Exponential Sums and Distinct Points on Arcs

Abstract

Suppose that some harmonic analysis arguments have been invoked to show that the indicator function of a set of residue classes modulo some integer has a large Fourier coefficient. To get information about the structure of the set of residue classes, we then need a certain type of complementary result. A solution to this problem was given by Gregory Freiman in 1961, when he proved a lemma which relates the value of an exponential sum with the distribution of summands in semi-circles of the unit circle in the complex plane. Since then, Freiman's Lemma has been extended by several authors. Rather than residue classes, one has considered the situation for finitely many arbitrary points on the unit circle. So far, Lev is the only author who has taken into consideration that the summands may be bounded away from each other, as is the case with residue classes. In this paper we extend Lev's result by lifting a recent result of ours to the case of the points being bounded away from each other.

Full text

Page 1

arXiv:0812.4653v1 [math.NT] 26 Dec 2008
EXPONENTIAL SUMS AND DISTINCT POINTS ON ARCS
ØYSTEIN J. RØDSETH
Dedicated in honour of Mel Nathanson’s 60thbirthday
(Contribution to the Festschrift)
Abstract. Suppose that some harmonic analysis arguments have been
invoked to show that the indicator function of a set of residue classes
modulo some integer has a large Fourier coefficient. To get informa-
tion about the structure of the set of residue classes, we then need a
certain type of complementary result. A solution to this problem was
given by Gregory Freiman in 1961, when he proved a lemma which
relates the value of an exponential sum with the distribution of sum-
mands in semi-circles of the unit circle in the complex plane. Since
then, Freiman’s Lemma has been extended by several authors. Rather
than residue classes, one has considered the situation for finitely many
arbitrary points on the unit circle. So far, Lev is the only author who
has taken into consideration that the summands may be bounded away
from each other, as is the case with residue classes. In this paper we
extend Lev’s result by lifting a recent result of ours to the case of the
points being bounded away from each other.
1. Introduction
In additive combinatorics, and in additive combinatorial number theory
in particular, situations of the following type are rather common. Let A be a
set of N residue classes modulo an integer m. Suppose that some harmonic
analysis arguments have been invoked to show that the indicator function
of A has a large Fourier coefficient; i.e.,
max
0?=x∈Z/mZ|?1A(x)| ≥ αN
for some α ∈ (0,1]. Following Green [1], or see Tao and Vu [13], we might
say that A has “Fourier bias”. Using this information one wishes to con-
clude that A has “combinatorial bias”; perhaps integer representatives of
Date: December 26, 2008.
2000 Mathematics Subject Classification. 11J71, 11K36, 11T23.
Key words and phrases. Exponential sums, unit circle, distribution, arcs.
1

Page 2

2Ø. J. RØDSETH
the residue classes in some affine image of A concentrates on some inter-
val. For this we then need tight upper bounds for the absolute values of
the exponential sums?1A(x). Now, the basic idea is that the absolute value
distributed. The case m prime was applied by Freiman [2, 3] in the proof of
his “2.4-theorem”. A proof of this theorem is presented in both Freiman’s
classical monograph [4] and in Mel Nathanson’s beautiful book [9].
Freiman [2, 3], Postnikova [10], Moran and Pollington [8], Lev [5, 6], and
Rødseth [11, 12] considered the situation in which one has finitely many ar-
bitrary points on the unit circle in the complex plane, rather than a subset
of Z/mZ. In possible applications the points on the unit circle will some-
times be bounded away from each other, like, for instance, if we are looking
at a subset of Z/mZ. If we add the condition that the points should be
bounded away from each other, we could hope for sharper results. Indeed,
by adding this assumption, Lev sharpened Freiman’s Lemma to (2) below.
Lev’s result [5, Theorem 2] seems, however, to be the only result in the
literature addressing this issue. In this paper we prove a result which ex-
tends both Lev’s result about points being bounded away from each other,
and our main result in [11, 12], where we did not take this property into
consideration.
First, we shall, however, present a version of Freiman’s Lemma.
include two different proofs in the hope of giving the reader an impression
of two recent techniques used in the search and study of results related to
Freiman’s Lemma.
of an exponential sum is small if the terms are, in some sense, uniformly
We
2. Three Theorems
In the following, n, N, κ, and k are non-negative or positive integers. We
write U for the unit circle in the complex plane; that is,
U := {z ∈ C : |z| = 1}.
An empty sum is taken as zero.
We now state Freiman’s Lemma [2, 3].
Theorem 1 (Freiman’s Lemma). Suppose that the complex numbers z1,...,zN∈
U have the property that any open semi-circle of U contains at most n of
them. Then
(1)|z1+ ··· + zN| ≤ 2n − N.

Page 3

EXPONENTIAL SUMS3
The complex numbers z1,...,zN are not necessarily all distinct. The
assumptions of the lemma imply that N ≤ 2n, and the result is sharp in the
range n ≤ N ≤ 2n. That is, for every n and N satisfying these inequalities,
there exist sequences z1,...,zN with zj∈ U, such that the hypotheses are
satisfied and |z1+···+zN| meets the bound 2n−N. In this sense, Freiman’s
Lemma is best possible. The result has, however, been extended by Moran
and Pollington [8], Lev [5, 6], and by Rødseth [11, 12].
The next theorem is Lev’s sharpening of Freiman’s Lemma at the ex-
pense of requiring the points to be bounded away from each other. Lev [5,
Theorem 2] proved the following theorem.
Theorem 2 (Lev). Let δ ∈ (0,π] satisfy nδ ≤ π. Suppose that the complex
numbers z1,...,zN∈ U have the following two properties:
(a) any open semi-circle of U contains at most n of them;
(ii) any open arc of U of length δ contains at most one of them.
Then
(2)|z1+ ··· + zN| ≤sin((2n − N)δ/2)
sin(δ/2)
.
So, by introducing the condition (ii), Lev reduced Freiman’s bound 2n−N
in (1) to the bound in (2), a refinement that, according to Lev, was crucial
in [7]. The result is sharp for n ≤ N ≤ 2n. By letting δ → 0+in Lev’s
result, we recover Freiman’s Lemma.
In this paper we shall prove the following theorem.
Theorem 3. Let δ,ϕ ∈ (0,π] satisfy nδ ≤ ϕ. Suppose that the complex
numbers z1,...,zN∈ U have the following two properties:
(i) any open arc of U of length ϕ contains at most n of them;
(ii) any open arc of U of length δ contains at most one of them.
Let N = κn + r, 1 ≤ r ≤ n, and assume that (κ + 1)ϕ ≤ 2π. Then we have
(3)
|z1+ ··· + zN| ≤sin(rδ/2)
sin(δ/2)sin(ϕ/2)
·sin((κ + 1)ϕ/2)
+sin((n − r)δ/2)
sin(δ/2)
·sin(κϕ/2)
sin(ϕ/2).
Notice that κ = ⌈N/n⌉ − 1. Also notice that we do need some condition
like (κ+1)ϕ ≤ 2π, to be certain that there is room enough on U to place N
points such that they satisfy the other conditions set in the theorem. This
condition can also be written as N ≤ ⌊2π/ϕ⌋n.
The restriction nδ ≤ ϕ is no problem. For if nδ ≥ ϕ, then (i) follows from
(ii) and can be omitted. Now, |z1+ ··· + zN| attains its maximum on any

Page 4

4Ø. J. RØDSETH
N-term geometric progression with ratio exp(iδ); hence
(4)|z1+ ··· + zN| ≤
??????
N−1
?
j=0
exp(ijδ)
??????
=sin(Nδ/2)
sin(δ/2)
.
The upper bound (3) is attained on the union of two finite 2-dimensional
geometric progressions on U, one consisting of geometric progressions with
ratio exp(iδ) and r terms each, centered around the points
exp((−κ + 2j)iϕ/2),j = 0,1,...,κ,
and the other consisting of geometric progressions with ratio exp(iδ) and
n − r terms each, centered around the points
exp((−κ + 1 + 2j)iϕ/2),j = 0,1,...,κ − 1.
This shows that Theorem 3 is sharp for N ≤ ⌊2π/ϕ⌋n.
Clearly, the bound (3) in Theorem 3 can be replaced by the weaker, but
smooth and nice, bound
|z1+ ··· + zN| ≤sin(nδ/2)
sin(δ/2)
·sin(ϕN/(2n))
sin(ϕ/2)
;
cf. Lev [6].
3. Two Proofs of Freiman’s Lemma
For real numbers α < β, the set of z ∈ U satisfying α < argz ≤ β for
some value of argz, is an open-closed arc. In Freiman’s Lemma one often
assumes that any open-closed (or closed-open) semi-circle of U contains at
most n of the points zj, instead of taking open semi-circles; cf. [4], [9]. It is,
however, easy to see that the two variants of the hypotheses are equivalent;
cf. Section 4.2 below.
Freiman’s proof of Theorem 1 was simplified by Postnikova [10], and it
is this proof we find in the books [4] and [9]. Here, we shall present two
other proofs in an attempt to give the reader a pleasant introduction to
two techniques recently employed in the quest for extensions of Freiman’s
Lemma. The two proofs are rather different. One proof can be characterized
as topological-combinatorial or as a perturbation method, and is due to Lev
(extracted from the proof of [5, Theorem 2]). The other proof uses properties
of a certain Fourier coefficient, and is independently due to Lev and the
present author.

Page 5

EXPONENTIAL SUMS5
3.1. First Proof. Assume that Freiman’s Lemma is false. We consider
the smallest N for which there exists an N-term sequence which satisfies
the hypotheses, but violates (1). Then N > 1. By considering open semi-
circles, the set of N-term sequences satisfying the hypotheses form a closed
subset of the compact topological space UN, and is itself compact. By the
continuity of the function z1,...,zN ?→ |S|, where S := z1+ ··· + zN,
we thus have that |S| attains a maximum value on some N-term sequence
Z := z1,...,zN, which satisfies the hypotheses. Then |S| > 2n − N. A
rotation of Z shows that we may assume that S is real and non-negative.
Suppose that there is a z ?= 1 in Z. By symmetry, we may assume that
Argz < 0, using the interval [−π,π) for the principal argument. Replacing
z by z exp(iε) for a small ε > 0, we get an increase in |S|. Thus the replace-
ment results in violation of the hypotheses in Freiman’s Lemma. The only
possibility is that the replacement produces an open semi-circle with more
than n points from Z; hence −z also belongs to Z.
We now remove ±z from Z. This gives us a sequence Z′satisfying the
hypotheses, and with parameters N′= N −2 and n′= n−1. Denoting the
sum of the terms of Z′by S′, we have by the minimality of N,
S = S′≤ 2n′− N′= 2n − N,
a contradiction. Thus all terms of Z are equal to 1, and N = S > 2n − N.
But a semi-circle containing 1 contains N terms from Z; hence n ≥ N, and
again we have a contradiction.
3.2. Second Proof. Assume that z1,...,zN satisfy the hypotheses. We
shall prove (1), and can without loss of generality assume that S is real and
non-negative. Let K(θ) denote the number of values j ∈ [1,N] such that
θ − π ≤ argzj< θ for some value of argzj. Then we have
(5)K(θ) + K(θ + π) = N,
so that N − n ≤ K(θ) ≤ n.
Moreover,
?π
−π
K(θ)sinθ dθ =
N
?
j=1
?argzj+π
argzj
sinθdθ = 2S,

Page 6

6Ø. J. RØDSETH
and we obtain
2S=
??0
−π
+
?π
?0
0
?
K(θ)sinθdθ
≤(N − n)
−π
sinθdθ + n
?π
0
sinθ dθ
=4n − 2N.
4. Proof of Theorem 3
We now turn to the proof of Theorem 3. In an attempt to make the proof
more readable, we split the proof up into several parts.
4.1. Notation. Set
ω = exp(iϕ/2) andρ = exp(iδ/2).
We often denote a sequence z1,...,zN ∈ UNby Z, and we write S =
z1+···+zN. If the sequence Z ∈ UNsatisfies the assumptions of Theorem 3
for a certain value of n, we say that Z is an (N,n)-admissible sequence.
We use the interval [−π,π) for the principal argument Argz of a non-zero
complex number z.
4.2. Arcs. An arc (u,v), where u,v ∈ U, consists of the points we pass in
moving counter-clockwise from u to v. Almost all arcs in this paper have
lengths at most 2π. In spite of the similarity of notation, an arc cannot be
confused with a real interval.
Consider the two statements
(i) any open arc of U of length ϕ contains at most n of the points zj;
(i′) any open-closed arc of U of length ϕ contains at most n of the points
zj.
Clearly, (i′) implies (i). On the other hand, if (i′) fails, then (i) fails. For if
an open-closed arc (u,uexp(iϕ)], u ∈ U, of U contains n+1 of the points zj,
then, for a sufficiently small real ε > 0, the open arc (uexp(iε),uexp(iϕ+iε))
contains the same n + 1 points. Thus (i) and (i′) are equivalent. (And the
reason is, of course, that we only consider finitely many points zj.)
4.3. Assumptions. The two bounds (3) and (4) coincide for nδ = ϕ. For
the proof of Theorem 3 we may therefore assume that nδ < ϕ. Moreover,
we observe that the right-hand side of (3) is continuous as a function of ϕ
on the real interval (nδ,2π/(κ + 1)]. Hence it suffices to prove the assertion
of Theorem 3 in the case ϕ < 2π/(κ + 1).

Page 7

EXPONENTIAL SUMS7
We shall prove Theorem 3 by contradiction. We therefore assume the
theorem false. Choose the smallest non-negative integer N for which there
exists an n such that (3) fails for some (N,n)-admissible sequence. Then
N > 1.
4.4. Geometric Progressions. A geometric progression ∆ in U with ratio
ρ2is called a δ-progression. If we write this progression as
(6) uρr−1−2j,j = 0,...,r − 1,
for some u ∈ U, then ∆ is a progression of length r, centered around u. The
point u may, or may not, belong to the progression. The point uρr−1is the
first and uρ−(r−1)is the end point (element, term) of the progression. If we
multiply each term of ∆ by v ∈ U, we get a new δ-progression denoted ∆v.
If all terms of ∆ belong to Z, we say that ∆ is in Z. The δ-progression (6)
is maximal in Z, if the progression is in Z, but neither uρr+1nor uρ−r−1
belongs to Z.
Notice that if ∆ is a δ-progression in Z of length r, then r ≤ n. The
reason is the inequality nδ < ϕ, and that an open arc of U of length ϕ
contains at most n points from Z.
4.5. Compactness. The arcs in both (i) and (ii) are open. Therefore, by
a standard compactness argument, |S| attains a maximum on the set of
(N,n)-admissible sequences. Let the maximum be attained at the (N,n)-
admissible sequence Z := z1,...,zN. A rotation of Z shows that we may
assume that the matching S = z1+ ··· + zNis real and non-negative.
4.6. Dispersion. Recall that N > 1 and N = κn + r, 1 ≤ r ≤ n. Suppose
that Z is a δ-progression. The length of this δ-progression is N.
If κ = 0, then
|z1+ ··· + zN| ≤sin(rδ/2)
sin(δ/2),
and (3) holds, contrary to hypothesis. Thus N ≥ n + 1, and there is a
closed arc of U of length nδ containing n + 1 of the points zj. But we have
nδ < ϕ, so this contradicts condition (i) of Theorem 3. Therefore Z is not
a δ-progression.
4.7. Perturbation. We have just seen that Z is not a δ-progression. This
implies that there exists at least one point z ∈ Z, z ?= 0, such that Argz > 0
and zρ−2?∈ Z, or Argz < 0 and zρ2?∈ Z. Choose such a z, as close to −1

Page 8

8Ø. J. RØDSETH
as possible. By symmetry, we can assume that Argz < 0. We split this case
into the two cases determined by (7) and (8) below.
4.7.1. Case I. First, we assume that there is an integer λ in the interval
1 ≤ λ ≤ κ satisfying
(7)−λϕ
2
≤ Argz < −(λ − 1)ϕ
2
.
The length of the shortest arc (z,z′) for z ?= z′∈ Z is at least δ. Since
zρ2?∈ Z, the arc (z,z′) has length greater than δ. We start the perturbation
procedure by performing a small counter-clockwise rotation of z along the
unit circle; that is, we replace z by z exp(iε) for a small ε > 0. If ε is
sufficiently small, then the length of the open arc (z exp(iε),z′) is still at
least δ, and the rotation of z does not disturb the truth of (ii).
Argz < 0, there will, however, be an increase in |S|. Thus the rotation
violates (i′). Therefore we have zω2∈ Z, and the open-closed arc (z,zω2]
contains exactly n terms from Z. If zω2ρ2∈ Z then the arc (zρ2,zω2ρ2]
contains n + 1 points of Z. Therefore zω2ρ2?∈ Z.
Next, we perform a small rotation of both the points z and zω2simulta-
neously and counter-clockwise along the unit circle. We have
Arg(z + zω2) = Argz +ϕ
Since
2,
and if λ ≥ 2, the rotation makes |S| increase. Hence zω4∈ Z. We also see
that zω4ρ2?∈ Z, and that the arc (zω2,zω4] contains exactly n terms from
Z.
Using the right inequality in (7), we find for 1 ≤ j ≤ λ, that
?j−1
ℓ=0
Arg
?
zω2ℓ
?
= Argz +(j − 1)ϕ
2
< −(λ − j)ϕ
2
≤ 0.
This shows that we may continue the perturbation process until we eventu-
ally obtain
zω2j∈ Zand zω2jρ2?∈ Zforj = 0,1,...,λ.
In addition, for j = 1,2,...,λ, each arc (zω2(j−1),zω2j] contains exactly n
points from Z.
By the left inequality in (7), we have
Arg(zω2λ) ≥ −Argz.
Using the definition of z, it follows that if we start at the point zω2λρ2,
which is in U but not in Z, and move counter-clockwise on U, then the first

Page 9

EXPONENTIAL SUMS9
point in Z we meet, is the end point of a maximal δ-progression in Z with
z as its first element.
The open-closed arc (z,zω2λ] contains λn points from Z, and in addition
we have r′≥ 1 points in the maximal δ-progression in Z with z as first
point. Thus we have
λn + r′= N = κn + r,1 ≤ r′,r ≤ n,
so that λ = κ and r′= r. This completes the first case (7).
4.7.2. Case II. Second, if z does not satisfy (7) for λ = κ, then
− π ≤ Argz < −κϕ
(8)
2,
and we can do one more step in the perturbation process. By the inequality
(κ + 1)ϕ < 2π, we have that the arc (z,zω2(κ+1)] contains exactly (κ + 1)n
points from Z. In addition, we have the point z itself. Thus we have found
(κ+1)n+1 points in Z. Since Z contains at most (κ+1)n points, we have
a contradiction. Thus there are no points in Z satisfying (8).
4.8. Primary Points. For the z ∈ Z defined at the beginning of Sec-
tion 4.7, we now know that for j = 1,2,...,κ, each open-closed arc (zω2(j−1),zω2j]
contains exactly n points from Z. In addition, there is the maximal δ-
progression ∆ of length r with z as first element. This accounts for all
elements in Z.
Thus we have
zω2j∈ Z,j = 0,1,...,κ.
Suppose that zρ−2∈ ∆. The open-closed arc (z,zω2] contains exactly n
points from Z. If zω2ρ−2?∈ Z, then the closed-open arc [zρ−2,zω2ρ−2)
contains n + 1 points from Z, contrary to hypotheses. Thus zω2ρ−2∈ Z.
Continuing in the same manner, we get (with a slight abuse of notation)
∆ω2⊆ Z. Repeating the argument, we ultimately obtain,
∆ω2j⊆ Z for j = 0,1,...,κ.
Notice that for j ≥ 1, the δ-progression ∆ω2jis not necessarily maximal.
We set
κ?
Then N1:= |Z1| = κr + r. We write S1for the sum of all elements in Z1,
and we want to find an upper bound for |S1|. (We may have N1= N, so we
Z1=
j=0
∆ω2j.

Page 10

10Ø. J. RØDSETH
cannot use (3).) We can, however, easily determine the exact value of |S1|,
??????
hypothesis. Thus we have κ ≥ 1 and 1 ≤ r < n.
(9)|S1| =
r−1
?
ℓ=0
ρ−2ℓ
κ
?
j=0
zω2j
??????
=sin(rδ/2)
sin(δ/2)
·sin((κ + 1)ϕ/2)
sin(ϕ/2)
.
If κ = 0 or r = n, then Z = Z1, and by (9), Eq. (3) holds, contrary to
4.9. Secondary Points. We set
Z2= Z \ Z1,
and put N2= |Z2|. Then N2= (κ − 1)(n − r) + (n − r). We have that Z2
is (N2,n − r)-admissible, and since N2< N, we can use (3). Writing S2for
the sum of the terms in Z2, we obtain
(10)|S2| ≤sin((n − r)δ/2)
sin(δ/2)
·sin(κϕ/2)
sin(ϕ/2).
We have
S ≤ |S1| + |S2|,
and (9) and (10) show that (3) holds; again we have reached a contradiction.
This concludes the proof of Theorem 3.
5. Closing Remarks
We can also state Theorem 3 in a slightly different way.
Theorem 4. Let δ,ϕ ∈ (0,π] satisfy nδ ≤ ϕ. Assume that the complex num-
bers z1,...,zN∈ U have the two properties (i) and (ii) stated in Theorem 3.
Then we have
(11)|z1+ ··· + zN| ≤sin((N − (k − 1)n)δ/2)
sin(δ/2)
+sin((kn − N)δ/2)
sin(δ/2)
·sin(kϕ/2)
sin(ϕ/2)
·sin((k − 1)ϕ/2)
sin(ϕ/2)
,
for any positive integer k ≤ 2π/ϕ.
Denote the right-hand side of (11) by Lk. We easily see that
Lk+1− Lk= 2sin((kn − N)δ/2)
sin(δ/2)
·sin(kϕ/2)
sin(ϕ/2)
?
cosnδ
2
− cosϕ
2
?
,
so that
L1≥ ... ≥ L⌈N/n⌉≤ L⌈N/n⌉+1≤ ... ≤ L⌊2π/ϕ⌋.

Page 11

EXPONENTIAL SUMS11
Thus Lkattains its minimum at k = ⌈N/n⌉ (and also at k = N/n + 1 if
n divides N). Therefore Theorem 3 and Theorem 4 are essentially one and
the same.
Now, by letting δ → 0+, we recover the main result of [11, 12].
The following direct extension of Theorem 2 is an immediate consequence
of Theorem 4.
Corollary 1. Let k ≥ 2 be an integer, and let δ ∈ (0,π] satisfy nδ ≤ 2π/k.
Suppose that the complex numbers z1,...,zN ∈ U have the following two
properties:
(b) any open arc of U of length 2π/k contains at most n of them;
(ii) any open arc of U of length δ contains at most one of them.
Then
|z1+ ··· + zN| ≤sin((kn − N)δ/2)
sin(δ/2)
.
If k is even, then Corollary 1 is an easy consequence of Theorem 2. This
does not seem to be the case for k odd. Corollary 1 is sharp for (k − 1)n ≤
N ≤ kn. By letting δ → 0+, we recover the result of Moran and Pollington
[8].
For applications, one would perhaps, although not always necessary, turn
the upper bound for |z1+ ··· + zN| into a lower bound for the number of
terms zjin at least one open arc of U of length ϕ.
In closing, let us consider such a lower bound for the residue class situation
mentioned in the introduction. Then we have an N-set A of residue classes
modulo an integer m > 1. Set δ = 2π/m, and put ϕ = 2π/k for some integer
k ≥ 2.
Let
?
By Corollary 1, there exist integers u,v satisfying u ≤ v < u + m/k such
that the image of the interval [u,v] under the canonical homomorphism
Z → Z/mZ contains n0elements of A, where
?
k
?1A(1) =
a∈A
exp(2πia/m).
n0≥
1
?
N +Arcsin(|?1A(1)|sin(π/m))
π/m
??
;
cf. [5, Corollary 2], [11, Section 6].

Page 12

12Ø. J. RØDSETH
References
[1] B. Green, Review MR2192089(2007a:11029) in MathSciNet.
[2] G. A. Freiman, Inverse problems of additive number theory. On the addition of sets
of residues with respect to a prime modulus (Russian). Doklady Akad. Nauk SSSR
141, 571–573 (1961)
[3] G. A. Freiman, Inverse problems of additive number theory. On the addition of sets of
residues with respect to a prime modulus. Soviet Math.-Doklady 2, 1520–1522 (1961)
[4] G. A. Freiman, Foundations of a Structural Theory of Set Addition. Translations
Math. Monographs, Vol. 37, American Math. Soc., Providence, R. I., 1973.
[5] V. F. Lev, Distribution of points on arcs. Integers 5 (2),#A11, 6 pp. (electronic)
(2005)
[6] V. F. Lev, More on points and arcs. Integers 7 (2), #A24, 3 pp. (electronic) (2007)
[7] V. F. Lev, Large sum-free sets in Z/pZ. Israel J. Math 154, 221–234 (2006)
[8] W. Moran, A. D. Pollington, On a result of Freiman. Preprint 1992.
[9] M. B. Nathanson, Additive Number Theory: Inverse Problems and the Geometry of
Sumsets. Graduate Texts Math. Vol. 165, Springer, New York, 1996.
[10] L. P. Postnikova, Fluctuations in the distribution of fractional parts. Doklady Akad.
Nauk SSSR, 161, 1282–1284 (1965)
[11] Ø. J. Rødseth, Distribution of points on the circle. J. Number Theory 127, (1),
127–135 (2007)
[12] Ø. J. Rødseth, Addendum to “Distribution of points on the circle”. J. Number Theory
128, (6), 1889–1892 (2008)
[13] T. Tao, V. Vu, Additive Combinatorics. Cambridge Studies Adv. Math., Vol. 105,
Cambridge Univ. Press, 2006.
Department of Mathematics, University of Bergen, Johs.Brunsgt. 12,
N-5008 Bergen, Norway
E-mail address: rodseth@math.uib.no
URL: http://math.uib.no/People/nmaoy