aa r X i v : . [ m a t h . N T ] J a n ON CERTAIN OTHER SETS OF INTEGERS
TOM SANDERS
Abstract.
We show that if A ⊂ { , . . . , N } contains no non-trivial three-term arithmeticprogressions then | A | = O ( N/ log / − o (1) N ). Introduction
In this paper we refine the quantitative aspects of Roth’s theorem on sets avoidingthree-term arithmetic progressions. Our main result is the following.
Theorem 1.1.
Suppose that A ⊂ { , . . . , N } contains no non-trivial three-term arithmeticprogressions. Then | A | = O ( N/ log / − o (1) N ) . The first non-trivial upper bound on the size of such sets was given by Roth [Rot52,Rot53], and there then followed refinements by Heath-Brown [HB87] and Szemer´edi [Sze90],and later Bourgain [Bou99, Bou08], culminating in the above with 2 / / Z / Z ) n .Salem and Spencer [SS42] showed that the surface of high-dimensional convex bodiescould be embedded in the integers to construct sets of size N − o (1) containing no three-term progressions, and Behrend [Beh46] noticed that spheres are a particularly good choice.Recently Elkin [Elk08] tweaked this further by thickening the spheres to produce the largestknown progression-free sets, and his argument was then considerably simplified by Greenand Wolf in the very short and readable note [GW10b].In the next section we establish our basic notation before we present a sketch of themain argument in §
3. The remainder of the paper then makes the sketch precise.2.
Basic notation: the Fourier transform and Bohr sets
Suppose that G is a finite abelian group. We write M ( G ) for the space of measures on G and given two measures µ, ν ∈ M ( G ) define their convolution µ ∗ ν to be the measureinduced by C ( G ) → C ( G ); f Z f ( x + y ) dµ ( x ) dν ( y ) . There is a privileged measure µ G ∈ M ( G ) on G called Haar probability measure, anddefined to be the measure assigning mass | G | − to each element of G .The significance of this measure is that it is the unique (up to scaling) translationinvariant measure on G : for x ∈ G and µ ∈ M ( G ) we define τ x ( µ ) to be the measureinduced by C ( G ) → C ( G ); f Z f ( y ) dµ ( y + x ) , and it is easy to see that τ x ( µ G ) = µ G for all x ∈ G .We use Haar measure to pass between the notion of function f ∈ L ( µ G ) and measure µ ∈ M ( G ). Indeed, since G is finite we shall often identify µ with dµ/dµ G , the Radon-Nikodym derivate of µ with respect to µ G . In light of this we can extend the notion oftranslation and convolution to L ( µ G ) with ease: given f ∈ L ( µ G ) and x ∈ G we definethe translation of f by x point-wise by τ x ( f )( y ) := d ( τ x ( f dµ G )) dµ G ) ( y ) = f ( x + y ) for all y ∈ G, and given f, g ∈ L ( µ G ) we define convolution of f and g point-wise by f ∗ g ( x ) := d (( f dµ G ) ∗ ( gdµ G )) dµ G ( x ) = Z f ( y ) g ( x − y ) dµ G ( y ) , and similarly for the convolution of f ∈ L ( µ G ) with µ ∈ M ( G ).Convolution operators can be written in a particularly simple form with respect to theFourier basis which we now recall. We write b G for the dual group, that is the finite Abeliangroup of homomorphisms γ : G → S , where S := { z ∈ C : | z | = 1 } . Given µ ∈ M ( G ) wedefine b µ ∈ ℓ ∞ ( b G ) by b µ ( γ ) := Z γdµ for all γ ∈ b G, and extend this to f ∈ L ( G ) by b f := [ f dµ G . It is easy to check that [ µ ∗ ν = b µ · b ν for all µ, ν ∈ M ( G ) and [ f ∗ g = b f · b g for all f, g ∈ L ( µ G ).It is also useful to have an inverse for the Fourier transform: given g ∈ ℓ ( b G ) we define g ∨ ∈ C ( G ) by g ∨ ( x ) = X γ ∈ b G g ( γ ) γ ( x ) for all x ∈ G, and note that ( g ∨ ) ∧ = g . It is now natural to define the convolution of two functions f, g ∈ ℓ ( b G ) to be f ∗ g ( γ ) = X γ ′ ∈ b G f ( γ ′ ) g ( γ − γ ′ ) for all γ ∈ b G, so that ( f ∗ g ) ∨ = f ∨ · g ∨ . N CERTAIN OTHER SETS OF INTEGERS 3
As is typical we shall establish the presence of non-trivial arithmetic progressions bycounting them, and to this end we introduce the tri-linear form T ( f, g, h ) := Z f ( x − y ) g ( y ) h ( x + y ) dµ G ( x ) dµ G ( y ) , which has a rather useful Fourier formulation: T ( f, g, h ) = X γ ∈ b G b f ( γ ) b g ( − γ ) b h ( γ ) . Following [Bou08] we use a slight generalization of the traditional notion of Bohr set, let-ting the width parameter vary according to the character. The advantage of this definitionis that the meet of two Bohr sets in the lattice of Bohr sets is then just their intersection.A set B is called a Bohr set if there is a frequency set
Γ of characters on G , and a widthfunction δ ∈ (0 , Γ such that B = { x ∈ G : | − γ ( x ) | δ γ for all γ ∈ Γ } . The size of the set Γ is called the rank of B and is denoted rk( B ). Technically the sameBohr set can be defined by different frequency sets and width functions; we make thestandard abuse that when we introduce a Bohr set we are implicitly fixing a frequency setand width function.There is a natural way of dilating Bohr sets which will be of particular use to us. Givensuch a B , and ρ ∈ R + we shall write B ρ for the Bohr set with frequency set Γ and widthfunction ρδ so that, in particular, B = B and more generally ( B ρ ) ρ ′ = B ρρ ′ .With these dilates we say that a Bohr set B ′ is a sub-Bohr set of another Bohr set B ,and write B ′ B , if B ′ ρ ⊂ B ρ for all ρ ∈ R + . Finally, we write β ρ for the probability measure induced on B ρ by µ G , and β for β .Throughout the paper C s will denote absolute, effective, but unspecified constants ofsize greater than 1 and c s will denote the same of size at most 1. Typically the constantswill be subscripted according to the result which they come from and superscripted withinarguments.We shall also find a variant of the big- O notation useful. Specifically, if X, Y > X = e O ( Y ) to mean X = O ( Y log O (1) Y ), and if X, Y X = e Ω( Y ) tomean X − = e O ( Y − ). 3. Sketching the argument
The following discussion is based on the approaches to Roth’s theorem developed byBourgain in [Bou99] and then [Bou08]. There have been a number of expositions of thefirst of these papers – the interested reader may wish to consult Tao and Vu [TV06] – andthe second paper contains quite a detailed heuristic discussion of the argument. Technically width function γ min { ρδ γ , } . TOM SANDERS
It should be remarked that the present section is not logically necessary for the rest ofthe paper, and the reader who is not already familiar with the basic ideas may find itdifficult to follow.We shall concentrate on the problem of showing that if G is a finite abelian group ofodd order and A ⊂ G has density α ∈ (0 ,
1] then T (1 A , A , A ) = exp( − e O ( α − C )) , ideally with as small a constant C as possible.It seems unlikely that the odd order condition is necessary in light of the work of Lev[Lev04], but removing it seems to add technicalities which obscure the underlying ideas.In any case the passage from finite abelian groups of odd order to { , . . . , N } is unaffectedand follows from the usual Fre˘ıman embedding.To begin recall that by the usual application of the inversion formula and the triangleinequality, if A has somewhat fewer than expected three-term progressions then(3.1) X b G = γ ∈ Spec Ω( α ) (1 A ) | c A ( γ ) | | c A ( − γ ) | = Ω( α ) . The basic approach.
Roth [Rot53] noted that the above implies, in particular, thatthe large spectrum is non-empty, and so we have a non-trivial character γ with | c A ( γ ) | =Ω( α ). This can be used to give a density increment on (a translate of) I := { x ∈ G : γ ( x ) ≈ } of the form α α (1 + Ω( α )).In [Bou99] Bourgain noted that approximate level sets of characters – sets like I –behave roughly like groups, and that crucially the preceding argument can be run relativeto anything which behaves in that way.The intersection of many approximate level sets is a Bohr set (hence their importance tous) and working through the previous for a set A of relative density α on a d -dimensionalBohr set B gives a new Bohr set B ′ with(3.2) d d + 1 and α α (1 + Ω( α )) . In general we only have a bound of the form(3.3) µ G ( B ′ ) > exp( − e O ( d )) µ G ( B ) , although for groups of bounded exponent this can be much improved. Indeed, if G is themodel group ( Z / Z ) n then µ G ( B ′ ) = Ω( µ G ( B )), and this is the reason for the difference inthe bounds in the model setting (see [Mes95]) and here.A density increment of the form in (3.2) cannot proceed for more that e O ( α − ) steps,and so there must be some point where A has about the expected number of three-termprogressions on a Bohr set B ′ of dimension d + e O ( α − ). It follows that µ G ( B ′ ) > exp( − e O ( d )) . . . exp( − e O ( d + α − )) µ G ( B )which tells us that(3.4) T (1 A , A , A ) = exp( − e O ( dα − + α − )) µ G ( B ) . N CERTAIN OTHER SETS OF INTEGERS 5
We are interested in the case d = 0 and B = G when this gives the bound in [Bou99].3.2. The energy increment approach.
The previous argument only uses the fact that(3.1) tells us the large spectrum is non-empty. It is natural to try to get a density incrementby noting that the Bohr set B with frequency set Spec Ω( α ) (1 A ) majorises that spectrum,and hence 1 A ∗ β has large L ( µ G )-mass.An energy increment technique motivated by the work of Heath-Brown [HB87] andSzemer´edi [Sze90] (essentially Lemma 7.2) now tells us that A has density α (1 + Ω(1)) on(a translate of) B . It remains to estimate the dimension of B . This is certainly at mostthe size of the largest dissociated subset of the large spectrum, and this can be boundedby Chang’s theorem [Cha02]; in this case it is e O ( α − ).All of this can, itself, be done for a set A of relative density α in a Bohr set B ofdimension d . In which case either we have many three-term progressions or else we have anew Bohr set with d d + e O ( α − ) and α α (1 + Ω(1)) . This process can be iterated at most O (log 2 α − ) times and so proceeding as before withthe lower bound (3.3) we conclude that(3.5) T (1 A , A , A ) = exp( − e O ( d + α − )) µ G ( B ) . This represents no improvement over (3.4) in our case of interest, but for large values of d it does and we shall find this fact useful later on.3.3. Comparing the approaches.
In the approach in § e O ( α − )different characters but we did so one at a time and every time we iterated we used (3.3)which cost us a factor exponential in the dimension of the Bohr set. In the approach in § e O ( α − ) characters but took almost all of them together meaning that weonly lost the factor from (3.3) a very small number of times.To get a better bound one should like to improve Chang’s theorem, but that is notpossible without additional information about A . (See [Gre03] for details.) However,suppose that we are in the extreme case of Chang’s theorem and there are many dissociatedcharacters in the large spectrum. They each lead to a different density increment of thetype in § The random sampling strategy.
In the first instance the argument proceeds byremoving the mass in (3.1) corresponding to Spec √ α ( A ) using the Heath-Brown-Szemer´edienergy increment technique as in § e O ( α − )on which we have a density increment of the form α α (1 + Ω(1)). Otherwise, by dyadic TOM SANDERS decomposition, there is some τ ∈ [Ω( α ) , √ α ] such that(3.6) X γ ∈ S τ | c A ( γ ) | = e Ω( τ − α ) , where S τ := Spec τ (1 A ) \ Spec τ (1 A ). We shall split into two cases depending on whether S τ contains a large dissociated set or not.Suppose that T is a dissociated subset of S τ , and let c λ be the phase of c A ( λ ). Byexpanding out we have(3.7) h A , Y λ ∈ T (1 + c λ Re λ ) i L ( µ G ) = α + X λ ∈ Λ | c A ( λ ) | + . . . . If the remaining terms really were an error term then we would have that the inner productis at least α (1 + Ω( τ | T | )).Now, a Riesz product p generated by a set Λ of characters is essentially invariant underconvolution with β , where B is the Bohr set with frequency set Λ. Thus h A , p i L ( µ G ) ≈ h A ∗ β, p i L ( µ G ) k A ∗ β k L ∞ ( µ G ) Z pdµ G . Of course, 1 A ∗ β ( x ) is just the density of A on x + B and so if the inner product on theleft is large and the integral of the Riesz product is 1 then we have a density increment.Since T is dissociated, the integral of our Riesz product is 1 and we would get the densityincrement α α (1 + Ω( τ | T | )) on a Bohr set of dimension O ( | T | ) if the remaining termsin (3.7) were error terms. Of course, they are not and to deal with this we sample from T .Let Λ be a subset of T with λ ∈ T chosen independently with probability θ . Then E h A , Y λ ∈ Λ (1 + c λ Re λ ) i L ( µ G ) = h A , Y λ ∈ T (1 + θc λ Re λ ) i L ( µ G ) > h A , exp( X λ ∈ T θc λ Re λ ) i L ( µ G ) (1 + O ( θ | T | )) > h A , X λ ∈ T θc λ Re λ i L ( µ G ) (1 + O ( θ | T | )) > α + θ | T | τ α/ θ cτ . Now we get a Bohr set of (expected) dimension O ( θ | T | ) on which we havedensity α (1 + Ω( τ θ | T | )).This can all be done for a set A of relative density α in a d -dimensional Bohr set, andtaking θ = cτ we have a dichotomy: if there are at least k dissociated elements in S τ thenwe get a new Bohr set with(3.9) d d + O ( τ k ) and α α (1 + Ω( τ k ));if there are at most k dissociated elements in S τ then we apply the Heath-Brown-Szemer´edienergy increment technique to get a Bohr set of dimension d + O ( k ) on which we havedensity e Ω( τ − α ) by equation (3.6). In this instance we apply (3.5) and are done. N CERTAIN OTHER SETS OF INTEGERS 7
Now, (3.9) can happen at most O ( τ − k − ) times which means that we have at most thatmany applications of (3.3), while the dimension of the Bohr set is at most O ( τ − k − ) .O ( τ k ) + e O ( α − ) = e O ( α − )since τ = Ω( α ). After that the iteration must terminate either with many progressions oran application of (3.5) and we find that for A ⊂ G of density α we have(3.10) T (1 A , A , A ) > exp( − e O ( τ − k − α − + k + τ α − )) . Of course τ ∈ [Ω( α ) , √ α ] so we can optimise with k = α − / to get the main result in[Bou08].3.5. Relaxing the dissociation.
To get the density increment from (3.8) we used thefact that T is dissociated so that the Riesz product has integral 1, but if the integral is atmost 1 + cθ | T | τ then the argument we used still works. Moreover, the random samplingprocess actually makes the integral closer to 1, and it is this which we shall exploit toimprove (3.10).We start with a decomposition of Bourgain from [Bou90] the existence of which is aneasy induction: for a parameter m ∈ N ,(i) either half of S τ has no dissociated set of size at least m ;(ii) or half of S τ is a disjoint union of dissociated sets of size m .In the former case we get a Bohr set of dimension m on which we have density Ω( τ − ) =Ω( α / ) by the Heath-Brown-Szemer´edi energy increment technique, and we are done afterapplying the bound (3.5). In the latter case write T for the union, and note by Rudin’sinequality and the triangle inequality that(3.11) 1 T ∗ · · · ∗ T (0 b G ) O ( r/m ) ( r − / | T | r − , where the convolution is r -fold. This sort of expression and related consequences have beenexploited extensively in the work of Shkredov [Shk08b, Shk09] and Shkredov and Yekhanin[SY10], and that work was one of the starting points for this paper.Let Λ be a subset of T with λ ∈ T chosen independently with probability θ . Now,suppose (for simplicity) that all the phases c λ from the previous subsection are 1. Thenwe have E Z Y λ ∈ Λ (1 + c λ Re λ ) dµ G θ Z X λ ∈ Λ c λ Re λdµ G + ∞ X r =2 θ r r ! 1 T ∗ · · · ∗ T (0 b G ) ∞ X r =2 mr | T | O (cid:18) θ | T |√ mr (cid:19) r since none of the characters is trivial. Of course if | T | > Cτ − √ m then we may take θ | T | = c ′ √ m whilst keeping θ cτ , and the upper bound is then just 1 + O ( θ | T | ).Assuming that this typical behaviour actually happens we get a density increment of α α (1 + Ω( τ √ m )) on a Bohr set of dimension O ( √ m ). For a suitable choice of m thiswill be an improvement on what we had before. TOM SANDERS
In the non-relative situation it is easy to see that T is as large as needed since it’s atleast half of S τ , which in turn has size Ω( τ − ) from (3.6). However, when we relativise toBohr sets this is not true any more. To solve this, if | T | Cτ − √ m then we shall arrangethings so that we can take a subset T ′ of size η | T | such that X λ ∈ T ′ | c A ( γ ) | = e Ω( ητ − α ) , and apply the usual Heath-Brown-Szemer´edi energy increment technique to get a Bohr setof dimension O ( ητ − √ m ) on which we have density e Ω( ητ − α ); we finish off this case when | T | is small by applying (3.5).Relativising all this as in the previous subsection we get that our iteration procedureterminates inside O ( τ − m − / ) steps and T (1 A , A , A ) > exp( − e O ( τ − m − / α − + m + ητ − √ m + η − τ α − )) . Of course τ ∈ [Ω( α ) , √ α ] so we can optimise with η = τ α − / and m = α − / to get ourmain theorem. 4. The large spectrum
In this section we shall be working with respect to a probability measure µ on G . Laterwe shall take µ to be a sort of approximate Haar measure on approximate groups, but forthe moment our analysis does not require this.Our object of study is the spectrum of functions with respect to µ : given a function f ∈ L ( µ ) and a parameter ǫ ∈ (0 ,
1] we define the ǫ -spectrum of f w.r.t. µ to be the setSpec ǫ ( f, µ ) := { γ ∈ b G : | ( f dµ ) ∧ ( γ ) | > ǫ k f k L ( µ ) } . This definition extends the usual one from the case µ = µ G . In that case there is a rathersimple estimate for the size of the spectrum which follows from Bessel’s inequality, and itwill be useful for us to have a relative version of this.Defining notions by what they do, rather than what they are is a fairly standard idea,but it has seen particular success in the development of approximate analogues of conceptsin additive combinatorics. This is spelt out particularly clearly by Gowers and Wolf in[GW10a], and the next definition (as well as similar later ones) is motivated by theirdiscussion of quadratic rank. We say that Λ ⊂ b G is K -orthogonal w.r.t. µ if Z | g ∨ | dµ (1 + K )(1 + k g k ℓ (Λ) ) for all g ∈ ℓ (Λ) . Lemma 4.1 (Monotonicity of orthogonality) . Suppose that µ ′ is another probability mea-sure, Λ is K -orthogonal w.r.t. µ , Λ ′ ⊂ Λ and K ′ > K . Then Λ ′ is K ′ -orthogonal w.r.t. µ ′ ∗ µ . There is some flexibility with η ; m is the critical parameter. N CERTAIN OTHER SETS OF INTEGERS 9
Proof.
The only part worthy of remark is the replacing of µ by µ ′ ∗ µ : this is immediate onnoting that τ − y (1 + g ∨ ) = 1 + h ∨ where h ( λ ) = λ ( y ) g ( λ ) and k h k ℓ (Λ) = k g k ℓ (Λ) . It followsthat Λ is K -orthogonal w.r.t. τ y ( µ ), and we get the result on integrating. (cid:3) We define the (
K, µ ) -relative size of Λ to be the size of the largest subset of Λ that is( K, µ )-orthogonal w.r.t. µ . By monotonicity the ( K ′ , µ ′ ∗ µ )-relative size dominates the( K, µ )-relative size; the (0 , µ G )-relative size of Λ is the size of the largest subset of Λ notcontaining { b G } .The Bessel bound mentioned earlier follows easily from this definition. Lemma 4.2 (The Bessel bound) . Suppose that f ∈ L ( µ ) is not identically zero and write L f := k f k L ( µ ) k f k − L ( µ ) . Then Spec ǫ ( f, µ ) has (1 , µ ) -relative size O ( ǫ − L − f ) .Proof. Suppose that Λ ⊂ Spec ǫ ( f, µ ) is (1 , µ )-orthogonal. Then1 + k g ∨ k L ( µ ) = 12 (cid:18)Z | − g ∨ | dµ + Z | g ∨ | dµ (cid:19) k g k ℓ (Λ) )for all g ∈ ℓ (Λ). It follows (by letting k g k ℓ (Λ) → ∞ ) that the map ℓ (Λ) → L ( µ ) hasnorm at most 2 / and so by duality the map L ( µ ) → ℓ (Λ) defined by f ( f dµ ) ∧ | Λ hasnorm at most 2 / , whence | Λ | ǫ k f k L ( µ ) X λ ∈ Λ | ( f dµ ) ∧ ( λ ) | k f k L ( µ ) . We get the result on rearranging. (cid:3)
In the case µ = µ G this yields the usual Parseval bound which pervades work in thearea. The bound was first understood in the context of Bessel’s inequality in the paper[GT08] of Green and Tao.Orthogonal random variables can be thought of as roughly pair-wise independent, andnaturally we can improve the above bound if we insist on a greater level of independence.The idea of doing this was developed by Chang in [Cha02] following on from work ofBourgain [Bou90].Given a set of characters Λ and a function ω : Λ → D := { z ∈ C : | z | } we define p ω, Λ := Y λ ∈ Λ (1 + Re( ω ( λ ) λ )) , and call such a function a Riesz product for
Λ. It is easy to see that all Riesz products arereal non-negative functions. They are at their most useful when they also have mass closeto 1: the set Λ is said to be K -dissociated w.r.t. µ if Z p ω, Λ dµ exp( K ) for all ω : Λ → D. If µ is positive definite then one sees by Plancherel’s theorem that the maximum above isattained for the constant function 1 and so we have the following lemma. Lemma 4.3 (Dissociativity for positive definite measures) . Suppose that Λ is a set ofcharacters and µ is positive definite. Then Λ is log R p , Λ dµ -dissociated w.r.t. µ . As before we also have a monotonicity result.
Lemma 4.4 (Monotonicity of dissociativity) . Suppose that µ ′ is another probability mea-sure, Λ is K -dissociated w.r.t. µ , Λ ′ ⊂ Λ and K ′ > K . Then Λ ′ is K ′ -dissociated w.r.t. µ ′ ∗ µ . The (
K, µ ) -relative entropy of a set Λ is the size of the largest subset of Λ that is K -dissociated w.r.t. µ . As before relative entropy is monotone; the (0 , µ G )-relative entropyof a set is the size of its largest dissociated (in the usual sense) subset.A slightly harder property of dissociated sets is that they satisfy a Chernoff-type esti-mate. The proof of this that we give here essentially localises [GR07, Proposition 3.4]. Lemma 4.5.
Suppose that Λ is K -dissociated w.r.t. µ . Then Z | exp( g ∨ ) | dµ exp( K + k g k ℓ (Λ) / for all g ∈ ℓ (Λ) . Proof.
Suppose that g ∈ ℓ (Λ) and begin by noting that Z | exp( g ∨ ) | dµ = Z exp(Re g ∨ ) dµ = Z Y λ ∈ Λ exp(Re( g ( λ ) λ )) dµ. Now we have the elementary inequality exp( ty ) cosh t + y sinh t whenever t ∈ R and − y
1, so Z | exp( g ∨ ) | dµ Z Y λ ∈ Λ: g ( λ ) =0 (cid:18) cosh | g ( λ ) | + Re( g ( λ ) λ ) | g ( λ ) | sinh | g ( λ ) | (cid:19) dµ, with the usual convention that t − sinh t is 1 if t = 0. We define ω ∈ ℓ ∞ (Λ) by ω ( λ ) := g ( λ ) sinh | g ( λ ) || g ( λ ) | cosh | g ( λ ) | so that we certainly have k ω k ℓ ∞ (Λ)
1, and hence p ω, Λ is a Riesz product and Z | exp( g ∨ ) | dµ Y λ ∈ Λ cosh | g ( λ ) | Z p ω, Λ dµ. The result follows since cosh x exp( x / (cid:3) It is, perhaps, instructive to compare the conclusion of this lemma with the definitionof being K -orthogonal w.r.t. µ .Green and Ruzsa used the above estimate (for K = 0) in [GR07] to prove Chang’s theo-rem for sets, but our argument will proceed along the more traditional lines of establishingRudin’s inequality (w.r.t. to µ ) and then applying duality so that it applies to generalfunctions. N CERTAIN OTHER SETS OF INTEGERS 11
Lemma 4.6 (The Chang bound) . Suppose that f ∈ L ( µ ) is not identically zero and write L f := k f k L ( µ ) k f k − L ( µ ) . Then Spec ǫ ( f, µ ) has (1 , µ ) -relative entropy O ( ǫ − log 2 L f ) .Proof. Suppose that Λ ⊂ Spec ǫ ( f, µ ) is K -dissociated w.r.t. µ . First we establish a relativeversion of Rudin’s inequality. Suppose that k ∈ N and g ∈ ℓ (Λ) has k g k ℓ (Λ) = 2 √ k . ThenLemma 4.5 tells us that k Re g ∨ k L k ( µ ) (2 k )! / k exp(1 / k + 4 k/ k ) = O ( √ k k g k ℓ (Λ) ) . However, since Re(( − i ) g ∨ ) = Im g ∨ we conclude that k g ∨ k L k ( µ ) = O ( √ k k g k ℓ (Λ) ) andhence that the map g g ∨ from ℓ (Λ) to L k ( µ ) has norm O ( √ k ).Now, as with the Bessel bound, by duality and convexity of the L p ( µ )-norms we see that | Λ | ǫ k f k L ( µ ) X λ ∈ Λ | ( f dµ ) ∧ ( λ ) | = O ( k k f k L k/ (2 k − ( µ ) ) = O ( k k f k L ( µ ) L /kf ) . The result follows on putting k = ⌈ log 2 L f ⌉ and rearranging. (cid:3) Although Chang’s theorem cannot be significantly improved, there are some small re-finements and discussions of their limitations in the work [Shk06, Shk07] and [Shk08a] ofShkredov. Our approach is largely insensitive to these developments and so we shall notconcern ourselves with them here.Instead of trying to improve Chang’s theorem we are going to show that if the largespectrum of a function contains many dissociated elements then we get correlation with aRiesz product. The result should be compared with Proposition (*) from [Bou08] whichhas a very readable proof in that paper.On a first reading one may wish to take µ = µ ′ . When we use the lemma we shall take µ ′ positive definite and such that µ ∗ µ ′ ≈ µ . It then follows that the integral on the rightof the second conclusion is roughly an upper bound for the constant of dissociativity of Λ ′′′ w.r.t. µ . (The reader may wish to look forwards to Lemma 7.3 to get some idea of whatwe need the lemma below to dovetail with.) Lemma 4.7.
Suppose µ ′ is another probability measure, A is an event with α := µ ( A ) > , Λ ⊂ Spec τ (1 A , µ ) is η -orthogonal w.r.t. µ ′ and has size k , and m > is a parameter. Thenat least one of the following is true: (i) there is a set Λ ′ of (1 , µ ′ ) -relative entropy O ( m log 2 k ) which contains at least k/ of the elements of Λ ; (ii) for any d with C log 2 τ − d c min { τ k, √ m } there is some set Λ ′′′ of O ( d ) characters and function ω ′ : Λ ′′′ → D such that h A , p ω ′ , Λ ′′′ i L ( µ ) > α (1 + τ d/ Z p , Λ ′′′ dµ ′ ; provided η c ′ τ .Proof. Let l := ⌈ m log 2 k ⌉ and decompose Λ as a disjoint union of sets Λ ′ , Λ , . . . , Λ r where (i) any subset of Λ ′ that is 1-dissociated w.r.t. µ ′ has size at most l ;(ii) every Λ i has size l and is 1-dissociated w.r.t. µ ′ .If | Λ ′ | is at least k/ ′′ := Λ ⊔ · · · ⊔ Λ r has size at least k/ ω : Λ ′′ → D be such that ω ( λ )(1 A dµ ) ∧ ( λ ) = | (1 A dµ ) ∧ ( λ ) | for all λ ∈ Λ ′′ ;and let ( X λ ) λ ∈ Λ ′′ be independent identically distributed random variables with P ( X λ = 1) = θ and P ( X λ = 0) = 1 − θ, where θ := d/k, possible by assuming d k . Since the X λ s are independent we have that(4.1) E p ω ′′ X, Λ ′′ ( x ) = Y λ ∈ Λ ′′ E (1 + X λ Re( ω ′′ ( λ ) λ ( x ))) = Y λ ∈ Λ ′′ (1 + θ Re( ω ′′ ( λ ) λ ( x ))) , for any ω ′′ : Λ ′′ → D . We write N for the event that at most 2 ed of the variables are 1and specialise to ω ′′ = ω so that again, by independence, E N p ωX, Λ ′′ ( x ) = E p ωX, Λ ′′ ( x ) − X r> ed θ r X S ⊂ Λ ′′ : | S | = r Y λ ∈ S Re( ω ( λ ) λ ( x )) > E p ωX, Λ ′′ ( x ) − X r> ed θ r (cid:18) | Λ ′′ | r (cid:19) > E p ωX, Λ ′′ ( x ) − exp( − Ω( d ))since (cid:0) nm (cid:1) ( en/m ) m for all non-negative integers n and m .Of course, 1 + z > exp( z − z ) whenever − / z /
2, and exp( z ) > z for all z so we have (since we may assume θ /
2) that E p ωX, Λ ′′ ( x ) > Y λ ∈ Λ ′′ exp( θ Re( ω ( λ ) λ ( x )) − θ )= exp( − θ k ) exp( θ Re ω ∨ ( x )) > exp( − θ k )(1 + θ Re ω ∨ ( x )) . Coupled with non-negativity of 1 A this means that E N h A , p ωX, Λ ′′ i L ( µ ) = h A , E p ωX, Λ ′′ i L ( µ ) − α exp( − Ω( d )) > exp( − θ k )( α + θ X λ ∈ Λ ′′ | (1 A dµ ) ∧ ( λ ) | ) − α exp( − Ω( d )) > (1 − θ k )( α + θτ αk/ − α exp( − Ω( d )) . (4.2)In the other direction we specialise to ω ′′ ≡ E p X, Λ ′′ ( x ) = Y λ ∈ Λ ′′ (1 + θ Re λ ( x )) exp( θ X λ ∈ Λ ′′ Re λ ( x )) = exp( θ Re 1 ∨ Λ ′′ ( x )) . However, exp( z ) z + z exp( z ) for all z ∈ R so that(4.3) E Z p X, Λ ′′ dµ ′ Z θ Re 1 ∨ Λ ′′ dµ ′ + Z ( θ Re 1 ∨ Λ ′′ ) exp( θ Re 1 ∨ Λ ′′ ) dµ ′ . N CERTAIN OTHER SETS OF INTEGERS 13
The second term on the right is estimated using the η -orthogonality of Λ. Indeed, supposethat | z | = 1, then 2 z Z Re 1 ∨ Λ ′′ dµ ′ = X λ ∈ Λ ′′ Z zλdµ ′ = X λ ∈ Λ ′′ (cid:18)Z | zλ | dµ ′ − (cid:19) ηk since | Λ ′′ | k . Thus, on suitable choice of z , we see that the relevant term in (4.3) isat most ηk in absolute value. To estimate the third term on the right in (4.3) we applyH¨older’s inequality with index q := 1 + log k and conjugate index q ′ to get that Z ( θ Re 1 ∨ Λ ′′ ) exp( θ Re 1 ∨ Λ ′′ ) dµ ′ θ (cid:18)Z | Re 1 ∨ Λ ′′ | q ′ dµ ′ (cid:19) /q ′ (cid:18)Z exp( θq Re 1 ∨ Λ ′′ ) dµ ′ (cid:19) /q . Now | Re 1 ∨ Λ ′′ | k and Z | Re 1 ∨ Λ ′′ | dµ ′ Z | ∨ Λ ′′ | dµ ′ k since Λ is certainly 1-orthogonal. It follows that (cid:18)Z | Re 1 ∨ Λ ′′ | q ′ dµ ′ (cid:19) /q ′ ((2 k ) .k q ′ − ) /q ′ = O ( k ) , and combining what we have so far gives E Z p X, Λ ′′ dµ ′ O ( ηθk ) + O θ k (cid:18)Z exp( θq Re 1 ∨ Λ ′′ ) dµ ′ (cid:19) /q ! . Recalling that each Λ i is 1-dissociated w.r.t. µ ′ and that Λ ′′ is their disjoint union it followsfrom Lemma 4.5 that Z exp( θq Re 1 ∨ Λ ′′ ) dµ ′ r Y i =1 (cid:18)Z exp( rqθ Re 1 ∨ Λ i ) dµ ′ (cid:19) /r r Y i =1 exp(1 /r ) exp( rq θ k Λ i k ℓ (Λ i ) / . Now, k Λ i k ℓ (Λ i ) = l and rl k , so E Z p X, Λ ′′ dµ ′ O ( θηk ) + O ( θ k exp( k qθ / l )) . Inserting this into (4.2) we get that E N h A , p ωX, Λ ′′ i L ( µ ) > α (cid:0) (1 − O ( ηd ) − O ( d k − exp( d q/ l )))(1 + τ d/ − exp( − Ω( d ))) E Z p X, Λ ′′ dµ ′ . It follows that there are absolute constants c, c ′ and C such that if η cτ, dk − exp( d q/ l ) c ′ τ and d > C log 2 τ − then E N h A , p ωX, Λ ′′ i L ( µ ) > E α (1 + τ d/ Z p X, Λ ′′ dµ ′ . By averaging there is some element such that the integrand on the left is at least that onthe right and the result follows on setting Λ ′′′ = { λ ∈ Λ ′′ : X λ = 1 } , and ω ′ = ω | Λ ′′′ . (cid:3) It is possible to optimise this lemma slightly more efficiently but we shall not concernourselves with that here. 5.
Basic properties of Bohr sets
In this section we collect the basic facts which we need about Bohr sets. Although thismaterial has all now become standard this is in no small part due to the expository workof Tao; [TV06, Chapter 4.4] is the recommended reference.To start with we have the following averaging argument c.f. [TV06, Lemma 4.20].
Lemma 5.1 (Size of Bohr sets) . Suppose that B is a Bohr set. Then µ G ( B ρ ) = exp( O (rk( B ))) µ G ( B ρ ) for all ρ ∈ R + and µ G ( B ) > exp( − O (rk( B ))) Y γ ∈ Γ δ γ . Given two Bohr sets B and B ′ we define their intersection to be the Bohr set withfrequency set Γ ∪ Γ ′ and width function δ ∧ δ ′ . It follows immediately from the previouslemma that Bohr sets are well behaved under intersections. Lemma 5.2 (Intersections of Bohr sets) . Suppose that B and B ′ are Bohr sets. Then rk( B ∧ B ′ ) rk( B ) + rk( B ′ ) and µ G ( B ∧ B ′ ) > exp( − O (rk( B ) + rk( B ′ ))) µ G ( B ) µ G ( B ′ ) . Proof.
The rank inequality is obvious. For the density note that( B ∧ B ′ ) ⊃ ( B / − B / ) ∩ ( B ′ / − B ′ / )by the triangle inequality. On the other hand µ G (( B / − B / ) ∩ ( B ′ / − B ′ / )) µ G ( B / ) µ G ( B ′ / )is at least Z B / ∗ B / B ′ / ∗ B ′ / dµ G = k B / ∗ B ′ / k L ( µ G ) > ( µ G ( B / ) µ G ( B ′ / )) , where the last inequality is Cauchy-Schwarz. The result now follows from the first case ofthe previous lemma. (cid:3) N CERTAIN OTHER SETS OF INTEGERS 15
It is the first condition of Lemma 5.1 which is the most important and informs thedefinition of the dimension of a Bohr set: a Bohr set B is said to be d -dimensional if µ G ( B ρ ) d µ G ( B ρ ) for all ρ ∈ (0 , , and it will be convenient to assume that it is always at least 1. Since any d -dimensionalBohr set is always d ′ -dimensional for all d ′ > d this is not a problem.Note, in particular, that by Lemma 5.1 a Bohr set of rank k has dimension O ( k ). It is,however, the dimension which is the important property of Bohr sets and the only reasonwe mention the rank is that it is sub-additive with respect to intersection, unlike dimension.Some Bohr sets behave better than others: a d -dimensional Bohr set is said to be C -regular if 11 + Cd | η | µ G ( B η ) µ G ( B ) Cd | η | for all η with | η | /Cd. Crucially, regular Bohr sets are plentiful:
Lemma 5.3 (Regular Bohr sets) . There is an absolute constant C R such that whenever B is a Bohr set, there is some λ ∈ [1 / , such that B λ is C R -regular. The result is proved by a covering argument due to Bourgain [Bou99]; for details onemay also consult [TV06, Lemma 4.25]. For the remainder of the paper we shall say regular for C R -regular. 6. Bohr sets as majorants
One of the key properties of Bohr sets is that they can be used to majorise sets ofcharacters. They are particularly good at this because if two sets are majorised by twodifferent Bohr sets then their sumset is majorised by (a dilation of) the intersection of theBohr sets.We begin by recalling a lemma which shows that a set majorised by a Bohr set isapproximately annihilated by a dilate, and if a set is approximately annihilated by a Bohrset then it is certainly majorised by it.
Lemma 6.1 (Majorising and annihilating) . Suppose that B is a regular d -dimensionalBohr set. Then { γ : | b β ( γ ) | > κ } ⊂ { γ : | − γ ( x ) | C ρκ − d for all x ∈ B ρ } , and { γ : | − γ ( x ) | η for all x ∈ B } ⊂ { γ : | b β ( γ ) | > − η } . Proof.
First, suppose that | b β ( γ ) | > κ and y ∈ B ρ . Then | − γ ( y ) | κ | Z γ ( x ) dβ − Z γ ( x + y ) dβ ( x ) | µ G ( B ρ \ B − ρ ) µ G ( B ) = O ( dρ )provided ρ /C R d . The first inclusion follows; the second is a trivial application of thetriangle inequality. (cid:3) The proof is from [GK09, Lemma 3.6] where the significance of this result seems to havefirst been clearly recognised.Of particular interest to us is the majorising of orthogonal and dissociated sets and inlight of the previous lemma that can be achieved by the following. Recall that if X is aset then P ( X ) denotes the power set , that is the set of all subsets of X . Lemma 6.2 (Annihilating orthogonal sets) . Suppose that B is a regular d -dimensionalBohr set and ∆ is a set of characters with ( η, β ) -relative size k . Then there is a set Λ , η -orthogonal w.r.t. β , and an order-preserving map φ : P (Λ) → P (∆) such that φ (Λ) = ∆ and for any Λ ′ ⊂ Λ and γ ∈ φ (Λ ′ ) we have | − γ ( x ) | C ( ν + ρd η − log 2 kη − ) for all x ∈ B ρ ∧ B ′ ν where B ′ is the Bohr set with constant width function and frequency set Λ ′ .Proof. Let L := ⌈ log k + 1) / η − ⌉ , the reason for which choice will become apparent,and define β + := β Lρ ′ ∗ β ρ ′ ∗ · · · ∗ β ρ ′ , where β ρ ′ occurs L times in the expression. By regularity (of B ) we can pick ρ ′ ∈ (Ω( η/dL ) ,
1] such that B ρ ′ is regular and µ G ( B Lρ ′ ) (1 + η/ µ G ( B ). On the otherhand we have the point-wise inequality µ G ( B ) β = 1 B B Lρ ′ ∗ β ρ ′ ∗ · · · ∗ β ρ ′ = µ G ( B Lρ ′ ) β + , whence β η/ β + . It follows that if Λ is η/ β + then Λ is η -orthogonal w.r.t. β , and hence Λ has size at most k . From now on all orthogonality willbe w.r.t. β + .We put η i := iη/ k + 1) and begin by defining a sequence of sets Λ , Λ , . . . iterativelysuch that Λ i is η i -orthogonal. We let Λ := ∅ which is easily seen to be 0-orthogonal. Now,suppose that we have defined Λ i as required. If there is some γ ∈ ∆ \ Λ i such that Λ i ∪ { γ } is η i +1 -orthogonal then let Λ i +1 := Λ i ∪ { γ } . Otherwise, terminate the iteration.Note that for all i k + 1, if the set Λ i is defined then it is certainly η/ | Λ i | k . However, if the iteration had continued for k + 1 steps then | Λ k +1 | > k .This contradiction means that there is some i k such that Λ := Λ i is η i -orthogonal andΛ i ∪ { γ } is not η i +1 -orthogonal for any γ ∈ ∆ \ Λ i .Now, we define φ in P (Λ) by φ (Λ ′ ) = { γ ∈ ∆ : | c β ρ ′ ( γ − λ ) | > / λ ∈ Λ ′ ∪ { b G }} . It is immediate that φ is order-preserving. Moreover we have the following claim. Claim. φ (Λ) = ∆ .Proof. If λ ∈ Λ then λ ∈ φ (Λ) since c β ρ ′ (0 b G ) = 1, so we need only consider γ ∈ ∆ \ Λ. Inthat case there is some g ∈ ℓ (Λ) and ν ∈ C such that Z | g ∨ + νγ | dβ + > (1 + η i +1 )(1 + k g k ℓ (Λ) + | ν | ) . N CERTAIN OTHER SETS OF INTEGERS 17
It follows from the η i -orthogonality of Λ that2 Re h g ∨ , νγ i L ( β + ) > (1 + η i +1 )(1 + k g k ℓ (Λ) + | ν | ) − (1 + η i )(1 + k g k ℓ (Λ) ) − | ν | > η k + 1) (1 + k g k ℓ (Λ) + | ν | ) > η ( k + 1) (1 + k g k ℓ (Λ) ) / | ν | . Thus |h g ∨ , γ i L ( β + ) | > η k + 1) (1 + k g k ℓ (Λ) ) / By Plancherel’s theorem this tells us that η k + 1) (1 + k g k ℓ (Λ) ) / (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c β + ( γ ) + X λ ∈ Λ g ( λ ) c β + ( γ − λ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (1 + k g k ℓ (Λ) ) sup λ ∈ Λ ∪{ b G } | c β ρ ′ ( γ − λ ) | L . By Cauchy-Schwarz we have 1 + k g k ℓ (Λ) √ k + 1(1 + k g k ℓ (Λ) ) / , and so it follows (giventhe size of L ) that there is some λ ∈ Λ ∪ { b G } such that | c β ρ ′ ( γ − λ ) | > / (cid:3) Now, suppose that Λ ′ ⊂ Λ and γ ∈ φ (Λ ′ ). It follows that there is some λ ∈ Λ ′ ∪ { b G } such that | c β ρ ′ ( γ − λ ) | > / | − γ ( x ) λ ( x ) | C ρ ′− ρd for all x ∈ B ρ . On the other hand by definition | − λ ( x ) | = O ( ν ) if x ∈ B ′ ν and so the result follows fromthe triangle inequality. (cid:3) The argument for dissociated sets is rather similar so we omit many of the details.
Lemma 6.3 (Annihilating dissociated sets) . Suppose that B is a regular d -dimensionalBohr set and ∆ is a set of characters with ( η, β ) -relative entropy k . Then there is a set Λ , η -dissociated w.r.t. β , such that for all γ ∈ ∆ we have | − γ ( x ) | C ( kν + ρd η − ( k + log 2 η − )) for all x ∈ B ρ ∧ B ′ ν where B ′ is the Bohr set with constant width function and frequency set Λ .Proof. By the same argument as at the start of the previous lemma, but with L = ⌈ log k k + 1) η − ⌉ , we get some i k , an η i -dissociated (w.r.t. β + ) set Λ of at most k characters, such that for all γ ∈ ∆ \ Λ there is a function ω : Λ → D and ν ∈ D such that Z p ω, Λ (1 + Re( νγ )) dβ + > exp( η i +1 ) . Now, suppose that γ ∈ ∆. If γ ∈ Λ then the conclusion is immediate, so we may assumethat γ ∈ ∆ \ Λ. Then, since Λ is η i -dissociated, we see that | Z p ω, Λ γdβ + | > exp( η i +1 ) − exp( η i ) > η k + 1) . Applying Plancherel’s theorem we get that η k + 1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X λ ∈ Span(Λ) d p ω, Λ ( λ ) c β + ( γ − λ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) k sup λ ∈ Span(Λ) | b β ρ ( γ − λ ) | L . Given the choice of L there is some λ ∈ Span(Λ) such that | c β ρ ′ ( γ − λ ) | > /
2. Theargument now proceeds as before on noting that | − λ ( x ) | = O ( kν ) if x ∈ B ′ ν by thetriangle inequality. (cid:3) It is also useful to record an immediate corollary of this.
Corollary 6.4.
Suppose that B is a regular d -dimensional Bohr set and ∆ is a set ofcharacters with ( η, β ) -relative entropy k . Then there is a Bohr set B ′ B with rk( B ′ ) rk( B ) + k and µ G ( B ′ ) > ( η/ dk ) O ( d ) (1 / k ) O ( k ) µ G ( B ) such that | − γ ( x ) | / for all x ∈ B ′ and γ ∈ ∆ . Getting a density increment
The basic dichotomy of the density increment strategy is driven by the following lemmawhich says that either we are in a situation where we have many three-term progressionsor else we have a concentration of Fourier mass.
Lemma 7.1.
Suppose that B is a regular d -dimensional Bohr set and A ⊂ B and A ′ ⊂ B ρ have density α > and α ′ > respectively. Then either (i) (Many three-term progressions) T (1 A , A ′ , A ) > α α ′ µ G ( B ) µ G ( B ρ ) / (Concentration of Fourier mass) X − γ ∈ Spec c α (1 A ′ ,β ρ ) | ((1 A − α )1 B ) ∧ ( γ ) | | (1 A ′ dβ ρ ) ∧ ( − γ ) | > c α α ′ µ G ( B ); provided ρ c ′ α/d .Proof. Write M : L ( β ) → L ( β ) for the linear operator taking f ∈ L ( β ) to x Z f ( x − y )1 A ′ ( y ) dβ ρ ( y )1 B ( x ) , and put T ( f ) := h M f, f i L ( β ) , N CERTAIN OTHER SETS OF INTEGERS 19 so that T (1 A , A ′ , A ) = µ G ( B ) µ G ( B ρ ) T (1 A ) . We begin by noting that T (1 A ) = T ((1 A − α )1 B ) + α h M A , B i L ( β ) + α h M B , A i L ( β ) − α h M B , B i L ( β ) . By regularity we have k M B − α ′ B k = O ( ρdα ′ µ G ( B ))provided ρ / C R d . Thus, by the triangle inequality we have T (1 A ) = T ((1 A − α )1 B ) + α α ′ + O ( αα ′ ρd ) . It follows that there is some absolute c > ρ cα/d then either we are in thefirst case of the lemma or else | T ((1 A − α )1 B ) | > α α ′ / X γ ∈ b G | ((1 A − α )1 B ) ∧ ( γ ) | | (1 A ′ dβ ρ ) ∧ ( − γ ) | > α α ′ µ G ( B ) / , from which it follows that we are in the second case of the lemma by the triangle inequality. (cid:3) The remaining two lemmas of the section encode the other half of the density incrementstrategy: having analysed the Fourier mass in some way, we need results which convertthis analysis into a density increment.First we have the Heath-Brown-Szemer´edi energy increment technique. There are moregeneral versions of the lemma suitable for other problems; the one here is adapted to ourcircumstance and, in particular, is the only place we use the fact that G has odd order. Lemma 7.2 (Energy to density lemma) . Suppose that B is a regular d -dimensional Bohrset, A ⊂ B has density α > , L is a set of characters such that X − γ ∈L | ((1 A − α )1 B ) ∧ ( γ ) | > Kα µ G ( B ) , and B ′ B ρ is a d ′ -dimensional Bohr set such that L ⊂ { γ : | − γ ( x ) | / for all x ∈ B ′ } . Then there is a regular Bohr set B ′′ with rk( B ′′ ) = rk( B ′ ) and µ G ( B ′′ ) > − O ( d ′ ) µ G ( B ′ ) such that k A ∗ β ′′ k L ∞ ( µ G ) > α (1 + c K ) provided ρ c ′ αK/d . Proof.
Put B ′′′ := − .B ′ / , and note that B ′′′ has the same rank and size as B ′ / (since G has odd order), and { γ : − γ ∈ L} ⊂ { γ : | − γ ( − x ) | / x ∈ B ′ } . Pick ν ∈ [1 / ,
1) such that B ′′ := B ′′′ ν is regular and so, since {− x : x ∈ B ′ } ⊃ { x : x ∈ B ′′ } , we conclude that { γ : − γ ∈ L} ⊂ { γ : | − γ ( x ) | / x ∈ B ′′ } . By the triangle inequality if γ is in this second set, then | c β ′′ ( γ ) | > /
2. Thus Plancherel’stheorem tells us that k ((1 A − α )1 B ) ∗ β ′′ k L ( µ G ) = X γ ∈ b G | ((1 A − α )1 B ) ∧ ( γ ) | | c β ′′ ( γ ) | > X − γ ∈L | ((1 A − α )1 B ) ∧ ( γ ) | > K α µ G ( B ) . But B ′′ ⊂ B ′ ⊂ B ρ , so by regularity we have that h f, B ∗ β ′′ ∗ β ′′ i L ( µ G ) = ( Z f dβ + O ( ρd k f k L ∞ ( β ) )) µ G ( B )for any f ∈ L ∞ ( β ) provided ρ / C R d . Thus k ((1 A − α )1 B ) ∗ β ′′ k L ( µ G ) = k A ∗ β ′′ k L ( µ G ) − α (( α + O ( ρd )) µ G ( B )+ α (1 + O ( ρd )) µ G ( B ) , whence k A ∗ β ′′ k L ( µ G ) > α µ G ( B ) + Kα µ G ( B ) / − O ( αρdµ G ( B )) . The result now follows provided ρ is sufficiently small on applying H¨older’s inequality anddividing by αµ G ( B ). (cid:3) The second lemma we need is used for leveraging of the second conclusion of Lemma4.7 and gives a way of passing between correlation with a Riesz product and a densityincrement on a Bohr set.
Lemma 7.3 (Riesz product correlation to density lemma) . Suppose that B is a regular d -dimensional Bohr set, A ⊂ B − ρ , Λ is a set of k characters and ω : Λ → D is such that h A , p ω, Λ i L ( β ) > α (1 + ǫ ) Z p , Λ dµ, where µ = β ρ ∗ β ρ . Then there is a regular Bohr set B ′ with rk( B ′ ) rk( B ) + k and µ G ( B ′ ) > ( ǫα/ k ) O ( k ) − O ( d ) µ G ( B ρ ) such that k A ∗ β ′ k L ∞ ( µ G ) > α (1 + ǫ/ . N CERTAIN OTHER SETS OF INTEGERS 21
Proof.
Let B ′′ be the Bohr set with frequency set Λ and width function equal to theconstant function 1. Let λ , . . . , λ k be an enumeration of Λ and put ω i : Λ → D ; λ ω λ if λ = λ j for some j > i λ = λ i ω λ λ ( y ) if λ = λ j for some j < i. With these definitions we have p ω, Λ ( x + y ) − p ω, Λ ( x ) = k X i =1 (Re( ω λ i λ i ( x + y )) − Re( ω λ i λ i ( x ))) p ω i , Λ ( x ) , but | Re( ω λ i λ i ( x + y )) − Re( ω λ i λ i ( x )) | = O ( ρ ′ )if y ∈ B ′′ ρ ′ . Since µ is positive definite we see that Λ is K -dissociated w.r.t. µ with K = log R p , Λ dµ . Thus, integrating the above, we get that Z | τ y ( p ω, Λ ) − p ω, Λ | dβ ∗ µ k X i =1 O ( ρ ′ ) . Z p ω i , Λ dβ ∗ µ = O ( ρ ′ k exp( K ))for all y ∈ B ′′ ρ ′ by monotonicity of dissociativity. Let ρ ′ = Ω( ǫα/k ) be such that the aboveerror term is at most ǫα exp( K ) / B ′ = ( B ′′ ρ ′ ∧ B ρ ) ν where ν ∈ [1 / ,
1) is such that B ′ is regular. It follows by the triangle inequality that Z | p ω, Λ ∗ β ′ − p ω, Λ | dβ ∗ µ ǫα exp( K ) / . On the other hand, since A ⊂ B − ρ we have that h A , p ω, Λ i L ( β ) = h A , p ω, Λ i L ( β ∗ µ ) , and so by the triangle inequality and hypothesis we have h A , p ω, Λ ∗ β ′ i L ( β ) > α (1 + ǫ/
2) exp( K ) . However, since A ⊂ B − ρ , we have that h A , p ω, Λ ∗ β ′ i L ( β ) = h A ∗ β ′ , p ω, Λ i L ( β ∗ µ ) . Finally, by monotonicity of dissociativity again we get that k A ∗ β ′ k L ∞ ( µ G ) exp( K ) > α (1 + ǫ/
2) exp( K ) , and the result follows. (cid:3) Roth’s theorem in high rank Bohr sets
From now on we assume that the underlying group has odd order so that we may applyLemma 7.2.The purpose of this section is to establish the following theorem which is the formalversion of the result discussed in § Theorem 8.1.
Suppose that B is a rank k Bohr set and A ⊂ B has density α > . Then T (1 A , A , A ) = exp( − e O ( k + α − )) µ G ( B ) . We shall prove this iteratively using the following lemma.
Lemma 8.2.
Suppose that B is a d -dimensional Bohr set such that B and B ρ are regular,and A ⊂ B and A ′ ⊂ B ρ have density α > and α ′ > respectively with α = Θ( α ′ ) .Then at least one of the following is true: (i) (Many three-term progressions) T (1 A , A ′ , A ) > α α ′ µ G ( B ) µ G ( B ρ ) / (Energy induced density increment) or there is a regular Bohr set B ′ with rk( B ′ ) rk( B ) + e O ( α − ) and µ G ( B ′ ) > exp( − e O ( d + α − )) µ G ( B ρ ) such that k A ∗ β ′ k L ∞ ( µ G ) > α (1 + c ) ;provided ρ c ′ α/d .Proof. By Lemma 7.1 (provided ρ c ′ α/d ) we are either in the first case of the lemmaor else(8.1) X − γ ∈ Spec c α (1 A ′ ,β ρ ) | ((1 A − α )1 B ) ∧ ( γ ) | > c α µ G ( B ) . By (the relative version of) the Chang bound (Lemma 4.6) we see that Spec c α (1 A ′ , β ρ )has (1 , β ρ )-relative entropy O ( α − log α ′− ) = e O ( α − ). Thus, by Corollary 6.4 there is aBohr set B ′′ B ρ withrk( B ′′ ) rk( B ) + e O ( α − ) and µ G ( B ′′ ) > exp( − e O ( d + α − )) µ G ( B ρ )such that | − γ ( x ) | / x ∈ B ′′ and γ ∈ Spec c α (1 A ′ , β ρ ) . It follows by Lemma 7.2 (with L := Spec c α (1 A ′ , β ρ ) provided ρ c ′ c α/d ) that weare in the second case of the lemma. (cid:3) This condition is unnecessary but it makes for simpler statements, particularly later on. It is also aslight abuse; to be formally correct the reader may wish to take, for example, 2 α ′ > α > α ′ / N CERTAIN OTHER SETS OF INTEGERS 23
Proof of Theorem 8.1.
We proceed iteratively to construct a sequence of regular Bohr sets B ( i ) of rank k i and dimension d i = O ( k i ). We write α i := k A ∗ β ( i ) k L ∞ ( µ G ) , put B (0) := B and suppose that we have defined B ( i ) . By regularity we have that(1 A ∗ β ( i ) ρ i + 1 A ∗ β ( i ) ρ i ρ ′ i ) ∗ β ( i ) ( x i ) = 2 α i + O ( ρ i k i )for some x i . Pick ρ i = Ω( α i /k i ) and then ρ ′ i = Ω( α i /k i ) such that B ( i ) ρ i and B ( i ) ρ ′ i ρ i areregular, (1 A ∗ β ( i ) ρ + 1 A ∗ β ( i ) ρ i ρ ′ i ∗ β ( i ) ( x i ) > α i (1 − c / , and ρ ′ i c ′ α i /d i . Thus there is some x ′ i such that1 A ∗ β ( i ) ρ ( x ′ i ) + 1 A ∗ β ( i ) ρ i ρ ′ i ( x ′ i ) > α i (1 − c / . If one of the two terms is at least α i (1 + c / B ( i +1) to be B ( i ) ρ i or B ( i ) ρ ′ i ρ i respectively. Otherwise we see that A i := ( x ′ i − A ) ∩ B ( i ) ρ i has β ( i ) ρ i ( A i ) > α i (1 − c / , and A ′ i := ( x ′ i − A ) ∩ B ( i ) ρ ′ i ρ i has β ( i ) ρ ′ i ρ i ( A ′ i ) > α i (1 − c / . By Lemma 8.2 we see that either we are done in that(8.2) T (1 A , A , A ) = Ω( α i µ G ( B ( i ) ρ i ) µ G ( B ( i ) ρ ′ i ρ i ))or there is a regular Bohr set B ( i +1) such that k i +1 k i + e O ( α − i ) ,µ G ( B ( i +1) ) > exp( − e O ( k i + α − i )) µ G ( B ( i ) ρ ′ i ρ i ) > exp( − e O ( k i + α − i )) µ G ( B ( i ) ) , and α i +1 > α i (1 + c / α i +1 > α i (1 + Ω(1)). This cannot happen for more than O (log 2 α − ) step,whence we are in (8.2) at some step i = O (log 2 α − ), where we shall have k i = k + e O ( α − ) and µ G ( B ( i ) ) > exp( − e O ( k + α − )) µ G ( B ) , from which the result follows. (cid:3) The main result
In this section we prove the following theorem which is the main result of the paper.
Theorem 9.1.
Suppose that A has density α > . Then T (1 A , A , A ) = exp( − e O ( α − / )) . To see how Theorem 1.1 follows simply note that if A ⊂ { , . . . , N } has density α and nonon-trivial three-term progressions then it can be embedded in Z / (4 N + 1) Z with densityΩ( α ) such that it also has no non-trivial three-term progressions there, but then(4 N + 1) > (4 N + 1) exp( − e O ( α − / )) , which can be rearranged to give the desired bound.As before we proceed iteratively; the following is the driver. Lemma 9.2.
Suppose that B is a d -dimensional Bohr set such that B and B ρ are regular,and A ⊂ B and A ′ ⊂ B ρ have density α > and α ′ > respectively with α = Θ( α ′ ) . Thenat least one of the following is true: (i) (Many three-term progressions) T (1 A , A ′ , A ) > α α ′ µ G ( B ) µ G ( B ρ ) / (Energy induced density increment from large coefficients) or there is a regularBohr set B ′ with rk( B ′ ) rk( B ) + e O ( α − ) and µ G ( B ′ ) > exp( − e O ( d + α − )) µ G ( B ρ ) such that k A ∗ β ′ k L ∞ ( µ G ) > α (1 + c ) ; (iii) (Terminal density increment) or there is a regular Bohr set B ′ with rk( B ′ ) rk( B ) + e O ( α − / ) and µ G ( B ′ ) > exp( − e O ( d + α − / )) µ G ( B ρ ) such that k A ∗ β ′ k L ∞ ( µ G ) = e Ω( α / ) ; (iv) (Density increment from many independent small coefficients) or there is a regularBohr set B ′ with rk( B ′ ) rk( B ) + e O ( α − / ) and µ G ( B ′ ) > exp( − e O ( d + α − / )) µ G ( B ρ ) such that k A ′ ∗ β ′ k L ∞ ( µ G ) > α ′ (1 + c α / ) ;provided ρ c ′ α/d .Proof. The argument is not difficult although it involves a number of somewhat technicaldetails. To start with (for the purpose of applying Lemma 7.3 later) we shall adjust theset A ′ slightly. Let ρ ′ = cα ′ /d for some absolute c > B ρρ ′ is regular and µ G ( B ρ \ B ρ (1 − ρ ′ ) ) α ′ / B ρ ), and put A ′′ := A ′ ∩ B ρ (1 − ρ ′ ) so that A ′′ ⊂ B ρ (1 − ρ ′ ) and β ρ ( A ′′ ) > α ′ (1 − α ′ / . N CERTAIN OTHER SETS OF INTEGERS 25
By Lemma 7.1 applied to A and A ′′ (provided ρ c ′ α/d ) we are either in the first caseof the lemma or else(9.1) X − γ ∈ Spec c α (1 A ′′ ,β ρ ) | ((1 A − α )1 B ) ∧ ( γ ) | | (1 A ′′ dβ ρ ) ∧ ( − γ ) | > c α α ′ µ G ( B ) . We assume that we are not in the first case and proceed with analysis of Spec c α (1 A ′′ , β ρ )depending on where the mass in (9.1) concentrates. Let ǫ := cα / where c is an absoluteconstant such that(9.2) C c − τ − c τ − / whenever τ c. First we split into two cases.
Case.
At least half the mass in (9.1) is supported by those γ with − γ ∈ Spec ǫ (1 A ′′ , β ρ ) .Analysis of case. We proceed as in the proof of Lemma 8.2: by (the relative version of)the Chang bound (Lemma 4.6) we see Spec ǫ (1 A ′′ , β ρ ) has (1 , β ρ )-relative entropy e O ( α − ).Thus, by Corollary 6.4 there is a Bohr set B ′′ B ρ withrk( B ′′ ) rk( B ) + e O ( α − ) and µ G ( B ′′ ) > exp( − e O ( d + α − )) µ G ( B ρ )such that | − γ ( x ) | / x ∈ B ′′ and γ ∈ Spec ǫ (1 A ′′ , β ρ ) . It follows by Lemma 7.2 (provided ρ c ′ c α/ d ) that we are in the second case of thelemma. (cid:3) Case.
At least half the mass in (9.1) is supported by those γ with − γ Spec ǫ (1 A ′′ , β ρ ) .Analysis of case. By averaging there is some τ ∈ [ c α, ǫ ] such that the set of γ such that − γ is a member of V := { γ : 2 τ α ′ > | (1 A ′′ dβ ρ ) ∧ ( γ ) | > τ α ′ } supports at least Ω(log − α − ) of the mass in (9.1), whence X − γ ∈V | ((1 A − α )1 B ) ∧ ( γ ) | = e Ω( τ − α ) µ G ( B ) . Note that by definition of A ′′ we have that | (1 A ′′ d ( β ρ ∗ β ρρ ′ )) ∧ ( γ ) | = | (1 A ′′ dβ ρ ) ∧ ( γ ) | , whence V ⊂
Spec τ (1 A ′′ , β ρ ∗ β ρρ ′ ). However, the Bessel bound (Lemma 4.2) tells us thatSpec τ (1 A ′′ , β ρ ∗ β ρρ ′ ) has (1 , β ρ ∗ β ρρ ′ )-relative size O ( α − ) and so by monotonicity V has( c ′ c α, β ρρ ′ )-relative size O ( α − ).Now, apply Lemma 6.2 to V and the Bohr set B ρρ ′ to get a set Λ of k characters,( c ′ c α )-orthogonal w.r.t. β ρρ ′ (and hence µ := β ρρ ′ ∗ β ρρ ′ ), and an order-preservingmap φ : P (Λ) → P ( V ) such that φ (Λ) = V and for all Λ ′ ⊂ Λ and γ ∈ φ (Λ ′ ) we have(9.3) | − γ ( x ) | / x ∈ B ρρ ′ ρ ′′ ∧ B ′′′ ν where B ′′′ is the Bohr set with constant width function 2 and frequency set Λ ′ , and ρ ′′ =Ω( α/d log 2 α − ) and ν = Ω(1).We define a sub-additive weight w on P (Λ) by w (Λ ′ ) = X − γ ∈ φ (Λ ′ ) | ((1 A − α )1 B ) ∧ ( γ ) | , and a sequence of sets Λ , . . . , Λ r as follows. We begin with Λ := Λ, and at stage i if thereis some Λ ′ ⊂ Λ i with | Λ ′ | > | Λ i | / w (Λ ′ ) w (Λ) / log 2 k then we put Λ i +1 := Λ i \ Λ ′ . Since | Λ i +1 | | Λ i | / k steps. At this point we have a set Λ i such that every subset of size atleast | Λ i | / w (Λ i ) / log 2 k . On the other hand w (Λ i ) > (cid:18) − k (cid:19) ⌈ log k ⌉ w (Λ) = e Ω( τ − α ) µ G ( B ) . Once again we split into two cases.
Case. Λ i is small: | Λ i | τ − α − / .Analysis of case. By averaging there is a subset Λ ′ ⊂ Λ i of size at most α − such that w (Λ ′ ) > ⌊ α − ⌋| Λ i | w (Λ i ) = e Ω( α / ) µ G ( B ) . It follows from (9.3) that there is a Bohr set B ′′ B withrk( B ′′ ) rk( B ) + α − and µ G ( B ′′ ) > exp( − e O ( d + α − )) µ G ( B ρ )such that | − γ ( x ) | / x ∈ B ′′ and γ ∈ φ (Λ ′ ) . We apply Lemma 7.2 (provided ρ c ′ α/d for some absolute c ′ ) and are certainly in thethird case of the lemma. (cid:3) Case. Λ i is large: | Λ i | > τ − α − / .Analysis of case. We apply Lemma 4.7 with the set A ′′ , measures β ρ and µ , and parameter m = α − / to get two possibilites. Case.
There is a set Λ ′ ⊂ Λ i with | Λ ′ | > | Λ i | / such that Λ ′ has (1 , µ ) -relative entropy e O ( α − / ) .Analysis of case. By monotonicity Λ ′ has (1 , β ρρ ′ )-relative entropy e O ( α − / ), and so byLemma 6.3 applied to Λ ′ and the Bohr set B ρρ ′ that there is a ρ ′′′ = ( α/ d ) O (1) and ν ′ = Ω(1 /k ) such that | − γ ( x ) | ν ′′ for all x ∈ ( B ρρ ′ ρ ′′′ ∧ B ′′′′ ν ′ ) ′′ ν and γ ∈ Λ ′ N CERTAIN OTHER SETS OF INTEGERS 27 where B ′′′′ is a Bohr set with constant width function 2 and a frequency set of size e O ( α − / ).It follows that B ′′′ B ρρ ′ ρ ′′′ ∧ B ′′′′ ν ′ , and inserting this in (9.3) we get that for all γ ∈ φ (Λ ′ ), | − γ ( x ) | / x ∈ B ρρ ′ min { ρ ′′ ,ρ ′′′ ν } ∧ B ν ′ ν . We conclude that there is a Bohr set B ′′ B ρ withrk( B ′′ ) rk( B ) + e O ( α − / ) and µ G ( B ′′ ) > exp( − e O ( d + α − / )) µ G ( B ρ )such that | − γ ( x ) | / x ∈ B ′′ and γ ∈ φ (Λ ′ ). On the other hand w (Λ ′ ) > w (Λ i )log 2 k = e Ω( τ − α ) µ G ( B ) = e Ω( α / ) µ G ( B ) . It follows from Lemma 7.2 that we are in the third case of the lemma provided ρ c ′′ α/d for some absolute c ′′ . (cid:3) Case.
There is a set of characters Λ ′′′ of O ( √ m ) characters and a function ω ′ : Λ ′′′ → D such that h A ′′ , p ω ′ , Λ ′′′ i L ( β ρ ) > α ′ (1 + τ √ m/ Z p , Λ ′′′ dµ. Analysis of case.
Apply Lemma 7.3 and we are in the final case of the lemma. (cid:3)(cid:3)(cid:3)
The proof is complete. (cid:3)
Proof of Theorem 9.1.
We proceed as before to construct a sequence of regular Bohr sets B ( i ) of rank k i and dimension d i = O ( k i ). We write α i := k A ∗ β ( i ) k L ∞ ( µ G ) , put B (0) := G and suppose that we have defined B ( i ) . By regularity we have that(1 A ∗ β ( i ) ρ i + 1 A ∗ β ( i ) ρ i ρ ′ i ) ∗ β ( i ) ( x i ) = 2 α i + O ( ρ i k i )for some x i . Pick ρ i = Ω( α / i /k i ) and then ρ ′ i = Ω( α / i /k i ) such that B ( i ) ρ i and B ( i ) ρ ′ i ρ i areregular, (1 A ∗ β ( i ) ρ + 1 A ∗ β ( i ) ρ i ρ ′ i ) ∗ β ( i ) ( x i ) > α i (1 − c α / i / , and ρ ′ i c ′ α i /d i . Thus there is some x ′ i such that1 A ∗ β ( i ) ρ ( x ′ i ) + 1 A ∗ β ( i ) ρ i ρ ′ i ( x ′ i ) > α i (1 − c α / i / . If one of the two terms is at least α i (1 + c α / i / B ( i +1) to be B ( i ) ρ i or B ( i ) ρ ′ i ρ i respectively. Otherwise we see that A i := ( x ′ i − A ) ∩ B ( i ) ρ i has β ( i ) ρ i ( A i ) > α i (1 − c α / i / , and A ′ i := ( x ′ i − A ) ∩ B ( i ) ρ ′ i ρ i has β ( i ) ρ ′ i ρ i ( A ′ i ) > α i (1 − c α / i / . By Lemma 9.2 we see that at least one of the following occurs:(i) we stop the iteration with T (1 A , A , A ) = Ω( α i µ G ( B ( i ) ρ i ) µ G ( B ( i ) ρ ′ i ρ i ));(ii) there is a regular Bohr set B ′ with rk( B ′ ) k i + e O ( α − / i ) and µ G ( B ′ ) > exp( − e O ( k i + α − / i )) µ G ( B ( i ) ρ ′ i ρ i ) > exp( − e O ( k i + α − / i )) µ G ( B ( i ) )such that k A ∗ β ′ k L ∞ ( µ G ) = e Ω( α / );(iii) there is a regular Bohr set B ( i +1) with k i +1 k i + e O ( α − / i ) and, µ G ( B ( i +1) ) > exp( − e O ( k i + α − / i )) µ G ( B ( i ) ρ ′ i ρ i ) > exp( − e O ( k i + α − / i )) µ G ( B ( i ) ) , such that α i +1 > α i (1 + c α / i / B ( i +1) with k i +1 k i + e O ( α − i ) and, µ G ( B ( i +1) ) > exp( − e O ( k i + α − i )) µ G ( B ( i ) ρ ′ i ρ i ) > exp( − e O ( k i + α − i )) µ G ( B ( i ) ) , such that α i +1 > α i (1 + c / T (1 A , A , A ) > exp( − e O ( k i + α − / i )) µ G ( B ( i ) ) . Otherwise we can be in the third case at most e O ( α − / ) times and in the fourth case O (log 2 α − ) times. It follows that the iteration must terminate at some step i = e O ( α − / ),where we shall have k i = e O ( α − ), and µ G ( B ( i ) ) > exp( − e O ( α − / )) µ G ( B ) , from which the result follows. (cid:3) Acknowledgement
The author should like to thank an anonymous referee for useful comments.
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Department of Pure Mathematics and Mathematical Statistics, University of Cam-bridge, Wilberforce Road, Cambridge CB3 0WB, England
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