A quantitative shrinking target result on Sturmian sequences for rotations

Let $R_\alpha$ be an irrational rotation of the circle, and code the orbit of any point $x$ by whether $R_\alpha^i(x)$ belongs to $[0,\alpha)$ or $[\alpha,1)$ -- this produces a Sturmian sequence. A point is undetermined at step $j$ if its coding up to time $j$ does not determine its coding at time $j+1$. We prove a pair of results on the asymptotic frequency of a point being undetermined, for full measure sets of $\alpha$ and $x$.

1. Introduction 1.1. Statement of the problem and main results. In this paper we study a shrinking target problem. Let α ∈ [0, 1), let R α : [0, 1) → [0, 1) be the rotation R α (x) = x+α (mod 1), and let λ denote Lebesgue measure. The following theorem, due to Weyl, is well known. That is, the asymptotic for the number of visits of the orbit of x to the target set B(y, ǫ) by step N is given by the sum of the size of the target over those N steps. The statement is written here in a slightly unusual way -the denominator is clearly N 2ǫ (assuming ǫ ≤ 1 2 and identifying [0, 1) with S 1 ). But it suggests the following sort of problem. Let {B i } be a sequence of measurable sets in [0, 1). What can be said about the behavior of N i=1 χ Bi (R i α x); in particular, is it asymptotic to N i=1 λ(B i )? This is, of course, an enormously varied problem. Cases which have generated significant interest are shrinking target problems, in which the B i form a decreasing, nested chain. By the Borel-Cantelli Lemma, the cases of real interest are when ∞ i=1 λ(B i ) = ∞. Several results on this problem for rotations and for interval exchange transformations are contained in [CC17], including the following.
Theorem 1.2. [CC17] For all α satisfying an explicit, full measure diophantine condition, and for any sequence {r i } such that ir i is non-increasing and ∞ i=1 r i = 1991 Mathematics Subject Classification. Primary: 37E10, 37A05, 37B10.
Key words and phrases. Shrinking target, Sturmian sequence, circle rotation, symbolic dynamics, continued fractions.
The first author is supported by NSF grants DMS-1004372, 135500, 1452762, the Sloan Foundation, a Warnock chair, and a Poincaré chair. ∞, and any y for almost every x.
In this paper we consider another shrinking target problem for rotations, but one whose targets arise in a very different way. Rather than being subject to some pre-determined analytic constraint (as for the sequence {r i }) the targets arise from the dynamics of the rotation itself.
Let α be given. Let P = {A 0 , A 1 } be the partition of [0, 1) given by A 0 = [0, α), A 1 = [α, 1). The bi-infinite sequences (c i (x)) i∈Z defined by c i (x) = j if R i α x ∈ A j are known as Sturmian sequences (see, e.g. [BFMS02], Ch. 6). These are sequences of minimal complexity, or with minimal block growth. They were introduced by Hedlund and Morse [MH40], and have been studied extensively.
For a sequence (c 0 , c 1 , . . .) (finite or infinite) of 0's and 1's, let C c0,c1,... = {x : is a coding for the orbit of x (or a portion thereof, if the sequence is finite). Let Σ be the set of finite codings c 0 , c 1 , . . . , c n which actually occur, i.e. for which C c0,...,cn = ∅. Let V j = {x : x ∈ C c0,...,cj and such that c 0 , . . . , c j , 0 and c 0 , . . . , c j , 1 ∈ Σ}. This is the set of 'undetermined' points at step j, that is, points whose coding up to step j does not determine the coding at step j + 1. The word c 0 , . . . , c j is also known as a right special word (see, e.g., [Lot02, §2.1.1].) We want to find asymptotics on how often a point is undetermined; specifically, we will prove for almost all x ∈ (0, 1).
As in [CC17], the full measure condition on α is a diophantine condition involving the continued fraction expansion of α. It will be stated explicitly in the proof.
To understand why Theorem A constitutes a shrinking target problem, consider the following. Let P j = ∨ j k=0 R k α P, the partition generated by P and its first j translates. For x ∈ X, denote by [[x]] j the atom of x in P j . The coding c 0 , . . . , c j determines only the atom as the image of this atom under one more rotation contains points in both A 0 and A 1 . We will denote [[1 − α]] j by U j -these are the shrinking targets which we are trying to hit. Note that U j = R j α (V j ). The logarithms in Theorem A indicate a weaker asymptotic result than in [CC17]. The stronger version is not true: does not exist for almost every x ∈ [0, 1).
Thus, Theorem A is in some sense the best one can hope for in this setting, an interesting contrast with the stronger results obtained for targets of the form B(y, r i ).
1.2. Notation and an outline of the paper. The key tool throughout the paper is the continued fraction expansion of α and its close relationship to the dynamics of the rotation by α. Throughout, α ∈ (0, 1) is assumed to be irrational. We write α = [0; a 1 , a 2 , a 3 , . . .] for the continued fraction expansion of α. Note that the elements a i of the continued fraction depend on α; we will at times write a i (α) to emphasize this dependence. The convergents to α are the rationals p k q k . The k th convergent is the best rational approximation to α with denominator ≤ q k . The q k can be computed by the recurrence relation q k+1 = a k+1 q k + q k−1 ; q 0 = 1, q 1 = a 1 .
We will prove Theorem B first, in Section 2. The almost sure existence of elements of the continued fraction expansion which are very large in relation to the preceding elements drives the argument.
Theorem A is proved in Section 3. There we prove a set of looser bounds on n i=1 a i and qn j=1 χ Vj (x) which hold for almost all α and which are sufficient for the statement of Theorem A.

Failure of a stronger convergence
We start with the proof of Theorem B. First, we prove the almost-sure existence of very large elements a n for the continued fraction expansion. We then use this to show that, for very long stretches of time certain points are undetermined more often than n j=1 λ(V j ) predicts. Proposition 2.1. For any C ∈ R and almost every α there exist infinitely many m such that We need a series of preliminary results to prove this. The following lemma appears in [Khi97, page 60].
Lemma 2.2. For any n, b 1 , ..., b n ∈ N we have From this it is an easy exercise to deduce: Let W n = α : n i=1 a i (α) < 10n log n .
Remark 2.5. The bound in Lemma 2.4 is not optimal, as is easily seen from the proof. We are only concerned to find some bound away from zero.
We are now ready to prove Proposition 2.1.
Proof of Proposition 2.1. Fix C > 0. Corollary 2.3 and the definition of W m−1 imply that From this we have that .
To complete the proof we need two lemmas. The first is a well known partial converse to the Borel-Cantelli Lemma for quasi-independent sets (as opposed to independent sets). Its proof is included for completeness.
Lemma 2.6. Let A i be measurable subsets of a space with probability measure λ.
for an arbitrary A * ⊂ W m−1 of the form above. Since a subcollection of the A i form a partition of W m−1 , by restricting the above estimate to that subcollection The result follows, using Lemma 2.4.
Applying these two lemmas we conclude that there is a positive measure set of α for which a m (α) ≥ C m−1 i=1 a i (α) infinitely often. If α is in this set, its image under the Gauss map is as well, so by the ergodicity of that map the set of such α in fact has full measure.
The following two lemmas on the shrinking targets U j are also needed to complete our proof of Theorem B. Recall that U j = R j α (V j ) and V j = {x : x ∈ C c0,...,cj and such that c 0 , . . . , c j , 0 and c 0 , . . . , c j , 1 ∈ Σ}.
These lemmas are proved using the partial fraction expansion of α. We will denote by {y} the value modulo 1 of a real number y and by y the distance from y to the nearest integer.
Proof. Note that r j α is smaller than or equal to iα for all i ≤ j.
Case 1: 0 < {r j α} < 1/2. As the convergents alternate in approximating α from above and below, 1/2 < {s j α} < 1. The only possible improvement in {r j α} as an upper bound for R α (U j ) would come from finding some l with lα < r j α . This is not possible for l ≤ j. Thus the upper endpoint of R α (U j ) is {r j α} as desired.
The lower bound on R α (U j ) given by {s j α} can be improved only by adding {r j α} some number of times, as r j is the only integer ≤ j with r j α < s j α . The lower endpoint will thus be of the form y = {s j α} + T {r j α} and will be found by taking T as large as possible such that the s j +T r j rotations required to produce this point do not exceed j; this number is t j .
We calculate that λ(U j ) = r j α + (1 − {s j α} − t j {r j α}) using the fact that in this case r j α = {r j α}. Since s j α = 1 − {s j α}, this simplifies to the desired result.
Case 2: 1/2 < {r j α} < 1. Then 0 < {s j α} < 1/2 and the lower endpoint of R α (U j ) is {r j α}. As before, the upper endpoint is of the form {s j α} − T (1 − {r j α}). The best such endpoint is found by taking T as large as possible, i.e. equal to t j .
Finally, we calculate again For use in the lemma below as well as later in the paper, we fix some notation. We will adopt interval notation ([n, m), etc.) to denote intervals of integers; context will make the distinction between these and subsets of the real interval [0, 1) clear. We Let J denote the collection of all the J i b 's. We note that J i b ⊂ I i and that these intervals are disjoint.
Further, let J i 2 = [q i−1 + q i , q i−1 + 2q i ) for all i, whether a i+1 ≥ 2 or not. If a i+1 = 1, J i 2 ⊂ I i+1 and it equals J i+1 1 , but we note that in any case {J i 2 } i∈N consists of pairwise disjoint intervals.
Lemma 2.10. For any J ∈ J , and for all l ∈ J, the sets V l = R −l α U l are pairwise disjoint.
We examine the two cases: b = 1 and b > 1.
, 1 < l − k < q i , and the endpoints of R α U are R qi α (0) and R qi−1+(b−1)qi α (0). The first time after q i−1 + (b − 1)q i that the orbit of 0 hits U is q i−1 + bq i . But (l − k) + q i < q i−1 + bq i since l − k < q i and b ≥ 2 and (l − k) + (q i−1 + (b − 1)q i ) < q i−1 + bq i since l − k < q i , so neither endpoint of R α U will return to R α U under R l−k α , again proving disjointness. Corollary 2.11. For all m, where U m and V m are intervals and K m → ∞ as m → ∞. Them, for any sufficiently large m, there exists a pair of points x * ∈ X m ∩ A and y * ∈ Y m ∩ A with |x * − y * | < δ.
Proof. Choose a positive ǫ satisfying ǫ < (.99)λ(A)δ 2+(.99)δ . This choice guarantees that ( 1 2 )(.99)(λ(A) − ǫ)δ > ǫ. Since A has finite measure, there is a finite, disjoint union of open intervals B = n i=1 I i such that λ(A∆B) < ǫ. By the equidistribution of points under R α and the fact that K m → ∞, we may pick M > 0 so large that for all m > M , for all i = 1, . . . , n, using our lower bound on the measures of X m and Y m . Further pick M so large that for m > M , the maximum separation between two adjacent points in {R k α 0} Km k=1 is < δ. Consider the intervals forming X m and Y m which are contained in I i . For each interval U which is a connected component of X m , let V U be its nearest neighbor to the right among the connected components of Y m . (Such a neighbor exists for all but possibly the last such U contained in I i . We may choose M so large that the number of X m intervals in I i is very large, making this exceptional subinterval's contribution to the argument below negligible.) Note that max x∈U,y∈VU |x − y| < δ. If a pair (x * , y * ) as desired does not exist, then for each pair (U, V U ), at least one of U, V U contains no points in A. Therefore, λ(I i \ A) > 1 2 (.99)λ(I i )δ. Thus, by our choice of ǫ. But this contradicts our choice of B, proving the lemma.
To simplify notation a bit, we set for all integers m: .
Where it exists, we set .
Note that, where it exists, lim m→∞ f m (x) = f (x) and f is measurable. In addition, by Fatou's Lemma, f will be integrable over the set where it is defined, since [0,1) f m dλ = 1 for all m. Therefore we can assume f takes only finite values.
We are now ready to prove Theorem B.
Proof of Theorem B. Fix C > 0 and apply Proposition 2.1 to find a full measure set of α satisfying equation 1 for infinitely many m. Fix any such m.
Note that by Lemma 2.10, this is a disjoint union, and using Lemma 2.8, In addition, if x ∈ W b , then it will belong to exactly one Choose any ρ ∈ (1/8, 1/4) in such a way that ρa m ∈ N (possible since a m is very large), and let X m = W ρam . We then estimate the measure of X m below using standard results on the convergents: Second, choose σ ∈ (1/16, 1/8) so that σa m ∈ N and is ≥ 2. Let Y m = W 1 \W σam . Then, as any y ∈ Y m will not belong to V j when j ∈ J m b for b ≥ σa m , Using Corollary 2.11, We can also estimate the measure of this set (recalling that a m is very large): Estimates here are certainly not precise; the key point is that X m and Y m have a positive lower bound on their measures which is independent of m. Let δ = 1 128 .
Using the results above, for all x ∈ X m and y ∈ Y m , Finally, using Corollary 2.11, By choosing C sufficiently large, and since ρ > σ, we have |f m (x) − f m (y)| ≥ D > 0 for all m such that equation 1 holds and all is within ǫ of f (x) . Then we may take x * ∈ X m ∩ G ∩ Z N0 and y * ∈ Y m ∩ G ∩ Z N0 with |x * − y * | < δ.
As both points are in G and |x * − y * | < δ, |f (x * ) − f (y * )| < ǫ. We conclude that |f m (x * ) − f m (y * )| < 3ǫ < D. But this contradicts our result above on the minimum difference between the values of f m at points in X m and Y m when a m satisfies 1. Therefore there is a set of full measure where the f m do not converge, completing the proof.

Proof of Theorem A
Towards Theorem A, we claim the following set of inequalities: There exists a positive constant C 1 such that for almost every α and x ∈ [0, 1), The middle inequality follows from almost the same proof as Corollary 2.11. We prove the other two inequalities in the following sequence of Lemmas. Lemma 3.1 specifies the full measure set of α for which we prove Theorem A.
Lemma 3.1. There exists a positive constant C 1 such that for almost every α, C 1 n(log n) 3 > n i=1 a i (α) for all sufficiently large n. Remark 3.2. Note that how large n must be for the given bound to hold does depend on n.
Proof. As in the proof of Lemma 2.4, set A n = {α : a i (α) < n 2 for all i ≤ n}. As before, An n i=1 a i (α)dλ(α) ≤ 5n log n (for n > 7). Note also that λ-a.e. α belongs to A n for all but finitely many n. It follows from Markov's inequality that Since almost every α belongs to A n for all but finitely many n, almost every α belongs to A 10 k for all but finitely many k. Then These measures form a summable sequence, so for a.e. α, 10 k i=1 a i (α) ≤ 10 k+1 log 10 k 2.1 for all but finitely many k.
We will give a lower bound on qn j=1 χ Vj (x) by bounding below the sum over the J i 2 . As we noted above, h i (x)dλ > 1/2.
Proof. As per Lemma 2.10, over j ∈ J i 2 , the V j are disjoint, so h i (x) ∈ {0, 1}. The length of the interval J i 2 is q i , and for j ∈ J i 2 , λ(V j ) = q i−1 α , using the description of R α (U j ) provided by Lemma 2.8. By Theorem 13 in [Khi97], q i−1 α > 1 qi−1+qi . We may then bound the integral from below by The following sequence of results prove that the random variables h i (x) are (approximately) independent.
Proposition 3.5. Fix k. For all i such that q i > k and any Proof. Fix k. Let i be so large that q i > k. By the previous lemma, the interval V k is hit by the left endpoints of the V l between λ(V k )|J i b | − 2 and λ(V k )|J i b | + 2 times. As the sets V l are disjoint and of the same measure over l ∈ J i b , this easily yields Translating to an inequality with multiplicative errors yields Proposition 3.5 asserts near independence of the events V k and ∪ l∈J i b V l . Using it for all k ∈ J j b ′ where j < i (which guarantees k < q i ) we get the following corollary. It relates to calculating the correlation between a point being undetermined in the intervals J j b ′ and J i b .
Corollary 3.6. For any k ∈ J i b ′ , and Proof. This follows from summing Proposition 3.5's inequalities over the disjoint sets V k for k ∈ J i b ′ . (The desire to compute this sum explains our preference for the formulation in terms of multiplicative bounds above.) Proof. First, As over J i 2 and over J j 2 the sets V l are disjoint, the integrand of the above has value 0 or 1 according to whether x ∈ ∪ l∈J i 2 V l ∩ ∪ l∈J j 2 V l . Thus, By Corollary 3.6, for l ∈ J i 2 we get To assess the value of the terms 1 ± 3 λ(V l )|J j 2 | consider an arbitrary l ∈ J i 2 . As U l = R α V l , using the description of R α U l given by Proposition 2.8 and [Khi97, Theorem 13], λ(V l ) > q i α > 1 qi−1+qi ≥ 1 qi+1 . From its description, |J j 2 | = q j . Using these two bounds, qj . Returning to our inequalities for h i h j , as the V l are disjoint over J j 2 or J i 2 we can translate back into integrals as so: These are the desired bounds on h i h j dλ.
The independence result we want is the following.
Summing the term on the right-hand side of the above inequality over all n yields a convergent series so by the Borel-Cantelli Lemma, for almost every x ∈ [0, 1), s (n−2) 2 (x) (n − 2) 2 → 0 as n → ∞.
The left-most inequality is Lemma 3.1 and the next is Corollary 2.11. It remains only to prove: Proof. This follows easily from Lemma 3.3 after noting that The inequalities collected above enable us to prove the main theorem: Proof of Theorem A. Consider the full measure set of α satisfying Lemma 3.1. For a given α, suppose n ∈ [q m , q m+1 ) is large enough for the bound in Lemma 3.1 to hold. Then we have the following for almost every x: χ Vj (x) < C 1 (m + 1)(log(m + 1)) 3 , λ(V j ) < C 1 (m + 1)(log(m + 1)) 3 .