MnLargeSymbols’164 MnLargeSymbols’171

# Existence of the carrying simplex for a $C^{1}$ retrotone map

Stephen Baigent
Department of Mathematics
University College London
Gower Street, London WC1E 6BT
United Kingdom of Great Britain and Northern Ireland
<EMAIL_ADDRESS>
and
Janusz Mierczyński
Faculty of Pure and Applied Mathematics
Wrocław University of Science and Technology
Wybrzeże Wyspiańskiego 27
PL 50-370 Wrocław
Poland
<EMAIL_ADDRESS>
###### Abstract

We present a dynamical approach to the study of decreasing, attracting
manifolds of retrotone maps commonly known as carrying simplices. Our approach
is novel in that it uses the radial representation of decreasing manifolds
over the probability simplex coupled with distances between these manifolds
measured by way of the Harnack and Hausdorff metrics. We establish Kuratowski
convergence of radial representations of decreasing manifolds to a unique
function which then provides the locally Lipschitz radial representation of
the carrying simplex.

## Introduction

There is a large class of maps widely used to study discrete-time population
dynamics that preserve all faces of the first orthant of Euclidean space that
are known as Kolmogorov maps. By varying the functional forms of these maps,
all the possible types of species interactions for populations where there is
no overlap of generations can be modelled, including predator-prey, mutualism
and competition to name a few. Here we are interested in a special subclass of
those Kolmogorov maps that feature only competition which, following recent
research trends, we call retrotone. While much of the existing research uses
competitive map where we use retrotone map \- indeed the distinction is not
crucial (necessary) for continuous time analogues - we prefer to use
’retrotone’ to describe maps that represent competitive interactions that also
have the specific feature that they preserve a convex cone backwards in time.
The Leslie–Gower map is an example of a competition model that is globally a
retrotone map [15], but not all competitive maps are retrotone, even on their
global attractor; the planar Ricker map discussed later in 3.1.3 is a classic
example of a map that represents competitive interactions, but is only
retrotone for a limited set of parameter values.

The focus of the present work is the carrying simplex which is a well-studied
feature of population models with competitive interactions [18, 21, 8, 23, 5,
9, 20]. The carrying simplex is a Lipschitz, codimension-one and compact
invariant manifold that attracts all nonzero points, and that is unordered,
which means that the carrying simplex is decreasing in each coordinate
direction. As we discuss later there are several ways of characterising the
carrying simplex, including as the set of nonzero points with globally defined
forward and backward orbits, or as the relative boundary of the global
attractor of bounded sets. The presence of the carrying simplex (which is
asymptotically complete) means that the limiting dynamics can be studied on
the carrying simplex which gives rise to a system of one fewer degree of
freedom. This has been exploited by a number of authors to study the global
dynamics of both continuous and discrete-time population models. For example,
[15, 14, 7] study global dynamics of maps with a carrying simplex by way of an
index theorem. Baigent and Hou [2, 3] utilise the carrying simplex in both
continuous and discrete time models to construct Lyapunov functions on forward
invariant sets, and Montes de Oca and Zeeman [17] use the carrying simplex
concept iteratively to reduce a continuous-time competition model to an easily
solved one-dimensional model.

That there exists such an invariant manifold under ecologically reasonable
assumptions is intuitive. We impose that the origin is unstable, which means
that small population densities grow; all the species cannot simultaneously go
extinct. We also require that the per-capita grow rate of a given species
decreases with increasing densities of all species (although we relax this to
non-positive for some densities later). This can be thought of as modelling
competition for resources. In differential equation models for competition,
these assumptions are typically sufficient for a carrying simplex to exist [8,
11], but not in the discrete time case. We follow other authors [23, 9, 20,
12] and add further conditions that render the map retrotone. Roughly speaking
the retrotone property, which can be imposed through spectral properties of
the derivative of the map [9, 20, 12], aside from modelling competitive
interactions, puts restrictions on the maximal change in total population
density over one generation, i.e. large radial changes under one application
of the map are not permitted. The retrotone properties of the map render the
carrying simplex unordered, so that it projects radially onto the probability
simplex. Hence on the carrying simplex, given the frequency of the species,
i.e. a point in the probability simplex, the radial coordinate, which is the
total population density, is determined.

We argue that the radial representation, in which phase space is the cartesian
product of the probability simplex and the positive real line, is a natural
coordinate choice, and it is the main set of coordinates that we use here. In
this description the carrying simplex is just the graph of a locally Lipschitz
function over the probability simplex. However, working with the radial
representation presents technical difficulties at the boundary of phase space
where derivatives can become unbounded. To resolve this issue we use the
Kuratowski metric to establish Hausdorff convergence of the unordered
manifolds generated by the graph transform approach. We generate one
increasing and one decreasing sequence of Kuratowski convergent sequences of
unordered manifolds and then we utilise the Harnack metric to show that the
two limits are identical and identified as the carrying simplex.

The paper is organized as follows. In section 1 we introduce our notation and
give important definitions, such as for cone-orderings, unordered and weakly
unordered sets, retrotone maps and weakly retrotone maps, and attractors of
various classes of sets. Section 2 is concerned with proof of the existence of
the carrying simplex. Subsection 2.1 deals with the dynamics of the map
restricted to the boundary, and is crucial for establishing the existence of a
bounded rectangle on which the map is retrotone/weakly retrotone and which
contains the unique compact attractor of bounded sets. That the map is
retrotone/weakly retrotone on the bounded rectangle is established in
subsection 2.2 and subsection 2.3 characterizes the compact attractor of
bounded sets of the map. In subsection 2.4 we construct the carrying
simplex/weak carrying simplex from sequences of unordered/weakly unordered
manifolds using the graph transform and Kuratowski convergence radial
coordinates, as well as give some of its properties. Attractor-repeller
pairings are used to establish further properties of the carrying simplex/weak
carrying simplex in subsections 2.5, and subsections 2.6 and 2.7 deal with
asymptotic completeness and a Lipschitz representation of the carrying
simplex. Lastly, in section 3 we discuss some examples that illustrate the
main ideas of the paper.

##  1 Notation and definitions

We will need the following notation and definitions.

$\lVert\cdot\rVert$ stands for the Euclidean norm in $\mathbb{R}^{d}$. For
$x,y\in^{d}$, $x\cdot y$ denotes the standard inner product
$\sum_{i=1}^{d}x_{i}y_{i}$ in $\mathbb{R}^{d}$. $\lVert\cdot\rVert_{1}$
denotes the $\ell_{1}$-norm in $\mathbb{R}^{d}$, and
$\lVert\cdot\rVert_{\infty}$ denotes the $\ell_{\infty}$-norm in
$\mathbb{R}^{d}$.

For $x\in\mathbb{R}^{n}$ and $A\subset\mathbb{R}^{n}$ we write

$\operatorname{dist}(x,A):=\inf\,\\{\,\lVert x-a\rVert:a\in A\,\\}.$

For nonempty compact $A,B\subset\mathbb{R}^{n}$ we denote by $d_{H}(A,B)$
their _Hausdorff distance_ ,

$d_{H}(A,B):=\max\\{\sup\,\\{\,\operatorname{dist}(a,B):a\in
A\,\\},\sup\,\\{\,\operatorname{dist}(b,A):b\in B\,\\}\\}.$

$C_{+}:=\\{x\in^{d}:x_{i}\geq 0,\ i=1,\ldots,d\\}$, and
$C_{++}:=\\{x\in^{d}:x_{i}>0,\ i=1,\ldots,d\\}$ will denote convex cones.
$C_{+}$ is often referred to as the _first orthant_ and $C_{++}$ is its
interior. $\partial C_{+}=C_{+}\setminus C_{++}$ is called the _boundary_ of
$C_{+}$ (indeed, it is the boundary of $C_{+}$ in $\mathbb{R}^{d}$).

$\mathbf{e}_{i}$ is the vector with a one at the $i$-th position and zeros
elsewhere and $\mathbf{e}=\sum_{i=1}^{d}\mathbf{e}_{i}$,
$\hat{\mathbf{e}}=\mathbf{e}/\sqrt{d}$. $\mathbb{N}=\\{0,1,2,3,\dots\\}$.

For a subset $I\subset\\{1,\ldots,d\\}$, let $(C_{I})_{+}:=\\{\,x\in
C_{+}:x_{i}=0$ for all $i\in\\{1,\ldots,d\\}\setminus I\,\\}$ denote a _$k$
-dimensional face_ of $C_{I}$, where $k=\operatorname{card}{I}$, and let
$(C_{I})_{++}:=\\{\,x\in(C_{I})_{+}:x_{i}>0$ for all $i\in I\,\\}$ denote the
relative interior of $(C_{I})_{+}$. A $1$-dimensional face is referred to as
an $i$-th axis, where $I=\\{i\\}$. Instead of $(C_{\\{i\\}})_{+}$, etc., we
write $(C_{i})_{+}$, etc. For $x\in C_{+}$ let $I(x)$ stand for the (unique)
subset of $\\{1,\ldots,d\\}$ such that $x\in(C_{I(x)})_{++}$. Often $I(x)$ is
called the _support_ of $x$.

For $A\subset C_{+}$, $\operatorname{Int}_{C_{+}}{A}$ stands for the relative
interior of $A$ in $C_{+}$, and $\operatorname{bd}_{C_{+}}{A}$ stands for the
relative boundary of $A$ in $C_{+}$. $\bar{A}$ denotes the closure of $A$.

$\Delta:=\\{x\in C_{+}:\sum_{i=1}^{d}x_{i}=1\\}$ denotes the standard
probability $(d-1)$-simplex. For $I\subset\\{1,\ldots,d\\}$,
$\Delta_{I}:=\Delta\cap(C_{I})_{+}$.

For $x,y\in C_{+}$, we write $x\leq y$ if $x_{i}\leq y_{i}$ for all
$i=1,\ldots,d$, and $x\ll y$ if $x_{i}<y_{i}$ for all $i=1,\ldots,d$. If
$x\leq y$ but $x\neq y$ we write $x<y$. The reverse relations are denoted by
$\geq,>,\gg$. Two points $x,y\in C_{+}$ are said to be _order-related_ if
either $x\leq y$ or $y\leq x$.

For $x,y\in C_{+}$ such that $x\leq y$ we define the order interval as

$[x,y]:=\\{\,z\in C_{+}:x\leq z\leq y\,\\}.$

We say $B\subset C_{+}$ is order convex if for any $x,y\in B$ with $x\leq y$
one has $[x,y]\subset B$.

Let $\emptyset\neq I\subset\\{1,\dots,d\\}$. For $x,y\in(C_{I})_{+}$, we write
$x\leq_{I}y$ if $x_{i}\leq y_{i}$ for all $i\in I$, and $x\ll_{I}y$ if
$x_{i}<y_{i}$ for all $i\in\\{1,\dots,d\\}$. If $x\leq_{I}y$ but $x\neq y$ we
write $x<_{I}y$. The reverse relations are denoted by
$\geq_{I},>_{I},\gg_{I}$.

For $I\subset\\{1,\dots,d\\}$, by $\pi_{I}$ we understand the orthogonal
projection of $C_{+}$ onto $(C_{I})_{+}$.

$\Pi$ denotes the orthogonal projection along $\mathbf{e}$ onto
$V:=\mathbf{e}^{\perp}$. $\Pi
u=u-(u\cdot\hat{\mathbf{e}})\hat{\mathbf{e}}=u-(u\cdot\mathbf{e})\mathbf{e}/d$.

We will consider maps of class $C^{k}(\Omega)$, where $k=1,2$, defined either
on $\Omega=C_{+}$ or on $\Omega=\Delta$. The standard definition for an
element of $C^{k}(\Omega)$ is that it can be extended to a $C^{k}$ mapping on
an open set containing $\Omega$. Indeed, since $C_{+}$ and $\Delta$ are
sufficiently regular, it follows from Whitney’s extension theorem (see, e.g.,
[19]) that the definitions are just the same as in the case of functions
defined on open sets, with derivatives replaced by one-sided derivatives where
necessary.

For a differentiable map $F$, the Jacobian matrix of $F$ at the point $x$ is
denoted by $DF(x)$. If $F$ is invertible with differentiable inverse $F^{-1}$
then by $DF^{-1}$ we mean the derivative of $F^{-1}$; this matrix is to be
distinguished from $(DF)^{-1}$, the matrix inverse of $DF$. For
$x\in\mathbb{R}^{d}$, $\operatorname{diag}[x]$ denotes the $d\times d$
diagonal matrix with diagonal elements $x_{1},\ldots,x_{d}$.

######  Definition 1.1.

A set $B\subset C_{+}$ is said to be _unordered_ if no two distinct points of
$B$ are ordered by the $<$ relation.

A set $B\subset C_{+}$ is said to be _weakly unordered_ if for any
$\emptyset\neq I\subset\\{1,\ldots,d\\}$ no two distinct points of
$B\cap(C_{I})_{++}$ are ordered by the $\ll_{I}$ relation.

The concept of a weakly unordered set (although not the name) was introduced
in [12, Remark 2.1(f)].

######  Example 1.2.

The standard probability simplex $\Delta\subset C_{+}$ is unordered.

######  Example 1.3.

Let $d>1$. For $a>0$ consider the set

$\displaystyle H(a)$ $\displaystyle:=\operatorname{bd}_{C_{+}}[0,a\mathbf{e}]$
$\displaystyle=\\{\,x\in C_{+}:\forall i\in\\{1,\ldots,d\\}\,0\leq x_{i}\leq
a,\ \exists k\in\\{1,\ldots,d\\}\text{ such that }x_{k}=a\,\\},$

where $a>0$. The set $H(a)$ is weakly unordered. To show this, suppose to the
contrary that there are a nonempty $I\subset\\{1,\dots,d\\}$ and $x,y\in
H\cap(C_{I})_{++}$ with $x\ll_{I}y$, which means that $x_{i}<y_{i}$ for all
$i\in I$ and $x_{j}=y_{j}=0$ for all $j\in\\{1,\dots,d\\}\setminus I$. But
then $x_{i}<a$ for all $1\leq i\leq d$, so $x$ cannot belong to $H(a)$.

On the other hand, $H(a)$ is not unordered: for example,
$a\mathbf{e}_{1}<a\mathbf{e}$ and both belong to $H(a)$.

######  Definition 1.4.

A map $F\colon C_{+}\to C_{+}$ is retrotone in a subset $B\subset C_{+}$, if,
for all $x,y\in B$ with $F(x)<F(y)$, one has that $x_{i}<y_{i}$ provided
$y_{i}>0$.

A map $F\colon C_{+}\to C_{+}$ is _weakly retrotone_ in $B\subset C_{+}$
provided for all $I\subset\\{1,\ldots,d\\}$ and any $x,y\in B$, if

$F(x)<F(y)\quad\text{and}\quad F_{i}(x)<F_{i}(y)\text{ for all }i\in I$

then

$x<y\quad\text{and}\quad x_{i}<y_{i}\text{ for all }i\in I.$

The name ‘retrotone’ in the context of competitive maps appeared first in [9]
and later in [20], [12] (however, notice that in [9] our ‘retrotone’ is called
‘strictly retrotone’). The term ‘weakly retrotone’ was introduced in [12].

From now on, we assume that $F\colon C_{+}\to C_{+}$ is a continuous map. We
will consider the dynamical system $(F^{n})_{n=0}^{\infty}$ on $C_{+}$, where
$F^{0}=\mathrm{Id}_{C_{+}}$.

We write $F^{n}=(F_{1}^{n},\ldots,F_{d}^{n})$.

For $x\in C_{+}$ we denote its forward orbit, $O^{+}(x)$, as

$O^{+}(x):=\\{\,F^{n}(x):n\in\mathbb{N}\,\\}.$

$A\subset C_{+}$ is forward invariant if $F(A)\subset A$, and invariant if
$F(A)=A$.

By a backward orbit of $x\in C_{+}$ we understand a set
$\\{\dots,x_{-n-1},x_{-n},\dots,\break x_{-2},x_{-1},x_{0}\\}$ such that
$x_{0}=x$ and $x_{-n}=F(x_{-n-1})$ for all $n\in\mathbb{N}$. A full orbit of
$x\in C_{+}$ is the union of a backward orbit of $x$ and the forward orbit,
$O^{+}(x)$.

In what follows, we use the terminology as in [22].

We say that $B\subset C_{+}$ attracts the set $A\subset C_{+}$ if for each
$\epsilon>0$ there is $n_{0}\in\mathbb{N}$ such that
$\operatorname{dist}(F^{n}(x),B)<\epsilon$ for all $n\geq n_{0}$ and all $x\in
A$.

For a set $A\subset C_{+}$ define its _$\omega$ -limit set_ as

$\omega(A):=\bigcap\limits_{k=0}^{\infty}\overline{\left(\bigcup\limits_{l=k}^{\infty}F^{l}(A)\right)}.$

The dynamical system $(F^{n})_{n=0}^{\infty}$ is said to be asymptotically
compact on $A\subset C_{+}$ if for any sequence $(n_{k})_{k=0}^{\infty}$,
$n_{k}\to\infty$, and any sequence $(x_{k})_{k=0}^{\infty}\subset A$, the
sequence $(F^{n_{k}}(x_{k}))_{k=0}^{\infty}$ has a convergent subsequence.

######  Lemma 1.5.

Assume that $(F^{n})_{n=0}^{\infty}$ is asymptotically compact on $A$. Then
$B$ attracts $A$ if and only if $\omega(A)\subset B$.

###### Proof.

See [22, Lemmas 2.6 and 2.8]. ∎

Let $B\subset C_{+}$ be forward invariant. By the compact attractor of bounded
sets in $B$ we mean a nonempty compact invariant set $\Gamma\subset B$ that
attracts any bounded $A\subset B$. Such a set is unique (see [22, Theorem 2.19
on p. 37]). By the compact attractor of neighbourhoods of compact sets in $B$
we mean a nonempty compact invariant set $\Gamma\subset B$ such that for any
compact $A\subset B$ there is a (relative) neighbourhood $U$ of $A$ in $B$
such that $\Gamma$ attracts $U$. Such a set is unique (see [22, Theorem 2.19
on p. 37]).

For $B=C_{+}$ we say simply compact attractor of bounded sets, which is the
same as the compact attractor of neighbourhoods of compact sets.

######  Proposition 1.6.

Let $B\subset C_{+}$ be forward invariant. The compact attractor of bounded
sets is characterized as the set of all $x\in B$ having bounded full orbits.

###### Proof.

See [22, Theorem 2.20 on p. 37]. ∎

A compact set $B\subset C_{+}$ is called a _uniform repeller_ in $C_{+}$ if
there is $\epsilon>0$ such that
$\liminf\limits_{n\to\infty}\operatorname{dist}(F^{n}(x),B)\geq\epsilon$ for
any $x\in C_{+}$. It is equivalent to the existence of a neighbourhood $U$ of
$B$ in $C_{+}$ such that for each $x\in U\setminus B$ there is
$n_{0}\in\mathbb{N}$ with the property that $F^{n}(x)\notin U$ for all $n\geq
n_{0}$, see [22, Remark 5.15 on p. 136].

We say that $F\colon C_{+}\to C_{+}$ is a _Kolmogorov map_ if
$F=\operatorname{diag}[\mbox{id}]f$ where $f\colon C_{+}\to C_{+}$.

######  Definition 1.7.

The _carrying simplex_ [resp.  _weak carrying simplex_] for a Kolmogorov map
$F\colon C_{+}\to C_{+}$ is a subset $\Sigma\subset C_{+}$ with the following
properties:

  1. (I)

$\Sigma$ is an unordered [resp. weakly unordered] subset of the compact
attractor of bounded sets $\Gamma$.

  2. (II)

$\Sigma$ is homeomorphic via radial projection to the $(d-1)$-dimensional
standard probability simplex $\Delta$.

  3. (III)

$F(\Sigma)=\Sigma$ and $F|_{\Sigma}\colon\Sigma\to\Sigma$ is a homeomorphism.

  4. (IV)

$\Sigma$ is the boundary (relative to $C_{+}$) of $\Gamma$ and $\Gamma$ is
order convex. In particular, $\Gamma=\\{\alpha
x:\alpha\in[0,1],x\in\Sigma\\}$.

  5. (V)

$\Sigma$ attracts any bounded $A\subset C_{+}$ with $0\notin\bar{A}$.

  6. (VI)

$\Gamma\setminus\Sigma=\\{\alpha x:\alpha\in[0,1),x\in\Sigma\\}$ is
characterized as the set of all those $x\in C_{+}$ that have a backward orbit
$\\{\ldots,x_{-2},x_{-1},x\\}$ with $\lim\limits_{n\to\infty}x_{-n}=0$.

  7. (VII)

$\Sigma$ is characterized as the set of all $x\in C_{+}$ having full orbits
that are bounded and bounded away from $0$.

  8. (VIII)

For any $x\in C_{+}\setminus\\{0\\}$ there is $y\in\Sigma$ such that
$\displaystyle\lim\limits_{n\to+\infty}\lVert F^{n}(x)-F^{n}(y)\rVert=0$ (this
property is called _asymptotic completeness_ or _asymptotic phase_) [resp. for
any $x\in\Gamma\setminus\\{0\\}$ there is $y\in\Sigma$ such that
$\displaystyle\lim\limits_{n\to+\infty}\lVert F^{n}(x)-F^{n}(y)\rVert=0$.

  9. (IX)

The inverse $(\Pi|_{\Sigma})^{-1}$ of the orthogonal projection of $\Sigma$
along $\mathbf{e}$ is Lipschitz continuous.

##  2 Existence of a carrying simplex

We split our proof of existence of a carrying simplex into two steps.

The definition of a carrying simplex given in 1.7 asks that the set $\Sigma$
attracts all of $C_{+}\setminus\\{0\\}$. In the first step we relax this
condition, replacing the unbounded rectangle $C_{+}$ by a bounded rectangle
$\Lambda$, so that $F$ restricted to $\Lambda$, $F|_{\Lambda}$, satisfies
definition 1.7 with instances of $C_{+}$ replaced by $\Lambda$. The main
result is Proposition 2.1 which shows a carrying simplex exists in $\Lambda$
under the assumptions AS1–AS4. We work this way because not all maps with a
carrying simplex are retrotone on all of $C_{+}$ (an example is given in
subsection 3.2).

In the second step we show that all points in $C_{+}$ eventually enter and
stay in $\Lambda$, so that the carrying simplex in $\Lambda$ is actually a
carrying simplex in $C_{+}$. In the second step the map is not required to be
retrotone outside $\Lambda$.

Recall that the spectral radius $\rho(P)$ of a square matrix $P$ is the
modulus of the eigenvalue with maximum modulus.

We make the following assumptions:

Let $F\colon C_{+}\to C_{+}$ be a Kolmogorov map
$F:=\operatorname{diag}[\mbox{id}]f$ where $f:C_{+}\to C_{+}$ satisfies

  * AS1

$f$ is of class $C^{1}$;

  * AS2

$f_{i}(\mathbf{e}_{i})=1$, $i=1,\ldots,d$;

  * AS3

$f(x)\gg 0$ and $Df(x)\leq 0$ with its diagonal terms negative, for all $x\in
C_{+}$ ;

  * AS4

the $d\times d$ matrix $Z(x)=\left(\left(-\frac{x_{i}}{f_{i}(x)}\frac{\partial
f_{i}(x)}{\partial x_{j}}\right)\right)$ has spectral radius $\rho(Z(x))<1$
for all $x\in\prod_{i=1}^{d}[0,1]\setminus\\{0\\}$.

Sometimes instead of AS3 we make the following stronger assumption:

  1. AS3′

$f(x)\gg 0$ and $Df(x)\ll 0$ for all $x\in C_{+}$.

######  Remark 2.1.

From the form $\operatorname{diag}[\mathrm{id}]f$ of $F$ in AS1 it follows
that for any $\emptyset\neq I\subset\\{1,\ldots,d\\}$ there holds
$F((C_{I})_{++})\subset(C_{I})_{++}$ and
$F^{-1}((C_{I})_{++})\subset(C_{I})_{++}$.

######  Remark 2.2.

In AS2, we assume that each axis has a fixed point of $F$ at $\mathbf{e}_{i}$,
but by rescaling we may deal with fixed point at $q_{i}\mathbf{e}_{i}$ for any
set of $q_{i}>0$, $i=1,\ldots,d$.

######  Remark 2.3.

Observe that AS4 is equivalent to saying that for any
$x\in\prod_{i=1}^{d}[0,1]\setminus\\{0\\}$ the
$\operatorname{card}{I(x)}\times\operatorname{card}{I(x)}$ matrix
$Z(x):=\left(-\frac{x_{i}}{f_{i}(x)}\frac{\partial f_{i}(x)}{\partial
x_{j}}\right)_{i,j\in I(x)}$ has spectral radius less than $1$.

######  Remark 2.4.

Regarding AS4, observe that for all $x\in C_{+}$ there holds

$DF(x)=\operatorname{diag}[f(x)](I-Z(x)).$ (2.1)

It follows from AS4, by the continuity of the spectral radius of a matrix,
that for any sufficiently small $\varkappa>0$ there holds $\rho(x)<1$ for all
$x\in\prod_{i=1}^{d}[0,1+\varkappa]\setminus\\{0\\}$.

We choose such a $\varkappa>0$, and put
$\Lambda:=\prod_{i=1}^{d}[0,1+\varkappa]$,
$\Lambda^{\prime}:=\Lambda\setminus\\{0\\}$.

######  Theorem 2.1.

Under the assumptions AS1–AS4 there exists a weak carrying simplex $\Sigma$.
If we assume additionally AS3′, $\Sigma$ is a carrying simplex.

The property of $\Sigma$ being unordered [weakly unordered] appears in all
papers dealing with the proofs of the existence of the CS. On the other hand,
$\Gamma$ is usually considered an attractor of points only. To the authors’
knowledge, the present paper is the first where the global attractor is proved
to be the attractor of bounded sets, although that was mentioned in [9]
without proof, as well as in [13]. The same holds for property (V).

Properties (II), (III) and (IV) appear in all papers (however, the order
convexity is not mentioned explicitly everywhere).

The characterization given in (VI) and (VII) appears occasionally explicitly
formulated, however usually it is an unstated consequence of the definition of
the carrying simplex as the upper (or the lower) boundary of the origin (or of
$\infty$). (VIII) appears in all papers, whereas (IX) is formulated and proved
in [8] only.

###  2.1 Restrictions of the dynamical system $(F^{n})_{n=0}^{\infty}$ to the
axes

For $F$ satisfying AS1–AS3 we define a map
$G_{i}\colon[0,\infty)\to[0,\infty)$, $i\in\\{1,\ldots,d\\}$, as
$G_{i}(s):=sg_{i}(s)$, where $g_{i}(s):=f_{i}(s\mathbf{e}_{i})$. The map
$G_{i}$ is nothing other than the dynamical rule $F$ restricted to the
invariant $i$-th axis.

$G^{n}_{i}$, for $n\in\mathbb{N}$, denotes the $n$-th iterate of $G_{i}$.
$(G_{i}^{n})_{n=0}^{\infty}$ is, up to renaming, the restriction of the
dynamical system $(F^{n})_{n=0}^{\infty}$ to the invariant set $(C_{i})_{+}$.

In the remainder of the present subsection, the terms from the theory of
dynamical systems, such as attractor, $\omega$-limit set, etc. will refer to
the dynamical system $(G_{i}^{n})_{n=0}^{\infty}$.

######  Lemma 2.2.

Under AS1–AS4, for $i\in\\{1,\ldots,d\\}$ the following holds.

  1. (i)

$1$ is the unique fixed point of $G_{i}$ on $(0,\infty)$.

  2. (ii)

$G_{i}(s)\begin{cases}\in(s,1)&\text{for }s\in(0,1),\\\ =s&\text{for }s=1\\\
\in(0,s)&\text{for }s\in(1,\infty).\end{cases}$

  3. (iii)

For any $s\in(0,\infty)$ the sequence $(G_{i}^{n}(s))$ converges, as
$n\to\infty$, to $1$ in an eventually monotone way.

  4. (iv)

For $s\in(1,1+\varkappa]$ there holds $G_{i}(s)\in(1,s)$, hence the sequence
$(G_{i}^{n}(s))$ strictly decreases to $1$.

###### Proof.

Fixed points of $G_{i}$ on $(0,\infty)$ are just the solutions of
$g_{i}(s)=1$. By AS2, $1$ is such a solution, and, by the second part of AS3,
the only one. This proves (i) as well as the second alternative (ii).

By the first part of AS3, $G_{i}(s)>0$ provided $s>0$. Furthermore, again by
$g_{i}(1)=1$ and the second part of AS3, $g_{i}(s)<1$, consequently
$G_{i}(s)<s$, for all $s>1$. This proves the third alternative (ii).

Now, let $s\in(0,1)$. It follows from AS4 that
$G^{\prime}_{i}(s)=sg^{\prime}_{i}(s)+g_{i}(s)>0$ for all $s\in(0,1]$.
Consequently, the function $G_{i}$ is strictly increasing on $[0,1]$. Further,
$g_{i}$ is strictly decreasing and $g_{i}(1)=1$, hence $g_{i}(0)>1$ on
$[0,1)$, which implies that $G_{i}(s)>s$ for $s\in(0,1)$. Since $G_{i}$ is
strictly increasing on $[0,1]$ and $G_{i}(1)=1$, $G_{i}(s)<1$ for $s\in(0,1)$.
This concludes the proof of (ii).

We proceed to proving (iii). If $s\in(0,1)$, the sequence $(G_{i}^{n}(s))$ is
strictly increasing and bounded, so it converges to the fixed point $1$. If
$s=1$, the sequence is constant. Assume that $s>1$. If $G_{i}^{n}(s)>1$ for
all $n$, the sequence $(G_{i}^{n}(s))$ is strictly decreasing and bounded, so
it converges to the fixed point $1$. If there is $n_{0}\geq 1$ such that
$G_{i}^{n_{0}}(s)=1$, then $G_{i}^{n}(s)=1$ for $n\geq n_{0}$. If there is
$n_{0}\geq 1$ such that $G_{i}^{n_{0}}(s)<1$, then the sequence is strictly
increasing, starting from $n_{0}$, to $1$.

(iv) follows since, by the definition of $\varkappa$,
$G^{\prime}_{i}(s)=sg^{\prime}_{i}(s)+g_{i}(s)>0$ for all
$s\in(0,1+\varkappa]$. ∎

The following result will be used several times, so we formulate it as a
separate lemma.

######  Lemma 2.3.

Assume AS1–AS4. Let $a\in[0,1+\varkappa]$. Then for each
$i\in\\{1,\ldots,d\\}$ and each $n\in\mathbb{N}$, $G_{i}^{n}$ is a strictly
increasing homeomorphism of $[0,a]$ onto $[0,G_{i}^{n}(a)]$.

###### Proof.

$G_{i}$ is strictly increasing on $[0,1+\varkappa]$ and
$G_{i}(1+\varkappa)<1+\varkappa$, so $G_{i}^{n}$ is strictly increasing on
$[0,1+\varkappa]$, and hence preserves the order $\leq$, for all
$n\in\mathbb{N}$. ∎

######  Proposition 2.4.

Assume AS1–AS4. Then for each $i\in\\{1,\ldots,d\\}$, the invariant set
$[0,1]$ is the compact attractor of bounded sets in $[0,\infty)$.

###### Proof.

By Lemma 2.2(iii), $[0,1]$ attracts any point $s\in[0,\infty)$ (in the
terminology of [22], $[0,1]$ is a compact attractor of points in
$[0,\infty)$). From Lemmas 2.3 and 2.2(iv) it follows that $[0,1]$ attracts
its open relative neighbourhood $[0,1+\varkappa)$. By [22, Theorem 2.31],
$[0,1]$ is the compact attractor of neighbourhoods of compact sets in
$[0,\infty)$, which is the same as the compact attractor of bounded sets in
$[0,\infty)$. ∎

###  2.2 Retrotonicity of the map $F$

Recall that a nonsingular $M$-matrix is a square matrix $P=\rho_{0}I-Z$ where
$Z$ is non-negative and $\rho_{0}$ exceeds the spectral radius of $Z$ and that
a $P$-matrix is a square matrix $P$ with positive principal minors. It is a
standard result that every nonsingular $M$-matrix is a $P$-matrix. Moreover,
every eigenvalue of a nonsingular $M$-matrix has positive real part and every
real eigenvalue of a $P$-matrix is positive [10].

In [6] Gale and Nikaidô proved an important result on the invertibility of
maps whose derivatives are $P$-matrices on rectangular subsets of
$\mathbb{R}^{d}$: If $\Omega\subset\mathbb{R}^{d}$ is a rectangle and
$F\colon\Omega\rightarrow^{d}$ is a continuously differentiable map such that
$DF(x)$ is a $P$-matrix for all $x\in\Omega$ then $F$ is one-to-one in
$\Omega$. If $\Omega$ is not rectangular, the map $F$ may not be one-to-one.

Our standing assumption in the present subsection is that $F$ satisfies
AS1–AS4. Recall that $\Lambda$ is chosen so that $\rho(Z(x))<1$ for all
$x\in\Lambda$.

######  Lemma 2.5.

  1. (i)

$F|_{\Lambda}$ a homeomorphism onto its image.

  2. (ii)

For $x\in\Lambda$, $(DF(x))_{ij}^{-1}\geq 0$ for all $i,j\in\\{1,\ldots,d\\}$;
moreover, for $x\in\Lambda^{\prime}$,

     * •

$(DF(x))^{-1}_{ii}>0$ for all $i\in I(x)$;

     * •

Under AS3′, $(DF(x))^{-1}_{ij}>0$ for all $i,j\in I(x)$.

###### Proof.

Let $x\in\Lambda^{\prime}$. Then $Z(x)$ is a non-negative matrix with
$\rho(Z(x))<1$. Including the origin where $P(0):=I$, $P(x):=I-Z(x)$ is an
$M$-matrix, and also a $P$-matrix for $x\in\Lambda$. Hence by Gale and
Nikaidô’s univalence theorem [6, Theorem 4], $F$ is a homeomorphism from
$\Lambda$ onto its image, showing (i). Since, by (2.1),
$DF=\operatorname{diag}[f](I-Z)$, $\rho(Z(x))<1$, $Z(x)_{ij}>0$ for $i,j\in
I(x)$, and $f(x)\gg 0$ we find that
$(DF(x))^{-1}=(I-Z(x))^{-1}\operatorname{diag}[f(x)]^{-1}=\left(\sum_{k=0}^{\infty}Z(x)^{k}\right)\operatorname{diag}[f(x)]^{-1}\geq
0$ for $x\in\Lambda$, and $(DF(x))^{-1}_{ii}>0$ for $i\in I(x)$ [under AS3′,
$(DF(x))^{-1}_{ij}>0$ for $i,j\in I(x)$], which shows (ii). ∎

######  Proposition 2.6.

  1. (1)

If $x,y\in\Lambda$ and $F(x)\leq F(y)$, then $x\leq y$ and $[F(x),F(y)]\subset
F([x,y])$, in particular $[0,F(x)]\subset F([0,x]),\;x\in\Lambda$;

  2. (2)

$F$ is weakly retrotone in $\Lambda$. If AS3′ holds, then $F$ is retrotone in
$\Lambda$.

###### Proof.

We start by proving (1). If $F(x)=F(y)$ then, by Lemma 2.5(i), $x=y$ and there
is nothing to prove. So assume $F(x)<F(y)$. Observe that $I(y)=I(F(y))$. We
can assume, for notational simplicity, that $I(y)=I(F(y))=\\{1,\dots,d\\}$,
that is, both $y$ and $F(y)$ are in $C_{++}$.

It follows from the Inverse Function Theorem that there is an open ball $B$
centred at $F(y)$ and contained in the interior of $F(\Lambda)$ such that for
each $\zeta\in B$ one has the following expression for
$(F|_{\Lambda})^{-1}(\zeta)$:

$y+\biggl{(}\int_{0}^{1}D(F|_{\Lambda})^{-1}(F(y)+t(\zeta-F(y)))\,\mathrm{d}t\biggr{)}\cdot(\zeta-F(y)).$

Take $z\in[F(x),F(y)]\setminus\\{F(y)\\}$, and consider the segment joining
$z$ to $F(y)$, parameterized by $t\in[0,1]$:

$J(t):=F(y)+t(z-F(y)).$

By Lemma 2.5(ii), for each $\zeta\in[F(x),F(y)]\cap B$, $\zeta\neq F(y)$,
there holds $(F|_{\Lambda})^{-1}(\zeta)<y$, hence
$(F|_{\Lambda})^{-1}(\zeta)\in\Lambda$. In particular, there is
$\theta\in(0,1]$ such that $(F|_{\Lambda})^{-1}(J(s))$ is given by the formula

$(F|_{\Lambda})^{-1}(J(s))=y+\biggl{(}\int_{0}^{s}D(F|_{\Lambda})^{-1}(F(y)+t(z-F(y)))\,\mathrm{d}t\biggr{)}\cdot(z-F(y)),$
(2.2)

hence is $<y$, for all $s\in(0,\theta)$.

Let $\tau\in(0,1]$ be the largest such $\theta$. As the mapping

$[0,\tau)\ni s\mapsto(F|_{\Lambda})^{-1}(J(s))\in[0,F(y)]$

is monotone and takes values in the compact set $[0,F(y)]$, there exists a
limit

$\lim\limits_{s\to\tau^{-}}(F|_{\Lambda})^{-1}(J(s))=:v.$

There holds
$F(v)=\lim\limits_{s\to\tau^{-}}F((F|_{\Lambda})^{-1}(J(s)))=J(\tau)$, so

$\displaystyle(F|_{\Lambda})^{-1}(J(\tau))$
$\displaystyle=y+\lim_{s\to\tau^{-}}\biggl{(}\int_{0}^{s}D(F|_{\Lambda})^{-1}(F(y)+t(z-F(y)))\,\mathrm{d}t\biggr{)}\cdot(z-F(y))$
$\displaystyle=y+\biggl{(}\int_{0}^{\tau}D(F|_{\Lambda})^{-1}(F(y)+t(z-F(y)))\,\mathrm{d}t\biggr{)}\cdot(z-F(y)).$

Suppose to the contrary that $\tau<1$. Then we can repeat our construction to
obtain the existence of $\tau_{1}\in(\tau,1]$ such that formula (2.2) holds
for $\zeta\in J((\tau,\tau_{1}))$, a contradiction. So,
$(F|_{\Lambda})^{-1}(z)<y$.

In a similar way we prove that $(F|_{\Lambda})^{-1}(z)>x$. This proves (1)

We have

$y=x+\biggl{(}\int_{0}^{1}D(F|_{\Lambda})^{-1}(F(x)+t(F(y)-F(x)))\,\mathrm{d}t\biggr{)}\cdot(F(y)-F(x)).$

Let $I:=\\{\,i\in\\{1,\ldots,d\\}:F_{i}(x)<F_{i}(y)\,\\}$. Whenever
$t\in(0,1]$ there holds $F_{i}(x)<F_{i}(x)+t(F_{i}(y)-F_{i}(x))$ for $i\in I$,
consequently, by Lemma 2.5(ii),

$\biggl{(}\int_{0}^{1}D(F|_{\Lambda})^{-1}(F(x)+t(F(y)-F(x)))\,\mathrm{d}t\biggr{)}_{ij}>0$

if $i=j\in I$ [under AS3′, if $i,j\in I$]. The required inequality follows in
a straightforward way. ∎

###  2.3 Existence of the compact attractor $\Gamma$ of bounded sets

Throughout the present subsection we assume AS1–AS4. At one place we assume
additionally AS3′.

######  Lemma 2.7.

Let $a>0$. Then
$F^{n}(\prod\limits_{i=1}^{d}[0,a])\subset\prod\limits_{i=1}^{d}G^{n}_{i}([0,a])$
for all $n\in\mathbb{N}$.

###### Proof.

We prove the lemma by induction on $n$. For $n=0$ we have
$[0,a]=G^{0}_{i}([0,a])$. Assume that the inclusion holds for some
$n\in\mathbb{N}$. Then, for $x\in\prod\limits_{i=1}^{d}[0,a]$,

$\displaystyle F_{i}^{n+1}(x)={}$ $\displaystyle F_{i}^{n}(x)f_{i}(F^{n}(x))$
$\displaystyle\leq{}$ $\displaystyle
F_{i}^{n}(x)f_{i}(F_{i}^{n}(x)\mathbf{e}_{i})$ (by AS3) $\displaystyle={}$
$\displaystyle F^{n}_{i}(x)g_{i}(F^{n}_{i}(x))=G_{i}(F^{n}_{i}(x))$ (by the
definitions of $g_{i}$ and $G_{i}$) $\displaystyle\in{}$ $\displaystyle
G_{i}(G^{n}_{i}([0,a]))$ $\displaystyle\text{ (by inductive hypothesis)}.$

Since, by Lemma 2.3, $G^{n+1}_{i}([0,a])$ equals the compact interval
$[0,G^{n+1}_{i}(a)]$, the membership of $F_{i}^{n+1}(x)$ in
$G^{n+1}_{i}([0,a])$ follows. ∎

######  Lemma 2.8.

  1. (i)

For $n\in\mathbb{N}$,
$F^{n}(\Lambda)\subset\prod\limits_{i=1}^{d}[0,G^{n}_{i}(1+\varkappa)]$. In
particular, $F(\Lambda)\subset\operatorname{Int}_{\,C_{+}}{\Lambda}$.

  2. (ii)

$F(\prod\limits_{i=1}^{d}[0,1])\subset\prod\limits_{i=1}^{d}[0,1]$. Under
AS3′, if
$x\in\prod\limits_{i=1}^{d}[0,1]\setminus\\{0,\mathbf{e}_{1},\dots,\mathbf{e}_{d}\\}$
then
$F(x)\in\operatorname{Int}_{\,C_{+}}\bigl{(}\prod\limits_{i=1}^{d}[0,1]\bigr{)}$.

###### Proof.

The first sentences in (i) and (ii) are direct consequences of Lemmas 2.7 and
2.3. The second sentence in (i) follows since, by Lemma 2.2(iv),
$G_{i}(1+\varkappa)<1+\varkappa$ for all $i$.

Assume AS3′, and let
$x\in\prod\limits_{i=1}^{d}[0,1]\setminus\\{0,\mathbf{e}_{1},\dots,\mathbf{e}_{d}\\}$.
If $i\in I(x)$ is such that $x_{i}<1$ then, by AS3′ and Lemma 2.2(ii),

$F_{i}(x)=x_{i}f_{i}(x)\leq x_{i}g_{i}(x_{i}\mathbf{e}_{i})=G_{i}(x_{i})<1.$

If $x_{i}=1$ then, as $x\neq\mathbf{e}_{k}$, $k=1,\dots,d$, there is some
$j\in I(x)\setminus\\{i\\}$ such that $x_{j}>0$, so we can apply AS3′ to
conclude that

$F_{i}(x)=x_{i}f_{i}(x)<x_{i}g_{i}(x_{i}\mathbf{e}_{i})=G_{i}(x_{i})\leq 1.$

∎

Put

$\Gamma:=\bigcap\limits_{n=0}^{\infty}F^{n}(\Lambda).$

Such a definition was given in [12, Lemma 5.4]. $\Gamma$, being the
intersection of a decreasing family of compact nonempty sets, is compact and
nonempty. Further, since $\Lambda\supset\ldots\supset F^{n}(\Lambda)\supset
F^{n+1}(\Lambda)\ldots\ $, there holds $\Gamma=\omega(\Lambda)$.

######  Lemma 2.9.

$\Gamma=\bigcap\limits_{n=0}^{\infty}F^{n}\bigl{(}\prod\limits_{i=1}^{d}[0,1]\bigr{)}$.
Moreover, $\mathbf{e}_{i}\in\Gamma$ for all $i\in\\{1,\ldots,d\\}$.

###### Proof.

Since $\prod\limits_{i=1}^{d}[0,1]\subset\Lambda$, the ‘$\supset$’ inclusion
is straightforward. To prove the other inclusion, observe first that it
follows from Lemmas 2.8(i) and 2.2(iii) that
$\Gamma\subset\prod\limits_{i=1}^{d}[0,1]$. Consequently,

$\Gamma=\omega(\Gamma)\subset\omega\Bigl{(}\prod\limits_{i=1}^{d}[0,1]\Bigr{)}=\bigcap\limits_{n=0}^{\infty}F^{n}\Bigl{(}\prod\limits_{i=1}^{d}[0,1]\Bigr{)},$

where the last equality holds since
$\prod\limits_{i=1}^{d}[0,1]\supset\ldots\supset
F^{n}\bigl{(}\prod\limits_{i=1}^{d}[0,1]\bigr{)}\supset
F^{n+1}\bigl{(}\prod\limits_{i=1}^{d}[0,1]\bigr{)}\ldots\ $.

$\mathbf{e}_{i}\in\Gamma$ because they are fixed points contained in
$\Lambda$. ∎

It should be remarked that in [21, Proposition 3.5] an analogue of $\Gamma$
was defined as
$\bigcap\limits_{n=0}^{\infty}F^{n}\bigl{(}\prod\limits_{i=1}^{d}[0,1]\bigr{)}$.

######  Lemma 2.10.

For a bounded $A\subset C_{+}$ there is $n_{0}$ such that
$F^{n}(A)\subset\Lambda$ for all $n\geq n_{0}$. In particular, $\Lambda$
attracts bounded sets $A\subset C_{+}$.

###### Proof.

Let $A\subset C_{+}$ be a bounded set and choose $a>1$ such that
$A\subset\prod\limits_{i=1}^{d}[0,a]$. Since, by Proposition 2.4, for each
$i\in\\{1,\ldots,d\\}$ in the dynamical system $(G_{i}^{n})$ the set $[0,1]$
attracts $[0,a]$, there exists $n_{0}\in\mathbb{N}$ such that for all $n\geq
n_{0}$ and all $i\in\\{1,\ldots,d\\}$ the set $G^{n}_{i}([0,a])$ is contained
in $[0,1+\varkappa)$. From Lemma 2.7 it follows that

$F^{n}\Bigl{(}\prod\limits_{i=1}^{d}[0,a]\Bigr{)}\subset\prod\limits_{i=1}^{d}G^{n}_{i}([0,a])\subset\Lambda$

for all $n\geq n_{0}$. ∎

######  Theorem 2.11.

  1. (i)

$\Gamma$ is the compact attractor of bounded sets in $C_{+}$.

  2. (ii)

$\Gamma$ is characterized as the set of those $x\in C_{+}$ for which there
exists a bounded full orbit.

###### Proof.

Let $A\subset C_{+}$ be bounded. By Lemma 2.10, $(F^{n})$ is asymptotically
compact on $A$; moreover, $\omega(A)\subset\omega(\Lambda)=\Gamma$. Therefore,
by Lemma 1.5, $\Gamma$ attracts $A$. This proves (i).

The characterization given in (ii) is a consequence of Proposition 1.6. ∎

A consequence of Theorem 2.11 is

######  Lemma 2.12.

For $x\in C_{+}$ the following are equivalent.

  1. (1)

$x\in\Gamma$.

  2. (2)

There is a full orbit of $x$, contained in $\Lambda$.

  3. (3)

There is a bounded backward orbit of $x$.

$\Gamma$ is the largest compact invariant set in $C_{+}$ (see [22, Theorem
2.19 on p. 37]). Further, since $\Gamma\subset\Lambda$ and, by Lemma 2.5,
$F|_{\Lambda}$ is a homeomorphism onto its image, $F|_{\Gamma}$ is a
homeomorphism onto $\Gamma$, so $((F|_{\Gamma})^{n})_{-\infty}^{\infty}$ is a
(two-sided) dynamical system on the compact metric space $\Gamma$ (and
$\Gamma$ is the largest subset of $C_{+}$ with that property).

In some papers ([9, 23]) $\Gamma$ is called the global attractor for $F$.

###  2.4 Construction of the carrying simplex

We assume AS1, AS2, AS3 and AS4. At some places we assume AS3′.

In the present subsection we write $F$ instead of $F|_{\Lambda}$.

Denote by $T$ the radial projection of $C_{+}\setminus\\{0\\}$ onto $\Delta$,
$T(x):=x/\lVert x\rVert_{1}$.

Let $\widehat{\mathcal{U}}$ [$\mathcal{U}$] stand for the set of bounded and
weakly unordered [unordered] hypersurfaces contained in $\Lambda$ that are
homeomorphic to the standard probability simplex $\Delta$ via radial
projection. In particular, a hypersurface $S\in\widehat{\mathcal{U}}$ is at a
positive distance from the origin, as the radial projection is not defined at
$0$.

It follows that for any $S\in\widehat{\mathcal{U}}$ the inverse
$(T|_{S})^{-1}$ can be written as

$\Delta\ni u\mapsto R(u)u,$

where $R\colon\Delta\to(0,\infty)$ is a continuous function (called the radial
representation of $S$). In other words,

$x=R(T(x))T(x),\quad x\in S.$

For $S\in\widehat{\mathcal{U}}$, its complement $C_{+}\setminus S$ is the
union of two disjoint sets, a bounded one, $S_{-}$, and an unbounded one,
$S_{+}$. One has $S_{-}=[0,1)S$, $S_{+}=(1,\infty)S$.

We write $\partial S$ for $S\cap\partial C_{+}$. $\partial
S=\operatorname{bd}_{C_{+}}S$.

######  Lemma 2.13.

If $S\in\widehat{\mathcal{U}}$ then $F(S)\in\widehat{\mathcal{U}}$. If
$S\in\mathcal{U}$ then $F(S)\in\mathcal{U}$. If AS3′ holds, $F$ maps
$\widehat{\mathcal{U}}$ into $\mathcal{U}$.

###### Proof.

Let $S\in\widehat{\mathcal{U}}$. We prove first that $F(S)$ is weakly
unordered. Indeed, suppose there are $\emptyset\neq I\subset\\{1,\ldots,d\\}$
and $x,y\in S\cap(C_{I})_{++}$ such that $F(x)\ll_{I}F(y)$. Then, by
Proposition 2.6(2), $x\ll_{I}y$, which is impossible.

Since $F$ is continuous on $C_{+}$ and $S\subset\Lambda$ is compact, $F(S)$ is
compact, and, by Lemma 2.8(i), $F(S)$ is a subset of $\Lambda^{\prime}$.
Define $\hat{F}$ on $\Lambda^{\prime}$ as $\hat{F}=F/\lVert F\rVert_{1}$, in
other words, $\hat{F}=T\circ F$. Then $\hat{F}$ is a continuous map on
$\Lambda^{\prime}$, and consequently the continuous map $\hat{F}|_{S}\colon
S\to\Delta$ is proper. $T|_{F(S)}$ is invertible at each element of $F(S)$,
since otherwise $F(S)$ would not be weakly unordered, which has been excluded
in the previous paragraph. Hence $\hat{F}|_{S}$ is locally invertible at each
$x\in S$, and we can apply [4, Cor. 2.3.6] to conclude that $\hat{F}|_{S}$ is
a homeomorphism onto $\Delta$. Therefore, $F(S)\in\widehat{\mathcal{U}}$.

Let $S\in\mathcal{U}$. If there are $x,y\in S$ such that $F(x)<F(y)$ then, by
Proposition 2.6(2), $x<y$, which contradicts the unorderedness of $S$

Assume that AS3′ holds, and let $S\in\widehat{\mathcal{U}}$. Suppose to the
contrary that $F(S)$ is not unordered, that is, there are $x,y\in S$ such that
$F(x)<F(y)$. By the competitivity of $F$, $x_{i}<y_{i}$ for all $i\in
I(y)=I(F(y))=:I$. If $I(x)=I$ then $x,y\in(C_{I})_{++}$, so $x\ll_{I}y$, a
contradiction. Suppose that $J:=I\setminus I(x)$ is nonempty. Since $S$ is
homeomorphic to $\Delta$ via radial projection, there is $\tilde{x}\in
S\cap(C_{I})_{++}$ sufficiently close to $x$ so that $\tilde{x}\ll_{I}y$,
again a contradiction. ∎

We introduce a partial order relation on $\widehat{\mathcal{U}}$. For
$S,S^{\prime}\in\widehat{\mathcal{U}}$, let $R,R^{\prime}$ denote their
respective radial representations. We write $S\preccurlyeq S^{\prime}$ if
$R(u)\leq R^{\prime}(u)$ for all $u\in\Delta$, $S\prec S^{\prime}$ if
$S\preccurlyeq S^{\prime}$ and $S\neq S^{\prime}$, and $S\llcurly S^{\prime}$
if $R(u)<R^{\prime}(u)$ for all $u\in\Delta$. It is straightforward that
$S\prec S^{\prime}$ if and only if $R(u)\leq R^{\prime}(u)$ for all
$u\in\Delta$ and there is $v\in\Delta$ with $R(v)<R^{\prime}(v)$.

Observe that $S\preccurlyeq S^{\prime}$ if and only if $S\subset
S^{\prime}\cup(S^{\prime})_{-}$ (or, which is the same, $S^{\prime}\subset
S\cup S_{+}$).

We assume the convention that $0\llcurly S$ for any
$S\in\widehat{\mathcal{U}}$.

For $S\preccurlyeq S^{\prime}$ we denote

$\langle S,S^{\prime}\rangle:=[1,\infty)S\cap[0,1]S^{\prime},$

and for $S\llcurly S^{\prime}$ we denote

$\llangle S,S^{\prime}\rrangle:=(1,\infty)S\cap[0,1)S^{\prime}.$

In other words, $\langle S,S^{\prime}\rangle=\\{\,\lambda
u:\lambda\in[R(u),R^{\prime}(u)],u\in\Delta\,\\}$, and $\llangle
S,S^{\prime}\rrangle=\\{\,\lambda
u:\lambda\in(R(u),R^{\prime}(u)),u\in\Delta\,\\}$. In particular, $\langle
0,S\rangle=\\{\,\lambda u:0\leq\lambda\leq R(u),u\in\Delta\,\\}=[0,1]S$.

Notice that $S\preccurlyeq S^{\prime}$ if and only if $S\subset\langle
0,S^{\prime}\rangle$.

######  Lemma 2.14.

For $S,S^{\prime}\in\widehat{\mathcal{U}}$ with $S\preccurlyeq S^{\prime}$,

$\langle S,S^{\prime}\rangle=\bigcup_{\begin{subarray}{c}x\in S\\\ y\in
S^{\prime}\end{subarray}}[x,y].$

###### Proof.

The ‘$\subset$’ inclusion is straightforward. Assume that $z\in[x,y]$ with
$x\in S$ and $y\in S^{\prime}$. There are $\xi\in S$ and $\eta\in S^{\prime}$
such that $z=\alpha\xi=\beta\eta$. We claim that $\alpha\geq 1$ and $\beta\leq
1$. Indeed, suppose that $\alpha<1$. Then $x\leq z\ll_{I}\xi$, where
$I:=I(z)=I(\xi)$, with both $x$ and $\xi$ in $S$. If $I(x)=I$ then
$x\ll_{I}\xi$, which contradicts the weak unorderedness of $S$. If not, we can
find $\tilde{x}\in S\cap(C_{++})_{I}$ so close to $x$ that
$\tilde{x}\ll_{I}\xi$, which again contradicts the weak unorderedness of $S$.
The case $\beta>1$ is excluded in much the same way. We have thus
$z\in[R(T(z)),R^{\prime}(T(z))]T(z)$, with $\xi=R(T(z))T(z)$ and
$\eta=R^{\prime}(T(z))T(z)$. ∎

As a consequence of Lemma 2.14, $\langle S,S^{\prime}\rangle$ is order convex.

######  Lemma 2.15.

For $S\in\widehat{\mathcal{U}}$, $F(\langle 0,S\rangle)=\langle
0,F(S)\rangle$.

###### Proof.

$\displaystyle\langle 0,F(S)\rangle$ $\displaystyle=\bigcup_{y\in F(S)}[0,y]$
(by Lemma 2.14) $\displaystyle=\bigcup_{x\in S}[0,F(x)]$
$\displaystyle\subset\bigcup_{x\in S}F([0,x])$ (by Proposition 2.6(1))
$\displaystyle=F\Bigl{(}\bigcup_{x\in S}[0,x]\Bigr{)}$
$\displaystyle=F(\langle 0,S\rangle)$ $\displaystyle\text{ (by
Lemma~{}\ref{lm:between})}.$

Suppose to the contrary that there is $z\in F(\langle
0,S\rangle)\setminus\langle 0,F(S)\rangle$. Let $y$ be the unique member of
$F(S)$ such that $y={\alpha}z$ with $\alpha>0$. By our choice of $z$, we have
$\alpha>1$ and $y\ll_{I}z$, where $I:=I(z)=I(y)$. Let $\eta\in S$ and
$\zeta\in\langle 0,S\rangle$ be such that $F(\eta)=y$ and $F(\zeta)=z$.
Proposition 2.6(2) yields $\eta\ll_{I}\zeta$. Let $x$ be the unique member of
$S$ such that $x={\beta}\zeta$ with $\beta\geq 1$. Then $x\in
S\cap(C_{I})_{++}$, and either $\zeta=x$ or $\zeta\ll_{I}x$. At any rate,
$\eta\ll_{I}x$, which contradicts the weak unorderedness of $S$. ∎

######  Lemma 2.16.

$F$ preserves the $\preccurlyeq$ and relations on $\widehat{\mathcal{U}}$.

###### Proof.

Let $S,S^{\prime}\in\widehat{\mathcal{U}}$, $S\preccurlyeq S^{\prime}$, which
means that $S\subset\langle 0,S^{\prime}\rangle$. Then, by Lemma 2.15,
$F(S)\subset F(\langle 0,S^{\prime}\rangle)=\langle 0,F(S^{\prime})\rangle$,
that is, $F(S)\preccurlyeq F(S^{\prime})$.

Assume that $S\llcurly S^{\prime}$. By the previous paragraph,
$F(S)\preccurlyeq F(S^{\prime})$, and by the fact that $F|_{\Lambda}$ is a
homeomorphism onto its image, $F(S)$ and $F(S^{\prime})$ are disjoint,
consequently $F(S)\llcurly F(S^{\prime})$. ∎

Let $\varepsilon\in(0,1)$ be small enough that $\min\limits_{1\leq i\leq
d}f_{i}(x)>1$ for all $x\in S_{0}:=\varepsilon\Delta$ (this is possible by
(AS2)–(AS3)). $S_{0}\subset\Lambda$, is unordered and projects
homeomorphically on $\Delta$, so it belongs to $\mathcal{U}$.

######  Proposition 2.17.

$\\{0\\}$ is a uniform repeller for $(F^{n})_{n=0}^{\infty}$.

###### Proof.

As $U$ in the definition of uniform repeller we take $\langle 0,S_{0}\rangle$.
It follows from the choice of $S_{0}$ and AS3 that there is $\delta>0$ such
that $f_{i}(x)\geq 1+\delta$, $i\in\\{1,\ldots,d\\}$, consequently $\lVert
F(x)\rVert_{1}\geq(1+\delta)\lVert x\rVert_{1}$, for all $x\in U$. So, we have
that there exists $n_{0}\in\mathbb{N}$ with the property that
$F^{n_{0}}(x)\notin U$ ($n_{0}$ is not larger than the least non-negative
integer $n$ such that $(1+\delta)^{n}\lVert x\rVert_{1}\geq\varepsilon$, where
$\varepsilon>0$ is as in the definition of $S_{0}$).

It remains to prove that $F^{n}(x)\notin U$ for all $n\geq n_{0}$. Suppose to
the contrary that there are $n>n_{0}$ such that $F^{n}(x)\in U$, and let
$n_{1}$ be their infimum. Then $F^{n_{1}}(x)\in U\subset\langle
0,S_{1}\rangle$, which is equal, by Lemma 2.15, to $F(\langle
0,S_{0}\rangle)$, so there is $\xi\in\langle 0,S_{0}\rangle$ with
$F(\xi)=F_{n_{1}}(x)$. Lemma 2.8 implies that $F^{n_{1}-1}(x)\in\Lambda$ and,
by Lemma 2.5(i), $F|_{\Lambda}$ is injective, so $F^{n_{1}-1}(x)=\xi$, which
belongs to $U$. This contradicts the choice of $n_{1}$. ∎

The sequence $S_{n}:=F^{n}(S_{0})\in\mathcal{U}$ by Lemma 2.13. Denote by
$R_{n}\colon\Delta\to(0,\infty)$ the radial representation of $S_{n}$. By our
choice of $S_{0}$, we have $S_{0}\llcurly S_{1}$. Lemma 2.16 implies
$S_{n}\llcurly S_{n+1}$ for all $n=1,2,\dots$. As a consequence, for each
$u\in\Delta$ the sequence $(R_{n}(u))_{n=0}^{\infty}$ is strictly increasing.
Let $R_{*}$ stand for its pointwise limit.

######  Definition 2.18.

$S_{*}$ is the _Kuratowski limit_ of the sequence $(S_{n})_{n=0}^{\infty}$ if
the following conditions are satisfied:

  1. (K1)

For each $x\in S_{*}$ there is a sequence $(x_{n})_{n=0}^{\infty}$ such that
$x_{n}\in S_{n}$ and $\lim\limits_{n\to\infty}x_{n}=x$.

  2. (K2)

For any sequence $(x_{n_{k}})_{k=1}^{\infty}$ with $x_{n_{k}}\in S_{n_{k}}$
and $n_{k}\underset{k\to\infty}{\longrightarrow}\infty$, if
$\lim\limits_{k\to\infty}x_{n_{k}}=x$ then $x\in S_{*}$.

See [1, Def. 4.4.13]. Recall that $d_{H}$ stands for the Hausdorff metric on
the family $2^{\Lambda}$ of nonempty compact subsets of $\Lambda$.
$(2^{\Lambda},d_{H})$ is a compact metric space (see [1, Thm. 4.4.15]).

######  Lemma 2.19.

$S_{*}=\\{\,R_{*}(u)u:u\in\Delta\,\\}$.

###### Proof.

Let $x=R_{*}(u)u$ for some $u\in\Delta$. There holds
$\lim\limits_{n\to\infty}R_{n}(u)u=x$ with $R_{n}(u)u\in S_{n}$, therefore, by
(K1), $R_{*}(u)u\in S_{*}$.

Let $(x_{n_{k}})_{k=1}^{\infty}$ with $x_{n_{k}}\in S_{n_{k}}$ and
$n_{k}\underset{k\to\infty}{\longrightarrow}\infty$ be a sequence convergent
to $x$. We can assume that $k\mapsto n_{k}$ is strictly increasing. As for any
$k\in\mathbb{N}$, $x_{n_{k+1}},x_{n_{k+2}},\ldots\in[1,\infty)S_{n_{k+1}}$, so
their limit, $x$ belongs to $[1,\infty)S_{n_{k+1}}\subset(1,\infty)S_{n_{k}}$.
In particular, the half-line originating at $0$ and passing through $x$
intersects $S_{n_{k}}$ at $R_{n_{k}}(u)u$ where $u:=T(x)$, and $R_{n_{k}}(u)u$
is between $0$ and $x$. As this holds for all $k$, we have
$R_{n_{1}}(u)<\ldots<R_{n_{k}}(u)<R_{n_{k+1}}(u)<\ldots<\lVert x\rVert_{1}$,
hence $R_{*}(u)=\sup\limits_{k\in\mathbb{N}}R_{n_{k}}(u)\leq\lVert
x\rVert_{1}$.

We claim that $x=R_{*}(u)u$. Suppose not. Put $I:=I(x)$,
$J:=\\{1,\ldots,d\\}\setminus I$. We have

$\bigl{(}R_{n_{k}}(u)u+(C_{++}\cup(C_{I})_{++})\bigr{)}\cap
S_{n_{k}}=\emptyset,\quad k\in\mathbb{N}.$

Indeed, $\bigl{(}R_{n_{k}}(u)u+(C_{I})_{++}\bigr{)}\cap S_{n_{k}}=\emptyset$
follows directly from the weak unordering of $S_{n_{k}}$, whereas, if we
suppose the existence of $\xi\in S_{n_{k}}\cap C_{++}$ with
$(R_{n_{k}}(u)u)_{i}<\xi_{i}$ for all $i\in I$ then we can find $\eta\in
S_{n_{k}}\cap C_{++}$ so close to $R_{n_{k}}(u)u$ that $\eta\ll\xi$, which
again contradicts the weak unorderedness of $S_{n_{k}}$. Since
$R_{n_{k}}(u)u\ll_{I}R_{*}(u)u$, there holds
$\bigl{(}R_{*}(u)u+(C_{++}\cup(C_{I})_{++})\bigr{)}\cap
S_{n_{k}}\subset\bigl{(}R_{n_{k}}(u)u+(C_{++}\cup(C_{I})_{++})\bigr{)}\cap
S_{n_{k}}=\emptyset$. Now,
$\bigl{(}R_{*}(u)u+(C_{++}\cup(C_{I})_{++})\bigr{)}$ is relatively open in
$C_{+}$ and contains $x$, hence $x$ cannot be approximated by $x_{n_{k}}\in
S_{n_{k}}$. ∎

######  Proposition 2.20.

The following properties hold:

  1. (1)

$d_{H}(S_{n},S_{*})\to 0$ as $n\to\infty$.

  2. (2)

The functions $R_{n}\colon\Delta\to(0,\infty)$ converge uniformly, as
$n\to\infty$, to (necessarily continuous) $R_{*}$.

###### Proof.

Part (1) is a consequence of the fact that for the compact $\Lambda$, the
Kuratowski convergence and the convergence in the Hausdorff metric are the
same, see [1, Prop. 4.4.15].

We proceed now to proving (2). Since $\Delta$ is compact, we use the fact that
uniform convergence is equivalent to continuous convergence: for any sequence
$(u_{n})_{n=1}^{\infty}\subset\Delta$ convergent to $u$ there holds
$\lim\limits_{n\to\infty}R_{n}(u_{n})=R_{*}(u)$. Indeed, let
$(R_{n_{k}}(u_{n_{k}}))_{k=1}^{\infty}$ be a subsequence such that
$\lim\limits_{k\to\infty}R_{n_{k}}(u_{n_{k}})u_{n_{k}}=x$. Then
$\lim\limits_{k\to\infty}T(R_{n_{k}}(u_{n_{k}})u_{n_{k}})=T(x)$, consequently
$T(x)=u$. Further, by (K2) in Definition 2.18, $x\in S_{*}$, so it follows
from Lemma 2.19 that $x=R_{*}(T(x))T(x)=R_{*}(u)u$; furthermore,
$\lim\limits_{k\to\infty}R_{n_{k}}(u_{n_{k}})=\lim\limits_{k\to\infty}\lVert
R_{n_{k}}(u_{n_{k}})u_{n_{k}}\rVert_{1}=\lVert x\rVert_{1}=R_{*}(u)$. Since
this holds for any convergent subsequence of $(R_{n}(u_{n}))_{n=1}^{\infty}$,
the statement follows.

To prove (2) we show first that $R_{*}\colon\Delta\to(0,\infty)$ is
continuous. Let $(u_{k})_{k=1}^{\infty}\in\Delta$ converge to $u\in\Delta$.
Let $(k_{l})_{l=1}^{\infty}$, $k_{l}\to\infty$ as $l\to\infty$, be such that
$R_{*}(u_{k_{l}})u_{k_{l}}\to x$ as $l\to\infty$. Then
$T(x)=\lim\limits_{l\to\infty}T(R_{*}(u_{k_{l}})u_{k_{l}})=\lim\limits_{l\to\infty}u_{k_{l}}=u$.
Since $x\in S_{*}$, it follows from Lemma 2.19 that $x=R_{*}(T(x))T(x)$, hence
$x=R_{*}(u)u$. Consequently, $R_{*}$ is continuous, and as it is the pointwise
limit of an increasing sequence of continuous functions, Dini’s theorem
implies that the convergence is uniform. ∎

######  Proposition 2.21.

$S_{*}$ is invariant under $F$, and $S_{*}\in\widehat{\mathcal{U}}$. Under
AS3′, $S_{*}\in\mathcal{U}$.

Recall that $F$ denotes here $F|_{\Lambda}$, so the invariance of $S_{*}$
means that, first, if $x\in S_{*}$ then $F(x)\in S_{*}$, and, second, if
$x\in\Lambda$ is such that $F(x)\in S_{*}$ then $x\in S_{*}$. For examples of
$x\in C_{+}\setminus\Lambda$ such that $F(x)\in S_{*}$ see subsubsection
3.1.3.

###### Proof.

$d_{H}(S_{n},S_{*})\to 0$, consequently $d_{H}(S_{n+1},S_{*})\to 0$, as
$n\to\infty$. But $S_{n+1}=F(S_{n})$, hence $d_{H}(F(S_{n}),S_{*})\to 0$ as
$n\to\infty$. On the other hand, $F(S_{n})$ converge to $F(S_{*})$ in the
Hausdorff metric, so $S_{*}=F(S_{*})$.

We know already that $S_{*}$ projects radially in a homeomorphic way to
$\Delta$. To show that $S_{*}\in\widehat{\mathcal{U}}$ it suffices to prove
that $S_{*}$ is weakly unordered. Suppose that $x,y\in S_{*}\cap(C_{I})_{++}$
are such that $x\ll_{I}y$. By the construction, there are $u,v\in\Delta_{I}$
such that $x=\lim\limits_{n\to\infty}R_{n}(u)u$ and
$y=\lim\limits_{n\to\infty}R_{n}(v)v$. Since the relation $\ll_{I}$ is
relatively open in $(C_{I})_{++}$, $R_{n}(u)u\ll_{I}R_{n}(v)v$ for $n$
sufficiently large. But $R_{n}(u)u,R_{n}(v)v\in S_{n}\cap(C_{I})_{++}$, which
contradicts the weak unorderedness of $S_{n}$.

The last sentence is a consequence of Lemma 2.13. ∎

Put $S^{0}:=\operatorname{bd}_{C_{+}}\Lambda=H(1+\varkappa)$ (see Example
1.3). We will consider a sequence $(S^{n})_{n=0}^{\infty}$, where
$S^{n}:=F^{n}(S^{0})$.

$S^{0}$, being a weakly unordered subset of $\Lambda$ and projecting radially
in a homeomorphic way on $\Delta$, belongs to $\widehat{\mathcal{U}}$, so
Lemma 2.13 states that $S^{1}\in\widehat{\mathcal{U}}$, and again, that
$S^{n}:=F^{n}(S^{0})\in\widehat{\mathcal{U}}$ for $n=1,2,\dots$.

By Lemma 2.8(i), we have $S^{1}\llcurly S^{0}$. By Lemma 2.16,
$S^{n+1}\llcurly S^{n}$ for all $n=1,2,\dots$. Denote by
$R^{n}\colon\Delta\to(0,\infty)$ the radial representation of $S^{n}$. Since
$S_{0}\llcurly S^{0}$, it follows from Lemma 2.16 that $S_{n}\llcurly
S^{n}\llcurly\ldots\llcurly S^{1}\llcurly S^{0}$ for all $n\in\mathbb{N}$,
consequently $S_{*}\llcurly S^{0}$. Again by Lemma 2.16, $S_{*}\llcurly S^{n}$
for all $n\in\mathbb{N}$, which means that
$R_{*}(u)<\ldots<R^{n+1}(u)<R^{n}(u)<\ldots<R^{0}(u)$ for all $u\in\Delta$.
Therefore there exists
$\lim\limits_{n\to\infty}R^{n}(u)=:R^{*}(u)\in[R_{*}(u),R^{0}(u))$. Put
$S^{*}$ to be the Kuratowski limit of the sequence $(S^{n})_{n=0}^{\infty}$.

There holds $S^{*}\subset[1,\infty)S_{*}$.

######  Lemma 2.22.

  1. (i)

$d_{H}(S^{n},S^{*})\to 0$ as $n\to\infty$.

  2. (ii)

$S^{*}$ is invariant under $F$.

###### Proof.

(i) is a consequence of the fact that for the compact $\Lambda$, the
Kuratowski convergence and the convergence in the Hausdorff metric are the
same, see [1, Prop. 4.4.15].

$d_{H}(S^{n},S^{*})\to 0$, consequently $d_{H}(S^{n+1},S^{*})\to 0$, as
$n\to\infty$. But $S^{n+1}=F(S^{n})$, hence $d_{H}(F(S^{n}),S^{*})\to 0$ as
$n\to\infty$. On the other hand, $F(S^{n})$ converge to $F(S^{*})$ in the
Hausdorff metric, so $S^{*}=F(S^{*})$. ∎

We introduce now the Harnack metric on $C_{+}$. For more, see [16].

For $x,y\in C_{+}$, $x\neq 0$, we define

$\lambda(x,y):=\sup\\{\,\lambda\geq 0:y-{\lambda}x\in C_{+}\,\\}\\\
=\min\left\\{\frac{y_{i}}{x_{i}}:1\leq i\leq d\text{ such that
}x_{i}>0\,\right\\}.$

We put $\lambda(x,y)=\infty$ if and only if $x=0$. Following [16], we call
$\lambda$ the order function on $C_{+}$.
$\mu(x,y):=\min\\{\lambda(x,y),\lambda(y,x)\\}$ is the symmetrized order
function on $C_{+}$.

We define on $C_{+}\setminus\\{0\\}$ the Harnack metric:

$h(x,y):=1-\mu(x,y).$

Let $x\leq y$ both belong to $(C_{I})_{++}$ for some $\emptyset\neq
I\subset\\{1,\dots,d\\}$. We have

$\mu(x,y)=\lambda(y,x)=\min\left\\{\frac{x_{i}}{y_{i}}:i\in I\,\right\\}.$

The following is straightforward.

######  Lemma 2.23.

Let $\lim\limits_{n\to\infty}x_{n}=x$, $\lim\limits_{n\to\infty}y_{n}=y$,
$x_{n}\leq y_{n}$ for all $n\in\mathbb{N}$, and
$x_{n},y_{n},x,y\in(C_{I})_{++}$ for some $\emptyset\neq
I\subset\\{1,\dots,d\\}$. Then
$h(x,y)=\lim\limits_{n\to\infty}h(x_{n},y_{n})$.

######  Proposition 2.24.

Let $\emptyset\neq I\subset\\{1,\ldots,d\\}$ and $x<y$ be such that
$x_{i}<y_{i}$ for $i\in I$. Then for each $\emptyset\neq J\subset I$ there
holds

$\mu(\pi_{J}(F(x)),\pi_{J}(F(y)))>\mu(\pi_{J}(x),\pi_{J}(y)).$

Consequently

$h(\pi_{J}(F(x)),\pi_{J}(F(y)))<h(\pi_{J}(x),\pi_{J}(y)).$

###### Proof.

For each $i\in J$ there holds $x_{i}<y_{i}$, so, by the second part of (AS3),

$\frac{F_{i}(x)}{F_{i}(y)}=\frac{x_{i}f_{i}(x)}{y_{i}f_{i}(y)}>\frac{x_{i}}{y_{i}}.$

Hence,

$\displaystyle\mu(\pi_{J}(x),\pi_{J}(y))$
$\displaystyle=\min\left\\{\frac{x_{i}}{y_{i}}:i\in J\,\right\\}$
$\displaystyle<\min\left\\{\frac{F_{i}(x)}{F_{i}(y)}:i\in
J\,\right\\}=\mu(\pi_{J}(F(x)),\pi_{J}(F(y))).$

∎

######  Theorem 2.25.

$S_{*}=S^{*}$.

###### Proof.

Suppose not. This means that there is $v\in\Delta$ such that the half-line
starting at $0$ and passing through $v$ intersects $S_{*}$ at $x$ and
intersects $S^{*}$ at $y$, where $x<y$. Put $I:=I(x)$. We have
$x,y\in(C_{I})_{++}$ and $x\ll_{I}y$.

By Proposition 2.21 and Lemma 2.22(ii), $S_{*}$ and $S^{*}$ are invariant.

From Proposition 2.6(2) it follows that $F^{-n}(x)\ll_{I}F^{-n}(y)$ for all
$n\in\mathbb{N}$.

We write

$\alpha(x):=\\{\,z\in\Lambda:\text{there is }n_{k}\to\infty\text{ such that
}F^{-n_{k}}(x)\to z\text{ as }k\to\infty\,\\},$

and similarly for $\alpha(y)$.

Fix $\xi\in\alpha(x)$. By passing to a subsequence, if necessary, we can
assume that $F^{-n_{k}}(y)$ converges to $\eta\in\alpha(y)$ for the same
sequence $n_{k}\to\infty$. There holds, by the closedeness of the $\leq$
relation, $\xi\leq\eta$. Put $\emptyset\neq J:=I(\xi)$. Then
$\xi=\pi_{J}(\xi)$ and $\pi_{J}(\xi),\pi_{J}(\eta)\in(C_{J})_{++}$. We have,
in view of Lemma 2.23 and Proposition 2.24,

$\displaystyle h(\pi_{J}(\xi),\pi_{J}(\eta))$
$\displaystyle=\lim_{k\to\infty}h(\pi_{J}(F^{-n_{k}}(x)),\pi_{J}(F^{-n_{k}}(y))$
$\displaystyle=\sup_{n\in\mathbb{N}}h(\pi_{J}(F^{-n}(x)),\pi_{J}(F^{-n}(y))>0,$

from which it follows that $\pi_{J}(\xi)<\pi_{J}(\eta)$. Let $K\neq\emptyset$
be the set of those $j\in J$ for which $\xi_{j}<v_{j}$. Proposition 2.6(2)
implies $F^{-1}(\xi)<F^{-1}(\eta)$ and $(F^{-1})_{j}(\xi)<(F^{-1})_{j}(\eta)$
for all $j\in K$. We have, by Lemma 2.23 and Proposition 2.24,

$\displaystyle h(\pi_{K}(\xi),\pi_{K}(v))$
$\displaystyle=\lim_{k\to\infty}h(\pi_{K}(F^{-n_{k}}(x)),\pi_{K}(F^{-n_{k}}(y))$
$\displaystyle=\sup_{n\in\mathbb{N}}h(\pi_{J}(F^{-n}(x)),\pi_{J}(F^{-n}(y))$
$\displaystyle=\lim_{k\to\infty}h(\pi_{K}(F^{-n_{k}-1}(x)),\pi_{K}(F^{-n_{k}-1}(y))$
$\displaystyle=h(\pi_{K}(F^{-1}(\xi)),\pi_{K}(F^{-1}(\eta))).$

But, again by Proposition 2.24,
$h(\pi_{K}(F^{-1}(\xi)),\pi_{K}(F^{-1}(\eta)))>h(\pi_{K}(\xi),\pi_{K}(\eta))$,
a contradiction. ∎

######  Proposition 2.26.

The functions $R^{n}\colon\Delta\to(0,\infty)$ converge uniformly, as
$n\to\infty$, to $R_{*}$.

###### Proof.

Since $\Delta$ is compact, it suffices to show that for any sequence
$(u_{n})_{n=1}^{\infty}\subset\Delta$ convergent to $u$ there holds
$\lim\limits_{n\to\infty}R^{n}(u_{n})=R_{*}(u)$. Indeed, let
$(R^{n_{k}}(u_{n_{k}}))_{k=1}^{\infty}$ be a subsequence such that
$\lim\limits_{k\to\infty}R^{n_{k}}(u_{n_{k}})u_{n_{k}}=x$. Then
$\lim\limits_{k\to\infty}T(R^{n_{k}}(u_{n_{k}})u_{n_{k}})=T(x)$, consequently
$T(x)=u$. Further, by the analogue of (K2) in Definition 2.18, $x\in S^{*}$,
which is, according to Theorem 2.25, equal to $S_{*}$. It follows from Lemma
2.19 that $x=R_{*}(T(x))T(x)=R_{*}(u)u$. Furthermore,
$\lim\limits_{k\to\infty}R^{n_{k}}(u_{n_{k}})=\lim\limits_{k\to\infty}\lVert
R^{n_{k}}(u_{n_{k}})u_{n_{k}}\rVert_{1}=\lVert x\rVert_{1}=R_{*}(u)$. Since
this holds for any convergent subsequence of $(R^{n}(u_{n}))_{n=1}^{\infty}$,
the statement follows.

It follows from Lemma 2.19 and Theorem 2.27 that $R^{n}$ converge pointwise in
a decreasing way to the (continuous by Proposition 2.20) function $R_{*}$, so,
by Dini’s theorem, the convergence is uniform. ∎

We denote $\Sigma:=S_{*}$.

Observe that, by Proposition 2.21 and Lemma 2.5(i), $\Sigma$ satisfies (I),
(II) and (III) in the definition of the carrying simplex [weak carrying
simplex].

######  Theorem 2.27.

$\Gamma=\langle 0,\Sigma\rangle$.

###### Proof.

$\displaystyle\omega(\Lambda)$
$\displaystyle=\bigcap_{n=0}^{\infty}\operatorname{cl}\Biggl{(}\bigcup_{k=n}^{\infty}F^{k}(\langle
0,S^{0}\rangle)\Biggr{)}$ $\displaystyle=\bigcap_{n=0}^{\infty}F^{n}(\langle
0,S^{0}\rangle)$ (by $F^{n+1}(\langle 0,S^{0}\rangle)\subset F^{n}(\langle
0,S^{0}\rangle)$) $\displaystyle=\bigcap_{n=0}^{\infty}\langle 0,S^{n}\rangle$
(by Lemma 2.15) $\displaystyle=\\{\,\alpha
v:v\in\Delta,0\leq\alpha\leq\inf_{n}R^{n}(v)\,\\}$ $\displaystyle=\\{\,\alpha
v:v\in\Delta,0\leq\alpha\leq R_{*}(v)\,\\}=\langle 0,\Sigma\rangle.$

∎

As it is straightforward that $\Sigma$ is the relative boundary of $\langle
0,\Sigma\rangle$ in $C_{+}$, the satisfaction of (IV) in the definition of
carrying simplex [weak carrying simplex] follows from Theorem 2.27 together
with Lemma 2.14.

###  2.5 Back to Dynamics on $C_{+}$

Assume AS1–AS4. From now on, $F$ is considered defined on the whole of $C_{+}$
again.

By Proposition 2.17, $\\{0\\}\subset\Gamma$ is a uniform repeller. According
to [22, Theorem 5.17 on p. 137], the compact attractor $\Gamma$ of
neighbourhoods of compact sets in $C_{+}$ is the union of pairwise disjoint
sets,

$\Gamma=\\{0\\}\cup E\cup H,$ (2.3)

with the following properties:

  * •

$H$ attracts any bounded $A\subset C_{+}$ with $0\notin\bar{A}$; further, if
$x\in C_{+}\setminus H$ has a bounded backward orbit
$\\{\,\ldots,x_{-n-1},x_{-n},\ldots,x_{-2},x_{-1},x\,\\}$ then
$\lim\limits_{n\to\infty}x_{-n}=0$;

  * •

$E$ consists of those $x\in\Gamma$ for which there exists a bounded full orbit
$\\{\,\ldots,x_{-n-1},x_{-n},\allowbreak\ldots,x_{-2},x_{-1},x,x_{1},x_{2},\ldots,x_{n},x_{n+1},\ldots\,\\}$
such that $\lim\limits_{n\to\infty}x_{-n}=0$ and
$\lim\limits_{n\to\infty}\operatorname{dist}(x_{n},H)=0$.

We claim that $E=(0,1)\Sigma$. Indeed, because $F|_{\Gamma}$ is a
homeomorphism of $\Gamma$ onto itself and $\Sigma$ is invariant,
$E\cap\Sigma=\emptyset$. As $0\notin E\subset\Gamma$, and
$\Gamma\setminus\\{0\\}$ is the disjoint union of $(0,1)\Sigma$ and $\Sigma$,
there holds $E=(0,1)\Sigma$ and $H=\Sigma$.

So we have the following classification of points in $x\in C_{+}$.

######  Theorem 2.28.

  1. (i)

If $x\in(0,1)\Sigma$, then $F^{n}(x)\in(0,1)\Sigma$ for all $n\in\mathbb{N}$
and $\lim\limits_{n\to\infty}\operatorname{dist}(F^{n}(x),\Sigma)=0$;
furthermore, there exists a backward orbit $\\{\ldots,x_{-2},x_{-1},x\\}$ with
$\lim\limits_{n\to\infty}x_{-n}=0$.

  2. (ii)

If $x\in\Sigma$, then $F^{n}(x)\in\Sigma$ for all $n\in\mathbb{N}$;
furthermore, there exists a backward orbit contained in $\Sigma$.

  3. (iii)

If $x\in(1,\infty)\Sigma$ then
$\lim\limits_{n\to\infty}\operatorname{dist}(F^{n}(x),\Sigma)=0$, and there
are the following possibilities:

     * •

$F^{n}(x)\in(1,\infty)\Sigma$ for all $n\in\mathbb{N}$;

     * •

there is $n_{0}\in\mathbb{N}\setminus\\{0\\}$ such that
$F^{n}(x)\in(1,\infty)\Sigma$ for $n=\\{0,\ldots,n_{0}-1\\}$ and
$F^{n}(x)\in\Sigma$ for $n\geq n_{0}$;

     * •

there is $n_{0}\in\mathbb{N}\setminus\\{0\\}$ such that
$F^{n}(x)\in(1,\infty)\Sigma$ for $n=\\{0,\ldots,n_{0}-1\\}$ and
$F^{n}(x)\in(0,1)\Sigma$ for $n\geq n_{0}$;

Further we have

######  Proposition 2.29.

$\Sigma$ attracts any bounded $A\subset C_{+}$ with $0\notin\bar{A}$. In other
words, $\Sigma$ is the compact attractor of neighbourhoods of compact sets in
$C_{+}\setminus\\{0\\}$.

We have thus obtained properties (V), (VI) and (VII) in the definition of the
carrying simplex [weak carrying simplex].

As a consequence of Proposition 2.29 we will show below that the carrying
simplex is the limit, whether in the Kuratowski sense or in the Hausdorff
metric, of the iterates of any $s_{0}\in\widehat{\mathcal{U}}$.

######  Proposition 2.30.

For $s_{0}\in\widehat{\mathcal{U}}$ put $s_{n}:=F^{n}(s_{0})$. Then the
following holds.

  1. (i)

$r_{n}$ converge uniformly to $R_{*}$, where $r_{n}$ denotes the radial
representation of $s_{n}$.

  2. (ii)

$\Sigma$ equals the Kuratowski limit of $(s_{n})_{n=0}^{\infty}$.

  3. (iii)

$d_{H}(s_{n},\Sigma)\to 0$ as $n\to\infty$.

###### Proof.

We start by noting that by Lemma 2.13, $s_{n}\in\widehat{\mathcal{U}}$, so
$r_{n}$ is well defined, for all $n$. From Proposition 2.29 it follows that
for each $k\in\mathbb{N}$ there is $n_{0}=n_{0}(k)\in\mathbb{N}$ such that
$s_{n}\subset\langle S_{k},S^{k}\rangle$ for all $n\geq n_{0}$. This means
that $R_{k}(u)\leq r_{n}(u)\leq R^{k}(u)$ for all $u\in\Delta$ and all $n\geq
n_{0}$. Since both $R_{k}$ and $R^{k}$ uniformly converge, as $k\to\infty$, to
$R_{*}$, we obtain (i).

It follows almost immediately from the above that the Kuratowski limit of
$(s_{n})_{n=0}^{\infty}$ is contained in $\Sigma$. On the other hand, for each
$u\in\Delta$ the point $R_{*}(u)u\in\Sigma$ equals the limit, as $n\to\infty$,
of $r_{n}(u)u\in s_{n}$, consequently $\Sigma$ is contained in the Kuratowski
limit of $(s_{n})_{n=0}^{\infty}$. This proves (ii).

(iii) is a consequence of the fact that in a compact metric space the
Kuratowski convergence and the convergence in the Hausdorff metric are the
same, see [1, Prop. 4.4.15]. ∎

Denote by $G_{*}$ the induced map on $\Delta$,

$G_{*}:=T\circ F\circ(R_{*}\,\mathrm{Id}).$ (2.4)

(2.4) gives conjugacy between $G_{*}$ and $F|_{\Sigma}$.

In the notation as in Proposition 2.30, define $G_{n}\colon\Delta\to\Delta$ as

$G_{n}:=T\circ F\circ(r_{n}\,\mathrm{Id}).$

Then it follows from Proposition 2.30(i) that $G_{n}$ converge uniformly to
$G_{*}$.

###  2.6 Asymptotic completeness

Let AS1–AS4 hold.

######  Proposition 2.31.

Assume additionally AS3′. For each $x\in C_{+}\setminus\\{0\\}$ there exists
$y\in\Sigma$ such that $\lim\limits_{n\to\infty}\lVert
F^{n}(x)-F^{n}(y)\rVert=0$.

###### Proof.

As the case $x\in\Sigma$ is obvious, in view of Lemma 2.10 and Theorem 2.28,
by replacing $x$ with some of its iterates, we can assume that either
$F^{n}(x)\in(0,1)\Sigma$ or $F^{n}(x)\in(1,\infty)\Sigma\cap\Lambda$, for all
$n\in\mathbb{N}$.

In the case $F^{n}(x)\in(0,1)\Sigma$ for all $n$, let
$A_{n}:=(F^{n}|_{\Lambda})^{-1}((F^{n}(x)+\mathbb{R}^{d}_{+})\cap\Sigma)$. The
sets $A_{n}$ are compact and, since $(F^{n}(x)+\mathbb{R}^{d}_{+})\cap\Sigma$
are nonempty, they are nonempty, too. We claim that $A_{n+1}\subset A_{n}$.
Indeed, let $\eta\in A_{n+1}$, which means that $F^{n+1}(\eta)\in\Sigma$ and
$F^{n+1}(x)\leq F^{n+1}(\eta)$. Proposition 2.6(1) gives that $F^{n}(x)\leq
F^{n}(\eta)$, that is,
$\eta\in(F^{n}|_{\Lambda})^{-1}((F^{n}(x)+\mathbb{R}^{d}_{+})$. The
intersection $A:=\bigcap\limits_{n=0}^{\infty}A_{n}$ is compact and nonempty.
Pick $y\in A$. By construction, $F^{n}(x)<F^{n}(y)$ for all $n\in\mathbb{N}$.
Suppose to the contrary that $\lVert F^{n}(x)-F^{n}(y)\rVert\not\to 0$ as
$n\to\infty$. Take a subsequence $n_{k}\to\infty$ such that
$\lim\limits_{k\to\infty}F^{n_{k}}(x)=u$ and
$\lim\limits_{k\to\infty}F^{n_{k}}(y)=v$ with $u\neq v$. Since the $\leq$
relation is preserved in the limit, we have $u\leq v$, consequently $u<v$. By
Theorem 2.28, both $u$ and $v$ belong to $\Sigma$, which contradicts the
unorderedness of $\Sigma$.

In the case $F^{n}(x)\in(1,\infty)\Sigma\cap\Lambda$ for all $n$, let
$A_{n}:=(F^{n}|_{\Lambda})^{-1}([0,F^{n}(x)]\cap\Sigma)$. The sets $A_{n}$ are
compact and nonempty. We claim that $A_{n+1}\subset A_{n}$. Indeed, let
$\eta\in A_{n+1}$, which means that $F^{n+1}(\eta)\in\Sigma$ and
$F^{n+1}(\eta)\leq F^{n+1}(x)$. Proposition 2.6(1) gives that $F^{n}(\eta)\leq
F^{n}(x)$, that is, $\eta\in(F^{n}|_{\Lambda})^{-1}([0,F^{n}(x)]$. The
intersection $A:=\bigcap\limits_{n=0}^{\infty}A_{n}$ is compact and nonempty.
Pick $y\in A$. By construction, $F^{n}(y)<F^{n}(x)$ for all $n\in\mathbb{N}$.
The rest of the proof goes as in the previous paragraph. ∎

######  Proposition 2.32.

For each $x\in\Gamma\setminus\\{0\\}$ there exists $y\in\Sigma$ such that
$\lim\limits_{n\to\infty}\lVert F^{n}(x)-F^{n}(y)\rVert=0$.

###### Proof.

Let a nonzero $x\in\Gamma\setminus\Sigma$. The sets $A_{n}$ are now defined as
$(F^{n}|_{\Lambda})^{-1}((F^{n}(x)+\mathbb{R}^{d}_{+})\cap\Sigma\cap(C_{I})_{+})$,
where $I:=I(x)$. As in the proof of Proposition 2.31 we obtain the existence
of $y\in\Sigma\cap(C_{I})_{+}$ such that $F^{n}(x)<F^{n}(y)$ for all
$n\in\mathbb{N}$. Take a subsequence $n_{k}\to\infty$ such that
$\lim\limits_{k\to\infty}F^{n_{k}}(x)=u$ and
$\lim\limits_{k\to\infty}F^{n_{k}}(y)=v$. By the closedness of the $\leq$
relation, $u\leq v$. For $j\in\\{1,\ldots,d\\}\setminus I$ we have
$u_{j}=v_{j}=0$.

Let $i\in I$. By Proposition 2.24,

$\lim_{k\to\infty}h(\pi_{i}(F^{n_{k}}(x)),\pi_{i}(F^{n_{k}}(y)))=\inf_{n\in\mathbb{N}}h(\pi_{i}(F^{n}(x)),\pi_{i}(F^{n}(y))).$
(2.5)

As a consequence, $0=u_{i}<v_{i}$ is impossible, that is, either
$u_{i}=v_{i}=0$ or $0<u_{i}\leq v_{i}$. Suppose $0<u_{i}<v_{i}$. It follows
from (2.5) with the help of Lemma 2.23 that

$\displaystyle h(\pi_{i}(u),\pi_{i}(v))$
$\displaystyle=\lim_{k\to\infty}h(\pi_{i}(F^{n_{k}}(x)),\pi_{i}(F^{n_{k}}(y)))$
$\displaystyle=\inf_{n\in\mathbb{N}}h(\pi_{i}(F^{n}(x)),\pi_{i}(F^{n}(y)))$
$\displaystyle=\lim_{k\to\infty}h(\pi_{i}(F^{n_{k}+1}(x)),\pi_{i}(F^{n_{k}+1}(y)))=h(\pi_{i}(F(u)),\pi_{i}(F(v))),$

But by Proposition 2.24,
$h(\pi_{i}(F(u)),\pi_{i}(F(v)))<h(\pi_{i}(u),\pi_{i}(v))$, a contradiction.
Since $u=v$ for any convergent subsequence, the statement of the proposition
holds. ∎

Therefore, (VIII) is satisfied.

###  2.7 Lipschitz Property

######  Lemma 2.33.

Assume AS1–AS4.

  1. (i)

$\Pi|_{\Sigma}$ is injective, consequently, a homeomorphism onto its image;

  2. (ii)

$\lVert\Pi x-\Pi y\rVert\geq\dfrac{1}{\sqrt{1+d}}\lVert x-y\rVert$ for any
$x,y\in\Sigma$.

###### Proof.

If $\Pi|_{\Sigma}$ were not injective, there would be $x,y\in\Sigma$ with
$y=x+{\alpha}\mathbf{e}$ for some $\alpha>0$. If $x\in C_{++}$ then $x\ll y$,
which contradicts the fact that $\Sigma$ is weakly unordered. If
$x\in\Sigma\cap\partial C_{+}$ then there is $\tilde{x}\in\Sigma\cap C_{++}$
so close to $x$ that $\tilde{x}\ll y$, again a contradiction.

In view of the linearity of $\Pi$ and the fact that no two points in $\Sigma$
are in the $\ll$ relation, we will prove (ii) if we show that for any nonzero
$u\in\mathbb{R}^{d}\setminus(C_{++}\cup(-C_{++}))$ there holds

$\frac{\lVert\Pi u\rVert}{\lVert u\rVert}\geq\dfrac{1}{\sqrt{1+d}}.$ (2.6)

Let $u\in\mathbb{R}^{d}$ be arbitrary nonzero. Again by linearity, we can
restrict ourselves to the case when $u=\pm\hat{\mathbf{e}}+v$, where $v\in V$.
If $\lVert v\rVert<\dfrac{1}{\sqrt{d}}$, then $\lVert
v\rVert_{\infty}\leq\lVert v\rVert<\dfrac{1}{\sqrt{d}}$, so $u\in
C_{++}\cup(-C_{++})$. Consequently, for
$u\in\mathbb{R}^{d}\setminus(C_{++}\cup(-C_{++}))$ we have, by the Pythagorean
theorem,

$\frac{\lVert\Pi u\rVert}{\lVert u\rVert}=\frac{\lVert
v\rVert}{\lVert\hat{\mathbf{e}}+v\rVert}=\frac{\lVert v\rVert}{\sqrt{1+\lVert
v\rVert^{2}}}\geq\dfrac{1}{\sqrt{1+d}}.$

∎

Property (IX) in the definition of carrying simplex [weak carrying simplex]
follows.

##  3 Examples

###  3.1 One-dimensional examples

Let $n=1$. Observe that our assumptions have now the following meaning:

  * AS11

$f$ is of class $C^{1}$ on $[0,\infty)$;

  * AS21

$f(1)=1$;

  * AS3${}^{\prime}_{1}$

$f(x)>0$ and $f^{\prime}(x)<0$ for all $x\geq 0$;

  * AS41

$F^{\prime}(x)=xf^{\prime}(x)+f(x)>0$ for $x\in(0,1]$.

####  3.1.1 Beverton–Holt

Take $f(x)=\frac{2}{1+x}$, so that $F(x)=\frac{2x}{1+x}$. The map $F$ has
fixed points only at $x=0,1$. AS11, AS21 are easily checked and $f(x)>0$ and
$f^{\prime}(x)=-\frac{2}{(1+x)^{2}}<0$ show AS3${}^{\prime}_{1}$ is satisfied
on $C_{+}=[0,\infty)$. Finally,

$Z(x)=-\frac{x}{f(x)}f^{\prime}(x)=\frac{x}{x+1}<1,\quad x\in[0,\infty),$

equivalently

$F^{\prime}(x)=-\frac{2x}{(1+x)^{2}}+\frac{2}{1+x}=\frac{2}{(1+x)^{2}}>0,\quad
x\in[0,\infty).$

$F$ is a homeomorphism from $[0,\infty)$ onto $[0,2)$ and we may take
$\Lambda=[0,1+\varkappa]$, where $\varkappa>0$ is arbitrary. The fixed point
$x=1$ is the carrying simplex and it is easy to check that it attracts any
$A\subset(0,\infty)$ with $0\notin\bar{A}$ (Figure 1 (a)).

Any $x\in(0,1)$ has a unique backward orbit
$\\{\ldots,x_{-n-1},x_{-n},\ldots,x_{-1},x\\}$ strictly decreasing, as
$n\to\infty$, to $0$. Its forward orbit strictly increases, as $n\to\infty$,
to $1$.

The forward orbit of any $x>1$ strictly decreases to $1$. No $x>1$ has an
(infinite) backward orbit, however for any $x\in(0,2)$ there are $n_{0}\geq 1$
and
$x=F(x_{-1})<x_{-1}=F(x_{-2})<\ldots<x_{-n_{0}+1}=F(x_{-n_{0}})<x_{-n_{0}}$.

####  3.1.2 Atkinson–Allen

Now take $f(x)=\lambda+\frac{2(1-\lambda)}{1+x}$ with $\lambda=1/2$, i.e.
$f(x)=\frac{1}{2}+\frac{1}{1+x}$. Then $F(x)=\frac{x}{2}+\frac{x}{1+x}$ has
fixed points at only $x=0,1$. Note that $f(x)>\lambda$ and
$f^{\prime}(x)=-\frac{1}{(x+1)^{2}}<0$ for $x\in C_{+}=[0,\infty)$. Hence $f$
satisfies AS11, AS21 and AS3${}^{\prime}_{1}$. Next,

$Z(x)=-\frac{x}{f(x)}f^{\prime}(x)=\frac{x}{(1+x)(1+\frac{(1+x)}{2})}<\frac{x}{(1+x)}<1,\quad
x\in[0,\infty),$

equivalently

$F^{\prime}(x)=-\frac{x}{(1+x)^{2}}+\frac{1}{2}+\frac{1}{1+x}=\frac{3+2x+x^{2}}{2(1+x)^{2}}>0,\quad
x\in[0,\infty),$

so AS41 is satisfied. In this example $F$ is a diffeomorphism from
$[0,\infty)$ onto $[0,\infty)$ and we may take $\Lambda=[0,1+\varkappa]$,
where $\varkappa>0$ is arbitrary. The carrying simplex is the fixed point
$x=1$, which attracts any bounded $A\subset(0,\infty)$ with $0\notin\bar{A}$.
(Figure 1 (b)).

Any $x\in(0,1)$ has a unique backward orbit
$\\{\ldots,x_{-n-1},x_{-n},\ldots,x_{-1},x\\}$ strictly decreasing, as
$n\to\infty$, to $0$. Its forward orbit strictly increases, as $n\to\infty$,
to $1$.

The forward orbit of any $x>1$ strictly decreases to $1$. Any $x>1$ has a
unique backward orbit $\\{\ldots,x_{-n-1},x_{-n},\ldots,x_{-1},x\\}$ strictly
increasing, as $n\to\infty$, to $\infty$.

####  3.1.3 Ricker

Take $f(x)=e^{\lambda(1-x)}$ with $\lambda>0$. Then $F(x)=xe^{\lambda(1-x)}$
has fixed points at only $x=0,1$. Note that $f(x)>0$ and
$f^{\prime}(x)=-\lambda e^{\lambda(1-x)}<0$ for $x\in C_{+}=[0,\infty)$. Hence
$f$ satisfies AS11, AS21 and AS3${}^{\prime}_{1}$.

Assume that $\lambda\in(0,1)$. To be more specific, take $\lambda=1/2$, i.e.
$f(x)=e^{(1-x)/2}$. Next,

$Z(x)=-\frac{x}{f(x)}f^{\prime}(x)=-\frac{x}{e^{(1-x)/2}}\Bigl{(}-\frac{1}{2}e^{(1-x)/2}\Bigr{)}=\frac{x}{2}<1,\quad
x\in[0,2),$

equivalently

$F^{\prime}(x)=-\frac{x}{2e^{(1-x)/2}}+e^{(1-x)/2}=\Bigl{(}1-\frac{x}{2}\Bigr{)}e^{(1-x)/2}>0,\quad
x\in[0,2),$

so AS41 is satisfied if we take $\Lambda=[0,1+\varkappa]$, where
$\varkappa\in(0,1)$. The function $F$ reaches a peak value of $2e^{-1/2}$ at
$x=2$, and $F$ restricted to $[0,2]$ is a homeomorphism onto $[0,2e^{-1/2}]$
(it is a diffeomorphism from $[0,2)$ onto $[0,2e^{-1/2})$). The carrying
simplex is the fixed point $x=1$, which attracts any $A\subset(0,\infty)$ with
$0\notin\bar{A}$.

Any $x\in(0,1)$ has a unique backward orbit
$\\{\ldots,x_{-n-1},x_{-n},\ldots,x_{-1},x\\}$ strictly decreasing, as
$n\to\infty$, to $0$. Its forward orbit strictly increases, as $n\to\infty$,
to $1$.

Put $x^{*}>1$ to be the only solution of $F(x^{*})=1$. For any $x\in(1,x^{*})$
its forward orbit strictly decreases, as $n\to\infty$, to $1$.
$F^{n}(x^{*})=1$ for all $n\geq 1$. If $x>x^{*}$ then $F(x)\in(0,1)$, so its
forward orbit eventually strictly increases to $1$.

Any $x\in(0,1)$ has two inverse images: the lower in $(0,x)$ and the upper in
$(x^{*},\infty)$. Any $x\in(1,2e^{-1/2})$ has two inverse images: the lower in
$(1,2)$ and the upper in $(2,x^{*})$ (Figure 1 (c)).

When $\lambda>1$ it is not possible to find a $\Lambda\ni 1$ for which AS41 is
satisfied, although for $\lambda<2$ the point $x=1$ attracts every point in
$(0,\infty)$ (Figure 1 (d)).

(a)  (b)

(c)  (d)

Figure 1: Examples for $d=1$.

###  3.2 Planar Ricker

Consider the following planar Ricker model:

$F(x,y)=(xe^{r(1-x-ay)},ye^{s(1-y-bx)})$ (3.1)

where $r,s>0$, $a,b\geq 0$. The map $F\colon C_{+}\rightarrow C_{+}$ where
$C_{+}=[0,\infty)^{2}$, always has $(0,0)$, $\mathbf{e}_{1}=(1,0)$ and
$\mathbf{e}_{2}=(0,1)$ as fixed points, so AS1 and AS2 are satisfied.

$DF(x,y)=\left(\begin{array}[]{cc}e^{r(1-x-ay)}(1-rx)&-arxe^{r(1-x-ay)}\\\
-sbye^{s(1-y-bx)}&e^{s(1-y-bx)}(1-sy)\end{array}\right)$

and $Z(x)=\left(\begin{array}[]{cc}rx&arx\\\ sby&sy\end{array}\right)$.

Under our assumptions AS3 is satisfied, and AS3′ holds provided $a>0$ and
$b>0$. The eigenvalues $\lambda$ of $Z(x,y)$ satisfy
$\lambda^{2}-(rx+sy)\lambda+rsxy(1-ab)=0$, so by the Jury condition the matrix
$Z(x,y)$ has spectral radius less than one when $(x,y)\in C_{+}$ satisfies

$rx+sy<1+rsxy(1-ab)<2.$ (3.2)

The bounded connected component $\Omega\subset C_{+}$ of the set defined by
(3.2) is a simply connected region whose closure contains the points
$(0,0),(1/r,0),(0,1/s)$. For the Ricker model to satisfy AS4 on $[0,1]^{2}$ we
require $[0,1]^{2}\subset\Omega$. Thus natural conditions are

$r,s<1\quad\text{and}\quad r+s<1+rs(1-ab)<2$

(since $\mathbf{e}_{1}$, $\mathbf{e}_{2}$, $(1,1)$ must belong to
$\bar{\Omega}$). Indeed, the former condition is redundant: $r+s<1+rs(1-ab)$
is equivalent to $(1-r)(1-s)>rsab$, from which it follows that $r$ and $s$ are
either both $>1$ or both $<1$, and $r+s<2$ gives $r,s<1$.

Since the boundary of $\Omega$ in $C_{+}$ is the graph of a decreasing
function we can take $\Lambda=[0,1+\varkappa]^{2}$, where $\varkappa>0$ is so
small that

$(1+\varkappa)(r+s)<1+(1+\varkappa)^{2}rs(1-ab)<2.$

In view of Proposition 2.6(2) we have the following.

######  Lemma 3.1.

The Ricker map (3.1) is weakly retrotone on $\Lambda$ if $r,s>0$, $a,b\geq 0$
and

$r+s<1+rs(1-ab)<2$ (3.3)

If $r,s,a,b>0$ and (3.3) holds then (3.1) is retrotone.

## Acknowledgements

J. Mierczyński is partially supported by the Faculty of Pure and Applied
Mathematics, Wrocław University of Science and Technology.

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