Density modulo 1 of sublacunary sequences: application of Peres-Schlag's arguments
aa r X i v : . [ m a t h . N T ] O c t Density modulo 1 of subla unary sequen es:appli ation of Peres-S hlag's argumentsMosh hevitin N.G. 1Abstra t. Let the sequen e { t n } ∞ n =1 of reals satisfy the ondition t n +1 t n > γn β , β < , γ > . Then the set { α ∈ [0 ,
1] : ∃ κ > ∀ n ∈ N || t n α || > κ n β log( n +1) } is un ountable. Moreover itsHausdor(cid:27) dimension is equal to 1. Consider the set of naturals of the form n m and let the sequen e s =1 , s =2 , s =3 , s =4 , s =6 , s = 8 , . . . performs this set as an in reasing sequen e. Then the set { α ∈ [0 ,
1] : ∃ κ > ∀ n ∈ N || s n α || > κ √ n log( n +1) } also has Hausdor(cid:27) dimension equal to 1. Theresults obtained use an original approa h due to Y. Peres and W. S hlag.1. Introdu tion. A sequen e { t j } , j = 1 , , , .. of positive real numbers is de(cid:28)ned to bela unary if for some M > one has t j +1 t j > M , ∀ j ∈ N . Erd(cid:4)os [1℄ onje tured that for any la unary sequen e there exists real α su h that the set of fra tionalparts { αt j } , j ∈ N is not dense in [0 , . This onje ture was proved by A. Pollington [2℄ and B. deMathan [3℄. Some quantitative improvements were due to Y. Katznelson [4℄, R. Akhunzhanov andN. Mosh hevitin [5℄ and A. Dubi kas [6℄. The best known quantitative estimate is due to Y. Peresand W. S hlag [7℄. The last authors proved that with some positive onstant γ > for any sequen e { t j } under onsideration there exists a real number α su h that || αt j || > γM log M , ∀ j ∈ N . Y. Peres and W. S hlag use an original approa h onne ted with the Lovasz lo al lemma.From another hand R. Akhunzhanov and N. Mosh hevitin in [8℄ generalized Pollington - deMathan's result to subla unary sequen es. For example for a sequen e { t j } under ondition t j +1 t j > γn β , ∀ j ∈ N , γ > , β ∈ (0 , / they proved the existen e of real irrational α su h that lim inf n →∞ (cid:0) || t n α || × n β (cid:1) > . Another appli ation from [8℄ deals with the sequen e of naturals of the form m n , m, n ∈ N ∪ { } .In the present paper we apply the arguments from [7℄ to improve the results from [8℄ mentionedabove.1 Resear h is supported by grants RFFI 06-01-00518, MD-3003.2006.1, NSh-1312.2006.1 and INTAS 03-51-50701. Results. Let t < t < ...t n < t n +1 < ... be a stri tly in reasing sequen e of reals and lim n →∞ t n = + ∞ . For a given sequen e { t n } we de(cid:28)ne the fun tion H ( n, τ ) = min (cid:26) k ∈ N : t n + k t n > τ (cid:27) . (1)Theorem 1. Let < η < . Consider a sequen e { h ( n ) } ∞ n =1 ⊂ N of natural numbers su h thatfor all natural n under ondition n > h ( n ) the fun tion n n − h ( n ) is in reasing and a de reasingsequen e { δ ( n ) } ∞ n =1 of positive real numbers. Let the sequen e { n k } Kk =0 of natural numbers is de(cid:28)nedto satisfy the ondition n k = n k +1 − h ( n k +1 ) (2)for k K − . Let our sequen es satisfy the following onditions (i),(ii) and (iii) below.(i) For any natural n under ondition n > h ( n ) the following inequality is valid h ( n ) > H ( n − h ( n ) , /δ ( n − h ( n ))) . (ii) For any k K − the following inequality is valid n k +1 − X v = n k +1 δ ( v ) (1 − η ) η . (iii) For k = 0 the following inequality is valid n X v =1 δ ( v ) − η . Then for the set A K = { α ∈ [0 ,
1] : || t n α || > δ ( n ) ∀ n n K } one has µ ( A K ) > η K +1 . Here µ ( · ) denotes the Lebesgue measure. Note that the sets A K are losed and nested: A K +1 ⊆ A K . Moreover if we have a natural number N we an onstru t a sequen e { n k } su h that n K = N , theequalities (2) are satis(cid:28)ed, n = n − h ( n ) > but n − h ( n ) . Hen e as a orollary of Theorem1 we immediately obtainTheorem 2. Let < η < . Consider a sequen e { h ( n ) } ∞ n =1 ⊂ N of natural numbers su h thatfor all natural n under ondition n > h ( n ) the fun tion n n − h ( n ) is in reasing and a de reasingsequen e { δ ( n ) } ∞ n =1 of positive reals. Let these sequen es satisfy the following onditions (i) fromTheorem 1 and the onditions (ii ′ ) (iii ′ ) below.(ii ′ ) For all natural numbers n under ondition n > h ( n ) the following inequality is valid n − X v = n − h ( n )+1 δ ( v ) (1 − η ) η . (iii ′ ) For all natural numbers n under ondition n h ( n ) the following inequality is valid n X v =1 δ ( v ) − η . A = { α ∈ [0 ,
1] : || t n α || > δ ( n ) ∀ n ∈ N } is nonempty.Theorem 3. Let the
onditions of theorem 2 be satis(cid:28)ed and an in(cid:28)nite sequen
e { n k } ∞ k =0 ofnaturals satis(cid:28)es the
ondition (2) for all natural k . Let the series ∞ X k =1 η k · (cid:18) t n k δ ( n k ) (cid:19) ν / (cid:18) t n k − δ ( n k − ) (cid:19) (3)
onverges for all ν < ν Then the set A from Theorem 2 has Hausdor(cid:27) dimension > ν .We give a
omplete proof of theorem 1 in Se
tions 3,4. In Se
tion 5 we give
omments to theproof of Theorem 3. In se
tion 6 we give some appli
ations of our results.4. Lemmata. For n > we de(cid:28)ne l n = (cid:22) log (cid:18) t n δ ( n ) (cid:19)(cid:23) . (4)From monotoni
ity of t n and δ ( n ) it follows that l n +1 > l n . Put E ( n, a ) = (cid:20) at n − δ ( n ) t n , at n + δ ( n ) t n (cid:21) Let A n be the union of dyadi
intervals of the form (cid:18) b l n , b + ε l n (cid:19) , b ∈ Z , ε ∈ { , } whi
h
overs the set [ a ⌈ t n ⌉ E ( n, a ) \ [0 , . So [ a ⌈ t n ⌉ E ( n, a ) \ [0 , ⊆ A n . De(cid:28)ne A cn = [0 , \ A n . Note that µ ( A n ) ( ⌈ t n ⌉ + 1) 2 δ ( n ) t n δ ( n ) and µ \ n n A cn ! > − n X n =1 δ ( n ) . (5)Lemma 1. Let n > h ( n ) . Let the
ondition (i) holds and µ \ j n − h ( n ) A cj > . Then µ \ j n − h ( n ) A cj \ A n δ ( n ) µ \ j n − h ( n ) A cj . (6)3roof. The set T j n − h ( n ) A cj an be
onsidered as a union \ j n − h ( n ) A cj = T [ ν =1 I ν (7)of the dyadi
intervals I ν = I ( n − h ( n )) ν of the form (cid:20) b l n − h ( n ) , b + 12 l n − h ( n ) (cid:21) , b ∈ Z where T > . Now the set A n ∩ I ν an be represented as a union A n \ I ν = W ν [ i =1 J i of intervals J i of the form (cid:20) b l n ) , b + 12 l n (cid:21) . Moreover W ν (cid:22)(cid:18) l n − h ( n ) + δ ( n )2 l n (cid:19) t n (cid:23) + 1 t n l n − h ( n ) + 2 . So µ ( A n ∩ I ν ) = W ν l n and µ \ j n − h ( n ) A cj \ A n T l n (cid:18) t n l n − h ( n ) + 2 (cid:19) = µ \ j n − h ( n ) A cj l n − h ( n ) l n (cid:18) t n l n − h ( n ) + 2 (cid:19) == µ \ j n − h ( n ) A cj (cid:18) t n l n + 2 · l n − h ( n ) l n (cid:19) . But t n l n δ ( n ) (8)from the de(cid:28)nition of l n (formula (4)). For the se
ond summand we have l n − h ( n ) l n · t n − h ( n ) t n · δ ( n ) δ ( n − h ( n )) δ ( n ) (9)from the
ondition (i) and the de(cid:28)nition (1) of the fun
tion H ( · , · ) .Now Lemma 1 follows from (8,9).For (cid:28)xed τ and v h ( τ ) de(cid:28)ne τ v = τ − h ( τ ) + v . Note that τ h ( τ ) = τ and τ = τ − h ( τ ) .Note that τ τ v τ. Lemma 2. Let the fun
tion n − h ( n ) is in
reasing and the
ondition (i) holds. Let for τ > h ( τ ) the following inequality is valid: µ \ j τ A cj ! > ηµ \ j τ − h ( τ ) A cj > (10)4ith some positive η .Then we have µ \ j τ A cj ! > − η τ − X v = τ δ ( v ) ! × µ \ j τ A cj ! . (11)Proof.We have µ \ j τ A cj ! = µ · · · \ j τ − h ( τ ) A cj \ A τ − h ( τ )+1 \ · · · \ A τ >> µ \ j τ − h ( τ ) A cj − h ( τ ) X v =1 µ A τ v \ \ j τ − h ( τ ) A cj . But as τ v τ from the monotoni
ity
ondition for n − h ( n ) we get τ − h ( τ ) > τ v − h ( τ v ) so \ j τ − h ( τ ) A cj ⊆ \ j τ v − h ( τ v ) A cj . (12)Now µ \ j τ A cj ! > µ \ j τ − h ( τ ) A cj − h ( τ ) X v =1 µ A τ v \ \ j τ v − h ( τ v ) A cj . We apply Lemma 1 for n = τ v , v = 1 , ..., h ( τ ) (it is possible as from (12) and µ (cid:16)T j τ − h ( τ ) A cj (cid:17) > it follows that µ (cid:16)T j τ v − h ( τ v ) A cj (cid:17) > for all v ) and obtain the inequality µ A τ v \ \ j τ v − h ( τ v ) A cj δ ( τ v ) µ \ j τ v − h ( τ v ) A cj . Now µ \ j τ A cj ! > µ \ j τ − h ( τ ) A cj − h ( τ ) X v =1 δ ( τ v ) × max v
4. Sket hed proof of of Theorem 3. In order to prove Theorem 3 one must do the following.In the proof of Theorem 1 instead of the inequality (6) of Lemma 1 one should prove µ (cid:16) I ( n − h ( n )) ν \ A n (cid:17) δ ( n ) µ (cid:0) I ( n − h ( n )) ν (cid:1) , where I ( n − h ( n )) ν is from partition (7). Then under the ondition µ (cid:16) I ( τ − h ( τ )) ν ′ ∩ A τ (cid:17) > ηµ (cid:0) I ( τ − h ( τ )) ν (cid:1) > one should prove instead of the inequality (11) of Lemma 2 the following inequality: µ I ( τ ) ν \ \ j τ A cj !! > − η τ − X v = τ δ ( v ) ! × µ (cid:0) I ( τ ) ν (cid:1) . It means that in ea h interval of the form I ( τ ) ν there exist not less than N = µ (cid:16) I ( τ ) ν T (cid:16)T j τ A cj (cid:17)(cid:17) µ (cid:16) I ( τ ) ν ′ (cid:17) > η l τ − l τ pairwise disjoint subintervals of the form I ( τ ) ν ′ . Then as in [8℄ one should take into a ount the onvergen e of (3) and apply the following well-known result:Theorem (Eggleston [9℄). Let for every k we have a set A k = R k F i =1 I k ( i ) where I k ( i ) are segmentsof real line of length | I k ( i ) | = ∆ k . Let ea h interval I k ( i ) has exa tly N k +1 > pairwise disjointsubintervals I k +1 ( i ′ ) of length ∆ k +1 from the set A k +1 . Let R k +1 = R k · N k +1 . Suppose <ν and forevery <ν<ν the series P ∞ k =2 ∆ k − ∆ k ( R k (∆ k ) ν ) − onverges. Then the set A = T ∞ k =1 A k has Hausdor(cid:27)dimension HD( A ) > ν .6. Examples. Note that the proof of Theorem 1 follows dire tly the arguments by Y.Peres andW. S hlag from [7℄. The author in [10℄ (following Peres-S hlag's arguments) established for la unarysequen e { t n } under ondition t j +1 t j > M , ∀ j ∈ N . the existen e of a real number α su h that || αt j || > M log M , ∀ j ∈ N . . We onsider some examples with subla unary sequen es below.A. Subla unary sequen es. Let { t n } ∞ n =1 satisfy the ondition t n +1 t n > γn β , β < , γ > . (13)6e take η < lose to 1 and h ( n ) = ⌊ c n β log( n + c ) ⌋ , δ ( n ) = (1 − β )(1 − η ) η c ( n + c ) β log( n + c ) , (14)Here large positive onstants c , c (depending on β and η ) should be de(cid:28)ned in the following way.In our situation under ondition n > h ( n ) for γ < γ one has t n t n − h ( n ) > n − Y j = n − h ( n ) (cid:18) γj β (cid:19) > exp n − X j = n − h ( n ) log (cid:18) γj β (cid:19) > exp (cid:18) ω h ( n ) n β (cid:19) > ( n + c ) ωc with ω = ω ( β, γ ) . Let c = c ( β, η ) be a large positive onstant su h that for all real y > we have y ωc > c y β log y (1 − β )(1 − η ) η . Then t n t n − h ( n ) > ( n + c ) ωc > c ( n + c ) β log( n + c )(1 − β )(1 − η ) η = 1 δ ( n ) > δ ( n − h ( n )) and the ondition (i ′ ) of Theorem 2 is satis(cid:28)ed.So we have c (cid:28)xed and then we de(cid:28)ne c . Let c = c ( β ) be a large positive onstant su h that max n ∈ N c log( n + c )( n + c ) − β , (15) h (cid:18) δ (0) (cid:19) = c c c β log c (1 − β )(1 − η ) η ! β log c c β log c − β + c ! δ (0) = c c β log c (1 − β )(1 − η ) η . (16) min y > (cid:18) (1 − β ) log( y + c ) − yy + c (cid:19) > (17)Then from (15) it follows that h ( n ) n + c and for n > h ( n ) we have n − X v = n − h ( n )+1 δ ( v ) (1 − β )(1 − η ) η c log( n − h ( n ) + c ) n − X v = n − h ( n )+1 v β (1 − η ) η × n − β − ( n − h ( n )) − β c log( n − h ( n ) + c ) (1 − η ) ηh ( n )2 c n β log( n − h ( n ) + c ) (1 − η ) η log( n + c )2 log( n − h ( n ) + c ) == (1 − η ) η × log( n + c )log( n + c ) + log(1 − h ( n ) n + c ) (1 − η ) η × log( n + c )log( n + c ) − log 2 (1 − η ) η . So the ondition (ii ′ ) of Theorem 2 is satis(cid:28)ed.Moreover for the value n = n ( β, c , c ) = max { n ∈ N : n h ( n ) } from (16) it follows that n δ (0) and the ondition (iii ′ ) of Theorem 2 is satis(cid:28)ed also.Also we must note that if y > and y > h ( y ) > c y β log( y + c ) then the fun tion y − c y β log( y + c ) is in reasing as from (17) it follows that ( y − c y β log( y + c )) ′ = 1 − βc y β − log( y + c ) − c y β y + c = c y β − (cid:18) (1 − β ) log( y + c ) − yy + c (cid:19) > . B = { α ∈ [0 ,
1] : ∃ κ > ∀ n ∈ N || t n α || > κ n β log( n + 1) } is nonempty (obviously, un ountable).Note that the set { n ∈ N : n h ( n ) } is (cid:28)nite. Hen e we an onstru t a sequen e of naturals { n k } satisfying (2).If it happens that in addition to (13) we have t n +1 t n γ n β (18)with some γ > γ then for the sequen e { n k } we get t n k t n k − n γ and k γ n − βk with positive γ , . Now η k · t νn k t n k − ≪ e γ n − βk · η k ≪ e γ η γ ) n − β (here all onstants γ j do not depend on η ) and for η lose to 1 the series (3) onverges. From Theorem3 it follows that the set B has Hausdor(cid:27) dimension equal to 1. We should note that it is possible to hoose fun tion h ( n ) (a tually in the same manner as it was done in [8℄) to satisfy the onditions ofTheorem 3 without additional assumption (18) on the rate of growth of the sequen e t n .We should note that it would be interesting to investigate winning properties of the onsideredsets (for the de(cid:28)nition of winning sets see [11℄,[12℄, for some partial results see [13℄).B. Subexponentional sequen es. Let { t n } ∞ n =1 satisfy the ondition γ exp( n β ) t n γ exp( n β ) , < β < , γ , > . (19)Then by the same reasons (as in example A) we have that the Hausdor(cid:27) dimension of the set { α ∈ [0 ,
1] : ∃ κ > ∀ n ∈ N || t n α || > κ n − β log( n + 1) } is equal to 1.C. F(cid:4)urstenberg's sequen e. Consider the set of naturals of the form n m and let thesequen e s =1 , s =2 , s =3 , s =4 , s =6 , s = 8 , . . . performs this set as an in reasing sequen e. F(cid:4)urstenberg [14℄ (see also [15℄) proved that for anyirrational α the set of fra tional parts { n m α } is dense in [0 , . Hen e lim inf n →∞ || s n α || =0 . We should note that we no nothing about the rate of onvergen e to zero here. Obviously for α = 1 / one has || s n / || > / . But / is a rational number.The sequen e { s n } satisfy (19) with β = 1 / . So from example B it follows that Hausdor(cid:27)dimension of the set { α ∈ [0 ,