Journal of Dynamics and Differential Equations

, Volume 26, Issue 3, pp 517–527

# Expansivity and Cone-fields in Metric Spaces

• Łukasz Struski
• Jacek Tabor
Open Access
Article

## Abstract

Due to the results of Lewowicz and Tolosa expansivity can be characterized with the aid of Lyapunov function. In this paper we study a similar problem for uniform expansivity and show that it can be described using generalized cone-fields on metric spaces. We say that a function $$f:X\rightarrow X$$ is uniformly expansive on a set $$\varLambda \subset X$$ if there exist $$\varepsilon >0$$ and $$\alpha \in (0,1)$$ such that for any two orbits $$\hbox {x}:\{-N,\ldots ,N\} \rightarrow \varLambda$$, $$\hbox {v}:\{-N,\ldots ,N\} \rightarrow X$$ of $$f$$ we have
\begin{aligned} \sup _{-N\le n\le N}d(\hbox {x}_n,\hbox {v}_n) \le \varepsilon \implies d(\hbox {x}_0,\hbox {v}_0) \le \alpha \sup _{-N\le n\le N}d(\hbox {x}_n,\hbox {v}_n). \end{aligned}
It occurs that a function is uniformly expansive iff there exists a generalized cone-field on $$X$$ such that $$f$$ is cone-hyperbolic.

## Keywords

Cone-field Hyperbolicity Expansive map Lyapunov function

37D20

## 1 Introduction

In 1892 Lyapunov [9] introduced the idea of Lyapunov functions to study stability of equilibria of differential equations. The Lyapunov approach allows to assess the stability of equilibrium points of a system without solving the differential equations that describe the system. This theory is widely used in qualitative theory of dynamical systems.

In Lewowicz [7, 8]proposed to use Lyapunov functions of two variables to study structural stability and similar concepts, such as topological stability and persistence. The method has been applied in particular to study hyperbolic diffeomorphisms on manifolds. For the survey of the results, methods and possible generalizations see [12].

Let us quote one of the most interesting results from [12]. Let $$f:M \rightarrow M$$ be a homeomorphism of a compact manifold $$M$$. For $$U:M \times M\rightarrow \mathbb {R}$$ we define
\begin{aligned} \varDelta _f U(x, y) := U(f(x), f(y))-U(x,y)\; \text { for }\; x,y\in M. \end{aligned}
We say that $$U$$ is a Lyapunov function for $$f$$ if it is continuous, vanishes on the diagonal, and $$\varDelta _f U(x, y)$$ is positive for $$(x, y)$$ on a neighborhood of the diagonal, $$x \ne y$$.

The following result characterizes expansive homeomorphisms in terms of Lyapunov functions.

### Theorem

[12, Theorem 3.2]. Let $$f$$ be a homeomorphism of a compact manifold $$M$$. The following conditions are equivalent:
1. i)

$$f$$ is expansive;

2. ii)

there exists a Lyapunov function for $$f$$.

The proof of this result for diffeomorphisms $$f$$ can be found in [7]; see Sect. 4 and Lemma 3.3 of that paper. Additional arguments required for the case of a homeomorphism are discussed in [6], Sect. 1]. See also [12], where Tolosa, basing on the results of Lewowicz, characterized the expansivity on metric spaces with the using Lyapunov functions.

In this paper we use a generalized notion of cone-fields on metric space to describe uniform expansivity. The notions of cone-fields and cone condition [4, 10] originally appeared in the late 60’s in the works of Alekseev, Anosov, Moser and Sinai. Recently, Sheldon Newhouse [10] obtained new conditions for dominated and hyperbolic splittings on compact invariant sets with the use of cone-fields. It is also worth mentioning that the notion of cone-field can be very useful in the study of hyperbolicity [1, 3, 4, 10].

Let us briefly describe the contents of this paper. In Sect. 2 we discuss the notion of uniform expansivity. We show that if $$f$$ is uniformly expansive then it is also expansive. In Sect. 3 we recall our generalization of cone-fields to metric spaces which we presented in paper [11] and show that the existence of hyperbolic cone fields guarantees uniform expansivity. In Sect. 4 we show how to construct functions $$c_s$$, $$c_u$$ for a uniformly expansive $$f$$ such that $$f$$ is cone-hyperbolic with respect to the cone-field $$(c_s,c_u)$$. The main result of the section can be summarized as follows:

### Main Result

[see Theorem 3]. Let $$X$$ be a metric space and let $$f:X{\rightharpoonup }X$$ be an L-bilipschitz map. Assume that $$\varLambda \subset X$$ is an invariant set for $$f$$ such that $$f$$ is uniformly expansive on $$\varLambda$$.

Then there exists a cone-field on $$\varLambda$$ such that $$f$$ is cone-hyperbolic on $$\varLambda$$.

## 2 Uniform Expansivity

First we define uniform expansivity of $$f$$ and show that this notion is stronger than the classical expansivity.

By a partial map from $$X$$ to $$Y$$ (written as $$f:X{\rightharpoonup }Y$$) we denote a function which domain is subset of $$X$$ [2, Chapter 2]. By $${\hbox {dom}}(f)$$ we denote the domain of a partial map $$f:X{\rightharpoonup }Y$$, and by $${\hbox {im}}(f)$$ we denote its inverse image. For a given $$f:X{\rightharpoonup }X$$ we say that a sequence $$\hbox {x}:I \rightarrow X$$ defined on a subinterval1 $$I$$ of $$\mathbb {Z}$$ is an orbit of  $$f$$ if
\begin{aligned} \hbox {x}_n\in {\hbox {dom}}(f)\ \text { and }\ \hbox {x}_{n+1}=f(\hbox {x}_n)\ \text { for }\ n\in I \text { such that } \;n+1\in I. \end{aligned}
We recall the classical definition of expansivity. We say that $$f:X{\rightharpoonup }X$$ is expansive on $$\varLambda \subset X$$ if there exists an $$\varepsilon >0$$ such that for any two orbits $$\hbox {x}:\mathbb {Z}\rightarrow \varLambda$$, $$\hbox {v}:\mathbb {Z}\rightarrow X$$ if $$\sup \limits _{n\in \mathbb {Z}}d(\hbox {x}_n,\hbox {v}_n) \le \varepsilon$$ then $$\hbox {x}=\hbox {v}$$.

### Definition 1

Let $$N \in \mathbb {N}$$, $$\varepsilon >0$$ and $$\alpha \in (0,1)$$ be given. We say that $$f:X{\rightharpoonup }X$$ is $$(N,\varepsilon ,\alpha )$$ -uniformly expansive on a set  $$\varLambda \subset X$$ if for any two orbits $$\hbox {x}:\{-N,\ldots ,N\} \rightarrow \varLambda$$, $$\hbox {v}:\{-N,\ldots ,N\} \rightarrow X$$ we have
\begin{aligned} d_{\sup }(\hbox {x},\hbox {v}) \le \varepsilon \implies d(\hbox {x}_0,\hbox {v}_0) \le \alpha d_{\sup }(\hbox {x},\hbox {v}), \end{aligned}
where
\begin{aligned} d_{\sup }(\hbox {x},\hbox {v}):=\sup _{-N\le n\le N}d(\hbox {x}_n,\hbox {v}_n). \end{aligned}

This notion is more useful because it does not need an infinite trajectory.

### Example 1

Consider a rotation of $$f:S^1\rightarrow S^1$$ by an angle $$\alpha$$. Then $$f$$ is an isometry, and therefore is not expansive, and consequently not $$(N,\varepsilon ,\alpha )$$-uniformly expansive on $$\varLambda =S^1$$.

### Example 2

Let us consider the function $$f:\mathbb {R}_+\ni x\mapsto x+\sqrt{x}\in \mathbb {R}_+$$. One can easily check that this function is expansive because its derivative at each point is strongly greater than $$1$$. On the other hand, $$f$$ is not uniformly expansive because for sufficiently large $$x$$ the derivative of the function at $$x$$ can become as close to $$1$$ as we want.

One can easily verify that uniform expansivity implies classical expansivity (this result can also be easily deduced from Theorem 1 below).

### Observation 1

[11, Observation 4.1] Let $$N \in \mathbb {N}$$, $$\varepsilon >0$$, $$\alpha \in (0,1)$$, $$\varLambda \subset X$$ and $$f:X{\rightharpoonup }X$$ be given. If $$f$$ is $$(N,\varepsilon ,\alpha )$$-uniformly expansive on $$\varLambda$$, then it is also expansive on $$\varLambda$$.

Given $$L\ge 1$$ and $$f:X{\rightharpoonup }Y$$ we call $$f$$ $$L$$ -bilipschitz if
\begin{aligned} L^{-1}d(x,y)\le d(f(x),f(y))\le Ld(x,y) \; \text { for }\; x,y\in {\hbox {dom}}{(f)}. \end{aligned}
(2.1)
Note that if a function $$f$$ is $$L$$-bilipschitz then it is injective.
For $$\delta >0$$ and a set $$A\subset X$$ we define the $$\delta$$ -neighbourhood of  $$A$$ as
\begin{aligned} A_{\delta }:=\bigcup _{x\in A}B(x,\delta ). \end{aligned}
Let an injective map $$f:X{\rightharpoonup }X$$ be given. We call $$A\subset {\hbox {dom}}(f)$$ an invariant set for  $$f$$ if $$f(x) \text { and } f^{-1}(x)\in A \text { for every } x\in A.$$

Now we show how to change the metric so that the function $$f$$ which is $$(N,\cdot ,\cdot )$$-uniformly expansive becomes $$(1,\cdot ,\cdot )$$-uniformly expansive.

### Theorem 1

Let $$f:X {\rightharpoonup }X$$ be an L-bilipschitz map for some $$L>1$$ and $$\alpha \in (0,1)$$. Let $$\varLambda \subset X$$ and $$\delta >0$$ be such that $$\varLambda _{\delta }\subset {\hbox {dom}}(f)\cap {\hbox {im}}(f)$$. We assume that $$\varLambda$$ is an invariant set for $$f$$ and that $$f$$ is $$(N,\delta ,\alpha )$$-uniformly expansive on $$\varLambda$$.

Then there exists a metric $$\rho$$ on $$\varLambda _{\delta L^{-N+1}}$$ such that
\begin{aligned} d(x,v) \le \rho (x,v) \le L^{N-1} d(x,v) \; \text { for } \; x,v\in \varLambda _{\delta L^{-N+1}}, \end{aligned}
(2.2)
that $$f$$ is $$(1,\delta L^{-N+1},\root N \of {\alpha })$$-uniformly expansive on $$\varLambda _{\delta L^{-N+1}}$$ and $$\max \{\alpha ^{-1/N}, L\}$$-bilipschitz map with respect to the metric $$\rho$$.

### Proof

Let $$\beta =\root N \of {\alpha }$$. We put
\begin{aligned} \rho (x,v):=\max _{k \in \{-N+1,\ldots ,N-1\}} \beta ^{|k|}d(f^k(x),f^k(v)) \; \text { for }\; x,v\in \varLambda _{\delta L^{-N+1}}. \end{aligned}
Inequalities (2.2) follow from the definition and (2.1). Note that for $$k\in \{-N+1, \ldots , N-1\}$$ we have
\begin{aligned} x,v\in \varLambda _{\delta L^{-N+1}} \Longrightarrow f^k(x), f^k(v)\in \varLambda _{\delta L^{-N+1+|k|}}. \end{aligned}
This means that $$\rho$$ is well defined on $$\varLambda _{\delta L^{-N+1}}$$.
First we show that $$f$$ is $$\max \{\beta ^{-1}, L\}$$-bilipschitz map with respect to the metric $$\rho$$. Since $$f$$ is $$L$$-bilipschitz in the metric $$d$$, we know that $$d(f^N(x),f^N(v))\le L d(f^{N-1}(x),f^{N-1}(v))$$ and finally we get
\begin{aligned} \begin{aligned} \rho (f(x),f(v))&=\max _{k \in \{-N+1,\ldots ,N-1\}} \beta ^{|k|}d(f^k(f(x)),f^k(f(v))) \\&=\max \{\beta ^{|-N+1|} d(f^{-N+2}(x),f^{-N+2}(v)), \ldots , \beta ^{N-1} d(f^N(x),f^N(v)) \}\\&=\max \{\beta ^{|-N+1|}\beta ^{-1}\beta d(f^{-N+2}(x),f^{-N+2}(v)), \ldots , \beta ^1\beta ^{-1} \beta d(x,v), \\& \beta ^0 \beta \beta ^{-1} d(f(x),f(v)), \ldots , \beta ^{N-2}\beta \beta ^{-1} d(f^{N-1}(x),f^{N-1}(v)),\\& \beta ^{N-1} d(f^N(x),f^N(v)) \}\\&=\max \{\beta \beta ^{|-N+2|}d(f^{-N+2}(x),f^{-N+2}(v)), \ldots , \beta \beta ^0 d(x,v), \\& \beta ^{-1}\beta ^1 d(f(x),f(v)), \ldots , \beta ^{-1}\beta ^{N-1} d(f^{N-1}(x),f^{N-1}(v)),\\& \beta ^{N-1} d(f^N(x),f^N(v)) \}\\&\le \max \{\beta \beta ^{|-N+2|}d(f^{-N+2}(x),f^{-N+2}(v)), \ldots , \beta \beta ^0 d(x,v), \\& \beta ^{-1}\beta ^1 d(f(x),f(v)), \ldots , \beta ^{-1}\beta ^{N-1} d(f^{N-1}(x),f^{N-1}(v)),\\& \beta ^{N-1} L d(f^{N-1}(x),f^{N-1}(v)) \}\\&\le \max \{\beta , \beta ^{-1}, L\}\cdot \rho (x,v)=\max \{\beta ^{-1}, L\}\cdot \rho (x,v). \end{aligned} \end{aligned}
Similarly, as for the opposite inequality, we know that $$L^{-1} d(f^{N-1}(x),f^{N-1}(v))\le d(f^N(x),f^N(v))$$ and $$L^{-1} d(f^{-N+1}(x),f^{-N+1}(v))\le d(f^{-N+2}(x),f^{-N+2}(v))$$. Hence
\begin{aligned} \begin{aligned} \rho (f(x),f(v))&=\max \{\beta \beta ^{|-N+2|}d(f^{-N+2}(x),f^{-N+2}(v)), \ldots , \beta \beta ^0 d(x,v), \\& \beta ^{-1}\beta ^1 d(f(x),f(v)), \ldots , \beta ^{-1}\beta ^{N-1} d(f^{N-1}(x),f^{N-1}(v)),\\& \beta ^{N-1} d(f^N(x),f^N(v)) \}\\&\ge \max \{\beta \beta ^{|-N+2|}d(f^{-N+2}(x),f^{-N+2}(v)), \ldots , \beta \beta ^0 d(x,v), \\& \beta ^{-1}\beta ^1 d(f(x),f(v)), \ldots , \beta ^{-1}\beta ^{N-1} d(f^{N-1}(x),f^{N-1}(v)),\\& \beta ^{N-1} L^{-1} d(f^{N-1}(x),f^{N-1}(v)) \}\\&\ge \min \{\beta , L^{-1}\}\cdot \rho (x,v). \end{aligned} \end{aligned}
Now we show that for $$x\in \varLambda$$ and $$v\in \varLambda _{\delta L^{-N+1}}$$ such that
\begin{aligned} \max \left\{ \rho (f^{-1}(x),f^{-1}(v)),\rho (x,v),\rho (f(x),f(v))\right\} \le \delta L^{-N+1} \end{aligned}
(2.3)
the following inequality holds:
\begin{aligned} \rho (x,v) \le \beta \max (\rho (f(x),f(v)),\rho (f^{-1}(x),f^{-1}(v))). \end{aligned}
We have to show that for $$k=-N+1,\ldots ,N-1$$
\begin{aligned}&\!\!\!\!\!\! \beta ^{|k|}d(f^k(x),f^k(v)) \le \beta \max \\&\!\!\! \times \left( \max _{k=-N+1,\ldots ,N-1}\beta ^{|k|}d(f^{k+1}(x),f^{k+1}(v)), \max _{k=-N+1,\ldots ,N-1}\beta ^{|k|}d(f^{k-1}(x),f^{k-1}(v))\!\right) . \end{aligned}
For $$k<0$$ or $$k>0$$ it is straightforward. Consider the case $$k=0$$. From (2.2) and (2.3) we get
\begin{aligned} \max \left\{ d(f^{-1}(x),f^{-1}(v)),d(x,v),d(f(x),f(v))\right\} \le \delta L^{-N+1}, \end{aligned}
which together with (2.1) implies that $$d(f^k(x),f^k(v))\le \delta$$ for $$k=-N,\dots ,N$$. By the uniform expansivity and the fact that $$\beta <1$$ we get
\begin{aligned}&d(x,v) \le \alpha \max _{|k| \le N}d(f^k(x),f^k(v)) \le \beta \max _{|k| \le N}(\beta ^{N-1}d(f^k(x),f^k(v)))\\&\le \beta \max \left( \max _{|k| \le N-1}\beta ^{|k|}d(f^{k+1}(x),f^{k+1}(v)), \max _{|k| \le N-1}\beta ^{|k|}d(f^{k-1}(x),f^{k-1}(v)) \right) . \end{aligned}

## 3 Cone-fields and Cone-hyperbolic Maps

In this section, for the convenience of the reader, we recall basic definitions concerning generalization of cone-fields to metric spaces (for more information and motivation see [5, 11]).

### Definition 2

[11, Definition 3.1] Let $$\delta >0$$ and $$\varLambda \subset X$$ be nonempty. We say that a pair of functions $$c_s,c_u:U \rightarrow \mathbb {R}_+$$ for some $$U\subset X\times X$$ forms a $$\delta$$ -cone-field on  $$\varLambda$$ if
\begin{aligned} \{x\}\times B(x,\delta )\subset U \; \text { for } x\in \varLambda . \end{aligned}
We put $$c(x,v):=\max \{c_s(x,v),c_u(x,v)\}$$. If there exists $$K>0$$ such that:
\begin{aligned}&\dfrac{1}{K}d(x,v) \le c(x,v) \le Kd(x,v) \text { for } (x,v) \in U \end{aligned}
then we call it $$(K,\delta )$$ -cone-field on  $$\varLambda$$ or uniform  $$\delta$$ -cone-field on  $$\varLambda$$.
For each point $$x\in \varLambda$$ we introduce unstable and stable cones by the formula
\begin{aligned}&C^u_x(\delta ):=\{v\in B(x,\delta ) \, : \, c_s(x,v)\le c_u(x,v)\}, \\&C^s_x(\delta ):=\{v\in B(x,\delta ) \, : \, c_s(x,v)\ge c_u(x,v)\}. \end{aligned}
We consider a partial map $$f:X \rightharpoonup Y$$ between metric spaces $$X$$ and $$Y$$ and $$\varLambda \subset {\hbox {dom}}(f)$$. Assume that $$X$$ is equipped with a uniform $$\delta$$-cone-field on $$\varLambda$$ and $$Y$$ is equipped with a uniform $$\delta$$-cone-field on a subset $$Z$$ of $$Y$$ such that $$f(\varLambda )\subset Z$$.
For every $$x \in {\hbox {dom}}(f)$$ we put
\begin{aligned} B_f(x,\delta ):=\{v \in B(x,\delta ) \cap {\hbox {dom}}(f): f(v) \in B(f(x),\delta )\}. \end{aligned}
Now we define $$u_x(f;\delta )$$ and $$s_x(f;\delta )$$, the expansion and the contraction rates of $$f$$, respectively. These rates are a modification of the classical definition from [10], but we do not assume that the function $$f$$ is invertible (for more information see [11]).

### Definition 3

[11, Definition 3.2] Let $$x \in {\hbox {dom}}(f)$$ and $$\delta >0$$ be given. We define
\begin{aligned} u_x(f;\delta )&:= \sup \{R \in [0,\infty ] \; | \; c(f(x),f(v)) \ge R c(x,v), v \in B_f(x,\delta ); v \in C^u_x(\delta )\},\\ s_x(f;\delta )&:= \inf \, \{R \in [0,\infty ] \; | \; c(f(x),f(v)) \le R c(x,v), v \in B_f(x,\delta ); f(v) \in C^s_{f(x)}(\delta )\}. \end{aligned}
Let $$u_\varLambda (f;\delta ) := \inf \limits _{x\in \varLambda }\{u_x(f;\delta )\}$$   and   $$s_\varLambda (f;\delta ) := \sup \limits _{x\in \varLambda }\{s_x(f;\delta )\}$$.

### Definition 4

We say that $$f$$ is $$\delta$$ -cone-hyperbolic on $$\varLambda$$ if
\begin{aligned} s_\varLambda (f;\delta )<1<u_\varLambda (f;\delta ). \end{aligned}

The next proposition is a simple analogue of [10, Lemma 1.1].

### Proposition 1

[11, Proposition 3.1] Every $$\delta$$-cone-hyperbolic is $$\delta$$-cone-invariant, i.e. for $$x\in \varLambda$$ and $$v\in B_f(x,\delta )$$ we have
\begin{aligned} v\in C^u_x(\delta ) \; \Longrightarrow \; f(v)\in C^u_{f(x)}(\delta ), \end{aligned}
and
\begin{aligned} f(v)\in C^s_{f(x)}(\delta ) \; \Longrightarrow \; v\in C^s_x(\delta ). \end{aligned}

### Theorem 2

[11, Theorem 4.1] Suppose that for $$K>0$$ and $$\delta >0$$ we are given a $$(K,\delta )$$-cone-field on $$\varLambda \subset X$$. Let $$f:\varLambda _{\delta } {\rightharpoonup }X$$ be $$\delta$$-cone-hyperbolic on $$\varLambda$$ and let $$\lambda >1$$ be chosen such that
\begin{aligned} s_\varLambda (f;\delta ) \le \lambda ^{-1}, \, u_\varLambda (f;\delta ) \ge \lambda . \end{aligned}
Then $$f$$ is $$(N,\delta ,K^2/\lambda ^N)$$-uniformly expansive on $$\varLambda$$ for every $$N \in \mathbb {N}$$, $$N > 2\log _{\lambda }K$$.

### Example 3

Let $$f:T^2\rightarrow T^2$$ be defined by $$f(x,y)=(2x+y,x+y),$$ where $$T^2=\mathbb {R}^2/\mathbb {Z}^2$$.

We know that $$f$$ is expansive (see [4, Sect. 1.8]). It is easy to show that
\begin{aligned} s_{T^2}(f;\delta ) \le \frac{3-\sqrt{5}}{2}<1, \quad u_{T^2}(f;\delta ) \ge \frac{3+\sqrt{5}}{2}>1. \end{aligned}
From Theorem 2 we conclude that $$f$$ is uniformly expansive on $$\varLambda =T^2$$.

## 4 Expansivity and Cone-fields

In this section we show that uniform expansiveness of $$f$$ on an invariant set $$\varLambda$$ lets us construct a cone-field on $$\varLambda$$ such that $$f$$ is cone-hyperbolic on $$\varLambda$$. In our reasoning we will need the notion of $$\varepsilon$$-quasiconvexity.

### Definition 5

Let $$I$$ be a subinterval of $$\mathbb {Z}$$, and let $$\varepsilon \ge 0$$ be fixed. We call a sequence $$\alpha :I\rightarrow \mathbb {R}\, \varepsilon$$ -quasiconvex if
\begin{aligned} \alpha _n\le \max \{\alpha _{n-1},\alpha _{n+1}\}-\varepsilon \; \text { for }\; n\in I:\; n-1,n+1\in I. \end{aligned}

Now we show some properties of $$\varepsilon$$-quasiconvex sequences, which will be used later.

### Observation 2

Let $$\varepsilon \ge 0$$ and $$\alpha :I\rightarrow \mathbb {R}$$ be an $$\varepsilon$$-quasiconvex sequence.

Then
1. i)
if $$m$$, $$m+2\in I$$ and $$\alpha _{m+1} > \alpha _m -\varepsilon$$ then
\begin{aligned} \alpha _{n+1} \ge \alpha _n + \varepsilon \text { for } n\ge m+1 \text { such that } n,n+1\in I. \end{aligned}
(4.1)

2. ii)
if $$m-1$$, $$m+1\in I$$ and $$\alpha _{m+1} < \alpha _m +\varepsilon$$ then
\begin{aligned} \alpha _{n+1} \le \alpha _n - \varepsilon \text { for } n< m \text { such that } n,n+1\in I. \end{aligned}
(4.2)

### Proof

The above statements are similar so we show the first one. The proof proceeds on induction. Suppose that $$m, m+2\in I$$ and $$\alpha _{m+1} > \alpha _m -\varepsilon$$. Since $$\alpha$$ is $$\varepsilon$$-quasiconvex,
\begin{aligned} \alpha _{m+1}\le \max \{\alpha _m, \alpha _{m+2}\} - \varepsilon =\max \{\alpha _m-\varepsilon , \alpha _{m+2} - \varepsilon \}. \end{aligned}
But $$\alpha _{m+1} > \alpha _m -\varepsilon$$, so we get
\begin{aligned} \alpha _{m+1}\le \alpha _{m+2} - \varepsilon , \end{aligned}
and hence
\begin{aligned} \alpha _{m+2}\ge \alpha _{m+1}+\varepsilon . \end{aligned}
It implies that (4.1) is valid for $$n=m+1$$. Suppose now that (4.1) holds for some $$n \ge m + 1$$, i.e. that $$n; n+1\in I$$ and $$\alpha _{n+1}\ge \alpha _n + \varepsilon$$. Assume additionally that $$n+2\in I$$. Then we get
\begin{aligned} \alpha _{n+1}\le \alpha _{n+2} - \varepsilon , \end{aligned}
thus
\begin{aligned} \alpha _{n+2}\ge \alpha _{n+1}+\varepsilon , \end{aligned}
which completes the proof.

The following proposition will be a basic tool in the proof of our main result, Theorem 3.

### Proposition 2

Let $$\varepsilon >0, L>1$$, $$\beta \in (0,1)$$ and let $$(Y, \rho )$$ be a metric space. Let $$\varLambda \subset Y$$ be given and $$f:Y{\rightharpoonup }Y$$ be an L-bilipschitz map such that $$\varLambda _\varepsilon \subset {\hbox {dom}}(f) \cap {\hbox {im}}(f)$$. Assume that $$\varLambda$$ is an invariant set for $$f$$ and that $$f$$ is $$(1,\varepsilon ,\beta )$$-uniformly expansive on $$\varLambda$$.

Then
\begin{aligned} \begin{array}{l} c_{s}(x,v) := \inf \{ \rho (f^k(x),f^k(v)) \, | \, k\in (-\infty ,0)\cap \mathbb {Z}: f^l(v) \in B(f^l(x),\varepsilon )\\ \quad \quad \text { for } l\in [k,0]\cap \mathbb {Z}\}, \\ c_{u}(x,v) := \inf \{ \rho (f^k(x),f^k(v)) \, | \, k\in [0,\infty )\cap \mathbb {Z}: f^l(v) \in B(f^l(x),\varepsilon )\\ \quad \quad \text { for } l\in [0,k]\cap \mathbb {Z}\}, \end{array} \end{aligned}
(4.3)
define an $$(L,\varepsilon /L)$$ cone-field on $$\varLambda$$. Moreover, $$f$$ is cone-hyperbolic on $$\varLambda$$ and
\begin{aligned} s_\varLambda (f;\varepsilon /L)\le \beta <\frac{1}{\beta }\le u_\varLambda (f;\varepsilon /L). \end{aligned}
(4.4)

### Proof

First we show that $$c_{s}(x,v)$$ and $$c_{u}(x,v)$$ defined above are $$(L,\varepsilon /L)$$ cone-field on $$\varLambda$$, i.e.
\begin{aligned} \dfrac{1}{L}\rho (x,v) \le c(x,v) \le L\rho (x,v) \, \text { for }\, (x,v) \in \big \{(x,v) \, :\, x\in \varLambda , v\in B(x,\varepsilon /L)\big \}, \end{aligned}
where $$c(x,v):=\max \{c_s(x,v),c_u(x,v)\}$$.

Choose an arbitrary point $$x\in \varLambda$$ and $$v\in B(x,\varepsilon /L)$$. We can assume that $$x\ne v$$, because the case $$x=v$$ is trivial ($$c_{s}(x,v) = c_{u}(x,v) = 0=\rho (x,v)$$).

Let $$I$$ be the biggest subinterval of $$\mathbb {Z}$$ containing 0 such that
\begin{aligned} \sup \{\rho (f^n(x),f^n(v))\, :\, n\in I\}\le \varepsilon . \end{aligned}
(4.5)
Since $$f$$ is $$L$$-bilipschitz, we know that $$f^{-1}(v) \in B(f^{-1}(x),\varepsilon )$$, and therefore $$\{-1,0\}\subset I$$. This yields $$c(x,v) < \infty$$.
Now we define a sequence $$\{a_n\}_{n\in I}\subset \mathbb {R}$$ by the formula
\begin{aligned} a_n:=\ln \rho (f^n(x),f^n(v)) \, \text { for } n\in I. \end{aligned}
(4.6)
Observe that $$a_n$$ is well-defined because $$\rho (f^n(x),f^n(v))>0$$ for all $$n\in I$$.
Let
\begin{aligned} I_-:=\{n\in I\, :\, n<0\} \; \text { and } \; I_+:=\{n\in I\, :\, n\ge 0\}. \end{aligned}
We have the following relations:
\begin{aligned} c_s(x,v)=\exp \left( \inf _{n\in I_-}\{a_n\}\right) \; \text { and }\; c_u(x,v)=\exp \left( \inf _{n\in I_+}\{a_n\}\right) , \end{aligned}
where we use the convention $$\exp (-\infty )=0$$.
We show that the sequence $$\{a_n\}$$ is $$\ln (1/\beta )$$-quasiconvex. Let $$n\in I$$ be such that $$n-1$$, $$n+1\in I$$. By (4.5) we observe that
\begin{aligned} \max \{\rho (f^{n-1}(x),f^{n-1}(v)),\rho (f^n(x),f^n(v)),\rho (f^{n+1}(x),f^{n+1}(v))\}\le \varepsilon . \end{aligned}
Consequently, by $$(1,\varepsilon ,\beta )$$-uniform expansivity of $$f$$ we get
\begin{aligned} \rho (f^n(x),f^n(v))\le \beta \max \{\rho (f^{n-1}(x),f^{n-1}(v)),\rho (f^{n+1}(x),f^{n+1}(v))\}, \end{aligned}
which implies that $$a_n\le \max \{a_{n-1},a_{n+1}\} - \ln (1/\beta )$$.
Now we consider two cases. If $$a_{-1} \le a_0$$ then by Observation 2 i) we get
\begin{aligned} a_{n+1}\ge a_n + \ln \frac{1}{\beta } \; \text { for }\; n\ge 0, n\in I, \end{aligned}
which yields
\begin{aligned} \inf _{n\in I_-}\{a_n\}\le a_{-1}\le a_0 = \inf _{n\in I_+}\{a_n\}, \end{aligned}
Hence
\begin{aligned} c_s(x,v)\le c_u(x,v) = c(x,v) = e^{a_0}=\rho (x,v). \end{aligned}
On the other hand if $$a_{-1} \ge a_0$$ then by Observation 2 ii) we get
\begin{aligned} a_{n+1}\le a_n - \ln \frac{1}{\beta } \; \text { for }\; n < -1, n\in I. \end{aligned}
Therefore
\begin{aligned} \inf _{n\in I_-}\{a_n\}=a_{-1}\ge a_0 \ge \inf _{n\in I_+}\{a_n\}, \end{aligned}
and consequently
\begin{aligned} c_u(x,v)\le c_s(x,v) = c(x,v)=e^{a_{-1}}=\rho (f^{-1}(x),f^{-1}(v)). \end{aligned}
Since $$f$$ is $$L$$-bilipschitz, we get that $$c_s,c_u$$ define an $$(L,\varepsilon /L)$$ cone-field on $$\varLambda$$.

Now we check that $$f$$ is cone-hyperbolic on $$\varLambda$$. Let us take $$x\in \varLambda$$ and $$v\in B_f(x,\varepsilon /L)$$ such that $$f(v) \in C^s_{f(x)}(\varepsilon /L)$$. We define the sequence $$\{a_n\}_{n\in I}$$ as in (4.6).

We show that $$a_0\ge a_1$$. Suppose that, on the contrary, $$a_0<a_1$$. By Observation 2 i) we get
\begin{aligned} a_{n+1}\ge a_n \; \text { for }\; n\ge 1, n\in I. \end{aligned}
Hence
\begin{aligned} \ln (c_u(f(x),f(v)))=\inf _{n\ge 1, n\in I}\{a_n\}= a_1 > a_0 \ge \inf _{n<1,n\in I}\{a_n\}=\ln (c_s(f(x),f(v))), \end{aligned}
which is a contradiction with $$f(v)\in C^s_{f(x)}(\varepsilon /L)$$. So we have $$a_1\le a_0$$. By the Observation 2 ii) we get
\begin{aligned} a_{n+1}\le a_{n}-\ln (1/\beta ) \text { for } n<0 \text { such that } n,n+1\in I. \end{aligned}
In particular,
\begin{aligned} a_0\le a_{-1}-\ln (1/\beta ). \end{aligned}
(4.7)
Consequently,
\begin{aligned}&c_u(f(x),f(v))=\exp \left( \inf _{n\ge 1, n\in I}\{a_n\}\right) \le \exp (a_1) \le \exp (a_0)\\&= \exp \left( \inf _{n<1,n\in I}\{a_n\}\right) = c_s(f(x),f(v)) = \underline{c(f(x),f(v))}\\&\mathop {\le }\limits ^{(4.7)} \beta \exp (a_{-1}) = \beta \exp \left( \inf _{n\in I_-}\{a_n\}\right) \le \underline{\beta c(x,v)}. \end{aligned}
Therefore
\begin{aligned} s_\varLambda (f;\varepsilon /L) = \sup \limits _{x\in \varLambda }\{s_x(f;\varepsilon /L)\}\le \beta < 1. \end{aligned}
Now we consider an $$x\in \varLambda$$ and $$v\in B_f(x,\varepsilon /L)$$ such that $$v\in C^u_x(\varepsilon /L)$$. We show that $$a_0\ge a_{-1}$$. Suppose the contrary, $$a_0<a_{-1}$$. By Observation 2 ii) we get
\begin{aligned} a_{n+1}\ge a_n \; \text { for }\; n<-1, n\in I. \end{aligned}
Hence
\begin{aligned} \inf _{n\in I_-}\{a_n\}= a_{-1} > a_0 \ge \inf _{n\in I_+}\{a_n\}, \end{aligned}
which is contradiction with $$v\in C^u_x(\varepsilon /L)$$. So we have $$a_0 \ge a_{-1}$$. By the Observation 2 i) we get
\begin{aligned} a_{n+1}\ge a_{n}+\ln (1/\beta ) \text { for } n\ge 0 \text { such that } n,n+1\in I. \end{aligned}
In particular,
\begin{aligned} a_1\ge a_0+\ln (1/\beta ). \end{aligned}
(4.8)
Finally
\begin{aligned}&c_s(f(x),f(v))=\exp \left( \inf _{n< 1, n\in I}\{a_n\}\right) \le \exp (a_0) \le \exp (a_1)\\&= \exp \left( \inf _{n\ge 1,n\in I}\{a_n\}\right) = c_u(f(x),f(v)) = c(f(x),f(v)), \end{aligned}
which yields
\begin{aligned} \exp (a_1) = \underline{c(f(x),f(v))} \mathop {\ge }\limits ^{4.8}\frac{1}{\beta } \exp (a_0) = \frac{1}{\beta } \exp \left( \inf _{n\in I_+}\{a_n\}\right) = \underline{\frac{1}{\beta } c(x,v)}. \end{aligned}
This shows that
\begin{aligned} u_\varLambda (f;\varepsilon /L) = \inf _{x\in \varLambda }\{u_x(f;\varepsilon /L)\}\ge \frac{1}{\beta }> 1. \end{aligned}
Therefore $$f$$ is cone-hyperbolic on $$\varLambda$$.

As a consequence of earlier results we obtain the following theorem.

### Theorem 3

Let $$\varepsilon >0, L>1, N\in \mathbb {N}$$, $$\alpha \in (0,1)$$ be fixed. Let $$(X,d)$$ be a metric space and $$\varLambda \subset X$$ be given. Let $$f:X{\rightharpoonup }X$$ be an L-bilipschitz map such that $$\varLambda _\varepsilon \subset {\hbox {dom}}(f)\cap {\hbox {im}}(f)$$. Assume that $$\varLambda$$ is an invariant set for $$f$$ and that $$f$$ is $$(N,\varepsilon ,\alpha )$$-uniformly expansive on $$\varLambda$$.

Then there exists an $$(\max \{\alpha ^{-1/N}L^{N-1}, L^N\},\min \{\varepsilon L^{-2N+1}\root N \of {\alpha }, \varepsilon L^{-2N} \})$$ cone-field on $$\varLambda$$ such that $$f$$ is cone-hyperbolic on $$\varLambda$$ and
\begin{aligned} s_\varLambda (f, \min \{\varepsilon L^{-2N+1}\root N \of {\alpha }, \varepsilon L^{-2N})\le \root N \of {\alpha } < \frac{1}{\root N \of {\alpha }} \le u_\varLambda (f, \min \{\varepsilon L^{-2N+1}\root N \of {\alpha }, \varepsilon L^{-2N}). \end{aligned}

### Proof

We will apply Proposition 2. By applying Theorem 1 (for $$\delta =\varepsilon$$) we obtain the metric $$\rho$$ which is equivalent to $$d$$ on $$U=\{x\, :\, d(x,\varLambda )<\varepsilon L^{-N+1}\}$$ and such that
1. i)

$$d(x,v)\le \rho (x,v)\le L^{N-1}d(x,v)$$ for $$x,v\in U$$,

2. ii)

$$f$$ is $$(1,\varepsilon L^{-N+1},\root N \of {\alpha })$$-uniformly expansive on $$U$$ with respect to the metric $$\rho$$,

3. iii)

$$f$$ is $$\max \{\alpha ^{-1/N}, L\}$$-bilipschitz map on $$U$$ with respect to the metric $$\rho$$.

Let $$\widetilde{Y}=\{y: d(y,\varLambda )< L^{-N+1}\varepsilon \}$$ and $$\widetilde{L}=\max \{\alpha ^{-1/N}, L\}$$. We use Proposition 2 (for $$\widetilde{\varepsilon }=\varepsilon L^{-N}$$, $$\widetilde{L}$$, $$\widetilde{\beta }=\root N \of {\alpha }$$, $$\widetilde{f}=f|_{\{x\,:\, d(x,\varLambda )<\varepsilon L^{-N}\}}$$) and construct functions $$\widetilde{c_s}$$, $$\widetilde{c_u}$$ which define an $$(\widetilde{L}, \widetilde{\delta })$$ cone-field on $$U$$ such that $$\widetilde{f}$$ is $$\widetilde{\delta }$$-cone-hyperbolic with respect to the metric $$\rho$$, where $$\widetilde{\delta }=\varepsilon L^{-N}/\widetilde{L}$$.

Now we need to “translate” the results from the metric $$\rho$$ to the original metric $$d$$. For clarity of notation we use the subscript $$(.)_d$$ to denote objects with respect to the metric $$d$$ and $$(.)_{\rho }$$ to denote objects with respect to the metric $$\rho$$.

By the definition of $$(\widetilde{L}, \widetilde{\delta })$$ cone-field on $$U$$ and i) we get
\begin{aligned}&\frac{1}{\widetilde{L}L^{N-1}} d(x,v)\le \frac{1}{\widetilde{L}} d(x,v)\le \frac{1}{\widetilde{L}}\rho (x,v)\le c(x,v)\le \widetilde{L}\rho (x,y)\\&\le \widetilde{L}L^{N-1}d(x,y) \text { for } (x,v)\in \big \{x\in U, v\in B(x,\widetilde{\delta })_\rho \}. \end{aligned}
From i) we have
\begin{aligned} B(x,\widetilde{\delta }/L^{N-1})_d \subset B(x,\widetilde{\delta })_\rho , \quad B_f(x,\widetilde{\delta }/L^{N-1})_d \subset B_f(x,\widetilde{\delta })_\rho , \end{aligned}
and
\begin{aligned} C_x^{u}(\widetilde{\delta }/L^{N-1})_d \subset C_x^{u}(\widetilde{\delta })_\rho , \quad C_x^{s}(\widetilde{\delta }/L^{N-1})_d \subset C_x^{s}(\widetilde{\delta })_\rho . \end{aligned}
Consequently, from Definition 3 for an arbitrary $$x\in U$$ we get
\begin{aligned} u_x(f;\widetilde{\delta })_\rho \le u_x(f;\widetilde{\delta }/L^{N-1})_d, \quad s_x(f;\widetilde{\delta })_\rho \ge s_x(f;\widetilde{\delta }/L^{N-1})_d. \end{aligned}
Hence
\begin{aligned} u_U(f;\widetilde{\delta })_\rho \le u_U(f;\widetilde{\delta }/L^{N-1})_d, \quad s_U(f;\widetilde{\delta })_\rho \ge s_U(f;\widetilde{\delta }/L^{N-1})_d. \end{aligned}
From the above inequalities and (4.4)
\begin{aligned} s_U(f;\widetilde{\delta })_\rho \le \widetilde{\beta }< 1 <\frac{1}{\widetilde{\beta }}\le u_U(f;\widetilde{\delta })_\rho , \end{aligned}
we obtain that $$f$$ is $$(\widetilde{\delta }/L^{N-1})$$-cone-hyperbolic in metric $$d$$ and
\begin{aligned} s_U(f;\widetilde{\delta }/L^{N-1})_d\le \widetilde{\beta }< 1 <\frac{1}{\widetilde{\beta }}\le u_U(f;\widetilde{\delta }/L^{N-1})_d. \end{aligned}
Finally we conclude that $$\widetilde{c_s}$$ and $$\widetilde{c_u}$$ are $$(\max \{\alpha ^{-1/N}L^{N-1}, L^N\},\widetilde{\delta }/L^{N-1} )$$-cone-field on $$\varLambda$$ such that $$f$$ is $$(\widetilde{\delta }/L^{N-1})$$-cone-hyperbolic on $$\varLambda$$ with respect to metric $$d$$.

## Footnotes

1. 1.

We say the $$I$$ is a subinterval of $$\mathbb {Z}$$ if $$[k, l]\cap \mathbb {Z}\subset I$$ for any $$k,l\in I$$.

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