Jean BOURGAIN Ciprian DEMETER

(Dedicated to Haim Brezis on the occasion of his 70th birthday)

1 The Theorem

Consider the truncated(elliptic)paraboloid in Rn,

For each cube Q in[0,1]n?1and g:Q → C,define the extension operator=EQas follows:

where e(z)=e2πiz,

and

This can be interpreted as the Fourier transformwhere the measure dσ is the lift of the Lebesgue measure from[0,1]n?1to the paraboloid.When Q=[0,1]n?1,we will sometimes write Eg for E[0,1]n?1g.

We will use the letters Q,q to denote cubes on the frequency side[0,1]n?1.We will use the letters B,Δ to denote cubes on the spatial side Rn.Throughout the whole paper we can and will implicitly assume that all cubes have side length in 2Z.This in particular will place(harmless)restrictions on various parameters such as δ,σ,R,that we will not bother to write down explicitly.Thanks to this assumption,we will be able to partition(rather than use finitely overlapping covers)large cubes into smaller cubes.Given a cube Q ? [0,1]n?1with side length l(Q)∈ 2?Nand α ∈ 2?Nsmaller than l(Q),we will denote by Partα(Q)the(unique)partition of Q by using cubes Qαof side length α.A similar notation will occasionally be used for spatial cubes B.

We will write B=B(cB,R)for the cube in Rncentered at cBand with side length l(BR)=R,and we will introduce the associated weight

The exponent 100n is chosen large enough to guarantee various integrability requirements.We will see that Theorem 1.1 remains true for any larger exponent E≥100n,and the implicit bounds will depend on E.This observation will allow us to run our induction argument,as explained in Section 3.

For a positive weight v:Rn→[0,∞)and for f:Rn→C,we define the weighted integral

For 2 ≤ p ≤ ∞ and δ∈ 4?N,let Dec(δ,p)=Decn(δ,p)be the smallest constant,such that the inequality

holds for every cube B ? Rnwith side length δ?1and every g:[0,1]n?1→ C.

The l2decoupling theorem proved in[3]reads as follows.We refer the reader to[3]for a few applications that motivate the theorem.

Theorem 1.1We have the following sharp(up to δ??losses)upper bound for Decn(δ,p):

if 2≤p≤.The implicit constant depends on?,p,n but not on δ.

We will present a rather detailed argument for this theorem.Essentially,we rewrite our original argument from[3]using a more streamlined approach.This approach has started to take shape in our subsequent papers on decouplings and has gotten to this final form in the joint work with Guth[4].One new feature of our argument compared to[3]is that we avoid the special interpolation from[3],that relies on wave-packet decomposition.Another one is that we use the multilinear Kakeya inequality,rather than the multilinear restriction theorem.The argument we describe here also clarifies various technical aspects of the theory,such as the role of the weights wBand the(essentially)locally constant behavior of Fourier transforms of measures supported on caps on the paraboloid.

We hope the argument will be accessible to experts outside the area of harmonic analysis.We believe this will serve as a warm up for the readers interested in understanding the proof of Vinogradov’s mean value theorem from[4].

A brief summary of the argument is presented in Section 3.The most important sections are the last two.The details from the remaining sections may be skipped at the first reading.

2 More Notation

Throughout this paper we will write A?υB to denote the fact that A≤CB,for a certain implicit constant C,that depends on the parameter υ.Typically,this parameter is either?,ν or K.The implicit constant will never depend on the scale δ,on the spatial cubes we integrate over,or on the function g.It will however most of the times depend on the degree n and on the Lebesgue index p.Since these can be thought of as being fixed parameters,we will in general not write?p,n.

We will denote by BRan arbitrary cube in Rnof side length l(BR)=R.We use the following two notations for averaged integrals:

Given a function η on Rnand a cube B=B(c,R)in Rn,we will use the rescaled version

|A|will refer to either the cardinality of A if A is finite,or to its Lebesgue measure if A has positive measure.

We will sometimes write?f,g?for the inner product?

3 A Brief Description of the Argument

We use two types of mechanisms to decouple.One is the L2decoupling(see Section 6).This is very basic,it relies just on Hilbert space orthogonality,but it is nevertheless very efficient.It decouples into frequency cubes whose side length is as small as permitted by the uncertainty principle,namely equal to the reciprocal of the side length of the spatial cube.The second mechanism is a multilinear decoupling that relies on the multilinear Kakeya inequality(see Theorem 9.2).Combining these with multiple iterations leads to the multiscale inequality(10.7).This inequality has a very simple form when 2≤p≤,and a short warm up argument is presented in the end of Section 10 to prove Theorem 1.1 in this range.

For the general case,the argument will go as follows.We will introduce a family of constants Decn(δ,p,ν,m),and will show in Section 8 that they dominate Decn(δ,p).On the other hand,in the last section,we use(10.7)to show that each Decn(δ,p,ν,m)can be controlled by a combination of powers of δ and some power of Decn(δ,p)(see(11.10)).This inequality represents an improvement over the trivial estimate Decn(δ,p,ν,m)?Decn(δ,p).By playing the two bounds(see(11.10)and(8.3))against each other,we arrive at the desired upper bound

An unfortunate technicality is the fact that we will need to work with the family of weights for a cube B=B(c,R)in Rn,

Here E ≥ 100n.For each such exponent E,we will let,as before,Decn(δ,p,E)denote the smallest constant that guarantees the following inequality for each g,B=Bδ?1:

All the quantities that will depend on weights will implicitly depend on E.This includes Decn(δ,p,ν,m),Dt(q,Br,g)and Ap(q,Br,s,g).Sometimes we will suppress the dependence on E,and will understand implicitly that the inequality is true for all E≥100n.The weight wB,Ewill always be the same on both sides of a given inequality.The implicit constants will depend on E but that is completely harmless.

We will prove Theorem 1.1 by using induction on the dimension n.We set a superficially stronger induction hypothesis,namely,we will assume that

for each 2≤p≤and each E ≥100(n?1).We will use this to prove

for each 2≤p≤and each E ≥ 100n.The reason for such a hypothesis is coming from inequality(8.8),which essentially uses the lower dimensional constant Decn?1(δ,p,F)to make a statement about Decn(δ,p,E).Larger dimensions demand higher values of E due to integrability requirements.

4 A Few Useful Inequalities

One technical challenge in the proof of Theorem 1.1 is to preserve the exponent E for the weights wBinvolved in various inequalities.

A key,easy to check property of the weights wB=wB,Ethat will be used extensively,is the following inequality:

valid for all cubes B with l(B)=R and all finitely overlapping covers B of B with cubes Δ of(fixed)side length 1≤R?≤R.The implicit constants in(4.1)will(harmlessly)depend on E,but crucially,they will be independent of R,R?.

We will find extremely useful the following simple result.

Lemma 4.1Let W be the collection of all weights,that is,positive,integrable functions on Rn.Fix R>0.Fix E.Let O1,O2:W →[0,∞]have the following four properties:

(W1)O1(1B)?O2(wB,E)for all cubes B?Rnof side length R.

(W2)O1(αu+βv)≤ αO1(u)+βO1(v)for each u,v ∈ W and α,β >0.

(W3)O2(αu+βv)≥ αO2(u)+βO2(v)for each u,v ∈ W and α,β >0.

(W4)If u≤v,then Oi(u)≤Oi(v).

Then

for each cube B with side length R.The implicit constant is independent of R,B and only depends on the implicit constant from(W1),on E and n.

We will sometimes be able to check a stronger assumption than(W1),where O2(wB)is replaced with O2(ηB)for some rapidly decreasing function η.

ProofLet B be a finitely overlapping cover of Rnwith cubes B?=B?(cB?,R).It suffices to note that

and

Remark 4.1It is rather immediate that for each f,

satisfies(W2)and(W4).Also,for fixed p ≥ 2 and fi,Minkowski’s inequality inshows that

satisfies(W3)and(W4).Most applications of Lemma 4.1 will use this type of operators.

We complete this section with the following reverse H¨older inequality.

Corollary 4.1For each q≥ p≥ 1,each cube Q ? [0,1]n?1with l(Q)=and each cube B in Rnwith l(B)=R,we have

with the implicit constant independent of R,Q,B and g.

ProofLet η be a positive smooth function on Rnsatisfying 1B(0,1)≤ ηB(0,1),such that the Fourier transform of η1pis supported on the cube B(0,1).We can thus write

Let θ be a Schwartz function which equals to 1 on the cube B(0,10).Since the Fourier transform ofis supported in the cube 3Q(having the same center as Q and side length three times as large),we have

and thus,by Young’s inequality,we can write

Here

Now,following the notation and ideas from the proof of Lemma 4.1,we may use the previous inequalities to write

Remark 4.2Note that there is a loss in regularity in(4.2),as the weight exponent ison the right-hand side.A simple example shows that this exponent is optimal.This will later cause some minor technicalities.In particular,it will force us to use the smaller weight wΔ,10E(as opposed to wΔ,E)in(8.1).This will in turn allow us to go from(11.8)to(11.9)by using(4.2)for indices p and 2.

5 An Equivalent Formulation

For δ<1 and Q ? [0,1]n?1,define the δ-neighborhood of Pn?1above Q to be

For each f:Rn→C and R?Rn,denote by fRthe Fourier restriction of f to R

In this section,we will make repeated use of the following inequalities,where BRwill refer to the cube centered at the origin in Rn,

and,when n=2,

We will need the following alternate form of decoupling,when we will derive inequality(8.8).

Theorem 5.1For each E≥100n,the following statement is true for each F≥Γn(E),where Γn(E)is a large enough constant depending on E and n.For p ≥ 2,each f:Rn→ C with Fourier transform supported in([0,1]n?1)and for each cube BR?Rn,we have

ProofTo simplify notation,we will show the computations when n=2.In this case,Γ2(E)=2E+2 will suffice.

Using Remark 4.1,it will suffice to prove

Due to translation/modulation invariance,we may assume BRto be centered at the origin.

A change of variables allows us to write

Next,combining this with the Taylor expansion

we can write for x∈BR,

where

Obviously,(5.3)leads to the following inequality:

It remains to prove that(note that we have F on the left-hand side and E on the right-hand side)

uniformly over j≥0.

An easy computation allows us to assume Q=[0,].Indeed,translating[u,u+]to[0,]on the frequency side will replace(x1,x2)with(x1+2ux2,x2)on the spatial side.Note that when 0≤u≤1,these shear transformations affect the weights wBonly negligibly.

We start by writing

Recall that

For x∈B(y,R),we write

and apply another Taylor expansion for e((? ξ2)(x2?y2))to arrive at

It now remains to prove

uniformly over j≥0.

where

ηis a Schwartz function equal to 1 on[?2,2]and supported in[?3,3],and

Let Mj(t)be a compactly supported Schwartz function which agrees with tjon[0,12]and satisfies the derivative bound

uniformly over j≥0 for each k≥0.

Note that we can also write

where

Applying H¨older’s inequality,we arrive at

It remains to show that

In fact,we will prove a slightly stronger inequality

An easy computation using(5.4)shows that for each s1,s2≥0,

Combining this with the fact thatis compactly supported inleads,via repeated integration by parts,to the following estimate for the Fourier transform:

where

and

Let IR=??,?and recall that BR=IR×IR.Using(5.7)and(5.1)(n=1),we may now write?

Recalling(5.5),we are left with proving that

We split the analysis into three cases.We will need F≥2E+2.

(a)|y2|≤R.In this case,

Using(5.1)with n=1 twice(first R?=then R?=R)and(5.2)with E1=E,E2=2,we get

as needed.

(b)|y2|～KR,with K ∈[1,]∩2N.In this case,

Using(5.1)twice(first R?=then R?=R),and(5.2)with E1=E,E2=3,we write

Note that summing over K ∈[1,]∩2Nleads to the desired estimate(5.8).

(c)|y2|～KR32with K ∈[1,∞)∩2N.In this case,

and so,by(5.1),we have

Next,combining this with(5.2)(E1=E,E2=E+2)and then with(5.1),we get

Note that summing over K ∈[1,∞)∩2Nleads to the desired estimate(5.8).

6 L2Decoupling

We will use Lemma 4.1 to prove a very simple but efficient decoupling.This exploits L2orthogonality and will allow us to decouple to the smallest possible scale,equal to the inverse of the radius of the cube.This process is illustrated by the following simple result.

Proposition 6.1(L2Decoupling)Let Q be a cube with l(Q)≥ R?1.Then for each cube BR?Rnwith side length R we have

ProofWe will prove that

Fix a positive Schwartz function η,such that the Fourier transform of √ηis supported in a small neighborhood of the origin,and such that η ≥ 1 on B(0,1).By invoking Lemma 4.1,we see that inequality(6.1)will follow once we check that

holds true for each cube B?with l(B?)=R.

Note that the Fourier transform ofwill be supported inside the R?1-neighborhood of the paraboloid above q,and that these neighborhoods are pairwise disjoint for two nonadjacent q.Since

(6.2)will now immediately follow from the L2orthogonality of the functions

7 Parabolic Rescaling

A nice property of the paraboloid Pn?1is the fact that each square-like cap on it can be stretched to the whole Pn?1via an affine transformation.Affine transformations interact well with the Fourier transform,and this facilitates a natural passage from the operator EQto E[0,1]n?1.

Proposition 7.1Let 0<δ≤σ<1 and p≥2.For each cube Q?[0,1]n?1with l(Q)=σ12 and each cube B ? Rnwith l(B)≥ δ?1,we have

The implicit constant is independent of δ,σ,Q,B.

ProofLet us first assume l(B)= δ?1.We will apply Lemma 4.1 to

(see Remark 4.1).It thus suffices to prove that

Assume Q=a+[0,]n?1with a=(a1,···,an?1).We will perform a parabolic rescaling via the affine transformation L=LQ,

A simple computation shows that for each cube?Q,we have

where=,gL=g?L.The image S of B under the affine transformation T=TQ

can be covered with a family F of pairwise disjoint cubes Δwith side length δ?1σ,such that we have the following double inequality,in the same spirit as(4.1):

The second inequality is very easy to guarantee for a proper covering,as l(B?)≤l(B).After a change of variables on the spatial side,we get(since QL=[0,1]n?1)

Using Minkowski’s inequality followed by(7.1),this is dominated by

By changing back to the original variables,this is easily seen to be the same as

This finishes the proof in the case l(B)= δ?1.

Let us next assume l(B)≥ δ?1.By invoking again Lemma 4.1(see Remark 4.1),it suffices to prove

Using(4.1)and Minkowski’s inequality,we may complete the argument as follows:

8 Linear Versus Multilinear Decoupling

Let π :Pn?1→ [0,1]n?1be the projection map.

Definition 8.1We say that the cubes Q1,···,Qn? [0,1]n?1are ν-transverse if the volume of the parallelepiped spanned by unit normals n(Pi)is greater than ν,for each choice of Pi∈Pn?1with π(Pi)∈ Qi.

For E ≥ 100n,2 ≤ p ≤ ∞,m ∈ N and 0< ν <1,we let Dec(δ,p,ν,m,E)=Decn(δ,p,ν,m,E)be the smallest constant,such that the inequality

holds for each cube B ? Rnwith l(B)= δ?1,each g:[0,1]n?1→ C and each ν-transverse cubes Qiwith equal side lengths μ satisfying μ ≥ δ2?m.Recall that Partμ?1(B)is the partition of B by using cubes Δ with l(Δ)= μ?1.The lower bound which we impose on the size of μ is a bit more severe than the minimal lower boundneeded in order to make sense of the quantity(Qi).This restriction can be ignored now and should only be paid attention to in the final argument from the last section,when we dominate(11.7)by(11.8).

Note that we use wΔ,10Erather than wΔ,Ein(8.1).This is done for purely technical reasons,as explained in Remark 4.2.

Since|EQig|can be thought of as being essentially constant on each Δ,the quantity

can be viewed as being comparable to

The former will be a preferred substitute for the latter due to purely technical reasons.

Several applications of H¨older’s inequality combined with(4.1)show that for each ν,m,

This inequality is too basic and will never be used.We will instead derive a stronger form of it in the last section(see(11.10)),which dominates Dec(δ,p,ν,m,E)by using a combination of powers of δ and some power of Dec(δ,p).

We will now prove and later use the following approximate reverse inequality.Recall the definition of Γn(E)from Theorem 5.1.

Theorem 8.1Let E≥100n.Assume that one of the following holds:

(i)n=2.

(ii)n ≥ 3 and Decn?1(δ,p,Γn?1(10E))

Then for each 0< ν ≤ 1,there is?(ν)=?(ν,p,E)withand Cν,m,such that for each m≥1,we have

for each R?ν,m1.

We next prove the case n=3 of the theorem and will then indicate the modifications needed for n≥4.The argument will also show how to deal with the case n=2.

Remark 8.1If P1,P2,P3∈P2,the volume of the parallelepiped spanned by the unit normals n(Pi)is comparable to the area of the triangle with vertices π(Pi).

The key step in the proof of Theorem 8.1 for n=3 is the following result.

Proposition 8.1Assume Dec2(δ,p,Γ2(10E))?δ??.Then there are constants C,C?,such that for each m≥1 and each R≥K2m,

ProofUsing Lemma 4.1(see Remark 4.1),it suffices to prove the inequality with the unweighted quantity?Eg?Lp(BR)on the left-hand side.Cover BRwith a family PartK(BR)of cubes BK?R3with side length K.

Let ψ :R3→C be a Schwartz function with Fourier transform equal to 1 on B(0,10).For each α ∈ PartK?1([0,1]2),define

Note that since Eαg=(Eαg)? ψKfor an appropriate modulation/dilation ψKof ψ,we have

This is a manifestation of the uncertainty principle that asserts that|Eαg|is essentially constant at scale K.Let α?= α?(K)∈ PartK?1([0,1]2)be a square that maximizes cα(BK).Define also

The number C will change its value from one line to the next one,but crucially,it will always be independent of K.

We will show that for each BK∈ PartK(BR)there exists a line L=L(BK)in the(ξ1,ξ2)plane,such that if

then for x∈BK,

To see this,we distinguish three scenarios.First,if there is no α ∈ Sbigwith dist(α,α?)≥then(8.4)suffices as

Otherwise,there is α??∈ Sbigwith dist(α??,α?)≥.The line L is determined by the centers of α1,α2,which are chosen to be furthest apart among all possible pairs in Sbig.Note that the distance between these centers is at least

Second,if there is α3∈ Sbig,such that α3intersects the complement of SL,then(8.5)suffices.To see this,note first that α3is forced to intersect the strip between α1and α2perpendicular to L.Thus,a triangle determined by any three points in αihas area?K?2.Combining this with Remark 8.1 shows that α1,α2,α3are K?2transverse for C large enough.

Third,if all α∈Sbigare inside SL,the sum of(8.4)and(8.6)will obviously suffice.

We now claim that(8.4)–(8.6)imply the following:

Only the third scenario above needs an explanation.Cover SLby pairwise disjoint rectangles U of dimensions K?1and K?12,with the long side parallel to L.To simplify notation,assume the equation of L is η=1 and that BK=[0,K]3.For each fixed y the Fourier transform of(x,z)?→ ESLg(x,y,z)is supported in the O(K?1)neighborhood of the parabola η = ξ2+1.Using Theorem 5.1 and our hypothesis Dec2(K?1,p,Γ2(10E))??K?,we can write

Next raise this inequality to the power p,integrate over y∈[0,K]and use

and Minkowski’s inequality to write

Note however that since we are dealing with the third scenario,the contribution of E[0,1]2SLg is small

Using the triangle inequality,we get

We conclude that(8.7)holds under the third scenario.The first two scenarios are quite immediate.

Using wBK,10E≤wBK,E,(8.7)further implies that

Finally,we raise(8.10)to the power p and sum over all BK∈PartK(BR),by invoking Minkowski’s inequality and(4.1),to get

An application of Lemma 4.1 finishes the proof.

Parabolic rescaling as in the proof of Proposition 7.1 leads to the following.The details are left to the reader.

Proposition 8.2Let τ? [0,1]2be a square with side length δ≥K2m?1.Assume

for all δ?<1.Then if R ≥ K2m,we have

The constants C?,C are independent of δ,R,τ,K.

We are now ready to prove Theorem 8.1 for n=3.Let K=.Let also R ≥ K2m=ν?2m?1.Iterate Proposition 8.2 starting with scale δ=1,until we reach scale δ=Each iteration lowers the scale of the square from δ to at leastThus we have to iterate O(logKR)times.We use the following immediate consequence of H¨older’s inequality:

Since

we get

The result in Theorem 8.1 now follows since C,C?do not depend on ν.

To summarize,the proof of Theorem 8.1 for n=3 relied on the hypothesis that the contribution coming from squares β living near a line is controlled by the negligible lower dimensional quantity Dec2(δ,p)=O(δ??).When n ≥ 4,the contribution from the cubes near a hyperplane H in[0,1]n?1will be similarly controlled by Decn?1(δ,p).That is because π?1(H)is a lower dimensional elliptic paraboloid whose principal curvatures are～1,uniformly over H.This paraboloid is an affine image of Pn?2,and can be analyzed by using parabolic rescaling.When n=2,there is no such lower dimensional contribution.

9 From Multilinear Kakeya to Multilinear Decouplings

We start by recalling the following multilinear Kakeya inequality due to Bennett,Carbery and Tao[1].We refer the reader to[6]for a short proof.

Theorem 9.1Let 0<ν<1.Consider n families Pjconsisting of tiles(rectangular boxes)P in Rnhaving the following properties:

(i)Each P has n?1 side lengths equal to R12and one side length equal to R which points in the direction of the unit vector vP.

(ii)vP1∧ ···∧vPn≥ ν for each Pi∈ Pi.

(iii)All tiles are subsets of a fixed cube B4Rwith side length 4R.

Then we have the following inequality:

for all functions Fjof the form

The implicit constant will not depend on R,cP,Pj.

We use this to prove the following key result.

Theorem 9.2Let p≥and δ<1.Consider n ν-transverse cubes Q1,···,Qn?[0,1]n?1.Let B be an arbitrary cube in Rnwith side length δ?2,and let B be the(unique)partition of B into cubes Δ of side length δ?1.Then for each g:[0,1]→ C we have

Moreover,the implicit constant is independent of g,δ,B.

Remark 9.1This result is part of a two-stage process.Note that,strictly speaking,this inequality is not a decoupling,since the size on the frequency cubes Qi,1remains unchanged.However,the size of the spatial cube increases from δ?1to δ?2,which will facilitate a subsequent decoupling,as we shall later see in Proposition 10.1.

ProofSince we can afford logarithmic losses in δ,it suffices to prove the inequality with the summation on both sides restricted to families of Qi,1for whichhave comparable size(within a multiplicative factor of 2)for each i.Indeed,the cubessatisfying(for some large enough C=O(1))can be easily dealt with by using the triangle inequality,since we automatically have

This leaves only log2(δ?O(1))sizes to consider.

Let us now assume that we have Nicubes Qi,1,withof comparable size.Since p≥,by H¨older’s inequality,(9.2)is at most

For each cube Q=Qi,1centered at cQ,we cover B with a family FQof pairwise disjoint,mutually parallel tiles TQ.They have n?1 short sides of length δ?1and one longer side of length δ?2,pointing in the direction of the normal N(cQ)to the paraboloid Pn?1at cQ.Moreover,we can assume these tiles to be inside the cube 4B.We let TQ(x)be the tile containing x,and we let 2TQbe the dilation of TQby a factor of 2 around its center.

Let us use q to abbreviateOur goal is to control the expression

We now define FQfor x∈ ∪TQ∈FQTQby

For any point x∈ Δ,we have Δ ?2TQ(x),and so we also have

Therefore,

Moreover,the functionis constant on each tile TQ∈FQ.Applying Theorem 9.1,we get the bound

It remains to check that for each Q=Qi,1,

Once this is established,it follows that(9.4)is dominated by

Recalling the restriction,we have made on Qi,1,(9.6)is comparable to

as desired.

To prove(9.5),we may assume,and thuswill be supported in[?δ,δ]n?1× [?δ2,δ2].Fix x=(x1,···,xn)with TQ(x) ∈ FQand let y ∈ 2TQ(x).Note that TQ(x)has sides parallel to the coordinate axes.In particular,y=x+y?with||<4δ?1for 1≤ j≤ n?1 and||<4δ?2.Then

Now,using Taylor series,we can write

Here Msnis the operator with Fourier multiplierwhere

We are able to insert the cuto ffbecause of our initial restriction on the Fourier support of EQg.

Plugging this estimate into(9.7),we obtain

Recalling the definition of FQand the fact that

we conclude that

Note that(t)agrees on[]with a compactly supported smooth functiondefined on R,with derivatives of any given order uniformly bounded over sn.It follows that

with implicit constant independent of sn,where

for all M>0.Letdenote the operator with multiplierWe can now write

where?denotes the convolution with respect to the last variable xn,and

Using this,one can easily check that

Combining this with(9.8)leads to the proof of(9.5)

The argument is now complete.

10 The Iteration Scheme

Let 0< ν<1.Throughout this section,we fix some 0< δ<1 and also n ν-transverse cubes Q1,···,Qn? [0,1]n?1with side length at least δ.

For a positive integer s,Bswill refer to cubes in Rnwith side length l(Bs)= δ?sand arbitrary centers.We will only encounter cubes B?Rnwith side length l(B)∈2N.This will allow us to perform decompositions by using cubes of smaller size in 2N.

The implicit constants will be independent of δ,g and the spatial cubes Qi.

Let t,p≥1 and consider the positive integers q≤s≤r.We define

To simplify notation,we will denote by Bs(Br)=Partδ?s(Br)the(unique)cover of Brwith cubes Bsof side length δ?s.Define

The letter A will remind us that we have an average.Note that when r=s,

For≤p,let 0≤κp≤ 1 satisfy

In other words,

Set also κp=0 for 2 ≤ p ≤

The next proposition will combine our main two decoupling devices,Theorem 9.2 and the L2decoupling.The result is a partial decoupling.Indeed,note that the term Ap(1,B2,1,g)in(10.1)involves frequency cubes of size δ,while the term Ap(2,B2,2,g)involves frequency cubes of smaller size δ2.Inequality(10.1)is only a partial decoupling in the range p>since the weight κpof the term Dp(1,B2,g)is nonzero.But this weight is zero when p ≤For these values of p,inequality(10.1)has the very simple form

This can be easily iterated and leads to a simpler proof of Theorem 1.1 in the range 2≤p≤See the discussion at the end of this section.

Proposition 10.1We have for each B2and p≥2,

ProofAssume first that p≥.By H¨older’s inequality,

Using this and Theorem 9.2,we can write

Using H¨older’s inequality,we can dominate this by

It suffices now to apply L2decoupling(Proposition 6.1)to the first term in(10.3).

We have “interpolated” between L2and Lp.We have used L2because(as explained in Section 6)this space facilitates the most efficient decoupling.Indeed,note that the term Ap(2,B2,2,g)on the right-hand side of(10.1)has cubes of side length δ2,which is as small as one can hope,given the size of the spatial cube B2.

Inequality(10.1)is easily seen to be true with κpreplaced with 1,by simply invoking(4.1)and the fact that D2?Dp.Consequently,it will be true for each exponent in the interval[κp,1].The example g=1Qwith l(Q)= δ shows that one can not consider exponents smaller than κp.The relevant thing about κpthat will be used in the final section is the fact that κp

The following sequence of propositions will allow us to rewrite(10.1)in a form that is more suitable for iteration.

Proposition 10.2We have for each cube BMwith M≥2 and p≥2,

The implicit constant is independent of M.

ProofRaising(10.1)to the power p,summing over all cubes B2∈B2(BM)and using H¨older’s inequality,we have

The only thing that needs to be verified is the inequality

This however immediately follows from Minkowski’s inequality(recall p ≥ 2)and the fact that

Proposition 10.3Let l,m∈N with l+1≤m.We have for each cube B2mand p≥2,

The implicit constant is independent of l,m.

ProofApply(10.4)with δ replaced by δ2land M=2m?l.

We can now iterate Proposition 10.3 to get the following immediate conclusion.

Proposition 10.4If m≥1 and p≥2,

The implicit constant is now allowed to depend on m,but this dependence will prove to be completely harmless.

We complete this section with a quick proof of

for 2≤p≤.This fact was first proved in[2].In this range κp=0 and(10.7)becomes a very satisfactory inequality

Combining this with(11.5),we may write

By invoking Cauchy–Schwarz,we can afford a rather trivial decoupling

Combining these two and substituting δ2m?→ δ we can write

Now choose m as large as desired to argue that

Finally,combine this with Theorem 8.1 by using induction on n to argue that

Now back to the case p>.As mentioned earlier,(10.7)is only a partial decoupling in this range.The argument for this case presented in the next section will go as follows.Assume the linear decoupling constant satisfies Decn(δ,p)～ δ?ηp.We will first apply parabolic rescaling to majorize the terms Dpin(10.7)by some powers of δ?ηp.Then we will combine(10.7)with a trivial decoupling(Cauchy–Schwarz)to derive an upper bound on the multilinear constant Decn(δ,p,ν)in terms of δ?ηp.We play this against Theorem 8.1,which produced a lower bound for Decn(δ,p,ν)involving δ?ηp.These will force ηpto be zero.

11 The Final Argument

In this section,we present the details for the proof of Theorem 1.1.Let E≥100n.By combining the triangle and Cauchy-Schwarz inequalities,we find that Decn(δ,p,E)? δ?Cpfor some Cplarge enough.For p ≥ 2,let ηp,n,E= ηp,E≥ 0 be the unique(finite)number,such that

and

We will use induction on n,as described at the end of Section 3.Assume either n=2 or n≥3,and in addition,assume that we have for E≥100(n?1)andWe need to prove ηp,n,E=0 for E ≥ 100n andNote that for such p,we automatically have that p is smaller than,the critical index for decouplings in Rn?1.In particular,if n ≥ 3,our induction hypothesis guarantees that

for each E≥100n and

Fixfor the rest of the proof(so in particular we have p ≤ 20).Fix also E≥100n.All quantities Apand Dpwill be implicitly assumed to be relative to this E.The casewill follow via a standard limiting argument explained in the end of the section.Note that for,we have

We start with the following rather immediate consequence of Proposition 10.4.

Theorem 11.1Consider n ν-transverse cubes Q1,···,Qn? [0,1]n?1with side length at least δ.Then for m ≥ 1 and p ≥ 2,we have

with the implicit constant independent of Qi.

ProofThis will follow from Proposition 10.4,once we make a few observations.

First,

This is a consequence of H¨older’s inequality,Minkowski’s inequality in lp2and(4.1)(like(10.5)very much).

Second,an application of Proposition 7.1 shows that

Finally,combine these with(11.1)and Proposition 10.4.

By replacing δ2mwith δ,we prefer to write the inequality in Theorem 11.1 as follows:

with the implicit constant independent of the cubes Qi.Here the assumption is l(Qi)≥ δ2?m.

Let B=B1be a cube in Rnwith l(B)= δ?1.Consider n ν-transverse cubes Q1,···,Qn?[0,1]n?1with side length μ ≥ δ2?m.Let,as before,Partμ?1(B)denote the partition of B by using cubes Δ with l(Δ)= μ?1.Denote also by Bm(B)the partition of B by using cubes Δmwith l(Δm)= δ?2?m.

We may write,first by combining Cauchy–Schwarz and(4.1),

and then by using Minkowski’s inequality and(4.2)(recall that p ≤ 20),we have

Invoking(11.6)and removing the normalization,we conclude that

By taking a supremum over all Qi,B,g as above,we deduce the following inequality,which is a stronger substitute for(8.2):

Combining this with Theorem 8.1(use(11.3))and(11.2),we may now write

for some sequence δlconverging to zero.This in turn forces

for each?,ν>0.Thus,letting?,ν → 0,we get

and by rearranging terms,we have

As this holds for each m ≥ 1,(11.4)will immediately force ηp=0.

Let us now show that ηpn=0 forLet B ? Rnbe a cube with l(B)= δ?1.Using inequality(4.2),for p

Combining this with H¨older’s inequality,we get

It suffices to note that q→1 as p→pn.

AcknowledgementsThe authors are grateful to Zane Li and Terry Tao for pointing out a few inaccuracies in an earlier version of this manuscript.

[1]Bennett,J.,Carbery,A.and Tao,T.,On the multilinear restriction and Kakeya conjectures,Acta Math.,196(2),2006,261–302.

[2]Bourgain,J.,Moment inequalities for trigonometric polynomials with spectrum in curved hypersurfaces,Israel J.Math.,193(1),2013,441–458.

[3]Bourgain,J.and Demeter,C.,The proof of the l2decoupling conjecture,Annals of Math.,182(1),2015,351–389.

[4]Bourgain,J.,Demeter,C.and Guth,L.,Proof of the main conjecture in Vinogradov’s mean value theorem for degrees higher than three,Ann.of Math.,184(2),2016,633–682.

[5]Bourgain,J.and Guth,L.,Bounds on oscillatory integral operators based on multilinear estimates,Geom.Funct.Anal.,21(6),2011,1239–1295.

[6]Guth,L.,A short proof of the multilinear Kakeya inequality,Math.Proc.Cambridge Philos.Soc.,158(1),2015,147–153.