On the recurrence rates of continued fractions

Chaos, Solitons and Fractals 114 (2018) 474–477

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Chaos, Solitons and Fractals Nonlinear Science, and Nonequilibrium and Complex Phenomena journal homepage: www.elsevier.com/locate/chaos

On the recurrence rates of continued fractionsR Chun Wei a, Min Wu b, Shuailing Wang b,∗ a b

School of Statistics and Mathematics, Zhongnan University of Economics and Law, Wuhan 430073, PR China Department of Mathematics, South China University of Technology, Guangzhou 510641, PR China

a r t i c l e

i n f o

a b s t r a c t Letting T be Gauss transformation on [0,1), we consider the first return time τ n (x) of x ∈ [0, 1) to the cylinder of order n containing x. We prove that the Hausdorff dimension of the set

Article history: Received 2 January 2018 Revised 10 July 2018 Accepted 23 July 2018



x ∈ [0, 1 ) : lim inf n→∞

log τn (x ) log τn (x ) = α , lim sup =γ ϕ (n ) ϕ (n ) n→∞



is either zero or one depending on ϕ , α and γ , where ϕ : N → R+ is a monotonically increasing function and 0 ≤ α ≤ γ ≤ +∞.

MSC: 28A80 11A55

© 2018 Elsevier Ltd. All rights reserved. Keywords: Continued fraction Recurrence rates Hausdorff dimension

1. Introduction Suppose (X, d) is a metric space. Let T: X → X be a measurepreserving transformation of the probability space (X, B, μ ). The classical Poincaré recurrence theorem implies that

lim inf d (T n x, x ) = 0, n→∞

μ-almost every x ∈ X (a.e. for short).

This means that almost all points are recurrent. Boshernitzan [3] improved the above equation to the following quantitative result

lim inf n1/α d (T n x, x ) < ∞, n→∞

μ-a.e.,

where α is the dimension of the space in some sense. The first return time is an important aspect used to characterize the behaviors of orbits in dynamical systems. The first return time of a point x ∈ X into a set E ⊂ X is defined as

τE (x ) = inf{k ≥ 1 : T k x ∈ E }, which has been well studied in the last decades. Denote τ r (x) as the first return time of the point x to the ball B(x, r). Barreira and Saussol [2] consider the limit of

log τr (x ) , − log r R

(1.1)

Supported by the NSFC (Nos. 11601161, 11771153, 11371148, 11811530269 and Guangdong Natural Science Foundation 2017A030310164). ∗ Corresponding author. E-mail addresses: [email protected] (C. Wei), [email protected] (M. Wu), [email protected] (S. Wang). https://doi.org/10.1016/j.chaos.2018.07.027 0960-0779/© 2018 Elsevier Ltd. All rights reserved.

which can provides a quantitative measure of the speed of recurrence of an orbit near to its starting point. There are also many results relate (1.1) to the local dimension of the invariant measure, see [1,5,6]. The quantitative recurrence is also related to the system of circle rotation and diophantine approximation (see [1,6]). Let ξ be a finite partition of X. For any x ∈ X, denote ξ n (x) as the intersection of elements of ξ , T −1 (ξ ), · · · , T −n+1 (ξ ) which contains x, usually called a cylinder of order n which contains x. If we replace the ball in (1.1) with ξ n (x), the limit is related to the entropy of the dynamical system. Ornstein and Weiss [12] proved that for a finite partition ξ of X, if there exists a T-invariant ergodic Borel probability measure μ, then

  log τξn (x ) (x ) μ x ∈ X : nlim = hμ ( ξ ) = 1 , →∞ n

(1.2)

where hμ (ξ ) is the measure-theoretic entropy of T with respect to the partition ξ . We would like to consider the size of the exceptional sets. In 2001, Feng and Wu [4] studied the exceptional sets of the one-sided shift space ( , σ ) with  = {0, 1, · · · , m − 1}N . Let ξ = {[0], [1], · · · , [m − 1]} be a partition of  , and τ n (x) denote the first return time of the point x to the cylinder of order n which contains x. They proved that



dimH

log τn (x ) log τn (x ) x ∈  : lim inf = α , lim sup =γ n→∞ n n n→∞



=1

for any 0 ≤ α ≤ γ ≤ +∞ (see also [14,15]), where dimH denotes the Hausdorff dimension of a given set. Instead of the exponential recurrence rate in [4], Peng [13] considered the polynomial rate.

C. Wei et al. / Chaos, Solitons and Fractals 114 (2018) 474–477

More recently, Kim and Li [11] studied the general rate, and they proved a quite strong theorem regarding the possible growth behavior of the recurrence time in a one-sided Bernoulli shift, with their results, constituting a significant improvement over previous results in this area. More precisely, let ϕ : N → R+ be a monotonically increasing function. They showed that for any 0 ≤ α ≤ γ ≤ +∞, the Hausdorff dimension of the set



log τn (x ) log τn (x ) x ∈  : lim inf = α , lim sup =γ n→∞ ϕ (n ) ϕ (n ) n→∞



is either zero or one depending on ϕ , α and γ . Comparing with the dimensional results in [4,11,13] where the space is always the symbolic space with finite alphabet, in this paper we consider the continued fraction dynamic system which induces a symbolic space with countable alphabet. The continued fraction expansion of real numbers is related to the Gauss transformation T: [0, 1) → [0, 1) defined by

T (0 ) := 0,

T (x ) =

1 − x

1 x

n→∞

log τn (x ) = 1, n

μ-a.e.

(1.3)

For the exceptional sets, in [14], the authors proved that



log τn (x ) log τn (x ) = α , lim sup =γ n n n→∞

dimH x ∈  : lim inf n→∞

 =1 (1.4)

for any 0 ≤ α ≤ γ ≤ +∞. The recurrence rate is asymptotically exponentially increasing. The corresponding results for the continued fractions of formal Laurent series were given by Hu et al. [7]. In this paper, we consider the general rate, as in [11]. Let ϕ : N → R+ be a monotonically increasing function. Denote

lim inf n→∞

Define ϕ

E α ,γ =

ϕ (n )

= δ,

log n

lim sup n→∞

ϕ (n ) log n

= λ.

x ∈ [0, 1 ) : lim inf n→∞

log τn (x ) log τn (x ) = α , lim sup =γ . ϕ (n ) ϕ (n ) n→∞

(1.5)

Then, the following theorem says that the Hausdorff dimension of ϕ the set Eα ,γ is either zero or one depending on ϕ , α and γ . Theorem 1.1. Let ϕ : N → R+ be a monotonically increasing function. Then for any 0 ≤ α ≤ γ ≤ +∞, ϕ dimH Eα ,γ =



1, 0,

if α ≥ λ1 and otherwise.

γ ≥ 1δ ;

1 0

= 0.

with the convention

= ∞ and

1 ∞

:= [a1 (x ), a2 (x ), · · · ],

1 a 2 ( x )+

..

.

Proposition 2.1 ([10]). Let In = In (a1 , a2 , · · · , an ) be a cylinder of order n. When n is odd, the sub-cylinders {In+1 (a1 , · · · , an , an+1 ) : an+1 ≥ 1} are distributed from the left endpoint of In to the right endpoint of In one after another as an+1 increases from 1 to ∞. When n is even, the sub-cylinders are distributed from the right endpoint of In to the left endpoint of In one after another as an+1 increases from 1 to ∞. Let qpn ((xx)) = [a1 (x ), a2 (x ), · · · , an (x )] be the n-convergent of x. n For any n ≥ 1, with the convention p−1 (x ) = 1, q−1 (x ) = 0 and p0 (x ) = 0, q0 (x ) = 1, we have qn (x ) = an (x )qn−1 (x ) + qn−2 (x ).

Let In (x) := In (a1 (x), a2 (x), , an (x)) be the cylinder of n containing x, then



In (x ) =

pn (x )+ pn−1 (x ) qn (x )+qn−1 (x ) pn (x )+ pn−1 (x ) pn (x ) , ], qn (x )+qn−1 (x ) qn (x )

[ qpnn ((xx)) ,

(

)

when n is even; when n is odd.

Where no confusion is likely to arise, we write an and qn in place of an (x) and qn (x) respectively for simplicity. Let |I| be the length of the interval I. The following lemma is well known. Lemma 2.2 ([8],[10]). For any n ≥ 1 and (a1 , a2 · · · , an ) ∈ Nn . Let qn = qn (a1 , a2 , , an ). Then

qn ≥ 2

n−1 2

,

n 

ak ≤ qn ≤

k=1

=

n 

( ak + 1 ),

|In (a1 , a2 · · · , an )|

k=1

1 . qn (qn + qn−1 )

Lemma 2.3 ([17]). For any n ≥ 1 and 1 ≤ k ≤ n,

qn (a1 , · · · , ak , ak+1 , · · · , an ) ak + 1 ≤ ≤ ak + 1 . 2 qn−1 (a1 , · · · , ak−1 , ak+1 , · · · , an ) Lemma 2.4 ([14]). If x, y ∈ [0, 1) satisfy ai (x ) = ai (y ) for all 1 ≤ i ≤ n but an+1 (x ) < an+1 (y ), then

1 2(an+2 (y ) + 1 )(an+1 (y ) + 1 )2 |In (a1 (y ), a2 (y ), · · · , an (y ))|.

|x − y| ≥

For any integer M ≥ 1, let BM be the set of points whose partial quotients do not exceed M, that is,

Lemma 2.5 ([9]). For any M ≥ 8,

This section is devoted to recalling some basic properties of continued fractions and fixing some notation. Let x ∈ (0, 1) and [a1 (x), a2 (x), ] be its continued fraction expansion, that is,

a1 ( x ) +

The following proposition concerns the position of a cylinder in [0,1).

Jarník [9] showed that

2. Preliminaries

1

In (a1 , a2 , · · · , an ) = {x ∈ [0, 1 ) : a1 (x ) = a1 , a2 (x ) = a2 , · · · , an (x ) = an }.

BM = {x ∈ [0, 1 ) : 1 ≤ an (x ) ≤ M for all n ≥ 1}.

Remark: Eq. (1.4) is a corollary of Theorem 1.1 with ϕ (n ) = n.

x=

n ≥ 1 and (a1 , a2 , · · · , an ) ∈ Nn . Denote a cylinder of order n by

From Proposition 2.1 and the above lemmas, we have





1 where an (x ) = T n−1

is called the nth partial quotient of x. Let (x )

pn (x ) = an (x ) pn−1 (x ) + pn−2 (x ),

for x ∈ (0, 1 ),

where · denotes the integer part of a real number. Denote by μ the unique T-invariant ergodic measure which is equivalent with the Lebesgue measure L, usually called the Gauss measure, and τ n (x) denotes the first return time of the point x to the cylinder of order n which contains x. From [14] and (1.1), we obtain

lim

475

(2.1)

1−

1 1 ≤ dimH BM ≤ 1 − . M log 2 8M log M

In order to prove our theorem, we also need the following lemma. Lemma 2.6 ([14]). For any ρ > 0, let N ≥ 4/ρ and

FN,M = {x ∈ BM : ai = M for 1 ≤ i ≤ N, akN+1 = a(k+1)N = 1 for all k ≥ 1}. Then dimH FN,M ≥ dimH BM − ρ .

476

C. Wei et al. / Chaos, Solitons and Fractals 114 (2018) 474–477

3. Proofs of Theorem 1.1

Therefore the desired set in (3.5) is a subset of the set A as in (3.3). From (3.4) we can obtain (3.5). 

3.1. Proof of the zero-dimensional part of Theorem 1.1

Proof of the zero-dimensional part of Theorem 1.1. Suppose α < 1 1 1 λ or γ < δ . The assumption α < λ implies α < +∞ and λ < +∞. Otherwise, for example, α = +∞, then λ1 > +∞, which is impossible. Similarly, the fact γ < 1δ indicates that γ < +∞ and δ < +∞. Since

Note that the system of continued fraction is the attractor of the conformal iterated function system (IFS) generated by {φi (x ) = 1/(i + x ) : i ∈ N}. Let be a finite or countable subset of N. Let  = {φi (x ) : i ∈ } and K be the attractor of . Suppose f : [0, 1] → R+ is a positive function with the tempered distortion property (the definition can be found in [16]). Given a finite word w1 w2 wn ∈ n , for each x ∈ In (w1 , , wn ), denote −1 S n f ( x ) = f ( x ) + f ( φw (x )) + · · · + f ((φw1 ◦ · · · ◦ φwn−1 )−1 (x )). 1

lim inf n→∞

and

lim inf n→∞

Put

log τn (x ) log τn (x ) ϕ (n ) ≤ lim inf · lim sup = αλ n→∞ log n ϕ (n ) log n n→∞ log τn (x ) log τn (x ) ϕ (n ) ≤ lim sup · lim inf = γ δ, n→∞ log n log n ϕ (n ) n→∞

we have

R( f ) = {x ∈ K : |x − T n x| < e−Sn f (x ) , i.o. n ∈ N}. −1 Since T (x ) = φw (x ) and T n (x ) = (φw1 ◦ · · · ◦ φwn )−1 (x ), 1 Theorem 1.1 in [16], we have

dimH R( f ) = inf{t ≥ 0 : P (−t (log |() | + f )) ≤ 0},

from

lim inf n→∞

log τn (x ) < 1. log n ϕ

(3.1)

From Lemma 3.2, we have dimH Eα ,γ = 0.



where P is the pressure function which is defined as

3.2. Proof of the one-dimensional part of Theorem 1.1

P (−t (log |(−1 ) | + f ))  1 = lim log sup (|(T n ) (x )|−1 e−Sn f (x ) )t . n→∞ n n x∈In (w )

Suppose that α ≥ λ1 and γ ≥ 1δ . In our proof, we will construct ϕ a subset of Eα ,γ with full Hausdorff dimension. We begin with the following lemma.

(3.2)

w∈

Since the set BM is the attractor of {φi }M , we have the followi=1 ing lemma. Lemma 3.1. Let  = {φi }M and R(b) = R( f ) with f (x ) = b log 2, i=1 b > 0, then

dimH R( f ) ≤ b−1 log2 M. Proof. Since f (x ) = b log 2, we have Sn f (x ) = bn log 2 for each x ∈ BM , and hence





R(b) = x ∈ BM : |T n (x ) − x| < 2−bn , i.o. n ∈ N . Note that because T is expanding, we have that for any w ∈ n , x ∈ In (w), |(Tn ) (x)| ≥ 1. From (3.2), we have

P (−t (log |(−1 ) | + b)) ≤ lim

n→∞

 1 log 2−bnt = log M − bt log 2. n n w∈

Taking t = log2 M, we obtain P (−(b−1 log2 M )(log |(−1 ) | + b)) ≤ 0. By (3.1) we have dimH R( f ) ≤ b−1 log2 M.  b−1



1/ ( α +ε )



, i.o. n ∈ N .

(3.3)

Then for any M ≥ 1, b > log M, we have A ∩ BM ⊂ R(b). Letting b → ∞ we obtain that dimH A ∩ BM = 0 by Lemma 3.1. Since A = ∪M≥1 (A ∩ BM ), we have

dimH A = 0.

(3.4)

Using (3.4), we can obtain the following lemma. Lemma 3.2. Let α < 1. Then



dimH x ∈ [0, 1 ) : lim inf n→∞

log τn (x ) ≤α log n

Proof. From the condition lim inf n→∞



log τn (x ) log n

= 0.

(3.5)

≤ α , we know that for

any 0 < ε < 1 − α , there exist infinitely many n s such that τn (x ) < nα +ε , which says n > τn (x )1/(α +ε ) . By the definition of τ n (x) and Lemma 2.2, we obtain that

|T τn (x) (x ) − x| ≤ |In (x )| ≤

1 1/ ( α +ε ) ≤ 21−n < 21−τn (x ) . qn ( x )2

(ii) lim

k→∞

k(nk +3 ) lk

= 0.

Denote

S ({nk }, {lk } ) = {x ∈ [0, 1 ) : τn (x ) = lk+1 for all n with nk < n ≤ nk+1 }.

Then dimH S({nk }, {lk } ) = 1. Proof. For any ρ > 0, let N > 4/ρ , then dimH FN,M ≥ dim BM − ρ , where FN, M is defined in Lemma 2.6. We will construct a map f: FN, M → S({nk }, {lk }) in the following and show that f −1 is (1 − ε )Hölder continuous on f(FN, M ) for any 0 < ε < 1. For any x = [a1 (x ), a2 (x ), · · · ] ∈ FN,M , we construct a sequence ξ (k) = (ξik )i≥1 ∈ NN as follows. Let k0 be the minimal integer such that lk0 > N and ξ (k0 −1 ) = · · · = ξ (1 ) = ξ (0 ) = a1 (x )a2 (x ) · · · . Assuming ξ (k−1 ) is defined (k ≥ k0 ), we obtain ξ (k) according to the inser) ) −1 ) tion of uk := 1ξ1(k−1 ) · · · ξn(kk−1 ) ηn(k−1 1 with ηn(k−1 = ξn(k+1 between +1 +1 k

the positions lk − 1 and lk of ξ (k−1 ) , that is,

For any positive numbers α , ε with α + ε < 1, denote

A = x ∈ [0, 1 ) : |T n (x ) − x| < 21−n

Lemma 3.3. Let {nk }k ≥ 1 and {lk }k ≥ 1 be two strictly increasing sequences of natural numbers satisfying the following conditions: (i) lk+1 ≥ lk + nk + 3;

k

k

ξ (k) = (ξi(k) )i≥1 = ξ1(k−1) · · · ξl(k−1−1) uk ξl(k−1) ξl(k+1−1) · · · . k

Then

we

obtain

a

k

{ξ (k) }

sequence

k

k≥1

with

) ξ1(k) · · · ξl(k−1 = k

−1 ) ξ1(k−1) · · · ξl(k−1 . Let ξ ∗ = (ξi∗ )i≥1 be the limit point of {ξ (k) }k ≥ 1 , k ∗ which says ξ is obtained by inserting the sequence of words {uk }k≥k0 in a1 (x)a2 (x). Put x∗ = [ξ1∗ , ξ2∗ , · · · ] as in (2.1).

Now we prove that x∗ ∈ S({nk }, {lk }). Firstly, for any nk < n ≤ nk+1 with nk > N, k ≥ 1, we have τn (x∗ ) ≤ lk+1 by the definition of ξ ∗ . It remains to be shown that τn (x∗ ) < lk+1 can not happen. In fact, from the structure of ξ ∗ , since ξ1∗ · · · ξn∗ begins with N consecutive M which cannot appear in any word ai+1 (x ) · · · ai+N (x ) with i ≥ N, then the only possible places where ξ1∗ · · · ξn∗ may appear are ) −1 ) ui for some i. However, from ηn(i−1 = ξn(i+1 and ni < n, we know +1 i

i

that ξ1∗ · · · ξn∗ cannot appear in ui for any 1 ≤ i ≤ k. So, τn (x∗ ) = lk+1 holds. Define the function f as f (x ) = x∗ for any x ∈ FN, M . Then f is injective and f(FN, M ) ⊂ S({nk }, {lk }). We consider f −1 on f(FN, M ) as f −1 (x∗ ) = x, that is, the words {uk } are removed from ξ ∗ . We claim

C. Wei et al. / Chaos, Solitons and Fractals 114 (2018) 474–477

that f −1 is (1 − ε )-Hölder continuous for any ε > 0. In fact, for any x∗ , y∗ ∈ In (x∗ ), where n is the largest integer such that y∗ ∈ In (x∗ ), there is an integer k such that lk < n ≤ lk+1 . From the definition of f −1 , we have x, y ∈ In (x ) where n is the largest integer such that y ∈ In (x ). Therefore, we have n ≥ lk+1 − ki=1 (ni + 3 ). From condition (ii), for any 0 < ε < 1, we obtain n ≥ (1 − n is large enough. So, by Lemma 2.3, we have

log 2 1 2 log(M+1 ) ε )n

when

have

≤ |In (x∗ )| ≤

1 q2n (x∗ )

and Lemma 2.4, we

Since |x − y| ≤ |In (x )| ≤ Lemma 2.2, we have

by

the

first

inequality

| x − y | ≤ C ( ε ) | x ∗ − y ∗ | 1 −ε .

of

(3.6)

(3.7)

is (1 − ε )-Hölder continuous on f(FN, M ). Therefore,

Proof of the one-dimensional part of Theorem 1.1. For any x ∈ S({nk }, {lk }), since τn (x ) = lk for all nk−1 + 1 ≤ n ≤ nk , we have

log τn (x ) log lk log τn (x ) = lim inf , lim sup ϕ (n ) ϕ (n ) k→∞ ϕ (nk ) n→∞ log lk = lim sup . k→∞ ϕ (nk−1 + 1 )

If we can choose two sequences {nk } and {lk } satisfying Lemma 3.3, which only depend on ϕ , α and γ , such that

lim inf k→∞

log lk

ϕ ( nk )

= α,

lim sup k→∞

log lk

ϕ (nk−1 + 1 )

= γ,

ψ (n ) log n

= δ,

lim sup n→∞

Define Fα ,γ =

ψ (n ) log n

= λ.



 x ∈ [0, 1 ) : lim inf n→∞

log κn (x ) log κn (x ) = α , lim sup =γ . ψ (n ) ψ (n ) n→∞

We can thus obtain the same 0–1 law for the Hausdorff dimension ψ of Fα ,γ as Theorem 1.1.

and the last inequality is obtained by Lemma 2.5 and 2.6. Letting ε → 0, N → ∞, ρ → 0, M → ∞, we obtain that dimH S({nk }, {lk } ) = 1. 

n→∞

 1 1 + , d1 ( x ) d1 (x )(d1 (x ) − 1 ) · · · dn−1 (x )(dn−1 (x ) − 1 )dn (x )

Jn (d1 , d2 , · · · , dn ) = {x ∈ (0, 1] : d1 (x ) = d1 , d2 (x ) = d2 , · · · , dn (x ) = dn }.

ψ

dimH S({nk }, {lk } ) ≥ dimH f (FN,M ) ≥ (1 − ε ) dimH FN,M 1 ≥ (1 − ε )(1 − − ρ ), M log 2

lim inf

∞

x=

n→∞

which says that there exists a constant C(ε ) such that

Then



1 , n ≥ 1. n+1 n ,

Then every x ∈ (0, 1] can be represented uniquely as the following infinite series:

lim inf

|x − y| −(1− log 2 2 )ε 2 n log(M+1 ) ≤ 2 2 −ε ( M + 1 ) 3 ( 1 −ε ) 2 , | x ∗ − y ∗ | 1 −ε

f −1

1

For any x ∈ (0, 1], let Jn (x) := Jn (d1 (x), d2 (x), , dn (x)) be the cylinder of n containing x, and κ n (x) the first return time of x to Jn (x). Let ψ : N → R+ be a monotonically increasing function. Denote

1 1 1 |In (x∗ )| ≥ 2 ( M + 1 )3 4(M + 1 )3 q2n (x∗ ) 1 1 ≥ . 4(M + 1 )3 2εn q2n (x )

|x∗ − y∗ | ≥

q−2 ( x ), n

when x ∈

with d1 (x ) = 1x + 1 and dn+1 (x ) = d1 (Sn x ). For any n ≥ 1 and (d1 , d2 , · · · , dn ) ∈ Nn with di ≥ 2 (1 ≤ i ≤ n), we can also define the cylinder of order n as

qn ( x ) 1 ≤ (M + 1 )(n−n ) ≤ 2 2 εn . qn ( x ) 1 2q2n (x∗ )



1 , n+1

n=2

∗

Combining with

S (x ) = n (n + 1 ) x −

477

(3.8)

ϕ

then S({nk }, {lk } ) ⊂ Eα ,γ . This completes the proof of the onedimensional part of Theorem 1.1. The choices of such {nk } and {lk } are guaranteed by applying the following key lemma.  Lemma 3.4 ([11]). Let ϕ : N → R+ be a monotonically increasing function. Then, for any 0 ≤ α ≤ γ ≤ ∞ with α ≥ λ1 and γ ≥ 1δ , we can choose two sequences {nk } and {lk } satisfying (i) and (ii) in Lemma 3.3 and (3.8). 4. Recurrence spectrum of Lüroth expansion The method presented in this paper can be applied to the Lüroth expansion. The Lüroth expansion is induced by the map S: (0, 1] → (0, 1] defined as

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