Uniform Estimates for Oscillatory Integral Operators with Polynomial

arXiv:1809.01300v1 [math.CA] 5 Sep 2018

Uniform Estimates for Oscillatory Integral Operators with Polynomial Phases Zuoshunhua Shi

∗

Abstract In this paper, we shall prove the uniform sharp Lp decay estimates for a class of oscillatory integral operators with polynomial phases. By this one-dimensional result, we can use the rotation method to obtain uniform sharp Lp estimates of certain higher-dimensional oscillatory integral operators.

Keywords Oscillatory integral operator, van der Corput’s Lemma, Uniform estimate Mathematics Subject Classification 47G10, 44A05.

1

Introduction

In this paper, we mainly consider the stability of certain oscillatory integral operators. The issue of stability for oscillatory integrals includes two major cases: (i) stable estimates under a small perturbation of a given function; see Karpushkin [14], Phong-Stein-Sturm [24], PhongSturm [25] and Greenblatt [8]; (ii) uniform estimates over a large class of phases satisfying certain nondegeneracy conditions; see Carbery-Christ-Wright [1], Carbery-Wright [2], ChristLi-Tao-Thiele [3], Phong-Stein-Sturm [26] and Gressman [12]. The second case of stability will be investigated for certain oscillatory integral operators with polynomial phases. Consider oscillatory integral operators of form Z ∞ eiλS(x,y) ϕ(x, y)f (y)dy, (1.1) Tλ f (x) = −∞

where λ is a real parameter and ϕ is a smooth cut-off function. For nondegenerate phases S, Hörmander [13] established the optimal L2 decay estimate; see also Phong-Stein [21, 22] and Phong-Stein-Sturm [26] for uniform L2 decay estimates. For general degenerate real-analytic phases S, Phong-Stein [22] established the relation between the decay rate of the L2 operator norm of Tλ and the Newton distance of the phase S; see Phong-Stein [20, 21] for earlier related results. Rychkov [28] extended this result to most smooth phases and full generalizations to smooth phases were established by Greenblatt [6]. For higher dimensional analogues, Tang [34] and Greenleaf-Pramanik-Tang [9] obtained sharp L2 decay estimates under certain genericity assumptions. Recently, Xiao [35] proved the full range of sharp Lp decay estimate for Tλ ; see also [36, 37, 31] for some related and earlier work. We also refer the reader to Greenleaf-Seeger [11] for a survey on degenerate oscillatory and Fourier integral operators. ∗

School of Mathematics and Statistics, Central South University, Changsha Hunan 410083, P.R. China. E-mail address: [email protected].

1

By imposing uniform positive lower bounds on certain mixed derivatives of the phase S, Carbery-Christ-Wright [1] established uniform sharp growth estimates for sublevel set operators associated with S. In [1], the authors also obtained uniform decay estimates for oscillatory integral operators with non-sharp decay exponents except for some especial cases; see also [2]. Up to logarithmic terms, Phong-Stein-Sturm [26] obtained uniform sharp Lp estimates for a class of multilinear sublevel set operators and oscillatory integral operators. In this paper, we shall remove the logarithmic terms in these estimates of [26] for Tλ . The phase S is a polynomial in R2 and takes the following form X S(x, y) = ak,l xk y l (1.2) kη+l=d

for two numbers η, d > 0. It is clear that S has nonvanishing partial derivatives ∂xk ∂yl S for positive integers k, l with ak,l 6= 0. Now we state our main result in this paper. Theorem 1.1 Assume S is a real-valued polynomial of form (1.2) in R2 . Let Tλ be an oscillatory integral operator as in (1.1). Then there exists a constant C, depending only on the cut-off ϕ and the degree deg(S) of the phase S, such that for all positive integers k, l satisfying kη + l = d, 1 1 k+l . (1.3) kTλ f kLp ≤ C|ak,l |− k+l |λ|− k+l kf kLp , p = k Our proof of the theorem relies on some uniform damping estimates related to Tλ . Roughly speaking, one of the damping estimates is a uniform L2 decay rate for oscillatory integral operators with damping factors related to the Hessian of the phase. Another damping estimate consists of uniform L1 → L1,∞ , HE1 → L1 and L1 → L1 boundedness for certain damped oscillatory integral operators. Here HE1 is a variant of the Hardy space H 1 associated with the phase S; see Phong-Stein [19], Pan [16], Greenleaf-Seeger [10] and Shi-Yan [31]. In this paper, the damped oscillatory integral operators with polynomial phases are of form Z eiλS(x,y) |D(x, y)|z ϕ(x, y)f (y)dy, (1.4) Wz f (x) = R

where D is the damping factor and z is the damping exponent. When the damping factor D ′′ , Phong-Stein [21, 22] proved the sharp L2 decay estimate for W is taken as the Hessian Sxy λ ′′ together with its higher derivatives; with the operator norm depending on upper bounds for Sxy see Seeger [30] for decay O(|λ|−1/2 ) with Re (z) > 1/2. As a consequence of our uniform L2 damping estimates, we are able to establish the stability of the result in Phong-Stein [22] for Wz with polynomial phases of form (1.2). More precisely, we have the following theorem. Theorem 1.2 Assume S is a real-valued polynomial of form (1.2). Let Wz be defined as in ′′ (x, y). Then there exists a constant C = C(deg(S)), depending only on (1.4) with D(x, y) = Sxy the degree of S, such that kWz f kL2 ≤ C(1 + |z|)2 M |λ|−1/2 kf kL2 for all z ∈ C with Re (z) = sup Ω

2 X k=0

1 2

and all ϕ ∈ C0∞ (R2 ) satisfying

k k δΩ,h (x) |∂yk ϕ(x, y)| + δΩ,v (y) |∂xk ϕ(x, y)| ≤ M, 2

where M is a positive number and Ω is a horizontally and vertically convex domain such that the cut-off ϕ is supported in Ω. Here δΩ,h (x) and δΩ,v (y) denote the length of the cross sections {y : (x, y) ∈ Ω} and {x : (x, y) ∈ Ω}, respectively. With the rotation method and a variant of Stein-Weiss interpolation with change of measures, we can apply Theorem 1.1 to obtain some uniform Lp estimates for higher dimensional oscillatory integral operators. In Section 6, we shall also discuss some examples related to a conjecture raised by Greenleaf-Pramanik-Tang in [9].

2

Preliminaries

In this section, we shall first present some basic notions concerning horizontally and vertically convex domains. With these useful notions, uniform L2 estimates will be established for nondegenerate oscillatory integral operators which are supported in horizontally (vertically) convex domains. These decay estimates are often known as the operator version of van der Corput lemma; see [21, 22, 26]. Finally, we also include an interpolation result with change of power weights. Definition 2.1 Let Ω ⊆ R2 . We say that Ω is horizontally convex if (x, z), (y, z) ∈ Ω imply (u, z) ∈ Ω for all x ≤ u ≤ y. Similarly, Ω is said to be vertically convex if (x, y), (x, z) ∈ Ω indicate (x, v) ∈ Ω for all y ≤ v ≤ z. We shall give a simple relation between horizontal convexity and the following curved trapezoid. Definition 2.2 Assume g and h are two monotone functions on [a, b] with g ≤ h. Then we call Ω = {(x, y) | a ≤ x ≤ b, g(x) ≤ y ≤ h(x)} a curved trapezoid. It is clear that a curved trapezoid is vertically convex. By the monotonicity of g, h, one can verify that a curved trapezoid is also horizontally convex. If the monotonicity assumption on g, h is dropped, the horizontal convexity for Ω is not generally true. But we have the following lemma. Lemma 2.1 Assume g, h are two continuous functions satisfying g ≤ h on [a, b]. If Ω = {(x, y) | a ≤ x ≤ b, g(x) ≤ y ≤ h(x)} is horizontally convex, then [a, b] can be divided into three subintervals I1 , I2 , I3 with disjoint interiors such that each domain Ωi = {(x, y) | x ∈ Ii , g(x) ≤ y ≤ h(x)} is a curved trapezoid. In other words, both g and h are monotone on each Ii . Proof. Choose a point c ∈ [a, b] such that h achieves its maximum at c. Then h must be increasing on [a, c] and decreasing on [c, d]. Otherwise, we can choose two points a ≤ x1 < x2 ≤ c such that h(x1 ) > h(x2 ). By the continuity of h, we can choose x3 ∈ [x2 , c] with h(x3 ) = h(x1 ). Since Ω is horizontally convex, we find that (u, z) ∈ Ω for all x1 ≤ u ≤ x3 if z = h(x1 ). But this contradicts the fact (x2 , z) ∈ / Ω. Thus h does not decrease on [a, c]. By a similar argument, we can prove that h is decreasing on [c, b]. 3

Assume g attains its minimum at some point d ∈ [a, b]. By the above argument, it follows that g is decreasing on [a, d] and is increasing on [d, b]. Without loss of generality, assume c ≤ d. Then Ii are obtained by taking a, c, d, b as the endpoints of these intervals. The proof is therefore complete. ✷ For horizontally and vertically convex domains, some notations will be used frequently. • δΩ,h (x): the length of the cross section {y : (x, y) ∈ Ω}. Here the subscript h means that δΩ,h is a function of the horizontal variable. • δΩ,v (y): the length of the cross section {x : (x, y) ∈ Ω}. The subscript v suggests that y is the vertical component. • Ω∗h : a horizontally expanded domain of Ω; see Definition 2.3. • Ω∗v : a vertically expanded domain of Ω; see Definition 2.3. • Ω∗ : an expanded domain of Ω of form Ω∗ = Ω∗h ∪ Ω∗v . • IΩ,h (x): the cross section {y : (x, y) ∈ Ω}. The subscript h means that IΩ,h is taken with respect to the horizontal component. • IΩ,v (y): the cross section {x : (x, y) ∈ Ω}. The subscript v means that IΩ,v is taken with respect to the vertical component. • a ∧ b: the minimum of two real numbers a and b. • a ∨ b: the maximum of a and b. Now we state the operator van der Corput lemma for nondegenerate oscillatory integral operators. The following lemma, with varying width of the “curve box”, was first established by Phong-Stein-Sturm in [26]. For convenience of readers, we also include a proof here. Lemma 2.2 Assume S is a real-valued polynomial in R2 . Let Tλ be defined as in (1.1) with supp (ϕ) contained in a curved trapezoid Ω = {(x, y) | a ≤ x ≤ b, g(x) ≤ y ≤ h(x)}. Suppose S satisfies the following conditions: (i) For some µ, A1 > 0, ′′ µ ≤ |Sxy (x, y)| ≤ A1 µ for all (x, y) ∈ Ω. (ii) We use δΩ,h (x) to denote the length of the cross-section IΩ,h (x) = {y | (x, y) ∈ Ω}. There exists a constant A2 > 0 such that 2 X

sup

k=0 (x,y)∈Ω

k δΩ,h (x) |∂yk ϕ(x, y)| ≤ A2 .

Then there exists a constant C = C(deg(S), A1 ) such that kTλ f kL2 ≤ CA2 |λµ|−1/2 kf kL2 . Proof. We first assume that g, h are increasing on [a, b]. We shall decompose Ω into a sequence of domains {Ωi } such that these domains satisfy the almost orthogonality property. Then it suffices to treat one Ti supported in Ωi , where Ti is defined as Tλ with insertion of the characteristic function χΩi into the cut-off of Tλ . More precisely, the decomposition procedure is described as follows. Let x0 = a. Find a point a ≤ x1 ≤ b such that g(x1 ) = h(x0 ). Then we choose x2 > x1 such that g(x2 ) = h(x1 ). Generally, if xk is chosen, then select one xk+1 > xk such that 4

g(xk+1 ) = h(xk ). This process will continue unless g(b) ≤ h(xN ) for some integer N . It is possible that this process will not be terminated in finite steps; for example, for those functions g, h satisfying g(b) = h(b) and g(x) < h(x) for all x ∈ [a, b), this decomposition process consists of infinite steps. Let Ωi = {(x, y) | xi ≤ x ≤ xi+1 , g(x) ≤ y ≤ h(x)}.PDefine Ti as Tλ in (1.1), but with the cut-off multiplied by χΩi . Then it is clear that T = Ti . For i, j satisfying |i − j| ≥ 2, ∗ ∗ Ti Tj = Ti Tj = 0. The reason is that if |i − j| ≥ 2 then the projections P of Ωi and ΩjPinto the x−axis (also y−axis) are disjoint up to a set of measure zero. Write T = i even Ti + i odd Ti . P P Observe that k i even Ti k ≤ sup i even kTi k and k i odd Ti k ≤ sup i odd kTi k. This observation implies kT k ≤ 2 sup i kTi k. Thus it suffices to estimate one Ti . For simplicity, we may assume a = xi and b = xi+1 . Under this assumption, we have h(a) ≥ g(b). With some abuse of notation, we also write Tλ instead of Ti . Now we use the classical T T ∗ method to prove our desired estimate. First, the kernel of Tλ Tλ∗ is given by Z eiλ[S(x,z)−S(y,z)] ϕ(x, z)ϕ(y, z)dz.

K(x, y) =

R

For z satisfying ϕ(x, z)ϕ(y, z) 6= 0, it follows from Assumption (i) that Z x ∂u ∂z S(u, z)du Φ(z) := ∂z [S(x, z) − S(y, z)] = y

satisfies |Φ(z)| ≥ µ|x − y|. For fixed x, y, define a differential operator D as Df (z) = (iλ)−1 [∂z S(x, z) − ∂z S(y, z)]−1 f ′ (z). Then its transpose D t is −f (z) ∂ D f (z) = . ∂z iλ(∂z S(x, z) − ∂z S(y, z)) t

By integration by parts, K(x, y) = =

Z

ZR

D 2 eiλ(S(x,z)−S(y,z)) ϕ(x, z)ϕ(y, z)dz

eiλ(S(x,z)−S(y,z)) (D t )2 (ϕ(x, z)ϕ(y, z)) dz.

R

Let ∆1 and ∆2 be two curved trapezoids defined by ∆1 = {(x, y) | a ≤ x ≤ b, h(a) ≤ y ≤ h(x)},

∆2 = {(x, y) | a ≤ x ≤ b, g(x) ≤ y ≤ h(a)}.

For z satisfying ϕ(x, z)ϕ(y, z) 6= 0, we see that z must belong to [g(x), h(x)] ∩ [g(y), h(y)] and that |Φ′ (z)| d δ−1 µ|x − y| 1 −1 dz Φ(z) = Φ(z)2 ≤ C (µ|x − y|)2 ≤ C (δµ|x − y|) , 5

where δ = (δ∆1 ,h (x) ∧ δ∆1 ,h (y)) ∨ (δ∆2 ,h (x) ∧ δ∆2 ,h (y)) and C = C(deg(S), A1 ). Similarly, we also have 2 ′′ ′ (z)2 d Φ (z) 1 Φ −2 −1 dz 2 Φ(z) = Φ(z)2 − 2 Φ(z)3 ≤ Cδ (µ|x − y|)

with the positive number δ defined as above. Observe that (D t )2 (ϕ(x, z)ϕ(y, z)) is a linear combination of the following terms m2 m3 1 dm1 1 1 d ∂ ϕ ∂ m4 ϕ (iλ)2 dz m1 Φ(z) dz m2 Φ(z) ∂z m3 ∂z m4 with nonnegative integers mi satisfying m1 + m2 + m3 + m4 = 2. Combining above results together with Assumption (ii), we have −2 h i−1 δ∆1 ,h (x) ∧ δ∆1 ,h (y) ∨ δ∆2 ,h (x) ∧ δ∆2 ,h (y) . |K(x, y)| ≤ CA22 |λ|µ|x − y|

Taking absolute value into the integral of the kernel K, we see that |K(x, y)| ≤ A22 δ∆1 ,h (x) ∧ δ∆1 ,h (y) + δ∆2 ,h (x) ∧ δ∆2 ,h (y) . These two inequalities imply

|K(x, y)| ≤ CA22

δ (1 + |λ|δµ|x − y|)2

with δ = (δ∆1 ,h (x) ∧ δ∆1 ,h (y)) ∨ (δ∆2 ,h (x) ∧ δ∆2 ,h (y)). Since g, h are increasing, it is clear that ( δ∆1 ,h (x), δ∆1 ,h (x) ∧ δ∆1 ,h (y) = δ∆1 ,h (y), and

( δ∆2 ,h (y), δ∆2 ,h (x) ∧ δ∆2 ,h (y) = δ∆2 ,h (x),

If x ≤ y, then |K(x, y)| ≤ CA22

if x ≤ y; if y ≤ x; if x ≤ y; if y ≤ x.

δ∆1 ,h (x) δ∆2 ,h (y) + . (1 + |λ|µδ∆1 ,h (x)|x − y|)2 (1 + |λ|µδ∆2 ,h (y)|x − y|)2

This implies Z Z Z |K(x, y)||f (y)g(x)|dxdy ≤ CA22 |λµ|−1 Mf (x)|g(x)|dx + |f (y)|Mg(y)dy x≤y

≤ CA22 |λµ|−1 kf k2 kgk2 ,

where M is the Hardy-Littlewood maximal operator. In the case y ≤ x, the same estimate is true for the kernel K. It follows immediately that kTλ k ≤ CA2 |λµ|−1/2 , 6

where the constant C depends only on deg(S) and A1 . By the same argument as above, we can prove the desired estimate if f, g are decreasing. There are two remaining cases: (i) f is increasing and g is decreasing; (ii) f is decreasing and g is increasing. In these two remaining cases, the proof is more direct since we need not decompose Ω into small pieces as above. In both cases (i) and (ii), the kernel K is bounded by δΩ,h (x) δΩ,h (y) 2 |K(x, y)| ≤ CA2 + . (1 + |λ|µδΩ,h (x)|x − y|)2 (1 + |λ|µδΩ,h (y)|x − y|)2 Invoking the maximal operator as above, we can also prove the desired estimate. The proof is therefore complete. ✷ Let Ω be a horizontally and vertically convex domain in R2 . If its horizontal and vertical cross-sections are closed intervals, then there exist two numbers a, b and functions g, h such that Ω can be written as Ω = {(x, y) ∈ R2 | a ≤ x ≤ b, g(x) ≤ y ≤ h(x)}.

(2.5)

Similarly, for some c, d ∈ R and functions u, v, the region Ω can also be given by Ω = {(x, y) ∈ R2 | c ≤ y ≤ d, u(y) ≤ x ≤ v(y)}.

(2.6)

Now we shall define an expanded domain Ω∗ for which δΩ∗ ,h and δΩ∗ ,v are suitably larger than δΩ,h and δΩ,v . Throughout the rest of this section, we always assume Ω is a horizontally and vertically convex domain in R2 . Definition 2.3 We say that Ω∗h is a horizontally expanded domain of Ω, if there exists a positive number ǫ > 0 and a nonnegative function η on [c, d] such that Ω∗h = {(x, y) | c ≤ y ≤ d, u(y) − η(y) ≤ x ≤ v(y) + η(y)} and for each y δΩ∗h ,v (y) ≥ (1 + 2ǫ)δΩ,v (y).

Similarly, Ω∗v is said to be a vertically expanded domain of Ω if there exists a number ǫ > 0 and a nonnegative function γ on [a, b] such that Ω∗v = {(x, y) | a ≤ x ≤ b, g(x) − γ(x) ≤ y ≤ g(x) + γ(x)} and there holds δΩ∗v ,h (x) ≥ (1 + 2ǫ)δΩ,h (x), x ∈ [a, b]. Definition 2.4 The set Ω∗ is said to be an expanded domain of Ω, if there exist horizontally and vertically expanded domains Ω∗h and Ω∗v , defined as in Definition 2.3, such that Ω∗ = Ω∗h ∪ Ω∗v . Definition 2.5 Let I be a bounded interval in R. We use I ∗ (B) to denote the concentric interval with length expanded by the factor B > 0. We now state a useful lemma concerning expanded intervals. Lemma 2.3 Assume I1 and I2 are two bounded intervals in R. Then if I1 ∩ I2 6= ∅, then we have |I1∗ (B) ∩ I2∗ (B)| ≥ (B − 1)|I1 | ∧ |I2 | for B > 1 with |I| being the length of the interval I. 7

Proof. Assume |I1 | ≤ |I2 | and I1 ∩ I2 = (a, b). For simplicity, write I1 as I1 = (c1 − δ1 , c1 + δ1 ). By Definition 2.5, we see that B−1 B−1 a− δ1 , b + δ1 ⊆ I1∗ (B) ∩ I2∗ (B) 2 2 for B > 1. Hence |I1∗ (B) ∩ I2∗ (B)| ≥ (B − 1)δ1 = (B − 1)|I1 |.

✷

With the above preliminaries, we can now present the oscillation estimate to treat the almost orthogonality between two oscillatory integral operators. It should be pointed out that operators considered here are supported on horizontally (vertically) convex domains. For operators supported on curved trapezoids, the corresponding oscillation estimate was obtained by Phong and Stein in [23]. (1)

(2)

Lemma 2.4 Let S be a real-valued polynomial in R2 with degree deg(S). Assume Tλ and Tλ are defined as Tλ in (1.1), but with the cut-off ϕ replaced by ϕ1 and ϕ2 respectively. Suppose that supp (ϕi ) ⊆ Ωi for two horizontally and vertically convex domains Ω1 and Ω2 . Assume that all of the following conditions are true. (i) For some µ, A > 0, there exist expanded domains Ω∗1 and Ω∗2 , defined as in Definition 2.4, such that ′′ µ ≤ |Sxy (x, y)| ≤ Aµ, (x, y) ∈ Ω∗1 .

(ii) For any horizontal line segment L joining one point in Ω∗1 and another one in Ω∗2 , the ′′ (x, y) does not change sign on L and sup |S ′′ (x, y)| ≤ Aµ. Hessian Sxy L xy (iii) There exists a positive number B such that for each y, we have IΩ∗2 ,v (y) ⊆ IΩ∗ ∗1 ,v (y, B), where IΩ∗i ,v (y) is the horizontal cross-section of Ω∗i at height y, i.e., IΩ∗i ,v (y) = {x : (x, y) ∈ Ω∗i }, and the interval IΩ∗ ∗ ,v (y, B) has the same center as IΩ∗1 ,v (y), but its length is B times as long as 1 that of IΩ∗1 ,v (y). (iv) For two positive numbers M1 and M2 , it is true that 2 X

sup (δΩi ,h (x))k |∂yk ϕi (x, y)| ≤ Mi ,

i = 1, 2.

k=0 Ωi

Then there exists a constant C, depending only on deg(S), A and the expanded factors ǫ appearing in the definition of Ω∗1 and Ω∗2 such that

(1) (2)∗

Tλ Tλ 2 2 ≤ CM1 M2 |λµ|−1 . L →L

(1)

(2)∗

Proof. Let K be the kernel associated with Tλ Tλ . Then K can be written as Z eiλ[S(x,z)−S(y,z)] ϕ1 (x, z)ϕ2 (y, z)dz. K(x, y) =

(2.7)

R

For those z such that ϕ1 (x, z)ϕ2 (y, z) 6= 0, we deduce form the assumptions (i), (ii) and (iii) that Φ(z) = ∂z S(x, z) − ∂z S(y, z) 8

satisfies

Z |Φ(z)| =

x y

∂u ∂z S(u, z)du ≥ Cµ|x − y|,

where the constant C depends only on the factor B in the assumption (iii) and ǫ1 , ǫ2 appearing in the definition of Ω∗1 , Ω∗2 . By integration by parts, we have Z eiλ[S(x,z)−S(y,z)] ϕ1 (x, z)ϕ2 (y, z)dz K(x, y) = ZR Dz2 eiλ(S(x,z)−S(y,z)) ϕ1 (x, z)ϕ2 (y, z)dz = ZR eiλ(S(x,z)−S(y,z)) (D t )2 (ϕ1 (x, z)ϕ2 (y, z)) dz, = R

where D is the differential operator Df (z) = [iλΦ′ (z)]−1 f ′ (z) and D t is its transpose, i.e., D t f (z) = −∂z [(iλΦ(z))−1 f (z)]. On the other hand, by the assumption (ii), we see that |Φ(z)| ≤ Aµ for z ∈ IΩ∗1 ,h (x) ∩ IΩ∗2 ,h (y). If IΩ1 ,h (x) ∩ IΩ2 ,h (y) = ∅, then K(x, y) = 0. Thus we need only consider those x, y for which IΩ1 ,h (x) ∩ IΩ2 ,h (y) 6= ∅. By Lemma 2.3, we have |IΩ∗1 ,h (x) ∩ IΩ∗2 ,h (y)| ≥ |IΩ∗1,v ,h (x) ∩ IΩ∗2,v ,h (y)| ≥ ǫ|IΩ1 ,h (x)| ∧ |IΩ2 ,h (y)|

with ǫ = 2ǫ1 ∧ ǫ2 . Here ǫ1 , ǫ2 are the expanded factors for Ω∗1 and Ω∗2 . It follows from the polynomial property that d Φ′ (z) Cµ|x − y||IΩ∗1 ,h (x) ∩ IΩ∗2 ,h (y)|−1 1 ≤ = − dz Φ(z) Φ2 (z) (µ|x − y|)2 ≤

Cµ|x − y| (δΩ1 ,h (x) ∧ δΩ2 ,h (y))−1 (µ|x − y|)2

for some constant C = C(deg(S), ǫ1 , ǫ2 ). Here the factor ǫi appears in the definition of Ω∗i,h . Likewise, it is also true that 2 ′ 2 ′′ d 1 ≤ |Φ (z)| + 2 Φ (z) ≤ C (µ|x − y|)−1 (δΩ ,h (x) ∧ δΩ ,h (y))−2 . 1 2 dz 2 Φ(z) (Φ(z))2 |Φ(z)|3

On the other hand, we see that

k k2 k3 k4 d 1 d ∂ ϕ1 ∂ ϕ2 1 1 1 X t 2 , → C− (D ) (ϕ1 (x, z)ϕ2 (y, z)) ≤ 2 k k k 1 2 |λ| dz Φ(z) dz Φ(z) ∂z k3 ∂z k4

→ − where the summation is taken over all k ∈ Z4 with nonnegative components ki satisfying k1 + k2 + k3 + k4 = 2. Combining above estimates together with Assumption (iv), we obtain |K(x, y)| ≤ CM1 M2 (1 + |λ|µδΩ1 ,h (x) ∧ δΩ2 ,h (y)|x − y|)−2 δΩ1 ,h (x) ∧ δΩ2 ,h (y). Note that

a∧b a b 2 ≤ 2 + (1 + |λ|µa ∧ b|x − y|) (1 + |λ|µa|x − y|) (1 + |λ|µb|x − y|)2 9

with a = δΩ1 ,h (x) and b = δΩ2 ,h (y). Thus for any f, g ∈ L2 , Z Z Z −1 |K(x, y)||f (y)g(x)|dxdy ≤ CM1 M2 |λµ| Mf (x)|g(x)|dy + |f (y)|Mg(y)dy R2

≤ CM1 M2 |λµ|−1 kf kL2 kgkL2 ,

where M is the Hardy-Littlewood maximal operator, and the constant C depends on deg(S), ǫ1 , ǫ2 and A, but not on λ, µ and f, g. Thus we complete the proof of the lemma. ✷ Remark 2.1 If we change the role of x and y in the lemma, we also have a similar estimate for (1)∗ (2) the L2 operator norm of Tλ Tλ . Now we shall point out that the bounds M1 , M2 appearing in the assumption (iv) can be slightly improved under certain additional conditions. Assume all assumptions in the lemma are true. Suppose ϕ1 and ϕ2 are also supported in e 1 and Ω e 2 , respectively. Then we can another two horizontally and vertically convex domains Ω e improve the bound Mi by replacing Ωi by Ωi ∩ Ωi in (iv) for i = 1, 2. For the proof of this improved result, we can follow the above argument line by line, but with δΩ∗i ,h replaced by δΩ∗ ∩Ωe i ,h . i

Now we shall present a uniform decay estimate for Tλ with non-sharp decay exponent. This result will be used in our proof of HE1 → L1 boundedness for damped oscillatory integral operators. Lemma 2.5 Let Tλ be defined as in (1.1) with S being a real-valued polynomial in R2 . For two positive integers j, k, assume |∂xj ∂yk S(x, y)| ≥ 1 on the unit square U := {|x| < 1/2, |y| < 1/2}. Then there exists a decay exponent δ > 0, depending only on the degree of S, such that we have for all ϕ ∈ C0∞ (U ) kTλ f kL2 ≤ C|λ|−δ kf kL2 , where the constant C depends only on deg(S) and ϕ. For the proof of this lemma with non-sharp δ, we refer the reader to Ricci-Stein [27] for the 1 and to Carbery-Christ-Wright [1] for a more general estimate above estimate with δ < 2 deg(S) −

1

1 2(j∨k) was proved by Phong(for smooth functions) with δ = 2(j∨k)2 j∧k . The maximal decay |λ| Stein-Sturm [26]. For further related topics, we refer the reader to see Christ-Li-Tao-Thiele [3] and Greenblatt [7] for uniform estimates under more general non-degeneracy concepts.

The following lemma is useful in the interpolation with change of power weights. Its formulation and proof are in many ways like the Stein-Weiss interpolation with change of measures; see Stein-Weiss [33], Pan-Sampson-Szeptycki [17] and Shi-Yan [31]. Lemma 2.6 Assume T is a sublinear operator defined for simple functions in R with Lebesgue measure. Suppose that there exist two constants A, B > 0 such that (i) kT f kL∞ ≤ Akf kL1 for all simple functions f; (ii) k|x|a T f kLp0 ≤ Bkf kLp0 , where a ∈ R, 1 < p0 < ∞ and a 6= −1/p0 .

10

Then for all 0 < θ < 1, there exists a constant C = C(a, p0 , θ) such that

1 θ

γ γ = −(1 − θ) + θa, = (1 − θ) + ,

|x| T f p ≤ CA1−θ B θ kf kLp , p p0 L where f is an arbitrary simple function.

Proof. By the assumption (i), it is easy to see that |x|−1 T is bounded from L1 (dx) into L1,∞ (dx). Since |x|a T is bounded on Lp0 (dx), we can define an operator Sf = |x|b T f and a measure dµ = |x|c dx such that S is bounded from L1 (dx) into L1,∞ (dµ) and maps Lp0 (dx) continuously into Lp0 (dµ). We shall now need a simple fact that |x|λ belongs to L1,∞ (|x|−1−λ dx) for any nonzero real number λ. By this fact, we see that S is bounded from L1 (dx) into L1,∞ (dµ) provided that b + c = −1 and b 6= 0. On the other hand, using assumption (ii), we see that S is bounded from Lp0 (dx) into Lp0 (dµ) if p0 b + c = p0 a. We solve this two equations for b and c, and then obtain b=

p0 a + 1 , p0 − 1

c = −1 −

p0 a + 1 . p0 − 1

Recall that we have assumed b 6= 0, i.e., a 6= −1/p0 . By the Marcinkiewicz interpolation theorem, we have a constant C = C(a, p0 , θ) such that k|x|b T f kLp (|x|c dx) ≤ CA1−θ B θ kf kLp (dx) with

1 p

=1−θ+

θ p0 .

Note that θ c c b+ =b+c 1−θ+ = (1 − θ)(b + c) + θ b + = −(1 − θ) + θa. p p0 p0

Then the desired inequality in the lemma follows immediately.

3

✷

Damped Oscillatory Integral Operators on L2

′′ can be written as Assume S is a real-valued polynomial of the form (1.2). Then its Hessian Sxy ′′ Sxy (x, y) = c0 xm y n

N Y (x − αi y η )

(3.8)

i=1

with c0 6= 0, α1 , α2 , · · · , αN ∈ C\{0} and η > 0. Since λ and S can be replaced by λc0 and S/c0 in (1.1), we may assume c0 = 1. By choosing s indices i1 < i2 < · · · < is , we define the corresponding damping factor D of the form D(x, y) = xm (x − αi1 y η )(x − αi2 y η ) · · · (x − αis y η ). Let Wz be the damped oscillatory integral operator Z eiλS(x,y) |D(x, y)|z ϕ(x, y)f (y)dy, Wz f (x) =

(3.9)

(3.10)

R

where z ∈ C lies in an appropriate strip. To estimate the L2 norm of Wz by Lemma 2.2 and Lemma 2.4, we also require that the damping function |D(x, y)|z should behave like a polynomial in x and y. However both D and |D|z are not polynomials generally. For this reason, we shall introduce the concept of polynomial type functions; see Phong-Stein [22, 23]. 11

Definition 3.1 Assume a function F is of class C N on an interval J. We say that F is of polynomial type with order N if there exists a constant C such that N X k=0

|I|k sup |F (k) (x)| ≤ C sup |F (x)| x∈I

x∈I

holds for all intervals I ⊆ J. A polynomial P in R of degree N is of polynomial type with order N on any bounded interval I. We also need the following two facts related to the concept of polynomial type functions. • Let F be a polynomial type function of order N on J. If 0 < µ ≤ |F (x)| ≤ Aµ for some µ, A > 0 and all x ∈ J, then |F |z is also of polynomial type with order N on J for any z ∈ C. • If F is of polynomial type with order N on some dyadic interval J ⊆ R+ , then, for any η > 0, the function G(x) = F (xη ) is also of polynomial type with order N on J 1/η , where the interval J 1/η is defined by J 1/η = {x1/η : x ∈ J}. Theorem 3.1 Let S and D be defined as above with η ≥ 1. If Wz is defined as in (3.10), then there exists a constant C, depending only on deg(S) and the cut-off ϕ, such that 

kWz f kL2 ≤ C(1 + |z|2 ) |λ|

Y

k ∈{i / 1 ,··· ,is }

for z ∈ C with real part

Re (z) =

−γ

|αk |

kf kL2 ,

γ=

1 2(n + (N − s)η + 1)

m + s − n − (N − s)η 1 · . 2(n + (N − s)η + 1) m + s

(3.11)

(3.12)

More precisely, the constant C can take the following form: C(deg(S)) sup Ω

2 X k=0

k k δΩ,h (x) |∂yk ϕ(x, y)| + δΩ,v (y) |∂xk ϕ(x, y)|

(3.13)

for all ϕ supported in a horizontally and vertically convex domain Ω. Here C(deg(S)) is a constant depending only on the degree of S. Remark 3.1 The assumption η ≥ 1 is crucial in our proof for technical reasons. If η < 1, we shall instead change the role of x and y. P Proof. Choose a smooth bump function Φ such that supp (Φ) ⊆ [1/2, 2] and j Φ(x/2j ) = 1 for x > 0. Define Wj,k as Wz by insertion of Φ(x/2j )Φ(y/2k ) into the cut-off. In other words, Z eiλS(x,y) |D(x, y)|z Φ(x/2j )Φ(y/2k )ϕ(x, y)f (y)dy. (3.14) Wj,k f (x) = R

Here we only consider the operator Wz in the first quadrant. Estimates in other quadrants can be treated similarly. We assume z has real part in (3.12) and |α1 | ≤ |α2 | ≤ · · · ≤ |αN | 12

throughout this proof. Step 1. All αi1 , αi2 , · · · , αis are real numbers. We first prove the theorem under the additional assumption αit ∈ R. One can see that our arguments are also applicable without essential change in presence of complex αit ; see Step 2. Case (i) |α1 |2(k−1)η ≥ 2j+2 . P Assume first m = 0. For fixed k, we define Wk = j Wj,k with the summation taken over all j satisfying Case (i). Then Wk is supported in the rectangle Rk : |x| ≤ |α1 |2(k−1)η−1 , 2k−1 ≤ y ≤ 2k+1 . For |k − l| ≥ 2, Wk Wl∗ = 0. Thus it suffices to estimate Wk∗ Wl to get the almost orthogonality. Since Wk∗ Wl and Wl∗ Wk have the same operator norm on L2 , we may assume k ≥ l. Observe that the Hessian of S has the uniform upper and lower bounds on an expanded rectangle Rk∗ of Rk . More precisely, we have ′′ (x, y)| |Sxy

kn

≈2

N Y i=1

!

|αi | 2N kη

on the expanded rectangle Rk∗ : |x| ≤ |α1 |2(k−1)η−1/2 , 2k−1−ǫ ≤ y ≤ 2k+1+ǫ with ǫ > 0 satisfying ǫη ≤ 41 . It is easily verified that |x − αi y η | ≈ |αi |2kη with bounds depending only on η. Moreover ′′ (x, y) does not change sign on all vertical line segments joining two points in R∗ and R∗ . We Sxy l k can apply Lemma 2.4 to get kWk∗ Wl k

"

N Y

kn

≤ C |λ|2

i=1

!

N kη

|αi | 2

#−1

ksη

2

s Y t=1

! Re (z)

lsη

|αit |

2

s Y

t=1

! Re (z)

|αit |

.

By the size estimate of for each Wk , we deduce from kWk∗ Wl k ≤ kWk∗ kkWl k that kWk∗ Wl k ≤ C

2ksη

s Y t=1

! Re (z)

|αit |

(|α1 |2kη )1/2 2k/2

2lsη

s Y t=1

! Re (z)

|αit |

(|α1 |2lη )1/2 2l/2 .

Taking a convex combination of the above two estimates, we obtain "

kWk∗ Wl k

kn

≤ C |λ|2

N Y

!

N kη

|αi | 2

i=1 kη (1−θ)/2 k(1−θ)/2

(|α1 |2 )

2

#−θ

ksη

2

s Y

! Re (z)

|αit |

t=1 lη (1−θ)/2 l(1−θ)/2

(|α1 |2 )

lsη

2

2

s Y t=1

.

! Re (z)

|αit |

Take θ = 2γ. We collect terms involving αi in the above inequality and obtain N Y i=1

!−θ

|αi |

s Y

t=1

!2 Re (z)

|αit |

|α1 |1−θ ≤

N Y i=1



≤  13

!−θ

|αi | Y

k ∈{i / 1 ,··· ,is }

s Y

|αit |

t=1 −θ

|αi |

!2 Re (z)+(1−θ)/s

,

(3.15)

where we have used the assumption |α1 | ≤ |α2 | ≤ · · · |αN | and the fact 2 Re (z) +

s − n − (N − s)η 1 n + (N − s)η 1 1−θ = · + · = θ. s n + (N − s)η + 1 s n + (N − s)η + 1 s

On the right side of (3.15), we add all exponents of 2k and obtain 1−θ 1−θ η+ 2 2 s − n − (N − s)η 1 η n + (N − s)η n + Nη + sη · · + · = − n + (N − s)η + 1 2[n + (N − s)η + 1] s 2 n + (N − s)η + 1 1 n + (N − s)η + · 2 n + (N − s)η + 1 1 sη 1 n + (N − s)η n + Nη + · + · = − n + (N − s)η + 1 2 n + (N − s)η + 1 2 n + (N − s)η + 1 1 n + Nη = − · . 2 n + (N − s)η + 1 −(n + N η)θ + sη Re (z) +

The exponent of 2l in (3.15) equals 1−θ 1−θ η+ 2 2 s − n − (N − s)η η n + (N − s)η 1 n + (N − s)η + · + · n + (N − s)η + 1 2 n + (N − s)η + 1 2 n + (N − s)η + 1 n + Nη . n + (N − s)η + 1

sη Re (z) + = =

η · 2 1 · 2

Then the inequality (3.15) becomes 

Y

kWk∗ Wl k ≤ C(1 + |z|2 )|λ|−θ 

k ∈{i / 1 ,··· ,is }

−θ

|αk |

2−|k−l|δ

with θ = 2γ and δ = (n + N η)/[2(n + (N − s)η + 1)] > 0. Now we address Case (i) with m > 0. Observe that Wj,k is supported in the rectangle Rj,k : 1 , we define the expanded x ∈ [2j−1 , 2j+1 ], y ∈ [2k−1 , 2k+1 ]. For some ǫ > 0 satisfying ǫ ≤ 4η rectangle ∗ Rj,k : 2j−3/2 ≤ x ≤ 2j+3/2 , 2k−1−ǫ ≤ y ≤ 2k+1+ǫ .

Then we can verify that all assumptions in Lemma 2.4 are true. Since Wj,k Wj∗′ ,k′ = 0 for ′ |k − k′ | ≥ 2, we shall view 2k and 2k as the same number. By Lemma 2.4, Wj,k Wj∗′ ,k′ satisfies, assuming j ≥ j ′ , kWj,k Wj∗′ ,k′ k

"

jm kn

≤ C |λ|2 "

′

2j m

2

N Y i=1

s Y

u=1

!

kN η

|αi | 2 !

′

|αiu | 2k sη 14

#−1 "

# Re (z)

.

jm

2

s Y

u=1

!

ksη

|αiu | 2

# Re (z)

×

Since kWj,k Wj∗′ ,k′ k ≤ kWj,k kkWj∗′ ,k′ k, it follows from size estimates for each operator that # Re (z) ! # Re (z) " ! " s s Y Y ′ ′ ′ ′ |αiu | 2k sη 2j /2 2k /2 . |αiu | 2ksη 2j/2 2k/2 2j m 2jm kWj,k Wj∗′ ,k′ k . u=1

u=1

We take a convex combination of the above two inequalities and obtain # Re (z) ! #−θ " ! " s N Y Y ksη jm kN η ∗ jm kn 2j(1−θ)/2 × |αiu | 2 2 |αi | 2 kWj,k Wj ′ ,k′ k . |λ|2 2 "

i=1

′

2k(1−θ)/2 2j m Put θ = 2γ. Since |k − of 2k is equal to τ

k′ |

s Y

u=1

!

u=1

′

|αiu | 2k sη

# Re (z)

′

′

2j (1−θ)/2 2k (1−θ)/2 .

(3.16)

≤ 1, we can identify k′ with k in this estimate. Then the exponent

:= −(n + N η)θ + 2sη Re (z) + 1 − θ n + Nη m + s − n − (N − s)η 1 n + (N − s)η = − + 2sη · + n + (N − s)η + 1 n + (N − s)η + 1 2(m + s) n + (N − s)η + 1 sη n + (N − s)η · ≤ 0. = − n + (N − s)η + 1 m + s

Recall that |α1 |2(k−1)η ≥ 2j+2 in Case (i). Inserting this estimate into (3.16) and collecting terms concerning αi , we obtain !2 Re (z) !−θ !− τ !2 Re (z) !−θ s N s s N sη Y Y Y Y Y −τ /η |αiu | |αi | |αiu | |αiu | |αi | |α1 | ≤ i=1

u=1

i=1

≤

N Y i=1



where we have used the equality 2 Re (z) −

τ sη

≤ 

!−θ

|αi |

Y

k ∈{i / 1 ,i2 ,··· ,is }

u=1 s Y

u=1

!θ

u=1

|αiu |

−θ

|αk |

,

m + s − n − (N − s)η 1 n + (N − s)η 1 · + · 2[n + (N − s)η + 1] m + s n + (N − s)η + 1 m + s 1 = θ. n + (N − s)η + 1

= 2· =

At the same time, the exponent of 2j becomes 1−θ τ − mθ + m Re (z) + η 2

1 n + (N − s)η sη m = − · · − + η n + (N − s)η + 1 m + s n + (N − s)η + 1 1 n + (N − s)η m + s − n − (N − s)η · + m· 2[n + (N − s)η + 1] m + s 2[n + (N − s)η + 1] n + (N − s)η s m = − · − 2[n + (N − s)η + 1] m + s 2[n + (N − s)η + 1] m s · (1 − θ) − θ. = − 2(m + s) 2 15

′

By direct calculation, we see that the exponent of 2j is m Re (z) +

1−θ 2

m + s − n − (N − s)η 1 n + (N − s)η · + 2[n + (N − s)η + 1] m + s 2[n + (N − s)η + 1] m s · (1 − θ) + θ. 2(m + s) 2

= m· =

Combining these estimates, we get 

kWj,k Wj∗′ ,k′ k . (1 + |z|2 )|λ|−θ 

Y

k ∈{i / 1 ,i2 ,··· ,is }

with δ defined by

δ=

−θ

|αk |

′

2−|j−j |δ

s m · (1 − θ) + · θ > 0. 2(m + s) 2

∗ W By the same argument, we can show that Wj,k j ′ ,k ′ satisfies a similar estimate. We omit the details here.

Case (ii) 2j ≥ |αN |2(k+1)η+2 . As in Case (i), the argument is slightly P different depending on whether n = 0. First consider the case n = 0. For each j, define Wj = k Wj,k with the summation taken over all k satisfying the condition in Case (ii). Then Wj is supported in the rectangle Rj : 2j−1 ≤ x ≤ 2j+1 , 0 ≤ y ≤ (|αN |−1 2j−2 )1/η . Since Wj∗ Wj ′ = 0 for |j − j| ≥ 2, it is enough to estimate Wj Wj∗′ . Observe that kWj Wj∗′ k = kWj ′ Wj∗ k. We may assume j ≥ j ′ in the following proof. It should be pointed out that the almost orthogonality estimate in Lemma 2.4 is not applicable here since D(x, y) is not a polynomial type function in y on 0 ≤ y ≤ (|αN |−1 2j−2 )1/η . However, if s = N then D is a polynomial in R2 . Assume first s < N . Since both D(x, y) and |D(x, y)|z are polynomial functions of order 2 with respect to x, we can apply Lemma 2.2 to obtain the following oscillation estimate −1/2 jm js Re (z) kWj k ≤ C(1 + |z|2 ) |λ|2jm 2jN 2 2 . By the Schur test, we obtain the size estimate kWj k ≤ C 2jm 2js

Re (z)

In the oscillation estimate, the exponent of 2j is −

2j/2 |αN |−1 2j−2

2η1

.

m+N m+N θ 1−θ + (m + s) Re (z) = − + (m + s)( − ) 2 2 2 2(m + s) 1−θ θ m+N − + (m + s) < 0 = − 2 2 2

because of s < N . On the other hand, the exponent of 2j in the size estimate is θ 1 1−θ 1 1 1 = (m + s) − + + (m + s) Re (z) + + 2 2η 2 2(m + s) 2 2η 1 θ θ > 0. = (m + s) + + 2 2 2η 16

Balancing the oscillation and size estimates, we will obtain 

X

X Y

Wj ≤

Wj ≤ C(1 + |z|2 ) |λ|

j

j

k ∈{i / 1 ,i2 ,··· ,is }

−γ

|αk |

kf kL2 .

Now we turn to the case when D(x, y) is a polynomial type function in y on [0, (|αN |−1 2j−2 )1/η ]; for example s = N . One can also verify the assumptions in Lemma 2.4. Hence Wj Wj∗′ satisfies Re (z) Re (z) ′ i−1 h ′ 2j m 2j s . 2jm 2js kWj Wj∗′ k ≤ C |λ|2jm 2jN

By the Schur test, kWj Wj∗′ k is bounded by

1 ′ Re (z) 1 ′ −2 2η ′ s Re (z) j ′ /2 2η −1 j−2 jm js j/2 −1 j j m j |αN | 2 2 kWj kkWj ′ k . 2 2 |αN | 2 2 2 2 . A convex combination of these two estimates yields kWj Wj∗′ k .

Re (z) Re (z) ′ i−θ ′ × 2j m 2j s 2jm 2js |λ|2jm 2jN 1−θ ′ 1−θ ′ 2η 2η 2j(1−θ)/2 |αN |−1 2j−2 2j (1−θ)/2 |αN |−1 2j −2 .

h

In the above estimate, the exponent of 2j equals

1−θ 1−θ + 2 2η m+N m + s − (N − s)η 1 = − + (m + s) · · + (N − s)η + 1 2[(N − s)η + 1] m + s (N − s) (N − s)η + 2[(N − s)η + 1] 2[(N − s)η + 1] m+N = − , 2[(N − s)η + 1] −(m + N )θ + (m + s) Re (z) +

′

and the exponent of 2j is given by 1−θ 1−θ + 2 2η 1 (N − s)η (N − s) m + s − (N − s)η · + + = (m + s) · 2[(N − s)η + 1] m + s 2[(N − s)η + 1] 2[(N − s)η + 1] m+N = . 2[(N − s)η + 1] (m + s) Re (z) +

On the other hand, we see that (1 − θ)/2η = (N − s)θ and |αN |−

1−θ 2η



≤

Y

k ∈{i / 1 ,··· ,is }

17

−θ

|αk |

.

(3.17)

All of these results imply that 

kWj Wj∗′ k ≤ C(1 + |z|2 )|λ|−θ 

Y

k ∈{i / 1 ,i2 ,··· ,is }

−θ

|αk |

′

2−|j−j |δ

with δ = 2[(Nm+N −s)η+1] ≥ 0. If δ = 0, then m = N = 0. Recall that we assumed n = 0. This ′′ = 1 and the L2 estimate (3.11) is a consequence of Plancherel theorem. Here implies that Sxy we need only consider the case δ > 0. Then the desired estimate in Case (ii) follows from the almost orthogonality principle. Now we turn to the estimate for n > 0 in Case (ii). For |k − k ′ | ≥ 2, Wj,k Wj∗′ ,k′ = 0. Assume |k − k′ | ≤ 1. The operator Wj,k is supported in the rectangle Rj,k : 2j−1 ≤ x ≤ 2j+1 , ′′ is comparable to a fixed value on an expanded region 2k−1 ≤ y ≤ 2k+1 . One can verify that Sxy ∗ which can be defined as in Case (i). Other assumptions in Lemma 2.4 are also true. Rj,k Assume j ≥ j ′ . By Lemma 2.4, there holds i−1 h Re (z) h j ′ m j ′ s i Re )(z) 2 2 . 2jm 2js kWj,k Wj∗′ ,k′ k ≤ C |λ|2jm 2kn 2jN The size estimate is, using kWj,k Wj∗′ ,k′ k ≤ kWj,k kkWj ′ ,k′ k, Re (z) j/2 k/2 h j ′ m j ′ s i Re )(z) j ′ /2 k′ /2 2 2 kWj,k Wj∗′ ,k′ k ≤ C 2jm 2js 2 2 . 2 2

We take a convex combination with θ = 2γ and obtain i−θ h Re (z) j(1−θ)/2 k(1−θ)/2 2jm 2js 2 2 × kWj,k Wj∗′ ,k′ k . |λ|2jm 2kn 2jN i Re (z) ′ h ′ ′ ′ 2j m 2j s 2j (1−θ)/2 2k (1−θ)/2 .

All terms and their exponents are given as follows: 1−θ 1−θ ′ 2j : − (m + N )θ + (m + s) Re (z) + , 2j : (m + s) Re (z) + ; 2 2 ′ 1−θ 1−θ , 2k : . 2k : − nθ + 2 2 ′ ′ ′ Since |k − k′ | ≤ 1 and 2k . (|αN |−1 2j )1/η , the product of two terms concerning 2k and 2k is ′ bounded by C(|αN |−1 2j )(−nθ+1−θ)/η . Then the new exponent of 2j becomes 1−θ 1 + (−nθ + 1 − θ) 2 η n (m + s)θ 1 − θ 1 − θ n − + − θ + θ + (N − s)θ = 2 2 2 η η m s = N+ θ. − 2 2 To obtain the almost orthogonality, we shall verify the exponent of 2j equals −(N + In fact, this is true. By direct computation, 1−θ −(m + N )θ + (m + s) Re (z) + 2 θ 1−θ 1−θ + = −(m + N )θ + (m + s) − 2 2 2 m s θ. − = − N+ 2 2 (m + s) Re (z) +

18

m 2

− s)θ.

On the other hand, the bound concerning αi is equal to (|αN |−1 )(−nθ+1−θ)/η . Observe that (−nθ+1−θ)/η = (N −s)θ. By the assumption |α1 | ≤ · · · ≤ |αN |, we see that (|αN |−1 )(−nθ+1−θ)/η Q −θ is less than or equal to . Combining this fact with above results, we obtain k ∈{i / 1 ,··· ,is } |αk | 

kWj,k Wj∗′ ,k′ k ≤ C |λ|

with δ given by

Y

k ∈{i / 1 ,··· ,ik }

−θ

|αk |

′

2−|k−k |δ

m s δ= N+ θ. − 2 2

∗ W A similar argument also shows that Wj,k j ′ ,k ′ satisfies the same estimate.

Case (iii) |α1 |2(k−1)η−2 < 2j < |αN |2(k+1)η+2 . For some large number N0 = N0 (η) > 0, consider the intervals Ii = (|αi |2−2N0 , |αi |22N0 ) with 1 ≤ i ≤ N . Since |α1 | ≤ |α2 | ≤ · · · |αN |, we see that Ii ∩Ii+1 = ∅ if and only if |αi+1 |/|αi | ≥ 24N0 . Let {t1 , t2 , · · · , ta } ⊆ {1, 2, · · · , N } consist of all indices i such that |αi+1 |/|αi | ≥ 24N0 . Of course, the set may very well be empty. We shall divide our proof into two subcases. Subcase (a) |αtr |2(k+1)η+2 ≤ 2j ≤ |αtr +1 |2(k−1)η−2 for some 1 ≤ r ≤ a. Define Ξ1 = {1, 2, · · · , tr } and Ξ2 = {tr + 1, tr + 2, · · · , N }. Then on the support of Wj,k , ′′ (x, y)| is comparable to |Sxy 

′′ |Sxy (x, y)| ≈ 2jm 2kn 

Y

t∈Ξ1

and the damping factor D(x, y) has size



|D(x, y)| ≈ 2jm 

Y

t∈Ξ1 ∩Θ



2j   

2j  

Y

t∈Ξ2

Y

t∈Ξ2 ∩Θ



|αt |2kη  

|αt |2kη 

where Θ = {i1 , i2 , · · · , is }. Since Wj,k Wj∗′ ,k′ = 0 for |k − k′ | ≥ 2, we assume now |k − k′ | ≤ 1 and j ≥ j ′ without loss of generality. By Lemma 2.4, there holds 



kWj,k Wj∗′ ,k′ k . |λ|2jm 2kn 2j|Ξ1 |  

Y

t∈Ξ2

j′m

2

j ′ |Ξ1 ∩Θ|

2

 

Y

−1 

|αt |2kη 

t∈Ξ2 ∩Θ

k′ η

|αt |2

19

2jm 2j|Ξ1 ∩Θ| 

 Re (z) 



.

Y

t∈Ξ2 ∩Θ

 Re (z)

|αt |2kη 

×

By the size estimate of each Wj,k , we have, using kWj,k Wj∗′ ,k′ k ≤ kWj,k kkWj∗′ ,k′ k, 



kWj,k Wj∗′ ,k′ k . 2jm 2j|Ξ1 ∩Θ| 

Y

t∈Ξ2 ∩Θ





2j ′ m 2j ′ |Ξ1 ∩Θ| 

Y

 Re (z)

|αt |2kη  ′

t∈Ξ2 ∩Θ

2j/2 2k/2 ×

 Re (z)

|αt |2k η 

′

′

2j /2 2k /2 .

With θ = 2γ, we take a convex combination of the oscillation and size estimates. This yields j ′ a′j ′ k ′ b′ ′ k

kWj,k Wj∗′ ,k′ k ≤ C|λ|−θ 2jaj 2kbk 2

2

,

(3.18)

where the above exponents are given as follows: aj = −(m + |Ξ1 |)θ + (m + |Ξ1 ∩ Θ|) Re (z) + a′j ′ = (m + |Ξ1 ∩ Θ|) Re (z) +

1−θ , 2

bk = −nθ − |Ξ2 |θη + |Ξ2 ∩ Θ| Re (z)η + b′k′ = |Ξ2 ∩ Θ| Re (z)η +

1−θ , 2

1−θ , 2

1−θ . 2

By the assumption |k − k′ | ≤ 1, we see that ′ ′

2kbk 2k bk′ ≤ C2k[2|Θ∩Ξ2 | Re (z)η+1−θ−|Ξ2 |ηθ−nθ] .

(3.19)

Define s1 = |Ξ1 ∩ Θ| and s2 = |Ξ2 ∩ Θ|. Then s = s1 + s2 . Using 1 − θ = [n + (N − s)η]θ, 2|Θ ∩ Ξ2 | Re (z)η + 1 − θ − |Ξ2 |ηθ − nθ = 2s2 η Re (z) + (1 − θ) − (N − tr )ηθ − nθ

= 2s2 η Re (z) + (1 − θ) − (N − s)ηθ − (s − tr )ηθ − nθ

= 2s2 η Re (z) + (tr − s)ηθ

= −η(θ − 2 Re (z))s2 + (tr − s1 )ηθ. ′

Since |αtr |2kη . 2j , 2j . |αtr +1 |2kη and |k − k′ | ≤ 1, the right side of (3.19) can be divided into two terms, and each term can treated as follows: k(tr −s1 )ηθ

2

.

1 j′ 2 |αtr |

(tr −s1 )θ

and

−kη(θ−2 Re (z))s2

2

.

1 |αtr +1 |

j

2

−(θ−2 Re (z))s2

. (3.20)

Substituting this estimate into (3.18), we obtain the new exponent of 2j : 1−θ 2 = −s2 θ + 2s2 Re (z) − (m + s1 )θ − (tr − s1 )θ + (m + s) Re (z) 1−θ −s2 Re (z) + 2 = −(m + s)θ + s2 Re (z) − (tr − s1 )θ + (m + s)θ/2.

−(θ − 2 Re (z))s2 + aj = −(θ − 2 Re (z))s2 − (m + tr )θ + (m + s1 ) Re (z) +

20

Note that −(m + s)θ/2 + s2 Re (z) = −(m + s) Re (z) − 1−θ 2 + s2 Re (z). It follows from the above equality that −(θ − 2 Re (z))s2 + aj

1−θ + s2 Re (z) − (tr − s1 )θ 2 1−θ = −(m + s1 ) Re (z) − − (tr − s1 )θ. 2 = −(m + s) Re (z) −

′

It is easy to see that the new exponent of 2j is 1−θ 2

(tr − s1 )θ + a′j ′ = (tr − s1 )θ + (m + |Ξ1 ∩ Θ|) Re (z) + = (tr − s1 )θ + (m + s1 ) Re (z) +

1−θ . 2

Combining the above estimates, we obtain ′

kWj,k Wj∗′ ,k′ k ≤ C(1 + |z|2 )2 |λ|−θ 2−|j−j |δ where 1−θ 2 m + s1 1−θ m + s1 ·θ− · (1 − θ) + = (tr − s1 )θ + 2 2(m + s) 2 s2 m + s1 ·θ+ (1 − θ). = (tr − s1 )θ + 2 2(m + s)

δ = (tr − s1 )θ + (m + s1 ) Re (z) +

Note that if {t1 , · · · , ta } = ∅ then we need only consider Subcase (b) below. Hence we can assume tr ≥Q1. It follows immediately that δ > 0. The constant C is bounded by a constant −θ multiple of ( k∈Θ / |αk |) . Its dependence on αi comes from (3.18) and (3.20), and we can verify this claim as follows,  

Y Ξ2

−θ 

|αt |



Y

Ξ2 ∩Θ

2 Re (z)

|αt |

1 |αtr |

(tr −s1 )θ

1 |αtr+1 |

−(θ−2 Re (z))s2

≤

Y

t∈Θ /

!−θ

|αt |

where we have used the following two inequalities: 

|αtr |−(tr −s1 )θ ≤ 

Y

t∈Ξ / 1 ∩Θ

−θ

|αt |



and |αtr +1 |(θ−2 Re (z))s2 ≤ 

Y

t∈Ξ2 ∩Θ

Therefore we have obtained the desired estimate in Subcase (a).

θ−2 Re (z)

|αt |

.

Subcase (b) |αtr +1 |2(k−1)η−2 < 2j < |αtr+1 |2(k+1)η+2 for some 0 ≤ r ≤ a. We use t0 to denote t0 = 0. Let G0 = {tr + 1, tr + 2, · · · , tr+1 } and Θ = {i1 , i2 , · · · , is }. σ0 Choose e0 ∈ G0 arbitrarily. For clarity, set e0 = tr + 1, the least number in G0 . Define Wj,k,l 0 η x−αe0 y . Here σ0 takes either + or −. as Wj,k , but with the cut-off multiplied by Φ σ0 2l0 21

Note that |αtr+1 |/|αtr +1 | ≤ 24N0 . By the almost orthogonality principle, there exists a constant C, depending on N0 , such that

X

Wj,k ≤ C sup kWj,k k,

j,k

where both the summation and the supremum are taken over all (j, k) satisfying Subcase (b). Associated with fixed j, k, l0 , we can decompose G0 into the following three subsets: G1,1 = {i ∈ G0 | |αi − αe0 |2kη ≥ 2l0 +N0 }

G1,2 = {i ∈ G0 | |αi − αe0 |2kη ≤ 2l0 −N0 }

G1,3 = {i ∈ G0 | 2l0 −N0 < |αi − αe0 |2kη < 2l0 +N0 }.

If G1,3 is empty, then our decomposition is finished. Otherwise, choose the least number e1 in x−αe1 y η σ0 ,σ1 σ0 σ0 G1,3 , and define Wj,k,l0,l1 as Wj,k,l0 by inserting Φ σ1 2l1 into the cut-off of Wj,k,l , where 0

σ1 = + or −. Since the number of l0 satisfying |αe1 − αe0 |2kη ≈ 2l0 is bounded by a constant C(N0 ), there exists a constant C, depending only on N0 , such that

X

σ0

σ0 ≤ C sup Wj,k,l

W

j,k,l0 , 0 l0

l0

where the summation and superemum are taken over all l0 satisfying |αe1 −αe0 |2kη ≈ 2l0 . Further decompose G1,3 as follows: G2,1 = {i ∈ G1,3 | |αi − αe1 |2kη ≥ 2l1 +N0 }

G2,2 = {i ∈ G1,3 | |αi − αe1 |2kη ≤ 2l1 −N0 }

G2,3 = {i ∈ G1,3 | 2l1 −N0 < |αi − αe1 |2kη < 2l1 +N0 }. If G2,3 is nonempty, we continue this decomposition procedure and obtain three subsets G3,1 , G3,2 , G3,3 of G2,3 . Generally, if Gi,1 , Gi,2 , Gi,3 are given with Gi,3 6= ∅, then we shall choose the least element ei of Gi,3 and decompose Gi,3 , for each fixed li , as follows: Gi+1,1 = {t ∈ Gi,3 | |αt − αei |2kη ≥ 2li +N0 }

Gi+1,2 = {t ∈ Gi,3 | |αt − αei |2kη ≤ 2li −N0 }

Gi+1,3 = {t ∈ Gi,3 | 2li −N0 < |αt − αei |2kη < 2li +N0 }.

Then we obtain three disjoint subsets Gi+1,1 , Gi+1,2 , Gi+1,3 . This process will continue until Gw,3 = ∅ for some w. We shall prove later that this process will terminate in finite steps. Although our construction of Gi,1 , Gi,2 , Gi,3 depends on li−1 , we may regard {Gi,1 , Gi,2 , Gi,3 } as a three-tuple satisfying three properties: (i) Gi,u

T

Gi,v = ∅ for u 6= v;

(ii) if Gi,3 is nonempty then so is Gi+1,2 ; (iii) Gi−1,3 = Gi,1 ∪ Gi,2 ∪ Gi,3 with G0,3 = G0 . 22

These properties imply |Gi+1,3 | ≤ |Gi,3 | − 1 provided that Gi,3 is nonempty. Thus the above decomposition process stops in finite steps. Since G0 is a finite set, |G0 | = tr+1 − tr ≤ N , the number of all three tuples (Gi,1 , Gi,2 , Gi,3 ), for each i ≥ 1, is bounded by a constant depending only on N . Therefore, in each step, we can divide the summation over li into a finite summation by restricting li in (Gi,1 , Gi,2 , Gi,3 ). In other words, li satisfies inequalities associated with Gi,u ; for example, inequalities associated with Gi,1 are |αt − αei |2kη ≥ 2li +N0 , where t ∈ Gi,1 and ei is the least member in Gi,2 . This σ0 ,σ1 ,··· ,σw−1 observation will simplify our proof of the decay estimate of Wj,k,l . 0 ,l1 ,··· ,lw−1 σ0 ,··· ,σw−1 σ0 ,··· ,σw−1 ′ Assume Wj,k,l0,··· ,lw−1 and Wj,k,l0,··· ,l′ are two operators such that lw−1 and lw−1 lie in the w−1

same three tuple (Gw−1,i )3i=1 . Now we shall verify the assumptions in Lemma 2.4. First observe σ0 ,··· ,σw−1 that Wj,k,l is supported in the intersection of Ω ⊇ supp (ϕ) and the following horizontally 0 ,··· ,lw−1 (also vertically) convex domain ( ) x y w−1 Y x − αe y η σ0 ,··· ,σw−1 t Φ σt Ωj,k,l0,··· ,lw−1 = Closure of (x, y) : Φ j Φ k 6= 0 . 2 2 2lt t=0

σ ,··· ,σ

∗

0 w−1 by Define an expanded region Ωj,k,l 0 ,··· ,lw−1

2j−1 − ǫ2j ≤ x ≤ 2j+1 + ǫ2j ,

2k−1 − ǫ2k ≤ y ≤ 2k+1 + ǫ2k ,

2lt −1 − ǫ2lt ≤ σt (x − αet y η ) ≤ 2lt +1 + ǫ2lt ,

0≤t≤w−1 σ ,··· ,σ

0 w−1 for some sufficiently small ǫ = ǫ(η) > 0. Observe that Wj,k,l = 0 unless lw−1 ≤ 0 ,··· ,lw−1 min{j, l0 , · · · , lw−1 } + C(η). One can verify that all conditions in Lemma 2.4 are true for σ0 ,··· ,σw−1 σ0 ,··· ,σw−1 Wj,k,l and Wj,k,l . ′ 0 ,··· ,lw−1 0 ,··· ,l w−1

σ ,··· ,σ

σ ,··· ,σ

∗

0 w−1 0 w−1 ′ k is bounded by a . By Lemma 2.4, kWj,k,l Wj,k,l Assume lw−1 ≥ lw−1 ′ 0 ,l1 ,··· ,lw−1 0 ,l1 ,··· ,lw−1 constant multiple of    −1 w N Y Y Y Y |λ|2jm 2kn 2jtr  2li−1  |αt − αei−1 |2kη |αt |2kη  ×

i=1



2jm 2j|Ξ1 ∩Θ|  

Y

t∈Gw,1 ∩Θ

w−1 Y i=1

 

t∈Gi,2

t∈Gi,1

Y

t∈Gi,1 ∩Θ

|αt − αew−1 |2kη

Y

|αt − αei−1 |2kη

Y

t∈Gw,2 ∩Θ

t∈Gi,2 ∩Θ



2lw−1  

Y

t∈Gw,1 ∩Θ

t=tr+1 +1



2 Re (z) Y 2li−1  |αt |2kη  × t∈Ξ2 ∩Θ


Y

t∈Gw,2 ∩Θ

′

 Re (z)

2lw−1 

where Ξ1 = {1, 2, · · · , tr }, Θ = {i1 , i2 · · · , is } and Ξ2 = {tr+1 + 1, · · · , N }. σ0 ,··· ,σw−1 σ0 ,··· ,σw−1 ∗ By size estimates for each operator, kWj,k,l Wj,k,l k is less than a constant ′ 0 ,l1 ,··· ,lw−1 0 ,l1 ,··· ,lw−1 multiple of   2 Re (z)  w−1 Y Y Y Y 2jm 2j|Ξ1 ∩Θ|  2li−1  |αt |2kη  |αt − αei−1 |2kη × i=1

t∈Gi,1 ∩Θ

t∈Gi,2 ∩Θ

23

t∈Ξ2 ∩Θ



Y



t∈Gw,1 ∩Θ


2lw−1 /2

Y

t∈Gw,2 ∩Θ



Y

2lw−1  

t∈Gw,1 ∩Θ

2lw−1 |αew−1 |2k(η−1)

1/2

|αt − αew−1 |2kη ′

′ lw−1 /2

2

Y

′ lw−1

2

t∈Gw,2 ∩Θ

2lw−1 |αew−1 |2k(η−1)

!1/2

 Re (z) 

×

,

where the last two terms are upper bounds for the measure of y x − αew−1 y η = 6 0 y | Φ k Φ σw−1 2 2u ′ with u ∈ {lw−1 , lw−1 }. A convex combination of the above oscillation and size estimates with θ = 2γ yields the following terms and their exponents:

2j :

2(m + |Ξ1 ∩ Θ|) Re (z) − (m + |Ξ1 |)θ,

2k :

− nθ − (|G1,1 | + |G2,1 | + · · · + |Gw,1 | + |Ξ2 |)ηθ+ 2 (|G1,1 ∩ Θ| + |G2,1 ∩ Θ| + · · · + |Gw,1 ∩ Θ| + |Ξ2 ∩ Θ|) η Re (z) − (η − 1)(1 − θ),

2l0 :

2|G1,2 ∩ Θ| Re (z) − |G1,2 |θ, · · · · · · ,

2lw−2 :

2|Gw−1,2 ∩ Θ| Re (z) − |Gw−1,2 |θ,

2lw−1 :

− |Gw,2 |θ + |Gw,2 ∩ Θ| Re (z) + 1 − θ,

′

2lw−1 :

|Gw,2 ∩ Θ| Re (z) + 1 − θ.

In this convex combination, the bound involving αi is  

Y

t∈G1,1

 

Y



|αt − αe0 | 

Y

t∈G2,1

t∈G1,1 ∩Θ



|αt − αe0 | 





|αt − αe1 | · · · 

Y

t∈G2,1 ∩Θ

Y

t∈Gw,1





|αt − αe1 | · · · 



|αt − αew−1 | 

Y

t∈Ξ2

Y

t∈Gw,1 ∩Θ

−θ

|αt |



|αt − αew−1 | 

Y

|αew−1 |−(1−θ) ×

t∈Ξ2 ∩Θ

2 Re (z)

|αt |

Define s1 = |Ξ1 ∩Θ|, s2 = |{tr +1, · · · , tr+1 }∩Θ| and s3 = |Ξ2 ∩Θ|. It is clear that s = s1 +s2 +s3 . Now we shall deal with the term 2kγk with γk given as above. The definition of Gi,1 implies that, for 1 ≤ t ≤ w, 2k(−|Gt,1 |θη+2|Gt,1 ∩Θ| Re (z)η)

−|Gt,1 |θ+2|Gt,1 ∩Θ| Re (z) lt−1 2  , .  min |αu − αet−1 | 

u∈Gt,1

since −|Gt,1 |θη + 2|Gt,1 ∩ Θ| Re (z)η ≤ 0. Our decomposition of G0 implies that there exists an absolute constant C such that li ≤ j + CN0 and li+1 ≤ li + CN0 for each i. With the restriction 24

.

of j, k in Subcase (b), it is true that k(−|Ξ2 |θη+2|Ξ2 ∩Θ| Re (z)η)

2

.

2lw−1 |αtr+1 |

−|Ξ2 |θ+2|Ξ2 ∩Θ| Re (z)

.

For the same reason as well as the assumption η ≥ 1, there also holds 2k[−nθ−(η−1)(1−θ)] .

2lw−1 |αtr+1 |

[− nη θ− η−1 (1−θ)] η

.

Since 2|Gi−1,2 ∩ Θ| Re (z) − |Gi−1,2 |θ ≤ 0, for 0 ≤ i ≤ w − 2, we also have 2li (2|Gi−1,2 ∩Θ| Re (z)−|Gi−1,2 |θ) . 2lw−1 (2|Gi−1,2 ∩Θ| Re (z)−|Gi−1,2 |θ) . σ ,··· ,σ

σ ,··· ,σ

0 w−1 0 w−1 Inserting these inequalities into the above estimate for kWj,k,l Wj,k,l ′ 0 ,l1 ,··· ,lw−1 0 ,l1 ,··· ,l

∗

w−1

that the resulting exponent of 2lw−1 equals

k, we see

! w X n 2(m + |Ξ1 ∩ Θ|) Re (z) − (m + |Ξ1 |)θ − θ − |Gi,1 | + |Ξ2 | θ + η i=1 ! w X (η − 1) 2 |Gi,1 ∩ Θ| + |Ξ2 ∩ Θ| Re (z) − (1 − θ) + η

i=1 w−1 X i=1

2|Gi,2 ∩ Θ| Re (z) − |Gi,2 |θ − |Gw,2 |θ + |Gw,2 ∩ Θ| Re (z) + 1 − θ

= (2m + 2s − |Gw,2 ∩ Θ|) Re (z) − (m + s)θ

= (m + s)θ − (1 − θ) − |Gw,2 ∩ Θ| Re (z) − (m + s)θ

= −|Gw,2 ∩ Θ| Re (z) − (1 − θ).

The new bound involving αi is bounded by a constant multiple of  −θ  2 Re (z) Y Y  |αt |  |αt | |αtr+1 |b , t∈Ξ2

t∈Ξ2 ∩Θ

where

η−1 n θ+ (1 − θ) η η n n + (N − s)η = −(1 − θ) + |Ξ2 ∩ Θ|θ + |Ξ2 \Θ|θ − 2|Ξ2 ∩ Θ| Re (z) + θ + (1 − θ) − θ η η = |Ξ2 ∩ Θ|θ + |Ξ2 \Θ|θ − 2|Ξ2 ∩ Θ| Re (z) − (N − s)θ, using N − s = |Ξ1 \Θ| + |G0 \Θ| + |Ξ2 \Θ|,

b = −(1 − θ) + |Ξ2 |θ − 2|Ξ2 ∩ Θ| Re (z) +

= |Ξ2 ∩ Θ|θ − 2|Ξ2 ∩ Θ| Re (z) − |Ξ1 \Θ|θ − |G0 \Θ|θ.

This implies  

Y

t∈Ξ2

−θ 

|αt |



Y

t∈Ξ2 ∩Θ

2 Re (z)

|αt |

25

b

|αtr+1 | ≤

Y

t∈Θ /

!−θ

|αt |

.

Now we have verified that

σ0 ,··· ,σw−1 σ0 ,··· ,σw−1 ∗

Wj,k,l0,l1 ,··· ,lw−1 Wj,k,l0,l1 ,··· ,l′ ≤ C w−1

|λ|

Y

k ∈Θ /

!−θ

|αk |

′

2−|lw−1 −lw−1 |δ

with δ = |Gw,2 ∩ Θ| Re (z) + 1 − θ. It remains to show δ > 0. Since N ≥ 1, we see that if 1 − θ = 0 then n = 0, s = N and hence δ = |Gw,2 |θ/2 > 0. On the other hand, if 1 − θ > 0 then δ=

aθ 2

+ 1−

a 2(m+s)

(1 − θ) > 0 with a = |Gw,2 ∩ Θ|. By the same argument as above, one σ ,··· ,σ

σ ,··· ,σ

∗

0 w−1 0 w−1 Wj,k,l can obtain a similar estimate for Wj,k,l ′ 0 ,l1 ,··· ,lw−1 0 ,l1 ,··· ,l

w−1

.

Step 2. Existence of complex αit Assume αit0 has nonzero imaginary part for some 1 ≤ t0 ≤ s. For the two cases 2j+2 ≤ |αit0 |2(k−1)η and 2j ≥ |αit0 |2(k+1)η+2 , the earlier arguments in Step 1 can apply without any change. So we need only consider the range 2j ≈ |αit0 |2kη . Note that |x − αit0 y η | ≈ |x − Re (αit0 )y η | + | Im (αit0 )|y η .

If | Re (αit0 )| ≤ | Im (αit0 )|, then |x − αit0 y η | ≈ |αit0 |2kη . The earlier arguments also produce the desired estimate. Assume now | Re (αit0 )| > | Im (αit0 )|. By balancing the oscillation and size estimates, we see that only Subcase (b) of Case (iii) requires a separate treatment. Indeed, if ′′ , is required, we shall insert a further decomposition, near the root x = αit0 y η of the Hessian Sxy x− Re (αi )yη t0 into Wj,k . With insertion of this cut-off, we have Φ σ 2l |x − αit0 y η | ≈ max{2l , | Im (αit0 )|2kη }.

By further decomposition as in Step 1, with Re (αit0 ) in place of αit0 , the Hessian is bounded from both below and above by the same bound up to a multiplicative constant. The previous proof applies without change and the desired estimate follows. It remains to show that all constants C appearing in above estimates have more precise form as in the theorem. Indeed, we can deduce from Remark 2.1 that all above constants C can take the following form C(deg(S)) sup Ω

2 X k=0

k k δΩ,h (x) |∂yk ϕ(x, y)| + δΩ,v (y) |∂xk ϕ(x, y)| .

Here C(deg(S)) is a constant depending only on the degree of S. Combining all above results, we have completed the proof of the theorem.

✷

As mentioned in Remark 3.1, the assumption η ≥ 1 is necessary in the proof of Theorem 3.1. Now we address the case η < 1. Let ν = η −1 > 1. Since S is a polynomial, (3.8) can be rewritten as M Y ′′ Sxy (x, y) = d0 xm y n (xν − βi y) (3.21) i=1

with d0 , βi ∈ C\{0}. Without loss of generality, we may assume c0 = 1. For some 0 ≤ s ≤ M , we first choose indices 1 ≤ i1 < · · · < is ≤ N . Then define the damping factor D as D(x, y) = xm

s Y t=1

26

(xν − βit y) .

Here we may take s = 0 and then define D(x, y) = xm . The following theorem can be regarded as a variant of Theorem 3.1. Theorem 3.2 Assume S is a real-valued polynomial such that its Hessian is given by (3.21) with ν ≥ 1. Let Wz be defined as in (3.10). Then there exists a constant C, depending only on deg(S) and ϕ, such that −γ  Y 1 , (3.22) |βk | kf kL2 , γ = kWz f kL2 ≤ C(1 + |z|2 ) |λ| 2(n + M − s + 1) k ∈{i / 1 ,··· ,is }

where z ∈ C has real part

Re (z) =

m + sν − n − (M − s) 1 · . 2(n + M − s + 1) m + sν

(3.23)

Remark 3.2 The proof of this theorem is the same as that of Theorem 3.1. For this reason, we omit the details here. It should be pointed out that the constant C in the above theorem can also take the form in Theorem 3.1. By the same argument as above, we can also prove Theorem 1.2 with uniformity on both the phases and the cut-off functions. We conclude this section with a L2 damping estimate for Wj,k in (3.14). Assume Θ = {i1 , i2 , · · · , is } = {tr + 1, tr + 2, · · · , tr+1 } where t1 , t2 , · · · , ta are defined as in Case (iii) of the proof of Theorem 3.1. In the case 2j ≈ |αis |2kη , we define the damping factor s − s Y s+2 x − αit y η , + D(x, y) = |λ||αis |−1 2−k(η−1) A

(3.24)

t=1

where A is a positive number given by



A = 2mj 2kn 2jtr 

N Y

t=tr+1 +1



|αt |2kη  .

(3.25)

Theorem 3.3 Assume m = 0 in (3.8) and {i1 , i2 , · · · , is } = {tr + 1, tr + 2, · · · , tr+1 } for some 0 ≤ r ≤ a − 1 and |αis |2(k−1)η−2 ≤ 2j ≤ |αis |2(k+1)η+2 . Let Wj,k be the damped oscillatory integral operator (3.14) with D(x, y) defined by (3.24). Then there exists a constant C of form (3.13) such that the decay estimate (3.11) is still true for Wj,k with the damping exponent z ∈ C having real part (3.12). Qs η ′′ behaves like A Proof. With 2j ≈ |αis |2kη , Sxy t=1 (x − αit y ) on the support of Wj,k . As in the proof of Theorem 3.1, we can decompose Wj,k as X σ ,σ ,··· ,σ 0 1 w−1 Wj,k = Wj,k,l . 0 ,l1 ,··· ,lw−1 σ ,σ ,··· ,σ

0 1 w−1 ′′ (also D) is bounded from both On the support of each operator Wj,k,l , the Hessian Sxy 0 ,l1 ,··· ,lw−1 below and above by the same bound up to a multiplicative constant.

27

Q σ0 ,σ1 ,··· ,σw−1 If D has size comparable to st=1 |x − αit y η | on the support of Wj,k,l , then all its 0 ,l1 ,··· ,lw−1 Qs η partial derivatives also have the same upper bounds as t=1 (x − αit y ). With this observation, the desired estimate can be proved by the same argument as in our proof of Theorem 3.1. σ0 ,σ1 ,··· ,σw−1 It remains to consider these operators Wj,k,l for which D has size 0 ,l1 ,··· ,lw−1 − s s+2 . |D(x, y)| ≈ |λ||αis |−1 2−k(η−1) A In other words, it suffices to prove that the desired estimate holds if s − s Y s+2 . |x − αit y η | ≤ |λ||αis |−1 2−k(η−1) A

(3.26)

t=1

Since αi1 , · · · , αis have equivalent sizes and y is restricted in the dyadic interval [2k−1 , 2k+1 ], we use the Schur test to obtain, by estimating the length of horizontal and vertical cross-sections σ0 ,σ1 ,··· ,σw−1 of the support of Wj,k,l , the following two size estimates: 0 ,l1 ,··· ,lw−1

X − 1 (1+s Re (z)) − 1 s+2 2

σ0 ,σ1 ,··· ,σw−1 −1 −k(η−1) k(η−1) W f ≤ C |λ||α | 2 A |α |2 kf kL2

is is j,k,l0,l1 ,··· ,lw−1 2 L

lw−1

and

X

σ0 ,σ1 ,··· ,σw−1 Wj,k,l f

0 ,l1 ,··· ,lw−1

L2

lw−1

− s Re (z) s+2 2j/2 2k/2 kf kL2 , ≤ C |λ||αis |−1 2−k(η−1) A

where the above summations are taken over all lw−1 such that the inequality (3.26) is true on σ0 ,σ1 ,··· ,σw−1 the support of Wj,k,l . Let γ be given by (3.11) and define θ as follows. 0 ,l1 ,··· ,lw−1 θ=

1 n + (N − s)η + 2 = + γ. 2[n + (N − s)η + 1] 2

s θ Note that γ = s+2 Re (z) + s+2 . A convex combination of the above estimates yields

X

1−θ −θ/2 −γ

σ0 ,σ1 ,··· ,σw−1 j/2 k/2 −1 −k(η−1) k(η−1) 2 2 ≤ C |λ||α kf kL2 , | 2 A |α Wj,k,l |2 f

is is 0 ,l1 ,··· ,lw−1 2 lw−1

L

where the exponents of 2j and 2k are given by 2j : aj = −tr γ + 21 (1 − θ),

2k : bk = [(η − 1) − n − (N − tr+1 )η]γ − (η − 1)θ/2 + (1 − θ)/2. By direct calculation, we have ηaj + bk = 0. In fact, using θ = s = tr+1 − tr , one can see that ηaj + bk equlas

= = = =

1 2

+ γ, 1 − θ =

1 2

− γ and

1 −tr γη + (1 − θ)η + [(η − 1) − n − (N − tr+1 )η]γ − (η − 1)θ/2 + (1 − θ)/2 2 1 1 1 1 1 1 [(η − 1) − n − (N − tr+1 )η − tr η]γ + · − γ · η − (η − 1) · +γ · + −γ · 2 2 2 2 2 2 1 1 1 1 1 1 1 1 −γ ·η− · +γ ·η+ · +γ + · −γ [(η − 1) − n − (N − s)η]γ + · 2 2 2 2 2 2 2 2 1 1 ηγ − − γη + 2 2 0. 28

Since sizes of αit are equivalent, the bound concerning αit in the resulting estimate is given by 

|αis |γ |αis |−tr γ  

≤ |αis |γ |αis |−tr γ  ≤

Y

t∈Θ /

!−γ

|αt |

N Y

t=tr+1 +1 N Y

t=tr+1 +1

−γ

|αt |

|αis |−θ/2 |αis |(1−θ)/2

−γ

|αt |

|αis |1/2−θ

with θ =

1 +γ 2

since Θ = {i1 , i2 , · · · , is } = {tr + 1, tr + 2, · · · , tr+1 }.

Combining above results, we have completed the proof of the theorem.

4

✷

Damped Oscillatory Integral Operators on HE1

In this section, we shall establish uniform HE1 → L1 estimates for damped oscillatory integral operators considered in Section 3. Generally, these operators are not bounded from HE1 into L1 . However, we can decompose them into three parts such that each operator has desired properties. ′′ can be written as Assume S is a real-valued polynomial in R2 and its Hessian Sxy ′′ Sxy (x, y) = xm y n

N Y i=1

x − αi y η

with α1 , α2 , · · · , αN ∈ C\{0} and η > 0. If N = 0, then S is a monomial up to pure x, y terms and the uniform estimate in this case is known; see for example [17] and [31]. For convenience, we also include a simple proof here. Without of loss of generality, we may assume m ≥ n. If m = 0, then Tλ is nondegenerate and the desired result follows immediately. Assume m > 0 and D(x, y) = xm . It is clear that Wz , defined by (3.10), is bounded from L1 into L1,∞ for Re (z) = −1/m. Combining this fact with Theorem 3.1, we can apply Lemma 2.6 to find that Tλ is bounded on Lp with p = (m + n + 2)/(m + 1). Moreover, the operator norm of Tλ on Lp satisfies 1 kTλ kLp →Lp ≤ C|λ|− m+n+2 with the constant C depending only on the cut-off ϕ and deg(S). Therefore we can assume N ≥ 1 throughout this section. Consider the following damped oscillatory integral operator Z eiλS(x,y) |D(x, y)|z ϕ(x, y)f (y)dy, Wz f (x) = R

where z ∈ C has fixed real part Re (z) = b. In this section, we shall further assume that the cut-off ϕ is supported in the unit cube P Q : |x|, |y| ≤ 1/2. Choose a bump function Φ ∈ C0∞ (R) such that (i) supp (Φ) ⊆ [1/2, 2]; (ii) j∈Z Φ(x/2j ) = 1 for all x > 0. Let Wj,k be defined as Wz , but with insertion of Φ(x/2j )Φ(y/2k ) into the cut-off of Wz . First choose s indices 29

i1 < i2 < · · · < is . Now we shall define the damping factor D as follows. 1 • If either m > 0 or max1≤t≤s |αit − αis | ≥ |αis |/4 is true, we take z ∈ C with Re (z) = − m+s and define s Y D(x, y) = xm x − αit y η . t=1

• If m = 0 and max1≤t≤s |αit − αis | < |αis |/4, we first choose a fixed large number N0 = N0 (η). Let {t1 , t2 , · · · , ta } consist of all indices 1 ≤ i ≤ N such that |αi+1 |/|αi | ≥ 24N0 ; see Case (iii) in our proof of Theorem 3.1. Since the sizes of αit are equivalent, for large N0 , we must have {i1 , i2 , · · · , is } ⊆ {tr + 1, tr + 2, · · · , tr+1 } for some 0 ≤ r ≤ a. Here we use the notation t0 = 0. In this case, it is more convenient to replace the set {i1 , i2 , · · · , is } by {tr + 1, tr + 2, · · · , tr+1 }. In other words, we shall set Θ := {i1 , i2 , · · · , is } = {tr + 1, tr + 2, · · · , tr+1 }.

(4.27)

With this revision, it should be pointed out that max1≤t≤s |αit − αis | < |αis |/4 may not still hold. But this does not affect our final results. Now we shall define damping factors Dj,k for each Wj,k as follows.  Qs η  if 2j ≥ |αis |2η(k+1)+2 or 2j ≤ |αi1 |2η(k−1)−2 ; t=1 x − αit y , − s Dj,k (x, y) = Q  |λ||αis |−1 2−k(η−1) A s+2 + s x − αit y η , otherwise; t=1

where A is defined as in (3.25).

As a variant of the classical Hardy space H 1 , we now shall define the space HE1 associated with the phase λS and the set Θ = {i1 , i2 , · · · , is } considered above; see Phong-Stein [19], Pan [16], Greenleaf-Seeger [10], Shi-Yan [31] and Xiao [35] for earlier work related to this space. Definition 4.1 Let Ik be the dyadic interval Ik = [2k−1 , 2k+1 ]. Associated with λS and the index set Θ, we say that a Lebesgue measurable function a is an atom in HE1 (Ik ) if there exists an interval I ⊆ Ik with the following three properties: (i) supp (a) ⊆ Ik ; (ii) |a(x)| ≤ |I|−1 , a.e. x ∈ I; (iii)

R

η

eiλS(αis cI ,y) a(y)dy = 0 with cI the center of I.

P The space HE1 (Ik ) consists of all L1 functions f which can be written as f = j λj aj , where P {aj } is a sequence of HE1 (Ik ) atoms and λj are complex numbers with |λj | < ∞. The norm 1 of f in HE (Ik ) is defined by   X  X kf kH 1 = inf |λj | : f = λj aj , aj atoms in HE1 (Ik ) . E   j

j

Now we state our main result in this section.

30

Theorem 4.1 Assume S, Wz and the cut-off Φ are given as above. Let Wj,k be defined as Wz with the cut-off multiplied by Φ( 2xj )Φ( 2yk ). If ηP≥ 1, then we can decompose Z2 into three disjoint subsets ∆1 , ∆2 and ∆3 such that each Wi = (j,k)∈∆i Wj,k , 1 ≤ i ≤ 3, satisfies kW1 f kL1,∞ ≤ Ckf kL1 ,

kW3 f kL1 ≤ Ckf kL1

and, for f ∈ HE1 (Ik ), kWj,k f kL1 ≤ C(1 + |z|)kf kH 1

with

E

(j, k) ∈ ∆2 ,

where the above constants C depend only on η and ϕ. ′′ as in (3.21). Similarly, Remark 4.1 As in Section 3, we can treat the case η < 1 by writing Sxy Q we shall define the damping factor D in the first case as D(x, y) = xm st=1 (xν − βit y) with 1 Re (z) = −1/(m + sν). While for the second case, we take z with Re (z) = − sν and define Dj,k by s Y Dj,k (x, y) = xν − βit y . t=1

We shall point out that this definition is slightly different from the case η ≥ 1. In fact, the following integral of Hilbert type is finite: Z |xν − βit y|−1/ν dx ≤ C(ν) |x|ν ≈|βis ||y|

provided that |βit − βis | < 41 |βis |. Hence, by Fubini’s theorem and H¨ older’s inequality, we see that Wj,k is bounded on L1 with a bound C(ν) for each (j, k) ∈ ∆2 . Other statements in the above theorem are also true in this case. Proof. According to the definition of D(x, y), we shall prove the theorem in two steps. Step 1. m > 0 or |αit − αis | ≥

|αis | 4

for some 1 ≤ t ≤ s.

Recall that each Wj,k is defined as Wz , but with the cut-off multiplied by Φ(x/2j )Φ(y/2k ). Now we divide the argument into several cases. Case (i) 2j > |αis |2η(k+1)+2 . P The set of all these (j, k) is denoted by ∆1 . Put W1 = (j,k)∈∆1 Wj,k . Then W1 satisfies the trivial estimate kW1 f kL1,∞ ≤ C sup |ϕ| kf kL1 . Case (ii) |αis |2η(k−1)−2 ≤ 2j ≤ |αis |2η(k+1)+2 . Let ∆2 be the set of these (j, k) and W2 =

P

∆2

Wj,k . Then we have

kW2 f kL1 ≤ C( sup |ϕ|)kf kL1 31

for some constant C. This can be verified by Fubini’s theorem. In fact, there holds Z Z Z |D(x, y)|−1/(m+s) |ϕ(x, y)||f (y)|dydx |W2 f (x)|dx ≤ C R

|αis ||y|η ≈|x|

R

≤ C sup |ϕ| ≤ C sup |ϕ|

Z

Z

Z

|x|≈|αis ||y|η

|y| |αis |/4 for some t. We shall point out that the above constants C depend only on m, s, η. Case (iii) 2j < |αis |2η(k−1)−2 . P Let ∆3 be the set of these pairs (j, k). Define W3 = (j,k)∈∆3 Wj,k . By Fubini’s theorem, we can prove that W3 is bounded on L1 with the operator norm less than a constant multiple of sup |ϕ|. By use of Fubini’s theorem and change of variables, we have Z Z Z 1 |D(x, y)|− m+s |ϕ(x, y)||f (y)|dydx |W3 f (x)|dx ≤ |αis y η |>2|x|

≤ C sup |ϕ|

Z

|y| 0 by b :=

w Y i=1

 

Y

t∈Gi,1



|αt − αei−1 |2kη  34

w−1 Y i=1

 

Y

t∈Gi,2



2lt−1  ,

then B = b 2lw−1 |Gw,2 | . It follows immediately that X

lw−1

σ ,··· ,σ

0 w−1 kWj,k,l k ≤ C 0 ,··· ,lw−1

≤ C

−1/s −1/2 X b2lw−1 |Gw,2 | |λ|Ab2lw−1 |Gw,2 |

lw−1

X

(|λ|Ab)−1/2 b−1/s 2−lw−1 |Gw,2 |(1/2+1/s)

lw−1

1/2 ≤ C |αis |−1 2−k(η−1) ,

where we have used the fact that lw−1 satisfies

−s/(s+2) |λ||αis |−1 2−k(η−1) A < b2lw−1 |Gw,2 | .

This proves our claim (4.30). Now we turn to address the HE1 → L1 estimate for Wj,k . Recall that we have assumed that the cut-off ϕ, appearing in the definition of Tλ , is supported in the unit rectangle |x| ≤ 1/2, |y| ≤ 1/2. Define a set G by G=

s [

{x : |x − αit cηI | ≤ C(η)|αis |2k(η−1) |I|}.

t=1

Then, by Hölder’s inequality, we can apply the L2 estimate (4.29) to obtain kWj,k akL1 (G) ≤ |G|1/2 kWj,k akL2 −1/2 ≤ C|G|1/2 |αis |2k(η−1) kakL2 −1/2 1/2 ≤ C |αis |2k(η−1) |I| |αis |2k(η−1) |I|−1/2 ≤ C.

Since G is the union of s intervals, it may not be an interval generally. However, we shall see that the interval (α1 cηI , αs cηI ) can be included in G such that kWj,k akL1 (G) ≤ C is still true. Let F be the set F = {x : |x − αis cηI | ≤ 2|αis − αi1 |cηI }. (4.33) As just claimed, there exists a constant C = C(η, ϕ) such that kWj,k akL1 (F ) ≤ C. For x ∈ Gc , |x − αit y η | ≈ |x − αit cηI |, y ∈ I, for all 1 ≤ t ≤ s. By this fact, we have Z

F \G

|Wj,k a(x)|dx ≤ C ≤ C

Z

s Y

F \G t=1

Z Y s

≤ C,

35

F t=1

|x − αit cηI |−1/s dx

|x − αit cηI |−1/s dx

(4.34)

where the last inequality is true since the singularities αi1 cηI , · · · , αis cηI in the above integral are suitably separated in F . More precisely, for x ∈ F , we have max{|x − αi1 cηI |, |x − αis cηI |} ≥ C|αis − αi1 |cηI .

(4.35)

By the definition (4.33) of F , (4.34) and (4.35), we have, for x ∈ F c ∩ Gc , min |x − αit y η | ≈ min |x − αit cηI | ≈ max |x − αit y η | ≈ max |x − αit y η |

1≤t≤s

1≤t≤s

1≤t≤s

1≤t≤s

(4.36)

and all these terms are not less than a constant multiple |αis − αi1 |cηI . Now we use the P ofσ0 ,··· ,σ σ0 ,··· ,σw−1 decomposition method in Section 3 and obtain Wj,k = Wj,k,l0,···w−1 ,lw−1 . Since Wj,k,l0 ,··· ,lw−1 is supported in the domain 2j−1 ≤ x ≤ 2j+1 , 2k−1 ≤ y ≤ 2k+1 , 2lt−1 −1 ≤ σt−1 (x − αet−1 y η ) ≤ 2lt−1 +1 , 1 ≤ t ≤ w, it follows from (4.36) that, for x ∈ F c ∩ Gc , we may assume 2lt−1 ≥ C max{|αis |2k(η−1) |I|, |αis − αi1 |2kη }, 1 ≤ t ≤ w. On the other hand, the sizes of |x − αit cηI | are equivalent up to constants depending only on η. In other words, we can assume l0 ≈ l1 ≈ · · · ≈ lw−1 now. With the above observations, we are now going to prove

X

σ0 ,··· ,σw−1 Wj,k,l0,··· ,lw−1 a 1 c c ≤ C (4.37)

L (F ∩G )

lw−1

for fixed l0 , · · · , lw−2 , where a is an atom in HE1 (Ik ). First we treat the simplest case w ≥ 2. For x ∈ F c ∩ Gc , it follows from (4.36) that Dj,k (x, y) ≥

s Y

t=1

|x − αit y η | ≈ |x − αe0 cηI |s .

x−αe0 y η σ0 ,··· ,σw−1 On the other hand, the cut-off function of Wj,k,l has the factor Φ σ . This 0 l 0 ,··· ,lw−1 20 implies Z Z X σ0 ,··· ,σw−1 |x − αe0 cηI |−1 dx ≤ C. Wj,k,l0,··· ,lw−1 a(x) dx ≤ C η F c ∩Gc

|x−αe0 cI |≈2l0

lw−1

This completes the proof of (4.37) for w ≥ 2. Now we turn to address the desired estimate (4.37) for w = 1. Before our proof of the estimate (4.37), we shall need the following inequality, which is reminiscent of the Hörmander condition for singular integrals (see Stein [32]), Z 1 (4.38) sup |K(x, y) − K(x, cI )|dx ≤ C(ϕ, η)(1 + |z|), Re (z) = − , s c c y∈I F ∩G

where the kernel K is defined by K(x, y) = ϕ(x, y)Φ

x y Φ k |D(x, y)|z . 2j 2 36

(4.39)

Let M be defined by

x y Φ k . 2j 2 z It is clear that K(x, y) = M (x, y)|D(x, y)| . By the mean value theorem, M (x, y) = ϕ(x, y)Φ

|M (x, y) − M (x, cI )| ≤ k∂y ϕk∞ kΦk2∞ |I| + kϕk∞ kΦk∞ kΦ′ k∞ k

|I| |I| ≤C k k 2 2

for y ∈ I and k ≤ 0. Recall that ϕ is supported in the unit rectangle |x| ≤ 1/2, |y| ≤ 1/2. Thus we need only consider j, k ≤ 0. For x ∈ F c ∩ Gc and y ∈ I, it follows from (4.36) that ! s Y s X 1 u |x − αis cηI |− s −δt |αit y η − αit cηI | |D(x, y)|z − |D(x, cI )|z ≤ C|z| ≤

u=1 t=1 C|z||x − αis cηI |−2 |αis |2k(η−1) |I|,

where δtu is the Kronecker symbol. Note also that the integral in (4.38) is taken over all x ∈ F c ∩ Gc and Φ(x/2j ) 6= 0. For this reason and 2j ≈ |αis |2kη , we can restrict the integration over C1 (η)|αis ||2k(η−1) |I| ≤ |x − αis ccI | ≤ C2 (η)|αis |2kη . With this observations, we see that k log( C2 the integral in (4.38) is bounded by a constant multiple of |I| |I| ) + 1. By the assumption 2k I ⊆ [2k−1 , 2k+1 ], this quantity is bounded above by a constant C independent of j, k, y, cI . Hence we have completed the proof of (4.38). With the Hömander condition (4.38), to obtain (4.37), it is enough to show that there exists a constant C such that Z Z 1 z j k iλS(x,y) a(y)dy dx ≤ C, Re (z) = − . (4.40) |D(x, cI )| Φ(x/2 )Φ(cI /2 )ϕ(x, cI ) e s c c I

F ∩G

Recall that supp (a) ⊆ I ⊆ that for x ∈ F c ∩ Gc ∩ Ij , Z eiλS(x,y) a(y)dy =

Ik = [2k−1 , 2k+1 ]. Using the cancellation property of a, we see

Z iλ[S(x,y)−S(x,cI )] iλ[S(αis cηI ,y)−S(αis cηI ,cI )] a(y)dy e −e I Z x Z y ∂u ∂v S(u, v)dudv ≤ C|λ| αis cη cI

I

I

≤ C|λ| A |x − αis cηI |s+1 |I|,

where A > 0 is defined by (3.25). Choose µ > 0 such that |λ|Aµs+1 |I| = 1. It is easily verified that (4.40) is true if integration is taken over the set of all x satisfying |x − αis cηI | ≤ µ and x ∈ F c ∩ Gc . Define γ0 = max{C(η)|αis |2k(η−1) |I|, C(η)|αis − αi1 |2kη , µ}. Write Z Z iλS(x,y) a(y)dy dx |K(x, cI )| e |x−αis cη |>γ0 I Z Z I η −1 iλS(x,y) a(y)dy dx |x − αis cI | e ≤ C1 η |≤C2 2j s cI

I

γ0 ≤|x−αi

≤ C1

log(C2 2j /γ0 ) Z

X l=0

1≤|x|γ0 I

I

l=0

Case (ii). At least one of αi1 , · · · , αis is complex.

The treatment of complex roots is similar to that of Step 2 in our proof of Theorem 3.1. Without loss of generality, we may assume that αi1 has nonzero imaginary part. If Re (αi1 ) ≤ Im (αi1 ), as shown at the beginning of our proof in this step, then Wj,k is bounded on L1 . So we need only take care of the case Re (αi1 ) > Im (αi1 ) in which the above arguments apply without essential change. ✷

5

Proof of the main result

In this section, we shall prove Theorem 1.1 by exploiting damping estimates established in Section 3 and Section 4. Proof. Throughout this section, we choose s indices {i1 , i2 , · · · , is } = {1, 2, · · · , s}. Then the damping factor D in Theorem 3.1 is equal to s Y (x − αt y η ). D(x, y) = x m

(5.41)

t=1

We first assume η ≥ 1 and m + s ≥ n + (N − s)η. As in the proof of Theorem 4.1, we shall divide our proof here into two steps. Step 1. Either m > 0 or |αit − αis | ≥

|αis | 4

is true for some 1 ≤ t ≤ s.

∞ Choose a bump function Φ as in the P previous jtwo sections. In other words, Φ ∈ C (R) satisfies (i) supp (Φ) ⊆ [1/2, 2] and (ii) j∈Z Φ(x/2 ) = 1 for x > 0. Define Wj,k by Z e y)|z ϕ(x, y)Φ x Φ y f (y)dy (5.42) eiλS(x,y) |D(x, Wj,k f (x) = 2j 2k R

e is slightly different from D and z ∈ C lies in the following strip: where the damping factor D −

m + s − n − (N − s)η 1 1 ≤ Re (z) ≤ γ := · . m+s 2[n + (N − s)η + 1] m + s 38

Case (i) 2j ≥ |αis |2(k+1)η+2 . e y) = xm+s . Let ∆1 be the set of allPthese (j, k) ∈ Z2 in Case (i). In this case, we define D(x, Then the operator W1 = (j,k)∈∆1 Wj,k satisfies 

kW1 f kL2 ≤ C(1+ |z|2 ) |λ|

Y

k ∈{i / 1 ,··· ,is }

and

−γ

|αk |

kf kL2 ,

Re (z) =

Re (z) = −

kW1 f kL1,∞ ≤ Ckf kL1 ,

m + s − n − (N − s)η 1 · , 2[n + (N − s)η + 1] m + s

1 , m+s

where both constants C depend only on deg(S) and the cut-off ϕ. The L2 decay estimate for W1 can be proved by the same argument as that of Theorem 3.1. Indeed, for each (j, k) ∈ ∆1 , e together with its partial derivatives has the same upper bounds, which it is easy to see that D were used in the proof of Theorem 3.1, as the damping factor D and its partial derivatives. Hence following the proof of Theorem 3.1, we can establish the above L2 decay estimate. The L1 → L1,∞ estimate for W1 is obvious for z ∈ C with Re (z) = −1/(m + s); see also Section 4. By Lemma 2.6, with T defined by W1 f = |x|(m+s)z Tλ f , a = (m + s)γ and p0 = 2, we obtain 

kW1 f kLp ≤ C(1 + |z|2 ) |λ|

Y

k ∈{i / 1 ,··· ,is }

where δ and p are given by δ=

−δ

|αk |

1 , m + s + n + (N − s)η + 2

p=

kf kLp

with

Re (z) = 0,

(5.43)

m + s + n + (N − s)η + 2 . m+s+1

(5.44)

Case (ii) |αis |2(k−1)η−2 < 2j < |αis |2(k+1)η+2 . The Pset of all (j, k) satisfying ˜Case (ii) is denoted by ∆2 . As in Section 4, we define W2 = ∆2 Wj,k . In this case, put D(x, y) = D(x, y). Then it is also true that 

kW2 f kL2 ≤ C(1 + |z|2 ) |λ|

Y

k ∈{i / 1 ,··· ,is }

−γ

|αk |

kf kL2 ,

γ=

1 2[n + (N − s)η + 1]

for z ∈ C with real part as in Theorem 3.1. On the other hand, as in Section 4, it is easy to prove W2 satisfies 1 . kW2 f kL1 ≤ Ckf kL1 , Re (z) = − m+s By the Stein complex interpolation theorem, for z with real part Re (z) = 0, we see that W2 also satisfies the estimate (5.43) for W1 . Case (iii) 2j ≤ |αis |2(k−1)η−2 . ∆3 to denote the set of all (j, k) satisfying Case (iii). Define W3 = P We use the notation e W with D(x, y) = D(x, y). As in the previous two cases, we can prove that W3 satisfies j,k ∆3 39

the same estimates as W2 in Case (ii). An interpolation yields the desired Lp decay estimate (5.43) with W3 in place of W1 . Step 2. m = 0 and |αit − αis |
0 given by

kn

A=2

N Y

t=t1 +1

|αt |2kη .

By Theorem 3.3, we have for each (j, k) ∈ ∆2 !−γ N Y kf kL2 , |αt | kWj,k f kL2 ≤ C(1 + |z|2 ) |λ| t=t1 +1

γ=

1 , 2(n + (N − t1 )η + 1)

with z having real part in Theorem 3.3. Here the constant C depends only on deg(S) and the cut-off ϕ as in (3.13). On the other hand, it follows from Theorem 4.1 that for each (j, k) ∈ ∆2 kWj,k f kL1 ≤ C(1 + |z|)kf kH 1 (Ik ) , E

Re (z) = −

1 , t1

where the constant C depends only deg(S) and the cut-off ϕ. By interpolation between the above L2 → L2 and HE1 → L1 estimates, we see that there exists a constant C = C(deg(S), ϕ) such that !−γ N Y kf kLp , Re (z) = 0, (5.45) |αt | sup kWj,k f kLp ≤ C(1 + | Im (z)|2 ) |λ| ∆2

t=t1 +1

40

with the supremum taken over all (j, k) in ∆2 and γ, p being given by γ=

1 , t1 + n + (N − t1 )η + 2

p=

t1 + n + (N − t1 )η + 2 . t1 + 1

With the above estimate (5.45), we are able to prove the corresponding Lp estimates for 1 ≤ s ≤ t1 satisfying m + s ≥ n + (N − s)η. However, we also need another Lp decay estimate corresponding to s = N for which {i1 , · · · , is } = {1, · · · , N }. The above arguments in Step 1 and this step are applicable without change. Indeed, if either m > 0 or sup 1≤t≤N |αt −αN | ≥ |αN | is true, we can follow the argument in Step 1 to obtain kWz f kLp ≤ C(1 + |z|2 )|λ|−γ kf kLp ,

γ=

1 , N +2

p = N + 2,

where C = C(deg(S), ϕ). If m = 0 and sup 1≤t≤N |αt − αN | < |αN |/4, we can use the same argument in this step to obtain the above estimate since t1 = N in this case. By a duality argument, with the role of x and y changed, we have 2

kWz f kLp ≤ C(1 + |z| ) |λ|

N Y t=1

!−γ

|αt |

kf kLp , γ =

1 , p = n + N η + 2. n + Nη + 2

(5.46)

Invoking the above estimates (5.45) and (5.46) with m = 0, we claim that for a constant C = C(deg(S), η) we have 2

kWz f kLp ≤ C(1 + |z| ) |λ|

N Y

t=s+1

!−γ

kf kLp

|αt |

(5.47)

with 1 ≤ s ≤ t1 satisfying s ≥ n + (N − s)η and γ, p given by γ=

1 , s + n + (N − s) + 2

p=

s + n + (N − s) + 2 . s+1

Now we are going to prove this claim. Choose θ ∈ [0, 1] such that t1 + 1 1 s+1 = ·θ+ · (1 − θ). s + n + (N − s)η + 2 t1 + n + (N − t1 )η + 2 n + Nη + 2 1 Set A = n + N η + 2 and B = η − 1. Subtracting − B1 = − η−1 from both sides of the above equality, we obtain

A+B A+B A+B = ·θ+ · (1 − θ). (A − Bs)B (A − Bt1 )B AB Hence we have

It follows that

1 1 1 = · θ + · (1 − θ). (A − Bs) (A − Bt1 ) A 1 1 − = (A − Bs) A − Bt1

1 1 − A (A − Bt1 )

41

· (1 − θ)

which implies

t1 − s t1 = · (1 − θ). A − Bs A By interpolation between the two Lp estimates (5.45) and (5.46), we get !−γ2 (1−θ) !−γ1 θ N N Y Y kf kLp , |αt | |αt | |λ| kWz f kLp ≤ C(1 + | Im (z)|2 ) |λ|

(5.48)

Re (z) = 0,

t=1

t=t1 +1

where p is given by

s + n + (N − s)η + 2 s+1 and we use γ1 and γ2 to denote the decay exponents in (5.45) and (5.46), respectively. With the θ considered above, we see that γ1 θ + γ2 (1 − θ) = 1/(A − Bs) and !−γ2 (1−θ) !−1/(A−Bs) t !−γ2 (1−θ) !−γ1 θ N N N 1 Y Y Y Y |αt | |αt | = |αt | |αt | p=

t=t1 +1

t=1

t=t1 +1

t=1

N Y

≈

t=t1 +1 N Y

≈

t=s+1

!−1/(A−Bs)

|αt |

!−1/(A−Bs)

|αt |

|αt1 |−t1 γ2 (1−θ)

,

where we have used the equality (5.48) and the assumption that α1 , · · · , αt1 have equivalent sizes up to constants depending only on η. This proves our claim (5.47). Combining all above results, we have established the desired Lp estimate for Tλ in (1.1) with η ≥ 1. Similarly, we can deal with the case η < 1 without essential change. Using the above interpolation methods, we are able to prove this decay estimate by invoking Theorem 3.2 and Remark 4.1. Now we turn to prove Theorem 1.1. First assume k ≥ l. We can choose an integer s such that one of the following two statements is true: (i) m + s + 1 = k and n + (N − s)η + 1 = l if η ≥ 1; (ii) m + sν + 1 = k and n + M − s + 1 = l with ν = η −1 if η < 1. Here the notations M and ν are defined as in Theorem 3.2. Choose {i1 , i2 , · · · , is } = {1, 2, . . . , s}. Without loss of generality, we assume that the coefficient c0 , appearing in (3.8) and (3.21), is equal to one. The reason is that c0 can be incorporated into the real parameter λ. Since |α1 | ≤ |α2 | ≤ · · · ≤ |αN | and X 1 1 αj1 αj2 · · · αjN−s ak,l = (−1)σ · · · k l 1≤j1 0 and d > 0, S can be written as X S(x, y) = Pi (x)Qi (y) ki +cli =d

with ki , li having the same meaning as in (i). Then there exists a constant C = C(deg(S), ϕ) such that kTλ f kLp (RnX ) is bounded by 1/p Z 1/p′ Z nY nX ′ − li n ′ ′ ′ − ki m −γ |Qi (y )| kf kLp (RnY ) , dσ(x ) dσ(y ) |Pi (x )| C|λ| S nY −1

S nX −1

where p and γ are given as follows: p=

ki m/nX + li n/nY , ki m/nX

γ=

1 . ki m/nX + li n/nY

Proof. We shall apply Theorem 1.1 to prove this theorem with the rotation method. By polar coordinates, we write x = ρx′ and y = ry ′ with ρ = |x| and r = |y|. Thus the phase S can be written as X S(x, y) = Pi (x′ )Qi (y ′ )ρki m r li n . ki +cli =d

We first calculate the Lp norm of Tλ f in the radial direction. Z ∞ |T f (x)|p ρnX −1 dρ 0 p Z ∞ Z Z ∞ ′ ,ry ′ ) iλS(ρx ′ ′ ′ n −1 ′ e f (ry )ϕ(ρx , ry )r Y drdσ(y ) ρnX −1 dρ = nY −1 S 0 0 p Z ∞ Z Z ∞ 1 1 1 1 1 iλS(ρ nX x′ ,r nY y ′ ) ′ ′ nX ′ nX ′ nY = C e ϕ(ρ x , r y )f (r y )drdσ(y ) dρ. 0

S nY −1

0

45

By Minkowski’s inequality and Hölder’s inequality, we can apply Theorem 6.2 to obtain 1/p Z ∞ |T f (ρx′ )|p ρnX −1 dρ 0

≤ C ≤ C

Z

ZS

nY −1

S nY −1 −δi

≤ C|λ|

Z

0

∞ Z ∞

iλS(ρ1/nX x′ ,r 1/nY y ′ )

e

ϕ(ρ

1 nX

′

x ,r

1 nX

′

y )f (r

1/nY

0

|λ|−δi |Pi (x′ )|−δi |Qi (y ′ )|−δi kf ((·)y ′ )kLp (rnY −1 dr) dσ(y ′ ) ′ −δi

|Pi (x )|

Z

n

S nY −1

Y ′ − li n

|Qi (y )|

1/p′ dσ(y ) kf kLp (RnY ) , ′

p 1/p dσ(y ′ ) y )dr dρ ′

δi =

1 . ki m/nX + li n/nY

In the second inequality, it should be pointed out that our application of Theorem 6.2 produces a constant C independent of x′ , y ′ . In fact, by Fourier expansion, this can be verified by Proposition 6.1. Taking the Lp norm of the above integrals over S nX −1 , we obtain the desired result. ✷ As a consequence of the above theorem, we obtain the following Lp boundedness of higher dimensional oscillatory integral operators without cut-off function. Theorem 6.4 Assume S is a real-valued polynomial satisfying all assumptions in Theorem 6.3. Let T (S, ϕ ≡ 1) be defined as Tλ in (6.49), but with λS replaced by S. If, for some i, Pi and Qi satisfy the integrability conditions: Z Z n n − Y − X |Qi (y ′ )| li n dσ(y ′ ) < ∞, |Pi (x′ )| ki m dσ(x′ ) < ∞ and S nY −1

S nX −1

then T is bounded from Lp (RnY ) into Lp (RnX ) with p defined as in Theorem 6.3. Now we discuss the necessity of degree gaps between polynomials Pi (also Qi ) in Theorem 6.3. Some examples show that the assumption (i) is necessary. On the other hand, these examples are related to a conjecture raised by A. Greenleaf, M. Pramanik and W. Tang p in [9]. We first 4 , let ρ = |x| = x21 + x22 + x23 + x24 present an example in (4 + 4) dimensions. For x, y ∈ R p and r = |y| = y12 + y22 + y32 + y42 . Consider the phase S(x, y) = (ρ4 − r 2 )36 . Then the decay estimates in Theorem 6.3 are not true for Tλ with this phase. Since Theorem 6.4 is a corollary of Theorem 6.3, it suffices to show that T is unbounded on certain Lebesgue spaces. In fact, we shall see that T , associated with phase S(x, y) = (ρ4 − r 2 )36 , is not bounded from Lp (R4 ) into itself for all 1 ≤ p ≤ ∞. Assume the converse. With change of variables in polar coordinates, we would see that the corresponding Lp boundedness is true in (1+1) dimension with phase √ S(u, v) = (u − v)36 . In other words, we would obtain a constant C < ∞ such that Z

0

∞ Z ∞ 0

√ i(u− v)36

e

p 1/p Z f (v)dv du ≤C

0

∞

p

|f (v)| dv

1/p

.

(6.53)

It is easy to see that this inequality is not true for p = 1, ∞. For 1 < p < ∞, let f be the characteristic function of the interval IM = (M 2 , M 2 + ǫ0 M ) with M ≥ 1 and 0 < ǫ0 ≪ 1. √ For u ∈ (M, M + ǫ0 ) and v ∈ IM , we have |u − v| ≤ Cǫ0 . For sufficiently small ǫ0 > 0, the left side of (6.53) is not less than a constant multiple of M . But the right side equals 4 2 36 (ǫ0 M )1/p . Therefore the above inequality is not P true for large M . Write S(x, y) = (ρ − r ) as the form in Theorem 6.3, i.e., S(x, y) = Pi (x)Qi (y). Then the degree gap between Qi 46

and Qi+1 is strictly less than nY = 4. This explains why we shall impose the assumptions |deg(Pi ) − deg(Pi+1 )| ≥ nX and |deg(Qi ) − deg(Qi+1 )| ≥ nY on the phase S. More generally, consider the phase S(x, y) = (ρ2m − r 2n )N with m, n ≥ 1 and N ≥ 2. Here ρ = |x| and r = |y|. If 2m = nX and 2n < nY , then Tλ does not satisfy the decay estimate in Theorem 6.3. For a real analytic function S ∈ C ω (Ω) withP Ω being a neighborhood of the origin in RnX +nY , we write it as the Taylor expansion S(x, y) = α,β cα,β xα y β . For our purpose, we may assume all pure x and y terms in this series vanish. Then the Newton polyhedron of S is the convex hull of all points (x, y) ∈ RnX × RnY satisfying x ≥ α, y ≥ β with cα,β 6= 0, where x ≥ α means xi ≥ αi for each i; so does y ≥ β. The Newton distance δ(S) is defined to be the infimum of all positive δ such that (δ, · · · , δ) ∈ RnX +nY belongs to N (S). Since the decay rate of kTλ k is invariant under linear transformations on RnX and RnY , Greenleaf, Pramanik and Tang [9] introduced the following modified Newton distance δmod (S) = sup δ S(Ax, By) : A ∈ GL(nX ), B ∈ GL(nY ) . It was conjectured in [9] that Tλ , with phase S, should satisfy the following decay estimate p 1 − (6.54) kTλ f kL2 ≤ C|λ| 2δmod (S) log(|λ|) kf kL2

for some p ≥ 0 and a constant C independent of λ. For homogeneous polynomial phase functions satisfying various genericity assumptions, for example homogeneous polynomial phases with either full rank or rank one Hessian away from the origin, Greenleaf, Pramanik and Tang [9] proved (6.54). Especially, their result is optimal in (2+2)-dimensions under genericity assumptions. For the phases considered above, the decay estimate (6.54) is not true. For example, S(x, y) = 4 (ρ − r 2 )36 , both the Newton distance and the modified one are equal to 12. Now we prove δmod (S) = 12. First observe that all terms in S(x, y) take the form xα y β with α1 + α2 + α3 + α4 β1 + β2 + β3 + β4 + = 36. 4 2

This implies that δ(S), δmod (S) ≥ 12. On the other hand, S(x, y) contains the following term 24 24 y12 + y22 + y32 + y42 x21 + x22 + x23 + x24

Q 12 which contains the monomial 4i=1 (x12 i yi ). This shows that δ(S) ≤ 12. Hence δ(S) = 12. Moreover, we can prove that S(Ax, By) contains also the monomial (x1 x2 x3 x4 )12 (y1 y2 y3 y4 )12 for arbitrary A, B ∈ GL(4). Now we show δmod (S) = 12 with a different method. One can verify that S(Ax, By) has the term (xT AT Ax)24 (y T B T By)24 which of course contains monomials 48 of form x48 i yj for 1 ≤ i, j ≤ 4. Then (48ei , 48ej ) belongs to N (S(Ax, By)). By a convex combination, we see that 12(e1 + e2 + e3 + e4 , e1 + e2 + e3 + e4 ) belongs to the Newton polyhedron of S(Ax, By). Here ei is the i − th coordinate unit vector in R4 . So δmod (S) = 12. If (6.54) were true, we would obtain p (6.55) kTλ f kL2 ≤ C|λ|−1/24 log |λ| kf kL2 for some p ≥ 0. Assume the smooth cut-off ϕ is radial in both variables x and y. For radial functions f , it follows from the above estimate that 2 !1/2 Z ∞ Z ∞ p √ 36 1/4 1/2 r) −1/24 iλ(ρ− dρ ϕ(ρ , r )f (r)dr log |λ| ≤ C|λ| kf kL2 e 0

0

47

1 1 for all f ∈ L2 (R+ ). Indeed, the optimal decay rate of the left side is 36 (< 24 ) for general f ∈ L2 . Choose a smooth cut-off ϕ which is positive near the origin. Then for some small 0 < ǫ0 < 1, we have ϕ(x, y) > 0 for |x|, |y| ≤ ǫ0 . For sufficiently large λ, we define f to be the characteristic p p p 1 function of ( ǫ20 , ǫ20 + ǫ20 |λ|−1/36 ). If we take ρ in the interval ( ǫ0 /2, ǫ0 /2 + ǫ0 /2|λ|− 36 ), then we see that the left side of the above inequality is not less than a constant multiple of |λ|−1/24 . 1 On the other hand, it is clear that kf kL2 = |λ|−1/72 . This implies that the decay rate σ ≤ 36 . 1 By change of variables, one has σ = 36 ; see [21, 22]. Hence the above inequality does not hold. Generally, we may consider the phase S(x, y) = (ρ2m − r 2n )N with m, n ≥ 1 and N ≥ 2. Here ρ and r denote the length of x and y, respectively. If 2m = nX and 2n < nY , then we have −1 which is less than N/2. Thus the decay rate 1/(2δmod (S)) > 1/N . δmod (S) = 2N nmX + nnY As above, we can prove the optimal L2 decay rate for Tλ (S, ϕ) is 1/N . This shows that the decay estimate (6.54) is not generally true.

Acknowledgements. This work was supported in part by the National Natural Science Foundation of China under Grant No. 11701573. The author would like to express his gratitude to Professor Xiaochun Li for his profitable discussions and valuable suggestions.

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