Shrinking targets and eventually always hitting points for interval maps

TL;DR

This paper investigates the measure and dimension of points that consistently hit shrinking targets in various interval maps, providing conditions for full or zero measure and calculating packing dimensions.

Contribution

It establishes necessary and sufficient conditions for the measure of the set of eventually always hitting points in several interval maps, including the doubling and Manneville-Pomeau maps.

Findings

Full measure conditions for the set of hitting points.

Zero measure conditions for the set of hitting points.

Dimension calculations for the set in the doubling map.

Abstract

We study shrinking target problems and the set $\mathcal{E}_{\text{ah}}$ of eventually always hitting points. These are the points whose first $n$ iterates will never have empty intersection with the $n$-th target for sufficiently large $n$. We derive necessary and sufficient conditions on the shrinking rate of the targets for $\mathcal{E}_{\text{ah}}$ to be of full or zero measure especially for some interval maps including the doubling map, some quadratic maps and the Manneville-Pomeau map. We also obtain results for the Gauss map and correspondingly for the maximal digits in continued fractions expansions. In the case of the doubling map we also compute the packing dimension of $\mathcal{E}_{\text{ah}}$ complementing already known results on the Hausdorff dimension of $\mathcal{E}_{\text{ah}}$.

Figures

Equations

$${\mathcal{A}}_{\text{i.o.}}={\mathcal{A}}_{\text{i.o.}}(\mathbf{B}):=\left\{x\in X:T^{n}x\in B_n\text{ for infinitely many }n\in \text{mathbbm}N\right\}.$$

$$\left\{x,T(x),T^2(x),\dots ,T^{m-1}(x)\right\}\cap B_m\ne \varnothing .$$

$${\mathcal{E}}_{\text{ah}}$$

$$=\bigcup _{n=1}^{\infty }\bigcap _{m=n}^{\infty }\bigcup _{k=0}^{m-1}{T}^{-k}(B_m).$$

$$\Lambda :=\bigcup _{k=0}^{\infty }{T}^{-k}\left(\bigcap _{i=1}^{\infty }{B}_i\right).$$

$$\operatorname{dim_{H}}\left(\left\{\,x\mathrel{}:\mathrel{}\hat{\nu}_{\beta}(x)\geq\hat{\nu}\,\right\}\right)=\Bigl{(}\frac{1-\hat{\nu}}{1+\hat{\nu}}\Bigr{)}^{2}$$

$$0<\operatorname{dim_{H}}\mathcal{E}_{\textup{ah}}=\Bigl{(}\frac{1-s}{1+s}\Bigr{)}^{2}<\operatorname{dim_{P}}\mathcal{E}_{\textup{ah}}=1-s<1$$

$$g_{\alpha }(x)=\{\begin{array}{ll}x(1+2^{\alpha }{x}^{\alpha })& \text{if }x\in [0,\frac{1}{2})\\ 2x-1& \text{if }x\in [\frac{1}{2},1).\end{array}$$

$$x=\frac{1}{a_1(x)+\frac{1}{a_2(x)+\frac{1}{a_3(x)+\frac{1}{\ddots }}}}$$

$$\mu\Bigl{(}\Bigl{\{}\,x\in[0,1]:\exists\,m_{0}\in\mathbbm{N}\,\forall m\geq m_{0}:\max_{1\leq k\leq m}a_{k}(x)\geq cm\,\Bigr{\}}\Bigr{)}=0.$$

$$\mu\biggl{(}\biggl{\{}\,x\in[0,1]:\exists\,m_{0}\in\mathbbm{N}\,\forall m\geq m_{0}:\max_{1\leq k\leq m}a_{k}(x)\geq\frac{cm}{(\log m)^{2}}\,\biggr{\}}\biggr{)}=1.$$

$${\mathcal{E}}_{\text{ah}}=\bigcup _{n=1}^{\infty }\bigcap _{m=n}^{\infty }\bigcup _{k=0}^{m-1}{T}^{-k}(B_m).$$

$${\mathcal{E}}_{\text{ah}}=\bigcup _{n=1}^{\infty }{A}_n$$

$$A_n=\bigcap _{m=n}^{\infty }{C}_m,C_m=\bigcup _{k=0}^{m-1}{T}^{-k}(B_m).$$

$$\mu \left({\mathcal{E}}_{\text{ah}}\right)=\underset{n\to \infty }{\lim }\mu (A_n).$$

$$\mu \left({\mathcal{E}}_{\text{ah}}△T^{-1}\left({\mathcal{E}}_{\text{ah}}\right)\right)=0.$$

$$T^{k-1}(T(x))=T^k(x)\in B_m\subseteq B_{m-1}.$$

$$\mu \left({\mathcal{E}}_{\text{ah}}\right)=\mu \left({\mathcal{E}}_{\text{ah}}\setminus \bigcap _{m\in \text{mathbbm}N}{B}_m\right)\le \mu \left(T^{-1}\left({\mathcal{E}}_{\text{ah}}\right)\right)=\mu \left({\mathcal{E}}_{\text{ah}}\right),$$

$$\mu \left(\bigcup _{n=1}^{\infty }{A}_n\right)=\underset{n\to \infty }{\lim }\mu (A_n)=\underset{n\to \infty }{\lim }\mu \left(\bigcap _{m=n}^{\infty }{C}_m\right)=1$$

$$\displaystyle\mu(C_{m})=\mu\left(\bigcup_{k=0}^{m-1}T^{-k}(B_{m})\right)\leq\sum_{k=0}^{m-1}\mu\bigl{(}T^{-k}(B_{m})\bigr{)}=m\mu(B_{m})$$

$${\tau }_E(x)=\inf \left\{i\in \text{mathbbm}N:T^{i}x\in E\right\}.$$

$$G_{E_m}(t):=\underset{m\to \infty }{\mathrm{limsup}}\mu \left(\left\{x\in X:{\tau }_{E_m}(x)\le \frac{t}{\mu (E_m)}\right\}\right).$$

$$\left\{x\in X:{\tau }_{B_{m_j}}(x)\le m_j\right\}\subset \left\{x\in X:{\tau }_{B_{m_j}}(x)\le \frac{c}{\mu (B_{m_j})}\right\}.$$

$${\mathcal{E}}_{\text{ah}}$$

$$=\bigcup _{n=1}^{\infty }\bigcap _{m=n}^{\infty }\left\{x\in X:{\tau }_{B_m}(x)<m\right\}$$

$$\subset \bigcup _{n=1}^{\infty }\bigcap _{m_j\ge n}\left\{x\in X:{\tau }_{B_{m_j}}(x)\le m_j\right\}\subset \bigcup _{n=1}^{\infty }\bigcap _{m_j\ge n}{\tilde{C}}_{m_j},$$

$${\tilde{C}}_{m_j}=\left\{x\in X:{\tau }_{B_{m_j}}(x)\le \frac{c}{\mu (B_{m_j})}\right\}.$$

$$\mu \left(\left\{x\in X:{\tau }_{B_{m_j}}(x)\le \frac{c}{\mu (B_{m_j})}\right\}\right)\to 1$$

$$G_{B_m}(t)=\underset{m\to \infty }{\lim }\mu \left(\left\{x\in X:{\tau }_{B_m}(x)\le \frac{t}{\mu (B_m)}\right\}\right)$$

$$\Vert f{\Vert }_1$$

$$\Vert f{\Vert }_{BV}$$

$$\biggl{|}\int f\circ T^{n}g\,\mathrm{d}\mu-\int f\,\mathrm{d}\mu\int g\,\mathrm{d}\mu\biggr{|}\leq\lVert f\rVert_{1}\lVert g\rVert_{BV}p(n)$$

Full text

We thank both Jörg Schmeling and Dmitry Kleinbock for interesting discussions and helpful comments and suggestions which improved the paper. We thank Jorge Freitas for answering our questions about Hitting Time Statistics for the Gauß map. We also acknowledge financial support by the Hamburg–Lund Funding Program 2018 which made several mutual research visits possible.

1. Introduction and setup

The term shrinking target problems in dynamical systems describes a class of questions which seek to understand the recurrence behaviour of typical orbits of a dynamical system. The standard ingredients of such questions are a measure-preserving dynamical system $(X,\mu,T)$, $T\colon X\to X$ and $\mu$ being a finite measure. Also, we have a sequence of subsets $\left\{B_{m}\right\}_{m=1}^{\infty}$ with $B_{m}\subset X$ and $\mu(B_{m})\to 0$. Recently shrinking target problems have also been investigated in the case when the measure is infinite, see [10].

In this paper we focus mostly on the case of finite measure and whenever this is the case we assume the measure to be normalized to a probability measure. If nothing else is stated this may be assumed to be the setting. A few of our results concern infinite measures and it will be stated explicitly whenever this is the case.

Throughout this paper $(X,T,\mu)$ will always denote a measure preserving system and $\mathbf{B}:=\left\{B_{m}\right\}_{m=1}^{\infty}$ will always denote a sequence of subsets of $X$ for which $\mu(B_{m})\to 0$. We refer to this as a sequence of shrinking targets. We call the sequence nested if $B_{m}\supset B_{m+1}$ for all $m$.

Classical questions in this area focus on the set of points in $X$, whose $n$’th iterate under $T$, lies in the set $B_{n}$ for infinitely many $n$. That is, given a sequence $\mathbf{B}=\left\{B_{m}\right\}_{m=1}^{\infty}$

[TABLE]

If $\sum\mu(B_{m})<\infty$ the Borel–Cantelli Lemma tells us that $\mu(\mathcal{A}_{\textup{i.o.}})=0$. If $\sum\mu(B_{m})=\infty$ the situation is more complicated since the Borel–Cantelli Lemma only guarantees $\mu(\mathcal{A}_{\textup{i.o.}})=1$ for independent events and this is usually not satisfied for dynamical systems. If we do have a sequence $\mathbf{B}$ for which $\mu(\mathcal{A}_{\textup{i.o.}}(\mathbf{B}))=1$ then we call $\mathbf{B}$ a Borel–Cantelli (BC) sequence. If we can prove that a large family $\mathcal{B}$ of sequences are all BC-sequences then we say that we have a dynamical Borel–Cantelli Lemma. In many cases such lemmas hold if the system satisfies some version of mixing which essentially acts as a replacement for independence. However, it is known that for any measure-preserving system we can find a sequence $\mathbf{B}$ satisfying $\sum\mu(B_{m})=\infty$ which is not BC for the system. We may even find such a sequence $\mathbf{B}$ which is nested [3, Proposition 1.6]. Hence there is no hope for $\mathcal{B}$ to be all sequences satisfying $\sum\mu(B_{m})=\infty$. It is therefore natural to look for the largest possible sub-families of $\mathcal{B}$ which consist only of BC-sequences. It turns out that sequences of balls with fixed center, and nested sequences of balls with fixed center are good and natural candidates. We say that $(X,\mu,T)$ has the shrinking target property (STP) if for any $x_{0}\in X$, every sequence of balls $B_{m}$ centered at $x_{0}$ satisfying $\sum\mu(B_{m})=\infty$ is a BC-sequence. We say that $(X,\mu,T)$ has the monotone shrinking target property (MSTP) if for any $x_{0}\in X$, every nested sequence of balls $B_{m}$ centered at $x_{0}$ satisfying $\sum\mu(B_{m})=\infty$ is a BC-sequence. Many interesting systems are known to have either the STP or MSTP property, see [1], [22], and references therein for examples. A more comprehensive introduction to dynamical Borel–Cantelli lemmas, including examples, can be found in [3].

In this paper we are interested in similar properties for a certain subset (aside for a set of measure 0) of $\mathcal{A}_{\textup{i.o.}}$ known as the set of points which are eventually always hitting. Due to the central importance of this concept we introduce it through a separate definition.

Definition 1** (Eventually always hitting).**

A point $x\in X$ is said to be eventually always hitting (EAH) for $\mathbf{B}=\left\{B_{m}\right\}_{m=1}^{\infty}$ under $T$ if there exists some $m_{0}(x)\in\mathbbm{N}$ such that for all $m\geq m_{0}(x)$ we have

[TABLE]

The set of all points in $X$ which are eventually always hitting for $\mathbf{B}$ under $T$ will be denoted $\mathcal{E}_{\textup{ah}}:=\mathcal{E}_{\textup{ah}}(\mathbf{B})$.

We remark that some authors study a slightly different version of eventually almost hitting, and require that for all $m\geq m_{0}(x)$ we have $T^{k}(x)\in B_{m}$ for some $k\leq m$, whereas we require $0\leq k<m$. For the results that we are discussing in this paper, it is unimportant which definition we use. The results are the same, with the same proofs, if we use the other definition instead.

Note that $\mathcal{E}_{\textup{ah}}$ may also be written as

[TABLE]

Set

[TABLE]

Then $\mu(\Lambda)=0$ since $\mu(B_{i})\to 0$, and we have that $\mathcal{E}_{\textup{ah}}\backslash\Lambda\subset\mathcal{A}_{\textup{i.o.}}$. In this sense, being eventually always hitting for $B_{m}$ is a stronger property than hitting $B_{m}$ infinitely often.

The term eventually always hitting was coined by Kelmer in [13] where this set was studied in the context of flows on hyperbolic manifolds. Kelmer proved necessary and sufficient conditions for the set of eventually always hitting points to be of full measure. Shortly afterwards Kelmer and Yu [14] extended the investigation to flows on homogeneous spaces. Also, Kleinbock and Wadleigh [16] studied the concept in the context of higher dimensional Diophantine approximations. Imposing a long-term independence property on the shrinking target system Kleinbock, Konstantoulas and Richter [15] recently obtained tight conditions on the shrinking rate of the targets so that $\mathcal{E}_{\textup{ah}}$ has measure zero or full measure. In particular, their assumptions are satisfied for specific choices of targets in product systems and Bernoulli shifts.

However, the concept had already been considered a few years earlier by Bugeaud and Liao [2] for a particular sequence of targets with exponential rate of shrinking in the setting of $\beta$-transformations $T_{\beta}(x)=\beta x-\lfloor\beta x\rfloor$ on $[0,1]$ for every $\beta>1$. For $x\in[0,1]$ they introduce the exponent $\hat{\nu}_{\beta}(x)$ as the supremum of real numbers $\hat{\nu}$ for which for every sufficiently large $N\in\mathbbm{N}$ the inequality $T^{n}_{\beta}(x)<\left(\beta^{N}\right)^{-\hat{\nu}}$ has a solution $1\leq n\leq N$. Note that this corresponds to $x$ satisfying $\{T_{\beta}^{n}(x)\}_{n=1}^{N}\cap B(0,\beta^{-\hat{\nu}N})\neq\emptyset$ for every sufficiently large $N$. Hence the set $\{x\in[0,1]:\hat{\nu}_{\beta}(x)\geq\hat{\nu}\}$ corresponds to $\mathcal{E}_{\textup{ah}}(B(0,\beta^{-\hat{\nu}N}))$ in our notation (aside from the discrepancy in definition mentioned above). They show that

[TABLE]

for every $\hat{\nu}\in[0,1]$, where $\operatorname{dim_{H}}$ is the Hausdorff dimension. Bugeaud and Liao also obtain analogous results in the setting of $b$-ary expansions.

In this paper we prove various results concerning the measure and also dimension of $\mathcal{E}_{\textup{ah}}$ in different settings. Our different results apply in various levels of generality, hence it would be complicated to state them all accurately in this introduction. Instead we illustrate our results through application to specific systems with simpler assumptions.

Note that throughout the paper $\log$ will denote the natural logarithm.

1.1. Main results for specific systems

Theorem 1** (The doubling map).**

Let $X=[0,1]$, let $T\colon X\to X$ be the doubling map $Tx=2x\mod 1$ and let $\mu$ denote the Lebesgue measure on $X$. Let $\left\{B_{m}\right\}_{m=1}^{\infty}$ denote a nested sequence of shrinking intervals with fixed center.

1. (a)

If $\mu(B_{m})\leq\frac{c}{m}$ for some $c\in\mathbbm{R}$, then $\mu(\mathcal{E}_{\textup{ah}})=0$. 2. (b)

If $\mu(B_{m})\geq\frac{c(\log m)^{2}}{m}$ for some $c>0$ sufficiently large, then $\mu(\mathcal{E}_{\textup{ah}})=1$.

Let $B_{m}=B(0,2^{-sm})$, $s\in\mathbbm{R}$ and $\operatorname{dim_{P}}$ denote the packing dimension.

1. (c)

If $s\in(0,1)$, then $\operatorname{dim_{P}}\mathcal{E}_{\textup{ah}}=1-s$. 2. (d)

If $s\geq 1$ then $\mathcal{E}_{\textup{ah}}$ is countable and in particular $\operatorname{dim_{P}}\mathcal{E}_{\textup{ah}}=0$.

We remark that we prove the statements (a) and (b) in the theorem above for many other dynamical systems, for instance piecewise expanding maps and some quadratic maps. See Section 4 for more details.

Note also that for the doubling map, and $B_{m}=B(0,2^{-sm})$, the result of Bugeaud and Liao [2] implies that \operatorname{dim_{H}}\mathcal{E}_{\textup{ah}}=\bigl{(}\frac{1-s}{1+s}\bigr{)}^{2}. Hence we have

[TABLE]

when $s\in(0,1)$.

Theorem 2** (The Manneville–Pomeau map).**

Let $X=[0,1]$, let $\alpha>0$ and let $g_{\alpha}\colon[0,1)\to[0,1)$ be the Manneville–Pomeau map given by

[TABLE]

Let $\mu_{\alpha}$ denote the absolutely continuous invariant measure for $([0,1),g_{\alpha})$, which is finite if and only if $\alpha\in(0,1)$. Let $\left\{B_{m}\right\}_{m=1}^{\infty}$ denote a nested sequence of intervals with a fixed center which is not [math].

If $\alpha\in(0,1)$ then we have the following results.

1. (a)

If $\mu_{\alpha}(B_{m})\leq\frac{c}{m}$ for some constant $c$, then $\mu_{\alpha}(\mathcal{E}_{\textup{ah}})=0$. 2. (b)

If $\mu_{\alpha}(B_{m})\geq\frac{c(\log m)^{2+\varepsilon}}{m}$ for some $\varepsilon>0$ and any $c>0$, then we have $\mu_{\alpha}(\mathcal{E}_{\textup{ah}})=1$.

If $\alpha\geq 1$ then we have the following results.

1. (c)

If $\alpha>1$ and $\mu_{\alpha}(B_{m})\leq\frac{c}{m^{\frac{1}{\alpha}+\varepsilon}}$ for some $\varepsilon>0$ and any $c>0$, then $\mu_{\alpha}(\mathcal{E}_{\textup{ah}})=0$.

If $\alpha=1$ and $\mu_{\alpha}(B_{m})\leq\frac{c}{m}$ for some $c>0$ then $\mu_{\alpha}(\mathcal{E}_{\textup{ah}})=0$. 2. (d)

If $\mu_{\alpha}(B_{m})\geq\frac{c}{m^{\frac{1}{\alpha}-\varepsilon}}$ for some $\varepsilon>0$ and any $c>0$, then $\mu_{\alpha}(\complement\mathcal{E}_{\textup{ah}})=0$.

We remark that the measure $\mu_{\alpha}$ is equivalent to Lebesgue measure on $[0,1]$. Hence the statements $\mu_{\alpha}(\mathcal{E}_{\textup{ah}})=0$ and $\mu_{\alpha}(\complement\mathcal{E}_{\textup{ah}})=0$ are equivalent to the corresponding statements involving the Lebesgue measure instead. Note also that in the case $\alpha\in(0,1)$ after normalising the measure $\mu_{\alpha}$ we can state $\mu_{\alpha}(\complement\mathcal{E}_{\textup{ah}})=0$ as the equivalent statement $\mu_{\alpha}(\mathcal{E}_{\textup{ah}})=1$. This is not possible if $\alpha\geq 1$, since the measure $\mu_{\alpha}$ is not finite in this case.

We may also consider the Gauß map defined by $G:(0,1]\to[0,1)$ given by $G(x)=\frac{1}{x}\mod 1$. $G$ admits an absolutely continuous invariant probability measure known as the Gauss measure which has density $\frac{1}{\log 2}\frac{1}{1+x}$. Statement (b) of Theorem 1 also holds true for the Gauß map and measure, while (a) of Theorem 1 holds true in this setting when $B_{m}:=[0,r_{m})$. This allows us to obtain a statement about the eventually always hitting property for maximal digits of continued fractions expansions. Recall that every point $x\in(0,1]$ can be written as a continued fraction, i.e.

[TABLE]

where the $a_{i}(x)$’s are generated by the algorithm $a_{1}(x)=\lfloor\frac{1}{x}\rfloor$ and $a_{j}(x)=a_{1}(G^{j-1}(x))$. In compact notation we write $x=[a_{1}(x),a_{2}(x),\dots]$.

Theorem 3** (Continued fractions).**

Let $\mu$ denote the Gauß measure.

1. (a)

For any $c>0$ we have

[TABLE] 2. (b)

If $c>0$ is sufficiently small, then

[TABLE]

1.2. Structure of the paper

In Section 2 we give some preliminary, general results concerning the set of eventually always hitting points that will prove useful later in the paper. In Section 3 we give necessary and sufficient conditions for $\mathcal{E}_{\textup{ah}}$ to be of full measure. In Section 4 we cite various results on mixing and hitting time statistics which in conjunction with the results of Section 3 allows us to deduce the conclusions of Theorem 1, 2 and 3 (except Theorem 1(c) and (d)). We also discuss further systems for which the results of Section 3 can be applied. Section 5 is dedicated to the proof of Theorem 1(c) and (d).

2. Preliminaries on eventually always hitting points

Recall that

[TABLE]

It will prove convenient to write

[TABLE]

where

[TABLE]

Note in particular that $A_{n}\subset A_{n+1}$ and hence

[TABLE]

The following lemma will prove very useful for studying the measure of $\mathcal{E}_{\textup{ah}}$ when the sets $B_{m}$ are assumed to be nested. It states that $\mathcal{E}_{\textup{ah}}$ is an almost invariant set under $T$. More precisely, this means that the symmetric difference between $\mathcal{E}_{\textup{ah}}$ and $T^{-1}(\mathcal{E}_{\textup{ah}})$ is a set of measure zero.

Lemma 1**.**

Let $(X,\mu,T)$ be a measure-preserving dynamical system ($\mu$ either finite or infinite) and let $\{B_{m}\}_{m=1}^{\infty}$ be a nested family of shrinking targets, i.e. $B_{m}\supset B_{m+1}$ and $\mu(B_{m})\to 0$. Then

[TABLE]

Hence, for ergodic transformations with respect to a finite $\mu$, $\mathcal{E}_{\textup{ah}}$ obeys a zero–one law. That is, either $\mu\left(\mathcal{E}_{\textup{ah}}\right)=0$ or $\mu\left(\mathcal{E}_{\textup{ah}}\right)=1$. For ergodic transformations with respect to an infinite $\mu$, $\mathcal{E}_{\textup{ah}}$ obeys a zero–infinity law.

Proof.

Let $x\in\mathcal{E}_{\textup{ah}}$. Then there is $m_{0}:=m_{0}(x)\in\mathbbm{N}$ such that for all $m\geq m_{0}$ there is $k\in\{0,1,\dots,m-1\}$ with $T^{k}(x)\in B_{m}$. Actually, if $x\notin\bigcap_{m\in\mathbbm{N}}B_{m}$ there is even $\tilde{m}_{0}:=\tilde{m}_{0}(x)\in\mathbbm{N}$ such that for all $m\geq\tilde{m}_{0}$ there is $k\in\{1,\dots,m-1\}$ with $T^{k}(x)\in B_{m}$ (because otherwise $x=T^{0}(x)$ would have to be in $B_{m}$ for all $m$ due to $x\in\mathcal{E}_{\textup{ah}}$ and the nesting property). Since the targets are nested, we also get for all $k\neq 0$ that

[TABLE]

Hence, $T(x)\in\mathcal{E}_{\textup{ah}}$. So, $\mathcal{E}_{\textup{ah}}\setminus\bigcap_{m\in\mathbbm{N}}B_{m}\subseteq T^{-1}\left(\mathcal{E}_{\textup{ah}}\right)$. Since $\mu\left(\bigcap_{m\in\mathbbm{N}}B_{m}\right)$ is a set of measure zero by $\mu(B_{m})\to 0$ and $T$ is measure-preserving, we have

[TABLE]

which yields the claim. ∎

In [13] Kelmer gave the following simple conditions for $\mathcal{E}_{\textup{ah}}$ to be of measure zero or one. We repeat the proof for completeness. Note that no assumption is made on the shape of the target sets.

Proposition 1**.**

Let $\left\{B_{m}\right\}_{m=1}^{\infty}$ denote a sequence of shrinking targets in $X$.

1. (c)

Let $\mu$ be a probability measure and assume that $\mu(\mathcal{E}_{\textup{ah}})=1$. Then there exists a sequence $c_{m}\to 1$ such that $\mu(B_{m})\geq\frac{c_{m}}{m}$. 2. (d)

Let $\mu$ be an infinite measure and assume that $\mu(\mathcal{E}_{\textup{ah}})=\infty$. Then there exists a sequence $c_{m}\to\infty$ such that $\mu(B_{m})\geq\frac{c_{m}}{m}$.

Proof.

(c) By the assumption we get that

[TABLE]

which implies that $\mu(C_{m})\to 1$ as $m\to\infty$. Now,

[TABLE]

where we used the $T$-invariance of $\mu$. Hence we see that since $\mu(C_{m})\to 1$, we have $\mu(B_{m})\geq\frac{c_{m}}{m}$ for some sequence $c_{m}\to 1$, because otherwise there is an $\varepsilon>0$ and infinitely many $m$ for which $\mu(B_{m})\leq\frac{1-\varepsilon}{m}$, which contradicts the estimates above.

(d) The proof goes exactly like for (c) with the obvious adaptations. ∎

For nested sequences $\left\{B_{m}\right\}_{m=1}^{\infty}$ and $T$ being ergodic we get the following sufficient condition for $\mu(\mathcal{E}_{\textup{ah}})=0$. The proof is just the negation of Proposition 1 followed by an application of Lemma 1.

Corollary 1**.**

Assume that $T$ is ergodic and that $\left\{B_{m}\right\}_{m=1}^{\infty}$ is a nested sequence of shrinking targets.

1. (e)

Let $\mu$ be a probability measure. If there exists a $c<1$ such that $\mu(B_{m})\leq\frac{c}{m}$ holds for infinitely many $m$, then $\mu(\mathcal{E}_{\textup{ah}})=0$. 2. (f)

Let $\mu$ be an infinite measure. If there exists a $c\in\mathbbm{R}$ such that $\mu(B_{m})\leq\frac{c}{m}$ holds for infinitely many $m$, then $\mu(\mathcal{E}_{\textup{ah}})=0$.

3. Necessary and sufficient conditions for $\mu(\mathcal{E}_{\textup{ah}})=1$

In this section we give proofs of various new necessary and sufficient conditions for $\mu(\mathcal{E}_{\textup{ah}})$ to be of measure zero or one.

3.1. Necessary conditions for $\boldsymbol{\mu(\mathcal{E}_{\textup{ah}})=1}$

We introduce some terminology and notation about hitting times in dynamical system. Given a set $E\subset X$, we denote by $\tau_{E}\colon X\to\mathbbm{N}$ the first hitting time to $E$ which is defined by

[TABLE]

Let $E_{m}\subset X$ denote a sequence of sets for which $\mu(E_{m})\to 0$ and define the function

[TABLE]

The next easy proposition gives a necessary condition for $\mathcal{E}_{\textup{ah}}$ to be of full measure when $G(t)<1$ for all $t\in\mathbbm{R}$.

Proposition 2**.**

Let $\left\{B_{m}\right\}_{m=1}^{\infty}$ denote a sequence of shrinking targets and assume that $G_{B_{m}}(t)<1$ for all $t\in\mathbbm{R}$. If $\mu(\mathcal{E}_{\textup{ah}})=1$, then for every $c\in\mathbbm{R}$ we have $\mu(B_{m})\geq\frac{c}{m}$ for all sufficiently large $m\in\mathbbm{N}$.

Proof.

To get a contradiction, fix $c\in\mathbbm{R}$, and assume that there is a sequence $m_{j}\to\infty$ such that $\mu(B_{m_{j}})\leq\frac{c}{m_{j}}$ for all $j$. We then have

[TABLE]

Using this inclusion we may rewrite as follows

[TABLE]

where

[TABLE]

Assuming that $\mu(\mathcal{E}_{\textup{ah}})=1$, we argue as in Proposition 1 that we must have $\mu(\tilde{C}_{m_{j}})\to 1$ which means that

[TABLE]

for $m_{j}\to\infty$. However, $\limsup_{j\to\infty}\mu(\tilde{C}_{m_{j}})\leq G_{B_{m}}(c)<1$ by assumption and hence we have a contradiction. Since this inequality is true for all $c\in\mathbbm{R}$ we get the desired conclusion. ∎

Again we get a sufficient condition for $\mu(\mathcal{E}_{\textup{ah}})=0$ under additional assumptions.

Corollary 2**.**

Assume that $T$ is ergodic and that $\left\{B_{m}\right\}_{m=1}^{\infty}$ is a nested sequence of shrinking targets and assume that $G_{B_{m}}(t)<1$ for all $t\in\mathbbm{R}$. If there exists $c\in\mathbbm{R}$ such that $\mu(B_{m})\leq\frac{c}{m}$ for infinitely many $m$, then $\mu(\mathcal{E}_{\textup{ah}})=0$.

The condition $G_{B_{m}}(t)<1$ is easily satisfied in many cases. It is interesting to note that often much more is known about $G_{B_{m}}(t)$. If we assume $B_{m}$ to be a nested sequence of shrinking balls with fixed center it is often known that

[TABLE]

exists and is non-degenerate which means that $G_{B_{m}}(t)$ takes at least one value different than 0 or 1. This property is known as the system having hitting time statistics (HTS) to $B_{m}$. Among these systems, many have exponential HTS to $B_{m}$ meaning that $G_{B_{m}}(t)=1-e^{-t}$. It is from the rich theory of HTS for dynamical systems that we borrow in order to prove Theorem 1(a), 2(a) and 3(a). We elaborate on this point in Section 4.

3.2. Sufficient conditions for $\boldsymbol{\mu(\mathcal{E}_{\textup{ah}})=1}$

Let $X\subset\mathbbm{R}$ in our probability measure preserving system $(X,T,\mu)$ and let $B_{m}=B(y_{m},r_{m})$ be a sequence of balls in $X$. We consider the $L^{1}$ and $BV$ norms of functions on $f\colon X\to\mathbbm{R}$, defined by

[TABLE]

where $\operatorname{var}f$ denotes the total variation of $f$.

We say that correlations decay as $p\colon\mathbbm{N}\to\mathbbm{R}$ for $L^{1}$ against $BV$, if

[TABLE]

holds for all $n$ and all functions $f$ and $g$ with $\lVert f\rVert_{1}$, $\lVert g\rVert_{BV}<\infty$.

Theorem 4**.**

Suppose that correlations decay as $p$ for $L^{1}$ against $BV$.

If $p$ satisfies

[TABLE]

for some $\tau>0$, then $\mu(\mathcal{E}_{\textup{ah}})=1$ provided

[TABLE]

for some $c>\tau^{-1}$ and all sufficiently large $m$. In particular $\mu(\mathcal{E}_{\textup{ah}})=1$ provided $\mu(B_{m})\geq\frac{c(\log m)^{2}h(m)}{m}$ for any $c>0$ and any function $h$ for which $h(m)\to\infty$ as $m\to\infty$.

If $p$ satisfies

[TABLE]

for some $t>0$, then $\mu(\mathcal{E}_{\textup{ah}})=1$ provided

[TABLE]

for any $c>0$ and $a<\frac{t}{1+t}$.

Note that do not require that the balls are nested in Theorem 4. We will need the following lemma.

Lemma 2**.**

Suppose that (2) holds for all $n$ and functions $f$ and $g$. If $f_{k}\colon X\to[0,\infty)$ and $n_{k}\in\mathbbm{N}$, then

[TABLE]

Proof.

Note that since $f_{k}\geq 0$, we have $\int f_{k}\,\mathrm{d}\mu=\lVert f_{k}\rVert_{1}$. By (2), we have

[TABLE]

and the first inequality follows by induction. The second inequality then follows trivially. ∎

Proof of Theorem 4.

Recall that

[TABLE]

We prove that $\mu(A_{n})\to 1$ as $n\to\infty$ which is equivalent to $\mu(\complement A_{n})\to 0$. We can write $\complement A_{n}$ as

[TABLE]

Hence, it is sufficient to prove that

[TABLE]

We will now bound $\mu(D_{m})$ from above. Let $\Delta_{m}>1$. We have that

[TABLE]

Then, we have by Lemma 2 that

[TABLE]

Since $B_{m}$ are balls, we have $\operatorname{var}1_{\complement B_{m}}\leq 2$ and $\lVert 1_{\complement B_{m}}\rVert_{1}\leq 1$. Hence we have $\lVert 1_{\complement B_{m}}\rVert_{BV}\leq 3$ and we get

[TABLE]

where the last inequality holds since $\log(1+x)\leq x$.

Assume now that $p(n)\leq Ce^{-\tau n}$. We prove that $\mu(\mathcal{E}_{\textup{ah}})=1$ if for all sufficiently large $m$, we have $\mu(B_{m})\geq\frac{c(\log m)^{2}}{m}$ where $c>\tau^{-1}$.

Take $\Delta_{m}=K\log m$ with any $K>\tau^{-1}$. Then

[TABLE]

so we can assume $(m/\Delta_{m})3p(\Delta_{m})<1$ if $m$ is large.

We assume that we have $\mu(B_{m})\geq\frac{c(\log m)^{2}}{m}$ for all $m$. The proof also works with obvious changes if this is only the case for all large enough $m$. We have

[TABLE]

Taken together, these two estimates give us the estimate

[TABLE]

which implies that $\mu(\tilde{D}_{m})$ is summable if $\frac{c}{K}>1$. Since we may take $K$ as close to $\tau^{-1}$ as we like, we can make $\frac{c}{K}>1$ as long as $c>\tau^{-1}$.

We now consider the case that $p(n)\leq C/n^{t}$ and $\mu(B_{m})\geq\frac{c}{m^{a}}$. Let $\Delta_{m}=m^{\frac{1}{1+t}}$. Then

[TABLE]

Hence

[TABLE]

where $\beta=\frac{t}{1+t}-a$. If $a<\frac{t}{1+t}$, then $\beta>0$ and $\mu(\tilde{D}_{m})$ is summable, which proves (5). ∎

We note that the sufficient condition obtained by Kelmer [13] in the setting of discrete time homogeneous flows acting on a finite volume quotient of $\mathbbm{H}^{n}$ is slightly better than what we get in our setting. More precisely, [13, Theorem 2] states that in the mentioned setting

[TABLE]

Inserting either $\mu(B_{m})\geq\frac{c(\log m)^{2}}{m}$ or $\mu(B_{m})\geq\frac{c}{m^{a}}$ in the above both result in convergent sums. A bound like $\mu(B_{m})\geq\frac{c(\log m)}{m}$ or lower would be a sharper bound compared to that of Kelmer. However, the settings are very different and it is not clear what an optimal bound would look like in either setting.

4. Application to examples

Many systems are known to have either polynomial or exponential decay of correlation for $L^{1}$ against $BV$ in the sense of (3) and (4). Examples of exponential decay includes $T$ being a piecewise expanding interval map and $\mu$ being a Gibbs measure, $T$ being a quadratic map for a Benedicks–Carleson parameter and $\mu$ being the absolutely continuous invariant measure, [18, 23]. Hence, for these systems, $\mu(\mathcal{E}_{\textup{ah}})=1$ whenever $\mu(B_{m})\geq\frac{c(\log m)^{2}}{m}$ for some $c>0$ sufficiently large.

Hitting time statistics is known for many interesting dynamical systems. Since HTS is often true for systems with sufficiently nice mixing properties, our necessary and sufficient conditions tend to hold for many of the same systems. Examples of HTS in dynamics include transitive Markov chains, Axiom A diffeomorphisms, uniformly expanding maps of the interval, non-uniformly hyperbolic maps, partially hyperbolic dynamical systems and toral automorphisms. Hence, for these systems $\mu(\mathcal{E}_{\textup{ah}})=0$ whenever $\mu(B_{m})\leq\frac{c}{m}$ for some $c\in\mathbbm{R}$. For a comprehensive overview of these results and references, see [19, Chapter 5]. Furthermore, [6, 7] establish a direct connection between HTS and so-called extreme value theory meaning that many HTS results can be obtained simply by translating known extreme value laws. See again [19] for an overview of such results.

In the following subsections we describe in more detail how Theorem 1, 2 and 3 are deduced.

4.1. The doubling map

HTS is known to hold for the doubling map. The precise statement is as follows. Let $r_{m}\in\mathbbm{R}$ be a sequence and set $E_{m}=B(y,r_{m})$, i.e. the ball with center $y\in[0,1]$ and radius $r_{m}$. For any sequence $r_{m}\to 0$ we have

[TABLE]

This explicit result can be deduced by applying [6, Theorem 2] (along with [7, p.7]) to [19, Corollary 4.2.11]. Hence Theorem 1(a) follows directly from Corollary 2.

The doubling map with Lebesgue measure is a well-known example of a piecewise expanding interval map with a Gibbs measure. Hence it is exponentially mixing for $L^{1}$ against $BV$ [18] and Theorem 1(b) follows directly from Theorem 4.

4.2. The Manneville–Pomeau map

Here we explain how Theorem 2 follows from our results. We begin with Theorem 2(a).

For the Manneville–Pomeau maps the following is known regarding hitting time statistics in the case when $\alpha<1$, that is when the invariant measure $\mu_{\alpha}$ is finite. Let $E_{m}=B(y,r_{m})$ as above, then for any sequence $r_{m}\to 0$ we have

[TABLE]

where $Dg_{\alpha}^{p}(y)$ denotes the derivative of $g_{\alpha}^{p}$ at the point $y$. This follows again by applying [6, Theorem 2] (along with [7, p.7]) to [8, Theorem 1 and 2]. Hence Theorem 2(a) follows directly from Corollary 2. Note that Theorem 2(a) actually also holds for balls centered at 0 as well.

We proceed to deducing Theorem 2(b), (c) and (d). Note first that the case $\alpha=1$ in Theorem 2(c) follows directly from Corollary 1(f).

The Manneville–Pomeau map is not known to have decay of correlations for $L^{1}$ against $BV$. However, through a technique known as inducing, explained below, we can obtain results almost as strong as if it had exponential decay of correlations with respect to said norms. We let $S\colon[\frac{1}{2},1)\to[\frac{1}{2},1)$ be the first return map of $g_{\alpha}$ to the interval $[\frac{1}{2},1)$. The structure of the map $S$ is illustrated in Figure 1.

Given a point $x\in[\frac{1}{2},1)$, there is then a sequence $R_{k}(x)$ such that

[TABLE]

The sequence $R_{k}(x)$ satisfies

[TABLE]

where $R$ is the return time $R(x)=\min\{\,n\geq 1:g_{\alpha}^{n}(x)\in[\frac{1}{2},1)\,\}$.

The return map $S$ is uniformly expanding and it follows by the paper of Rychlik [20] that it has exponential decay of correlations for $L^{1}$ against $BV$. Hence we may apply Theorem 4 to $S$.

The absolutely continuous invariant measure $\mu_{\alpha}$ of $g_{\alpha}$ is finite on $[\frac{1}{2},1)$ and we write $\tilde{\mu}_{\alpha}$ for the normalized measure, i.e. $\tilde{\mu}_{\alpha}([\frac{1}{2},1))=1$. The measure $\mu_{\alpha}$ is finite on $[0,1)$ if and only if $\alpha\in(0,1)$. In fact, the density $h$ of the measure $\mu_{\alpha}$ is a decreasing and positive function, and it satisfies $h(x)\sim x^{-\alpha}$ when $x$ is close to zero [21].

Using the first return map enables us to estimate $\tilde{\mu}_{\alpha}(\mathcal{E}_{\textup{ah}}(\mathbf{B})\cap[\frac{1}{2},1))$ for $\mathbf{B}=\left\{B(y,r_{m})\right\}_{m=1}^{\infty}$ with center $y\in[\frac{1}{2},1)$ depending on the rate of shrinking of $r_{m}$. At the end of the section we argue why this is sufficient. For now set $B_{m}=B(y,r_{m})$ and assume that $y\in[\frac{1}{2},1)$. By using the fact that $g_{\alpha}^{n}(x)\in B_{m}$ can only happen along the subsequence $n_{k}:=R_{k}(x)$, a short argument gives the inclusions

[TABLE]

where

[TABLE]

and

[TABLE]

The small argument verifying (6) is left to the reader.

Assume first that $\mu_{\alpha}$ is finite, i.e. $\alpha\in(0,1)$. We will use (6) together with the following estimate of $R_{n}$ from [10, Theorem 2.19]. There exists a constant $C>1$ such that the set of $x$ for which the inequality,

[TABLE]

does not hold is of arbitrarily small measure if $n$ is sufficiently large. Let $N_{0}$ be so large that the set of $x$ for which (7) holds for $n>N_{0}$ is of measure at least $1-\delta$ and call this set $D_{\delta}$. For $x\in D_{\delta}$ we may apply (7) and we find that the condition $\tilde{\mu}_{\alpha}(B_{m})\geq\frac{c(\log m)^{2+\varepsilon}}{m}$ implies that $\tilde{\mu}_{\alpha}(B_{R_{n}(x)})\geq\frac{c(\log n)^{2+\varepsilon}}{Cn}>\frac{c_{0}(\log n)^{2}}{n}$ for large $n$. So by picking $C_{m}:=B(y,r_{m})$ with $\tilde{\mu}_{\alpha}(C_{m})=\frac{c(\log m)^{2+\varepsilon}}{m}$ we get $\tilde{A}_{1}\subset A_{1}$ where

[TABLE]

The first set in the intersection is $\mathcal{E}_{\textup{ah}}(C_{m})$ for the system $([\frac{1}{2},1),S,\tilde{\mu}_{\alpha})$ which has measure one by Theorem 4. This shows that $\tilde{\mu}_{\alpha}(\tilde{A}_{1})\geq 1-\delta$ and since $\delta$ is arbitrary we get $\tilde{\mu}_{\alpha}(A_{1})=1$. This implies that $\tilde{\mu}_{\alpha}(\mathcal{E}_{\textup{ah}}\cap[\frac{1}{2},1))=1$.

Assume now that $\alpha>1$ in which case $\mu_{\alpha}$ is infinite. In this case we use (6) together with a different estimate of $R_{n}$ which also originates from [10, Theorem 2.19]. In this case we have that for any $\kappa>0$, the set of $x$ for which the estimate

[TABLE]

does not hold is of arbitrarily small measure if $n$ is sufficiently large. Let again $N_{0}$ be so large that the set $D_{\delta}$ of $x$ for which (8) holds for $n>N_{0}$ is of measure at least $1-\delta$. For $x\in D_{\delta}$ we have that if there exists an $\varepsilon>0$ such that $\tilde{\mu}_{\alpha}(B_{m})\leq\frac{c}{m^{\frac{1}{\alpha}+\varepsilon}}$ for any $c>0$, then

[TABLE]

for sufficiently large $n$ when we choose $\kappa$ sufficiently small compared to $\varepsilon$. As before we may pick $C_{m}:=B(y,r_{m})$ such that $\mu(C_{m})=\frac{c_{1}}{m}$. As before we can define a set $\tilde{A_{2}}$ to which we can apply Corollary 1 and by the same reasoning we conclude $\tilde{\mu}_{\alpha}(\mathcal{E}_{\textup{ah}}\cap[\frac{1}{2},1))=0$.

The proof that $\tilde{\mu}_{\alpha}(B_{m})\geq\frac{c}{m^{\frac{1}{\alpha}-\varepsilon}}$ implies $\tilde{\mu}_{\alpha}(\complement\mathcal{E}_{\textup{ah}}\cap[\frac{1}{2},1))=0$ is similar to the previous two cases and we omit the details.

We now argue why our results for $\tilde{\mu}_{\alpha}(\mathcal{E}_{\textup{ah}}(\mathbf{B})\cap[\frac{1}{2},1))$ imply the general statement. Note first that if $y\in[\frac{1}{2},1)$ and $\tilde{\mu}_{\alpha}(\mathcal{E}_{\textup{ah}}(\mathbf{B})\cap[\frac{1}{2},1))=0$ then $\mu_{\alpha}(\complement\mathcal{E}_{\textup{ah}}(\mathbf{B}))>0$. By Lemma 1 we then have $\mu_{\alpha}(\mathcal{E}_{\textup{ah}}(\mathbf{B}))=0$. The identical argument works when $\mathcal{E}_{\textup{ah}}$ is replaced by $\complement\mathcal{E}_{\textup{ah}}$.

To go from $y\in[0,\frac{1}{2})$ to $y\in[0,1)$ requires only a little more consideration. Notice first that each assumption on $\mu(B_{m})$ in Theorem 2(b), (c) and (d) is invariant under multiplication by a constant. Assume for example the setting of Theorem 2(c), i.e. $\alpha\geq 1$ and $\mu(B_{m})\leq\frac{c}{m^{\frac{1}{\alpha}+\varepsilon}}$ for some $\varepsilon>0$ and any $c>0$. Assume that $y\in[0,\frac{1}{2})$. Let $k_{0}$ denote the smallest number such that $g_{\alpha}^{k_{0}}(y)\in[\frac{1}{2},1)$. There exists a $K>0$ such that for all $m\in\mathbbm{N}$ we have $\mu_{\alpha}(g_{\alpha}^{k_{0}}(B(y,r_{m})))\leq K\mu_{\alpha}(B(y,r_{m}))$. This is an easy consequence of l’Hôpitals rule applied to the function $f(r):=\mu_{\alpha}(g_{\alpha}(B(y,r)))/\mu_{\alpha}(B(y,r))$. Pick $\tilde{B}_{m}$ to be the smallest ball with center $g_{\alpha}^{k_{0}}(y)$ such that $g_{\alpha}^{k_{0}}(B(y,r_{m}))\subset\tilde{B}_{m}$. Then $\tilde{B}_{m}$ also satisfies the assumption of Theorem (c) and we know that $\mu_{\alpha}(\mathcal{E}_{\textup{ah}}(\tilde{B}_{m}))=0$ from the arguments above.

We argue that $\mathcal{E}_{\textup{ah}}(B_{m})\subset\mathcal{E}_{\textup{ah}}(\tilde{B}_{m})$. Assume that $x\in\mathcal{E}_{\textup{ah}}(B_{m})$, i.e. $\exists\,m_{0}\,\forall\,m\geq m_{0}\,\exists\,k<m:g_{\alpha}^{k}(x)\in B_{m}$. But if $g_{\alpha}^{k}(x)\in B_{m}$ then $g_{\alpha}^{k+k_{0}}(x)\in\tilde{B}_{m}$. Hence $\forall\,m\geq m_{0}+k_{0}\,\exists\,k<m$ such that $g_{\alpha}^{k}(x)\in\tilde{B}_{m}$, i.e. $x\in\mathcal{E}_{\textup{ah}}(\tilde{B}_{m})$. The cases Theorem 2(b) and (d) follow by analogue arguments. This completes the proof of Theorem 2.

4.3. The Gauß map

In this section, we will consider the Gauß map $G\colon(0,1]\to[0,1)$ defined by $G(x)=\frac{1}{x}\mod 1$. This is a piecewise expanding map with infinitely many branches. There is a unique measure which is an invariant probability measure and absolutely continuous with respect to the Lebesgue measure. We denote this measure by $\mu$, and its density with respect to Lebesgue measure is given by $h(x)=\frac{1}{\log 2}\frac{1}{1+x}$.

HTS for the Gauß map is known for $B_{m}:=[0,r_{m})$, i.e. the interval with fixed left endpoint being 0 and right endpoint shrinking towards 0. More precisely, for $E_{m}=[0,r_{m}]$ we have,

[TABLE]

This follows from a classic result of Doeblin111To be precise, Doeblins proof contained a gap which was repaired by Iosifescu [12] in 1977, but not before Galambos [9] had proven a special case 1972 which is sufficient for our purposes. [4] which may easily be translated into the above, see for example [11, Section 5]. Hence for this choice of targets, Corollary 2 applies. In particular, we have $\mu(\mathcal{E}_{\textup{ah}})=0$ provided $\mu([0,r_{m}))\leq\frac{c_{0}}{m}$ for some constant $c_{0}>0$.

In order to show that we may apply Theorem 4 we need the following short argument. Define $g:[0,1]\to[0,1]$ by $g(x)=\frac{1}{|G^{\prime}(x)|}=x^{2}$ whenever $x$ is not 0 or of the form $x=\frac{1}{n}$ for some $n\in\mathbbm{N}$. In the remaining points, we let $g(x)=0$. Obviously, $g$ is then of bounded variation, since the heights of the jumps at $\frac{1}{n}$ are summable.

We have $\lVert g\rVert_{\infty}=1$, but $\lVert g\cdot g\circ G\rVert_{\infty}<1$, since $g(x)\to 1$ only if $x\to 1$, and $x=1$ is not a fixed point under $G$. This implies that $G^{2}$ together with $g_{2}=g\cdot g\circ G$ satisfies the assumptions of Rychlik [20]. Hence we may conclude [20, Theorem 1] that if $P\colon BV\to BV$ denotes the transfer operator associated to $G$, then, since $G$ is mixing, $P$ can be written in the form $P=Q+R$, where $Q$ is the projection on the invariant density $h$, $R$ has a spectral radius which is strictly less than $1$ and $QR=RQ=0$.

Let $f\in L^{1}$ and $g\in BV$. Let $c=-\int g\,\mathrm{d}\mu$, so that $\int(g+c)\,\mathrm{d}\mu=0$. Then

[TABLE]

By the choice of $c$, we have $Q((g+c)h)=0$. Hence

[TABLE]

Since $R$ has a spectral radius strictly less than $1$, there are positive constants $C_{1}$ and $\tau$ such that $\lVert R^{n}((g+c)h)\rVert_{BV}\leq Ce^{-\tau n}\lVert(g+c)h\rVert_{BV}$. In particular

[TABLE]

From this, it follows that

[TABLE]

Hence, with $C=\frac{C_{2}}{2\log 2}$ we have

[TABLE]

In conclusion, we may apply Theorem 4 to the Gauß map. In particular, we have $\mu(\mathcal{E}_{\textup{ah}})=1$ provided $\mu(B_{m})\geq\frac{c_{0}(\log m)^{2}}{m}$ for some constant $c_{0}>\tau^{-1}$.

In order to obtain Theorem 3 we now use the fact that when $r_{m}$ is small and $G^{j-1}(x)\in[0,r_{m})$ then $a_{j}(x)\sim\frac{1}{G^{j-1}(x)}$. An easy calculation shows that in the definition of $\mathcal{E}_{\textup{ah}}([0,r_{m}))$ for the Gauß map we can replace $\left\{G^{k}(x)\right\}_{k=0}^{m-1}(x)\cap[0,r_{m})\neq\emptyset$ with $\left\{a_{k}(x)\right\}_{k=1}^{m}\cap\left[\frac{1}{r_{m}},\infty\right)\neq\emptyset$. The only thing left to do now is to compute the bounds on $r_{m}$ when $\mu([0,r_{m}))\leq\frac{c_{0}}{m}$ and $\mu([0,r_{m}))\geq\frac{c_{0}(\log m)^{2}}{m}$. Using the density of the Gauß measure we get that

[TABLE]

This leads to the following conclusions. If $r_{m}\leq e^{\frac{c_{0}\log 2}{m}}-1$ for some $c_{0}>0$, then

[TABLE]

and if $r_{m}\geq e^{\frac{c_{0}\log 2(\log m)^{2}}{m}}-1$ for some $c_{0}>0$ sufficiently large, then

[TABLE]

Since $\frac{c_{0}\log 2}{m}\leq e^{\frac{c_{0}\log 2}{m}}-1\leq\frac{2c_{0}\log 2}{m}$ if $m$ is large and since $c_{0}>0$ arbitrary, we can conclude that (9) holds provided $r_{m}\leq\frac{1}{cm}$ for some $c>0$. Letting $b_{m}=r_{m}^{-1}$, we have that

[TABLE]

if $b_{m}\geq cm$ for some $c>0$. This is statement Theorem 3(a).

Similarly, we may conclude that (10) holds provided $r_{m}\geq\frac{(\log m)^{2}}{cm}$ for some sufficiently small $c>0$. Letting $b_{m}=r_{m}^{-1}$ we then have

[TABLE]

if $b_{m}\leq\frac{cm}{(\log m)^{2}}$ for some sufficiently small $c>0$. This is Theorem 3(b).

5. Results on the Hausdorff and packing dimension

In this section we consider the case when $T$ is the doubling map $[0,1)\to[0,1)$, defined by $T(x)=2x\mod 1$, and $\mu$ is Lebesgue measure. If we put $\Sigma=\{0,1\}^{\mathbbm{N}}$, then every $x\in[0,1)$ can be coded by a sequence $(x_{i})=(d_{i}(x))_{i=0}^{\infty}$ such that

[TABLE]

Note that in the symbolic setting the doubling map becomes the left shift on $\Sigma$, i.e. for $d(x)=x_{0},x_{1},\dots$ we have $T(d(x))=x_{1},x_{2},\dots$. Given a finite sequence $x_{0},x_{1},\ldots,x_{m}$, we let

[TABLE]

Then $C(x_{0},x_{1},\ldots,x_{m})$ is an interval of length $2^{-(m+1)}$.

We recall the construction of the packing dimension. Let $F\subset\mathbbm{R}^{d}$ and let $\delta>0$. A collection $\left\{B_{i}\right\}$ of disjoint balls of radii at most $\delta$ with centres in $F$ is called a $\delta$-packing collection for $F$. For $s\geq 0$, let

[TABLE]

and

[TABLE]

We then define the $s$-dimensional packing measure by

[TABLE]

where the infimum is taken over all countable covers of $F$. Finally, the packing dimension of $F$ is defined by

[TABLE]

In the hierarchy of dimensions the packing dimension falls between the Hausdorff dimension and the upper box-counting dimension in the sense that

[TABLE]

Theorem 5**.**

Suppose $B_{m}=B(0,2^{-sm})$, where $s\in(0,1)$. Then,

[TABLE]

If $s\geq 1$ then $\operatorname{dim_{P}}\mathcal{E}_{\textup{ah}}=0$. Indeed, $\mathcal{E}_{\textup{ah}}$ is a countable set in this case.

Proof.

If $s=1$, then we will prove that $\mathcal{E}_{\textup{ah}}$ consists only of those $x\in[0,1)$, such that $T^{n}(x)=0$ for some $n$. Hence $\mathcal{E}_{\textup{ah}}$ is the set of all finite words concatenated with an infinite tail of zeroes from which it is clear that $\mathcal{E}_{\textup{ah}}$ is countable. Then $\mathcal{E}_{\textup{ah}}$ must also be countable for all $s>1$. Assume $x\in\mathcal{E}_{\textup{ah}}$. Then $x\in A_{n}$ for some $n$ which means that $\left\{T^{i}x\right\}_{i=0}^{m-1}\cap B_{m}\neq\emptyset$ for all $m\geq n$. Then somewhere from digit number 0 to digit number $m-1$, a block of $m$ zeroes starts. Regardless of where this block starts it will overlap with the digit at place $m-1$ and hence the digit on place $m-1$ is [math]. Since this is true for all $m\geq n$ we have shown that $d_{m}(x)=0$ for all $m\geq n-1$.

From now on, we assume that $s<1$. We will first prove that $\operatorname{dim_{P}}\mathcal{E}_{\textup{ah}}\leq 1-s$. Since $\operatorname{dim_{P}}$ is countably stable, $\mathcal{E}_{\textup{ah}}=\bigcup A_{n}$ and $\operatorname{dim_{P}}\leq\operatorname{dim_{B}}$, it is enough to prove that $\operatorname{dim_{B}}A_{n}\leq 1-s$ for all $n$, where $\operatorname{dim_{B}}$ denotes the box dimension. If there is an $n$ such that $T^{n}(x)=0$, then $x\in A_{n}$. As in the introduction, let $\Lambda$ denote the set of all such points which, as discussed above, is countable and may be disregarded, since the packing dimension of any countable set is 0. Set $A_{n}^{\prime}:=A_{n}\backslash\Lambda$.

We first prove that $x\in A_{n}^{\prime}$ if and only if $(d_{i}(x))_{i=0}^{\infty}$ has blocks of $n_{j}+1$ zeroes starting at position $m_{j-1}$ for all $j\geq 1$ where $(n_{j})_{j=0}^{\infty}$ and $(m_{j})_{j=0}^{\infty}$ are strictly increasing sequences satisfying

1. a)

$m_{0}<n$. 2. b)

$n_{j}\geq sm_{j}$. 3. c)

$m_{j-1}<(1-s)m_{j}$.

Suppose first that $x\in A_{n}^{\prime}$. Then starting somewhere not later than at position $n-1$, the sequence $(d_{i}(x))_{i=0}^{\infty}$ contains a block of at least $sn$ zeroes222Since $sn$ is typically non-integer we should, in principle, be more diligent and write $\lceil sn\rceil$. However, to improve readability we let the relevant ceiling and floor functions be implicitly understood throughout the proof. The outcome is invariant under this abuse of notation.. We let $n_{1}$ be the length of the largest such block, and let $m_{0}<n$ be the position of its first zero. This satisfies a).

The above property is true for all $m\geq n$, i.e. starting somewhere not later than at position $m-1$ a block of at least $sm$ zeroes start. This allows us to define sequences $(n_{j})_{j=1}^{\infty}$ and $(m_{j})_{j=0}^{\infty}$ as follows. The numbers $n_{1}$ and $m_{0}$ are already defined. Suppose that $n_{j}$ and $m_{j-1}$ are defined. Then we let $m_{j}$ be the position of the first digit of the leftmost block of at least $n_{j}+1$ consequtive zeroes in the sequence $(d_{i}(x))_{i=0}^{\infty}$ and we let $n_{j+1}$ be the maximal number of zeroes in such a block. Note that, if we make any change in the digits $d_{i}(x)$ outside of the blocks of $n_{j}$ zeroes described above, then we get a new sequence $d_{i}(y)$ with $y\in A_{n}^{\prime}$.

Since $x\in A_{n}^{\prime}$ and $n_{j}$ was chosen to be maximal, the block of length $n_{j}$ must be long enough to ensure that $\left\{T^{i}x\right\}_{i=0}^{m_{j}-1}\cap B_{m_{j}}\neq\emptyset$, i.e. we always have $n_{j}\geq sm_{j}$ and b) is satisfied. Furthermore, again due to $n_{j}$ being maximal and since $x\notin\Lambda$, the blocks of zeroes are separated, and we have $m_{j-1}+n_{j}<m_{j}$. Hence $m_{j-1}<(1-s)m_{j}$ and c) is satisfied.

Conversely it is clear that any $x\in[0,1]$ for which a), b) and c) holds true for $(d_{i}(x))_{i=0}^{\infty}$ is an element of $A_{n}^{\prime}$.

Let now $N$ be fixed and take $k$ such that $m_{k-1}<N\leq m_{k}$. From $m_{j-1}<(1-s)m_{j}$ we obtain that

[TABLE]

Hence

[TABLE]

The numbers of zeroes in $(d_{i}(x))_{i=0}^{\infty}$ between digit number $m_{0}$ and digit number $N$ is at least $s(N-m_{0})$. This is the case, since the finite sequence $d_{m_{0}}(x),\ldots,d_{N}(x)$ can be cut into $k$ sequences starting at $d_{m_{j}}(x)$, $j=0,1,\ldots k-1$ and each of these subsequences contains at least a quotient of $s$ zeroes.

We will now cover the set $A_{n}^{\prime}$ by intervals of the form

[TABLE]

The sequences $x_{0},x_{1},\ldots,x_{N}$ that we need to consider are only those that can be obtained from a sequence $(n_{j})_{j=1}^{k}$ and $(m_{j})_{j=0}^{k}$ satisfying the inequalities a), b) and c) and therefore also with $k\leq 2+\frac{\log N}{-\log(1-s)}$.

The sequence $(m_{j})_{j=0}^{k}$ can be chosen in at most $N^{k+1}$ different ways, and once $(m_{j})_{j=0}^{k}$ is chosen, we can choose $(n_{j})_{j=1}^{k}$ in at most $N^{k}$ different way, hence in total at most $N^{2k+1}$ different ways to choose the sequences. (These are very rough estimates, but sufficient for our purpose.)

Once the sequences $(n_{j})_{j=1}^{k}$ and $(m_{j})_{j=0}^{k}$ are chosen, we have specified a certain number of zeroes, while the other digits in the sequence $x_{0},x_{1},\ldots,x_{N}$ remain free. There are at most $m_{0}+(1-s)(N-m_{0})\leq(1-s)N+n$ digits that are free, and hence once the sequences $(n_{j})_{j=1}^{k}$ and $(m_{j})_{j=0}^{k}$ are chosen, we may choose the sequence $x_{0},x_{1},\ldots,x_{N}$ in at most $2^{(1-s)N+n}$ ways.

In total, the number of sequences $x_{0},x_{1},\ldots,x_{N}$ that we need in order to cover $A_{n}^{\prime}$ with the sets $C(x_{0},x_{1},\ldots x_{N})$, are not more than

[TABLE]

Since the sets $C(x_{0},x_{1},\ldots x_{N})$ have diameter $2^{-(N+1)}$ we get from the definition of box dimension that

[TABLE]

The last equality follows since $N^{\log N}$ grows with $N$ slower than any exponential.

We now finish by proving that $\operatorname{dim_{P}}\mathcal{E}_{\textup{ah}}\geq 1-s$. Let $(m_{j})_{j=1}^{\infty}$ be a strictly increasing sequence of natural numbers. Using this sequence, we will construct a subset of $\mathcal{E}_{\textup{ah}}$ and prove that the packing dimension of this subset is $1-s$ if the sequence $(m_{j})_{j=1}^{\infty}$ is chosen such that

[TABLE]

For example, we could choose the sequence inductively as $m_{k}:=k\sum_{j=1}^{k-1}m_{j}$, however, the explicit choice is irrelevant. Let $F$ consist of those $x\in[0,1)$ such that

[TABLE]

for all $j\geq 0$. Then $F\subset\mathcal{E}_{\textup{ah}}$.

We let $\mu$ be the probability measure on $F$ defined by

[TABLE]

if $C(x_{0},x_{1},\ldots,x_{n})$ intersects $F$ where $N(n)$ is the number of intervals $C(x_{0},x_{1},\ldots,x_{n})$ that have non-empty intersection with $F$. Otherwise we assign measure 0. The upper pointwise dimension of $\mu$ at $x\in F$ is defined by

[TABLE]

Let $r_{n}=2^{-n-2}$. Then $B(x,r_{n})$ is contained in the cylinder $C(x_{0},x_{1},\ldots,x_{n})$ and one of the neigbouring cylinders. The measure of the neighbouring cylinder is either zero or equal to that of the cylinder $C(x_{0},x_{1},\ldots,x_{n})$. Hence

[TABLE]

and since $\log r_{n}<0$ we have

[TABLE]

It therefore follows that

[TABLE]

We will use this inequality to show that $\overline{d}_{\mu}(x)\geq 1-s$ for any $x\in F$. By [5, Proposition 10.1], this implies that $\operatorname{dim_{P}}F\geq 1-s$.

We have that

[TABLE]

and

[TABLE]

Hence

[TABLE]

Since $m_{k}$ are chosen so that (11) holds, we have

[TABLE]

which implies that $\operatorname{dim_{P}}\mathcal{E}_{\textup{ah}}\geq\operatorname{dim_{P}}F\geq 1-s$. ∎