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On Fourier coefficients of sets with small doubling

I.D. Shkredov

Annotation.

Let AA be a subset of a finite abelian group such that AA has a small difference set AAA-A and the density of AA is small. We prove that, counter–intuitively, the smallness (in terms of |AA||A-A|) of the Fourier coefficients of AA guarantees that AA is correlated with a large Bohr set. Our bounds on the size and the dimension of the resulting Bohr set are close to exact.

1 Introduction

Let 𝐆{\mathbf{G}} be a finite abelian group and 𝐆^\widehat{{\mathbf{G}}} its dual group. For any function f:𝐆f:{\mathbf{G}}\to\mathbb{C} and χ𝐆^\chi\in\widehat{{\mathbf{G}}} define the Fourier transform of ff at χ\chi by the formula

f^(χ)=g𝐆f(g)χ(g)¯.\widehat{f}(\chi)=\sum_{g\in{\mathbf{G}}}f(g)\overline{\chi(g)}\,. (1)

This paper considers the case when the function ff has a special form, namely, ff the characteristic function of a set A𝐆A\subseteq{\mathbf{G}} and we want to study the quantity

(A):=maxχχ0|A^(χ)|,\mathcal{M}(A):=\max_{\chi\neq\chi_{0}}|\widehat{A}(\chi)|\,, (2)

where by χ0\chi_{0} we have denoted the principal character of 𝐆^\widehat{{\mathbf{G}}} and we use the same capital letter to denote a set A𝐆A\subseteq{\mathbf{G}} and its characteristic function A:𝐆{0,1}A:{\mathbf{G}}\to\{0,1\}. It is well–known that (A){\mathcal{M}}(A) is directly related to the uniform distribution properties of the set AA (see, e.g., [13], [28]). Moreover, we impose an additional property on the set AA, namely, that it is a set with small doubling, i.e. we consider sets AA for which the ratio |AA|/|A||A-A|/|A| is small. Recall that given two sets A,B𝐆A,B\subseteq{\mathbf{G}} the sumset of AA and BB is defined as

A+B:={a+b:aA,bB}.A+B:=\{a+b~{}:~{}a\in{A},\,b\in{B}\}\,.

In a similar way we define the difference sets AAA-A and the interated sumsets, e.g., 2AA2A-A is A+AAA+A-A. In terms of the difference sets |AA|:=K|A||A-A|:=K|A| it is easy to see that

2(A)(1o(1))|A|2K{\mathcal{M}}^{2}(A)\geqslant(1-o(1))\cdot\frac{|A|^{2}}{K} (3)

and estimate (3) gives us a simple lower bound for the quantity (A){\mathcal{M}}(A). In general, bound (3) is tight, but the situation changes dramatically depending on the density δ:=|A|/|𝐆|\delta:=|A|/|{\mathbf{G}}| of AA. For example, in [11, Lemma 4.1] (also, see [24] and [22, Propostion 6.1]) it was proved that

2(A)(1o(1))|A|2,{\mathcal{M}}^{2}(A)\geqslant(1-o(1))\cdot|A|^{2}\,, (4)

provided δ=exp(Ω(K))\delta=\exp(-\Omega(K)). Thus, if the density is exponentially small relative to KK, then (A){\mathcal{M}}(A) is close to its maximum value |A||A|. On the other hand, it is easy to see we always have δK1\delta\leqslant K^{-1} and for a random set A𝐆A\subseteq{\mathbf{G}} one has δK1\delta\gg K^{-1}. Therefore, one of the reasonable questions here is the following. Assume that δKd\delta\sim K^{-d}, where d>1d>1, that is, the dependence of δ\delta on KK is polynomial. What non–trivial properties do the Fourier coefficients of set AA have in this case? Such a problem arises naturally in connection with the famous Freiman 3k43k-4 theorem in 𝔽p\mathbb{F}_{p}, see, e.g., [7], [17], [18]. In particular, in [18, Theorem 6] it was proved that if δK3\delta\ll K^{-3}, then

2(A)|A|2K(1+κ0),{\mathcal{M}}^{2}(A)\geqslant\frac{|A|^{2}}{K}(1+\kappa_{0})\,, (5)

where κ0>0\kappa_{0}>0 is an absolute constant. Thus, a non-trivial lower bound for (A){\mathcal{M}}(A) exists in any abelian group, and hence the Fourier coefficients of sets with small doubling have some interesting properties. Other results on the quantity (A){\mathcal{M}}(A) and its relation to sets AxA_{x} (see Section 2 below) were obtained in [27]. The main result of this paper (all required definitions can be found in Sections 2, 4) concerns an even broader regime δK2\delta\ll K^{-2} and gives us a new structural property of sets with small doubling and small Fourier coefficients.

Theorem 1

Let 𝐆{\mathbf{G}} be a finite abelian group, A𝐆A\subseteq{\mathbf{G}} be a set, |A|=δ|𝐆||A|=\delta|{\mathbf{G}}|, |AA|=K|A||A-A|=K|A|, and 1MK1\leqslant M\leqslant K be a parameter. Suppose that

100K2δ1.100K^{2}\delta\leqslant 1\,. (6)

Then either there is x0x\neq 0 such that

|A^(x)|2>M|A|2K,|\widehat{A}(x)|^{2}>\frac{M|A|^{2}}{K}\,, (7)

or for any BAB\subseteq A, |B||A|=δ|𝐆||B|\gg|A|=\delta|{\mathbf{G}}| there exists a regular Bohr set {\mathcal{B}}_{*} and z𝐆z\in{\mathbf{G}} such that

|B(+z)|||8M,|B\cap({\mathcal{B}}_{*}+z)|\geqslant\frac{|{\mathcal{B}}_{*}|}{8M}\,, (8)

and

dim()M2(log(δ1K)+log2M),{\rm dim}({\mathcal{B}}_{*})\ll M^{2}\left(\log(\delta^{-1}K)+\log^{2}M\right)\,, (9)

as well as

|||𝐆|exp(O(dim()log(Mdim()))).|{\mathcal{B}}_{*}|\gg|{\mathbf{G}}|\cdot\exp(-O({\rm dim}({\mathcal{B}}_{*})\log(M{\rm dim}({\mathcal{B}}_{*}))))\,. (10)

Let us make a few remarks regarding Theorem 10. First of all, the bounds (8)—(10) hold for any dense subset of our set AA, which means that AA has a very rigid structure (for example, see the second part of Corollary 9 below). Secondly, the dependencies on parameters in (8)—(10) have polynomial nature, which distinguishes them from the best modern results on the structure of sets with small doubling, see [22], [23]. Also, it appears that Theorem 10 does not depend on the recent breakthrough progress concerning Polynomial Freiman–Ruzsa Conjecture [8], [9]. Thirdly, the inversion of bound (7) guarantees the existence of a large intersection of AA and a translation of a Bohr set {\mathcal{B}}_{*}, see inequality (8). This is quite surprising, because usually we have the opposite picture: small Fourier coefficients help to avoid such structural objects as Bohr sets. Finally, our proof drastically differs from the arguments of [11], [24], [22], which give us inequality (4). Instead of using almost periodicity of multiple convolutions [6] and the polynomial growth of sumsets, we apply some observations from the higher energies method, see papers [25], [27]. In particular, our structural subset of AAA-A has a different nature (it resembles some steps of the proof of [8], [9] and even older constructions of Schoen, see [26, Examples 5,6]). Let us consider the following motivating example (see, e.g., [26, Example 5]).

Example 2

(H+ΛH+\Lambda sets). Let 𝐆=𝔽2n{\mathbf{G}}=\mathbb{F}_{2}^{n}, H𝐆H\leqslant{\mathbf{G}} be the space spanned by the first k<nk<n coordinate vectors, ΛH\Lambda\subseteq H^{\perp} be a basis, |Λ|=2K|\Lambda|=2K, KK\to\infty, and A:=H+ΛA:=H+\Lambda. In particular, |A|=|H||Λ||A|=|H||\Lambda| and |AA|=(1+o(1))K|A||A-A|=(1+o(1))\cdot K|A|. Further for sHs\in H one has As=AA_{s}=A, but it is easy to check that for s(AA)Hs\in(A-A)\setminus H each AsA_{s} is a disjoint union of two shifts of HH. Hence

𝖤(A)|A|3/KsH|As|2s(AA)H|As|2\mathsf{E}(A)\sim|A|^{3}/K\sim\sum_{s\in H}|A_{s}|^{2}\sim\sum_{s\in(A-A)\setminus H}|A_{s}|^{2}

(the existence of two “dual” subsets of AAA-A where 𝖤(A)\mathsf{E}(A) is achieved is in fact a general result, see [2] and also [26]), but for k>2k>2 the higher energies are supported on the set of measure zero (namely, HH) in the sense that

𝖤k(A):=s|As|k=(1+ok(1))sH|As|k.\mathsf{E}_{k}(A):=\sum_{s}|A_{s}|^{k}=(1+o_{k}(1))\cdot\sum_{s\in H}|A_{s}|^{k}\,. (11)

Moreover if one considers the function φk(s)=|As|k,\varphi_{k}(s)=|A_{s}|^{k}\,, then it is easy to see that

φ^k(χ):=s|As|kχ(s)=(1+ok(1))sH|As|kχ(s)=(1+ok(1))|H|kH^(χ),\widehat{\varphi}_{k}(\chi):=\sum_{s}|A_{s}|^{k}\chi(s)=(1+o_{k}(1))\cdot\sum_{s\in H}|A_{s}|^{k}\chi(s)=(1+o_{k}(1))\cdot|H|^{k}\widehat{H}(\chi)\,,

where χ\chi is an arbitrary additive character on 𝐆{\mathbf{G}}. Thus, as kk goes to infinity, the Fourier transform φ^k(χ)\widehat{\varphi}_{k}(\chi) becomes very regular, and this gives us a new way to extract the structure piece from sets with small doubling.

Roughly speaking, the same Example 2 shows that Theorem 10 is tight (also, see more rigorous Example 20 below). Indeed, it is easy to see that A^=H^Λ^\widehat{A}=\widehat{H}\cdot\widehat{\Lambda} and assuming that Λ\Lambda is a sufficiently small set and 2(Λ)|Λ|K{\mathcal{M}}^{2}(\Lambda)\ll|\Lambda|\ll K, we get 2(A)=|H|22(Λ)|A|2/K{\mathcal{M}}^{2}(A)=|H|^{2}{\mathcal{M}}^{2}(\Lambda)\ll|A|^{2}/K. Thus all conditions of Theorem 10 hold but the obvious subspace in AAA-A is of course HH and we have codim(H)=log(δ1K){\rm codim}(H)=\log(\delta^{-1}K), which coincides with (9) up to multiple constants.

2 Definitions and notation

Let 𝐆{\mathbf{G}} be a finite abelian group and we denote the cardinality of 𝐆{\mathbf{G}} by NN. Given a set A𝐆A\subseteq{\mathbf{G}} and a positive integer kk, let us put

Δk(A):={(a,a,,a):aA}𝐆k.\Delta_{k}(A):=\{(a,a,\dots,a)~{}:~{}a\in A\}\subseteq{\mathbf{G}}^{k}\,.

Now we have

AA:={ab:a,bA}={x𝐆:A(A+x)},A-A:=\{a-b~{}:~{}a,b\in A\}=\{x\in{\mathbf{G}}~{}:~{}A\cap(A+x)\neq\emptyset\}\,, (12)

and for xAAx\in A-A we denote by AxA_{x} the intersection A(A+x)A\cap(A+x). The useful inclusion of Katz–Koester [14] is the following

B+Ax(A+B)x.B+A_{x}\subseteq(A+B)_{x}\,. (13)

A natural generalization of the last formula in (12) is the set

{(x1,,xk)𝐆k:A(A+x1)(A+xk)}=AkΔk(A),\{(x_{1},\dots,x_{k})\in{\mathbf{G}}^{k}~{}:~{}A\cap(A+x_{1})\cap\dots\cap(A+x_{k})\neq\emptyset\}=A^{k}-\Delta_{k}(A)\,, (14)

which is called the higher difference set (see [25]). For any two sets A,B𝐆A,B\subseteq{\mathbf{G}} the additive energy of AA and BB is defined by

𝖤(A,B)=𝖤(A,B)=|{(a1,a2,b1,b2)A×A×B×B:a1b1=a2b2}|.\mathsf{E}(A,B)=\mathsf{E}(A,B)=|\{(a_{1},a_{2},b_{1},b_{2})\in A\times A\times B\times B~{}:~{}a_{1}-b_{1}=a_{2}-b_{2}\}|\,.

If A=BA=B, then we simply write 𝖤(A)\mathsf{E}(A) for 𝖤(A,A)\mathsf{E}(A,A). Also, one can define the higher energy (see [25] and another formula (19) below)

𝖤k(A)=|{(a1,,ak,a1,,ak)A2k:a1a1==akak}|.\mathsf{E}_{k}(A)=|\{(a_{1},\dots,a_{k},a^{\prime}_{1},\dots,a^{\prime}_{k})\in A^{2k}~{}:~{}a_{1}-a^{\prime}_{1}=\dots=a_{k}-a^{\prime}_{k}\}|\,. (15)

Let 𝐆^\widehat{{\mathbf{G}}} be its dual group. For any function f:𝐆f:{\mathbf{G}}\to\mathbb{C} and χ𝐆^\chi\in\widehat{{\mathbf{G}}} we define its Fourier transform using the formula (1). The Parseval identity is

Ng𝐆|f(g)|2=χ𝐆^|f^(χ)|2.N\sum_{g\in{\mathbf{G}}}|f(g)|^{2}=\sum_{\chi\in\widehat{{\mathbf{G}}}}\big{|}\widehat{f}(\chi)\big{|}^{2}\,. (16)

Given a function ff and ε(0,1]\varepsilon\in(0,1] define the spectrum of ff as

Specε(f)={χ𝐆^:|f^(χ)|εf1}.{\rm Spec\,}_{\varepsilon}(f)=\{\chi\in\widehat{{\mathbf{G}}}~{}:~{}|\widehat{f}(\chi)|\geqslant\varepsilon\|f\|_{1}\}\,. (17)

We need Chang’s lemma [5].

Lemma 3

Let f:𝐆f:{\mathbf{G}}\to\mathbb{C} be a function and ε(0,1]\varepsilon\in(0,1] be a parameter. Then

dim(Specε(f))ε2logf22Nf12.{\rm dim}({\rm Spec\,}_{\varepsilon}(f))\ll\varepsilon^{-2}\log\frac{\|f\|_{2}^{2}N}{\|f\|_{1}^{2}}\,.

By measure we mean any non–negative function μ\mu on 𝐆{\mathbf{G}} such that g𝐆μ(g)=1\sum_{g\in{\mathbf{G}}}\mu(g)=1. If f,g:𝐆f,g:{\mathbf{G}}\to\mathbb{C} are some functions, then

(fg)(x):=y𝐆f(y)g(xy) and (fg)(x):=y𝐆f(y)g(y+x).(f*g)(x):=\sum_{y\in{\mathbf{G}}}f(y)g(x-y)\quad\mbox{ and }\quad(f\circ g)(x):=\sum_{y\in{\mathbf{G}}}f(y)g(y+x)\,.

One has

fg^=f^g^.\widehat{f*g}=\widehat{f}\cdot\widehat{g}\,. (18)

Having a function f:𝐆f:{\mathbf{G}}\to\mathbb{C} and a positive integer k>1k>1, we write f(k)(x)=(fff)(x)f^{(k)}(x)=(f\circ f\circ\dots\circ f)(x), where the convolution \circ is taken k1k-1 times. For example, A(4)(0)=𝖤(A)A^{(4)}(0)=\mathsf{E}(A) and for k2k\geqslant 2 one has

𝖤k(A)=x(AA)k(x)=x1,,xk1(Ak1Δk1(A))2(x1,,xk1).\mathsf{E}_{k}(A)=\sum_{x}(A\circ A)^{k}(x)=\sum_{x_{1},\dots,x_{k-1}}(A^{k-1}\circ\Delta_{k-1}(A))^{2}(x_{1},\dots,x_{k-1})\,. (19)

A finite set Λ𝐆\Lambda\subseteq{\mathbf{G}} is called dissociated if any equality of the form

λΛελλ=0, where ελ{0,±1},λΛ\sum_{\lambda\in\Lambda}\varepsilon_{\lambda}\lambda=0\,,\quad\quad\mbox{ where }\quad\quad\varepsilon_{\lambda}\in\{0,\pm 1\}\,,\quad\quad\forall\lambda\in\Lambda

implies ελ=0\varepsilon_{\lambda}=0 for all λΛ\lambda\in\Lambda. Let dim(A){\rm dim}(A) be the size of the largest dissociated subset of AA and we call dim(A){\rm dim}(A) the additive dimension of AA. Let us denote all combinations of the form {λΛελλ}ελ{0,±1}\{\sum_{\lambda\in\Lambda}\varepsilon_{\lambda}\lambda\}_{\varepsilon_{\lambda}\in\{0,\pm 1\}} as Span(Λ){\rm Span\,}(\Lambda).


We need the generalized triangle inequality [25, Theorem 7].

Lemma 4

Let k1,k2k_{1},k_{2} be positive integers, W𝐆k1W\subseteq{\mathbf{G}}^{k_{1}}, Y𝐆k2Y\subseteq{\mathbf{G}}^{k_{2}} and X,Z𝐆X,Z\subseteq{\mathbf{G}}. Then

|W×X||YΔk2(Z)||W×Y×ZΔk1+k2+1(X)|.|W\times X||Y-\Delta_{k_{2}}(Z)|\leqslant|W\times Y\times Z-\Delta_{k_{1}+k_{2}+1}(X)|\,. (20)

The signs \ll and \gg are the usual Vinogradov symbols. All logarithms are to base ee.

3 The model case

In this section we follow the well–known logic (see [10]) and first consider the simplest case 𝐆=𝔽2n{\mathbf{G}}=\mathbb{F}_{2}^{n}, where the technical difficulties are minimal, and the case of general groups will be considered later.

We need a generalization of a result from [27, Remark 6] (also see the proof of [27, Theorem 4]). For the convenience of the reader, we recall our argument.

Lemma 5

Let A,B𝐆A,B\subseteq{\mathbf{G}} be sets, |AA|=K|A||A-A|=K|A| and k2k\geqslant 2 be an integer. Then

𝖤k(B)𝖤k(A,A+B)|A|2k+1|B|2kK.\mathsf{E}_{k}(B)\mathsf{E}^{k}(A,A+B)\geqslant\frac{|A|^{2k+1}|B|^{2k}}{K}\,. (21)

P r o o f.  Let S=A+BS=A+B. In view of inclusion (13), we have

𝖤(A,S)=x|Ax||Sx|x|Ax||B+Ax|.\mathsf{E}(A,S)=\sum_{x}|A_{x}||S_{x}|\geqslant\sum_{x}|A_{x}||B+A_{x}|\,.

Using Lemma 20, we see that for any Z𝐆lZ\subseteq{\mathbf{G}}^{l} the following holds

|Z|𝒟k:=|Z||Bk1Δk1(B)||BZ|k|Z|\mathcal{D}_{k}:=|Z||B^{k-1}-\Delta_{k-1}(B)|\leqslant|B-Z|^{k}

and therefore by the Hölder inequality one has

𝖤(A,S)𝒟k1/kx|Ax|1+1/k𝒟k1/k(x|Ax|)1+1/k|AA|1/k.\mathsf{E}(A,S)\geqslant\mathcal{D}^{1/k}_{k}\sum_{x}|A_{x}|^{1+1/k}\geqslant\mathcal{D}^{1/k}_{k}\left(\sum_{x}|A_{x}|\right)^{1+1/k}|A-A|^{-1/k}\,. (22)

Thus

𝖤k(A,S)𝒟k|A|2k+1K1.\mathsf{E}^{k}(A,S)\geqslant\mathcal{D}_{k}|A|^{2k+1}K^{-1}\,. (23)

On the other hand, applying the Cauchy–Schwarz inequality and formula (19), we derive

|B|2k=(y𝐆k1(Bk1Δk1(B)(y))2𝒟ky𝐆k1(Bk1Δk1(B)2(y)=𝒟k𝖤k(B).|B|^{2k}=\left(\sum_{y\in{\mathbf{G}}^{k-1}}(B^{k-1}\circ\Delta_{k-1}(B)(y)\right)^{2}\leqslant\mathcal{D}_{k}\sum_{y\in{\mathbf{G}}^{k-1}}(B^{k-1}\circ\Delta_{k-1}(B)^{2}(y)=\mathcal{D}_{k}\mathsf{E}_{k}(B)\,. (24)

Combining (23) and (24), one obtains

𝖤k(B)𝖤k(A,S)𝖤k(B)𝒟k|A|2k+1K1|B|2k|A|2k+1K1\mathsf{E}_{k}(B)\mathsf{E}^{k}(A,S)\geqslant\mathsf{E}_{k}(B)\mathcal{D}_{k}|A|^{2k+1}K^{-1}\geqslant|B|^{2k}|A|^{2k+1}K^{-1} (25)

as required. \hfill\Box


Now we are ready to prove our main technical proposition.

Proposition 6

Let 𝐆=𝔽2n{\mathbf{G}}=\mathbb{F}_{2}^{n}, A,B𝐆A,B\subseteq{\mathbf{G}} be sets, |A|=δN|A|=\delta N, |B|=ω|A||B|=\omega|A|, |A+B|=K|A||A+B|=K|A|, |AA|=K|A||A-A|=K^{\prime}|A|, and 0<MK0<M\leqslant K, 0<MK0<M^{\prime}\leqslant K^{\prime}, κ>0\kappa>0, ζ(0,1)\zeta\in(0,1), 1<TM(M+κ)ω11<T\leqslant M^{\prime}(M+\kappa)\omega^{-1} be some parameters. Suppose that

2(A)M|A|2K,𝖤(B)M|B|3K,{\mathcal{M}}^{2}(A)\leqslant\frac{M|A|^{2}}{K},\,\quad\quad\mathsf{E}(B)\leqslant\frac{M^{\prime}|B|^{3}}{K^{\prime}}\,, (26)

and

|A+B|2κ|A|N.|A+B|^{2}\leqslant\kappa|A|N\,. (27)

Then there is 𝐆\mathcal{L}\leqslant{\mathbf{G}} (\mathcal{L} depends on BB only) and z𝐆z\in{\mathbf{G}} such that

|B(+z)|(1ζ)ω||T(M+κ),|B\cap(\mathcal{L}+z)|\geqslant\frac{(1-\zeta)\omega|\mathcal{L}|}{T(M+\kappa)}\,,

and

codim()(ωζ)2T2(M+κ)2(log(δ1K)+logT(M(M+κ)ω1)log((M+κ)ω1)).{\rm codim}(\mathcal{L})\ll(\omega\zeta)^{-2}T^{2}(M+\kappa)^{2}\cdot\left(\log(\delta^{-1}K^{\prime})+\log_{T}(M^{\prime}(M+\kappa)\omega^{-1})\cdot\log((M+\kappa)\omega^{-1})\right)\,.

P r o o f.  Let a=|A|a=|A|, b=|B|=βN=ωδNb=|B|=\beta N=\omega\delta N, S=A+BS=A+B and 𝖤k=𝖤k(B)\mathsf{E}_{k}=\mathsf{E}_{k}(B). Thanks to the condition (27) and Parseval identity (16), we have

𝖤(A,S)a2|A+B|2N+Ma3(M+κ)a3.\mathsf{E}(A,S)\leqslant\frac{a^{2}|A+B|^{2}}{N}+Ma^{3}\leqslant(M+\kappa)a^{3}\,.

Using Lemma 21, we obtain for all integers k2k\geqslant 2 that

𝖤k+1b2k+2K(M+κ)k+1ak=bk+2ωkK(M+κ)k+1.\mathsf{E}_{k+1}\geqslant\frac{b^{2k+2}}{K^{\prime}(M+\kappa)^{k+1}a^{k}}=\frac{b^{k+2}\omega^{k}}{K^{\prime}(M+\kappa)^{k+1}}\,. (28)

Suppose that for any k2k\geqslant 2 the following holds

𝖤k+1b𝖤kM,\mathsf{E}_{k+1}\leqslant\frac{b\mathsf{E}_{k}}{M_{*}}\,,

where M1M_{*}\geqslant 1 is a parameter. Then thanks to (26), we derive

𝖤k+1bk1𝖤2Mk1Mbk+2KMk1,\mathsf{E}_{k+1}\leqslant\frac{b^{k-1}\mathsf{E}_{2}}{M^{k-1}_{*}}\leqslant\frac{M^{\prime}b^{k+2}}{K^{\prime}M^{k-1}_{*}}\,, (29)

and using (28), we have

M(M+κ)k+1ωkMk1.M^{\prime}(M+\kappa)^{k+1}\geqslant\omega^{k}M^{k-1}_{*}\,.

Put M=ω1(M+κ)TM_{*}=\omega^{-1}(M+\kappa)T. It gives us

M(M+κ)2ωTk1M^{\prime}(M+\kappa)^{2}\geqslant\omega T^{k-1}

and we obtain a contradiction if kk0:=10logT(M(M+κ)ω1)+10k\geqslant k_{0}:=\lceil 10\log_{T}(M^{\prime}(M+\kappa)\omega^{-1})\rceil+10, say. Thus there is 2kk02\leqslant k\leqslant k_{0} such that

𝖤k+1b𝖤kM.\mathsf{E}_{k+1}\geqslant\frac{b\mathsf{E}_{k}}{M_{*}}\,. (30)

Consider the function φ(x)=|Bx|k\varphi(x)=|B_{x}|^{k}. Clearly, φ^0\widehat{\varphi}\geqslant 0 (use, for example, formula (18)) and φ1=𝖤k\|\varphi\|_{1}=\mathsf{E}_{k}. In terms of Fourier transform we can rewrite inequality (30) as

𝖤k+1=1Nξ|B^(ξ)|2φ^(ξ)b𝖤kM=ωb𝖤kT(M+κ).\mathsf{E}_{k+1}=\frac{1}{N}\sum_{\xi}|\widehat{B}(\xi)|^{2}\widehat{\varphi}(\xi)\geqslant\frac{b\mathsf{E}_{k}}{M_{*}}=\frac{\omega b\mathsf{E}_{k}}{T(M+\kappa)}\,.

Now we use the parameter ζ\zeta and derive

1NξSpecζ/M(φ)|B^(ξ)|2φ^(ξ)(1ζ)ωb𝖤kT(M+κ).\frac{1}{N}\sum_{\xi\in{\rm Spec\,}_{\zeta/M_{*}}(\varphi)}|\widehat{B}(\xi)|^{2}\widehat{\varphi}(\xi)\geqslant\frac{(1-\zeta)\omega b\mathsf{E}_{k}}{T(M+\kappa)}\,. (31)

Let ΛSpecζ/M(φ)\Lambda\subseteq{\rm Spec\,}_{\zeta/M_{*}}(\varphi) be a dissociated set such that |Λ|=dim(Specζ/M(φ))|\Lambda|={\rm dim}({\rm Spec\,}_{\zeta/M_{*}}(\varphi)). Put =SpanΛ\mathcal{L}^{\perp}={\rm Span\,}\Lambda. Then

(1ζ)ωb𝖤kT(M+κ)1Nξ|B^(ξ)|2φ^(ξ)𝖤kNξ|B^(ξ)|2=𝖤k||x(BB)(x)(x).\frac{(1-\zeta)\omega b\mathsf{E}_{k}}{T(M+\kappa)}\leqslant\frac{1}{N}\sum_{\xi\in\mathcal{L}^{\perp}}|\widehat{B}(\xi)|^{2}\widehat{\varphi}(\xi)\leqslant\frac{\mathsf{E}_{k}}{N}\sum_{\xi\in\mathcal{L}^{\perp}}|\widehat{B}(\xi)|^{2}=\frac{\mathsf{E}_{k}}{|\mathcal{L}|}\sum_{x}(B\circ B)(x)\mathcal{L}(x)\,. (32)

By the pigeonhole principle there is zBz\in B such that

|B(+z)|(1ζ)ω||T(M+κ).|B\cap(\mathcal{L}+z)|\geqslant\frac{(1-\zeta)\omega|\mathcal{L}|}{T(M+\kappa)}\,. (33)

Using the Chang Lemma 3 and estimate (28), we derive

codim()ζ2M2log(φ22Nφ12)(ωζ)2T2(M+κ)2log(bkN𝖤k){\rm codim}(\mathcal{L})\ll\zeta^{-2}M^{2}_{*}\log\left(\frac{\|\varphi\|_{2}^{2}N}{\|\varphi\|^{2}_{1}}\right)\ll(\omega\zeta)^{-2}T^{2}(M+\kappa)^{2}\log\left(\frac{b^{k}N}{\mathsf{E}_{k}}\right) (34)
(ωζ)2T2(M+κ)2log(K(M+κ)kβωk1).\ll(\omega\zeta)^{-2}T^{2}(M+\kappa)^{2}\log\left(\frac{K^{\prime}(M+\kappa)^{k}}{\beta\omega^{k-1}}\right)\,. (35)

Recalling that kk0k\leqslant k_{0}, we finally obtain

codim()(ωζ)2T2(M+κ)2{\rm codim}(\mathcal{L})\ll(\omega\zeta)^{-2}T^{2}(M+\kappa)^{2}\cdot
(log(δ1K)+logT(M(M+κ)ω1)log((M+κ)ω1)).\left(\log(\delta^{-1}K^{\prime})+\log_{T}(M^{\prime}(M+\kappa)\omega^{-1})\cdot\log((M+\kappa)\omega^{-1})\right)\,. (36)

This completes the proof. \hfill\Box

Remark 7

Suppose that A=BA=B. Then in terms of Proposition 6 one has

𝖤(A)|A|4N+2(A)|A||A|4N+M|A|3KM|A|3K(1+δKM)M|A|3K(1+κKM).\mathsf{E}(A)\leqslant\frac{|A|^{4}}{N}+{\mathcal{M}}^{2}(A)|A|\leqslant\frac{|A|^{4}}{N}+\frac{M|A|^{3}}{K}\leqslant\frac{M|A|^{3}}{K}\left(1+\frac{\delta K}{M}\right)\leqslant\frac{M|A|^{3}}{K}\left(1+\frac{\kappa}{KM}\right)\,.

Thus M(1+κKM)MM^{\prime}\leqslant\left(1+\frac{\kappa}{KM}\right)M.

We derive some consequences of Proposition 6. Let us start with the case when (A){\mathcal{M}}(A) is really small, namely, 2(A)(2ε)|A|2/K{\mathcal{M}}^{2}(A)\leqslant(2-\varepsilon)|A|^{2}/K.

Corollary 8

Let 𝐆=𝔽2n{\mathbf{G}}=\mathbb{F}_{2}^{n}, A𝐆A\subseteq{\mathbf{G}} be a set, |A|=δN|A|=\delta N, |AA|=K|A||A-A|=K|A|, and ε(0,1)\varepsilon\in(0,1) be a parameter. Suppose that

100K2δε.100K^{2}\delta\leqslant\varepsilon\,.

Then either there is x0x\neq 0 such that

|A^(x)|2(2ε)|A|2K,|\widehat{A}(x)|^{2}\geqslant\frac{(2-\varepsilon)|A|^{2}}{K}\,,

or there exists 𝐆\mathcal{L}\leqslant{\mathbf{G}} with AA\mathcal{L}\subseteq A-A and

codim()ε2log(δ1K)+ε3.{\rm codim}(\mathcal{L})\ll\varepsilon^{-2}\log(\delta^{-1}K)+\varepsilon^{-3}\,.

P r o o f.  We apply Proposition 6 with B=AB=-A, ω=1\omega=1, M=2εM=2-\varepsilon, M=(1+κKM)MM+κM^{\prime}=\left(1+\frac{\kappa}{KM}\right)M\leqslant M+\kappa (see Remark 7), T=1+κT=1+\kappa, and κ=ζ=ε/100\kappa=\zeta=\varepsilon/100, say. Thus we find 𝐆\mathcal{L}\leqslant{\mathbf{G}} and z𝐆z\in{\mathbf{G}} such that

|A(+z)|(1ζ)||T(M+κ)||(12+ε8),|A\cap(\mathcal{L}+z)|\geqslant\frac{(1-\zeta)|\mathcal{L}|}{T(M+\kappa)}\geqslant|\mathcal{L}|\left(\frac{1}{2}+\frac{\varepsilon}{8}\right)\,, (37)

and

codim()ζ2T2(M+κ)2(log(δ1K)+logT(M(M+κ))log(M+κ)){\rm codim}(\mathcal{L})\ll\zeta^{-2}T^{2}(M+\kappa)^{2}\cdot\left(\log(\delta^{-1}K)+\log_{T}(M^{\prime}(M+\kappa))\cdot\log(M+\kappa)\right)
ε2log(δ1K)+ε3.\ll\varepsilon^{-2}\log(\delta^{-1}K)+\varepsilon^{-3}\,.

The inequality (37) implies that AA\mathcal{L}\subseteq A-A. This completes the proof. \hfill\Box


Now we are ready to obtain our structural result for sets with small doubling and small (A){\mathcal{M}}(A). Given two sets A,B𝐆A,B\subseteq{\mathbf{G}} we write ABA\dotplus B if |A+B|=|A||B||A+B|=|A||B|.

Corollary 9

Let 𝐆=𝔽2n{\mathbf{G}}=\mathbb{F}_{2}^{n}, A𝐆A\subseteq{\mathbf{G}} be a set, |A|=δN|A|=\delta N, |AA|=K|A||A-A|=K|A|, and 1MK1\leqslant M\leqslant K be a parameter. Suppose that

100K2|A|N.100K^{2}|A|\leqslant N\,.

Then either there is x0x\neq 0 such that

|A^(x)|2>M|A|2K,|\widehat{A}(x)|^{2}>\frac{M|A|^{2}}{K}\,, (38)

or for any BAB\subseteq A, |B|=βN|B|=\beta N there exist H𝐆H\leqslant{\mathbf{G}}, z𝐆z\in{\mathbf{G}} with H3B+zH\subseteq 3B+z and

codim(H)(δβ1M)2(log(δ1K)+log2(δβ1M)).{\rm codim}(H)\ll(\delta\beta^{-1}M)^{2}\left(\log(\delta^{-1}K)+\log^{2}(\delta\beta^{-1}M)\right)\,.

In the last case one can find Λ𝐆/H\Lambda\subseteq{\mathbf{G}}/H such that

|A(ΛH)||A|16M and |Λ||H|16M|A|.|A\cap(\Lambda\dotplus H)|\geqslant\frac{|A|}{16M}\quad\quad\mbox{ and }\quad\quad|\Lambda||H|\leqslant 16M|A|\,.

P r o o f.  If 2(A)>M|A|2/K{\mathcal{M}}^{2}(A)>M|A|^{2}/K, then there is nothing to prove. Otherwise, 2(A)M|A|2/K{\mathcal{M}}^{2}(A)\leqslant M|A|^{2}/K and since |AB||AA|=K|A||A-B|\leqslant|A-A|=K|A|, it follows that condition (27) of Proposition 6 takes place. We apply this proposition with κ=1\kappa=1, T=2T=2, ζ=1/8\zeta=1/8, |B|=βN:=ω|A||B|=\beta N:=\omega|A| and M=2MM^{\prime}=2M (see Remark 7) to find 𝐆\mathcal{L}\leqslant{\mathbf{G}} such that

codim()(ωζ)2T2(M+κ)2(log(δ1K)+logT(M(M+κ)ω1)log((M+κ)ω1)){\rm codim}(\mathcal{L})\ll(\omega\zeta)^{-2}T^{2}(M+\kappa)^{2}\cdot\left(\log(\delta^{-1}K)+\log_{T}(M^{\prime}(M+\kappa)\omega^{-1})\cdot\log((M+\kappa)\omega^{-1})\right)
(δβ1M)2(log(δ1K)+log2(δβ1M)),\ll(\delta\beta^{-1}M)^{2}\left(\log(\delta^{-1}K)+\log^{2}(\delta\beta^{-1}M)\right)\,,

and (see (32) or (37))

|B(+z)|(1ζ)||T(M+κ)||8M.|B\cap(\mathcal{L}+z)|\geqslant\frac{(1-\zeta)|\mathcal{L}|}{T(M+\kappa)}\geqslant\frac{|\mathcal{L}|}{8M}\,. (39)

Put B=B(+z)B^{\prime}=B\cap(\mathcal{L}+z). It remains to apply the Kelley–Meka bound (see [16] and [3, Theorem 3]) and find HH\leqslant\mathcal{L}, z𝐆z\in{\mathbf{G}} such that H3B+zH\subseteq 3B^{\prime}+z and

codim(H)codim()+O(logO(1)M)(δβ1M)2(log(δ1K)+log2(δβ1M)).{\rm codim}(H)\leqslant{\rm codim}(\mathcal{L})+O(\log^{O(1)}M)\ll(\delta\beta^{-1}M)^{2}\left(\log(\delta^{-1}K)+\log^{2}(\delta\beta^{-1}M)\right)\,.

Returning to (32) with B=AB=A, we see that

x(AA)(x)(x)|||A|8M.\sum_{x}(A\circ A)(x)\mathcal{L}(x)\geqslant\frac{|\mathcal{L}||A|}{8M}\,.

Put A=λ𝐆/(A(+λ))A=\bigsqcup_{\lambda\in{\mathbf{G}}/\mathcal{L}}(A\cap(\mathcal{L}+\lambda)). Then the last inequality is equivalent to

2λ𝐆/:|Aλ|||/16M|Aλ|2λ𝐆/|Aλ|2|||A|8M.2\sum_{\lambda\in{\mathbf{G}}/\mathcal{L}~{}:~{}|A_{\lambda}|\geqslant|\mathcal{L}|/16M}|A_{\lambda}|^{2}\geqslant\sum_{\lambda\in{\mathbf{G}}/\mathcal{L}}|A_{\lambda}|^{2}\geqslant\frac{|\mathcal{L}||A|}{8M}\,.

Let Λ={λ𝐆/:|Aλ|||/16M}\Lambda=\{\lambda\in{\mathbf{G}}/\mathcal{L}~{}:~{}|A_{\lambda}|\geqslant|\mathcal{L}|/16M\}. Putting A=A(Λ)A_{*}=A\cap(\Lambda\dotplus\mathcal{L}), we see that |A||A|/16M|A_{*}|\geqslant|A|/16M and |Λ|||16M|A||\Lambda||\mathcal{L}|\leqslant 16M|A|. This completes the proof. \hfill\Box


The following consequence of Proposition 6 allows us to have a “regularization” of any set AA in terms of the doubling constant of AA and its density, see inequality (40) below. Given a set A𝐆A\subseteq{\mathbf{G}} put K[A]:=|AA|/|A|K[A]:=|A-A|/|A|.

Corollary 10

Let 𝐆=𝔽2n{\mathbf{G}}=\mathbb{F}_{2}^{n}, A𝐆A\subseteq{\mathbf{G}} be a set, |A|=δN|A|=\delta N. Then there is H𝐆H\leqslant{\mathbf{G}} and z𝐆z\in{\mathbf{G}} such that

codim(H)δ2log3(1/δ).{\rm codim}(H)\ll\delta^{-2}\log^{3}(1/\delta)\,.

and for A~=A(H+z)\tilde{A}=A\cap(H+z) one has δ~=|A~|/|H|δ\tilde{\delta}=|\tilde{A}|/|H|\geqslant\delta, K~=K[A~]\tilde{K}=K[\tilde{A}] and

100K~2δ~>1.100\tilde{K}^{2}\tilde{\delta}>1\,. (40)

P r o o f.  Let K=K[A]K=K[A]. If 100K2δ>1100K^{2}\delta>1, then there is nothing to prove. Otherwise, we apply Proposition 6 with B=AB=-A, ω=1\omega=1, κ=1\kappa=1, T=2T=2, ζ=1/8\zeta=1/8 and M=2MM^{\prime}=2M as we did in the proof of Corollary 9. Here MM is defined as (A)2=M|A|2/K{\mathcal{M}}(A)^{2}=M|A|^{2}/K, where K=K[A]K=K[A]. Exactly as in (39) we find H1𝐆H_{1}\leqslant{\mathbf{G}} and z1𝐆z_{1}\in{\mathbf{G}} such that

|A(H1+z1)||H1|8M,|A\cap(H_{1}+z_{1})|\geqslant\frac{|H_{1}|}{8M}\,, (41)

and

codim(H1)M2(log(δ1K)+log2M).{\rm codim}(H_{1})\ll M^{2}\left(\log(\delta^{-1}K)+\log^{2}M\right)\,.

Now suppose that M1/(16δ)M\leqslant 1/(16\delta) and therefore codim(H1)δ2log2(δ1){\rm codim}(H_{1})\ll\delta^{-2}\log^{2}(\delta^{-1}). Then (41) shows that the AA has density 2δ2\delta inside H1+z1H_{1}+z_{1}. On the other hand, if M>1/(16δ)M>1/(16\delta) then there is x0x\neq 0 such that

|A^(x)||A|(M/K)1/2=22δ|A|.|\widehat{A}(x)|\geqslant|A|\cdot(M/K)^{1/2}=2^{-2}\delta|A|\,.

Here we have used the trivial fact that Kδ1K\leqslant\delta^{-1}. In this case we can use classical density increment (see [19] or [28]) and find H1𝐆H^{\prime}_{1}\leqslant{\mathbf{G}}, codim(H1)=1{\rm codim}(H^{\prime}_{1})=1, z1𝐆z^{\prime}_{1}\in{\mathbf{G}} such that

|A(H1+z1)|δ(1+23δ)|H1|.|A\cap(H^{\prime}_{1}+z^{\prime}_{1})|\geqslant\delta(1+2^{-3}\delta)|H^{\prime}_{1}|\,. (42)

After that, one can repeat our dichotomy for A1:=A(H1+z1)A_{1}:=A\cap(H_{1}+z_{1}) or A1:=A(H1+z1)A^{\prime}_{1}:=A\cap(H^{\prime}_{1}+z^{\prime}_{1}). Using (41) or (42), we see that the algorithm must stop after a finite number of steps and the resulting codimension is big–O of

log(1/δ)δ2log2(δ1)+δ1=O(δ2log3(1/δ)).\log(1/\delta)\cdot\delta^{-2}\log^{2}(\delta^{-1})+\delta^{-1}=O(\delta^{-2}\log^{3}(1/\delta))\,.

This completes the proof. \hfill\Box

4 General case

Now we are ready to consider the case of an arbitrary finite abelian group 𝐆{\mathbf{G}}. The main tool here is the so–called Bohr sets, and we recall all the necessary definitions and properties of this object. Bohr sets were introduced to additive number theory by Ruzsa [20] and Bourgain [4] was the first who used Fourier analysis on Bohr sets to improve the estimate in Roth’s theorem [19]. Sanders (see, e.g., [21], [22]) developed the theory of Bohr sets proving many important theorems, see for example Lemma 17 below.

Definition 11

Let Γ\Gamma be a subset of 𝐆^\widehat{{\mathbf{G}}}, |Γ|=d|\Gamma|=d, and ε=(ε1,,εd)(0,1]d\varepsilon=(\varepsilon_{1},\dots,\varepsilon_{d})\in(0,1]^{d}. Define the Bohr set =(Γ,ε){\mathcal{B}}={\mathcal{B}}(\Gamma,\varepsilon) by

(Γ,ε)={n𝐆|γjn<εj for all γjΓ},{\mathcal{B}}(\Gamma,\varepsilon)=\{n\in{\mathbf{G}}~{}|~{}\|\gamma_{j}\cdot n\|<\varepsilon_{j}\mbox{ for all }\gamma_{j}\in\Gamma\}\,,

where x=|argx|/2π.\|x\|=|\arg x|/2\pi.

The number d=|Γ|d=|\Gamma| is called dimension of {\mathcal{B}} and is denoted by dimB{\rm dim}B. If M=+nM={\mathcal{B}}+n, n𝐆n\in{\mathbf{G}} is a translation of {\mathcal{B}}, then, by definition, put dim(M)=dim(){\rm dim}(M)={\rm dim}({\mathcal{B}}). The intersection {\mathcal{B}}\wedge{\mathcal{B}}^{\prime} of two Bohr sets =(Γ,ε){\mathcal{B}}={\mathcal{B}}(\Gamma,\varepsilon) and =(Γ,ε){\mathcal{B}}^{\prime}={\mathcal{B}}(\Gamma^{\prime},\varepsilon^{\prime}) is the Bohr set with the generating set ΓΓ\Gamma\cup\Gamma^{\prime} and new vector ε~\tilde{\varepsilon} equals min{εj,εj}\min\{\varepsilon_{j},\varepsilon^{\prime}_{j}\}. Furthermore, if =(Γ,ε){\mathcal{B}}={\mathcal{B}}(\Gamma,\varepsilon) and ρ>0\rho>0 then by ρ{\mathcal{B}}_{\rho} we mean (Γ,ρε).{\mathcal{B}}(\Gamma,\rho\varepsilon).

Definition 12

A Bohr set =(Γ,ε){\mathcal{B}}={\mathcal{B}}(\Gamma,\varepsilon) is called regular, if for every η,d|η|1/100\eta,\,d|\eta|\leqslant 1/100 we have

(1100d|η|)|1|<|1+η|<(1+100d|η|)|1|.(1-100d|\eta|)|{\mathcal{B}}_{1}|<{|{\mathcal{B}}_{1+\eta}|}<(1+100d|\eta|)|{\mathcal{B}}_{1}|\,. (43)

We formulate a sequence of basic properties of Bohr, which will be used later.

Lemma 13

Let (Γ,ε){\mathcal{B}}(\Gamma,\varepsilon) be a Bohr set. Then there exists ε1\varepsilon_{1} such that ε2<ε1<ε\frac{\varepsilon}{2}<\varepsilon_{1}<\varepsilon and (Γ,ε1){\mathcal{B}}(\Gamma,\varepsilon_{1}) is regular.

Lemma 14

Let (Γ,ε){\mathcal{B}}(\Gamma,\varepsilon) be a Bohr set. Then

|(Γ,ε)|N2j=1dεj.|{\mathcal{B}}(\Gamma,\varepsilon)|\geqslant\frac{N}{2}\prod_{j=1}^{d}\varepsilon_{j}\,.
Lemma 15

Let (Γ,ε){\mathcal{B}}(\Gamma,\varepsilon) be a Bohr set. Then

|(Γ,ε)|8|Γ|+1|(Γ,ε/2)|.|{\mathcal{B}}(\Gamma,\varepsilon)|\leqslant 8^{|\Gamma|+1}|{\mathcal{B}}(\Gamma,\varepsilon/2)|\,.
Lemma 16

Suppose that (1),,(k){\mathcal{B}}^{(1)},\dots,{\mathcal{B}}^{(k)} is a sequence of Bohr sets. Then

|i=1k(i)|Ni=1k|1/2(i)|N.|\bigwedge_{i=1}^{k}{\mathcal{B}}^{(i)}|\geqslant N\cdot\prod_{i=1}^{k}\frac{|{\mathcal{B}}^{(i)}_{1/2}|}{N}\,.

Recall a local version of Chang’s lemma [5], see [22, Lemma 5.3] and [21, Lemmas 4.6, 6.3].

Lemma 17

Let ε,ν,ρ(0,1]\varepsilon,\nu,\rho\in(0,1] be positive real numbers. Suppose that {\mathcal{B}} is a regular Bohr set and f:f:{\mathcal{B}}\to\mathbb{C}. Then there is a set Λ\Lambda of size O(ε2log(f22||/f12))O(\varepsilon^{-2}\log(\|f\|_{2}^{2}|{\mathcal{B}}|/\|f\|^{2}_{1})) such that for any γSpecε(f)\gamma\in{\rm Spec\,}_{\varepsilon}(f) we have

|1γ(x)||Λ|(ν+ρdim2(B))xρν,|1-\gamma(x)|\ll|\Lambda|(\nu+\rho{\rm dim}^{2}(B))\quad\quad\forall x\in{\mathcal{B}}_{\rho}\wedge{\mathcal{B}}^{\prime}_{\nu}\,,

where =(Λ,1/2){\mathcal{B}}^{\prime}={\mathcal{B}}(\Lambda,1/2).

Now we are ready to obtain an analogue of Proposition 6.

Proposition 18

Let 𝐆{\mathbf{G}} be a finite abelian group, A,B𝐆A,B\subseteq{\mathbf{G}} be sets, |B|=ω|A||B|=\omega|A|, |A+B|=K|A||A+B|=K|A|, |AA|=K|A||A-A|=K^{\prime}|A|, and 0<MK0<M\leqslant K, 0<MK0<M^{\prime}\leqslant K^{\prime}, κ>0\kappa>0, ζ(0,1/2)\zeta\in(0,1/2), 1<TM(M+κ)ω11<T\leqslant M^{\prime}(M+\kappa)\omega^{-1} be some parameters. Suppose that

2(A)M|A|2K,𝖤(B)M|A|3K,{\mathcal{M}}^{2}(A)\leqslant\frac{M|A|^{2}}{K},\,\quad\quad\mathsf{E}(B)\leqslant\frac{M^{\prime}|A|^{3}}{K^{\prime}}\,, (44)

and

|A+B|2κ|A|N.|A+B|^{2}\leqslant\kappa|A|N\,. (45)

Then there is a regular Bohr set =(Γ,ε){\mathcal{B}}_{*}={\mathcal{B}}(\Gamma,\varepsilon) ({\mathcal{B}}_{*} depends on BB only) and z𝐆z\in{\mathbf{G}} such that

|B(+z)|(12ζ)||T(M+κ),|B\cap({\mathcal{B}}_{*}+z)|\geqslant\frac{(1-2\zeta)|{\mathcal{B}}_{*}|}{T(M+\kappa)}\,,

and

dim()(ωζ)2T2(M+κ)2(log(δ1K)+logT(M(M+κ)ω1)log((M+κ)ω1)),{\rm dim}({\mathcal{B}}_{*})\ll(\omega\zeta)^{-2}T^{2}(M+\kappa)^{2}\cdot\left(\log(\delta^{-1}K)+\log_{T}(M^{\prime}(M+\kappa)\omega^{-1})\cdot\log((M+\kappa)\omega^{-1})\right)\,, (46)

as well as

||Nexp(O(dim()log((ωζ)1(M+κ)Tdim()))).|{\mathcal{B}}_{*}|\gg N\cdot\exp(-O({\rm dim}({\mathcal{B}}_{*})\log((\omega\zeta)^{-1}(M+\kappa)T{\rm dim}({\mathcal{B}}_{*}))))\,. (47)

P r o o f.  We use the same argument and the notation of the proof of Proposition 6. The argument before inequality (31) does not depend on a group, therefore we have

1NξSpecζ/M(φ)|B^(ξ)|2φ^(ξ)(1ζ)ωb𝖤kT(M+κ).\frac{1}{N}\sum_{\xi\in{\rm Spec\,}_{\zeta/M_{*}}(\varphi)}|\widehat{B}(\xi)|^{2}\widehat{\varphi}(\xi)\geqslant\frac{(1-\zeta)\omega b\mathsf{E}_{k}}{T(M+\kappa)}\,. (48)

Applying Lemma 17 (combining with the triangle inequality) with =𝐆{\mathcal{B}}={\mathbf{G}} (hence dim()=1{\rm dim}({\mathcal{B}})=1), f=φf=\varphi, and ρ=ν=cζ/(M|Λ|)\rho=\nu=c\zeta/(M_{*}|\Lambda|), where c>0c>0 is a sufficiently small absolute constant, we see that for any ξSpecζ/M(φ)\xi\in{\rm Spec\,}_{\zeta/M_{*}}(\varphi) one has

Specζ/M(φ)(ξ)||2|^(ξ)|2(1+ζ),{\rm Spec\,}_{\zeta/M_{*}}(\varphi)(\xi)\leqslant|{\mathcal{B}}_{*}|^{-2}|\widehat{{\mathcal{B}}_{*}}(\xi)|^{2}(1+\zeta)\,,

where =ρν{\mathcal{B}}_{*}={\mathcal{B}}_{\rho}\wedge{\mathcal{B}}^{\prime}_{\nu}. Thus (48) gives us

1Nξ|B^(ξ)|2|^(ξ)|2(12ζ)ωb||2T(M+κ).\frac{1}{N}\sum_{\xi}|\widehat{B}(\xi)|^{2}|\widehat{{\mathcal{B}}_{*}}(\xi)|^{2}\geqslant\frac{(1-2\zeta)\omega b|{\mathcal{B}}_{*}|^{2}}{T(M+\kappa)}\,.

This is equivalent to

x(BB)(x)()(x)(12ζ)ωb||2T(M+κ).\sum_{x}(B\circ B)(x)({\mathcal{B}}_{*}\circ{\mathcal{B}}_{*})(x)\geqslant\frac{(1-2\zeta)\omega b|{\mathcal{B}}_{*}|^{2}}{T(M+\kappa)}\,.

By the pigeonhole principle there is z𝐆z\in{\mathbf{G}} such that

|B(+z)|(12ζ)ω||T(M+κ).|B\cap({\mathcal{B}}_{*}+z)|\geqslant\frac{(1-2\zeta)\omega|{\mathcal{B}}_{*}|}{T(M+\kappa)}\,.

Now

dim()=|Λ|(ωζ)2T2(M+κ)2(log(δ1K)+logT(M(M+κ)ω1)log((M+κ)ω1)),{\rm dim}({\mathcal{B}}_{*})=|\Lambda|\ll(\omega\zeta)^{-2}T^{2}(M+\kappa)^{2}\cdot\left(\log(\delta^{-1}K)+\log_{T}(M^{\prime}(M+\kappa)\omega^{-1})\cdot\log((M+\kappa)\omega^{-1})\right)\,,

see computations in (34)—(36). Applying Lemmas 14, 16, we see that

||N(ζ/(M|Λ|))O(|Λ|)Nexp(O(dim()log((ωζ)1(M+κ)Tdim()))).|{\mathcal{B}}_{*}|\gg N\cdot(\zeta/(M_{*}|\Lambda|))^{O(|\Lambda|)}\gg N\cdot\exp(-O({\rm dim}({\mathcal{B}}_{*})\log((\omega\zeta)^{-1}(M+\kappa)T{\rm dim}({\mathcal{B}}_{*}))))\,.

Finally, in view of Lemma 13 one can assume that {\mathcal{B}}_{*} is a regular Bohr set. This completes the proof. \hfill\Box


Proposition 47 immediately implies an analogue of Corollary 9.

Corollary 19

Let 𝐆{\mathbf{G}} be a finite abelian group, A𝐆A\subseteq{\mathbf{G}} be a set, |A|=δN|A|=\delta N, |AA|=K|A||A-A|=K|A|, and 1MK1\leqslant M\leqslant K be a parameter. Suppose that

100K2|A|N.100K^{2}|A|\leqslant N\,. (49)

Then either there is x0x\neq 0 such that

|A^(x)|2>M|A|2K,|\widehat{A}(x)|^{2}>\frac{M|A|^{2}}{K}\,, (50)

or for any BAB\subseteq A or BAB\subseteq-A, |B|=βN|B|=\beta N there exists a regular Bohr set =(Γ,ε){\mathcal{B}}_{*}={\mathcal{B}}(\Gamma,\varepsilon) and z𝐆z\in{\mathbf{G}} such that

|B(+z)|||8M,|B\cap({\mathcal{B}}_{*}+z)|\geqslant\frac{|{\mathcal{B}}_{*}|}{8M}\,,

and the bounds

dim()(δβ1M)2(log(δ1K)+log2(δβ1M)),{\rm dim}({\mathcal{B}}_{*})\ll(\delta\beta^{-1}M)^{2}\left(\log(\delta^{-1}K)+\log^{2}(\delta\beta^{-1}M)\right)\,, (51)
||Nexp(O(dim()log(δβ1Mdim())))|{\mathcal{B}}_{*}|\gg N\cdot\exp(-O({\rm dim}({\mathcal{B}}_{*})\log(\delta\beta^{-1}M{\rm dim}({\mathcal{B}}_{*})))) (52)

take place.

Now we present a construction showing that the estimates in Corollary 19 are close to exact.

Example 20

Let pp be a prime number, d=4d=4 and 𝐆{\mathbf{G}} be the cyclic group 𝐆=𝔽pd{\mathbf{G}}=\mathbb{F}^{*}_{p^{d}}. Put A={ind(g+j)}j=0,1,,p1A=\{\mathrm{ind}(g+j)\}_{j=0,1,\dots,p-1}, where gg is a fixed element that generates 𝐆{\mathbf{G}}. Then by Katz’s result (see [15, Theorem 1] and the proof of [1, Lemma 1]) all non–zero Fourier coefficients of AA are bounded by (d1)p=3|A|(d-1)\sqrt{p}=3\sqrt{|A|}. Clearly, K=|AA|/|A||A|K=|A-A|/|A|\leqslant|A| and hence

2(A)(d1)2|A|(d1)2|A|2K|A|2K.{\mathcal{M}}^{2}(A)\leqslant(d-1)^{2}|A|\leqslant\frac{(d-1)^{2}|A|^{2}}{K}\ll\frac{|A|^{2}}{K}\,. (53)

In other words, the set AA has small Fourier coefficients. Also,

K2δ|A|3N1=o(1)K^{2}\delta\leqslant|A|^{3}N^{-1}=o(1)

and, therefore, condition (49) is satisfied for large NN. Now, if there exists a regular Bohr set =(Γ,ε){\mathcal{B}}={\mathcal{B}}(\Gamma,\varepsilon) and z𝐆z\in{\mathbf{G}} such that |A(+z)||||A\cap({\mathcal{B}}+z)|\gg|{\mathcal{B}}|, then |||A|N1/4|{\mathcal{B}}|\ll|A|\ll N^{1/4} (if one believes in GRH [13], then it is possible to obtain even better upper bounds for the cardinality of the intersection A(+z)A\cap({\mathcal{B}}+z), see the argument of [12]). But then estimate (52) gives us

dim()logNloglogNlog(δ1K)loglog(δ1K),{\rm dim}({\mathcal{B}})\gg\frac{\log N}{\log\log N}\geqslant\frac{\log(\delta^{-1}K)}{\log\log(\delta^{-1}K)}\,,

and this coincides with (51) up to double logarithm.

Similarly, it is easy to prove an analogue of Corollary 8 (and we leave the derivation of the analogue of Corollary 40 to the interested reader).

Corollary 21

Let 𝐆{\mathbf{G}} be a finite abelian group, A𝐆A\subseteq{\mathbf{G}} be a set, |A|=δN|A|=\delta N, |AA|=K|A||A-A|=K|A|, and ε(0,1)\varepsilon\in(0,1) be a parameter. Suppose that

100K2δε.100K^{2}\delta\leqslant\varepsilon\,.

Then either there is x0x\neq 0 such that

|A^(x)|2(2ε)|A|2K,|\widehat{A}(x)|^{2}\geqslant\frac{(2-\varepsilon)|A|^{2}}{K}\,,

or there is a regular Bohr set =(Γ,ε){\mathcal{B}}_{*}={\mathcal{B}}(\Gamma,\varepsilon) and z𝐆z\in{\mathbf{G}} such that AA{\mathcal{B}}_{*}\subseteq A-A and

dim()ε2log(δ1K)+ε3,{\rm dim}({\mathcal{B}}_{*})\ll\varepsilon^{-2}\log(\delta^{-1}K)+\varepsilon^{-3}\,, (54)

as well as

||Nexp(O(dim()log(ε1dim()))).|{\mathcal{B}}_{*}|\gg N\cdot\exp(-O({\rm dim}({\mathcal{B}}_{*})\cdot\log(\varepsilon^{-1}{\rm dim}({\mathcal{B}}_{*}))))\,. (55)

P r o o f.  We apply the argument of Proposition 47 with B=AB=-A, ω=1\omega=1, M=2εM=2-\varepsilon, M=(1+κKM)MM+κM^{\prime}=\left(1+\frac{\kappa}{KM}\right)M\leqslant M+\kappa (see Remark 7), T=1+κT=1+\kappa, and κ=ζ=ε/200\kappa=\zeta=\varepsilon/200, say. Thus we find a Bohr set 𝐆{\mathcal{B}}_{*}\subseteq{\mathbf{G}} and z𝐆z\in{\mathbf{G}} such that

(Aμ)(z)(12ζ)||T(M+κ)(12+ε8),(A*\mu)(z)\geqslant\frac{(1-2\zeta)|\mathcal{L}|}{T(M+\kappa)}\geqslant\left(\frac{1}{2}+\frac{\varepsilon}{8}\right)\,, (56)

where μ\mu is any measure on {\mathcal{B}}_{*}. Also, we have

d:=dim()ζ2T2(M+κ)2(log(δ1K)+logT(M(M+κ))log(M+κ))d:={\rm dim}({\mathcal{B}}_{*})\ll\zeta^{-2}T^{2}(M+\kappa)^{2}\cdot\left(\log(\delta^{-1}K)+\log_{T}(M^{\prime}(M+\kappa))\cdot\log(M+\kappa)\right)
ε2log(δ1K)+ε3,\ll\varepsilon^{-2}\log(\delta^{-1}K)+\varepsilon^{-3}\,,

and thanks to Lemmas 14, 16, one has

||N(ζ/(M|Λ|))O(|Λ|)Nexp(O(dim()log(ε1dim()))).|{\mathcal{B}}_{*}|\gg N\cdot(\zeta/(M_{*}|\Lambda|))^{O(|\Lambda|)}\gg N\cdot\exp(-O({\rm dim}({\mathcal{B}}_{*})\cdot\log(\varepsilon^{-1}{\rm dim}({\mathcal{B}}_{*}))))\,.

Now we follow the argument of [22, Lemma 9.2]. Namely, put t=100ε1dt=\lceil 100\varepsilon^{-1}d\rceil and η=1/2t\eta=1/2t and consider the sequence of Bohr sets

()1/2()1/2+η()1/2+tη=.({\mathcal{B}}_{*})_{1/2}\subseteq({\mathcal{B}}_{*})_{1/2+\eta}\subseteq\dots\subseteq({\mathcal{B}}_{*})_{1/2+t\eta}={\mathcal{B}}_{*}\,.

Applying Lemma 15, we see that there is j[t]j\in[t] such that

|′′|:=|()1/2+jη|8(d+1)/t|()1/2+(j1)η|(1+ε/4)|()1/2+(j1)η|:=(1+ε/4)||.|{\mathcal{B}}^{\prime\prime}|:=|({\mathcal{B}}_{*})_{1/2+j\eta}|\leqslant 8^{(d+1)/t}|({\mathcal{B}}_{*})_{1/2+(j-1)\eta}|\leqslant(1+\varepsilon/4)|({\mathcal{B}}_{*})_{1/2+(j-1)\eta}|:=(1+\varepsilon/4)|{\mathcal{B}}^{\prime}|\,. (57)

Consider the measure μ(x)=(x)+′′(x)||+|′′|\mu(x)=\frac{{\mathcal{B}}^{\prime}(x)+{\mathcal{B}}^{\prime\prime}(x)}{|{\mathcal{B}}^{\prime}|+|{\mathcal{B}}^{\prime\prime}|}, 𝗌𝗎𝗉𝗉(μ)\mathsf{supp}(\mu)\subseteq{\mathcal{B}}_{*}. Then inequality (56) gives us

|A(+z)|+|A(′′+z)|(12+ε8)(||+|′′|).|A\cap({\mathcal{B}}^{\prime}+z)|+|A\cap({\mathcal{B}}^{\prime\prime}+z)|\geqslant\left(\frac{1}{2}+\frac{\varepsilon}{8}\right)\cdot(|{\mathcal{B}}^{\prime}|+|{\mathcal{B}}^{\prime\prime}|)\,. (58)

One the other hand, for any x()ηx\in({\mathcal{B}}_{*})_{\eta}, we have

(AA)(x)((A(+z))(A(′′+z)))(x)(A\circ A)(x)\geqslant((A\cap({\mathcal{B}}^{\prime}+z))\circ(A\cap({\mathcal{B}}^{\prime\prime}+z)))(x)
|A(+z)|+|A(′′+z)||(A(+z+x))(A(′′+z))|\geqslant|A\cap({\mathcal{B}}^{\prime}+z)|+|A\cap({\mathcal{B}}^{\prime\prime}+z)|-|(A\cap({\mathcal{B}}^{\prime}+z+x))\bigcup(A\cap({\mathcal{B}}^{\prime\prime}+z))|
|A(+z)|+|A(′′+z)||A(′′+z)||A(+z)|+|A(′′+z)||′′|.\geqslant|A\cap({\mathcal{B}}^{\prime}+z)|+|A\cap({\mathcal{B}}^{\prime\prime}+z)|-|A\cap({\mathcal{B}}^{\prime\prime}+z)|\geqslant|A\cap({\mathcal{B}}^{\prime}+z)|+|A\cap({\mathcal{B}}^{\prime\prime}+z)|-|{\mathcal{B}}^{\prime\prime}|\,.

Using formulae (57), (58), we get

(AA)(x)(12+ε8)(||+|′′|)|′′|=(12+ε8)||(12ε8)(1+ε4)||(A\circ A)(x)\geqslant\left(\frac{1}{2}+\frac{\varepsilon}{8}\right)\cdot(|{\mathcal{B}}^{\prime}|+|{\mathcal{B}}^{\prime\prime}|)-|{\mathcal{B}}^{\prime\prime}|=\left(\frac{1}{2}+\frac{\varepsilon}{8}\right)|{\mathcal{B}}^{\prime}|-\left(\frac{1}{2}-\frac{\varepsilon}{8}\right)\left(1+\frac{\varepsilon}{4}\right)|{\mathcal{B}}^{\prime}|
ε||8>0.\geqslant\frac{\varepsilon|{\mathcal{B}}^{\prime}|}{8}>0\,. (59)

The inequality (59) implies that ()ηAA({\mathcal{B}}_{*})_{\eta}\subseteq A-A. We see that (54), (55) take place (use Lemmas 14, 16 again) and thanks to Lemma 13 one can assume that we have deal with a regular Bohr set. This completes the proof. \hfill\Box

Problem 22

In Example 20 we constructed a set A𝐆A\subseteq{\mathbf{G}}, |A|=δ|𝐆||A|=\delta|{\mathbf{G}}|, |AA|=K|A||A-A|=K|A| such that for d>1d>1 one has

Kd1δ1,K^{d-1}\delta\sim 1\,, (60)

and

2(A)(d1)2|A|2K,{\mathcal{M}}^{2}(A)\leqslant\frac{(d-1)^{2}|A|^{2}}{K}\,,

see Example 53. On the other hand, we always have a universal lower bound (5), provided d4d\geqslant 4. Given d>1d>1 and a set AA such that (60) takes place, what are the proper upper/lower bounds for (A){\mathcal{M}}(A)?

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I.D. Shkredov
ilya.shkredov@gmail.com