A Note on the Measure-Theoretic Pressure in Mean Metric

Abstract

In this paper, we apply the max metric $d_{n, q}$ and the mean metric ${\overline{d}}_{n, q}$ to the definitions of measure-theoretic pressure and topological pressure. In fact, we generalize Katok’s entropy formula to measure-theoretic pressure by the mean metric.

1 Introductions

Entropy is one of the measurements of the complexity of dynamical systems. Measure-theoretic entropy and topological entropy are important determinants of complexity in dynamical system. The relationship between these two quantities is the well-known variational principle. In 1958, Kolmogorov firstly introduced the measure-theoretic entropy of a measure-preserving dynamical system as an isomorphic invariant [11, 12]. Later Adler et al. introduced the topological entropy of a topological dynamical system by open covers as a conjugate invariant [1]. Equivalent definitions based on separated and spanning set were independently given by Bowen [2] and Dinaburg [5].

Topological pressure is a generalization of topological entropy. The theories of topological pressure, varitional principle and equilibrium states play a prominent role in statistical mechanics, ergodic theory and dynamical systems [10, 14]. Since the works of Bowen [3] and Ruelle [16], the topological pressure turned into a basic tool of the dimension theory in dynamical systems.

In the investigation of the Sarnak’s conjecture, Huang, Wang and Ye [7] introduced the measure complexity of an invariant measure $\mu$ similar to the one introduced by Katok [13], by using the mean metric instead of the Bowen metric (for discussion and results related to mean metric, see also [18]). In 2015, by using separated sets, Gröger and Jäger [9] gave a definition of topological entropy of the whole system in mean metrics, and proved that the topological entropy defined by mean metrics is equivalent to the classical topological entropy. Huang et al. [7] introduced the notion of measure complexity for a topological dynamical system. Then, they [8] established Katok’s entropy formula for ergodic measures in the case of mean metrics. In fact, this formula [8] showed that one can replace the Bowen metrics with the mean metrics when studying the measure complexity for a topological dynamical system. In this paper, inspired by Huang et al. [8], we apply two metrics: the max metric $d_{n, q}$ and the mean metric ${\overline{d}}_{n, q}$ to the definitions of measure-theoretic pressure and topological pressure. In fact, we generalize Katok’s entropy formula to measure-theoretic pressure by the mean metric.

2 Preliminaries and Main Results

Throughout this paper, by a topological dynamical system (t.d.s. for short) we mean a pair (X, T), where X is a compact metric space with a metric d and T is a homeomorphism from X to itself. Let $C(X, {\mathbb {R}})$ denote the set of all continuous functions of X. Given a t.d.s. (X, T), let M(X) denote the collection of all Borel probability measures on measurable space $\left( X, {\mathcal {B}}_X\right)$, where ${\mathcal {B}}_X$ denote the Borel $\sigma$-algebra on X. The set of all T-invariant Borel probability measures in M(X) will be denoted by M(X, T). The set of all T-invariant ergodic measures in M(X, T) will be denoted by $M^e(X, T)$. It is noted that each $\mu \in M(X, T)$ induces a measure preserving system (m.p.s. for short) $\left( X, {\mathcal {B}}_X, \mu , T\right)$.

Let (X, T) be a t.d.s. with a metric d. For $q, n \in {\mathbb {N}}$, we consider the max metric

$$\begin{aligned} d_{n, q}(x, y)=\max _{0 \le i \le n-1} d\left( T^{q i} x, T^{q i} y\right) \end{aligned}$$

for any $x, y \in X$. For $\varphi \in C(X,{\mathbb {R}})$,$\mu \in M(X, T)$ and $\varepsilon >0$, set

$$\begin{aligned} S_{n,q}\varphi (x):=\varphi (x)+\varphi (T^{q}x)+...+\varphi (T^{q(n-1)}x), \end{aligned}$$

and

$$\begin{aligned} \textrm{Var}(\varphi , \varepsilon ):=\sup \{|\varphi (x)-\varphi (y)|: d(x, y)<\varepsilon \}. \end{aligned}$$

Let

$$\begin{aligned} S_{d_{n, q}}(\varphi ,\mu , \varepsilon )=\inf \left\{ \sum _{i=1}^{k} e^{S_{n,q}\varphi (x_i)}, \text{ s.t. } \mu \left( \bigcup _{i=1}^k B_{d_{n, q}}\left( x_i, \varepsilon \right) \right) >1-\varepsilon \right\} \end{aligned}$$

where

$$\begin{aligned} B_{d_{n, q}}(x, \varepsilon )=\left\{ y \in X: d_{n, q}(x, y)<\varepsilon \right\} \end{aligned}$$

for any $x \in X$. We define

$$\begin{aligned} S_n^*(\varphi ,\mu , \varepsilon )=\inf _{q \ge 1} S_{d_{n, q}}(\varphi ,\mu , \varepsilon ). \end{aligned}$$

When $q=1$, [9] establish measure-theoretic pressure version of Katok’s entropy formula as follows:

Theorem 2.1

[9] Let (X, T) be a t.d.s. If $\mu \in M(X, T)$ is ergodic, then

$$\begin{aligned} h_\mu (T)+\int \varphi \textrm{d} \mu =\lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log S_{{d}_{n,1}}(\varphi ,\mu , \varepsilon )=\lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log S_{{d}_{n,1}}(\varphi ,\mu , \varepsilon ). \end{aligned}$$

For $\mu \in M(X, T)$ and $q, n \in {\mathbb {N}}$, we consider the mean metric

$$\begin{aligned} {\overline{d}}_{n, q}(x, y)=\frac{1}{n} \sum _{i=0}^{n-1} d\left( T^{q i} x, T^{q i} y\right) \end{aligned}$$

for any $x, y \in X$. For $\varepsilon >0$ and $\varphi \in C(X,{\mathbb {R}})$, let

$$\begin{aligned} S_{{\overline{d}}_{n, q}}(\varphi ,\mu , \varepsilon )=\inf \left\{ \sum _{i=1}^{k} e^{S_{n,q}\varphi (x_i)} \text{ s.t. } \mu \left( \bigcup _{i=1}^k B_{{\overline{d}}_{n, q}}\left( x_i, \varepsilon \right) \right) >1-\varepsilon \right\} , \end{aligned}$$

where

$$\begin{aligned} B_{{\overline{d}}_{n, q}}(x, \varepsilon )=\left\{ y \in X: {\overline{d}}_{n, q}(x, y)<\varepsilon \right\} \end{aligned}$$

for any $x \in X$. We define

$$\begin{aligned} {\overline{S}}_n^*(\varphi ,\mu , \varepsilon )=\inf _{q \ge 1} S_{{\overline{d}}_{n, q}}(\varphi , \mu ,\varepsilon ). \end{aligned}$$

When $q=1$, [6] established measure-theoretic pressure version of Katok’s entropy formula in mean metrics as follows:

Theorem 2.2

[6] Let (X, T) be a t.d.s. If $\mu \in M(X, T)$ is ergodic, then

$$\begin{aligned} h_\mu (T)+\int \varphi \textrm{d} \mu =\lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log S_{{\overline{d}}_{n,1}}(\varphi ,\mu , \varepsilon )=\lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log S_{{\overline{d}}_{n,1}}(\varphi ,\mu , \varepsilon ). \end{aligned}$$

Next, we need to define the topological pressure:

Let (X, T) be a t.d.s., for $q, n \in {\mathbb {N}}$ and $\varphi \in C(X,{\mathbb {R}})$, we consider

$$\begin{aligned} P_n(T, \varphi , \varepsilon )=\inf \left\{ \sum _{x_i \in F } e^{S_{n,1}\varphi (x_i)} \mid F \text{ is } \text{ a } (n, \varepsilon ) \text{ spanning } \text{ set } \text{ for } X\right\} \text{. } \end{aligned}$$

For $\varepsilon > 0$ and $\varphi \in C(X,{\mathbb {R}})$,put

$$\begin{aligned} P(T, \varphi , \varepsilon )=\limsup _{n \rightarrow \infty } \frac{1}{n} \log P_n(T,\varphi , \varepsilon ). \end{aligned}$$

When $\varepsilon \rightarrow 0$, we can get

$$\begin{aligned} P_{\textrm{top}}(T, \varphi ) = \lim _{\varepsilon \rightarrow 0} P(T, \varphi , \varepsilon ). \end{aligned}$$

Also, the variation principle [19] reveals the basic relationship between topological pressure and measure-theoretic pressure: let (X, T) is a TDS and $\varphi \in C(X,{\mathbb {R}})$, then

$$\begin{aligned} P_{\textrm{top}}(T, \varphi )=\sup \left\{ h_\mu (T)+\int \varphi \mathrm {~d} \mu : \mu \in M(X, T)\right\} . \end{aligned}$$

We give the computation formula of the measure-theoretic pressure of an ergodic measure preserving system (resp., the topological pressure of a topological dynamical system) by the two metrics $d_{n, q}$ and ${\overline{d}}_{n, q}$.

Our main results are as follows:

Theorem 2.3

Let (X, T) be a t.d.s.. Then for any $\mu \in M^e(X, T)$ and $\varphi \in C(X,{\mathbb {R}})$, we have

$$\begin{aligned} \int \varphi \mathrm {~d} \mu +h_\mu (T)=\lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\mu ,\varphi , \varepsilon )=\lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\mu , \varphi , \varepsilon ) \end{aligned}$$

and

$$\begin{aligned} \int \varphi \mathrm {~d} \mu +h_\mu (T)=\lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\mu , \varphi , \varepsilon )=\lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\mu , \varphi , \varepsilon ). \end{aligned}$$

Let (X, T) be a t.d.s. with a metric d. For $q, n \in {\mathbb {N}}$ and $\varepsilon >0$,

$$\begin{aligned} S_{{\overline{d}}_{n, q}}(\varphi ,\varepsilon )=\inf \left\{ \sum _{i=1}^{k} e^{S_{n,q}(x_i)} \text{ s.t. } \bigcup _{i=1}^k B_{{\overline{d}}_{n, q}}\left( x_i, \varepsilon \right) =X\right\} . \end{aligned}$$

We define

$$\begin{aligned} {\overline{S}}_n^*(\varepsilon )=\inf _{q \ge 1} S_{{\overline{d}}_{n, q}}(\varphi ,\varepsilon ). \end{aligned}$$

Similarly, we can define $S_{d_{n, q}}(\varphi ,\varepsilon ), S_n^*(\varphi ,\varepsilon )$ by changing ${\overline{d}}_{n, q}$ into $d_{n, q}$.

Theorem 2.4

Let (X, T) be a t.d.s and $\varphi \in C(X,{\mathbb {R}})$. Then

$$\begin{aligned} P_{\textrm{top}}(T,\varphi )=\lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\varphi ,\varepsilon )=\lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\varphi ,\varepsilon ) \end{aligned}$$

and

$$\begin{aligned} P_{\textrm{top}}(T,\varphi )=\lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\varepsilon )=\lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\varepsilon ). \end{aligned}$$

3 Proof of Theorem 2.3

Before the proof of Theorem 2.3, we need a lemma which similar to Lemma 3.1 in [6] as follows. In order to prove the Lemma 3.1, we need the relationship between the ${\overline{d}}_{n,q}$-ball $B_{{\overline{d}}_{n,q}}(x, \varepsilon )$ where the mistake function $g: {\mathbb {N}} \times (0,1] \rightarrow {\mathbb {R}}$ by $g(n, \varepsilon ):=n \varepsilon$ for any $\varepsilon \in (0,1], n \in {\mathbb {N}}$. The function g is called a mistake function. Note that g is a special mistake function and the definition of the function g is different from the definition of the mistake function in [17] or [20].

Definition 3.1

Let $0<\varepsilon \le 1$ and $n \ge 1$. The mistake dynamical ball $B_{d_{n,q}}(g; x, \varepsilon )$ centered at x with radius $\varepsilon$ and length n associated to function g is defined as follows:

$$\begin{aligned} \begin{aligned} B_{d_{n,q}}(g; x, \varepsilon )&=\left\{ y \in X: \#\left\{ 0 \le i \le n-1: d\left( T^{iq} x, T^{iq} y\right)<\varepsilon \right\}>n-g(n, \varepsilon )\right\} \\&=\left\{ y \in X: \frac{\#\left\{ 0 \le i \le n-1: d\left( T^{iq} x, T^{iq} y\right) <\varepsilon \right\} }{n}>1-\frac{g(n, \varepsilon )}{n}\right\} . \end{aligned} \end{aligned}$$

Lemma 3.1

For any $x \in X, n \in {\mathbb {N}}$ and $\varepsilon >0$, we have

$$\begin{aligned} B_{{\overline{d}}_{n,q}}(x, \varepsilon ) \subset B_{d_{n,q} }(g; x, \sqrt{\varepsilon }) \end{aligned}$$

where $g=n\sqrt{\varepsilon }.$

Proof

For any $q \in {\mathbb {N}}$, Set

$$\begin{aligned} \begin{aligned}&I_1=\left\{ 0 \le i \le n-1: d\left( T^{iq} x, T^{iq} y\right) <\sqrt{\varepsilon }\right\} , \\&I_2=\left\{ 0 \le i \le n-1: d\left( T^{iq} x, T^{iq}y\right) \ge \sqrt{\varepsilon }\right\} . \end{aligned} \end{aligned}$$

Noting that

$$\begin{aligned} \begin{aligned} \frac{1}{n} \sum _{i=0}^{n-1} d\left( T^{iq} x, T^{iq} y\right)&=\frac{\sum _{i \in I_1} d\left( T^{iq} x, T^{iq} y\right) +\sum _{i \in I_2} d\left( T^{iq} x, T^{iq} y\right) }{n} \\&\ge \frac{\sum _{i \in I_1} d\left( T^{iq} x, T^{iq} y\right) +\sqrt{\varepsilon } \# I_2}{n} \\&\ge \frac{\sqrt{\varepsilon }}{n} \# I_2. \end{aligned} \end{aligned}$$

If $y \in B_{{\overline{d}}_{n,q}}(x, \varepsilon )$, i.e., ${\overline{d}}_{n,q}(x, y)<\varepsilon$, we have

$$\begin{aligned} \# I_2<n \sqrt{\varepsilon }. \end{aligned}$$

Thus, we obtain

$$\begin{aligned} \frac{\#\left\{ 0 \le i \le n-1: d\left( T^{iq} x, T^{iq} y\right) \ge \sqrt{\varepsilon }\right\} }{n}<\sqrt{\varepsilon }, \end{aligned}$$

Then $y \in B_{d_{n,q} }(g; x, \sqrt{\varepsilon })$,

$$\begin{aligned} \frac{\#\left\{ 0 \le i \le n-1: d\left( T^{iq} x, T^{iq} y\right) <\sqrt{\varepsilon }\right\} }{n}>1-\sqrt{\varepsilon }=1-\frac{g(n, \sqrt{\varepsilon })}{n}. \end{aligned}$$

Therefore, we have $B_{{\overline{d}}_{n,q}}(x, \varepsilon ) \subset B_{d_{n,q} }(g; x, \sqrt{\varepsilon })$. $\square$

Next, we prove the Theorem 2.3.

Proof

Let (X, T) be a t.d.s. First we show that for any $\mu \in M^e(X, T)$

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_{n}^*(\varphi ,\mu , \varepsilon ) \le h_\mu (T)+\int \varphi \mathrm {~d} \mu . \end{aligned}$$

Fix $\varepsilon >0$ and $\mu \in M^e(X, T)$. Recall that $d_{n,1}(x, y)=\max \limits _{0 \le i \le n-1} d\left( T^i x, T^i y\right)$. It is obvious that ${\overline{d}}_{n, 1}(x, y) \le d_{n,1}(x, y)$ and $B_{d_{n,1}}(x, \varepsilon ) \subset B_{{\overline{d}}_{n, 1}}(x, \varepsilon )$. For any $\tau > 0$ and n large enough, by Theorem 2.1 there exists a finite subset $F_n$ of X such that

$$\begin{aligned} \sum \limits _{x_i\in F_{n}}e^{S_{n,1}\varphi (x_i)}\le e^{n({h_{\mu }(T)+\int \varphi \mathrm {~d} \mu +\tau })} \end{aligned}$$

and an open subset $K_n$ of X with $\mu \left( K_n\right) >1-\varepsilon$ such that for any $x \in K_n$ there is $y \in F_n$ with $d_{n,1}(x, y)<\varepsilon$, which implies ${\overline{d}}_{n, 1}(x, y) \le d_{n,1}(x, y)<\varepsilon$. This implies that

$$\begin{aligned} S_{{\overline{d}}_{n, 1}}(\varphi ,\mu , \varepsilon ) \le \sum \limits _{x_i\in F_{n}}e^{S_{n,1}\varphi (x_i)}\le e^{n({h_{\mu }(T)+\int \varphi \mathrm {~d} \mu +\tau })}, \end{aligned}$$

and hence we have

$$\begin{aligned} \begin{aligned} \lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\varphi ,\mu , \varepsilon )&\le \lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log S_{{\overline{d}}_{n, 1}}(\varphi ,\mu , \varepsilon )\\&\le h_\mu (T)+\int \varphi \mathrm {~d} \mu +\tau . \end{aligned} \end{aligned}$$

Let $\tau \rightarrow 0$, we finish this proof.

Now we are to prove that for any $\mu \in M^e(X, T)$

$$\begin{aligned} h_\mu (T)+\int \varphi \mathrm {~d} \mu \le \lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\varphi ,\mu , \varepsilon ). \end{aligned}$$

Given a Borel partition $\eta =\left\{ A_1, \ldots , A_k\right\}$ of X and $0<\delta <1$, it is sufficient to show

$$\begin{aligned} h_\mu (T, \eta )+ \int \varphi \mathrm {~d} \mu \le a+2 \delta , \end{aligned}$$

where $a=\lim \limits _{\varepsilon \rightarrow 0} \liminf \limits _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\varphi ,\mu , \varepsilon )$. Take $0<\tau <\frac{1}{2}$ with

$$\begin{aligned} -2 \tau \log 2 \tau -(1-2 \tau ) \log (1-2 \tau )+4 \tau \log (k+1)+2\tau ||\varphi ||<\delta . \end{aligned}$$

By [19], Lemma 4.15 and Corollary 4.12.1], it is easy to see that there is a Borel partition $\xi =\left\{ B_1, \ldots , B_k, B_{k+1}\right\}$ of X associated with $\eta$ such that $B_i \subset A_i$ is closed for $1 \le i \le k$ and $B_{k+1}=X \backslash K$ with $K=\bigcup _{i=1}^k B_i, \mu \left( B_{k+1}\right) <\tau ^2$ and

$$\begin{aligned} h_\mu (T, \eta )+\int \varphi \mathrm {~d} \mu \le h_\mu (T, \xi )+\delta +\int \varphi \mathrm {~d} \mu . \end{aligned}$$

(3.1)

It is clear that $B_i \cap B_j=\emptyset , 1 \le i<j \le k$ and $\mu (K)>1-\tau ^2$. Let $b=\min \limits _{1 \le i<j \le k} d\left( B_i, B_j\right)$. Then $b>0$.

Given $\gamma >0,0<\varepsilon <\min \left\{ 1-2 \tau , \frac{1}{2} b(1-2 \tau )\right\}$. For $q, n \in {\mathbb {N}}$ and $\gamma \ge 0$, by the definition of $S_{{\overline{d}}_{n, q}}(\varphi ,\mu , \varepsilon )$ there are $x_1, \ldots , x_{m(n, q)} \in X$ such that $\mu \left( \bigcup _{i=1}^{m(n, q)} B_{{\overline{d}}_{n, q}}\left( x_i, \varepsilon \right) \right) >1-\varepsilon$, where

$$\begin{aligned} \sum \limits _{i=1}^{m(n,q)}e^{S_{n,q}\varphi (x_i)}\le S_{{\overline{d}}_{n, q}}(\varphi ,\mu , \varepsilon )+\gamma . \end{aligned}$$

(3.2)

Let $F_{n, q}=\bigcup _{i=1}^{m(n, q)} B_{{\overline{d}}_{n, q}}\left( x_i, \varepsilon \right) .$ Then $\mu \left( F_{n, q}\right) >1-\varepsilon$. For $x \in X$, let $K_q(x)=\left\{ i \in {\mathbb {Z}}_{+}: T^{q i} x \in K\right\}$ and

$E_{n, q}=\left\{ x \in X: \frac{\left| K_q(x) \cap [0, n-1]\right| }{n} \le 1-\tau \right\}$. Note that

$$\begin{aligned} \int _X \frac{\left| K_q(x) \cap [0, n-1]\right| }{n} \textrm{d} \mu (x)=\int _X \frac{1}{n} \sum _{i=0}^{n-1} 1_K\left( T^{q i} x\right) \textrm{d} \mu (x)=\mu (K)>1-\tau ^2. \end{aligned}$$

We have

$$\begin{aligned} (1-\tau ) \mu \left( E_{n, q}\right) +1-\mu \left( E_{n, q}\right) \ge \int _X \frac{\left| K_q(x) \cap [0, n-1]\right| }{n} \textrm{d} \mu (x)>1-\tau ^2, \end{aligned}$$

which implies that $\mu \left( E_{n, q}\right) <\tau$. Put $W_{n, q}=\left( K \cap F_{n, q}\right) \backslash E_{n, q}$. Then

$$\begin{aligned} \mu \left( W_{n, q}\right)>1-\tau ^2-\tau -\varepsilon >1-2 \tau -\varepsilon \end{aligned}$$

and for $z \in W_{n, q}$

$$\begin{aligned} \frac{\left| K_q(z) \cap [0, n-1]\right| }{n}>1-\tau . \end{aligned}$$

For a given $1 \le i \le m(n, q)$, let

$$\begin{aligned} \xi _{n, q}(i)=\left\{ U \in \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi : U \cap B_{{\overline{d}}_{n, q}}\left( x_i, \varepsilon \right) \cap W_{n, q} \ne \emptyset \right\} . \end{aligned}$$

We claim

$$\begin{aligned} \left| \xi _{n, q}(i)\right| \le C_n^{[n c(\varepsilon )]} \cdot (k+1)^{[n c(\varepsilon )]} \end{aligned}$$

(3.3)

where $c(\varepsilon )=2 \frac{\varepsilon }{b}+2 \tau$ and $[n c(\varepsilon )]$ is the integer part of $n c(\varepsilon )$. To see this, let $U_1, U_2 \in \xi _{n, q}(i)$. Take $x \in U_1 \cap B_{{\overline{d}}_{n, q}}\left( x_i, \varepsilon \right) \cap W_{n, q}$ and $y \in U_2 \cap B_{{\overline{d}}_{n, q}}\left( x_i, \varepsilon \right) \cap W_{n, q}$. Then ${\overline{d}}_{n, q}(x, y)<2 \varepsilon$ and $\left| K_q(x) \cap K_q(y) \cap [0, n-1]\right| >(1-2 \tau ) n$. This implies that

$$\begin{aligned} \left| \left\{ 0 \le j \le n-1: T^{q j} x \in B_{k_j}, T^{q j} y \in B_{l_j}, 1 \le k_j \ne l_j \le k+1\right\} \right| \le [n c(\varepsilon )]. \end{aligned}$$

Hence (3.3) holds. It remains to show $h_\mu (T, \xi )+\int \varphi \textrm{d}\mu \le a+\delta$. If $C \in \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \cap W_{n,q}$,We define

$$\begin{aligned} \alpha (C):=\sup \{(S_{n,q}\varphi )(x):~ x \in C \}. \end{aligned}$$

For each $C \in \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \cap W_{n,q}$, we can choose some $x_C \in \bar{C}$, so that $S_{n,q}\varphi (x_C)=\alpha (C)$. For $q, n \in {\mathbb {N}}$, we know that

$$\begin{aligned} \begin{aligned}&H_\mu \left( \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \right) +\int S_{n,q} \varphi \mathrm {~d} \mu \\ \le&H_\mu \left( \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \vee \left\{ W_{n, q}, X \backslash W_{n, q}\right\} \right) +\int _X S_{n, q} \varphi \textrm{d} \mu \\ \le&-\mu (W_{n,q}) \sum _{U\in \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi } \frac{\mu (U \cap W_{n,q})}{\mu (W_{n,q})}\log \frac{\mu (U \cap W_{n,q})}{\mu (W_{n,q})}\\&-\mu (X \backslash W_{n, q}) \sum _{U\in \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi } \frac{\mu (U \cap (X \backslash W_{n, q}))}{\mu (X \backslash W_{n,q})}\log \frac{\mu (U \cap (X \backslash W_{n, q}))}{\mu ((X \backslash W_{n, q}))} \\ {}&+\int _{W_{n,q}} S_{n,q}\varphi (x) \mathrm {~d} \mu +\int _{X \backslash W_{n, q}} S_{n,q}\varphi (x) \mathrm {~d} \mu -\mu \left( W_{n, q}\right) \log \mu \left( W_{n, q}\right) \\&-\left( 1-\mu \left( W_{n, q}\right) \right) \log \left( 1-\mu \left( W_{n, q}\right) \right) \\ \le&\mu (W_{n,q}) H_{\mu _{W_{n,q}}}( \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \cap W_{n,q} ) + \int _{W_{n, q}} S_{n,q}\varphi (x ) \textrm{d} \mu \\ {}&+\mu (X \backslash W_{n, q}) \log \left| \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \right| +\mu \left( X \backslash W_{n, q}\right) n||\varphi ||+\log 2 \\ \le&H_{\mu _{W_{n,q}}}( \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \cap W_{n,q} ) +\int S_{n,q}\varphi (x) d \mu _{W_{n, q}} \\ {}&+\mu (X \backslash W_{n, q}) \log \left| \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \right| +\mu \left( X \backslash W_{n, q}\right) n||\varphi ||+\log 2 \\ \le&\sum \limits _{C \in \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \cap W_{n,q} } \mu _{W_{n, q}}(C)[-\log (\mu _{W_{n, q}}(C)+\alpha (C)]\\ {}&+\mu \left( X \backslash W_{n, q}\right) (\log (k+1)^n+ n||\varphi ||)+\log 2 \\ \le&\log \sum \limits _{C \in \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \cap W_{n,q} }e^{\alpha (C)} +n(2\tau +\varepsilon )\left( \log (k+1) +||\varphi ||\right) +\log 2 \\ \le&\log \sum \limits _{0 \le i \le m(n,q)} \sum \limits _{C\in \xi _{n, q}(i)}e^{\alpha (C)} +n(2\tau +\varepsilon )\left( \log (k+1)+||\varphi ||\right) +\log 2. \end{aligned} \end{aligned}$$

For any $C \in \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \cap W_{n, q}$, then $C \subset \bigcup _{i=1}^{m(n, q)} B_{{\overline{d}}_{n, q}}\left( x_i, \varepsilon \right)$, there exists $1 \le i \le m(n, q)$ such that ${\overline{d}}_{n, q}\left( x_C, x_i\right) \le \varepsilon <2 \varepsilon$. Hence, by Lemma 3.1, $x_C \in B_{d_{n, q}}\left( g; x_i, \sqrt{2 \varepsilon }\right)$ with $g=n \sqrt{2 \varepsilon }$ and

$$\begin{aligned} \alpha (C)=S_{n, q} \varphi \left( x_C\right) \le S_{n, q} \varphi \left( x_i\right) +n {\text {Var}}(\varphi , \sqrt{2 \varepsilon })+n \sqrt{2 \varepsilon }\Vert \varphi \Vert . \end{aligned}$$

Then

$$\begin{aligned}&H_\mu \left( \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \right) +\int _X S_{n, q} \varphi \textrm{d} \mu \\&\qquad \le \log \sum _{0 \le i \le m(n, q)} \sum _{C \in \xi _{n, q}(i)} e^{S_{n, q} \varphi \left( x_i\right) +n {\text {Var}}(\varphi , \sqrt{2 \varepsilon })+n \sqrt{2 \varepsilon }\Vert \varphi \Vert }\\&\qquad \quad +n(2 \tau +\varepsilon )(\log (k+1)+\Vert \varphi \Vert )+\log 2. \end{aligned}$$

Combining this with 3.3, we have

$$\begin{aligned} \begin{aligned}&H_\mu \left( \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \right) + n\int \varphi \mathrm {~d} \mu \\ =&H_\mu \left( \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \right) +\int S_{n,q} \varphi \mathrm {~d} \mu \\ \le&\log \left( C_n^{[n c(\varepsilon )]} \cdot (k+1)^{[n c(\varepsilon )]} \cdot \sum \limits _{0 \le i \le m(n,q)}e^{ S_{n,q}\varphi (x_i)+n\textrm{Var}(\varphi , \sqrt{2\varepsilon })+n\sqrt{2\varepsilon }\Vert \varphi \Vert }\right) \\ {}&+n(2\tau +\varepsilon )\left( \log (k+1)+||\varphi ||\right) +\log 2. \end{aligned} \end{aligned}$$

Hence,

$$\begin{aligned} \begin{aligned}&\dfrac{1}{n}H_\mu \left( \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \right) +\int \varphi \mathrm {~d} \mu \\ \le&\dfrac{1}{n}\log \left( \sum \limits _{0 \le i \le m(n,q)}e^{ S_{n,q}\varphi (x_i)}\right) + \dfrac{1}{n}\log C_n^{[n c(\varepsilon )]} +\dfrac{1}{n}\log (k+1)^{[n c(\varepsilon )]}\\&+ \textrm{Var}(\varphi , \sqrt{2\varepsilon }) +\sqrt{2\varepsilon }\Vert \varphi \Vert + (2\tau +\varepsilon )\left( \log (k+1)+||\varphi ||\right) +\frac{\log 2}{n}\\ \le&\dfrac{1}{n}\log \left( S_{{\overline{d}}_{n, q}}(\varphi ,\mu , \varepsilon )+\gamma \right) +\dfrac{1}{n}\log C_n^{[n c(\varepsilon )]} +\dfrac{1}{n}\log (k+1)^{[n c(\varepsilon )]}\\&+ \textrm{Var}(\varphi , \sqrt{2\varepsilon }) +\sqrt{2\varepsilon }\Vert \varphi \Vert + (2\tau +\varepsilon )\left( \log (k+1)+||\varphi ||\right) +\frac{\log 2}{n}. \quad \quad \text {(By}(3.2)) \end{aligned} \end{aligned}$$

Since q is arbitrary, we have

$$\begin{aligned} \begin{aligned}&\dfrac{1}{n}H_\mu \left( \xi \vee T^{-q} \xi \vee \cdots \vee T^{-q(n-1)} \xi \right) +\int \varphi \mathrm {~d} \mu \\ \le&\dfrac{1}{n}\log \left( {\overline{S}}^*_{n}(\varphi ,\mu , \varepsilon )+\gamma \right) +\dfrac{1}{n}\log C_n^{[n c(\varepsilon )]} +\dfrac{1}{n}\log (k+1)^{[n c(\varepsilon )]}\\&+ \textrm{Var}(\varphi , \sqrt{2\varepsilon }) +\sqrt{2\varepsilon }\Vert \varphi \Vert + (2\tau +\varepsilon )\left( \log (k+1)+||\varphi ||\right) +\frac{\log 2}{n}. \end{aligned} \end{aligned}$$

Combing with the Stirling’s formula which implies that

$$\begin{aligned} \Delta (\varepsilon ):=\lim _{n \rightarrow +\infty } \frac{1}{n} \log C_n^{[n c(\varepsilon )]}=-c(\varepsilon ) \log c(\varepsilon )-(1-c(\varepsilon )) \log (1-c(\varepsilon )), \end{aligned}$$

one has

$$\begin{aligned} h_{\mu }(T,\xi )+\int \varphi \mathrm {~d} \mu \le&\liminf \limits _{n\rightarrow \infty } \dfrac{1}{n}\log {\overline{S}}^*_{n}(\varphi ,\mu , \varepsilon )+\Delta (\varepsilon )+c(\varepsilon )\log (k+1) \\&+ \textrm{Var}(\varphi , \sqrt{2\varepsilon }) +\sqrt{2\varepsilon }\Vert \varphi \Vert + (2\tau +\varepsilon )\left( \log (k+1)+||\varphi ||\right) . \end{aligned}$$

By taking $\varepsilon \rightarrow 0$, we can have

$$\begin{aligned}{} & {} h_{\mu }(T,\xi )+\int \varphi \mathrm {~d} \mu \le a-2 \tau \log 2 \tau -(1-2 \tau ) \log (1-2 \tau )+4 \tau \log (k+1)\\{} & {} \quad +2\tau ||\varphi ||<a+\delta . \end{aligned}$$

Hence by (3.1), $h_\mu (T, \eta )+\int \varphi \mathrm {~d} \mu <a+2 \delta$. This finishes the proof of the first formula of Theorem 2.3.

Next we are going to prove the second formula

$$\begin{aligned} h_\mu (T)+\int \varphi \mathrm {~d} \mu =\lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\mu , \varepsilon )=\lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\mu , \varepsilon ). \end{aligned}$$

Fix $\varepsilon >0$ and $\mu \in M^e(X, T)$. Recall that for any $x, y\in X, {\overline{d}}_{n, q}(x, y) \le d_{n, q}(x, y)$ and then

$$\begin{aligned} B_{d_{n, q}}(x, \varepsilon ) \subset B_{{\overline{d}}_{n, q}}(x, \varepsilon ). \end{aligned}$$

Hence, $S_{{\overline{d}}_{n, q}}(\varphi ,\mu , \varepsilon ) \le S_{d_{n, q}}(\varphi ,\mu , \varepsilon ).$ Thus ${\overline{S}}_n^*(\varphi ,\mu , \varepsilon ) \le S_n^*(\varphi ,\mu , \varepsilon )$. Furthermore, we have

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\mu , \varepsilon ) \ge \lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\varphi ,\mu , \varepsilon )=h_\mu (T)+\int \varphi \mathrm {~d} \mu . \end{aligned}$$

Now we proceed to prove the inverse inequality: for any $\mu \in M^e(X, T)$,

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\mu , \varepsilon ) \le h_\mu (T)+\int \varphi \mathrm {~d} \mu . \end{aligned}$$

Fix $\varepsilon >0$ and $\mu \in M^e(X, T)$. Recall that $d_{n, 1}(x, y)=d_n(x, y)$. Hence by Theorem 2.1 we have

$$\begin{aligned} \begin{aligned} \lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\mu , \varepsilon )&\le \lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log S_{d_{n, 1}}(\varphi ,\mu , \varepsilon ) =h_\mu (T)+\int \varphi \mathrm {~d} \mu . \end{aligned} \end{aligned}$$

So we are done. $\square$

4 Proof of Theorem 2.4

Proof

Let (X, T) be a t.d.s. with a metric d. First we show that $\lim \limits _{\varepsilon \rightarrow 0} \limsup \limits _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\varphi ,\varepsilon ) \le P_{\textrm{top}}(T,\varphi )$. For anty $\tau >0.$ we can choose $\varepsilon$ small enough and n large enough, By the definition of Topological pressure, there exists a finite subset $F_n$ of X with $\bigcup _{x_i\in F_n} B_{d_{n,1}}\left( x_i, \varepsilon \right) =X$ such that

$$\begin{aligned} \sum \limits _{x_i\in F_{n}}e^{S_{n,1}\varphi (x_i)}\le e^{n(P_{\textrm{top}}(T,\varphi )+\tau )} , \end{aligned}$$

(4.1)

Fix the above $\varepsilon >0$. For any $x, y\in X$, it is clear that

$$\begin{aligned} {\overline{d}}_{n, 1}(x, y) \le d_{n,1}(x, y) \text{ and } B_{d_{n,1}}(x, \varepsilon ) \subset B_{{\overline{d}}_{n, 1}}(x, \varepsilon ). \end{aligned}$$

Recall that for $n \in {\mathbb {N}}$,

$$\begin{aligned} S_{{\overline{d}}_{n,q}}(\varphi , \varepsilon )=\inf \left\{ \sum \limits _{i=1}^k e^{S_{n,q}\varphi (x_i)} \text{ s.t. } \bigcup _{i=1}^k B_{d_{n,q}}\left( x_i, \varepsilon \right) =X\right\} ,\\ {\overline{S}}_{n}^*(\varphi , \varepsilon )=\inf _{q \ge 1} S_{{\overline{d}}_{n,q}}(\varphi , \varepsilon ). \end{aligned}$$

Since for any $x \in X$ there is $y \in F_n$ with ${\overline{d}}_{n, 1}(x, y) \le d_n(x, y)<\varepsilon$. Connecting with (4.1), one has

$$\begin{aligned} {\overline{S}}_n^*(\varphi , \varepsilon )\le S_{{\overline{d}}_{n, 1}}(\varphi , \varepsilon ) \le \sum \limits _{x_i\in F_{n}}e^{S_{n,1}\varphi (x_i)}\le e^{n(P_{\textrm{top}}(T,\varphi )+\tau )}. \end{aligned}$$

Hence by taking $\varepsilon \rightarrow 0$ we have

$$\begin{aligned} \begin{aligned} \lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\varphi , \varepsilon ) \le P_{\textrm{top}}(T,\varphi )+\tau . \end{aligned} \end{aligned}$$

Let $\tau \rightarrow 0$, we finish this prove.

Now we are going to prove that

$$\begin{aligned} P_{\textrm{top}}(T,\varphi ) \le \lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\varphi ,\varepsilon ). \end{aligned}$$

By the varitional principle [19] it is sufficient to show that for any $\mu \in M^e(X, T)$

$$\begin{aligned} h_\mu (T)+\int \varphi \mathrm {~d} \mu \le \lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\varphi ,\varepsilon ) \end{aligned}$$

It is easy to see that for $\varepsilon >0$ and $n \in {\mathbb {N}}$,

$$\begin{aligned} {\overline{S}}_n^*(\varphi ,\mu , \varepsilon ) \le {\overline{S}}_n^*(\varphi ,\varepsilon ). \end{aligned}$$

Then by Theorem 2.3,

$$\begin{aligned} h_\mu (T)+\int \varphi \mathrm {~d} \mu =\lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\varphi ,\mu , \varepsilon ) \le \lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log {\overline{S}}_n^*(\varphi ,\varepsilon ). \end{aligned}$$

This finishes the proof of the first formula of Theorem 2.4.

Finally we are going to prove the second formula:

$$\begin{aligned} P_{\textrm{top}}(T,\varphi )=\lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\varepsilon )=\lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\varepsilon ). \end{aligned}$$

To show $P_{\textrm{top}}(T,\varphi ) \le \lim \limits _{\varepsilon \rightarrow 0} \liminf \limits _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\varepsilon )$, by the variation principle it is sufficient to show that for any $\mu \in M^e(X, T)$

$$\begin{aligned} h_\mu (T)+\int \varphi \mathrm {~d} \mu \le \lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\varepsilon ). \end{aligned}$$

Note that for $\varepsilon >0$ and $n \in {\mathbb {N}}$, $S_n^*(\varphi ,\mu , \varepsilon ) \le S_n^*(\varphi ,\varepsilon ).$ Then by Theorem 2.3,

$$\begin{aligned} h_\mu (T)+\int \varphi \mathrm {~d} \mu =\lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\mu , \varepsilon ) \le \lim _{\varepsilon \rightarrow 0} \liminf _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\varepsilon ). \end{aligned}$$

Now we proceed to the proof of the inverse inequality:

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\varepsilon ) \le P_{\textrm{top}}(T,\varphi ). \end{aligned}$$

It is not hard to see that

$$\begin{aligned} \begin{aligned} \lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log S_n^*(\varphi ,\varepsilon )&\le \lim _{\varepsilon \rightarrow 0} \limsup _{n \rightarrow \infty } \frac{1}{n} \log S_{d_{n, 1}}(\varphi ,\varepsilon ) \le P_{\textrm{top}}(T,\varphi ). \end{aligned} \end{aligned}$$

This finishes the proof of Theorem 2.4. $\square$