On multiplicatively badly approximable numbers

The Littlewood Conjecture states that liminf_{q\to \infty} q . ||qx|| . ||qy|| = 0 for all pairs (x,y) of real numbers. We show that with the additional factor of log q . loglog q the statement is false. Indeed, our main result implies that the set of (x,y) for which liminf_{q\to\infty} q . log q . loglog q . ||qx|| . ||qy||>0 is of full dimension.


Introduction
The famous Littlewood conjecture (LC) states that for any pair of real numbers (α, β) where • denotes the distance to the nearest integer.Equivalently, the set is empty.This problem was conjectured in 1930's and it is still open.For recent progress concerning this fundamental problem see [4,6] and references therein.It is easily seen that ( 1) holds for all α ∈ R and β ∈ R outside the set Bad of badly approximable numbers defined as follows Bad := {α ∈ R : lim inf q→∞ q||qα|| > 0}.
In attempt to understand what should be a proper analogue of badly approximable points in multiplicative case several authors investigated the following set (we will follow the notation introduced in [2]).For λ 0 let In other words, Mad λ is a modification of the set in (2) such that the corresponding condition is weakened by (log q) λ .More generally, given a function f : N → R + , define the set In [2] the author and Velani conjectured that Conjecture A (BV) Mad λ = ∅ for any λ < 1, dim(Mad λ ) = 2 for any λ 1 where dim(•) denotes the Hausdorff dimension.If true this conjecture implies that the proper multiplicative analogue of the set Bad is Mad 1 .Note that LC is equivalent to the statement that Mad 0 is empty.Therefore BV conjecture implies LC.Regarding the first part of BV conjecture all that is known to date is the remarkable result of Einsiedler, Katok and Lindenstrauss [4] which states that dim Mad 0 = 0. On the other hand according to the second part the best known result is due to Bugeaud and Moschevitin [3].It states that dim Mad 2 = 2.So we have a gap 0 λ < 2 where the behavior of Mad λ is completely unknown.
In this paper we will address the second part of the BV conjecture.In particular, we will show that dim Mad(f ) = 2 if f (q) = log q • log log q.
It is worth mentioning that the 'mixed' analogue of this result was achieved recently by author and Velani.It was proven that the set has full Hausdorff dimension.All the details can be found in [2].

Simultaneous and dual variants of Mad
It is well known that Littlewood conjecture has an equivalent formulation in terms of linear forms.In other words, (1) is equivalent to the statement that lim inf where |x| * := max{|x|, 1}.However it is not known if (3) can be reformulated in the same manner.In other words, define the sets and Mad λ L := Mad L (log λ q).Then Mad(f ) and Mad L (f ) are not necessarily the same.However as it will be shown in the next sections these sets are closely related to each other.For consistency in further discussion we will use the notation Mad λ P and Mad P (f ) instead of Mad λ and Mad(f ) respectively.It will reflect the fact that in one case we deal with points and in another case we deal with lines.
It appears that instead of investigating Mad P (f ) and Mad L (f ) independently it is easier to deal with them simultaneously.In particular, we prove the following result:

Main result
For convenience, we define the 'modified logarithm' function log * : R → R as follows log * x := 1 for x < e; log x for x e.

From now on
f (q) := log * q • log * log q.
The key to establishing Theorem 1 is to investigate the intersection of the sets Mad P (f ) and Mad L (f ) along fixed vertical lines in the (x, y)-plane.With this in mind, let L x denote the line parallel to the y-axis passing through the point (x, 0).
The following constitutes our main theorem.
Theorem 2 For any θ ∈ Bad Since by Jarník (1928) the Hausdorff dimension of Bad is one, Theorem 1 can be easily derived from Theorem 2 with the help of the following general result that relates the dimension of a set to the dimensions of parallel sections, enables us to establish the complementary lower bound estimate -see [5, pg. 99].
Proposition Let F be a subset of R 2 and let E be a subset of the x-axis.If dim(F ∩L x ) t for all x ∈ E, then dim F t + dim E.
Indeed, let F = Mad P (f ) ∩ Mad L (f ) and E = Bad.In view of dim(Bad) = 1 and Theorem 2, one gets dim Mad P (f ) ∩ Mad L (f ) 2. Since Mad P (f ) ∩ Mad L (f ) ⊂ R 2 , the upper bound statement for the dimension is trivial.Therefore the main ingredient in establishing Theorem 1 is Theorem 2.
Regarding the proof of Theorem 2 we will use ideas similar to those in [2] which firstly appeared in joint work of author, Pollington and Velani [1].However the technical details in this paper are substantially more complicated than those in [2].
Denote by ∆(P, δ) the following segment on L θ : Given a line with integer coeffitients denote by the height of L the value Denote by ∆(L, δ) the following segment on L θ : .
Given constants c > 0 and Q > 0 define the auxiliary sets: and For convenience we will omit the parameter Q where it is irrelevant and write Mad P (f, c) and Mad L (f, c) for Mad P (f, c, Q) and Mad L (f, c, Q) respectively.So it suffices to prove that the set Mad P (f, c) ∩ Mad L (f, c) ∩ L θ has full Hausdorff dimension for some positive constant c.
Geometrically, the set Mad P (f, c) consists of points that avoid the "neighborhood" of each rational point P = (p/q, r/q) defined by the inequality This "neighborhood" of P will remove the interval ∆(P, cf (q) −1 ) from L θ .Without loss of generality we can assume that |qθ − p| = ||qθ||.Otherwise we just replace the point P by P ′ := (p ′ /q, r/q) such that |qθ − p ′ | = ||qθ||.Then ∆(P ′ ) ⊃ ∆(P ) and the "neighborhood" of P will not remove anything more than one of P ′ .
Similarly one can show that the set Mad L (f, c) consists of points that avoid the "neighborhood" of each line L(A, B, C) defined by where the coefficients A, B, C satisfy (A, B) <> (0, 0) and gcd(A, B, C) = 1.For B = 0 it leads to the following inequality: Take c < inf q∈N q||qθ||.Then this inequality is not true for x = θ, in other words the "neighborhood" of the line do not remove anything from L θ .Therefore it is sufficient to consider the lines L(A, B, C) with B = 0, so the coefficients (A, B, C) will satisfy (5).Then the "neighborhood" of L(A, B, C) will remove the interval ∆(L, cf (|A| * |B| * ) −1 ) from L θ .

Cantor sets
In the proof we will use the general Cantor framework firstly introduced in [2].Here we reproduce the definitions and facts which will be used in later discussion.For more details we refer to the paper [2].
Let I be a closed interval in R. Let R := (R n ) with n ∈ Z 0 be a sequence of natural numbers and r := (r m,n ) with m, n ∈ Z 0 , m n be a two parameter sequence of nonnegative real numbers.
The construction.We start by subdividing the interval I into R 0 closed intervals I 1 of equal length and denote by I 1 the collection of such intervals.Thus, Next, we remove at most r 0,0 intervals I 1 from I 1 .Note that we do not specify which intervals should be removed but just give an upper bound on the number of intervals to be removed.Denote by J 1 the resulting collection.Thus, For obvious reasons, intervals in J 1 will be referred to as (level one) survivors.It will be convenient to define J 0 := {I}.In general, for n 0, given a collection J n we construct a nested collection J n+1 of closed intervals J n+1 using the following two operations.Splitting procedure.We subdivide each interval J n ∈ J n into R n closed sub-intervals I n+1 of equal length and denote by I n+1 the collection of such intervals.Thus, Removing procedure.For each interval J n ∈ J n we remove at most r n,n intervals I n+1 ∈ I n+1 that lie within J n .Note that the number of intervals I n+1 removed is allowed to vary amongst the intervals in J n .Next, for each interval J n−1 ∈ J n−1 we additionally remove at most r n−1,n intervals I n+1 ∈ I n+1 that lie within J n−1 .In general, for each interval we additionally remove at most r n−k,n intervals I n+1 ∈ I n+1 that lie within J n−k .Then the collection J n+1 consists of all intervals I n+1 ∈ I n+1 that survive after all these removing procedures for k = 1, 2, . . ., n.Thus, the total number of survivors is at most Finally, having constructed the nested collections J n of closed intervals we consider the limit set Any set K(I, R, r) which can be achieved by the procedure described will be referred to as a (I, R, r) Cantor set.
Of course in general it can happen that for some choice of parameters R and r and some choice of removed intervals in removing procedure the (I, R, r) Cantor set becomes empty.However the next result shows that with some additional conditions on the parameters the Hausdorff dimension of this set is bounded below.
Here we use the convention that the product term in (8) is one when k = 0 and by definition log Rn 2 := log 2/ log R n .The proof of Theorem BV4 is presented in [2, Theorem 4].

Duality between points and lines
The next two propositions show that there is a 'kind' of duality between rational points P (p, r, q) and lines L(A, B, C).It will play a crucial role in our proof.

This together with (A
The assumption (P 2 ) θ ∈ ∆(P 1 , δ) is equivalent to Finally by the triangle inequality we find that By combining all these inequalities together we get that Now we are ready to calculate the bound .
Then the first inclusion in (9) follows immediately.For the second one we just use calculated estimate for |B|.Also by combining the bounds for |A| and |B| we get an estimate for the height H(L): This completes the proof of Proposition 1. ⊠ Before we start the proof of Proposition 2 let's establish some basic facts regarding the point of intersection of two lines . These facts will be of use in further discussion as well.An intersection L 1 ∩ L 2 is a rational point P (p, r, q) which is the solution of the following system of equations which leads to the following equalities Therefore we get that where Proof of Proposition 2. By (13) an upper bound for q is given by An upper bound for |qθ − p| can be derived from ( 14) and the assumption L 2 ∩ L θ ∈ ∆(P, δ): Finally we get the required bounds and To get the last inclusion in (11) we just use calculated bound for q.This completes the proof of Proposition 2. ⊠ As we will see the duality between points and lines will appear throughout the whole paper.
3 Proof of Theorem 2 In other words, for any positive integer q the following inequality is satisfied Let R e 9 c −1 (θ) be an integer.Choose constants c and c 1 sufficiently small such that they satisfy the following inequalities and Finally choose the parameter where The goal is to construct a (I, R, r) Cantor type set K c with properly chosen parameters I, R and r so that K c is a subset of Mad P (f, c, Q) ∩ Mad L (f, c, Q).Then we use Theorem BV4 to estimate its Hausdorff dimension.Let I be any interval of length c 1 contained within the unit interval {θ} × [0, 1] ⊂ L θ .Define J 0 := {I}.We are going to construct, by induction on n, a collection J n of closed intervals J n such that J n is nested in J n−1 ; that is, each interval J n in J n is contained in some interval J n−1 in J n−1 .The length of an interval J n will be given by Moreover, each interval J n in J n will satisfy the conditions that and In particular, we put By construction, conditions (19) and (20) ensure that The aim of the rest of the paper is to show that K c is in fact a (I, R, r) Cantor set with R = (R n ) given by and r = (r m,n ) given by Then Theorem 2 will follow from Theorem BV4.Indeed for n < 3 the condition (8) is obviously satisfied.For n 3 and R 2 7 we have that the l.h.s. of (8 R n 4 = r.h.s. of (8) .
Therefore Theorem BV4 implies that which completes the proof of Theorem 2.
3.2 Basic construction.Splitting into collections C P (n, l, k) and C L (n, l, k) Now we will describe the procedure of constructing the collections J n .For n = 0, we trivially have that ( 19), (20) are satisfied for the sole interval I ∈ J 0 .The point is that by the choice of Q there are neither points nor lines satisfying the height condition Q < H(P ), H(L) < c(θ).
Then we construct J i , i = 1, 2, 3 by just subdividing each intervals of equal length.Again for the same reason the conditions ( 19) and (20) are satisfied for any J i ∈ J i , i = 1, 2, 3. Note that In general, given J n satisfying ( 19) and (20) we wish to construct a nested collection J n+1 of intervals J n+1 for which (19) and (20) are satisfied with n replaced by n + 1.By definition, any interval J n in J n avoids intervals ∆(P, cf −1 (q)) and ∆(L, cf −1 (|A| * |B| * )) arising from points and lines with height bounded above by c(θ)R n−1 F (n − 1).Since any 'new' interval J n+1 is to be nested in some J n , it is enough to show that J n+1 avoids intervals ∆(P, cf −1 (q)) and ∆(L, cf −1 (|A| * |B| * )) arising from points and lines with height satisfying Denote by C P (n) the collection of all rational points satisfying this height condition.Formally and it is precisely this collection of rationals that comes into play when constructing J n+1 from J n .By analogy for 'lines' let We now proceed with the construction.Assume that n 3. We subdivide each J n in J n into R n = [R(n + 1) log * (n + 1)] closed intervals I n+1 of length Denote by I n+1 the collection of such intervals.In view of the nested requirement, the collection J n+1 which we are attempting to construct will be a sub-collection of I n+1 .In other words, the intervals I n+1 represent possible candidates for J n+1 .The goal now is simple -it is to remove those 'bad' intervals I n+1 from I n+1 for which or So we define Consider the rational point P (p, r, q) ∈ C P (n).Note that since q 2 q 2 ||qθ|| = H(q) cR n−1 F (n − 1), we have that for sufficiently large R. We use Stirling formula to show that for n 3, Therefore the left hand side of (26) is bigger than Note that for any line L(A, B, C) ∈ C L (n) we have the analogous bound For l ∈ Z we split C P (n) into sub-collections In view of (23) we have that where c 3 := (log R + 2)/ log 2 is an absolute constant independent on n and l.
Additionally with k ∈ Z we split the collection C P (n, l) into sub-collections C ′ P (n, l, k) such that Take any P (p, r, q) ∈ C ′ P (n, l, k).In view of (16) the value k should be nonnegative.On the other hand one can get an upper bound for k by (23): The upshot is that for fixed n, l the number of classes C ′ P (n, l, k) is at most c 3 n log * n.Note that within the collection C ′ P (n, l, k) we have very sharp control of the height H(P ).Then by ( 28) and (31) we also have very sharp control on the value q as well, namely Concerning the collection C L (n) we also partition it into sub-collections.Firstly we partition it into sub-collections C L (n, l) such that Then we split One can check that l and k satisfy the same conditions ( 30) and (32) as in the case of points.
Note that within each collection we have very good control of all point and line parameters.
The procedure of removing "bad" intervals from I n+1 will be as follows.We will firstly remove all intervals I n+1 ∈ I n+1 such that there exists a point P ∈ C P (n, 0) which satisfy I n+1 ∩ ∆(P, cf −1 (q)) = ∅ or there exists a line a line L ∈ C L (n, 0) which satisfy I n+1 ∩ ∆(L, cf −1 (|A| * |B| * )) = ∅.Then we repeat this removing procedure for collections C P (n, 1) and C L (n, 1), . . ., C P (n, c 2 log * n) and C L (n, c 2 log * n) in exactly this order.
We will use lexicographical order for pairs in Z 2 .That is, we say that (a, b) then such a point will not remove anything more than was removed by a line L. Therefore such a point can be ignored.The same is true if there exists a point P ′ (p ′ , r ′ , q ′ ) ∈ C P (n ′ , l ′ ) such that H(P ′ ) < H(P ) and ∆(P, cf −1 (q)) ⊂ ∆(P ′ , cf −1 (q ′ )).
Therefore instead of collection C ′ P (n, l, k) we can work with By the same procedure we construct the collection C L (n, l, k) from C ′ L (n, l, k).Note that by the construction of C P (n, l, k) there exists at most one point P (p, r, q) ∈ C P (n, l, k) with given second coordinate r/q.

Blocks of intervals B P (J) and B L (J)
Take the maximal possible constant c 2 > 0 such that Fix the triple (n, l, k) and consider an arbitrary interval . Then for any P (p, r, q) ∈ C P (n, l, k) we have |∆(P, cf −1 (q))| < |J|.Indeed this is true because The last inequality is true provided c 2 c(θ) 2c which in turn is true by the second inequality of ( 17) and (36).One can easily check that the same fact is true for any ∆(L, . Then all rational points P (p, r, q) ∈ C P (n, l, k) such that ∆(P, cf −1 (q)) ∩ J = ∅ lie on a single line.
Proof.Consider an arbitrary point P (p, r, q) ∈ C P (n, l, k).Then Suppose we have three points P i (p i , r i , q i ) ∈ C P (n, l, k), i = 1, 2, 3 such that ∆(P i , cf −1 (q i )) ∩ J = ∅ and they do not lie on a single line.Then they form a triangle which has the area at least area(△P 1 P 2 P 3 ) 1 2q 1 q 2 q 3 (33) .
On the other hand the first coordinates p i /q i of the points P i should satisfy (37) and their second coordinates r i /q i should lie within the interval of length |J| + |∆(P i , cf −1 (q i ))| 2|J|.Therefore we have the following upper bound for the area of triangle △P 1 P 2 P 3 : .
Finally by (36) we get that the last value is bounded above by which is impossible.So we get a contradiction.⊠ So given interval J of length c 2 2 −l R n−1 F −1 (n − 1) if we have at least two points P ∈ C P (n, l, k) as in Lemma 1 then all the points with such property will lie on a single line L. We denote this line by L J .If there is at most one point P ∈ C p (n, l, k) as in Lemma 1 then we just say that L J is undefined.
Note that L J can not be horizontal because by the construction of C P (n, l, k) there is only one point P (p, r, q) ∈ C P (n, l, k) with given second coordinate r/q.L J can not be vertical too.Otherwise its equation can be written as x = C/A, gcd(A, C) = 1.Then by the construction of θ we have that which together with (37) gives us Then by defitnition of L J there exist two points P 1 (p 1 , r 1 , q 1 ), P 2 (p 2 , r 2 , q 2 ) with |r 1 /q 1 − r 2 /q 2 | < 2|J|.However So we get a contradiction.The statement of Lemma 1 can be strengthened if we have more than two points P ∈ C P (n, l, k) such that ∆(P, cf −1 (q)) ∩ J = ∅.
Lemma 2 Let J be an interval on L θ of length |J| = c 2 2 −l R −n+1 F −1 (n − 1).Assume that there exists a line L J .Consider the sequence of consecutive intervals M i ⊂ L θ , i ∈ N, |M i | = |J|, M 1 := J and bottom end of M i coincides with the top end of M i+1 .Define the set P(J, m) := P ∈ C P (n, l, k) : P ∈ L J and ∆(P, cf −1 (q)) ∩ Then all rational points P ∈ C P (n, l, k) such that Remark 1.Since the number of points P ∈ C P (n, l, k), P ∈ L J is finite, the value m P (J) is correctly defined.Indeed since by assumption #P(J, 1) 2, m + 1 → ∞ and #P(J, m) is bounded then m(J) exists and is finite.
Remark 2. We define the block of intervals We will work with it as with one unit.If for some interval J the line L J is undefined then we define m(J) := 1 and B P (J) := J.So now m(J) and B P (J) are well defined for all intervals J of length c 2 2 −l R −n+1 F −1 (n − 1).
Proof.Is similar to the proof of Lemma 1.Let P(J, m(J)) = (P i (p i , r i , q i )) where the sequence r i /q i is ordered in ascending order.Assume that there is a point P (p, r, q) ∈ C P (n, l, k) such that P ∈ L J and ∆(P, cf −1 (q)) ∩ B P (J) = ∅.Then the triangle ∆(P P 1 P m(J)+1 ) is splitted into m P (J) disjoint triangles ∆(P P i P i+1 ), 1 i m P (J) each of which has the area area(∆(P On the other hand the first coordinates of the points P 1 , . . ., P m P (J)+1 and P satisfy (37) and their second coordinates lie within the interval of length at most (m P (J) + 1)|J|.Therefore we have the following estimate for the area of the triangle m P (J) 2 3l−3k+4 R 3(n−1) F 3 (n − 1) area(△P P 1 P m P (J)+1 ) 2 9 (m P (J) + 1)c 2 c(θ) .
which is impossible since the l.h.s of this inequality is bigger than its r.h.s.⊠ Lemmas 1 and 2 have their full analogues for lines L ∈ C L (n, l, k).However the proofs areslightly different.We will formulate them in the next two lemmata.
Proof.We will use the following well-known fact.Let us have three planar lines L i (A i , B i , C i ), i = 1, 2, 3 defined by equations A i x − B i y + C i = 0. Then they intersect in one point (probably at infinity) if and only if det Suppose that there are three lines L 1 , L 2 , L 3 ∈ C L (n, l, k) which do not intersect at one point but their thickenings intersect J. Then det On the other hand we can make a vertical shifts of L 1 , L 2 , L 3 to the distances 3 such that they will intersect at one point on J.By vertically shifting a line to the distance ǫ we change its C-coordinate by the value Bǫ.Therefore we have det However the latter determinant is bounded above by < 1 We get a contradiction.⊠ So given interval J of length c 2 2 −l R n−1 F −1 (n − 1) if we have at least two lines from C L (n, l, k) as in Lemma 3 then all lines with such property will intersect at one rational point P .We denote this point by P J .If there is at most one line from C L (n, l, k) as in Lemma 3 then we just say that P J is undefined.
The next Lemma is a "line" analogue of Lemma 2. Note that for any line L(A, B, C) ∈ C L (n, l, k) which go through P M (p, r, q) the distance can differ by factor at most 16 from the same distance for line L 2 .Therefore for all L ∈ L M , However since L M ⊂ C L (n, l, k) we get by the construction of C L (n, l, k) that L M has to be empty -a contradiction.Therefore the Case 2L is impossible.Now consider the Case 3L.Let's order all the lines in in such a way that the sequence of the second coordinates of L i ∩ L θ is increasing.Then we have On the other hand by ( 14) and ( 35) the smallest difference between two consecutive By combining the upper and lower bounds for

Final step of the proof
Let n 3. Fix an interval J n−3 ∈ J n−3 .We will firstly estimate the quantity such that the bottom endpoint of M i coincides with the top endpoint of M i+1 (1 i K − 1).We start by constructing blocks from intervals M 1 , . . ., M K .Define B 1 := B P (M n 1 ), B 2 := B P (M n 2 ), . . ., B t := B P (M nt ) in such a way that n 1 := 1 and the bottom endpoint of B P (M n i ) coincides with the top endpoint of B P (M n i+1 ).By Lemma 2 for any 1 i < t we have 45) Now let's consider the last block B t .The problem is that this block is not necessarily included in J n−3 so we need to treat it independently.As it was discussed in Section 3.4, we have two possible cases.In Case 1P we have that for any i n t #{P (p, r, q) ∈ C P (n, l, k) : ∆(P, cf −1 (q)) ∩ M i = ∅} 2 2 .
Now estimate the number of intervals I n+1 ∈ I n+1 which are removed by ∆(P, cf −1 (q)) where P is some interval from C P (n, l, k).The upshot is that for any interval J n−3 ∈ J n−3 the number of 'bad' intervals I n+1 ∈ I n+1 which are to be removed is bounded by r n−3,n .Therefore the desired set K c is indeed a (I, R, r) Cantor type set.The proof is complete.Regarding the sets of the form Mad P (f ) ∩ Mad L (f ) ∩ L θ , Theorem BV5 enables us to show that for any finite family θ 1 , . . ., θ n of badly approximable numbers one can find α ∈ R such the following inclusion holds simultaneously for all 1 i n: (α, θ i ) ∈ Mad P (f ) ∩ Mad L (f ).
Moreover the set of such numbers α is of full Hausdorff dimension.The proof is based on intersecting the corresponding Cantor type sets K c (i) associated with each set Mad P (f, c) ∩ Mad L (f, c) ∩ L θ i for c sufficiently small and then on applying Theorem BV4 to the intersection.We will leave the details to the enthusiastic reader.
We also believe that the same fact will be true for countable collection {θ i } of badly approximable numbers.However it can not be proven with existing technique.

Lemma 4
Let J be an interval on L θ of length |J| = c 2 2 −l R −n+1 F −1 (n − 1).Assume that there exists a point P J .Consider the sequence of consecutive intervalsM i ⊂ L θ , i ∈ N, |M i | = |J|,M 1 := J and bottom end of M i coincides with the top end of M i+1 .Define the set L(J, m) := L ∈ C L (n, l, k) : P J ∈ L and ∆(L, cf −1 (|A| * |B| * )) ∩ m i=1 M i = ∅ and the value m L (J) := max{m ∈ N | #L(J, m) m + 1}.

4
Final remark.In the proof of Theorem 2 we showed that Mad P (f, c) ∩ Mad L (f, c) ∩ L θ contains (I, R, r) Cantor type set.It allows us to use Theorem 5 from [2]: Theorem (BV5) For each integer 1 i k, suppose we are given a Cantor set K(I, R, r i ).Then k i=1 K(I, R, r i ) is a (I, R, r) Cantor set where r := (r m,n ) with r m,n := k i=1 r (i) m,n .