A note on preimage entropy

Abstract

Cheng and Newhouse (Ergod. Th. & Dynam. Sys. 25 (2005), 1091–1113) proved a variational principle for topological preimage entropy $h_{\mathrm {pre}}(f)$:

$$ \begin{align*} h_{\mathrm{pre}}(f)=\sup_{\mu\in\mathcal{M}(X,f)}h_{\mathrm{pre},\mu}(f). \end{align*} $$

Unfortunately, we show in this note that this variational principle is not true.

1 Introduction

Let $(X,f)$ be a topological dynamical system (t.d.s. for short), that is, $(X,d)$ is a compact metric space and $f:X\rightarrow X$ is a continuous self-map. Preimage entropies were introduced and studied by Langevin and Przytycki [6], Hurley [5], Nitecki and Przytycki [7], and Fiebig, Fiebig and Nitecki [3]. These quantities give relevant information of how ‘non-invertible’ a system is. Among these entropy-like invariants, there are two kinds of pointwise preimage entropies:

$$ \begin{align*} h_m(f)=\lim_{{\epsilon}\to0}\limsup_{n\to\infty}\frac{1}{n}\log\sup_{x\in X}s(n,{\epsilon},f^{-n}(x)), \end{align*} $$

$$ \begin{align*} h_p(f)=\sup_{x\in X}\lim_{{\epsilon}\to0}\limsup_{n\to\infty}\frac{1}{n}\log s(n,{\epsilon},f^{-n}(x)), \end{align*} $$

where $s(n,{\epsilon },Z)$ (or $s(n,{\epsilon },Z,f)$ ) denotes the largest cardinality of any $(n,{\epsilon })$ -separated set of $Z\subset X$ . An important question is: can one introduce the counterpart of $h_m(f)$ or $h_p(f)$ from the measure-theoretic point of view, and obtain a variational principle relating them?

The first progress on this research was made by Cheng and Newhouse [1]. They defined a new notion of topological preimage entropy:

$$ \begin{align*} h_{\mathrm{pre}}(f)=\lim_{{\epsilon}\to0}\limsup_{n\to\infty}\frac{1}{n}\log\sup_{x\in X,k\geq n} s(n,{\epsilon},f^{-k}(x)). \end{align*} $$

On the measure-theoretic side, they defined a corresponding measure-theoretic preimage entropy:

$$ \begin{align*} h_{\mathrm{pre},\mu}(f)=\sup_{{\alpha}}h_{\mathrm{pre},\mu}(f,{\alpha}), \end{align*} $$

where ${\alpha }$ ranges over all finite partitions of X,

$$ \begin{align*} h_{\mathrm{pre},\mu}(f,{\alpha})=h_\mu(f,{\alpha}|\mathcal{B}^-) =\limsup_{n\to\infty}\frac{1}{n}H_\mu({\alpha}_0^{n-1}|\mathcal{B}^-), \end{align*} $$

and $\mathcal {B}^-$ is the infinite past $\sigma $ -algebra $\bigcap _{n\geq 0}f^{-n}\mathcal {B}$ related to the Borel $\sigma $ -algebra $\mathcal {B}$ . In addition, they stated a variational principle:

(1.1) $$ \begin{align} h_{\mathrm{pre}}(f)=\sup_{\mu\in\mathcal{M}(X,f)}h_{\mathrm{pre},\mu}(f), \end{align} $$

where $\mathcal {M}(X,f)$ denotes the set of all f-invariant Borel probability measures on X.

Recently, Wu and Zhu [9] developed a variational principle for $h_m(f)$ under the condition of uniform separation of preimages. They introduced a new version of pointwise metric preimage entropy:

$$ \begin{align*} h_{m,\mu}(f)=\sup_{{\alpha}}h_{m,\mu}(f,{\alpha}), \end{align*} $$

where ${\alpha }$ ranges over all finite partitions of X and

$$ \begin{align*} h_{m,\mu}(f,{\alpha})=\limsup_{n\to\infty}\frac{1}{n}H_\mu({\alpha}_0^{n-1}|f^{-n}\mathcal{B}). \end{align*} $$

For f with uniform separation of preimages, the authors [9] established the following variational principle relating $h_{m,\mu }(f)$ and $h_m(f)$ :

$$ \begin{align*} h_m(f)=\sup_{\mu\in\mathcal{M}(X,f)}h_{m,\mu}(f). \end{align*} $$

In fact, it was shown in [10, Proposition 3.1] that

$$ \begin{align*} h_{m,\mu}(f)=h_{\mathrm{pre},\mu}(f) \quad\text{for any }\mu\in\mathcal{M}(X,f). \end{align*} $$

For related definitions of topological and measure-theoretic entropies, we refer to the books [2, 4, 8].

In this note, we shall give an example to show that

$$ \begin{align*} h_{\mathrm{pre}}(f)>\sup_{\mu\in\mathcal{M}(X,f)}h_{\mathrm{pre},\mu}(f). \end{align*} $$

So the variational principle in equation (1.1) is not true.

2 Main result

In this section, we will state and prove our main result.

Lemma 2.1. Let $A=\{0, 1, 2\}$ and endow $A^{\mathbb {N}\times \mathbb {N}}$ with the product topology of the discrete topology on A. Denote by $f: A^{\mathbb {N}\times \mathbb {N}}\to A^{\mathbb {N}\times \mathbb {N}}$ the left shift map on rows; that is,

$$ \begin{align*} (f x)_{m,i} = x_{m,i+1} \quad\text{for }m, i\in \mathbb{N}. \end{align*} $$

For each array $x=(x_{m,i})_{m,i\geq 0}$ , denote by $i_0(x)$ the minimal $i\geq 0$ such that $x_{0,i}=0$ . If such an i does not exist, then we set $i_0(x)=\infty $ . Let $X\subset A^{\mathbb {N}\times \mathbb {N}}$ consist of arrays such that:

1. (1) for all $i\geq i_0(x)$ and all $m\geq 0$ , we have $x_{m,i}=0$ ;

2. (2) for all $0\leq i< i_0(x)$ and all $m\geq 0$ , we have $x_{m,i}\in \{1,2\}$ and if both $m\geq 1$ and $i\geq 1$ , then $x_{m,i}=x_{m-1,i-1}$ .

For the t.d.s $(X,f)$ , we have $h_{\mathrm {pre}}(f)\geq \log 2$ and $h_{\mathrm {pre},\mu }(f)=0$ for any $\mu \in \mathcal {M}(X,f)$ .

Proof. For $0\leq n\leq \infty $ , let $A_n$ denote the set of points $x\in X$ with $i_0(x)=n$ and $\textbf {0}$ denote the array consisting of just zeros. Then we have the following observations.

1. (1) $A_0=\{\textbf {0}\}$ and the element $\textbf {0}$ has infinitely many preimages.

2. (2) Any element $x\in X\setminus A_0$ has exactly one preimage.

3. (3) $(A_\infty ,f)$ is an invertible subsystem.

Let ${\epsilon }_0>0$ be so small that $x,y\in X$ with $x_{0,0}\neq y_{0,0}$ implies $d(x,y)\geq {\epsilon }_0$ . Note that if we just observe the zero-row of $f^{-n}(\textbf {0})$ , we will see elements starting with any block of any length $0\leq k\leq n$ over $1,2$ (followed by zeros). So we have

$$ \begin{align*}s(n,{\epsilon}_0,f^{-n}(\textbf{0}))\geq\sum_{k=0}^n2^k=2^{n+1}-1.\end{align*} $$

Hence,

$$ \begin{align*} h_{\mathrm{pre}}(f)&=\lim_{{\epsilon}\to0}\limsup_{n\to\infty}\frac{1}{n}\log\sup_{x\in X,k\geq n} s(n,{\epsilon},f^{-k}(x))\\ &\geq\limsup_{n\to\infty}\frac{1}{n}\log s(n,{\epsilon}_0,f^{-n}(\textbf{0}))\\ &\geq\log2. \end{align*} $$

Now we pass to evaluating the measure-theoretic preimage entropy. Notice that for each $0<n<\infty $ , the set $A_n$ is visited by any orbit at most once implying that $\mu (A_n)=0$ for any $\mu \in \mathcal {M}(X,f)$ . So, any invariant measure $\mu $ is supported by $A_0\cup A_\infty $ . Fix $\mu \in \mathcal {M}(X,f)$ . Without loss of generality, we may assume that $\mu (A_0)>0$ and $\mu (A_\infty )>0$ . Consider the conditional measures

$$ \begin{align*} \mu_{A_0}(\cdot)=\frac{\mu(\cdot\cap A_0)}{\mu_{A_0}}\quad\text{and}\quad \mu_{A_\infty}(\cdot)=\frac{\mu(\cdot\cap A_\infty)}{\mu_{A_\infty}}. \end{align*} $$

It is easy to verify that both $\mu _{A_0}$ and $\mu _{A_\infty }$ are invariant and $\mu =\mu (A_0)\mu _{A_0}+\mu (A_\infty )\mu _{A_\infty }$ . By the affinity of measurable conditional entropy (see, for example, [2, Theorem 2.5.1], [1, Theorem 2.3] or [9, Proposition 2.12]), we have

$$ \begin{align*} h_{\mathrm{pre},\mu}(f)&=\mu(A_0)h_{\mathrm{pre},\mu_{A_0}}(f) +\mu(A_\infty)h_{\mathrm{pre},\mu_{A_\infty}}(f)\\ &\leq\mu(A_0)h_{\mu_{A_0}}(f) +\mu(A_\infty)h_{\mathrm{pre},\mu_{A_\infty}}(f)\\ &=0.\\[-3.2pc] \end{align*} $$

By Lemma 2.1, we can get our main result.

Theorem 2.2. There exists a t.d.s. $(X,f)$ such that

$$ \begin{align*} 0=\sup_{\mu\in\mathcal{M}(X,f)}h_{\mathrm{pre},\mu}(f)<\log2\leq h_{\mathrm{pre}}(f). \end{align*} $$

Thus, the Cheng–Newhouse variational principle in equation (1.1) fails.

3 Another definition of preimage entropy

In [1], the authors show that $h_{\mathrm {pre}}(f)$ can also be defined as

(3.1) $$ \begin{align} h_{\mathrm{pre}}(f)=\lim_{{\epsilon}\to0}\limsup_{n\to\infty}\frac{1}{n}\log\sup_{x\in X,k\geq1}s(n,{\epsilon},f^{-k}x). \end{align} $$

This result is based on their variational principle in equation (1.1). Now we shall give a topological proof of equation (3.1). In fact, it is a consequence of the following result.

For $Z\subset X$ , let $r(n,{\epsilon },Z)$ denote the smallest cardinality of any $(n,{\epsilon })$ -spanning set of $Z\subset X$ . It is clear that the above topological notions of entropies defined by separated sets can also be defined by spanning sets.

Theorem 3.1. Let $f:X\to X$ be a continuous map. Then,

$$ \begin{align*} h_{\mathrm{pre}}(f)&=\lim_{{\epsilon}\to0}\limsup_{n\to\infty}\frac{1}{n}\log\sup_{x\in X}s(n,{\epsilon},P_x)\\ &=\lim_{{\epsilon}\to0}\limsup_{n\to\infty}\frac{1}{n}\log\sup_{x\in X}r(n,{\epsilon},P_x), \end{align*} $$

where

$$ \begin{align*} P_x=\bigcup_{j\geq0}f^{-j}f^jx. \end{align*} $$

Proof. Fix $y\in X$ , $n\in \mathbb {N}$ and $k\geq n$ . If $f^{-k}y\neq \emptyset $ , then pick $x\in f^{-k}y$ . So,

$$ \begin{align*} r(n,{\epsilon},f^{-k}y)=r(n,{\epsilon},f^{-k}f^kx)\leq r(n,{\epsilon},P_x), \end{align*} $$

which implies

$$ \begin{align*} h_{\mathrm{pre}}(f)\leq\lim_{{\epsilon}\to0}\limsup_{n\to\infty}\frac{1}{n}\log\sup_{x\in X}r(n,{\epsilon},P_x). \end{align*} $$

Next, we show the remaining inequality. Fix $s>h_{\mathrm {pre}}(f)$ . For any ${\epsilon }>0$ , there exists $N\in \mathbb {N}$ such that

$$ \begin{align*} r(n,{\epsilon},f^{-k}x)\leq e^{sn} \end{align*} $$

for all $x\in X$ , $n\geq N$ and $k\geq n$ .

Fix $x\in X$ , $n\geq N$ . For $k\geq n$ , let $E_k\subset X$ be an $(n,{\epsilon })$ -spanning set of $f^{-k}f^kx$ with $\#E_k=r(n,{\epsilon },f^{-k}f^kx)\leq e^{sn}$ . Let $K(X)$ be the space of non-empty closed subsets of X equipped with the Hausdorff metric. Then we have $E_k\in K(X)$ . As X is compact, $K(X)$ is also compact. So there exists a subsequence $\{k_j\}_{j\geq 1}$ such that $E_{k_j}\rightarrow E(j\to \infty )$ . Then we have $\#E\leq e^{sn}$ .

We claim that

$$ \begin{align*} P_x\subset\bigcup_{z\in E}B_n(z,2{\epsilon}). \end{align*} $$

To see this, pick $y\in P_x$ . Then there exists $J\geq n$ such that for any $j\geq J$ , one has

$$ \begin{align*} y\in f^{-k_j}f^{k_j}x\subset\bigcup_{z\in E_{k_j}}B_n(z,{\epsilon}). \end{align*} $$

Furthermore, we can pick $z_{k_j}\in E_{k_j}$ to get

$$ \begin{align*} d_n(y,z_{k_j})<{\epsilon} \quad\text{for all }j\geq J. \end{align*} $$

Without loss of generality, we assume that $\lim _{j\to \infty }z_{k_j}=z$ . Then it is easy to see that $z\in E$ and

$$ \begin{align*} d_n(y,z)\leq{\epsilon}. \end{align*} $$

So the claim is true. Hence, we have $r(n,2{\epsilon },P_x)\leq \#E\leq e^{sn}$ , from which one can get

$$ \begin{align*} \lim_{{\epsilon}\to0}\limsup_{n\to\infty}\frac{1}{n}\log\sup_{x\in X}r(n,{\epsilon},P_x)\leq s. \end{align*} $$

By the choice of s, we obtain the reversed inequality.