Littlewood and Duffin--Schaeffer-Type Problems in Diophantine

LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS IN DIOPHANTINE APPROXIMATION

SAM CHOW AND NICLAS TECHNAU

Dedicated to Andy Pollington

Abstract. Gallagher’s theorem describes the multiplicative diophantine approximation rate of a typical vector. We establish a fully-inhomogeneous version of Gallagher’s theorem, a diophantine fibre refinement, and a sharp and unexpected threshold for Liouville fibres. Along the way, we prove an inhomogeneous version of the Duffin–Schaeffer conjecture for a class of non-monotonic approximation functions.

Table of Contents

1. Introduction 2

1.1. Main results 6

1.2. Key ideas and further results 9

1.3. Open problems 14

1.4. Organisation and notation 17

2. Preliminaries 18

2.1. Continued fractions 18 arXiv:2010.09069v1 [math.NT] 18 Oct 2020 2.2. Ostrowski expansions 20

2.3. Bohr sets 23

2.4. Measure theory 25

2.5. Real analysis 25

2.6. Geometry of numbers 27

2020 Mathematics Subject Classiﬁcation. 11J83, 11J54, 11H06, 52C07, 11J70. Key words and phrases. Metric diophantine approximation, geometry of numbers, additive combinatorics, continued fractions. 1 2 SAM CHOW AND NICLAS TECHNAU

2.7. Primes and sieves 30

3. A fully-inhomogeneous version of Gallagher’s theorem 32

3.1. Notation and reduction steps 32

3.2. Divergence of the series 34

3.3. Overlap estimates, localised Bohr sets, and the small-GCD regime 40

3.4. Large GCDs 47

3.5. A convergence statement 52

4. Liouville ﬁbres 54

4.1. A special case 54

4.2. Diophantine second shift 56

4.3. Liouville second shift 61

4.4. Rational second shift 65

5. Obstructions on Liouville ﬁbres 65

Appendix A. Pathology 70

References 72

1. Introduction This manuscript concerns two fundamental problems in diophantine approximation. We introduce a method to tackle, in a general context, inhomogeneous versions of Littlewood’s conjecture which are metric in at least one parameter. Further, we prove an inhomogeneous version of the Duﬃn–Schaeﬀer conjecture for a relatively large class of functions.

Let us begin by explaining the link to Littlewood’s conjecture, and defer elaborating on the inhomogeneous Duffin–Schaeffer conjecture to Section 1.2. Around 1930, Littlewood raised the question of whether planar badly approx- 2 imable vectors exist in a multiplicative sense: That is, if for all (α1, α2) ∈ R , we have lim inf nknα1k · knα2k = 0, (1.1) n→∞ where k·k denotes the distance to the nearest integer. Until the time of writing, finding non-trivial examples of (α1, α2) satisfying (1.1), barring rather special LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 3

cases, evades the best eﬀorts of the mathematical√ √ community. For instance, the problem remains open even for (α1, α2) = ( 2, 3). Here non-trivial means that α1, α2 lie in the set

Bad = {β ∈ R : ∃c>0 nknβk > c for all n ∈ N} of badly approximable numbers. This set has Lebesgue measure zero, by the Borel–Cantelli lemma, but full Hausdorﬀ dimension, by the Jarn´ık–Besicovitch theorem [12, Theorem 3.2].

Remark 1.1. By Dirichlet’s approximation theorem, the inequality nknβk < 1 holds inﬁnitely often for any β ∈ R. So badly approximable numbers are precisely those numbers for which Dirichlet’s approximation theorem is optimal, up to a constant.

The study of the measure theory and fractal geometry centring around (1.1) has turned out to be a fruitful endeavour. Indeed, it has led to various exciting developments in homogeneous dynamics and in diophantine approximation. We presently expound upon this.

Homogeneous dynamics. There is a classical correspondence—the Dani correspondence—between the diophantine properties of a given vector in Rk, and the dynamical properties of the associated orbit of a lattice in the space

SLk+1(R)/SLk+1(Z), under the group of diagonal matrices in R(k+1)×(k+1) having determinant 1. By virtue of this, Littlewood’s conjecture is closely linked to a conjecture of Mar- gulis [43, Conjecture 1]. If true, Margulis’ conjecture would imply Littlewood’s conjecture.

With this dynamical perspective, Einsiedler, Katok, and Lindenstrauss [24] established, inter alia, the striking result that the set of putative counterexamples to Littlewood’s conjecture (1.1) has Hausdorﬀ dimension zero. A crucial ingredient was a deep result of Ratner.

From a similar departure point, Shapira [50] established a measure-theoretic, uniform version of an inhomogeneous Littlewood-type problem, solving an old problem of Cassels. To state it, we stress that ‘for almost all’ (and similarly for ‘almost every’) is in this manuscript always means with respect to the Lebesgue measure on the ambient space, expressing that the complement of the set under consideration has Lebesgue measure zero. Shapira proved that for almost all (α, β) ∈ R2 the relation lim inf nknα − γk · knβ − δk = 0 n→∞ holds for any γ, δ ∈ R. Gorodnik and Vishe [31] improved this by a factor of (log log log log log n)λ, for some constant λ > 0. 4 SAM CHOW AND NICLAS TECHNAU

While the above results suggest that Littlewood’s conjecture might be correct, there is an indication against it: Adiceam, Nesharim, and Lunnon [1] proved that a certain function ﬁeld analogue of the Littlewood conjecture is false.

In what follows, the word metric is used to mean ‘measure-theoretic’, as is cus- tomary in this area. The goal of metric number theory is to classify behaviour up to exceptional sets of measure zero.

Metric multiplicative diophantine approximation. Alongside the theory of homogeneous dynamics linked to Littlewood’s conjecture, there have been signif- icant advances towards the corresponding metric theory. The ﬁrst systematic result in this direction is a famous theorem of Gallagher [29]:1 For any non- k increasing ψ : N → R>0 and almost all (α1, . . . , αk) ∈ R , we have

knα1k · · · knαkk < ψ(n) (1.2) infinitely often if the relevant series of measures diverges, that is, if X ψ(n)(log n)k−1 = ∞. (1.3) n>1 2 Consequently (1.1) holds true for almost every (α1, α2) ∈ R by a log-squared margin: 2 lim inf n(log n) knα1k · knα2k = 0. n→∞ Continuing this line of research, Pollington and Velani [47] showed the following fibre statement, by exploiting the Fourier decay property of a certain fractal measure: If α1 ∈ Bad then there is a set of numbers α2 ∈ Bad, of full Hausdorff dimension, such that n log nknα1k · knα2k < 1 holds for infinitely many n ∈ N.

A further ﬁbre statement concerning (1.1) was established by Beresnevich, Haynes, and Velani in [11, Theorem 2.4]. To state it, we recall that Liouville numbers are irrational real numbers α such that for any w > 0 the inequality knαk < n−w holds inﬁnitely often. We denote the set of Liouville numbers by L.

1In fact Gallagher stated the theorem in a slightly weaker form, namely that k lim inf n(log n) knα1k · · · knαkk = 0 n→∞ for almost all (α1, . . . , αk), but cognoscenti will recall that the proof carries through com- fortably to give the divergence part of the version above, and that the convergence part was already known, see [14, Remark 1.2]. Gallagher’s theorem is commonly referred to in one of three forms: the weakened divergence part, the full-strength divergence part, and the complete Lebesgue theory. We will refer to all three versions as Gallagher’s theorem. LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 5

Theorem 1.2 ([11, Theorem 2.4]). Let α1 ∈ R \ (Q ∪ L) and γ ∈ R. If the Duﬃn–Schaeﬀer conjecture is true, then for almost all α2 ∈ R we have 2 lim inf n(log n) knα1 − γk · knα2k = 0. n→∞

When this theorem was proved, the Duffin–Schaeffer conjecture was still open. The former was then proved, without appealing to the Duffin–Schaeffer conjecture, by the first named author [18] in a stronger form:

Theorem 1.3 ([18]). Let α1, γ ∈ R and assume that α1 ∈ R \ (Q ∪ L). If ψ : N → R>0 is non-increasing and the series X ψ(n) log n (1.4) n>1 diverges, then for almost all α2 ∈ R there exist inﬁnitely many n ∈ N such that knα1 − γk · knα2k < ψ(n).

The proof used the structural theory of Bohr sets (see Section 1.2), as well as continued fractions and the geometry of numbers, in a crucial way. This combinatorial–geometric method was further developed by the authors [20] to prove higher-dimensional results, removing the reliance on continued fractions. Instead, a more versatile framework from the geometry of numbers was brought to bear on the problem. Another interesting facet of the approach is the application of diophantine transference inequalities [13, 16, 17, 19, 30, 38] to deal with the inhomogeneous shifts.

By ﬁxing α1 above, one considers pairs (α1, α2) which lie on a vertical line in the plane. With Yang, the ﬁrst named author showed in [21] that if L0 is an arbitrary line in the plane then for almost all (α1, α2) ∈ L0 we have 2 lim inf n(log n) knα1k · knα2k = 0. n→∞

Subject to a generic diophantine assumption on L0, it was also shown there that if ψ : N → [0, ∞) is non-increasing and X ψ(n) log n < ∞ n>1 then for almost all (α1, α2) ∈ L0 the inequality

knα1k · knα2k < ψ(n) has at most ﬁnitely many solutions n ∈ N. The divergence statement was attained via an eﬀective asymptotic equidistribution theorem for unipotent orbits in SL3(R)/SL3(Z), whilst the convergence statement involved the correspondence between Bohr sets and generalised arithmetic progressions. All of this sits within the broader context of metric diophantine approximation on manifolds, for which there is a vast literature [7, 9, 35, 36, 39, 56]. 6 SAM CHOW AND NICLAS TECHNAU

A common feature of these results is that they are homogeneous in the metric parameter, i.e. they involve knα2k but not knα2 − γk with a general parameter γ. Even a weak inhomogeneous version of Gallagher’s theorem, akin to Shapira’s [50, Theorem 1.2], remains completely open, despite numerous at- tempts. In light of this, Beresnevich, Haynes, and Velani [11, Problem 2.3] posed the following problem:

Problem 1.4 (A fully-inhomogeneous version of Gallagher’s theorem on vertical planar lines, weak form). Let α1, γ1, γ2 ∈ R, and suppose that α1 6∈ L∪Q. Prove that 2 lim inf n (log n) knα1 − γ1k · knα2 − γ2k = 0 for almost all α2 ∈ . n→∞ R

They write in the paragraph leading up to [11, Problem 2.3] concerning this problem that it “currently seems well out of reach”. Nevertheless, a stronger conjecture was put forth by the ﬁrst named author [18, Conjecture 1.6]:

Conjecture 1.5 (A fully-inhomogeneous version of Gallagher’s theorem on vertical planar lines, strong form). Let α1, γ1, γ2 be as in Problem 1.4. Suppose ψ : N → R>0 is non-increasing and that the series (1.4) diverges. Then for almost all α2 ∈ R there exist inﬁnitely many n ∈ N such that

knα1 − γ1k · knα2 − γ2k < ψ(n).

This manuscript resolves Problem 1.4 a fortiori by proving Conjecture 1.5. Additionally, our methods are capable of deducing a higher-dimensional generalisation, as conjectured by the authors in [20, Conjecture 1.7]. Furthermore, we resolve Conjecture 2.1 of Beresnevich, Haynes, and Velani [11]:

Conjecture 1.6 (A fully-inhomogeneous version of Gallagher’s theorem in the plane). Let γ1, γ2, ∈ R, and let ψ : N → R>0 be a non-increasing function such 2 that the series (1.4) diverges. Then for almost all (α1, α2) ∈ R the inequality

knα1 − γ1k · knα2 − γ2k < ψ(n) holds inﬁnitely often.

Note that Conjecture 1.5 implies Conjecture 1.6, since L ∪ Q has Lebesgue measure zero. We now formulate our results in greater detail. 1.1. Main results. Recall that the multiplicative exponent ω×(α) of a vector d α = (α1, . . . , αd) ∈ R is the supremum of all w > 0 such that −w knα1k · · · knαdk < n inﬁnitely often. The property of ω×(α) being ﬁnite can be interpreted as a higher-dimensional generalisation of being an irrational, non-Liouville number. LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 7

Theorem 1.7 (A fully-inhomogeneous version of Gallagher’s theorem on ver- k−1 tical lines, strong form). Let k > 2. Fix α = (α1, . . . , αk−1) ∈ R and γ1, . . . , γk ∈ R. For k = 2, suppose that α1 is an irrational, non-Liouville number, and for k > 3 suppose that k − 1 ω×(α) < . (1.5) k − 2 If ψ : N → R>0 is non-increasing and satisﬁes (1.3), then for almost all αk ∈ R there are inﬁnitely many n ∈ N for which

knα1 − γ1k · · · knαk − γkk < ψ(n). (1.6)

Remark 1.8. The set of α ∈ Rk−1 for which k − 1 ω×(α) > k − 2 has Lebesgue measure zero and, stronger still, has Hausdorﬀ dimension strictly less than k − 1. The former follows from the Borel–Cantelli lemma and the latter from the work of Hussain and Simmons [37]. The set of Liouville numbers has Lebesgue measure zero and, stronger still, has Hausdorﬀ dimension 0.

Theorem 1.7 is precisely [20, Conjecture 1.7], and we have the following note- worthy special cases.

Corollary 1.9. Conjectures 1.5 and 1.6 are true.

We have the following generalisation of Conjecture 1.6, which includes its com- plementary convergence theory.

Corollary 1.10 (A fully-inhomogeneous version of Gallagher’s theorem). Let γ1, . . . , γk ∈ R, and let ψ : N → (0, ∞) be a non-increasing function. Write × × k W = W (ψ, γ1, . . . , γk) for the set of (α1, . . . , αk) ∈ [0, 1] such that (1.6) has inﬁnitely many solutions n ∈ N. Then  ∞  X k−1 1, if ψ(n)(log n) = ∞ ×  n=1 µk(W ) = ∞ X 0, if ψ(n)(log n)k−1 < ∞,  n=1

where µk denotes k-dimensional Lebesgue measure.

The divergence part of Corollary 1.10 is a consequence of Theorem 1.7 and Remark 1.8. The convergence part requires only classical techniques, and will be proved in Section 3.5. Theorem 1.7 also resolves Problem 1.4 in the following stronger and more general form: 8 SAM CHOW AND NICLAS TECHNAU

k−1 Corollary 1.11. Let k > 2. Fix a ﬁbre vector α = (α1, . . . , αk−1) ∈ R , and shifts γ1, . . . , γk ∈ R. For k = 2, suppose that α1 is an irrational, non-Liouville number, and for k > 3 assume (1.5). Then for almost all αk ∈ R there are inﬁnitely many n ∈ N for which 1 knα − γ k · · · knα − γ k < . 1 1 k k n(log n)k log log n

When γk = 0, the results above follow from [18]. In the planar case, we go beyond the scope of Problem 1.4. Indeed, we also solve it on ﬁbres (α1, R) where α1 is a Liouville number:

Theorem 1.12. Let α1 ∈ L, and let γ1, γ2 ∈ R. Then, for almost all α2 ∈ R, we have 2 lim inf n(log n) knα1 − γ1k · knα2 − γ2k = 0. n→∞

In view of Theorem 1.7, we see that this result holds for any irrational α1, Liouville or not. To be clear, we obtain the following statement.

Theorem 1.13. Let α1 ∈ R \ Q, and let γ1, γ2 ∈ R. Then, for almost all α2 ∈ R, we have 2 lim inf n(log n) knα1 − γ1k · knα2 − γ2k = 0. n→∞

However, if α1 ∈ Q, then one can easily construct a counterexample by choosing any γ1 ∈/ α1Z and applying Sz¨usz’stheorem [51], see Theorem 1.16. In this sense, and in the sense described in the next two paragraphs, Theorem 1.13 is deﬁnitive.

The analysis on Liouville ﬁbres is delicate, owing to the erratic behaviour of the arising sums of reciprocals of fractional parts [11]. In light of our earlier discussion on Problem 1.4 and Conjecture 1.5, one might expect Theorem 1.12 not to be sharp. Surprisingly, the result is sharp, as we now detail.

Let us ask for a strengthening of Theorem 1.12 by considering an approximation function with a faster decay, say 1 ψ (n) = , (1.7) ξ n(log n)2ξ(n) where ξ : N → [1, ∞) is an unbounded and non-decreasing function. Then the strengthened statement of Theorem 1.12, with ψ(n) = ψξ(n), is false:

Theorem 1.14. Let ξ : N → [1, ∞) be non-decreasing and unbounded. Then there are continuum many pairs (α1, γ1) ∈ L × R such that for any γ2 ∈ R and LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 9 almost all α2 ∈ R the inequality

knα1 − γ1k · knα2 − γ2k < ψξ(n) has at most ﬁnitely many solutions n ∈ N.

Remark 1.15. (1) One could regard this as a result ‘in the opposite direction’ to Littlewood’s conjecture, though there are several diﬀerences. A volume heuristic, and the works of Peck [45] and Pollington–Velani [47], suggest that (1.1) can be strengthened by roughly a logarithm— see the discussion in [5]—and meanwhile Badziahin [4] has shown that it cannot be strengthened much further than that. For this reason, we consider that results in the opposite direction to Littlewood’s conjecture are worthy of further study.

(2) In the course of our proof, we explicitly construct the pairs (α1, γ1). This feature is often not present in results of this ﬂavour.

To conclude this discussion, Theorem 1.13 is sharp, and has no unnecessary restrictions on α1, γ1, and γ2.

By work of Beresnevich and Velani [14, Section 1], as well as Hussain and Simmons [37], ‘fractal’ Hausdorﬀ measures are known to be insensitive to the multiplicative nature of these types of problems. Fix k > 2. For ψ : N → R>0 k non-increasing with limn→∞ ψ(n) = 0, and γ = (γ1, . . . , γk) ∈ R , denote by × k Wk (ψ, γ) the set of (α1, . . . , αk) ∈ R satisfying (1.6) for inﬁnitely many n. k Further, denote by Wk(ψ, γ) the set of (α1, . . . , αk) ∈ [0, 1] for which

max(knα1 − γ1k,..., knαk − γkk) < ψ(n) has infinitely many solutions n ∈ N. By [37, Corollary 1.4] and [12, Theorem 6.1], for γ ∈ Rk we have the Hausdorff measure identity s × s−(k−1) H (Wk (ψ, γ)) = H (W1(ψ, γ)) (k − 1 < s < k). (1.8) We interpret from this that multiplicatively approximating k reals using the same denominator is no different to approximating one of the k numbers, except possibly for a set of zero Hausdorff s-measure. This behaviour differs greatly from that of the Lebesgue case s = k, where there are extra logarithms for the multiplicative problem. As explained in [14, 37], in the remaining s × ranges for s the Hausdorff theory trivialises: if s > k then H (Wk (ψ), γ) = 0, s × irrespective of ψ, whereas if s 6 k − 1 then H (Wk (ψ, γ)) = ∞. 1.2. Key ideas and further results.

A fully-inhomogeneous ﬁbre reﬁnement of Gallagher’s theorem. Owing to the robustness of our method, much of the argument for Theorem 1.7 is transparent already in the planar case k = 2. As it is simpler from a technical point of 10 SAM CHOW AND NICLAS TECHNAU

view, we shall outline the proof only in this case, and indicate in passing how to generalise to higher dimensions. To begin, let us isolate the metric parameter α = α2 in (1.6) on the left hand side of the inequality. As k · k is 1-periodic, it suffices if we show ψ(n) knα − γk < Φ(n) := , γ := γ2, (1.9) knα1 − γ1k holds for almost every α ∈ [0, 1] infinitely often. If Φ were a non-increasing function, then we could utilise Szüsz’sextension of Khintchine’s theorem, which grants a sharp description of the inhomogeneous approximation rate of a generic real number:

Theorem 1.16 (Sz¨usz[51]). If Ψ: N → R>0 is non-increasing and γ ∈ R, then the Lebesgue measure of the set of α ∈ [0, 1] for which knα − γk < Ψ(n) (1.10) holds inﬁnitely often is 1 (resp. 0) if X Ψ(n) (1.11) n>1 diverges (resp. converges).

Since Φ is very much not monotonic, deducing (1.9) infinitely often for almost P all α, from the divergence of n Φ(n), is far more demanding. In fact, it is known that this naive condition is insufficient, as Duffin and Schaeffer [23] pointed out: There exists Ψ : N → R>0 such that (1.11) diverges but for γ = 0 and almost all α ∈ [0, 1] the inequality (1.10) holds at most finitely often. This was generalised by Ram´ırezin [48].

To circumvent their counterexamples, Duffin and Schaeffer restricted attention to reduced fractions, and correspondingly imposed the condition that the series X ϕ(n) Ψ(n) (1.12) n n>1 diverges, where ϕ is Euler’s totient function. The Duffin–Schaeffer conjecture was a major open problem in diophantine approximation since the 1940s. Over the course of the nearly eight decades, various partial results towards the Duffin–Schaeffer conjecture were obtained, by

• Erd˝os[25] in 1970, Vaaler [55] in 1978

• Pollington–Vaughan [46] in 1990

• Harman [33] in 1990, Haynes–Pollington–Velani [34] in 2012, LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 11

and several other authors [2, 3, 10]. Recently, Koukoulopoulos and Maynard (2019) broke through with a complete proof of the Duﬃn–Schaeﬀer conjecture:

Theorem 1.17 (Koukoulopoulos–Maynard [41]). If Ψ: N → R>0 is such that (1.12) diverges then, for almost all α ∈ [0, 1], the inequality |nα − a| < Ψ(n) holds for inﬁnitely many coprime a, n ∈ N.

A natural generalisation would be an inhomogeneous version of the Duﬃn– Schaeﬀer conjecture:

Conjecture 1.18 (Inhomogeneous Duffin–Schaeffer, see Ram´ırez[48]). Let γ ∈ R. If Ψ: N → R>0 is such that (1.12) diverges, then for almost all α ∈ [0, 1] the inequality |nα − a − γ| < Ψ(n) holds for infinitely many coprime a, n ∈ N.

For us it is enough to establish a similar result for a concrete class of functions of the shape (1.9). We consider sets An that are roughly of the form {α ∈ [0, 1] : knα − γk < Φ(n)}. By standard probabilistic arguments, it suﬃces to show that the measures of the sets An are not summable, and that the sets are quasi-independent in an averaged sense. The latter property is, as always, the crux of the matter, and involves overlap estimates that quantify how large µ(An ∩ Am) is compared to µ(An)µ(Am) on average.

To this end, we may confine our analysis to a reasonably large set G of ‘good’ indices n. To simplify matters, we decompose N into dyadic ranges, wherein n N, and in addition the oscillating factor knα1 − γ1k has a fixed order of magnitude. Sets of such integers n are essentially Bohr sets B = Bγ1 (N; ρ ) := {n ∈ : |n| N, knα − γ k ρ }, (1.13) α1 1 Z 6 1 1 6 1 which appear in many areas of mathematics. A novelty of this paper is to show how to handle the overlap estimates via congruences in Bohr sets. Here the structural theory from our previous work [20], constructing the correspondence between Bohr sets and generalised arithmetic progressions—a central pillar of additive combinatorics [52]—in the present context, plays a pivotal role. Previously there was progress made in this direction by Tao–Vu [53], and by the first named author [18]. Further, it will be helpful to group m, n according to the size of the greatest common divisor d of m and n.

We handle the overlap estimates by averaging over indices m, n from different Bohr sets of the shape (1.13). After summing over different dyadic ranges, we 12 SAM CHOW AND NICLAS TECHNAU can then infer the required quasi-independence on average. In the course of our analysis, we need to count solutions to congruences in generalised arithmetic progressions that are essentially Bohr sets. The range in which d is large requires extra care: For γ∈ / L∪Q, a repulsion stemming from this diophantine assumption enables us to treat this challenging case. For γ ∈ L ∪ Q, an additional argument enables us to crack this devilish final case; the idea is to introduce a counterpart to reduced fractions, which we call ‘shift-reduced’.

0 Deﬁnition 1.19. Let γ ∈ R, η ∈ (0, 1), and n ∈ N. Denote by ct/qt the continued fraction convergent of γ for which t is maximal satisfying 0 η qt 6 n . (1.14) 0 The pair (a, n) ∈ Z × N is (γ, η)-shift-reduced if (qta + ct, n) = 1.

To our knowledge, this notion has not hitherto appeared in the diophantine approximation literature.

Remark 1.20. Note that γ can be a rational in the above deﬁnition, in which 0 case the sequence (qt)t terminates. Indeed, in the case γ = 0, we recover the 0 traditional notion of reduced fractions: Letting ct = 0 and qt = 1, the fraction a/n is reduced if and only if the pair (a, n) is (0, 1/2)-shift-reduced. Moreover, η 0 if γ ∈ Q and n is greater than or equal to the denominator of γ, then γ = ct/qt. We provide some background on continued fractions in Section 2.1.

Our definition of shift-reduced fractions is sensitive to the shift γ. This fi- nesse slightly complicates matters, because we lose measure by not using all fractions. However, by sieve theory we are able to show that shift-reduced fractions are at least as prevalent as reduced fractions, and we are then able to establish the divergence of the relevant series. We suspect that the idea of using shift-reduced fractions will be useful for other arithmetic problems, including perhaps a version of the inhomogeneous Duffin–Schaeffer conjecture, as we will shortly discuss further.

Yet all of this combined yields only that (1.6) holds infinitely often on a set of positive measure. In the absence of a zero–one law for inhomogeneous diophantine approximation, we need to carefully ‘localise’ the overlap estimates to indeed deduce that (1.6) holds for a set of αk of full measure; this machinery, though not confined to the realm of metric diophantine approximation, is well- explained in the monograph of Beresnevich, Dickinson, and Velani [8]. This subtlety complicates the analysis non-trivially, as it requires us to keep hold of a factor of µ(I) throughout. In the regime of m, n in which d is essentially constant it turns out to be surprisingly difficult to do so. To avoid this scenario, we introduce artificial powers of 4 as divisors, which guarantees us that d is not too small. This is an unorthodox manoeuvre, but one that we found useful in practice, and one that may find other uses. LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 13

In the course of our proof, we establish the following weakened version of Con- jecture 1.18 for a class of non-monotonic approximating functions generalising Φ as deﬁned above.

Conjecture 1.21 (Weak inhomogeneous Duffin–Schaeffer conjecture). Let γ ∈ R. If Ψ: N → R>0 is such that (1.12) diverges, then for almost all α ∈ [0, 1] the inequality knα − γk < Ψ(n) has infinitely many solutions n ∈ N.

Theorem 1.22 (Special case of weak inhomogeneous Duﬃn–Schaeﬀer). Let k−1 k > 2. Fix α = (α1, . . . , αk−1) ∈ R and γ1, . . . , γk ∈ R. For k = 2, suppose that α1 is an irrational, non-Liouville number, and for k > 3 assume (1.5). Let ψ : N → R>0 be a non-increasing function satisfying (1.3), and let ψ(n) Φ(n) = (n ∈ N). (1.15) knα1 − γ1k · · · knαk−1 − γk−1k Then Conjecture 1.21 holds for Ψ = Φ.

We wonder if there is a sharp dichotomy along the lines of Conjecture 1.21.

Question 1.23 (Inhomogeneous Duffin–Schaeffer dichotomy). Let γ ∈ R and Ψ: N → R>0. Does there exist η ∈ (0, 1) with the following property? Denote by W(Ψ; γ, η) the set of α ∈ [0, 1] such that |nα − γ − a| < Ψ(n), (a, n) is (γ, η)-shift-reduced has infinitely many solutions (a, n) ∈ Z × N. Then  X ϕγ,η(n) 1, if Ψ(n) = ∞  n µ(W(Ψ; γ, η)) = n>1 X ϕγ,η(n) 0, if Ψ(n) < ∞,  n n>1 where

ϕγ,η(n) = #{a ∈ {1, 2, . . . , n} :(a, n) is (γ, η)-shift-reduced}.

Remark 1.24. It is not clear at this stage whether η should need to depend on γ or Ψ. Moreover, it could be that the property holds for all η ∈ (0, η0), for some η0 ∈ (0, 1]. This question has the appeal of a matching convergence theory, unlike Conjecture 1.21.

We are also able to answer Question 1.23 positively for Ψ = Φ, subject to natural assumptions: 14 SAM CHOW AND NICLAS TECHNAU

Theorem 1.25 (Special case of inhomogeneous Duﬃn–Schaeﬀer dichotomy). k−1 Let k > 2. Fix α = (α1, . . . , αk−1) ∈ R and γ1, . . . , γk ∈ R. For k = 2, suppose that α1 is an irrational, non-Liouville number, and for k > 3 assume (1.5). Let ψ : N → R>0 be a non-increasing function satisfying (1.3), and let Φ be as in (1.15). Then, with γ = γk and the notation of Question 1.23, there exists η ∈ (0, 1) such that ∞ X ϕγ,η(n) Φ(n) = ∞ (1.16) n n=1 and µ(W(Φ; γ, η)) = 1. (1.17) In particular, Question 1.23 has a positive answer for Ψ = Φ.

Our method comes close to directly establishing Theorem 1.25. However, it misses a pathological case where an auxiliary function exceeds 1/2 inﬁnitely often, causing the relevant intervals to overlap and their union to have smaller measure. We are able to circumvent this by ad-hoc means, and provide the details in an appendix.

Inhomogeneous, non-monotonic diophantine approximation is also discussed in Harman’s book [32, Chapter 3], as well as in recent work of Yu [57, 58].

Our method is robust with respect to the dimension k. If k > 3, the combinatorics for controlling the overlap estimates is relatively similar to the planar case, k = 2. The notable differences are that there are more dyadic ranges to sum over and the choice of cutoff parameters needs to be adapted. An inhomogeneous version of Gallagher’s theorem on Liouville fibres. For The- orem 1.12, the overall structure of our proof parallels that of the Duffin– Schaeffer theorem [32, Theorem 2.3], which is a special case of the Duffin– Schaeffer conjecture. This naturally leads us to count solutions to congruences in Bohr sets. In this setting, the latter are rather sparse in the set of positive integers, which presents new difficulties. The partial quotients, see Section 2, grow extremely rapidly infinitely many times, and we use this together with classical continued fraction analysis to deal with the combinatorial aspects of the overlap estimates. Liouville fibres: sharpness. The prove the sharpness result, Theorem 1.14, we construct pairs (α, γ) ∈ L × R in such a way as to keep knα − γk away from zero. We achieve this via the Ostrowski expansion [11, Section 3], by choosing each Ostrowski coefficient of γ with respect to α to be roughly half times the corresponding partial quotient of α, and by choosing α to have extremely rapidly-growing partial quotients. 1.3. Open problems. We have already discussed a few open questions. Here are some that we have yet to cover. LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 15

Relaxing the diophantine condition. We expect that the assumption (1.5) in Theorem 1.7 can be somewhat relaxed. This likely requires diﬀerent methods.

Convergence theory. It would be desirable to advance the convergence theory complementing Theorem 1.7, especially for k > 3. For k = 2, progress has been made by Beresnevich, Haynes, and Velani [11].

By partial summation and the Borel–Cantelli lemma, proving the desired convergence statement can be reduced to showing that X 1 N(log N)k−1. knα1 − γ1k · · · knαk−1 − γk−1k n6N In the case k = 2 and γ1 = 0, for a generic choice of α1 ∈ R this bound is false, see [11, Example 1.1]. However, the logarithmically averaged sums

γ X 1 Sα(N) = , γ = (γ1, . . . , γk−1) nknα1 − γ1k · · · knαk−1 − γk−1k n6N could be better behaved, and accurately bounding these would lead to a similar outcome. On probabilistic grounds, we expect that γ k Sα(N) α,γ (log N) . Subject to a generic diophantine condition on α, it is less diﬃcult to prove matching lower bounds via dyadic pigeonholing and estimates for the cardinality of a Bohr set, see [20, Lemma 3.1]. So the task is to determine the order γ of magnitude of Sα(N). Beresnevich, Haynes and Velani [11, Theorem 1.4] γ 2 showed that if k = 2 and γ ∈ R then Sα(N) α,γ (log N) for almost every α ∈ R. With this in mind, we pose the following problem which, if resolved in a suﬃciently positive manner, would entail a coherent convergence theory.

k−1 k−1 Problem 1.26. Let k > 3, and let Ck be the set of (α, γ) ∈ R × R for which γ k Sα(N) α,γ (log N) . Is the set Ck non-empty? What is its Lebesgue measure?

0 3 Empirical evidence mildly supports the assertion that Sα(N) α (log N) holds for generic values of α = (α1, α2). We randomly generated

α1 ≈ 0.957363115715396, α2 ≈ 0.3049448415027476. With S0 (H) H = 106, c = α ≈ 1.73475, (log H)3 0 3 we plotted Sα(N) and c(log N) against N for N = 2, 3,...,H, see Figure 1. There are ‘jumps’ when knα1k · knα2k is very small, but these do not appear 0 to aﬀect the order of magnitude of Sα(N). For further discussion, we refer the reader to the article by Lˆeand Vaaler [42]. 16 SAM CHOW AND NICLAS TECHNAU

0 3 Figure 1. Sα(N) and c(log N) against N

Dual approximation. There are natural dual versions of Gallagher’s theorem. Loosely speaking, in the dual framework one studies how close a given vector is to a hyperplane of bounded height, as the height bound increases. In fact, we intend to address the next problem in a future work.

k−1 Conjecture 1.27. Let k > 2, and (α1, . . . , αk−1) ∈ R . Then for almost all k αk ∈ R there exist inﬁnitely many (n1, . . . , nk) ∈ Z such that 1 kn α + ··· + n α k < , 1 1 k k H(n)(log H(n))k + + + where H(n) = H(n1, . . . , nk) = n1 ··· nk and n = max(|n|, 2).

This is also [20, Conjecture 1.9], which comes with some further discussion.

The dual convergence theory is also of interest, that is, to show that the convergence of the series in (1.3) implies that for almost almost all αk the inequality kn1α1 + ··· + nkαkk < ψ(H(n)) holds at most finitely often. As with the usual multiplicative approximation problems described above, the convergence theory in the dual setting is very much open. There is a discussion of the relevant sums, as well as a reference to a possible departure point, in the paragraphs surrounding Conjecture 1.1 of Beresnevich, Haynes, and Velani [11]. 1.3.1. A Hausdorff dimension problem. In Theorem 1.14, we did not estimate 2 the Hausdorff dimension of the set of pairs (α1, γ1) ∈ R such that for any γ2 ∈ R and almost all α2 ∈ R the inequality

knα1 − γ1k · knα2 − γ2k < ψξ(n) LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 17 has at most finitely many solutions n ∈ N. It would be of interest to do so, for ξ increasing slowly to infinity, or even for ξ(n) = log log n. This problem can be further simplified by fixing γ2 = 0.

1.4. Organisation and notation.

Organisation of the manuscript. In Section 2, we collect tools and technical lemmata that would otherwise disrupt the course of the main arguments. In Section 3, we prove Theorems 1.7 and 1.22 together, followed by the convergence part of Corollary 1.10. Thereafter, we prove Theorems 1.12 and 1.14 in Sections 4 and 5, respectively. Appendix A describes how the proof of Theorem 1.22 can be adapted to give Theorem 1.25.

Notation. We use the Vinogradov and Bachmann–Landau notations: For functions f and positive-valued functions g, we write f g or f = O(g) if there exists a constant C such that |f(x)| 6 Cg(x) for all values of x under consideration. We write f g or f = Θ(g) if f g f. Throughout this manuscript, the implied constants are allowed to depend on:

• The approximation function ψ

k−1 • A ﬁxed vector α ∈ R , and γ1, . . . , γk ∈ R

• A constant C > 2, which in turn only needs to depend on α, γ1, . . . , γk, specifying the ranges [Cj,Cj+1] to which we localise various parameters

• A small positive constant ε0, which only needs to depend on α, γ1, . . . , γk.

These dependencies shall usually not be indicated by a subscript. To be ex- plicit, we do consider the dimension k of the ambient space to be data of the vector α and hence shall not indicate its dependency. If any other dependence occurs, we record this using an appropriate subscript. If S is a set, we denote the cardinality of S by |S| or #S. The symbol p is reserved for primes. The pronumeral N denotes a positive integer, suﬃciently large in terms of α, γ1, . . . , γk, the approximation function ψ, and a bounded interval I that will arise in the course of some of the proofs. For a vector α ∈ Rk−1, we abbreviate the product of its coordinates to

Π(α) = α1 ··· αk−1. (1.18) When x ∈ R, we write kxk for the distance from x to the nearest integer. Furthermore, µ denotes one-dimensional Lebesgue measure. Given S ⊆ R, we write S6X for {x ∈ S : x 6 X}.

Finally, we often have to deal with expressions such as X 1 , (log n) log log n n6N 18 SAM CHOW AND NICLAS TECHNAU

which are not always well-defined because of finitely many small n ∈ N. To deal with this, we write ln(x) for the natural logarithm of a positive real number x, and put log(x) = max(ln(x), 1) to ensure that these logarithms and their iterates are indeed well-defined and positive.

Funding and acknowledgements. SC was supported by EPSRC Fellowship Grant EP/S00226X/2. NT was supported by the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme (Grant agreement No. 786758), as well as by the Austrian Science Fund (FWF) grant J 4464-N, and is grateful to Rajula Srivastava for her eternal encouragement. We thank Sanju Velani for his enthusiasm towards this topic, Andy Pollington for many enthralling conversations, Victor Beresnevich for further encouragement, and Zeev Rudnick for comments on a draft.

2. Preliminaries In this subsection, we gather together a panoply of tools. We begin with the theory of continued fractions, before continuing to that of Bohr sets. Then we present some standard measure theory and real analysis. The latter will enable us to introduce the artiﬁcial divisors to which we alluded earlier, at essentially no cost. After that we discuss some estimates from the geometry of numbers, whose raison d’ˆetreis to count solutions to congruences in generalised arithmetic progressions. We then review some basic prime number theory and sieve theory, culminating in the fundamental lemma of sieve theory, which we will later use to count shift-reduced fractions.

2.1. Continued fractions. The material here is standard; see for instance [15, Chapter 1, Section 2]. For each irrational number α there exists a unique sequence of integers a0, a1, a2,..., the partial quotients of α, such that aj > 1 for j > 1 and 1 α = a + . 0 1 a1 + a2 + ... For j > 0, let [a0; a1, . . . , aj] denote the truncation the above inﬁnite continued fraction at place j, and let pj ∈ Z and qj ∈ N be the coprime integers satisfying pj [a0; a1, . . . , aj] = . qj This rational number is the called the j-th convergent to α. Continued fractions enjoy an impressive portfolio of beautiful properties. One particular feature for which we have ample need is the following recursion: For j > 1, we have

qj+1 = aj+1qj + qj−1 and pj+1 = aj+1pj + pj−1. (2.1) The initial values are

p0 = a0, q0 = 1, p1 = a1a0 + 1, q1 = a1. LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 19

For each j > 0, the quantity

Dj = qjα − pj (2.2) satisﬁes the bound 1 |D | q 1. (2.3) 2 6 j j+1 6 Furthermore, it is well-known that the signs Dj/|Dj| are alternating. For an irrational number α ∈ R, the diophantine exponent of α is −w ω(α) = sup{w > 0 : kqαk < q for inﬁnitely many q > 1}. Note that ω(α) = ω×(α). We require the following well-known fact character- ising diophantine numbers in terms of the growth of the consecutive continued fraction denominators [11, Lemma 1.1]:

Lemma 2.1. Let α ∈ R\Q, and (qk)k be the sequence of its continued fraction denominators. Then log q ω(α) = lim sup k+1 . k→∞ log qk In particular α 6∈ L if and only if log qk+1 log qk.

Continued fractions have been used to prove a highly aesthetic result called the three gap theorem. For α ∈ R \ Q and m ∈ N, this asserts that there are at most three distinct gaps di+1 − di, where

{d1, . . . , dm} = {iα − biαc : 1 6 i 6 m} and 0 = d0 < ··· < dm+1 = 1. Many ﬁnd this surprising at ﬁrst. The sizes of the gaps can be computed and described in terms of the continued fraction expansion. We will need only the size of the largest gap. The following theorem combines parts of Theorem 1 and Corollary 1 of [44], and adopts the typical convention that q−1 = 0.

Theorem 2.2. Let α ∈ R \ Q and m ∈ N. Then:

(a) There is a unique representation

m = rqk + qk−1 + s, for some

k > 0, 1 6 r 6 ak+1, 0 6 s 6 qk − 1.

(b) If s < qk − 1 then ( |Dk+1| + |Dk|, if r = ak+1 max{di+1 − di : 0 6 i 6 m} = |Dk+1| + (ak+1 − r + 1)|Dk|, if r < ak+1. 20 SAM CHOW AND NICLAS TECHNAU

If s = qk − 1 then ( |Dk|, if r = ak+1 max{di+1 − di : 0 6 i 6 m} = |Dk+1| + (ak+1 − r)|Dk|, if r < ak+1.

Rational numbers also have continued fraction expansions, however they are ﬁnite, taking the form 1 a + . 0 1 a + 1 a + 2 .. . 1 + at

The partial quotients a0 ∈ Z and a1, . . . , at ∈ N, as well as the convergents pj/qj (0 6 j 6 t), are deﬁned in the same way is in the irrational case, except that we impose the additional constraint at > 1 to be sure that the expansion is unique.

In the next subsection, we ﬁx an irrational number α and describe an expansion, the Ostrowski expansion, that allows us to accurately read oﬀ for each n ∈ N the size of knαk, and more generally knα − γk for γ ∈ R.

2.2. Ostrowski expansions. Let n ∈ N, and let K > 0 be such that qK 6 n < qK+1. (2.4) It is known [49, p. 24] that there exists a uniquely determined sequence of non-negative integers ck = ck(n) such that X n = ck+1qk, k>0 satisfying ck+1 = 0 for all k > K, as well as the following additional constraints:

0 6 c1 < a1, 0 6 ck+1 6 ak+1 (k ∈ N), if ck+1 = ak+1 then ck = 0. The structure of the set of integers whose initial Ostrowski digits are prescribed is well-understood; this set is referred to as a cylinder set. We require information concerning the size of the gap between consecutive elements, which we retrieve from [11, Lemma 5.1].

Lemma 2.3 (Gaps lemma). Let m > 0, and let A (d1, . . . , dm+1) denote the set of positive integers whose initial Ostrowski digits are d1, . . . , dm+1. Let n1 < n2 < . . . be the elements of the set A (d1, . . . , dm+1), and let i ∈ N. If LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 21

dm+1 > 0 then ni+1 − ni > qm+1, and if dm+1 = 0 then ni+1 − ni ∈ {qm+1, qm}. Further, if dm+1 = 0 and ni+1 −ni = qm then cm+2 (ni) = am+2 and the gap ni+1 −ni is preceded by am+2 gaps of size qm+1.

With reference to (2.2), we will apply the theory of this subsection to pairs (α, γ), where X γ = bk+1Dk, k>0 such that a a a 64, k b k (k 1), a = 0. (2.5) k > 4 6 k 6 2 > 0 Before proceeding in earnest, we conﬁrm some technical conditions that will put us into a standard setting.

Lemma 2.4. If we have (2.5), then 1 0 < α < , 0 γ < 1 − α, 64 6 and knα − γk= 6 0 (n ∈ N). (2.6)

Proof. The ﬁrst inequality follows from a0 = 0 and a1 > 64.

We compute that

ak+1 1 ak+1/4 1 |bk+1Dk| 6 6 , |bk+1Dk| > > (k > 0). qk+1 qk 2qk+1 16qk As ak+1 > 64 for all k > 0, we see from these inequalities that |bk+1Dk| is a monotonically decreasing sequence which converges to zero as k → ∞. Using that the signs of the Dk alternate and that D0 = {α} > 0, we conclude that bk+1Dk + bk+2Dk+1 > 0 if k is even and bk+1Dk + bk+2Dk+1 6 0 if k is odd. Therefore X 0 6 (b2k+1D2k + b2k+2D2k+1) = γ k>0 supplies the lower bound in the second inequality. For the upper bound, note that X b1 γ = b1D0 + (b2k+2D2k+1 + b2k+3D2k+2) 6 b1D0 6 6 1/2 < 1 − α. a1 k>0

Finally, we turn our attention towards (2.6). Observe that X nα − γ = (ck+1qkα − bk+1(qkα − pk)) ∈ Σ + Z, k>0 22 SAM CHOW AND NICLAS TECHNAU

where X Σ = δk+1Dk, δk+1 = ck+1 − bk+1 (k > 0). (2.7) k>0 Using (2.3) and (2.1), as well as (2.5), we compute that 3 X 3 X 1 |Σ| > |D0| − ak+1|Dk| > {α} − , 4 4 qk k>1 k>1 where {α} denotes the fractional part of α, and

X 1 1 X −r 64 6 64 = . qk q1 63q1 k>1 r>0 Since 1 65 < a + 1 = q + 1 q , {α} 1 1 6 64 1 we have 64 16 |Σ| > − q−1 > 0. 65 21 1 Moreover X X 1 1 1 |Σ| 6 |D0| + ak+1|Dk| 6 {α} + 6 + < 1. qk 64 63 k>1 k>1 Verily we have (2.6).

It turns out that we can quantify the size of knα − γk in terms of the quantity Σ from (2.7), as the next lemma details. The assumption (2.5) simpliﬁes several technicalities. The following combines [11, Lemmata 4.3, 4.4, and 4.5], in this special case.

Lemma 2.5. If we have (2.5), then knα − γk = kΣk = min(|Σ| , 1 − |Σ|).

Furthermore, let m = m(n) be the smallest i > 0 such that δi+1 6= 0, and let K be as in (2.4). Then we have the following estimates for |Σ| and 1 − |Σ|.

(1) |Σ| = (|δm+1| − 1) |Dm| + um+2 |Dm+1| + um+3 |Dm+2| + Υ,

where um+2, um+3,Υ are non-negative real numbers constrained by

um+2 am+2, um+3 am+3,Υ |Dm+2|.

(2) ˜ 1 − |Σ| = u1 |D0| + u2 |D1| + Υ, ˜ where u1, u2, Υ are non-negative, and constrained by ˜ u1 a1, u2 a2, Υ |D1|. LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 23

2.3. Bohr sets. For α, γ ∈ Rk−1, we have ample need for bounds on the cardinality of Bohr sets γ B = Bα(N; ρ) := {n ∈ Z : |n| 6 N, knαi − γik 6 ρi (1 6 i 6 k − 1)}, (2.8) that are sharp up to multiplication by absolute constants. Further, it turns out to be crucial to have precise control over the ranges in which the Bohr sets have a suﬃciently nice structure. If the width parameters δi and the length parameters Ni are in a suitable regime, then the Bohr set B is enveloped—eﬃciently, as we detail soon—in a k-dimensional generalised arithmetic progression

P(b; A1,...,Ak; N1,...,Nk) = {b + A1n1 + ··· + Aknk : |ni| 6 Ni}, (2.9) where b, A1,...,Ak,N1,...,Nk ∈ N. The thresholds for the admissible regimes depend naturally on the diophantine properties of α.

Inside B, we will ﬁnd a large (asymmetric) generalised arithmetic progression + P (b; A1,...,Ak; N1,...,Nk) = {b+A1n1 +···+Aknk : 1 6 ni 6 Ni}. (2.10) + This is proper if for each n ∈ P (b, A1,...,Ak,N1,...,Nk) there is a unique k vector (n1, . . . , nk) ∈ N for which n = b + A1n1 + ··· + Aknk, ni 6 Ni (1 6 i 6 k). Throughout this subsection we operate under the assumption (1.5), which in the case k = 2 simply means that α is irrational and non-Liouville. Set 1 k − 2 η(α) = − ∈ (0, 1]. ω×(α) k − 1 In [20, Section 3], we exploited the strict positivity of η(α) to describe regimes in which the Bohr sets (2.8) contain and are contained in generalised arithmetic progressions of the expected size. We presently outline the key ﬁndings from that investigation, as far as they are needed here.

Lemma 2.6 (Inner structure). Let ϑ > 1. Then there exists ε˜ > 0 with the following property. If ε ∈ (0, ε˜] is ﬁxed, and N is large in terms of ϑ, ε, α, and −ε N 6 ρi 6 1 (1 6 i 6 k − 1), (2.11) then there exists a proper generalised arithmetic progression + P = P (b; A1,...,Ak; N1,...,Nk) contained in B, for which √ ϑε ε N |P| ρ1 ··· ρk−1N, min Ni > N ,N 6 b 6 , i6k 10 and gcd(A1,...,Ak) = 1.

Proof. Observe that the statement gets weaker as ε decreases, in the sense that if it holds for ε =ε ˜ then it holds for any ε ∈ (0, ε˜]. It is almost identical to 24 SAM CHOW AND NICLAS TECHNAU the statement in [20, Lemma 3.1], but there the variable ϑ was equal to 1. By replacingε ˜ byε/ϑ ˜ , thereby shrinking the range of admissible values of ε, we obtain the statement here.

Next, we quantify the range of the width vector ρ for which the Bohr sets are eﬃciently contained in generalised arithmetic progressions. The outer structure lemma as stated in [20, Lemma 3.2] is homogeneous and fails to record the feature that gcd(A1,...,Ak) = 1. Here we require an inhomogeneous version as well as the latter feature, and fortunately we can extract this additional information from the proof of [20, Lemma 3.2], as we now explain.

Lemma 2.7 (Outer structure). Let ϑ > 1. Then there exists ε˜ > 0 with the following property. If ε ∈ (0, ε˜] is ﬁxed and N is suﬃciently large in terms of ε, ϑ, and we have (2.11), then there exists a generalised arithmetic progression

P = P(b; A1,...,Ak; N1,...,Nk), γ containing Bα(N; ρ), for which ϑε min Ni > N , |P| ρ1 ··· ρk−1N, gcd(A1,...,Ak) = 1. i6k

γ Proof. Lemma 2.6 implies that the Bohr set Bα(N; ρ) is non-empty, so choose γ γ b ∈ Bα(N; ρ) arbitrarily. For any n ∈ Bα(N; ρ), the triangle inequality yields 0 n − b ∈ Bα(N; 2ρ). Now [20, Lemma 3.2] assures us that

n − b ∈ P(0; A1,...,Ak; N1,...,Nk).

The coprimality property gcd(A1,...,Ak) = 1 comes out of the proof, but was not recorded in [20, Lemma 3.2] because was not needed there. Similarly, the ε inequality mini6k Ni > N arises in the proof. By replacingε ˜ byε/ϑ ˜ , thereby shrinking the range of admissible values of ε, we are able to bootstrap the ϑε inequality to mini6k Ni > N .

Finally, recall that Beresnevich, Haynes, and Velani [11, Lemma 6.1] determined a convenient condition under which a rank 1 Bohr set has the expected size.

Lemma 2.8. Let α ∈ R \ Q, let (qk)k be its sequence of denominators of convergents of continued fractions, and let δ ∈ (0, kq2αk/2). If there exists ` ∈ Z such that 1 q N, 2δ 6 ` 6 then 0 δN − 1 6 #Bα(N; δ) 6 32δN. LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 25

2.4. Measure theory. To bound from below the measure of the limit superior sets of interest, we deploy the ‘divergence’ Borel–Cantelli lemma [32, Lemma 2.3].

Lemma 2.9. Let E1, E2,... be a sequence of Borel subsets of [0, 1] such that ∞ X µ(En) = ∞, n=1

and let E = lim sup En. Then n→∞ !2 X µ(En) n6N µ(E) > lim sup X . N→∞ µ (En ∩ Em) n,m6N

In many applications in metric diophantine approximation, it suﬃces to establish that a limit superior set E has positive measure in order to conclude that it has full measure: Often one of the classical zero–one laws, such as Cassels’ or Gallagher’s [32, Section 2.2], rules out the possibility that µ(E) ∈ (0, 1). However, for our purposes no such zero–one law is available. To establish full measure, we turn instead to another device, which follows from Lebesgue’s density theorem. Intuitively, the next lemma—sometimes referred to as Knopp’s lemma—states that any set whose local densities are uniformly and positively bounded from below must have full measure. This is a special case of [8, Proposition 1].

Lemma 2.10. If E ⊆ [0, 1] is Borel set and µ(E ∩ I) µ(I) for any interval I ⊆ [0, 1], then µ(E) = 1.

2.5. Real analysis. The next lemma will later enable us to eschew a certain small-GCD regime. In the ﬁrst instance, it asserts that if ψ(n) log n is not summable then neither is ψ(an) log n for any ﬁxed a ∈ N. In fact, we can allow a to increase very slowly with n.

Lemma 2.11. Let d ∈ N. Let ψ : N → R>0 be non-increasing such that X ψ(n)(log n)d n>1 diverges. Then, there exists a strictly increasing sequence (Ki)i of positive integers satisfying Ki > exp(exp i), for all i > 2, such that K1 = 1, Ki = o(Ki+1), and: 26 SAM CHOW AND NICLAS TECHNAU

(1) The map f deﬁned by

f(n) = i (Ki 6 n < Ki+1) satisﬁes f(n) log log n.

(2) If ψˆ(n) = ψ(4f(n)n), for all n, then the series X ψˆ(n)(log n)d n>1 diverges.

Proof. We construct (Ki)i recursively. Write X d X ˆ d Sψ(N) = ψ(n)(log n) ,Sψˆ(N) = ψ(n)(log n) . n6N n6N Set K1 = 1, let i > 1, and suppose that Ki has already been constructed. If a ∈ N and N > Na, where Na = N(a, ψ) is large, then 2 X X d Sψ(aN) = Sψ(a ) + ψ (aj + r) (log (aj + r)) a6j Ki + Na, for some i ∈ N. Supposing we choose Ki+1 = N + 1, then as i > f(j) for any j 6 N, we have i f(j) ˆ ψ(aj) = ψ(4 j) 6 ψ(4 j) = ψ(j)(j 6 N), and so −1 −i i Sψˆ(N) a Sψ(aN) = 4 Sψ(4 N). We deﬁne Ki+1 to be one more than the smallest positive integer

N > exp(exp(2Ki + N4i )) −i i for which 4 Sψ(4 N) > i. By construction, we have

Sψˆ(Ki+1) i, Ki+1 > exp(exp(i + 1)). The latter implies that f(n) 6 log log n for n > K2, completing the proof.

The following lemma helps us with dyadic pigeonholing.

Lemma 2.12. Let h : N → R>0 be non-increasing, and ﬁx C > 2, κ > 0, as well a positive integer J0 1. Then, for N ∈ N and J = blog N/ log Cc, we have J X X h(n)(log n)κ jκCjh(Cj). (2.12) J C 0 6n6N j=J0 LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 27

Proof. Since h is non-increasing, for j = 1, 2,...,J we have X κ j+1 j+1 κ h(n)(log n) > C (1 − 1/2)h(C )(j log C) Cj

j=J0 and J X X h(n)(log n)κ Cjh(Cj)(j log C)κ. J C 0 6n6N j=J0

2.6. Geometry of numbers. The following lattice point counting theorem originates from the work of Davenport [22], see also [6] and [54, p. 244]. Our precise statement follows from [6, Lemmata 2.1 and 2.2].

Theorem 2.13 (Davenport). Let d, h ∈ N, and let S be a compact subset of Rd. Assume that the two following conditions are met:

(i) Any line intersects S in a set of points which, if non-empty, comprises at most h intervals.

(ii) The ﬁrst condition holds, with j in place of d, for the projection of S onto any j-dimensional subspace.

Let λ1 6 ··· 6 λd be the successive minima, with respect to the Euclidean unit ball, of a (full-rank) lattice Λ in Rd. Then d−1 vol(S) X Vj(S) |S ∩ Λ| − d,h , det Λ λ ··· λ j=0 1 j where Vj(S) is the supremum of the j-dimensional volumes of the projections of S onto any j-dimensional subspace. We adopt the convention that V0(S) = 1. 28 SAM CHOW AND NICLAS TECHNAU

We have not encountered a reference for the following classical result, so we provide a proof.

Lemma 2.14. Let k ∈ N, and let A1,...,Ak, d ∈ N with

gcd(A1,...,Ak, d) = 1. Then the congruence

A1n1 + ··· + Aknk ≡ 0 mod d (2.13) deﬁnes a full-rank lattice Λ of determinant d.

Proof. We proceed by induction on k. For k = 1, a basis is given by {d}, so the lattice has rank 1 and determinant d.

Now suppose k > 2, and that the conclusion holds with k − 1 in place of k. (1) (k−1) k−1 0 Put g = gcd(Ak, d), and let b ,..., b ∈ Z be a basis for the lattice Λ deﬁned by A1n1 + ··· + Ak−1nk−1 ≡ 0 mod g. We claim that (1) (k−1) {(b , f1),..., (b , fk−1), (0,..., 0, d/g)} ⊂ Λ is a basis for Λ, where (A ,...,A ) · b(j) f = −A /g 1 k−1 (1 j k − 1), j k g 6 6 wherein Ak/g denotes an integer which is inverse to Ak/g modulo d/g. As these vectors are independent over R, our task is to show that their Z-span is 0 0 Λ. Let n = (n1, . . . , nk) ∈ Λ. Then n := (n1, . . . , nk−1) ∈ Λ , so there exist c , . . . , c ∈ such that n0 = P c b(j). Now 1 k−1 Z j6k−1 j ! X A1n1 + ··· + Ak−1nk−1 (−A /g) n − c f ≡ k k j j g j6k−1 (j) X (A1,...,Ak−1) · b − c j g j6k−1 A n + ··· + A n (A ,...,A ) · n0 ≡ 1 1 k−1 k−1 − 1 k−1 g g ≡ 0 mod d/g, so X nk ≡ cjfj mod d/g, j6k−1 and ﬁnally X (j) n = ck(0,..., 0, d/g) + cj(b , fj) j6k−1 LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 29

for some ck ∈ Z. This conﬁrms our asserted basis for Λ, and so the determinant is indeed g · d/g = d.

We need to be able to count elements of a generalised arithmetic progression divisible by a given positive integer d. The previous two facts enable us to accurately do so, provided that d is not too large.

Lemma 2.15.

(i) Let d ∈ N, and let P be generalised arithmetic progression given by (2.9), where gcd(A1,...,Ak) = 1. Then

#{n ∈ P : d | n} −1 −1 d + (min Ni) . (2.14) N1 ··· Nk i6k

(ii) Let P be a proper, asymmetric generalised arithmetic progression given by (2.10), where gcd(A1,...,Ak) = 1, and let d ∈ N. Then

#{n ∈ P : d | n} −1 = d + O(1/ min Ni). (2.15) N1 ··· Nk i6k

Proof. (i) The quantity #{n ∈ P : d | n} is bounded above by the number of integer solutions

(n1, . . . , nk) ∈ [−N1,N1] × · · · × [−Nk,Nk] to b + A1n1 + ··· + Aknk ≡ 0 mod d. We may assume that this has a solution ∗ ∗ (n1, . . . , nk) ∈ [−N1,N1] × · · · × [−Nk,Nk], and then 0 0 ∗ ∗ (n1, . . . , nk) = (n1, . . . , nk) − (n1, . . . , nk)

lies in [−2N1, 2N1] × · · · × [−2Nk, 2Nk] and satisﬁes 0 0 A1n1 + ··· + Aknk ≡ 0 mod d. By Lemma 2.14, this deﬁnes a full-rank lattice of determinant d, and this is a sublattice of Zn so the successive minima are greater than or equal to 1. By Theorem 2.13, we now have k 2 N1 ··· Nk #{n ∈ P : d | n} 6 + Ok(N1 ··· Nk/ min Ni), d i6k giving (2.14).

(ii) As P is proper, the quantity #{n ∈ P : d | n} counts integer solutions

(n1, . . . , nk) ∈ [1,N1] × · · · × [1,Nk] 30 SAM CHOW AND NICLAS TECHNAU

to b + A1n1 + ··· + Aknk ≡ 0 mod d. ∗ ∗ Since gcd(A1,...,Ak) = 1, there exist integers n1, . . . , nk such that ∗ ∗ b + A1n1 + ··· + Aknk = 0. Now 0 0 ∗ ∗ (n1, . . . , nk) = (n1, . . . , nk) − (n1, . . . , nk) ∗ ∗ ∗ ∗ lies in [1 − n1,N1 − n1] × · · · × [1 − n1,Nk − n1] and satisﬁes 0 0 A1n1 + ··· + Aknk ≡ 0 mod d. By Lemma 2.14, this deﬁnes a full-rank lattice of determinant d, and this is a sublattice of Zn so the successive minima are greater than or equal to 1. By Theorem 2.13, we now have

(N1 − 1) ··· (Nk − 1) #{n ∈ P : d | n} = + Ok(N1 ··· Nk/ min Ni), d i6k giving (2.15).

2.7. Primes and sieves. We require one of Mertens’ three famous, classical estimates [40, Theorem 3.4(c)].

Theorem 2.16 (Mertens’ third theorem). For x > 2, we have Y 1 e−γ 1 1 − = 1 + O , p log x log x p6x where γ is the Euler–Mascheroni constant.

Sieve theory is a powerful collection of techniques used to study prime numbers. Let P be set of primes, and let

A = (an)16n6x be a ﬁnite sequence of non-negative real numbers. The main object of interest is the sifting function X S(A, z) = an, (n,P (z))=1 where Y P (z) = p, p∈P p 2. For example, if J ⊆ [1, x] ∩ Z and an = 1 for n ∈ J and an = 0 for n∈ / J , then S(A, z) counts elements of J not divisible by any prime p ∈ P for which p < z. LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 31

Evaluating the sifting function using the inclusion–exclusion principle yields X S(A, z) = µ(d)Ad(x), d|P (z) where X Ad(x) = an (d ∈ N). n6x n≡0 mod d This gives rise to the main term XV (z), where

• X is typically chosen to approximate A1(x)

• The density function, g, is a multiplicative arithmetic function satisfying 0 6 g(p) < 1 (p ∈ P), and g(p) is typically chosen to approximate Ap(x)/X

• Y V (z) = (1 − g(p)). p∈P p

However, if z is large then P (z) could have many prime factors, and the error terms may combine to overwhelm the main term. This impasse stood for a long time before Viggo Brun was able to overcome it in many situations. Brun’s idea was to approximate µ by a function of smaller support, thereby reducing the number of error terms. One particularly useful outcome is the fundamental lemma, which involves the following additional objects:

• Remainders

rd = Ad(x) − g(d)X (d ∈ N)

• The dimension κ > 0, typically chosen to approximate a suitable average of g(p)p over p ∈ P

• The level D > z, and the sifting variable s = log D/ log z.

The result, for which the lower bound is stated below, is taken from Opera de Cribro [28, Theorem 6.9]. In principle there is a great deal of ﬂexibility, however in practice there is often a natural choice of parameters that reﬂects the nature of the problem, and any less principled choice tends to produce weaker information. 32 SAM CHOW AND NICLAS TECHNAU

Theorem 2.17 (Fundamental lemma of sieve theory). Let κ > 0, z > 2, 9κ+1 D > z , K > 1, and assume that Y log z κ (1 − g(p))−1 K (2 w < z). (2.16) 6 log w 6 w6p XV (z)(1 − e K ) − |rd|. d|P (z) d

3. A fully-inhomogeneous version of Gallagher’s theorem In this section, we prove Theorem 1.7, along the way establishing Theorem 1.22. At the end we prove the convergence part of Corollary 1.10. The reasoning presented here is sensitive to the diophantine nature of the shift γk. We begin by introducing some notation, and reducing the statements of Theorems 1.7 and 1.22 to proving statistical properties of certain sets. 3.1. Notation and reduction steps. For ease of exposition, we use the abbreviations α = αk, γ = γk throughout the present section. With f as in Lemma 2.11, specialising d = k−1 therein, we set ψˆ(n) = ψ(ˆn), wheren ˆ = 4f(n)n. In light of (1.3), we then have ∞ X ψ(ˆn)(log n)k−1 = ∞. (3.1) n=1

For ε0 > 0, we deﬁne ˆ −4ε0 −2ε0 G = {h ∈ N : h 6 khαi − γik 6 h (1 6 i 6 k − 1)}, ˆ G = {n ∈ N :n ˆ ∈ G}, and 1 G = n ∈ G : ψ(ˆn) . > n(log n)k+1

Remark 3.1. The constant ε0 is sufficiently small depending on the diophantine nature of the fibre vector α and the final shift γ. Since, throughout this section, we operate under the assumption (1.5) for k > 3 and for k = 2 that α∈ / (L∪Q), the constant ε0 will always be small enough so that the structural theory of Bohr sets applies. In particular, we will have ε0 6 ε˜ when Lemmata 2.6 and −1 2.7 are applied with ϑ = 20k. We will also always assume that ε0 < (99k) .

Let us now also introduce a parameter η = η(γ) ∈ (0, 1). We shall in due course be more speciﬁc about ε0 and η, if γ is diophantine, rational, or Liouville. The reader seeking these details instantly may consult (3.36), (3.42), and (3.44). LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 33

For n ∈ N, let ψ(ˆn) Ψ(n) = 1G(n), knαˆ 1 − γ1k · · · knαˆ k−1 − γk−1k where 1G is the indicator function of G. Fix a non-empty interval I ⊆ [0, 1] and γ ∈ R, and for n ∈ N let ( a + γ ∈ nˆI, ) I,γ En := α ∈ [0, 1] : ∃a ∈ Z s.t. |nαˆ − γ − a| < Ψ(n), . (3.2) (a, nˆ) is (γ, η)-shift-reduced We call these (localised) approximation sets. Note that if n is large in terms of I then I,γ µ(En ) µ(I)Ψ(n). (3.3) Observe that G = {n ∈ N : µ(En) > 0}. We will often suppress the dependence on γ and I in the notation by writing I,γ En in place of En .

We say that γ is admissible if there exists η ∈ (0, 1) such that for any interval I ⊆ [0, 1] there are inﬁnitely positive integers X for which the following two properties hold:

(1) We have X X k−1 µ(En) µ(I) ψ(ˆn)(log n) . (3.4) n6X n6X

(2) The sets En are quasi-independent on average for those X, i.e. !2 X X k−1 µ(En ∩ Em) µ(I) ψ(ˆn)(log n) . (3.5) m,n6X n6X

By (3.1), the right hand side of (3.4) is unbounded as a function of X. It is worth stressing that the implicit constants in (3.4) and (3.5) are only allowed to depend on α, γ1, . . . , γk, and are uniform in I.

Now we show that Theorems 1.22 and 1.7 can be reduced to showing that every γ ∈ R is admissible. We assume throughout this section that Ψ(n) 6 1/2 (n large), (3.6) as we may because otherwise these two theorems are trivial.

Remark 3.2. Unless otherwise specified, ‘(sufficiently) large’ means large in terms of α, γ1, . . . , γk, ψ, I. Similarly, unless otherwise specified, a positive real number is ‘(sufficiently) small’ if it is small in terms of α, γ1, . . . , γk, ψ, I. 34 SAM CHOW AND NICLAS TECHNAU

Proposition 3.3. If every γ ∈ R is admissible, then Theorems 1.22 and 1.7 are true.

Proof. Let γ ∈ R, and let η be as in the deﬁnition of admissibility. We ﬁrst show that W(Ψ; γ, η) := lim supn→∞ An has full measure, where |nαˆ − γ − a| < Ψ(n), A = α ∈ [0, 1] : ∃a ∈ s.t. . n Z (a, nˆ) is (γ, η)-shift-reduced Fix a non-empty subinterval I0 of [0, 1], and let I be its dilation by 1/2 about its I,γ 0 centre. Observe using the triangle inequality that if n > n0(I) then En ⊆ I . I,γ I,γ Let R = lim sup En . Inserting (3.4) and (3.5) into Lemma 2.9, we obtain n→∞ 0 I,γ 0 µ(W(Ψ; γ, η) ∩ I ) > µ(R ) µ(I) µ(I ), where the implied constant is independent of I0. Lemma 2.10 now grants us that W(Ψ; γ, η) has full measure in [0, 1]. As Ψ(n) 6 Φ(ˆn) for all n, we obtain Theorem 1.22. By 1-periodicity of k · k, we thereby obtain Theorem 1.7.

For the remainder of this section, we establish that any γ ∈ R is admissible. In fact, we will verify a fortiori that these properties of the approximation sets hold for all suﬃciently large values of X. Moreover, the ﬁrst property will hold for all η ∈ (0, 1).

3.2. Divergence of the series. Let γ ∈ R and η ∈ (0, 1), and let X ∈ N be large. Let ε0 be small, as in Remark 3.1. In this subsection, we establish the property (3.4). We begin by estimating µ(En).

Lemma 3.4. Let n ∈ G be large. Then ϕ(ˆn) µ(I)Ψ(n) µ(E ) µ(I)Ψ(n). (3.7) nˆ n

Proof. The enunciated upper bound follows at once by observing that En is contained in O(ˆnµ(I)) many intervals of length 2Ψ(n)/nˆ. On the other hand, it contains X ϕ0(ˆn) := 1 a+γ∈nˆI 0 (qta+ct,nˆ)=1 many open intervals of length 2Ψ(n)/nˆ and, by (3.6), these are disjoint. We proceed to show that ϕ0(ˆn) µ(I)ϕ(ˆn), (3.8) which would complete the proof. We will ﬁnd that the implied constant in (3.8) is absolute. LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 35

To infer (3.8), we apply Theorem 2.17 with the following speciﬁcations. The 0 set P of relevant primes is the set of primes that dividen ˆ but not qt, so that Y P (z) = p,

p∈P 1 is an absolute constant, and s = 10(1 + log K),D = zs. The dimension condition (2.16) follows from Mertens’ third theorem (Theorem 2.16). With Y V (z) = (1 − p−1),

p∈P (1 − e K )XV (z) − |rd|, (m,P (z))=1 d|P (z) d

0 For d < D dividing P (z), we have (d, qt) = 1, so X −1 am = d X + O(1), m≡0 mod d ergo rd 1. Therefore X O(1) |rd| D = (log n) , d|P (z) d (1 − e )XV (z) + o(XV (z)) XV (z). (m,P (z))=1 Finally, the union bound gives X X X log n a X p−1 = o(XV (z)), m z log z (m,P (x)/P (z))>1 p|n p>z 36 SAM CHOW AND NICLAS TECHNAU

and so X ϕ0(ˆn) = am XV (z) µ(I)ϕ(ˆn). (m,P (x))=1

Let C > 4 be a constant that is large in terms of the implied constants in Lemmata 2.6 and 2.7. To estimate P ϕ(ˆn) µ(I)Ψ(n), it will be useful to n6X nˆ gather all n on a scale N such that, additionally, the knαˆ i−γik are in prescribed C-adic ranges. Here and in the following, we assume for a purely technical reason that C is a perfect square, and let N be large in terms of C and the other constants. Deﬁne ˆ ˆ N < h 6 CN,d Bloc(N; ρ) = h ∈ N : ρi < khαi − γik 6 Cρi (i = 1, . . . , k − 1) k−1 ! ˆ [ = B(CNd; Cρ) \ B(N; Cρ) ∪ B(CNd; ρi) , (3.9) i=1

where ρi = (ρi,1, . . . , ρi,k−1), and ρi,j is deﬁned to be ρj if i = j and Cρj otherwise. For this section ρ denotes a parameter in the hyperrectangle W(N) := [Nˆ −4.1ε0 , Nˆ −1.9ε0 ]k−1. (3.10) Similarly we will have a large parameter M 6 N, and δ will denote a parameter in W(M). We also write ˆ Bloc(N; ρ) = {n ∈ N :n ˆ ∈ Bloc(N; ρ)}. (3.11) By the construction of f, there is a uniquely determined integer u = u(N) with f(n) ∈ {u, u + 1} (N 6 n 6 CN). (3.12) Furthermore, since f(n) log log n, we know that if n N thenn/n ˆ = N o(1). We have ample use for this estimate, and shall use it without further mention.

We proceed to study these localised Bohr sets, ﬁrst deriving a cardinality estimate in the range of interest. Heuristically, one might expect that each of ˆ ˆ the Θ(N) many integers in the interval [N, CNd] lies in {nˆ : n ∈ Bloc(N; ρ)} with probability roughly Π(ρ)/4f(N), recalling the notation (1.18). For the ranges occurring implicitly in the set G, this heuristic correctly predicts the order of magnitude of #Bloc(N; ρ):

Lemma 3.5. We have

#Bloc(N; ρ) Π(ρ)N, uniformly for ρ ∈ W(N). LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 37

Proof. In light of (3.12), we have f(N) f(n) ν ν+2 4 , 4 ∈ {2 , 2 } (n ∈ Bloc(N; ρ)), (3.13) for some positive integer ν, where ν O(log log N) O(1) 2 6 4 = (log N) .

For the upper bound, each n ∈ Bloc(N; ρ) gives rise to a diﬀerent element of ˆ ν Bloc(N; ρ) that is divisible by 2 . We can count the latter using Lemmata 2.7 and 2.15, noting that ˆ 20kε0 N1 ··· Nk N · Π(ρ), min Ni > N i6k therein. We obtain ˆ ν ˆ f(N) #Bloc(N; ρ) N · Π(ρ)/2 N · Π(ρ)/4 = N · Π(ρ).

For√ the lower bound,√ we divide the interval (N,CN] into two subintervals (N, CN] and ( CN,CN], and observe that f must be constant√ on at least f(n) ν one of these√ subintervals. We assume that 4 = 2 on (N, CN]; the cases involving ( CN,CN] and/or 2ν+2 can be dealt with in the same manner. A lower bound is then given by the number of elements of √ ˆ ν ν Bloc(N; ρ) ∩ (2 N, C2 N] √ k−1 √ ! \ ˆ [ \ = B( CN; Cρ) \ B(N; Cρ) ∪ B( CN;(ρi,1, . . . , ρi,k−1)) i=1 that are divisible by 2ν, which we can estimate using Lemmata 2.6, 2.7, and √ \ 2.15. The point is that C is large, so the count for B( CN; Cρ) dominates.

Before we demonstrate (3.4), we prepare one more lemma. Since (3.4) requires us to determine the order of magnitude of P µ(E ), it is helpful to know n6N n that the totient weights on the left hand side of (3.7) ‘average well’ in suitable C-adic ranges, i.e. that for n in a localised Bohr set Bloc(N; ρ), the average of ϕ(ˆn)/nˆ has order of magnitude 1.

Remark 3.6. A similar strategy for proving this point was employed by the ﬁrst named author in [18, Lemma 3.1], and then by both authors in [20, Lemma 4.1]. The additional diﬃculty here is a mild one: Roughly speaking, we need to average only over elements of the Bohr sets which are divisible by an appropriate power of four.

Lemma 3.7 (Good averaging). We have X ϕ(ˆn) #B (N; ρ), nˆ loc n∈Bloc(N;ρ) 38 SAM CHOW AND NICLAS TECHNAU

uniformly for all ρ ∈ W(N).

Proof. By the AM–GM inequality, we have 1 ! #B (N;ρ) τ 1 X ϕ(ˆn) Y ϕ(ˆn) loc Y 1 p > = 1 − , #Bloc(N; ρ) nˆ nˆ p n∈B (N;ρ) n∈P loc p6CNd where #{n ∈ Bloc(N; ρ): p | nˆ} τp = . #Bloc(N; ρ) Now   1 X ϕ(ˆn) X τp − ln   , #Bloc(N; ρ) nˆ p n∈B (N;ρ) loc p6CNd so as τ2 6 1 it suﬃces to show that −ε0 τp p (3 6 p 6 CNd).

Suppose 3 6 p 6 CNd. By Lemma 3.5, we have #{h ∈ B(CNd; Cρ): h ≡ 0 mod 2νp} τ , p ρN where ν is as in (3.13) and ρ = Π(ρ). We can estimate the numerator via Lemmata 2.7 and 2.15, noting that

20kε0 N1 ··· Nk CNd · ρ, min Ni > N i6k therein. We obtain

ν ν −1 −20kε0 −ε0 τp 2 ((2 p) + N ) p .

We can now estimate the normalised-totient-weighted sums of measures, thereby attaining the main result of this subsection. For j ∈ N, deﬁne j j+1 j j+1 Dj = (C ,C ] ∩ G, Gj = (C ,C ] ∩ G. (3.14)

Lemma 3.8. The sets En satisfy (3.4) for all suﬃciently large X.

k−1 Proof. For j large, observe that Dj is contained in a union of O(j ) sets

j −t1 −tk−1 Bloc(C ;(C ,...,C )) satisfying the hypotheses of Lemma 3.5, and so

X 1 k−1 j Tj := j C . (3.15) knαˆ 1 − γ1k · · · knαˆ k−1 − γk−1k n∈Dj LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 39

j Let J be the largest integer j such that C 6 X. By (3.3), we have X X j µ(En) µ(I)Ψ(n) 6 µ(I)ψ(Cc)Tj Cj

X j X k−1 j j ψ(Cc)Tj 1 + j C ψ(Cc). j6J j6J By (2.12), we now have

−1 X X k−1 j j X k−1 µ(I) µ(En) 1 + j C ψ(Cc) ψ(ˆn)(log n) , n6X j6J n6X recalling that X is large and recalling from (3.1) that the right hand side diverges as X → ∞. We have established the upper bound in (3.4).

For the lower bound, we begin by applying (3.7) to give X X ϕ(ˆn) µ(E ) µ(I) Ψ(n), n nˆ n6X C1

We proceed to lower-bound the contribution from n ∈ Dj, for j large. To this j end, consider the localised Bohr sets Bloc(C ; ρ(t)), where ρ(t) = (C−t1 ,...,C−tk−1 ). j For n ∈ Bloc(C ; ρ(t)), where t = (t1, . . . , tk−1) satisﬁes

2.1ε0(j + 1) 6 tr 6 3.9ε0j (1 6 r 6 k − 1), we deduce that −4ε −t 1−t −2ε nˆ 0 6 C r 6 C r 6 nˆ 0 (1 6 r 6 k − 1).

Hence n ∈ Dj, and so we see that j Bloc(C ; ρ(t)) ⊆ Dj. 40 SAM CHOW AND NICLAS TECHNAU

Next, observe that X ϕ(ˆn) ψ(ˆn) X X ϕ(ˆn) ψ(C[j+1) , nˆ knαˆ − γ k · · · knαˆ − γ k nρˆ (t) 1 1 k−1 k−1 j n∈Dj t n∈Bloc(C ;ρ(t)) where t runs through all the integer vectors as above and ρ(t) = Π(ρ(t)). j j Lemma 3.5 implies #Bloc(C ; ρ(t)) C ρ(t), and so Lemma 3.7 assures us that X ϕ(ˆn) Cj, nρˆ (t) j n∈Bloc(C ;ρ(t)) uniformly for each of the Θ(jk−1) many choices of t. Whence X ϕ(ˆn) ψ(ˆn) ψ(C[j+1)Cj+1(j + 1)k−1. nˆ knαˆ 1 − γ1k · · · knαˆ k−1 − γk−1k n∈Dj log X Summing over large j 6 log C and invoking Lemma 2.12 furnishes (3.16).

Finally, by (3.15) we have X ϕ(ˆn) ψ(ˆn) nˆ knαˆ 1 − γ1k · · · knαˆ k−1 − γk−1k n∈G\G ∞ X X ψ(ˆn) 6 knαˆ 1 − γ1k · · · knαˆ k−1 − γk−1k j=1 n∈Dj \G ∞ ∞ X jk−1Cj X 1 + 1 + j−2 1, Cj(j log C)k+1 j=1 j=1 which is (3.17).

Having established (3.4), our ﬁnal task for this section is to prove (3.5) for some ε0 > 0 and η ∈ (0, 1) depending on γ.

3.3. Overlap estimates, localised Bohr sets, and the small-GCD regime. In the present subsection, we reduce the task of proving (3.5) to demonstrat- ing a uniform estimate in localised Bohr sets, see Lemma 3.9. Thereafter, we recast the desired estimate in terms of counting solutions to diophantine inequalities in localised Bohr sets, see Lemma 3.10. At the end of this subsection, we establish such a counting result in a regime where the arising GCDs are relatively small, see Proposition 3.11.

Let N > M, where M is large. The next lemma asserts that if we have a good bound on I,γ X I,γ I,γ R (M,N; ρ, δ) := µ(En ∩ Em ),

n∈Bloc(N;ρ) m∈Bloc(M;δ) n6=m LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 41

suﬃciently uniform in ρ and δ, then we can deduce (3.5). As before, we shall drop the dependency on γ and I for most of the time, and simply write R(M,N; ρ, δ) instead.

Lemma 3.9. If R(M,N; ρ, δ) µ(I)MNψ(Mˆ )ψ(Nˆ), (3.18) uniformly for M large and N > M, ρ ∈ W(N), and δ ∈ W(M), then (3.5) holds.

j Proof. Let J be the least integer j such that C > N, and recall (3.14). Then X X X µ(En ∩ Em) 1 + µ(En ∩ Em). m,n6X j6i6J n∈Gi m∈Gj

Let X0 ∈ N be a large constant. By Lemma 3.8, the contribution from j < X0 is bounded by a constant times X X k−1 µ(En) µ(I) ψ(ˆn)(log n) . n6X n6X Thus, recalling from (3.1) that the completed series diverges, it remains to show that !2 X X X k−1 µ(En ∩ Em) µ(I) ψ(ˆn)(log n) . (3.19) X06j6i6J n∈Gi n6X m∈Gj

Presently, we ﬁx i, j with X0 6 j 6 i 6 J. Consider the vectors ρ(t) = (C−t1 ,...,C−tk−1 ), δ(`) = (C−`1 ,...,C−`k−1 ), (3.20) wherein ranges for the exponents will be prescribed shortly. Let us account for the contribution of the diagonal ﬁrst. By (3.1), the summands for which n = m contribute !2 X X X k−1 X k−1 µ(En) µ(I) ψ(ˆn)(log n) µ(I) ψ(ˆn)(log n) . X06i6J n∈Gi n6X n6X

Next, we account for the oﬀ-diagonal contribution. To this end, we approxi- mately decompose X S(i, j) := µ(En ∩ Em)

n∈Gi m∈Gj n6=m 42 SAM CHOW AND NICLAS TECHNAU

into sums of the shape X S(i, j; t, `) := µ(En ∩ Em). i n∈Bloc(C ;ρ(t)) j m∈Bloc(C ;δ(`)) n6=m Speciﬁcally, the sum S(i, j) is bounded above by the sum of S(i, j; t, `) over the integer vectors t, ` within the ranges

1.9ε0i 6 t1, . . . , tk−1 6 4.1ε0i, 1.9ε0j 6 `1, . . . , `k−1 6 4.1ε0j. By the assumed estimate (3.18), we have S(i, j; t, `) µ(I)CiCjψˆ(Ci)ψˆ(Cj), uniformly in t and ` as above. Summing over the O(ik−1) many choices for t and the O(jk−1) many choices for `, we see that S(i, j) µ(I)ik−1Cijk−1Cjψˆ(Ci)ψˆ(Cj). Summing over i and j gives 2 X X X k−1 j ˆ j µ(En ∩ Em) µ(I) j C ψ(C ) . X06j6i6J n∈Gi X06j6J m∈Gj Now Lemma 2.12 delivers (3.19), completing the proof.

Denote by N1,...,Nk the length parameters, arising from Lemma 2.7, associ- ˆ ated to the outer structure of Bloc(N; ρ). Speciﬁcally, we apply the lemma to B(CNd; Cρ), with ε = ε0 and ϑ = 20k. By symmetry, we may assume that N1 = min{Ni : 1 6 i 6 k}, (3.21) and we do so in order to simplify notation. Note that

20kε0 N1 > N . Let M be large, let N > M, and let ψ(Nˆ) ψ(Mˆ ) ∆ = CMd + CNd . (3.22) Π(ρ) Π(δ)

Importantly, if there exists m ∈ Bloc(M; δ) ∩ G then we have a lower bound on ∆ of the strength ψ(Mˆ ) ψ(m ˆ ) CNd ∆ CNd CNd > Π(δ) > Π(δ) > Π(δ)m(log m)k+1 CNd 1 1 . > CM Π(δ)(log(CM))k+1 > Π(δ)(log(CM))k+1 For δ ∈ W(M), recalling (3.10), we ﬁnd that for all suﬃciently large M we have ∆ > M (k−1)ε0 > 1. (3.23) LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 43

0 Let (q`)` be the sequence of continued fraction denominators of γ. We estimate the contribution to the quantity R(M,N; ρ, δ) from the small-GCD regime (including the diagonal), i.e. X Rgcd6(M,N; ρ, δ) := µ(En ∩ Em), (3.24) n∈Bloc(N;ρ) m∈Bloc(M;δ) 0 gcd(ˆn,mˆ )6max{3∆/kq2γk,N1} and the contribution from large-GCD regime, i.e. X Rgcd>(M,N; ρ, δ) := µ(En ∩ Em), (3.25)

n∈Bloc(N;ρ) m∈Bloc(M;δ) 0 gcd(ˆn,mˆ )>max{3∆/kq2γk,N1} n6=m

separately. When bounding these quantities, we may assume that Bloc(M; δ) intersects G, and so we have (3.23).

Let us write ρ = Π(ρ), δ = Π(δ).

As a ﬁrst step towards estimating Rgcd6(M,N; ρ, δ), we record a useful relation between the size of Rgcd6(M,N; ρ, δ) and the number of solutions to a diophantine inequality with various constraints.

Lemma 3.10. Let M be large, let N > M, and let ρ ∈ W(N), and δ ∈ W(M). 2 2 Denote by Dgcd6 the number of quadruples (n, m, a, b) ∈ N × Z for which a + γ b + γ , ∈ I, n ∈ B (N; ρ) ∩ G, m ∈ B (M; δ) ∩ G, (3.26) nˆ mˆ loc loc as well as 0 gcd(ˆn, mˆ ) 6 max(3∆/kq2γk,N1) (3.27) and |(ˆn − mˆ )γ − (ma ˆ − nbˆ )| 6 ∆, (3.28) and ﬁnally (a, nˆ), (b, mˆ ) are (γ, η)-shift-reduced. (3.29) 2 2 Furthermore, let Dgcd> be the number of quadruples (n, m, a, b) ∈ N ×Z which satisfy (3.26) , (3.28), and (3.29), but instead of (3.27) the reversed inequality 0 gcd(ˆn, mˆ ) > max(3∆/kq2γk,N1) and the constraint n 6= m. If

Dgcd6 = O(µ(I)ρNδM∆), (3.30) ˆ ˆ then Rgcd6(M,N; ρ, δ) µ(I)MNψ(M)ψ(N). Moreover, if we have

Dgcd> = O(µ(I)ρNδM∆), (3.31) ˆ ˆ then Rgcd>(M,N; ρ, δ) µ(I)MNψ(M)ψ(N).

Proof. We detail the proof only in the case of Dgcd6 since the case Dgcd> can be dealt with in the same way. We begin by observing that each En (resp. Em) 44 SAM CHOW AND NICLAS TECHNAU

is contained in a union of ﬁnitely many intervals a + γ − Ψ(n) a + γ + Ψ(n) I = , n,a nˆ nˆ

(resp. Im,b), and so

µ(En ∩ Em) 6 max{min(µ(In,a), µ(Im,b)) : a, b ∈ Z}· #{(a, b): In,a ∩ Im,b 6= ∅}. The first factor is bounded above by Ψ(n) Ψ(m) ψ(Nˆ) ψ(Mˆ ) 2 min , 6 2 min , . nˆ mˆ ρNˆ δMˆ To bound the second factor, we have to count how often the centre (a + γ)/nˆ of In,a is ‘sufficiently close’ to the centre (b + γ)/mˆ of an interval Im,b. Here sufficiently close means that the distance of the centres is less than sum of the radii of the intervals, i.e. b + γ a + γ Ψ(n) Ψ(m) − + . mˆ nˆ 6 nˆ mˆ Multiplying bym ˆ nˆ, we see that |nˆ(b + γ) − mˆ (a + γ)| 6 CMd Ψ(n) + CNdΨ(m) 6 ∆, and the number of integer solutions (n, m, a, b) to the above inequality, subject

to our constraints, is at most Dgcd6. Thus, if (3.30) holds then ψ(Nˆ) ψ(Mˆ ) Rgcd (M,N; ρ, δ) min , µ(I)ρNδM∆. 6 ρNˆ δMˆ

Next, observe that ψ(Nˆ) ψ(Mˆ ) ∆ 6 2CMd CNd max , . ρCNd δCMd It follows from (3.12) that Nb CNd and Mc CMd , and so ψ(Nˆ) ψ(Mˆ ) ψ(Nˆ) ψ(Mˆ ) min , min , . ρNˆ δMˆ ρCNd δCMd

The upshot is that Rgcd6(M,N; ρ, δ) is at most a constant times ψ(Nˆ) ψ(Mˆ ) ψ(Nˆ) ψ(Mˆ ) min , ρNδMµ(I)CMd CNd max , . (3.32) ρCNd δCMd ρCNd δCMd Applying the simple identity ψ(Nˆ) ψ(Mˆ ) ψ(Nˆ) ψ(Mˆ ) ψ(Nˆ) ψ(Mˆ ) min , max , = , ρCNd δCMd ρCNd δCMd ρCNd δCMd ˆ ˆ we see that (3.32) simpliﬁes to µ(I)MNψ(M)ψ(N). LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 45

Next, we establish overlap estimates in the regime in which the GCDs are relatively small. Here congruence considerations and the structural theory of Bohr sets will be decisive; a welcome circumstance is that the bound works for any ﬁnal shift γ, irrespective of its diophantine nature.

Proposition 3.11. Let M be large, and let N > M. Then ˆ ˆ Rgcd6(M,N; ρ, δ) µ(I)MNψ(M)ψ(N), uniformly for ρ ∈ W(N) and δ ∈ W(M).

Proof. By Lemma 3.10, it remains to verify (3.30). We distinguish several cases according to the size of the of the greatest common divisor d := (ˆn, mˆ ). Put nˆ = dx andm ˆ = dy. As f is non-decreasing, we have f(n) f(m) f(M) −1 d = (ˆn, mˆ ) = (4 n, 4 m) > 4 > µ(I) , where the last inequality comes from M being large in terms of I. We rewrite the overlap inequality (3.28) as |(x − y)γ − (ya − xb)| 6 ∆/d. (3.33)

0 Recall that (q`)` is the sequence of continued fraction denominators of γ.

−1 0 Case 1: µ(I) 6 d 6 3∆/kq2γk

By Lemma 3.5, there are O(ρNδM) possible choices of n ∈ Bloc(N; ρ) and m ∈ Bloc(M; δ). Now suppose we are given n and m, which then uniquely determine d, x, y.

Case 1a: ∆ < n/ˆ 4

Since x and y are coprime, the inequality (3.33) restricts the integer a to one of at most 2∆/d + 1 residue classes modulo x. Furthermore, the constraint a + γ ∈ nˆI places the integer a in an interval of length dxµ(I). Within this interval, and given r ∈ Z/xZ, there can be at most dµ(I) + 1 integers a ≡ r mod x. Thus, given n and m, there are at most (dµ(I) + 1)(2∆/d + 1) µ(I)∆ possibilities for a. Finally, the inequality (3.33) constrains the integer b to an interval of length at most 2∆/nˆ < 1/2, so b (if it exists at all) is determined by the other variables.

Case 1b: ∆ > n/ˆ 4 46 SAM CHOW AND NICLAS TECHNAU

We may suppose that (3.28) has a solution (a, b) = (a0, b0) with a0 + γ ∈ nI. By the triangle inequality, any solution to (3.28) for which a + γ ∈ nI satisﬁes 0 0 0 |maˆ − nbˆ | 6 2∆, a ∈ nˆ(I − I), (3.34) where 0 0 a = a − a0, b = b − b0, I − I = {r1 − r2 : r1, r2 ∈ I}. Denote by R the closure of the set of (a0, b0) ∈ R2 satisfying (3.34). We apply Theorem 2.13 to the region R and the lattice Z2. Observe using the triangle inequality that R is contained in the rectangle 2∆ (a0, b0) ∈ 2 : a0 ∈ nˆ(I − I), |b0| + 2µ(I)m ˆ , R 6 nˆ and in particular the projection of R onto any line has length at most 4∆ 2µ(I)ˆn + + 4µ(I)m ˆ nµˆ (I), nˆ ˆ ˆ where for the ﬁnal inequality we have used that ψ(N), ψ(M) 6 ψ(1) 1, as well as that ρ ∈ W(N) and δ ∈ W(M). The area of R is O(∆µ(I)), and we glean that the number of solutions is O(∆µ(I) +nµ ˆ (I) + 1) = O(∆µ(I)).

We conclude that there are O(µ(I)ρNδM∆) solutions in total coming from Case 1, uniformly in ρ ∈ W(N) and δ ∈ W(M).

0 Case 2: 3∆/kq2γk 6 d 6 N1

By Lemma 3.5, there are O(δM) possibilities for m ∈ Bloc(M; δ). Next, we o(1) choose d ∈ [∆,N1] dividing m, in one of M ways. By Lemma 2.7, we have

dx =n ˆ = A1n1 + ··· + Aknk + s,

where |ni| 6 Ni (1 6 i 6 k) and ˆ N1 ··· Nk ρN.

As d 6 N1, wherein we recall (3.21), Lemma 2.15 assures us that the congruence A1n1 + ··· + Aknk + s ≡ 0 mod d ˆ has O(ρN/d) solutions n1, . . . , nk, and these variables determine x.

Now suppose that we are given d, x, y ∈ N with d > ∆ and gcd(x, y) = 1. Then, as d > ∆, by the same argument as in Case 1a there are at most (dµ(I) + 1)(2∆/d + 1) dµ(I) many choices for the pair (a, b) ∈ Z2 subject to (3.33) as well as a + γ ∈ nˆI. In view of (3.23), the total is again O(µ(I)ρNδM∆).

We have established (3.30), completing the proof. LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 47

3.4. Large GCDs. Let M be large, N > M, ρ ∈ W(N), and δ ∈ W(M). On account of the preceding subsections, our ﬁnal sub-task for this section is to establish (3.31). We will choose ε0 and η according to how well γ can be approximated by rationals.

3.4.1. Diophantine final shift. Throughout the present subsubsection, the final shift γ ∈ R is diophantine, i.e. there exists λ > 2 such that −λ/2 knγk n (n ∈ N). (3.35) This is equivalent to γ being irrational and non-Liouville. We fix such a value of λ throughout the current subsubsection.

The approximation sets are as in (3.2), however now we are more speciﬁc about ε0 and η. Letε ˜ be the minimum of its values when Lemmata 2.6 and 2.7 are applied with ϑ = 20k, and ﬁx a small positive real number ε0 such that 2 −2 1000k ε0 < min{λ , ε˜}. (3.36) In this diophantine case, the value of η is of no importance, and we arbitrarily choose η = 1/2.

4 Recall that Dgcd> denotes the number of quadruples (m, n, a, b) ∈ N satisfying a + γ b + γ , ∈ I, n ∈ B (N; ρ) ∩ G, m ∈ B (M; δ) ∩ G \ {n}, nˆ mˆ loc loc 0 as well as d := gcd(ˆn, mˆ ) > max{3∆/kq2γk,N1}, (3.28), and (3.29). Put nˆ = dx,m ˆ = dy, and note that (3.28) entails (3.33) and hence k(x − y)γk < ∆/d and |x − y| < CN/d.d (3.37)

Combining the diophantine assumption (3.35) with (3.37) yields ∆/d |x − y|−λ (N/dˆ )−λ, and hence 1/(1+λ) λ/(1+λ)+o(1) d 6 ∆ N . (3.38) We enlarge the Bohr set implicit in (3.37) to n CNd Nˆ −1/λ ∆o B0 = u : kuγk + , 6 d 6 d d observing that u := |x − y| ∈ B0.

We claim that 1− 1 N λ ∆N #B0 N o(1) + . (3.39) 6 d d2 ˆ ˆ This is clearly true if d N, so let us now assume that d 6 cN for some small constant c > 0. By (3.35), we have ω(γ) 6 λ/2, 48 SAM CHOW AND NICLAS TECHNAU

so by Lemma 2.1 there exists ` ∈ N such that ˆ 1/λ 0 ˆ (N/d) 6 q` 6 N/d. Thus, we may apply Lemma 2.8 to the enlarged Bohr set B0, reaping (3.39).

We begin by choosing m ∈ Bloc(M; δ), and by Lemma 3.5 there are at most 0 O(δM) choices. Next, we choose d | m with d > max{3∆/kq2γk,N1}, and by the standard divisor function bound there are M o(1) possible choices of d.

Given an element u from the Bohr set B0 as well as a choice of y, the value of x is then determined in at most two ways. Furthermore, we claim that for each choice of x, y the number of possible choices of a, b is O(dµ(I)). To see this, let v be the integer closest to (x − y)γ, and note from (3.33) that a, b must satisfy ya − xb = v. All integer solutions to this linear diophantine equation have the form a = a0 + tx, b = b0 + ty, for a speciﬁc solution a0, b0 and a parameter t ∈ Z. Therefore the number of such a, b constrained by a + γ ∈ nˆI is nµˆ (I) O + 1 = O(dµ(I)). x Hence, by (3.39), the number of choices for u, a, b is at most N o(1)µ(I)(d1/λN 1−1/λ + ∆N/d)

which, by (3.38) and the inequality d > N1, is at most o(1) 1/(λ+λ2) 1/(1+λ)+1−1/λ N µ(I)(∆ N + ∆N/N1). (3.40)

The contribution to Dgcd> coming from the second term in (3.40) is at most

o(1) 1+o(1)−20kε0 δMN µ(I)∆N/N1 6 µ(I)δMN ∆ 6 µ(I)ρδMN∆.

Finally, the contribution to Dgcd> corresponding to the first term on the right hand side of (3.40) is at most 1 ∆ 2 δMµ(I) λ+λ N 1+o(1). N This quantity being at most µ(I)ρδMN∆ is equivalent to 1−1/(λ+λ2) 1/(λ+λ2) o(1) ρ∆ N > N . (3.41) To confirm (3.41), we first recall from (3.23) that ∆ > 1. Next, the fact that ρ ∈ W(N), together with the bound (3.36) on ε0, give −5kε −1/(9λ2) ρ > N 0 > N . These considerations bestow (3.41) and thence (3.31). LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 49

We now prove Theorems 1.7 and 1.22 in the remaining situation γ ∈ Q ∪ L, recalling that it suﬃces to establish (3.31). The underpinning mechanisms are easier to grasp when γ is rational, so we begin with this case.

3.4.2. Rational ﬁnal shift. Throughout this subsubsection we assume that γ is a rational number, given in lowest terms by γ = c0/d0. We continue to use the notation introduced in the previous subsubsection, and alert the reader to any deviation that we make. Letε ˜ be the minimum of its values when Lemmata 2.6 and 2.7 are applied with ϑ = 20k, and ﬁx a small positive real number −1 ε0 < min{ε,˜ (99k) }.

In this case η is again of little importance, and so we once more choose η = 1/2 0 arbitrarily. Observe that if n is large then ct = c0 and qt = d0, whereupon I,γ En = En is the set of α ∈ [0, 1] for which there is an integer a satisfying a + γ |nαˆ − γ − a| < Ψ(n), ∈ I, and (d a + c , nˆ) = 1. (3.42) nˆ 0 0

The overlap inequality (3.28) is equivalent to

|c0(ˆn − mˆ ) − d0(ma ˆ − nbˆ )| 6 d0∆. Dividing the above inequality by d gives ∆d |c (x − y) − d (ya − xb)| 0 . 0 0 6 d

Assume for a contradiction that c0(x − y) = d0(ya − xb). Then

x(c0 + d0b) = y(c0 + d0a). As n 6= m, we have max{x, y} > 2. Let us assume for simplicity that x > 2; a similar argument handles the case y > 2. Let p be a prime divisor of x. By Euclid’s lemma p divides y or c0 + d0a. The former is excluded by the coprimality of x and y, and the latter by the shift-reduction. This contradiction means that ∆d 1 |c (x − y) − d (ya − xb)| 0 , 6 0 0 6 d so d 6 d0∆.

On the other hand, we have d > ∆ by assumption. So let us ﬁx a non-zero integer h ∈ [−d0, d0] and count integer quadruples (n, m, a, b) with n 6= m satisfying c0(x − y) − d0(ya − xb) = h (3.43) as well as the constraints (3.26). There are O(ρNδM) many viable choices for the two integers n 6= m. Then by (3.43) there are O(µ(I)d) = O(∆µ(I)) many choices for the numerators a, b. Summing this bound over the O(1) many choices for h veriﬁes (3.31). 50 SAM CHOW AND NICLAS TECHNAU

It remains to consider the case in which γ is a Liouville number. Informally, this is a careful interpolation between the diophantine and rational cases.

3.4.3. Liouville final shift. Throughout the present subsubsection, we fix a Liouville number γ. In a nutshell, we need to find an way to balance between the regimes in which γ behaves like a rational number, and those in which γ behaves like a diophantine number. This is incarnated in the definition of the approximation sets, and shift-reduced fractions play a crucial role.

Fix a non-empty interval I in [0, 1], and a large positive constant C. Further, let η = 9(k − 1)ε0. (3.44)

Here, as before, the positive constant ε0 is at most the lower value ofε ˜ when Lemmata 2.6 and 2.7 are applied with ϑ = 20k, and moreover ε < (99k)−1. I,γ Recall that the approximation set En = En is the set of α ∈ [0, 1] for which there exists a ∈ Z satisfying |nαˆ − γ − a| < Ψ(n) (3.45) and a + γ ∈ I, and the pair (a, nˆ) is (γ, η) − shift-reduced. (3.46) nˆ As in the previous subsubsections, we write d = gcd(ˆn, mˆ ), dx =n ˆ, dy =m ˆ .

0 Recall that qt is the greatest continued fraction denominator of γ not exceeding η n , and that ct is the corresponding numerator. We separate our argument 0 according to the size of the subsequent denominator qt+1.

0 Case 1: qt+1 > 10CN/dd

This inequality and the continued fraction approximation (2.3) yield

ct d γ − . q0 6 0 t 10qtCNd Moreover, the inequality (3.28) can be written in the form |nˆ(b + γ) − mˆ (a + γ)| 6 ∆, or equivalently q0∆ |x(q0b + q0γ) − y(q0a + q0γ)| t . t t t t 6 d Note that

Recall from (3.29) that the pairs (a, nˆ) and (b, mˆ ) are (γ, η)-shift-reduced, from 0 0 which we now deduce that x(qtb+ct)−y(qta+ct) is a non-zero integer. Indeed, 0 0 suppose for a contradiction that x(qtb + ct) = y(qta + ct). As n 6= m, we have max{x, y} > 2. Let us assume for simplicity that x > 2; a similar argument handles the case y > 2. Let p be a prime divisor of x. Then p divides y or 0 qta + ct. The former is excluded by the coprimality of x and y, and the latter by the shift-reduction. This contradiction means that q0∆ 1 1 |x(q0b + c ) − y(q0a + c )| t + , 6 t t t t 6 d 5 0 and consequently d 6 2qt∆. The upshot is that we are in the rather special 0 scenario that ∆ 6 d 6 2qt∆.

0 0 0 Setting h = x(qtb+ct)−y(qta+ct), we see that there are O(qt) many realisable values for the integer h. Now we are in the position to derive an acceptable bound on the contribution from this case to Dgcd>. There are O(δM) many o(1) options for m and at most N many ways to choose d > N1 dividingm ˆ . It follows from Lemmata 2.7 and 2.15 that there are 1 1 ρN 1+o(1) O ρCNd + 6 d N1 N1 0 many choices for n ∈ Bloc(N; ρ) divisible by d. For each choice of h qt, the 0 parameters a, b are determined up to O(µ(I)d) = O(µ(I)qt∆) many possibilities. By summing over all choices of h, we obtain 1+o(1) ρN 0 2 Dgcd> = O δM (qt) µ(I)∆ . N1

0 2 o(1) 2η+o(1) 20(k−1)ε0 Since (qt) N 6 N < N 6 N1, this bound is acceptable.

0 Case 2: qt+1 < 10CN/dd

0 η By deﬁnition, we have qt+1 > N , and therefore d < 10CNNd −η. (3.47) Now, akin to the proof from Subsubsection 3.4.1, we work with an enlarged Bohr set, namely B := {u 6 10CN/dd : kuγk 6 L}, where L = L(d) = max(∆/d, N −η).

0 Note that qt+1 ∈ [1/(2L), 10CN/dd ], where the lower bound comes from max- imality in the deﬁnition of shift-reduction. The existence of a continued fraction denominator in this range is a key technical ingredient. As M is large and 0 0 d > 3∆/kq2γk, we have L < kq2γk/2. Therefore Lemma 2.8 is applicable, and hence N 1+o(1) #B L(d). d 52 SAM CHOW AND NICLAS TECHNAU

0 ˆ First choose d > max(3∆/kq2γk,N1). There are O(M/d) many conceivable options for m divisible by d. Then u = |x − y| lies in B, and therefore admits at most #B possibilities, and then x is determined up at most two choices. The number of viable choices of a, b is then O(dµ(I)). So, for a ﬁxed choice of d, the number of valid possibilities for (m, n, a, b) is at most M N L(d) N o(1) L(d)dµ(I) = N o(1)µ(I)MN . d d d Upon summing over d, we infer that Dgcd> is at most X L(d) N o(1)µ(I)MN . d max(∆,N1)

To conclude, it suﬃces to show that for any d in the range of interest we have o(1) N L(d) 6 ∆δρ. (3.48) If L(d) = N −η then −4.2(k−1)ε −4.2(k−1)ε −8.4(k−1)ε o(1)−η o(1) ρδ > N 0 N 0 = N 0 > N = N L(d). o(1) Since ∆ > 1, from (3.23), we conclude that N L(d) 6 δρ∆. If on the other hand L(d) = ∆/d, then 20kε0 d > N1 > N and therefore o(1) −1 o(1) o(1)−20kε L(d)N /∆ = d N 6 N 0 6 δρ. We have (3.48) in both cases, completing the proofs of Theorems 1.7 and 1.22.

3.5. A convergence statement. In this subsection, we prove the convergence side of Corollary 1.10. Assume that ∞ X ψ(n)(log n)k−1 < ∞. (3.49) n=1 Replacing ψ(n) by max(ψ(n), n−2), we may suppose that −2 ψ(n) > n (n ∈ N). × We wish to show that µk(W ) = 0. By 1-periodicity, we may assume that

−1 < γ1, . . . , γk 6 0.

We abbreviate α˜ = (α1, . . . , αk) and α = (α1, . . . , αk−1) for the remainder of × this section. Observe that W = lim sup An, where n→∞ k An = {α˜ ∈ [0, 1] : knα1 − γ1k · · · knαk − γkk < ψ(n)} (n ∈ N). k For a = (a1, . . . , ak) ∈ Z , let k An,a = α˜ ∈ [0, 1] : |nαk − γk − ak| < min(1,Q(α)) , LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 53

where ψ(n) Q(α) = Q(α; a1, . . . , ak−1) = Y . |nαi − γi − ai| i6k−1

Lemma 3.12. For n ∈ N, we have [ An = An,a. 06a1,...,ak6n+1

k Proof. First suppose α˜ ∈ An,a, for some a ∈ {0, 1, . . . , n + 1} . Then Y Y knαi − γik 6 |nαi − γi − ai| < ψ(n), i6k i6k [ and so α˜ ∈ An. Therefore An,a ⊆ An. 06a1,...,ak6n+1

−1/2 6 −γi − 1/2 6 ai 6 n − γi + 1/2 6 n + 3/2 (1 6 i 6 k), [ and therefore An ⊆ An,a. 06a1,...,ak6n+1

By the union bound, if n ∈ N then n+1 X µk(An) 6 µk(An,a). a1,...,ak=0 Further, if n ∈ N and 0 6 a1, . . . , ak 6 n + 1 then Z −1 −1 µk(An,a) n min(1,Q(α)) dα 6 n (I1 + I2), [0,1]k−1 where k−1 −k I1 = µk−1(R), R = {α ∈ [0, 1] : min |nαi − γi − ai| 6 n }, 16i6k−1 and Z I2 = Q(α) dα. [0,1]k−1\R

Here µk−1 denotes (k − 1)-dimensional Lebesgue measure.

−1−k k−1 We have I1 n by inspection. To estimate I2, we cover [0, 1] \R by O((log n)k−1) dyadically-restricted regions k−1 {α ∈ [0, 1] : δi < |nαi − γi − ai| 6 2δi (1 6 i 6 k − 1)}. 54 SAM CHOW AND NICLAS TECHNAU

The integral of Q over such a region is O(ψ(n)/nk−1), and so ψ(n)(log n)k−1 I . 2 nk−1

−2 Recalling that ψ(n) > n , we therefore have I + I ψ(n)(log n)k−1 µ (A ) 1 2 (0 a , . . . , a n + 1), k n,a n nk 6 1 k 6 and ﬁnally k−1 µk(An) ψ(n)(log n) . In view of (3.49), we now have ∞ X µk(An) < ∞, n=1 × and the ﬁrst Borel–Cantelli lemma completes the proof that µk(W ) = 0.

4. Liouville fibres In this section, we establish Theorem 1.12.

4.1. A special case. For expository purposes, we begin with the ‘partially homogeneous’ case γ2 = 0. Here is it simplest to invoke the resolution of the Duffin–Schaeffer conjecture by Koukoulopoulos and Maynard, though this is by no means an essential ingredient. By Theorem 1.17, it suffices to show that ∞ X ϕ(n) 1 · = ∞, n n(log n)2knα − γk n=1

where α = α1 and γ = γ1.

9 By Lemma 2.1, there are inﬁnitely many positive integers such that qt > qt−1, where the q` are continued fraction denominators of α. Let T be a sparse, 9 inﬁnite set of integers t > 9 for which qt > qt−1.

Given t ∈ T , we begin by using Theorem 2.2, with

m = qt = atqt−1 + qt−2 + 0,

to ﬁnd a small positive integer bt such that kbtα − γk is small. Recall that this concerns the gaps di+1 − di, where

{d1, . . . , dm} = {iα − biαc : 0 6 i 6 m}, 0 = d0 < ··· < dm+1 = 1. By (2.3), Theorem 2.2 tells us that

max{di+1 − di : 0 6 i 6 m} 6 |Dt| + |Dt−1| 6 2/qt, where we have employed the standard notation (2.2). For some i ∈ {0, 1, . . . , m}, the fractional part of γ lies in [di, di+1], and for some bt ∈ {1, 2, . . . , m} the fractional part of btα is either di or di+1. The upshot is that we have a positive integer bt 6 qt such that kbtα − γk 6 2/qt. LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 55

For t ∈ T , denote by Dt the set of integers of the form

n = bt + qt−1x + qty, where x, y ∈ Z satisfy 1/4 1/3 2/3 3/4 qt 6 y 6 qt , qt 6 x 6 qt . If n is as above then x n qty, log n log qt, knα − γk . qt Indeed, for the final estimate, observe using (2.3) that 2 x y kbtα − γk 6 , kxqt−1αk , kyqtαk , qt qt qt and apply the triangle inequality. The sets Dt (t ∈ T ) are disjoint, since T is sparse. Moreover, for n ∈ T the representation above is unique, since 3/4 qt−1qt < qt. Fix t ∈ T , and let X ϕ(n) S = v , t n n n∈Dt −1 2 where vn = n(log n) knα − γk. As T is infinite, it suffices to prove that St 1. Note that in this section our implicit constants do not depend on t.

As a ﬁrst step, we compute that X −2 X −1 vn (log qt) (xy) 1.

n∈Dt 1/4 1/3 qt 6y6qt 2/3 3/4 qt 6x6qt Let C = P v −1, and for n ∈ D let w = C v v , so that we have t n∈Dt n t n t n n P w = 1. The weighted AM–GM inequality gives n∈Dt n

X Y τp St wnϕ(n)/n > (1 − 1/p) , n∈Dt p where X τp = wn.

n∈Dt n≡0 mod p Now X X − ln St 6 O(1) − τp ln(1 − 1/p) 1 + τp/p, p p so it remains to show that X τp/p < ∞. (4.1) p

Let p be prime, and let X,Y be parameters in the ranges 1/4 1/3 2/3 3/4 qt 6 Y 6 qt /2, qt 6 X 6 qt /2. 56 SAM CHOW AND NICLAS TECHNAU

Denote by Np(X,Y ) the number of integer solutions (x, y) ∈ [X, 2X] × [Y, 2Y ] to b + qt−1x + qty ≡ 0 mod p. Let us assume that this congruence has a solution in [X, 2X] × [Y, 2Y ], forcing 2 p 6 qt . Then, by Lemma 2.15, we have −1/8 Np(X,Y ) XY/p + X + Y XY p . (4.2) If X 6 x 6 2X,Y 6 y 6 2Y, n = b + qt−1x + qty −1 2 then wn XY (log qt) . Using (4.2), and summing over X,Y that are powers of 2 or the endpoints of their allowed ranges, gives −1/8 τp p , and in particular (4.1).

4.2. Diophantine second shift. In this subsection, we prove Theorem 1.12 in the case that γ2 is diophantine. Let λ > 2 satisfy −λ/2 knγ2k n (n ∈ N). (4.3) Then λ > 2ω(γ2), where ∞ −w ω(γ2) = sup{w > 0 : ∃ q ∈ N kqγ2k < q }. Fix a constant c1 > 0, and for n ∈ N let c1 Ψ(n) = 2 ∈ (0, +∞], n(log n) knα1 − γ1k recalling our notational conventions from Section 1.4. By 1-periodicity of k · k, our task is to show that for almost all α2 ∈ [0, 1] the inequality

knα2 − γ2k < Ψ(n) has inﬁnitely many solutions n ∈ N. Let I be a non-empty subinterval of [0, 1]. Let θ3, θ4 be constants satisfying −3 θ3 = 1 − λ < θ4 < 1.

Let T0 be large in terms of I, and let T be a sparse, inﬁnite set of integers 1−θ4 t > T0 for which qt > qt−1.

For t ∈ T , let Dt be the set of integers of the form

n = bt + qt−1x1 + qty1 (4.4) with 1/4 1/3 θ3 θ4 qt 6 y1 6 qt , qt 6 x1 6 qt , where bt, qt−1, and qt are as in the previous subsection. Let t ∈ T , and let θ4 n ∈ Dt. The representation above is unique, since qt > qt qt−1. Moreover, we have 2 x1 2 y1 x1 knα1 − γ1k > kx1qt−1α1k − − y1kqtα1k > − − > , qt 2qt qt qt+1 3qt LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 57

so n qty1, log n log qt, knα1 − γ1k x1/qt, and in particular

qt −1 −2 1 Ψ(n) (qty1) (log qt) 2 . x1 x1y1(log qt)

For n ∈ N and a ∈ Z, deﬁne [ An,a = {β ∈ I : |nβ − γ2 − a| < Ψ(n)}, An = An,a, a∈Z and observe that if nµ(I) > 1 then

µ(An) µ(I)Ψ(n). Let F (1),F (2),... be a sequence of powers of 4, non-increasing, satisfying F (t) 6 log log t for all t, and such that X F (t)−1 = ∞. t∈T

For t ∈ T , let Gt = {n ∈ Dt : n ≡ 0 mod F (t)}. We will show that if t, s ∈ T then X µ(I) µ(A ) (4.5) n F (t) n∈Gt and X X µ(I) µ(A ∩ A ) . (4.6) n m F (t)F (s) n∈Gt m∈Gs m

n,m∈XN n∈XN n∈XN !2 X µ(An) .

n∈XN At that stage Lemmata 2.9 and 2.10 would give

µ(lim sup{β ∈ [0, 1] : knβ − γ2k < Ψ(n)}) = 1, n∈X 58 SAM CHOW AND NICLAS TECHNAU

which would ﬁnish the proof. The upshot is that it remains to establish (4.5) and (4.6).

Let t ∈ T , and let X and Y be parameters in the ranges

θ3 θ4 1/4 1/3 qt 6 X 6 qt /2, qt 6 Y 6 qt /2. (4.7) Then −1 X −2 −1 X µ(I) µ(An) (log qt) (XY ) 1 n=bt+qt−1x1+qty1 X

It remains to prove (4.6). Writing

n = bt + qt−1x1 + qty1, m = bs + qs−1x2 + qsy2, (4.8) where

1/4 1/3 θ3 θ4 1/4 1/3 θ3 θ4 qt 6 y1 6 qt , qt 6 x1 6 qt , qs 6 y2 6 qs , qs 6 x2 6 qs , (4.9) we have Ψ(n) Ψ(m) µ(A ∩ A ) min , n,a m,b n m qt qs min 2 , 2 x1(qty1 log qt) x2(qsy2 log qs) for a, b ∈ Z.

Let X1,Y1,X2,Y2 be parameters in the ranges

1/4 1/3 θ3 θ4 qt 6 Y1 6 qt /2, qt 6 X1 6 qt /2, 1/4 1/3 θ3 θ4 qs 6 Y2 6 qs /2, qs 6 X2 6 qs /2.

Suppose An,a ∩ Am,b is non-empty, and that we have (4.8) and

Xi 6 xi 6 2Xi,Yi 6 yi 6 2Yi (i = 1, 2). (4.10) Then dist(a + γ2, nI) < 1/2, dist(b + γ2, mI) < 1/2, (4.11)

and there exists β ∈ An,a ∩ Am,b, so that

|nβ − γ2 − a| < Ψ(n), |mβ − γ2 − b| < Ψ(m). The triangle inequality gives

|(n − m)γ2 − (ma − nb)| < mΨ(n) + nΨ(m) < ∆, (4.12) LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 59

where qsY2 qtY1 ∆ 2 + 2 . X1Y1(log qt) X2Y2(log qs) Note that q q 2 t s 1−θ4−o(1) 1−θ4−o(1) ∆ 2 2 > qt qs . (4.13) X1X2(log qt) (log qs)

Given t, s ∈ T , denote by N(t, s) the number of solutions 2 (n, m, a, b) ∈ Gt × Gs × Z to the diophantine system given by (4.11), (4.12) and n > m for which we have (4.8) for some x1, x2, y1, y2 in the ranges (4.10). Our goal is to show that −1 −1 N(t, s) X1X2Y1Y2∆µ(I)F (t) F (s) . (4.14)

Note that the number of solutions with n = m is at most X1Y1, which is negli- gible. Assuming for the time being that we can achieve (4.14), the contribution to P P µ(A ∩ A ) from x , x , y , y in these dyadic ranges is n∈Dt m∈Ds n m 1 2 1 2 X1X2Y1Y2∆µ(I) qt qs O min 2 , 2 , F (t)F (s) X1(qtY1 log qt) X2(qsY2 log qs) which is µ(I) O 2 2 . F (t)F (s)(log qt) (log qs) Summing over X1,X2,Y1,Y2 that are powers of 2 or endpoints of the prescribed ranges, we would thereby deduce (4.6). The upshot is that it remains to prove (4.14).

We partition our solutions according to the value of d = gcd(m, n), and write n = dx, m = dy, so that gcd(x, y) = 1 and x > y. Then (4.12) becomes

|(x − y)γ2 − (ya − xb)| < ∆/d. (4.15) −1 0 Note that d > min{F (t),F (s)} > µ(I) . Let (q`)` be the sequence of continued fraction denominators of γ2.

−1 0 Case 1: µ(I) < d < 3∆/kq2γ2k

0 −1 −1 Choose m, n with gcd(m, n) < 3∆/kq2γ2k in O(F (t) F (s) X1X2Y1Y2) ways, via Lemma 2.15. Then there are O(∆/d) possibilities for h = ya − xb, by (4.15). Given h, the value of a is determined modulo x, and lies in an interval of length O(dxµ(I)), so there are O(dµ(I) + 1) possibilities for a, whereupon b is determined. As d > µ(I)−1, the contribution to N(t, s) from this case is −1 −1 O(X1X2Y1Y2∆µ(I)F (t) F (s) ).

0 Case 2: 3∆/kq2γ2k 6 d < Y1 60 SAM CHOW AND NICLAS TECHNAU

0 Choose m in O(X2Y2) ways, and then choose d | m with d ∈ [3∆/kq2γ2k,Y1) o(1) in Y2 ways. Then, by Lemma 2.15, choose n ∈ Dt such that n ≡ 0 mod d in O(X1Y1/d) ways. Finally there are dµ(I) possibilities for a and b so, using (4.13), the contribution to N(t, s) from this case is bounded above by 1+o(1) −1 −1 X2Y2 X1Y1µ(I) X1Y1X2Y2∆µ(I)F (t) F (s) .

0 Case 3: d > max{Y1, 3∆/kq2γ2k}

0 o(1) Choose m in O(X2Y2) ways and d > max{Y1, 3∆/kq2γ2k} dividing m in qt ways. Put N = qtY1. Set x − y = u ∈ N and ya − xb = v ∈ Z, so that u 6 3N/d, |uγ2 − v| < ∆/d. (4.16) From our choice of λ, we have ∆/d (N/d)−λ, so d (∆N λ)1/(1+λ). We relax the inequalities above, giving −1/λ u 6 CN/d, kuγ2k 6 (N/d) + ∆/d, (4.17) where C is a large, positive constant.

We claim that (4.17) has O((N/d)((N/d)−1/λ + ∆/d)) solutions u ∈ N. This is clearly true if d N, so let us now assume that d 6 cN for some small constant c > 0. By (4.3), we have ω(γ2) 6 λ/2, so by Lemma 2.1 there exists ` ∈ N such that 1/λ 0 (N/d) 6 q` 6 CN/d. Thus, we may apply Lemma 2.8 to the Bohr set deﬁned by (4.17), establishing the claim.

Consequently, there are O((N/d)((N/d)−1/λ + ∆/d)) pairs (u, v) ∈ N × Z satisfying (4.16). Hence, given m and d as above, the number of possibilities for u, v, a, b is at most a constant times (N/d)((N/d)−1/λ + ∆/d)d = N 1−(1/λ)d1/λ + N∆/d 1/(λ+λ2) 1−(1/λ)+1/(1+λ) ∆ N + N∆/Y1. Recalling that −3 θ3 1−λ 1/4 N = qtY1,X1 > qt = qt ,Y1 > qt , LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 61

we ﬁnd that the contribution to N(t, s) from this case is O(N1(t, s)+N2(t, s)), where o(1) 1/(λ+λ2) 1−1/(λ+λ2)+o(1) N1(t, s) X2Y2qt (∆/N) N X2Y2∆qt Y1 −1 −1 X1Y1X2Y2∆µ(I)F (t) F (s) and o(1) 1+o(1) N2(t, s) X2Y2qt N∆/Y1 = X2Y2qt ∆ −1 −1 X1Y1X2Y2∆µ(I)F (t) F (s) .

We have considered all possible size ranges for d, conﬁrming (4.14). We conclude that Theorem 1.12 holds in the case that γ2 is diophantine.

4.3. Liouville second shift. Next, we prove Theorem 1.12 in the case that γ2 ∈ L. This is a more sophisticated variant of the proof given in the previous subsection that works whenever γ2 ∈ R \ Q. As we discuss in the next subsection, a simpler version of it works when γ2 ∈ Q.

Fix a constant c1 > 0, and for n ∈ N let c1 Ψ(n) = 2 ∈ (0, +∞]. n(log n) knα1 − γ1k

By 1-periodicity of k · k, it suﬃces to show that for almost all α2 ∈ [0, 1] the inequality knα2 − γ2k < Ψ(n) has inﬁnitely many solutions n ∈ N. Indeed, the latter would imply that 2 lim inf n(log n) knα1 − γ1k · knα2 − γ2k c1, n→∞ 6 and c1 > 0 is arbitrary.

Let I be a non-empty subinterval of [0, 1], let T0 be large in terms of I, and 9 let T be a sparse, inﬁnite set of integers t > T0 for which qt > qt−1, where again q1, q2,... are the denominators of the continued fraction convergents to the Liouville number α1.

Fix λ > 2. For t ∈ T , we deﬁne Dt as in the previous subsection. The sets Dt (t ∈ T ) are disjoint, because T is sparse. For t ∈ T and n ∈ Dt, the representation (4.4) is unique, and moreover

x1 1 n qty1, log n log qt, knα1 − γ1k , Ψ(n) 2 . qt x1y1(log qt)

We also deﬁne Gt via Dt as in the previous subsection, for t ∈ T , using the arithmetic function F . 62 SAM CHOW AND NICLAS TECHNAU

0 For t ∈ T , let ct0 /qt0 be the continued fraction convergent to γ2 for which 0 qt0 < qt is maximal. For n ∈ Dt and a ∈ Z, we again deﬁne

An,a = {β ∈ I : |nβ − γ2 − a| < Ψ(n)}, but now we let An be the union of An,a over integers a for which 0 (qt0 a + ct0 , n) = 1. As in the previous subsection, it remains to establish (4.5) and (4.6).

Lemma 4.1. Let n ∈ Dt. Then ϕ(n) µ(I)Ψ(n) µ(A ) µ(I)Ψ(n). n n

Proof. For the upper bound, observe that there are O(nµ(I)) integers a for which An,a is non-empty, and for each of these µ(An,a) Ψ(n)/n. For the lower bound, it suﬃces to show that

0 ϕ(n) #{a ∈ nI :(q 0 a + c 0 , n) = 1} µ(I). t t n This follows routinely from the fundamental lemma of sieve theory, in the same way as (3.8).

Let t ∈ T , and let X and Y be parameters in the ranges (4.7). Then −1 X µ(I) µ(An)

n=bt+qt−1x1+qty1 n≡0 mod F (t) X

−2 −1 X ϕ(bt + qt−1x1 + qty1) (log qt) (XY ) . bt + qt−1x1 + qty1 X

We claim that X ϕ(bt + qt−1x1 + qty1) S := XY/F (t). bt + qt−1x1 + qty1 X

U = {n = bt + qt−1x1 + qty : X < x1 6 2X, Y < y1 6 2Y, n ≡ 0 mod F (t)}, and apply the AM–GM inequality to give !1/|U| Y ϕ(n) Y Y S |U| = |U| (1 − 1/p)1/|U|. > n n∈U p n∈U n≡0 mod p LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 63

By Lemma 2.15, we have |U| XY/F (t), so for the claim it suﬃces to show that Y Y S0 := (1 − 1/p)1/|U| 1. p n∈U n≡0 mod p Next, observe that 0 X X − ln(S ) = − τp ln(1 − 1/p) τp/p, p p where −1 τp = |U| #{n ∈ U : n ≡ 0 mod p}, so for the claim it remains to prove that X τp/p 1. (4.18) p>3 Let p > 3, and note that p - F (t) because F (t) is a power of 4. As τp = 0 for 2 2 p > qt , let us also assume that p 6 qt . By Lemma 2.15, we have 1 1 1 1 τ F (t) + + + Y −1/2 p−1/16, p pF (t) X Y p giving (4.18).

Thus, we have the claim, and so −1 X −2 −1 µ(I) µ(An) (log qt) F (t) .

n=bt+qt−1x1+qty1 n≡0 mod F (t) X

It remains to prove (4.6). To this end, it again suﬃces to prove (4.14), where X1,Y1,X2,Y2 are parameters in the ranges

1/4 1/3 θ3 θ4 qt 6 Y1 6 qt /2, qt 6 X1 6 qt /2, 1/4 1/3 θ3 θ4 qs 6 Y2 6 qs /2, qs 6 X2 6 qs /2, but now in the count N(t, s) we impose the additional restrictions 0 0 (qt0 a + ct0 , n) = 1, (qs0 b + cs0 , m) = 1. (4.19) Cases 1 and 2 from the previous subsection are unaﬀected, so our task is to count solutions for which 0 d = (m, n) > max{Y1, 3∆/kq2γ2k} (Case 3). As T is sparse, this is only possible if s = t. 64 SAM CHOW AND NICLAS TECHNAU

Put N = qtY1. Let us again write n = dx and m = dy, so that x > y and (x, y) = 1. Let C be a large, positive constant.

1/λ Case 3a: γ2 has a continued fraction denominator in [(N/d) , CN/d], or d > N

In the case the proof from the previous subsection carries through, for in this case we may apply Lemma 2.8 therein.

1/λ Case 3b: γ2 has no continued fraction denominator in [(N/d) , CN/d], and d < N

In this case, as 0 1/λ qt0+1 > qt > qtY1/d = N/d > (N/d) , we must have 0 qt0+1 > CN/d, 0 0 where qt0+1 is the continued fraction denominator of γ2 subsequent to qt0 . Therefore 0 0 −1 |qt0 γ2 − ct0 | < (qt0+1) < d/(CN). As 0 0 0 |(n − m)qt0 γ2 − qt0 (ma − nb)| < qt0 ∆, the triangle inequality now confers 0 0 |(n − m)ct0 − qt0 (ma − nb)| < qt0 ∆ + d/2. We thus have 0 0 0 1 6 |x(qt0 b + ct0 ) − y(qt0 a + ct0 )| < qt0 ∆/d + 1/2, owing to the coprimality restrictions (4.19) that we have thrust upon the problem. Whence 0 0 0 d < 2qt0 ∆, ct0 (x − y) ≡ O(qt0 ∆/d) mod qt0 . (4.20)

Recall that in this case we have s = t. We begin our count by choosing d so that 0 0 max{Y1, 3∆/kq2γ2k} 6 d < 2qt0 ∆. Next, we choose x in O(qtY1/d) ways. Then

y qtY2/d 0 0 lies in one of O(qt0 ∆/d) residue classes modulo qt0 , according to (4.20), so there are 0 qt0 ∆ qtY2 O 0 + 1 d dqt0 possibilities for y. After that, the variable a is then determined modulo x from (4.15), and so there are at most dµ(I) 6 d possibilities for a and then −3 θ3 1−λ ﬁnally b is uniquely determined. Recalling that X1,X2 > qt = qt and LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 65

1/4 Y1,Y2 > qt , our total count from this case is at most a constant times 0 X qtY1qt0 ∆ qtY2 0 + 1 6 N1(t, s) + N2(t, s), d dq 0 max{Y ,3∆/kq0 γ k} d<2q0 ∆ t 1 2 2 6 t0 where 2 X q Y1Y2∆ X1Y1X2Y2∆µ(I) N (t, s) = t q2∆Y 1 d2 t 2 F (t)F (s) d>Y1 and 0 X qtY1qt0 ∆ 0 0 2+o(1) N (t, s) = q Y q 0 (log q 0 )∆ q Y ∆ 2 d t 1 t t 6 t 1 d<2q0 ∆ t0 X Y X Y ∆µ(I) 1 1 2 2 . F (t)F (s)

We have considered all cases, conﬁrming (4.14). We conclude that Theorem 1.12 holds in the case that γ2 ∈ L.

4.4. Rational second shift. Finally, we prove Theorem 1.12 in the case that γ2 ∈ Q. Let γ2 = c0/d0, where c0 ∈ Z and d0 ∈ N are ﬁxed and coprime. We 0 follow the previous subsection, but this time we replace qt0 and ct0 by d0 and c0, respectively, for all t ∈ T , and the proof carries through.

We have covered all possibilities for γ2, completing the proof of Theorem 1.12.

5. Obstructions on Liouville fibres Our proof of Theorem 1.14, via an Ostrowski expansion construction, rests upon the following technical lemma. In the following lemma and its proof, the implied constants are allowed to depend on α.

Lemma 5.1. Suppose α, γ satisfy (2.5). Let m(n) denote the least i > 0 such that δi+1(n) 6= 0. Deﬁne

Wu,d = {n ∈ N : m(n) = u, |δu+1(n)| = d} , whenever 1 6 d 6 au+1 − bu+1, as well as X 1 S = . u,d n(log n)2 knα − γk n∈Wu,d

Then min Wu,d qu, uniformly in d. Moreover, we have  1 , if d > bu+1  d log qu+1 S 1 , if d = b u,d log qu u+1  qu+1 1  2 + , if d < bu+1. (bu+1−d)qu(log((bu+1−d)qu)) d d log qu+1 66 SAM CHOW AND NICLAS TECHNAU

Proof. For n ∈ Wu,d, Lemma 2.5 implies that

knα − γk min ((d − 1) |Du| + au+2 |Du+1| , a1 |D0| + a2|D1|)

= (d − 1) |Du| + au+2 |Du+1| . By (2.3), we have d − 1 (d − 1) |Du| qu+1 and au+2 au+2 1 au+2 |Du+1| = . qu+2 au+2qu+1 + qu qu+1 Therefore d knα − γk (n ∈ Wu,d). (5.1) qu+1 Using the notation of Lemma 2.3, observe that

Wu,d = A (b1, . . . , bu, bu+1 + d) ∪ A (b1, . . . , bu, bu+1 − d) , where A (b1, . . . , bu, bu+1 + d) is understood to be empty if bu+1 + d > au+1, and A (b1, . . . , bu, bu+1 − d) is empty if bu+1 − d < 0.

Observe that + − Su,d = Su,d + Su,d, where X 1 S± = . u,d n(log n)2knα − γk n∈A(b1,...,bu,bu+1±d)

Case 1: n ∈ A(b1, . . . , bu, bu+1 + d)

Then X X n > bk+1qk + (bu+1 + d) qu + ck+1qk > bu+1qu qu+1. 06ku

From Lemma 2.3, applied with m = u and du+1 = bu+1 + d > 0, any two th distinct elements of A (b1, . . . , bu, bu+1 + d) diﬀer by at least qu+1. So the r smallest element nr of A (b1, . . . , bu, bu+1 + d) satisﬁes

nr > min A (b1, . . . , bu, bu+1 + d) + (r − 1)qu+1 rqu+1, uniformly in d. Combining this with (5.1), we deduce that X 1 S+ u,d 2 d rqu+1(log(rqu+1)) r>1 qu+1 ! 1 X 1 X 1 = 2 + 2 d r(log(rqu+1)) r(log(rqu+1)) rqu+1 ! 1 X 1 X 1 6 2 + 2 . d r(log qu+1) r(log r) rqu+1 LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 67

The ﬁrst sum is O(1/ log qu+1), and the second sum is bounded by a constant times Z ∞ X j 1 dx 1 e j 2 2 . e j log qu+1 x log qu+1 j>log qu+1 Therefore + 1 Su,d . d log qu+1

Case 2: n ∈ A(b1, . . . , bu, bu+1 − d)

For the aforementioned set to be non-empty, following our earlier convention, we must have 1 6 d 6 bu+1. We distinguish two sub-cases.

Case 2a: 1 6 d < bu+1

Note that X X n > bk+1qk + (bu+1 − d) qu + ck+1qk > (bu+1 − d) qu. 06ku th Let n0 = min Wu,d, and for r > 1 denote by nr the r smallest element of Wu,d \{n0}. It follows from Lemma 2.3 that

nr > rqu+1 + (bu+1 − d) qu (r > 0). Together with (5.1), this gives X 1 S− u,d 2 d (rqu+1 + (bu+1 − d) qu)(log(rqu+1 + (bu+1 − d) qu)) r>0 qu+1 qu+1 X 1 + . 6 2 2 d (bu+1 − d) qu(log((bu+1 − d) qu)) d rqu+1(log(rqu+1)) r>1 qu+1 As in Case 1, we have X 1 1 , 2 d rqu+1(log(rqu+1)) d log qu+1 r>1 qu+1 and so

− qu+1 1 Su,d 2 + . (bu+1 − d) qu(log((bu+1 − d) qu)) d d log qu+1

Case 2b: d = bu+1

Note that X n > bk+1qk auqu−1 qu. 06k

th Let n0 = min Wu,d, and for r > 1 denote by nr the r smallest element of Wu,d \{n0}. By (5.1), we have − Su,d T1 + T2, where X 1 Tj = (j = 1, 2). 2 bu+1 nj+2r(log nj+2r) r>0 qu+1 Let j ∈ {1, 2}. We infer from Lemma 2.3 that if r > 0 then nj+2r rqu+1 +qu. Whence X 1 Tj 2 bu+1 (rqu+1 + qu)(log(rqu+1 + qu)) r>0 qu+1 1 X 1 + 2 bu+1 2 qu(log qu) r(log(rqu+1)) bu+1 qu+1 r>1 1 1 1 2 + . (log qu) log qu+1 log qu

We are now in the position to prove the main result of this section.

∞ Proof of Theorem 1.14. Let A = (an)n=1 be a sequence in 64N, suﬃciently rapidly-increasing that

• The sequence deﬁned by

q0 = 1, q1 = a1, qu+1 = au+1qu + qu−1 (u > 1) satisﬁes X 1 1 + < ∞ (5.2) log qu ξ(qu) u>0

• au+1 > qu!(u > 0).

Let V be the collection of all such sequences A = (ai)i. For A ∈ V, deﬁne N α(A) := [0; a1, a2,...]. For σ = (σi)i ∈ {0, 1} and A = (ai)i ∈ V, we deﬁne a sequence (bi(A, σ))i by ai bi(A, σ) = (i ∈ N), 21+σi as well as a real number X γ(A, σ) := bk+1(A, σ)Dk(A), k>0 th where Dk(A) = qkα(A) − pk and pk/qk is the k convergent to α(A). Note that we have (2.5), so by Lemma 2.4 we have γ(A, σ) ∈ [0, 1 − α(A)) and knα(A) − γ(A, σ)k= 6 0 (n ∈ N). LITTLEWOOD AND DUFFIN–SCHAEFFER-TYPE PROBLEMS 69

There are continuum many α(A), and they are Liouville by Lemma 2.1. For the rest of the proof, we ﬁx a pair (α(A), γ(A, σ)), where A ∈ V and σ ∈ {0, 1}N, and abbreviate α = α(A), γ = γ(A, σ).

Fix γ2 ∈ R, and consider α1 = α and γ1 = γ. By the Borel–Cantelli lemma, it remains to prove that X ψξ(n) < ∞, (5.3) knα − γk n>1 where ψξ is as in (1.7). To this end, observe that

X ψξ(n) X X 1 X 1 6 max Su,d + max Su,bu+1 , knα − γk n∈Wu,d ξ(n) n∈Wu,bu+1 ξ(n) n>1 u>0 d6=bu+1 u>0

where here and henceforth d 6= bu+1 means that

d ∈ {1, . . . , au+1 − bu+1}\{bu+1}. Since ξ non-decreasing and unbounded, we infer from Lemma 5.1 that 1 1 1 max = , n∈Wu,d ξ(n) ξ(min Wu,d) ξ(qu−1)

where here q−1 = 1. By Lemma 5.1, we now have

X ψξ(n) T + T + T , knα − γk 1 2 3 n>1 where X 1 1 T1 = , ξ(qu−1) log qu u>0 X 1 X 1 T2 = , ξ(qu−1) d log qu+1 u>0 d6=bu+1 X 1 X qu+1 T3 = 2 . ξ(qu−1) (bu+1 − d) qu(log((bu+1 − d) qu)) d u>0 d

We see from (5.2) that T1 < ∞. The convergence of T2 also follows straight- forwardly from (5.2), since

X 1 log au+1 X 1 T2 6 < ∞. ξ(qu−1) log qu+1 ξ(qu−1) u>0 u>0

Our ﬁnal task is to establish the convergence of the series deﬁning T3. We begin with the observation that X qu+1 X 1 T3 6 2 ξ(qu−1)qu d(bu+1 − d)(log(bu+1 − d)) u>0 d0 d

The inner sum is at most Xu + Yu, where X 1 Xu = 2 d(bu+1 − d)(log(bu+1 − d)) d6bu+1/2 and X 1 Yu = 2 > Xu. t(bu+1 − t)(log t) t6bu+1/2 Finally, we have

X bu+1 X 1 X 1 X 1 T3 Yu 2 , ξ(qu−1) ξ(qu−1) t(log t) ξ(qu−1) u>0 u>0 t>1 u>0 which converges.

Appendix A. Pathology In this appendix, we establish Theorem 1.25. If we have (3.6) then the proof of Theorem 1.22 prevails. We also need to consider the pathological situation in which Ψ(n) > 1/2 for inﬁnitely many n ∈ N. We will slightly alter this dichotomy.

Let c∗ be a small, positive constant. Our implicit constants will not depend on c∗ unless otherwise stated. The ﬁrst idea is to restrict the support of Ψ to ϕ(ˆn) G∗ = n ∈ G : c∗ , nˆ > ∗ where G and η = η(γ) are as in Section 3. That is, we introduce Ψ = Ψ1G∗ and ( a + γ ∈ nˆI, ) ∗ ∗ En = α ∈ [0, 1] : ∃a ∈ Z s.t. |nαˆ − γ − a| < Ψ (n), . (a, nˆ) is (γ, η)-shift-reduced

∗ Case: Ψ (n) 6 1/2 for large n

We commence by discussing (1.17). Our modiﬁcation can only reduce the left hand side of (3.5) so, by the reasoning of Proposition 3.3, it remains to justify ∗ (3.4) with En in place of En. The upper bound is immediate from the inequality ∗ µ(En) 6 µ(En), leaving us to deal with the lower bound. By (3.7), we have

X ∗ X ϕ(ˆn) ψ(ˆn) µ(En) µ(I) , (A.1) ∗ nˆ knαˆ 1 − γ1k · · · knαˆ k−1 − γk−1k n6X n∈G C1

Let N be large, and let us now specialise I = [0, 1]. By (3.1) and the above, we have ∞ X ∗ µ(En) = ∞. n=1 By (3.8), we have ϕ (ˆn) ϕ(ˆn) γ,η c∗ (n ∈ G∗, n > N), nˆ nˆ > and note also that ∗ µ(En) 6 µ(En) Ψ(n) 6 Φ(ˆn)(n > N). Therefore ∞ X ϕγ,η(ˆn) X X Φ(ˆn) c∗ µ(E ∗) = c∗ µ(E ∗) = ∞, nˆ n n n=1 n∈G∗ n>N n>N which implies (1.16). Having established (1.16) and (1.17), we have completed the proof of the theorem in this case.

Case: Ψ∗(n) > 1/2 inﬁnitely often

Let N be large, and put ∗ S = {n ∈ N :Ψ (n) > 1/2, n > N}. 72 SAM CHOW AND NICLAS TECHNAU

The inequalities ϕ (ˆn) γ,η c∗, Φ(ˆn) Ψ∗(n) > 1/2 (n ∈ S), nˆ > together with the inﬁnitude of S, yield

X ϕγ,η(ˆn) Φ(ˆn) = ∞, nˆ n∈S which implies (1.16).

For n ∈ , deﬁne N ( 1/2, if n ∈ S Ψ†(n) = 0, if n∈ / S. and ( a + γ ∈ nˆI, ) † † En = α ∈ [0, 1] : ∃a ∈ Z s.t. |nαˆ − γ − a| < Ψ (n), . (a, nˆ) is (γ, η)-shift-reduced Let I be a non-empty subinterval of [0, 1]. Henceforth, our implied constants will be allowed to depend on c∗ but not I. By (3.8), we have ϕ (ˆn) µ(E †) 0 µ(I)(n ∈ S), n nˆ and clearly † † † µ(En ∩ Em) 6 µ(En) µ(I)(n, m ∈ S). Applying Lemmata 2.9 and 2.10, as in Proposition 3.3, furnishes µ lim sup An = 1, n→∞ where † A = α ∈ [0, 1] : ∃a ∈ s.t. |nαˆ − γ − a| < Ψ (n), . n Z (a, nˆ) is (γ, η)-shift-reduced † This implies (1.17), since Φ(ˆn) > Ψ (n) for all n, and completes the proof of Theorem 1.25.

Remark A.1. In the context of Remark 1.24, our proof works for any suﬃciently small η, by scaling ε0 accordingly.

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Mathematics Institute, Zeeman Building, University of Warwick, Coventry CV4 7AL, United Kingdom

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School of Mathematical Sciences, Tel Aviv University, Tel Aviv 69978, Israel

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Department of Mathematics, University of Wisconsin, 480 Lincoln Drive, Madison, WI, 53706, USA

Email address: [email protected]