On the complexity function for sequences which are not uniformly recurrent

TL;DR

This paper investigates the complexity functions of non-minimal transitive subshifts, establishing growth bounds and structural restrictions, and extends known results on minimal systems to more general transitive cases.

Contribution

It proves new growth bounds for complexity functions of non-minimal transitive subshifts and extends Boshernitzan's results on minimal systems to transitive systems.

Findings

Non-minimal transitive subshifts have complexity growth exceeding 1.5n infinitely often.

Certain transitive subshifts with specific complexity bounds must be minimal.

Restrictions imply unique ergodicity for some non-minimal transitive subshifts.

Abstract

We prove that every non-minimal transitive subshift $X$ satisfying a mild aperiodicity condition satisfies $\limsup c_n(X) - 1.5n = \infty$, and give a class of examples which shows that the threshold of $1.5n$ cannot be increased. As a corollary, we show that any transitive $X$ satisfying $\limsup c_n(X) - n = \infty$ and $\limsup c_n(X) - 1.5n < \infty$ must be minimal. We also prove some restrictions on the structure of transitive non-minimal $X$ satisfying $\liminf c_n(X) - 2n = -\infty$, which imply unique ergodicity (for a periodic measure) as a corollary, which extends a result of Boshernitzan from the minimal case to the more general transitive case.

Equations

$$x=0^{\infty }.10^{g_1}10^{g_2}10^{g_3}10^{g_4}1\dots$$

$$g_i=n_k\text{ if i is the product of 2k and an odd natural number where k≥0.}$$

$$x=0^{\infty }.10^{n_0}10^{n_1}10^{n_0}10^{n_2}10^{n_0}10^{n_1}10^{n_0}10^{n_3}\dots$$

$$w(k)=0^{n_k}10^{g_1}10^{g_2}1\cdots 10^{g_{2^k-1}}10^{n_k}=0^{n_k}10^{n_0}10^{n_1}10^{n_0}10^{n_2}\cdots 0^{n_2}10^{n_0}10^{n_1}10^{n_0}10^{n_k}.$$

$$|w(k)|=2^k+2n_k+\sum _{j=1}^k{2}^{j-1}{n}_{k-j}=2^k+2n_k+\sum _{j=0}^{k-1}{2}^{k-j-1}{n}_j.$$

$$c_{n+1}(X)=c_n(X)+1+\left|\{n\}\cap R\right|.$$

$$c_n(X)=n+1+\left|\{1,2,\dots ,n-1\}\cap R\right|$$

$$0.5|w(k)|-n_k+|R\cap \{1,\dots ,|w(k-1)|\}|=2^{k-1}+\sum _{j=0}^{k-1}{2}^{k-2-j}{n}_j+|R\cap \{1,\dots ,|w(k-1)|\}|.$$

$$\left|\{1,2,\dots ,n-1\}\cap R\right|$$

$$<0.5(|w(k-1)|+1)+f(|w(k-1)|+1)\le 0.5n+f(n).$$

$$\left|\{1,2,\dots ,n-1\}\cap R\right|$$

$$=0.5n+\left(0.5n-n_k+\left|\{1,2,\dots ,|w(k-1)|\}\cap R\right|\right)$$

$$<0.5n+\left(0.5|w(k)|-n_k+\left|\{1,2,\dots ,|w(k-1)|\}\cap R\right|\right)$$

$$<0.5n+f(n_k)\le 0.5n+f(n).$$

Full text

The second author gratefully acknowledges the support of NSF grant DMS-1500685.

1. Introduction and definitions

In this work, we describe some simple connections between the recurrence properties of a two-sided sequence $x$ and the so-called word complexity function $c_{n}(x)$ which measures the number of words of length $n$ appearing in $x$. One of the most fundamental results of this sort is the Morse-Hedlund theorem, which has slightly different statements in the one- and two-sided cases (see [6]).

Theorem 1.1**.**

(Morse-Hedlund Theorem)* Suppose that $A$ is a finite alphabet, $x\in A^{\mathbb{N}}$ or $x\in A^{\mathbb{Z}}$, and there exists $n$ such that the number of $n$-letter subwords of $x$ is less than or equal to $n$. Then, if $x\in A^{\mathbb{Z}}$, $x$ must be periodic, and if $x\in A^{\mathbb{N}}$, then $x$ must be eventually periodic.*

One way to view this theorem is that it yields a lower bound on $c_{n}(x)$; if $x$ is two-sided and not periodic, then $c_{n}(x)\geq n+1$ for all $n$. It is well-known that this bound is sharp; there exist aperiodic sequences called Sturmian sequences (see Chapter 6 of [4] for an introduction) for which $c_{n}(x)=n+1$ for all $n$. There are also other examples in the literature ([1]) with $1<\liminf(c_{n}(x)/n)<\limsup(c_{n}(x)/n)<1+\epsilon$ for arbitrarily small $\epsilon$. All of these examples are uniformly recurrent sequences, meaning that for every subword $w$, there exists $N$ so that every $N$-letter subword contains $w$. Equivalently, a sequence is uniformly recurrent whenever the shift map acting on its orbit closure forms a minimal topological dynamical system.

There are also fairly simple examples of sequences which are not uniformly recurrent and yet have $c_{n}(x)<n+k$ for all $n$ and some constant $k$, given by any $x$ which is not periodic but is eventually periodic in both directions. For example, if $x=\ldots 1212344444\ldots$ then $c_{n}(x)=n+3$ for all $n\geq 1$, and $x$ is clearly not uniformly recurrent since the word $1234$ occurs just once.

This leads to a natural question: must a sequence with complexity function “close to $n$” be either uniformly recurrent or eventually periodic in both directions? Our main result shows that this is indeed the case.

Theorem 1.2**.**

If $x$ is not uniformly recurrent and it is not true that $x$ is eventually periodic in both directions, then $\limsup(c_{n}(x)-1.5n)=\infty$.

This gives a large gap in the complexity functions achievable by sequences which are not uniformly recurrent; any such complexity is either below $n+k$ for all $n$ and some constant $k$, or has a subsequence along which $c_{n}(x)-1.5n$ approaches infinity. In particular, this means that some interesting examples from the literature ([1], [5]) with $1<\limsup(c_{n}(x)/n)<1.5$ can only be achieved by uniformly recurrent sequences. We also show that Theorem 1.2 is tight in the sense that the threshold of $1.5n$ cannot be meaningfully increased.

Theorem 1.3**.**

For any nondecreasing $g:\mathbb{N}\rightarrow\mathbb{R}$ with $\lim g(n)=\infty$, there exists an $x$ which is not uniformly recurrent where $c_{n}(x)<1.5n+g(n)$ for sufficiently large $n$.

Our proof is an analysis by cases, and in most of the cases, the much stronger bound $\liminf(c_{n}(x)-2n)>-\infty$ holds. We can then prove a fairly strong structure on those $x$ for which it does not.

Theorem 1.4**.**

*If $x$ is not uniformly recurrent, it is not true that $x$ is eventually periodic in both directions, and $\liminf(c_{n}(x)-2n)=-\infty$, then there exist a constant $k$ and periodic orbit $M$ with the following property: for every $N$, there exists $m>N$ so that every $(3m+k)$-letter subword of $x$ contains an $m$-letter subword of a point in $M$. *

Informally, the conclusion of Theorem 1.4 says that $x$ can be partitioned, at arbitrarily large “scales,” into long (possibly infinite on one side) pieces of the periodic orbit $M$ and pieces not in $M$ which are not much longer. Unsurprisingly, this structure is quite similar to the structure of the examples proving Theorem 1.3, as we will see in Section 3.

Theorem 1.4 implies a useful corollary which extends a result of Boshernitzan. He proved in [2] that if $X$ is minimal and $\liminf(c_{n}(X)-2n)=-\infty$, then $X$ is uniquely ergodic, i.e. there is only one shift-invariant Borel probability measure on $X$. The following result uses the same complexity hypothesis, but applies to non-minimal transitive systems.

Theorem 1.5**.**

If $X=\overline{\mathcal{O}(x)}$, $x$ is not uniformly recurrent, it is not true that $x$ is eventually periodic in both directions, and $\liminf(c_{n}(X)-2n)=-\infty$, then $X$ is uniquely ergodic, with unique shift-invariant measure supported on a periodic orbit.

(We would like to note that the proof in [2] could theoretically be applied to transitive systems with very few changes, and so the main new content in our result is the triviality of the measure in the non-minimal case.)

Cyr and Kra ([3]) recently generalized a different result of Boshernitzan’s, proving that under no assumption on $X$ whatsoever, for any $k\in\mathbb{N}$, $\liminf(c_{n}(X)/n)<k$ implies that $X$ has fewer than $k$ nonatomic shift-invariant measures which have so-called generic points. Theorem 1.5 applies only to the case $k=2$ and assumes transitivity of $X$ and some aperiodicity of $x$, but uses a weaker complexity hypothesis and implies that $X$ cannot have multiple shift-invariant measures at all, rather than only forbidding multiple nonatomic shift-invariant measures.

2. Definitions

Let $A$ denote a finite set, which we will refer to as our alphabet.

Definition 2.1**.**

A bi-infinite sequence $x\in A^{\mathbb{Z}}$ is periodic if there exists $n\neq 0$ so that $x(k)=x(k+n)$ for all $k\in\mathbb{Z}$. A one-sided sequence $x\in A^{\mathbb{N}}$ is eventually periodic if there exist $n,N\in\mathbb{N}$ so that $x(k)=x(k+n)$ for all $k>N$; the definition is analogous for $x\in A^{-\mathbb{N}}$. A bi-infinite sequence $x$ is eventually periodic in both directions if $x(0)x(1)x(2)\ldots$ and $\ldots x(-2)x(-1)$ are each eventually periodic.

Definition 2.2**.**

A subshift $X$ on an alphabet $A$ is any subset of $A^{\mathbb{Z}}$ which is invariant under the left shift map $\sigma$ and closed in the product topology.

Definition 2.3**.**

A subshift $X$ is transitive if it can be written as $\overline{\mathcal{O}(x)}$ for some $x\in A^{\mathbb{Z}}$, where $\mathcal{O}(x):=\{\sigma^{n}x\ :\ n\in\mathbb{Z}\}$.

Definition 2.4**.**

A subshift $X$ is minimal if it contains no proper nonempty subshift; equivalently, if $X=\overline{\mathcal{O}(x)}$ for all $x\in X$.

A routine application of Zorn’s Lemma shows that every nonempty subshift contains a nonempty minimal subshift.

Definition 2.5**.**

A word over $A$ is a member of $A^{n}$ for some $n\in\mathbb{N}$, which we call the length of $w$ and denote by $|w|$. A word $w$ is called a subword of a longer word or infinite or bi-infinite sequence $u$ if there exists $i$ so that $u(i+j)=w(j)$ for all $1\leq j\leq|w|$.

Definition 2.6**.**

A sequence $x\in A^{\mathbb{Z}}$ is recurrent if every subword of $x$ appears infinitely many times within $x$, and uniformly recurrent if, for every $w\in W(x)$, there exists $N$ so that every $N$-letter subword of $x$ contains $w$ as a subword.

Definition 2.7**.**

For any words $v\in A^{n}$ and $w\in A^{m}$, we define the concatenation $vw$ to be the word in $A^{n+m}$ whose first $n$ letters are the letters forming $v$ and whose next $m$ letters are the letters forming $w$.

Definition 2.8**.**

For a word $u\in A^{n}$, if $u$ can be written as the concatenation of two words $u=vw$ then we say that $v$ is a prefix of $u$ and that $w$ is a suffix of $u$.

Definition 2.9**.**

For any infinite or bi-infinite sequence $x$, we denote by $W(x)$ the set of all subwords of $x$ and, for any $n\in\mathbb{N}$, define $W_{n}(x)=W(x)\cap A^{n}$, the set of subwords of $x$ with length $n$. For a subshift $X$, we define $W(X)=\bigcup_{x\in X}W(x)$ and $W_{n}(X)=\bigcup_{x\in X}W_{n}(x)$.

Definition 2.10**.**

For any infinite or bi-infinite sequence $x$, $c_{n}(x):=|W_{n}(x)|$ is the word complexity function of $x$; for a subshift $X$, $c_{n}(X)$ is similarly defined.

Definition 2.11**.**

A word $w$ is right-special within a subshift $X$ if there exist $a\neq b\in A$ so that $wa,wb\in W(X)$.

We note that for every subshift $X$ and $w\in W(X)$, there exists at least one letter $a$ so that $wa\in W(X)$. Therefore, for any $n$, $c_{n+1}(X)-c_{n}(X)$ is greater than or equal to the number of right-special words in $W_{n}(X)$.

Definition 2.12**.**

A sliding block code with window size $k$ is a function $\phi$ defined on a subshift $X$ where $\phi(X)$ is a subshift and $(\phi(x))(i)$ depends only on $x(i)x(i+1)\ldots x(i+k-1)$ for all $x\in X$ and $i\in\mathbb{Z}$.

For a sliding block code $\phi$ with window size $k$, even though $\phi$ technically is defined on $X$, it induces an obvious action on words in $W_{n}(X)$ for $n\geq k$; for any such $w$, one can define $\phi(w)\in W_{n-k+1}(\phi(X))$ to be $(\phi(x))(0)\ldots(\phi(x))(n-k+1)$ for any $x$ with $x(0)\ldots x(n-1)=w$. (This is independent of choice of $x$ by the definition of sliding block code.) This induces a surjection from $W_{n}(X)$ to $W_{n-k+1}(\phi(X))$, and so for any such $\phi$ and $n\geq k$, $c_{n}(X)\geq c_{n-k+1}(\phi(X))$.

3. Proofs

3.1. Proof of Theorem 1.2

Throughout, $x$ will represent a bi-infinite sequence and $X$ will represent its orbit closure, $X=\overline{\mathcal{O}(x)}$. Note that then $W_{n}(X)$ is just the set of words of length $n$ appearing as subwords of $x$, and $c_{n}(X)$ is the number of such words, i.e., $W_{n}(X)=W_{n}(x)$ and $c_{n}(X)=c_{n}(x)$. We assume throughout that $x$ is not uniformly recurrent and that it is not true that $x$ is eventually periodic in both directions, and will now break into various cases and give lower bounds on $c_{n}(x)$ in each.

3.1.1. $x$ is non-recurrent

Lemma 3.1**.**

If $x$ is non-recurrent and it is not true that $x$ is eventually periodic in both directions, then there exists a constant $k$ so that $c_{n}(X)>2n-k$ for all $n$.

Proof.

Since $x$ is not recurrent, there exists a word $v$ which appears in $x$ only finitely many times. We can then write $x=\ell wr$ where $w$ contains all occurrences of $v$ in $x$; then $w$ occurs only once in $x$. Then $\ell$ and $r$ do not contain $w$, and one of $\ell$ or $r$ is not eventually periodic. We treat only the $r$ case here, as the $\ell$ case is similar. Since $r$ is not eventually periodic, by Theorem 1.1, it contains at least $n+1$ distinct $n$-letter subwords for every $n$, and none of these contain $w$ as a subword. In addition, $x=\ell wr$ contains $n-|w|+1$ subwords of length $n$ which contain $w$, which are all distinct since they contain $w$ exactly once at different locations. Therefore, $c_{n}(X)\geq(n+1)+(n-|w|+1)=2n-|w|+2$ for all $n$, completing the proof. ∎

3.1.2. $x$ is recurrent and not uniformly recurrent

In this case, since $x$ is not uniformly recurrent, $X$ is not minimal. Then $X$ must properly contain some minimal subshift.

Lemma 3.2**.**

If $X$ properly contains an infinite minimal subshift $M$, then there exists a constant $k$ so that $c_{n}(X)>2n-k$ for all $n$.

Proof.

Suppose that $X$, $M$ are as in the theorem. Since $\overline{\mathcal{O}(x)}=X\neq M$, $x$ contains a subword not in $W(M)$, let’s call it $w$. By shifting $x$ if necessary, we may assume that $w=x(0)\ldots x(|w|-1)$. By recurrence, $x$ contains infinitely many occurrences of $w$. However, since $\overline{\mathcal{O}(x)}=X\supset M$, $x$ contains arbitrarily long subwords in $W(M)$, none of which may contain $w$. Choose any $n\geq|w|$, and consider a subword of $x$ of length $n$ which does not contain $w$; take it to be $x(k)\ldots x(k+n-1)$, and for now assume that $k>0$. Now, the word $x(0)\ldots x(k+n-1)$ contains $w$ at least once (as a prefix), so we may define the rightmost occurrence of $w$ within it; say this happens at $x(j)\ldots x(j+|w|-1)$. Note that since $x(k)\ldots x(k+n-1)$ contains no occurrences of $w$, we know that $j<k$.

Finally, consider the $n$-letter subwords of $x$ defined by $u_{i}=x(i)\ldots x(i+n-1)$, where $j-n+|w|\leq i\leq j$. Each contains the occurrence of $w$ at $x(j)\ldots x(j+|w|-1)$, and no occurrence of $w$ to the right, by definition of $j$. Therefore, all are distinct, and so $x$ contains $n-|w|+1$ subwords of length $n$, which each contain $w$.

On the other hand, since $M$ is an infinite minimal subshift, it is aperiodic, and so by Theorem 1.1, $W_{n}(M)$ contains at least $n+1$ subwords of length $n$, none of which contain $w$ since $w\notin W(M)$. Since $M\subset X$, $W_{n}(M)\subset W_{n}(X)$, and so $c_{n}(X)>(n+1)+(n-|w|+1)=2n-|w|+2$ for all $n\geq|w|$, completing the proof when $k>0$. Since the complexity function is unaffected by reflecting $x$ (and $w$) about the origin, the same holds when $k<0$, completing the proof.

∎

We now only need treat the case where $X$ contains only finite minimal subshifts (i.e. periodic orbits), and will first deal with the case where it contains more than one.

Lemma 3.3**.**

If $X$ contains two minimal subshifts and it is not true that $x$ is eventually periodic in both directions, then there exists a constant $k$ so that $c_{n}(X)>2n-k$ for all $n$.

Proof.

Denote by $M$ and $M^{\prime}$ minimal subshifts of $X$; by definition of minimality, $M$ and $M^{\prime}$ are disjoint. If either is infinite, then we are done by Lemma 3.2. So, assume that both are finite, and therefore periodic orbits. Choose $k$ so that $W_{k}(M)$ and $W_{k}(M^{\prime})$ are disjoint, and their union is strictly contained in $W_{k}(X)$. Since $X=\overline{\mathcal{O}(x)}$, $x$ contains arbitrarily long words in $W(M)$ and $W(M^{\prime})$. As above, we may assume without loss of generality that $x$ is not eventually periodic to the right. Then, for all $n\geq k$, there exists $\ell$ so that $x(\ell)\ldots x(\ell+n-1)\in W(M)$ and $x(\ell+n-k+1)\ldots x(\ell+n)\notin W_{k}(M)$. Similarly, there exists $m$ so that $x(m)\ldots x(m+n-1)\in W(M^{\prime})$ and $x(m+n-k+1)\ldots x(m+n)\notin W_{k}(M^{\prime})$.

Define the $n$-letter words $u_{i}=x(\ell+i)\ldots x(\ell+i+n-1)$ and $v_{j}=x(m+j)\ldots x(m+j+n-1)$ for $0<i,j\leq n-k$; clearly all are in $W_{n}(X)$. In each $u_{i}$, the leftmost $k$-letter word not in $W_{k}(M)$ is $u_{i}(n-k-i+2)\ldots u_{i}(n-i+1)=x(m+n-k+1)\ldots x(m+n)$, and so all $u_{i}$ are distinct. The same argument (using $W_{k}(M^{\prime})$) shows that all $v_{j}$ are distinct. Finally, all $u_{i}$ begin with a word in $W_{k}(M)$ and all $v_{j}$ begin with a word in $W_{k}(M^{\prime})$, and so the sets $\{u_{i}\}$ and $\{v_{j}\}$ are also disjoint. Therefore, $c_{n}(X)\geq 2n-2k$ for $n>k$, completing the proof.

∎

The remaining case is that $x$ is recurrent and that $X$ properly contains a periodic orbit $M$, which is the only minimal subshift contained in $X$. For simplicity, we assume that $M$ is a single fixed point, which we may do via the following lemma.

Lemma 3.4**.**

Suppose that $x$ is recurrent and $X=\overline{\mathcal{O}(x)}$ strictly contains a periodic orbit $M$ which is the only minimal subshift contained in $X$. Then there is a sliding block code $\phi$ with the following properties: $\phi(X)$ has alphabet $\{0,1\}$, $\phi(X)$ strictly contains the unique minimal subshift $\{0^{\infty}\}$, and $\phi(w)=0^{i}$ implies that $w\in W(M)$.

Proof.

Choose such $X$ and $M$, and choose any $k$ greater than the period $p$ of $M$. Define $\phi$ as follows: for every $i$, $(\phi(x))(i)=0$ if $x(i)\ldots x(i+k-1)\in W_{k}(M)$, and $1$ otherwise. Trivially $\phi(X)$ has alphabet $\{0,1\}$. If $\phi(w)=0^{i}$, then $w$ has period $p$ (since all words in $W_{k}(M)$ have period $p$) and begins with a $k$-letter word in $W(M)$, and is therefore itself in $W(M)$. Since $X\supsetneq M$, $\phi(X)$ contains points other than $0^{\infty}$. Finally, if $\phi(X)$ contained a minimal subshift not equal to $\{0^{\infty}\}$, then it would be disjoint from $\{0^{\infty}\}$, and so its preimage would contain a minimal subshift of $X$ other than $M$, a contradiction.

∎

3.1.3. $x$ is recurrent, $x\in\{0,1\}^{\mathbb{Z}}$, $M=\{0^{\infty}\}$ is the only minimal subsystem of $X$

In this case, $x$ must contain infinitely many $1$s (by recurrence) and must contain $0^{n}$ as a subword for every $n$ (since $0^{\infty}\in\overline{\mathcal{O}(x)}=X$). We will need the following slightly stronger fact.

Lemma 3.5**.**

For $x$ satisfying the conditions of this section, and for all $n$, $0^{n}1$ and $10^{n}$ are subwords of $x$.

Proof.

Choose any $n$. We know already that $0^{n}$ is a subword of $x$. If neither $0^{n}1$ nor $10^{n}$ were subwords of $x$, then every occurrence of $0^{n}$ in $x$ would force [math]s on both sides, implying $x=0^{\infty}$, a contradiction. Therefore, either $0^{n}1$ or $10^{n}$ is a subword of $x$; assume without loss of generality that it is the former. Then by recurrence, $0^{n}1$ appears twice as a subword of $x$, implying that $x$ contains a subword of the form $0^{n}1w0^{n}1$. Remove the terminal $1$, and consider the rightmost $1$ in the remaining word; it must be followed by $0^{n}$, and so $x$ also (in addition to $0^{n}1$) contains $10^{n}$ as a subword. Since $n$ was arbitrary, this completes the proof.

∎

By Lemma 3.5, for every $n$ there exists a one-sided sequence $y_{n}$ beginning with $1$ so that $0^{n}y_{n}$ appears in $x$. By compactness, there exists a limit point $y$ of the $y_{n}$ (which begins with $1$), and then since $X$ is closed, $0^{\infty}y\in X$. Similarly, there exists a one-sided sequence $z$ ending with $1$ so that $z0^{\infty}\in X$. We first treat the case whether either $y$ or $z$ is not unique.

Theorem 3.6**.**

For $x$ satisfying the conditions of this section, if there exist either $y\neq y^{\prime}\in\{0,1\}^{\mathbb{N}}$ beginning with $1$ for which $0^{\infty}y,0^{\infty}y^{\prime}\in X$ or $z\neq z^{\prime}\in\{0,1\}^{-\mathbb{N}}$ ending with $1$ for which $z0^{\infty},z^{\prime}0^{\infty}\in X$, then there exists $k$ so that $c_{n}(X)>2n-k$ for all $n$.

Proof.

We prove only the statement for $y,y^{\prime}$, as the corresponding proof for $z,z^{\prime}$ is trivially similar. Assume that such $y,y^{\prime}$ exist. Since $y\neq y^{\prime}$, there exists $k$ so that $y(k)\neq y^{\prime}(k)$.

For any $n\geq k$, define the $n$-letter words $u_{i}=0^{i}y(1)\ldots y(n-i)$ and $v_{i}=0^{i}y^{\prime}(1)\ldots y^{\prime}(n-i)$ for $0\leq i\leq n-k$. First, note that $u_{i}$ and $v_{i}$ both begin with $0^{i}1$ for every $i$, and since $0^{i}1$ is never a prefix of $0^{j}1$ for $i\neq j$, the sets $\{u_{i},v_{i}\}$ and $\{u_{j},v_{j}\}$ are disjoint whenever $i\neq j$. Finally, for every $i$, $u_{i}(i+k)=y(k)\neq y^{\prime}(k)=v_{i}(i+k)$, so $u_{i}\neq v_{i}$. This yields $2n-2k+2$ words in $W_{n}(X)$ for $n\geq k$, completing the proof. ∎

We from now on assume that $y$ and $z$ are unique sequences beginning with $1$ and ending with $1$ respectively which satisfy $0^{\infty}y,z0^{\infty}\in X$.

Theorem 3.7**.**

For $x$ satisfying the conditions of the section, if either $y$ or $z$ contains only finitely many $1$s, then there exists $k$ so that $c_{n}(X)>2n-k$ for every $n$.

Proof.

We again treat only the $y$ case, as the proof for the $z$ case is similar. Suppose that $y$ contains only finitely many $1$s. Then, we can write $y=w0^{\infty}$ for some $w$ beginning and ending with $1$, and $0^{\infty}y=0^{\infty}w0^{\infty}\in X$. By recurrence, $x$ contains a subword $v$ which contains more than $|w|$ $1$s.

Again, by recurrence $x$ contains $v$ infinitely many times. Also, $x$ contains the subword $0^{n}$ for all $n$, which never contains $v$. Therefore, for every $n$, there exists a word $u$ of length $n$ so that either $vu$ or $uv$ is a subword of $x$ and contains $v$ only once as a subword. We treat only the former case, as the latter is similar, and so suppose that $x(k)\ldots x(k+n+|v|-1)=vu$.

For any $n\geq\max(|v|,|w|)$, consider the $n$-letter subwords of $x$ given by $t_{j}=x(j)\ldots x(j+n-1)$ for $k-n+|v|\leq j\leq k$. The rightmost occurrence of $v$ within $t_{j}$ begins at the $(k-j+1)$th letter of $t_{j}$, and so all $t_{j}$ are distinct. This yields $n-|v|+1$ words in $W_{n}(X)$ which each contain $v$. On the other hand, we can define $u_{i}=0^{i}w0^{n-|w|-i}$ for $0\leq i\leq n-|w|$, each of which is contained in $0^{\infty}w0^{\infty}\in X$. Each $u_{i}$ contains $w$ exactly once, beginning at the $(i+1)$th letter, and so all are distinct. In addition, each $u_{i}$ contains at most $|w|$ $1$s, and so none contains $v$, meaning no $t_{j}$ and $u_{i}$ can be equal.

Therefore, for $n\geq\max(|v|,|w|)$, $c_{n}(X)\geq(n-|v|+1)+(n-|w|+1)=2n-|v|-|w|+2$, completing the proof.

∎

Theorem 3.8**.**

For $x$ satisfying the conditions of the section, if the lengths of runs of [math]s in $y$ or $z$ are bounded, then there exists $k$ so that $c_{n}(X)>2n-k$ for all $k$.

Proof.

As usual, we treat only the $y$ case since the $z$ case is similar. Suppose that there exists $k$ so that $0^{k}$ is not a subword of $y$. Then, for any $n>k$, consider the $n$-letter words $u_{i}=0^{i}y(1)\ldots y(n-i)$, $k\leq i<n$, and $v_{j}=z(-j)\ldots z(-1)0^{n-j}$, $0<j\leq n-k$. Each $u_{i}$ begins with $0^{i}1$, and $0^{i}1$ is never a prefix of $0^{i^{\prime}}1$ for $i\neq i^{\prime}$, so all $u_{i}$ are distinct; a similar argument shows that all $v_{j}$ are distinct. In addition, all $v_{j}$ end with $0^{k}$, and all $u_{i}$ either have final $k$ letters containing $y(1)=1$ or end with a $k$-letter subword of $y$, and in either case do not end with $0^{k}$. Therefore, no $u_{i}$ and $v_{j}$ can be equal, and so $c_{n}(X)\geq 2n-k$, completing the proof. ∎

We finally arrive at the only case in which $\liminf(c_{n}(x)-2n)$ may be $-\infty$: $y$ and $z$ contain infinitely many $1$s and arbitrarily long runs of [math]s. In this case, we instead prove the weaker bound from the conclusion of Theorem 1.2, and interestingly only require the stated hypotheses on $y$.

Theorem 3.9**.**

For $x$ satisfying the conditions of the section, if $y$ contains infinitely many $1$s and contains $0^{n}$ as a subword for all $n$, then $\limsup(c_{n}(x)-1.5n)=\infty$.

Proof.

For every $k$, choose $m\geq 2k$ so that $y(1)\ldots y(m)$ ends with $1$ and contains exactly $2k$ $1$s; note that $y(1)\ldots y(m)$ does not contain $0^{m-2k+1}$ as a subword. Then, choose $\ell$ so that $y(1)\ldots y(\ell)0^{m-2k+1}$ is a prefix of $y$ and contains $0^{m-2k+1}$ only at the end, i.e. $y(\ell+1)\ldots y(\ell+m-2k+1)$ is the first occurrence of $0^{m-2k+1}$ in $y$. Clearly, $\ell\geq m$ since $y(1)\ldots y(m)$ ends with $1$ and did not contain $0^{m-2k+1}$. Also, by definition of $\ell$, $y(1)\ldots y(\ell)0^{m-2k}=y(1)\ldots y(\ell+m-2k)$ does not contain $0^{m-2k+1}$.

Now, consider the $(\ell+m-2k)$-letter words defined by $u_{i}=0^{i}y(1)\ldots y(\ell+m-2k-i)$, $0\leq i<\ell+m-2k$, and $v_{j}=z(-j)\ldots z(-1)0^{\ell+m-2k-j}$, $0\leq j<\ell$. Again, since each $u_{i}$ begins with $0^{i}1$, all $u_{i}$ are distinct; similarly, all $v_{j}$ are distinct. In addition, all $v_{j}$ end with $0^{m-2k+1}$, and all $u_{i}$ either have final $m-2k+1$ letters containing $y(1)=1$ or end with a subword of $y(1)\ldots y(\ell+m-2k)$, and in either case do not end with $0^{m-2k+1}$. Therefore, no $u_{i}$ and $v_{j}$ can be equal, and so $c_{\ell+m-2k}(X)\geq(\ell+m-2k)+\ell=2\ell+m-2k$.

Recall that $\ell\geq m$; therefore, $2\ell+m-2k\geq 1.5\ell+1.5m-2k=1.5(\ell+m-2k)+k$. In other words, for $n=\ell+m-2k$, $c_{n}(X)\geq 1.5n+k$. Since $k$ was arbitrary, $c_{n}(x)-1.5n$ is unbounded from above, completing the proof.

∎

We are now prepared to combine the results from the previous subsections to prove Theorem 1.2.

Proof of Theorem 1.2.

We assume that $X$ is not minimal and that it is not the case that $x$ is eventually periodic in both directions. By Lemma 3.1, if $x$ is non-recurrent, then $\liminf(c_{n}(X)-2n)>-\infty$, implying that $\limsup(c_{n}(X)-1.5n)=\infty$. By Lemmas 3.2 and 3.3, if $X$ contains either two minimal subsystems or an infinite minimal subsystem, then $\liminf(c_{n}(X)-2n)>-\infty$, implying that $\limsup(c_{n}(X)-1.5n)=\infty$.

So, we can assume that $x$ is recurrent and that $X$ properly contains a unique minimal subsystem, which is finite. Take the sliding block code $\phi$ (with window size $k$) guaranteed by Lemma 3.4. If we define $y=\phi(x)$ and $Y=\phi(X)$, then $Y=\overline{\mathcal{O}(y)}$ has alphabet $\{0,1\}$, strictly contains the unique minimal subshift $\{0^{\infty}\}$, and (since $\phi$ has window size $k$) satisfies $c_{n}(X)\geq c_{n-k+1}(Y)$ for all $n$.

By Theorems 3.6, 3.7, 3.8, and 3.9, $\limsup(c_{n}(Y)-1.5n)=\infty$, and since $c_{n}(X)\geq c_{n-k+1}(Y)$ for all $n$, it must be the case that $\limsup(c_{n}(X)-1.5n)=\infty$, completing the proof.

∎

3.2. Proof of Theorem 1.3

Fix any nondecreasing unbounded $g:\mathbb{N}\rightarrow\mathbb{R}$. Clearly there exist $N$ and a nondecreasing unbounded $f:\mathbb{N}\rightarrow\mathbb{N}$ so that $f(n)+1\leq g(n)$ for all $n>N$.

We will construct a point $x\in\{0,1\}^{\mathbb{Z}}$ of the following form

[TABLE]

where all $g_{i}\geq 1$. We will refer to these numbers $\{g_{i}\}$ as the gaps (between 1s in $x$). Next we describe how the gaps are defined.

We will construct an increasing sequence of natural numbers $n_{1}<n_{2}<n_{3}<\cdots$, and for every $i$ define

[TABLE]

As such, $x$ will have the form

[TABLE]

Our goal is to show that the natural numbers $n_{0}<n_{1}<n_{2}<\cdots$ may be chosen so that $c_{n}(x)<1.5n+1+f(n)$ for all $n\in\mathbb{N}$; since $1.5n+1+f(n)\leq 1.5n+g(n)$ for $n>N$, we will then be done.

We will establish this by consideration of right-special words occurring in $x$ of various lengths. First note that for any $j\geq 1$, $0^{j}$ is a right-special word: both $0^{j+1}$ and $0^{j}1$ appear in $x$.

Set $w(0)=0^{n_{0}}10^{n_{0}}$, and for $k\geq 1$, let $w(k)$ be the unique word in $x$ of the form $w=0^{n_{k}}1u10^{n_{k}}$ where $u$ has no occurrence of $0^{n_{k}}$. The uniqueness of $w(k)$ can be seen from noting that a gap of $n_{k}$ or longer must correspond to the $i$th gap in $x$ where $i$ is multiple of $2^{k}$. By the construction of $x$, the sequence of gaps that occur between any two consecutive gaps of $n_{k}$ or more are always the same and are equal to $g_{1},g_{2},\ldots,g_{2^{k}-1}$. Thus

[TABLE]

We make a series of claims about the words $w(k)$ for all $k\geq 0$.

Claim 1: Every $w(k)$ is right-special. Because there are gaps of exactly $n_{k}$, $w(k)1$ occurs in $x$, and because there are gaps larger than $n_{k}$, $w(k)0$ occurs in $x$.

Claim 2: Neither $0w(k)$ nor $1w(k)$ are right-special. Given two consecutive multiples of $2^{k}$, one is the product of an odd natural number and $2^{k}$ and the other is a multiple of $2^{k+1}$. Therefore, given two consecutive gaps of $n_{k}$ or longer, one is exactly $n_{k}$ and the other is strictly more than $n_{k}$. Therefore, neither $0w(k)0$ nor $1w(k)1$ occur in $x$, but both $0w(k)1$ and $1w(k)0$ occur in $x$.

Claim 3: Any right-special word $w$ is a suffix of $w(k)$ for some $k>0$. Clearly, $w=0^{n}$ is a suffix of $w(k)$ for $k$ large enough that $n_{k}>n$. Now assume $w$ is a right-special word of the form $u10^{n}$ for some word $u$ and some $n\geq 0$. Then $w1$ occurs in $x$, meaning that $n=n_{k}$ for some $k\geq 0$. Therefore, $w=u10^{n_{k}}$. If $|w|\leq|w(k)|$ then $w$ is a suffix of $w(k)$ by the uniqueness of $w(k)$. Now assume $|w|>|w(k)|$. Then again by the uniqueness of $w(k)$, $w=vw(k)$ for some word $v$ with $|v|>1$. But since any suffix of a right-special word is right-special, this implies that either $0w(k)$ or $1w(k)$ is right-special, contradicting Claim 2.

Next we give a recursive formula for $|w(k)|$. Between any two consecutive multiples of $2^{k}$, for $1\leq j\leq k$, there are $2^{j-1}$ odd multiples of $2^{k-j}$. Therefore, in $w(k)$ we have two runs of $n_{k}$ 0s, $2^{k}$ 1s, and $2^{j-1}$ gaps of $n_{k-j}$ for $0<j\leq k$. This gives

[TABLE]

In order to analyze $c_{n}(x)$, we consider the number of right-special words of length $n$. For all $n\geq 1$, we have $0^{n}$ and for any $n\in(n_{k},|w(k)|]$, we have a suffix of $w(k)$ that contains at least one 1. In what follows, we will always recursively choose the sequence $\{n_{k}\}$ so that $n_{k}>|w(k-1)|$, implying that the intervals $(n_{k},|w(k)|]$ are pairwise disjoint. Therefore, for some values of $n$ we will have exactly one right-special word ($0^{n}$), and for $n$ which are in $(n_{k},|w(k)|]$ for some $k$, we have exactly two right-special words ($0^{n}$ and the suffix of $w(k)$ of length $n$).

Let $R=\mathbb{N}\cap\bigcup_{k\geq 0}(n_{k},|w(k)|]$. For $n\in R$ there are two right-special words of length $n$, and for $n\not\in R$ there is just one right-special word of length $n$. This gives us the recursion formula

[TABLE]

From this, and the fact that $c_{1}(X)=2$ it follows that

[TABLE]

for all $n\geq 1$.

It remains to show that the sequence $n_{0}<n_{1}<n_{2}<\cdots$ can be chosen so that $\left|\{1,2,\ldots,n-1\}\cap R\right|<0.5n+f(n)$ for all $n\in\mathbb{N}$. First, we choose $n_{0}=1$, meaning that $|w(0)|=2n_{0}+1=3$. Then clearly $\left|\{1,2,\ldots,n-1\}\cap R\right|\leq 2<0.5n+f(n)$ for $n\leq 3=|w(0)|$.

3.2.1. Choice of $n_{k}$, $k\geq 1$

Suppose $n_{0},n_{1},\ldots,n_{k-1}$ have been chosen so that

$\left|\{1,2,\ldots,n-1\}\cap R\right|<0.5n+f(n)$ for all $n\leq|w(k-1)|$. Choose $n_{k}$ so that $f(n_{k})$ is greater than

[TABLE]

For $n\in(|w(k-1)|,n_{k})$, we have

[TABLE]

For $n\in[n_{k},|w(k)|)$ we have

[TABLE]

(The second-to-last inequality came from (2).) We’ve shown that $\left|\{1,2,\ldots,n-1\}\cap R\right|<0.5n+f(n)$ for all $n$, and so by (1), $c_{n}(X)<1.5n+f(n)+1$ for all $n$. Since $f(n)+1\leq g(n)$ for $n>N$, this means that $c_{n}(X)<1.5n+g(n)$ for $n>N$, completing the proof.

3.3. Proofs of Theorems 1.4 and 1.5

Proof of Theorem 1.4.

We first note that by Lemmas 3.1, 3.2, and 3.3, $\liminf(c_{n}(X)-2n)=-\infty$ implies that $x$ is recurrent and that $X$ properly contains a periodic orbit $M$, which is the unique minimal subshift contained in $X$. We will for now assume that $M=\{0^{\infty}\}$ and that $X$ has alphabet $\{0,1\}$, and will then extend to the general case by Lemma 3.4.

By Lemma 3.5, $0^{n}1\in W(X)$ for all $n$, and so $0^{n}$ is right-special for all $n$. Therefore, for any $n$ where there is another right-special word in $W_{n}(X)$, $c_{n+1}(X)-c_{n}(X)\geq 2$.

For any $N$, choose $n$ so that $c_{n}(X)-2n\leq-N$, and define $S=\{j<n\ :\ 0^{j}$ is the only right-special word in $W_{j}(X)\}$. Then $\displaystyle c_{n}(X)=\sum_{j=0}^{n-1}(c_{j+1}(X)-c_{j}(X))\geq|S|+2(n-|S|)$, and so $|S|\geq N$. Define $m$ to be the maximal element of $S$; then $0^{m}$ is the only right-special word in $W_{m}(X)$, $m\geq N$, and $\displaystyle c_{m}(X)=c_{n}(X)-\sum_{j=m}^{n-1}(c_{j+1}(X)-c_{j}(X))\leq 2n-2(n-m)=2m$.

We claim that every word in $W_{3m}(X)$ contains $0^{m}$, and so that every subword of length $3m$ of $x$ contains $0^{m}$. Suppose for a contradiction that this is false, i.e. that there is $y\in X$ where $y(1)\ldots y(3m)$ does not contain $0^{m}$. Since $c_{m}(X)\leq 2m$, there exist $1\leq i<j\leq 2m$ so that $y(i)\ldots y(i+m-1)=y(j)\ldots y(j+m-1)$. Also, all $m$-letter words $y(k)\ldots y(k+m-1)$ for $i\leq k\leq j$ are not $0^{m}$ and so, since $m\in S$, are not right-special, i.e. there is only one letter that can follow each of them in a point of $X$. This means that $y(i)y(i+1)\ldots$ is in fact periodic with period $j-i$. Since $y(i)\ldots y(j+m-1)$ does not contain $0^{m}$ (as a subword of $y(1)\ldots y(3m)$), $y(i)y(i+1)\ldots$ cannot contain $0^{m}$, a contradiction to $0^{\infty}$ being the only minimal subsystem of $X$. This means that the original claim was true, completing the proof in the case $M=\{0^{\infty}\}$.

Now suppose that $M$ is an arbitrary periodic orbit, and take the sliding block code $\phi$ (with window size $k$) guaranteed by Lemma 3.4. As before, define $y=\phi(x)$ and $Y=\phi(X)$; then $Y$ has alphabet $\{0,1\}$, strictly contains the unique minimal subshift $\{0^{\infty}\}$, and satisfies $c_{n}(X)\geq c_{n-k+1}(Y)$ for all $n$, implying that $\liminf(c_{n}(Y)-2n)=-\infty$.

From the above proof, for all $N$, there exists $m\geq N$ so that every $3m$-letter subword of $y$ contains $0^{m}$. Every $(3m+3k)$-letter subword of $x$ has image under $\phi$ which is a $(3m+2k)$-letter subword of $y$, and therefore contains $0^{m}$. By definition of $\phi$, $x$ contains an $(m+k)$-letter word at the corresponding location which is in $W(M)$; since $m+k\geq m\geq N$, the proof is complete.

∎

The proof of Theorem 1.5 uses a few basic notions from ergodic theory, which we briefly and informally summarize here. Firstly, a (shift-invariant Borel probability) measure $\mu$ is called ergodic if every measurable set $A$ with $A=\sigma A$ has $\mu(A)\in\{0,1\}$. Ergodic measures are valuable because of the pointwise ergodic theorem, which says that $\mu$-almost every point $x$ in $X$ is generic for $\mu$, which means that for every $f\in C(X)$, $\displaystyle\frac{1}{n}\sum_{i=0}^{n-1}f(\sigma^{i}x)\rightarrow\int f\ d\mu$. For the purposes of the proof below, we need only a very simple application of the ergodic theorem: for any generic point for $\mu$, the frequency of [math] symbols is equal to $\mu([0])$, where $[0]$ is the set of $z\in X$ containing a [math] at the origin. Finally, the ergodic decomposition theorem states that every (shift-invariant Borel probability) measure is a sort of generalized convex combination of ergodic measures. Again, we need only a very simple corollary: if a subshift has only one ergodic measure, then it has only one (shift-invariant Borel probability) measure. For a more detailed introduction to ergodic theory, see [7].

Proof of Theorem 1.5.

Assume that $x$ is not uniformly recurrent, it is not the case that $x$ is periodic in both directions, and $\liminf(c_{n}(X)-2n)=-\infty$. Then as above, $X$ properly contains a periodic orbit $M$, which is the unique minimal subshift contained in $X$. We again first treat the case where $M=\{0^{\infty}\}$. Assume for a contradiction that $X$ has a (shift-invariant Borel probability) measure $\mu$ not equal to $\delta_{0^{\infty}}$. Since $\delta_{0^{\infty}}$ is obviously ergodic, by ergodic decomposition we may assume without loss of generality that $\mu$ is ergodic.

By ergodicity, $\mu(0^{\infty})=0$, and so there exists $j$ so that $\mu([0^{j}])<1/6$. Since $\mu$ is ergodic, by the pointwise ergodic theorem there is a point $z$ which is generic for $\mu$. Note that the only periodic orbit in $X$ is $\{0^{\infty}\}$, and so $z$ cannot be eventually periodic in both directions, since then it would be generic for $\delta_{0^{\infty}}$. In addition, note that $\overline{\mathcal{O}(z)}$ must contain $M=\{0^{\infty}\}$ (since $M$ is the unique minimal subshift contained in $X$), and so $z$ is not uniformly recurrent.

Finally, since $\liminf(c_{n}(X)-2n)=-\infty$, we know that $\liminf(c_{n}(z)-2n)=-\infty$, and so $z$ satisfies the hypotheses of Theorem 1.4. We apply that theorem with $N=2j$ to find $m\geq 2j$ for which every $3m$-letter subword of $z$ contains $0^{m}$. Then, the frequency of occurrences of $0^{j}$ in $z$ is at least $\frac{m-j}{3m}$, which is greater than or equal to $1/6$ since $m\geq 2j$. By genericity of $z$ for $\mu$, $\mu([0^{j}])\geq 1/6$, contradicting the definition of $j$ and so the existence of $\mu$.

Now, suppose that $M$ is an arbitrary periodic orbit, define the sliding block code $\phi$ (with window size $k$) guaranteed by Lemma 3.4, and again define $y=\phi(x)$ and $Y=\phi(X)$. As usual, $Y$ has alphabet $\{0,1\}$, strictly contains the unique minimal subshift $\{0^{\infty}\}$, and satisfies $c_{n}(X)\geq c_{n-k+1}(Y)$ for all $n$. We note that $y$ cannot be eventually periodic in both directions; if it were, then it would have to begin and end with infinitely many [math]s, which would imply that $x$ was eventually periodic in both directions, a contradiction.

Finally, since $c_{n}(X)\geq c_{n-k+1}(Y)$ for all $n$, $\liminf(c_{n}(Y)-2n)=-\infty$. So, by the proof above in the $M=\{0^{\infty}\}$ case, $Y$ has unique invariant measure $\delta_{0^{\infty}}$. Any invariant measure $\nu$ in $X$ then must have pushforward $\delta_{0^{\infty}}$ under $\phi$, and so must have $\nu(M)=1$. It is easily checked that there is only one such $\nu$, namely the measure equidistributed over the points of $M$.

∎