Decaying and non-decaying badly approximable numbers
aa r X i v : . [ m a t h . N T ] A p r DECAYING AND NON-DECAYING BADLY APPROXIMABLE NUMBERS
RYAN BRODERICK, LIOR FISHMAN, AND DAVID SIMMONS
Abstract.
We call a badly approximable number decaying if, roughly, the Lagrange constants of integermultiples of that number decay as fast as possible. In this terminology, a question of Y. Bugeaud (’15)asks to find the Hausdorff dimension of the set of decaying badly approximable numbers, and also of theset of badly approximable numbers which are not decaying. We answer both questions, showing that theHausdorff dimensions of both sets are equal to one. Part of our proof utilizes a game which combines theBanach–Mazur game and Schmidt’s game, first introduced in Fishman, Reams, and Simmons (preprint’15). Introduction
Fix d ∈ N . For each x ∈ R d , write L ( x ) = lim inf q →∞ q /d d ( q x , Z d )= sup { c ≥ k x − p /q k ≥ c/q /d for all but finitely many ( p , q ) ∈ Z d × N } , where k · k denotes some norm on R d (e.g. the max norm). The point x is called well approximable if L ( x ) = 0, and badly approximable if L ( x ) >
0. If x is badly approximable and i/j ∈ Q , we can get a trivialbound on L (cid:0) ij x (cid:1) in terms of L ( x ): if L ( x ) > c , then for all but finitely many ( p , q ) ∈ Z d × N (cid:13)(cid:13)(cid:13)(cid:13) ij x − p q (cid:13)(cid:13)(cid:13)(cid:13) = ij (cid:13)(cid:13)(cid:13)(cid:13) x − j p iq (cid:13)(cid:13)(cid:13)(cid:13) ≥ ij c ( iq ) /d = ci /d j q and thus(1.1) L (cid:18) ij x (cid:19) ≥ i /d j L ( x ) . The special case j = 1 gives the bound(1.2) L ( i x ) ≥ L ( x ) /i /d . Recently, Y. Bugeaud [3] has investigated the question of whether the bound (1.2) is “optimal”. Precisely,call a badly approximable point x decaying if there exists C > i ∈ N , L ( i x ) ≤ C/i /d . In this terminology, the main result of [3] states that certain algebraic badly approximable points aredecaying. Bugeaud also asks the following question:
Question 1.1 ([3, Problem 4.4]) . What is the Hausdorff dimension of the set of decaying badly approx-imable points? of the set of badly approximable points which are not decaying?The latter question has been recently answered in dimension 1 by a preprint of D. Badziahin and S.Harrap [1, Theorem 15]. In this paper, we give a different proof of Badziahin–Harrap’s result, which isvalid in higher dimensions as well. Our proof shows that the set of badly approximable points which are The condition “badly approximable” has been omitted in the statement of [3, Problem 4.4], but the question obviouslymakes no sense without it. In the current version of their paper they do not mention this connection; we informed them of it by private communicationwhile writing this paper. not decaying is hyperplane winning, which shows that its intersection with certain “nice” fractals has fullHausdorff dimension; see § Acknowledgements.
The second-named author was supported in part by the Simons Foundation grant
Main results.
In our first main theorem, we find a full dimension set of badly approximable pointswhich are not only non-decaying, but non-decaying when multiplied by any chosen sequence of rationalnumbers, with as little decay in the Lagrange values as desired. (The full dimension set depends on thechosen sequence and decay rate.) Precisely:
Theorem 1.2.
Let i j , i j , . . . ∈ Q be a sequence of distinct rational numbers, and let g : N → (0 , ∞ ) satisfy g ( k ) → ∞ as k → ∞ . Then the set (1.3) (cid:26) x ∈ BA d : lim sup k →∞ g ( k ) L (cid:18) i k j k x (cid:19) = ∞ (cid:27) is hyperplane winning. This theorem is quite similar to [1, Theorem 15], with three main differences: • We allow any sequence of rationals (cid:0) i k j k (cid:1) k , rather than just the sequence of all natural numbers. • We prove the theorem in any dimension, rather than just dimension 1. • We prove that the set in question is hyperplane winning, whereas Badziahin and Harrap show thatit is Cantor-winning. These concepts share many similar properties and both imply full Hausdorffdimension, but the former is stronger [1, Theorem 12] and implies strong C incompressibility onsufficiently regular fractals (see [2]).The special case i k j k = k , g ( k ) = k /d yields an answer to the latter part of Question 1.1, namely it showsthat the set of badly approximable points which are not decaying is hyperplane winning, and therefore offull Hausdorff dimension.We remark that this theorem also answers a related question. Namely, fix an irrational i/j ∈ Q andconsider the sequence i k /j k = ( i/j ) k . By (1.1) we have L (cid:0) ( i/j ) k x (cid:1) ≥ i /d j ) k L ( x ), so it is natural to askabout the size of the set of x for which(1.4) lim k →∞ ( i /d j ) k L (cid:0) ( i/j ) k x (cid:1) = ∞ . Taking i k /j k = ( i/j ) k and any g ( k ) ∈ o (cid:0) ( i /d j ) k (cid:1) , Theorem 1.2 implies that for a full-dimension set of x ∈ BA d there exists a sequence ( k n ) n such that L (cid:0) ( i/j ) k n x (cid:1) ≥ g ( k n ) for all n . Fix such an x ; applying(1.1) one then obtains L (cid:0) ( i/j ) k x (cid:1) ≥ (cid:0) g ( k n )( i /d j ) k − k n (cid:1) − for all k n ≤ k < k n +1 . It follows easily fromthis that (1.4) holds for this x . Thus, the set of x satisfying (1.4) has full Hausdorff dimension in R d . Asa concrete example, there is a full-dimension set of x ∈ R such that L (2 n x ) ∈ ω (2 − n ).On the other hand, in dimension one we show that the set of x for which L (cid:0) ij x (cid:1) decays quickly also hasfull dimension, providing a complement to both of the above results. Namely, we obtain full dimension forthe set of x for which lim k →∞ ( ij ) k L (cid:0) ( ij ) k x (cid:1) < ∞ , as well as full dimension of the set of decaying badlyapproximable numbers, via the following theorem: Theorem 1.3.
The set n x ∈ BA : L (cid:16) ij x (cid:17) ≤ ij for all reduced ij ∈ Q o has full Hausdorff dimension in R . Since the set in question is contained in the set of decaying badly approximable numbers, Theorem 1.3answers the first part of Question 1.1 by showing that the set of decaying badly approximable points hasfull dimension in R . ECAYING AND NON-DECAYING BADLY APPROXIMABLE NUMBERS 3 Games
In this section we describe two variants of Schmidt’s game which will be used in the following sections.Both games are played by two players, whom we will call Alice and Bob. The hyperplane absolute game (or just hyperplane game ) was introduced in [2] and involves a parameter 0 < β < / S ⊆ R d . Bob begins by choosing a ball B = B ( x , ρ ) ∈ R d . Inductively, if B i = B ( x i , ρ i ) havebeen chosen for i = 1 , . . . , k , Alice chooses a hyperplane L k +1 and a number 0 < ε k +1 ≤ β , determininga set A k +1 = N ( L , ε k +1 ρ k ) = { x ∈ R d : d ( x , L ) ≤ ε k +1 ρ k } . Bob then must choose a ball B k +1 = B ( x k +1 , ρ k +1 ) ⊆ B k \ A k +1 satisfying ρ k +1 = βρ k . The sets B ⊇ B ⊇ B ⊇ . . . are closed and nestedwith radii tending to zero, so the intersection B ∞ = T ∞ k =0 B k is a singleton. The unique point x suchthat B ∞ = { x } is called the result of the game. If x ∈ S , Alice is declared the winner; otherwise, Bobwins. If Alice has a strategy to win this game regardless of how Bob plays for each 0 < β < /
3, then S is called hyperplane absolute winning , or just hyperplane winning . Hyperplane winning sets intersectany sufficiently regular fractal in a set of full dimension, and the class of hyperplane winning sets isclosed under diffeomorphisms and countable intersections. Combining these three properties one obtains strong C -incompressibility of hyperplane winning sets on “nice” fractals (see [2] for precise definition anddiscussion): Theorem 2.1 ([2, Corollary 5.4]) . Let S be a hyperplane winning subset of R d , let K be the support ofan absolutely decaying and Ahlfors regular measure, and let f i : U → R d ( i ∈ N ) be nonsingular C maps.Then dim ∞ \ i =1 f − i ( S ) ∩ K ! = dim( K ) . In particular, dim( S ) = d . Many self-similar fractals are supports of absolutely decaying and Ahlfors regular measures so we obtainfor example that when S is a hyperplane winning subset of R and C is the Cantor ternary set, there is aset of dimension dim( C ) consisting of real numbers x ∈ C such that x, x , x , . . . are all in S .The Banach–Mazur–Schmidt game , or
BMS game , was introduced in [4]. It is played on a completemetric space (
X, d ) with a target set S and parameter 0 < β < B = B ( x , ρ ). Now suppose B , B , . . . , B k have been chosen and write B i = B ( x i , ρ i ).Alice then chooses a ball A k +1 = B ( y k +1 , βρ k ) satisfying d ( x k , y k +1 ) + βρ k ≤ ρ k . (Note that this implies A k +1 ⊆ B k .) Bob then chooses B k +1 = B ( x k +1 , ρ k +1 ) satisfying d ( x k +1 , y k +1 ) + ρ k +1 ≤ βρ k . Here Bob isallowed to choose 0 < ρ k +1 ≤ ρ k arbitrarily. Again, the balls B i ( i ∈ N ) are closed and nested with radiitending to zero, so the intersection B ∞ = T k B k is a singleton. Again, the result x is the unique pointsuch that B ∞ = { x } , and we call Alice the winner if and only if x ∈ S . If Alice has a strategy to winthis game regardless of how Bob plays, we say that S is β -BMS winning . If S is β -BMS winning for some0 < β < S is BMS winning . Informally, the BMS game is a two-player game in whichthe first player (Bob) chooses balls according to the relatively lax rules of the Banach–Mazur game, whilethe second player (Alice) chooses according to the stricter rules of Schmidt’s game. Accordingly, the classof BMS winning sets is very restrictive and has strong geometric properties. To make this precise, we needthe following definition:
Definition 2.2.
Given β >
0, a set E ⊆ X is said to be uniformly β -porous if there exists r > B ( x, r ) ⊆ X with r ≤ r , there exists an open ball B ◦ ( y, βr ) ⊆ B ( x, r ) such that B ◦ ( y, βr ) ∩ E = ∅ . Theorem 2.3 ([4, Theorem 2.2]) . Let ( X, d ) be a complete, separable metric space and fix < β < .Then a Borel set S ⊆ X is β -BMS winning if and only if X \ T ∞ i =1 S i is a countable union of uniformly β -porous sets. The original hyperplane game only requires Bob’s radius to satisfy ρ k +1 ≥ βρ k rather than equality, making the gamediscussed here the “modified hyperplane game”. However, the hyperplane game is equivalent to the modified hyperplanegame (this follows from appropriately modifying the proof of [5, Proposition 4.5]), so this distinction is not important. RYAN BRODERICK, LIOR FISHMAN, AND DAVID SIMMONS
When (
X, d ) is given additional structure, this theorem yields an estimate on the Hausdorff measure ofa BMS winning set. Specifically, we consider metric spaces (
X, d ) which support a measure of the followingtype:
Definition 2.4.
A Borel measure on a metric space (
X, d ) is called
Ahlfors s -regular if there exist C > ρ > x ∈ X and 0 < ρ ≤ ρ we have(2.1) 1 C ρ s ≤ µ ( B ( x, ρ )) ≤ Cρ s . We say µ is Ahlfors regular if it is Ahlfors s -regular for some s > X, d ) supports some Ahlfors s -regular Borel measure, then the s -dimensional Hausdorff measureon X is Ahlfors s -regular as well [8, Theorem 8.5]. It follows that dim X = s . On a metric space ( X, d )supporting such a measure, BMS winning sets are very large in the sense that their complement has positivecodimension:
Proposition 2.5 ([4, Corollary 2.5]) . If X is Ahlfors s -regular and S ⊆ X is BMS winning, then dim( X \ S ) < s . It follows from this that any BMS winning set on a metric space supporting an Ahlfors s -regular measuremust be conull with respect to the s -dimensional Hausdorff measure.3. Proof of Theorem 1.2
We will need the following lemma due to Davenport (see [7, Lemma 4]) for a version which implies theone stated here):
Lemma 3.1 (Simplex Lemma) . Let d ∈ N , r > , and x ∈ R d . Then the rational points in B ( x , r ) withdenominator at most (2 d ! r ) − dd +1 all lie in a single hyperplane.Proof of Theorem 1.2. Fix 0 < β < /
3. Let ρ > B . Since g ( k ) → ∞ , after extractinga subsequence we can assume without loss of generality that β k +1 / (4 d !) ≥ g ( k ) − / for all k . (Takinga subsequence makes the set (1.3) smaller, so if we can show that (1.3) is hyperplane winning for thesubsequence, then it is hyperplane winning for the whole sequence as well.) Write ε k = g ( k ) − / . Alicewill play the game in such a way that she guarantees that if x denotes the result of the game, i.e. theunique point so that T j ≥ B j = { x } , then for each k ∈ N , (cid:13)(cid:13)(cid:13) i k j k x − p q (cid:13)(cid:13)(cid:13) ≥ ε k q − /d for all sufficientlylarge q . She will do this by partitioning N into infinitely many arithmetic progressions P k = 2 k − + 2 k N ( k ∈ N ) and using her turns corresponding to indices in P k to ensure that the condition holds for k . (Thisargument mimics the original proof of the countable intersection stability of Schmidt’s winning property.)This ensures that L (cid:16) i k j k x (cid:17) is larger than ε k .Fix k ∈ N and ℓ ∈ P k , and we will describe Alice’s strategy for the turn corresponding to the index ℓ .Let Q m = (cid:18) d ! i k j k β ℓ ρ (cid:19) − dd +1 and let Q m = (cid:26) j k p i k q : Q m − ≤ q < Q m (cid:27) . Note that every (reduced) j k p i k q with q sufficiently large is in some Q m (where “sufficiently large” depends on k ). Since ℓ ∈ P k , we have ℓ = 2 k − + 2 k m for some m ≥
0. Given B ℓ = B k − +2 k m , we consider Q m ∩ B ℓ ,where 2 B ℓ denotes the ball with the same center as B ℓ and twice the radius. Applying Lemma 3.1 with r = 2( i k /j k ) β ℓ ρ , we see that all rationals in a given ball of radius r of the form p q with Q m − ≤ q < Q m lieon a single hyperplane. Now, the map x j k i k x sends balls of radius 2( i k /j k ) β ℓ ρ to balls of radius 2 β ℓ ρ and also sends hyperplanes to hyperplanes, so it follows that Q m ∩ B ℓ is contained in a single hyperplane L . Alice will choose A ℓ to be N ( L , β ℓ +1 ρ ). Then for any x ∈ A ℓ and any j k p i k q ∈ Q m ∩ B ℓ , we have (cid:13)(cid:13)(cid:13)(cid:13) x − j k p i k q (cid:13)(cid:13)(cid:13)(cid:13) ≥ β ℓ +1 ρ = β k +1 i k /j k ( i k /j k ) β k − +2 k ( m − ρ = β k +1 i k /j k Q − (1+1 /d ) m − d ! ≥ β k +1 d ! 1 i k /j k q − (1+1 /d ) ≥ ε k i k /j k q − (1+1 /d ) . On the other hand, for any j k p i k q ∈ Q m \ B ℓ we have (cid:13)(cid:13)(cid:13)(cid:13) x − j k p i k q (cid:13)(cid:13)(cid:13)(cid:13) ≥ β ℓ ρ ≥ β ℓ +1 ρ ≥ ε k i k /j k q − (1+1 /d ) . Thus, if Alice plays according to this strategy and x denotes the result of the game, then for each k ∈ N , (cid:13)(cid:13)(cid:13) i k j k x − p q (cid:13)(cid:13)(cid:13) ≥ ε k q − (1+1 /d ) for all ( p , q ) ∈ Z d × N . (cid:3) Proof of Theorem 1.3
Let F M ⊆ R be the set of real numbers with partial quotients at most M . Recall that the set of badlyapproximable numbers is S M ∈ N F M . We will prove that our set has full dimension in R by showing thatit has full dimension within each F M . To this end we prove the following: Proposition 4.1.
For each integer M ≥ and each reduced ij ∈ Q , the set { x ∈ F M : L (cid:16) ij x (cid:17) ≤ ij } is aBMS winning set on F M . Before proving this, we show how it can be used to deduce Theorem 1.3.
Proof of Theorem 1.3.
Fix M ∈ N . One can easily check that F M ∩ [0 ,
1] is the limit set of the con-tracting family of self-conformal maps { f i } Mi =1 on [ a, b ], where f i ( x ) = i + x , a = [0; M, , M, , . . . ] and b = [0; 1 , M, , M, . . . ], and that this system satisfies the open set condition. It follows (see [9, Lemma3.14]) that the δ M -dimensional Hausdorff measure H δ M is Ahlfors regular, where δ M = dim( F M ). ByProposition 4.1 the set (cid:26) x ∈ F M : L (cid:18) ij x (cid:19) ≤ ij for all i, j ∈ N (cid:27) is a countable intersection of BMS winning sets, hence conull with respect to H δ M by Proposition 2.5. Inparticular, dim (cid:26) x ∈ R : L (cid:18) ij x (cid:19) ≤ ij for all i, j ∈ N (cid:27) ≥ δ M . Since dim (cid:0) S M F M (cid:1) = dim(BA ) = 1, we have lim M →∞ δ M = 1, so the theorem follows. (cid:3) The main idea of the proof of Proposition 4.1 is as follows: Call an approximation p/q of x “good” if | x − p/q | < /q . If i/j ∈ Q is fixed and ( jp ) / ( iq ) is a good approximation of x , then | x − ( jp ) / ( iq ) | < / ( iq ) ,and rearranging gives | ( i/j ) x − p/q | < / ( ijq ). Thus, if x has infinitely many good approximations whosedenominator is a multiple of i and whose numerator is a multiple of j , then L (cid:0) ij x (cid:1) ≤ ij . Alice will directthe play of the game to numbers x with this property by using her turns to choose digits of the continuedfraction expansion in such a way that infinitely many truncated expansions (i.e. convergents), which aregood approximations to x , have numerator and denominator in the required sets. To show that this ispossible we first prove the following lemma: Lemma 4.2.
For any i, j ∈ N and M ≥ , there exists T = T ( i, j, M ) ∈ N such that for any n ∈ N andany a , a , . . . , a n ∈ N , there exist ≤ t ≤ T and a n +1 , . . . , a n + t ∈ { , . . . , M } such that [0; a , . . . , a n + t ] isa rational with numerator in j N and denominator in i N . RYAN BRODERICK, LIOR FISHMAN, AND DAVID SIMMONS
Proof.
Recall [6, Theorem 1] that if x = [0; a , a , . . . ], then the approximations obtained by truncatingthe continued fraction expansion of x , p n q n = [0; a , a , . . . , a n ], satisfy the recurrence relation(4.1) p n q n = a n p n − + p n − a n q n − + q n − · In other words, (cid:20) p n − q n − p n q n (cid:21) = (cid:20) a n (cid:21) (cid:20) p n − q n − p n − q n − (cid:21) . Now it is well-known (e.g. [6, Theorem 2]) that(4.2) det (cid:20) p n − q n − p n q n (cid:21) = ( − n . Fix i , j , and M , and let a , . . . , a n ∈ { , . . . , M } . We may assume i and j are coprime (i.e. that thefraction i/j is reduced). We want to choose a n +1 , a n +2 , . . . , a n + t in such a way that (cid:20) p n + t − q n + t − p n + t q n + t (cid:21) = (cid:20) a n + t (cid:21) (cid:20) a n + t − (cid:21) · · · (cid:20) a n +1 (cid:21) (cid:20) p n − q n − p n q n (cid:21) ≡ (cid:20) m m j i (cid:21) (mod ij ) , where m , m ∈ Z /ij Z satisfy m j + m i ≡ ( − n (mod ij ) (such numbers exist since gcd( i, j ) = 1). Itwill suffice to show that the semigroup H generated by elements of the form g a = (cid:20) a (cid:21) (1 ≤ a ≤ M )is equal to G = SL ± (2 , Z /ij Z ). (Since G is finite, if H = G then there exists T such that every elementof G is the product of at most T elements of the form g a .) Since G is finite, for any g ∈ H there exist1 ≤ k < k such that g k = g k and therefore g − = g k − k − ∈ H , so H is equal to the group generatedby the elements g a . In particular, we have (cid:20) − a
11 0 (cid:21) = (cid:20) a (cid:21) − = g − a ∈ H for all 1 ≤ a ≤ M .Recall (e.g. [11, p.139]) that (cid:20) (cid:21) and (cid:20) − (cid:21) generate SL(2 , Z ). Since (cid:20) (cid:21) = (cid:20) − (cid:21) (cid:20) (cid:21) ∈ H (cid:20) (cid:21) = (cid:20) (cid:21) (cid:20) − (cid:21) ∈ H (cid:20) − (cid:21) = (cid:20) (cid:21) (cid:20) − (cid:21) (cid:20) (cid:21) = (cid:20) (cid:21) (cid:20) (cid:21) − (cid:20) (cid:21) ∈ H, it follows that SL(2 , Z /ij Z ) ⊆ H . Since det( g a ) = − a , we have G = SL ± (2 , Z /ij Z ) = H , whichcompletes the proof. (cid:3) Proof of Proposition 4.1.
A modification of the argument of [10, Theorem 1.2] shows that the BMS game isinvariant under quasisymmetric homeomorphisms, and in particular under the coding map π : { , . . . , M } N → F M defined by the formula π ( a , a , . . . ) = [0; a , a , . . . ]. When the BMS game is played on the space { , . . . , M } N , the gameplay is equivalent to the following: on each of Bob’s moves, he chooses a word inthe alphabet E = { , . . . , M } which is an extension of Alice’s most recent word, and on each of Alice’smoves, she chooses a word in the alphabet E which is an extension of Bob’s most recent word by T letters,where T ∈ N is fixed at the start of the game. The result of the game is the unique infinite word ω whichextends all of these words, so Alice’s goal is to make the number x = π ( ω ) satisfy L (cid:0) ij x (cid:1) ≤ ij . Now let T be given by Lemma 4.2, and let Alice’s strategy be given as follows: if Bob’s most recent word is the move a , . . . , a n , then Alice’s next move will be the word a , . . . , a n + T , where a n +1 , . . . , a n + t are given by Lemma4.2, and a n + t +1 , . . . , a n + T are arbitrary. When Alice uses this strategy, each of her moves determines a ECAYING AND NON-DECAYING BADLY APPROXIMABLE NUMBERS 7 new convergent of the continued fraction expansion of the final word whose numerator is in j N and whosedenominator is in i N .Now, suppose p k q k is a convergent with p k ∈ j N and q k ∈ i N , say p k = jp and q k = iq . Then (cid:12)(cid:12)(cid:12)(cid:12) x − jpiq (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) x − p k q k (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) p k +1 q k +1 − p k q k (cid:12)(cid:12)(cid:12)(cid:12) = 1 q k +1 q k ≤ i q and hence (cid:12)(cid:12)(cid:12)(cid:12) ij x − pq (cid:12)(cid:12)(cid:12)(cid:12) = ij (cid:12)(cid:12)(cid:12)(cid:12) x − jpiq (cid:12)(cid:12)(cid:12)(cid:12) ≤ iji q = 1 ij q · Since this holds for infinitely many fractions pq , it follows that L (cid:16) ij x (cid:17) ≤ ij . (cid:3) References
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340 Rowland Hall (Bldg.
E-mail address : broderir @ uci.edu University of North Texas, Department of Mathematics, 1155 Union Circle
E-mail address : lior.fishman @ unt.edu University of York, Department of Mathematics, Heslington, York YO10 5DD, UK
E-mail address : David.Simmons @ york.ac.uk URL ::