ASYMPTOTIC DIOPHANTINE APPROXIMATION: THE MULTIPLICATIVE CASE MARTIN WIDMER ABSTRACT Let α and β be irrational real numbers and 0 < ε < 1/30 We prove a precise estimate for the number of positive integers q Q that satisfy qα qβ < ε If we choose ε as a function of Q we get asymptotics as Q gets large, provided εq grows quickly enough in terms of the (multiplicative) Diophantine type of (α, β), eg, if (α, β) is a counterexample to Littlewood s conjecture then we only need that εq tends to infinity Our result yields a new upper bound on sums of reciprocals of products of fractional parts, and sheds some light on a recent question of Lê and Vaaler 1 INTRODUCTION Let α and β be irrational real numbers, and let be the distance to the nearest integer Littlewood s conjecture asserts that lim inf q qα qβ = 0 q We assume that φ : [1, ) (0, 1/4] is a a non-increasing 1 function (depending on (α, β)) such that (11) q qα qβ φ(q) for all positive integers q Note that φ can be chosen to be constant if and only if the pair (α, β) is a counterexample to Littlewood s conjecture The condition (11) has been considered in various forms, eg, by Badziahin [1] He takes a function f : N (0, ) and considers the set { } (1) Mad( f ) = (α, β) R ; lim inf f (q) q qα qβ > 0 q Special cases of these sets already appeared in [] If we assume that 1/ f is also nonincreasing then (α, β) lies in Mad( f ) if and only if (11) holds true with a φ satisfying 1/ f (q) α,β φ(q) α,β 1/ f (q) Throughout this article, let Q 1, ε > 0, and T > 0 be real numbers, and assume (13) ε/t 1/e, Date: March 6, 018 010 Mathematics Subject Classification Primary 11J13; 11J5; 11J54 Secondary 11H46; 37A17 Key words and phrases Diophantine approximation, reciprocals of fractional parts, flows, counting, Littlewood s conjecture 1 By non-increasing we mean that x, y [1, ) and x y implies that φ(x) φ(y) 1
MARTIN WIDMER where e denotes the base of the (natural) logarithm We consider the finite set { M α,β (ε, T, Q) = (p 1, p, q) Z 3 ; p 1 + qα p + qβ < ε, max{ p 1 + qα, p + qβ } T, } 0 < q Q Theorem 11 Suppose that (11) and (13) hold, and set C 1 = 3 8 Then we have ( ( T ) ( T M α,β(ε, T, Q) 4εQ log + 1) C ε 1 (1 + T) )( ) εq /3 log ε We shall see that the main term is just the volume of the set Z defined in Section The constant C 1 could easily be improved Choosing T = 1/ we have M α,β (ε, 1/, Q) = {q Z; qα qβ < ε, 0 < q Q} which is of particular interest, and hence we state this case of Theorem 11 as a corollary Corollary 11 Suppose that (11) holds, that 0 < ε 1/(e), and set C = 4C 1 = 4 3 8 Then we have M α,β (ε, 1/, Q) 4εQ(1 log(4ε)) ( ) εq /3 C log(ε) If we choose a value ε = ε(q) 1/(e) for each value of Q, and we let Q tend to infinity then we get asymptotics for M α,β (ε, 1/, Q) provided 1/ = o( ε(q)q) Let us write log + Q = max{1, log Q} A result of Gallagher [5] implies that for f (q) = (log + q) λ the set Mad( f ) has full Lebesgue measure if λ > and measure zero when λ Hence, if ε(q) (log + Q) λ /Q with λ > then the asymptotics are given by the main term in Corollary 11 for almost every pair (α, β) R Bugeaud and Moshchevitin [3] showed that when λ = the set Mad( f ) still has full Hausdorff dimension This was substantially improved by Badziahin [1] who showed that even with f (q) = (log + q)(log + (log + q)) the set Mad( f ) has full Hausdorff dimension We now discuss an application of Corollary 11 In [6] Lê and Vaaler showed that Q(log + Q) Q q=1 ( qα qβ ) 1 Motivated by this they raised the question whether there exist real irrational numbers α, β such that (14) Q q=1 ( qα qβ ) 1 α,β Q(log + Q) Lê and Vaaler showed that (14) holds for (α, β) provided the latter is a counterexample to Littlewood s conjecture, ie, provided one can choose φ from (11) to be a constant function We show that α,β 1/(log + Q) suffices Corollary 1 Suppose that (11) holds, and set C 3 = 1 and C 4 = 3 3 Then we have ( ( )) Q ( ) ( qα qβ ) 1 Q Q C 3 Q log + C 4 log Q q=1 With respect to the Lebesgue measure
ASYMPTOTIC DIOPHANTINE APPROXIMATION: THE MULTIPLICATIVE CASE 3 Einsiedler, Katok and Lindenstrauss [4] showed that the set of counterexamples to Littlewood s conjecture has Hausdorff dimension zero, and it is widely believed that no such counterexample exists at all On the other hand there is evidence for the existence of pairs (α, β) with α,β 1/(log + Q) In fact, Badziahin and Velani [, (L)] (see also [1, Conjecture 1]) conjectured that the set of these pairs has full Hausdorff dimension Unfortunately, it is not known whether such a pair (α, β) really exists and so we cannot unconditionally answer Lê and Vaaler s question PREREQUISITES Lemma 1 If εq < then the stated inequality in Theorem 11 holds true Proof Suppose (p 1, p, q) M α,β (ε, T, Q) Hence, 1 q Q and qα qβ p 1 + qα p + qβ < ε On the other hand by (11), and using the monotonicity of φ, we have qα qβ φ(q)/q /Q Thus, if εq < then M α,β (ε, T, Q) = 0 It remains to show that the main term is covered by the error term As T /ε e we have log(t /ε) + 1 < log(t /ε) Using that εq < 1/4 we see that 4εQ < (C 1 /)(εq/) /3 This shows that the main term is bounded by the error term, and this proves the lemma (1) For the proof of Theorem 11 we thus can and will assume that εq For a vector x in R n we write x for the Euclidean length of x Let Λ be a lattice of rank n in R n We define the first successive minimum λ 1 (Λ) of Λ as the shortest Euclidean length of a non-zero lattice vector λ 1 = inf{ x ; x Λ, x = 0} From now on suppose n, M 1 is also an integer, and let L be a non-negative real We say that a set S is in Lip(n, M, L) if S is a subset of R n, and if there are M maps ι 1,, ι M : [0, 1] n 1 R n satisfying a Lipschitz condition ι i (x) ι i (y) L x y for x, y [0, 1] n 1, i = 1,, M such that S is covered by the images of the maps ι i We will apply the following counting result which is an immediate consequence of [7, Theorem 54] Lemma Let Λ be a lattice of rank n in R n with first successive minimum λ 1 Let S be a set in R n such that the boundary S of S is in Lip(n, M, L) Then S is measurable, and moreover, ( ( ) ) VolS Λ S L n 1 det Λ D nm 1 +, where D n = n n Next we introduce the sets and the lattice Clearly, () λ 1 H = {(x, y) R ; xy < ε, x T, y T}, Z = H (0, Q], Λ = (1, 0, 0)Z + (0, 1, 0)Z + (α, β, 1)Z M α,β (ε, T, Q) = Λ Z
4 MARTIN WIDMER Instead of working with Z it is more convenient to decompose Z into four identically shaped parts Z j and two rectangles R j We set H 1 = {(x, y) R ; xy < ε, 0 < x T, 0 < y T}, H = {(x, y) R ; xy < ε, T x < 0, 0 < y T}, H 3 = {(x, y) R ; xy < ε, 0 < x T, T y < 0}, H 4 = {(x, y) R ; xy < ε, T x < 0, T y < 0} Furthermore, we put for 1 j 4 Z j = H j (0, Q], so that we have the following partition Due to the irrationality of α and β we have Hence, R 1 = [ T, T] {0} (0, Q], R = {0} [ T, T] (0, Q], Z = Z 1 Z Z 3 Z 4 R 1 R Λ R 1 = Λ R = T + 1 < (T + 1) 4 Λ Z Λ Z j < 4(T + 1) j=1 Using the automorphisms defined by τ 1 (x, y, z) = (x, y, z), τ (x, y, z) = ( x, y, z), τ 3 (x, y, z) = (x, y, z), and τ 4 (x, y, z) = ( x, y, z) we have τ j Z j = Z 1 Setting for 1 j 4 we find (3) M α,β(ε, T, Q) Λ j = τ j (Λ), 4 j=1 Λ j Z 1 < 4(T + 1) Unfortunately, our set Z 1 is increasingly distorted when approaching the coordinateaxes After the trivial decomposition of Z we shall now consider a less obvious decomposition of our new counting domain Z 1 3 PARTITIONING THE COUNTING DOMAIN First let us decompose H 1 into three disjoint pieces Set Hence, we have (31) x = {(x, y); 0< y < (ε/t )x, 0 < x T}, y = {(x, y); (T /ε)x y T, 0 < x < ε/t}, S = {(x, y); 0< (ε/t )x y < (T /ε)x, xy < ε} Λ j Z 1 = Λ j x (0, Q] + Λ j y (0, Q] + Λ j S (0, Q] The sets x and y are long and thin triangles, distorted only in x-direction or y-direction respectively The set S is more troublesome and requires a further decomposition into about log(ε/t ) pieces Recall that by hypothesis 0 < ε/t 1/e Let ν [1/e, 1/e] be maximal such that N = log(ε/t )/log ν is an integer Hence, (3) 1 N log(ε/t )
ASYMPTOTIC DIOPHANTINE APPROXIMATION: THE MULTIPLICATIVE CASE 5 Decompose S into the N pieces S N+1,, S N, where Then we have the following partition (33) Note that (34) S i = {(x, y); 0 < ν i x y < ν i 1 x, xy < ε} S = A straightforward calculation yields εν ε/ν Vol (S 0 ) = Hence, N+1 i N S i S 0 [0, ε] [0, ε/ν] [0, 3 ε] ε ε + νε x dx ε = ε log ν V = Vol 3 (S 0 (0, Q]) = ε Q log ν Thus (35) εq V εq 4 APPLYING FLOWS In this section we construct certain elements of the diagonal flow on R 3 that transform our distorted sets into sets of small diameter We introduce the following automorphisms of R g i (x, y) = (ν i/ x, ν i/ y) Then we have for N + 1 i N g i S i = S 0 We extend g i to an automorphism of R 3 G i (x, y, z) = (ν i/ x, ν i/ y, z), so that G i (S i (0, Q]) = S 0 (0, Q] Next we introduce a further automorphism of R 3 where Let us write Then we have Combining (34) and (35) we get (41) Similarly, we find (4) (43) G θ (x, y, z) = (θx, θy, θ z), θ = V1/3 ε ϕ i = G θ G i ϕ i (S i (0, Q]) = θs 0 (0, θ Q] ϕ i (S i (0, Q]) = θs 0 (0, θ Q] [0, 3θ ε] (0, θ Q] [0, 3V 1/3 ] 3 ϕ N ( x (0, Q]) [0, 3V 1/3 ] 3, ϕ N+1 ( y (0, Q]) [0, 3V 1/3 ] 3
6 MARTIN WIDMER Lemma 41 For N + 1 i N the boundary of ϕ i (S i (0, Q]), ϕ N ( x (0, Q]) and ϕ N+1 ( y (0, Q]) lies in Lip(3, M, L) where M = 5, L = C L V 1/3 and C L = 1 Proof The boundary of the set ϕ i (S i (0, Q]) = θs 0 (0, θ Q] can be covered by 4 planes and the set {(x, θ ε/x, z); θ νε x θ ε, 0 z Qθ } For the Jacobian J of the parameterising map (t 1, t ) (at 1 + b, θ ε (at 1 + b), ct ) with a = θ ε(1 ν), b = θ νε, c = Q/θ, and domain [0, 1] we get for its l -operator norm J 4V 1/3 which yields the required Lipschitz condition thanks to the Mean- Value Theorem Hence, we are left with the linear pieces of the boundary Clearly, a subset of a plane with diameter no larger than d can be parameterised by a single affine map with domain [0, 1] and Lipschitz constant d Thus, it suffices to show that the diameter of ϕ i (S i (0, Q]) is 6V 1/3 But the latter holds due to (41) Finally, the boundary of the set ϕ N ( x (0, Q]) and of the set ϕ N+1 ( y (0, Q]) can each be covered by 5 planes Moreover, by (4) and (43) their diameter is also 6V 1/3 This proves the lemma 5 CONTROLLING THE ORBITS Our transformations of the previous section have brought our distorted sets into nice shapes Unfortunately, they transform our lattices Λ j in a less favourable manner Indeed, the corresponding orbit of Λ j escapes to infinity, ie, the fist successive minimum gets arbitrarily small However, the rate of escape is controllable and sufficiently slow Lemma 51 For 1 j 4, N + 1 i N, and Q 1 we have λ 1 (ϕ i Λ j ) min{1, 1/(T)} 1/3 Proof Let v Λ j be an arbitrary non-zero lattice point Then there exist ǫ 1 and ǫ in { 1, 1} and p 1, p, q Z, not all zero, such that v = (ǫ 1 (p 1 + qα), ǫ (p + qβ), q) First suppose q = 0 Then by the inequality of arithmetic and geometric means we have ϕ i v 3( p 1 + qα p + qβ q ) /3 Using our hypothesis (11) we get p 1 + qα p + qβ q φ( q ) If q Q we conclude, by the monotonicity of φ, that φ( q ), and hence ϕ i v 1/3 If, on the other hand, q > Q then, looking only at the last coordinate, and using (1), we find ϕ i v > θ Q (εq) 1/3 1/3 Suppose now that q = 0 Then p 1 and p are not both zero, and hence ϕ i v max{θν i/ p 1, θν i/ p } θν i / θν N/ Recall that ν N/ = ε/t, and θ = V 1/3 / ε 1/3( Q/ ε ) 1/3 Hence, This proves the lemma θν N/ 1 T (εq)1/3 1 T 1/3
ASYMPTOTIC DIOPHANTINE APPROXIMATION: THE MULTIPLICATIVE CASE 7 6 PROOF OF THEOREM 11 Let 1 j 4 Decomposing the set Z 1 using (31) and (33) and then applying the automorphisms ϕ i yields Λ j Z 1 = Λ j x (0, Q] + Λ j y (0, Q] + = ϕ N Λ j ϕ N ( x (0, Q]) + ϕ N+1 Λ j ϕ N+1 ( y (0, Q]) + N i= N+1 ϕ i Λ j ϕ i (S i (0, Q]) N i= N+1 Λ j S i (0, Q] Note that det ϕ i Λ j = 1 Applying Lemma to each summand, using Lemma 41, and collecting the main terms and the error terms terms yields Λ j Z 1 Vol 3 (Z 1 ) ( ) D 3 MC V /3 (61) L 1 + λ 1 (ϕ i Λ j ) N i= N+1 Then, applying Lemma 51, we see that the right hand-side of (61) is bounded by ( ) 4D 3 MC L max{1, T} N 1 + V/3 /3 Using that by (35) V εq, and then again that εq we conclude that the latter is bounded by ( ) εq /3 8D 3 MC L (1 + T) N Putting C 5 = 8D 3 MC L = 8 318 5 1, and recalling that N log(t /ε) we conclude that Λ j Z 1 Vol 3 (Z 1 ) ( T C 5 (1 + T) )( ) εq /3 log ε By virtue of inequality (3), we get M α,β (ε, T, Q) 4Vol 3 (Z 1 ) 5C 5 (1 + T) log Finally, we note that 5C 5 < 3 8 = C 1 and Vol 3 (Z 1 ) = εq ( log and this completes the proof of Theorem 11 We have Q q=1 ( qα qβ ) 1 = ( T ε ( T ε ) ) + 1, 7 PROOF OF COROLLARY 1 )( ) εq /3 k+1 {q; 1 q Q, k 1 qα qβ < k } k+1 {q; 1 q Q, qα qβ < k } k+1 M α,β ( k, 1/, Q)
8 MARTIN WIDMER Moreover, in the proof of Lemma 1 we have seen that M α,β (ε, T, Q) = when ε < /Q We apply this with ε = k and T = 1/ Hence, (71) k+1 M α,β ( k, 1/, Q) = From Corollary 11 we get for integers k 5 (7) log (Q/) k+1 M α,β ( k, 1/, Q) log (Q/) < 4 5 Q + k=5 M α,β ( k, 1/, Q) 4(log )Qk k + C (log )k k+1 M α,β ( k, 1/, Q) ( ) /3 k Q Combining (71) and (7) Corollary 1 follows from a straightforward calculation using the trivial estimates K k K and K kxk Kx K+1 /(x 1) (where x > 1) and that 1/4 ACKNOWLEDGEMENTS This article was initiated during a visit at the University of York, and, in parts, motivated by a question of Sanju Velani I am very grateful to Victor Beresnevich, Alan Haynes and Sanju Velani for many interesting and stimulating discussions and their encouragement I also would like to thank Thái Hoàng Lê for fruitful discussions, drawing my attention to [], and for pointing out an error in an early draft of the manuscript REFERENCES 1 D Badziahin, On multiplicatively badly approximable numbers, Mathematika 59, no1 (013), 31 55 D A Badziahin and S Velani, Multiplicatively badly approximable numbers and generalised cantor sets, Adv Math 8, no5 (011), 766 796 3 Y Bugeaud and N Moshchevitin, Badly approximable numbers and Littlewood-type problems, Math Proc Cambridge Phil Soc 150 (011), 15 6 4 M Einsiedler, A Katok, and E Lindenstrauss, Invariant measures and the set of exceptions to Littlewood s conjecture, Ann of Math () 164 (006), 513 560 5 P Gallagher, Metric simultaneous diophantine aproximations, J London Math Soc 37 (196), 387 390 6 T H Lê and J D Vaaler, Sums of products of fractional parts, to appear in Proc London Math Soc 7 M Widmer, Counting primitive points of bounded height, Trans Amer Math Soc 36 (010), 4793 489 DEPARTMENT OF MATHEMATICS, ROYAL HOLLOWAY, UNIVERSITY OF LONDON, TW0 0EX EGHAM, UK E-mail address: martinwidmer@rhulacuk