About the Measure of Large Values of the Modulo of Trigonometric Sum

Gregory A. Freiman and Alexander A. Yudin

Abstract We study the connection between the additive structure of a finite

set A ⊂ ℤ and the measure of large values of a trigonometric sum.

Keywords: Inverse additive problems, trigonometric sum.

1 Introduction

Let ℤ be the ring of integer rational numbers, ℝ - the field of real numbers,

𝕋 = ℝ/ℤ, e(α) = e^{2πiα}, i = \sqrt{−1}, [a, b] = \{a, a + 1, …, b\}, a, b ∈ ℤ, a < b .

For A ⊂ ℤ, |A| = card A - denotes the number of elements of a set A.

S_{A}(α) = ∑_{a∈A} e(αa), α ∈ 𝕋,

Let

E(A, λ|A|) = {α ∈ 𝕋 : |S_{A}(α)| ≥ λ|A|} .

In this paper, we study the following problem: find the set A for which mes E(A, λ|A|) has the maximal value assuming that |A| is given and A is a part of a short interval.

The problem has a rather long history (see [1], [2], [3] and [4]). The connection of this theme with problems of probability theory, harmonic analysis and coding theory is shown in [4], [5], [6] and [7]. Results in this direction you can also find in [9], [10] and [11]. There exists very strong connections of this field with Roth three element progression problem ([3], pp. 140-142).

In this paper, we shall prove the following

Theorem 1 Let A ⊂ [−N, N], |A| = k ≥ N + 1, N - sufficiently large integer.

Then

max_{A} mesE(A, λ|A|) = μ^{∗}(|A|) .

where λ = \frac{2\sqrt{2}}{π} = 0.90032^{1} is attained if and only if A is an arithmetic progression.

Let us stress that the conditions of Theorem 1 lead to the fact that the difference of arithmetic progression A is equal to 1 and only in the case k = N + 1 is equal to 1 or 2.

An example: if |A| = N + 1 then

μ^{∗}(N + 1) = mes \{α ∈ 𝕋 : | ∑_{a=0}^{N} e(αa)| ≥ \frac{2\sqrt{2}}{π}(N + 1)\} = \frac{2θ}{N + 1} + O(\frac{1}{N^{3}} ),

where θ is a solution of the equation

\frac{sin πθ}{πθ} = \frac{2\sqrt{2}}{π}

so that

πθ = 0.775

Let us review existing results in more detail and explain existing and possible applications.

The problem of finding the maximal measure μ_{max} = sup_{A} μ for sets of α for which |S_{A}(α)| is bigger than some given number which is less than the trivial estimate equal to k and of finding the sets A with this condition was first formulated in [3], page 144.

1The result stated here for λ = 0.90032 is, in fact, valid for λ = 0.75. This last result was obtained with the help of computations too massive to be included in this paper.

The case when

|S_{A}(α)| ≥ (1 − ε)|A|

and ε = o(1) was studied in [4] by A. Yudin.

The case when ε is some positive constant was studied in [7] by A. Besser. Nevertheless, the constant achieved appeared to be very small (ε = \frac{1}{20000})

and further progress connected with too large technical difficulties.

That’s why in [9] it was proposed to add an additional condition and to study only those sets included in a segment of not very big length. This approach was illustrated in [1] by a simple example. We explain now that an advance in this direction includes important applications giving a strong justification for the result of this paper.

For the set A which satisfies |2A| < \frac{10}{3}|A| − 5 was shown in [18] that we were led to a study of trigonometric sums for which the condition

|S_{A}(α)| ≥ 0.56|A|  (1)

was valid for some specially chosen α. We see that the constant in (1) and the constant 0.75 in Theorem 1 are already very close (near).

As to the sets in the short intervals or, in other words, sets with a large density, we may mention that most problems for sum-tree sets are of that nature.

Now we will describe the connection with problems of information transfer. The code word (ε_{0}, ε_{1}, …, ε_{n−1}), where ∀i ε_{i} ∈ {+1, −1} is transmitted with the aid of a signal

F_{n}(t) = ∑_{j=0}^{n−1} ε_{i}e(jt) .

Technical conditions ask for the value

max_{t∈𝕋}|F_{n}(t)|

be as small as possible.

We have

F_{n}(t) = 2 ∑_{k:ε_{k}=1} e(kt) − D_{n}(t) ,

where D_{n}(t) = ∑_{k=0}^{n−1} e(kt) denotes Dirichlets’ kernel. The behavior of D_{n}(t) is known. So, the study of |F_{n}(z)| leads us to the study of a trigonometric sum

∑_{k:ε_{n}=1} e(kt) .  (2)

We know that “almost all” code words

(ε_{0}, ε_{1}, …, ε_{n−1})

have almost the same number of +1 and −1. Thus the number of summands in (2) is near n/2 and the conditions of Theorem 1 are satisfied. See [15] for the connection of this work with information theory problems.

In this paragraph we will introduce some definitions and formulate and prove lemmas needed for the proof of Theorem 1. In §2 we will complete the proof of Theorem 1.

Zamechanie.Tak kak utvergdenie Theorem 1 , kasaetsya meru mnogestva, to pri dokazatel’stve neravenstv o velechine meru togo ili inogo mnogestva mu mogem prenebrech’ lyubum mnogestvom meru 0.

Let N ∈ ℤ, N > 0 and α ∈ [0, 1).

Definition. The segment

[β, γ] ⊂ 𝕋 ,

where

β, γ ∈ 𝕋 = [0, 1)(mod 1)

is the set of numbers ρ, for which

ρ ∈ [β, γ)(mod 1) ,

if

∃β_{0} ∈ β + ℤ , γ_{0} ∈ γ + ℤ , β_{0} < γ_{0} ≤ β_{0} + 1

and

β_{0} ≤ ρ ≤ γ_{0} .

Example 1 The interval [\frac{1}{3}, \frac{2}{3}] (mod 1) is defined by a set of numbers ρ ∈ ℝ for which

1/3 ≤ ρ ≤ 2/3, ρ ∈ ℝ .

Example 2 The interval [\frac{4}{5}, \frac{1}{5}] (mod 1) is defined by a set

\{\frac{4}{5} ≤ ρ ≤ 1\} ⋃ \{0 < ρ ≤ \frac{1}{5}\} , ρ ∈ ℝ .

Lemma 1 Let α ∈ 𝕋 and

L = {e(αx) : x ∈ [a, b], a, b ∈ ℤ, a < b} ,
A ⊂ [a, b], |A| = m, 2 ≤ m ≤ b − a + 1 .

Let A_{0} ⊆ [a, b] be an extremal set i.e. , for which |A_{0}| = m, A_{0} ⊂ L we have

max_{A⊂[a,b],}_{|A|=m} | ∑_{a∈A} e(αa)| = | ∑_{a∈A_{0}} e(αa)| .

Then A_{0} is a set such that a minimal segment [β, γ] containing the set αA_{0} does not include points of the set α([a, b]\A_{0}). For almost all points α of the set 𝕋 .

Lemma 1 is obvious from a geometrical point of view. Let m be an integer for which 1 ≤ m ≤ N. Take all subsets of L of cardinality m. It is clear that the subset of L the sum of vectors of which has the maximal length - is all the vectors of L from a certain angle.

Note that A_{0} exists because |A| is finite.

The case A_{0} = [a, b] is trivial. Look at the set

E = {e(αx) − e(αx^{∗}) : x ∈ A_{0}, x^{∗} ∈ [a, b]\A_{0}} .

Let us suppose, first, that there exist two vectors, x and x^{∗}, x ∈ A_{0}, x^{∗} ∈ [a, b]\A_{0} such that the projection of e(αx)−e(αx^{∗}) on the vector S_{A_{0}}(α) is positive, i.e. the angle between these two vectors is less than \frac{π}{2} (we identify here ℂ with ℝ^{2} in the usual way).

We have for A′ = (A ∪ {x^{∗}})\{x}, and we have

|S_{A′}(α)| > |S_{A}(α) for S_{A′}(α) = S_{A_{0}}(α) − (e(αx) − e(αx^{∗})) ,

contradiction.

We have to study now the case in which projection of all e(αx) − e(αx^{∗}) on S_{A_{0}}(α) are positive. Then the projection of any e(αx) on S_{A_{0}}(x) is bigger than a projection of an e(αx^{∗}). It means that the angle between e(αx^{∗}) and S_{A_{0}}(α) is larger than the angle between any e(αx) and S_{A_{0}}(α). In other words, all vectors e(αx^{∗}) are situated outside of the angle where all vectors e(αx) are situated. ■ We will now need the following definitions and notations.

z = (z_{1}, …, z_{k}) ∈ ℂ^{k} ,

∥z∥ = (∑_{j=1}^{k} |z_{j}|^{2})^{1/2} ,

(u, v) = ∑_{j=1}^{k} u_{j}\overline{v_{j}} ,

\overline{a + bi} = a − bi

e_{A} = (x_{−N}, …, x_{0}, …, x_{N}) , where x_{i} = \begin{cases}1 & if i ∈ A \\ −1 & if i ∉ A \end{cases}

e_{α} = (e(−Nα), e(−(N − 1)α), …, e(Nα)) .

Note that e_{0} = \underbrace{(1, 1, …, 1)}_{2N+1} and let K_{H}(α) = ∑_{k=−N}^{N} cos 2πkα.

Lemma 2 Let H = 2N + 1. In the conditions of Theorem 1 we have

1) E (A, \frac{2\sqrt{2}}{π}|A|) ⊂ \{α ⊆ 𝕋 : |D_{H}(α)| ≥ \frac{H}{60}\} ⋃ \{α ∈ 𝕋 : |D_{H}(2α)| ≥ \frac{H}{60}\} .

2) For every j = 1, 2, …, 2N we have \frac{j}{H} ∉ E (A, \frac{2\sqrt{2}}{π}|A|).

Proof of Lemma 2. Let

e_{A} = c_{0}e_{0} + c_{α}e_{α} + c_{−α}e_{−α} + e_{⊥} ,  (3)

where e_{⊥} ∈ ℂ^{H}, which is orthogonal to vectors e_{0}, e_{α}, e_{−α}.

Using scalar multiplication of both sides of (3) on each of the vectors e_{0}, e_{α}, e_{−α} we will obtain for the numbers c_{0}, c_{α}, c_{−α} the following system of equations

\begin{cases}2|A| − H = c_{0}H + c_{α}K_{H}(α) + c_{−α}K_{H}(α) \\ 2\overline{S}_{A}(α) − K_{H}(α) = c_{0}K_{H}(α) + c_{α}H + C_{−α}K_{H}(2α) \\ 2S_{A}(α) − K_{H}(α) = c_{0}K_{H}(α) + c_{α}K_{H}(2α) + c_{−α}H \end{cases}  (4)

Let

G = (\begin{matrix}H & K_{H}(α) & K_{H}(α) \\ K_{H}(α) & H & K_{H}(2α) \\ K_{H}(α) & K_{H}(2α) & H \end{matrix})

be the matrix of (4). A simple computation shows that

det G = H^{3} + 2K_{H}^{2}(α)K(2α) − H(2K_{H}^{2}(α) + K_{H}^{2}(2α))
= H(H^{2} − K_{H}^{2}(2α)) − 2K_{H}^{2}(α)(H − K_{H}(2α)) =
= (H − K_{H}(2α))(H(H + K_{H}(2α)) − 2K_{H}^{2}(α)) .

Let us now find those α ∈ 𝕋 for which the matrix G is non-singular. If det G = 0 we get

H − K_{H}(2α) = 0  (5)

or

H(H + K_{H}(2α)) − 2K_{H}^{2}(α) = 0 .  (6)

From equation (5) it follows that

α = α_{0} = 0  (7)

or

α = α_{1} = \frac{1}{2} .

We now analyze equation (6) and show that it has only one solution (7). We have

H^{2} + HK_{H}(2α) − 2K_{H}^{2}(α) ≥ H^{2} − H|K_{H}(2α)| − 2K_{H}^{2}(α) .

In a neighborhood of α = \frac{1}{2} we have |K_{H}(α)| ≤ C, and K_{H}(2α) ≥ −\frac{2}{3π}H. Thus, near α = \frac{1}{2}, equation (6) has no solutions.

If α^{∗} ≠ 0, |α^{∗}| is small and α^{∗} is the solution of (6) then the vectors e_{0}, e_{α^{∗}}, e_{−α^{∗}} are linearly dependent, i.e. for each j ∈ [−N, N]

e_{0} = x_{α^{∗}}e_{α^{∗}} + x_{−α^{∗}}e_{−α^{∗}}
1 = x_{α^{∗}}e(jα^{∗}) + x_{−α^{∗}}e(−jα^{∗}) ,

and this is impossible.

Thus the matrix of system (4) is not degenerate everywhere, excepting the case α ∈ {0, 1/2}. This means that if α ∉ {0, 1/2} the system (4) has a solution for c_{0}, c_{α}, c_{−α} and only one. Because we are interested in a value of a measure of E (A, \frac{2\sqrt{2}}{π}|A|) elimination of two points does not matter.

We now find the eigenvalues of matrix G. Solving the characteristic equations

|\begin{matrix}H − λ & K_{H}(α) & K_{H}(α) \\ K_{H}(α) & H − λ & K_{H}(2α) \\ K_{H}(α) & K_{H}(2α) & H − λ \end{matrix}| = 0

we obtain three solutions (eigenvalues)

λ_{1} = H − K_{H}(2α)

λ_{2} = H + \frac{1}{2}(K_{H}(2α) + \sqrt{K_{H}^{2}(2α) + 8K_{H}^{2}(α)})  (8)

λ_{3} = H − \frac{1}{2}( − K_{H}(2α) + \sqrt{K_{H}^{2}(2α) + 8K_{H}^{2}(α)})

Solutions of the system (4) are

(c_{0}, c_{α}, c_{−α})^{t} = G^{−1}(2|A| − H, 2\overline{S}_{A}(α) − K_{H}(α), 2S_{A}(α) − K_{H}(α))^{t}

or c⃗ = G^{−1}h⃗. From (3) and (4) we get

(h⃗, c⃗) = (Gc⃗, c⃗) = ∥e_{A}∥^{2} − ∥e_{⊥}∥^{2} ≤ H  (9)

and

|(h⃗, c⃗)| = |(h⃗, G^{−1}h⃗)| ≥ λ_{min}(G^{−1})∥h⃗∥^{2}  (10)

where λ_{min}(G^{−1})-minimal eigenvalue of the matrix G^{−1}. From (9) and (10) we get

λ_{min}(G^{−1})∥h⃗∥^{2} ≤ H

or

∥h⃗∥^{2} < \frac{H}{λ_{min}(G^{−1})} = λ_{max}(G)H ,  (11)

where λ_{max}(G) is the maximum eigenvalue of the matrix G.

The inequality (10) and the equality λ_{min}(G^{−1}) = λ_{max}(G)^{−1} are valid because of extremal properties of hermitian matrices (see [16]).

The inequality (11) may be rewritten in the following way:

(2|A| − H)^{2} + 2|2S_{A}(α) − K_{H}(α)|^{2} ≤ λ_{max}(G)H  (12)

We shall now prove the second conclusion of Lemma 2.

Let α = \frac{j}{H}, j ≠ 0. Then

K_{H}(α) = K_{H}(2α) = 0, λ_{1} = λ_{2} = λ_{3} = H .

From (12) and |S_{A}(α)| > \frac{3}{4}|A| we get that

(2|A| − H)^{2} + 8|S_{A}(α)|^{2} ≤ H^{2} ⇒

(2|A| − H)^{2} + 8 (\frac{3|A|}{4})^{2} ≤ H^{2} ⇒

4|A|^{2} − 4|A|H + \frac{9|A|^{2}}{2} ≤ 0

and thus

|A| ≤ \frac{8(2N + 1)}{17} ,

which contradicts |A| ≥ N + 1. We get |S_{A}(α)| < \frac{2\sqrt{2}}{π}|A| for all α = \frac{j}{H}, j ≠ 0.

In order to prove the first assertion of Lemma 2 we shall estimate λ_{max}(G) using (8). We get

λ_{max}(G) ≤ H + 2 max(|D_{H}(α)|, |D_{H}(2α)|) .

From (12), we get

(2|A| − H)^{2} + 2|2S_{A}(α) − K_{H}(α)|^{2} ≤ H(H + 2 max(|D_{H}(α)|, |D_{H}(2α)|) .

Let α ∈ E (A, \frac{2\sqrt{2}}{π}|A|). If

2|S_{A}(α)| − |K_{H}(α)| < 0

then

|D_{H}(α)| > 2|S_{A}(α)| > 2\frac{2\sqrt{2}}{π}|A| = \frac{4\sqrt{2}}{π}|A| ≥ \frac{2\sqrt{2}}{π}H ,

and thus α ∈ (−\frac{1}{H} , \frac{1}{H} ), i.e. α satisfies the condition 1 of Lemma 2.

Now let’s assume

2|S_{A}(α)| − |K_{H}(α)| > 0

Then using |2S_{A}(α) − K_{H}(α)| ≥ 2|S_{A}(α)| − |D_{H}(α)| and (12), we get

(2|A| − H)^{2} + 2(2|S_{A}(α)| − |D_{H}(α)|)^{2}
≤ H(H + 2 max(|D_{H}(α)|, |D_{H}(2α)|) .

It follows from here that

max(|D_{H}(α)|, |D_{H}(2α)|) ≥ \frac{H}{60} .

■

Lemma 2 shows that the values α in the sums of the form

∑_{a∈[h_{1},h_{2}]} e(αa)

are small and we can look at these sums as integral sums and replace them on integrals with an error which is sufficiently small.

2

The proof of Theorem 1 follows from steps A, B, C below.

A. E (A, \frac{2\sqrt{2}}{π}|A|) ⊂ (−\frac{1}{H} , \frac{1}{H} ), then the set is an arithmetic progression with a difference one.

B. The set E (A, \frac{2\sqrt{2}}{π}|A|) is included in a neighborhood of 0, defined in a Lemma 2; then E (A, \frac{2\sqrt{2}}{π}|A|) ⊂ (−\frac{1}{H} , \frac{1}{H} )

C. We assume that there is α such that α ∈ E (A, \frac{2\sqrt{2}}{π}|A|) and α belongs to a neighborhood of 1/2.

Case A. The condition in A is such that we look on trigonometric sums for which

α ∈ (−\frac{1}{H} , \frac{1}{H} ) .

Choose α^{∗} ∈ E (A, \frac{2\sqrt{2}}{π}|A|), where A is any set of cardinality |A| = m ≥ N + 1. According to Lemma 1 max_{A,|A|=m}|S_{A}(α^{∗})| is achieved if the set A is an arithmetic progression with a difference equal to 1. Denote it by A_{0}.

We see that for almost all α ∈ E(A, \frac{2\sqrt{2}}{π}|A|) and for every set A such that |A| = |A_{0}| we have

|S_{A_{0}}(α)| ≥ |S_{A}(α)| .

We conclude that in this case mesE (A_{0}, \frac{2\sqrt{2}}{π}|A_{0}|) ≥ mesE (A, \frac{2\sqrt{2}}{π}|A|).

Case B. We will show first that if

\frac{1}{H} < |α| ≤ \frac{100}{H}

then a.e. we have

|S_{A}(α)| < \frac{2\sqrt{2}}{π}|A| .  (13)

In fact we will prove (13) only in case α ∈ (\frac{1}{H} , \frac{2}{H} ). This case is the most difficult and other cases may be studied analogously and they are much simpler.

The use of Lemma 1 shows that |S_{A}(α)| in the case we study achieves its maximum value if A is a union of two segments:

A = [−N, −(k + 1)] ∪ [k + 1, N] .

Thus

|S_{A}(α)| ≤ 2| ∑_{s=k+1}^{N} e(αs)| = 2 |\frac{sin π(N − k)α}{sin πα}| .

According to the condition of Theorem 1 we have |A| ≥ N + 1 and therefore |A| = 2(N − k) ≥ N + 1, 1 − \frac{k}{N} > \frac{1}{2}.

In our case α ∈ (\frac{1}{H} , \frac{2}{H} ) = (\frac{1}{2N+1}, \frac{2}{2N+1}). Take α = \frac{θ}{N} and we will get

\frac{1}{2} < θ < 1 .

Now we have

\frac{|S_{A}(α)|}{|A|} = \frac{2}{|A|}|\frac{sin π (1 − \frac{k}{N} ) θ}{sin π \frac{θ}{N}} | =

=\frac{| sin π (1 − \frac{k}{N} ) θ|}{N (1 − \frac{k}{N} ) | sin π \frac{θ}{N} |} = \frac{sin π (1 − \frac{k}{N} ) θ}{π (1 − \frac{k}{N} ) θ} (1 + O (\frac{1}{N} ))

Because of \frac{1}{2} < θ < 1 and \frac{1}{2} < (1 − \frac{k}{N} ) < 1 we have to find

max_{\frac{1}{4} <x<1} \frac{sin πx}{πx}

A simple computation shows that for every \frac{π}{4} ≤ x ≤ π we have:

\frac{sin x}{x} ≤ \frac{sin π/4}{π/4} = \frac{2\sqrt{2}}{π}

Thus, for every α ∈ (\frac{1}{H} , \frac{2}{H} ) and N sufficiently large we have:

|S_{A}(α)| ≤ \frac{2\sqrt{2}}{π}|A| .

It is clear that both parts of this inequality may be equal for a set of zero measure and thus (13) is proved. Case C. Let α = \frac{1}{2} + β, |β| ≤ \frac{100}{N} and

|S_{A} (\frac{1}{2} + β) | ≥ \frac{2\sqrt{2}}{π}|A|, |A| ≥ N + 1 .

We will show that in this case

|β| ≤ \frac{0.13}{N} .

Using Lemma 1 we can change the set A on a set which is a union of a finite number of segments.

Take

A_{0} = {a ∈ A : a ≡ 0(mod 2)}, A_{1} = {a ∈ A : a ≡ 1(mod 2)}

and

2πγ = argS_{A} (\frac{1}{2} + β) .

We have

|S_{A}(\frac{1}{2} + β)| = S_{A}(\frac{1}{2} + β)e^{−2πiγ} = ∑_{a∈A_{0}} cos 2π(aβ − γ) − ∑_{a∈A_{1}} cos 2π(aβ − γ) .

Summing these two last sums on all even and odd numbers of the segment [−N, N] we will get the following estimate.

|S_{A} (\frac{1}{2} + β) | ≤ \frac{1}{2} ∑_{x∈[−N,N]} | cos 2π(βx − γ)| + O(1) = (1/2)Σ_{1} + O(1) .

According to Lemma 2, the value of β is small and we can approximate Σ_{1}, with the aid of an integral

Σ_{1} = ∫_{−N}^{N} | cos 2π(βx − γ)|dx = \frac{1}{β} ∫_{−Nβ−γ}^{Nβ−γ} | cos 2πt|dt .

Let us use the Cauchy inequality

Σ_{1} ≤ \frac{1}{β} \sqrt{2Nβ}\sqrt{Nβ + \frac{1}{4π}}

and thus

|S_{A} (\frac{1}{2} + β) | ≤ \frac{1}{2} \frac{1}{β} \sqrt{2Nβ}\sqrt{Nβ + \frac{1}{4π}} + O(1) .

Using the estimate

|S_{A} (\frac{1}{2} + β) | ≥ \frac{2\sqrt{2}}{π}|A| ≥ \frac{2\sqrt{2}}{π}(N + 1) ,

we obtain

|β| ≤ \frac{0.13}{N} .

Let A be an extremal set,

A ⊆ [−N, N]

for which the trigonometric sum is large in a neighborhood of α = 1/2. Then

|S_{A}(α)| ≥ \frac{2\sqrt{2}}{π}|A|,  (14)

where α = 1/2 + β. As was shown earlier

|β| ≤ \frac{0.13}{N}.  (15)

Lemma 2, [1] , pp.35 shows that the cardinality of the set of points {αa}, a ∈ A which belong to a suitable segment of the length \frac{1}{2} in ℝ/ℤ is no less than 0.95|A|.

Because α = 1/2+β and β is small in view of (15) we see that those a ∈ A for which {αa} ∈ γ are or all even or all odd. Suppose, for example, that they are even. If |S_{A}(β + \frac{1}{2})| ≥ \frac{2\sqrt{2}}{π}|A|, then from (15) we get |S_{A}(β)| ≥ \frac{2\sqrt{2}}{π}|A|. These inequalities lead us to an equality

|S_{A}(β)|^{2} + |S_{A}(β + \frac{1}{2})|^{2} = |S_{A_{0}}(β) + S_{A_{1}}(β)|^{2} + |S_{A_{0}}(β) − S_{A_{1}}(β)|^{2} (16)
= 2|S_{A_{0}}(β)|^{2} + 2|S_{A_{1}}(β)|^{2} ≥ \frac{16}{π^{2}}|A|^{2}  (17)

We will estimate now the measure of the set of solutions of inequality (14).

Let us stress once more that our conditions (14) imply that the maximal values of |S_{A_{0}}(β)| and |S_{A_{1}}(β)| are achieved as it follows from Lemma 1 only in a case when A_{0} consists of an interval of even numbers and A_{1} consists of an interval of odd numbers.

We see that if (16) takes place then A = A_{0} ∪ A_{1} and |A_{0}| ≥ 0.95|A|. We have

|S_{A}(α)| = |S_{A}(1/2 + β)| = | ∑_{a∈A_{0}} e^{2πiβa} − ∑_{a∈A_{1}} e^{2πiβa}| = |S_{A_{0}}(β) − S_{A_{1}}(β)|.

The modulus of the sum

S_{A_{0}}(α) + S_{A_{1}}(α)

for every α will grow it A_{0} will be to build an interval. In fact, the norm of the sum of two vectors can only grow, it will grow the norm the bigger of summands.

Then

|S_{A_{0}}(β)| ≤ | ∑_{0≤a≤|A_{0}|−1} e^{2πiβ2a}|

and

|S_{A_{1}}(β)| ≤ | ∑_{0≤a≤|A_{1}|−1} e^{2πiβ2a}|.

We see that the measure of solutions of the inequality (14) is not greater then the measure of solutions of the inequality

\frac{sin^{2} 2πβ|A_{0}|}{sin^{2} 2πβ} + \frac{sin^{2} 2πβ|A_{1}|}{sin^{2} 2πβ} ≥ \frac{8}{π^{2}} |A|^{2}.

And let now stress that the function

sin^{2} 2πβ|A_{0}| + sin^{2} 2πβ|A_{1}|

under the conditions (15), |A_{0}| ≥ 0.95|A|, has a maximal value only if |A_{0}| = |A_{1}|. So, the measure of the set of solutions of inequality (14) is not greater than

mes\{β : |\frac{sin 2πβ|A|}{sin 2πβ}| ≥ \frac{2\sqrt{2}}{π}|A|\},

and it is the measure of the set

\{α : | ∑_{0≤a≤N} e^{2πiαa}| ≥ \frac{2\sqrt{2}}{π}|A|\},

i.e. of arithmetic progression of the length N + 1. ■.

References

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[17] R. Courant and D. Hilbert. Methoden der mathematischen Physik, Ersteband, Berlin, Verlag von Julius Springer ,I,(1931).

[18] G.A. Freiman. Inverse problems of additive number theory VII. Additions of finite sets IV. The method of trigonometric sums. Izv. Vyss. Ucebn Zaved. Mathematika 6(31), (Russian), (1962), 131–144.

Gregory A. Freiman  Alexander A. Yudin
Department of Pure Mathematics  Department of Mathematics
Faculty of Exact Sciences  Vladimir Pedagogical University
School of Mathematical Sciences  600024,Vladimir,
Tel Aviv University, Tel Aviv 69978, Israel st.Stroitelej,h.11,Russia