On Harrap s conjecture in Diophantine approximation

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1 On Harrap s conjecture in Diophantine approximation by Nikolay Moshchevitin Abstract. We prove a conjecture due to Stephen Harrap on inhomogeneous linear Diophantine approximation related to BADαβ sets.. The result. Let α β 0 α + β =. Let denote the distance to the nearest integer. Consider the set arxiv: v [math.nt] Apr 202 BADαβ = {θ θ 2 R 2 : inf q Z + maxq α qθ q β qθ 2 > 0}. For Θ = θ θ 2 R 2 consider the set BAD Θ αβ = {η η 2 R 2 : inf q Z + maxq α qθ η q β qθ 2 η > 0}. In [2] it was proved that if Θ BADαβ then the set BAD Θ αβ has full Hausdorff dimension in R 2. Moreover it was noted there that in this case it is possible to establish the winning property of the set BAD Θ αβ. It was conjectured in [2] that the set BAD Θ αβ should be a set of full Hausdorff dimension without the additional assumption Θ BADαβ. In the case α = β = /2 this is true it follows from Khintchine a and Jarnik s approach [3 4]. In the present note we give a solution to the problem fro. For simplicity reason we restrict ourselves by the case α = 2/3β = /3. Theorem. For any Θ the set BAD Θ 2/3/3 is non-empty. Moreover it has full Hausdorff dimension. 2. Best approximations. We may suppose that θ θ 2 are linearly independent over Z. Otherwise the result is obvious. We define m = m m 2 Z 2 \{0} to be a best approximation vector if in the parallelepiped {x 0 x x 2 R 3 : max x x 2 2 max m m 2 2 x 0 +θ x +θ 2 x 2 θ m +θ 2 m 2 } there is no integer points different from the points 000±m 0.m m 2 where m 0 Z is defined from θ m +θ 2 m 2 = m 0 +θ m +θ 2 m 2. All the best approximation vectors should be arranged in the infinite sequence m ν = m ν. m ν2 ν = M ν = max m ν m µ.2 2 form an increasing sequence M ν+ > M ν and the values ζ ν = θ m ν +θ 2 m 2ν form a decreasing sequence ζ ν+ < ζ ν. From the Minkowski convex body theorem it follows that ζ ν Mν+ 3. research is supported by RFBR grant No a and by the grant of Russian Government project. G

2 Note that if m ν is a best approximation vector then in any parallelepiped of the form {x 0 x x 2 R 3 : max2 x ξ 4 x 2 ξ 2 2 M 2 ν x 0 +θ x +θ 2 x 2 ξ 0 ζ ν 2 } 2 with ξ 0 ξ ξ 2 R 3 there exists not more than one integer point. As the set {x 0 x x 2 R 3 : max x x 2 2 2M ν 2 x 0 +θ x +θ 2 x 2 ζ ν }\ {x 0 x x 2 R 3 : max x x 2 2 M 2 ν x 0 +θ x +θ 2 x 2 ζ ν } may be partitioned into 2 28 sets of the form 2 and the lattice Z 3 is 0-symmetric we see that M ν+28 2M ν 3 for every value of ν. There are two kinds of best approximation vectors. For some best approximation vectors we have M ν = max m ν m µ.2 2 = m ν. 4 For other best approximation vectors we have M ν = max m ν m µ.2 2 = m ν2. 5 All the best approximation vectors satisfying 4 form the sequence m [] ν = m [] ν.m [] ν2 ν = M ν [] = max m [] ν m[] µ.2 2 = m [] ν form an increasing sequence. All the best approximation vectors satisfying 5 form the sequence ν = ν. ν2 ν = M ν [2] = max ν µ.2 2 = ν2 form an increasing sequence. From 3 we see that M [] ν+28 2M[] ν M[2] ν+28 2M[2] ν Linear form bounded from zero. In this section we prove that there exists a vector η = η η 2 satisfying inf ν η m ν +η 2 m ν2 > 0. 7 Standard transference argument see [] Chapter V in view of shows that for such η we have η BAD Θ 2/3/3. Moreover standard argument with resonance sets see [2 5] shows that the set of η satisfying 7 is a set of full Hausdorff dimension. 2

3 To get 7 it is enough to show that Let R be a large positive integer. Put inf η m [j] ν +η 2m [j] ν2 > 0 j = 2. 8 ν δ = R3. 9 We will discribe the inductive process. Its base is trivial. Suppose that for a positive integer n we have a rectangle B R 2 of a form [ B = b b + δ ] [ b R 2n 2 b 2 + δ ]. R n We suppose that for all η B and for all best approximation vectors and for all best approximation vectors m [] ν M [] ν R n ν M [2] ν R n we have η m [j] ν +η 2 m [j] ν2 > ε with ε = δ R. 0 By dividing the segments [ ] [ ] b b + δ R 2n b2 b 2 + δ R into R 2 and R equal parts correspondingly we n get a partition of B into R 3 smaller rectangles B of the form [ ] [ B = b b δ + b R 2b 2n+ 2 + δ ]. R n+ We should show that there exist many subrectangles of the form such that for all the points from these rectanges we have 0 for all the best approximation vectors m [] ν Rn < M [] ν R n+ 2 ν Rn < M [2] ν R n+. 3 That will be enough.. We will deal with the best approximation vectors 2. For a single vector m [] ν from 2 consider the collection of parallel lines L [] ν = l ν [] c l ν [] c = {x x 2 R 2 : m [] νx +m [] ν.2x 2 = c}. c Z Consider a rectangleb B of the form which has a point ξ ξ 2 with ξ m [] ν.+ξ 2 m [] with some c Z. Let H ν [] be the number of such subrectangles. Fix x 2 and consider the one-dimensional section with x 2 fixed. Then the point m [] ν2 c < ε ν2 x 2+c m [] ν belongs to the line l ν [] c. The section of the subrectangle B under the consideration must completely lie in the segment Jc of the form [ m [] ν2x 2 +c m [] ν ε m [] ν δ R 2n+ k[] δ R n+x 2 3 m [] ν2x 2 +c m [] ν + ε m [] ν + x 2 ] δ R + k[] δ 2n+ R n+x 2

4 where k [] = m[] ν2 m [] ν m [] ν. R n we use 4 and the lower bound from 2. For the length [] ν of such a segment Jc we have the upper bound [] ν 2ε+2δR 2 +2δR. R 2n We suppose R to be large enough so by 9 we have m [] ν [] ν > δ We see that each section of rectangle B with fixed value of x 2 can intersect not more than just one segment Jc. Now H ν [] [] ν δ/r n 2ε δ 2 /R 3n+ δ R3 +3R 2 = 5R 2. The number of vectors m [] ν satisfying 2 is 56log 2 R due to 6. We come to the following conclusion. The total number of dangerous subrectangles B B which are killed by the lines corresponding to the vectors 2 is less than ν satisfy 2 R 2n. H [] ν 600R 2 log 2 R. 2. We will deal with the best approximation vectors 3. For a single vector ν from 3 consider the collection of parallel lines L [2] ν = c Z l ν [2] c l[2] ν c = {x x 2 R 2 : ν x + ν.2 x 2 = c}. Consider a rectangleb B of the form which has a point ξ ξ 2 with ξ ν. +ξ 2 ν2 c < ε with some c Z. Let H [2] ν be the number of such subrectangles. Fix x and consider the one-dimensional section with x fixed. Then the point x m[2] ν x +c ν2 belongs to the line l ν [2] c. The section of the subrectangle B under the consideration must completely lie in the segment [ where x m[2] νx +c ν2 ε ν2 δ R k[2] δ n+ R 2n+ x m[2] νx +c ν2 k [2] = m[2] ν m[2] ν2 R n+ ν2 ] + ε ν2 + δ R + k[2] δ n+ R 2n+ we use 5 and the upper bound from 3. For the length [2] ν of such a segment we have the upper bound [2] ν 2ε+4δR. R n 4

5 So by 9 we have Now H ν [2] [2] ν2 [2] ν > δ R n. ν δ/r 2n 2ε δ 2 /R 3n+ δ R3 +4R 2 = 6R 2. The number of vectors ν satisfying 3 is 28log 2 R due to 6. We come to the following conclusion. The total number of dangerous subrectangles B B which are killed by the lines corresponding to the vectors 3 is less than ν satisfy 3 H [2] ν 400R 2 log 2 R. Combining together the couclusions from. and 2. we see that the total number of bad subrectangles is less than 000R 2 log 2 R. The whole number of subrectangles is R 3. So there exist at R 3 000R 2 log 2 R subrectangles B for which the desired property is satisfied for the n + -th step. It is enough to prove the existence of η. The full Hausdorff dimension of the set of such η s follows from the analysis of the resonance sets L [j] ν. References [] J.W.S. Cassels An introduction to Diophantine approximations Cambridge Univ. Press 957. [2] S. Harrap Twisted inhomogeneous Diophantine approximation and badly approximable sets Acta Arithmetica ; preprint available at arxiv: v4. [3] A. Khintchine Über die angenäherte Auflösung Linearer Gleichungen in ganzen Zahlen. // Acta Arithmetica [4] V. Jarník On lineárnich nehomogennich diofantických aproximacich Rozpravy II. Třidy České Akademie Ročnik LI Čislo [5] S. Kristensen R. Thorn S. Velani Diophantine approximation and badly approximable sets Adv. Math no