On a mixed Littlewood conjecture in Diophantine approximation

Transcription

1 ACTA ARITHMETICA (2007) On a mixed Littlewood conjecture in Diohantine aroximation by Yann Bugeaud (Strasbourg), Michael Drmota (Wien) and Bernard de Mathan (Bordeaux) 1. Introduction. A famous oen roblem in simultaneous Diohantine aroximation is the Littlewood conjecture [9]. It claims that, for every given air (α, β) of real numbers, we have (1) lim inf q q qα qβ = 0, where denotes the distance to the nearest integer. The first significant contribution to this question goes back to Cassels and Swinnerton-Dyer [3] who showed that (1) holds when α and β belong to the same cubic field. Further exlicit examles of airs (α, β) of real numbers satisfying (1) have been given in [10, 1]. Desite some recent remarkable rogress [13, 5] the Littlewood conjecture remains an oen roblem. Recently, de Mathan and Teulié [12] roosed a mixed Littlewood conjecture that can be stated as follows. Let D = (d k ) k Z be a sequence of integers greater than or equal to 2. Set e 0 = 1 and, for any n 1, e n = d k. For q Z, set 0<k n w D (q) = su{n N : q e n Z}, q D = 1/e wd (q) = inf{1/e n : q e n Z}. When D is the constant sequence equal to, where is a rime number, then D is the usual -adic absolute value, normalized by = 1. In analogy with the Littlewood conjecture, de Mathan and Teulié asked whether (2) lim inf q q qα q D = Mathematics Subject Classification: 11J04, 11J61. Research of Y. Bugeaud suorted by the Austrian Science Fundation FWF, grant M822-N12. [107] c Instytut Matematyczny PAN, 2007

2 108 Y. Bugeaud et al. for every real number α. They roved that (2) holds for every quadratic irrationality α when the sequence D is bounded. In the resent aer, we focus on the articular case when D is the constant sequence equal to a rime number. Thus, we investigate the following conjecture. Mixed Littlewood Conjecture. For every real number α and every rime number, we have (3) lim inf q q qα q = 0. Obviously, this holds if α is rational or has unbounded artial quotients. Thus, we only consider the case when α is an element of the set Bad of badly aroximable numbers, where Bad = {α R : inf q qα > 0}. q 1 We are concerned with the following question: Problem 1. Is there any α in Bad which is irrational and not quadratic, and such that, for any rime number, the air (α, ) satisfies (3)? As briefly outlined on. 231 of [12], the answer to Problem 1 is ositive when α lies in a subset of Bad with Hausdorff dimension 1; see also [6] for a stronger result. Nevertheless, these aroaches do not rovide any new exlicit examles of airs (α, ) satisfying (3) with α in Bad. The urose of the resent note is recisely to construct exlicitly uncountably many real numbers α in Bad such that the air (α, ) satisfies (3) for any rime number. We further extend the roblem osed by de Mathan and Teulié, by considering an inhomogeneous mixed Littlewood conjecture. We ask whether (4) lim inf q q qα q y = 0 for any real number α, any rime number and any y in Z, the ring of -adic integers. It turns out that our methods allow us to establish (4) for a wide class of real numbers α in Bad. Our roofs heavily deend on -adic analysis, and our key tool is the -adic logarithm function. It is not clear to us whether our Theorem 1 has a real analogue, or an analogue for formal ower series over a finite field. 2. Results. Our main result shows that (3) holds for any real number α whose sequence of artial quotients is, in some sense, quasi-eriodic. Theorem 1. Let α be in Bad and write α = [a 0 ; a 1, a 2,...].

3 Mixed Littlewood conjecture 109 Let T 1 be an integer and b 1,...,b T be ositive integers. If there exist two sequences (m k ) k 1 and (h k ) k 1 of ositive integers with (h k ) k 1 being unbounded and then a mk +j+nt = b j for every j = 1,...,T and every n = 0,...,h k 1, (5) lim inf q q qα q y = 0 for every rime number and every y Z. If, furthermore, there exists a constant C such that (6) m k Ch k for k 1, then (7) lim inf q q log q qα q y <, for every rime number and every y Z. It is worth rehrasing the assumtion of Theorem 1 by using the terminology of combinatorics on words. Let α = [a 0 ; a 1, a 2,...] be in Bad and view its sequence of artial quotients as the infinite word a = a 0 a 1 a 2... on the alhabet A = {1,...,M}, where M is an uer bound for the a i s. Theorem 1 asserts that if there exists a finite, non-emty word B on the alhabet A such that, for every k 1, the concatenation of k coies of B occurs in the word a, then (5) holds for every rime number and every y in Z. Consequently, Theorem 1 rovides an uncountable, exlicit class of badly aroximable real numbers for which the mixed Littlewood conjecture, and even the inhomogeneous mixed Littlewood conjecture, is true. We further mention that if the real number α satisfies the assumtion of Theorem 1 with sequences (m k ) k 1 and (h k ) k 1 such that (6) holds, then α is either a quadratic irrationality, or a transcendental number. This follows from Theorem 3.2 of [2]. We dislay an immediate consequence of the roof of Theorem 1. Theorem 2. Let α be a quadratic real number. Let be a rime number. For any y in Z, there exist a ositive constant c(α,, y), deending only on α, and y, and arbitrarily large ositive integers q with and qα c(α,, y)/q (8) q y c(α,, y)/log q. In articular, (7) holds. The case y = 0 was already established in Théorème 2.1 of [12]. Notice that the latter result actually covers the more general case of a bounded

4 110 Y. Bugeaud et al. sequence D of ositive integers. In view of [11], we cannot relace log q in (8) by (log q) λ with λ very large. Theorem 2 also holds for an arbitrary bounded sequence D when y is an integer. Using some of the ideas occurring in the roof of Theorem 1, we get another uncountable, exlicit class of badly aroximable real numbers for which the mixed Littlewood conjecture, and even the inhomogeneous mixed Littlewood conjecture, is true. Theorem 3. Let α = [a 0 ; a 1, a 2,...] be in Bad. If for every integer h 1 there exists an integer T such that a j+nt = a j for every j = 1,...,T and every n = 0,...,h, then, for any rime number and any integer y, lim inf q q qα q y <. More generally, Theorem 3 holds for an arbitrary (bounded) sequence D. Its analogue in the function field case can also be easily established. Using arguments from [13], we deduce that (4) holds for many real numbers α with bounded artial quotients. Theorem 4. The set of real numbers α with bounded artial quotients for which lim inf q q qα q y = 0 for every rime number and every y in Z has Hausdorff dimension 1. A stronger result holds when y = 0. Namely, Einsiedler and Kleinbock [6] established that the set of α for which (3) does not hold has Hausdorff dimension 0. Presumably, their aroach could be modified to get an analogous result for (4), but this is not clear to us. In view of Theorem 1, we would like to address the following roblem. Problem 2. Let ε be a ositive real number. Find a real number α in Bad, a rime number, and a rational integer y such that (9) lim inf q q1+ε qα q y <. It follows from the -adic analogue of the Schmidt subsace theorem, established by Schlickewei [15], that any real number α satisfying (9) must be transcendental. Aarently, our construction does not allow us to tackle Problem 2. Nevertheless, by using the folding lemma recalled in Section 6, it is ossible to give a ositive answer to Problem 2.

5 Mixed Littlewood conjecture 111 Theorem 5. For any rime number, there exists a real number α in Bad such that lim inf q q2 qα q 1. Since (9) cannot hold with ε > 1 and α in Bad, Theorem 5 is best ossible. We end this section with a third question. Problem 3. Given α in Bad, is there a rime number such that (3) holds for the air (α, )? Aarently, there is no contribution towards Problem 3, which seems to be quite difficult. 3. Proof of Theorem 1 Preliminaries to the roof. Let (q n ) n 1 be the sequence of the denominators of the convergents of α. These numbers satisfy the recurrence relation q n = a n q n 1 + q n 2 with q 0 = 1 and q 1 = 0. For n 0, set ( Q n = q n q n 1 We can write ( ) a n 1 Q n = P n Q n 1 where P n =. 1 0 Set ( ) b j 1 M j =, j = 1,...,T, M = M T...M Relacing if necessary b 1,...,b T by b 1,...,b T, b 1,...,b T, we may relace T by 2T, and thus we can suose detm = 1. The matrix M is diagonalizable and its eigenvalues are quadratic units ω > 1 and 1/ω. We have N Q(ω)/Q ω = 1. Set a = Tr Q(ω)/Q ω. Then a > 2 is a ositive integer, and ω is a root of the olynomial X 2 ax + 1. We have Q mk +nt = MQ mk +(n 1)T for every 1 n h k, hence Q mk +nt = M n Q mk. The Cayley Hamilton theorem imlies that, for 2 n h k, we have Q mk +nt = M 2 Q mk +(n 2)T = (am I)Q mk +(n 2)T ) = aq mk +(n 1)T Q mk +(n 2)T. Hence the sequence (q mk +nt) n 0 satisfies (10) q mk +nt = aq mk +(n 1)T q mk +(n 2)T when 2 n h k..

6 112 Y. Bugeaud et al. Note that Q mk +1+nT = M Q mk +1+(n 1)T for 0 < n < h k, where M = M 1 M T...M 2 = M 1 MM1 1. Hence, M has the same characteristic olynomial as M, and the sequence (q mk +1+nT) n 0 also satisfies the recurrence q mk +1+nT = aq mk +1+(n 1)T q mk +1+(n 2)T when 2 n < h k. Consequently, we may relace m k by m k + 1. Since q mk and q mk +1 are corime, we can, without any loss of generality, suose that does not divide q mk. We have (11) q mk +nt = A k ω n + B k ω n for 0 n h k, where A k and B k are given by (11) with n = 0 and n = 1: (12) (13) A k = ωq m k +T q mk ω 2, 1 B k = ω q m k +T ωq mk ω 2. 1 We also denote by ω a zero of the olynomial X 2 ax + 1 in the algebraic closure of Q. We still denote by the -adic value extended to this field. As ω is a unit, we have ω = 1. The formulæ (12) and (13) hold in Q. Since ω and ω 1 are conjugate in Q(ω), and so are A k and B k, we can change A k into B k by changing the embedding of ω in Q (ω) it is convenient to make this embedding deend uon k. We can thus suose that A k B k (i.e., q mk +T ωq mk ωq mk +T q mk ). Since we suose that A k +B k = q mk = 1, we then have 1 A k 1/ ω 2 1. The field Q (ω) is comlete. The ball G = {x Q (ω) : x 1 < 1/( 1) } is a subgrou of finite index in the multilicative grou {x Q (ω) : x =1}. Hence, relacing again T by lt, therefore ω by ω l, where l is a suitable ositive integer, we may also suose that ω 1 < 1/( 1). In what follows, we shall make use of the -adic logarithm function, which is defined on the multilicative grou {x Q (ω) : x 1 < 1} in Q (ω) by log x = ( 1) n 1 (x 1) n /n. We have n=1 log xy = log x + log y for x, y in {x Q (ω) : x 1 < 1}, and, for every x, y in G, log x = x 1

7 and Mixed Littlewood conjecture 113 log x log y = log x y = x y 1 = x y. The constants imlied in the symbols and occurring below will only deend uon b 1,...,b T and on. The roofs in the cases y = 0 and y 0 are rather different. First, we deal with the case y = 0, which is slightly more difficult. An auxiliary result for the case y = 0. Kee the above notations and set either ε k = ω 2 1 q mk +T ωq mk or ε k = 1/2 ω 2 1 q mk +T ωq mk, in such a way that ε k < 1 is an integral ower of. (14) (15) Lemma 1. There exist integers x k and y k, with y k = 1, such that Furthermore, x k q mk + y k q mk +T ε k, max{ x k, y k } ε 1/2 k. (16) x k q mk + y k q mk +T ω 2 1 x k + y k ω (max{ x k, y k }) 2. and To rove Lemma 1, we need the following version of Liouville s lemma. Lemma 2. For any integers X and Y, not both zero, Proof. We have X + Y ω (max{ X, Y }) 2. N Q(ω)/Q (X + Y ω) max{ X, Y } 2 N Q(ω)/Q (X + Y ω) = (X + Y ω)(x + Y/ω) X + Y ω. As N Q(ω)/Q (X + Y ω) is a non-zero integer, we get N Q(ω)/Q (X + Y ω) N Q(ω)/Q (X + Y ω) 1. It then follows that X + Y ω max{ X, Y } 2 1, as claimed. Proof of Lemma 1. Let s be a ositive integer. By the igeonhole rincile, there exist integers x k,s and y k,s, not both zero, such that (17) x k,s q mk + y k,s q mk +T ε k s, (18) max{ x k,s, y k,s } ε 1/2 k s/2. First, we rove that there exists a non-negative integer S, deending only on ω, such that if s > S, then y k,s cannot be divisible by s. Indeed, let

8 114 Y. Bugeaud et al. σ be a ositive integer with 0 σ s such that y k,s is divisible by σ. Then x k,s as well is divisible by σ. Indeed, we have y k,s q mk +T σ and x k,s q mk + y k,s q mk +T σ, hence x k,s q mk σ. Since we have assumed that q mk = 1, we get x k,s σ. Setting x k,s = σ x k,s and y k,s = σ y k,s, we deduce from (17) that (19) and from (18) that x k,s + y k,s q mk +T q mk ε k s+σ (20) max{ x k,s, y k,s } ε 1/2 k s/2 σ. Hence, writing (21) x k,s + y k,s ω = x k,s + y k,s q mk +T q mk + y k,s ( ω q m k +T q mk and noticing that (19) imlies that x k,s + q y mk +T k,s q mk < q mk +T ωq mk, we get Then by (20), If σ = s, we thus get x k,s + y k,s ω q mk +T ωq mk ε k. x k,s + y k,s ω max{ x k,s, y k,s }2 s 2σ. x k,s + y k,s ω max{ x k,s, y k,s }2 s, and it follows from Lemma 2 that we must have s 1, i.e., s S, where S 0 is an integral constant, deending only uon ω. Thus, if s > S, then y k,s cannot be divisible by s. Take s = S+1. If we define σ by y k,s = σ, then σ < s. Set x k = σ x k,s and y k = σ y k,s, so that (22) y k = 1. By (19) and (20), as s = S + 1, we deduce (14) and (15). To conclude, let us show (16). Indeed, as ε k < q mk +T ωq mk, we deduce from (22), (14) and (21) that x k + y k ω = q mk +T ωq mk. Accordingly, (14) and (15) lead to the desired conclusion. Comletion of the roof for the case y = 0. For any k 1, let x k and y k be the integers given by Lemma 1. For any n 0, set (23) Q k,n = x k q mk +nt + y k q mk +(n+1)t. ),

9 Mixed Littlewood conjecture 115 Our aim is to rove that there exists an integer n(k) h k 1 such that Q k,n(k) 0, and inf Q k,n(k) Q k,n(k) Q k,n(k) α = 0. k 1 First, we note that, for every 0 n < h k, we have We thus get Q k,n max{ x k, y k }q mk +nt, Q k,n α max( x k, y k )/q mk +nt. (24) Q k,n Q k,n α max{ x k, y k } 2. Further, with A k and B k being defined as in (12) and (13), we can write where Q k,n = C k ω n + D k ω n C k = A k (x k + y k ω), D k = B k (x k + y k ω 1 ). Since A k 1, we deduce from Lemma 1 that C k +D k = x k q mk +y k q mk +T ω 2 1 x k +y k ω ω 2 1 C k. Hence, C k + D k < C k, and thus C k = D k. We then get (25) 1 + D k /C k ω 2 1 < 1/( 1). Write Q k,n = C k ω n (D k /C k + ω 2n ). Since C k x k + y k ω (max{ x k, y k }) 2, we thus get and, by (24), Q k,n (max{ x k, y k }) 2 D k /C k + ω 2n (26) Q k,n Q k,n Q k,n α D k /C k + ω 2n. Now, inequality (25) enables us to use the -adic logarithm in the domain G where log x log y = x y. As D k /C k + ω 2n = log( D k /C k ) 2nlog ω, we may also write (26) as Q k,n Q k,n Q k,n α log( D k /C k ) 2log ω n. Now let us show that (log( D k /C k ))/(2log ω) lies in Q. This is trivial if ω Q. In the case where ω Q, there exists a unique Q -automorhism σ of Q (ω), different from the identity. We have σ(ω) = 1/ω, and σ is isometrical. We have σ(log ω) = log(σ(ω)) = log ω, and σ( D k /C k ) = C k /D k, since σ(c k ) = D k. We thus have σ(log( D k /C k )) = log( C k /D k ) = log( D k /C k ).

10 116 Y. Bugeaud et al. Hence, ( ) log( Dk /C k ) σ 2log ω = log( D k/c k ), 2log ω and so the element log( D k /C k )/(2log ω) of Q (ω) lies in Q. Further, it lies in Z, since log( D k /C k ) = D k /C k + 1 ω 2 1 = 2log ω, by (25). If t k is a ositive integer with 1 2 h k < t k 1 2 h k, then there exists an integer n, with 0 n < t k, such that log( D k /C k ) n 2log ω t k. Relacing if necessary n by n + t k, we may ensure that log( D k /C k ) 2nlog ω, hence D k /C k + ω 2n 0. Taking either n(k) = n or n(k) = n + t k, we have thus constructed an integer n(k) satisfying 0 n(k) < h k, Q k,n(k) 0 and (27) Q k,n(k) Q k,n(k) Q k,n(k) α t k 1 h k. Since the sequence (h k ) k 1 is unbounded, we have inf Q k,n(k) Q k,n(k) Q k,n(k) α = 0. k 1 It remains for us to rove that the estimate (7) holds when (6) is satisfied. Note that as the artial quotients of α are bounded, by (23) we have hence log Q k,n(k) log max{ x k, y k } + m k + n(k), log Q k,n(k) log max{ x k, y k } + m k + h k. Now, by (15) and Lemma 2, we have max{ x k, y k } q mk +T ωq mk 1/2 q mk. Thus, log max{ x k, y k } m k and so log Q k,n(k) m k + h k. Accordingly, if (6) is satisfied, then log Q k,n(k) h k, and (7) follows from (27). The case y 0. We shall rove that for any non-zero y in Z, there exists an infinite set Q of ositive integers Q satisfying (28) Q Qα 1

11 Mixed Littlewood conjecture 117 and (29) lim inf Q Q Q y = 0. Note that if (29) is true for y, then it is true for My, where M is any ositive integer. Indeed, we may relace Q by MQ, while reserving (28). Thus it is enough to rove (29) when y = 1. Actually we shall construct a set of ositive integers Q, with (28), such that the -adic toological closure of this set contains the unit circumference y 1 = 1. For this urose, it is enough to construct a set of Q, with (28), whose -adic closure contains the ball y 1 λ for some ositive integer λ. Indeed, if we relace the set of Q by the set of µq, where µ runs through the integers 0 < µ < λ with µ = 1, the -adic closure of this last set will contain the unit circumference y 1 = 1. First, suose that 2. Let (m k ) k 1 and (h k ) k 1 be as in the statement of the theorem. Define A k and B k by (12) and (13). Recall that we suose that q mk = 1 and A k B k, hence A k 1. We may also suose that A k B k ω 1. Indeed, we may relace q mk by q mk +T or q mk +2T, reserving the condition q mk = 1, since (10) ensures that if q mk is not divisible by, then neither is q mk +T or q mk +2T. Now, if we relace q mk by q mk +T (res. q mk +2T), then A k is relaced by A k ω (res. A k ω 2 ), and B k is relaced by B k ω 1 (res. B k ω 2 ). If A k B k < ω 1, then A k ω B k ω 1 = A k (ω ω 1 ) + (A k B k )ω 1 = A k (ω ω 1 ) ω 2 1 = ω 1 in view of the roerties of the logarithm function, since ω 1 < 1/( 1). In the same way, if A k B k < ω 1, then A k ω 2 B k ω 2 ω 1. Accordingly, by these changes, we may suose that q mk = 1 and (30) A k B k ω 1. Let λ be the ositive integer such that ω 1 2 = λ+1. As q mk = 1 there exists an integer L k with 0 < L k < λ and L k = 1 such that (31) L k (A k + B k ) 1 mod λ. We shall rove that the -adic closure of the set comosed by the integers L k q mk +nt for k 1 and 0 n h k contains the ball y 1 λ. Note that, L k being bounded, the integers L k q mk +nt satisfy (28), and thus the result will be roved. Set L k A k = A k and L kb k = B k. For 0 n h k, we have L k q mk +nt = A k ωn + B k ω n.

12 118 Y. Bugeaud et al. Consider the ma ϕ k from the ball {x : x 1 < 1} of Q (ω) into Q (ω) such that ϕ k (x) = A k x + B k x 1. Let y Q with (32) y 1 λ. We shall find x Q (ω) such that ϕ k (x) = y. We must have A k x2 yx+b k = 0. Accordingly we take (33) x = y + y 2 4A k B k 2A k However, we must make recise the meaning of the symbol. We define the function over the ball {z : z 1 1 } in Q, with values in the same ball, by the equality log z = 1 2 log z (that is to say, z = ex log z 2, where exu = n=0 un /n! for u 1 ). Therefore we have (34) z 1 = z 1. Now, by (31) and (32), we have (35) y 2 4A k B k (A k B k )2 = y 2 (A k + B k )2 λ, and, by (30) and A k B k 2 λ+1, we get (36) y 2 4A k B k (A k 1 B k )2 1. Moreover the number (y 2 4A k B k )/(A k B k )2 lies in Q, since in the case where ω Q, it is invariant under the above automorhism σ. Inequality (36) allows us to define the number (y 2 4A k B k )/(A k B k )2 in Q, and we ut in (33) y 2 4A k B k = (A k B k ) y 2 4A k B k (A k B k )2. Thus x (see (33)) is well defined in Q (ω). Further, let us rove that (37) x 1 ω 1. Indeed, writing y 2 4A k B k (A k B k ) = A k B k y 2 4A k B k (A k 1 B k )2, we deduce from (34) that y 2 4A k B k (A k B k ) = y2 4A k B k (A k B k )2 A k B k,

13 and from (35) that Mixed Littlewood conjecture 119 y 2 4A k B k (A k B k ) λ A k B k. Since A k B k λ/2, it follows that y 2 4A k B k (A k B k ) λ/2. As y (A k + B k ) λ, we thus get y + y 2 4A k B k 2A k λ/2, which leads to (37) since A k 1. Lastly we rove that log x/log ω belongs to Q. Indeed, in the case where ω Q, we have σ( y 2 4A ) k B k y 2 4A k (A k = B k B k )2 (A k B k )2 since this number lies in Q. Hence, as σ(a k B k ) = B k A k, we get σ( y 2 4A k B k ) = y 2 4A k B k. We thus have ( y + y 2 4A ) k σ B k 2A = y y 2 4A k B k k 2B k. It follows immediately that (38) σ(x) = 1/x and as ω also satisfies (38), we conclude as above that log x/log ω belongs to Q. Moreover, as log x/log ω = x 1 / ω 1, we conclude from (37) that log x/log ω belongs to Z. Accordingly, given a ositive integer N, there exists an integer n, with 0 n < N, such that log x log ω n N, i.e., log x log ω n N ω 1, and thus x ω n N ω 1. As B k A k 1/ ω 1, we thus get ϕ k (x) ϕ k (ω n ) N. Now ϕ k (x) = y, and if 0 n h k, then ϕ k (ω n ) = L k q mk +nt. Let us select k such that h k N ; we have thus found an integer n, with 0 n h k, such that (39) L k q mk +nt y N. Accordingly, the -adic closure of the set of L k q mk +nt with k 1 and 0 n h k contains the ball y 1 λ.

14 120 Y. Bugeaud et al. There are some minor changes when = 2. First note that, using the logarithm function over Q 2 for z 1 1/4, we get z = 1 2 z 1 2. The function is then defined over the ball z 1 2 1/8 in Q 2, and satisfies z 1 2 = 2 z 1 2. Also note that if A k B k 2 < 1, then A k 2 = B k 2 = 2, since A k + B k 2 = 1. Reasoning as above, we may thus suose that A k B k ω 1 2. The number λ is then determined by 2 λ = 1 16 ω 1 2 2, and L k is determined by L k q mk 1 mod 2 λ and 0 < L k < 2 λ. Then the 2-adic closure of the set of integers L k q mk +nt contains the ball {y : y λ }, and (39) holds. In order to obtain (7), it is enough to note that log L k q mk +nt m k +n, and thus, if condition (6) is satisfied, then log L k q mk +nt h k. We may choose the integer N above such that h k / < N h k, i.e., N h k, and from (39), we get (7). 4. Proof of Theorem 3. Let α be as in the statement of Theorem 3. Let be a rime number. First, notice that if ω is any quadratic unit, then the index of the multilicative grou G = {x Q (ω) : x 1 < 1/( 1) } in the unit ball {x Q (ω) : x 1 = 1} is a divisor of 2 ( 2 1). Hence, ω 2 ( 2 1) 1 < 1/( 1). Accordingly, in the statement of Theorem 3, if we relace T by 2 ( 2 1)T, we can suose, as in the roof of Theorem 1, that the eigenvalues ω T, 1/ω T of the matrix M = M T...M 1 are quadratic units with ω 2 ( 2 1) T 1 < 1/( 1). Thus, in order to rove Theorem 3 for y = 0 it is enough to consider Q n = q nt 1 for 0 n h. This sequence satisfies (10), hence we can write since Q 0 = 0. We have Q n = A(ω n T ω n T ) A 1/ ω 2 T 1. Hence, for 0 < s log h/log, we have Q s ω2s T 1 ωt 2 1 = s. Taking s such that h/ < s h, we get Q s 1/h. This roves the result, in view of (28).

15 Mixed Littlewood conjecture 121 Let y be a non-zero integer. Consider Q n = yq nt = Eω n T +Fω n T, where Since Q 0 = y, we have as above max{ E, F } 1/ ω 2 T 1. Q 2 s y ω2s T 1 ωt 2 1 which again roves the result by (28). = s, Remarks on the roofs. Although Theorem 2 for y = 0 is a consequence of Theorem 1, and was already roved in [12], we would like to note that in this case, the roof of (7) is very simle, and the use of the -adic logarithm function is actually not necessary. Indeed, the assumtions of Theorem 1 are satisfied for some ositive integer m, and we have a m+j+nt = b j for every non-negative integer n and for any j = 1,...,T. For any k 1, we set m k = m, and we choose the integers x k = q m+t and y k = q m. Then, as in the roof of Theorem 1, we consider the integers Q n = q m+t q m+nt q m q m+(n+1)t. They satisfy (28) since m is now a fixed number. In this case, as Q 0 = 0, we have C k = D k, that is to say, we can write Q n = C(ω n ω n ) with C 0 since q m+t ωq m 0. Hence, Q n 0 for n > 0, and Q n ω 2n 1. Therefore, we only have to check that ω 2s 1 s ω 2 1. This follows from an elementary induction by writing ω 2s = 1+u s and using Newton s formula. As log Q n n, we thus see that the integers Q s satisfy (28) with log Q s s and Q s s. This rovides the estimate (7). The same holds when y 0 is an integer. Indeed, by the Bézout theorem, q m and q m+1 being corime, we can take integers x and y such that x q m + y q m+1 = y (where m 0 is chosen in a such way that a m+j+nt = b j for every non-negative integer n and for any j = 1,...,T). Since both the sequences (q m+nt ) n 0 and (q m+1+nt ) n 0 satisfy (10), for every non-negative integer n we can write x q m+nt + y q m+1+nt = Eω n + Fω n where E and F are in Q (ω) with E 1/ ω 2 1 and F 1/ ω 2 1. The integers Q n = x q m+nt + y q m+1+nt satisfy (28). Since ω 2s 1 s ω 2 1, we see that for n = 2 s we have Q 2 s Q 0 s, that is to say, Q 2 s y s. Further, it is easy to see that Theorems 2 and 3 remain valid uon relacing in (8) the -adic absolute value by D, if we take y Z.

16 122 Y. Bugeaud et al. In the other arts of the roof, the role layed by the -adic logarithm function seems deeer. For instance, it is easy to rove the analogue of Theorem 2 for formal ower series over a finite field, when y is a olynomial. But we do not know whether Theorem 1 has an analogue in this setting. The crucial oint is the use of the -adic logarithm function, which we lose in the formal ower series case. 5. Proof of Theorem 4. Let be a rime number and y be an integer. Let µ denote the Kaufman measure, whose existence has been roved in [7]. The measure µ is suorted on the set Bad. For any n 1, set q n := n +y. For any real number α we have q n q n α q n y = q n q n α n (1 + y ) q n α. Thus, (4) holds as soon as α satisfies (40) inf n 1 q nα = 0. Using the exonential growth of the sequence (q n ) n 1 and a result of Davenort, Erdős, and LeVeque [4], as exlained in Section 4 of [13], we infer that (40) holds for µ-almost all α. Hence, µ-almost all α satisfy (4) for any rime and any integer y. Arguing then as on. 294 of [13], we conclude that the Hausdorff dimension of the set of real numbers α in Bad for which (4) holds for any rime and any integer y is equal to 1, as claimed. 6. Proof of Theorem 5. Our roof is very much insired by [8]. It rests on the folding lemma, recalled below. Lemma F. If n /q n = [a 0 ; a 1, a 2,..., a n ] with a n 2, then n q n + ( 1)n q 2 n = [a 0 ; a 1, a 2,..., a n 1, a n + 1, a n 1, a n 1,...,a 2, a 1 ]. Lemma F follows from Proositions 2 and 3 of [14]. Recall that [0; a 1,...,a n, 1, 1] = [0; a 1,..., a n, 2] for any ositive integers a 1,...,a n. We dislay an immediate consequence of Lemma F. Lemma 3. If a/m = [0; 1, 1, a 3,..., a h 1, a h ] with h 4 and a h 2, then ma + ( 1) h m 2 = [0; 1, 1, a 3,...,a h 1, a h + 1, a h 1, a h 1,..., a 3, 2]. Lemma 3 is the main tool for the roof of Theorem 5. Let be a rime number. Observe that in the oen real interval with endoints [0; 1, 1, 1, 2] and [0; 1, 1, 1, 3] there are rational numbers whose denominator is a ower of. Consequently, there exist ositive integers a and b, with 1 a b

17 Mixed Littlewood conjecture 123 and a corime with, such that a b = [0; 1, 1, a 3,...,a h 1, a h ] with h 4 and a h 2. Set M = max{a 3,...,a h 1, a h + 1}. By reeated alication of Lemma 3, we see that, for any j 2, the continued fraction of the rational number α j := a b + ( 1)h 2b 1 22 b 1 2j b reads [0; 1, 1,...,2] and has all its artial quotients bounded by M. Set α = lim α j = a j b + ( 1)h 2b 1 2j b. j 2 By construction, all the artial quotients of α are less than or equal to M, hence α is in Bad. Furthermore, it is easily checked that, for any j 2, we have 2jb 2jb α j+1b. This imlies that as asserted. lim inf q q2 qα q 1, References [1] B. Adamczewski and Y. Bugeaud, On the Littlewood conjecture in simultaneous Diohantine aroximation, J. London Math. Soc. 73 (2006), [2],, On the Maillet Baker continued fractions, J. Reine Angew. Math., to aear. [3] J. W. S. Cassels and H. P. F. Swinnerton-Dyer, On the roduct of three homogeneous linear forms and indefinite ternary quadratic forms, Philos. Trans. Roy. Soc. London Ser. A 248 (1955), [4] H. Davenort, P. Erdős and W. J. LeVeque, On Weyl s criterion for uniform distribution, Michigan Math. J. 10 (1963), [5] M. Einsiedler, A. Katok and E. Lindenstrauss, Invariant measures and the set of excetions to the Littlewood conjecture, Ann. of Math. 164 (2006), [6] M. Einsiedler and D. Kleinbock, Measure rigidity and -adic Littlewood-tye roblems, Comos. Math. 143 (2007), [7] R. Kaufman, Continued fractions and Fourier transforms, Mathematika 27 (1980), [8] T. Komatsu, On a Zaremba s conjecture for owers, Sarajevo J. Math. 1 (2005), [9] J. E. Littlewood, Some Problems in Real and Comlex analysis, D. C. Heath and Raytheon Education Co., Lexington, MA, 1968.

18 124 Y. Bugeaud et al. [10] B. de Mathan, Conjecture de Littlewood et récurrences linéaires, J. Théor. Nombres Bordeaux 13 (2003), [11], On a mixed Littlewood conjecture for quadratic numbers, ibid. 17 (2005), [12] B. de Mathan et O. Teulié, Problèmes diohantiens simultanés, Monatsh. Math. 143 (2004), [13] A. D. Pollington and S. Velani, On a roblem in simultaneous Diohantine aroximation: Littlewood s conjecture, Acta Math. 185 (2000), [14] A. J. van der Poorten and J. Shallit, Folded continued fractions, J. Number Theory 40 (1992), [15] H. P. Schlickewei, The -adic Thue Siegel Roth Schmidt theorem, Arch. Math. (Basel) 29 (1977), Mathématiques Université Louis Pasteur 7, rue René Descartes Strasbourg Cedex, France bugeaud@math.u-strasbg.fr Institut für Diskrete Mathematik und Geometrie TU Wien A-1040 Wien, Austria michael.drmota@tuwien.ac.at Institut de Mathématiques Université Bordeaux I 351, cours de la Libération Talence Cedex, France demathan@math.u-bordeaux1.fr Received on and in revised form on (5208)