# Sufficient conditions for rescaling expansivity

A. Rojas  Instituto de Matemática, Universidade Federal do Rio de Janeiro, P.
O. Box 68530, 21945-970 Rio de Janeiro, Brazil<EMAIL_ADDRESS>,  X. Wen
LMIB, Institute of Artificial Intelligence and School of Mathematical Science,
Beihang University, Beijing, China<EMAIL_ADDRESS>and  Y. Yang
Department of Mathematics, Liaoming University, China<EMAIL_ADDRESS>
###### Abstract.

We demonstrate that any k*-expansive vector field on a closed manifold
exhibits rescaling expansiveness. This enhances the principal outcome outlined
in [3]. The verification of this assertion hinges on the introduction and
exploration of the novel concept termed as a "singular-expansive flow," which
will be thoroughly examined.

###### Key words and phrases:

Singular-expansive, k*-expansive, Rescaling expansive

###### 2020 Mathematics Subject Classification:

Primary 37C10; Secondary 37B05

AR was partially supported by Basic Science Research Program through the NRF
funded by the Ministry of Education (Grant Number: 2022R1l1A3053628) and CNPq-
Brazil grant No 307776/2019-0. XW was partially supported by NSFC12071018 and
the Fundamental Research Funds for the Central Universities. YY was partially
supported by the National Natural Science Foundation of China (Grant No.
12101281).

##  1\. Introduction

The concept of expansive flows, introduced by Bowen and Walters [10] as an
extension of the notion for homeomorphisms [16], forms the basis of a vast
theory. However, this theory excludes significant examples such as the
geometric Lorenz attractor [1, 11] and the Cherry flow [15]. Various attempts
have been made by different authors to incorporate these examples into the
expansive theory. For instance, Komuro [13] defined k*-expansive flows and
established the k*-expansiveness of the geometric Lorenz attractor. Oka [14]
demonstrated the equivalence between k*-expansivity and expansivity for
nonsingular flows. Araujo et al. [2] extended Komuro’s work by proving the
k*-expansiveness of every singular-hyperbolic attractor in three-dimensional
differentiable flows. Several authors have also revealed intriguing properties
of k*-expansive flows, as evident in [4, 5, 8]. More recently, Wen and Wen
[17] introduced rescaling expansive flows and demonstrated the rescaling
expansiveness of multisingular hyperbolic flows [7]. Artigue [3] contributed
by establishing a sufficient condition, termed efficiency, for a k*-expansive
flow to be rescaling expansive. Notably, this condition holds true, for
instance, when the fixed points of the flow are hyperbolic. Consequently, it
was concluded that $C^{1}$ generic k*-expansive vector fields on closed
manifolds are rescaling expansive.

In this paper, we advance Artigue’s work [3] by proving that every
k*-expansive vector field on a closed manifold exhibits rescaling
expansiveness. This proof hinges on a novel notion of expansivity for flows
termed singular-expansivity. Our analysis delves into the dynamics of
singular-expansive flows on metric spaces. Specifically, we establish that the
set of periodic orbits in a singular-expansive flow is countable. Moreover, if
the singular set is dynamically isolated, the set of periodic orbits with a
specified period is finite. We also demonstrate the existence of singular-
expansive flows possessing the shadowing property but lacking expansiveness.
Additionally, we explore the notion of singular-equicontinuous flows,
extending the classical concept of equicontinuous flows [6]. Notably, we prove
the existence of flows that are singular-expansive and singular-equicontinuous
yet not equicontinuous. Furthermore, we establish that the Bowen entropy of
nonsingular points vanishes for singular-equicontinuous flows. Let us now
precisely state our main result.

Denote by $M$ a closed manifold i.e. a compact connected boundaryless manifold
endowed with a Riemannian metric $\|\cdot\|$. In this section we let $d$
denote the distance in $M$ induced by $\|\cdot\|$. Denote by $B_{a}(x)$ the
$a$-ball with center at $x$. The exponential map of $M$ will be denoted by
$exp$.

Let $V$ be a vector field of $M$ (all vector fields will be assumed to be
$C^{1}$). Denote by $V_{t}(x)$ the solution curve of the ODE $\dot{x}=V(x)$
with initial condition $x\in M$. This produces a one-parameter family of
diffeomorphisms $\\{V_{t}:M\to M\\}_{t\in\mathbb{R}}$. We say that
$\Lambda\subset M$ is invariant if $V_{t}(\Lambda)=\Lambda$ for every
$t\in\mathbb{R}$.

######  Definition 1 ([17]).

We say that $V$ is rescaling expansive on $\Lambda\subset M$ if for every
$\epsilon>0$ there is $\delta>0$ such that if $x,y\in\Lambda$ satisfy

$d(V_{t}(x),V_{s(t)}(y))\leq\delta\|V(V_{t}(x))\|$

for every $t\in\mathbb{R}$ and some increasing homeomorphism
$s:\mathbb{R}\to\mathbb{R}$, then $V_{s(0)}(y)\in
V_{[-\epsilon,\epsilon]}(x)$. If $V$ is rescaling expansive on $M$, we just
say that $V$ is a rescaling expansive vector field.

Actually this is not the original definition [17] but an equivalent one [18].
Next we recall the notion of k*-expansive vector field based on Komuro [13].

######  Definition 2.

We say that $V$ is k*-expansive on $\Lambda$ if for every $\epsilon>0$ there
is $\delta>0$ such that if $x,y\in\Lambda$ satisfy
$d(V_{t}(x),V_{s(t)}(y))\leq\delta$ for every $t\in\mathbb{R}$ and some
increasing homeomorphism $s:\mathbb{R}\to\mathbb{R}$ fixing $0$, then
$V_{s(t_{0})}(y)\in V_{[t_{0}-\epsilon,t_{0}+\epsilon]}(x)$ for some
$t_{0}\in\mathbb{R}$. If $V$ is k*-expansive on $X$, we just say that $V$ is
k*-expansive.

With these definitions we can state our main result.

######  Theorem 1.

Let $V$ be a vector field of a closed manifold $M$. If $V$ is k*-expansive on
a compact invariant set $\Lambda\subset M$, then $V$ is rescaling expansive on
$\Lambda$.

The proof of this theorem is based on the following definition. Denote by
$Sing(V)=\\{x\in X:V(x)=0\\}$ the singular set of a vector field $V$. The
distance between $z\in M$ and $A\subset M$ is defined by $dist(z,A)=\inf_{a\in
A}d(z,a)$.

######  Definition 3.

We say that $V$ is singular-expansive on $\Lambda\subset M$ if for every
$\epsilon>0$ there is $\delta>0$ such that if $x,y\in\Lambda$ satisfy
$d(V_{t}(x),V_{s(t)}(y))\leq\delta dist(V_{t}(x),Sing(V))$ for every
$t\in\mathbb{R}$ and some increasing homeomorphism
$s:\mathbb{R}\to\mathbb{R}$, then $V_{s(t_{0})}(y)\in
V_{[t_{0}-\epsilon,t_{0}+\epsilon]}(x)$ for some $t_{0}\in\mathbb{R}$. If $V$
is singular-expansive on $X$, we just say that $V$ is a singular-expansive
flow.

Theorem 1 is clearly a direct consequence of the following two propositions.

######  Proposition 1.

Let $V$ be a vector field of a closed manifold $M$. If $V$ is k*-expansive on
$\Lambda\subset M$, then $V$ is singular-expansive on $\Lambda$.

######  Proposition 2.

Let $V$ be a vector field of a closed manifold $M$. If $V$ is singular-
expansive on a compact invariant set $\Lambda\subset M$, then $V$ is rescaling
expansive on $\Lambda$.

This paper is organized as follows. In Section 2 we will prove Proposition 2.
In Section 3 we will extend the notion of singular-expansivity from vector
fields on closed manifolds to flows on metric spaces and prove Proposition 1.
In Section 4 we state some topological properties of the singular-expansive
flows. These properties will be proved in Section 5.

The first author thanks Beihang University, Beijing, Peoples Republic of
China, for its kindly hospitality during the preparation of this work.

##  2\. Proof of Proposition 2

We first recall a flowbox theorem given in [17]. Let $V$ be a $C^{1}$ vector
field on a closed manifold $M$. Let $x\in M\setminus Sing(V)$. Let

$N_{x}(r)=\\{v\in T_{x}M:v\bot X(x),\|v\|<r\\},$ $U_{x}(r)=\\{v+tX(x):v\in
N_{x}(r\|V(x)\|),t\in\mathbb{R},|t|<r\\}.$

Denote by

$F_{x}:U_{x}\to M,\ \ \ \ F_{x}(v+tX(x))=\varphi_{t}(\exp_{x}(v)).$

The conorm (or mininorm) of a linear operator $L$ is defined by

$m(L)=\inf_{\|v\|=1}\|L(v)\|.$

######  Lemma 1 (Proposition 2.2 in [17]).

Let $V$ be a $C^{1}$ vector field on $M$. There exists a constants $r_{0}>0$
such that for any $x\in M\setminus Sing(V)$, $F_{x}:U_{x}(r_{0})\to M$ is an
embedding and $\|D_{p}F_{x}\|<3$ and $m(D_{p}F_{x})>1/3$ for any $p\in
U_{x}(r_{0})$.

Since $M$ is compact, there is $a>0$ such that

$\|D_{p}\exp_{x}\|<3/2\ \ \ \text{ and }\ \ \ \ \|(D_{p}\exp_{x})^{-1}\|<3/2$

for all $x\in M$ and $p\in T_{x}M$ with $\|p\|<a$. Assume $V$ is a $C^{1}$
vector field on $M$. Also by the compactness of $M$, we can find $L>0$ such
that for any $x\in M$, the vector field

$\overline{V}=(\exp^{-1}_{x})_{*}(V|_{B_{a}(x)})$

is Lipschitz with Lipschitz constant $L$. We call $L$ the local Lipschitz
constant of $V$. The following is an easy lemma.

######  Lemma 2.

Let $V$ be a $C^{1}$ vector field on $M$. There is $c>0$ such that for any
$x\in M\setminus Sing(V)$, if $d(y,x)<c\|V(x)\|$, then

$\frac{1}{2}\|V(x)\|\leq\|V(y)\|\leq 2\|V(x)\|.$

###### Proof.

Since there is an upper bound of $\|V(x)\|$ by the compactness of $M$, we can
find $c>0$ such that $c\|V(x)\|<a$ for all $x\in M$ at first. Let $L$ be a
local Lipschitz constant of $V$, we choose $c>0$ such that $c\|V(x)\|<a$ and
$c\leq\frac{1}{4L}$. Then for any $x\in M\setminus Sing(V)$ and any $y\in M$
with $d(x,y)<c\|V(x)\|$, denoting by
$\overline{V}=(\exp^{-1}_{x})_{*}(V|_{B_{a}(x)})$, we have

$\|\overline{V}(\exp_{x}^{-1}(y))-\overline{V}(\exp_{x}^{-1}(x))\|\leq
Ld(y,x)\leq Lc\|V(x)\|\leq\frac{1}{4}\|V(x)\|.$

Note that $\overline{V}(\exp_{x}^{-1}(x))=V(x)$. Thus we have

$\frac{3}{4}\|V(x)\|\leq\|\overline{V}(\exp_{x}^{-1}(y))\|\leq\frac{5}{4}\|V(x)\|.$

Hence

$\|V(y)\|=\|(D_{exp_{x}^{-1}(y)}\exp_{x})^{-1}(\overline{V}(\exp_{x}^{-1}(y)))\|\leq\frac{3}{2}\cdot\frac{5}{4}\|V(x)\|<2\|V(x)\|,$
$\|V(y)\|=\|(D_{exp_{x}^{-1}(y)}\exp_{x})^{-1}(\overline{V}(\exp_{x}^{-1}(y)))\|\geq\frac{2}{3}\cdot\frac{3}{4}\|V(x)\|\geq\frac{1}{2}\|V(x)\|.$

This ends proof of the lemma. ∎

######  Lemma 3 (Lemma 2.3 in [17]).

For every $C^{1}$ vector field $V$ of a closed manifold $M$ there is $r_{0}>0$
such that

  1. (1)

if $x\in M\setminus Sing(V)$, $0<\delta<r_{0}/3$ and $t\in\mathbb{R}$ satisfy
$V_{[0,t]}(x)\subset B(x,\delta\|V(x)\|)$, then $|t|<3\delta$.

  2. (2)

if $x\in M\setminus Sing(V)$, $0<\delta<r_{0}/3$, $|t|<r_{0}$ and $V_{t}(x)\in
B(x,\delta\|V(x)\|)$, then $|t|<3\delta$.

Note that the constants $r_{0}$ in Lemma 1 and Lemma 3 are same. The following
Lemma follows the idea of Lemma 2.4 of [17].

######  Lemma 4.

Let $V$ be a $C^{1}$ vector field on $M$ and $r_{0}$ be given as in Lemma 3.
There is $\delta_{0}>0$ such that for any $x\in M\setminus Sing(V),y\in M$ and
any increasing homeomorphism $h:\mathbb{R}\to\mathbb{R}$, if

$d(V_{t}(x),V_{h(t)}(y))<\delta_{0}\|V(V_{t}(x))\|$

for all $t\in\mathbb{R}$, then $|h(t)-h(0)-t|<r_{0}/2$ for all
$t\in(-r_{0},r_{0})$.

###### Proof.

Let $L$ be a local Lipschitz constant of $V$ and $c$ be the constant given in
Lemma 2. By the continuous dependence of solutions with respect to initial
conditions, we know that $d(V_{t}(x),V_{t}(y))\leq e^{L|t|}d(x,y)$ for any
$x,y\in M$ and $t\in\mathbb{R}$. Then we know that

$e^{-L|t|}\leq\displaystyle\frac{\|V(V_{t}(x))\|}{\|V(x)\|}\leq e^{L|t|}$

for any $x\in M\setminus Sing(V)$ and $t\in\mathbb{R}$.

Now we choose $\delta_{0}$ such that the following properties are satisfied:

$e^{2Lr_{0}}\delta_{0}<c,\ \ 6(1+2e^{2Lr_{0}})\delta_{0}<r_{0}/2,\ \
3(e^{Lr_{0}}+e^{2Lr_{0}})\delta_{0}<r_{0}/2,$
$9(e^{Lr_{0}}+e^{2Lr_{0}})\delta_{0}<c,\ \
6[3+9(e^{Lr_{0}}+e^{2Lr_{0}})]\delta_{0}<r_{0}/2.$

Assume that $x\in M\setminus Sing(V),y\in M$ and homeomorphism
$h:\mathbb{R}\to\mathbb{R}$ satisfy

$d(V_{t}(x),V_{h(t)}(y))<\delta_{0}\|V(V_{t}(x))\|$

for all $t\in\mathbb{R}$. We show that $|h(t)-h(0)-t|<r_{0}/2$ for any
$t\in(-r_{0},r_{0})$

Without loss of generality, we assume that $t\in(0,r_{0})$. First we consider
the case of $h(t)-h(0)\leq t$. In this case we have

$d(V_{h(t)}(y),V_{t+h(0)}(y))\leq
d(V_{h(t)}(y),V_{t}(x))+d(V_{t}(x),V_{t+h(0)}(y))$
$<\delta_{0}\|V(V_{t}(x))\|+e^{Lt}d(x,V_{h(0)}(y))<\delta_{0}\|V(V_{t}(x))\|+\delta_{0}e^{Lt}\|V(x)\|$
$\leq\delta_{0}\|V(V_{t}(x))\|+\delta_{0}e^{2Lt}\|V(V_{t}(x))\|<(1+e^{2Lr_{0}})\delta_{0}\|V(V_{t}(x))\|.$

In the proof of above inequalities, we have already shown that

(1)
$d(V_{t}(x),V_{t+h(0)}(y))<\delta_{0}e^{2Lt}\|V(V_{t}(x))\|<\delta_{0}e^{2Lr_{0}}\|V(V_{t}(x))\|.$

By the assumption $e^{2Lr_{0}}\delta_{0}<c$ we know that
$d(V_{t}(x),V_{t+h(0)}(y))<c\|V(V_{t}(x))\|$, hence $\|V(V_{t}(x))\|\leq
2\|V(V_{t+h(0)}(y))\|$. Then we have

$d(V_{h(t)}(y),V_{t+h(0)}(y))<(1+e^{2Lr_{0}})\delta_{0}\|V(V_{t}(x))\|<2(1+e^{2Lr_{0}})\delta_{0}\|V(V_{t+h(0)}(y))\|.$

By the assumption that $6(1+2e^{2Lr_{0}})\delta_{0}<r_{0}/2$ we know
$2(1+e^{2Lr_{0}})\delta_{0}<\frac{r_{0}}{3}$. Note that

$-r_{0}<-t<h(t)-(t+h(0))\leq 0.$

We can get that $|h(t)-(t+h(0))|<6(1+e^{2Lr_{0}})\delta_{0}<r_{0}/2$ by the
second item of Lemma 3.

Now let us consider the case of $h(t)-h(0)>t$. Since $h$ is an increasing
homeomorphism, there exists $s\in(0,t)$ such that $h(s)=h(0)+t$. Then

$d(V_{s}(x),V_{t}(x))\leq d(V_{s}(x),V_{h(0)+t}(y))+d(V_{h(0)+t}(y),V_{t}(x))$
$=d(V_{s}(x),V_{h(s)}(y))+d(V_{t+h(0)}(y),V_{t}(x))\overset{\eqref{for}}{<}\delta_{0}\|V(V_{s}(x))\|+\delta_{0}e^{2Lt}\|V(V_{t}(x))\|$
$\leq\delta_{0}e^{L(t-s)}\|V(V_{t}(x))\|+\delta_{0}e^{2Lt}\|V(V_{t}(x))\|<(e^{Lr_{0}}+e^{2Lr_{0}})\delta_{0}\|V(V_{t}(x))\|.$

By the assumption that $3(e^{Lr_{0}}+e^{2Lr_{0}})\delta_{0}<r_{0}/2$ we know
that $(e^{Lr_{0}}+e^{2Lr_{0}})\delta_{0}<\frac{r_{0}}{3}$. Noting that
$0<t-s<t<r_{0}$, we have

$t-s<3(e^{Lr_{0}}+e^{2Lr_{0}})\delta_{0}<r_{0}/2$

from the second item of Lemma 3. By Lemma 1 we know that for any
$\tau^{\prime}\in[s,t]$

$d(V_{\tau^{\prime}}(x),V_{t}(x))=d(F_{V_{t}(x)}((\tau^{\prime}-t)V(V_{t}(x))),F_{V_{t}(x)}(0))<3|\tau^{\prime}-t|\|V(V_{t}(x))\|$
$=3(t-\tau^{\prime})\|V(V_{t}(x))\|\leq
3(t-s)\|V(V_{t}(x))\|<9(e^{Lr_{0}}+e^{2Lr_{0}})\delta_{0}\|V(V_{t}(x))\|.$

By the assumption that $9(e^{Lr_{0}}+e^{2Lr_{0}})\delta_{0}<c$ we know

$\|V(V_{\tau^{\prime}}(x))\|\leq 2\|V(V_{t}(x))\|.$

For any $\tau\in[h(0)+t,h(t)]=[h(s),h(t)]$, we can find
$\tau^{\prime}\in[s,t]$ such that $h(\tau^{\prime})=\tau$, then we have

$d(V_{\tau}(y),V_{h(t)}(y))\leq
d(V_{\tau}(y),V_{\tau^{\prime}}(x))+d(V_{\tau^{\prime}}(x),V_{t}(x))+d(V_{t}(x),V_{h(t)}(y))$
$<\delta_{0}\|V(V_{\tau^{\prime}}(x))\|+9(e^{Lr_{0}}+e^{2Lr_{0}})\delta_{0}\|V(V_{t}(x))\|+\delta_{0}\|V(V_{t}(x))\|.$
$\leq
2\delta_{0}\|V(V_{t}(x))\|+9(e^{Lr_{0}}+e^{2Lr_{0}})\delta_{0}\|V(V_{t}(x))\|+\delta_{0}\|V(V_{t}(x))\|$
$=[3+9(e^{Lr_{0}}+e^{2Lr_{0}})]\delta_{0}\|V(V_{t}(x))\|.$

Noting that
$d(V_{h(t)}(y),V_{t}(x))<\delta_{0}\|V(V_{t}(x))\|<c\|V(V_{t}(x))\|$, we have
$\|V(V_{t}(x))\|\leq 2\|V(V_{h(t)}(y))\|$ from Lemma 2. Hence we have

$d(V_{\tau}(y),V_{h(t)}(y))<2[3+9(e^{Lr_{0}}+e^{2Lr_{0}})]\delta_{0}\|V(V_{h(t)}(y))\|$

for all $\tau\in[h(0)+t,h(t)]$. This means that

$V_{[0,h(0)+t-h(t)]}(V_{h(t)}(y))\subset
B(V_{h(t)}(y),2[3+9(e^{Lr_{0}}+e^{2Lr_{0}})]\delta_{0}\|V(V_{h(t)}(y))\|)$

By the choice of $\delta_{0}$ such that
$6[3+9(e^{Lr_{0}}+e^{2Lr_{0}})]\delta_{0}<r_{0}/2$, we know
$2[3+9(e^{Lr_{0}}+e^{2Lr_{0}})]\delta_{0}<\frac{r_{0}}{3}$. By the first part
of Lemma 3 we know
$|h(t)-h(0)-t|<6[3+9(e^{Lr_{0}}+e^{2Lr_{0}})]\delta_{0}<r_{0}/2$. This ends
the proof of the lemma. ∎

######  Lemma 5.

Let $V$ be a $C^{1}$ vector field on $M$. There exists
$\delta_{0}<\frac{r_{0}}{3}$ such that for any $\delta\in(0,\delta_{0})$ and
any $x\in M\setminus Sing(V)$ and any $y\in M$ with a increasing homeomorphism
$h:\mathbb{R}\to\mathbb{R}$, if

$d(V_{t}(x),V_{h(t)}(y))<\delta\|V(V_{t}(x))\|$

for all $t\in\mathbb{R}$ and $V_{h(0)}(y)\in V_{(-r_{0},r_{0})}(x)$, then

$V_{h(t)}(y)\in V_{(-3\delta,3\delta)}(V_{t}(x))\subset
V_{(-r_{0},r_{0})}(V_{t}(x))$

for any $t\in(-r_{0},r_{0})$.

###### Proof.

Let $\delta_{0}$ be chosen as in Lemma 4. Without loss of generality, we can
assume that $6\delta_{0}<r_{0}$. Assume that there are
$\delta\in(0,\delta_{0})$ and $x\in M\setminus Sing(V)$ and $y\in M$ with an
increasing homeomorphism $h:\mathbb{R}\to\mathbb{R}$ such that

$d(V_{t}(x),V_{h(t)}(y))<\delta\|V(V_{t}(x))\|$

for all $t\in\mathbb{R}$ and $V_{h(0)}(y)\in V_{(-r_{0},r_{0})}(x)$. Denote by
$V_{t_{0}}(x)=V_{h(0)}(y)$. Note that $d(V_{h(0)}(y),x)<\delta\|V(x)\|$, hence
by Lemma 1 we have

$\|t_{0}V(x)\|=\|F_{x}^{-1}(V_{t_{0}}(x))\|<3d(V_{t_{0}}(x),x)<3\delta\|V(x)\|,$

thus $|t_{0}|<3\delta$. Let any $t\in(-r_{0},r_{0})$ be given. It is easy to
see that

$V_{h(t)}(y)=V_{h(t)-h(0)}(V_{h(0)}(y))=V_{h(t)-h(0)}(V_{t_{0}}(x))=V_{h(t)-h(0)-t+t_{0}}(V_{t}(x)).$

By Lemma 4 we know that $|h(t)-h(0)-t|<r_{0}/2$, hence we have
$|h(t)-h(0)-t+t_{0}|<\frac{r_{0}}{2}+|t_{0}|<\frac{r_{0}}{2}+3\delta_{0}<r_{0}$.
By the fact that

$d(V_{h(t)}(y),V_{t}(x))<\delta\|V(V_{t}(x))\|$

we know $d(V_{h(t)-h(0)-t+t_{0}}(V_{t}(x)),V_{t}(x))<\delta\|V(V_{t}(x))\|$.
From the second part of Lemma 3 we know $|h(t)-h(0)-t+t_{0}|<3\delta$. This
says that

$V_{h(t)}(y)=V_{h(t)-h(0)-t+t_{0}}(V_{t}(x))\in
V_{(-3\delta,3\delta)}(V_{t}(x))$

and ends the proof of the lemma. ∎

######  Lemma 6.

Let $V$ be a $C^{1}$ vector field on $M$. There exists $\delta_{0}$ such that
for any $\delta\in(0,\delta_{0})$ and any $x\in M\setminus Sing(V)$ and any
$y\in M$ with a increasing homeomorphism $h:\mathbb{R}\to\mathbb{R}$, if

$d(V_{t}(x),V_{h(t)}(y))<\delta\|V(V_{t}(x))\|$

for all $t\in\mathbb{R}$ and $V_{h(0)}(y)\in V_{(-r_{0},r_{0})}(x)$, then

$V_{h(t)}(y)\in V_{(-3\delta,3\delta)}(V_{t}(x))$

for any $t\in\mathbb{R}$.

###### Proof.

Let $\delta_{0}$ be the constant chosen in Lemma 5 and
$\delta\in(0,\delta_{0}]$ be given. If $x\in M\setminus Sing(V)$ and $y\in M$
with a increasing homeomorphism $h:\mathbb{R}\to\mathbb{R}$ satisfy

$d(V_{t}(x),V_{h(t)}(y))<\delta\|V(V_{t}(x))\|$

for all $t\in\mathbb{R}$ and $V_{h(0)}(y)\in V_{(-r_{0},r_{0})}(x)$. Then by
Lemma 5 we know that

$V_{h(\frac{r_{0}}{2})}(y)\in V_{(-r_{0},r_{0})}(V_{\frac{r_{0}}{2}}x).$

Let
$\tilde{h}(t)=h(t+\frac{r_{0}}{2})-h(\frac{r_{0}}{2}),\tilde{x}=V_{\frac{r_{0}}{2}}x,\tilde{y}=V_{h(\frac{r_{0}}{2})}(y)$,
then

$d(V_{t}(\tilde{x}),V_{\tilde{h}(t)}(\tilde{y}))=d(V_{t+\frac{r_{0}}{2}}(x),V_{h(t+\frac{r_{0}}{2})}(y))<\delta\|V(V_{t+\frac{r_{0}}{2}}(x))\|=\delta\|V(V_{t}(\tilde{x}))\|$

for all $t\in\mathbb{R}$. And we also have $V_{\tilde{h}(0)}(\tilde{y})\in
V_{(-r_{0},r_{0})}(\tilde{x})$. Hence for any $t\in(-r_{0},r_{0})$ we have

$V_{\tilde{h}(t)}(\tilde{y})\in V_{(-3\delta,3\delta)}(V_{t}(\tilde{x})),$

and then

$V_{h(t+\frac{r_{0}}{2})}(y)\in
V_{(-3\delta,3\delta)}(V_{t+\frac{r_{0}}{2}}(x)).$

This prove that $V_{h(t)}(y)\in V_{(-3\delta,3\delta)}(V_{t}(x))$ for any
$t\in[0,\frac{3}{2}r_{0}]$. Similarly we prove the same thing holds as
$t\in[-\frac{3}{2}r_{0},0]$. By induction we can prove that for any
$n\in\mathbb{N}$ we have $V_{h(t)}(y)\in V_{(-3\delta,3\delta)}(V_{t}(x))$ for
any $t\in[-\frac{n}{2}r_{0},\frac{n}{2}r_{0}]$. Hence $V_{h(t)}(y)\in
V_{(-3\delta,3\delta)}(V_{t}(x))$ for all $t\in\mathbb{R}$. ∎

###### Proof of Proposition 2.

Let $V$ be a $C^{1}$ vector field and $\Lambda$ be a compact invariant set of
$V$. Assume that $V$ is singular-expansive on $\Lambda$. Since $V$ is $C^{1}$
(in particular Lipschitz), there is $B>0$ such that

$\|V(z)\|\leq Bdist(z,Sing(V)),\quad\quad\forall z\in M.$

Fix $\epsilon>0$. Without loss of generality, we assume that $\epsilon<r_{0}$.
For this $\epsilon$ we let $\delta^{\prime}>0$ be given by the singular-
expansivity of $V$ on $\Lambda$. Define
$\delta=\min\\{\frac{\delta^{\prime}}{B},\delta_{0},\frac{\epsilon}{3}\\}$
where $\delta_{0}$ was given as in Lemma 6. Suppose that

$d(V_{t}(x),V_{s(t)}(y))\leq\delta\|V(V_{t}(x))\|$

for every $t\in\mathbb{R}$ and $x\in\Lambda\setminus Sing(V)$, $y\in\Lambda$
and an increasing homeomorphism $s:\mathbb{R}\to\mathbb{R}$.

Then, $x,y\in\Lambda$ satisfy

$\delta\|V(V_{t}(x))\|\leq\frac{\delta^{\prime}}{B}\|V(V_{t}(x))\|\leq\delta^{\prime}dist(V_{t}(x),Sing(V)),\quad\quad\forall
t\in\mathbb{R}$

and so

$d(V_{t}(x),V_{s(t)}(y))\leq\delta^{\prime}dist(V_{t}(x),Sing(V)),\quad\quad\forall
t\in\mathbb{R}.$

It follows from singular expansivity that
$V_{s(t_{0})}(y)=V_{\tau}(V_{t_{0}}(x))$ for some
$\tau\in[-\epsilon,\epsilon]$ and $t_{0}\in\mathbb{R}$. Now let
$\tilde{x}=V_{t_{0}}(x),\tilde{y}=V_{s(t_{0})}(y),\tilde{s}(t)=s(t+t_{0})-s(t_{0})$,
then we have

$d(V_{t}(\tilde{x}),V_{\tilde{s}(t)}(\tilde{y}))=d(V_{t+t_{0}}(x),V_{s(t+t_{0})}(y))\leq\delta\|V(V_{t+t_{0}}(x))\|=\delta\|V(V_{t}(\tilde{x}))\|,$

and $V_{\tilde{s}(0)}(\tilde{y})=V_{\tau}(\tilde{x})\in
V_{(-r_{0},r_{0})}(\tilde{x})$. By Lemma 6 we have

$V_{\tilde{s}(t)}(\tilde{y})\in V_{(-3\delta,3\delta)}(V_{t}(\tilde{x}))$

for all $t\in\mathbb{R}$. Thus

$V_{{s}(t)}(y)\in V_{(-3\delta,3\delta)}(V_{t}({x}))\subset
V_{[-\epsilon,\epsilon]}(V_{t}(x))$

for all $t\in\mathbb{R}$. Therefore, $V$ is rescaling expansive on $\Lambda$
and the proof follows. ∎

##  3\. Singular-expansive flows and Proof of Proposition 1

In this section we will transform the notion of singular-expansivity from
vector fields on closed manifolds (Definition 3) to flows on compact metric
spaces. In particular, we will prove Proposition 1. Previously, we present
some basic definitions and facts.

Let $X$ be a metric space. A flow of $X$ is a continuous map
$\phi:\mathbb{R}\times X\to X$ satisfying $\phi(0,x)=x$ and
$\phi(t+s,x)=\phi(t,\phi(s,x))$ for every $x\in X$ and $t,s\in\mathbb{R}$.
Denote by $\phi_{t}(x)=\phi(t,x)$ the time $t$-map. If $I\subset\mathbb{R}$ we
denote $\phi_{I}(x)=\\{\phi_{t}(x):t\in I\\}$. We define
$\phi_{[a,b]}(x)=\phi_{[b,a]}(x)$ for $a\geq b$. Recall that a singularity of
a flow $\phi$ is a point $\sigma\in X$ such that $\phi_{t}(\sigma)=\sigma$ for
every $t\in\mathbb{R}$. Denote by $Sing(\phi)$ the set of singularities of
$\phi$. Notice that the singularities of a vector field match with the
singularities of its corresponding flow.

For the sake of comparison we will present the current notions of expansivity
for flows on metric spaces. The first one is the classical notion of expansive
flow by Bowen and Walters [10].

######  Definition 4.

We say that $\phi$ is expansive on $\Lambda\subset X$ if for every
$\epsilon>0$ there is $\delta>0$ such that if $x,y\in\Lambda$ satisfy
$d(\phi_{t}(x),\phi_{s(t)}(y))\leq\delta$ for every $t\in\mathbb{R}$ and some
continuous map $s:\mathbb{R}\to\mathbb{R}$ fixing $0$, then
$y\in\phi_{[-\epsilon,\epsilon]}(x)$. If $\phi$ is expansive on $X$, we just
say that $\phi$ is expansive.

The second is the notion of k*-expansive flow by Komuro [13].

######  Definition 5.

We say that $\phi$ is k*-expansive on $\Lambda$ if for every $\epsilon>0$
there is $\delta>0$ such that if $x,y\in\Lambda$ satisfy
$d(\phi_{t}(x),\phi_{s(t)}(y))\leq\delta$ for every $t\in\mathbb{R}$ and some
increasing homeomorphism $s:\mathbb{R}\to\mathbb{R}$ fixing $0$, then
$\phi_{s(t_{0})}(y)\in\phi_{[t_{0}-\epsilon,t_{0}+\epsilon]}(x)$ for some
$t_{0}\in\mathbb{R}$. If $\phi$ is k*-expansive on $X$, we just say that
$\phi$ is k*-expansive.

Now, by replacing the vector field $V$ by a general flow $\phi$ in Definition
3 we obtain the following definition.

######  Definition 6.

We say that $\phi$ is singular-expansive on $\Lambda$ if for every
$\epsilon>0$ there is $\delta>0$ such that if $x,y\in\Lambda$ satisfy
$d(\phi_{t}(x),\phi_{s(t)}(y))\leq\delta dist(\phi_{t}(x),Sing(\phi))$ for
every $t\in\mathbb{R}$ and some increasing homeomorphism
$s:\mathbb{R}\to\mathbb{R}$, then
$\phi_{s(t_{0})}(y)\in\phi_{[t_{0}-\epsilon,t_{0}+\epsilon]}(x)$ for some
$t_{0}\in\mathbb{R}$. If $\phi$ is singular-expansive on $X$, we just say that
$\phi$ is a singular-expansive flow.

We can therefore reformulate Proposition 2 by saying that if the flow of a
vector field $V$ of a closed manifold $M$ is singular-expansive on
$\Lambda\subset M$, then such a flow is also rescaling expansive on $\Lambda$.
Therefore, the property of being singular-expansive is stronger than rescaling
expansive.

On the other hand, there are singular-expansive flows which are not expansive:
take for instance the trivial flow namely the one for which every point is a
singularity (it is clearly singular-expansive but not expansive unless the
space is finite). In particular, singular-expansive flows which are not
k*-expansive also exist.

The next result asserts that every k*-expansive flow is singular-expansive.

######  Theorem 2.

Let $\phi$ be a flow of a compact metric space $X$. If $\phi$ is k*-expansive
on $\Lambda\subset X$, then $\phi$ is singular-expansive on $\Lambda$.

###### Proof.

Fix $\epsilon>0$ and let $\delta>0$ be given by the k*-expansivity of $\phi$
on $\Lambda$. Take $x,y\in X$ and an increasing homeomorphism
$s:\mathbb{R}\to\mathbb{R}$ such that
$d(\phi_{t}(x),\phi_{s(t)}(y))\leq\frac{\delta}{diam(X)}dist(\phi_{t}(x),Sing(\phi))$
for every $t\in\mathbb{R}$. Since $dist(z,Sing(\phi))\leq diam(X)$ for every
$z\in X$, we get $d(\phi_{t}(x),\phi_{s(t)}(y)\leq\delta$ for every
$t\in\mathbb{R}$. Then, by k*-expansivity and the choice of $\delta$ we get
$\phi_{s(t_{0})}(y)\in\phi_{[t_{0}-\epsilon,t_{0}+\epsilon]}(x)$ proving the
result. ∎

###### Proof of Proposition 1.

Just apply Theorem 2 to the flow generated by $V$. ∎

We finish this section by proving that both the rescaling and singular-
expansivity are equivalent when the vector field is invertible at the
singularities. More precisely, the following result holds.

######  Theorem 3.

Let $V$ be a $C^{1}$ vector field of a closed manifold $M$ such that
$DV(\sigma)$ is invertible for every $\sigma\in Sing(V)$. Then, $V$ is
singular-expansive on $\Lambda\subset M$ if and only if it is rescaling
expansive on $\Lambda$.

###### Proof.

Clearly the following property implies that $V$ is singular-expansive on
$\Lambda$:

  1. (*)

For every $\epsilon>0$ there is $\delta>0$ such that if $x,y\in\Lambda$
satisfy

$d(V_{t}(x),V_{s(t)}(y))\leq\delta dist(V_{t}(x),Sing(V))$

for every $t\in\mathbb{R}$ and some increasing homeomorphism
$s:\mathbb{R}\to\mathbb{R}$, then $V_{s(0)}(y)\in
V_{(-\epsilon,\epsilon)}(x)$.

On the other hand, by Proposition 2, the singular-expansivity of $V$ on
$\Lambda$ implies the rescaling expansivity of $V$ on $\Lambda$. It remains to
prove that the rescaling expansivity of $V$ on $\Lambda$ implies (*).

First note that by Lemma 3.9 in [3] there is $C>0$ such that

$dist(z,Sing(V))\leq C\|V(z)\|,\quad\quad\forall z\in M.$

Fix $\epsilon>0$ and let $\delta^{\prime}$ be given by rescaling-expansivity.
Let $\delta=\frac{\delta^{\prime}}{C}$ and suppose that

$d(V_{t}(x),V_{s(t)}(y))\leq\delta dist(V_{t}(x),Sing(V))$

for every $t\in\mathbb{R}$ and some increasing homeomorphism
$s:\mathbb{R}\to\mathbb{R}$. Then,

$\delta
dist(V_{t}(x),Sing(V))\leq\frac{\delta^{\prime}}{C}dist(V_{t}(x),Sing(V))\leq\delta^{\prime}\|V(V_{t}(x))\|,\quad\quad\forall
t\in\mathbb{R}$

hence $d(V_{t}(x),V_{s(t)}(y))\leq\delta^{\prime}\|V(V_{t}(x))\|,$ for every
$t\in\mathbb{R}$ and so $V_{s(0)}(y)=V_{t}(x)$ for some
$t\in[-\epsilon,\epsilon]$. This completes the proof. ∎

These results motivate the study of the singular-expansive flows. In the next
section we will present our results in this direction.

##  4\. Properties of singular-expansive flows

In the sequel we will study the topological properties of the singular-
expansive flows. This requires some notations.

Let $\phi$ be a flow of a compact metric space $X$. We say that $x\in X$ is a
periodic point of $\phi$ if $x\notin Sing(\phi)$ but there is $t>0$ such that
$\phi_{t}(x)=x$. The minimal of such $t^{\prime}s$ is called the period of
$x$. We say that $x\in X$ is nonwandering if $U\cap(\bigcap_{t\geq
T}\phi_{t}(U))\neq\emptyset$ for every $T>0$ and every neighborhood $U$ of
$x$. The set of nonwandering points is denoted by $\Omega(\phi)$.

We say that $K\subset X$ is invariant if $\phi_{t}(K)=K$ for every
$t\in\mathbb{R}$. In such a case we denote by $\phi|_{K}$ the flow restricted
to $K$. We say that $K$ is dynamically isolated if there is a neighborhood $U$
of $K$ (called isolated block) such that

$K=\bigcap_{t\in\mathbb{R}}\phi_{t}(U).$

######  Theorem 4.

The following properties hold for every singular-expansive flow $\phi$ of a
compact metric space $X$.

  1. (1)

The set of periodic orbits of $\phi$ is countable.

  2. (2)

If $Sing(\phi)=\emptyset$ or consists of finitely many isolated points of $X$,
then $\phi$ is expansive.

  3. (3)

If $Sing(\phi)$ is dynamically isolated, the set of periodic orbits with
period $\tau\in[0,t]$ is finite ($\forall t>0$).

Item (3) above is false without the hypothesis that $Sing(\phi)$ is
dynamically isolated. Based on Subsection 3.5 in [3] we obtain the following
counterexample.

######  Example 1.

There is a compact metric space exhibiting a singular-expansive flow with
infinitely many periodic orbits of period $2\pi$.

###### Proof.

Define

$X=\\{(0,0)\\}\cup\bigcup_{n\in\mathbb{N}}\\{z\in\mathbb{R}^{2}:\|z\|=e^{-n}\\}.$

We have that $X$ is a compact metric space if equipped with the Euclidean
metric. Consider the flow $\phi$ on $X$ obtained by restricting (to $X$) the
flow of the vector field in $\mathbb{R}^{2}$ defined by $V(x,y)=(-y,x)$.
Proposition 3.21 in [3] implies that $V$ is rescaling expansive on $X$. Since
$DV(0,0)$ is invertible, we conclude from Theorem 3 that $\phi$ singular-
expansive. On the other hand, by direct integration we obtain

$\phi_{t}(z)=(-y\sin t+x\cos t,x\sin t+y\cos y),\quad\quad\forall z=(x,y)\in
X,\quad t\in\mathbb{R}.$

Then, $X\setminus Sing(\phi)=X\setminus\\{(0,0)\\}$ consists of infinitely
many periodic orbits of period $2\pi$. ∎

On the other hand, we say that $\phi$ is equicontinuous when for every
$\epsilon>0$ there is $\delta>0$ such that if $x,y\in X$ and
$d(x,y)\leq\delta$, then $d(\phi_{t}(x),\phi_{t}(y))\leq\epsilon$ for every
$t\in\mathbb{R}$. The equicontinuous flows have been widely studied in the
literature [6]. Their singular version are given below.

######  Definition 7.

A flow $\phi$ of a metric space $X$ is singular-equicontinuous if for every
$\epsilon>0$ there is $\delta>0$ such that if $x,y\in X$ and $d(x,y)\leq\delta
dist(x,Sing(\phi))$, then $d(\phi_{t}(x),\phi_{t}(y))\leq\epsilon$ for every
$t\in\mathbb{R}$.

Every equicontinuous flow $\phi$ on a compact metric space is singular-
equicontinuous. Indeed, take $\epsilon>0$ and let $\delta>0$ be given by the
equicontinuity of $\phi$. If $\delta^{\prime}=\frac{\delta}{diam(X)}$ and
$d(x,y)\leq\delta^{\prime}dist(x,Sing(\phi))$, then $d(x,y)\leq\delta$ and so
$d(\phi_{t}(x),\phi_{t}(y))\leq\epsilon$ for every $t\in\mathbb{R}$.
Therefore, $\phi$ is singular-equicontinuous.

The converse is true if there are no singularities. More precisely, every
singular-equicontinuous flow without singularities on a compact metric space
is equicontinuous. It follows that flows which are both singular-expansive and
singular-equicontinuous do exist (e.g. the trivial flow). A different example
will be reported below.

First recall that if $\delta,T>0$ a $(\delta,T)$-pseudo orbit of a flow $\phi$
is a sequence $(x_{i},t_{i})_{i\in\mathbb{Z}}$ formed by points $x_{i}\in X$
and times $t_{i}\in\mathbb{R}$ such that $t_{i}\geq T$ and
$d(\phi_{t_{i}}(x_{i}),x_{i+1})\leq\delta$ for every $i\in\mathbb{Z}$. Given
$\epsilon>0$ we define $Rep(\epsilon)$ as the set of increasing homeomorphism
$s:\mathbb{R}\to\mathbb{R}$ such that

$\left|\frac{s(t)-s(r)}{t-r}-1\right|\leq\epsilon\quad(\forall t\neq r).$

We say that $(x_{i},t_{i})_{i\in\mathbb{Z}}$ can be $\epsilon$-shadowed if
there are $x\in X$ and $s\in Rep(\epsilon)$ such that

$d(\phi_{s(t)}(x),\phi_{t-S(i)}(x_{i}))\leq\epsilon,\quad\quad\forall
i\in\mathbb{Z},\quad\forall S(i)\leq t<S(i+1)$

where

$S(i)=\left\\{\begin{matrix}t_{0}+t_{1}+\cdots t_{i-1}&\hbox{ if }i>0,\\\
0&\hbox{ if }i=0,\\\ -t_{1}-t_{2}-\cdots-t_{i}&\hbox{ if
}i<0.\end{matrix}\right.$

Given $\Lambda\subset X$ we say that $\phi$ has the shadowing property on
$\Lambda$ if for every $\epsilon>0$ there is $\delta>0$ such that every
$(\delta,1)$-pseudo orbit $(x_{i},t_{i})_{i\in\mathbb{Z}}$ on $\Lambda$ (i.e.
$x_{i}\in\Lambda$ for $i\in\mathbb{Z}$) can be $\epsilon$-shadowed. If $\phi$
has the shadowing property on $X$ we just say that $\phi$ has the shadowing
property. This definition was abbreviated to SPOTP (strong pseudo-orbit
tracing property) in Komuro [12]. Examples of flows with this property are the
Anosov ones [15].

######  Theorem 5.

There is a compact metric space exhibiting a flow with the shadowing property
which is singular-expansive, singular-equicontinuous but not equicontinuous.

An interesting property of the equicontinuous flows is that they have zero
topological entropy. Recall the topological entropy of a flow $\phi$ of a
compact metric space $X$ defined by $h(\phi)=h(\phi,X)$ where

$h(\phi,K)=\lim_{\epsilon\to 0}\limsup_{t\to\infty}\frac{1}{t}\ln
r(t,\epsilon)$

for $K\subset X$ compact, and $r(t,\epsilon)$ is the minimal cardinality of
$F\subset K$ satisfying

$K\subset\bigcup_{x\in F}\\{y\in
X:d(\phi_{s}(x),\phi_{s}(y))\leq\epsilon,\forall 0\leq s\leq t\\}$

($F$ is then called $(t,\epsilon)$-spanning set). We would like to prove the
same property for the singular-equicontinuous flows. Instead, we will consider
the Bowen entropy [9] of the nonsingular points namely

$h^{*}(\phi)=\sup\\{h(\phi,K):K\subset X\setminus Sing(\phi)\mbox{ is
compact}\\}$

In general $h^{*}(\phi)\leq h(\phi)$ hence $h^{*}(\phi)=0$ if $\phi$ is
equicontinuous. Since $\phi$ restricted to $Sing(\phi)$ is equicontinuous,
$h(\phi,Sing(\phi))=0$ we expect $h^{*}(\phi)=h(\phi)$. These facts motivate
the result below.

######  Theorem 6.

Let $\phi$ be a flow of compact metric spaces.

  1. (1)

If $\Omega(\phi)\setminus Sing(\phi)$ is closed, then $h^{*}(\phi)=h(\phi)$.

  2. (2)

If $\phi$ is singular-equicontinuous, then $h^{*}(\phi)=0$.

##  5\. Proof of theorems 4 to 6

The next lemma is closely related to Lemma 2.2 in [3].

######  Lemma 7.

Let $\phi$ be a flow of a compact metric space $X$. If $Sing(\phi)$ is
dynamically isolated, then there is $\beta_{0}>0$ such that
$diam(\phi_{\mathbb{R}}(x))\geq\beta_{0}$ for every $x\in X\setminus
Sing(\phi)$.

###### Proof.

Otherwise, there is a sequence $x_{n}\in X\setminus Sing(\phi)$ such that
$diam(\phi_{\mathbb{R}}(x_{n}))\to 0$ as $n\to\infty$. Let $U$ be an isolating
block of $Sing(\phi)$. Since $diam(\phi_{\mathbb{R}}(x_{n}))\to 0$ and $X$ is
compact, we can assume that $x_{n}\to\sigma$ for some $\sigma\in Sing(\phi)$.
Then, $\phi_{\mathbb{R}}(x_{n})\subset U$ for $n$ large so
$x_{n}\in\bigcap_{t\in\mathbb{R}}\phi_{t}(U)=Sing(\phi)$ for $n$ large which
is absurd. This completes the proof. ∎

The next lemma proves that the set of singularities of a singular-expansive
flow is dynamically isolated.

######  Lemma 8.

Let $\phi$ be a flow of a compact metric space $X$. If $\phi$ is expansive on
$X\setminus Sing(\phi)$, then $Sing(\phi)$ is dynamically isolated.

###### Proof.

Suppose that $Sing(\phi)$ is not dynamically isolated. We have three cases to
be considered:

  1. (1)

There is a sequence of periodic orbits
$\\{\gamma_{1},\gamma_{2},\cdots,\gamma_{n},\cdots\\}$ accumulating on
$Sing(\phi)$ whose periods are bounded away from $0$.

  2. (2)

There is a sequence of periodic orbits
$\\{\gamma_{1},\gamma_{2},\cdots,\gamma_{n},\cdots\\}$ accumulating on
$Sing(\phi)$ whose periods tent to $0$ as $n\to\infty$.

  3. (3)

There is a neighborhood $U_{0}$ of $Sing(\phi)$ such that no periodic orbit of
$\phi$ is entirely contained in $U_{0}$.

In Case (1) we take $\epsilon_{0}=\min\\{\frac{\tau}{4},1\\}$ where $\tau$ is
a positive lower bound of the periods of the periodic orbits $\gamma_{n}$
($n=1,2,\cdots$). Fix $\delta>0$. For this $\delta$ there exists a
neighborhood $U_{\delta}$ of $Sing(\phi)$ such that $d(\phi_{t}(z),z)<\delta$
for every $z\in U_{\delta}$ and every $t\in[0,2]$. Since $\gamma_{n}$
accumulates on $Sing(\phi)$, one has $\gamma_{n}\subset U_{\delta}$ for some
$n\in\mathbb{N}$. Take $z\in\gamma_{n}$ and $y=\phi_{2\epsilon_{0}}(z)$. It
follows from the choice of $\epsilon_{0}$ that $2\epsilon_{0}\in[0,2]$. Then,
$d(\phi_{t}(z),\phi_{t}(y))=d(\phi_{t}(z),\phi_{2\epsilon_{0}}(\phi_{t}(z)))<\delta$
for every $t\in\mathbb{R}$. Again the choice of $\epsilon_{0}$ implies
$2\epsilon_{0}<\tau$ hence $y\notin\phi_{[-\epsilon_{0},\epsilon_{0}]}(z)$.

In Case (2) we take $\epsilon_{0}=1$. In this case have that
$\gamma_{n}\to\sigma$ for some $\sigma\in Sing(\phi)$. Given $\delta>0$ we can
choose two different periodic orbits $\gamma_{n}$ and $\gamma_{m}$ with
$d_{H}(\gamma_{n},\gamma_{m})<\delta$ ($d_{H}$ here is the Hausdorff
distance). Taking $z\in\gamma_{n}$ and $y\in\gamma_{m}$ we have
$d(\phi_{t}(z),\phi_{t}(y))<\delta$ for every $t\in\mathbb{R}$. Since
$\gamma_{n}$ and $\gamma_{m}$ are different orbits, we also have
$y\notin\phi_{[-\epsilon_{0},\epsilon_{0}]}(x)$.

In Case (3) we take $\epsilon_{0}=1$ once more. Take $\delta>0$. As before
there is a neighborhood $U_{\delta}$ of $Sing(\phi)$ such that
$d(\phi_{t}(z),z)<\delta$ for every $z\in U_{\delta}$ and every $t\in[0,2]$.
Since $Sing(\phi)$ is not dynamically isolated, there is $x\in U\setminus
Sing(\phi)$ such that the full orbit of $x$ contained in $U_{\delta}$. The
assumption in Case (3) implies that the orbit of $x$ is not periodic. Take
$y=\phi_{2}(x)$. Then, $d(\phi_{t}(x),\phi_{t}(y))<\delta$ for every
$t\in\mathbb{R}$. Since $x$ is not periodic, we have
$y\notin\phi_{[-1,1]}(x)$.

Taking $h$ as the identity of $\mathbb{R}$ in all these cases we obtain that
$\phi$ is not expansive on $X\setminus Sing(\phi)$. This completes the proof.
∎

We also need the following lemma.

######  Lemma 9.

The properties below hold for every flow $\phi$ of a compact metric space $X$:

  1. (1)

If $\phi$ is expansive on $X\setminus Sing(\phi)$, then $\phi$ is singular-
expansive.

  2. (2)

If $\phi$ is singular-expansive and $Sing(\phi)$ is open, then $\phi$ is
expansive on $X\setminus Sing(\phi)$.

###### Proof.

First we prove Item (1). By Lemma 8 we have that $Sing(\phi)$ is dynamically
isolated. Take an isolating block $U$ of $Sing(\phi)$. Fix $\Delta>0$ such
that

(2) $\\{z\in X:dist(z,Sing(\phi))\leq\Delta\\}\subset U.$

Let $\epsilon>0$ and set $\epsilon^{\prime}=\min\\{\epsilon,\Delta\\}$. For
this $\epsilon^{\prime}$ we let $\delta^{\prime}>0$ be given by the
expansivity of $\phi$ on $X\setminus Sing(\phi)$. Define
$\delta=\frac{\min\\{\delta^{\prime},\Delta\\}}{diam(X)}$ and let $x,y\in X$
such that

$d(\phi_{t}(x),\phi_{s(t)}(y))\leq\delta dist(\phi_{t}(x),Sing(\phi))$

for every $t\in\mathbb{R}$ and some increasing homeomorphism
$s:\mathbb{R}\to\mathbb{R}$. Define $\hat{s}(t)=s(t)-s(0)$ for
$t\in\mathbb{R}$ and $\hat{y}=\phi_{s(0)}(y)$. Then,
$\hat{s}:\mathbb{R}\to\mathbb{R}$ is a continuous map fixing $0$ such that

$d(\phi_{t}(x),\phi_{\hat{s}(t)}(\hat{y}))\leq\delta
dist(\phi_{t}(x),Sing(\phi)),\quad\quad\forall t\in\mathbb{R}.$

If $x\in Sing(\phi)$, $\phi_{t}(x)=x\in Sing(\phi)$ for $t\in\mathbb{R}$ hence
$\hat{y}=x$ and so $\phi_{s(0)}(y)=\phi_{t}(x)$ with
$t=0\in[-\epsilon,\epsilon]$.

On the other hand,

$\delta
dist(z,Sing(\phi))=\min\\{\delta^{\prime},\Delta\\}\frac{dist(z,Sing(\phi))}{diam(X)}\leq\min\\{\delta^{\prime},\Delta\\},\quad\quad\forall
z\in X.$

Therefore,

$d(\phi_{t}(x),\phi_{\hat{s}(t)}(\hat{y}))\leq\min\\{\delta^{\prime},\Delta\\},\quad\quad\forall
t\in\mathbb{R}.$

If $y\in Sing(\phi)$, then $d(\phi_{t}(x),y)\leq\Delta$ for every
$t\in\mathbb{R}$. This and the inclusion (2) imply
$x\in\bigcap_{t\in\mathbb{R}}\phi_{t}(U)=Sing(\phi)$ and then
$\phi_{s(0)}(y)=\phi_{t}(x)$ with $t\in[-\epsilon,\epsilon]$ as before.
Therefore, we can assume $x,y\in X\setminus Sing(\phi)$. Since

$d(\phi_{t}(x),\phi_{\hat{s}(t)}(\hat{y}))\leq\min\\{\delta^{\prime},\Delta\\}\leq\delta^{\prime},\quad\quad\forall
t\in\mathbb{R},$

we conclude from the expansivity on $X\setminus Sing(\phi)$ that
$\phi_{s(0)}(y)=\hat{y}=\phi_{t}(x)$ for some $t\in[-\epsilon,\epsilon]$
proving Item (1).

To prove Item (2) we assume that $\phi$ is singular-expansive and that
$Sing(\phi)$ is open. Then, $X\setminus Sing(\phi)$ is closed and $\phi$ has
no singularities there so by Item (ii) of Theorem 1 in [10] we just need to
consider increasing homeomorphisms fixing $0$ to prove the expansivity of
$\phi$ on $X\setminus Sing(\phi)$. Since $Sing(\phi)$ and $X\setminus
Sing(\phi)$ are closed disjoint, there is $c>0$ such that
$\inf\\{dist(z,Sing(\phi)):z\in X\setminus Sing(\phi)\\}\geq c.$ Now, let
$\epsilon>0$ and consider $\delta^{\prime}$ from the singular-expansivity of
$\phi$ for this $\epsilon$. Define $\delta=\delta^{\prime}c$ and take $x,y\in
X\setminus Sing(\phi)$ such that $d(\phi_{t}(x),\phi_{s(t)}(y))\leq\delta,$
for every $t\in\mathbb{R}$ and some increasing homeomorphism
$s:\mathbb{R}\to\mathbb{R}$ fixing $0$. Since $x\in X\setminus Sing(\phi)$
which is invariant, $\phi_{t}(x)\in X\setminus Sing(\phi)$ for every
$t\in\mathbb{R}$. Then, $c\leq dist(\phi_{t}(x),Sing(\phi))$ for every
$t\in\mathbb{R}$ and then

$d(\phi_{t}(x),\phi_{s(t)}(y))\leq\delta=\delta^{\prime}c\leq\delta^{\prime}dist(\phi_{t}(x),Sing(\phi)),\quad\quad\forall
t\in\mathbb{R}.$

Therefore, $y=\phi_{s(0)}(y)=\phi_{t}(x)$ for some $t\in[-\epsilon,\epsilon]$
proving that $\phi$ is expansive on $X\setminus Sing(\phi)$. This completes
the proof. ∎

######  Example 2.

It is natural to ask if we can remove the hypothesis that $Sing(\phi)$ is open
in Item (2) of Lemma 9. However, this is false and a counterexample is given
by the geometric Lorenz attractor.

We also need the result below below.

######  Lemma 10.

Let $\phi$ be a singular-expansive flow of a compact metric space $X$. Then,
$\phi$ is expansive on every nonsingular compact invariant set of $\phi$.

###### Proof.

Let $\Lambda$ be a nonsingular compact invariant set of $\phi$. We assert that
$\phi|_{\Lambda}$ is singular-expansive.

Fix $\epsilon>0$ and let $\delta^{\prime}$ be given by the singular-
expansivity of $\phi$ for this $\epsilon$. Since $\Lambda$ is compact and
nonsingular, there exists $\delta^{\prime}>0$ such that if $a,b\in\Lambda$ and
$d(a,b)\leq\delta^{\prime}diam(X)$, then $d(a,b)\leq\delta
dist(a,Sing(\phi))$.

Now suppose that $x,y\in\Lambda$ and
$d(\phi_{t}(x),\phi_{s(t)}(y))\leq\delta^{\prime}dist(\phi_{t}(x),Sing(\phi|_{\Lambda}))$
for every $t\in\mathbb{R}$ and some increasing homeomorphism
$s:\mathbb{R}\to\mathbb{R}$. Since $\phi$ is nonsingular,
$Sing(\phi|_{\Lambda})=\emptyset$ hence
$dist(\phi_{t}(x),Sing(\phi|_{\Lambda}))=dist(\phi_{t}(x),\emptyset)=diam(X)$
for every $t\in\mathbb{R}$. Then,
$d(\phi_{t}(x),\phi_{s(t)}(y))\leq\delta^{\prime}diam(X)$ for every
$t\in\mathbb{R}$. Taking $a=\phi_{t}(x)$ and $b=\phi_{s(t)}(y)$ we get
$d(\phi_{t}(x),\phi_{s(t)}(y))\leq\delta dist(\phi_{t}(x),Sing(\phi))$ for all
$t\in\mathbb{R}$. and so
$\phi_{s(t_{0})}(y)\in\phi_{[t_{0}-\epsilon,t_{0}+\epsilon]}(x)$ for some
$t_{0}\in\mathbb{R}$ proving the assertion.

On the other hand, since $\Lambda$ is nonsingular,
$Sing(\phi|_{\Lambda})=\emptyset$ and so $Sing(\phi|_{\Lambda})$ is open.
Then, $\phi|_{\Lambda}$ is expansive by Lemma 9 proving the result. ∎

This corollary motivates the question if conversely every flow which is
expansive on every nonsingular compact invariant set is singular-expansive.
But the answer is negative by the following example.

######  Example 3.

There is a compact metric space exhibiting a flow which not singular-expansive
but expansive on every nonsingular compact invariant set.

###### Proof.

Following the ideas of Example 1 we define

$X=\\{(0,0)\\}\cup\bigcup_{n\in\mathbb{N}}\\{z\in\mathbb{R}^{2}:\|z\|=n^{-1}\\}.$

Again $X$ is a compact metric space if equipped with the Euclidean metric.
Once more we consider the flow $\phi$ on $X$ obtained by restricting that of
the vector field in $\mathbb{R}^{2}$ defined by $V(x,y)=(-y,x)$ on $X$. As in
Example 1 we have that

$\phi_{t}(z)=(-y\sin t+x\cos t,x\sin t+y\cos y),\quad\quad\forall z=(x,y)\in
X,\quad t\in\mathbb{R}.$

Then, $Sing(\phi)=\\{(0,0)\\}$ and $\phi_{t}$ is a linear isometry for every
$t\in\mathbb{R}$. Now take $\epsilon=1$. Define the sequences
$z_{n},z_{n}^{\prime}\in X$ by $z_{n}=(\frac{1}{n},0)$ and
$z_{n}^{\prime}=(\frac{1}{n+1},0)$ for $n\in\mathbb{N}$. Notice that $z_{n}$
and $z_{n}^{\prime}$ belong to different circles of $X$ and so their orbits
are different. From this we obtain $z_{n}^{\prime}\notin\phi_{[-1,1]}(z_{n})$.
On the other hand,

$d(\phi_{t}(z_{n}),\phi_{t}(z_{n}^{\prime}))=\|z_{n}-z_{n}^{\prime}\|=\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}<\frac{1}{n^{2}},$

and $dist(\phi_{t}(z_{n}),Sing(\phi))=\|z_{n}-(0,0)\|=\frac{1}{n}$ so

$d(\phi_{t}(z_{n}),\phi_{t}(z_{n}^{\prime}))<\frac{1}{n}dist(\phi_{t}(z_{n}),Sing(\phi)),\quad\quad\forall
t\in\mathbb{R}.$

Since $z_{n}$ and $z_{n}^{\prime}$ belong to different circles, $\phi$ is not
singular-expansive. Finally, since every nonsingular compact invariant set
consists of finitely many periodic orbits, one has that $\phi$ is expansive on
all such sets. This completes the proof. ∎

The lemma below will be used to prove Theorem 5.

######  Lemma 11.

Let $\phi$ be a flow of a compact metric space $X$. If for every $\epsilon>0$
there is $\delta>0$ such that $B[x,\delta
dist(x,Sing(\phi))]\subset\phi_{[-\epsilon,\epsilon]}(x)$ for every $x\in X$,
then $\phi$ is both singular-expansive and singular-equicontinuous.

###### Proof.

Let $\epsilon>0$ and $\delta>0$ be given by the condition. If
$s:\mathbb{R}\to\mathbb{R}$ is an increasing homeomorphism satisfying
$d(\phi_{t}(x),\phi_{s(t)}(y))\leq\delta dist(\phi_{t}(x),Sing(\phi))$ for all
$t\in\mathbb{R}$, then $\phi_{s(0)}(y)\in B(x,\delta dist(x,Sing(\phi)))$ and
so $\phi_{s(0)}(y)\in\phi_{[-\epsilon,\epsilon]}(x)$ by the condition.
Therefore, $\phi$ is singular-expansive.

To prove that $\phi$ is singular-equicontinuous, take $\epsilon>0$ and
$\eta>0$ such that if $y=\phi_{s}(x)$ with $|s|\leq\eta$, then
$d(\phi_{t}(x),\phi_{t}(y))\leq\epsilon$ for every $t\in\mathbb{R}$ (c.f. p.
181 in [10]). For this $\eta$ we take $\delta>0$ given by the condition.
Therefore, if $d(x,y)\leq\delta dist(x,Sing(\phi))$, that is $y\in B[x,\delta
dist(x,Sing(\phi))]$, then $y=\phi_{s}(x)$ for some $|s|\leq\eta$ thus
$d(\phi_{t}(x),\phi_{t}(y))\leq\epsilon$ for every $t\in\mathbb{R}$ proving
the result. ∎

###### Proof of Theorem 4.

Let $\phi$ be a singular-expansive flow of a compact metric space $X$. Given
$\delta>0$ we denote by $U_{\delta}(Sing(\phi))$ the open $\delta$-ball around
$Sing(\phi)$. Define

(3) $X_{\delta}=\bigcap_{t\in\mathbb{R}}\phi_{t}(X\setminus
U_{\delta}(Sing(\phi))).$

It follows that $X_{\delta}$ is a compact invariant set without singularities
of $\phi$. Since $\phi$ is singular-expansive, $\phi$ is expansive on
$X_{\delta}$ by Lemma 10. On the other hand, as is well known [10], the set of
periodic orbits of an expansive flow is countable. Since the periodic orbits
of $\phi$ are contained in $\bigcup_{n\in\mathbb{N}}X_{\frac{1}{n}}$, we
conclude that the set of periodic orbits of $X$ is countable. This proves Item
(1).

To prove Item (2) we see that if $Sing(\phi)=\emptyset$ or consists of
finitely many isolated points of $X$, then $Sing(\phi)$ is open. Therefore,
$\phi$ is expansive on $X\setminus Sing(\phi)$ by Item (2) of Lemma 9. Since
$X\setminus Sing(\phi)$ and $Sing(\phi)$ are closed disjoint, we conclude that
$\phi$ is expansive. This proves Item (2).

To prove Item (3) we further assume that $Sing(\phi)$ is isolated. Fix $t>0$
and suppose by contradiction that $\phi$ has infinitely many distinct periodic
orbits $O_{n}$ with period $t_{n}\leq t$. If
$inf_{n\in\mathbb{N}}dist(O_{n},Sing(\phi))>0$, then
$\bigcup_{n\in\mathbb{N}}O_{n}\subset X_{\delta}$ for some $\delta>0$ which is
a contradiction since $\phi$ is expansive on $X_{\delta}$ (see [10]). Then, we
can assume that there is a sequence $x_{n}\in O_{n}$ and $\sigma\in
Sing(\phi)$ such that $x_{n}\to\sigma$. Since $\sigma\in Sing(\phi)$ and the
period of $O_{n}$ is bounded by $t$, we have that the whole $O_{n}\to\sigma$
with respect to the Hausdorff metric of compact subsets of $X$. In particular,
$diam(O_{n})\to 0$ contradicting Lemma 7. This completes the proof. ∎

###### Proof of Theorem 5.

Let $X=[0,1]$ be the unit interval endowed with the Euclidean metric. For
every $\lambda\in\mathbb{R}$ we define $\phi:\mathbb{R}\times X\to X$ by

$\phi_{t}(x)=\frac{xe^{\lambda t}}{1+x(e^{\lambda t}-1)},\quad\quad\forall
0\leq x\leq 1,t\in\mathbb{R}.$

It is not difficult to see that $\phi$ is a flow of $X$. If $\lambda=0$, then
$\phi$ is the trivial flow. If $\lambda\neq 0$, then $Sing(\phi)=\\{0,1\\}$
and the remainder orbits go from $0$ to $1$ or viceversa depending on whether
$\lambda>0$ or $\lambda<0$. Then, $\phi$ is Morse-Smale and so it has the
shadowing property but is not equicontinuous. We shall prove that if
$\lambda=1$, then $\phi$ satisfies the condition in Lemma 11.

First note that $dist(z,Sing(\phi))=\min\\{z,1-z\\}$ for every $z\in X.$ It
follows that $dist(z,Sing(\phi))=z$ (if $z\leq\frac{1}{2}$) or $1-z$
(otherwise).

Take $\epsilon>0$ and $0<\delta<\frac{1}{2}$ such that

$\ln\left(\frac{1+\delta}{1-\delta}\right)\leq\epsilon.$

We will show that

$|x-y|\leq\delta\min\\{x,1-x\\}\quad\Rightarrow\quad
y\in\phi_{[-\epsilon,\epsilon]}(x).$

Notice that the left-hand side of the above implication is equivalent to

$x-\delta\min\\{x,1-x\\}\leq y\leq x+\delta\min\\{x,1-x\\}.$

Since $\delta<\frac{1}{2}$, one has $y\in X\setminus Sing(\phi)$.

Assume $x\leq\frac{1}{2}$. It follows that

$(1-\delta)\leq\frac{y}{x}\leq(1+\delta)\quad\mbox{ and
}\quad\frac{1-x}{1-x(1-\delta)}\leq\frac{1-x}{1-y}\leq\frac{1-x}{1-x(1+\delta)}.$

Since $0<x\leq\frac{1}{2}$, $0<\frac{x}{1-x}\leq 1$ and then

$\frac{1-x}{1-x(1-\delta)}=\frac{(1-x)}{(1-x)+x\delta}=\frac{1}{1+\left(\frac{x}{1-x}\right)\delta}\geq\frac{1}{1+\delta}.$

Likewise,

$\frac{1-x}{1-x(1+\delta)}\leq\frac{1}{1-\delta}$

so

$\frac{1}{1+\delta}\leq\frac{1-x}{1-y}\leq\frac{1}{1-\delta}$

thus

$\ln\left(\frac{1-\delta}{1+\delta}\right)\leq\ln\left(\frac{y}{x}\cdot\frac{1-x}{1-y}\right)\leq\ln\left(\frac{1+\delta}{1-\delta}\right).$

On the other hand, it follows from the definition of $\phi$ that the equation
$\phi_{t}(x)=y$ is solved by

$t=\ln\left(\frac{y}{x}\cdot\frac{1-x}{1-y}\right).$

Then, the choice of $\delta$ implies

$-\epsilon\leq t\leq\epsilon$

yielding $y\in\phi_{[-\epsilon,\epsilon]}(x)$ for $x\leq\frac{1}{2}$.
Interchanging the roles of $x$ and $y$ above by $1-x$ and $1-y$ respectively
we get $y\in\phi_{[-\epsilon,\epsilon]}(x)$ when $x>\frac{1}{2}$ too.
Therefore, $\phi$ satisfies the condition in Lemma 11, and so, it is both
singular-expansive and singular-equicontinuous. This completes the proof. ∎

###### Proof of Theorem 6.

First suppose that $\Omega(\phi)\setminus Sing(\phi)$ is closed. Then,
$K=\Omega(\phi)\setminus Sing(\phi)$ is compact contained in $X\setminus
Sing(\phi)$ hence $h(\phi,K)\leq h^{*}(\phi)$. On the other hand, by well-
known properties of the topological entropy (p. 403 in [9]),
$h(\phi)=h(\phi,X)=h(\phi,K\cup
Sing(\phi))\leq\max\\{h(\phi,K),h(\phi,Sing(\phi))\\}=h(\phi,K)\leq
h^{*}(\phi)$. Since $h^{*}(\phi)\leq h(\phi)$, we are done.

Now suppose that $\phi$ is singular-equicontinuous. Take $K\subset X\setminus
Sing(\phi)$ compact and $\epsilon>0$. For this $\epsilon$ let $\delta$ be
given by the singular-equicontinuity of $\phi$. Since $K\cap
Sing(\phi)=\emptyset$, $\delta dist(x,Sing(\phi))>0$ for every $x\in K$. Then,
since $K$ is compact, there is $F\subset K$ finite such that

$K\subset\bigcup_{x\in F}B[x,\delta dist(x,Sing(\phi))].$

Now take $y\in K$ and $t>0$. Then,

$d(x,y)\leq\delta dist(x,Sing(\phi))$

for some $x\in F$, and so, by singular-equicontinuity,

$d(\phi_{l}(x),\phi_{l}(y))\leq\epsilon\quad\quad\forall 0\leq l\leq t.$

It follows that $F$ is $(t,\epsilon)$-spanning for every $t>0$ thus
$r(t,\epsilon)\leq car(F)$ the cardinality of $F$ for every $t>0$. Therefore,

$\limsup_{t\to\infty}\frac{1}{t}\ln
r(t,\epsilon)\leq\limsup_{t\to\infty}\frac{\ln
car(F)}{t}=0,\quad\quad\forall\epsilon>0.$

Then, $h(\phi,K)=0$ for every compact subset $K\subset X\setminus Sing(\phi)$
so $h^{*}(\phi)=0$. ∎

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