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arXiv:1012.0675v1 [math.NT] 3 Dec 2010
Multiplicative zero-one laws and
metric number theory
Victor Beresnevich∗
Alan Haynes†
Sanju Velani‡
Abstract
We develop the classical theory of Diophantine approximation without
assuming monotonicity or convexity. A complete ‘multiplicative’ zero-one
law is established akin to the ‘simultaneous’ zero-one laws of Cassels and
Gallagher. As a consequence we are able to establish the analogue of the
Duffin-Schaeffer theorem within the multiplicative setup. The key ingredient
is the rather simple but nevertheless versatile ‘cross fibering principle’. In a
nutshell it enables us to ‘lift’ zero-one laws to higher dimensions.
Keywords: Zero-one laws, metric Diophantine approximation
Subject classification: 11J13, 11J83, 11K60
1Introduction
The theory of multiplicative Diophantine approximation is concerned with the set
S×
n(ψ) := {(x1,...,xn) ∈ [0,1]n:
n
?
i=1
?qxi? < ψ(q) for i.m. q ∈ N},
where ?qx? = min{|qx−p| : p ∈ Z}, ‘i.m.’ means ‘infinitely many’ and ψ : N → R≥0
is a a non-negative function. For obvious reasons the function ψ is often referred
to as an approximating function. For convenience, we work within the unit cube
[0,1]nrather than Rn; it makes full measure results easier to state and avoids
ambiguity. In fact, this is not at all restrictive since the set under consideration is
invariant under translation by integer vectors.
Multiplicative Diophantine approximation is currently an active area of re-
search. In particular, the long standing conjecture of Littlewood that states that
S×
references within. In this paper we will address the multiplicative analogue of yet
another long standing classical problem; namely, the Duffin-Schaeffer conjecture.
2(q ?→ εq−1) = R for any ε > 0 has attracted much attention – see [1, 16, 18] and
∗EPSRC Advanced Research Fellow, grant EP/C54076X/1
†Research supported by EPSRC grant EP/F027028/1
‡Research supported by EPSRC grants EP/E061613/1 and EP/F027028/1
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Given q ∈ N and x ∈ R, let
?qx?′:= min{|qx − p| : p ∈ Z, (p,q) = 1},
and consider the standard simultaneous sets
Dn(ψ) := {(x1,...,xn) ∈ [0,1]n:?
max
1≤i≤n?qxi?′?n< ψ(q) for i.m. q ∈ N}
and
Sn(ψ) := {(x1,...,xn) ∈ [0,1]n:?
An elegant measure theoretic property of these sets is that they are always of zero
or full Lebesgue measure | . | irrespective of the dimension or the approximating
function. Formally, for n ≥ 1 and any non-negative function ψ : N → R≥0
max
1≤i≤n?qxi??n< ψ(q) for i.m. q ∈ N}.
|Sn(ψ)| ∈ {0,1}
and
|Dn(ψ)| ∈ {0,1}.(1)
The former zero-one law is due to Cassels [7] while the latter is due to Gallagher
[10] when n = 1 and Vilchinski [19] for n arbitrary. By making use of a refined
version of Cassels’ zero-one law, Gallagher [12] proved that for n ≥ 2
|Sn(ψ)| = 1if
∞
?
q=1
ψ(q) = ∞ .(2)
Remark. Regarding the above statement and indeed the statements and conjec-
tures below, by making use of the Borel-Cantelli Lemma from probability theory,
it is straightforward to establish the complementary convergent results; i.e. if the
sum in question converges then the set in question is of zero measure.
The case that n = 1 is excluded from the statement given by (2) since it is false.
Indeed, Duffin & Schaeffer [8] gave a counterexample and formulated an alternative
appropriate statement. The Duffin-Schaeffer conjecture1states that
|Dn(ψ)| = 1if
∞
?
q=1
?ϕ(q)
q
?n
ψ(q) = ∞ ,(3)
where ϕ is the Euler phi function. The consequence of the zero-one law for Dn(ψ)
is that it reduces the Duffin-Schaeffer conjecture to showing that |Dn(ψ)| > 0.
Using this fact the conjecture has been established in the case n ≥ 2 by Pollington
& Vaughan [15]. Although various partial results have been obtained in the case
n = 1, the full conjecture represents a key unsolved problem in number theory.
For background and recent developments regarding this fundamental problem see
1To be precise Duffin and Schaeffer stated their conjecture for n = 1. The higher dimensional
version is attributed to Sprindˇ zuk – see [17, pg63].
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[2, 8, 13, 14]. However, it is worth highlighting the Duffin-Schaeffer theorem which
states that (3) holds whenever
limsup
Q→∞
?Q
q=1
?
?ϕ(q)
q
?
ψ(q)
??Q
?
q=1
ψ(q)
?−1
> 0 .
Note that this condition implies that the convergence/divergence properties of the
sums in (2) and (3) are equivalent.
As already mentioned, the purpose of this paper is to consider the multiplica-
tive setup and in particular, the multiplicative analogue of the Duffin-Schaeffer
conjecture. With this in mind, it is natural to define the set
D×
n(ψ) := {(x1,...,xn) ∈ [0,1]n:
n
?
i=1
?qxi?′< ψ(q) for i.m. q ∈ N}.
The ultimate goal is to prove the following two statements.
Conjecture 1 Let n ≥ 2 and ψ : N → R≥0be a non-negative function. Then
|S×
n(ψ)| = 1
if
∞
?
q=1
ψ(q)logn−1q = ∞. (4)
Conjecture 2 Let n ≥ 1 and ψ : N → R≥0be a non-negative function. Then
|D×
n(ψ)| = 1
if
∞
?
q=1
?ϕ(q)
q
?n
ψ(q)logn−1q = ∞.(5)
In view of the Duffin-Schaeffer counterexample it is necessary to exclude n = 1
from the statement of Conjecture 1. Clearly, the Duffin-Schaeffer conjecture and
Conjecture 2 coincide when n = 1.
Remark. For n ≥ 2, the results of Gallagher and Pollington & Vaughan establish
the analogues of the above conjectures for the standard simultaneous sets Sn(ψ)
and Dn(ψ).
1.1 The story so far: convexity versus monotonicity
Throughout this section, assume that n ≥ 2. Geometrically, the multiplicative sets
S×
n(ψ) consist of points in the unit cube that lie within infinitely many
‘hyperbolic’ domains
n(ψ) and D×
H = H(ψ,p,q) := {x ∈ Rn:?n
centered around rational points p/q where p = (p1,...,pn) ∈ Znand q ∈ N. In
the case of D×
n(ψ) we impose the additional co-primeness condition (pi,q) = 1 on
i=1|xi− pi/q| < ψ(q)/qn}
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the rational points. The approximating function ψ governs the size of the domains
H. In the case of the standard simultaneous sets Sn(ψ) and Dn(ψ) the domains H
are replaced by the ‘cubical’ domains
C = C(ψ,p,q) :=
?
x ∈ Rn:?
max
1≤i≤n|xi− pi/q|?n< ψ(q)/qn
?
.
The significant difference between the standard and multiplicative situation
is that the domains C are convex while the domains H are non-convex. It is this
difference that lies behind the fact that Conjectures 1 & 2 are still open whilst their
standard simultaneous counterparts have been established – recall we assuming
that n ≥ 2. In short, without imposing additional assumptions, convexity is vital
in the methods employed by Gallagher and Pollington & Vaughan to establish (2)
and (3) respectively. Indeed, their methods can be refined and adapted to deal
with limsup sets arising from more general convex domains but convexity itself
seems to be unremovable – see [13, Chp.3] and references within. However, the
landscape is completely different if we impose the additional assumption that the
approximating function ψ is monotonic. For instance we can then overcome the fact
that the domains H associated with the sets S×
and Conjectures 1 & 2 correspond to a well known theorem of Gallagher [11].
In fact, Gallagher considers limsup sets arising from more general domains but
monotonicity plays a crucial role in his approach and seems to be unremovable.
Note that for monotonic ψ the convergence/divergence properties of the sums
appearing in (4) and (5) are equivalent and since S×
Conjecture 2 implies Conjecture 1.
The upshot is that the current body of metrical results for limsup sets re-
quires that either the approximating domains are convex or that the approximating
function is monotonic. We stress that this includes existing zero-one laws.
n(ψ) and D×
n(ψ) are non-convex
n(ψ) ⊃ D×
n(ψ) it follows that
1.2 Statement of results
Our first theorem is the multiplicative analogue of the Cassels-Gallagher zero-one
law. It reduces Conjectures 1 & 2 to showing that the corresponding sets are of
positive measure. In principal, it is easier to prove positive measure statements
than full measure statements. More to the point, there is a well established mech-
anism in place to obtain lower bounds for the measure of limsup sets – see §4
below or [3, §8] for a more comprehensive account.
Theorem 1 Let n ≥ 1 and ψ : N → R≥0be a non-negative function. Then
|S×
n(ψ)| ∈ {0,1}
and
|D×
n(ψ)| ∈ {0,1}.
The proof will rely on the general technique developed in §2 which we refer to as
the cross fibering principle. Given its simplicity, we suspect that it may well have
applications elsewhere in one form or another.
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The following theorem represents our ‘direct’ contributions to Conjectures 1
& 2 and is the complete multiplicative analogue of the Duffin-Schaeffer theorem.
Theorem 2 Let n ≥ 1 and ψ : N → R≥0be a non-negative function. Then
|S×
n(ψ)| = 1 = |D×
n(ψ)|
if
∞
?
q=1
ψ(q)logn−1q = ∞
and
limsup
Q→∞
?Q
q=1
?
?ϕ(q)
q
?n
ψ(q)logn−1q
??Q
?
q=1
ψ(q)logn−1q
?−1
> 0 . (6)
Note that the ‘additional’ assumption (6) implies that the convergence/divergence
properties of the sums within Conjectures 1 & 2 are equivalent.
Remark. Theorem 2 enables us to establish the complete analogue of Gallagher’s
multiplicative theorem [11] within the framework of the ‘p-adic Littlewood Con-
jecture’ – see §4.1.
2 Cross Fibering Principle
Let X and Y be two non-empty sets. Let S ⊂ X × Y . Given x ∈ X, the set
Sx:= {y : (x,y) ∈ S} ⊂ Y
will be called a fiber of S through x. Similarly, given y ∈ Y , the set
Sy:= {x : (x,y) ∈ S} ⊂ X
will be called a fiber of S through y. Given a measure µ over X, we will say that
A ⊂ X is µ-trivial if A is either null or full with respect to µ; that is
µ(A) = 0orµ(X \ A) = 0.
It is an immediate consequence of Fubini’s theorem (see below) that
S is µ × ν-trivial=⇒
µ-almost every fiber Sxis ν-trivial,(7)
and likewise
S is µ × ν-trivial=⇒
ν-almost every fiber Syis µ-trivial. (8)
Neither of these implications can be reversed in their own right. However, if the
right hand side statements are combined together then we actually have a criterion
which we will refer to as the cross fibering principle.
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Theorem 3 Let µ be a σ-finite measure over X, ν be a σ-finite measure over Y
and S ⊂ X × Y be a µ × ν-measurable set. Then
S is µ × ν-trivial
⇐⇒
µ-almost every fiber Sxis ν-trivial
&
ν-almost every fiber Syis µ-trivial.
(9)
The proof of this theorem will make use of the following general form of
Fubini’s theorem which can be found in [5, pg.233] and [9, §2.6.2].
Fubini’s Theorem Let µ be a σ-finite measure over X and ν be a σ-finite mea-
sure over Y . Then µ × ν is a regular measure over X × Y such that
(i) If A is a µ-measurable set and B is a ν-measurable set then A × B is a
µ × ν-measurable set and
(µ × ν)(A × B) = µ(A) · ν(B).
(ii) If S is a µ × ν-measurable set, then
Sy
is µ-measurable for ν-almost all y,
Sx
is ν-measurable for µ-almost all x,
the functions
X → R : x ?→ ν(Sx)
and
Y → R : y ?→ µ(Sy)(10)
are integrable and
(µ × ν)(S) =
?
µ(Sy)dν =
?
ν(Sx)dµ.(11)
2.1Proof of Theorem 3
The measures µ and ν are σ-finite. Thus, without loss of generality we can assume
that the measures are finite and indeed that they are probability measures; that
is
µ(X) = 1 = ν(Y ).
Necessity (=⇒). Without loss of generality, we can assume that (µ × ν)(S) = 0
since otherwise we can replace S by its complement X \ S. Therefore, both the
integrals appearing in (11) vanish. Note that the integrals themselves are obtained
by integrating the non-negative functions (10). The upshot is that these functions
vanish almost everywhere with respect to the appropriate measures which in turn
implies the right hand side of (9).
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Sufficiency (⇐=). Let˜ X be the set of x ∈ X such that Sxis ν-measurable and
trivial. Similarly, let˜Y be the set of y ∈ Y such that Syis µ-measurable and trivial.
In view of part (ii) of Fubini’s theorem and the right hand side of (9) we have that
both˜ X and˜Y are sets of full measure; that is µ(X \˜ X) = 0 and ν(Y \˜Y ) = 0.
In particular,˜ X is µ-measurable and˜Y is ν-measurable. Now partition˜ X and˜Y
into two disjoint subsets as follows:
X0:= {x ∈˜ X : ν(Sx) = 0},Y0:= {y ∈˜Y : µ(Sy) = 0},
X1:=˜ X \ X0= {x ∈˜ X : ν(Sx) = 1},Y1:=˜Y \ Y0= {y ∈˜Y : ν(Sy) = 1}.
Let XAdenote the characteristic function of a set A. By definition and part (ii) of
Fubini’s theorem, the functions (10) almost everywhere coincide with the functions
XX1and XY1. Since the functions (10) are integrable, the functions XX1and XY1
are also integrable and so it follows that the sets X1 and Y1 are respectively µ
and ν-measurable. This together with the fact that˜ X and˜Y are respectively µ
and ν-measurable, implies that X0=˜ X \ X1is µ-measurable and Y0=˜Y \ Y1is
ν-measurable.
Obviously µ(X0)+µ(X1) = 1 and ν(Y0)+ν(Y1) = 1. Let us assume that the
sets Xiand Yiare non-trivial. In other words,
0 < µ(Xi) < 1and0 < ν(Yi) < 1fori = 0,1.(12)
By part (i) of Fubini’s theorem, the set M := X0× Y1is µ × ν-measurable. Now
consider the set S∩M and observe that My= X0if y ∈ Y1and My= ∅ otherwise.
Therefore, on using the first equality of (11) we obtain that
(µ × ν)(S ∩ M) =
?
µ(Sy∩ My)dν =
?
µ(Sy∩ X0) XY1(y)dν.(13)
By definition, for y ∈ Y1 the set Syis full in X and thus is full in X0. As a
consequence, we have that µ(Sy∩ X0) = µ(X0) for y ∈ Y1. Therefore, (12) and
(13) imply that
(µ × ν)(S ∩ M) =
?
µ(X0)XY1(y)dν = µ(X0)ν(Y1) > 0. (14)
On the other hand, observe that Mx= Y1if x ∈ X0and Mx= ∅ otherwise. Then,
on using the second equality of (11) we obtain that
(µ × ν)(S ∩ M) =
?
ν(Sx∩ Mx)dµ =
?
ν(Sx∩ Y1)XX0(x)dµ.(15)
By definition, for x ∈ X0 the set Sx is null and so ν(Sx∩ Y1) = 0 for x ∈ X0.
Therefore, (15) implies that
(µ × ν)(S ∩ M) =
?
0dµ = 0.
This contradicts (14). Therefore at least one of the sets Xiand Yimust be trivial.
This together with (11) implies that S is trivial and thereby completes the proof.
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3Proof of Theorem 1
The proof is by induction. Consider the set S×
S×
is one-dimensional Lebesgue measure on X := [0,1] .
Now assume that n > 1 and that Theorem 1 is true for all dimensions k < n.
Given a k-tuple (x1,...,xk) ∈ [0,1]k, consider the function
n(ψ). When n = 1, we have that
1(ψ) is µ-trivial where µ
1(ψ) = S1(ψ) and Cassels’ zero-one law implies that S×
ψ(x1,...,xk)(q) :=
ψ(q)
?qx1?...?qxk?.
Here we adopt the convention that α/0 := +∞ if α > 0 and that α/0 := 0 if
α = 0. With reference to §2, let Y := [0,1]n−1and let ν be (n − 1)-dimensional
Lebesgue measure on Y . Furthermore, let S := S×
that for any x1 ∈ X the fiber Sx1is equal to the set S×
for any (x2,...,xn) ∈ Y the fiber S(x2,...,xn)is equal to the set S×
In view of the induction hypothesis, we have that Sx1is µ-trivial and S(x2,...,xn)
is ν-trivial. Therefore, by Theorem 3 it follows that S is µ × ν-trivial. In other
words, the n-dimensional Lebesgue measure of S×
establishes Theorem 1 for the set S×
n(ψ).
n(ψ). Then it is readily verified
1(ψ(x1)) and similarly
n−1(ψ(x2,...,xn)).
n(ψ) is either zero or one. This
Apart from obvious notational changes, the proof for the set D×
the same as above except for that fact that when n = 1 we appeal to Gallagher’s
zero-one law rather than Cassels’ zero-one law.
n(ψ) is exactly
3.1A multiplicative zero-one law for linear forms
In what follows m ≥ 1 and n ≥ 1 are integers. Given a ‘multi-variable’ approxi-
mating function Ψ : Zn→ R≥0, let S×
that
Π(qX + p) < Ψ(q)
n,m(Ψ) denote the set of X ∈ [0,1]mnsuch
(16)
holds for infinitely many (p,q) ∈ Zn× Zm? {0}. Here Π(y) :=?n
as a row vector. Thus, qX ∈ Rnrepresents a system of n real linear forms in m
variables. Naturally, let D×
X ∈ [0,1]mnfor which (16) holds infinitely often with the additional co-primeness
condition (pi,q) = 1 for all 1 ≤ i ≤ n. Clearly, when m = 1 and Ψ(q) = ψ(|q|) the
sets S×
The following statement is the natural generalisation of Theorem 1 to the
linear forms framework. It also gives a positive answer to Question 4 raised in [4].
i=1|yi| for a
vector y = (y1,...,yn) ∈ Rn, X is regarded as an m×n matrix and q is regarded
m,n(Ψ) denote the subset of S×
m,n(Ψ) corresponding to
m,n(Ψ) and S×
n(ψ) coincide as do the sets D×
m,n(Ψ) and D×
n(ψ).
Theorem 4 Let m,n ≥ 1 and Ψ : Zn→ R≥0be a non-negative function. Then
|S×
m,n(Ψ)| ∈ {0,1}
and
|D×
m,n(Ψ)| ∈ {0,1}.
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In view of the linear forms version of the Cassels-Gallagher zero-one law estab-
lished in [4], the proof of Theorem 4 is pretty much the same as the proof of
Theorem 1 with obvious modification. More specifically, all that is required from
[4] is Theorem 1 with n = 1.
4 Proof of Theorem 2
To begin with, observe that S×
theorem for D×
n(ψ). In view of Theorem 1, we are done if we can show that
n(ψ) ⊃ D×
n(ψ) and therefore is suffices to prove the
|D×
n(ψ)| > 0. (17)
With reference to §1.1, given q ∈ N let
H(ψ,q) := [0,1]n
∩
?
p=(p1,...,pn)∈Zn:
(pi,q)=1
H(ψ,p,q)
Then, by definition
D×
n(ψ) = limsup
q→∞
H(ψ,q).
The following lemma provides a mechanism for establishing lower bounds for the
measure of limsup sets. The statement is a generalisation of the divergent part
of the standard Borel-Cantelli lemma in probability theory, see for example [17,
Lemma 5].
Lemma 1 Let (Ω,A,µ) be a probability space and {Ek} ⊆ A be a sequence of sets
such that?∞
Ek) ≥ limsup
Q→∞
k=1µ(Ek) = ∞. Then
µ(limsup
k→∞
??Q
?Q
s=1µ(Es)
?2
s,t=1µ(Es∩ Et)
.
In view of Lemma 1, the desired statement (17) will follow on showing that
the sets H(ψ,q) are pairwise quasi-independent on average and that the sum of
their measures diverges. It is easily verified that2
|H(ψ,q)| ≍
?ϕ(q)
q
?n
ψ(q)logn−1q (18)
and thus (6) together with the divergent sum hypothesis implies that
∞
?
q=1
|H(ψ,q)| = ∞.
2The Vinogradov symbols ≪ and ≫ indicate an inequality with an unspecified positive mul-
tiplicative constant. If a ≪ b and a ≫ b we write a ≍ b, and the quantities a and b are said to
be comparable.
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Regarding pairwise quasi-independence on average, Lemma 2 in [11] implies that
|H(ψ,q) ∩ H(ψ,r)| ≪ ψ(q)logn−1q ψ(r)logn−1r ifq ?= r.
Hence, for Q sufficiently large it follows that
Q
?
q,r=1
|H(ψ,q) ∩ H(ψ,r)| ≪
? Q
?
q=1
ψ(q)logn−1q
?2
(6)&(18)
≪
? Q
?
q=1
|H(ψ,q)|
?2
.
This thereby completes the proof of Theorem 2.
4.1 An application to p-adic approximation
Theorems 1 & 2 settle the conjecture and problem stated in [6, §4.5] regarding the
multiplicative set S×
n(ψ). In particular, as a consequence of Theorem 2 we are able
to prove the following generalisation of the main result appearing in [6]. In short
the statement corresponds to the complete analogue of Gallagher’s multiplicative
theorem [11] within the framework of the ‘p-adic Littlewood Conjecture’ – for
further details see [1, 6] and references within.
Theorem 5 Let p1,...,pkbe distinct prime numbers and f1,...,fk: R≥0→ R≥0
be positive functions. Furthermore, let ψ : N → R≥0be a non-negative decreasing
function. Then, for almost every (α1,...,αn) ∈ Rnthe inequality
f1(|q|p1)···fk(|q|pk)?qα1?···?qαn? ≤ ψ(q) ,
has infinitely many solutions q ∈ N if
∞
?
q=1
ψ(q)
f1(|q|p1)···fk(|q|pk)
logn−1q = ∞.
Armed with Theorem 2, the proof is a straightforward adaptation of the ideas
used to establish the n = 1 case [6, Theorem 2].
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Victor V. Beresnevich: Department of Mathematics, University of York,
Heslington, York, YO10 5DD, England.
e-mail: vb8@york.ac.uk
Alan K. Haynes: Department of Mathematics, University of York,
Heslington, York, YO10 5DD, England.
e-mail: akh502@york.ac.uk
Sanju L. Velani: Department of Mathematics, University of York,
Heslington, York, YO10 5DD, England.
e-mail: slv3@york.ac.uk
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