∑ α β 1 6 c x y α,β |x|n |a(y)|dydx M dQ <|x−xQ |1,|β|=l ∫ ∫ ∑ 1 |cα,β ||x||α| |y|l n |a(y)|dydx 6 |x| M dQ <|x−xQ |1,|β|=l ∫ ∫ ∑ 6 |cα,β ||x||α|−n |y|l |a(y)|dydx ∫

M dQ <|x−xQ |
∫ .

1 |Q| ,

|cα,β ||x||α|−n |dQ |l |a(y)|dydx.

Q |α|>1,|β|=l

∫ M dQ <|x−xQ |
∫

∑

we have

∫

6

iQ(x,y)

Q |α|>1,|β|=l

∫

M dQ <|x−xQ |
Since |xQ | < 2dQ and ∥a∥∞ 6

iP (x,y)

∑

|cα,β ||x||α|−n |dQ |l |a(y)|dydx

Q |α|>1,|β|=l

∑

|x|62r |α|>1,|β|=l

|cα,β ||x||α|−n |dQ |l dx

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∑

. |dQ |l

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|cα,β ||r||α| .

|α|>1,|β|=l

There must exist (α0 , β0 ) such that |α0 | > 1, |β0 | = l and |dQ |l/|α0 | |cα0 ,β0 |1/|α0 | =

max

|α|>1,|β|=l

|dQ |l/|α| |cα,β |1/|α| .

Set r−1 = |dQ |l/|α0 | |cα0 ,β0 |1/|α0 | . Then, it is obvious that I9 6 C. On the other hand, by inductive hypothesis, I10 6 C. Thus, we complete the argument of I7 . For I8 , we have ∫ ∫ iP (x,y) I8 = K(x, y)a(y)dy dx e |x−xQ |>max{M dQ ,r}

∫ iP (x,y) (K(x, y) − K(x, xQ ))a(y)dy dx 6 e |x−xQ |>max{M dQ ,r} ∫ ∫ + |K(x, xQ )| eiP (x,y) a(y)dy dx ∫

|x−xQ |>max{M dQ ,r}

=: I11 + I12 . From (3.2), it is easy to verify I11 6 C. Given |xQ | 6 2dQ , we have |K(x, xQ )| . Therefore, ∫ ∫ iP (x,y) I12 6 |K(x, xQ )| e a(y)dy dx

1 |x|n +|xQ |n

≈

1 |x−xQ |n .

|x−xQ |>r

∫ 1 eiP (x,y) a(y)dy dx n |x−xQ |>r |x − xQ | ∫ +∞ ∫ ∑ 1 eiP (x,y) a(y)dy dx = n |x − xQ | j=0 Rj ∫ +∞ ∫ ∑ 1 χRj (x) eiP (x,y) a(y)dy dx = n |x − xQ | j=0 Rj ∫ +∞ ∫ ∑ 1 iP (x,y) dx, χ (x) e a(y)dy . Rj jn n 2 r j=0 Rj ∫

.

where Rj = {x ∈ Rn : 2j r 6 |x − xQ | < 2j+1 r}. Set x = xQ + 2j ru, y = xQ + dQ v and Pj (u, v) = P (xQ + 2j ru, xQ + dQ v). Then, ∫ 1 iP (x,y) dx χ (x) e a(y)dy Rj jn n r j j+1 r 2 j=0 2 r6|x−xQ |<2 ∫ +∞ ∫ ∑ iPj (u,v) n χ du. = ˆ (u) e d a(x + d v)dv Q Q Q Rj

I12 .

+∞ ∫ ∑

j=0

16|u|<2

Suppose φ ∈ C0∞ (Rn ) and φ(v) ≡ 1

for |v| 6 1,

Define an operator Lj by

φ(v) ≡ 0 for

|v| > 2.

∫ Lj f (u) = χ ˆRj (u)

Then, I12 .

eiPj (u,v) φ(v)f (v)dv.

+∞ ∫ ∑ j=0

16|u|<2

|Lj (b)(u)|du,

(3.6)

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where b(v) = dnQ a(xQ + dQ v) is an atom associated with the unit cube centered at the origin. Set ∫ ∗ Lj (u, w) = Ker(Lj Lj ) = χ ˆRj (u)χ ˆRj (w) eiPj (u,v)−iPj (w,v) |φ(v)|2 dv. Since Pj (u, v) − Pj (w, v) ∑ ˜ w, v), = cα,β [(xQ + 2j ru)α − (xQ + 2j rw)α ](xQ + dQ v)β + Q(u, |α|>1,|β|=l

from Lemma 3.3, we have −1/l ∑ j α j α l |Lj (u, w)| 6 χ ˆRj (u)χ ˆRj (w) cα,β0 [(xQ + 2 ru) − (xQ + 2 rw) ]dQ |α|>1

[( )α ( )α ] −1/l ∑ xQ xQ j |α| =χ ˆRj (u)χ ˆRj (w) cα,β0 |2 r| +u +w dlQ − j j 2 r 2 r |α|>1

∑ [( )α ( )α ] −1/l cα,β0 dlQ xQ xQ j|α| ˆRj (u)χ ˆRj (w) 2 +u − +w =χ j j |α|/|α | |α|l/|α | 0 0 2 r 2 r |cα0 ,β0 | |dQ | |α|>1 ∑ )α ( )α ] −1/l [( xQ xQ ˆRj (u)χ ˆRj (w) + u − + w := χ bα,β0 2j|α| . j j 2 r 2 r |α|>1

On the other hand, it is obvious that |Lj (u, w)| 6 C. For a large number N , we have ∑ )α ( )α ] −1/N l [( xQ xQ j|α| |Lj (u, w)| 6 C χ ˆRj (u)χ ˆRj (w) +u − +w . bα,β0 2 j j 2 r 2 r

(3.7)

|α|>1

Now we figure out the coefficient of the term uα0 in the right-hand side of (3.7) and denote it by Aα0 . Thus, ( )α−α0 ∑ xQ 2j|α| Cd,α bα,β0 j Aα0 = 2j|α0 | bα0 ,β0 + , 2 r |α|>|α0 |+1

where Cd,α is a constant that depends only on the degree of P and α. From Lemma 3.4, we have ∫ sup |Lj (u, w)|du 6 C(Aα0 )−1/N l . w

Rn

Because |bα0 ,β0 | = 1, |bα,β0 | 6 1 and r > M dQ >

M 2 |xQ |,

we can choose a sufficiently large M such that

2j|α0 |−1 6 |Aα0 | 6 3 · 2j|α0 |−1 . Thus, we can obtain

∫ sup w

Rn

|Lj (u, w)|du 6 C2−jθ ,

where θ is a positive constant independent of the coefficients of P . The same method can be applied to ∫ supu Rn |Lj (u, w)| and leads to the same estimate. By Schur’s lemma, we have ∥Lj ∥2 6 C2−jθ . Now, we come back to (3.6). By H¨older’s inequality, I12 6

+∞ ∑ j=0

C∥Lj (b)∥ 6

+∞ ∑ j=0

C∥Lj ∥2 ∥b∥2 6 C.

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Case II.

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|xQ | > 2dQ . In this case, we decompose the integral into two parts ∫ ∫ ∫ |T P a|dx = |T P a|dx + |T P a|dx |x−xQ |6M |xQ |

|x−xQ |>M |xQ |

=: I13 + I14 . We shall get I13 6 C from the analog of I3 . On the other hand, I14 is similar to I6 , and following the same pattern to deal with I6 would yield I14 6 C. Thus, we complete our proof.

4

Optimality of decay rates and examples

The optimality of decay rates can be derived from the proof of Theorem 4.1 in [3]; therefore, it is omitted here. Next, we give an example to demonstrate our main result. Let n = 2, d = 6, and S(x, y) = 15 (x51 y1 + x1 y15 + x1 x42 y2 + x1 y14 y2 + x41 x2 y1 + x2 y1 y24 + x52 y2 + x2 y25 ). Then, the Hessian matrix of S(x, y) is ′′ = Sxy

(

) . 4

x41 + y14 x42 + y14 x41 + y24 x42 + y2

Hence, ′′ ∥HS ∥Sxy

1/(d−2)

= [(x41 + y14 )2 + (x41 + y24 )2 + (x42 + y14 )2 + (x42 + y24 )2 ]1/8 .

′′ The equation above can be regarded as a composition of three different simple norms. Then, ∥Sxy ∥HS 2 2 is obviously a norm in R × R . Thus, this example satisfies the decay estimate in Theorem A. If we let n = 2, d = 6, and S(x, y) = 51 (x51 y1 + x1 y15 + x52 y2 + x2 y25 ), then the Hessian matrix of S(x, y) is 1/(d−2)

′′ Sxy

( =

) . 4

x41 + y14

0

0

x42 + y2

This is the simplest case because the related oscillatory integral operator can be separated variables. By iterating the one-dimensional result of [13], we can show that ∥Tλ ∥p 6 Cλ−1/3

for

6/5 6 p 6 6.

Thus, d/(d − n) < p < d/n is not necessary to guarantee the sharp decay. Acknowledgements This work was supported by National Natural Science Foundation of China (Grant Nos. 11471309, 11271162 and 11561062).

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