Two Erdos problems on lacunary sequences: Chromatic number and Diophantine approximation

Abstract

Let ${n_k}$ be an increasing lacunary sequence, i.e., $n_{k+1}/n_k>1+r$ for some $r>0$. In 1987, P. Erdos asked for the chromatic number of a graph $G$ on the integers, where two integers $a,b$ are connected by an edge iff their difference $|a-b|$ is in the sequence ${n_k}$. Y. Katznelson found a connection to a Diophantine approximation problem (also due to Erdos): the existence of $x$ in $(0,1)$ such that all the multiples $n_j x$ are at least distance $\delta(x)>0$ from the set of integers. Katznelson bounded the chromatic number of $G$ by $Cr^{-2}|\log r|$. We apply the Lov\'asz local lemma to establish that $\delta(x)>cr|\log r|^{-1}$ for some $x$, which implies that the chromatic number of $G$ is at most $Cr^{-1} |\log r|$. This is sharp up to the logarithmic factor.

Full text

arXiv:0706.0223v1 [math.CO] 1 Jun 2007
TWO ERD
˝
OS PROBLEMS ON LACUNARY SEQUENCES:
CHROMATIC NUMBER AND DIOPHANTINE APPROXIMATION
YUVAL PERES AND WILHELM SCHLAG
Abstract. Let {n
k
} be an increasing lacunary sequence, i.e., n
k+1
/n
k
> 1 + ǫ for some
ǫ > 0. In 19 87, P. Erd˝os asked for the chromatic number χ(G) of a graph G with vertex
set Z, where two integers x, y ∈ Z are connected by an edge iff their differe nce |x − y| is
in the se quence {n
k
}. Y. Katznelso n found a connection to a Diophantine approximation
problem (also due to Erd˝os): the exis tence of θ ∈ (0, 1) such that all the multiples n
j
θ are
at least dista nce δ(θ) > 0 from Z. Katznelso n showed that χ(G) ≤ Cǫ
−2
| log ǫ|. We apply
the Lov´asz loca l lemma to establish that δ(θ) > cǫ| log ǫ|
−1
for some θ, which implies that
χ(G) < Cǫ
−1
| log ǫ|. This is sharp up to the logarithmic factor.
1. Introduction
The chromatic number χ(G) of a graph G is the minimal number of colors that can be a ssigned
to the vertices of G so that no edge connects two vertices of the same color. In 1987, Paul
Erd˝os posed the following problem
1
:
Problem A: Let ǫ > 0 be fixed and suppose S = {n
j
}
∞
j=1
is a sequence of positive integers
such that n
j+1
> (1 + ǫ)n
j
for all j ≥ 1. Define a graph G = G(S) with vertex set Z (the
integers) by letting the pair (n, m) be an edge iff | n − m| ∈ S. Is the c hromatic number χ(G)
finite?
There is a related Diophantine approximation problem, po sed earlier by Erd˝os [5]:
Problem B: Let ǫ > 0 and S be as above. Is there a number θ ∈ (0 , 1) so that the sequence
{n
j
θ}
∞
j=1
is not den s e modulo 1 ?
1991 Mathematics Subject Classification. Primary: 05C15 Secondary: 42A55, 11B05.
Key words and phrases. Lacunary sequence, chromatic number, diophantine approximation.
Research of Peres was partially supported by NSF grant DMS-0605166. Research of Schlag was par tially
supported by NSF grant DMS-0 617854.
1
according to Y. Katznelson [10].
1

2 YUVAL PERES AND WILHELM SCHLAG
The relation b etween Problem A and Problem B was discovered by Katznelson [10]: Let δ > 0
and θ ∈ (0, 1) be such that inf
j
kθn
j
k > δ, where k · k denotes the distance to the closest
integer. Partition the circle T = [0, 1) into k = ⌈δ
−1
⌉ disjoint intervals I
1
, . . . , I
k
of length
1
k
≤ δ. Let G be the graph from Problem A and color the vertex n ∈ Z with color j iff
nθ ∈ I
j
(mod 1). Clearly, any two vertices connected by an edge must have different colors.
Therefore, χ(G) ≤ k = ⌈δ
−1
⌉. From this, one can easily deduce that χ(G) < ∞ for any
lacunary sequence with ratio at least 1 + ǫ by partitioning into several subsequences (see the
end of the introduction); however, the bound obtained this way grows exponentially in ǫ
−1
.
Problem B was solved by de Mathan [11] and Pollington [15] and their proofs provide bo unds
on χ(G) that grow polynomially in ǫ
−1
. More precisely, they show that there exists θ ∈ (0, 1)
such that
inf
j≥1
kθn
j
k > cǫ
4
| log ǫ|
−1
where c > 0 is some constant. This bo und was improved by Katznelson [10], who showed
that there exists a θ such that
inf
j≥1
kθn
j
k > cǫ
2
| log ǫ|
−1
. (1.1)
Akhunzhanov and Moshchevitin [1] removed the logarithmic factor on t he right hand side of
(1.1), see also Dubicka s [4].
We can now state the main result of this note.
Theorem 1.1. Suppose S = {n
j
} satisfies n
j+1
/n
j
≥ 1 + ǫ, where 0 < ǫ < 1/4. Then there
exists θ ∈ (0, 1) such that
inf
j≥1
kθn
j
k > cǫ| log ǫ|
−1
, (1.2)
where c > 0 is a universal constant. Therefore, the graph G = G(S) described in Problem A
satisfies χ(G) ≤ 1 + c
−1
ǫ
−1
| log ǫ|.
Up to the | log ǫ|
−1
factor, (1.2) cannot be improved. Indeed, let n
j
= j for j = 1, 2, . . . , ⌊ǫ
−1
⌋
and continue this as a lacunary sequence with ratio 1 + ǫ. It is clear that χ(G) > ⌊ǫ
−1
⌋ in this
case, so that the p ower of ǫ in (1.2) cannot be decreased.
In order to prove Theorem 1.1, we use the Lov´asz local lemma from probabilistic combi-
natorics, see [6] or [2, Chap. 5]. Loosely speaking, given events A
1
, A
2
, . . . in a probability
space, this lemma bounds P

T
N
j=1
A
c
j

from below if the events A
j
have small probability and

TWO ERD
˝
OS PROBLEMS 3
each A
i
is almost independent of most of the others. See the following section for a precise
statement. Theorem 1.1 is established in Section 3.
Finally, we recall how the aforementioned relation between problems A and B yields an
easy proof that χ(G) < ∞ for any ǫ > 0. See [10], [19, Chap. 5] and [17] for variants of this
argument. First suppose that n
j+1
/n
j
> 4 for all j. In this case,
∞
\
j=1
n
θ ∈ T : kθn
j
k >
1
4
o
6= ∅. (1.3)
Indeed, fix some j and notice that the set {θ ∈ T : kθn
j
k >
1
4
} is the union of the middle
halves of intervals [
ℓ
n
j
,
ℓ+1
n
j
] where ℓ = 0, 1, . . . , n
j
− 1. Since n
j+1
> 4n
j
, each such middle half
contains an entire interva l of the for m [
ℓ
′
n
j+1
,
ℓ
′
+1
n
j+1
]. Iterating this yields a sequence of nested
interva ls and establishes (1.3) .
Now suppose just that n
j+1
/n
j
> 1 + ǫ > 1. Pick K = ⌈2ǫ
−1
⌉ so that (1 + ǫ)
K
> 4. Divide
the given sequence S into K subsequences {n
Kj+r
}
∞
j=0
, with r = 1, 2, . . . , K. Applying (1.3)
to each such subsequence yields (θ
1
, . . . , θ
K
) ∈ T
K
so that
inf
j≥0
kn
Kj+r
θ
r
k ≥
1
4
for all r = 1, 2, . . . , K.
Coloring each integer m according to which quarter of the unit interval mθ
r
falls into for
1 ≤ r ≤ K, shows that χ(G) ≤ 4
K
. Observe that as ǫ → 0, this bound grows exponentially
in
1
ǫ
.
2. A one-sided version of the local lemma
The following lemma is the variant of the Lov´asz local lemma [6] that we apply to Problem B
above. Since it is not exactly stated in this form (neither in terms of the hypotheses nor the
conclusion) in [6] or [2], we provide a proof for the reader’s convenience. We stress, however,
that it is a simple adaptation of the argument given in chapter 5 of [2].
Lemma 2.1. Let {A
j
}
N
j=1
be events in a p robability space (Ω, F, P) and let {x
j
}
N
j=1
be a
sequence of numbers in (0, 1). Assume that for every i ≤ N, there is an integ er 0 ≤ m(i) < i
so that
P

A
i



\
j<m(i)
A
c
j

≤ x
i
i−1
Y
j=m(i)
(1 − x
j
). (2.1)

4 YUVAL PERES AND WILHELM SCHLAG
Then for any integer n ∈ [1, N], we hav e
P

n
\
i=1
A
c
i

≥
n
Y
ℓ=1
(1 − x
ℓ
). (2.2)
Proof. Denote B
1
= Ω and B
ℓ
= A
c
1
∩ . . . ∩ A
c
ℓ−1
for ℓ > 1. We claim that for each ℓ ≥ 1,
P(A
ℓ
|B
ℓ
) ≤ x
ℓ
. (2.3)
Since
P

n
\
i=1
A
c
i

=
n
Y
ℓ=1

1 − P(A
ℓ
|B
ℓ
)

,
(2.3) implies (2.2). The claim (2.3) is verified inductively:
P

A
ℓ



B
ℓ

=
P

A
ℓ
∩ B
ℓ



B
m(ℓ)

P

B
ℓ



B
m(ℓ)

≤
P

A
ℓ



B
m(ℓ)

P

B
ℓ



B
m(ℓ)

. (2.4)
The denominator in the rightmost fraction can be written as a product
P

B
ℓ



B
m(ℓ)

=
ℓ−1
Y
j=m(ℓ)

1 − P(A
j
|B
j
)

. (2.5)
By the inductive hypothesis, this product is at least
Q
ℓ−1
j=m(ℓ)
(1−x
j
), whereas the numerator in
the right-hand side of (2.4) is at most x
ℓ
Q
ℓ−1
j=m(ℓ)
(1 − x
j
) by (2 .1). This finishes the proof. 
3. Rotation orbits sampled along a lacunary sequence
In this section we present our quantita t ive result on problem B, which extends Theorem 1.1
and is also a pplicable to unions of lacunary sequences. Observe that if S = {n
j
} is lacunary
with ratio 1 + ǫ, then it satisfies the hypothesis of the next theorem with M = ⌈ǫ
−1
⌉. More
generally, if S
1
, . . . , S
ℓ
are lacunary with ratios 1 + ǫ
1
, . . . , 1 + ǫ
ℓ
respectively, then their union
satisfies the hypothesis with M =
P
ℓ
i=1
⌈ǫ
−1
i
⌉. We shall assume that M ≥ 4.
Theorem 3.1. Suppose S = {n
j
} satisfies n
j+M
> 2n
j
for all j. Define
E
j
=

θ ∈ T : kn
j
θk <
c
0
M log
2
M

(3.1)

TWO ERD
˝
OS PROBLEMS 5
for j ≥ 1. If 240 c
0
≤ 1, then
∞
\
j=1
E
c
j
6= ∅. (3.2)
Proof. Set
δ =
c
0
M log
2
M
. (3.3)
For each j = 1, 2, . . . define an integer ℓ
j
by
2
−ℓ
j
−1
<
2δ
n
j
≤ 2
−ℓ
j
. (3.4)
Let A
j
be the union of a ll the open dyadic intervals of size 2
−ℓ
j
that intersect E
j
. Observe
that E
j
is the union of n
j
interva ls of length
2δ
n
j
, and each one of them is covered by at most
two dyadic intervals of length 2
−ℓ
j
. Therefore,
P(A
j
) ≤ 2 · 2
−ℓ
j
n
j
≤ 8δ. (3.5)
where P is Lebesgue measure on [0, 1]. D efine
h = ⌈C
1
log
2
M⌉M, (3.6)
where C
1
≥ 5 is a constant to be determined. Our goa l is to apply Lemma 2.1 with m(i) = i−h
and x
i
= x = h
−1
for all i ∈ Z
+
. To verify (2.1), fix some i > h. Then
\
j<i−h
A
c
j
=
[
s
I
s
with dyadic intervals I
s
of length |I
s
| = 2
−ℓ
i−h−1
. Hence
P

A
i
∩
\
j<i−h
A
c
j

=
X
s
P(A
i
∩ I
s
) ≤
X
s
(1 + |I
s
|n
i
)2
1−ℓ
i
≤ P

\
j<i−h
A
c
j
h
2 n
i
2
−ℓ
i
+ 2
1+ℓ
i−h−1
−ℓ
i
i
≤ P

\
j<i−h
A
c
j
h
8δ + 4
n
i−h−1
n
i
i
. (3.7)
To pass to (3.7), one uses (3.4). By (3.6),
n
i−h−1
n
i
≤ 2
−C
1
log
2
M
= M
−C
1
. (3.8)

6 YUVAL PERES AND WILHELM SCHLAG
Inserting this bound into (3.7) yields
P

A
i



\
j<i−h
A
c
j

≤ 12δ, provided that (3.9)
c
0
M
C
1
M log
2
M
≥ 1, which is the same as M
−C
1
≤ δ. (3.10)
By (3.5), the estimate (3.9) holds also if i ≤ h. In order to satisfy (2.1) , we need to ensure
that
12δ ≤ x(1 − x)
h
. (3.11)
Since x = h
−1
≤ 1/16, we have (1 − x)
h
≥ 1/3. Thus (3.11) will be satisfied if
36δ ≤ x = h
−1
. (3.12)
By (3.3) and (3.6), hδ = ⌈C
1
log
2
M⌉ c
0
/ log
2
M ≤
10
9
C
1
c
0
, since M ≥ 4 and C
1
≥ 5. There-
fore, (3.12) will hold if
40C
1
c
0
≤ 1 . (3.13)
Take C
1
= 6 and c
0
=
1
240
; then (3.13) and (3.10) are both satisfied. By Lemma 2.1,
P

n
\
i=1
A
c
i

≥

1 − x

n
. (3.14)
Since each of the A
c
j
is compact, and A
c
j
⊂ E
c
j
, the intersection ∩
∞
j=1
E
c
j
is nonempty, as
claimed. 
4. Intersective sets
Let S = {n
j
} with n
j+1
≥ (1 + ǫ)n
j
be a la cunary sequence of positive integers with ratio
1 + ǫ > 1. Denote M = ⌈ǫ
−1
⌉. We have shown that there is a coloring of the graph G with at
most CM log M colors. Let A
max
be a set of integers of the same color with upper density
D
∗
(A
max
) = lim sup
N→∞
card(A
max
∩ [−N, N])
2N + 1
>
c
M log M
,
where c is a constant. By the definition of coloring,
(A
max
− A
max
) ∩ S = ∅.
A set H is called intersective if
(A − A) ∩ H 6= ∅

TWO ERD
˝
OS PROBLEMS 7
for any A ⊂ Z with D
∗
(A) > 0. Intersective sets are precisely the Poincar´e sets considered
by F¨urstenberg [8], see [3] and [13]. Generally speaking, it is not a simple matter to decide
whether a given set is intersective or not. F¨urstenberg [7] and S´ark¨ozy [18] showed that the
squares (and more generally, the set {P (n)}
n∈Z
where P is a polynomial over Z that vanishes
at some integer) are intersective. In particular, intersective sets can have zero density. On
the other hand, the previous discussion shows that any finite union of lacunary sequences is
not intersective. There are some related concepts of intersectivity which we briefly recall; for
a nice introduction to this subject see chapter 2 in Montgomery’s book [12]. A set of integers
H is called a van der Corput set, if for any sequence {x
j
}
j
of numbers with the property
that {x
j+h
− x
j
}
j
is uniformly distributed modulo 1 for all h ∈ H, one has that {x
j
}
j
is
uniformly distributed modulo 1. It was shown by Kamae and Mendes–France [9] that any
van der Corput set is intersective. There is a more quantitative version of this fact, due to
Ruzsa [16]: Define
γ
H
= inf
Z
1
0
T (x) dx where T (x) = a
0
+
X
h∈H
a
h
cos(2πhx) ≥ 0 and T (0) = 1,
T being a trigonometric polynomial. It it is known that H is a van der Corput set iff γ
H
= 0 ,
see [16] and [12]. Also, let
δ
H
= sup{D
∗
(A) : (A − A) ∩ H = ∅}.
By definition, H is intersective iff δ
H
= 0 . It was shown in [16] that δ
H
≤ γ
H
. In view of this
fact, our Theorem 1.1 has the following consequence.
Corollary 4.1. Let S = {n
j
}
j
be lacunary with ratio 1 + ǫ, and suppose T is a non negative
trigonometric polynomial of the form
T (x) = a
0
+
X
j
a
j
cos(2πn
j
x) with T (0) = 1.
Then for some universal constant c > 0,
Z
1
0
T (x) dx = a
0
> cǫ | log ǫ|
−1
. (4.1)
If one could show that the bound given by (4.1) is optimal, then it would follow that the
| log ǫ|
−1
factor in (1.2) cannot be removed.

8 YUVAL PERES AND WILHELM SCHLAG
Remark. The first author heard Y. Katznelson present his proof of (1.1) in a lecture at
Stanford in 19 91, but that proof only appeared ten years later [10]. The proof of our main
result, Theorem 1.1, was obtained in 1999 and presented in a lecture [14] at the IAS, Princeton
in 2000. We thank J. Gr ytczuk for urging us t o publish this result and providing references
for recent work on related problems.
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http://www.math.ias.edu∼warfield/abstr2000.html#peres
[15] Pollington, A. D. On the density of the sequence {n
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TWO ERD
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OS PROBLEMS 9
[17] Ruzsa, I. Z., Tuza, Z. and Voigt, M. Distance Graphs with Finite Chromatic Number. Jour. Combin.
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Soc., Providence, RI (2000).
Microsoft Research, Redmond, WA and University of California, Berkeley, CA
E-mail address: peres@microsoft.com
Department of Mathematics, University of Chicago.
E-mail address: schlag@math.uchicago.edu