# Endpoint Estimates for Oscillatory Singular Integrals with Hölder Class Kernels

Authors: Hussain Al-Qassem; Leslie Cheng; Yibiao Pan
Journal: Journal of Function Spaces
 (2019)

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## Abstract

We prove the uniformL1→L1,∞ and HE1→L1 boundedness of oscillatory singular integral operators whose kernels are the products of an oscillatory factor with bilinear phase and a Calderón-Zygmund kernel K(x,y) satisfying a Hölder condition. This Hölder condition appreciably weakens the C1 condition imposed in existing literature.


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## Body

## 1. Introduction

Letn∈N. We shall consider the following oscillatory singular integral operator:(1)TB,K:f→p.v.∫RneiBx,yKx,yfydywhere B(·,·) is a real-valued bilinear form. In past studies of this type of operators, K(x,y) is typically assumed to be a Calderón-Zygmund kernel satisfying a C1 condition away from the diagonal Δ=x,x:x∈Rn, i.e., there exists an A>0 such that(i)
for allx,y∈Rn×Rn∖Δ,(2)Kx,y≤Ax-yn;(ii)
K(x,y)∈C1((Rn×Rn)∖Δ), and for (x,y)∈(Rn×Rn)∖Δ(3)∇xKx,y+∇yKx,y≤Ax-yn+1;(iii)
(4)ToL2Rn→L2Rn≤Awhere(5)Tofx=p.v.∫RnKx,yfydy.Under conditions (i), (ii), and (iii), Phong and Stein proved theLp boundedness of TB,K for 1<p<∞ ([1]). The Lp result of Phong and Stein was then extended to operators with polynomial phases by Ricci and Stein ([2]), under the same conditions (i), (ii), and (iii) on K(x,y), while the weak (1,1) boundedness of such operators was subsequently established by Chanillo and Christ in [3] (for all polynomial phase functions, bilinear or otherwise).TheC1 property of K in condition (ii) was instrumental when van der Corput’s lemma, a standard tool in the treatment of oscillatory integrals, was used in past studies, including the seminal papers cited above. There has been widespread interest in finding out what happens when the C1 kernel K(x,y) is replaced by a “rougher” kernel. Many interesting results have been obtained for kernels that are homogeneous and of convolutional type but lack smoothness (i.e., K(x,y)=|x-y|-nΩ((x-y)/|x-y|)). See, for example, [4–6].In this paper we are interested in general kernelsK(x,y) for which condition (ii) is replaced by the following weaker condition of Hölder type:( i i ) ′ there exists a δ>0 such that(6)Kx,y-Kx′,y≤Ax-x′δx-y+x′-yn+δwhenever |x-x′|<(1/2)max{|x-y|,|x′-y|} and(7)Kx,y-Kx,y′≤Ay-y′δx-y+x-y′n+δwhenever |y-y′|<(1/2)max{|x-y|,|x-y′|}.In a recent paper, we were able to prove the following uniformLp boundedness of TB,K for 1<p<∞:Theorem 1 (see [7]).
LetA,δ>0 and TB,K be given as in (1). Suppose that K(x,y) satisfies (i), (ii)′, and (iii). Then, for 1<p<∞, there exists a positive Cp which may depend on p, n, δ, and A but is independent of the bilinear form B(·,·), such that(8)TB,KfLpRn≤CpfLpRnfor all f∈Lp(Rn).In this paper we shall investigate the endpoint casep=1 and obtain both the weak type (1,1) and Hardy space bounds. We begin with the weak (1,1) result.Theorem 2.
LetA,δ>0 and TB,K be given as in (1). Suppose that K(x,y) satisfies (i), (ii)′, and (iii). Then, TB,K is of weak type (1,1), i.e., there exists a positive C such that(9)x∈Rn:TB,Kfx>α≤Cα-1fL1Rnfor all f∈L1(Rn) and α>0. Moreover, while the constant C may depend on n, δ, and A, it is otherwise independent of B(·,·) and K(·,·).In the statement above, we used|S| to denote the Lebesgue measure of a measurable set S.In order to describe our result on Hardy spaces, letHE1(Rn) be the Hardy space introduced by Phong and Stein in [1] as a variant of the standard Hardy space H1(Rn) suitable for the study of oscillatory singular integrals. It was proved there that under conditions (i), (ii), and (iii) TB,K is a bounded operator from HE1(Rn) to L1(Rn). As an improvement over their result, we have the following.Theorem 3.
Under conditions (i), (ii)′, and (iii),TB,K is a bounded operator from HE1(Rn) to L1(Rn). Moreover, the bound for the operator norm may depend on n, δ, and A but is otherwise independent of B(·,·) and K(·,·).It is already well-known that analogous results do not hold forHp when p<1.The proof of the weak type (1,1) estimate will appear in Sections2 and 3. It follows a L1→L2 strategy pioneered by C. Fefferman in [8] (see also [3, 4, 9, 10]). The proof of the Hardy space estimate will be given in Section 4.We now close this section by posing the following natural question which should be of interest to many working in this field: are theLp and endpoint results for oscillatory singular integrals in Theorems 1–3 still true when the bilinear phase functions are replaced by general polynomials in x,y with real coefficients?


## 2. Basic Reductions for theL1→L1,∞ Bounds

We shall begin the proof of Theorem2 with a few reductions. Since the one-dimensional case is relatively easier, we shall focus our attention on n≥2. For x=(x1,…,xn),y=(y1,…,yn)∈Rn, B(x,y) can be expressed as(10)Bx,y=∑j=1n∑k=1nbjkxjyk.Ifbjk=0 for all j,k∈{1,…,n}, then TB,K=To. It is well-known that, under the conditions (2), (4), and (6)-(7), To is bounded from from L1(Rn) to L1,∞(Rn) as well as from Lp(Rn) to Lp(Rn) for 1<p<∞. Thus, from this point on, we may assume that bjk≠0 holds for at least one pair (j,k). Let(11)b=maxbjk:1≤j,k≤n.If we letρ denote the dilation operator(12)g·→g·b,then(13)ρ∘TB,K=TρB,b-n/2ρK∘ρ.Since b-n/2ρK satisfies (i), (ii)′, and (iii) with the same constants A and δ as K, it suffices to establish (9) under the additional assumption that, for some k0∈{1,…,n}, |b1k0|=b=1 (after reindexing the variables if necessary). Clearly, we may also assume that δ≤1.For any cubeQ in Rn, let l(Q) and xQ denote its sidelength and center, respectively. For any β>0, we let βQ denote the cube that has the same center as Q and sidelength βl(Q). Also, let jQ=log2(l(Q)).Letf∈L1(Rn) and α>0. Then there is a collection of dyadic cubes F with disjoint interiors such that the following are satisfied:(14)fL∞Ωc≤αwhereΩ=⋃Q∈FQ;(15)1Q∫Qf≤Cα∀Q∈F;(16)∑Q∈FQ≤Cα-1fL1Rn;(17)jQ-jQ′≤CwheneverQ,Q′∈FanddistQ,Q′≤2lQ.Fors∈N∪{0}, let Ks(x,y)=K(x,y)χ[2s,∞)(|x-y|). Let K~(x,y)=K(x,y)-K0(x,y),(18)G=Q∈F:lQ≥2and(19)H=F∖G=Q∈F:lQ≤1.Then,(20)x∈Rn:TB,Kfx>α⊆⋃j=15Ujwhere(21)U1=x∈Rn:TB,K0fχΩcx>α4,U2=⋃Q∈F4Q,(22)U3=x∈⋃Q∈F4Qc:∑Q∈GTB,K0fχQx>α4,(23)U4=x∈⋃Q∈F4Qc:∑Q∈HTB,K0fχQx>α4,and(24)U5=x∈Rn:TB,K~fx>α4.It follows from Theorem1 and a standard argument that (8) remains valid when K is replaced by K0 or K~. Thus, by (14)-(15),(25)U1≤α4-2TB,K0fχΩcL2Rn≤Cα-2∫Ωcf2≤Cα-1fL1Rn,and(26)U2=4n∑Q∈FQ≤Cα-1fL1Rn.The set U5 can be treated by a finite overlapping argument. Let Q0=(-1/2,1/2]n. For each h∈Zn, let h+Q0={h+y:y∈Q0} and(27)fhy=eiBy,y-hχh+Q0yfy.Clearly,(28)suppTB,K~χh+Q0f⊆h+3Q0.By (2) and b=1, for x∈h+3Q0,(29)TB,K~χh+Q0fx-eiBx,hT0fhx≤C∫x-y≤1χh+Q0yfyx-yn-1dy+∫1≤x-y≤2nχh+Q0yfyx-yndy≤Cg0∗χh+Q0fx,where g0(x)=|x|-n+1χ[0,2n](|x|). Thus,(30)U5≤∑h∈Znx∈Rn:TB,K~χh+Q0fx>4·10n-1α≤∑h∈Znx∈Rn:T0fhx>8·10n-1α+x∈Rn:g0∗χh+Q0fx>8C·10n-1α≤C∑h∈Znα-1fhL1Rn+α-1g0L1Rnχh+Q0fL1Rn≤Cα-1fL1Rn.The proof of Theorem 2 has thus been reduced to the verification of the following:(31)Uj≤Cα-1fL1Rnforj=3,4.


## 3. Estimates forU3 and U4

For two nonnegative integerss and m, let(32)Lsmx,y=∫RneiBz,x-yKsz-xKmz-y¯dz.Fora∈Rn and r>0, let Γ(a,r)={x∈Rn:|x-a|≥r}. Let(33)Px=∑k=1nb1kxk.Observe that(34)12∑k=1nxk≤Px+∑k≠k0xk≤2∑k=1nxk.Lemma 4.
Lets,m∈N∪{0} such that s≥m. Suppose that K satisfies (2) and (6)-(7).(i)
For anyx,y∈Rn,(35)Lsmx,y≤C2-sn1+s-m(ii)
For anyx,y∈Rn,(36)Lsmx,y≤Cx-y-nln⁡2+2-mx-y(iii)
For anyx,y∈Rn satisfying |x-y|>2s+1 and |P(x-y)|>4π,(37)Lsmx,y≤CPx-y-δx-y-n.Proof.
We shall omit the arguments for (i) and (ii) because they utilize (2) only and therefore can be found in [3] (see page 152).
Suppose thatx,y∈Rn, |x-y|>2s+1 and |P(x-y)|>4π. For any z=(z1,z2,…,zn)=(z1,z~), let z±=(z1±θ,z~) where θ=π/P(x-y). Thus,(38)Ls,mx,y≤∫Rn-1∫Reiz1Px-yKsz,xKmz,y¯dz1dz~=12∫Rn-1∫Reiz1Px-yKsz,xKmz,y¯-Ksz+,xKmz+,y¯dz1dz~≤M1x,y+M2x,y+M3x,y+M4x,ywhere(39)M1x,y=∫Γx,2s∩Γy,2mKz,x-Kz+,xKz,ydz,M2x,y=∫Γx,2sΔΓx-,2s∩Γy,2mKz+,xKz,ydz,M3x,y=∫Γx-,2s∩Γy,2mKz,y-Kz+,yKz+,xdz,M4x,y=∫Γy,2mΔΓy-,2m∩Γx-,2sKz+,yKz+,xdz.When z∈Γ(x,2s), we have(40)z-x≥2s≥2θ=2z-z+.It follows from (2) and (6) that(41)M1x,y≤Cθδ∫Γx,2s∩Γy,2mdzz-xn+δz-yn≤Cθδ∫Rndz1+z-xn+δ1+z-yn≤Cθδ∫z-y≥x-y/2dz1+z-xn+δz-yn+∫z-x≥x-y/2≥z-ydzz-xn1+z-xδ1+z-yn≤Cθδx-y-n∫Rndz1+zn+δ=CPx-y-δx-y-n.When z∈Γ(x-,2s),(42)z-x≤z-x-+θ≤2z-x-=2z+-x.Thus,(43)M3x,y≤Cθδ∫Rndz1+z-yn+δ1+z-xn.It follows from the the arguments for M1(x,y) that(44)M3x,y≤CPx-y-δx-y-n.When z∈Γ(x,2s)ΔΓ(x-,2s), we have(45)34z-x≤z+-x≤43z-xand(46)342s≤z-x,z+-x≤432s.Thus,(47)z-x≤432s≤23x-yand(48)z-y≥x-y-z-x≥13x-y.Therefore,(49)M2x,y≤C∫Γx,2sΔΓx-,2sz+-x-nz-y-ndz≤C2-nsx-y-nΓx,2sΔΓx-,2s≤C2-nsx-y-n2sn-1θ≤CPx-y-1x-y-n.Similarly,(50)M4x,y≤C2-nmx-y-n2mn-1θ≤CPx-y-1x-y-n.The proof of Lemma 4 is now complete.Letd(S1,S2) denote the distance between two sets S1 and S2. For Q∈G, let(51)GQ=Q′∈G:lQ′≤lQ.Lemma 5.
IfQ∈G and x∈Q, then(52)∑Q′∈GQ∫Q′LjQ,jQ′x,yfydy≤Cα.Proof.
LetQ∈G and(53)G1Q=Q′∈GQ:dQ′,Q≤2lQ.For each x∈Q, let(54)G2Q,x=Q′∈G∖G1Q:infy∈Q′Px-y≤8nlQ′;(55)G3Q,x=Q′∈G∖G1Q:infy∈Q′Px-y>8nlQ′.It follows from (15), (17), and Lemma 4(i) that the cardinality of G1(Q) is bounded by a constant and(56)∑Q′∈G1Q∫Q′LjQ,jQ′x,yfydy(57)≤C∑Q′∈G1Q1+lQ-lQ′Q-1∫Q′fydy≤Cα.When Q′∈G∖G1(Q),(58)supy∈Q′ln⁡2+2-jQ′x-yx-yn≤Cinfy∈Q′ln⁡2+2-jQ′x-yx-yn.Also, for any Q′∈G2(Q,x) and y∈Q′,(59)Px-y≤infv∈Q′Px-v+nlQ′≤9nlQ′.Thus, by Lemma 4(ii), (15), and (58),(60)∑Q′∈G2Q,x∫Q′LjQ,jQ′x,yfydy≤Cα∑Q′∈G2Q,x∫Q′x-y-nln⁡2+2-jQ′x-ydy≤Cα∑m=1jQ∫y∈Rn:Px-y≤9n2mlQ+x-y-nln⁡2+2-mx-ydy=Cα∑m=1jQ∫u∈Rn:Pu≤9n2jQ-m+u-nln⁡2+udu≤Cα∑m=1jQ∫Rn-1∫v1≤9n2jQ-m+v~+v1-nln⁡2+v~+v1dv1dv~≤Cα∑m=1jQ1+jQ-m2-jQ-m≤Cα.For any Q′∈G3(Q,x),(61)supy∈Q′Px-y≤infy∈Q′Px-y+nlQ′≤2infy∈Q′Px-y.It follows from Lemma 4(iii), (15), (58), and (61) that(62)∑Q′∈G3Q,x∫Q′LjQ,jQ′x,yfydy≤Cα∑Q′∈G3Q,x∫Q′Px-y-δx-y-ndy≤Cα∫Rn1+Pu-δ1+u-ndu≤Cα∫Rn1+v1-δ1+v-ndv≤Cα.Lemma 5 is proved.ForQ∈H, let H(Q)={Q′∈H:l(Q′)≤l(Q)}.Lemma 6.
IfQ∈H and x∈Q, then(63)∑Q′∈HQ∫Q′L0,0x,yfydy≤Cα.Proof.
The argument is very similar to the proof of the previous lemma. We will point out the differences but omit most of the details.
LetQ∈H and x∈Q. Let(64)H1Q=Q′∈HQ:dQ′,Q≤2,H2Q,x=Q′∈H∖H1Q:infy∈Q′Px-y≤4π;H3Q,x=Q′∈H∖H1Q:infy∈Q′Px-y>4π.While there is no uniform bound on the cardinality of H1(Q) (unlike G1(Q)), by using |L0,0(x,y)|≤C, we still have(65)∑Q′∈H1Q∫Q′L0,0x,yfydy≤Cα∑Q′∈H1QQ′≤Cα.By |L0,0(x,y)|≤C|x-y|-nln⁡2+x-y,(66)∑Q′∈H2Q,x∫Q′L0,0x,yfydy≤Cα∫u∈Rn:Pu≤4π+n1+u-nln⁡2+udu≤Cα.Finally, H3(Q,x) can be treated the same as G3(Q,x), which finishes the proof of Lemma 6.We now employ a well-knownL2→L1 technique to obtain the desired estimates for |U3| and |U4| (see, for example, [3, 8–11]). By Lemma 5,(67)U3≤Cα-2∑Q∈GTB,K0fχQL2⋃Q∈F4Qc2≤Cα-2∑Q∈G∫Q∑Q′∈GQLjQ,jQ′x,yfydyfxdx≤Cα-1∑Q∈G∫Qfxdx≤Cα-1fL1Rn.By Lemma 6,(68)U4≤Cα-2∑Q∈H∫Q∑Q′∈HQL0,0x,yfydyfxdx≤Cα-1∑Q∈H∫Qfxdx≤Cα-1fL1Rn.The proof of Theorem 2 is now complete.


## 4.HE1→L1 Boundedness

We shall begin by recalling the definition of the function spaceHE1(Rn).Definition 7.
A measurable functiona(·) on Rn is called an atom if there exists a cube Q such that supp(a)⊆Q, a∞≤|Q|-1 and(69)∫QeiBxQ,yaydy=0.A function f is in HE1(Rn) if there exist a sequence {λj} in C and a sequence of atoms {aj} such that(70)f=∑jλjaj.The HE1 norm of f is the infimum of ∑jλj over all possible expressions of f described in (70).In order to prove Theorem3, it suffices to establish that, for every atom a(·),(71)TB,KaL1Rn≤C.By using a dilation (as in Section 2) as well as a translation, we can further reduce the task to the verification of (71) under the assumption that, for some j0∈{1,…,n} and h>0,(72)max⁡bjk:1≤j,k≤n=bj01=1,suppa⊆-h,h-n,a∞≤2h-nand∫-h,hnaydy=0.Let η=max{2nh,h-1}. Then(73)TB,KaL1Rn≤I1+I2+I3+I4where(74)I1=∫x≤2nhTB,Kaxdx,I2=∫x≥2nh∫RneiBx,yKx,y-Kx,0aydydx,I3=∫2nh≤x≤ηKx,0∫RneiBx,yaydydx,I4=∫x≥ηKx,0∫RneiBx,yaydydx.By Theorem 1,(75)I1≤Chn/2TB,KaL2Rn≤Chn/2aL2Rn≤C.By (7),(76)I2≤∫x≥2nh∫y∈-h,hnyδaydyxn+δdx≤ChδaL1Rn∫x≥2nhx-n-δdx≤C.By the vanishing integral property of a(·),(77)I3=∫2nh≤x≤ηKx,0∫y∈-h,hneiBx,y-1aydydx≤C∫2nh≤x≤ηx-n+1∫y∈-h,hnyaydydx≤CaL1Rnhη-2nh≤C.Let F1 denote the partial Fourier transform in the first variable. Then,(78)I4≤C∫Rnη+x-n∫y~∈Rn-1F1a·,y~-∑j=1nbj1xjdy~dx≤C∫y~∈Rn-1∫u∈Rnη+u-nF1a·,y~u1dudy~≤C∫y~∈Rn-1∫u~∈Rn-1∫Rdu1η+u2n1/2∫RF1a·,y~u12du11/2du~dy~≤C∫u~∈Rn-1η+u~-n+1/2du~∫y~∈Rn-1∫Ray2dy11/2dy~≤Cηh-1/2≤C.The proof of Theorem 3 is now complete.



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