Quantitative recurrence for beta expansion. Wang BaoWei

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Introduction Further study Quantitative recurrence for beta expansion Huazhong University of Science and Technology July 8 2010

Introduction Further study Contents 1 Introduction Background Beta expansion 2 Known results and our formulations Proofs 3 Further study

Introduction FurtherBackground study Beta expansion 1 Introduction Background Beta expansion 2 3 Further study

Introduction FurtherBackground study Background Beta expansion A well known theory on recurrence is Poincaré s recurrence theorem. Theorem (Poincaré s Recurrence Theorem) Let T : X X be a measurable transformation on a probability space (X, B, µ). Let B B with µ(b) > 0. Then for almost all points x B, the orbit {T n (x)} n 0 returns to B infinitely often.

Introduction FurtherBackground study Background Beta expansion A well known theory on recurrence is Poincaré s recurrence theorem. Theorem (Poincaré s Recurrence Theorem) Let T : X X be a measurable transformation on a probability space (X, B, µ). Let B B with µ(b) > 0. Then for almost all points x B, the orbit {T n (x)} n 0 returns to B infinitely often. However, this information is only of qualitative nature and it is only concerned with a fixed target B.

Introduction FurtherBackground study Problems Beta expansion Problem 1 With which frequency an orbit visits a given set of positive measure; and

Introduction FurtherBackground study Problems Beta expansion Problem 1 With which frequency an orbit visits a given set of positive measure; and Problem 2 With which rate a given point returns to an arbitrarily small neighborhood of itself.

Introduction FurtherBackground study Problems Beta expansion Problem 1 With which frequency an orbit visits a given set of positive measure; and Problem 2 With which rate a given point returns to an arbitrarily small neighborhood of itself. Problem 3 What happens if the ball B shrinks with time and more generally if the ball also moves around with time?

Introduction FurtherBackground study Beta expansion Beta expansion Let β > 1 be a real number. The beta transformation with respect to β is given as T β (x) = βx(mod 1). Hence, any x [0, 1) can be expanded uniquely as an finite or infinite series with the form x = ɛ 1(x, β) β + ɛ 2(x, β) β 2 +, where ɛ n (x, β) = [βt n 1 β (x)] are called the digits of x and the ( ) sequence ɛ 1 (x, β), ɛ 2 (x, β), is called the sequence of the beta expansion of x with respect to the base β.

Introduction FurtherBackground study Admissible sequence Beta expansion Definition An n-block (ɛ 1, ɛ 2,, ɛ n ) is called admissible in base β if there exists x [0, 1] such that ɛ k (x, β) = ɛ k for all 1 k n. An infinite sequence (ɛ 1, ɛ 2,, ɛ n, ) is admissible in base β if (ɛ 1, ɛ 2,, ɛ k ) is admissible in base β for all k 1.

Introduction FurtherBackground study Admissible sequence Beta expansion Definition An n-block (ɛ 1, ɛ 2,, ɛ n ) is called admissible in base β if there exists x [0, 1] such that ɛ k (x, β) = ɛ k for all 1 k n. An infinite sequence (ɛ 1, ɛ 2,, ɛ n, ) is admissible in base β if (ɛ 1, ɛ 2,, ɛ k ) is admissible in base β for all k 1. Denote by Σ β all infinite admissible sequence in the beta expansion with respect to the base β.

Introduction FurtherBackground study Beta expansion Characterizations on admissible sequence Theorem (W. Parry) Let β > 1 be a real number and ɛ(1, β) be the β-expansion of 1. We denote by w an infinite sequence of positive integer. (1) If ɛ(1, β) is infinite, w Σ β if and only if σ k (w) < lex ɛ(1, β), for all k 0. (2)If ɛ(1, β) is finite, i.e., ɛ(1, β) = (ɛ 1 (1, β),, ɛ n (1, β), 0 ) with ɛ n (1, β) 0. Then w Σ β if and only if σ k (w) < lex ɛ (1, β), for all k 0. where ɛ (1, β) = ( ɛ 1 (1, β), ɛ 2 (1, β),, ɛ n 1 (1, β), (ɛ n (1, β) 1)) is a purely periodic sequence.

Introduction FurtherKnown study results and our formulations Proof of the result 1 Introduction 2 Known results and our formulations Proofs 3 Further study

Introduction FurtherKnown study results and our formulations Known results Proof of the result Quantitative recurrence properties to general settings. Theorem (Boshernitzan) Let (X, µ, T, d) be a measure dynamical system with a metric d. Assume that, for some α > 0, the Hausdorff α-measure H α is σ- finite on the space X. Then for µ-almost all x X, lim inf n n 1 α d(t n x, x) <. If, moreover, H α (X) = 0, then for µ-almost all x X, lim inf n n 1 α d(t n x, x) = 0.

Introduction FurtherKnown study results and our formulations Proof of the result When applied to beta transformation, one has, almost surely, lim inf n nd(t n x, x) <.

Introduction FurtherKnown study results and our formulations Proof of the result Quantitative recurrence properties to beta transformations. Theorem (W. Philipp) Let φ be a positive function defined on natural numbers N. The set { } R β (φ) := x [0, 1] : Tβ n x B(x, φ(n) 1 ), infinitely often n N is a full or null set if and only if the series n 1 φ 1 (n) diverges or not.

Introduction FurtherKnown study results and our formulations Our formulations Proof of the result Given a general function φ : N R +, find the Hausdorff dimension of the set { } x [0, 1] : Tβ n x x < φ(n) 1, infinitely often n N.

Introduction FurtherKnown study results and our formulations Our formulations Proof of the result Given a general function φ : N R +, find the Hausdorff dimension of the set { } x [0, 1] : Tβ n x x < φ(n) 1, infinitely often n N. Theorem (Main result) For any β > 1, dim H R β (φ) = 1, where b = lim inf 1 + b n log β φ(n). n

Introduction FurtherKnown study results and our formulations Difficulties Proof of the result In studying the metric theory related to beta transformation, one of the big obstacle lies in estimating the length of a cylinder I β (ɛ 1,, ɛ n ) = {x [0, 1] : ɛ k (x, β) = ɛ k, 1 k n} for general β > 1.

Introduction FurtherKnown study results and our formulations Our strategy Proof of the result Principle As far as Haudorff dimension or metric theory is concerned, we can discard some bad points without affecting the final results.

Introduction FurtherKnown study results and our formulations Our strategy Proof of the result Principle As far as Haudorff dimension or metric theory is concerned, we can discard some bad points without affecting the final results. As a result, we confine our attention to good points: {x = (ɛ 1,, ɛ n, ) : (ɛ 1,, ɛ n, ) Σ βn }, where β N is defined as the solution to the equation 1 = ɛ 1(1, β) β N + ɛ 2(1, β) β 2 N + + ɛ N(1, β) βn N.

Introduction FurtherKnown study results and our formulations Proof of the result By the criterion for admissible sequence, one can know easy that Σ βn Σ β. More importantly, for any admissible block (ɛ 1,, ɛ n ) Σ βn, (ɛ 1,, ɛ n, 0 N, 1) Σ βn, so is in Σ β.

Introduction FurtherKnown study results and our formulations Proof of the result By the criterion for admissible sequence, one can know easy that Σ βn Σ β. More importantly, for any admissible block (ɛ 1,, ɛ n ) Σ βn, (ɛ 1,, ɛ n, 0 N, 1) Σ βn, so is in Σ β. As a consequence, Proposition For any admissible block (ɛ 1,, ɛ n ) Σ βn, then it holds the following length estimation I β (ɛ 1,, ɛ n ) 1 β n.

Introduction FurtherKnown study results and our formulations Cantor subset Proof of the result It is easy to see that if the β-expansion of T n x and that of x coincide in a long block from the very beginning, one can make sure that T n x and x are close enough. So, our Cantor subset is defined as follows.

Introduction FurtherKnown study results and our formulations Cantor subset Proof of the result It is easy to see that if the β-expansion of T n x and that of x coincide in a long block from the very beginning, one can make sure that T n x and x are close enough. So, our Cantor subset is defined as follows. Level 1. E 1 = I β ((ɛ 1,, ɛ n1 0 N ) w1+1 0 N ) ɛ 1,,ɛ n1 Σ n 1 β N where w 1 is some rational with w 1 (n 1 + N) N such that β (n 1+N)w 1 < φ(n 1 ) 1, but β (n 1+N)w 1 +1 φ(n 1 ) 1. Then it follows that, for any x I β ((ɛ 1,, ɛ n1 0 N ) w 1+1 0 N ), T n 1 β x x < 1 φ(n 1 ). (2.1)

Introduction FurtherKnown study results and our formulations Proof of the result Level 2. Write m 1 = (w 1 + 1)(n 1 + N), t 2 = n 2 m 1. For each I β = I β (ɛ 1,, ɛ m1 ), set E 2 (I β ) = I β ((ɛ 1,, ɛ 0 N n2 ) w2+1 0 N ), ɛ m1 +1,,ɛ n2 Σ t 2 βn where w 2 is some rational with w 2 (n 2 + N) N such that β (n 2+N)w 2 < φ(n 1 2 ), but β (n 2+N)w 2 +1 φ(n 1 2 ).

Introduction FurtherKnown study results and our formulations Proof of the result Level 2. Write m 1 = (w 1 + 1)(n 1 + N), t 2 = n 2 m 1. For each I β = I β (ɛ 1,, ɛ m1 ), set E 2 (I β ) = I β ((ɛ 1,, ɛ 0 N n2 ) w2+1 0 N ), ɛ m1 +1,,ɛ n2 Σ t 2 βn where w 2 is some rational with w 2 (n 2 + N) N such that β (n 2+N)w 2 < φ(n 1 2 ), but β (n 2+N)w 2 +1 φ(n 1 2 ). Then finally, define E 2 = I β (ɛ 1,,ɛ m1 ) E 1 E 2 (I β (ɛ 1,, ɛ m1 )).

Introduction Further study 1 Introduction 2 3 Further study

Introduction Further study Further study Size of the level set defined by the Birkhoff ergodic related to beta expansion. Theorem (Fan, Feng, Wu) Let (Σ A, T ) be a topologically mixing subshift of finite type on an alphabet consisting of m symbols and let Φ : Σ A R d be a continuous function. For any possible limit of the ergodic limit α of lim n 1 n n 1 j=0 Φ(T j x), one has n 1 1 ) { h top (x Σ A : lim Φ(T j x) = α = sup h µ : n n j=0 } Φdµ = α where h µ denotes the entropy of µ and h top denotes the topological entropy.c d.

Introduction Further study If we relate above system with beta shift, it deals with the β with finite expansion of the unit 1. Then how about the case for general β. Question 1 Given β > 1. Denote by Σ β the collection of all infinite admissible sequence in the beta expansion in base β. Let Φ : Σ β R be a continuous function. How about the size of the set x [0, 1] : lim n n 1 1 n j=0 Φ(T j (ɛ(x, β))) = α, where ɛ(x, β) denotes the sequence of the β-expansion of x.

Introduction Further study Shrinking target problems for beta transformation. Question 2 Given β > 1 and x 0 [0, 1). Let Φ be a positive function defined on N. Find the Hausdorff dimension of the set { } x [0, 1] : T n x x 0 < Φ(n), for infinite many n N.

Introduction Further study Thanks THANK YOU!