Max-plus convexity in Riesz spaces

TL;DR

This paper explores max-plus convexity within Riesz spaces, establishing algebraic definitions, associated norms, and geodesic distances, and extends fixed point and selection theorems to this setting, showing hyperspaces are Absolute Retracts.

Contribution

It introduces a new algebraic framework for max-plus convexity in Riesz spaces and extends classical convexity results to this non-linear context without assuming topological structures.

Findings

Max-plus convex sets are geodesically closed under certain norms.

Fixed point and continuous selection theorems are extended to max-plus convex sets.

Hyperspaces of compact max-plus convex sets are Absolute Retracts.

Abstract

We study max-plus convexity in an Archimedean Riesz space $E$ with an order unit $\un$; the definition of max-plus convex sets is algebraic and we do not assume that $E$ has an {\it a priori} given topological structure. To the given unit $\un$ one can associate two equivalent norms $\norm\cdot\norm_{\un}$ and $\norm\cdot\norm_{\hun}$ on $E$; the distance ${\sf D}_{\hun}$ on $E$ associated to $\norm\cdot\norm_{\hun}$ is a geodesic distance for which max-plus convex sets in $E$ are geodesically closed sets. Under suitable assumptions, we establish max-plus versions of some fixed points and continuous selection theorems that are well known for linear convex sets and we show that hyperspaces of compact max-plus convex sets are Absolute Retracts.

Equations

$$\forall x,y,z\in E\forall t\in {\mathbb{R}}_{+}x⩽y\Rightarrow (tx+z)⩽(ty+z)$$

$$\{\begin{array}{ll}(1)& \forall x,y\in Ex⩽y\iff (y-x)\in E_{+}\\ (2)& \forall x,y\in E_{+}\forall t\in {\mathbb{R}}_{+}(tx+y)\in E_{+}\end{array}$$

$$\forall \hat{x}\in \hat{E}\hat{x}=\sup \{x\in E:x⩽\hat{x}\}=\inf \{x\in E:\hat{x}⩽x\}$$

$$\forall x\in E_{+}\exists n\in \mathbb{N}\text{ such that }x⩽n\mathbf{u}$$

$$\forall x,y\in E|x|⩽|y|\Rightarrow \Vert x\Vert \le \Vert y\Vert$$

$$\forall x,y\in E_{+}\Vert x\vee y\Vert =\max \{\Vert x\Vert ,\Vert y\Vert \}$$

$$\Vert x{\Vert }_{\mathbf{u}}=\inf \{t\in {\mathbb{R}}_{\star }:|x|⩽t\mathbf{u}\}$$

$$\Vert x{\Vert }_{\mathbf{u}}=\underset{\omega \in \Omega }{\max }\frac{|x(\omega )|}{\mathbf{u}(\omega )}.$$

$$(\mu\vee\nu)(A)=\sup\big{\{}\mu(B)+\nu(A\setminus\!B):B\subseteq A\text{ and }B\in\mathcal{F}\big{\}}.$$

$$\{\begin{array}{l}{\mathbf{p}}_{\mathbf{u}}(x)=\inf \{t\in \mathbb{R}:x⩽t\mathbf{u}\}\\ {\mathbf{q}}_{\mathbf{u}}(x)=\sup \{t\in \mathbb{R}:t\mathbf{u}⩽x\}\end{array}$$

$${\mathbf{p}}_{\mathbf{u}}(x)=-{\mathbf{q}}_{\mathbf{u}}(-x)$$

$$\forall x,y\in Ex⩽y\Rightarrow \{\begin{array}{l}{\mathbf{p}}_{\mathbf{u}}(x)\le {\mathbf{p}}_{\mathbf{u}}(y)\\ {\mathbf{q}}_{\mathbf{u}}(x)\le {\mathbf{q}}_{\mathbf{u}}(y)\end{array}$$

$$\forall x\in E{\mathbf{q}}_{\mathbf{u}}(x)\mathbf{u}⩽x⩽{\mathbf{p}}_{\mathbf{u}}(x)\mathbf{u}$$

$$\forall x,y\in E\{\begin{array}{l}(1){\mathbf{p}}_{\mathbf{u}}(x\vee y)=\max \{{\mathbf{p}}_{\mathbf{u}}(x),{\mathbf{p}}_{\mathbf{u}}(y)\}\\ (2){\mathbf{q}}_{\mathbf{u}}(x\wedge y)=\min \{{\mathbf{q}}_{\mathbf{u}}(x),\mathbf{q}(y)\}\end{array}$$

$$\forall x\in E{\mathbf{p}}_{\mathbf{u}}(x^{+})=\max \{0,{\mathbf{p}}_{\mathbf{u}}(x)\}\text{ and }{\mathbf{p}}_{\mathbf{u}}(x^{-})=\max \{0,{\mathbf{p}}_{\mathbf{u}}(-x)\}$$

$$\forall x,y\in E\text{ and }\forall s\in {\mathbb{R}}_{+}\{\begin{array}{l}(1){\mathbf{p}}_{\mathbf{u}}(x+y)\le {\mathbf{p}}_{\mathbf{u}}(x)+{\mathbf{p}}_{\mathbf{u}}(y)\text{ and }{\mathbf{p}}_{\mathbf{u}}(sx)=s{\mathbf{p}}_{\mathbf{u}}(x)\\ (2){\mathbf{q}}_{\mathbf{u}}(x)+{\mathbf{q}}_{\mathbf{u}}(y)\le {\mathbf{q}}_{\mathbf{u}}(x+y)\text{ and }{\mathbf{q}}_{\mathbf{u}}(sx)=s{\mathbf{q}}_{\mathbf{u}}(x)\end{array}$$

$$x\mapsto {\mathbf{p}}_{\mathbf{u}}(x^{+})+{\mathbf{p}}_{\mathbf{u}}(x^{-})=\Vert x{\Vert }_{\mathbf{hu}}$$

$$\forall x\in E\Vert x{\Vert }_{\mathbf{hu}}=\Vert x^{+}{\Vert }_{\mathbf{hu}}+\Vert x^{-}{\Vert }_{\mathbf{hu}}=\Vert x^{+}{\Vert }_{\mathbf{u}}+\Vert x^{-}{\Vert }_{\mathbf{u}}$$

$$\Vert x{\Vert }_{\mathbf{hu}}=\max \{x_i^{+}:i\in [n]\}+\max \{x_i^{-}:i\in [n]\}={\mathbf{p}}_{\mathbf{u}}(x^{+})+{\mathbf{p}}_{\mathbf{u}}(x^{-})$$

$$\forall (x,y,t)\in E\times E\times \mathbb{R}\{\begin{array}{l}x\oplus y\text{ for }x\vee y\text{ and}\\ t\odot x\text{ for }x+t\mathbf{u}.\end{array}$$

$$\{\begin{array}{ll}(1)& t\odot (x\oplus y)=(t\odot x)\oplus (t\odot y)\\ (2)& (t_1\oplus t_2)\odot x=(t_1\odot x)\oplus (t_2\odot x)\\ (3)& t_2\odot (t_1\odot x)=(t_1\odot t_2)\odot x\\ (4)& 1\odot x=x\end{array}$$

$$\{\begin{array}{ll}(1)\forall x\in E\cup \{\perp \}& \perp \oplus x=x\oplus \perp =x\\ (2)\forall x\in E\cup \{\perp \}& -\infty \odot x=\perp \\ (3)\forall t\in \mathbb{R}\cup \{-\infty \}& t\odot \perp =\perp \end{array}$$

$$[[S\cup \{x\}]]=\bigcup _{y\in [[S]]}[[x,y]]$$

$$\begin{cases}(a)\,C\text{ is max-plus convex}.\\ (b)\,\forall t\leq 0\quad\forall x,y\in C\quad x\vee(y+t\boldsymbol{u})\in C\\ (c)\,\big{[}\,\forall x_{1},x_{2}\in C\quad x_{1}\vee x_{2}\in C\big{]}\text{ and }\big{[}\,\forall x_{1},x_{2}\in C\text{ such that }x_{1}\leqslant x_{2}\quad\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}\subset C\,\big{]}.\end{cases}$$

$${\mathsf{D}}_{\mathbf{hu}}(x_1,x_2)={\mathsf{D}}_{\mathbf{hu}}(x_1,x_1\vee x_2)+{\mathsf{D}}_{\mathbf{hu}}(x_2,x_1\vee x_2).$$

$${\mathsf{D}}_{\mathbf{hu}}(x_1,x_2)={\mathsf{D}}_{\mathbf{hu}}(x_1,z)+{\mathsf{D}}_{\mathbf{hu}}(z,x_2).$$

$${\mathsf{D}}_{\mathbf{hu}}(x_1,x_2)={\mathsf{D}}_{\mathbf{hu}}(x_1,z)+{\mathsf{D}}_{\mathbf{hu}}(z,x_2)$$

$${\mathsf{D}}_{\mathbf{hu}}(x_1,x_2)={\mathsf{D}}_{\mathbf{hu}}(x_1,z)+{\mathsf{D}}_{\mathbf{hu}}(z,x_1\vee x_2)+{\mathsf{D}}_{\mathbf{hu}}(x_2,x_1\vee x_2).$$

$${\mathsf{D}}_{\mathbf{hu}}(z,z^{\prime })=\{\begin{array}{ll}{\mathsf{D}}_{\mathbf{hu}}(z,x_1\vee x_2)-{\mathsf{D}}_{\mathbf{hu}}(z^{\prime },x_1\vee x_2)& \text{if }{x}_i⩽z⩽z^{\prime }⩽x_1\vee x_2\\ {\mathsf{D}}_{\mathbf{hu}}(z,x_1\vee x_2)+{\mathsf{D}}_{\mathbf{hu}}(z^{\prime },x_1\vee x_2)& \text{if }{x}_1⩽z⩽x_1\vee x_2\text{ and }{x}_2⩽z^{\prime }⩽x_1\vee x_2\end{array}$$

$$t\le {\mathbf{q}}_{\mathbf{u}}(z-x_2)$$

Full text

horvath@univ-perp.fr

MSC: 14T99, 46A40, 54H25, 54C55, 54C65, 62P20.

1. Introduction

To keep the size of this paper reasonable, the definition of max-plus convexity in arbitrary Riesz spaces with respect to a given unit $\boldsymbol{u}$ is given in Section 4 with barely no justification as to why one should be interested in max-plus convexity. If the Riesz space in question is $\mathbb{R}^{n}$ and $\boldsymbol{u}=(1,\cdots,1)$ then the max-plus convex sets with respect to $\boldsymbol{u}$ are the usual max-plus convex subsets of $\mathbb{R}^{n}$. The usual finite dimensional max-plus convexity lives in \big{(}\mathbb{R}\cup\{-\infty\}\big{)}^{n}, the extension to arbitrary Archimedean Riesz spaces with a unit that is presented here is therefore not a full generalization of the finite dimensional framework which does not mean that such a thing could not be done. The reader looking for motivations and applications is referred to [15], [20], [23], [26]. Section 2 is mainly about basic concepts and a few exemples. Section 3 is about two norms, denoted here by $\|\cdot\|_{\boldsymbol{u}}$ and $\|\cdot\|_{\boldsymbol{hu}}$, which can be associated to a given unit $\boldsymbol{u}$ in an Archimedean Riesz space $E$. If $E=\mathbb{R}^{n}$ and $\boldsymbol{u}=(1,\cdots,1)$ then $\|(x_{1},\cdots,x_{n})\|_{\boldsymbol{u}}=\max|x_{i}|$ and $\|(x_{1},\cdots,x_{n})\|_{\boldsymbol{hu}}=\max x_{i}^{+}+\max x_{i}^{-}$, the so called Hilbert affine norm. In Section 4 one will find the definition of max-plus convex sets in an Archimedean Riesz space, with respect to a given unit, and some of their basic properties the most important one being the Kakutani Property, also known in the standard linear framework as the Algebraic Hahn-Banach Property. Section 5 shows that ${\sf D}_{\boldsymbol{hu}}$, the metric associated to $\|\cdot\|_{\boldsymbol{hu}}$, is a geodesic distance on the Riesz space $E$ with respect to which the geodesically closed sets are precisely the max-plus convex sets, with respect to the given unit $\boldsymbol{u}$. That max-plus convexity in $\mathbb{R}^{n}$ should be a geodesic structure with respect to an appropriate metric is a not so rescent idea; it had been discussed years ago with Walter Briec from Université de Perpignan and it is Stefan Gaubert, in a discussion with the author at École Polytechnique, who hinted at the fact that the Hilbert affine metric should be the appropriate metric; shortly thereafter, in a private communication [14] mailed to the author, Stefan Gaubert proved that this is indeed the case. The proof given here differs somewhat from Gaubert’s straightforward coordinatewise proof in $\mathbb{R}^{n}$ but would have been impossible without that proof. Section 6 deals with the basic topological properties of max-plus convex sets and hyperspaces of compact max-plus sets; Section 7 is about infinite dimensional max-plus versions of some standard results: Ky Fan Best Approximation and conequently Schauder’s Fixed Point Theorem, Kakutani’s Fixed Point Theorem for upper semicontinuous maps, Michael’s Selection Theorem, Dugundji’s Extension Theorem, and consequently the fact that max-plus convex sets are Absolute Retracts. Max-plus convexity in $\mathbb{R}^{S}$ and hyperspaces of compact max-plus convex subsets of $\mathbb{R}^{S}$ have been studied by L. Bazylevych, D. Repovs and M. Zarichnyi in [3]. Hyperspaces of max-plus compact convex sets in $\mathcal{C}(X)$, where $X$ is a compact metrizable topological space, have been studied by L. Bazylevych and M. Zarichnyi in [2]. In Section 6 one can find a few remarks on hyperspaces of compact max-plus convex sets in a Riesz space .

2. Preliminaries and a few examples

We will denote by $\mathbb{R}_{+}$ the set of positive real numbers and by $\mathbb{R}_{++}$ the set of strictly positive real numbers.

A Riesz space, or a vector lattice, is a real vector space $E$ endowed with a partial order $\leqslant$ that is compatible with the linear structure, that is

[TABLE]

and such that all pairs $\{x,y\}$ of elements of $E$ have a least upper bound for which we will use the standard notation $x\vee y$.

The positive cone is $E_{+}=\{x\in E:0\leqslant x\}$ which has the following properties;

[TABLE]

All pairs $\{x,y\}$ of elements of $E$ have a greatest lower bound, for which we will use the standard notation $x\wedge y$; one easily sees that x\wedge y=-\big{(}(-x)\vee(-y)\big{)}.

A Riesz space $E$ is Archimedean if, whenever $x$ and $y$ are two elements of $E$ such that, for all $n\in\mathbb{N}$, $ny\leqslant x$, one has $y\leqslant 0$.

A Riesz space $E$ is Dedekind complete (respectively, Dedekind $\boldsymbol{\sigma}$-complete) if every non-empty (respectively, countable) subset $S$ of $E$ which has an upper bound has a least upper bound.

Since $S$ has an upper bound if and only if $-S$ has a lower bound one can replace in the definition of Dedekind complete (resp. $\sigma$-completness) “upper bound” by “lower bound”. A Dedekind $\sigma$-complete Riesz space is Archimedean.

Every Archimedean Riesz space has a Dedekind completion, more precisely: there exists a Dedekind complete Riesz space $\hat{E}$ containing $E$ as a vector sublattice such that

[TABLE]

A strong order unit of a Riesz space $E$ is an element $\boldsymbol{u}\in E_{+}$ such that

[TABLE]

Since strong units are the only kind of units we will consider we will drop the adjective “strong”.

A Riesz norm on a Riesz space $E$ is a norm such that,

[TABLE]

For a Riesz norm one has, for all $x\in E$, $\|\boldsymbol{|}x\boldsymbol{|}\|=\|x\|$. 111$x^{+}=0\vee x$, $x^{-}=-(0\wedge x)=0\vee(-x)$ and $\boldsymbol{|}x\boldsymbol{|}=x\vee(-x)=x^{+}+x^{-}=x^{+}\vee x^{-}$. Also, $\boldsymbol{|}tx\boldsymbol{|}=|t|\,\boldsymbol{|}x\boldsymbol{|}$, $\boldsymbol{|}x+y\boldsymbol{|}\leqslant\boldsymbol{|}x\boldsymbol{|}+\boldsymbol{|}y\boldsymbol{|}$, $\boldsymbol{|}x\boldsymbol{|}\leqslant y$ if and only if $-y\leqslant x\leqslant y$, $(x+y)^{+}\leqslant x^{+}+y^{+}$, $(x+y)^{-}\leqslant x^{-}+y^{-}$, $(x\vee y)^{+}=x^{+}\vee y^{+}$, $(x\vee y)^{-}=x^{-}\wedge y^{-}$.

A Riesz space equipped with a Riesz norm is a normed lattice. A normed lattice is Archimedean and the lattice operations are uniformly continuous.

Let $S$ be a subset of a Riesz space $E$ for which there exist $x_{1},x_{2}\in E$ such that, for all $x\in S$, $x_{1}\leqslant x\leqslant x_{2}$ ($S$ is an order bounded set); if $\|\cdot\|$ is a Riesz norm on $E$ then, $x\in S$, $\|x_{1}\|\leqslant\|x\|\leqslant\|x_{2}\|$. That is, in a normed lattice an order bounded set is norm-bounded.

Any two complete lattice norms on a given Riesz space $E$ are equivalent, page 352 or [12] Proposition 25 A.

An M**-norm** on a Riesz space $E$ is a Riesz norm such that,

[TABLE]

If $\|\cdot\|$ is an M-norm on $E$ and if $0\leqslant x\leqslant y$ then $\|x\|\leq\|y\|$ ( from $x\vee y=y$, $\|y\|=\max\{\|x\|,\|y\|\}$). If $\|\cdot\|$ is an M-norm and a Riesz norm on $E$ then, for all $x,y\in E$, $\|x\vee y\|\leq\max\{\|x\|,\|y\|\}$.

An AM**-space** is a Riesz space equipped with a complete norm which is an M-norm and a Riesz norm.

An AM**-space with a unit** (resp. an M**-space with a unit**) is an AM-space $E$ (resp. an M-space ) with a unit $\boldsymbol{u}$ such that $\|\boldsymbol{u}\|=1$ in which case the unit ball is $\{x\in E:-\boldsymbol{u}\leqslant x\leqslant\boldsymbol{u}\}$.

There is a standard way to associate to each given unit $\boldsymbol{u}$ on an Archimedean Riesz space $E$ an M-norm $\|\cdot\|_{\boldsymbol{u}}$ on $E$, namely :

[TABLE]

That norm $\|\cdot\|_{\boldsymbol{u}}$ is a Riesz norm is evident; that it is also an M-norm is well known, it can also be seen from Lemma 2 below.

If $\boldsymbol{u}_{1}$ and $\boldsymbol{u}_{2}$ are two units on an Archimedean Riesz space $E$ then, the norms $\|\cdot\|_{\boldsymbol{u}_{1}}$ and $\|\cdot\|_{\boldsymbol{u}_{2}}$ are equivalent since there exists two natural numbers $n_{1}$ and $n_{2}$ such that $\boldsymbol{u}_{1}\leqslant n_{1}\boldsymbol{u}_{2}$ and $\boldsymbol{u}_{2}\leqslant n_{2}\boldsymbol{u}_{1}$.

In an Archimedean Riesz space $E$, possibly without a unit, take an arbitrary element $\boldsymbol{u}\in E_{+}\!\setminus\!\{0\}$ and let

$E_{\boldsymbol{u}}=\{x\in E:\exists n\in\mathbb{N}\,\,\boldsymbol{|}x\boldsymbol{|}\leqslant n\boldsymbol{u}\}$ (the principle ideal spanned by $\boldsymbol{u}$). Then \big{(}E_{\boldsymbol{u}},\|\cdot\|_{\boldsymbol{u}}\big{)} is an M-space with unit. If $\boldsymbol{u}$ is a unit of $E$ then $E_{\boldsymbol{u}}$ is $E$ itself. Furthermore, if $(E,\|\cdot\|)$ is a complete normed lattice or if $E$ is Dedekind $\sigma$-complete then, for all $\boldsymbol{u}\in E$, \big{(}E_{\boldsymbol{u}},\|\cdot\|_{\boldsymbol{u}}\big{)} is an AM-space with unit; details can be found [12], Lemma 25I and Lemma 25J.

If the Riesz space $E$ is equipped with a complete Riesz norm $\|\cdot\|$ and if $\boldsymbol{u}$ is a unit in $E$ then $\|\cdot\|$ and $\|\cdot\|_{\boldsymbol{u}}$ are complete lattice norms on $E$; they are therefore equivalent.

If $\Omega$ is a compact topological space then the space $\mathcal{C}(\Omega)$ of continuous real valued functions on $\Omega$ is a Riesz space, the fucntion $\boldsymbol{1}$ identically equal to $1$ is a unit and, for all $x\in\mathcal{C}(\Omega)$, $\|x\|_{\boldsymbol{1}}=\sup_{\omega\in\Omega}|x(\omega)|=\|x\|_{\infty}$; \big{(}\mathcal{C}(\Omega),\|\cdot\|_{\infty}\big{)} is of course an AM-space with unit.

The Bohnenblust-Kakutani-Krein Representation Theorem says that ${\sf AM}$-spaces with a unit are isomorphic, as normed Riesz spaces222Riesz spaces $E_{1}$ and $E_{2}$ are isomorphic if there exists a linear isomorphism which is also a lattice isomorphism. , to spaces of continuous functions on a compact topological space $\Omega$; see [32], Chapter 17, &121.

(BKK RepresentationTheorem) An AM-space with unit is Riesz isomorphic and norm isomorphic to a space $\mathcal{C}(\Omega)$ equipped with the $\sup$-norm, for an appropriate compact Hausdorff space $\Omega$.

Now, consider an Archimedean Riesz space $E$ with unit $\boldsymbol{u}$ and let $\hat{E}$ be its Dedekind completion; by the preceding discussion, \big{(}\hat{E}_{\boldsymbol{u}},\|\cdot\|_{\boldsymbol{u}}\big{)} is an AM-space with unit; by the BBK-Representation Theorem there exits a compact topological space $\Omega$ and a Riesz space isomorphism $\Phi:\hat{E}_{\boldsymbol{u}}\to\mathcal{C}(\Omega)$ such that, for all $\hat{x}\in\hat{E}_{\boldsymbol{u}}$, $\|\hat{x}\|_{\boldsymbol{u}}=\|\Phi(\hat{x})\|_{\infty}$ and $\Phi(\boldsymbol{u})=\boldsymbol{1}$; since $E$ is itself a Riesz subspace of $\hat{E}_{\boldsymbol{u}}$ and $\boldsymbol{u}\in E$; $E$ can be identified with the Riesz subspace $\Phi(E)=\tilde{E}$ of $\mathcal{C}(\Omega)$. In conclusion,

An Archimedean Riesz $E$ space with unit $\boldsymbol{u}$ can be identified with a Riesz subspace of a space $\mathcal{C}(\Omega)$ where $\Omega$ is a compact Hausdorff space that is, there is a Riesz subspace ${\tilde{E}}$ of $\mathcal{C}(\Omega)$ such that $\boldsymbol{1}\in{\tilde{E}}$ and there exists a Riesz space isomorphism $\Phi:E\to{\tilde{E}}$ such that, for all $x\in E$, $\|x\|_{\boldsymbol{u}}=\|\Phi(x)\|_{\infty}$.

The closure $\bar{E}$ of the Riesz space $\tilde{E}$ in $\mathcal{C}(\Omega)$ is an AM-space, and since $\boldsymbol{1}\in{\tilde{E}}\subseteq\bar{E}$ it is an AM-space with unit, since $\boldsymbol{1}$ is a unit in $\mathcal{C}(\Omega)$, which shows that an Archimedean Riesz space possessing a unit can be embedded as a dense Riesz subspace of an AM-space with unit and therefore as dense Riesz subspace of some $\mathcal{C}(\Omega^{\prime})$ containing $\boldsymbol{1}$, where $\Omega^{\prime}$ is some compact topological space. If the Riesz space $\tilde{E}$ separates the points of $\Omega$ then $\bar{E}=\mathcal{C}(\Omega)$, by the Stone-Weierstrass Theorem.

We conclude this section with a few simple examples of the previous constructions.

For all sets $S$, the space $\mathcal{F}(S)=\mathbb{R}^{S}$ of arbitrary real valued functions on $S$ with pointwise operations, is an Archimedean Riesz space, without a unit if $S$ is not a finite set; $\mathcal{F}_{\star}(S)$, the space of bounded real valued functions on $S$, is an Archimedean Riesz space with a unit: $\boldsymbol{u}=(1,1,1,\ldots)$.

Let $\boldsymbol{u}\in\mathcal{F}(S)$ be a positive function which is not identically [math]. For all $x\in\mathcal{F}(S)$ let $Z(x)=\{\omega\in S:x(\omega)=0\}$. Let $E=\mathcal{F}(S)$; then $x\in E_{\boldsymbol{u}}$ if $Z(x)\subset Z(\boldsymbol{u})$ and $\displaystyle{\sup_{\omega\not\in Z(\boldsymbol{u})}\frac{|x(\omega)|}{\boldsymbol{u}(\omega)}<\infty}$ in which case $\|x\|_{\boldsymbol{u}}$ is this supremum. If $\boldsymbol{u}:S\to\mathbb{R}$ is identically equal to one the $E_{\boldsymbol{u}}=\mathcal{F}_{\star}(S)$.

If $\Omega$ is a compact topological space then ${\mathcal{C}}(\Omega)$ is Dedekind complete (resp. Dedekind $\sigma$-complete) if and only if the closure of every open set (resp. of every $F_{\sigma}$ open set) is open. If $\boldsymbol{u}\in\mathcal{C}(\Omega)$ is a unit, that is a strictly positive function, then

[TABLE]

The classical sequence spaces $l_{p}$ are Riesz subspaces of $\mathbb{R}^{\mathbb{N}}$; the sequence whose terms are all equal to $1$ is a unit of $l_{\infty}$.

Let \big{(}X,\mathcal{B},\mu\big{)} be a measured space where the measure $\mu$ is finite; {\mathcal{M}}\big{(}X,\mathcal{B},\mu\big{)}, the space of almost $\mu$- everywhere finite real valued functions, with the usual identification of almost $\mu$-everywhere equal functions, is a Dedekind complete Riesz space. Also, still under the hypothesis that $\mu$ is a finite measure, the spaces L_{p}\big{(}X,\mathcal{B},\mu\big{)}, $1\leq p\leq\infty$ are Dedekind complete. Details can be found in [25] page $126-127$.

Given a measured space $\left(\Omega,\mathcal{B},\mu\right)$, $L_{\infty}\left(\Omega,\mathcal{B},\mu\right)$ is also, with for its usual norm, an AM-space: the constant map $\boldsymbol{u}(\omega)=1$ is a unit.

Let $\Omega$ be a non empty set, $\mathcal{F}$ a field of subsets of $\Omega$ and ${\sf ba}(\Omega,\mathcal{F})$ the family of bounded charges333$\emptyset\in\mathcal{F}$ on $\mathcal{F}$ and $\Omega\in\mathcal{F}$ ; if $A,B\in\mathcal{F}$ then $A\cup B\in\mathcal{F}$ and $A\setminus\!B\in\mathcal{F}$. An element of ${\sf ba}(\Omega,\mathcal{F})$ is a map $\mu:\mathcal{F}\to\mathbb{R}$ which is additive ($\mu(A\cup B)=\mu(A)+\mu(B)$ if $A\cap B=\emptyset$) and such that $\mu(\emptyset)=0$ and $\sup|\mu(A)|<\infty$. on $\mathcal{F}$; ${\sf ba}(\Omega,\mathcal{F})$ is a real vector space and the relation $\mu\leqslant\nu$ if, for all $A\in\mathcal{F}$, $\mu(A)\leq\nu(A)$, is a partial order on ${\sf ba}(\Omega,\mathcal{F})$ compatible with the vector space structure; endowed with that partial order ${\sf ba}(\Omega,\mathcal{F})$ is a Riesz space, the maximum of two elements $\mu$ and $\nu$ of ${\sf ba}(\Omega,\mathcal{F})$ is given by, for all $A\in\mathcal{F}$,

[TABLE]

Furthermore, ${\sf ba}(\Omega,\mathcal{F})$ is Dedekind complete and $\mu\mapsto\boldsymbol{|}\mu\boldsymbol{|}(\Omega)$ is a complete Riesz norm on ${\sf ba}(\Omega,\mathcal{F})$ but not an M-norm444By definition of the absolute value of an element of a Riesz space, $\boldsymbol{|}\mu\boldsymbol{|}=\mu^{+}+\mu^{-}$; an explicit formula for the absolute value of a bounded charge $\mu$ is \boldsymbol{|}\mu(A)\boldsymbol{|}=\sup_{\mathcal{R}}\big{\{}\sum_{F\in\mathcal{R}}|\mu(F)|:\mathcal{R}\text{ is a finite partition of }A\text{ by elements of }\mathcal{F}\big{\}}.; by the discussion above, if $\boldsymbol{u}\in{\sf ba}(\Omega,\mathcal{F})$ is a positive charge then ${\sf ba}(\Omega,\mathcal{F})_{\boldsymbol{u}}$ equipped with the norm $\|\cdot\|_{\boldsymbol{u}}$ is an AM-space.

A charge $\mu\in{\sf ba}(\Omega,\mathcal{F})$ is a measure if, for all countable family $\{F_{n}:n\in\mathbb{N}\}$ of pairwise disjoint elements of $\mathcal{F}$ whose union belongs to $\mathcal{F}$, one has $\mu(\cup_{n}F_{n})=\sum_{n}\mu(F_{n})$; the set of elements of ${\sf ba}(\Omega,\mathcal{F})$ that are measures is a Dedekind-complete Riesz subspace of ${\sf ba}(\Omega,\mathcal{F})$ which is also a Banach sublattice. Details can be found in [4].

We assume throughout that $\boldsymbol{E}$ is an Archimedean Riesz space and that $\boldsymbol{u}$ is a unit of $\boldsymbol{E}$.

3. Max-plus norms in Archimedean spaces

For all $x\in E$ let

[TABLE]

One has

[TABLE]

and

[TABLE]

Since $\boldsymbol{u}$ is a unit, $\{t\in\mathbb{R}:x\leqslant t\boldsymbol{u}\}\neq\emptyset$ from which $\boldsymbol{p}_{\boldsymbol{u}}(x)<\infty$. If $\boldsymbol{p}_{\boldsymbol{u}}(x)$ were $-\infty$ then we would have, for all $t\in\mathbb{R}$, $t\boldsymbol{u}\leqslant(-x)$ and, since $E$ is Archimedean, we would have $\boldsymbol{u}\leqslant 0$; which is not the case. In conclusion, $\boldsymbol{p}_{\boldsymbol{u}}(x)$ is a real number and, by (9), so is $\boldsymbol{q}_{\boldsymbol{u}}(x)$.

Lemma 1**.**

[TABLE]

from which $x=0$ if and only if $\boldsymbol{q}_{\boldsymbol{u}}(x)=\boldsymbol{p}_{\boldsymbol{u}}(x)=0$.

Proof.

If we show that $x\leqslant\boldsymbol{p}_{\boldsymbol{u}}(x)\boldsymbol{u}$ then the other the inequality will follow from (1).

If $\boldsymbol{p}_{\boldsymbol{u}}(x)<t$ then $x\leqslant t\boldsymbol{u}$ and therefore, for all $n\in\mathbb{N}_{\star}$, $x\leqslant(\boldsymbol{p}_{\boldsymbol{u}}(x)+\frac{1}{n})\boldsymbol{u}$ that is $n(x-\boldsymbol{p}_{\boldsymbol{u}}(x)\boldsymbol{u})\leqslant\boldsymbol{u}$ and since $E$ is Archimedean, $x-\boldsymbol{p}_{\boldsymbol{u}}(x)\boldsymbol{u}\leqslant 0$. $\Box$

Call a $\boldsymbol{u}$-box in $E$ any order interval of the form $[s\boldsymbol{u},t\boldsymbol{u}]=\{y\in E:s\boldsymbol{u}\leqslant y\leqslant t\boldsymbol{u}\}$, where $s\leq t$ are real numbers; then, by definition, $[\boldsymbol{q}_{\boldsymbol{u}}(x),\boldsymbol{p}_{\boldsymbol{u}}(x)]$ is the smallest $\boldsymbol{u}$-box containing $x$.

Lemma 2**.**

[TABLE]

From which we have

[TABLE]

and also $x=0$ if and only if $\boldsymbol{p}_{\boldsymbol{u}}(x^{+})=\boldsymbol{p}_{\boldsymbol{u}}(x^{-})=0$.

Proof.

From $x\leqslant\boldsymbol{p}_{\boldsymbol{u}}(x)\boldsymbol{u}$ and $y\leqslant\boldsymbol{p}_{\boldsymbol{u}}(y)\boldsymbol{u}$ we have $x\vee y\leqslant\max\{\boldsymbol{p}_{\boldsymbol{u}}(x),\boldsymbol{p}_{\boldsymbol{u}}(y)\}\boldsymbol{u}$ and therefore $\boldsymbol{p}_{\boldsymbol{u}}(x\vee y)\leq\max\{\boldsymbol{p}_{\boldsymbol{u}}(x),\boldsymbol{p}_{\boldsymbol{u}}(y)\}$.

From $x\leqslant x\vee y$ and $y\leqslant x\vee y$ we have $\boldsymbol{p}_{\boldsymbol{u}}(x)\leqslant\boldsymbol{p}_{\boldsymbol{u}}(x\vee y)$ and $\boldsymbol{p}_{\boldsymbol{u}}(y)\leqslant\boldsymbol{p}_{\boldsymbol{u}}(x\vee y)$ from which we have $\max\{\boldsymbol{p}_{\boldsymbol{u}}(x),\boldsymbol{p}_{\boldsymbol{u}}(y)\}\leq\boldsymbol{p}_{\boldsymbol{u}}(x\vee y)$.

From (9) and x\wedge y=-\big{(}(-x)\vee(-y)\big{)} we have $\boldsymbol{q}_{\boldsymbol{u}}(x\wedge y)=\min\{\boldsymbol{q}_{\boldsymbol{u}}(x),\boldsymbol{q}_{\boldsymbol{u}}(y)\}$.

From $x^{+}=0\vee x$ we have $\boldsymbol{p}_{\boldsymbol{u}}(x^{+})=\max\{\boldsymbol{p}_{\boldsymbol{u}}(0),\boldsymbol{p}_{\boldsymbol{u}}(x)\}$ and, from $x^{-}=0\vee(-x)$ we have

$\boldsymbol{p}_{\boldsymbol{u}}(x^{-})=\boldsymbol{p}_{\boldsymbol{u}}(0\vee(-x))=\max\{0,\boldsymbol{p}_{\boldsymbol{u}}(-x)\}$.

If $\boldsymbol{p}_{\boldsymbol{u}}(x^{+})=0=\boldsymbol{p}_{\boldsymbol{u}}(x^{-})$ then $\boldsymbol{p}_{\boldsymbol{u}}(x)\leq 0$ and $\boldsymbol{p}_{\boldsymbol{u}}(-x)\leq 0$; from $x\leqslant\boldsymbol{p}_{\boldsymbol{u}}(x)\boldsymbol{u}$, $-x\leqslant\boldsymbol{p}_{\boldsymbol{u}}(-x)\boldsymbol{u}$ and $\boldsymbol{u}\in E_{+}$ we have $x\in E_{+}$ and $-x\in E_{+}$ and therefore $x=0$. $\Box$

Lemma 3**.**

[TABLE]

and, if $s<0$ then $\boldsymbol{p}_{\boldsymbol{u}}(sx)=s\boldsymbol{q}_{\boldsymbol{u}}(x)$.

Proof.

From $x\leqslant\boldsymbol{p}_{\boldsymbol{u}}(x)\boldsymbol{u}$ and $y\leqslant\boldsymbol{p}_{\boldsymbol{u}}(y)\boldsymbol{u}$ we have $x+y\leqslant(\boldsymbol{p}_{\boldsymbol{u}}(x)+\boldsymbol{p}_{\boldsymbol{u}}(y))\boldsymbol{u}$ from which the first part of $(1)$ follows. For the second part, there is nothing to prove if $s=0$; if $s>0$ then $sx\leqslant(s\boldsymbol{p}_{\boldsymbol{u}}(x))\boldsymbol{u}$ from which we have $\boldsymbol{p}_{\boldsymbol{u}}(sx)\leq s\boldsymbol{p}_{\boldsymbol{u}}(x)$.

From $sx\leq\boldsymbol{p}_{\boldsymbol{u}}(sx)\boldsymbol{u}$ we have $x\leq(s^{-1}\boldsymbol{p}_{\boldsymbol{u}}(sx))\boldsymbol{u}$ and therefore, $\boldsymbol{p}_{\boldsymbol{u}}(x)\leq s^{-1}\boldsymbol{p}_{\boldsymbol{u}}(sx)$ that is $s\boldsymbol{p}_{\boldsymbol{u}}(x)\leq\boldsymbol{p}_{\boldsymbol{u}}(sx)$.

The second part is a consequence of $\boldsymbol{q}_{\boldsymbol{u}}(x)=-\boldsymbol{p}_{\boldsymbol{u}}(-x)$.

If $s<0$ then \boldsymbol{p}_{\boldsymbol{u}}(sx)=\boldsymbol{p}_{\boldsymbol{u}}\big{(}(-s)(-x)\big{)}=|s|\boldsymbol{p}_{\boldsymbol{u}}(-x)=(-|s|)(-\boldsymbol{p}_{\boldsymbol{u}}(-x))=s\boldsymbol{q}_{\boldsymbol{u}}(x). $\Box$

Theorem 1**.**

The map from $E$ to $\mathbb{R}$ given by

[TABLE]

is a norm on $E$. Furthermore, for all $x,y\in E$, $\|x\vee y\|_{\boldsymbol{hu}}\leq\max\{\|x\|_{\boldsymbol{hu}},\|y\|_{\boldsymbol{hu}}\}$ with equality if both $x$ and $y$ are in $E_{+}$ and

[TABLE]

Proof.

From $x^{+}\in E_{+}$ and $x^{-}\in E_{+}$ we have $\boldsymbol{p}_{\boldsymbol{u}}(x^{+})\geq 0$ and $\boldsymbol{p}_{\boldsymbol{u}}(x^{-})\geq 0$ and therefore $\|x\|_{\boldsymbol{hu}}\geq 0$. If $\|x\|_{\boldsymbol{hu}}=0$ then $\boldsymbol{p}_{\boldsymbol{u}}(x^{+})=\boldsymbol{p}_{\boldsymbol{u}}(x^{-})=0$ and by Lemma 2, $x=0$. Clearly, $\|0\|_{\boldsymbol{hu}}=0$.

If $s>0$ the $(sx)^{+}=sx^{+}$ and $(sx)^{-}=sx^{-}$; from Lemma 3 we obtain $\|sx\|_{\boldsymbol{hu}}=s\|x\|_{\boldsymbol{hu}}$.

If $s<0$ then $sx=|s|(-x)$ from which $\|sx\|_{\boldsymbol{hu}}=|s|\,\|-x\|_{\boldsymbol{hu}}$ and, $\|-x\|_{\boldsymbol{hu}}=\boldsymbol{p}_{\boldsymbol{u}}((-x)^{+})+\boldsymbol{p}_{\boldsymbol{u}}((-x)^{-})=\boldsymbol{p}_{\boldsymbol{u}}(x^{-})+\boldsymbol{p}_{\boldsymbol{u}}(x^{+})=\|x\|_{\boldsymbol{hu}}$.

$\|x+y\|_{\boldsymbol{hu}}\leq\|x\|_{\boldsymbol{hu}}+\|y\|_{\boldsymbol{hu}}$ follows From $(x+y)^{+}\leqslant x^{+}+y^{+}$, $(x+y)^{-}\leqslant x^{-}+y^{-}$ and Lemma 3 we have \|x+y\|_{\boldsymbol{hu}}=\boldsymbol{p}_{\boldsymbol{u}}\big{(}\big{(}x+y)^{+}\big{)}+\boldsymbol{p}_{\boldsymbol{u}}\big{(}(x+y)^{-}\big{)}\leqslant\boldsymbol{p}_{\boldsymbol{u}}(x^{+})+\boldsymbol{p}_{\boldsymbol{u}}(y^{+})+\boldsymbol{p}_{\boldsymbol{u}}(x^{-})+\boldsymbol{p}_{\boldsymbol{u}}(y^{-})=\|x\|_{\boldsymbol{hu}}+\|y\|_{\boldsymbol{hu}}.

Without loss of generality let us assume that $\boldsymbol{p}_{\boldsymbol{u}}(x^{+})\leq\boldsymbol{p}_{\boldsymbol{u}}(y^{+})$. From $(x\vee y)^{+}=x^{+}\vee y^{+}$ and $(x\vee y)^{-}=x^{-}\wedge y^{-}$ and from $(1)$ of Lemma 2 we have $\|x\vee y\|_{\boldsymbol{hu}}\leq\max\{\boldsymbol{p}_{\boldsymbol{u}}(x^{+}),\boldsymbol{p}_{\boldsymbol{u}}(y^{+})\}+\min\{\boldsymbol{p}_{\boldsymbol{u}}(x^{-})+\boldsymbol{p}_{\boldsymbol{u}}(y^{-})\}\leq\boldsymbol{p}_{\boldsymbol{u}}(y^{+})+\boldsymbol{p}_{\boldsymbol{u}}(y^{-})$. Finally, if $x$ and $y$ are both in $E_{+}$ then $x\vee y\in E_{+}$. $\Box$

We will call $\|\cdot\|_{\boldsymbol{hu}}$ the max-plus norm on $E$ associated to the unit $\boldsymbol{u}$.

From Theorem 1, the max-plus norm $\|\cdot\|_{\boldsymbol{hu}}$ is an M-norm, but it is not a Riesz norm as can be seen by taking $E=\mathbb{R}^{2}$, $\boldsymbol{u}=(1,1)$ and $x=(-1,1)$ for which we have $\|x\|_{\boldsymbol{hu}}=2$ and $\|\boldsymbol{|}x\boldsymbol{|}\|_{\boldsymbol{hu}}=1$.

More generally, we always have $\|\boldsymbol{|}x\boldsymbol{|}\|_{\boldsymbol{hu}}\leq\|x\|_{\boldsymbol{hu}}$ since $\|\boldsymbol{|}x\boldsymbol{|}\|_{\boldsymbol{hu}}=\|x^{+}+x^{-}\|_{\boldsymbol{hu}}\leq\|x^{+}\|_{\boldsymbol{hu}}+\|x^{-}\|_{\boldsymbol{hu}}=\boldsymbol{p}_{\boldsymbol{u}}(x^{+})+\boldsymbol{p}_{\boldsymbol{u}}(x^{-})$.

Corollary 1**.**

*The map $(x,y)\mapsto{\sf D}_{\boldsymbol{hu}}(x,y)=\max\{0,\boldsymbol{p}_{\boldsymbol{u}}(x-y)\}-\min\{0,\boldsymbol{q}_{\boldsymbol{u}}(x-y)\}$

 

$\phantom{AAAAAAAAAAAAAAAAAAAAAAAAA}=\max\{0,\boldsymbol{p}_{\boldsymbol{u}}(x-y)\}+\max\{0,\boldsymbol{p}_{\boldsymbol{u}}(y-x)\}$

is a metric on $E$.*

We will call ${\sf D}_{\boldsymbol{hu}}$ the max-plus distance on $E$ associated to the unit $\boldsymbol{u}$.

For $E=\mathbb{R}^{n}$, $\boldsymbol{u}=(1,\cdots,1)$ and $x=(x_{1},\cdots,x_{n})$ we have555$[n]=\{1,\cdots,n\}$

$\begin{cases}\boldsymbol{p}_{\boldsymbol{u}}(x)=\inf\{t\in\mathbb{R}:\forall i\in[n]\,\,x_{i}\leq t\}=\max\{x_{i}:i\in[n]\}\\ \boldsymbol{q}_{\boldsymbol{u}}(x)=\sup\{t\in\mathbb{R}:\forall i\in[n]\,\,t\leq x_{i}\}=\min\{x_{i}:i\in[n]\}\end{cases}$

From $\boldsymbol{p}_{\boldsymbol{u}}(-x)=-\min\{x_{i}:i\in[n]\}=\max\{-x_{i}:i\in[n]\}$ we finally have

$\begin{cases}\max\{0,\boldsymbol{p}_{\boldsymbol{u}}(x)\}=\max\{x^{+}_{i}:i\in[n]\}=\boldsymbol{p}_{\boldsymbol{u}}(x^{+})\\ \max\{0,\boldsymbol{p}_{\boldsymbol{u}}(-x)\}=\max\{x^{-}_{i}:i\in[n]\}=\boldsymbol{p}_{\boldsymbol{u}}(x^{-})\end{cases}$

which gives

[TABLE]

Put simply, if $x\not\in\mathbb{R}^{n}_{+}$ or if $-x\not\in\mathbb{R}^{n}_{+}$ then $\|x\|_{\boldsymbol{hu}}$ is the difference between the largest and the smallest coordinate of $x$; if $x\in\mathbb{R}^{n}_{+}$, $\|x\|_{\boldsymbol{hu}}$ is the largest coordinate of $x$ and, if $-x\in\mathbb{R}^{n}_{+}$, $\|x\|_{\boldsymbol{hu}}$ is the absolute value of the smallest coordinate of $x$ that is, in this last two case, $\|x\|_{\boldsymbol{hu}}=\|x\|_{\infty}=\max_{i\in[n]}|x_{i}|$. The norm $\|\cdot\|_{\boldsymbol{hu}}$ on $\mathbb{R}^{n}$ associated to the unit $\boldsymbol{u}=(1,\cdots,1)$ is the Hilbert affine norm of S. Gaubert.

Let \boldsymbol{U}\big{(}E\big{)} be the set of units of the Riesz space $E$. The set of units of $\mathbb{R}^{n}$ is $\mathbb{R}^{n}_{++}=\{x\in\mathbb{R}^{n}_{+}:0<\min_{i\in[n]}x_{i}\}$.

Lemma 4**.**

For all \boldsymbol{u}_{1},\boldsymbol{u}_{2}\in\boldsymbol{U}\big{(}E\big{)} the norms $\|\cdot\|_{\boldsymbol{hu}_{i}}$ and $\|\cdot\|_{\boldsymbol{u}_{j}}$, $i,j\in\{1,2\}$ are equivalent.

Proof.

First, take $\boldsymbol{u}_{1}=\boldsymbol{u}_{2}=\boldsymbol{u}$. From the definitions, \|x\|_{\boldsymbol{u}}=\boldsymbol{p}\big{(}\boldsymbol{|}x\boldsymbol{|}\big{)} and therefore $\|x\|_{\boldsymbol{u}}=\|x\|_{\boldsymbol{hu}}$ for all $x\in E_{+}$ which gives $\|x\|_{\boldsymbol{hu}}=\|x^{+}\|_{\boldsymbol{u}}+\|x^{-}\|_{\boldsymbol{u}}\leq 2\|x\|_{\boldsymbol{u}}$.

From $\|x\|_{\boldsymbol{u}}=\|\boldsymbol{|}x\boldsymbol{|}\|_{\boldsymbol{u}}$ we get $\|x\|_{\boldsymbol{u}}=\|x^{+}+x^{-}\|_{\boldsymbol{u}}\leq\|x^{+}\|_{\boldsymbol{u}}+\|x^{-}\|_{\boldsymbol{u}}=\boldsymbol{p}(x^{+})+\boldsymbol{p}(x^{-})=\|x\|_{\boldsymbol{hu}}$.

To complete the proof, notice that, for \boldsymbol{u}_{1},\boldsymbol{u}_{2}\in\boldsymbol{U}\big{(}E\big{)}, the norms $\|\cdot\|_{\boldsymbol{hu}_{1}}$ and $\|\cdot\|_{\boldsymbol{hu}_{2}}$ are equivalent since $n_{1}\boldsymbol{u}_{2}\leqslant\boldsymbol{u}_{1}\leqslant n_{2}\boldsymbol{u}_{1}$ for some non zero whole numbers $n_{1}$ and $n_{2}$. $\Box$

From Lemma 4 one has that $\|\cdot\|_{\boldsymbol{hu}}$ is complete for a given \boldsymbol{u}\in\boldsymbol{U}\big{(}E\big{)} if and only if $\|\cdot\|_{\boldsymbol{hu}}$ is complete for all \boldsymbol{u}\in\boldsymbol{U}\big{(}E\big{)} if and only if $\|\cdot\|_{\boldsymbol{u}}$ is complete for a given \boldsymbol{u}\in\boldsymbol{U}\big{(}E\big{)} if and only if $\|\cdot\|_{\boldsymbol{u}}$ is complete for all \boldsymbol{u}\in\boldsymbol{U}\big{(}E\big{)}.

Lemma 5**.**

For all \boldsymbol{u}\in\boldsymbol{U}\big{(}E\big{)} the lattice operations $\vee$, $\wedge$ are uniformly continuous with respect to the metric ${\sf D}_{\boldsymbol{hu}}$. Furthermore, if $K\subset E$ is ${\sf D}_{\boldsymbol{hu}}$-compact then there exists $t_{1},t_{2}\in\mathbb{R}$ such that, for all $x\in K$, $t_{1}\boldsymbol{u}\leqslant x\leqslant t_{2}\boldsymbol{u}$.

Proof.

The norms $\|\cdot\|_{\boldsymbol{u}}$ and $\|\cdot\|_{\boldsymbol{hu}}$ are equivalent and $\|\cdot\|_{\boldsymbol{u}}$ is a lattice norm, which implies uniform continuity of the lattice operations with respect to $\|\cdot\|_{\boldsymbol{u}}$. $\Box$

4. Max-plus convexity in Archimedean Riesz spaces with a unit

Given a Riesz space $E$ and a unit $\boldsymbol{u}$ of $E$ one can, as in the now standard finite dimensional case $E=\mathbb{R}^{n}$, introduce on $E$ two operations, max-plus addition $\oplus$ and max-plus multiplication $\odot$ by real numbers:

[TABLE]

It would have been more appropriate to write $t\odot_{\boldsymbol{u}}x$ for $x+t\boldsymbol{u}$ since this “tropical multiplication” depends on the chosen unit $\boldsymbol{u}$.

Furthermore, writing ${\sf 1}$ for the real number [math], $(\mathbb{R},\odot,{\sf 1})$ is a group (the additive group of real numbers written multiplicatively) and $(t,x)\to t\odot x$ is an action of the group $(\mathbb{R},\odot,{\sf 1})$ on $E$ with the following properties:

[TABLE]

The notation makes everything look very familiar; the peculiarity here is that the “sum” is idempotent: $x\oplus x=x$.

In the finite dimensional case one can enlarge the set of scalars to $\mathbb{R}\cup\{-\infty\}$; tropical addition and multiplication are extended to $\mathbb{R}\cup\{-\infty\}$ in the obvious way: $(-\infty)\oplus x=x\oplus(-\infty)=x$ and $(-\infty)\odot x=-\infty$. Writing ${\sf 0}$ for $-\infty$ this becomes ${\sf 0}\oplus x=x\oplus{\sf 0}=x$ and ${\sf 0}\odot x={\sf 0}$.

E=\big{(}\mathbb{R}^{n}\cup\{-\infty\}\big{)}^{n} with pointwise $\oplus$ and $\odot$ operations, is an indempotent semimodule over the idempotent semi-field $\mathbb{R}\cup\{-\infty\}$; this is the standard max-plus semi-module $\mathbb{R}_{max+}^{n}$.

Such a coordinatewise procedure that turns an arbitrary Riesz space with unit into an idempotent semi-module over the semi-field \big{(}{\mathbb{R}}\cup\{-\infty\},\oplus,\odot,{\sf 0},{\sf 1}) is not readily available. By adding a single element to $E$ which becomes by decree the smallest element, the Riesz space $E$ is embedded in a semi-module over $\mathbb{R}\cup\{-\infty\}$ of which it is a max-plus convex subset(the definition of max-plus convexity is given below).

To an arbitrary Riesz space $E$ add a smallest element, let us call it $\boldsymbol{\bot}$, and one can extend scalar multiplication to \odot:\mathbb{R}\cup\{-\infty\}\times\big{(}E\cup\{\boldsymbol{\bot}\}\big{)}\to\big{(}E\cup\{\boldsymbol{\bot}\}\big{)} and the tropical sum to \big{(}E\cup\{\boldsymbol{\bot}\}\big{)}\times\big{(}E\cup\{\boldsymbol{\bot}\}\big{)} in such a way that $E\cup\{\boldsymbol{\bot}\}$ becomes an idempotent semimodule over the totally ordered semi-field $\mathbb{R}\cup\{-\infty\}$:

[TABLE]

The max-plus convex hull of a non empty subset $S\subset E\cup\{\boldsymbol{\bot}\}$ is the set of elements of $E\cup\{\boldsymbol{\bot}\}$ which can be written as $(x_{1}+t_{1}\boldsymbol{u})\vee\cdots\vee(x_{m}+t_{m}\boldsymbol{u})$ with $\{x_{1},\cdots,x_{m}\}\subset S$, $\{t_{1},\cdots,t_{m}\}\subset\mathbb{R}\cup\{-\infty\}$ and $\max\{t_{1},\cdots,t_{m}\}=0$666In tropical notation this becomes $(t_{1}\odot x_{1})\oplus\cdots\oplus(t_{m}\odot x_{m})$ with $t_{1}\oplus\cdots\oplus t_{m}={\sf 1}$ (recall here ${\sf 1}$ is the usual $0\in\mathbb{R}$ and that $\odot$ depends on a fixed unit $\boldsymbol{u}$ : $t\odot x=x+t\boldsymbol{u}$) exactly as if it were a usual affine combination.. We will use the notation $\boldsymbol{[\![}S\boldsymbol{]\!]}_{\boldsymbol{u}}$ for the max-plus convex hull of $S$ with respect to the unit $\boldsymbol{u}$; we set $\boldsymbol{[\![}\emptyset\boldsymbol{]\!]}_{\boldsymbol{u}}=\emptyset$. Whenever a single fixed unit $\boldsymbol{u}$ is under consideration we drop the index $\boldsymbol{u}$.

If $S\subset E$ then $\boldsymbol{[\![}S\boldsymbol{]\!]}_{\boldsymbol{u}}$ is a subset of $E$; it is the set of of elements of $E$ which can be written as

$(x_{1}+t_{1}\boldsymbol{u})\vee\cdots\vee(x_{m}+t_{m}\boldsymbol{u})$ with $\{x_{1},\cdots,x_{m}\}\subset S$, $\{t_{1},\cdots,t_{m}\}\subset\mathbb{R}$ and $\max\{t_{1},\cdots,t_{m}\}=0$ since, if $t_{i}=-\infty$ then $x_{i}+t_{i}\boldsymbol{u}=\boldsymbol{\bot}$ and, since one the coefficients is [math], let us say $t_{j}=0$, we have $(x_{i}+t_{i}\boldsymbol{u})\vee(x_{j}+t_{j}\boldsymbol{u})=\boldsymbol{\bot}\vee x_{j}=x_{j}$.

A subset $C$ of $E\cup\{-\infty\}$ is said to be max-plus convex (with respect to $\boldsymbol{u}$) if $C=\boldsymbol{[\![}C\boldsymbol{]\!]}_{\boldsymbol{u}}$.

If $S=\{x_{1},\cdots,x_{m}\}$ is a finite set we will write $\boldsymbol{[\![}x_{1},\cdots,x_{m}\boldsymbol{]\!]}$ for $\boldsymbol{[\![}S\boldsymbol{]\!]}$. The max-plus convex hull of two points $x_{1}$ and $x_{2}$ is $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}=\{x_{1}\vee(x_{2}+t\boldsymbol{u}):t\leq 0\}\cup\{(x_{1}+t\boldsymbol{u})\vee x_{2}:t\leq 0\}$; if $x_{1}\neq x_{2}$, the set $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$ will be called a max-plus segment. Since $\boldsymbol{u}$ is a unit, there exists $s\geq 0$ such that $(x_{2}-x_{1})\leqslant s\boldsymbol{u}$ that is $x_{2}+(-s)\boldsymbol{u}\leqslant x_{1}$ which shows that $x_{1}\in\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$, and similarly for $x_{2}$.

Given a non empty set $S$ we will denote by $\langle S\rangle$ the family of non empty finite subsets of $S$. A set of the form $\boldsymbol{[\![}S\boldsymbol{]\!]}$ with $S$ finite is a max-plus polytope (with respect to $\boldsymbol{u}$); this definition makes the empty set into a polytope.

Max-plus convex subsets of $E\cup\{\boldsymbol{\bot}\}$ will be rarely referred to. Unless otherwise specified, “max-plus convex set” will mean “max-plus convex set of $E$”. What matters here, is that $E$ is a max-plus convex subset of the $(\mathbb{R}\cup\{-\infty\})$-idempotent semimodule $E\cup\{\boldsymbol{\bot}\}$ and therefore, the max-plus convex subsets of $E$ are exactly the max-plus convex subsets of $E\cup\{\boldsymbol{\bot}\}$ that are contained in $E$.

The proof of Lemma 6 below is left to the reader; that of Lemma 7 can here be done by hand or one can go to Lemma 2.1.5 in [19].

Lemma 6**.**

*Given a Riesz space $E$ a unit $\boldsymbol{u}$ of $E$ the following properties hold:

$(1)$ $S\subset\boldsymbol{[\![}S\boldsymbol{]\!]}$.

$(2)$ If $S_{1}\subset S_{2}$ then $\boldsymbol{[\![}S_{1}\boldsymbol{]\!]}\subset\boldsymbol{[\![}S_{2}\boldsymbol{]\!]}$.

$(3)$ $\displaystyle{\boldsymbol{[\![}S\boldsymbol{]\!]}=\bigcup_{A\in\langle S\rangle}\boldsymbol{[\![}A\boldsymbol{]\!]}}$ where $\langle S\rangle$ denotes the set of non empty finite subsets of $S$.

$(4)$ $\boldsymbol{[\![}\,\boldsymbol{[\![}S\boldsymbol{]\!]}\,\boldsymbol{]\!]}=\boldsymbol{[\![}S\boldsymbol{]\!]}$.*

Lemma 7**.**

For all finite subset $S$ of $E$ and for all $x\in E$,

[TABLE]

Lemma 8**.**

*(1) For all $x_{1},x_{2}\in E$, $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}=\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}\cup\boldsymbol{[\![}x_{2},x_{1}\vee x_{2}\boldsymbol{]\!]}$ and $\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}\cap\boldsymbol{[\![}x_{2},x_{1}\vee x_{2}\boldsymbol{]\!]}=\{x_{1}\vee x_{2}\}$.

(2) If $w\in\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$ then either $w\vee x_{2}=x_{1}\vee x_{2}$ or $w\vee x_{2}=w$.

(3) If $x_{1}$ and $x_{2}$ are comparable then $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$ is, with respect to the partial order of $E$, a totally ordered subset.*

Proof.

(1) If $z=x_{1}\vee(x_{2}+s\boldsymbol{u})$ with $s\leq 0$ then, from $(x_{1}+s\boldsymbol{u})\vee(x_{2}+s\boldsymbol{u})=(x_{1}\vee x_{2})+s\boldsymbol{u}$ and $x_{1}+s\boldsymbol{u}\leqslant x_{1}$ we have x_{1}\vee\big{(}(x_{1}\vee x_{2})+s\boldsymbol{u}\big{)}=x_{1}\vee(x_{1}+s\boldsymbol{u})\vee(x_{2}+s\boldsymbol{u})=x_{1}\vee(x_{2}+s\boldsymbol{u})=z which shows that $z\in\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}$; similarly, if $z=(x_{1}+s\boldsymbol{u})\vee x_{2}$ with $s\leq 0$ then $z\in\boldsymbol{[\![}x_{2},x_{1}\vee x_{2}\boldsymbol{]\!]}$.

From $\{x_{i},x_{1}\vee x_{2}\}\subset\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$ and from Lemma 6 we have $\boldsymbol{[\![}x_{i},x_{1}\vee x_{2}\boldsymbol{]\!]}\subset\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$.

Take $z\in\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}\cap\boldsymbol{[\![}x_{2},x_{1}\vee x_{2}\boldsymbol{]\!]}$ ; if $z=tx_{i}\vee(x_{1}\vee x_{2})$ with $t\leq 0$ then $z=x_{1}\vee x_{2}$. If $z=x_{1}\vee s(x_{1}\vee x_{2})=x_{2}\vee t(x_{1}\vee x_{2})$ with $s,t\leq 0$ then $z=z\vee z=x_{1}\vee_{2}\vee s(x_{1}\vee x_{2})\vee s(x_{1}\vee x_{2})=x_{1}\vee x_{2}$. This completes the proof of (1).

(2) If $w\in\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}$ then $x_{1}\leqslant w\leqslant x_{1}\vee x_{2}$ and therefore $x_{1}\vee x_{2}\leqslant w\vee x_{2}\leqslant x_{1}\vee x_{2}\vee x_{2}=x_{1}\vee x_{2}$.

If $w\in\boldsymbol{[\![}x_{2},x_{1}\vee x_{2}\boldsymbol{]\!]}$ then $x_{2}\leqslant w\leqslant x_{1}\vee x_{2}$ and therefore $w\vee x_{2}=w$.

(3) Assume that $x_{1}\leqslant x_{2}$. An arbitrary element of $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$ is either of the form $(x_{1}+s\boldsymbol{u})\vee x_{2}$ or $x_{1}\vee(x_{2}+s\boldsymbol{u})$ with $s\leq 0$; since $(x_{1}+s\boldsymbol{u})\vee x_{2}=x_{2}$ and $x_{1}\leqslant x_{1}\vee(x_{2}+s\boldsymbol{u})\leq x_{1}\vee x_{2}=x_{2}$, $x_{1}$ is the smallest element of $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$ and $x_{2}$ is its largest element. Take $z$ and $z^{\prime}$ in $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$; we can assume that $z=x_{1}\vee(x_{2}+s\boldsymbol{u})$ and $z^{\prime}=x_{1}\vee(x_{2}+s^{\prime}\boldsymbol{u})$ with $s\leq 0$ and $s^{\prime}\leq 0$. We have either $s\leq s^{\prime}$ of $s^{\prime}\leq s$ from which it follows that either $z\leqslant z^{\prime}$ or $z^{\prime}\leqslant z$. $\Box$

Lemma 9**.**

$(1)$* A singleton is max-plus convex.

$(2)$ An arbitrary intersection of max-plus convex sets is max-plus convex.

$(3)$ $\boldsymbol{[\![}S\boldsymbol{]\!]}$ is the smallest max-plus convex set containing $S$.

$(4)$ $C$ is max-plus convex if and only if, for all $x_{1},x_{2}\in C$, $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}\subset C$.

$(5)$ The following statements are equivalent:*

[TABLE]

Proof.

Only $(4)$ needs to be checked, $(1),(2)$ and $(3)$ are direct consequences of the definitions. Let us see that $C$ is max-plus convex if for all $x_{1},x_{2}\in C$, $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}\subset C$; the reverse implication is a consequence of $(2)$ and $(4)$ of Lemma 6. By $(3)$ of Lemma 6 we have to show that, for all $A\in\langle C\rangle$, $\boldsymbol{[\![}A\boldsymbol{]\!]}\subset C$ which can be proved using Lemma 7 and an obvious induction on the cardinality of the set $A$.

The equivalence of $(4)$ and $(5)$ follows from $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}=\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}\cup\boldsymbol{[\![}x_{2},x_{1}\vee x_{2}\boldsymbol{]\!]}$. $\Box$

A max-plus half-space of a convex subset $C$ of $E$ is a max-plus convex set $D\subset C$ such that $C\setminus\!D$ is also max-plus convex. A max-plus convex subset $C$ of $E$ has the Kakutani-Property if, for arbitrary disjoint max-plus convex subsets $C_{1}$ and $C_{2}$ there exists a half-space $D$ of $C$ such that $C_{1}\subset D$ and $C_{2}\cap D=\emptyset$.

Lemma 10**.**

Max-plus convex subsets of $E$ have the Kakutani-Property.

Proof.

From 3.0.12 in [19]. $\Box$

We conclude this section by showing that the max-plus metric is additive on max-plus segments,

Proposition 1 below.

Lemma 11**.**

For $x_{1},x_{2}\in E$

[TABLE]

Proof.

If $x_{1}$ and $x_{2}$ are comparable then either $x_{1}=x_{1}\vee x_{2}$ or $x_{2}=x_{1}\vee x_{2}$ in which case there is nothing to prove. If $\boldsymbol{p}_{\boldsymbol{u}}(x_{1}-x_{2})\leq 0$ then $x_{1}\leqslant x_{2}$ and $x_{1}$ and $x_{2}$ are comparable, and similarly if $\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-x_{1})\leq 0$. We can therefore assume that $0\leq\boldsymbol{p}_{\boldsymbol{u}}(x_{1}-x_{2})$ and $0\leq\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-x_{1})$.

We now have ${\sf D}_{\boldsymbol{hu}}(x_{1},x_{2})=\max\{0,\boldsymbol{p}_{\boldsymbol{u}}(x_{1}-x_{2})\}+\max\{0,\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-x_{1})\}=\boldsymbol{p}_{\boldsymbol{u}}(x_{1}-x_{2})+\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-x_{1})$ and we have to see that $\boldsymbol{p}_{\boldsymbol{u}}(x_{1}-x_{2})+\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-x_{1})\geq\|x_{1}-x_{2}\vee x_{1}\|_{\boldsymbol{u}}+\|x_{2}-x_{2}\vee x_{1}\|_{\boldsymbol{u}}$.

From $x_{1}\leqslant\boldsymbol{p}_{\boldsymbol{u}}(x_{1}-x_{2})\boldsymbol{u}+x_{2}$ and $\boldsymbol{p}_{\boldsymbol{u}}(x_{1}-x_{2})\boldsymbol{u}\in E_{+}$ we have $x_{1}\vee x_{2}\leqslant\boldsymbol{p}_{\boldsymbol{u}}(x_{1}-x_{2})\boldsymbol{u}+x_{2}$ and, since $\|\cdot\|_{\boldsymbol{u}}$ is a Riesz norm and $\|\boldsymbol{u}\|_{\boldsymbol{u}}=1$, $\|x_{1}\vee x_{2}-x_{2}\|_{\boldsymbol{u}}\leq\boldsymbol{p}_{\boldsymbol{u}}(x_{1}-x_{2})$. Similarly, $\|x_{1}\vee x_{2}-x_{1}\|_{\boldsymbol{u}}\leq\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-x_{1})$. $\Box$

Lemma 12**.**

*If $x_{1}\leq x_{2}$ then, for all $z\in\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$, ${\sf D}_{\boldsymbol{hu}}(z,x_{2})=\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-z)=\|x_{2}-z\|_{\boldsymbol{u}}$,

z=x_{1}\vee\big{(}x_{2}-\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-z)\boldsymbol{u}\big{)} and $\boldsymbol{q}_{\boldsymbol{u}}(z-x_{2})=\sup\{s\leq 0:z=x_{1}\vee(x_{2}+s\boldsymbol{u})\}$.*

Proof.

From $z\in\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$ we have either $z=(x_{1}+s\boldsymbol{u})\vee x_{2}$ or $z=x_{1}\vee(x_{2}+s\boldsymbol{u})$ with $s\leq 0$. In the first case, from $x_{1}\leq x_{2}$, we have $z=x_{2}$ in which case $\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-z)=\boldsymbol{p}_{\boldsymbol{u}}(0)=0=\boldsymbol{q}_{\boldsymbol{u}}(0)$ and x_{1}\vee\big{(}x_{2}-\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-z)\boldsymbol{u}\big{)}=x_{1}\vee x_{2}=x_{2}=z.

If $z=x_{1}\vee(x_{2}+s\boldsymbol{u})$ with $s\leq 0$ then $x_{1}\vee z=z$ and $x_{2}\vee z=x_{2}$, which shows that $x_{1}\leqslant z\leqslant x_{2}$.

From $0\leqslant x_{2}-z$ we have $\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-z)=\inf\{t\geq 0:x_{2}-z\leqslant t\boldsymbol{u}\}={\sf D}_{\boldsymbol{hu}}(z,x_{2})=\|x_{2}-z\|_{\boldsymbol{u}}$.

From $x_{2}+s\boldsymbol{u}\leqslant x_{1}\vee(x_{2}+s\boldsymbol{u})=x_{1}\vee z=z$ we have $\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-z)\leq(-s)$ from which we obtain

$x_{2}+s\boldsymbol{u}\leqslant x_{2}-\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-z)\boldsymbol{u}$ and z=x_{1}\vee(x_{2}+s\boldsymbol{u})\leqslant x_{1}\vee\big{(}x_{2}-\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-z)\boldsymbol{u}\big{)}.

From Lemma 1 we have $x_{2}-z\leqslant\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-z)\boldsymbol{u}$ that is, $x_{2}-\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-z)\boldsymbol{u}\leqslant z$ and, since $x_{1}\leqslant z$,

x_{1}\vee\big{(}x_{2}-\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-z)\boldsymbol{u}\big{)}\leqslant z.

We have shown that z=x_{1}\vee\big{(}x_{2}-\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-z)\boldsymbol{u}\big{)} and that $s\leq-\boldsymbol{p}_{\boldsymbol{u}}(x_{2}-z)=\boldsymbol{q}_{\boldsymbol{u}}(z-x_{2})$. $\Box$

Lemma 13**.**

If $x_{1}$ and $x_{2}$ are two comparable elements of $E$ then, for all $z\in\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$,

[TABLE]

Proof.

From Lemma 12 we can write $z=x_{1}\vee(x_{2}+s\boldsymbol{u})$ with $s=-{\sf D}_{\boldsymbol{hu}}(z,x_{2})$.

From $x_{1}\leqslant z$, ${\sf D}_{\boldsymbol{hu}}(x_{1},z)=\|z-x_{1}\|_{\boldsymbol{u}}=\inf\{t\geq 0:z-x_{1}\leq t\boldsymbol{u}\}=\inf\{t\geq 0:x_{1}\vee(x_{2}+s\boldsymbol{u})\leq x_{1}+t\boldsymbol{u}\}$. If $t\geq 0$ and $(x_{2}+s\boldsymbol{u})\leq x_{1}+t\boldsymbol{u}$ then $x_{1}\vee(x_{2}+s\boldsymbol{u})\leq(x_{1}+t\boldsymbol{u})\vee x_{1}=(x_{1}+t\boldsymbol{u})$ which shows that ${\sf D}_{\boldsymbol{hu}}(x_{1},z)\leq\inf\{t\geq 0:(x_{2}-x_{1})\leqslant(t-s)\boldsymbol{u}\}={\sf D}_{\boldsymbol{hu}}(x_{1},x_{2})+s={\sf D}_{\boldsymbol{hu}}(x_{1},x_{2})-{\sf D}_{\boldsymbol{hu}}(z,x_{2})$. $\Box$

Proposition 1**.**

For all $x_{1},x_{2}\in E$ an for all $z\in\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$

[TABLE]

Proof.

Let $z\in\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$. By Lemma (8) either $w\in\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}$ or $z\in\boldsymbol{[\![}x_{2},x_{1}\vee x_{2}\boldsymbol{]\!]}$. Without loss of generality assume that $z\in\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}$.

By Lemma 13 we have ${\sf D}_{\boldsymbol{hu}}(x_{1},x_{1}\vee x_{2})={\sf D}_{\boldsymbol{hu}}(x_{1},z)+{\sf D}_{\boldsymbol{hu}}(z,x_{1}\vee x_{2})$ and, by Lemma 11,

${\sf D}_{\boldsymbol{hu}}(x_{1},x_{2})={\sf D}_{\boldsymbol{hu}}(x_{1},x_{1}\vee x_{2})+{\sf D}_{\boldsymbol{hu}}(x_{2},x_{1}\vee x_{2})$. Therefore

[TABLE]

From Lemma 8 we have ${\sf D}_{\boldsymbol{hu}}(z,x_{1}\vee x_{2})={\sf D}_{\boldsymbol{hu}}(z,z\vee x_{2})$ and consequently ${\sf D}_{\boldsymbol{hu}}(x_{1},x_{2})={\sf D}_{\boldsymbol{hu}}(x_{1},z)+{\sf D}_{\boldsymbol{hu}}(z,z\vee x_{2})+{\sf D}_{\boldsymbol{hu}}(x_{2},z\vee x_{2})$ which yields, by Lemma 11, ${\sf D}_{\boldsymbol{hu}}(x_{1},x_{2})={\sf D}_{\boldsymbol{hu}}(x_{1},z)+{\sf D}_{\boldsymbol{hu}}(z,x_{2})$. $\Box$

5. Geodesics

Given a metric space $(X,D)$, let us say that non empty subset $Z\subset X$ is a geodesic, with respect to the metric $D$, if there exists an onto map $\theta$ from a closed interval $[a,b]\subset\mathbb{R}$ to $Z$ such that, for all $t_{1},t_{2}\in[a,b]$, D\big{(}\theta(t_{1}),\theta(t_{2})\big{)}=\mid t_{1}-t_{2}\mid in which case we will say that $\theta:[a,b]\to X$ is a parametrized geodesic from $\theta(a)$ to $\theta(b)$.

The metric space $(X,D)$ is a geodesic space if, for all pair $(x_{1},x_{2})\in X\times X$, there exists a parametrized geodesic $\theta:[a,b]\to X$ from $x_{1}$ to $x_{2}$. A geodesic structure on a geodesic metric space $(X,D)$ is a family \Theta=\big{(}\theta_{(x_{1},x_{2})},[a_{(x_{1},x_{2})},b_{(x_{1},x_{2})}]\big{)}_{(x_{1},x_{2})\in X\times X} where \big{(}\theta_{(x_{1},x_{2})},[a_{(x_{1},x_{2})},b_{(x_{1},x_{2})}]\big{)} is a parametrized geodesic from $x_{1}$ to $x_{2}$ ; the geodesic \theta_{(x_{1},x_{2})}\big{(}[a_{(x_{1},x_{2})},b_{(x_{1},x_{2})}\big{)}\subset X will be denoted by $\boldsymbol{\theta}(x_{1},x_{2})$.

Given a geodesic structure $\Theta$ on a metric space $(X,D)$, we will say that a subset $C\subset X$ is geodesically convex (with respect to $\Theta$) if, for all $(x_{1},x_{2})\in X\times X$, $\boldsymbol{\theta}(x_{1},x_{2})\subset C$; the empty set is convex by default.

If $\theta:[a,b]\to X$ is a parametrized geodesic from $x_{1}$ to $x_{2}$ then, for all $s\in\mathbb{R}$, $t\mapsto\theta(t-s)$ is a parametrized geodesic $\tilde{\theta}$ on $[a+s,b+s]$ from $x_{1}$ to $x_{2}$ such that $\boldsymbol{\theta}(x_{1},x_{2})=\boldsymbol{\tilde{\theta}}(x_{1},x_{2})$ ; since $D(x_{1},x_{2})=b-a$, the map $\tilde{\theta}:[0,D(x_{1},x_{2})]\to X$ defined by $\tilde{\theta}(t)=\theta(t+a)$ is a parametrized geodesic from from $x_{1}$ to $x_{2}$ with $\boldsymbol{\theta}(x_{1},x_{2})=\boldsymbol{\tilde{\theta}}(x_{1},x_{2})$; let us call $\tilde{\theta}:[0,D(x_{1},x_{2})]\to X$ the standard parametrized geodesic associated to the parametrized geodesic $\theta:[a,b]\to X$.

Given a parametrized geodesic $\theta:[a,b]\to X$, from $x_{1}$ to $x_{2}$, the affinely parametrized geodesic

$\hat{\theta}:[0,1]\to X$ associated to $\theta$ is \hat{\theta}(t)=\theta\big{(}(1-t)a+tb\big{)}=\theta\big{(}a+tD(x_{1},x_{2})\big{)}.

For all $t_{1},t_{2}\in[0,1]$, D\big{(}\hat{\theta}(t_{1}),\hat{\theta}(t_{2})\big{)}=\mid t_{1}-t_{2}\mid D(x_{1},x_{2}).

We show that \big{(}E,{\sf D}_{\boldsymbol{hu}}\big{)} is a geodesic metric space, more precisely, there is a geodesic strucuture \Gamma=\big{(}\gamma_{\boldsymbol{u},(x_{1},x_{2})},[\alpha_{\boldsymbol{u},(x_{1},x_{2})},\beta_{\boldsymbol{u},(x_{1},x_{2})}]\big{)} on $E$ for which the geodesics are precisely the max-plus segments and therefore, the max-plus convex sets are precisely the geodesicaly convex sets. To avoid cumbersome double subscripts we will write $\gamma_{\boldsymbol{u}}(x_{1},x_{2},t)$ for $\gamma_{\boldsymbol{u}(x_{1},x_{2})}(t)$ and similarly for $\alpha$ and $\beta$. The explicit form of $\gamma_{\boldsymbol{u}}(x_{1},x_{2},t)$ is given in Theorem 2; for $E=\mathbb{R}^{n}$ and $\boldsymbol{u}=(1,\cdots,1)$ this is due to Stefan Gaubert, [14].

Lemma 14**.**

Two arbitrary points $x_{1},x_{2}\in E$ being given, any two other points of $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$ can be labelled $z$ and $z^{\prime}$ in such a way one of the following two assertions holds:

[TABLE]

Proof.

From Lemma 8, there are two cases to consider: both $z$ and $z^{\prime}$ are in $\boldsymbol{[\![}x_{i},x_{1}\vee x_{2}\boldsymbol{]\!]}$ for the same $i$, one of $z$ and $z^{\prime}$ is in $\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}$ and the other is in $\boldsymbol{[\![}x_{2},x_{1}\vee x_{2}\boldsymbol{]\!]}$.

Let us assume that $z$ and $z^{\prime}$ are both in $\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}$ which, by Lemma 8, is a totally ordered set with respect the partial order of $E$. If the two points $z$ and $z^{\prime}$ are labeled such that $x_{1}\leqslant z\leqslant z^{\prime}\leqslant x_{1}\vee x_{2}$ then $z^{\prime}\in\boldsymbol{[\![}z,x_{1}\vee x_{2}\boldsymbol{]\!]}$ and, from Proposition 1, ${\sf D}_{\boldsymbol{hu}}(z,z^{\prime})+{\sf D}_{\boldsymbol{hu}}(z^{\prime},x_{1}\vee x_{2})={\sf D}_{\boldsymbol{hu}}(z,x_{1}\vee x_{2})$

For the second case, label the points such that $z\in\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}$ and $z^{\prime}\in\boldsymbol{[\![}x_{2},x_{1}\vee x_{2}\boldsymbol{]\!]}$ that is:

$x_{1}\leqslant z\leqslant x_{1}\vee x_{2}$ and $x_{2}\leqslant z^{\prime}\leqslant x_{1}\vee x_{2}$ from which we have $x_{1}\vee x_{2}=z\vee z^{\prime}$. Proposition 1 yields ${\sf D}_{\boldsymbol{hu}}(z,z^{\prime})={\sf D}_{\boldsymbol{hu}}(z,z\vee z^{\prime})+{\sf D}_{\boldsymbol{hu}}(z^{\prime},z\vee z^{\prime})$. $\Box$

Lemma 15**.**

If $x_{1}\leqslant x_{2}$ then the map $t\mapsto\gamma(t,x_{1},x_{2})=x_{1}\vee(x_{2}+t\boldsymbol{u})$ is one to one and onto from the interval $[\boldsymbol{q}_{\boldsymbol{u}}(x_{1}-x_{2}),0]$ to $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$. Furthermore, given $z\in\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$, the unique $s\in[\boldsymbol{q}_{\boldsymbol{u}}(x_{1}-x_{2}),0]$ such that $z=x_{1}\vee(x_{2}+s\boldsymbol{u})$ is $-{\sf D}_{\boldsymbol{hu}}(z,x_{2})$.

Proof.

The points $x_{1}$ and $x_{2}$ being fixed, write $\gamma(t)$ for $\gamma(t,x_{1},x_{2})$. If $z\in\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$ then either $z=(x_{1}+t\boldsymbol{u})\vee x_{2}$ or $z=x_{1}\vee(x_{2}+t\boldsymbol{u})$ with $t\leq 0$ ; in the first case $z=x_{2}=x_{1}\vee x_{2}=x_{1}\vee(x_{2}+0\boldsymbol{u})$. This shows that $\gamma$ is onto.

To see that $\gamma$ is one to one on $[\boldsymbol{q}_{\boldsymbol{u}}(x_{1}-x_{2}),0]$ let $z=x_{1}\vee(x_{2}+s\boldsymbol{u})$, $z^{\prime}=x_{1}\vee(x_{2}+t\boldsymbol{u})$ with $\boldsymbol{q}_{\boldsymbol{u}}(x_{1}-x_{2})\leq s\leq t\leq 0$ and assume that $z=z^{\prime}$. We show that $s=t$.

First, $x_{2}+t\boldsymbol{u}\leqslant x_{1}\vee(x_{2}+t\boldsymbol{u})=z^{\prime}=z$ from which we have $t\boldsymbol{u}\leqslant z-x_{2}$ and therefore, from the definition of $\boldsymbol{q}_{\boldsymbol{u}}$,

[TABLE]

Now, z-x_{2}=\big{[}x_{1}\vee(x_{2}+s\boldsymbol{u})\big{]}-x_{2}=(x_{1}-x_{2})\vee\big{(}(x_{2}+s\boldsymbol{u})-x_{2}\big{)}=(x_{1}-x_{2})\vee(s\boldsymbol{u}).

To complete the proof, we identify $E$ with a Riesz subspace of the space $\mathcal{C}(\Omega)$ containing the constant function $\boldsymbol{1}$, where $\Omega$ is a compact Hausdorff topological space, and we set $\boldsymbol{u}=\boldsymbol{1}$.

[TABLE]

and therefore,

[TABLE]

We have shown that $\boldsymbol{q}_{\boldsymbol{u}}(z-x_{2})\leq s$ which, with (18), gives $s=t$.

The last part follows from Lemma 12.$\Box$

For $x_{1},x_{2}\in E$ let $\alpha_{\boldsymbol{u}}(x_{1},x_{2})=\min\{0,\boldsymbol{q}_{\boldsymbol{u}}(x_{1}-x_{2})\}$ and $\beta_{\boldsymbol{u}}(x_{1},x_{2})=\max\{0,\boldsymbol{p}_{\boldsymbol{u}}(x_{1}-x_{2})\}$ Then $\beta_{\boldsymbol{u}}(x_{1},x_{2})=-\alpha_{\boldsymbol{u}}(x_{2},x_{1})$, $\alpha_{\boldsymbol{u}}(x_{1},x_{2})\leq\beta_{\boldsymbol{u}}(x_{1},x_{2})$ and ${\sf D}_{\boldsymbol{hu}}(x_{1},x_{2})=\beta_{\boldsymbol{u}}(x_{1},x_{2})-\alpha_{\boldsymbol{u}}(x_{1},x_{2})$.

Lemma 16**.**

If $x_{1}$ and $x_{2}$ are two elements of $E$ and $s,t$ are real numbers such that $\alpha_{\boldsymbol{u}}(x_{1},x_{2})\leq t\leq 0\leq s\leq\beta_{\boldsymbol{u}}(x_{1},x_{2})$ then, for $z=x_{1}\vee(x_{2}+t\boldsymbol{u})\in\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}$ and $z^{\prime}=(x_{1}-s\boldsymbol{u})\vee x_{2}\in\boldsymbol{[\![}x_{2},x_{1}\vee x_{2}\boldsymbol{]\!]}$ one has

[TABLE]

Proof.

We have z=x_{1}\vee(x_{2}+t\boldsymbol{u})\leqslant x_{1}\vee\big{(}(x_{1}\vee x_{2})+t\boldsymbol{u}\big{)}=x_{1}\vee[(x_{1}+t\boldsymbol{u})\vee(x_{2}+t\boldsymbol{u})]=[x_{1}\vee(x_{1}+t\boldsymbol{u})]\vee(x_{2}+t\boldsymbol{u})=x_{1}\vee(x_{2}+t\boldsymbol{u}).

In conclusion, z=x_{1}\vee\big{(}(x_{1}\vee x_{2})+t\boldsymbol{u}\big{)}.

Let us see that $\min\{0,\boldsymbol{q}_{\boldsymbol{u}}(x_{1}-x_{2})\}=\boldsymbol{q}_{\boldsymbol{u}}(x_{1}-x_{1}\vee x_{2})$.

By definition $\boldsymbol{q}_{\boldsymbol{u}}(x_{1}-x_{2})=\sup\{t:t\boldsymbol{u}\leqslant x_{1}-x_{2}\}$. If $t\leq 0$ and $t\boldsymbol{u}\leqslant x_{1}-x_{2}$ then $x_{2}\leqslant x_{1}-t\boldsymbol{u}$ and $x_{1}\vee x_{2}\leqslant(x_{1}-t\boldsymbol{u})\vee x_{1}=x_{1}-t\boldsymbol{u}$ which shows that $\min\{0,\boldsymbol{q}_{\boldsymbol{u}}(x_{1}-x_{2})\}\leq\boldsymbol{q}_{\boldsymbol{u}}(x_{1}-x_{1}\vee x_{2})$.

Reciprocally, if $t\leq\boldsymbol{q}_{\boldsymbol{u}}(x_{1}-x_{1}\vee x_{2})$ then $t\leq 0$ and $t\boldsymbol{u}\leqslant(x_{1}-x_{1}\vee x_{2})\leqslant x_{1}-x_{2}$ and therefore,

$t\leq\min\{0,\boldsymbol{q}_{\boldsymbol{u}}(x_{1}-x_{2})\}$.

The conclusion follows from Lemma 15 and $\beta_{\boldsymbol{u}}(x_{1},x_{2})=-\alpha_{\boldsymbol{u}}(x_{2},x_{1})$. $\Box$

Lemma 17**.**

The map $\eta_{\boldsymbol{u}}$ defined on $E\times E\times\mathbb{R}$ by

[TABLE]

is uniformly continuous with respect to the product topology induced by the metric ${\sf D}_{\boldsymbol{hu}}$ on $E$. Furthermore,

[TABLE]

Proof.

Since $\|\cdot\|_{\boldsymbol{u}}$ is a Riesz norm, the lattice operations are uniformly continuous with respect to $\|\cdot\|_{\boldsymbol{u}}$ from which it follows that $\eta_{\boldsymbol{u}}$ is uniformly continuous with respect to $\|\cdot\|_{\boldsymbol{u}}$ and therefore with respect ${\sf D}_{\boldsymbol{hu}}$, since the norms $\|\cdot\|_{\boldsymbol{u}}$ and $\|\cdot\|_{\boldsymbol{hu}}$ are equivalent. The verification of (23) is straightforward. $\Box$

Theorem 2**.**

For all $x_{1},x_{2}\in E$ and $t\in\mathbb{R}$ let $\gamma_{\boldsymbol{u}}$ be the restriction of $\eta_{\boldsymbol{u}}$ to $[\alpha_{\boldsymbol{u}}(x_{1},x_{2}),\beta_{\boldsymbol{u}}(x_{1},x_{2})]$ that is,

[TABLE]

Then, for all $x_{1},x_{2}\in E$ and for all $t_{1},t_{2}\in[\alpha_{\boldsymbol{u}}(x_{1},x_{2}),\beta_{\boldsymbol{u}}(x_{1},x_{2})]$

[TABLE]

In other words, \Gamma_{\boldsymbol{u}}=\big{(}\gamma_{\boldsymbol{u}}(x_{1},x_{2},-),[\alpha(x_{1},x_{2}),\beta(x_{1},x_{2})]\big{)}_{(x_{1},x_{2})\in X\times X} is a geodesic structure on \big{(}E,{\sf D}_{\boldsymbol{u}}\big{)} for which the geodesics are precisely the max-plus segments.

Proof.

Let $z=\gamma_{\boldsymbol{u}}(x_{1},x_{2},t)$ and $z^{\prime}=\gamma_{\boldsymbol{u}}(x_{1},x_{2},s)$. From Lemma 8, there are three cases to consider:

[TABLE]

In all cases, the conclusion follows from Lemma 16 and Lemma 14. $\Box$

Corollary 2**.**

A subset $C$ of $E$ is max-plus convex if and only if it is a geodesically convex set with respect to the geodesic structure $\Gamma_{\boldsymbol{u}}$.

Let $\hat{\gamma}_{\boldsymbol{u}}:E\times E\times[0,1]\to E$ be the affinely parametrized geodesic associated to $\gamma_{\boldsymbol{u}}$ that is,

[TABLE]

then

[TABLE]

If one defines the midpoint map $\mu_{\boldsymbol{u}}:E\times E\to E$ by $\mu_{\boldsymbol{u}}(x_{1},x_{2})=\hat{\gamma}_{\boldsymbol{u}}(x_{1},x_{2},1/2)$ then:

[TABLE]

A closed subset $C$ on a topological vector space is convex (in the usual affine structure) if and only if, for all $x_{1},x_{2}\in C$, $\displaystyle{\frac{1}{2}x_{1}+\frac{1}{2}x_{2}}\in C$.

Proposition 2 below characterizes closed max-plus convex sets as semilattices containing the (max-plus) midpoints of any pair of their points.

First, notice that, if $x_{1}\leqslant x_{2}$ then $-\alpha_{\boldsymbol{u}}(x_{1},x_{2})={\sf D}_{\boldsymbol{u}}(x_{1},x_{2})=\|x_{2}-x_{1}\|_{\boldsymbol{u}}=\beta_{\boldsymbol{u}}(x_{2},x_{1})$ and

$x_{1}\leqslant\mu_{\boldsymbol{u}}(x_{1},x_{2})\leqslant x_{2}$ since \mu_{\boldsymbol{u}}(x_{1},x_{2})=x_{1}\vee\big{(}x_{2}+(\alpha_{\boldsymbol{u}}(x_{1},x_{2})/2)\boldsymbol{u}\big{)}.

Lemma 18**.**

With respect to the metric topology associated to ${\sf D}_{\boldsymbol{u}}$ on $E$, the affinely parametrized geodesic $\hat{\gamma}_{\boldsymbol{u}}:E\times E\times[0,1]\to E$ is continuous on $E\times E\times[0,1]$.

Proof.

The restriction of \eta_{\boldsymbol{u}}\big{(}x_{1},x_{2},\alpha_{\boldsymbol{u}}(x_{1},x_{2})+t{\sf D}_{\boldsymbol{u}}(x_{1},x_{2})\big{)} to $E\times E\times[0,1]$ is $\hat{\gamma}_{\boldsymbol{u}}(x_{1},x_{2},t)$ and, by Lemma 17, $\eta_{\boldsymbol{u}}:E\times E\times\mathbb{R}\to E$ is continuous. Showing that $(x_{1},x_{2})\mapsto\alpha_{\boldsymbol{u}}(x_{1},x_{2})$ is continuous on $E\times E$ will complete the proof. We show that $(x_{1},x_{2})\mapsto\beta_{\boldsymbol{u}}(x_{1},x_{2})$ is uniformly continuous on $E\times E$ with respect to the topology of the norm $\|\cdot\|_{\boldsymbol{u}}$, which is a Riesz norm on $E$, which implies that $x\mapsto x^{+}$ is uniformly continuous on $E$, and therefore $x\mapsto\|x^{+}\|_{\boldsymbol{u}}$ is uniformly continuous. The conclusion follows from $\|x^{+}\|_{\boldsymbol{u}}=\boldsymbol{p}_{\boldsymbol{u}}(x^{+})=\max\{0,\boldsymbol{p}_{\boldsymbol{u}}(x)\}=\beta_{\boldsymbol{u}}(x,0)$ and $\beta_{\boldsymbol{u}}(x_{1},x_{2})=\beta_{\boldsymbol{u}}(x_{1}-x_{2},0)$. $\Box$

Corollary 3**.**

The midpoint map $\mu_{\boldsymbol{u}}:E\times E\to E$ is continuous.

Proposition 2**.**

For all non empty closed subset $C$ of $E$ the first and the last of the three assertions below are equivalent; if $C$ is complete then the three assertions are equivalent.

(A)* $C$ is max-plus convex.*

(B)* $\forall x_{0},x_{1}\in C\quad\mu_{\boldsymbol{u}}(x_{0},x_{1})\in C.$*

(C)* $\forall x_{0},x_{1}\in C\quad\begin{cases}(1)&x_{0}\vee x_{1}\in C\text{ and }\\ (2)&\text{ if }x_{0}\leqslant x_{1}\text{ then }\mu_{\boldsymbol{u}}(x_{0},x_{1})\in C.\end{cases}$*

Proof.

Clearly, $(A)$ implies $(B)$ and $(C)$. We show that $(C)$ implies $(A)$.

We have to see that, for all $x_{0},x_{1}\in C$, $\boldsymbol{[\![}x_{0},x_{1}\boldsymbol{]\!]}\subset C$. From Lemma 9 and hypothesis $(1)$, it is sufficient to show that $\boldsymbol{[\![}x_{0},x_{1}\boldsymbol{]\!]}\subset C$ whenever $x_{0},x_{1}\in C$ and $x_{0}\leqslant x_{1}$.

Let $D_{n}=\{(k/2^{n}):0\leq k\leq n\}$. If $x_{0},x_{1}\in C$ and $x_{0}\leqslant x_{1}$ then, by the second hypothesis, $x_{1/2}=\mu_{\boldsymbol{u}}(x_{0},x_{1})\in C$ and $x_{0}\leqslant x_{1/2}\leqslant x_{1}$.

Let $n\geq 1$ and assume that, for each $m\leq n$ we have a sequence of points $S_{m}=\{x_{m,t}:t\in D_{m}\}\subset\boldsymbol{[\![}x_{0},x_{1}\boldsymbol{]\!]}$ such that:

$(a)$ If $t\in D_{m-1}$ then $x_{m-1,t}=x_{m,t}$

$(b)$ If $t,t^{\prime}\in D_{m}$ and $t\leq t^{\prime}$ then $x_{m,t}\leqslant x_{m,t^{\prime}}$

$(c)$ If ${\sf D}_{\boldsymbol{hu}}(x_{n,k/2^{n}},x_{n,(k+1)/2^{n}})=2^{-n}{\sf D}_{\boldsymbol{hu}}(x_{0},x_{1})$.

To construct $S_{n+1}$ such that $(a),(b)$ and $(c)$ hold consider $t=(k/2^{n+1})$ and $t\not\in D_{n}$; there are then $t^{\prime},t^{\prime\prime}\in D_{n}$ such that $t=(t^{\prime}+t^{\prime\prime})/2$; let $x_{n+1,t}=\mu_{\boldsymbol{u}}(x_{n,t^{\prime}},x_{n,t^{\prime\prime}})$.

Now, let $S=\{x_{t}:t\in D\}=\cup_{n}S_{n}$ where $D\subset[0,1]$ is the set of dyadic numbers. For all given $n$, $S_{n}$ is a linearly ordered subset of $\boldsymbol{[\![}x_{0},x_{1}\boldsymbol{]\!]}$. Take $x\in\boldsymbol{[\![}x_{0},x_{1}\boldsymbol{]\!]}$; if $x=x_{0}$ then ${\sf D}_{\boldsymbol{hu}}(x,x_{n,1/2^{n}})=(1/2^{n}){\sf D}_{\boldsymbol{hu}}(x_{0},x_{1})$; if $x=x_{1}$ then ${\sf D}_{\boldsymbol{hu}}(x,x_{n,2^{n-1}/2^{n}})=(1/2^{n}){\sf D}_{\boldsymbol{hu}}(x_{0},x_{1})$. If $x$ is neither $x_{0}$ nor $x_{1}$, nor a point of $S_{n}$, there are two points $x_{n,k/2^{n}}$ and $x_{n,(k+1)/2^{n}}$ such that $x_{n,k/2^{n}}\leqslant x\leqslant x_{n,(k+1)/2^{n}}$, since, by Lemma 8, $\boldsymbol{[\![}x_{0},x_{1}\boldsymbol{]\!]}$ is linearly ordered by the restriction of the partial of $E$. We have $\boldsymbol{[\![}x_{0},x_{1}\boldsymbol{]\!]}=\boldsymbol{[\![}x_{0},x_{n,k/2^{n}}\boldsymbol{]\!]}\cup\boldsymbol{[\![}x_{n,k/2^{n}},x_{n,(k+1)/2^{n}}\boldsymbol{]\!]}\cup\boldsymbol{[\![}x_{n,(k+1)/2^{n}},x_{1}\boldsymbol{]\!]}$ and since $x\not\in S_{n}$, $x$ belongs to only one these max-plus segments, which is $\boldsymbol{[\![}x_{n,k/2^{n}},x_{n,(k+1)/2^{n}}\boldsymbol{]\!]}$. From $(1/2^{n})={\sf D}_{\boldsymbol{hu}}(x_{n,k/2^{n}},x_{n,(k+1)/2^{n}})={\sf D}_{\boldsymbol{hu}}(x_{n,k/2^{n}},x)+{\sf D}_{\boldsymbol{hu}}(x,x_{n,(k+1)/2^{n}})$ we have ${\sf D}_{\boldsymbol{hu}}(x_{n,k/2^{n}},x)\leq(1/2^{n})$.

We have shown that $S$ is a dense subset of $\boldsymbol{[\![}x_{0},x_{1}\boldsymbol{]\!]}$ and that $S\subset C$. Since $C$ is closed we have $\boldsymbol{[\![}x_{0},x_{1}\boldsymbol{]\!]}\subset C$.

Assuming that $C$ is complete - with respect to ${\sf D}_{\boldsymbol{hu}}$ - we show that $(B)$ implies $(A)$.

Since $(C,{\sf D}_{\boldsymbol{hu}},\mu_{\boldsymbol{u}})$ is a complete midpoint space in the sense of [18] there exists an affinely parametrized geodesic $\varphi:C\times C\times[0,1]\to C$ such, for all $(x_{0},x_{1})\in C\times C$, $\mu_{\boldsymbol{u}}(x_{0},x_{1})=\varphi(x_{0},x_{1},1/2)$ which is obtained by dyadic approximation starting from $\mu_{\boldsymbol{u}}(x_{1},x_{2})$, Lemma 3.0.1 and its proof in [18]; the restriction of $\hat{\gamma}_{\boldsymbol{u}}$ to $C\times C\times[0,1]$ is also an affinely parametrized geodesic. Furthermore, for all $(x_{0},x_{1})\in C\times C$, $\hat{\gamma}_{\boldsymbol{u}}(x_{0},x_{1},1/2)=\mu_{\boldsymbol{u}}(x_{0},x_{1})=\varphi(x_{0},x_{1},1/2)$. By dyadic approximation we have, $(x_{0},x_{1})\in C\times C\times[0,1]$, $\hat{\gamma}_{\boldsymbol{u}}(x_{0},x_{1},t)=\varphi(x_{0},x_{1},t)\in C$ that is $\boldsymbol{[\![}x_{0},x_{1}\boldsymbol{]\!]}\subset C$. $\Box$

6. On the topology of max-plus convex sets

In this section, the topology on a given Riesz space $E$ is the metric topology ${\sf D}_{\boldsymbol{hu}}$ associated to a given unit $\boldsymbol{u}$ of $E$. Either of the first proposition or the first lemma of this section shows that in the metric space \big{(}E,{\sf D}_{\boldsymbol{hu}}\big{)} open and closed balls are max-plus convex, and therefore geodesically convex ; consequently, the topology is locally max-plus convex. A max-plus convex set is an absolute retract from which we have the max-plus version of the classical Kakutani Fixed Point Theorem. There are also max-plus versions of some classical continuous (approximate) selection theorems for upper semicontinuous multivalued maps (here simply called “maps”).

Lemma 19**.**

Balls, open or closed with respect to $\|\cdot\|_{\boldsymbol{hu}}$, are max-plus convex.

Proof.

(A) Let B=B_{\boldsymbol{hu}}(0,\delta)=\big{\{}x\in E:\|x\|_{\boldsymbol{hu}}<\delta\big{\}}; from Theorem 1 we have $x_{1}\vee x_{2}\in B$ if $\{x_{1},x_{2}\}\subset B$.

Since $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}=\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}\,\cup\,\boldsymbol{[\![}x_{1},x_{1}\vee x_{2}\boldsymbol{]\!]}$ it is sufficient to prove that $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}\subset B$ whenever $x_{1},x_{2}\in B$ and $x_{1}$ and $x_{2}$ are comparable.

(1) Assume that $x_{1}\leqslant x_{2}$.

If $t\leq 0$ then $x_{1}+t_{1}\boldsymbol{u}\leq x_{2}$ from which we have $(x_{1}+t_{1}\boldsymbol{u})\vee x_{2}=x_{2}\in B$.

(2) Assume that $x_{2}\leqslant x_{1}$.

We have to see that, if $t\leq 0$ then $(x_{1}+t\boldsymbol{u})\vee x_{2}\in B$ or, equivalently, that $(x_{1}-s\boldsymbol{u})\vee x_{2}\in B$ if $s\geq 0$.

From $s\geq 0$ and $x_{2}\leqslant x_{1}$ we have $(x_{1}-s\boldsymbol{u})\vee x_{2}\leq x_{1}\vee x_{2}=x_{1}$ and therefore \boldsymbol{p}_{\boldsymbol{u}}\big{(}(x_{1}-s\boldsymbol{u})\vee x_{2}\big{)}\leq\boldsymbol{p}_{\boldsymbol{u}}(x_{1}) and

[TABLE]

From $(s\boldsymbol{u}-x_{1})\wedge(-x_{2})\wedge(-x_{1})=(s\boldsymbol{u}-x_{1})\wedge(-x_{1})\leq(-x_{1})$ and $-[(x_{1}-s\boldsymbol{u})\vee x_{2}]=(s\boldsymbol{u}-x_{1})\wedge(-x_{2})$ we have $-[(x_{1}-s\boldsymbol{u})\vee x_{2}]\leqslant(-x_{1})$ from which, \boldsymbol{p}_{\boldsymbol{u}}\big{(}-[(x_{1}-s\boldsymbol{u})\vee x_{2}]\big{)}\leq\boldsymbol{p}_{\boldsymbol{u}}(-x_{1}) and

[TABLE]

Adding the inequalities from (28) and (29) gives $\|(x_{1}-s\boldsymbol{u})\vee x_{2}]\|_{\boldsymbol{hu}}\leq\|x_{1}\|_{\boldsymbol{hu}}$.

We have shown that $B$ is max-plus convex; the same procedure shows that closed balls centered at [math] are max-plus convex;

(B) We show that arbitrary balls are max-plus convex. Let $B=\{x\in E:{\sf D}_{\boldsymbol{hu}}(x_{0},x)\leq\delta\}$ be a ball centered at $x_{0}$ and let $x_{1}$ and $x_{2}$ be two points of $B$; from $x_{i}-x_{0}\in(B-x_{0})$ and from the first part of the proof, we have, for all $t\leq 0$, \big{[}(x_{1}-x_{0})+t\boldsymbol{u}\big{]}\vee(x_{2}-x_{0})\in(B-x_{0}) and therefore, $(x_{1}+t\boldsymbol{u})\vee x_{2}\in B$. $\Box$

Proposition 3**.**

For all $(y,x_{1},x_{2})\in E\times E$ and for all $x\in\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$

[TABLE]

and, more generally, for all non empty set $S\subset E$ and for all $y\in E$

[TABLE]

Proof.

Let $r=\max\left\{{\sf D}_{\boldsymbol{hu}}(y,x_{1}),{\sf D}_{\boldsymbol{hu}}(y,x_{2})\right\}$ and let $B$ be the closed ball - with respect to ${\sf D}_{\boldsymbol{hu}}$ - of radius $r$ centered at $y$; from $x_{i}\in B$ and Lemma 19 we have $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}\subset B$ which establishes (30).

Assume $S=\{x_{1},\cdots,x_{m}\}$ with $m>2$ and let $S_{1}=\{x_{1},\cdots,x_{m-1}\}$ and proceed by induction: if $x\in\boldsymbol{[\![}S_{1}\boldsymbol{]\!]}$ then {\sf D}_{\boldsymbol{hu}}\big{(}y,x\big{)}\leq\max_{1\leq i\leq m-1}{\sf D}_{\boldsymbol{hu}}\big{(}y,x_{i}\big{)}; if $x\in\boldsymbol{[\![}S\boldsymbol{]\!]}$ then, by Lemma 7, $x\in\boldsymbol{[\![}x^{\prime},x_{m}\boldsymbol{]\!]}$ with $x^{\prime}\in\boldsymbol{[\![}S_{1}\boldsymbol{]\!]}$ and, by (30), {\sf D}_{\boldsymbol{hu}}\big{(}y,x\big{)}\leq\max\{{\sf D}_{\boldsymbol{hu}}\big{(}y,x^{\prime}\big{)},{\sf D}_{\boldsymbol{hu}}\big{(}y,x_{m}\big{)}\}. We have shown that (31) holds for finite sets.

For the general case, from $\displaystyle{\boldsymbol{[\![}S\boldsymbol{]\!]}=\bigcup_{A\in\langle S\rangle}\boldsymbol{[\![}A\boldsymbol{]\!]}}$, $(3)$ of Lemma 6, we have $x\in\boldsymbol{[\![}A\boldsymbol{]\!]}$ for some finite set $A\subset S$ from which the conclusion follows. $\Box$

Corollary 4**.**

The diameter with respect to the metric ${\sf D}_{\boldsymbol{hu}}$ of $\boldsymbol{[\![}S\boldsymbol{]\!]}$ is $\max\left\{{\sf D}_{\boldsymbol{hu}}(x,y):(x,y)\in S\times S\right\}$ which is the diameter of $S$.

Lemma 19 and Proposition 3 are equivalent, they say that the metric ${\sf D}_{\boldsymbol{hu}}$ is (max-plus) quasiconvex. One can obviously define a max-plus quasiconvex function $f:C\to\mathbb{R}$, where $C$ is a max-plus convex subset of $E$, by the property $\max_{x\in\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}}f(x)=\max\{f(x_{1}),f(x_{2})\}$ and show that, for such functions, $\max_{x\in\boldsymbol{[\![}S\boldsymbol{]\!]}}f(x)=\max_{x\in S}f(x)$.

Given two subsets $S_{1},S_{2}$ of $E$ let $S_{1}\vee S_{2}=\{v_{1}\vee v_{2}:(v_{1},v_{2})\in S_{1}\times S_{2}\}$ and let $S_{i}+t\boldsymbol{u}=\{v_{i}+t\boldsymbol{u}:v_{i}\in S_{i}\}$.

Lemma 20**.**

If $(E,\|\cdot\|_{\boldsymbol{u}})$ is complete then,

[TABLE]

where $B_{\boldsymbol{u}}(x,\delta)$ is the open ball with respect to $\|\cdot\|_{\boldsymbol{u}}$ of radius $\delta$ centered at $x$.

Proof.

We can assume that $(E,\|\cdot\|_{\boldsymbol{u}})$ is $\mathcal{C}(\Omega)$, for some compact topological space $\Omega$, and that $\boldsymbol{u}:\Omega\to\mathbb{R}$ is the constant function $1$; $\|\cdot\|_{\boldsymbol{u}}$ is then the sup-norm $\|x\|_{\infty}=\max_{\omega\in\Omega}|x(\omega)|$; we simply write $B(x,\delta)$ for the open ball with respect to the sup-norm. If $\|x_{i}-y_{i}\|_{\infty}<\delta$ then, for all $\omega\in\Omega$, $x_{i}(\omega)-\delta<y_{i}(\omega)<x_{i}(\omega)+\delta$ and therefore $\max\{x_{1}(\omega),x_{2}(\omega)\}-\delta<\max\{y_{1}(\omega),y_{2}(\omega)\}<\max\{x_{1}(\omega),x_{2}(\omega)\}+\delta$ which shows that $B(x_{1},\delta)\vee B(x_{2},\delta)\subseteq B(x_{1}\vee x_{2},\delta)$.

To prove the other inclusion let $\Omega_{1}=\{\omega\in\Omega:x_{2}(\omega)\leq x_{1}(\omega)\}$ and similarly for $\Omega_{2}$.

If $y\in B_{\boldsymbol{u}}(x_{1}\vee x_{2},\delta)$ then

[TABLE]

By the Tietze-Urysohn’s Theorem, there exists a continuous function $v_{i}:\Omega\to]-\delta,\delta[$ such that,

[TABLE]

Let $z_{i}=v_{i}+x_{i}$ and notice that $\begin{cases}\forall\omega\in\Omega_{i}\quad z_{i}(\omega)=y(\omega)\\ \forall\omega\in\Omega\quad x_{i}(\omega)-\delta<z_{i}(\omega)<x_{i}(\omega)+\delta\end{cases}$

from which we have $z_{i}\in B(x_{i},\delta)$ and $y\leq\max\{z_{1},z_{2}\}$. Let $y_{i}=\min\{y,z_{i}\}$ then $y_{i}\in B(x_{i},\delta)$ and $y=\max\{y_{1},y_{2}\}$. $\Box$

Proposition 4**.**

If $\|\cdot\|_{\boldsymbol{u}}$ is a complete norm on $E$ then the interior of a max-plus convex subset of $E$ is max-plus convex.

Proof.

Let $C\subseteq E$ be a max-plus convex set whose interior $\stackrel{{\scriptstyle\circ}}{{C}}$ is not empty; take $x_{1},x_{2}\in\stackrel{{\scriptstyle\circ}}{{C}}$ and $\delta>0$ such that $B_{\boldsymbol{u}}(x_{i},\delta)\subset C$. For all $t\leq 0$, B_{\boldsymbol{u}}(x_{1},\delta)\vee\big{(}B_{\boldsymbol{u}}(x_{2},\delta)+t\boldsymbol{u}\big{)}\subset C and $B_{\boldsymbol{u}}(x_{2},\delta)+t\boldsymbol{u}=B_{\boldsymbol{u}}(x_{2}+t\boldsymbol{u},\delta)$ therefore, $B_{\boldsymbol{u}}(x_{1},\delta)\vee B_{\boldsymbol{u}}(x_{2}+t\boldsymbol{u},\delta)\subset C$ which, from Lemma 20, gives B_{\boldsymbol{u}}\big{(}x_{1}\vee(x_{2}+t\boldsymbol{u}),\delta)\subseteq C. $\Box$

Given $\delta>0$ and a non empty set $S\subseteq E$ let $U_{\delta}(S)=\{x\in E:\inf_{y\in S}\|x-y\|_{\boldsymbol{u}}<\delta\}$ and

$V_{\delta}(S)=\{x\in E:\inf_{y\in S}\|x-y\|_{\boldsymbol{hu}}<\delta\}$

Lemma 21**.**

For all non empty subset $E$ of $E$ one has $V_{\delta}(S)\subset U_{\delta}(S)\subset V_{2\delta}(S)$ and, for all max-plus convex set $C\subset E$ and for all $\delta>0$, $U_{\delta}(C)$ is max-plus convex. As a consequence, the closure of a max-plus convex set is max-plus convex.

Proof.

The first part follows from $\|\cdot\|_{\boldsymbol{u}}\leq\|\cdot\|_{\boldsymbol{hu}}\leq 2\|\cdot\|_{\boldsymbol{u}}$.

Given a non empty max-plus set $C\subset E$, take $y_{i}\in U_{\delta}(C)$ and $x_{i}\in C$, $i=1,2$, and $\eta>0$ such that $\|x_{i}-y_{i}\|_{\boldsymbol{u}}\leq\eta<\delta$. From $\boldsymbol{|}x_{i}-y_{i}\boldsymbol{|}\leqslant\eta\boldsymbol{u}$ we have $x_{i}-\eta\boldsymbol{u}\leqslant y_{i}\leqslant x_{i}+\eta\boldsymbol{u}$. For all $t\in\mathbb{R}$, $x_{2}+t\boldsymbol{u}-\eta\boldsymbol{u}\leqslant y_{2}+t\boldsymbol{u}\leqslant x_{2}+t\boldsymbol{u}+\eta\boldsymbol{u}$. From $x_{2}+t\boldsymbol{u}-\eta\boldsymbol{u}\leqslant y_{2}+t\boldsymbol{u}$ and $x_{1}-\eta\boldsymbol{u}\leqslant y_{1}$ we have $(x_{1}-\eta\boldsymbol{u})\vee(x_{2}+t\boldsymbol{u}-\eta\boldsymbol{u})\leqslant y_{1}\vee(y_{2}+t\boldsymbol{u})$ that is $x_{1}\vee(x_{2}+t\boldsymbol{u})-\eta\boldsymbol{u}\leqslant y_{1}\vee(y_{2}+t\boldsymbol{u})$. Similarly, from $(y_{2}+t\boldsymbol{u})\leqslant(x_{2}+t\boldsymbol{u}+\eta\boldsymbol{u})$ and $y_{1}\leqslant x_{1}+\eta\boldsymbol{u}$ we have $y_{1}\vee(y_{2}+t\boldsymbol{u})\leqslant x_{1}\vee(x_{2}+t\boldsymbol{u})+\eta\boldsymbol{u}$. We have shown that $\boldsymbol{|}y_{1}\vee(y_{2}+t\boldsymbol{u})-x_{1}\vee(x_{2}+t\boldsymbol{u})\boldsymbol{|}\leqslant\eta\boldsymbol{u}$. If $t\leq 0$ then $x_{1}\vee(x_{2}+t\boldsymbol{u})\in C$ and $\|y_{1}\vee(y_{2}+t\boldsymbol{u})-x_{1}\vee(x_{2}+t\boldsymbol{u})\|_{\boldsymbol{u}}\leq\eta$ which shows that $y_{1}\vee(y_{2}+t\boldsymbol{u})\in U_{\delta}(C)$.

The last part follows from $\overline{C}=\cap_{\delta>0}U_{\delta}(C)$. $\Box$

Proposition 5**.**

$(1)$* If $x_{1}\neq x_{2}$ then $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$ is a topological arc in the metric space \big{(}E,{\sf D}_{\boldsymbol{u}}\big{)}.

$(2)$ For all finite sets $\{x_{1},\cdots,x_{m}\}\subset E$, $\boldsymbol{[\![}x_{1},\cdots,x_{m}\boldsymbol{]\!]}$ is a compact subset of the metric space \big{(}E,{\sf D}_{\boldsymbol{u}}\big{)}.*

Proof.

$(1)$ If $x_{1}\neq x_{2}$ then $\alpha_{\boldsymbol{u}}<\beta_{\boldsymbol{u}}$ and $\gamma_{\boldsymbol{u}}:[\alpha_{\boldsymbol{u}}(x_{1},x_{2}),\beta_{\boldsymbol{u}}(x_{1},x_{2})]\to\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$ is a homeomorphism.

$(2)$ By induction on $m$. We can assume that $m\geq 3$. By the induction hypothesis $K=\boldsymbol{[\![}x_{1},\cdots,x_{m-1}\boldsymbol{]\!]}$ is compact. Take $t_{1},t_{2}\in\mathbb{R}$ such that, for all $x\in K$, $t_{1}\boldsymbol{u}\leqslant x-x_{m}\leqslant t_{2}\boldsymbol{u}$, Lemma 5. We have, for all $x\in K$, $t_{1}\leq\boldsymbol{q}_{\boldsymbol{u}}(x-x_{m})$ and $\boldsymbol{p}_{\boldsymbol{u}}(x-x_{m})\leq t_{2}$; without loss of generality, we can assume that $t_{1}<0<t_{2}$ which yields, for all $x\in K$, $t_{1}\leq\alpha(x,x_{m})\leq\beta(x,x_{m})\leq t_{2}$.

On $K\times[t_{1},t_{2}]$ define $\gamma_{\star}$ by $\gamma_{\star}(x,t)=\gamma_{\boldsymbol{u}}(x,x_{m},t)$; from $[\alpha(x,x_{m}),\beta(x,x_{m})]\subseteq[t_{1},t_{2}]$ we have, for all $x\in K$, \gamma_{\star}\big{(}x,[t_{1},t_{2}]\big{)}=\boldsymbol{[\![}x,x_{m}\boldsymbol{]\!]} and, from Lemma 7, we have \gamma_{\star}\Big{(}K\times[t_{1},t_{2}]\Big{)}=\boldsymbol{[\![}x_{1},\cdots,x_{m}\boldsymbol{]\!]}.

To complete the proof, notice that by the definition of $\gamma_{\boldsymbol{u}}$, (24), and Lemma 5, $\gamma_{\star}$ is continuous. $\Box$

Lemma 22**.**

Non empty max-plus convex susbets of $E$ are contractible.

Proof.

By Lemma 18, $\hat{\gamma}:E\times E\times[0,1]\to E$ is continuous ; if $C$ be a max-plus subset of $E$ then \hat{\gamma}\big{(}C\times C\times[0,1]\big{)}=C futhermore, for all $(x,y)\in C\times C$, $\hat{\gamma}(x,y,0)=x$ and $\hat{\gamma}(x,y,1)=y$. $\Box$

Proposition 6**.**

If $C\subset E$ is a max-plus convex subset of $E$ which is complete with respect to ${\sf D}_{\boldsymbol{hu}}$ then, for all compact subsets $K\subset C$, $\boldsymbol{[\![}K\boldsymbol{]\!]}$ is compact.

Proof.

From Proposition 5 and Theorem 3.8 in [30]. $\Box$

Given a metric space $(X,D)$ let ${\sf Bd}(X)$ (respectively, ${\sf Comp}(X)$), be the family of non empty bounded (respectively, compact) subsets of $X$.

For $S\subset X$ and $\delta>0$ let $\mathcal{N}_{\delta}(S)=\{x\in X:D(x,S)<\delta\}$; for $X=E$ and $D$ the distance associated to $\|\cdot\|_{\boldsymbol{u}}$ (respectively, $\|\cdot\|_{\boldsymbol{hu}}$) this is $U_{\delta}(S)$ (respectively, $V_{\delta}(S)$). Recall that the Hausdorff metric on ${\sf Bd}(X)$ associated to the metric $D$ on $X$ is given by \mathcal{H}_{D}(S_{1},S_{2})=\inf\{\delta>0:S_{1}\subset\mathcal{N}_{\delta}(S_{2})\text{ and }S_{2}\subset\mathcal{N}_{\delta}(S_{1})\}=\max\big{\{}\sup_{x\in S_{1}}D(x,S_{2}),\sup_{x\in S_{2}}D(x,S_{1})\big{\}}. For $X=E$ and $D$ the distance associated to $\|\cdot\|_{\boldsymbol{u}}$ (respectively, the distance ${\sf D}_{\boldsymbol{hu}}$) the corresponding Hausdorff metric will be written $\mathcal{H}_{\boldsymbol{u}}$ (respectively, $\mathcal{H}_{\boldsymbol{hu}}$). From the first part of Lemma 21, $\mathcal{H}_{\boldsymbol{u}}\leq\mathcal{H}_{\boldsymbol{hu}}\leq 2\mathcal{H}_{\boldsymbol{u}}$.

Proposition 7**.**

With respect to either of the equivalent metrics $\mathcal{H}_{\boldsymbol{hu}}$ or $\mathcal{H}_{\boldsymbol{u}}$, the convex hull operator $S\mapsto\boldsymbol{[\![}S\boldsymbol{]\!]}$ is Lipschitz continuous from ${\sf Bd}(E)$ to itself.

Proof.

First, from Corollary 4, $\boldsymbol{[\![}S\boldsymbol{]\!]}$ is bounded if $S$ is bounded. Choose $\delta>\mathcal{H}_{\boldsymbol{hu}}(S_{1},S_{2})$ then $S_{1}\subset V_{\delta}(S_{2})\subset U_{\delta}(S_{2})$; from $S_{2}\subset\boldsymbol{[\![}S_{2}\boldsymbol{]\!]}$ we have $\boldsymbol{[\![}S_{1}\boldsymbol{]\!]}\subset\boldsymbol{[\![}\,U_{\delta}(\boldsymbol{[\![}S_{2}\boldsymbol{]\!]})\,\boldsymbol{]\!]}$ and finally, from Lemma 21, $\boldsymbol{[\![}\,U_{\delta}(\boldsymbol{[\![}S_{2}\boldsymbol{]\!]})\,\boldsymbol{]\!]}=U_{\delta}(\boldsymbol{[\![}S_{2}\boldsymbol{]\!]})\subset V_{2\delta}(\boldsymbol{[\![}S_{2}\boldsymbol{]\!]})$. We have shown that $\boldsymbol{[\![}S_{1}\boldsymbol{]\!]}\subset V_{2\delta}(\boldsymbol{[\![}S_{2}\boldsymbol{]\!]})$ and similarly $\boldsymbol{[\![}S_{2}\boldsymbol{]\!]}\subset V_{2\delta}(\boldsymbol{[\![}S_{1}\boldsymbol{]\!]})$ from which $\mathcal{H}_{\boldsymbol{hu}}(\boldsymbol{[\![}S_{1}\boldsymbol{]\!]},\boldsymbol{[\![}S_{2}\boldsymbol{]\!]})\leq 2\mathcal{H}_{\boldsymbol{hu}}(S_{1},S_{2})$. $\Box$

Given a non empty max-plus convex $C$ subset of $E$ let ${\sf MPCC}_{\boldsymbol{u}}(C)$ be the family of non empty max-plus convex and compact subsets of $C$;

Corollary 5**.**

If $C$ is a non empty compact max-plus convex subset of $E$ then, ${\sf MPCC}_{\boldsymbol{u}}(C)$ is an absolute retract.

Proof.

Since $C$ is compact it is, with respect to either of the metrics associated to the norms $\|\cdot\|_{\boldsymbol{u}}$ or $\|\cdot\|_{\boldsymbol{hu}}$ a Peano continuum777A compact, connected and locally connected metric space therefore, by Wojdislawki’s Theorem [31], ${\sf Comp}(C)$ is an absolute retract; by Proposition 6 and Proposition 7, $K\mapsto\boldsymbol{[\![}K\boldsymbol{]\!]}$ is a continuous retraction from ${\sf Comp}(C)$ to $\sf{\sf MPCC}_{\boldsymbol{u}}(C)$ which is consequently an absolute retract. $\Box$

Corollary 6**.**

A non empty compact and max-plus convex subset is an absolute retract.

Proof.

Let $C$ be a non empty compact and maxplus convex subset of $E$. The map $x\mapsto\{x\}$ is continuous from $C$ to ${\sf CCMP}_{\boldsymbol{u}}(C)$. An arbitrary non empty max-plus $K$ has a unique maximal element $\bigvee K$ and, for all $x\in C$, $\bigvee\{x\}=x$; the map $K\mapsto\bigvee K$ is continous and onto from ${\sf MPCC}_{\boldsymbol{u}}(C)$ to $C$. In conclusion, $C$ is a retract of ${\sf MPCC}_{\boldsymbol{u}}(C)$. $\Box$

We will see in the next section that, in Corollary 6, the compactness assumption is superfluous, Theorem 6, but the proof of that result is somewhat more involved.

If $C$ is a compact max-plus convex set $\mathbb{R}^{n}$ ($\boldsymbol{u}=(1,1\cdots,1)$), it has been shown by L. Bazylevych, M. Zarichnyi [2] that ${\sf MPP}_{m}(C)$, the space of (non enmpty) max-plus polytopes of the form $\boldsymbol{[\![}S\boldsymbol{]\!]}$ with $S\subset C$ and of cardinality at most $m$, is an absolute retract. In that same paper they also show that, for a compact and metrizable space $\Omega$, whose unit is taken to be the constant map $\omega\mapsto 1$, {\sf MPCC}\big{(}\mathcal{C}(\Omega)\big{)} is an absolute retract homeomorphic to $l_{2}$.

The compact topological space $\Omega$ of the BBK Representation Theorem is unique up to homeomorphism; and it can be realized as a closed subspace of the unit ball of the dual space endowed with the weak topology (hence its compactness, by Alaoglu’s Theorem); also, if \big{(}E,\|\cdot\|_{\boldsymbol{u}}\big{)} is separable then the unit ball of the dual space is metrizable in the weak topology. The BBK Representation Theorem combined with the theorem of L. Bazylevych and M. Zarichnyi from [2] cited above yields the following proposition.

Proposition 8**.**

Let $E$ be a Riesz space with unit $\boldsymbol{u}$. If \big{(}E,\boldsymbol{u}\big{)} is complete and separable then ${\sf MPCC}_{\boldsymbol{u}}(E)$, the hyperspace of max-plus non empty compact subsets of $E$, is an absolute retract.

7. Fixed points and selections

From Proposition 3 one has a max-plus version of Fan’s Best Approximation Theorem, (A) of Proposition 9 below, from which one has the max-plus version of Brouwer’s Fixed Point Theorem. For the proof, in the context of geodesic spaces, the reader is referred to [18], and to [11] page 146 for the proof of the original Fan’s Theorem in normed spaces.

Proposition 9**.**

*Let $f:C\to E$ be a continuous function, with respect to ${\sf D}_{\boldsymbol{hu}}$, defined on a compact max-plus convex subset of $E$. If the metric ${\sf D}_{\boldsymbol{hu}}$ is complete then the following hold:

(A) (Fan’s Best Approximation Theorem in max-plus) There exists $x_{0}\in C$ such that, for all $y\in C$, ${\sf D}_{\boldsymbol{hu}}(x_{0},f(x_{0}))\leq{\sf D}_{\boldsymbol{hu}}(y,f(x_{0}))$.

(B) (Fan’s Fixed Point Criteria in max-plus) For $f$ to have a fixed point it is sufficient that, for all $x\in C$ such that $x\neq f(x)$, $C\cap\boldsymbol{[\![}x,f(x)\boldsymbol{]\!]}$ contains a point other than $x$.*

One could relax the completness assumption on the metric ${\sf D}_{\boldsymbol{hu}}$ by assuming that $C$ is contained in a complete max-plus convex subset $X$ of $E$ and that $f(C)\subset X$. If $f$ takes its values in $C$ itself then, either (A) or (B) of Proposition 9, implies that $f$ has a fixed point. This “Brouwer’s Fixed Point Theorem in max-plus” follows also from the more general “Kakutani’s Fixed Point Theorem in max-plus”, Theorem 7 below.

Recall that the metric ${\sf D}_{\boldsymbol{hu}}$ is equivalent to the metric associated to the norm $\|\cdot\|_{\boldsymbol{u}}$ and if there is on $E$ a complete Riesz norm then, for all $\boldsymbol{u}\in E_{+}$, $E_{\boldsymbol{u}}$ equipped with the norm $\|\cdot\|_{\boldsymbol{u}}$ is complete, [12] page 65.

Theorem 3** (Michael’s Selection Theorem).**

If $X\subset E$ is a max-plus convex subset of $E$ which is complete with respect to ${\sf D}_{\boldsymbol{hu}}$ then, all lower semicontinuous maps $\Gamma:Y\to X$ with non empty closed max-plus convex values defined a paracompact space $Y$ have a continuous selection.

Furthermore, if $A\subset Y$ is a closed set then any continuous selection of the resriction of $\Gamma$ to $A$ extends to a continuous selection of $\Gamma$.

Proof.

From Theorem 3.4 in [17]. $\Box$

Theorem 4** (Approximate selections for usc maps).**

Let $X$ be a non empty max-plus convex subset of $E$, $Y$ a paracompact topological space and $\Gamma:Y\to X$ an upper semicontinuous map with non empty max-plus convex values. Then, for all $\delta>0$, there exists a continuous map $f:Y\to X$ such that, for all $y\in Y$, $f(y)\in V_{\delta}(\Gamma y)$.

Furthermore, if the values of $\Gamma$ are max-plus convex and compact then any neighborhood $\Theta\subset Y\times X$ of the graph of $\Gamma$ contains the graph of a continuous map $f:Y\to X$.

Proof.

From Theorem 3.5 in [17]. $\Box$

Theorem 5** (Dungundji’s Extension Theorem).**

Let $A\subset X$ be non empty closed subset of an arbitrary metric space $(X,d)$ and $f:A\to C$ a continuous map from $A$ to an arbitrary non empty max-plus convex subset of $E$. Then, there exists a continuous maps $\hat{f}:X\to\boldsymbol{[\![}f(A)\boldsymbol{]\!]}$ whose restriction to $A$ is $f$.

Proof.

From Theorem 4.1 in [17]. $\Box$

Theorem 5 says that max-plus convex subsets (with respect to a given unit $\boldsymbol{u}$) of a Riesz space $E$ are, with respect to the topology induced by either of the norms $\|\cdot\|_{\boldsymbol{u}}$ or $\|\cdot\|_{\boldsymbol{hu}}$, absolute extensors for the class of metric spaces.

Theorem 6**.**

An arbitrary non empty max-plus convex subset of $E$, equipped with the metric topology associated to ${\sf D}_{\boldsymbol{hu}}$, is an absolute retract.

Proof.

A metrizable absolute extensor for the class of metric spaces is an absolute retract. $\Box$

A map (single valued or multivalued) $\Gamma:X\to Y$, where $Y$ is a toplological space, is a compact map if there is a compact set $K\subset Y$ such that $\Gamma(Y)\subset K$. A set $X$ has the fixed point property for a given class $\mathcal{M}$ of maps $\Gamma:X\to X$ if, for all $\Gamma\in\mathcal{M}$ there exists $x\in X$ such that $x\in\Gamma x$.

Theorem 7** (Kakutani-Fan - Himmelberg’s Theorem).**

An arbitrary non empty max-plus convex subset of $E$ has the fixed point property for upper semicontinuous compact maps with closed max-plus convex non empty values.

Proof.

From Proposition 5, Lemma 10 and Theorem 5.2 in [17]. $\Box$

Since a max-plus convex set is an absolute retract, Theorem 7 follows from the much harder Eilenberg-Montgomery Fixed Point Theorem, Corollary 7.4 in [11] or from (iv) of Corollary 7.5 on the same page of which the following statement is a particular instance :

If $X$ is an absolute retract then, arbitrary compact upper semicontinous maps $S:X\to X$ with contactible values have a fixed point.

From the results of this section one can derive tropical versions of some classical results from mathematical economics, existence of Nash equilibria, existence of equilibria for abstract economies, existence of maximal elements for a preference relation. The question of their relevance in economics or game theory is left open.

8. A few drawings in $\mathbb{R}^{2}$

[TABLE]

The closed unit ${\sf D}_{\boldsymbol{hu}}$-ball in $\mathbb{R}^{2}$ about $\ast$; it is $\boldsymbol{[\![}x_{1},x_{2},x_{3}\boldsymbol{]\!]}$.

The next example shows that the metric ${\sf D}_{\boldsymbol{hu}}$ does not have the unique extension property.

[TABLE]

Three possible geodesic extensions of the geodesic $\boldsymbol{[\![}x_{1},x_{2}\boldsymbol{]\!]}$

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$\boldsymbol{[\![}x_{1},x_{3}\boldsymbol{]\!]}$, $\boldsymbol{[\![}x_{1},x_{3}^{\prime}\boldsymbol{]\!]}$, $\boldsymbol{[\![}x_{1},x_{3}^{\prime\prime}\boldsymbol{]\!]}$ are geodesics.

8.1. A few max-plus polytopes in $\mathbb{R}^{2}$ and a max-plus convex set that is not a polytope

In each case the max-plus polytope in question is the max-plus convex hull of the points labeled $x_{1},x_{2},x_{2},\ldots$

[TABLE]

This example shows that in $\mathbb{R}^{2}$ a max-plus convex polytopes with three extreme points (none of the three points is in the max-plus convex hull of the other two) can have empty interior and have topological dimension equal to $1$.

[TABLE]

[TABLE]

In $\mathbb{R}^{n}$ a max-plus segment is piecewise linear, it is made of at most $n$ affine segments. A max-plus polytope is a contractible finite union of affine convex polytopes, and consequently an absolute retract; it is also a contractible simplicial complex. What is the structure of max-plus polytopes in infinite dimensional Riesz spaces?